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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeJul 10th 2012
    • CommentRowNumber2.
    • CommentAuthorDavid_Corfield
    • CommentTimeJul 10th 2012

    If we have

    π n(O)\pi_n(O) = Z/2, Z/2, 0, Z, 0, 0, 0, Z for n = 0, …, 7 mod 8,

    and SpSp is half a cycle out, don’t we have

    π n(Sp)\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=3n= 3 the first level to co-kill?

    Hmm, but elsewhere I see π 1(Sp(2n))=Z\pi_1(Sp(2 n)) = Z, so a double cover is fine. Is it a questioning of stabilizing the Sp(2n)Sp(2 n)? Is the universal cover of Sp(2n)Sp(2 n) of any interest?

    (By the way, any reason for using \mathcal{B} on the spin structure page?)

    • CommentRowNumber3.
    • CommentAuthorUrs
    • CommentTimeJul 10th 2012

    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 π 1(Sp(n))=\pi_1(Sp(n)) = \mathbb{Z}

    Yes, the maximal compact subgroup of Sp(n) is U(n)U(n), and so both have the same homotopy groups.

    • CommentRowNumber4.
    • CommentAuthorDavid_Corfield
    • CommentTimeJul 10th 2012
    • (edited Jul 10th 2012)

    By half a cycle, I meant this result:

    π k(O)=π k+4(Sp)\pi_k(O)=\pi_{k+4}(\operatorname{Sp})

    π k(Sp)=π k+4(O),k=0,1,\pi_k(\operatorname{Sp})=\pi_{k+4}(O) ,k=0,1,\dots

    What am I getting wrong here? p. 38 of this has π 1(Sp(n))=0\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, C), Sp(2n,R)Sp(2n, R) and Sp(n)Sp(n)? But according to this Wikipedia page, only Sp(2n,R)Sp(2n, R) has nontrivial fundamental group.

    OK, so you were talking about the latter.

    • CommentRowNumber5.
    • CommentAuthorUrs
    • CommentTimeJul 10th 2012

    Right, I should better disentangle my notation. I have been speaking about Sp(2n,)Sp(2n,\mathbb{R}) here throughout. It certainly has π 1=\pi_1 = \mathbb{Z}.

    Let me try to make the notation in the entries more consistent…

    • CommentRowNumber6.
    • CommentAuthorUrs
    • CommentTimeJul 10th 2012

    So I went through a couple of entries and wrote out






    and so on everywhere. I hope I caught them all, not to leave a mess.

    • CommentRowNumber7.
    • CommentAuthorzskoda
    • CommentTimeJul 10th 2012

    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.

    • CommentRowNumber8.
    • CommentAuthorUrs
    • CommentTimeJul 10th 2012

    I’d think that deserves an entry of its own!