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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeJul 10th 2012
   • PermaLink
   Author: Urs
   Format: MarkdownItexstub for _[[metaplectic structure]]_

   stub for metaplectic structure

2. • CommentRowNumber2.
   • CommentAuthorDavid_Corfield
   • CommentTimeJul 10th 2012
   • PermaLink
   Author: David_Corfield
   Format: MarkdownItexIf we [have](http://ncatlab.org/nlab/show/string+group#definition_14) >$\pi_n(O)$ = 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?)

   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?)

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeJul 10th 2012
   • PermaLink
   Author: Urs
   Format: MarkdownItex> 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)](http://ncatlab.org/nlab/show/symplectic+group#MaximalCompactSubgroup) is $U(n)$, and so both have [the same homotopy groups](http://ncatlab.org/nlab/show/symplectic+group#HomotopyGroups).

   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.

4. • CommentRowNumber4.
   • CommentAuthorDavid_Corfield
   • CommentTimeJul 10th 2012
   • (edited Jul 10th 2012)
   • PermaLink
   Author: David_Corfield
   Format: MarkdownItexBy half a cycle, I meant this [result](http://en.wikipedia.org/wiki/Bott_periodicity_theorem#Context_and_significance): >$\pi_k(O)=\pi_{k+4}(\operatorname{Sp})$ >$\pi_k(\operatorname{Sp})=\pi_{k+4}(O) ,k=0,1,\dots$ What am I getting wrong here? p. 38 of [this](http://felix.physics.sunysb.edu/~abanov/Teaching/Spring2009/Notes/abanov-cpA1-upload.pdf) 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](http://en.wikipedia.org/wiki/Symplectic_group), only $Sp(2n, R)$ has nontrivial fundamental group. OK, so you were talking about the latter.

   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.

5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeJul 10th 2012
   • PermaLink
   Author: Urs
   Format: MarkdownItexRight, 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...

   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…

6. • CommentRowNumber6.
   • CommentAuthorUrs
   • CommentTimeJul 10th 2012
   • PermaLink
   Author: Urs
   Format: MarkdownItexSo 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.

   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.

7. • CommentRowNumber7.
   • CommentAuthorzskoda
   • CommentTimeJul 10th 2012
   • PermaLink
   Author: zskoda
   Format: MarkdownItexUrs, 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.

   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.

8. • CommentRowNumber8.
   • CommentAuthorUrs
   • CommentTimeJul 10th 2012
   • PermaLink
   Author: Urs
   Format: MarkdownItexI'd think that deserves an entry of its own!

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