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

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?

• An easy example: A product of uncountably many infinite discrete spaces is not paracompact. Therefore $\mathbb{Z}^\mathbb{R}$ is a non-paracompact group. – Johannes Hahn Jul 20 '18 at 14:15
• Avoid writing a question in the title without repeating it... this makes confusion, especially if you add a second question below. I edited accordingly. – YCor Jul 20 '18 at 14:17
• @YCor Thanks, that's a good point about re-stating the question. – Tim Campion Jul 20 '18 at 14:21
• @JohannesHahn Thanks, I wish I had realized there were such easy examples! – Tim Campion Jul 20 '18 at 14:22

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.
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.