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stub for metaplectic structure
If we have
$\pi_n(O)$ = Z/2, Z/2, 0, Z, 0, 0, 0, Z for n = 0, …, 7 mod 8,
and $Sp$ is half a cycle out, don’t we have
$\pi_n(Sp)$ = 0, 0, 0, Z, Z/2, Z/2, 0, Z for n = 0, …, 7 mod 8?
So where does the double cover come in? Why isn’t $n= 3$ the first level to co-kill?
Hmm, but elsewhere I see $\pi_1(Sp(2 n)) = Z$, so a double cover is fine. Is it a questioning of stabilizing the $Sp(2 n)$? Is the universal cover of $Sp(2 n)$ of any interest?
(By the way, any reason for using \mathcal{B} on the spin structure page?)
Sp is half a cycle out,
Wait, what do you mean by that? Maybe there is some wrong assumption going into this statement. But maybe I am missing something, let me know what you have in mind.
but elsewhere I see $\pi_1(Sp(n)) = \mathbb{Z}$
Yes, the maximal compact subgroup of Sp(n) is $U(n)$, and so both have the same homotopy groups.
By half a cycle, I meant this result:
$\pi_k(O)=\pi_{k+4}(\operatorname{Sp})$
$\pi_k(\operatorname{Sp})=\pi_{k+4}(O) ,k=0,1,\dots$
What am I getting wrong here? p. 38 of this has $\pi_1(Sp(n)) = 0$ in a table called “Homotopy groups of symplectic groups”
Oh, am I getting confused between $Sp(2n, C)$, $Sp(2n, R)$ and $Sp(n)$? But according to this Wikipedia page, only $Sp(2n, R)$ has nontrivial fundamental group.
OK, so you were talking about the latter.
Right, I should better disentangle my notation. I have been speaking about $Sp(2n,\mathbb{R})$ here throughout. It certainly has $\pi_1 = \mathbb{Z}$.
Let me try to make the notation in the entries more consistent…
So I went through a couple of entries and wrote out
Sp(2n,\mathbb{R})
and
Mp(2n\mathbb{R})
and
Ml(n,\mathbb{R})
and so on everywhere. I hope I caught them all, not to leave a mess.
Urs, I know some references on metaplectic representation, which I would be glad to add to a relevant entry. I am not sure what is your big plan: to consider it under metaplectic group, under metaplectic structure or to create a new entry metaplectic representation. I would be glad to do create it if this fits your picture.
I’d think that deserves an entry of its own!
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