Math 120A: Lecture notes

Definition 1.1 (Smoothness classes). A function is of class $C^0$ if it is continuous. For $k \ge 1$, a function of one or several variables is of class $C^k$ if all of its partial derivatives of order at most $k$ exist and are continuous. It is of class $C^\infty$ (or smooth) if it is of class $C^k$ for every $k$. A vector-valued function is of class $C^k$ if each of its components is.

Lemma 1.2 (Product rules). Let $f, g$ be differentiable functions from an interval to $\mathbb{R}^3$. Then $$\frac{d}{dt}\langle f, g \rangle = \Big\langle \frac{df}{dt}, g \Big\rangle + \Big\langle f, \frac{dg}{dt} \Big\rangle \qquad\text{and}\qquad \frac{d}{dt}(f \times g) = \frac{df}{dt} \times g + f \times \frac{dg}{dt}.$$ In particular, if $|f(t)|$ is constant, then $f'(t)$ is perpendicular to $f(t)$ for every $t$.

Proof. The two formulas follow from the usual product rule applied to each term of $\langle f, g \rangle = \sum f^i g^i$ and of each component of $f \times g$. For the last statement, if $\langle f, f \rangle$ is constant, then $0 = \frac{d}{dt}\langle f, f\rangle = 2\langle f', f \rangle$. ∎

Theorem 1.3 (Inverse function theorem in one variable). Let $g \colon (c, d) \to \mathbb{R}$ be of class $C^k$, $k \ge 1$, with $g'(r) \ne 0$ for all $r$. Then $g$ is either strictly increasing or strictly decreasing, its image is an open interval $(a, b)$, and the inverse function $h \colon (a, b) \to (c, d)$ is also of class $C^k$, with $$h'(t) = \frac{1}{g'(h(t))}.$$

Proof. Since $g'$ is continuous and never zero on an interval, it has constant sign by the intermediate value theorem; say $g' > 0$. By the mean value theorem $g$ is strictly increasing, so it is one-to-one, and its image is an interval, which is open because $g$ has no maximum or minimum on $(c, d)$. The inverse $h$ is continuous (an increasing bijection between intervals is continuous). Differentiating $g(h(t)) = t$, which is justified by a standard argument with difference quotients, gives $g'(h(t))\,h'(t) = 1$. The formula $h' = 1/(g' \circ h)$ shows that if $h$ is $C^j$ for some $j < k$, then $h'$ is $C^j$, so $h$ is $C^{j+1}$; by induction, $h$ is $C^k$. ∎

Definition 1.4 (Regular curve). A regular curve in $\mathbb{R}^3$ is a function $\alpha \colon (a, b) \to \mathbb{R}^3$ of class $C^k$ for some $k \ge 1$, such that $\alpha'(t) \ne 0$ for all $t$. Here $(a, b)$ is an open interval, and we allow $a = -\infty$ or $b = \infty$.

Definition 1.5 (Velocity, speed and unit tangent). Let $\alpha$ be a regular curve. The velocity vector of $\alpha$ at time $t$ is $\alpha'(t)$, and the speed is $|\alpha'(t)|$. The unit tangent vector is $$T(t) = \frac{\alpha'(t)}{|\alpha'(t)|}.$$ The tangent line to $\alpha$ at $t = t_0$ is the line $\{\alpha(t_0) + \lambda T(t_0) : \lambda \in \mathbb{R}\}$.

Example 1.6 (Straight lines). For fixed vectors $u$ and $v$ in $\mathbb{R}^3$, the curve $\alpha(t) = u + tv$ has $\alpha'(t) = v$. It is regular exactly when $v \ne 0$. Its image is the straight line through $u$ in the direction $v$, its unit tangent is the constant vector $T = v/|v|$, and its tangent line at every point is the line itself.

Example 1.7 (Same image, not regular). The curve $\alpha(t) = (t^3, 0, 0)$ has image the whole $x$-axis, the same as the regular curve $\beta(t) = (t, 0, 0)$. But $\alpha'(t) = (3t^2, 0, 0)$ vanishes at $t = 0$, so $\alpha$ is not regular: the particle comes to a stop at the origin and then moves on. On the other hand, $\gamma(t) = (t^3 + t, 0, 0)$ has $\gamma'(t) = (3t^2 + 1, 0, 0) \ne 0$, so it is regular, again with the same image.

Example 1.8 (Graphs). If $g \colon (a, b) \to \mathbb{R}$ is differentiable, the curve $\alpha(t) = (t, g(t), 0)$ traces the graph of $g$ in the $xy$-plane. Since $\alpha'(t) = (1, g'(t), 0) \ne 0$, it is always regular. The tangent line at $t_0$ is $\{(t_0 + \lambda, g(t_0) + \lambda g'(t_0), 0)\}$, which is the familiar tangent line to a graph from calculus.

Example 1.9 (Circular helix). Let $r > 0$ and $h$ be constants, and let $\alpha(t) = (r\cos t, r\sin t, ht)$. Then $\alpha'(t) = (-r\sin t, r\cos t, h)$, which is never zero, and $|\alpha'(t)| = \sqrt{r^2 + h^2}$. The curve winds around the cylinder $x^2 + y^2 = r^2$, rising by $2\pi h$ in each turn. If $h > 0$ the helix is right-handed (it turns like an ordinary screw), and if $h < 0$ it is left-handed. If $h = 0$ it is a circle of radius $r$ in the $xy$-plane.

Example 1.10 (A smooth curve with a cusp). Let $\alpha(t) = (t^2, t^3, 0)$. Each component is a polynomial, so $\alpha$ is $C^\infty$. But $\alpha'(t) = (2t, 3t^2, 0)$ is zero at $t = 0$, so $\alpha$ is not regular. Its image is the set of points with $y^2 = x^3$, which has a sharp point (a cusp) at the origin: as $t$ increases through $0$, the particle comes in along the lower branch, stops at the origin, and leaves along the upper branch. (Compare Homework 1, Problem 3.)

Example 1.11 (A smooth curve with a corner). Let $\varphi(t) = e^{-1/t^2}$ for $t > 0$ and $\varphi(t) = 0$ for $t \le 0$. One can show that $\varphi$ is $C^\infty$, with all derivatives equal to $0$ at $t = 0$. The curve $\alpha(t) = (\varphi(t) - \varphi(-t),\ \varphi(t) + \varphi(-t),\ 0)$ is therefore $C^\infty$. For $t > 0$ it is $(\varphi(t), \varphi(t), 0)$, on the line $y = x$; for $t < 0$ it is $(-\varphi(-t), \varphi(-t), 0)$, on the line $y = -x$. So its image is the graph of $y = |x|$ near the origin, which has a corner. The particle slows down to speed zero at the corner, turns, and speeds up again; this is why the curve can be smooth. Requiring regularity rules out such behavior, and guarantees a well-defined tangent direction at every point.

Definition 1.12 (Reparametrization). Let $\alpha \colon (a, b) \to \mathbb{R}^3$ be a regular curve. A reparametrization of $\alpha$ is a one-to-one, onto function $g \colon (c, d) \to (a, b)$ such that both $g$ and its inverse $h \colon (a, b) \to (c, d)$ are of class $C^k$ for some $k \ge 1$. The new curve $\beta = \alpha \circ g \colon (c, d) \to \mathbb{R}^3$ is also called a reparametrization of $\alpha$.

Lemma 1.13 (Checking reparametrizations). A function $g \colon (c, d) \to (a, b)$ is a reparametrization if and only if $g$ is of class $C^k$ for some $k \ge 1$, $g$ is onto, and $g'(r) \ne 0$ for all $r$. In that case $\alpha \circ g$ is a regular curve whenever $\alpha$ is.

Proof. If $g$ is a reparametrization with inverse $h$, then differentiating $h(g(r)) = r$ gives $h'(g(r))\,g'(r) = 1$, so $g'(r) \ne 0$. Conversely, if $g' \ne 0$ and $g$ is onto, then Theorem 1.3 says that $g$ is one-to-one with a $C^k$ inverse. For the last statement, by the chain rule $(\alpha \circ g)'(r) = \alpha'(g(r))\,g'(r)$, which is a product of a nonzero vector and a nonzero number. ∎

Example 1.14 (A reparametrization). The function $g \colon (0, 1) \to (1, 2)$, $g(r) = 1 + r^2$, is onto, $C^\infty$, and $g'(r) = 2r > 0$ on $(0, 1)$. So it is a reparametrization; its inverse is $h(t) = \sqrt{t - 1}$, which is $C^\infty$ on $(1, 2)$.

Example 1.15 (Not a reparametrization). The function $g(r) = r^3$ is a one-to-one, onto, $C^\infty$ map from $\mathbb{R}$ to $\mathbb{R}$, but its inverse $h(t) = t^{1/3}$ is not differentiable at $0$, and indeed $g'(0) = 0$. So $g$ is not a reparametrization. This is exactly what happened in Example 1.7: composing the regular curve $(t, 0, 0)$ with $g$ produced the non-regular curve $(t^3, 0, 0)$.

Proposition 1.16 (The unit tangent is geometric). Let $\alpha \colon (a, b) \to \mathbb{R}^3$ be a regular curve, let $g \colon (c, d) \to (a, b)$ be a reparametrization, and let $\beta = \alpha \circ g$. Let $T$ and $S$ be the unit tangents of $\alpha$ and $\beta$. If $t_0 = g(r_0)$, then $S(r_0) = T(t_0)$ if $g'(r_0) > 0$, and $S(r_0) = -T(t_0)$ if $g'(r_0) < 0$.

Proof. By the chain rule, $\beta'(r_0) = \alpha'(t_0)\,g'(r_0)$. Dividing by $|\beta'(r_0)| = |\alpha'(t_0)|\,|g'(r_0)|$ gives $S(r_0) = T(t_0)\,g'(r_0)/|g'(r_0)| = \pm T(t_0)$. ∎

Definition 2.1 (Curve segment). A regular curve segment is a function $\alpha \colon [a, b] \to \mathbb{R}^3$ that is the restriction of a regular curve defined on some open interval containing $[a, b]$.

Definition 2.2 (Length). The length of a regular curve segment $\alpha \colon [a, b] \to \mathbb{R}^3$ is $$L(\alpha) = \int_a^b |\alpha'(t)|\,dt = \int_a^b \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2}\;dt,$$ where $\alpha(t) = (x(t), y(t), z(t))$.

Proposition 2.3 (Length does not depend on the parametrization). Let $\alpha \colon [a, b] \to \mathbb{R}^3$ be a regular curve segment and let $g \colon [c, d] \to [a, b]$ be a reparametrization (the restriction of a reparametrization defined on slightly larger open intervals). Then $\beta = \alpha \circ g$ has the same length as $\alpha$.

Proof. By the chain rule, $|\beta'(r)| = |\alpha'(g(r))|\,|g'(r)|$. Suppose first that $g' > 0$, so $g(c) = a$ and $g(d) = b$. By the substitution $t = g(r)$, $$L(\beta) = \int_c^d |\alpha'(g(r))|\,g'(r)\,dr = \int_a^b |\alpha'(t)|\,dt = L(\alpha).$$ If $g' < 0$, then $|g'| = -g'$, $g(c) = b$ and $g(d) = a$, and the same substitution gives $L(\beta) = -\int_b^a |\alpha'(t)|\,dt = L(\alpha)$. ∎

Definition 2.4 (Arc length). The arc length of $\alpha$ measured from $t_0$ is the function $$s = h(t) = \int_{t_0}^t |\alpha'(u)|\,du, \qquad a < t < b.$$

Theorem 2.5 (Arc length is a reparametrization). Let $\alpha \colon (a, b) \to \mathbb{R}^3$ be a regular curve of class $C^k$, and let $h$ be its arc length measured from $t_0$. Then $h$ is one-to-one with image an open interval $(c, d)$, and its inverse $g \colon (c, d) \to (a, b)$ is an orientation-preserving reparametrization. The reparametrized curve $\beta = \alpha \circ g$ satisfies $|\beta'(s)| = 1$ for all $s$.

Proof. By the fundamental theorem of calculus, $h'(t) = |\alpha'(t)|$, which is positive because $\alpha$ is regular, and which is of class $C^{k-1}$ because $\alpha'$ is $C^{k-1}$ and nonzero (the length function $v \mapsto |v|$ is smooth away from $0$). So $h$ is of class $C^k$ with positive derivative. By Theorem 1.3, $h$ is one-to-one onto an open interval $(c, d)$ and its inverse $g$ is $C^k$ with $g' = 1/(h' \circ g) > 0$, so $g$ is an orientation-preserving reparametrization. Finally, by the chain rule, $$|\beta'(s)| = |\alpha'(g(s))|\,g'(s) = \frac{|\alpha'(g(s))|}{h'(g(s))} = 1.$$ ∎

Definition 2.6 (Unit speed curve). A regular curve $\beta$ is a unit speed curve, or is parametrized by arc length, if $|\beta'(s)| = 1$ for all $s$.

Corollary 2.7 (Arc length formulas). If $\alpha$ is a regular curve and $s = h(t)$ is its arc length, then

1. $s = \int_{t_0}^t |\alpha'(u)|\,du$;
2. $\dfrac{ds}{dt} = \Big|\dfrac{d\alpha}{dt}\Big|$;
3. $\dfrac{d\alpha}{dt} = \dfrac{ds}{dt}\,T$;
4. $T = \dfrac{d\alpha}{ds}$, where $\alpha$ is now regarded as a function of $s$.

Example 2.8 (Straight line). For $\alpha(t) = u + tv$ with $v \ne 0$, we have $|\alpha'(t)| = |v|$, so $s = h(t) = |v|\,t$ (measuring from $t_0 = 0$), $t = g(s) = s/|v|$, and the unit speed reparametrization is $\beta(s) = u + s\,v/|v|$.

Example 2.9 (Circle). For $\alpha(t) = (r\cos t, r\sin t, 0)$ with $r > 0$, we have $\alpha'(t) = (-r\sin t, r\cos t, 0)$ and $|\alpha'(t)| = r$. So $s = rt$, and the unit speed parametrization of the circle of radius $r$ is $$\beta(s) = \Big(r\cos\frac{s}{r},\ r\sin\frac{s}{r},\ 0\Big).$$ Its unit tangent is $T(s) = \big(-\sin\frac{s}{r}, \cos\frac{s}{r}, 0\big)$.

Example 2.10 (Helix). For the helix $\alpha(t) = (r\cos t, r\sin t, ht)$ of Example 1.9, the speed is the constant $\sqrt{r^2 + h^2}$. It is convenient to write $$\omega = \frac{1}{\sqrt{r^2 + h^2}}.$$ Then $s = t/\omega$, and the unit speed parametrization is $\beta(s) = (r\cos\omega s,\ r\sin\omega s,\ h\omega s)$. (Check the details in Homework 1, Problem 6.) We will use this formula repeatedly.

Example 2.11 (Ellipse). For $\alpha(t) = (2\sin t, \cos t, 0)$, whose image is the ellipse $x^2/4 + y^2 = 1$, we get $\alpha'(t) = (2\cos t, -\sin t, 0)$ and $$|\alpha'(t)| = \sqrt{4\cos^2 t + \sin^2 t} = \sqrt{1 + 3\cos^2 t}.$$ The integral $\int \sqrt{1 + 3\cos^2 t}\,dt$ is an elliptic integral; it has no antiderivative expressible using elementary functions. (Historically, this is where the name "elliptic integral" comes from.)

Example 2.12 (Parabola). For $\alpha(t) = (t, t^2/2, 0)$, we get $|\alpha'(t)| = \sqrt{1 + t^2}$, and measuring from $t_0 = 0$ (substitute $t = \tan\theta$ and integrate $\sec^3\theta$ by parts), $$s = \int_0^t \sqrt{1 + u^2}\,du = \frac{1}{2}\Big(t\sqrt{1 + t^2} + \ln\big(t + \sqrt{1 + t^2}\big)\Big).$$ So $s$ is an explicit function of $t$, but there is no way to solve this equation for $t$ in terms of elementary functions. The unit speed parametrization exists by Theorem 2.5, but we cannot write it down.

Definition 3.1 (Curvature). The curvature of a unit speed curve $\alpha$ is $\kappa(s) = |T'(s)| = |\alpha''(s)|$.

Example 3.2 (Straight line). For a unit speed line $\alpha(s) = u + s v$ with $|v| = 1$, the unit tangent $T = v$ is constant, so $\kappa \equiv 0$.

Example 3.3 (Circle). For the unit speed circle $\alpha(s) = (r\cos\frac{s}{r}, r\sin\frac{s}{r}, 0)$ of Example 2.9, $$T(s) = \Big(-\sin\frac{s}{r},\ \cos\frac{s}{r},\ 0\Big), \qquad T'(s) = \frac{1}{r}\Big(-\cos\frac{s}{r},\ -\sin\frac{s}{r},\ 0\Big),$$ so $\kappa = 1/r$. A circle of radius $r$ has constant curvature $1/r$; the smaller the circle, the larger the curvature, as we wanted.

Proposition 3.4 (Zero curvature means straight). If a unit speed curve $\alpha$ has $\kappa \equiv 0$ on an interval, then on that interval $\alpha$ is a segment of a straight line.

Proof. If $\kappa = |T'| = 0$ on an interval, then $T$ is a constant unit vector $v$ there. Integrating $\alpha'(s) = v$ gives $\alpha(s) = sv + x_0$ for a constant vector $x_0$, which parametrizes a straight line. ∎

Definition 3.5 (Principal normal and binormal). Let $\alpha$ be a unit speed curve and suppose $\kappa(s) \ne 0$. The principal normal vector of $\alpha$ at $s$ is $$N(s) = \frac{T'(s)}{\kappa(s)},$$ and the binormal vector is $B(s) = T(s) \times N(s)$.

Lemma 3.6 (The Frenet frame). If $\kappa(s) \ne 0$, then $T(s), N(s), B(s)$ are mutually perpendicular unit vectors, and they form a right-handed basis of $\mathbb{R}^3$.

Proof. $T$ is a unit vector, so by Lemma 1.2 its derivative $T'$ is perpendicular to $T$; hence $N = T'/\kappa$ is a unit vector perpendicular to $T$. The cross product of two perpendicular unit vectors is a unit vector perpendicular to both, so $B = T \times N$ is a unit vector perpendicular to $T$ and $N$. Three mutually perpendicular unit vectors are linearly independent, so they form a basis, and the basis $(T, N, T \times N)$ is right-handed by the definition of the cross product, just as $(e_1, e_2, e_3 = e_1 \times e_2)$ is. ∎

Definition 3.7 (Torsion). Let $\alpha$ be a unit speed curve with $\kappa \ne 0$. The torsion of $\alpha$ is $$\tau(s) = -\langle B'(s), N(s) \rangle,$$ so that $B'(s) = -\tau(s)\,N(s)$.

Definition 3.8 (Frenet-Serret apparatus). The Frenet-Serret apparatus of a unit speed curve with $\kappa \ne 0$ is the collection $\{\kappa, \tau, T, N, B\}$. For a regular curve $\beta(t)$ that is not unit speed, we define its Frenet-Serret apparatus at time $t$ to be that of its unit speed reparametrization at the corresponding value $s$ of arc length.

Example 3.9 (Circle). For the circle of radius $r$, from Example 3.3, $$N(s) = \Big(-\cos\frac{s}{r},\ -\sin\frac{s}{r},\ 0\Big),$$ which points from $\alpha(s)$ toward the center of the circle. Then $B = T \times N = (0, 0, 1)$ is constant, so $B' = 0$ and $\tau = 0$. The osculating plane is the plane of the circle at every point.

