mathtools: A fixed point Brouwer's invariance of domain theorems, and Hilbert's fifth problem

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We recall Brouwer’s famous fixed point theorem:

Theorem 1 (Brouwer fixed point theorem) Let ${f: B^n \rightarrow B^n}$ be a continuous function on the unit ball ${B^n := \{ x \in {\bf R}^n: \|x\| \leq 1 \}}$ in a Euclidean space ${{\bf R}^n}$ . Then ${f}$ has at least one fixed point, thus there exists ${x \in B^n}$ with ${f(x)=x}$ .

This theorem has many proofs, most of which revolve (either explicitly or implicitly) around the notion of the degree of a continuous map ${f: S^{n-1} \rightarrow S^{n-1}}$ of the unit sphere ${S^{n-1} := \{ x \in {\bf R}^n: \|x\|=1\}}$ to itself, and more precisely around the stability of degree with respect to homotopy. (Indeed, one can view the Brouwer fixed point theorem as an assertion that some non-trivial degree-like invariant must exist, or more abstractly that the homotopy group ${\pi_{n-1}(S^{n-1})}$ is non-trivial.)

One of the many applications of this result is to prove Brouwer’s invariance of domain theorem:

Theorem 2 (Brouwer invariance of domain theorem) Let ${U}$ be an open subset of ${{\bf R}^n}$ , and let ${f: U \rightarrow {\bf R}^n}$ be a continuous injective map. Then ${f(U)}$ is also open.

This theorem in turn has an important corollary:

Corollary 3 (Topological invariance of dimension) If ${n > m}$ , and ${U}$ is a non-empty open subset of ${{\bf R}^n}$ , then there is no continuous injective mapping from ${U}$ to ${{\bf R}^m}$ . In particular, ${{\bf R}^n}$ and ${{\bf R}^m}$ are not homeomorphic.

This corollary is intuitively obvious, but note that topological intuition is not always rigorous. For instance, it is intuitively plausible that there should be no continuous surjection from ${{\bf R}^m}$ to ${{\bf R}^n}$ for ${n>m}$ , but such surjections always exist, thanks to variants of the Peano curve construction.

Theorem 2 or Corollary 3 can be proven by simple ad hoc means for small values of ${n}$ or ${m}$ (for instance, by noting that removing a point from ${{\bf R}^n}$ will disconnect ${{\bf R}^n}$ when ${n=1}$ , but not for ${n>1}$ ), but I do not know of any proof of these results in general dimension that does not require algebraic topology machinery that is at least as sophisticated as the Brouwer fixed point theorem. (Lebesgue, for instance, famously failed to establish the above corollary rigorously, although he did end up discovering the important concept of Lebesgue covering dimension as a result of his efforts.)

Nowadays, the invariance of domain theorem is usually proven using the machinery of singular homology. In this post I would like to record a short proof of Theorem 2 using Theorem 1 that I discovered in a paper of Kulpa, which avoids any use of algebraic topology tools beyond the fixed point theorem, though it is more ad hoc in its approach than the systematic singular homology approach.

Remark 1 A heuristic explanation as to why the Brouwer fixed point theorem is more or less a necessary ingredient in the proof of the invariance of domain theorem is that a counterexample to the former result could conceivably be used to create a counterexample to the latter one. Indeed, if the Brouwer fixed point theorem failed, then (as is well known) one would be able to find a continuous function ${F: B^n \rightarrow S^{n-1}}$ that was the identity on ${S^{n-1}}$ (indeed, one could take ${F(x)}$ to be the first point in which the ray from ${f(x)}$ through ${x}$ hits ${S^{n-1}}$ ). If one then considered the function ${G: B^n \rightarrow {\bf R}^n}$ defined by ${G(x) := (1+\|x\|) F(x)}$ , then this would be a continuous function which avoids the interior of ${B^n}$ , but which maps the origin ${0}$ to a point on the sphere ${S^{n-1}}$ (and maps ${S^{n-1}}$ to the dilate ${2 \cdot S^{n-1}}$ ). This could conceivably be a counterexample to Theorem 2, except that ${G}$ is not necessarily injective. I do not know if there is a more rigorous way to formulate this connection.

