1. Definition. Setting the function
equal to a positive constant gives an implicit definition of a surface in R3. This follows by a well known criterion, since the partial derivatives of km with respect to x, y, z are simultaneously zero only at the origin. For the constant 3a^4, we call the resulting surface Kummer's surface M(a) of size a. For simplicity, we take a=1 and write M=M(1). Any other choice of a just gives a scale model of M.
(Note: In algebraic geometry, a "surface" is allowed to have singularities that are forbidden to the--everywhere smooth--surfaces of differential geometry. In particular, the term "Kummer's surface" is applied much more broadly in algebraic geometry.)
Here is what Kummer's surface looks like:
The six "tubes" here, of course, actually extend out to infinity along the x, y, z axes in both directions. One can see from the formula for km that their cross-sections shrink down toward limit 0.
2. Symmetries.The figure exhibits visually the symmetries that can be deduced immediately from the formula, namely, invariance (a) under any permutation of the Euclidean coordinates x, y, z and (b) under changes of sign of any of these coordinates. In fact, these operations generate a group of 6·8=48 mappings, a typical map sending (x,y,z) to (y,-z,x).
These mappings are orthogonal transformations of R3 and
hence are isometries of the whole space R3. Since they carry M to
itself, their restrictions to M are congruences, that is, isometries of M that
preserve not only the intrinsic geometry of M, but also its shape in R3.
In fact, these 48 congruences are the only isometries of any kind from M to M (to be
proved).
3.Drawing the Surface.
We use Mathematica's ContourPlot3D. This command makes serious
demands on a computer, since (for example) to give a reasonable idea of
Kummer's surface, it would have to check points densely throughout a quite large
3-dimensional box to see which satisfy the equation km=3. Thus we construct the
picture above in pieces, starting with its central portion:
ContourPlot3D[km,{x,-2,2},{y,-2,2},
{z,-2,2},Contours->{3},Axes->True,
PlotPoints->{4,5},ImageSize->216]
We strip away the box and axes to get the official center:
ctr=Show[%,Boxed->False,Axes->False]
.
Now six tubular extensions are needed. For example, the one running in the
positive x direction is given by:
xp=ContourPlot3D[km,{x,2,6},{y,-1,1},
{z,-1,1},
Contours->{3},Boxed->False, PlotPoints->{3,5},
DisplayFunction->Identity]
The latter option prevents immediate display -- but the picture will appear in a moment.
After the other five extensions are similarly ordered, the whole surface is drawn by:
Show[{ctr,xp,xm,yp,ym,zp,zm},
ViewPoint->{1.3,2.6,1.7},
ImageSize->544, DisplayFunction->$DisplayFunction]
4. The Unit Points of M.
Evidently, the permutations of x,y,z leave the point (1,1,1) of M fixed.
Since these maps are isometries, it can be deduced without computation that
p=(1,1,1) is a critical point of the Gaussian curvature K of M. We verify this
computationally in the next section.
Any geometrical assertion about (1,1,1) must hold for the other points reachable by isometries mentioned above. Since only the sign-change isometries move (1,1,1), the full set of eight unit points is:
(1,1,1), (-1,1,1), (1,-1,1), (1,1,-1), (-1,-1,1), (1,-1,-1), (1,-1,-1), (-1,-1,-1).
The boxed figure of the central part of M suggests that at the point (1,-1,1)--recall Mathematica looks from y<0--the Gaussian curvature K is zero, but that at nearby points, K is probably negative. Indeed the large figure of M suggests that K<0 holds everywhere except at the unit points. So it is time to compute the curvature of M.
5. A Maximal Patch.
We could instruct Mathematica to compute Gaussian curvature K of M using
formulas appropriate for an implicitly defined surface, but this approach turns out to
be complicated. It is much simpler to define a largest possible Monge patch in M--
because of the symmetries of M, this one patch will be all we need.
Solving km=3 for z, and choosing the positive sign, we arrive at the Monge patch
mon[x_,y_] := {x,y,Sqrt[(3-x^2*y^2)/(x^2+y^2)]]
The formula for this patch shows that it is well defined on the region
D: 3-x^2 y^2>0 bounded by the four branches of the hyperbolas x*y=±
3. D is just the orthogonal projection of M onto the xy plane, and
the patch covers all points of M that lie above D. Roughly speaking, this is
the top half of M, including the entire+z tube. Note that the origin is not in D,
since no points of M lie above it--in fact, the vertical line up from the origin is
the axis of the +z tube.