Example 3.10 (Circular helix). Consider the unit speed helix $\alpha(s) = (r\cos\omega s,\ r\sin\omega s,\ h\omega s)$ with $r > 0$ and $\omega = 1/\sqrt{r^2 + h^2}$, from Example 2.10. Then $$T(s) = \omega\,(-r\sin\omega s,\ r\cos\omega s,\ h), \qquad T'(s) = -\omega^2 r\,(\cos\omega s,\ \sin\omega s,\ 0).$$ So the curvature is the constant $$\kappa = \omega^2 r = \frac{r}{r^2 + h^2},$$ and $N(s) = (-\cos\omega s, -\sin\omega s, 0)$, which points horizontally from the curve toward the axis of the cylinder. Next, $$B = T \times N = \omega\,(h\sin\omega s,\ -h\cos\omega s,\ r), \qquad B'(s) = \omega^2 h\,(\cos\omega s,\ \sin\omega s,\ 0) = -\omega^2 h\,N(s).$$ So the torsion is the constant $$\tau = -\langle B', N \rangle = \omega^2 h = \frac{h}{r^2 + h^2}.$$ For a right-handed helix ($h > 0$) the torsion is positive, for a left-handed helix it is negative, and when $h = 0$ we recover the circle, with $\kappa = 1/r$ and $\tau = 0$. Notice that both the curvature and the torsion of a helix are constant, but the helix is not a circle. In Lecture 5 we will see that the helices (together with circles) are the only curves with constant curvature and torsion.

Example 3.11 (The osculating plane can jump). Let $\varphi(t) = e^{-1/t^2}$ for $t \ne 0$ and $\varphi(0) = 0$; it is $C^\infty$, with all derivatives zero at $t = 0$. Define $\alpha(t) = (t, \varphi(t), 0)$ for $t \le 0$ and $\alpha(t) = (t, 0, \varphi(t))$ for $t > 0$. This curve is $C^\infty$ and regular, since its first component is $t$. For $t < 0$ it lies in the $xy$-plane, and for $t > 0$ in the $xz$-plane, so its osculating plane is the $xy$-plane for $t < 0$ and the $xz$-plane for $t > 0$ (wherever $\kappa \ne 0$). At $t = 0$ the curvature is zero and the osculating plane changes abruptly. (See Homework 2, Problem 9.)

Lemma 4.1 (Expansion in an orthonormal basis). If $e_1, e_2, e_3$ is an orthonormal basis of $\mathbb{R}^3$, then every vector $v$ satisfies $v = \langle v, e_1 \rangle e_1 + \langle v, e_2 \rangle e_2 + \langle v, e_3 \rangle e_3$.

Proof. Write $v = c^1 e_1 + c^2 e_2 + c^3 e_3$, which is possible because the $e_i$ form a basis. Taking the inner product with $e_j$ and using $\langle e_i, e_j \rangle = 0$ for $i \ne j$ and $\langle e_j, e_j \rangle = 1$ gives $\langle v, e_j \rangle = c^j$. ∎

Theorem 4.2 (Frenet-Serret equations). Let $\alpha$ be a unit speed curve with $\kappa \ne 0$ and Frenet-Serret apparatus $\{\kappa, \tau, T, N, B\}$. Then $$T' = \kappa N, \qquad N' = -\kappa T + \tau B, \qquad B' = -\tau N.$$

Proof. The first and third equations hold by definition, as just recalled. For the second, apply Lemma 4.1 to $v = N'$ with the orthonormal basis $T, N, B$: $$N' = \langle N', T \rangle T + \langle N', N \rangle N + \langle N', B \rangle B.$$ We compute the three coefficients by differentiating the relations among $T$, $N$, $B$. Differentiating $\langle N, T \rangle = 0$ gives $\langle N', T \rangle = -\langle N, T' \rangle = -\langle N, \kappa N \rangle = -\kappa$. Since $N$ is a unit vector, $\langle N', N \rangle = 0$. Differentiating $\langle N, B \rangle = 0$ gives $\langle N', B \rangle = -\langle N, B' \rangle = -\langle N, -\tau N \rangle = \tau$. Hence $N' = -\kappa T + \tau B$. ∎

Corollary 4.3 (Plane curves have zero torsion). Let $\alpha$ be a unit speed curve with $\kappa \ne 0$. The following are equivalent:

1. $\alpha$ lies in a plane;
2. $B$ is a constant vector;
3. $\tau \equiv 0$.

Proof. Statements 2 and 3 are equivalent because $B' = -\tau N$ and $N \ne 0$.

Suppose $B$ is constant, and let $x_0 = \alpha(0)$. Then $$\frac{d}{ds}\langle \alpha(s) - x_0, B \rangle = \langle \alpha'(s), B \rangle = \langle T, B \rangle = 0,$$ so $\langle \alpha(s) - x_0, B \rangle$ is constant, and it is $0$ at $s = 0$. Hence $\langle \alpha(s) - x_0, B \rangle = 0$ for all $s$, which says that $\alpha$ lies in the plane through $x_0$ perpendicular to $B$.

That a plane curve has constant $B$ is Homework 2, Problem 3: the idea is that $T = \alpha'$ and $N = \alpha''/\kappa$ are both parallel to the plane containing the curve, so $B = T \times N$ is a unit normal to that plane. ∎

Proposition 4.4 (Parallel osculating planes). A unit speed curve with $\kappa \ne 0$ lies in a plane if and only if all of its osculating planes are parallel.

Proof. If $\alpha$ lies in a plane, that plane is its osculating plane at every point (by Corollary 4.3, $B$ is constant and perpendicular to it), so all osculating planes coincide. Conversely, if all osculating planes are parallel, then their unit normals $B(s)$ are all $\pm$ the same vector; since $B$ is continuous, it cannot jump between the two, so $B$ is constant, and $\alpha$ lies in a plane by Corollary 4.3. ∎

Proposition 4.5 (Tangent lines through a point). A unit speed curve is a straight line if and only if there is a point $x_0$ such that every tangent line of the curve passes through $x_0$.

Proof. If $\alpha$ is a straight line, any point $x_0$ on it works. Conversely, suppose every tangent line passes through $x_0$. Then $\alpha(s) - x_0 = \lambda(s) T(s)$ for some function $\lambda$, which is differentiable since $\lambda = \langle \alpha - x_0, T \rangle$. If $\kappa \equiv 0$ the curve is a line by Proposition 3.4, so suppose $\kappa(s_0) \ne 0$ at some point; then $\kappa \ne 0$ on an interval around $s_0$. Differentiating on that interval, $$T = \alpha' = \lambda' T + \lambda \kappa N.$$ Since $T$ and $N$ are linearly independent, $\lambda' = 1$ and $\lambda\kappa = 0$, so $\lambda = 0$ on the interval. But then $\lambda' = 0$, a contradiction. So $\kappa \equiv 0$ and $\alpha$ is a line. ∎

Definition 4.6 (General helix). A regular curve $\alpha$ is a (general) helix if there is a fixed unit vector $u$ such that $\langle T, u \rangle$ is constant. The line through the origin in the direction $u$ is the axis of the helix.

Theorem 4.7 (Lancret's theorem). A unit speed curve with $\kappa \ne 0$ is a helix if and only if $\tau/\kappa$ is constant.

Proof. Suppose $\alpha$ is a helix with axis $u$, so $\langle T, u \rangle = \cos\theta$ for a constant angle $\theta$. If $\cos\theta = \pm 1$, then $T = \pm u$ is constant and $\kappa = 0$, which we have excluded; so $\sin\theta \ne 0$. Differentiating gives $0 = \langle T', u \rangle = \kappa\langle N, u \rangle$, so $\langle N, u \rangle = 0$, and by Lemma 4.1 $u = \cos\theta\, T + b\, B$ with $b = \langle u, B \rangle$. Since $|u| = 1$, $b = \pm\sin\theta$; replacing $\theta$ by $-\theta$ if necessary, we may assume $u = \cos\theta\,T + \sin\theta\,B$, where $\theta$ is constant (by continuity, $b$ cannot jump between $\sin\theta$ and $-\sin\theta$). Differentiating once more, $$0 = u' = \cos\theta\,\kappa N - \sin\theta\,\tau N = (\kappa\cos\theta - \tau\sin\theta)\,N.$$ So $\tau/\kappa = \cot\theta$ is constant.

Conversely, suppose $\tau = c\kappa$ for a constant $c$. Choose $\theta \in (0, \pi)$ with $\cot\theta = c$, and set $u = \cos\theta\,T + \sin\theta\,B$. Then $u$ is a unit vector, and $$u' = \cos\theta\,\kappa N - \sin\theta\,\tau N = \kappa\sin\theta\,(\cot\theta - c)\,N = 0,$$ so $u$ is constant. Since $\langle T, u \rangle = \cos\theta$ is constant, $\alpha$ is a helix with axis $u$. ∎

Theorem 5.1 (Existence and uniqueness for ordinary differential equations (Picard)). Let $A(x, t)$ be a function with values in $\mathbb{R}^n$, defined and continuous for $x \in \mathbb{R}^n$ and $t$ in an open interval $I$ containing $t_0$. Suppose that $A$ has bounded partial derivatives with respect to the coordinates of $x$ on each set of the form $\mathbb{R}^n \times J$ with $J$ a closed bounded subinterval of $I$. Then for every $c \in \mathbb{R}^n$ there is a unique differentiable function $x \colon I \to \mathbb{R}^n$ with $$\frac{dx}{dt} = A(x(t), t), \qquad x(t_0) = c.$$

Theorem 5.2 (Fundamental theorem of curves). Let $I$ be an open interval containing $0$, let $\bar\kappa \colon I \to \mathbb{R}$ be a $C^1$ function with $\bar\kappa(s) > 0$ for all $s$, and let $\bar\tau \colon I \to \mathbb{R}$ be continuous. Let $x_0$ be a point of $\mathbb{R}^3$ and let $D, E, F$ be a right-handed orthonormal basis of $\mathbb{R}^3$. Then there is a unique unit speed curve $\alpha \colon I \to \mathbb{R}^3$ of class $C^3$ such that

1. $\alpha(0) = x_0$ and $T(0) = D$, $N(0) = E$, $B(0) = F$;
2. $\kappa(s) = \bar\kappa(s)$ and $\tau(s) = \bar\tau(s)$ for all $s$.

Proof. The idea is to first construct the moving frame, by solving the Frenet-Serret equations as a differential equation, and then to obtain the curve by integrating its unit tangent.

Step 1: the frame. Consider the linear system for three unknown vector functions $u_1, u_2, u_3$: $$u_1' = \bar\kappa\,u_2, \qquad u_2' = -\bar\kappa\,u_1 + \bar\tau\,u_3, \qquad u_3' = -\bar\tau\,u_2,$$ with initial values $u_1(0) = D$, $u_2(0) = E$, $u_3(0) = F$. By Theorem 5.1 it has a unique solution on all of $I$. Write the system as $u_i' = \sum_j a_{ij} u_j$, where $(a_{ij})$ is the skew-symmetric matrix with $a_{12} = \bar\kappa$, $a_{23} = \bar\tau$.

Step 2: the frame is orthonormal. Let $p_{ij} = \langle u_i, u_j \rangle$. Then $$p_{ij}' = \langle u_i', u_j \rangle + \langle u_i, u_j' \rangle = \sum_k a_{ik}\,p_{kj} + \sum_k a_{jk}\,p_{ik},$$ which is again a linear system, for the nine functions $p_{ij}$, with initial values $p_{ij}(0) = \delta_{ij}$ (that is, $1$ if $i = j$ and $0$ otherwise) because $D, E, F$ are orthonormal. The constant functions $p_{ij} = \delta_{ij}$ also solve this system, since then the right side is $a_{ij} + a_{ji} = 0$ by skew-symmetry. By uniqueness, $p_{ij}(s) = \delta_{ij}$ for all $s$: the vectors $u_1(s), u_2(s), u_3(s)$ are orthonormal. Moreover, the triple product $[u_1, u_2, u_3] = \langle u_1 \times u_2, u_3 \rangle$ is then always $\pm 1$, it is continuous, and it equals $[D, E, F] = 1$ at $s = 0$, so it is always $1$: the frame is right-handed, and $u_3 = u_1 \times u_2$.

Step 3: the curve. Define $\alpha(s) = x_0 + \int_0^s u_1(\sigma)\,d\sigma$. Then $\alpha(0) = x_0$ and $\alpha' = u_1$, which is a unit vector, so $\alpha$ is a regular unit speed curve with $T = u_1$. Since $u_1' = \bar\kappa u_2$ and $\bar\kappa$ is $C^1$ and the $u_j$ are differentiable, $\alpha'' = u_1'$ is differentiable, and $\alpha''' = \bar\kappa' u_2 + \bar\kappa(-\bar\kappa u_1 + \bar\tau u_3)$ is continuous. So $\alpha$ is $C^3$. (This is where we used that $\bar\kappa$ is $C^1$.)

Step 4: the curvature and torsion. We have $T' = u_1' = \bar\kappa u_2$ with $\bar\kappa > 0$ and $|u_2| = 1$, so $\kappa = |T'| = \bar\kappa$ and $N = T'/\kappa = u_2$. Then $B = T \times N = u_1 \times u_2 = u_3$, and $B' = u_3' = -\bar\tau u_2 = -\bar\tau N$, so $\tau = \bar\tau$. The initial conditions hold by construction.

Step 5: uniqueness. If $\beta$ is another curve with the required properties, then by Theorem 4.2 its Frenet frame satisfies the same linear system with the same initial values as $(u_1, u_2, u_3)$, so by uniqueness in Theorem 5.1 its frame equals $(u_1, u_2, u_3)$. In particular $\beta' = u_1 = \alpha'$, and $\beta(0) = \alpha(0)$, so $\beta = \alpha$. ∎

Corollary 5.3 (Curves are determined by curvature and torsion). Two unit speed curves defined on the same interval, with the same curvature $\kappa > 0$ and the same torsion, differ by a rigid motion.

Proof. Let $\alpha$ and $\beta$ be such curves. Choose the rigid motion $M(x) = Rx + b$ that takes $\beta(0)$ to $\alpha(0)$ and the frame $(T_\beta(0), N_\beta(0), B_\beta(0))$ to $(T_\alpha(0), N_\alpha(0), B_\alpha(0))$: $R$ is the orthogonal matrix taking one right-handed orthonormal basis to the other, which has determinant $1$, and $b = \alpha(0) - R\beta(0)$. Then $M \circ \beta$ is a unit speed curve with the same curvature, torsion, starting point and starting frame as $\alpha$, so $M \circ \beta = \alpha$ by the uniqueness part of Theorem 5.2. ∎

Corollary 5.4 (Constant curvature and torsion). A unit speed curve with constant curvature $\kappa > 0$ and constant torsion $\tau$ is, up to a rigid motion, the circular helix $$\alpha(s) = (r\cos\omega s,\ r\sin\omega s,\ h\omega s), \qquad r = \frac{\kappa}{\kappa^2 + \tau^2}, \quad h = \frac{\tau}{\kappa^2 + \tau^2}, \quad \omega = \frac{1}{\sqrt{r^2 + h^2}}.$$ In particular, if $\tau = 0$, it is a circle of radius $1/\kappa$.

Proof. With $r$ and $h$ as given, $r^2 + h^2 = 1/(\kappa^2 + \tau^2)$, so $\omega^2 = \kappa^2 + \tau^2$. By Example 3.10, the circular helix has curvature $\omega^2 r = \kappa$ and torsion $\omega^2 h = \tau$. By Corollary 5.3, every unit speed curve with these constant curvature and torsion differs from it by a rigid motion. ∎

Example 5.5 (A curve with curvature proportional to arc length). Take $\kappa(s) = s$ for $s > 0$ and $\tau = 0$. Then $\theta(s) = s^2/2$, and the plane curve is $$\alpha(s) = \Big(\int_0^s \cos\frac{\sigma^2}{2}\,d\sigma,\ \int_0^s \sin\frac{\sigma^2}{2}\,d\sigma,\ 0\Big).$$ These integrals (Fresnel integrals) cannot be computed in elementary terms, but the shape is easy to describe: the curvature grows linearly, so the curve winds around in ever tighter circles, spiraling in toward a limit point. This curve is called the clothoid or Euler spiral, and is used to design highway and railway curves, because a car driving along it turns its steering wheel at a constant rate.

Proposition 6.1 (Frenet-Serret apparatus of a non-unit speed curve). Let $\beta(t)$ be a regular curve with speed $v = |\dot\beta|$. Then:

1. $T = \dot\beta / v$;
2. $\dot\beta \times \ddot\beta = v^3\kappa B$, so $\kappa = \dfrac{|\dot\beta \times \ddot\beta|}{v^3}$;
3. if $\kappa \ne 0$, then $B = \dfrac{\dot\beta \times \ddot\beta}{|\dot\beta \times \ddot\beta|}$ and $N = B \times T$;
4. if $\kappa \ne 0$, then $\tau = \dfrac{[\dot\beta, \ddot\beta, \dddot\beta]}{|\dot\beta \times \ddot\beta|^2}$, where $[a, b, c] = \langle a \times b, c \rangle$ is the triple product.

Proof. We differentiate $\beta$ three times with respect to $t$, each time using $\dot X = vX'$ and the Frenet-Serret equations.

First, $\dot\beta = \dot s\,\alpha' = v\,T$, which gives (1).

Second, differentiating $\dot\beta = vT$, $$\ddot\beta = \dot v\,T + v\,\dot T = \dot v\,T + v^2\,T' = \dot v\,T + v^2\kappa\,N.$$ (The first term is the acceleration along the curve, due to speeding up; the second is the acceleration perpendicular to it, due to turning.) Since $T \times T = 0$ and $T \times N = B$, $$\dot\beta \times \ddot\beta = vT \times (\dot v\,T + v^2\kappa N) = v^3\kappa\,B.$$ Taking lengths gives $\kappa = |\dot\beta \times \ddot\beta|/v^3$, which is (2). If $\kappa \ne 0$, dividing $\dot\beta \times \ddot\beta$ by its length gives $B$, and $N = B \times T$ because $(T, N, B)$ is a right-handed orthonormal basis (just as $e_2 = e_3 \times e_1$). This is (3).

Third, differentiating $\ddot\beta = \dot v\,T + v^2\kappa\,N$ and using $\dot T = v\kappa N$ and $\dot N = vN' = v(-\kappa T + \tau B)$, $$\dddot\beta = \ddot v\,T + \dot v v\kappa\,N + \frac{d(v^2\kappa)}{dt}\,N + v^3\kappa(-\kappa T + \tau B) = (\ddot v - v^3\kappa^2)\,T + \big(\dot v v\kappa + \tfrac{d}{dt}(v^2\kappa)\big)\,N + v^3\kappa\tau\,B.$$ We only need the $B$ component: since $\dot\beta \times \ddot\beta = v^3\kappa B$, $$[\dot\beta, \ddot\beta, \dddot\beta] = \langle v^3\kappa B, \dddot\beta \rangle = v^3\kappa \cdot v^3\kappa\tau = |\dot\beta \times \ddot\beta|^2\,\tau,$$ which gives (4). ∎

Corollary 6.2 (A test for plane curves). A regular curve $\beta$ with $\dot\beta \times \ddot\beta \ne 0$ lies in a plane if and only if $[\dot\beta, \ddot\beta, \dddot\beta] \equiv 0$.

Proof. The condition $\dot\beta \times \ddot\beta \ne 0$ means $\kappa \ne 0$, and then $[\dot\beta, \ddot\beta, \dddot\beta] \equiv 0$ means $\tau \equiv 0$, which by Corollary 4.3 means the curve lies in a plane. ∎

Example 6.3 (The helix again). For $\beta(t) = (r\cos t, r\sin t, ht)$, $$\dot\beta = (-r\sin t, r\cos t, h), \qquad \ddot\beta = (-r\cos t, -r\sin t, 0), \qquad \dddot\beta = (r\sin t, -r\cos t, 0).$$ Then $v = \sqrt{r^2 + h^2}$ and $\dot\beta \times \ddot\beta = (rh\sin t,\ -rh\cos t,\ r^2)$, with length $r\sqrt{r^2 + h^2}$. So $$\kappa = \frac{r\sqrt{r^2 + h^2}}{(r^2 + h^2)^{3/2}} = \frac{r}{r^2 + h^2}, \qquad \tau = \frac{\langle (rh\sin t, -rh\cos t, r^2), (r\sin t, -r\cos t, 0) \rangle}{r^2(r^2 + h^2)} = \frac{r^2 h}{r^2(r^2 + h^2)} = \frac{h}{r^2 + h^2},$$ in agreement with Example 3.10, but without reparametrizing.