The reason I was looking for a proof of the invariance of domain theorem was that it comes up in the very last stage of the solution to Hilbert’s fifth problem, namely to establish the following fact:

Theorem 4 (Hilbert’s fifth problem) Every locally Euclidean group is isomorphic to a Lie group.

Recall that a locally Euclidean group is a topological group which is locally homeomorphic to an open subset of a Euclidean space ${{\bf R}^n}$ , i.e. it is a continuous manifold. Note in contrast that a Lie group is a topological group which is locally diffeomorphic to an open subset of ${{\bf R}^n}$ , it is a smooth manifold. Thus, Hilbert’s fifth problem is a manifestation of the “rigidity” of algebraic structure (in this case, group structure), which turns weak regularity (continuity) into strong regularity (smoothness).

It is plausible that something like Corollary 3 would need to be invoked in order to solve Hilbert’s fifth problem. After all, if Euclidean spaces ${{\bf R}^n}$ , ${{\bf R}^m}$ of different dimension were homeomorphic to each other, then the property of being locally Euclidean loses a lot of meaning, and would thus not be a particularly powerful hypothesis. Note also that it is clear that two Lie groups can only be isomorphic if they have the same dimension, so in view of Theorem 4, it becomes plausible that two Euclidean spaces can only be homeomorphic if they have the same dimension, although I do not know of a way to rigorously deduce this claim from Theorem 4.

Interestingly, Corollary 3 is the only place where algebraic topology enters into the solution of Hilbert’s fifth problem (although its cousin, point-set topology, is used all over the place). There are results closely related to Theorem 4, such as the Gleason-Yamabe theorem mentioned in a recent post, which do not use the notion of being locally Euclidean, and do not require algebraic topological methods in their proof. Indeed, one can deduce Theorem 4 from the Gleason-Yamabe theorem and invariance of domain; we sketch a proof of this (following Montgomery and Zippin) below the fold.

— 1. Invariance of domain —

Now we prove Theorem 2. By rescaling and translation invariance, it will suffice to show the following claim:

Theorem 5 (Invariance of domain, again) Let ${f: B^n \rightarrow {\bf R}^n}$ be an continuous injective map. Then ${f(0)}$ lies in the interior of ${f(B^n)}$ .

Let ${f}$ be as in Theorem 5. The map ${f: B^n \rightarrow f(B^n)}$ is a continuous bijection between compact Hausdorff spaces and is thus a homeomorphism. In particular, the inverse map ${f^{-1}: f(B^n) \rightarrow B^n}$ is continuous. Using the Tietze extension theorem, we can find a continuous function ${G: {\bf R}^n \rightarrow {\bf R}^n}$ that extends ${f^{-1}}$ .

The function ${G}$ has a zero on ${f(B^n)}$ , namely at ${f(0)}$ . We can use the Brouwer fixed point theorem to show that this zero is stable:

Lemma 6 (Stability of zero) Let ${\tilde G: f(B^n) \rightarrow {\bf R}^n}$ be a continuous function such that ${\|G(y)-\tilde G(y)\| \leq 1}$ for all ${x \in f(B^n)}$ . Then ${\tilde G}$ has at least one zero (i.e. there is an ${y \in f(B^n)}$ such that ${\tilde G(y)=0}$ ).

Proof: Apply Theorem 1 to the function

$\displaystyle x \mapsto x - \tilde G(f(x)) = G(f(x)) - \tilde G(f(x)).$

$\Box$

Now suppose that Theorem 5 failed, so that ${f(0)}$ is a not an interior point of ${f(B^n)}$ . We will use this to locate a small perturbation of ${G}$ that no longer has a zero on ${f(B^n)}$ , contradicting Lemma 6.