In view of the symmetries of M, it can be covered entirely by patches congruent to mon, so this patch will suffice to describe much of the geometry of Kummer's surface.
6. Gaussian Curvature of M.
We denote by kkm the Gaussian curvature of M as expressed in terms of the
patch mon. Explicitly, the value of kkm at a point of D is the Gaussian
curvature of M at the point mon(x,y). There are standard elementary
formulas for curvature in this patch case. However, we will use the general
command superK, whose construction and storage in a file "superK.m" is
described for both Mathematica and
Maple. It is installed in the current page by
<<superK.m
Then
kkm=Simplify[superK[mon][x,y]]
The formula for kkm is not trivial, so to get some understanding of kkm, we plot it, with domain the largest square in D centered at the origin. The formulas for the boundary of D show at once that the corners of the square are (c,c), (-c,c), (c,-c), (c,c), where c is the fourth root of 3:
c=3^(1/4)
The command N[c] gives 1.316 as the approximate numerical value of c. Then the graph of kkm is produced by
spider=Plot3D[kkm,{x,-c,c},{y,-c,c},
ViewPoint->{1,-2,1},
PlotPoints->30] 
This picture is full of information. First, it indicates that the unit points are maximum points (local, at least). Second, it shows that on the tubes, curvature is strictly negative and increases toward limiting value 0 as one travels out all the way. The latter assertion is shown by the hill in the center of the figure. (Recall that the top of the hill is not in the patch, since the origin is not in its domain D.)
7. Proof that the Unit Points are Maxima.
Again, general formulas for the derivatives of kkm are not very informative. By symmetry, we only need these derivatives of kkm at (1,1) in D, where mon(1,1)=(1,1,1).
Curvature is zero there, as shown by:
kkm/.{x->1,y->1}
The first derivatives at (1,1) are both zero as shown by
{D[kkm,x]/.{x->1,y->1},D[kkm,x]/.{x->1,y->1}}
So (1,1) is a critical point of curvature.
The second derivatives at (1,1) are arranged into a Hessian matrix by the command
MatrixForm[{{D[kkm,x,x],D[kkm,x,y]},
{D[kkm,y,x],D[kkm,y,y]}}]/{x->1,y->1}
The command
Eigenvalues[%]
gives -3,-1 as its eigenvalues, so the Hessian matrix is negative definite. Consequently, mon(1,1)=(1,1,1) is a maximum point of curvature-- and so are the other unit points.
8. Minima of Curvature.
We assert that somewhere on Kummer's surface the Gaussian curvature
of M must take on a minimum value. Proof: Chopping off the ends of the six tubes
where K is arbitrarily near zero leaves a compact region--on it, K must take on a
minimum.
It is not obvious how to find where the minimum value occurs. For example, it is a struggle to go in the natural theoretical direction by using the command
Solve[{D[kkm,x]==0,D[kkm,y]==0},{x,y}]
So we abandon computation for graphics, and redraw the spider with a larger domain--ignoring points from outside the region D.
Plot3D[kkm,{x,-3,3},{y,-3,3},
ViewPoint->{1.3,-2.4,1.5}, PlotPoints->35]
Ignore outlying pieces! Here we see four minimum points, whose coordinates differ only in sign. Consider the one near x=1.3, y=0. Reducing the domain above to smaller regions, such as {x,1.2,1.4},{y,1.2,1.4}, makes the y=0 value look good. The plot
Plot[kkm/.{y->0},{x,1.26,1.37}]
shows the behavior of curvature down the z+ tube and out the x+ tube. Then taking ever smaller domains, starting with, say,
Plot[kkm/.{y->0},{x,1.26,1.37}]
clearly indicates that the minimum is at (c,0), where, as above, c~1.316... is the fourth root of 3. We verify this computationally by the same pattern as for the maximum points. The command shows that both derivatives are zero, so (c,0) is a critical point.
