# Is every group object in TopMan a Lie group?

Recall that a Lie group is a group object in the category of C manifolds.

If I have a group object in the category of topological manifolds, can I necessarily equip it with a smooth structure so that all the group operations are smooth? If so, how unique is this structure?

Is a continuous group-homomorphism between two Lie groups necessarily smooth?

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This is exactly Hilbert's fifth problem, which is summarized reasonably in Wikipedia. en.wikipedia.org/wiki/Hilbert%27s_fifth_problem – Greg Kuperberg Dec 13 '09 at 20:06
The best there is, if I recall correctly, is that a locally compact topological group such that the identity has a neighborhood which does not contain a subgroup can be smoothed to an analytic manifold in a unique way so that it becomes an analytic Lie group. There was an article in the AMS Notices recently about Gleason, with interesting details about the history of this result – Mariano Suárez-Alvarez Dec 13 '09 at 20:11
What I had in mind is that we should freeze or community-wiki questions that exactly match Wikipedia pages. MO should go beyond Wikipedia. – Greg Kuperberg Dec 13 '09 at 20:15
@Greg: That's fair. I wouldn't have known to google for "Hilbert's fifth problem", though, and there is not a link to it from the Wikipedia page on Lie groups. I almost closed the question as "no longer relevant", but it seems that we can only make actions once every 30 seconds, and by then Pete had responded. – Theo Johnson-Freyd Dec 13 '09 at 20:17
I think I disagree with the idea that if it's on Wikipedia then it's not a good MO question. Wikipedia is a currently voluminous, only mostly correct, and potentially limitless repository of mathematical knowledge. The idea of MO, I think, is that what you are wondering about might be easy or -- better yet -- already known to some other mathematician. I think Theo's question is fine. – Pete L. Clark Dec 13 '09 at 22:23

As Greg Kuperberg indicated in the comments, this is Hilbert's 5th Problem. The answer is yes, a theorem of Gleason, Montgomery and Zippin from the 1950's.

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Well, I think I should accept your answer, since Greg didn't actually leave it :) – Theo Johnson-Freyd Dec 13 '09 at 20:14

I just wanted to add that there is a fairly easy proof for your final question: Is every continuous homomorphism between Lie groups actually smooth?

The theorem we need is the closed subgroup theorem (also called the Cartan Theorem): If H is a topologically closed subgroup of a Lie group G, then H is actually an embedded Lie subgroup.

Granting this, one proves all continuous homomorphisms are smooth as follows:

Given Lie groups H and G with $f:H\rightarrow G$ a continuous homomorphism, consider the subgroup $K$ of $H\times G$ given by the graph of $f$. The graph is a closed subset of $H\times G$ precisely because $f$ is continuous, and hence, by the closed subgroup theorem, the graph is an embedded smooth submanifold of $H\times G$. Thus, the restriction of the two canonical projection maps $\pi_1:H\times G\rightarrow H$ and $\pi_2:H\times G\rightarrow G$ are smooth when restricted to K.

Now, $\pi_1$ restricted to $K$ is clearly* a diffeomorphism onto $H$, and hence has a smooth inverse and so is smooth. But then we find that $f = \pi_2\circ \pi_1^{-1}$ is a composition of smooth maps, and hence is smooth. (To be clearer, the $\pi_1^{-1}$ means the inverse of $\pi_1:K\rightarrow H$.)

*- (Edited in due to comments). One knows by Sard's theorem that there is a point $p\in K$ such that $d_p \pi_1$ is invertible (of full rank). I claim that this implies that for all $q\in K$, $d_q \pi_1$ is invertible. The point is that $\pi_1$ is group homomorphism, which is the same as saying $\pi_1\circ L_{qp^{-1}} = L_{\pi_1(qp^{-1})}\circ \pi_1$, where $L_g$ denotes left multiplication by $g: L_g(h) = gh$. Taking the differentials at p on each side of this equation and using the chain rule, one finds

$$d_q \pi_1 \circ d_p L_{qp^{-1}} = d_{\pi_1(p)}L_{\pi_1(qp^{-1})}\circ d_p \pi_1.$$

The fact that $L_g$ is a diffeomorphism (with inverse $L_{g^{-1}}$) implies that $dL$ is invertible at any point, and hence we see that

$$d_q\pi_1 = d_{\pi_1(p)}L_{\pi_1{qp^{-1}}}\circ d_p \pi_1\circ d_pL_{pq^{-1}};$$ i.e., that $d_q \pi_1$ is a composition of invertible maps, and hence is itself invertible.

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I think one needs a bit more argument to see that you get a diffeomorphism from $K$ to $H$: something like Sard's theorem to make sure the derivative bijects. – Fran Burstall Dec 13 '09 at 22:05
I'll clarify it.... – Jason DeVito Dec 13 '09 at 22:40
I just wanted to add that this also addresses the uniqueness question asked: A topological group has a unique smooth structure. For if we give G two smooth structures, then the identity map from G to itself is continuous, and hence by the above smooth. Since the identity map is its own inverse, this implies the identity map is a diffeomorphism between the two smooth structures. – Jason DeVito Jan 2 '10 at 20:02
To clarify my above comment, I mean IF a topological group has a smooth structure, then it's unique. – Jason DeVito Mar 3 '10 at 1:01