Example 6.4 (A twisted cubic). Let $\beta(t) = (t,\ t^2/2,\ t^3/3)$. Then $$\dot\beta = (1, t, t^2), \qquad \ddot\beta = (0, 1, 2t), \qquad \dddot\beta = (0, 0, 2),$$ so $v = \sqrt{1 + t^2 + t^4}$ and $$\dot\beta \times \ddot\beta = (t \cdot 2t - t^2 \cdot 1,\ t^2 \cdot 0 - 1 \cdot 2t,\ 1 \cdot 1 - t \cdot 0) = (t^2,\ -2t,\ 1).$$ Hence $$\kappa = \frac{\sqrt{t^4 + 4t^2 + 1}}{(1 + t^2 + t^4)^{3/2}}, \qquad [\dot\beta, \ddot\beta, \dddot\beta] = \langle (t^2, -2t, 1), (0, 0, 2) \rangle = 2, \qquad \tau = \frac{2}{t^4 + 4t^2 + 1}.$$ The unit vectors are $$T = \frac{(1, t, t^2)}{\sqrt{1 + t^2 + t^4}}, \qquad B = \frac{(t^2, -2t, 1)}{\sqrt{t^4 + 4t^2 + 1}},$$ and $N = B \times T$. At $t = 0$ this gives $\kappa = 1$, $\tau = 2$, $T = (1, 0, 0)$, $N = (0, 1, 0)$ and $B = (0, 0, 1)$. The torsion is never zero, so this curve does not lie in any plane, even though each pair of its coordinates satisfies a simple algebraic relation.

Example 6.5 (A curve that is planar in disguise). Let $\beta(t) = (t,\ t^2 + t,\ t^2 - t)$. Then $\dot\beta = (1, 2t + 1, 2t - 1)$, $\ddot\beta = (0, 2, 2)$, and $\dddot\beta = 0$, so $[\dot\beta, \ddot\beta, \dddot\beta] = 0$, while $\dot\beta \times \ddot\beta = (4,\ -2,\ 2) \ne 0$. By Corollary 6.2 the curve lies in a plane; indeed $B$ is parallel to $(2, -1, 1)$ and $\langle \beta(t), (2, -1, 1) \rangle = 2t - t^2 - t + t^2 - t = 0$, so $\beta$ lies in the plane $2x - y + z = 0$.

Proposition 6.6 (Frenet-Serret equations with respect to $t$). For a regular curve $\beta(t)$ with speed $v$ and $\kappa \ne 0$, $$\dot T = v\kappa\,N, \qquad \dot N = -v\kappa\,T + v\tau\,B, \qquad \dot B = -v\tau\,N.$$

Proof. Each equation is $\dot X = vX'$ combined with the corresponding equation of Theorem 4.2. ∎

Definition 7.1 (Line integral). Let $\alpha(t) = (x(t), y(t))$, $a \le t \le b$, be a $C^1$ curve segment with image $C$, and let $f$, $g$ be continuous functions of $(x, y)$. The line integral of $f\,dx + g\,dy$ along $C$ is $$\int_C f\,dx + g\,dy = \int_a^b \Big[f(x(t), y(t))\,\frac{dx}{dt} + g(x(t), y(t))\,\frac{dy}{dt}\Big]\,dt.$$

Example 7.2 (Two line integrals over the circle). Let $C$ be the unit circle traversed counterclockwise, $\alpha(\theta) = (\cos\theta, \sin\theta)$, $0 \le \theta \le 2\pi$. Then $dx/d\theta = -\sin\theta$ and $dy/d\theta = \cos\theta$, so $$\int_C y\,dx + x\,dy = \int_0^{2\pi} (-\sin^2\theta + \cos^2\theta)\,d\theta = \int_0^{2\pi} \cos 2\theta\,d\theta = 0,$$ $$\int_C x\,dy - y\,dx = \int_0^{2\pi} (\cos^2\theta + \sin^2\theta)\,d\theta = 2\pi.$$

Theorem 7.3 (Green's theorem). Let $C$ be a positively oriented simple closed curve made up of finitely many $C^1$ curve segments, and let $R$ be the region inside $C$. If $f$ and $g$ have continuous partial derivatives on a neighborhood of $R$ together with $C$, then $$\int_C f\,dx + g\,dy = \iint_R \Big(\frac{\partial g}{\partial x} - \frac{\partial f}{\partial y}\Big)\,dx\,dy.$$

Corollary 7.4 (Area as a line integral). If $C$ is a positively oriented simple closed curve made up of finitely many $C^1$ segments, enclosing a region of area $A$, then $$A = \frac{1}{2}\int_C x\,dy - y\,dx.$$

Proof. Apply Green's theorem with $f = -y/2$ and $g = x/2$: then $\partial g/\partial x - \partial f/\partial y = \tfrac12 + \tfrac12 = 1$, so the right side is $\iint_R 1\,dx\,dy = A$. ∎

Definition 7.5 (Normal and signed curvature of a plane curve). Let $\alpha(s)$ be a unit speed plane curve with unit tangent $T(s)$. Its normal $n(s)$ is the unit vector obtained by rotating $T(s)$ counterclockwise by $90^\circ$: if $T = (a, b)$ then $n = (-b, a)$. The signed curvature (or plane curvature) $k(s)$ is defined by $$T'(s) = k(s)\,n(s).$$

Lemma 7.6 (Plane Frenet equations). For a unit speed plane curve, $T' = k\,n$ and $n' = -k\,T$. If $\alpha(s) = (x(s), y(s))$, then $T = (x', y')$, $n = (-y', x')$, and $k = x'y'' - x''y'$.

Proof. The first equation is the definition of $k$. For the second, $n'$ is perpendicular to $n$, so $n' = \langle n', T \rangle T$, and $\langle n', T \rangle = -\langle n, T' \rangle = -k$. The formulas for $T$ and $n$ are the definitions. Finally $T' = (x'', y'')$, so $k = \langle T', n \rangle = -x''y' + y''x'$. ∎

Example 7.7 (Circles in both directions). The circle $\alpha(s) = (r\cos\frac{s}{r}, r\sin\frac{s}{r})$, traversed counterclockwise, has $T = (-\sin\frac{s}{r}, \cos\frac{s}{r})$, $n = (-\cos\frac{s}{r}, -\sin\frac{s}{r})$ pointing toward the center, and $T' = \frac{1}{r}\,n$, so $k = 1/r$. The same circle traversed clockwise, $\beta(s) = (r\cos\frac{s}{r}, -r\sin\frac{s}{r})$, has $T = (-\sin\frac{s}{r}, -\cos\frac{s}{r})$, $n = (\cos\frac{s}{r}, -\sin\frac{s}{r})$ pointing away from the center, and $k = -1/r$. So the sign of $k$ depends on the direction of travel, while $\kappa = 1/r$ does not.

Proposition 7.8 (Angle functions). Let $\alpha \colon [0, L] \to \mathbb{R}^2$ be a unit speed $C^2$ plane curve, and let $\theta_0$ be an angle with $T(0) = (\cos\theta_0, \sin\theta_0)$. Then the function $$\theta(s) = \theta_0 + \int_0^s k(\sigma)\,d\sigma$$ satisfies $T(s) = (\cos\theta(s), \sin\theta(s))$ for all $s$. It is the unique continuous function with this property and with $\theta(0) = \theta_0$, and $\theta'(s) = k(s)$.

Proof. Let $u(s) = (\cos\theta(s), \sin\theta(s))$ and let $Ju = (-\sin\theta, \cos\theta)$ be $u$ rotated by $90^\circ$; note that $n = JT$ and $\langle Jv, w \rangle = -\langle v, Jw \rangle$ for all vectors $v$, $w$. Then $u' = \theta' Ju = k\,Ju$, and $$\frac{d}{ds}\langle T, u \rangle = \langle k\,JT, u \rangle + \langle T, k\,Ju \rangle = k\big(\langle JT, u \rangle + \langle T, Ju \rangle\big) = 0.$$ Since $\langle T(0), u(0) \rangle = 1$, we get $\langle T, u \rangle \equiv 1$, and two unit vectors with inner product $1$ are equal: $T = u$.

For uniqueness, if $\tilde\theta$ is another continuous function with the same property, then $(\tilde\theta - \theta)/2\pi$ is a continuous function with integer values, hence constant, and it is $0$ at $s = 0$. ∎

Definition 8.1 (Rotation index). The rotation index of a unit speed closed plane curve $\alpha$ with period $L$ is the integer $$i_\alpha = \frac{\theta(L) - \theta(0)}{2\pi} = \frac{1}{2\pi}\int_0^L k(s)\,ds.$$

Example 8.2 (Some rotation indices).

1. A circle traversed once counterclockwise has $k = 1/r$ and $L = 2\pi r$, so $i_\alpha = \frac{1}{2\pi}\cdot\frac{1}{r}\cdot 2\pi r = 1$. Traversed clockwise, $k = -1/r$ and $i_\alpha = -1$.
2. A figure eight, traced so that one loop is traversed counterclockwise and the other clockwise, has rotation index $0$: the tangent turns once counterclockwise around one loop and once clockwise around the other.
3. A curve that goes around a large loop counterclockwise and, along the way, makes a small extra loop also counterclockwise (like a limaçon, the curve $r = \frac12 + \cos t$ in polar coordinates) has rotation index $2$.

Theorem 8.3 (Hopf's rotation index theorem (Umlaufsatz)). The rotation index of a simple closed plane curve is $+1$ or $-1$. It is $+1$ if the curve is positively oriented (traversed counterclockwise) and $-1$ otherwise.

Lemma 8.4 (Continuous angle functions on a triangle). Let $\Delta$ be a closed triangle in the plane and let $e \colon \Delta \to S^1$ be a continuous function to the unit circle $S^1 = \{(a, b) : a^2 + b^2 = 1\}$. Fix a point $p_0 \in \Delta$ and an angle $\theta_0$ with $e(p_0) = (\cos\theta_0, \sin\theta_0)$. Then there is a unique continuous function $\theta \colon \Delta \to \mathbb{R}$ with $\theta(p_0) = \theta_0$ and $e(p) = (\cos\theta(p), \sin\theta(p))$ for all $p \in \Delta$.

Proof sketch. Since $\Delta$ is compact, $e$ is uniformly continuous, so there is $\delta > 0$ such that $e(p)$ and $e(q)$ are not opposite (in fact are within an angle less than $\pi/2$ of each other) whenever $|p - q| < \delta$. For each $p \in \Delta$, the segment from $p_0$ to $p$ lies in $\Delta$ (a triangle is convex), and along this segment there is exactly one continuous angle function starting at $\theta_0$: cut the segment into subsegments of length less than $\delta$, on each of which $e$ varies by less than $\pi/2$, so that on each subsegment the angle can be chosen continuously, and match the choices at the ends of consecutive subsegments. Define $\theta(p)$ to be its value at the end $p$. To see that $\theta$ is continuous, compare the segments from $p_0$ to two nearby points $p$ and $q$: corresponding points on the two segments are less than $\delta$ apart, so the two angle functions stay within $\pi/2$ of each other all the way along, and hence $\theta(p)$ is the value of the angle of $e(p)$ that is closest to $\theta(q)$. Uniqueness follows as before: the difference of two such functions is a continuous function with values in $2\pi\mathbb{Z}$ on a connected set, so it is constant. ∎

Proof of Theorem 8.3. Let $\alpha$ be a simple closed unit speed curve with period $L$. Rotating and translating the plane does not change the rotation index, and neither does changing the starting point of the parametrization. So we may assume that the starting point $\alpha(0)$ is a lowest point of the curve, that is, the $y$-coordinate of $\alpha(s)$ is smallest at $s = 0$ (a minimum exists because the curve is the continuous image of the closed interval $[0, L]$). Then $y'(0) = 0$, so $T(0) = (\pm 1, 0)$. Reversing the direction of travel changes the sign of the rotation index, so we may assume $T(0) = (1, 0)$, and we must show that $i_\alpha = 1$.

Consider the triangle $\Delta = \{(u, v) : 0 \le u \le v \le L\}$, with corners $A = (0, 0)$, $B = (0, L)$ and $C = (L, L)$. Define $e \colon \Delta \to S^1$ by $$e(u, v) = \frac{\alpha(v) - \alpha(u)}{|\alpha(v) - \alpha(u)|} \quad\text{if } u < v \text{ and } (u, v) \ne (0, L), \qquad e(u, u) = T(u), \qquad e(0, L) = -T(0).$$ The denominator is nonzero because $\alpha$ is simple. The function $e$ is continuous: away from the diagonal and the corner $B$ this is clear; near a diagonal point $(s, s)$, the chord direction from $\alpha(u)$ to $\alpha(v)$ approaches the tangent direction $T(s)$ by the definition of the derivative; and near $B$, where $u$ is close to $0$ and $v$ close to $L$, the point $\alpha(v)$ is close to $\alpha(L) = \alpha(0)$ but just behind it, so the chord from $\alpha(u)$ to $\alpha(v)$ points approximately in the direction $-T(0)$.

By Lemma 8.4 there is a continuous angle function $\theta(u, v)$ for $e$ with $\theta(0, 0) = 0$ (recall $e(0, 0) = T(0) = (1, 0)$). Along the diagonal, $s \mapsto \theta(s, s)$ is a continuous angle function for $T(s)$, so by the definition of the rotation index, $$2\pi\,i_\alpha = \theta(L, L) - \theta(0, 0) = \theta(C) - \theta(A).$$ We compute this difference the other way around the triangle, from $A$ to $B$ to $C$.

From $A$ to $B$: for $0 < v < L$, $e(0, v)$ is the direction from the lowest point $\alpha(0)$ to $\alpha(v)$, which has $y$-component $\ge 0$. So along the side $AB$, the vector $e$ stays in the closed upper half of the circle. A continuous angle function starting at $\theta(A) = 0$ whose values are always angles of vectors in the upper half-circle must stay in the interval $[0, \pi]$ (to leave it, it would have to pass through an angle of a vector in the open lower half-circle). At the end, $e(B) = -T(0) = (-1, 0)$, whose angles in $[0, \pi]$ are only $\pi$. So $\theta(B) = \pi$.

From $B$ to $C$: for $0 < u < L$, $e(u, L)$ is the direction from $\alpha(u)$ to $\alpha(L) = \alpha(0)$, the lowest point, so it has $y$-component $\le 0$. By the same reasoning, along the side $BC$ the angle stays in $[\pi, 2\pi]$, and at the end $e(C) = T(L) = T(0) = (1, 0)$, so $\theta(C) = 2\pi$.

Hence $2\pi\,i_\alpha = \theta(C) - \theta(A) = 2\pi$, and $i_\alpha = 1$.

Finally, in our normalization, with $\alpha(0)$ a lowest point and $T(0) = (1, 0)$, the curve leaves its lowest point moving to the right; the inside of the curve lies above the lowest point, hence on the left, so the curve is positively oriented. This shows that positively oriented simple closed curves have rotation index $1$, and so negatively oriented ones have rotation index $-1$. ∎

Definition 8.5 (Piecewise regular curve). A piecewise regular curve is a continuous function $\alpha \colon [a, b] \to \mathbb{R}^2$ together with points $a = s_0 < s_1 < \dots < s_m = b$ such that the restriction of $\alpha$ to each interval $[s_{j-1}, s_j]$ is a regular curve segment.

Definition 8.6 (Rotation index of a piecewise regular closed curve). For a closed piecewise regular curve, let $\delta\theta_j = \int_{s_j}^{s_{j+1}} k\,ds$ be the total turning of the tangent along the $j$-th smooth piece, and let $\Delta\theta_j$ be the exterior angles at the corners. The rotation index is $$i_\alpha = \frac{1}{2\pi}\Big(\sum_j \delta\theta_j + \sum_j \Delta\theta_j\Big).$$

Example 8.7 (A triangle). Let $\alpha$ be the triangle with vertices $(0, 0)$, $(2, 0)$, $(1, 1)$, traversed in that order (counterclockwise). The sides are straight, so every $\delta\theta_j = 0$. The exterior angles at the three vertices are $\pi$ minus the interior angles, namely $\frac{3\pi}{4}$, $\frac{3\pi}{4}$ and $\frac{\pi}{2}$, whose sum is $2\pi$. So the rotation index is $1$. More generally, the exterior angles of any convex polygon add up to $2\pi$; this is equivalent to the familiar fact that the interior angles of an $m$-gon add up to $(m - 2)\pi$.

Proposition 8.8 (Hopf's theorem for piecewise regular curves). A simple closed piecewise regular plane curve has rotation index $\pm 1$, with $+1$ if it is positively oriented.

Theorem 9.1 (Isoperimetric inequality). Let $\alpha$ be a simple closed plane curve of length $L$, made up of finitely many $C^1$ curve segments, and let $A$ be the area of the region it bounds. Then $$L^2 \ge 4\pi A,$$ and equality holds if and only if $\alpha$ is a circle.

Lemma 9.2 (Arithmetic and geometric means). For real numbers $a, b \ge 0$, $\sqrt{ab} \le \frac12(a + b)$, with equality if and only if $a = b$.

Lemma 9.3 (Cauchy-Schwarz inequality). For vectors $v$ and $w$, $|\langle v, w \rangle| \le |v|\,|w|$, with equality if and only if one of $v$, $w$ is a multiple of the other.

Proof. If $w = 0$ there is nothing to prove. Otherwise, for every real $\lambda$, $0 \le |v - \lambda w|^2 = |v|^2 - 2\lambda\langle v, w \rangle + \lambda^2|w|^2$. Taking $\lambda = \langle v, w \rangle/|w|^2$ gives $0 \le |v|^2 - \langle v, w \rangle^2/|w|^2$, which is the inequality. Equality means $|v - \lambda w| = 0$ for this $\lambda$, that is, $v = \lambda w$. ∎

Proof of Theorem 9.1. Parametrize $\alpha$ by arc length, $\alpha(s) = (x(s), y(s))$ for $0 \le s \le L$, positively oriented; length and area do not depend on the parametrization or direction, so this is no loss of generality.

Step 1: the comparison circle. The function $x(s)$ is continuous on the closed interval $[0, L]$, so it attains a maximum and a minimum. The vertical lines through these extreme values are tangent to the curve (at a maximum or minimum of $x$ we have $x' = 0$, so the tangent is vertical), and the whole curve lies in the strip between them. Translate the plane horizontally so that the strip is centered on the $y$-axis: the lines are $x = r$ and $x = -r$, where $2r > 0$ is the width of the strip. Change the starting point of the arc length so that $x(0) = r$, and let $s_1$ be a value with $x(s_1) = -r$.

Let $D$ be the disk of radius $r$ centered at the origin; it fits in the same strip, touching both lines. Define the function $$w(s) = \begin{cases} \sqrt{r^2 - x(s)^2} & 0 \le s \le s_1, \\ -\sqrt{r^2 - x(s)^2} & s_1 \le s \le L. \end{cases}$$ Then $(x(s), w(s))$ always lies on the circle of radius $r$: for $0 \le s \le s_1$ on its upper half, moving from $(r, 0)$ to $(-r, 0)$, and for $s_1 \le s \le L$ on its lower half, moving back to $(r, 0)$. (It may move back and forth along the way, and it need not be a regular curve, but this does not matter.)

