Reference: D. Blanusa, "Über die Einbettung hyperbolischer Räume in Euclidische Räume," Monatsch. Math. 55, 217-229.
The hyperbolic plane will be given in hyperbolic
polar
coordinates
u=r, v=
in R2,
with the
geometry specified by E=1, F=0, G=G(u)=cosh2u. (As
in the ordinary Euclidean case, where G=1, the behavior of polar
coordinates at the origin does not present a problem, and the
choice
of
G(u) as above yields curvature K=-1.) Thus what's needed is
a
one-to-one differentiable map x from the right halfplane D into
R6 such that
xu
xu = 1,
xu
xv = 0,
xv xv = cosh2(u).
We give
the general form of the map x in terms of certain functions of u
only that will be defined later: three differentiable
functions int, f1, f2 and two step functions
1,
2.
x(u,v)
= (int(u), v, f1(u)
cos(v 1(u)),
f1(u) sin(v 1(u)),
f2(u) cos(v 2(u)),
f2(u) cos(v 2(u)).
Now compute E(x), F(x), G(x) in turn, by machine or by hand. The results will suggest how to define the five functions above.
Since 1 and
2 are step functions,
their derivatives are zero, and we find
E(x)(u,v) = int'(u)2 + f1'(u)2 + f2'(u)2.
Thus E(x)=1 requires
int'(u) = (1 - f1'(u)2 - f2'(u)2)1/2.
The function int can be suitably defined by an integral provided
(*) f1'(u)2 + f2'(u)2 < 1.
Next, F(x)=0 is immediate
since the derivatives of
1 and
2 are zero.
Finally,
G(x)(u,v) = 1 +
f1(u)2
1(u)2 +
f2(u)2
2(u)2.
Thus to get (**) G(x)(u,v) =cosh2(u), as required, the sum of the last two terms must be sinh2(u).
> A function k is defined by
k(t) = exp(-1/sin2
t)
sin(t).
The exponential here, a close
relative of the well known
exp(-1/t2), levels off the zeros
of
sin(t), as the graph shows.
Evidently, k
has the same periodicity and symmetries as
sin(t).
> Functions
1 and 2
are defined by
integrating k. Let A be the integral of k from 0 to 1.
Then let
1(u)=(1/A)
0u+1 k(t) dt,
and
2(u)=
(1/A)0u
k(t) dt.
Thus 1(u)=
2(u+1). Evidently
these functions are
nonnegative and periodic of period 2.
> Step functions
1 and
2 are
defined
by
1(u) = exp(2[(u+1)/2] + 3)
and
2(u) =
exp(2[u/2] + 4),
where [t] denotes the greatest integer in t.
> Functions f1 and f2 are defined by
f1(u) =
(1)
1/2 /
1(u)) sinh u,
and
f2(u) =
(2)
1/2 /
2(u)) sinh u
There is
nothing mysterious about the step functions. In
f1 and
f2, they are smoothed out by the
exponential
exp(-1/sin2 t) in the
definition of k.
Their role is to ensure that the derivatives of
f1 and
f2 are small enough so that the
inequality (*) will
hold.
f1'(u) < 1/
2
and
f2'(u) < 1/2.
For this we need an upper bound B for the
absolute values of
the derivatives of
(1)
1/2
and
(2)
1/2.
Because
1(u) =
2(u+1), it will suffice to
work with the latter. Since
2(u+2) =2(u),
we need only consider 2 on the interval [0,2],
and
since
2(2-u)
=2(u)
we can
further reduce to [0,1].
By the definition of 2,
((2)1/2)'
=
k(u)/(2A (2)1/2).
The accompanying figure shows the graph of this function, plotted numerically. Evidently, 3 is a satisfactory value for B. (There is no need for a close bound; 300 would do as well.)
Now we estimate |f1'|. By the definition of f1,
f1' = ((1)1/2)' /1) sinh u +
((1)1/2)/1) cosh u.
The values
of
1 lie in [0,1],
so using the bound B=3 gives
|f1'| <
(3/1)
sinh u +
(1/1)
cosh u <
(4/1)
eu.
Now 2[(u+1)/2]+1 > u, so the definition of
1
gives 1(u) >
eu+2.
Hence
|f1'| < 4/e2 <
1/2.
The same holds for f2; thus the proof of (*) is complete.
Proof of (**). Substituting the definitions of f1 and f2 into the formula for G(x)(u,v) in Section 1 gives
G(x)(u,v) = 1 + (1(u)
+
2(u))
sinh2(u).
So to get cosh2(u), we must
show that
1 +
2 = 1.
Using the fact that
k(u+1) = -k(u), we find
A 1(u) = 0u+1 k(t) dt =
01 k(t) dt +
1u+1
k(t) dt =
A + 0
u k(t+1)
dt = A -
0u k(t) dt =
A - A 2(u).
QED
Finally, for the map x to be an imbedding, it must also be one-to-one. But if x(u1,v1)=x(u2, v2), then since the second component of x is just v, we have v1=v2. And (*) shows that the first component of x is strictly increasing, so u1=u2.
Remark. It is not known whether the hyperbolic plane H can be isometrically imbedded in R5--or even in R4 (though the latter seems most unlikely).