# Is a left topological group which is a manifold a topological group?

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.

• 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.
– YCor
Feb 22, 2014 at 7:48
• 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? Feb 22, 2014 at 13:30
• 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).
– YCor
Feb 22, 2014 at 13:33
• Well, I sort of anticipated that you were primarily interested by the connected case, so I left the above as a comment.
– YCor
Feb 22, 2014 at 14:28
• (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. Feb 22, 2014 at 15:05

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 $(x,y)$.
• 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 Feb 24, 2014 at 18:11
• @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.
• 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). Feb 25, 2014 at 1:05