{D[kkm,x]/.{x->c,y->0},D[kkm,x]/.{x->c,y->0}}
Then the command
MatrixForm[{{D[kkm,x,x],D[kkm,x,y]},
{D[kkm,y,x],D[kkm,y,y]}}]/{x->c,y->0}
yields a diagonal matrix with entries 4/3 and 14/3. Thus (c,0) is a minimum point of kkm=K(mon) on D. Since mon(c,0)=(c,0,c), it follows from the symmetries of M that there are twelve minimum points of K on M, four in each coordinate plane:
(0,c,c), (0,-c,c), (0,c,-c), (0,-c,-c) (c,0,c), (-c,0,c), (c,0,-c), (-c,0,-c) (c,c,0), (-c,c,0), (c,-c,0), (-c,-c,0)
The same minimum value is taken on at each of these point, namely -1/3. This is readily found by hand from the formula for mon or by computer from
kkm /.{x->c,y->0}
9. Monkey Saddles in Kummer's Surface.
A close look at pictures of M above suggests that the point (1,-1,1) may be the
center of a monkey saddle--where three hills and three valleys meet.
To take a closer look, using the equivalent point p=(1,1,1), we refer the
nearby surface to the tangent plane to M at p. Either by symmetry or by
explicitly using the gradient of km, we find that the unitvector
(1,1,1)/3 at p is normal to the surface at p. Hence the
height of the nearby points--above or below--the tangent plane at p is given
in terms of the patch mon by the function
ht=(mon[x,y]-{1,1,1}).{1,1,1}/Sqrt[3]
In plotting this function, we stretch the normal direction slightly to emphasize the shape of the surface.
Plot3D[ht, {x,.8,1.2}, {y,.8,1.2},
PlotPoints->20, PlotRange->All,
BoxRatios->{1,1,1.2}]
Here the topography should reveal a monkey saddle. And as usual the other unit points are geometrically the same.
10. Geometry of Kummer's Surface M(a)
The Gaussian curvature K takes on its maximum value 0 at the eight points
(±a,±a,±a), where the sign are independently chosen. It takes on its minimum
value -1/(3a²) at the twelve points am, where m
ranges through the twelve minimum points for M=M(1) found in Sec. 8.
Proof. The mapping scalar multiplication by a is similarity map (or homothety) of R3 that sends M to M(a). Hence for a point x in M, the curvature K(ax) of M(a) at the point ax is K(x)/a2. This is true for any homothety,and can be verified by direct computation in the case at hand. Thus the maximum points of K on M map to maximum points on M(a), and similarly for the minimum points.
The 48 congruences of M are the restrictions of isometries of R3, and evidently they carry every M(a) to itself, so their restrictions to M(a) are congruences of M(a).
To complete the story of the isometries of M(a), we require this fact--for which it suffices to consider only M=M(1):
Lemma. Every isometry of M that leaves (1,1,1) fixed is one of the 6 congruences given by permutations of x,y,z.
Outline of Proof. Any isometry permutes unit points since they are the maximum points of curvature. Thus an isometry F that leaves (1,1,1) fixed must permute the three unit points (-1,1,1), (1,-1,1), (1,1-1) that are nearest to (1,1,1). There is a unique shortest geodesic from (1,1,1) to each of these three. For instance, the shortest route from (1,1,1) to (-1,1,1) is along the curve sliced from M by the plane y=z. This a consequence of the symmetry sending (x,y,z) to (x,z,y), and is visually clear in the large picture of M above.
Then F must permute these three geodesics and hence permute their initial velocities. But there is a unique congruence C that produces the same permutation of unit points, hence of geodesics, hence of initial velocities. Thus F and C have the same effect on the tangent space to M at (1,1,1). By a standard theorem, it follows that F=C.
Corollary. The 48 conguences of M are the only isometries of M -- and similarly for M(a).
Proof. Let H be an arbitrary isometry of M. It must carry (1,1,1) to one of the 8 unit points, say,u. Then there is evidently a congruence C--given by a suitable permutation of coords and change of signs--that carries u back to (1,1,1). Thus the composite map C H is an isometry of M that leaves (1,1,1) fixed. By the Lemma, this is a congruence, say B. But then C H=B, and consequently, H=C-1 B, which is again a congruence.
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