Step 2: the area of the disk. We compute $-\int y\,dx$ along this path on the circle. On the first part, $w = \sqrt{r^2 - x^2}$ is a function of $x$ alone, so by substituting $x = x(s)$, $$-\int_0^{s_1} w(s)\,x'(s)\,ds = -\int_{r}^{-r} \sqrt{r^2 - x^2}\,dx = \int_{-r}^{r} \sqrt{r^2 - x^2}\,dx = \frac{\pi r^2}{2},$$ the area of a half-disk. On the second part, similarly, $$-\int_{s_1}^{L} w(s)\,x'(s)\,ds = \int_{-r}^{r} \sqrt{r^2 - x^2}\,dx = \frac{\pi r^2}{2}.$$ (Any back-and-forth motion cancels out, because the integrand depends only on $x$.) So $$\pi r^2 = -\int_0^L w\,x'\,ds.$$

Step 3: the main estimate. By the area formula for $\alpha$, $A = \int_C x\,dy = \int_0^L x\,y'\,ds$. Adding, $$A + \pi r^2 = \int_0^L \big(x\,y' - w\,x'\big)\,ds = \int_0^L \big\langle (x, w),\ (y', -x') \big\rangle\,ds.$$ By the Cauchy-Schwarz inequality, the integrand is at most $|(x, w)|\,|(y', -x')|$. Here $|(x, w)| = r$ because $(x, w)$ lies on the circle, and $|(y', -x')| = |\alpha'| = 1$ because $\alpha$ has unit speed. So $$A + \pi r^2 \le \int_0^L r\,ds = rL.$$

Step 4: conclusion. By Lemma 9.2 and Step 3, $$\sqrt{A \cdot \pi r^2} \le \frac{A + \pi r^2}{2} \le \frac{rL}{2}.$$ Squaring and dividing by $\pi r^2$ gives $A \le L^2/4\pi$, that is, $L^2 \ge 4\pi A$.

Step 5: the equality case. Suppose $L^2 = 4\pi A$. Then all the inequalities in Step 4 are equalities. Equality in the arithmetic-geometric mean inequality gives $A = \pi r^2$, and then equality in $A + \pi r^2 \le rL$ gives $2\pi r^2 = rL$, so $$r = \frac{L}{2\pi}.$$ The width $2r$ of the strip was measured in the horizontal direction, but we could have used parallel tangent lines in any direction. So the curve has the same width $L/\pi$ in every direction.

Equality in Step 3 also means that $\int_0^L \big(r - \langle (x, w), (y', -x') \rangle\big)\,ds = 0$, where the integrand is continuous (on each smooth piece) and nonnegative, so it is zero everywhere. Thus we have equality in the Cauchy-Schwarz inequality at every $s$, with a nonnegative multiple: $(x, w) = r\,(y', -x')$. In particular $$x = r\,y'.$$ Now repeat the argument using two horizontal tangent lines instead of vertical ones, that is, after rotating the coordinate system by $90^\circ$ (which preserves the orientation). If the horizontal strip is centered on the line $y = d$, the rotated coordinates are $\tilde x = y - d$, $\tilde y = -x$, and the same reasoning gives $\tilde x = r\,\tilde y'$ with the same $r = L/2\pi$, that is, $$y - d = -r\,x'.$$ Therefore $$x^2 + (y - d)^2 = r^2\big((y')^2 + (x')^2\big) = r^2,$$ so the curve lies on the circle of radius $r$ centered at $(0, d)$. Being a simple closed curve, it is that circle. Conversely, we saw that circles give equality. ∎

Definition 10.1 (Open set). A subset $U$ of $\mathbb{R}^2$ is open if for every point $(a, b) \in U$ there is a number $\varepsilon > 0$ such that every point $(u^1, u^2)$ with $(u^1 - a)^2 + (u^2 - b)^2 < \varepsilon^2$ is also in $U$.

Example 10.2 (Open and non-open sets).

1. The open disk $\{(x, y) : x^2 + y^2 < 1\}$ is open: for a point at distance $d < 1$ from the origin, take $\varepsilon = 1 - d$.
2. The closed disk $\{x^2 + y^2 \le 1\}$ is not open: no disk around the boundary point $(1, 0)$ stays inside it.
3. The outside of the circle, $\{x^2 + y^2 > 1\}$, and the open rectangle $\{1 < x < 2,\ 3 < y < 5\}$ are open.
4. The half-plane $\{x \ge 0\}$ is not open, but $\{x > 0\}$ is.
5. The whole plane $\mathbb{R}^2$ is open, and so is any union of open sets.

Definition 10.3 (Coordinate patch). A $C^k$ coordinate patch, or simple surface, is a one-to-one $C^k$ function $x \colon U \to \mathbb{R}^3$, $k \ge 1$, where $U$ is an open subset of $\mathbb{R}^2$ with coordinates $(u^1, u^2)$, such that $$\frac{\partial x}{\partial u^1} \times \frac{\partial x}{\partial u^2} \ne 0 \quad\text{at every point of } U.$$

Definition 10.4 (Tangent plane and unit normal). Let $x \colon U \to \mathbb{R}^3$ be a coordinate patch and $(a, b) \in U$. The tangent plane to $x$ at $P = x(a, b)$ is the plane through $P$ spanned by $x_1(a, b)$ and $x_2(a, b)$, that is, the plane through $P$ perpendicular to $x_1 \times x_2$. The unit normal at $P$ is $$n(a, b) = \frac{x_1 \times x_2}{|x_1 \times x_2|}.$$

Example 10.5 (A plane). For vectors $p$, $v$, $w$ with $v$ and $w$ linearly independent, $x(u^1, u^2) = p + u^1 v + u^2 w$, defined on all of $\mathbb{R}^2$, is a coordinate patch. Here $x_1 = v$ and $x_2 = w$ are constant, $x_1 \times x_2 = v \times w \ne 0$, and the image is the plane through $p$ spanned by $v$ and $w$, which is its own tangent plane at every point.

Example 10.6 (Graphs (Monge patches)). Let $f(u^1, u^2)$ be a $C^k$ function on an open set $U$ and let $$x(u^1, u^2) = (u^1,\ u^2,\ f(u^1, u^2)).$$ This is one-to-one, because the first two coordinates of $x(u^1, u^2)$ are $u^1$ and $u^2$. Writing $f_1 = \partial f/\partial u^1$ and $f_2 = \partial f/\partial u^2$, $$x_1 = (1, 0, f_1), \qquad x_2 = (0, 1, f_2), \qquad x_1 \times x_2 = (-f_1, -f_2, 1) \ne 0.$$ So every graph of a $C^k$ function is a coordinate patch, with unit normal $n = (-f_1, -f_2, 1)/\sqrt{1 + f_1^2 + f_2^2}$. Such a patch is called a Monge patch. The tangent plane at $(a, b, f(a, b))$ is $z - f(a, b) = f_1(a, b)(x - a) + f_2(a, b)(y - b)$, the tangent plane to a graph from multivariable calculus.

Example 10.7 (Upper hemisphere). On the open unit disk $U = \{(u^1)^2 + (u^2)^2 < 1\}$, let $f = \sqrt{1 - (u^1)^2 - (u^2)^2}$. The Monge patch $x(u^1, u^2) = (u^1, u^2, \sqrt{1 - (u^1)^2 - (u^2)^2})$ has image the upper half of the unit sphere $S^2$ (without the equator). We need $U$ open: on the closed disk, $f$ is not differentiable at the boundary. Here $$x_1 = \Big(1, 0, \frac{-u^1}{f}\Big), \qquad x_2 = \Big(0, 1, \frac{-u^2}{f}\Big), \qquad n = (u^1,\ u^2,\ f) = x(u^1, u^2),$$ so the unit normal at a point of the sphere is the point itself, the outward radial direction, as we expect. At $(u^1, u^2) = (\tfrac12, \tfrac12)$, the point is $P = (\tfrac12, \tfrac12, \tfrac{\sqrt2}{2})$ and the tangent plane is $\langle X - P, P \rangle = 0$, that is, $\frac12(x - \frac12) + \frac12(y - \frac12) + \frac{\sqrt2}{2}(z - \frac{\sqrt2}{2}) = 0$.

Example 10.8 (Spherical coordinates). Let $U = \{(\theta, \phi) : -\pi < \theta < \pi,\ -\pi/2 < \phi < \pi/2\}$ and $$x(\theta, \phi) = (\cos\theta\cos\phi,\ \sin\theta\cos\phi,\ \sin\phi).$$ Here $\theta$ is longitude and $\phi$ is latitude. The $\theta$-curves are circles of latitude (horizontal circles, which are great circles only for $\phi = 0$), and the $\phi$-curves are meridians (half great circles from pole to pole). We compute $$x_\theta = (-\sin\theta\cos\phi,\ \cos\theta\cos\phi,\ 0), \qquad x_\phi = (-\cos\theta\sin\phi,\ -\sin\theta\sin\phi,\ \cos\phi),$$ $$x_\theta \times x_\phi = \cos\phi\,(\cos\theta\cos\phi,\ \sin\theta\cos\phi,\ \sin\phi) = \cos\phi\;x(\theta, \phi).$$ Since $\cos\phi > 0$ on $U$, this is never zero, and $n = x(\theta, \phi)$ is again the outward radial vector. The patch is one-to-one on $U$ (latitude and longitude are determined by the point when $-\pi < \theta < \pi$), and its image is the sphere minus the two poles and minus the $180^\circ$ meridian $\{(x, 0, z) : x \le 0\}$. We had to remove these: at the poles $\phi = \pm\pi/2$ we would have $x_\theta = 0$, and allowing $\theta = \pm\pi$ would make the patch not one-to-one.

Example 10.9 (Cylinder). Let $x(\theta, z) = (r\cos\theta,\ r\sin\theta,\ z)$ for $-\pi < \theta < \pi$ and $z \in \mathbb{R}$, with $r > 0$. Then $x_\theta = (-r\sin\theta, r\cos\theta, 0)$, $x_z = (0, 0, 1)$, $x_\theta \times x_z = r(\cos\theta, \sin\theta, 0) \ne 0$, and $n = (\cos\theta, \sin\theta, 0)$ points horizontally outward. The image is the cylinder $x^2 + y^2 = r^2$ minus the line $\{(-r, 0, z)\}$.

Example 10.10 (A smooth map that is not a patch). The map $x(u^1, u^2) = ((u^1)^3, (u^2)^3, u^1 u^2)$ on $\mathbb{R}^2$ is $C^\infty$ and one-to-one (because $t \mapsto t^3$ is one-to-one), but $x_1 = (3(u^1)^2, 0, u^2)$ and $x_2 = (0, 3(u^2)^2, u^1)$ give $x_1 \times x_2 = (-3(u^2)^3, -3(u^1)^3, 9(u^1 u^2)^2)$, which vanishes at the origin. So it is not a coordinate patch on all of $\mathbb{R}^2$. Restricted to the open set $\{u^1 > 0, u^2 > 0\}$, it is.

Example 10.11 (Surfaces of revolution). Let $\alpha(t) = (r(t), z(t))$ be a regular, one-to-one plane curve with $r(t) > 0$, and rotate it about the $z$-axis. The resulting surface is parametrized by $$x(t, \theta) = (r(t)\cos\theta,\ r(t)\sin\theta,\ z(t)), \qquad -\pi < \theta < \pi.$$ The $t$-curves (copies of the original curve) are the meridians and the $\theta$-curves (horizontal circles) are the circles of latitude. One checks that $x$ is a coordinate patch and computes $x_1$, $x_2$, $n$ in Homework 4, Problem 2. The sphere, the cylinder and the torus are all surfaces of revolution (Homework 4, Problem 3).

Example 10.12 (The saddle $z = xy$). The Monge patch $x(u^1, u^2) = (u^1, u^2, u^1 u^2)$ has $x_1 = (1, 0, u^2)$, $x_2 = (0, 1, u^1)$ and $x_1 \times x_2 = (-u^2, -u^1, 1)$. At $(u^1, u^2) = (1, 2)$, the point is $(1, 2, 2)$, the normal direction is $(-2, -1, 1)$, and the tangent plane is $-2(x - 1) - (y - 2) + (z - 2) = 0$. This surface contains two families of straight lines: for fixed $u^2 = b$ the $u^1$-curve $(u^1, b, b\,u^1)$ is a line, and similarly for fixed $u^1$. For instance, the parameter curves through $(1, 2, 2)$ are the lines $(t, 2, 2t)$ and $(1, t, t)$, and both lie in the tangent plane, which therefore cuts the surface along two lines. So, unlike for a sphere, the tangent plane of a saddle can meet the surface in more than one point near the point of tangency.

Definition 11.1 (Coordinate transformation). Let $U$ and $V$ be open subsets of $\mathbb{R}^2$, with coordinates $(u^1, u^2)$ on $U$ and $(v^1, v^2)$ on $V$. A $C^k$ coordinate transformation is a one-to-one, onto $C^k$ function $f \colon V \to U$ whose inverse $g \colon U \to V$ is also $C^k$.

Definition 11.2 (Jacobian). The Jacobian matrix of $f$ is the $2 \times 2$ matrix of partial derivatives $$J(f) = \left(\frac{\partial u^i}{\partial v^\alpha}\right) = \begin{pmatrix} \dfrac{\partial u^1}{\partial v^1} & \dfrac{\partial u^1}{\partial v^2} \\[1ex] \dfrac{\partial u^2}{\partial v^1} & \dfrac{\partial u^2}{\partial v^2} \end{pmatrix},$$ a matrix-valued function on $V$.

Example 11.3 (Polar coordinates). Let $V = \{(r, \theta) : r > 0,\ -\pi < \theta < \pi\}$ and let $U$ be the plane minus the ray $\{(x, 0) : x \le 0\}$. Then $f(r, \theta) = (r\cos\theta, r\sin\theta)$ is one-to-one and onto $U$, with $$J(f) = \begin{pmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{pmatrix}, \qquad \det J(f) = r > 0.$$ So it is a coordinate transformation. Its inverse is $g(x, y) = (\sqrt{x^2 + y^2},\ \theta(x, y))$, where $\theta(x, y)$ is the angle of $(x, y)$ in $(-\pi, \pi)$.

Lemma 11.4 (Composing with a coordinate transformation). Let $x \colon U \to \mathbb{R}^3$ be a coordinate patch and $f \colon V \to U$ a coordinate transformation. Then $y = x \circ f \colon V \to \mathbb{R}^3$ is a coordinate patch with the same image as $x$, and $$y_\alpha = \sum_{i=1}^2 x_i\,\frac{\partial u^i}{\partial v^\alpha} \quad (\alpha = 1, 2), \qquad y_1 \times y_2 = \det J(f)\;(x_1 \times x_2).$$

Proof. The map $y$ is $C^k$ as a composition of $C^k$ maps, it is one-to-one since $x$ and $f$ are, and its image is $x(f(V)) = x(U)$ since $f$ is onto. The formula for $y_\alpha = \partial y/\partial v^\alpha$ is the chain rule. Then, using $x_1 \times x_1 = x_2 \times x_2 = 0$ and $x_2 \times x_1 = -x_1 \times x_2$, $$y_1 \times y_2 = \Big(x_1\frac{\partial u^1}{\partial v^1} + x_2\frac{\partial u^2}{\partial v^1}\Big) \times \Big(x_1\frac{\partial u^1}{\partial v^2} + x_2\frac{\partial u^2}{\partial v^2}\Big) = \Big(\frac{\partial u^1}{\partial v^1}\frac{\partial u^2}{\partial v^2} - \frac{\partial u^2}{\partial v^1}\frac{\partial u^1}{\partial v^2}\Big)\,x_1 \times x_2,$$ and the factor in parentheses is $\det J(f)$. Since $\det J(f) \ne 0$ and $x_1 \times x_2 \ne 0$, also $y_1 \times y_2 \ne 0$, so $y$ is regular. ∎

Proposition 11.5 (The tangent plane does not depend on the patch). Let $x$ be a coordinate patch, $f$ a coordinate transformation and $y = x \circ f$. At a point $P = x(f(c)) = y(c)$:

1. the tangent plane to $x$ at $P$ equals the tangent plane to $y$ at $P$;
2. the unit normals of $x$ and $y$ at $P$ are equal if $\det J(f) > 0$ at $c$, and opposite if $\det J(f) < 0$.

Proof. By Lemma 11.4, $y_1 \times y_2$ is a nonzero multiple of $x_1 \times x_2$, so the two planes through $P$ perpendicular to these vectors coincide, and dividing by lengths gives $n_y = \pm n_x$ with the sign of $\det J(f)$. ∎

Example 11.6 (The hemisphere in two ways). Let $U = \{(u^1, u^2) : (u^1)^2 + (u^2)^2 < 1,\ u^1 < 0\}$ and $V = \{(v^1, v^2) : (v^1)^2 + (v^2)^2 < 1,\ v^2 > 0\}$. On $U$ take the Monge patch $x(u^1, u^2) = (u^1, u^2, \sqrt{1 - (u^1)^2 - (u^2)^2})$, whose image is the part of the sphere with $x < 0$, $z > 0$. On $V$ take the patch $y(v^1, v^2) = (-\sqrt{1 - (v^1)^2 - (v^2)^2},\ v^1,\ v^2)$, which describes the left hemisphere $x < 0$ as a graph over the $yz$-plane; restricted to $v^2 > 0$ its image is the same quarter-sphere. The coordinate transformation relating them is $f \colon V \to U$, $$f(v^1, v^2) = \Big(-\sqrt{1 - (v^1)^2 - (v^2)^2},\ v^1\Big),$$ which is one-to-one and onto with inverse $g(u^1, u^2) = (u^2, \sqrt{1 - (u^1)^2 - (u^2)^2})$, and indeed $x \circ f = y$. Its Jacobian is $$J(f) = \begin{pmatrix} \dfrac{v^1}{\sqrt{1 - (v^1)^2 - (v^2)^2}} & \dfrac{v^2}{\sqrt{1 - (v^1)^2 - (v^2)^2}} \\[1ex] 1 & 0 \end{pmatrix}, \qquad \det J(f) = \frac{-v^2}{\sqrt{1 - (v^1)^2 - (v^2)^2}} < 0.$$ So the two patches have opposite unit normals: one points out of the sphere and the other into it.

Definition 11.7 (Tangent vector). Let $x \colon U \to \mathbb{R}^3$ be a coordinate patch and $P = x(a, b)$. A vector $X$ is tangent to $x$ at $P$ if it is the velocity vector of some curve in the patch through $P$: that is, there are $C^1$ functions $\gamma^1(t), \gamma^2(t)$ defined for $|t| < \varepsilon$, with $(\gamma^1(t), \gamma^2(t)) \in U$ and $(\gamma^1(0), \gamma^2(0)) = (a, b)$, such that $X$ is the velocity at $t = 0$ of the curve $\gamma(t) = x(\gamma^1(t), \gamma^2(t))$.

Proposition 11.8 (The tangent vectors form a plane). The set of tangent vectors to a coordinate patch $x$ at $P = x(a, b)$ is a two-dimensional vector space with basis $x_1(a, b), x_2(a, b)$. Every tangent vector can be written uniquely as $$X = X^1 x_1 + X^2 x_2 = \sum_i X^i x_i,$$ and, drawn starting at $P$, the tangent vectors fill out exactly the tangent plane at $P$.

Definition 12.1 (Neighborhoods in a subset). Let $M$ be a subset of $\mathbb{R}^3$, $P \in M$ and $\varepsilon > 0$. The $\varepsilon$-neighborhood of $P$ in $M$ is the set of points of $M$ at distance less than $\varepsilon$ from $P$, that is, the intersection of $M$ with the open ball of radius $\varepsilon$ about $P$. A subset $W$ of $M$ is open in $M$ if every point of $W$ has an $\varepsilon$-neighborhood in $M$ contained in $W$.

Definition 12.2 (Continuity on a subset). A function $g \colon M \to \mathbb{R}^2$ defined on a subset $M \subset \mathbb{R}^3$ is continuous at $P \in M$ if for every $\varepsilon > 0$ there is a $\delta > 0$ such that $|g(Q) - g(P)| < \varepsilon$ for all $Q \in M$ with $|Q - P| < \delta$. It is continuous if it is continuous at every point of $M$.

Definition 12.3 (Proper patch). A coordinate patch $x \colon U \to \mathbb{R}^3$ is proper if its inverse $x^{-1} \colon x(U) \to U$ is continuous.