We turn to the details. Let ${\epsilon > 0}$ be a small number to be chosen later. By continuity of ${G}$ , we see (if ${\epsilon}$ is small enough) that we have ${\|G(y)\| \leq 0.1}$ whenever ${y \in {\bf R}^n}$ and ${\|y-f(0)\| \leq 2\epsilon}$ .

On the other hand, since ${f(0)}$ is not an interior point of ${f(B^n)}$ , there exists a point ${c \in {\bf R}^n}$ with ${\|c-f(0)\| \leq \epsilon}$ that lies outside ${f(B^n)}$ . By translating ${f}$ if necessary, we may take ${c=0}$ ; thus ${f(B^n)}$ avoids zero, ${\|f(0)\| \leq \epsilon}$ , and we have $\displaystyle \|G(y)\| \leq 0.1 \hbox{ whenever } \|y\| \leq \epsilon. \ \ \ \ \ (1)$

Let ${\Sigma}$ denote the set ${\Sigma := \Sigma_1 \cup \Sigma_2}$ , where

$\displaystyle \Sigma_1 := \{ y \in f(B^n): \|y\| \geq \epsilon \}$

and $\displaystyle \Sigma_2 := \{ y \in {\bf R}^n: \|y\| = \epsilon \}.$

By construction, ${\Sigma}$ is compact but does not contain ${f(0)}$ . It is a perturbation of ${f(B^n)}$ , and crucially there is a continuous map ${\Phi: f(B^n) \rightarrow \Sigma}$ defined by setting $\displaystyle \Phi(y) := \max( \frac{\epsilon}{\|y\|}, 1 ) y. \ \ \ \ \ (2)$

Note that ${\Phi}$ is continuous and well-defined since ${f(B^n)}$ avoids zero.

By construction, ${G}$ is non-zero on ${\Sigma_1}$ ; since ${\Sigma_1}$ is compact, ${G}$ is bounded from below on ${\Sigma_1}$ by some ${0 . By the Weierstrass approximation theorem, we can find a polynomial <IMG class=latex title=$ such that $\displaystyle \| P(y) - G(y) \| for all <IMG class=latex title=$ ; in particular, ${P}$ does not vanish on ${\Sigma_1}$ . At present, it is possible that ${P}$ vanishes on ${\Sigma_2}$ . But as ${P}$ is smooth and ${\Sigma_2}$ has measure zero, ${P(\Sigma_2)}$ also has measure zero; so by shifting ${P}$ by a small generic constant we may assume without loss of generality that ${P}$ also does not vanish on ${\Sigma_2}$ . (If one wishes, one can use an algebraic geometry argument here instead of a measure-theoretic one, noting that ${P(\Sigma_2)}$ lies in an algebraic hypersurface and can thus be generically avoided by perturbation. A purely topological way to avoid zeroes in ${\Sigma_2}$ is also given in Kulpa’s paper.)

Now consider the function ${\tilde G: f(B^n) \rightarrow {\bf R}^n}$ defined by

$\displaystyle \tilde G(y) := P( \Phi( y ) ).$

This is a continuous function that is never zero. From (3), (2) we have $\displaystyle \| G(y) - \tilde G(y) \| whenever <IMG class=latex title=$ is such that ${\|y\| > \epsilon}$ . On the other hand, if ${\|y\| \leq \epsilon}$ , then from (2), (1) we have $\displaystyle \| G(y) \|, \| G( \Phi(y) ) \| \leq 0.1$

and hence by (3) and the triangle inequality $\displaystyle \| G(y) - \tilde G(y) \| \leq 0.2 + \delta.$

Thus in all cases we have

$\displaystyle \| G(y) - \tilde G(y) \| \leq 0.2 + \delta \leq 0.3$

for all ${y \in f(B^n)}$ . But this, combined with the non-vanishing nature of ${\tilde G}$ , contradicts Lemma 6.

