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If $G$ is a paracompact topological group, then is $G \times G$ paracompact?

This question is raised by Gepner and Henriques (first paragraph of 2.2). Of course, this is not true for arbitrary paracompact spaces, as shown by the Sorgenfrey plane.

Actually -- what's an example of a non-paracompact group?

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    $\begingroup$ An easy example: A product of uncountably many infinite discrete spaces is not paracompact. Therefore $\mathbb{Z}^\mathbb{R}$ is a non-paracompact group. $\endgroup$
    – Johannes Hahn
    Commented Jul 20, 2018 at 14:15
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    $\begingroup$ Avoid writing a question in the title without repeating it... this makes confusion, especially if you add a second question below. I edited accordingly. $\endgroup$
    – YCor
    Commented Jul 20, 2018 at 14:17
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    $\begingroup$ @YCor Thanks, that's a good point about re-stating the question. $\endgroup$
    – Tim Campion
    Commented Jul 20, 2018 at 14:21
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    $\begingroup$ @JohannesHahn Thanks, I wish I had realized there were such easy examples! $\endgroup$
    – Tim Campion
    Commented Jul 20, 2018 at 14:22

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I am at a topology conference today, and among the many good topologists here is Jan van Mill, a leading expert on topological groups. I ran your question by him, thinking he might know the answer off the top of his head. He did -- the answer is that if $G$ is a paracompact topological group, then $G \times G$ need not be paracompact.

The construction of such a group can be found in

Yinhe Peng and Liuzhen Wu, "A Lindelöf group with non-normal square" (link)

Their main theorem is that there is a Hausdorff group $G$ such that $G$ is Lindelöf but $G \times G$ is not. Every Hausdorff group is regular, every regular Lindelöf space is paracompact, and every Hausdorff paracompact space is normal. Thus their group $G$ is paracompact, while $G \times G$ is not.

More information on Peng and Wu's work can be found in these slides of Yinhe Peng's.

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  • $\begingroup$ Thanks! I see— you explain why G is paracompact. On the other hand from the title I see that GxG is not normal. But it is Hausdorff and every paracompact Hausdorff space is normal. So GxG is not paracompact. $\endgroup$
    – Tim Campion
    Commented Jul 20, 2018 at 17:23
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    $\begingroup$ I just realized that Gepner and Henriques (my motivating context) are working in compactly-generated spaces, which might change the answer. Perhaps I will ask this as a separate question... $\endgroup$
    – Tim Campion
    Commented Jul 20, 2018 at 17:44

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