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I am sorry if my question is stupid (or very hard) or common knowledge, or should be placed at math.stackexchage.com. As long as a math student read the definition of Lie group, several natural questions appear instantly:

Is it true, that any compact manifold admits Lie group structure? (NO)

Is it true, that there exists compact manifold that admits different Lie group structures? (YES)

Answers to these questions can be found here - Lie Groups and Manifolds

But I was not able to find the answer for third most natural question: Which connected compact manifolds admits unique Lie group structure? (As pointed out by YCor, without the connectedness assumption, there are trivial non-uniqueness examples.)

Thanks a lot for your answers!

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    $\begingroup$ If it's not connected, there are trivial non-uniqueness examples. Probably in the connected case there's indeed uniqueness (non-isomorphic connected compact Lie groups are non-homotopy-equivalent). The classification of compact groups is known, so what more do you want? $\endgroup$ – YCor May 7 '17 at 21:04
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    $\begingroup$ @YCor, Ok, I've edited the question, if you can provide working argument in connected case, please post your answer and I will accept it. $\endgroup$ – Hedgehog May 8 '17 at 3:47
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    $\begingroup$ @YCor what about the examples in mathoverflow.net/questions/67030 ? $\endgroup$ – Yemon Choi May 8 '17 at 3:59
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    $\begingroup$ So I'd suggest to specify the question to the following more specific: for which compact connected Lie groups $G$ is the group structure unique up to conjugation by a homeomorphism (equivalently, every Lie group homeomorphic to $G$ is isomorphic to $G$ as a topological/Lie group)? The answer is yes when $G$ is simply connected and in general seems to depend on $G$. Yemon Choi links to the example of two non-isomorphic semi-direct products $SU(2)\ltimes SO(3)$, namely with trivial and nontrivial action, yielding non-isomorphic $SU(2)\times SO(3)$ and $SO(4)$. $\endgroup$ – YCor May 8 '17 at 9:18
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    $\begingroup$ I think you ticked an answer, but your question is not answered. $\endgroup$ – YCor May 10 '17 at 11:30
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Your question seems to be of a tall order!

There are many,vmany necessary conditions involving the entire spectrum of algebraic topology such as homotopy and cohomology. This 3rd edition book by Kakl H. Hofmann and Sidney A. Morris on compact groups is full of these. For instance if you asked: which compact connected n-dimensional manifold can support an abelian Lie group, one would tell you "only $\mathbb{T}^n =(\mathbb{R}/\mathbb{Z})^n$" but without the qualification "unique". There are tons of discontinuous automorphisms of $\mathbb{T}$, and so you have millions of group topologies (all isomorphic!).

By contrast, the $3$-dimensional real projective space supports only one compact Lie group topology; namely, that of $SO(3)$. This has to do with the automatic continuity of algebraic automorphisms of certain compact connected Lie groups (see loc.cit., Corollary 5.66 and Exercise E5.21 following it, page 168). For more stuff see (just as a for instance) 8.59, 8.83, 9.59, A3.90, A3.92.

Life is not so simple that we can hope for a positive answer to your question as it stands. However, information in the area is around so that as soon as you pose a question of a more specific nature on the topic of compact connected manifolds and Lie groups, there is hope for an answer.

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