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So I was working through some exercises on Lie groups and I was wondering if the group isomorphisms carry any of the differentiable structure with them.

Explicitly, if two groups, $G$ and $H$ are isomorphic and $G$ is a Lie group of dimension n can we conclude anything about the other group?

The group isomorphism, $\phi$, is a bijective function and so we know that the set theoretic structure of the spaces are the same, so we can put a topology on $H$ using the images of open sets in $G$ under the isomorphism (so the new topology will still be Hausdorff and second countable). This automatically makes $\phi$ into a homeomorphism and so we can then define charts using the charts of $G$ composed with $\phi$ and these will be $C^{\infty}$-compatible since we have just stuck a bijection and it's inverse into the middle when checking two charts are compatible. Then I believe that multiplication and inverses in $H$ are now smooth since when checking smoothness by composing with charts we have again just inserted a bijection and it's inverse into the composition, which will not alter the result and so won't change smoothness.

I think that what I have come up with is correct but I wanted to check that I haven't missed something.

EHH
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  • So even though this is not a useful thing to do in general we can in fact use it to define a smooth structure on a group which doesn't have an obvious one. Such as the group of affine transformations which are isomorphic to a certain group of matrices which inherit nice smooth structure from euclidean space. – EHH Apr 10 '14 at 13:56

2 Answers2

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Your construction looks like it will work to the extent that it gives you a new topology on $H$. However, notice that this doesn't have to have anything to do with the topology you would a priori expect for $H$.

For example, $\mathbb R$ and $\mathbb C$ under addition are both Lie groups but of different dimension. However they are isomorphic as groups, since both have dimension $2^{\aleph_0}$ as vector spaces over $\mathbb Q$.

hmakholm left over Monica
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  • But my argument might in fact work then since even if the smooth structure and topology aren't the obvious one, it still makes the other group into a Lie group (if I haven't gone wrong somewhere). – EHH Apr 10 '14 at 13:39
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    @EdwardHart: It is right as far as it goes. Going from $\mathbb R$ to $\mathbb C$ it will give you an alternative topology on $\mathbb C$ that makes it look like the real line -- but which doesn't have anything to do with the standard topology on $\mathbb C$. – hmakholm left over Monica Apr 10 '14 at 13:40
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    @EdwardHart "is a Lie group" is not a property of the group itself, but a property of the group along with a topology on the group. – Tobias Kildetoft Apr 10 '14 at 13:40
  • @TobiasKildetoft ah yes of course! Thanks I'd overlooked that. – EHH Apr 10 '14 at 13:43
  • This doesn't directly answer the question, but it's also worth noting that if $G$ and $H$ are Lie groups and $\phi\colon G\to H$ is a continuous group isomorphism, then $G$ and $H$ are isomorphic as Lie groups, because continuous homomorphisms between Lie groups are smooth. (A proof can be found in just about any book on Lie groups.) – Jack Lee Apr 10 '14 at 17:06
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If you do not have any structure on the group $H$, and have an isomorphism $G\to H$, then yes you can shift any information of $G$ on $H$ via this isomorphism. It is more or less given by the definition of the structure that you put.

Basically, you are doing nothing, and are in fact working on $G$, which has a structure, and just give a different "name" to the elements of the groups, via the identification $G\to H$. You can even do this with only a bijection to a set, and then create by this way the group structure on the other set.

Jérémy Blanc
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  • So basically the answer to the question is no since in any situation we would have already determined the topological and smooth structure on both groups and there is no reason to expect that an isomorphism has any smoothness to it. – EHH Apr 10 '14 at 13:48
  • In your question, it is not said that there is a structure on $H$, and the title seems to say the converse. If one is given, then of course you are right (and it can be checked by looking at the other answer). If you do not give anything (which is what I have read), then you get a new structure, but as I said it is quite tautological. – Jérémy Blanc Apr 10 '14 at 13:59
  • I was originally not considering a structure on $H$ but then realised that for common examples we are really interested in the obvious structures and topologies. But I have now managed to use it to place a structure on a group which doesn't have an obvious one, but is isomorphic to a matrix group which does and so it is useful in this situation. Thanks for your help! – EHH Apr 10 '14 at 14:02