Example 12.4 (A patch that is not proper). Take the open rectangle $U = (0, 2) \times (0, 1)$ and bend it into a surface shaped like the figure "6" times an interval: map the $u^1$-direction along a plane curve that starts at a point $Q$, makes a loop, and comes back to end at a point $P$ in the middle of the curve, approaching it without reaching it, and let $u^2$ be the height. (For instance, let $c(t)$, $0 < t < 2$, be a one-to-one regular plane curve with $\lim_{t \to 2} c(t) = c(1)$, and set $x(u^1, u^2) = (c(u^1), u^2)$.) This is a one-to-one regular patch. But points of the surface near $x(1, \tfrac12)$ include points $x(u^1, \tfrac12)$ with $u^1$ close to $2$, which are far from $(1, \tfrac12)$ in $U$. So $x^{-1}$ is not continuous at $x(1, \frac12)$, and $x$ is not proper. Its image looks like it touches itself along a line, and we do not want to count it as a surface near that line.

Definition 12.5 (Surface). A $C^k$ surface in $\mathbb{R}^3$ is a subset $M \subset \mathbb{R}^3$ such that:

1. every point $P \in M$ lies in the image of some proper $C^k$ coordinate patch $x \colon U \to \mathbb{R}^3$ whose image $x(U)$ is contained in $M$ and contains an $\varepsilon$-neighborhood of $P$ in $M$ for some $\varepsilon > 0$;
2. any two such patches are compatible: if $x \colon U \to M$ and $y \colon V \to M$ are two of them, with overlapping images, then

$$y^{-1} \circ x \colon x^{-1}\big(x(U) \cap y(V)\big) \to y^{-1}\big(x(U) \cap y(V)\big)$$ is a $C^k$ coordinate transformation.

Definition 12.6 (Tangent space of a surface). Let $M$ be a surface and $P \in M$. A vector $X$ is tangent to $M$ at $P$ if it is tangent to some (hence every) coordinate patch of $M$ containing $P$. The set of these vectors is the tangent space $T_pM$ (or $T_PM$).

Example 12.7 (The sphere as a surface). Let $S^2 = \{(x, y, z) : x^2 + y^2 + z^2 = 1\}$ and let $D$ be the open unit disk in the plane. The sphere is covered by six hemispheres, each the graph of a function over $D$: $$x^\pm(u^1, u^2) = \big(u^1, u^2, \pm\sqrt{1 - (u^1)^2 - (u^2)^2}\big), \qquad y^\pm(v^1, v^2) = \big(v^1, \pm\sqrt{1 - (v^1)^2 - (v^2)^2}, v^2\big),$$ $$z^\pm(w^1, w^2) = \big(\pm\sqrt{1 - (w^1)^2 - (w^2)^2}, w^1, w^2\big),$$ with images the upper, lower, right, left, front and back open hemispheres. Each is a Monge patch (in suitable coordinates), and each is proper, because its inverse simply forgets one coordinate: for example $(x^+)^{-1}(x, y, z) = (x, y)$, which is continuous. Every point of $S^2$ has some nonzero coordinate, so lies in one of the six open hemispheres.

To check compatibility we must compute the coordinate transformation on each overlap. For instance, $x^+(D) \cap y^+(D)$ is the quarter-sphere $\{y > 0, z > 0\}$; its preimage under $x^+$ is $\{(u^1, u^2) \in D : u^2 > 0\}$, and $$(y^+)^{-1} \circ x^+(u^1, u^2) = (y^+)^{-1}\big(u^1, u^2, \sqrt{1 - (u^1)^2 - (u^2)^2}\big) = \big(u^1, \sqrt{1 - (u^1)^2 - (u^2)^2}\big).$$ This is $C^\infty$ on its domain (the square root is of a positive number), one-to-one, with $C^\infty$ inverse $(v^1, v^2) \mapsto (v^1, \sqrt{1 - (v^1)^2 - (v^2)^2})$. The other overlaps are similar (compare Example 11.6). So $S^2$ is a $C^\infty$ surface. In Homework 5, Problem 3 you show that two patches suffice: the upper hemisphere and the inverse of stereographic projection.

Example 12.8 (The cylinder). Let $M = \{(x, y, z) : x^2 + y^2 = 1\}$. The patches $$x(\theta, t) = (\cos\theta, \sin\theta, t), \quad -\tfrac{3\pi}{4} < \theta < \tfrac{3\pi}{4}, \qquad y(\phi, t) = (\cos\phi, \sin\phi, t), \quad \tfrac{\pi}{4} < \phi < \tfrac{7\pi}{4},$$ with $t \in \mathbb{R}$, are proper and cover $M$. Their overlap consists of two pieces: on the piece where $\pi/4 < \theta < 3\pi/4$, the transformation is $(\theta, t) \mapsto (\phi, t) = (\theta, t)$, and on the piece where $-3\pi/4 < \theta < -\pi/4$ it is $(\theta, t) \mapsto (\theta + 2\pi, t)$. Both are coordinate transformations, so $M$ is a surface. Note that the overlap of two patches need not be connected.

Theorem 12.9 (Regular level sets are surfaces). Let $F \colon \mathbb{R}^3 \to \mathbb{R}$ be a $C^k$ function, $c$ a number, and $M = \{(x, y, z) : F(x, y, z) = c\}$. If the gradient $(F_x, F_y, F_z)$ is nonzero at every point of $M$, then $M$ is a $C^k$ surface. In fact, near a point where $F_z \ne 0$, $M$ is the graph of a function $z = h(x, y)$, and so is described by a Monge patch; similarly near points where $F_x \ne 0$ or $F_y \ne 0$.

Example 12.10 (Quadric surfaces). For $a, b, c > 0$ the following are surfaces, since the gradient of the defining function vanishes only at the origin, which is not on the set:

1. the ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$, including the sphere;
2. the hyperboloid of one sheet $\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1$;
3. the hyperboloid of two sheets $\frac{x^2}{a^2} - \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1$.

The hyperbolic paraboloid $x^2 - y^2 - z = 0$ (a saddle) is also a surface: the gradient $(2x, -2y, -1)$ is never zero. On the other hand the cone $x^2 + y^2 - z^2 = 0$ is not covered by the theorem, because the gradient vanishes at the origin, which is on the cone; and indeed the cone is not a surface near its vertex.

Definition 13.1 (First fundamental form). The first fundamental form of a surface $M$ at $P$ is the inner product $\langle X, Y \rangle$ on $T_PM$, restricted from $\mathbb{R}^3$. If $x \colon U \to M$ is a coordinate patch, its metric coefficients are the functions $$g_{ij}(u^1, u^2) = \langle x_i(u^1, u^2), x_j(u^1, u^2) \rangle, \qquad i, j = 1, 2,$$ on $U$. We also write $g = \det(g_{ij})$, and $(g^{ij})$ for the inverse matrix of $(g_{ij})$.

Lemma 13.2 (Facts about the metric coefficients). For a coordinate patch $x$:

1. $g = g_{11}g_{22} - g_{12}^2 = |x_1 \times x_2|^2 > 0$;
2. $g^{11} = g_{22}/g$, $g^{12} = g^{21} = -g_{12}/g$, $g^{22} = g_{11}/g$;
3. $\sum_k g_{ik}\,g^{kj} = \delta_i^j$.

Proof. For (1), if $\theta$ is the angle between $x_1$ and $x_2$, then $|x_1 \times x_2|^2 = |x_1|^2|x_2|^2\sin^2\theta = |x_1|^2|x_2|^2 - \langle x_1, x_2 \rangle^2 = g_{11}g_{22} - g_{12}^2$, and this is positive by the regularity condition. Statement (2) is the formula for the inverse of a $2 \times 2$ matrix, and (3) says that the product of a matrix and its inverse is the identity matrix. ∎

Example 13.3 (Distance travelled on the Earth). A ship's GPS records its longitude $\theta(t)$ and latitude $\phi(t)$. How far has it travelled? Using the spherical coordinates of Example 10.8 on a sphere of radius $1$, we have $$x_\theta = (-\sin\theta\cos\phi,\ \cos\theta\cos\phi,\ 0), \qquad x_\phi = (-\cos\theta\sin\phi,\ -\sin\theta\sin\phi,\ \cos\phi),$$ so $g_{11} = \cos^2\phi$, $g_{12} = 0$, $g_{22} = 1$: $$(g_{ij}) = \begin{pmatrix} \cos^2\phi & 0 \\ 0 & 1 \end{pmatrix}, \qquad ds^2 = \cos^2\phi\,d\theta^2 + d\phi^2.$$ The distance travelled between times $t_0$ and $t_1$ is $\int_{t_0}^{t_1}\sqrt{\cos^2\phi(t)\,\theta'(t)^2 + \phi'(t)^2}\,dt$. For example, sailing along the circle of latitude $\phi = \phi_0$ through a change of longitude $\Delta\theta$ covers a distance $|\Delta\theta|\cos\phi_0$: degrees of longitude get shorter toward the poles. (On the Earth, multiply by its radius.)

Example 13.4 (Graphs). For the Monge patch $x(u^1, u^2) = (u^1, u^2, f(u^1, u^2))$ of Example 10.6, $x_1 = (1, 0, f_1)$ and $x_2 = (0, 1, f_2)$, so $$(g_{ij}) = \begin{pmatrix} 1 + f_1^2 & f_1 f_2 \\ f_1 f_2 & 1 + f_2^2 \end{pmatrix}, \qquad g = 1 + f_1^2 + f_2^2.$$ For the saddle $f = u^1 u^2$ this gives $g_{11} = 1 + (u^2)^2$, $g_{12} = u^1 u^2$, $g_{22} = 1 + (u^1)^2$. Here $g_{12} \ne 0$ away from the axes, so the parameter curves do not meet at right angles. For the upper hemisphere, $f = \sqrt{1 - (u^1)^2 - (u^2)^2}$, $$(g_{ij}) = \frac{1}{1 - (u^1)^2 - (u^2)^2}\begin{pmatrix} 1 - (u^2)^2 & u^1 u^2 \\ u^1 u^2 & 1 - (u^1)^2 \end{pmatrix}.$$

Example 13.5 (Plane and cylinder). For the plane $x(u^1, u^2) = (u^1, u^2, 0)$ we get $(g_{ij}) = I$, the identity matrix. For the cylinder of radius $r$, $x(\theta, z) = (r\cos\theta, r\sin\theta, z)$, we get $g_{11} = r^2$, $g_{12} = 0$, $g_{22} = 1$. If instead we use the arc length $u = r\theta$ along the circle as the first coordinate, $y(u, z) = (r\cos\frac{u}{r}, r\sin\frac{u}{r}, z)$, then $y_u = (-\sin\frac{u}{r}, \cos\frac{u}{r}, 0)$ and $y_z = (0, 0, 1)$, so again $(g_{ij}) = I$.

Definition 14.1 (Normal and geodesic curvature). For a unit speed curve $\gamma$ on a surface $M$, the normal curvature $\kappa_n$ and the geodesic curvature $\kappa_g$ are defined by $$\gamma'' = T' = \kappa_n\,n + \kappa_g\,S, \qquad\text{that is,}\qquad \kappa_n = \langle \gamma'', n \rangle, \quad \kappa_g = \langle \gamma'', S \rangle.$$

Example 14.2 (Curves in a plane). Let $M$ be the $xy$-plane with $n = (0, 0, 1)$. For a unit speed curve in $M$, $\gamma''$ lies in the plane, so $\kappa_n = 0$. And $S = n \times T$ is $T$ rotated counterclockwise by $90^\circ$, which is the normal $n$ of plane curves in Definition 7.5, so $\kappa_g$ is the signed plane curvature $k$ (Homework 6, Problem 6). So for curves in a plane, all the bending is within the surface.

Example 14.3 (A great circle). On the unit sphere with outward normal $n(P) = P$, consider the equator $\gamma(s) = (\cos s, \sin s, 0)$. Then $\gamma'' = -(\cos s, \sin s, 0) = -n$, so $\kappa_n = -1$ and $\kappa_g = 0$. All of the bending of the equator is forced by the sphere; within the sphere, the equator goes straight. The same is true for every great circle, by symmetry.

Definition 14.4 (Second fundamental form coefficients and Christoffel symbols). For a coordinate patch $x$, define functions $L_{ij}$ and $\Gamma^k_{ij}$ ($i, j, k = 1, 2$) on $U$ by $$x_{ij} = L_{ij}\,n + \sum_k \Gamma^k_{ij}\,x_k.$$ The $L_{ij}$ are the coefficients of the second fundamental form, and the $\Gamma^k_{ij}$ are the Christoffel symbols of $x$.

Proposition 14.5 (Gauss's formulas). For a coordinate patch $x$:

1. $L_{ij} = \langle x_{ij}, n \rangle$ and $\Gamma^k_{ij} = \sum_l \langle x_{ij}, x_l \rangle\,g^{lk}$;
2. for a unit speed curve $\gamma(s) = x(\gamma^1(s), \gamma^2(s))$,

$$\kappa_n = \sum_{i, j} L_{ij}\,(\gamma^i)'(\gamma^j)', \qquad \kappa_g\,S = \sum_k \Big((\gamma^k)'' + \sum_{i, j}\Gamma^k_{ij}\,(\gamma^i)'(\gamma^j)'\Big)\,x_k.$$

Proof. (1) Since $x_1$, $x_2$ are tangent and $n$ is a unit normal, taking the inner product of the definition of $L_{ij}$ and $\Gamma^k_{ij}$ with $n$ gives $\langle x_{ij}, n \rangle = L_{ij}$. Taking the inner product with $x_l$ instead gives $$\langle x_{ij}, x_l \rangle = \sum_k \Gamma^k_{ij}\,g_{kl}.$$ This is a system of linear equations for the $\Gamma^k_{ij}$. Multiplying by $g^{lm}$ and summing over $l$, using $\sum_l g_{kl}\,g^{lm} = \delta^m_k$ (Lemma 13.2), gives $\sum_l \langle x_{ij}, x_l \rangle\,g^{lm} = \Gamma^m_{ij}$. ("Raising the index $l$.")

(2) Substitute the definition of $L_{ij}$ and $\Gamma^k_{ij}$ into the formula for $\gamma''$ above: $$\gamma'' = \Big(\sum_{i, j} L_{ij}\,(\gamma^i)'(\gamma^j)'\Big)\,n + \sum_k \Big((\gamma^k)'' + \sum_{i, j} \Gamma^k_{ij}\,(\gamma^i)'(\gamma^j)'\Big)\,x_k.$$ The first term is normal to $M$ and the second is tangent. Comparing with $\gamma'' = \kappa_n n + \kappa_g S$, where $\kappa_n n$ is normal and $\kappa_g S$ is tangent, gives both formulas. ∎

Example 14.6 (Gauss's formulas for the cylinder). Let $x(u^1, u^2) = (r\cos u^1,\ r\sin u^1,\ u^2)$ be the cylinder of radius $r$. Then $$x_1 = (-r\sin u^1, r\cos u^1, 0), \quad x_2 = (0, 0, 1), \quad n = (\cos u^1, \sin u^1, 0),$$ with $g_{11} = r^2$, $g_{12} = 0$, $g_{22} = 1$. The second derivatives are $x_{11} = (-r\cos u^1, -r\sin u^1, 0) = -r\,n$ and $x_{12} = x_{21} = x_{22} = 0$. Hence $$L_{11} = -r, \qquad L_{12} = L_{22} = 0, \qquad \Gamma^k_{ij} = 0 \text{ for all } i, j, k.$$ Now consider the helix $(r\cos\omega s,\ r\sin\omega s,\ h\omega s)$ with $\omega = 1/\sqrt{r^2 + h^2}$. In this patch it is $\gamma^1(s) = \omega s$, $\gamma^2(s) = h\omega s$, a straight line in the $(u^1, u^2)$-plane, traversed at constant speed. So $(\gamma^k)'' = 0$, and since all $\Gamma^k_{ij}$ vanish, $\kappa_g = 0$. And $$\kappa_n = L_{11}\,((\gamma^1)')^2 = -r\omega^2 = -\frac{r}{r^2 + h^2}.$$ Compare with Example 3.10: $\kappa = r/(r^2 + h^2)$, so all of the curvature of the helix is normal curvature, and $N = -n$ points toward the axis. Within the cylinder, the helix does not turn at all. This makes sense if we unroll the cylinder into a plane: the helix becomes a straight line.

Proposition 15.1 (Christoffel symbols in terms of the metric). For a coordinate patch with metric coefficients $g_{ij}$, $$\Gamma^k_{ij} = \frac{1}{2}\sum_l g^{kl}\Big(\frac{\partial g_{il}}{\partial u^j} - \frac{\partial g_{ij}}{\partial u^l} + \frac{\partial g_{jl}}{\partial u^i}\Big).$$

Proof. The proof uses a classical trick, a cyclic permutation of the indices. By Proposition 14.5, $\Gamma^k_{ij} = \sum_l \langle x_{ij}, x_l \rangle\,g^{lk}$, so it suffices to express $\langle x_{ij}, x_l \rangle$ in terms of derivatives of the $g$'s. Differentiating $g_{ij} = \langle x_i, x_j \rangle$ with the product rule, $$\frac{\partial g_{ij}}{\partial u^l} = \langle x_{il}, x_j \rangle + \langle x_i, x_{jl} \rangle.$$ Write this identity three times, with the indices permuted: $$\frac{\partial g_{il}}{\partial u^j} = \langle x_{ij}, x_l \rangle + \langle x_i, x_{lj} \rangle, \qquad \frac{\partial g_{ij}}{\partial u^l} = \langle x_{il}, x_j \rangle + \langle x_i, x_{jl} \rangle, \qquad \frac{\partial g_{jl}}{\partial u^i} = \langle x_{ji}, x_l \rangle + \langle x_j, x_{li} \rangle.$$ Add the first and third and subtract the second. Using $x_{lj} = x_{jl}$, $x_{ji} = x_{ij}$ and $x_{li} = x_{il}$, the terms $\langle x_i, x_{jl} \rangle$ and $\langle x_j, x_{il} \rangle$ cancel, and what remains is $$\frac{\partial g_{il}}{\partial u^j} - \frac{\partial g_{ij}}{\partial u^l} + \frac{\partial g_{jl}}{\partial u^i} = 2\,\langle x_{ij}, x_l \rangle.$$ Multiply by $\frac12 g^{lk}$ and sum over $l$. ∎

Proposition 15.2 (Geodesic curvature is intrinsic). The geodesic curvature of a unit speed curve $\gamma(s) = x(\gamma^1(s), \gamma^2(s))$ is $$\kappa_g = \sqrt{g}\,\Big[(\gamma^1)'\Big((\gamma^2)'' + \sum_{i,j}\Gamma^2_{ij}(\gamma^i)'(\gamma^j)'\Big) - (\gamma^2)'\Big((\gamma^1)'' + \sum_{i,j}\Gamma^1_{ij}(\gamma^i)'(\gamma^j)'\Big)\Big],$$ where $g = \det(g_{ij})$. In particular it depends only on the metric coefficients and the curve in coordinates.

Proof. Write $\kappa_g S = \sum_k A^k x_k$, with $A^k = (\gamma^k)'' + \sum_{i,j}\Gamma^k_{ij}(\gamma^i)'(\gamma^j)'$, and $T = \sum_j (\gamma^j)' x_j$. Since $(T, S, n)$ is right-handed orthonormal, $\kappa_g = \langle \kappa_g S, S \rangle = \langle \kappa_g S, n \times T \rangle = \langle n, T \times \kappa_g S \rangle$, using the cyclic symmetry of the triple product. Now $$T \times \kappa_g S = \sum_{j, k} (\gamma^j)' A^k\,x_j \times x_k = \big((\gamma^1)'A^2 - (\gamma^2)'A^1\big)\,x_1 \times x_2,$$ and $\langle n, x_1 \times x_2 \rangle = |x_1 \times x_2| = \sqrt g$ (Lemma 13.2). Every ingredient is intrinsic. ∎

Example 15.3 (Christoffel symbols of a graph). For the Monge patch $x(u^1, u^2) = (u^1, u^2, f(u^1, u^2))$, write $f_{ij}$ for second partial derivatives and $D = 1 + f_1^2 + f_2^2$. Then $x_{ij} = (0, 0, f_{ij})$ and $n = (-f_1, -f_2, 1)/\sqrt D$, so $$L_{ij} = \langle x_{ij}, n \rangle = \frac{f_{ij}}{\sqrt{D}}.$$ For the Christoffel symbols we compute extrinsically: $\langle x_{ij}, x_l \rangle = f_{ij}f_l$, and with $(g^{lk}) = \frac{1}{D}\begin{pmatrix} 1 + f_2^2 & -f_1f_2 \\ -f_1f_2 & 1 + f_1^2 \end{pmatrix}$, $$\Gamma^k_{ij} = \sum_l f_{ij}f_l\,g^{lk} = \frac{f_{ij}f_k}{D}.$$ (For instance $\Gamma^1_{ij} = \frac{f_{ij}}{D}\big(f_1(1 + f_2^2) - f_2 f_1 f_2\big) = \frac{f_{ij}f_1}{D}$.) Computing the same symbols from Proposition 15.1 is possible but longer; often it is easiest to use whichever of the two methods is more convenient.

