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First let me fix some notation:

Let $O(n)$ be the group of $n \times n$ real matrices $T$ which are "orthogonal", $U(n)$ be the group of $n \times n$ complex matrices $T$ which are "unitary" and $Sp(n)$ be the group of $n \times n$ quaternionic matrices $T$ which are "symplectic" (in all three cases $T^hT=TT^h=I$).

Let $Sp(2n,F)$ be the group of $2n \times 2n$ matrices that preserve a non-degenerate skew-symmetric bilinear form on $F^{2n}$, where $F$ is the field of real $\mathbb{R}$, complex $\mathbb{C}$ or quaternion $\mathbb{H}$ numbers (skew-field in the case of quaternions).

The following are true:

$O(2n) \cap Sp(2n,\mathbb{R}) = U(n)$

$U(2n) \cap Sp(2n,\mathbb{C}) = Sp(n)$

So my question is about the next logical step. Clearly both $Sp(2n)$ and $Sp(2n,\mathbb{H})$ are groups acting on $\mathbb{H}^{2n}$ but do they intersect to a non-empty group? In other words what is $X(n)$ below (if anything)?

$Sp(2n) \cap Sp(2n,\mathbb{H}) = X(n)$?

PS 1 This is a question I naturally asked myself after reading Baez's "Symplectic, Quaternionic, Fermionic" blog posting: http://math.ucr.edu/home/baez/symplectic.html

PS 2 By writing $X(n)$ instead of $X(2n)$ above I am hinting something related to Octonions but I don't want to scare off anyone.

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It is not that they are accepted to be true: they are true! –  Mariano Suárez-Alvarez Jan 26 '10 at 17:44
    
Thanks, I edited "accepted" away. –  Marios Jan 26 '10 at 17:50

1 Answer 1

You can't find an answer, because the question is false. There is no non-degenerated skew-symmetric bilinearform on $\mathbb{H}^{2n}$. Assume $B: \mathbb{H}^{k}\times\mathbb{H}^k \to \mathbb{H}$ is skew-symmetric and not zero, then $B(v,w)\neq 0$ for some $v,w\in\mathbb{H}^k$. Then $baB(v,w)=bB(av,w)=B(av,bw)=aB(v,bw)=abB(v,w) \implies ba=ab$. That's a contradiction. The same argument works if scalars go on the right or if you consider the space as right vector space. Even if the bilinearity is interpreted as $B(av,bw)=aB(v,w)b$, there is a contradiction:$B(v,w)ba=B(v,bw)a=B(v,abw)=B(v,w)ab \implies ab=ba$

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Notice that you have proved that there is no non-zero bilinear form, as skew-symmetry plays no role. –  Mariano Suárez-Alvarez Jan 26 '10 at 20:09
    
I fixed a LaTeX typo. –  Theo Johnson-Freyd Jan 26 '10 at 20:28
    
So, your argument seems to prove that there is no non-zero bilinear form on $\mathbb H^k$, independent of skew symmetry. And the problem is that asking the form to be instead sesquilinear, or whatever, rules out the possibility of demanding it be skew-symmetric. –  Theo Johnson-Freyd Jan 26 '10 at 22:30
    
This is indeed the case. You can have hermitian and skewhermitian forms on $\mathbb{H}^n$. The corresponding invariance groups are $\mathrm{Sp}(n)$ and $\mathrm{SO}^*(2n)$, at least in some notation. This is described in, say, Rossmann's Lie groups: an introduction through linear groups. –  José Figueroa-O'Farrill Jan 26 '10 at 23:04

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