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Recall that a (real) topological vector space is a real vector space ${V = (V, 0, +, \cdot)}$ equipped with a topology ${{\mathcal F}}$ that makes the vector space operations ${+: V \times V \rightarrow V}$ and ${\cdot: {\bf R} \times V \rightarrow V}$ continuous. One often restricts attention to Hausdorff topological spaces; in practice, this is not a severe restriction because it turns out that any topological vector space can be made Hausdorff by quotienting out the closure ${\overline{\{0\}}}$ of the origin ${\{0\}}$ . One can also discuss complex topological vector spaces, and the theory is not significantly different; but for sake of exposition we shall restrict attention here to the real case.

An obvious example of a topological vector space is a finite-dimensional vector space such as ${{\bf R}^n}$ with the usual topology. Of course, there are plenty of infinite-dimensional topological vector spaces also, such as infinite-dimensional normed vector spaces (with the strong, weak, or weak-* topologies) or Frechet spaces.

One way to distinguish the finite and infinite dimensional topological spaces is via local compactness. Recall that a topological space is locally compact if every point in that space has a compact neighbourhood. From the Heine-Borel theorem, all finite-dimensional vector spaces (with the usual topology) are locally compact. In infinite dimensions, one can trivially make a vector space locally compact by giving it a trivial topology, but once one restricts to the Hausdorff case, it seems impossible to make a space locally compact. For instance, in an infinite-dimensional normed vector space ${V}$ with the strong topology, an iteration of the Riesz lemma shows that the closed unit ball ${B}$ in that space contains an infinite sequence with no convergent subsequence, which (by the Heine-Borel theorem) implies that ${V}$ cannot be locally compact. If one gives ${V}$ the weak-* topology instead, then ${B}$ is now compact by the Banach-Alaoglu theorem, but is no longer a neighbourhood of the identity in this topology. In fact, we have the following result:

Theorem 1 Every locally compact Hausdorff topological vector space is finite-dimensional.

The first proof of this theorem that I am aware of is by André Weil. There is also a related result:

Theorem 2 Every finite-dimensional Hausdorff topological vector space has the usual topology.

As a corollary, every locally compact Hausdorff topological vector space is in fact isomorphic to ${{\bf R}^n}$ with the usual topology for some ${n}$ . This can be viewed as a very special case of the theorem of Gleason, which is a key component of the solution to Hilbert’s fifth problem, that a locally compact group ${G}$ with no small subgroups (in the sense that there is a neighbourhood of the identity that contains no non-trivial subgroups) is necessarily isomorphic to a Lie group. Indeed, Theorem 1 is in fact used in the proof of Gleason’s theorem (the rough idea being to first locate a “tangent space” to ${G}$ at the origin, with the tangent vectors described by “one-parameter subgroups” of ${G}$ , and show that this space is a locally compact Hausdorff topological space, and hence finite dimensional by Theorem 1).

Theorem 2 may seem devoid of content, but it does contain some subtleties, as it hinges crucially on the joint continuity of the vector space operations ${+: V \times V \rightarrow V}$ and ${\cdot: {\bf R} \times V \rightarrow V}$ , and not just on the separate continuity in each coordinate. Consider for instance the one-dimensional vector space ${{\bf R}}$ with the co-compact topology (a non-empty set is open iff its complement is compact in the usual topology). In this topology, the space is ${T_1}$ (though not Hausdorff), the scalar multiplication map ${\cdot: {\bf R} \times {\bf R} \rightarrow {\bf R}}$ is jointly continuous, and the addition map ${+: {\bf R} \times {\bf R} \rightarrow {\bf R}}$ is continuous in each coordinate (i.e. translations are continuous), but not jointly continuous; for instance, the set ${\{ (x,y) \in {\bf R}: x+y \not \in [0,1]\}}$ does not contain a non-trivial Cartesian product of two sets that are open in the co-compact topology. So this is not a counterexample to Theorem 2. Similarly for the cocountable or cofinite topologies on ${{\bf R}}$ (the latter topology, incidentally, is the same as the Zariski topology on ${{\bf R}}$ ).

Another near-counterexample comes from the topology of ${{\bf R}}$ inherited by pulling back the usual topology on the unit circle ${{\bf R}/{\bf Z}}$ . Admittedly, this pullback topology is not quite Hausdorff, but the addition map ${+: {\bf R} \times {\bf R} \rightarrow {\bf R}}$ is jointly continuous. On the other hand, the scalar multiplication map ${\cdot: {\bf R} \times {\bf R} \rightarrow {\bf R}}$ is not continuous at all. A slight variant of this topology comes from pulling back the usual topology on the torus ${({\bf R}/{\bf Z})^2}$ under the map ${x \mapsto (x,\alpha x)}$ for some irrational ${\alpha}$ ; this restores the Hausdorff property, and addition is still jointly continuous, but multiplication remains discontinuous.

As some final examples, consider ${{\bf R}}$ with the discrete topology; here, the topology is Hausdorff, addition is jointly continuous, and every dilation is continuous, but multiplication is not jointly continuous. If one instead gives ${{\bf R}}$ the half-open topology, then again the topology is Hausdorff and addition is jointly continuous, but scalar multiplication is only jointly continuous once one restricts the scalar to be non-negative.

