Hilbert’s Theorem, via moving frames
Abstract
We present a proof that the hyperbolic plane cannot be isometrically immersed in Euclidean -space by a map. Ideas from many topics in (essentially) undergraduate mathematics are applied; the use of moving frames and connection forms to express the geometry simplifies the outline of the proof, compared to, say, using coordinate patches and Christoffel symbols. The key transition is from expressions in terms of the principal directions on the immersed surface (which give access to the Gaussian curvature) to expressions in terms of the asymptotic directions (which yield a coordinate system and give access to surface area).
1 Outline of the proof
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•
Assume for simplicity that is a isometric embedding (immersions are no more difficult). In this section, “the surface” will refer to the image of .
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Construct an orthonormal frame field on which is mapped by to unit vectors in the principal directions on the surface (maximizing and minimizing normal curvature).
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Construct a frame field on which is mapped by to linearly independent unit vectors in asymptotic directions (where the normal curvature equals zero).
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Express the connection -form (for ) first in terms of the -forms dual to and the principal curvatures [Lemma 1], then in terms of the same forms and the angle between either of the asymptotic directions and [Lemma 2], and finally change basis to the -forms dual to [equation (23)]. (The Gaussian curvature on being identically equal to is used in the proof of Lemma 2.)
- •
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•
Define to be the (oriented) angle between and ; hence, is the area distortion factor when mapping from the -plane to . Show that [equation (26)].
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This approach to Hilbert’s theorem was inspired by the argument given by Rubens Leão in the Appendix of [dC83]. Citations will be supplied for equations that come from undergraduate differential geometry and/or from tensor analysis, but proofs will usually be omitted. For differential geometry, [O’N06] and [Shi18, Chapter 3, Section 3] use moving frames; [MP77] and [dC16] may also be helpful. Basic facts about differential forms in (exterior differentiation, wedge product, pullback) can be found in [dC94, Chapter 1] and in [Spi79a, Chapter 7].
2 Proof of the main theorem
Theorem 1.
There does not exist a isometric immersion of the hyperbolic plane into Euclidean -space.
Before starting the proof, we make a foundational definition and a remark on notation.
Definition 1.
A moving frame on an open subset of is an ordered -tuple of smooth (i.e., ) vector fields , such that the vectors at each point form an orthonormal basis for the tangent space .
The previous definition agrees with [dC83, page 118]; see also [dC94, page 77] and [O’N06, Chapter II, Section 1]. Elsewhere, such as in [Spi79b, page 285], the vectors in a moving frame need only be linearly independent (and span).
Remark.
We will use lower subscripts for contravariant objects such as vectors, and upper subscripts for their components with respect to a basis (as in the preceding definition). Similarly, we will use upper subscripts for covariant objects such as differential forms, and lower subscripts for their components with respect to a basis. Summation signs are needed only a few times, and they are not omitted, as they would be in “Einstein summation convention” [Spi79a, pages 50–51, pages 155–158].
Proof of Theorem 1.
The Poincaré disk will represent the hyperbolic plane; define (see the Appendix for the Riemannian metric and other details). Suppose that is a isometric immersion.
2.1 Construct vector fields and -forms on
For all , there is a positive number , such that is an embedding, when restricted to the open disk of radius in the hyperbolic metric. The tangent space to the surface at can inherit an orientation from , via , so there is a preferred unit normal vector at , which can be continuously (indeed, smoothly) extended to a unit normal vector field .
At every point , the map , is a self-adjoint linear transformation of [dC16, page 142]; note that and are both names for the orthogonal complement in of . The eigenvalues are equal to the principal curvatures (let denote the positive eigenvalue, and the negative eigenvalue), and the corresponding (one-dimensional, orthogonal) eigenspaces give the principal directions [dC16, page 146]. At this stage, we are using the fact that the Gaussian curvature (product of the principal curvatures) is negative at every point of , but not (yet) the fact that it is constant and equal to .
Via , we can transfer these eigenspaces back to form two smooth line fields on each . Whenever , the line fields for and will agree at , since they both describe the extrinsic geometry of at . Hence we can merge all of the “local” line fields into two “global” line fields on . There is a two-sheeted covering space of , consisting of all the unit vectors in , and since is simply-connected, this covering space is not connected. In other words, there are two unit vector fields on which everywhere point along . We define to be the result of rotating a quarter-turn counter-clockwise (i.e., radians).
This allows us to define and on each , so we now have a (positively oriented) moving frame on . As in [dC94, page 82], we can extend this moving frame to a small open subset of by displacing frames normally by a sufficiently small amount (choose such that by is a diffeomorphism). See Figure 2 for an illustration.
Since are linear transformations on at every point of (in other words, they are -valued -forms, as defined in [Spi79a, page 546], while are -valued -forms), we can define the connection -forms by
(1) |
for all vector fields on [dC94, page 78] [Spi79b, page 287]. The notation follows [Spi79b, page 287] and corresponds to in [dC94, page 78].
On each we can also define -forms dual to the frame, that is,
(2) |
(using the Kronecker delta). These -forms on are related to each other by the Cartan structure equations, which follow.