Example 15.4 (A circle on a cone). Let $M$ be the cone $x(r, \theta) = (r\cos\theta,\ r\sin\theta,\ r)$, $r > 0$, $-\pi < \theta < \pi$, which makes an angle of $45^\circ$ with its axis. Then $x_r = (\cos\theta, \sin\theta, 1)$ and $x_\theta = (-r\sin\theta, r\cos\theta, 0)$, so $$g_{11} = 2, \qquad g_{12} = 0, \qquad g_{22} = r^2.$$ The coordinates are orthogonal and only $g_{22}$ is non-constant, depending on $u^1 = r$; by the formulas above, the nonzero Christoffel symbols are $$\Gamma^1_{22} = -\frac{\partial_r g_{22}}{2g_{11}} = -\frac{r}{2}, \qquad \Gamma^2_{12} = \Gamma^2_{21} = \frac{\partial_r g_{22}}{2g_{22}} = \frac{1}{r}.$$ Consider the horizontal circle $r = r_0$, parametrized by arc length: $\gamma^1(s) = r_0$, $\gamma^2(s) = s/r_0$, since $|x_\theta| = r_0$. Then $(\gamma^1)' = 0$, $(\gamma^2)' = 1/r_0$, and all second derivatives vanish, so $$\kappa_g S = \Gamma^1_{22}\,((\gamma^2)')^2\,x_r + \Gamma^2_{22}\,((\gamma^2)')^2\,x_\theta = -\frac{r_0}{2}\cdot\frac{1}{r_0^2}\,x_r = -\frac{1}{2r_0}\,x_r.$$ Since $|x_r| = \sqrt2$, this gives $|\kappa_g| = \frac{1}{\sqrt2\,r_0}$, and $S = \pm x_r/\sqrt2$ points along the cone's generating line. (The sign depends on the choice of $n$.) We can check this extrinsically: the circle has radius $r_0$, so $\kappa = 1/r_0$, and $\kappa^2 = \kappa_n^2 + \kappa_g^2$ gives $|\kappa_n| = \frac{1}{\sqrt2\,r_0}$ as well. Half of the bending of the circle (in the sense of squares) is within the cone and half is forced by the cone. If we cut the cone along a line and unroll it flat, the circle becomes an arc of a circle in the plane of radius $\sqrt2\,r_0$ (its distance from the vertex along the cone), whose curvature $1/(\sqrt2\,r_0)$ is exactly $|\kappa_g|$, as it must be, since unrolling does not change lengths or angles.

Example 15.5 (A great circle in the Monge hemisphere). In the upper hemisphere $x(r, s) = (r, s, \sqrt{1 - r^2 - s^2})$, consider the great semicircle $\gamma(t) = (\sin t, 0, \cos t)$, $-\pi/2 < t < \pi/2$, which is unit speed, with $\gamma^1(t) = \sin t$, $\gamma^2(t) = 0$. Then $T = (\cos t, 0, -\sin t)$, $T' = -(\sin t, 0, \cos t) = -n$ (since $n$ is the outward radial vector on the sphere). So $\kappa_n = -1$, $\kappa_g = 0$. In the Monge coordinates this curve is not a straight line traversed at constant speed: $(\gamma^1)'' = -\sin t \ne 0$. The Christoffel symbols exactly cancel this, as they must since $\kappa_g = 0$. So, unlike for the cylinder, whether a curve is "straight" in the surface is not visible by looking at it in coordinates; it requires the Christoffel symbols.

Definition 16.1 (Geodesic). A geodesic on a surface $M$ is a unit speed curve in $M$ whose geodesic curvature is zero everywhere: $\kappa_g \equiv 0$.

Proposition 16.2 (Geodesics have normal acceleration). A unit speed curve $\gamma$ in $M$ is a geodesic if and only if its acceleration $\gamma''$ is everywhere normal to $M$, that is, a multiple of $n$.

Proof. By definition, $\gamma'' = \kappa_n n + \kappa_g S$ with $S$ tangent to $M$. So $\gamma''$ is a multiple of $n$ exactly when $\kappa_g = 0$. ∎

Example 16.3 (Great circles). Let $\gamma(s)$ be a great circle on the unit sphere $S^2$, parametrized by arc length: $\gamma(s) = \cos s\,a + \sin s\,b$ for orthonormal vectors $a$, $b$. Then $\gamma'' = -\gamma$, and the unit normal of the sphere at $\gamma(s)$ is $\pm\gamma(s)$. So $\gamma''$ is normal, and every great circle is a geodesic. In the next lecture we will see that there are no other geodesics on the sphere.

Example 16.4 (Straight lines in a surface). If a surface contains a straight line, traversed at unit speed, then $\gamma'' = 0$, which is (trivially) normal. So straight lines in a surface are geodesics. For example, the lines on the cylinder parallel to its axis, and the lines in the saddle $z = xy$ found in Example 10.12, are geodesics.

Proposition 16.5 (The geodesic equations). Let $x$ be a coordinate patch with Christoffel symbols $\Gamma^k_{ij}$. A unit speed curve $\gamma(s) = x(\gamma^1(s), \gamma^2(s))$ is a geodesic if and only if $$\frac{d^2\gamma^k}{ds^2} + \sum_{i, j} \Gamma^k_{ij}\big(\gamma^1(s), \gamma^2(s)\big)\,\frac{d\gamma^i}{ds}\,\frac{d\gamma^j}{ds} = 0 \qquad\text{for } k = 1, 2.$$

Proof. By Gauss's formula (Proposition 14.5), $\kappa_g S = \sum_k A^k x_k$, where $A^k$ is the left side of the equation. Since $x_1$, $x_2$ are linearly independent, $\kappa_g = 0$ if and only if $A^1 = A^2 = 0$. ∎

Example 16.6 (A geodesic on a saddle). Let $M$ be the Monge patch $x(u, v) = (u, v, u^2 - v^2)$, and consider the parabola $\gamma(t) = (t, 0, t^2)$, the intersection of $M$ with the plane $y = 0$. It is not parametrized by arc length, so we use Proposition 16.2 together with the chain rule: if $s$ is arc length, then $\frac{d\gamma}{ds} = \dot\gamma\,\frac{dt}{ds}$ and $$\frac{d^2\gamma}{ds^2} = \ddot\gamma\,\Big(\frac{dt}{ds}\Big)^2 + \dot\gamma\,\frac{d^2t}{ds^2}.$$ So $\frac{d^2\gamma}{ds^2}$ is a combination of $\dot\gamma$ and $\ddot\gamma$, and it is perpendicular to $\dot\gamma$ because the curve has unit speed in $s$. If $n$, $\dot\gamma$, $\ddot\gamma$ are linearly dependent, then (since $n$ and $\dot\gamma$ are independent) $\ddot\gamma$ lies in the plane spanned by $n$ and $\dot\gamma$, hence so does $\frac{d^2\gamma}{ds^2}$; being perpendicular to $\dot\gamma$, it is then a multiple of $n$, and the curve is a geodesic by Proposition 16.2. (Conversely, for a geodesic the three vectors are dependent: in other words, a unit speed curve is a geodesic if and only if $[n, T, T'] = 0$.) Here $\dot\gamma = (1, 0, 2t)$, $\ddot\gamma = (0, 0, 2)$, and at $\gamma(t)$ the normal is a multiple of $x_u \times x_v = (-2u, 2v, 1) = (-2t, 0, 1)$. All three vectors have second coordinate $0$, so they lie in a plane and are dependent. Hence $\gamma$, parametrized by arc length, is a geodesic. The same argument shows that the parabola $(0, t, -t^2)$ is a geodesic. In both cases the curve lies in a plane of symmetry of the surface; compare Homework 7, Problem 2.

Theorem 16.7 (Existence and uniqueness of geodesics). Let $P$ be a point of a surface $M$ and $X$ a unit tangent vector to $M$ at $P$. Then for some $\varepsilon > 0$ there is a geodesic $\gamma \colon (-\varepsilon, \varepsilon) \to M$ with $\gamma(0) = P$ and $\gamma'(0) = X$, and any two such geodesics agree on the intersection of their domains.

Proof. Choose a patch $x$ around $P$ with $x(a, b) = P$, and write $X = X^1 x_1 + X^2 x_2$. We want to solve the geodesic equations with the initial conditions $(\gamma^1(0), \gamma^2(0)) = (a, b)$ and $((\gamma^1)'(0), (\gamma^2)'(0)) = (X^1, X^2)$.

To use Theorem 5.1, which is about first-order equations, introduce the four unknowns $w^1 = \gamma^1$, $w^2 = \gamma^2$, $w^3 = (\gamma^1)'$, $w^4 = (\gamma^2)'$. The geodesic equations become the first-order system $$(w^1)' = w^3, \qquad (w^2)' = w^4, \qquad (w^{k+2})' = -\sum_{i, j} \Gamma^k_{ij}(w^1, w^2)\,w^{i+2}\,w^{j+2} \quad (k = 1, 2),$$ with initial values $(a, b, X^1, X^2)$. The right side is $C^1$ (the $\Gamma^k_{ij}$ involve first derivatives of the $g_{ij}$, so they are $C^1$ if $x$ is $C^3$), so its partial derivatives are bounded near the initial point, and Theorem 5.1 (in its local form) gives a unique solution for $s$ near $0$.

It remains to check that the resulting curve $\gamma(s) = x(\gamma^1(s), \gamma^2(s))$ has unit speed, since that was part of the definition of a geodesic. The computation in the proof of Proposition 14.5 did not use unit speed: for any curve in the patch, $$\gamma'' = \Big(\sum_{i,j} L_{ij}(\gamma^i)'(\gamma^j)'\Big)\,n + \sum_k \Big((\gamma^k)'' + \sum_{i,j}\Gamma^k_{ij}(\gamma^i)'(\gamma^j)'\Big)\,x_k.$$ For our solution the second sum vanishes, so $\gamma''$ is normal to $M$, and in particular perpendicular to $\gamma'$. Hence $\frac{d}{ds}|\gamma'|^2 = 2\langle \gamma'', \gamma' \rangle = 0$, so the speed is constant, and it is $1$ at $s = 0$ because $X$ is a unit vector. So $\gamma$ is a unit speed curve with $\kappa_g = 0$, a geodesic.

Uniqueness: any geodesic with the same initial point and velocity satisfies the same system with the same initial values, so it agrees with $\gamma$ near $0$; a connectedness argument as in the uniqueness part of Picard's theorem extends this to the common domain. ∎

Example 16.8 (A geodesic that stops). Let $M$ be the plane with the single point $(2, 0)$ removed. The geodesic starting at $P = (0, 0)$ in the direction $X = (1, 0)$ is $\gamma(s) = (s, 0)$, but it is only defined for $s < 2$: it would have to pass through the missing point. A surface on which every geodesic extends to all $s \in \mathbb{R}$ is called complete.

Example 17.1 (All geodesics of the sphere). By Example 16.3, every great circle on $S^2$ is a geodesic. Conversely, let $\gamma$ be any geodesic with $\gamma(0) = P$ and $\gamma'(0) = X$. The great circle $\beta(s) = \cos s\,P + \sin s\,X$ also has $\beta(0) = P$ and $\beta'(0) = X$ (note that $X \perp P$ because $X$ is tangent to the sphere at $P$, so $P$, $X$ are orthonormal). By the uniqueness in Theorem 16.7, $\gamma = \beta$. So the geodesics of the sphere are exactly the great circles, traversed at unit speed.

Example 17.2 (All geodesics of the cylinder). Use the patch $y(u, z) = (r\cos\frac{u}{r}, r\sin\frac{u}{r}, z)$ of Example 13.5, with $(g_{ij}) = I$. All Christoffel symbols vanish (they are computed from derivatives of the $g_{ij}$), so the geodesic equations are $u'' = z'' = 0$: geodesics are the curves $u = as + b$, $z = cs + d$ with $a^2 + c^2 = 1$. On the cylinder these are

• vertical lines ($a = 0$),
• horizontal circles ($c = 0$), and
• circular helices (both $a, c \ne 0$),

traversed at unit speed. (Since the patch misses one vertical line, use a second patch rotated around the axis to cover it; the answer is the same.) This matches our intuition from unrolling: a cylinder rolled out flat becomes a plane, and its geodesics become straight lines.

So through two points of a cylinder on the same vertical line there are infinitely many geodesics: the vertical segment, and helices that wind around once, twice, and so on, before arriving. Through two points on the same horizontal circle there are two arcs of that circle and infinitely many helices.

Theorem 17.3 (Shortest curves are geodesics). Let $\gamma \colon [a, b] \to M$ be a unit speed curve in a surface $M$ from $P = \gamma(a)$ to $Q = \gamma(b)$. If $\gamma$ is the shortest curve in $M$ from $P$ to $Q$, then $\gamma$ is a geodesic.

Proof of Theorem 17.3. Suppose that $\kappa_g(s_0) \ne 0$ for some $s_0 \in (a, b)$; we derive a contradiction. By continuity, $\kappa_g \ne 0$ on some interval $[c, d]$ around $s_0$, and we may take this interval so small that $\gamma([c, d])$ lies in a single coordinate patch $x$. Write $\gamma(s) = x(\gamma^1(s), \gamma^2(s))$ there. Let $\lambda$ be the bump function above, and write the tangent vector field $\lambda\kappa_g S$ along $\gamma$ in terms of the patch: $$\lambda(s)\,\kappa_g(s)\,S(s) = \sum_k v^k(s)\,x_k.$$ For small $|t|$, define the curves $$\gamma_t(s) = x\big(\gamma^1(s) + t v^1(s),\ \gamma^2(s) + t v^2(s)\big) \quad (c \le s \le d), \qquad \gamma_t(s) = \gamma(s) \text{ otherwise}.$$ Each $\gamma_t$ joins $P$ to $Q$ in $M$ (it agrees with $\gamma$ outside $(c, d)$, and the $v^k$ vanish at $c$ and $d$), and $\gamma_0 = \gamma$. For small $|t|$ these curves are regular, though not unit speed. Write $\alpha(s, t) = \gamma_t(s)$. By the chain rule, $$\frac{\partial\alpha}{\partial t}(s, 0) = \sum_k v^k(s)\,x_k = \lambda\kappa_g S, \qquad \frac{\partial\alpha}{\partial s}(s, 0) = \gamma'(s) = T, \qquad \frac{\partial^2\alpha}{\partial s^2}(s, 0) = \gamma''(s) = \kappa_n n + \kappa_g S.$$ Let $L(t) = \int_a^b \big|\frac{\partial\alpha}{\partial s}\big|\,ds$ be the length of $\gamma_t$. Since $\gamma = \gamma_0$ is the shortest curve, $L$ has a minimum at $t = 0$, so $L'(0) = 0$. On the other hand, only $s \in [c, d]$ contributes to the change in length, and differentiating under the integral sign, $$L'(0) = \int_c^d \frac{\partial}{\partial t}\Big\langle \frac{\partial\alpha}{\partial s}, \frac{\partial\alpha}{\partial s} \Big\rangle^{1/2}\Big|_{t=0} ds = \int_c^d \frac{\big\langle \frac{\partial^2\alpha}{\partial t\,\partial s}, \frac{\partial\alpha}{\partial s} \big\rangle}{\big|\frac{\partial\alpha}{\partial s}\big|}\Big|_{t=0} ds = \int_c^d \Big\langle \frac{\partial^2\alpha}{\partial s\,\partial t}, \frac{\partial\alpha}{\partial s} \Big\rangle\Big|_{t=0} ds,$$ using $|\partial\alpha/\partial s| = 1$ at $t = 0$ and the equality of mixed partial derivatives. By the product rule, the integrand is $$\frac{\partial}{\partial s}\Big\langle \frac{\partial\alpha}{\partial t}, \frac{\partial\alpha}{\partial s} \Big\rangle - \Big\langle \frac{\partial\alpha}{\partial t}, \frac{\partial^2\alpha}{\partial s^2} \Big\rangle.$$ The first term integrates to $\big\langle \frac{\partial\alpha}{\partial t}, \frac{\partial\alpha}{\partial s} \big\rangle$ evaluated between $s = c$ and $s = d$, which is $0$ because $\partial\alpha/\partial t = \lambda\kappa_g S$ vanishes there. Hence $$0 = L'(0) = -\int_c^d \big\langle \lambda\kappa_g S,\ \kappa_n n + \kappa_g S \big\rangle\,ds = -\int_c^d \lambda\,\kappa_g^2\,ds.$$ But $\lambda > 0$ and $\kappa_g^2 > 0$ on $(c, d)$, so the last integral is strictly negative. This contradiction shows that $\kappa_g \equiv 0$. ∎

Example 17.4 (Long geodesics on the sphere). Let $P$ and $Q$ be points of $S^2$ that are neither equal nor antipodal. There is exactly one great circle through them, and it is divided by $P$ and $Q$ into two arcs of different lengths. Both arcs are geodesics, but only the shorter one is the shortest curve from $P$ to $Q$. Also, between antipodal points such as the north and south poles there are infinitely many geodesics of the same length (all the meridians), so even the shortest geodesic need not be unique.

Definition 18.1 (Vector field along a curve). Let $\gamma \colon [a, b] \to M$ be a curve in a surface $M$. A vector field along $\gamma$ is a function $X$ that assigns to each $t \in [a, b]$ a tangent vector $X(t) \in T_{\gamma(t)}M$. It is differentiable if $X \colon [a, b] \to \mathbb{R}^3$ is differentiable.

Definition 18.2 (Parallel vector field). A differentiable vector field $X$ along $\gamma$ is parallel along $\gamma$ if $dX/dt$ is normal to $M$ for every $t$.

Example 18.3 (Parallel vectors in a plane). Let $M$ be the $xy$-plane and $X(t) = (A(t), B(t), 0)$ a vector field along a curve in $M$. Since $dX/dt = (A', B', 0)$ is tangent to $M$, it is normal only if it is $0$. So $X$ is parallel exactly when $A$ and $B$ are constant: this is the usual notion of parallel vectors, and it does not depend on the curve.

Example 18.4 (Along the equator). On the unit sphere, let $\gamma(t) = (\cos t, \sin t, 0)$ be the equator and let $X(t) = (0, 0, 1)$, the unit vector pointing north. It is tangent to the sphere at every point of the equator, and $dX/dt = 0$ is (trivially) normal, so $X$ is parallel along the equator.

Example 18.5 (Not parallel along a latitude). Along the circle of latitude at height $\frac{\sqrt2}{2}$, $\gamma(t) = (\frac{\sqrt2}{2}\cos t, \frac{\sqrt2}{2}\sin t, \frac{\sqrt2}{2})$, the unit vector pointing north (toward the north pole, along the meridian) is $X(t) = (-\frac{\sqrt2}{2}\cos t, -\frac{\sqrt2}{2}\sin t, \frac{\sqrt2}{2})$. Its derivative $dX/dt = (\frac{\sqrt2}{2}\sin t, -\frac{\sqrt2}{2}\cos t, 0)$ is horizontal and tangent to the circle, so it is tangent to the sphere and nonzero. So "always pointing north" is not parallel along this latitude: walking around the circle while pointing north, you would feel yourself turning the arrow.