Below the fold, I record the textbook proof of Theorem 2 and Theorem 1. There is nothing particularly original in this presentation, but I wanted to record it here for my own future reference, and perhaps these results will also be of interest to some other readers.

— 1. Proof of Theorem 2 —

Let ${V}$ be a finite-dimensional Hausdorff topological space, with topology ${{\mathcal F}}$ . We need to show that every set which is open in the usual topology, is open in ${{\mathcal F}}$ , and conversely.

Let ${v_1,\ldots,v_n}$ be a basis for the finite-dimensional space ${V}$ . From the continuity of the vector space operations, we easily verify that the linear map ${T: {\bf R}^n \rightarrow V}$ given by

$\displaystyle T( x_1,\ldots,x_n) := x_1 v_1 + \ldots + x_n v_n$

is continuous. From this, we see that any set which is open in ${{\mathcal F}}$ , is also open in the usual topology.

Now we show conversely that every set which is open in the usual topology, is open in ${{\mathcal F}}$ . It suffices to show that there is a bounded open neighbourhood of the origin in ${{\mathcal F}}$ , since one can then translate and dilate this open neighbourhood to obtain a (sub-)base for the usual topology. (Here, “bounded” refers to the usual sense of the term, for instance with respect to an arbitrarily selected norm on ${V}$ (note that on a finite-dimensional space, all norms are equivalent).)

We use ${T}$ to identify ${V}$ (as a vector space) with ${{\bf R}^n}$ . As ${T}$ is continuous, every set which is compact in the usual topology, is compact in ${{\mathcal F}}$ . In particular, the unit sphere ${S^{n-1} := \{x \in {\bf R}^n: \|x\|=1\}}$ (in, say, the Euclidean norm ${\| \|}$ on ${{\bf R}^n}$ ) is compact in ${{\mathcal F}}$ . Using this and the Hausdorff assumption on ${{\mathcal F}}$ , we can find an open neighbourhood ${U}$ of the origin in ${F}$ which is disjoint from ${S^{n-1}}$ .

At present, ${U}$ need not be bounded (note that we are not assuming ${V}$ to be locally connected a priori). However, we can fix this as follows. Firstly, using the joint continuity of the addition map (and of the negation map ${x \mapsto -x}$ , which is a specialisation of the scalar multiplication map), we can find an open neighbourhood ${U'}$ of the origin such that the difference-set ${U'-U' := \{x-y: x,y \in U' \}}$ is contained in ${U}$ . Next, using the joint continuity of the scalar multiplication map, one can find another open neighbourhood ${U''}$ of the origin and an open interval ${(1-\epsilon,1+\epsilon)}$ around ${1}$ such that the product set ${(1-\epsilon,1+\epsilon) \cdot U'' := \{ t x: t \in (1-\epsilon,1+\epsilon); x \in U'' \}}$ is contained in ${U'}$ . Combining the two statements, we see that for every ${x \in U''}$ , we have ${sx \in U}$ for every ${s \in (-\epsilon,\epsilon)}$ . In particular, since ${U}$ avoids the unit sphere ${S^{n-1}}$ , ${U''}$ must avoid the region ${\{ x \in {\bf R}^n: \|x\| > 1/\epsilon \}}$ and is thus bounded, as required.

Remark 1 Note how the joint continuity of both scalar multiplication and addition was needed in the proof of Theorem 2. In view of the near-counterexamples of this theorem, this dependence on the joint continuity hypotheses is necessary.

Corollary 3 In a Hausdorff topological space ${V}$ , every finite-dimensional subspace ${W}$ is closed.

Proof: It suffices to show that every vector ${x \in V \backslash W}$ is in the exterior of ${W}$ . But this follows from Theorem 2 after restricting to the finite-dimensional space spanned by ${W}$ and ${x}$ . $\Box$

— 2. Proof of Theorem 1 —

Let ${V}$ be a locally compact Hausdorff space, thus there exists a compact neighbourhood ${K}$ of the origin. Then the dilate ${\frac{1}{2} K}$ is also a neighbourhood of the origin, and so by compactness ${K}$ can be covered by finitely many translates of ${\frac{1}{2} K}$ , thus

$\displaystyle K \subset S + \frac{1}{2} K$

for some finite set ${S}$ . If we let ${W}$ be the finite-dimensional vector space generated by ${S}$ , we conclude that $\displaystyle K \subset W + \frac{1}{2} K.$

Iterating this we have $\displaystyle K \subset W + 2^{-n} K$

for any ${n \geq 1}$ . On the other hand, if ${U}$ is a neighbourhood of the origin, then for every ${x \in V}$ we see that ${2^{-n} x \in U}$ for sufficiently large ${n}$ . By compactness of ${K}$ (and continuity of the scalar multiplication map at zero), we conclude that ${2^{-n} K \subset U}$ for some sufficiently large ${n}$ , and thus $\displaystyle K \subset W + U$

for any neighbourhood ${U}$ of the origin; thus ${K}$ is in the closure of ${W}$ . By Corollary 3, we conclude that $\displaystyle K \subset W.$

But ${K}$ is a neighbourhood of the origin, thus for every ${x \in V}$ we have ${2^{-n} x \in K}$ for all sufficiently large ${n}$ , and thus ${x \in 2^n W = W}$ . Thus ${V=W}$ , and the claim follows.

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Tuesday, 31 May 2011

Locally compact topological vector spaces