Proposition 1.
Proof.

Now, on each , we can define the -forms , (noting that pulls back to the zero -form) and , . The reader can easily check that are dual to .
Remark.
Since are globally defined on , the moving frames at , arising respectively from and , will agree. Hence the -forms and , defined above, will agree on . Consequently, as we did above with the line fields, we merge the “local” ’s and ’s into “global” smooth -forms that are defined on .
Furthermore, since the “unbarred” versions of equations (3) and (4) can be justified by basic facts about pullbacks and wedge products in each , Proposition 1 implies the following equations on (or more precisely, in the exterior algebra of differential forms on ):
(5) |
(6) |
Also, by definition, in each ,
(7) |
[BL10, page 179], [MP77, pages 125–129], [Spi79b, pages 103–106]. From the first equation we have, using (1), and ; similarly, the second equation implies and . We conclude, after pullback to , with , that
(8) |
Remark.
To keep the signs straight, it is useful to keep in mind the example of the unit sphere (with one point removed, so that there exists a moving frame), oriented with outward normal. Then equals at , so is the identity map. The normal curvatures at each point all equal (negative since curves on the sphere bend away from the normal vector). So for this surface the principal curvatures both equal , which is consistent with equation (7).
2.2 Expressions for in terms of
We now turn to the proofs of two key lemmas, which relate to the principal curvatures and to the angle between the asymptotic directions and .
Lemma 1.
With , , and defined as above,
(9) |
Proof.
To verify this equation of -forms, it suffices to show the following two equations:
(10) |
We rewrite in two ways, using equations (5), (6), and (8), obtaining
In other words, . Evaluating both sides of this equation on the pair (,), we obtain
which justifies the first half of (10). Similarly, rewriting ,
In other words, . Again evaluating the left-hand and right-hand sides at (,),
which justifies the second half of (10). ∎
For all , the normal curvatures at can be described by , defined in terms of the Euclidean inner product for vectors in [dC94, page 87]. In essence, this is the second fundamental form for the immersed surface near , pulled back to a quadratic form on . Using Euler’s theorem [dC16, page 147] [MP77, page 129], and (for the first time) making use of the fact that Gaussian curvature equals everywhere on , we have normal curvature zero when
for a unit vector that makes an angle with . We conclude that , so set , and define the asymptotic directions
(11) |
It follows that
(12) |
Lemma 2.
With , , and defined as above,
(13) |
2.3 Use to construct a global coordinate map
We now turn our attention to the properties of the asymptotic vector fields .
Lemma 3.
Let be the -forms which are dual to . Then and .
Proof.
It is not hard to use the definitions of and in (11) to verify that
(16) |
Next, we see that
(17) |
and similarly
(18) |
We next show that the vector fields and on each generate a global flow (i.e., each integral curve is defined for all ). We will use the following special case of [Cod61, Chapter 6, §6, Theorem 2].
Theorem 2.
Let be positive numbers, let , let , and let . If
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1.
is a continuous function, and
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2.
there exists such that for all , , and
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3.
there exists such that for all , ,
then the initial-value problem has a solution
where . ∎
Note that does not depend on , the Lipschitz constant.
Proposition 2.
If is a vector field on which everywhere has unit length with respect to the hyperbolic metric, then every maximal integral curve of is defined for all .
Proof.
Preparing to apply Theorem 2, we will think of the unit-hyperbolic-length vector field as a vector field on the open unit disk in , and now measure length with respect to the Euclidean metric. Set and . Next, set , and set , so that , and define . By the definition of the hyperbolic metric on (see Appendix), we can take (for any choice of ). Hence, .
The partial derivatives of with respect to the standard coordinates of are continuous on the compact set , and hence are bounded, so we conclude that the Lipschitz constant exists. By Theorem 2, we have a solution .
Now fix an arbitrary point . There exists a hyperbolic isometry which maps to (see equation (30)). Use to map the vector field to a “recentered” vector field on , then apply the previous reasoning to , obtaining an integral curve such that . The curve will be the desired integral curve through .
Since is a smooth vector field on the smooth manifold , and every point of can flow forward and backward along for at least one unit of time, it follows from the Uniform Time Lemma [Lee13, Lemma 9.15, page 216] that every integral curve can be extended to an integral curve defined on . ∎
Corollary 1.
All integral curves of and , the unit asymptotic directions, are defined for all time. ∎
Proposition 3.
There is a diffeomorphism from onto , such that
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1.
is a (global) coordinate chart for , and
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2.
, .
Proof.
Working in , with , define . is well-defined, using Green’s theorem, since and are closed and any two paths from to bound a region in . By definition, the differential of at will map to , for . Since and are everywhere linearly independent, the differential is everywhere nonsingular, so is a local diffeomorphism (using a version of the Inverse Function theorem; see [Lee13, Theorem C.34] ).
In addition, is surjective, since in order to find a point in which maps to , it suffices to start at at time zero, follow an integral curve for (forward if , backward if ) for time , then similarly follow an integral curve for for time , using Corollary 1. By the definition of , along the first part of the curve, the second coordinate of the output will not change, and vice versa.
