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If $(G_i,\varphi_i)$ is a inverse system of Hausdorff locally compact groups, then $\varprojlim G_i $ is locally compact?

I have looked for information about it, or any additional condition for the conclusion holds.

I don't know if I added the condition that $G_i$ are compact for almost everywhere it's enough.

Thanks you all for any suggestion or any counterexample

Luis Antonio Sanchez
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  • You write a "direct system" but use the notation for "inverse" limit. Which is it, direct or inverse? – Moishe Kohan Nov 02 '21 at 07:31
  • I've edited the question – Luis Antonio Sanchez Nov 02 '21 at 07:36
  • If it were true, arbitrary countable (unrestricted) direct products of locally compact groups would be locally compact. But actually, a countable direct product $\prod_n G_n$ is locally compact if and only if $G_n$ is compact for all but finitely many $n$. Example: $\mathbf{Z}^\mathbf{N}$ is not locally compact. – YCor Nov 09 '21 at 10:17

1 Answers1

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This is unfortunately false.

We can write an infinite dimensional $\mathbb{R}$ vector space as an inverse limit of finite dimensional $\mathbb{R}$ vector spaces (see here, or PaulFrost's comment below). Now since finite dimensional vector spaces are locally compact, but infinite dimensional spaces aren't (see here, for instance), we see inverse limits cannot preserve local compactness.

You ask about additional conditions to make the conclusion hold, such as the condition that the $G_n$s are "compact almost everywhere":

If we replace local compactness by compactness, then the theorem becomes true (because of Tychonoff's theorem). In fact, if all but finitely many of the $G_n$s are compact, and the remaining finitely many are merely locally compact, then the inverse limit is locally compact (so a version of your "almost everywhere" criterion works!).

To see why, recall $\varprojlim G_n$ is a certain (closed) subgroup of $\prod G_n$. Since the product of finitely many locally compact spaces is locally compact (see here, say), we see $\prod G_n$ is still locally compact, and thus $\varprojlim G_n$, which is a closed subspace, is locally compact too (see here).

I hope this helps ^_^

HallaSurvivor
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