— 2. Hilbert’s fifth problem —

We now sketch how invariance of domain can be used to establish the solution to Hilbert’s fifth problem (as formulated in Theorem 4). Our main tool is the Gleason-Yamabe theorem, in the form stated in Theorem 4 of this previous post:

Theorem 7 (Gleason-Yamabe theorem) Every locally compact Hausdorff group has an open subgroup that is the projective limit of Lie groups.

A locally Euclidean group is of course locally compact Hausdorff; it is also first countable, and hence (by the Birkhoff-Kakutani theorem, discussed in this post) is also metrisable. Because of this, it is not difficult to show that the open subgroup given by the Gleason-Yamabe theorem in this case can be obtained as a projective limit

$\displaystyle G := \lim_{\stackrel{\leftarrow}{n}} G_n$

of a countable sequence ${(G_n)_{n \in {\bf N}}}$ of Lie groups, with continuous homomorphisms from ${G_{n+1}}$ to ${G_n}$ for each ${n}$ .

Using Cartan’s theorem to obtain the smoothness of outer automorphisms on Lie groups as in previous posts, we see that to show that a topological group is a Lie group, it suffices to locate an open subgroup that is a Lie group. In view of these reductions, it suffices to show

Proposition 8 Let ${G}$ be a locally Euclidean group that is the inverse limit $\displaystyle G := \lim_{\stackrel{\leftarrow}{n}} G_n$
of Lie groups. Then ${G}$ is itself isomorphic to a Lie group.

We now prove the proposition. We first observe by shrinking the ${G_n}$ if necessary (and using Cartan’s theorem that locally compact subgroups of a Lie group are still Lie) we may assume that all the projection maps in the inverse limit are surjective; indeed, by local compactness we may take ${G_n = G/H_n}$ where ${H_n}$ are a sequence of compact normal subgroups converging to the identity.

As continuous homomorphisms of Lie groups are automatically smooth (as proven in this previous post), the projection maps from ${G_{n+1}}$ to ${G_n}$ induce an associated projection map from ${{\mathfrak g}_{n+1}}$ to ${{\mathfrak g}_n}$ at the Lie algebra level. As the former maps are surjective, the latter maps are also. In particular, the dimensions of the finite-dimensional Lie algebras ${{\mathfrak g}_n}$ are non-decreasing in ${n}$ .

Another consequence of surjectivity is that every tangent vector in ${{\mathfrak g}_n}$ can be lifted up to ${{\mathfrak g}_{n+1}}$ . Continuing this process and passing to the inverse limit, it is possible to show that every one-parameter subgroup in one of the factor groups ${G_n}$ can be lifted up (locally, at least) to a one-parameter subgroup of ${G}$ . In fact, one can show that a neighbourhood of the origin in ${{\mathfrak g}_n}$ can be lifted up into ${G}$ , to provide a continuous injection of an open subset of that Lie algebra into ${G}$ .

Now we crucially use the invariance of domain (Corollary 3) to conclude that the dimensions of the Lie algebras ${{\mathfrak g}_n}$ must be bounded in ${n}$ (indeed, they cannot exceed the dimension of the locally Euclidean group ${G}$ ). As these dimensions are non-decreasing, they must therefore be eventually constant; by discarding finitely many of the factor groups, we may thus assume that all the ${G_n}$ have the same dimension. In particular, the kernels of the projection maps from ${G_{n+1}}$ to ${G_n}$ are zero-dimensional Lie groups and are thus discrete. As ${G_n = G/H_n}$ , this means that the ${H_n/H_{n+1}}$ are also discrete, thus ${H_{n+1}}$ is an open subgroup of ${H_n}$ , and so (by compactness of ${H_n}$ ) has finite index. As the ${H_n}$ converge to the trivial group as ${n \rightarrow \infty}$ , we conclude that each of the ${H_n}$ are profinite, and in particular are totally disconnected. Thus, for any ${n}$ , we have a short exact sequence

