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Let $G$ be a left topological group, i.e. a topological space with group operation such that left multiplication $L_g : x \mapsto gx$ is continuous (but right multiplication and inversion are not required to be). Assume also $G$ to be a topological manifold. Does this imply that $G$ is a topological group?

EDIT 1: as pointed out by Yves Cornulier, the aswer is no if $G$ is not required to be connected. Hence, I would like to know if the statement still holds when $G$ is assumed to be connected.

EDIT 2: as discussed in the comments, we are also assuming $G$ to be paracompact, with countable atlas say.

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    $\begingroup$ If you start from a topological group $H$ with a closed subgroup $F$, define a topology on $H$ for which the left cosets $gF$ are open and have the original topology from $F$; thus $H_F$ is homeomorphic to the product of $F$ and a discrete set (in bijection with $G/F$). For this topology, the left translations of $H$ are still continuous. If $G=SL_2(\mathbf{R})$ and $H$ is a closed 1-parameter subgroup, still, the unit component is not normal and thus this fails to be a topological group. $\endgroup$
    – YCor
    Commented Feb 22, 2014 at 7:48
  • $\begingroup$ If you make the coset of a a one-parameter subgroup open (and hence also closed), how can the group still be a topological manifold? $\endgroup$
    – Maurizio Monge
    Commented Feb 22, 2014 at 13:30
  • $\begingroup$ I'm not sure I understand your question since as I understand it, the answer is in my comment: the resulting topology on $H$ is homeomorphic to the product of $F$ with a discrete set; thus if $F$ is a topological manifold, so is $H_F$ (which is $H$ endowed with this topology, I forgot to say). $\endgroup$
    – YCor
    Commented Feb 22, 2014 at 13:33
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    $\begingroup$ Well, I sort of anticipated that you were primarily interested by the connected case, so I left the above as a comment. $\endgroup$
    – YCor
    Commented Feb 22, 2014 at 14:28
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    $\begingroup$ (2) Here is a nice theorem: if $G$ is a group and a complete metric space, and if multiplication is separately continuous, then it is a topological group (multiplication is jointly continuous and inversion is continuous). So (in case your manifold is complete in some metric) it will be enough to show right multiplication is continuous. $\endgroup$
    – Gerald Edgar
    Commented Feb 22, 2014 at 15:05

1 Answer 1

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Here is a counter-example with $G$ homeomorphic to $\mathbb R^2$. Let $f:\mathbb R\to\mathbb R$ be a discontinous additive homomorphism (constructed using a Hamel basis of $\mathbb R$ over $\mathbb Q$). Define a group operation $*$ on $\mathbb R^2$ by $$ (x,y)*(x',y') = (x+x'e^{f(y)},y+y') . $$ This groups is a semidirect product of $\mathbb R$ and $\mathbb R$ with respect to the action where $y\in\mathbb R$ acts on $\mathbb R$ by multiplication to $e^{f(y)}$. The formula is continuous in $(x',y')$ but not in $(x,y)$.

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    $\begingroup$ It is interesting to point out that the same example cannot work with compact semisimple Lie groups (replacing $xe^{f(y)}$ with the conjugacy $yxy^{-1}$, that sends as well $G\rightarrow{}Inn(G)\subseteq{}Aut(G)$), because in this case any automorphism is automatically continuous, as pointed out in mathoverflow.net/a/40700/3680 $\endgroup$
    – Maurizio Monge
    Commented Feb 24, 2014 at 18:11
  • $\begingroup$ @Maurizio: what do you mean by "cannot work in compact semisimple Lie groups"? the conclusion would rather be that it cannot work with a semidirect product $H\rtimes K$ with $K$ compact semisimple. A further question would be whether there is an example with $G$ connected metrizable manifold, but not homeomorphic to any Lie group. $\endgroup$
    – YCor
    Commented Feb 24, 2014 at 21:55
  • $\begingroup$ That the idea does not work "as is" with $H=K$, where I was looking for a non-continuous $H\rightarrow{}K$ that could then be composed with the canonical $K\rightarrow{}Inn(K)$. But actually I just realized that this can work very easily with $S^1$ and $SO(2)$, considering an isomorphical embedding $S^1\rightarrow{}SO(2)$ and a non-continuous homomorphism $S^1\rightarrow{}S^1$ (which exists). In this way we easily have a non-continuous homomorphism $S^1\rightarrow{}Aut(SO(2))$, and a similar example (that is also compact). $\endgroup$
    – Maurizio Monge
    Commented Feb 25, 2014 at 1:05

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