With respect to the orthonormal bases in and in , any local inverse for has Jacobian matrix
(19) |
(using equation (11)) and the singular values of this matrix are easily calculated to be and . From this we can conclude that all local inverses are Lipschitz maps with constant . Now let be a continuous path, and let be a point in such that . We claim that has a unique lift to ; that is, there is a unique path such that and such that .
Let . This is an open subset of , since when lifts as far as , a local diffeomorphism from an open subset of to an open neighborhood of will allow the lift to be (uniquely) extended a little further. It is also true that is a closed subset of , since if lifts, then is uniformly continuous on . Using the Lipschitz bound in the previous paragraph, is also uniformly continuous, and hence (see Appendix for details) has a unique extension to . Finally, since is connected, and is a non-empty subset that is both open and closed, we conclude that , hence lifts uniquely.
Since is path-connected, and is simply-connected, the following lemma applies to the map .
Lemma 4 ([Lim03, Proposition 6.12]).
Let be a local homeomorphism with the unique-path-lifting property. If is path-connected and is simply-connected, then is a homeomorphism.
Proof.
(Sketch) Construct a path between any two points in that maps to the same place. Use to map the path, continuously deform the mapped path to a constant path, and then lift the homotopy, to show that the initial and final points of the original path are equal. ∎
Since is now known to be a homeomorphism (and in particular is injective), and has already been shown to be a local diffeomorphism, it follows that is a diffeomorphism. ∎
Corollary 2.
Let be defined by mapping the point to the point in obtained by starting at , following an integral path for the vector field for time , then following an integral path for the vector field for time (moving backward when these numbers are negative). Then .
Proof.
By construction, is the identity map on . Proposition 3 implies that is injective, so implies . ∎
Since Proposition 3 has shown that is a coordinate mapping for , we can define coordinate functions on such that and
(20) |
(By a common abuse of notation, we let and denote the functions on that return, respectively, the first and second coordinates of a point.)
2.4 Area computations
Adding and subtracting the equations in (16), we have
(21) |
Therefore, using Lemma 2,
(22) |
We define , the angle between and (using the capital letter to avoid confusion with the 1-forms ). Now (22) becomes
(23) |
which means that (using (16) and to show the final equality)
(24) |
where represents the “area form” on (since is orthonormal). At the same time,
(25) |
on , which we can transfer to the -plane, after defining , as
(27) |
By Proposition 3 and the definition of , is the factor by which area is distorted, when maps the -plane to . Hence, the area of the hyperbolic plane is given by
(28) |
and since , is bounded above by . On the other hand, the hyperbolic plane has infinite area, as computed below (in the Poincaré model; see the Appendix for the Riemannian metric):
(29) |
Assuming the existence of has led to a contradiction, hence there does not exist an isometric immersion of into , concluding the proof of Theorem 1. ∎
Corollary 3.
If is a Riemannian -manifold for which the metric is complete and has constant curvature , then there does not exist a isometric immersion of into .
Proof.
Corollary 4.
If is a Riemannian -manifold for which the metric is complete and has constant curvature (), then there does not exist a isometric immersion of into .
Proof.
Suppose that is such an isometric immersion, and let be the dilation . Since the effect of the dilation is to divide all normal curvatures by , the image of the composition has constant curvature everywhere, which contradicts Corollary 3 (if the original inner product at every point on is multiplied by , the new Riemannian metric will still be complete, and will be an isometric immersion). ∎
3 Brief History
Hilbert’s original proof [Hil01] assumed that the surface in was real-analytic (and had constant negative curvature ). However, the outline of his argument will work if is smooth enough to ensure that the PDE in equation (26) makes sense. Tracing back the definition of , we can see that if , then the normal vector is , and the eigenvalues of are , as is . So exists and is continuous on . With more sophisticated techniques, Efimov proved that there is no isometric immersion of the hyperbolic plane; see [Mil72] for an exposition and more history. On the other hand, by work of Nash and Kuiper, there does exist a isometric embedding of the hyperbolic plane in [Nas54] [Kui55].
4 Appendix: Metric properties of
Let . is the Poincaré disk model for , with Riemannian metric [MP77, page 179], [Rat94, Theorem 4.5.5]. inherits an orientation from the usual orientation on . Furthermore, the corresponding distance function on is [Rat94, Theorem 4.5.1]
where and denotes the Euclidean length of the vector .
Proofs of the following properties can be found in many textbooks, including [Rat94].
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1.
The metrics and are invariant under Möbius transformations which leave invariant.
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2.
The -metric topology on equals the topology that the inherits as a subset of (with the usual topology).
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3.
Every open -ball centered at a point in equals (as a set) an open Euclidean disk whose closure is contained in .
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4.
The metric space is complete [Rat94, Theorem 8.5.1].
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5.
After identifying with the complex number , the maps
(30)
Furthermore, if and is uniformly continuous, then there is a unique extension of to a (uniformly) continuous map . This is a very special case of an exercise in [Mun00, Chapter 7]. The proof hinges on the fact that every sequence in converging to is a Cauchy sequence and will be mapped by to a Cauchy sequence in .
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