$\displaystyle 0 \rightarrow H_n \rightarrow G \rightarrow G_n \rightarrow 0$

describing ${G}$ as the extension of a Lie group ${G_n}$ by a totally disconnected compact group ${H_n}$ . Furthermore, we can lift a small neighbourhood of the identity of ${G_n}$ to ${G}$ in a continuously injective manner. This neighbourhood is connected, and acts on the normal subgroup ${H_n}$ by conjugation. But ${H_n}$ is totally disconnected, and so all orbits of this action must be points; thus, the lift of this neighbourhood commutes with ${H_n}$ . As a consequence, one can show that ${G}$ has the local topological (and group) structure of the direct product ${H_n \times G_n}$ . But ${G}$ is also locally Euclidean; as ${H_n}$ is totally disconnected, these two statements are only compatible if ${H_n}$ is discrete. Thus the Lie group ${G_n}$ is an open subgroup of ${G}$ , and we are done.

Remark 2 The above argument shows that any metrisable group which is “finite-dimensional” in the sense that it does not contain continuous injective images of non-trivial open sets of Euclidean spaces ${{\bf R}^n}$ of arbitrarily large dimension, has the local structure (as a topological space) of the product of a Euclidean space and a totally disconnected space. Of course, direct products of Lie groups and totally disconnected groups provide one such example of this claim. A more non-trivial example is given by the solenoid groups, a simple example of which is the (additive) group $\displaystyle G := ({\bf Z}_p \times {\bf R}) / {\bf Z}^\Delta$
where ${{\bf Z}^\delta := \{ (n,n): n \in {\bf Z} \}}$ is the diagonally embedded copy of the integers. This is a compact Hausdorff group that has a short exact sequence $\displaystyle 0 \rightarrow {\bf Z}_p \rightarrow G \rightarrow {\bf R}/{\bf Z} \rightarrow 0$
and has the local structure of ${{\bf Z}_p \times {\bf R}/{\bf Z}}$ , although ${{\bf R}/{\bf Z}}$ does not embed into this group and so this is not a direct or semi-direct product. (Topologically, it can be viewed as the set ${{\bf Z}_p \times [0,1]}$ after identifying ${(x,0)}$ with ${(x+1,1)}$ for every ${p}$ -adic integer ${x \in {\bf Z}_p}$ .) It has topological dimension one, in the sense that it contains continuous injective images of open subsets of ${{\bf R}^n}$ if and only if ${n \leq 1}$ . It is also the projective limit of the Lie groups $\displaystyle G_n := (({\bf Z}/p^n{\bf Z}) \times {\bf R})/{\bf Z}^\Delta,$
each of which is isomorphic to the circle ${{\bf R}/{\bf Z}}$ , but in a manner which becomes increasingly “twisted” as ${n}$ increases (which helps explain the terminology “solenoid”). It is an instructive exercise to verify that ${G}$ is connected, but not path connected or locally connected. Thus we see that connected locally compact groups can contain some significant non-Lie behaviour, even in the abelian setting.

Remark 3 Of course, it is possible for locally compact groups to be infinite-dimensional; a simple example is the infinite-dimensional torus ${({\bf R}/{\bf Z})^{\bf N}}$ , which is compact, abelian, metrisable, and locally connected, but infinite dimensional. (It will still be an inverse limit of Lie groups, though.)

Remark 4 The above analysis also gives another purely topological characterisation of Lie groups; a topological group is Lie if and only if it is locally compact, Hausdorff, first-countable, locally connected, and finite-dimensional. It is interesting to note that this characterisation barely uses the real numbers ${{\bf R}}$ , which are of course fundamental in defining the smooth structure of a Lie group; the only remaining reference to ${{\bf R}}$ comes through the notion of finite dimensionality. It is also possible, using dimension theory, to obtain alternate characterisations of finite dimensionality (e.g. finite Lebesgue covering dimension) that avoid explicit mention of the real line, thus capturing the concept of a Lie group using only the concepts of point-set topology (and the concept of a group, of course).

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Wednesday 20 July 2011

A fixed point Brouwer's invariance of domain theorems, and Hilbert's fifth problem

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