A smooth manifold $G$ (of dimension $m$) with a group structure given by a multiplication map $\mu: G \times G \to G, \, (g,h) \mapsto gh$ is called a Lie group if both the multiplication $\mu$ and the inversion $\iota:G \to G, \, g \mapsto g^{-1}$ are smooth maps.

It is well-known that smoothness of the multiplication already implies the smoothness of the inversion, which is proven via the inverse function theorem. Smoothness of $\mu$ moreover implies that the left and right multiplication maps $$L_g: G \to G, \, h \mapsto gh \, \text{ and } \, R_g: G \to G, \, h \mapsto hg^{-1} \quad (g \in G)$$ are smooth by pre-composing $\mu$ with suitable smooth maps $G \to G \times G$.

My question is about the converse of the previous statement: Does smoothness of the maps $L_g$ and $R_g$ for all $g \in G$ also imply that $G$ is a Lie group, i.e. that the multiplication $\mu$ is smooth as well? If yes, does it also suffice to assume that $L_g$ is smooth for every $g \in G$?

  • $\begingroup$ @Matematleta The assumption is not that $\eta: G \times G \to G, \, (g,h) \mapsto gh^{-1}$ is smooth, only that the map $\eta(-, h): G \to G$ is smooth for every $h \in G$. $\endgroup$ Apr 29, 2021 at 15:37
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    $\begingroup$ This is like asking if a function with partial derivatives is differentiable. As we all know, if the function and its partial derivatives are continuous, then the function is differentiable. But you're not even assuming continuity, so .... it will take extra work on topological groups here. $\endgroup$ Apr 29, 2021 at 16:12
  • $\begingroup$ @TedShifrin So if we use a definition of smooth functions that always assumes continuity, then this should be true by just translating to the "classical" case with charts? $\endgroup$ Apr 29, 2021 at 16:27
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    $\begingroup$ No, I meant that if you assume that $\mu$ is continuous, then smoothness will follow from smoothness in each variable separately. But I am not an expert in special things that happen on topological groups. $\endgroup$ Apr 29, 2021 at 16:31
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    $\begingroup$ @TedShifrin Ah, thanks for the clarification. There are definitely groups where continuity of multiplication in each variable does not imply global continuity, see here, but I have not seen an example where $G$ is a smooth manifold. $\endgroup$ Apr 29, 2021 at 16:43

1 Answer 1


It is proven in

Ellis, Robert, Locally compact transformation groups, Duke Math. J. 24, 119-125 (1957). ZBL0079.16602.

that a locally compact Hausdorff topological space equipped with a group structure where both left and right multiplications are continuous, is actually a topological group, i.e. the multiplication is jointly continuous. Now, you can use the result that a topological manifold which is also a topological group admits a unique Lie group structure.

  • $\begingroup$ Thanks for the answer. Can anything be said if we only assume the left multiplications $L_g$, $g \in G$, to be continuous/smooth? $\endgroup$ Apr 29, 2021 at 21:56
  • $\begingroup$ @SebastianSpindler: let me think about this. $\endgroup$ Apr 29, 2021 at 22:39
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    $\begingroup$ More precisely, a top. manifold that is also a top. group has a unique smooth structure for which the law is smooth. I'm not sure how you eventually conclure. Question: assuming that left and right multiplications are smooth, why is then this smooth structure equal to the original smooth structure? $\endgroup$
    – YCor
    May 4, 2021 at 23:11
  • $\begingroup$ @YCor One can at least deduce that the smooth structures agree after showing smoothness of $\mu$ with the argument Ted Shifrin gave in the comments, but I'm also not sure how to show that the structures agree without showing smoothness of $\mu$ first. $\endgroup$ May 7, 2021 at 23:32

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