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

13 June, 2011 in expository, math.GN, math.GR | Tags: , , , | by | 24 comments

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.

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