Proposition 18.6 (Parallel transport equations). Let $\gamma(t) = x(\gamma^1(t), \gamma^2(t))$ be a regular curve in a patch and $X(t) = \sum_k X^k(t)\,x_k$ a differentiable vector field along it. Then $X$ is parallel along $\gamma$ if and only if $$\frac{dX^k}{dt} + \sum_{i, j} \Gamma^k_{ij}\,X^i\,\frac{d\gamma^j}{dt} = 0, \qquad k = 1, 2.$$

Proof. By the product rule and the chain rule, $$\frac{dX}{dt} = \sum_k \frac{dX^k}{dt}\,x_k + \sum_{i, j} X^i\,x_{ij}\,\frac{d\gamma^j}{dt}.$$ Substituting $x_{ij} = L_{ij}n + \sum_k \Gamma^k_{ij}x_k$, $$\frac{dX}{dt} = \Big(\sum_{i,j} L_{ij}X^i\frac{d\gamma^j}{dt}\Big)\,n + \sum_k \Big(\frac{dX^k}{dt} + \sum_{i,j}\Gamma^k_{ij}X^i\frac{d\gamma^j}{dt}\Big)\,x_k.$$ This is normal to $M$ exactly when the coefficients of $x_1$ and $x_2$ vanish. ∎

Theorem 18.7 (Existence of parallel transport). Let $\gamma \colon [a, b] \to M$ be a regular $C^2$ curve and $X_0 \in T_{\gamma(a)}M$. There is a unique vector field $X$ parallel along $\gamma$ with $X(a) = X_0$.

Proof. In a patch, the equations of Proposition 18.6 form a linear system of first-order differential equations for $X^1(t)$, $X^2(t)$, with continuous coefficients $\Gamma^k_{ij}(\gamma(t))\,d\gamma^j/dt$. By Theorem 5.1 (for linear systems) it has a unique solution on the whole interval with the given initial value. If $\gamma$ does not lie in one patch, cut $[a, b]$ into finitely many pieces each lying in a patch, and solve on each piece in turn, using the final value on one piece as the initial value on the next. ∎

Proposition 18.8 (Parallel transport preserves lengths and angles). If $X$ and $Y$ are parallel along $\gamma$, then $\langle X(t), Y(t) \rangle$ is constant. In particular $|X(t)|$ is constant, and the angle between $X(t)$ and $Y(t)$ is constant.

Proof. $\frac{d}{dt}\langle X, Y \rangle = \langle \frac{dX}{dt}, Y \rangle + \langle X, \frac{dY}{dt} \rangle = 0$, because $dX/dt$ and $dY/dt$ are normal to $M$ while $X$ and $Y$ are tangent. The length is $\sqrt{\langle X, X \rangle}$ and the cosine of the angle is $\langle X, Y \rangle/(|X||Y|)$. ∎

Proposition 18.9 (Geodesics have parallel tangents). A unit speed curve $\gamma$ is a geodesic if and only if its unit tangent $T$ is parallel along $\gamma$. In that case, a unit vector field $X$ along $\gamma$ is parallel if and only if the angle between $X$ and $T$ is constant.

Proof. $dT/ds = \gamma''$, which is normal exactly when $\gamma$ is a geodesic (Proposition 16.2). Suppose $\gamma$ is a geodesic. Then $S = n \times T$ is also parallel: $\frac{dS}{ds} = \frac{dn}{ds} \times T + n \times T'$, where $n \times T' = 0$ because $T'$ is a multiple of $n$, and $\frac{dn}{ds}$ is perpendicular to $n$ (as $n$ is a unit vector), so $\frac{dn}{ds} = aT + bS$ for some $a$, $b$, and $\frac{dn}{ds} \times T = b\,S \times T = -b\,n$ is normal. (The converse is Homework 7, Problem 9.) So $T$ and $S$ are parallel and form an orthonormal basis of the tangent plane. A unit vector field can be written $X = \cos\theta(s)\,T + \sin\theta(s)\,S$, and $dX/ds = \theta'(-\sin\theta\,T + \cos\theta\,S) + (\text{normal})$, which is normal exactly when $\theta' = 0$. ∎

Example 18.10 (Around a circle of latitude). On the unit sphere use the coordinates $x(\theta, \phi) = (\cos\theta\cos\phi, \sin\theta\cos\phi, \sin\phi)$, with $g_{11} = \cos^2\phi$, $g_{12} = 0$, $g_{22} = 1$. By the formulas for orthogonal coordinates in Lecture 15, the only nonzero Christoffel symbols are $$\Gamma^1_{12} = \Gamma^1_{21} = -\tan\phi, \qquad \Gamma^2_{11} = \sin\phi\cos\phi.$$ Let $0 < \phi_0 < \pi/2$ and let $\gamma(t) = x(t, \phi_0)$, $0 \le t \le 2\pi$, the circle of latitude at latitude $\phi_0$; here $d\gamma^1/dt = 1$ and $d\gamma^2/dt = 0$. The parallel transport equations for $X = X^1 x_\theta + X^2 x_\phi$ become $$\frac{dX^1}{dt} - \tan\phi_0\,X^2 = 0, \qquad \frac{dX^2}{dt} + \sin\phi_0\cos\phi_0\,X^1 = 0.$$ Start with the unit vector pointing north, $X(0) = x_\phi$, so $X^1(0) = 0$, $X^2(0) = 1$. Writing $c = \sin\phi_0$, the solution is $$X^1(t) = \frac{\sin(ct)}{\cos\phi_0}, \qquad X^2(t) = \cos(ct),$$ as one checks by differentiating. Since $x_\theta/\cos\phi_0$ and $x_\phi$ are orthonormal (pointing east and north), $X(t)$ is the unit vector making an angle $ct$ with north, turned toward the east. After going once around the circle, at $t = 2\pi$, we return to the starting point, but the vector has turned through the angle $2\pi\sin\phi_0$. Only on the equator, $\phi_0 = 0$, does it come back unchanged.

There is a beautiful interpretation: the polar cap above this latitude has area $2\pi(1 - \sin\phi_0)$, so the vector has turned by $2\pi$ minus the area of the cap. Equivalently, measured as a rotation, it has fallen behind a full turn by exactly the area enclosed.

Example 18.11 (Around a geodesic triangle). On the unit sphere, start at the north pole $N$ with a tangent vector $X$ pointing along a meridian toward a point $A$ on the equator. Transport it along the meridian from $N$ to $A$, then along the equator from $A$ to a point $B$, then back along a meridian from $B$ to $N$; all three sides are geodesics. Along each side, by Proposition 18.9, the vector keeps a constant angle with the direction of travel. It arrives at $A$ pointing south, along the direction of travel. Along the equator it keeps a right angle with the direction of travel, so it keeps pointing south, and arrives at $B$ pointing south. Along the meridian from $B$ to $N$ the direction of travel is north, so the vector, pointing south, makes the constant angle $\pi$ with it; it arrives at $N$ pointing opposite to the direction of travel, that is, along the meridian from $N$ toward $B$. It started pointing from $N$ toward $A$, so it has rotated by the angle $\theta$ between the two meridians at $N$. This angle $\theta$ is also the area of the triangle $NAB$, since the triangle is the fraction $\theta/2\pi$ of the northern hemisphere, whose area is $2\pi$.

Definition 19.1 (Directional derivative). Let $f$ be a differentiable function defined on $M$ near $P$, let $X \in T_PM$, and let $\alpha(t)$ be a curve in $M$ with $\alpha(0) = P$ and $\alpha'(0) = X$. The directional derivative of $f$ in the direction $X$ is $$Xf = \frac{d}{dt}\Big|_{t=0} f(\alpha(t)).$$ If $f = (f^1, f^2, f^3)$ has values in $\mathbb{R}^3$, then $Xf = (Xf^1, Xf^2, Xf^3)$.

Proposition 19.2 (Directional derivatives in coordinates). Let $x$ be a patch with $x(a, b) = P$, and $X = \sum_i X^i x_i$. Then $$Xf = \sum_i X^i\,\frac{\partial (f \circ x)}{\partial u^i}(a, b).$$ In particular $Xf$ does not depend on the choice of the curve $\alpha$, and it is linear in $X$: $(X + Y)f = Xf + Yf$ and $(rX)f = r(Xf)$.

Proof. Write $\alpha(t) = x(\alpha^1(t), \alpha^2(t))$. Then $f(\alpha(t)) = (f \circ x)(\alpha^1(t), \alpha^2(t))$, and by the chain rule its derivative at $t = 0$ is $\sum_i \frac{\partial (f \circ x)}{\partial u^i}\,\frac{d\alpha^i}{dt}(0)$. Since $\alpha'(0) = \sum_i \frac{d\alpha^i}{dt}(0)\,x_i = X$, we have $\frac{d\alpha^i}{dt}(0) = X^i$. The right side depends only on the components $X^i$, and is linear in them. ∎

Definition 19.3 (Weingarten map). The Weingarten map (or shape operator) of $M$ at $P$ is the function $L \colon T_PM \to \mathbb{R}^3$ defined by $$L(X) = -Xn.$$

Proposition 19.4 (The Weingarten map is a linear map of the tangent plane).

1. $L$ is a linear map from $T_PM$ to $T_PM$.
2. In a patch, $L(x_i) = -n_i$, where $n_i = \partial n/\partial u^i$, and if we write $L(x_i) = \sum_j L^j_i\,x_j$, then

$$L_{ik} = \langle L(x_i), x_k \rangle = \sum_j L^j_i\,g_{jk} \qquad\text{and}\qquad L^j_i = \sum_k L_{ik}\,g^{kj},$$ where $L_{ik} = \langle x_{ik}, n \rangle$ are the coefficients of Definition 14.4.

Proof. Linearity follows from Proposition 19.2. To see that $L(X)$ is tangent, note that $\langle n, n \rangle = 1$, so $0 = X\langle n, n \rangle = 2\langle Xn, n \rangle$ (the directional derivative obeys the product rule, because it is an ordinary derivative along a curve). So $Xn \perp n$, that is, $Xn \in T_PM$.

For (2), $L(x_i) = -x_i n = -\partial n/\partial u^i$. Differentiating $\langle n, x_k \rangle = 0$ with respect to $u^i$ gives $\langle n_i, x_k \rangle + \langle n, x_{ki} \rangle = 0$, so $\langle L(x_i), x_k \rangle = \langle x_{ki}, n \rangle = L_{ik}$, using $x_{ki} = x_{ik}$. Substituting $L(x_i) = \sum_j L^j_i x_j$ gives $L_{ik} = \sum_j L^j_i g_{jk}$, and multiplying by $g^{km}$ and summing over $k$ gives $\sum_k L_{ik}g^{km} = L^m_i$. ∎

Definition 19.5 (Second fundamental form). The second fundamental form of $M$ at $P$ is the bilinear form on $T_PM$ $$\mathrm{II}(X, Y) = \langle L(X), Y \rangle.$$ In a patch, $\mathrm{II}(x_i, x_j) = L_{ij}$, so $\mathrm{II}(X, Y) = \sum_{i,j} L_{ij}\,X^i Y^j$.

Example 19.6 (The plane). For a plane, $n$ is constant, so $L(X) = -Xn = 0$ for every $X$: the Weingarten map and the second fundamental form vanish. Equivalently, every $x_{ij}$ is tangent to the plane, so $L_{ij} = 0$.

Example 19.7 (The sphere of radius $r$). Let $M$ be the sphere of radius $r$ centered at the origin, with outward normal $n(P) = P/r$. If $\alpha$ is a curve in $M$ with $\alpha(0) = P$ and $\alpha'(0) = X$, then $n(\alpha(t)) = \alpha(t)/r$, so $$L(X) = -\frac{d}{dt}\Big|_{t=0}\frac{\alpha(t)}{r} = -\frac{1}{r}X.$$ So $L = -\frac1r\,\mathrm{Id}$ is a multiple of the identity, and $\mathrm{II}(X, X) = -\frac1r|X|^2$: every curve on the sphere has normal curvature $-1/r$, whatever its direction. With the inward normal, $L = \frac1r\mathrm{Id}$. (Checking this in a coordinate patch for the unit sphere is Homework 8, Problem 1.)

Example 19.8 (The cone). Take the cone $x(r, \theta) = (r\cos\theta, r\sin\theta, r)$ of Example 15.4, with $g_{11} = 2$, $g_{12} = 0$, $g_{22} = r^2$. Then $x_r \times x_\theta = (-r\cos\theta, -r\sin\theta, r)$, so $n = \frac{1}{\sqrt2}(-\cos\theta, -\sin\theta, 1)$, which points up and toward the axis. The second derivatives are $$x_{rr} = 0, \qquad x_{r\theta} = (-\sin\theta, \cos\theta, 0), \qquad x_{\theta\theta} = (-r\cos\theta, -r\sin\theta, 0),$$ so $L_{11} = 0$, $L_{12} = 0$, $L_{22} = \frac{r}{\sqrt2}$. Raising an index, $$(L^j_i) = \begin{pmatrix} 0 & 0 \\ 0 & \dfrac{1}{\sqrt2\,r} \end{pmatrix}.$$ So $L(x_r) = 0$: moving along a ruling of the cone, the normal does not change at all. And $L(x_\theta) = \frac{1}{\sqrt2\,r}x_\theta$. The normal curvature of the horizontal circle $r = r_0$ is $\mathrm{II}(T, T)$ with $T = x_\theta/r_0$, namely $\frac{1}{\sqrt2\,r_0}$, agreeing in absolute value with Example 15.4.

Example 19.9 (A graph at a critical point). Let $x(u^1, u^2) = (u^1, u^2, f(u^1, u^2))$ be a Monge patch with $f(0, 0) = 0$ and $f_1(0, 0) = f_2(0, 0) = 0$, so that the tangent plane at the origin is horizontal. By Example 13.4 and Example 15.3, at the origin $g_{ij} = \delta_{ij}$, $n = (0, 0, 1)$ and $L_{ij} = f_{ij}$. Hence at the origin $$(L^j_i) = (L_{ij}) = \begin{pmatrix} f_{11} & f_{12} \\ f_{12} & f_{22} \end{pmatrix},$$ the Hessian matrix of $f$. So the Weingarten map at a point where the tangent plane is horizontal is given by the second derivatives of the height function, which describe the shape of the surface near that point to second order. Since any point of any surface can be moved to the origin with a horizontal tangent plane by a rigid motion, and any surface is locally a graph over its tangent plane, this example is typical.

Proposition 20.1 (Meusnier's theorem). Let $\gamma$ be a unit speed curve in $M$ with $\gamma(0) = P$ and $\gamma'(0) = X$. Then $\kappa_n(0) = \mathrm{II}(X, X)$. In particular, all curves through $P$ tangent to the line through $X$, traversed in either direction, have the same normal curvature at $P$, since $\mathrm{II}(-X, -X) = \mathrm{II}(X, X)$.

Proof. This is the formula $\kappa_n = \mathrm{II}(T, T)$ from Lecture 19, evaluated at $s = 0$. ∎

Lemma 20.2 (Critical directions are eigenvectors). If a unit vector $X \in T_PM$ is a maximum or minimum of $\mathrm{II}(X, X)$ on the unit circle, then $X$ is an eigenvector of $L$: $L(X) = \lambda X$, and the eigenvalue is $\lambda = \mathrm{II}(X, X)$.

Proof. In a patch, write $X = \sum X^i x_i$. We extremize $F(X^1, X^2) = \sum L_{ij}X^iX^j$ subject to $G(X^1, X^2) = \sum g_{ij}X^iX^j = 1$. At an extremum, there is a number $\lambda$ with $\partial F/\partial X^k = \lambda\,\partial G/\partial X^k$ for $k = 1, 2$, which (using the symmetry of $L_{ij}$ and $g_{ij}$) says $$\sum_j L_{kj}X^j = \lambda\sum_j g_{kj}X^j, \qquad k = 1, 2.$$ Multiplying by $Y^k$ and summing, $\mathrm{II}(Y, X) = \lambda\langle Y, X \rangle$ for every $Y \in T_PM$, that is, $\langle Y, L(X) - \lambda X \rangle = 0$ for all $Y$. Taking $Y = L(X) - \lambda X$ shows $L(X) = \lambda X$. Then $\mathrm{II}(X, X) = \langle \lambda X, X \rangle = \lambda$. ∎

Lemma 20.3 (Eigenvectors of a self-adjoint map). Let $L$ be a self-adjoint linear map of a two-dimensional inner product space. If $X$ is a unit eigenvector of $L$ and $Y$ is a unit vector perpendicular to $X$, then $Y$ is also an eigenvector.

Proof. $\langle L(Y), X \rangle = \langle Y, L(X) \rangle = \lambda\langle Y, X \rangle = 0$, so $L(Y)$ is perpendicular to $X$. In two dimensions this means $L(Y)$ is a multiple of $Y$. ∎

Theorem 20.4 (Principal curvatures and directions). At each point $P$ of $M$ there is an orthonormal basis $X_1, X_2$ of $T_PM$ consisting of eigenvectors of $L$, with eigenvalues $\kappa_1 \ge \kappa_2$. The number $\kappa_1$ is the maximum and $\kappa_2$ the minimum of the normal curvature $\mathrm{II}(X, X)$ over unit vectors $X \in T_PM$.

Proof. Let $X_1$ be a unit vector where $\mathrm{II}(X, X)$ attains its maximum $\kappa_1$. By Lemma 20.2 it is an eigenvector with eigenvalue $\kappa_1$, and by Lemma 20.3 a unit vector $X_2$ perpendicular to it is an eigenvector, with some eigenvalue $\kappa_2$. Any unit vector is $X = \cos\theta\,X_1 + \sin\theta\,X_2$, and then $$\mathrm{II}(X, X) = \langle L(X), X \rangle = \kappa_1\cos^2\theta + \kappa_2\sin^2\theta.$$ Since $\kappa_1$ is the maximum, $\kappa_2 \le \kappa_1$, and the formula shows that the minimum is $\kappa_2$. ∎

Definition 20.5 (Principal curvatures, umbilics, lines of curvature). The eigenvalues $\kappa_1 \ge \kappa_2$ of $L$ at $P$ are the principal curvatures of $M$ at $P$, and the corresponding unit eigenvectors are principal directions. A point where $\kappa_1 = \kappa_2$ is an umbilic. A curve in $M$ whose tangent vector is a principal direction at each of its points is a line of curvature.

Corollary 20.6 (Euler's formula). If $X_1, X_2$ are orthonormal principal directions at $P$ with principal curvatures $\kappa_1, \kappa_2$, and $X$ is a unit vector making angle $\theta$ with $X_1$, then $$\mathrm{II}(X, X) = \kappa_1\cos^2\theta + \kappa_2\sin^2\theta.$$

Example 20.7 (Plane and sphere). By Example 19.6 and Example 19.7, every point of a plane is an umbilic with $\kappa_1 = \kappa_2 = 0$, and every point of a sphere of radius $r$ is an umbilic with $\kappa_1 = \kappa_2 = \mp 1/r$. Every curve on a plane or sphere is a line of curvature. (It can be shown that a connected surface consisting entirely of umbilics is part of a plane or a sphere.)

Example 20.8 (The cone). By Example 19.8, at a point of the cone at distance $r$ from the axis (in the $xy$-direction), the principal curvatures are $\frac{1}{\sqrt2\,r}$ and $0$, with principal directions along the horizontal circle and along the ruling. The rulings and the horizontal circles are lines of curvature. By Euler's formula, a curve leaving the point at angle $\theta$ to the horizontal circle has normal curvature $\frac{\cos^2\theta}{\sqrt2\,r}$.

Example 20.9 (The saddle at the origin). For $z = xy$, the patch $x(u^1, u^2) = (u^1, u^2, u^1u^2)$ has a horizontal tangent plane at the origin, and by Example 19.9, $L$ is given there by the Hessian $$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},$$ with eigenvalues $\kappa_1 = 1$, $\kappa_2 = -1$ and principal directions $\frac{1}{\sqrt2}(1, 1, 0)$ and $\frac{1}{\sqrt2}(1, -1, 0)$. So the surface bends upward (toward $n = (0, 0, 1)$) along the diagonal $y = x$, where it contains the parabola $z = x^2$, and downward along $y = -x$, where $z = -x^2$. In the coordinate directions $X = (1, 0, 0)$ and $(0, 1, 0)$, at angle $\theta = \pm\pi/4$ to the principal directions, Euler's formula gives normal curvature $\frac12 - \frac12 = 0$; indeed the surface contains the coordinate axes, which are straight lines.

