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Let $G$ be a smooth manifold and a group, whose product $G\times G\to G$ is "separately smooth," meaning $L_g:h\mapsto gh$ and $R_g:h\mapsto hg$ are smooth for each $g\in G$. Must $G$ (with the given product) be a Lie group? In other words, must this separately smooth product be jointly smooth? (Note that a jointly smooth product implies a smooth inverse, as shown here.)

The same question is asked here, but the answer there is insufficient and perhaps even inaccurate. It shows that $G$ can be equipped with a product turning it into a Lie group, but not necessarily that the given product makes $G$ into a Lie group (as is pointed out in the comments). The purpose of my question is to clarify this important detail.

WillG
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    Where's the gap in the following reasoning? By Ellis, Robert, Locally compact transformation groups, Duke Math. J. 24, 119-125 (1957), the product map is continuous. Therefore, thanks to the argument by Ted Shifrin in the comments to https://math.stackexchange.com/q/4120965 the product map is smooth, isn't it? – Ben Oct 06 '22 at 11:13
  • @Ben I don't fully understand that argument. Is it claiming that any function $F:X\times Y\to Z$ (without any group structure) that is (a) jointly continuous and (b) smooth in each variable separately must be jointly smooth? I know that continuity of partial derivatives implies differentiability, as Ted Shifrin points out, but I still don't see how the above claim follows. – WillG Oct 06 '22 at 13:59
  • I see. To me, at least at the moment, it seems tautological after passing to compatible (thanks to continuity) charts. But maybe I'm wrong or just confused. – Ben Oct 06 '22 at 15:11
  • By the answer in your link, multiplication is jointly continuous. By the solution to Hilbert's fifth problem, it is (jointly) smooth. – Cronus Sep 02 '23 at 16:55

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