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I recently went to a talk where the speaker mentioned that

$$ 0 \rightarrow SU(n) \rightarrow SU(n+1) \rightarrow S^{2n + 1} \rightarrow 0$$

was exact. I think it's well-known that this sequence is a fibration, but is it an exact sequence too?

For this to the the case we would need to put a group structure on $S^{2n + 1}$ and define a group homomorphism from $SU(n+1)$ to it... but I don't know how to do this. There is a bijection between $S^{2n+1}$ and cosets $SU(n+1)/SU(n)$, but since $SU(n)$ is not normal in $SU(n+1)$, $SU(n+1)/SU(n)$ does not have a group structure. Hence my source of confusion.

Was the speaker correct, or did they mean something else by 'exact'?

Sebastiano
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Cookie Monster
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1 Answers1

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This is not a standard usage of "exact", and looks like a misstatement. In particular, $S^{2n+1}$ does not admit any topological group structure if $n>1$. I imagine they really just meant that it's a fiber sequence.

Eric Wofsey
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  • What's a good reference to learn about when topological spaces admit topological group structures? In particular, what's a method of proving that S^n does not admit a topological group structure when n > 1? – Cookie Monster Nov 03 '20 at 22:45
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    Typically, you use some sort of invariant from algebraic topology. For instance, the cohomology ring of a topological group has a canonical Hopf algebra structure, and so you can prove many spaces don't admit topological group structures by showing their cohomology rings don't admit Hopf algebra structures. – Eric Wofsey Nov 03 '20 at 22:49
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    For $S^{2n+1}$, you have to use a more sophisticated approach. For instance, supposing $S^{2n+1}$ were a topological group, you can compute what the cohomology ring of its classifying space would be using the Serre spectral sequence, and then you can use Steenrod powers to show no space can exist with that cohomology ring. – Eric Wofsey Nov 03 '20 at 22:50
  • I know some basics of cohomology, but I didn't know about the Hopf algebra result. This looks like something exciting to explore, thanks! – Cookie Monster Nov 03 '20 at 22:54
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    @CookieMonster This question might be of interest. – Kajelad Nov 03 '20 at 23:02