Definition 21.1 (Gaussian and mean curvature). The Gaussian curvature and the mean curvature of $M$ at $P$ are $$K = \det L = \kappa_1\kappa_2, \qquad H = \tfrac12\operatorname{tr} L = \tfrac12(\kappa_1 + \kappa_2).$$

Proposition 21.2 ($K$ and $H$ in a patch). In a coordinate patch, $$K = \det(L^j_i) = \frac{\det(L_{ij})}{\det(g_{ij})} = \frac{L_{11}L_{22} - L_{12}^2}{g_{11}g_{22} - g_{12}^2},$$ $$H = \frac12\sum_i L^i_i = \frac12\sum_{i, j} L_{ij}\,g^{ij} = \frac{L_{11}g_{22} - 2L_{12}g_{12} + L_{22}g_{11}}{2(g_{11}g_{22} - g_{12}^2)},$$ and the principal curvatures are $\kappa_{1,2} = H \pm \sqrt{H^2 - K}$.

Proof. The matrix of $L$ in the basis $x_1, x_2$ is $(L^j_i) = (g^{ij})(L_{ij})$ (Proposition 19.4), so its determinant is $\det(L_{ij})/\det(g_{ij})$, and its trace is $\sum_{i,k} L_{ik}g^{ki}$. Substituting $g^{11} = g_{22}/g$, $g^{12} = -g_{12}/g$, $g^{22} = g_{11}/g$ gives the explicit formula for $H$. Finally, $\kappa_1$ and $\kappa_2$ are the roots of $\lambda^2 - 2H\lambda + K = 0$. ∎

Example 21.3 (Graphs). For a Monge patch $x(u^1, u^2) = (u^1, u^2, f(u^1, u^2))$, with $D = 1 + f_1^2 + f_2^2$, we found $g_{ij} = \delta_{ij} + f_if_j$, $\det(g_{ij}) = D$ and $L_{ij} = f_{ij}/\sqrt D$ (Example 13.4, Example 15.3). Hence $$K = \frac{f_{11}f_{22} - f_{12}^2}{(1 + f_1^2 + f_2^2)^2}, \qquad H = \frac{(1 + f_2^2)f_{11} - 2f_1f_2f_{12} + (1 + f_1^2)f_{22}}{2(1 + f_1^2 + f_2^2)^{3/2}}.$$ For the saddle $f = u^1u^2$ this gives $$K = -\frac{1}{(1 + (u^1)^2 + (u^2)^2)^2}, \qquad H = -\frac{u^1u^2}{(1 + (u^1)^2 + (u^2)^2)^{3/2}}.$$ The Gaussian curvature is negative everywhere, most negative at the origin ($K = -1$), and tends to $0$ far away, where the surface flattens out.

Example 21.4 (The cone). For the cone of Example 19.8, $\kappa_1 = \frac{1}{\sqrt2\,r}$ and $\kappa_2 = 0$, so $K = 0$ and $H = \frac{1}{2\sqrt2\,r}$.

Definition 21.5 (Elliptic, hyperbolic, parabolic and planar points). A point $P$ of $M$ is elliptic if $K > 0$, hyperbolic if $K < 0$, parabolic if $K = 0$ but $L \ne 0$, and planar if $L = 0$.

Definition 22.1 (Area). Let $x \colon U \to M$ be a coordinate patch and $R$ a region in $x(U)$ (a set to which the integral below applies, such as the image of a disk or a rectangle). The area of $R$ is $$A(R) = \iint_{x^{-1}(R)} \sqrt{g}\;du^1\,du^2 = \iint_{x^{-1}(R)} |x_1 \times x_2|\;du^1\,du^2.$$ The expression $dA = \sqrt g\,du^1du^2$ is the area element. More generally, the integral of a function $f$ over $R$ is $\iint_R f\,dA = \iint_{x^{-1}(R)} f(x(u^1, u^2))\,\sqrt g\,du^1du^2$.

Example 22.2 (The sphere). In spherical coordinates (Example 13.3), $g_{11} = \cos^2\phi$, $g_{12} = 0$, $g_{22} = 1$, so $\sqrt g = \cos\phi$, and the area of the part of the unit sphere covered by the patch is $$\int_{-\pi/2}^{\pi/2}\int_{-\pi}^{\pi} \cos\phi\;d\theta\,d\phi = 2\pi\,[\sin\phi]_{-\pi/2}^{\pi/2} = 4\pi.$$ The missing poles and meridian have zero area, so the area of the unit sphere is $4\pi$. Similarly, the polar cap above latitude $\phi_0$ has area $\int_{\phi_0}^{\pi/2}\int_{-\pi}^{\pi}\cos\phi\,d\theta\,d\phi = 2\pi(1 - \sin\phi_0)$, as used in Example 18.10.

Example 22.3 (The torus). For the torus $x(u^1, u^2) = ((2 + \cos u^1)\cos u^2, (2 + \cos u^1)\sin u^2, \sin u^1)$ of Homework 4, Problem 1, one computes $x_1 = (-\sin u^1\cos u^2, -\sin u^1\sin u^2, \cos u^1)$ and $x_2 = (2 + \cos u^1)(-\sin u^2, \cos u^2, 0)$, so $g_{11} = 1$, $g_{12} = 0$, $g_{22} = (2 + \cos u^1)^2$ and $\sqrt g = 2 + \cos u^1$. The area of the torus is $$\int_{-\pi}^{\pi}\int_{-\pi}^{\pi} (2 + \cos u^1)\,du^1\,du^2 = 2\pi \cdot 4\pi = 8\pi^2.$$ This is the length $2\pi$ of the small circle times the length $4\pi$ of the circle traced by its center, an instance of a theorem of Pappus.

Proposition 22.4 (Rodrigues' formula). A unit speed curve $\gamma$ in $M$ is a line of curvature if and only if $$\frac{dn}{ds} = -\lambda(s)\,T(s)$$ for some function $\lambda$, where $n(s)$ is the unit normal of $M$ at $\gamma(s)$. In that case $\lambda(s)$ is the principal curvature in the direction of $T(s)$.

Proof. By the definition of the directional derivative, $\frac{dn}{ds} = Tn = -L(T)$. So the condition says $L(T) = \lambda T$, that is, $T$ is an eigenvector of $L$, a principal direction, with eigenvalue $\lambda$. ∎

Definition 22.5 (Gauss map). The Gauss map (or normal spherical image) of a patch $x$ is the map $\nu$ that sends a point $P$ of the patch to the point $n(P)$ of the unit sphere $S^2$.

Proposition 22.6 (Gaussian curvature as an area ratio). For a patch $x$, $n_1 \times n_2 = K\,\sqrt{g}\;n$. Consequently, if $R$ is a small region around $P$ and $\nu(R)$ its image under the Gauss map, then $$|K(P)| = \lim_{R \to P}\frac{\text{area of } \nu(R)}{\text{area of } R},$$ where the limit is taken over regions shrinking to $P$, and areas on the sphere are counted with multiplicity.

Proof. The formula $n_1 \times n_2 = K\sqrt g\,n$ is Homework 9, Problem 8; it follows from the Weingarten equations. The map $(u^1, u^2) \mapsto n(u^1, u^2)$ parametrizes the image $\nu(R)$ (possibly not regularly, and possibly covering parts several times), so by the same reasoning as for the area of a patch, the area of $\nu(R)$ is $\iint |n_1 \times n_2|\,du^1du^2 = \iint |K|\sqrt g\,du^1du^2$, while the area of $R$ is $\iint \sqrt g\,du^1du^2$. If $K$ is continuous, the ratio of these integrals tends to $|K(P)|$ as $R$ shrinks to $P$. ∎

Definition 23.1 (Riemann curvature tensor). The Riemann curvature tensor of a patch is the collection of sixteen functions $$R^l_{ijk} = \partial_j\Gamma^l_{ik} - \partial_k\Gamma^l_{ij} + \sum_p\big(\Gamma^p_{ik}\Gamma^l_{pj} - \Gamma^p_{ij}\Gamma^l_{pk}\big), \qquad 1 \le i, j, k, l \le 2.$$

Theorem 23.2 (Gauss and Codazzi-Mainardi equations). For a $C^3$ coordinate patch and all $i, j, k, l$:

1. (Gauss's equations) $R^l_{ijk} = L_{ik}L^l_j - L_{ij}L^l_k$;
2. (Codazzi-Mainardi equations) $\partial_k L_{ij} - \partial_j L_{ik} = \sum_p\big(\Gamma^p_{ik}L_{pj} - \Gamma^p_{ij}L_{pk}\big)$.

Proof. Differentiate Gauss's formula $x_{ij} = L_{ij}n + \sum_l \Gamma^l_{ij}x_l$ with respect to $u^k$, using the product rule: $$x_{ijk} = (\partial_k L_{ij})\,n + L_{ij}\,n_k + \sum_l (\partial_k\Gamma^l_{ij})\,x_l + \sum_p \Gamma^p_{ij}\,x_{pk}.$$ Now substitute the Weingarten equation $n_k = -\sum_l L^l_k x_l$ and Gauss's formula $x_{pk} = L_{pk}n + \sum_l \Gamma^l_{pk}x_l$, and collect the normal and tangential parts: $$x_{ijk} = \Big(\partial_k L_{ij} + \sum_p \Gamma^p_{ij}L_{pk}\Big)\,n + \sum_l\Big(\partial_k\Gamma^l_{ij} - L_{ij}L^l_k + \sum_p \Gamma^p_{ij}\Gamma^l_{pk}\Big)\,x_l.$$ Since $x$ is $C^3$, $x_{ijk} = x_{ikj}$: the left side does not change when we swap $j$ and $k$. The right side, however, is not visibly symmetric in $j$ and $k$. Swapping $j$ and $k$ in it and comparing the coefficients of the basis vectors $n$, $x_1$, $x_2$ gives two sets of equations.

The coefficients of $n$ give $$\partial_k L_{ij} + \sum_p \Gamma^p_{ij}L_{pk} = \partial_j L_{ik} + \sum_p \Gamma^p_{ik}L_{pj},$$ which is the Codazzi-Mainardi equation. The coefficients of $x_l$ give $$\partial_k\Gamma^l_{ij} - L_{ij}L^l_k + \sum_p \Gamma^p_{ij}\Gamma^l_{pk} = \partial_j\Gamma^l_{ik} - L_{ik}L^l_j + \sum_p \Gamma^p_{ik}\Gamma^l_{pj},$$ and rearranging, $$\partial_j\Gamma^l_{ik} - \partial_k\Gamma^l_{ij} + \sum_p\big(\Gamma^p_{ik}\Gamma^l_{pj} - \Gamma^p_{ij}\Gamma^l_{pk}\big) = L_{ik}L^l_j - L_{ij}L^l_k.$$ The left side is $R^l_{ijk}$ by definition. This is Gauss's equation. ∎

Example 23.3 (Checking the equations on the sphere). For the unit sphere with outward normal, $L = -\mathrm{Id}$, so $L^l_j = -\delta^l_j$ and $L_{ij} = -g_{ij}$ (because $L_{ij} = \sum_k L^k_i g_{kj}$). Gauss's equations predict $$R^l_{ijk} = (-g_{ik})(-\delta^l_j) - (-g_{ij})(-\delta^l_k) = g_{ik}\,\delta^l_j - g_{ij}\,\delta^l_k.$$ In particular, with $i = k = 2$, $j = l = 1$ in spherical coordinates, $R^1_{212} = g_{22} = 1$. We will compute $R^1_{212}$ intrinsically in Example 24.2 and find the same answer, as it must be. The Codazzi-Mainardi equations reduce, since $L_{ij} = -g_{ij}$, to $\partial_k g_{ij} - \partial_j g_{ik} = \sum_p(\Gamma^p_{ik}g_{pj} - \Gamma^p_{ij}g_{pk})$, which is a consequence of the formula for the Christoffel symbols in terms of the metric.

Theorem 23.4 (Fundamental theorem of surfaces (stated without proof)). Let $U$ be an open disk (or any simply connected open set) in the plane, and let $g_{ij}$ and $L_{ij}$ be functions on $U$, with $g_{ij}$ of class $C^2$ and $L_{ij}$ of class $C^1$, both symmetric in $i$ and $j$, such that $g_{11} > 0$ and $\det(g_{ij}) > 0$. If they satisfy the Gauss and Codazzi-Mainardi equations (with $\Gamma^l_{ij}$, $R^l_{ijk}$ and $L^l_j$ computed from the $g_{ij}$ and $L_{ij}$ by the usual formulas), then there is a coordinate patch $x \colon U \to \mathbb{R}^3$ with first and second fundamental form coefficients $g_{ij}$ and $L_{ij}$. It is unique up to a rigid motion of $\mathbb{R}^3$.

Theorem 24.1 (Theorema Egregium (Gauss, 1827)). The Gaussian curvature of a surface is intrinsic. More precisely, in any coordinate patch, $$K = \frac{R_{2121}}{\det(g_{ij})} = \frac{\sum_l R^l_{212}\,g_{l1}}{g_{11}g_{22} - g_{12}^2},$$ which is computed from the metric coefficients $g_{ij}$ and their first and second derivatives.

Proof. By Gauss's equations in the lowered form $R_{ijkm} = L_{ik}L_{jm} - L_{ij}L_{km}$ (Theorem 23.2), with $i = 2$, $j = 1$, $k = 2$, $m = 1$, $$R_{2121} = L_{22}L_{11} - L_{21}L_{21} = \det(L_{ij}).$$ By Proposition 21.2, $K = \det(L_{ij})/\det(g_{ij})$. The left side $R_{2121} = \sum_l R^l_{212}\,g_{l1}$ is intrinsic, since the Riemann tensor is computed from the Christoffel symbols, which are computed from the metric (Proposition 15.1). ∎

Example 24.2 (The sphere, intrinsically). Use the coordinates $u^1 = \theta$, $u^2 = \phi$ on the unit sphere, with $g_{11} = \cos^2\phi$, $g_{12} = 0$, $g_{22} = 1$. In Example 18.10 we found the nonzero Christoffel symbols $$\Gamma^1_{12} = \Gamma^1_{21} = -\tan\phi, \qquad \Gamma^2_{11} = \sin\phi\cos\phi.$$ Since $g_{21} = 0$, we need only $R^1_{212}$: $$R^1_{212} = \partial_1\Gamma^1_{22} - \partial_2\Gamma^1_{21} + \sum_p\big(\Gamma^p_{22}\Gamma^1_{p1} - \Gamma^p_{21}\Gamma^1_{p2}\big).$$ Here $\Gamma^1_{22} = 0$, $\partial_2\Gamma^1_{21} = \frac{d}{d\phi}(-\tan\phi) = -\sec^2\phi$, all $\Gamma^p_{22}$ vanish, and in the last sum only $p = 1$ contributes, $\Gamma^1_{21}\Gamma^1_{12} = \tan^2\phi$. So $$R^1_{212} = 0 + \sec^2\phi + 0 - \tan^2\phi = 1, \qquad K = \frac{R^1_{212}\,g_{11}}{\det(g_{ij})} = \frac{\cos^2\phi}{\cos^2\phi} = 1.$$ This agrees with the extrinsic computation $K = \kappa_1\kappa_2 = (-1)^2 = 1$ from Example 20.7, and with Example 23.3.

Example 24.3 (The cone, intrinsically). For the cone $x(r, \theta) = (r\cos\theta, r\sin\theta, r)$ with $u^1 = r$, $u^2 = \theta$, $g_{11} = 2$, $g_{12} = 0$, $g_{22} = r^2$, the nonzero Christoffel symbols are $\Gamma^1_{22} = -r/2$ and $\Gamma^2_{12} = \Gamma^2_{21} = 1/r$ (Example 15.4). Then $$R^1_{212} = \partial_r\Big(-\frac r2\Big) - 0 + \big(\Gamma^1_{22}\Gamma^1_{11} + \Gamma^2_{22}\Gamma^1_{21}\big) - \big(\Gamma^1_{21}\Gamma^1_{12} + \Gamma^2_{21}\Gamma^1_{22}\big) = -\frac12 + 0 - \Big(0 + \frac1r\cdot\Big(-\frac r2\Big)\Big) = 0,$$ so $K = 0$, agreeing with Example 21.4.

Definition 24.4 (Isometry and local isometry). A map $f \colon M \to N$ between surfaces is differentiable if, in coordinate patches $x$ of $M$ and $y$ of $N$, the map $y^{-1} \circ f \circ x$ is differentiable. A one-to-one, onto, differentiable map $f \colon M \to N$ is an isometry if for every curve $\gamma$ in $M$, the curve $f \circ \gamma$ in $N$ has the same length as $\gamma$. Two surfaces $M$ and $N$ are locally isometric if every point of $M$ has a neighborhood that is isometric to an open subset of $N$, and vice versa.

Proposition 24.5 (Local isometries and the first fundamental form). Two surfaces $M$ and $N$ are locally isometric if and only if around every point of $M$ there is a patch $x \colon U \to M$, and a patch $y \colon U \to N$ on the same domain, such that $x$ and $y$ have the same metric coefficients $g_{ij}$; and similarly with the roles of $M$ and $N$ exchanged.

Proof sketch. If $x$ and $y$ have the same metric coefficients, then $f = y \circ x^{-1}$ is an isometry from $x(U)$ to $y(U)$: for a curve $\gamma(t) = x(\gamma^1(t), \gamma^2(t))$, the curve $f \circ \gamma = y(\gamma^1(t), \gamma^2(t))$ has the same length, since both lengths are $\int \sqrt{\sum g_{ij}\dot\gamma^i\dot\gamma^j}\,dt$ with the same $g_{ij}$. Conversely, if $f$ is an isometry from $x(U)$ to an open subset of $N$, set $y = f \circ x$. Applying the length condition to the parameter curves shows that $y_i$ has the same length as $x_i$, and applying it to the curves $t \mapsto x(a + t, b + t)$ shows that $|x_1 + x_2| = |y_1 + y_2|$; together these give $g_{ij} = h_{ij}$, where $h_{ij}$ are the metric coefficients of $y$. ∎

Corollary 24.6 (Local isometries preserve Gaussian curvature). If $f$ is a local isometry from $M$ to $N$, then the Gaussian curvatures of $M$ at $P$ and of $N$ at $f(P)$ are equal.

Example 24.7 (The catenoid and the helicoid). The catenoid is the surface of revolution $x(u, \theta) = (\cosh u\cos\theta,\ \cosh u\sin\theta,\ u)$, obtained by rotating the curve $r = \cosh z$ about the $z$-axis; one computes $g_{11} = g_{22} = \cosh^2 u$ and $g_{12} = 0$. The helicoid is the surface $y(v, \phi) = (v\cos\phi,\ v\sin\phi,\ \phi)$, swept out by a horizontal line rotating as it rises, like a spiral staircase; here $h_{11} = 1$, $h_{12} = 0$, $h_{22} = 1 + v^2$. These look like very different surfaces. But substituting $v = \sinh u$, $\phi = \theta$ into the helicoid gives a patch $\tilde y(u, \theta) = y(\sinh u, \theta)$ with $\tilde y_u = \cosh u\,y_v$ and $\tilde y_\theta = y_\phi$, so its metric coefficients are $\cosh^2 u \cdot 1 = \cosh^2 u$, $0$, and $1 + \sinh^2 u = \cosh^2 u$: exactly those of the catenoid. So the catenoid and the helicoid are locally isometric. One can bend a piece of a catenoid into a piece of a helicoid without stretching it, and the meridians of the catenoid become the straight lines of the helicoid. By Corollary 24.6 they have the same Gaussian curvature at corresponding points; both, as it happens, are minimal surfaces.

Corollary 24.8 (No perfect map of the Earth). No open subset of a sphere is isometric to an open subset of the plane.

Proof. A sphere of radius $r$ has Gaussian curvature $1/r^2 \ne 0$ at every point, and the plane has Gaussian curvature $0$. A local isometry would preserve $K$ by Corollary 24.6. ∎