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I would like to know the properties of orthogonal matrices and symplectic matrices in terms of the forms they preserve. Could someone please add and/or correct, maybe give some refs/examples?

AFAIK, given a quadratic form q on a vector space V over a field F, there is an associated orthogonal group O(2n) ,a subgroup of GL(n,F), which preserve q; if F is the reals O(2n) preserves q= inner-product and norm (since in R, the norm is induced by the inner-product). Symplectic matrices only preserve symplectic forms, i.e., bilinear,antisymmetric,non-degenerate forms.

Are there relations between these groups; do they overlap, intersect, etc?

I am interested mostly in the case where the field is Z/2.

Thanks

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take a look at artin's geometric algebra –  yoyo Apr 10 '11 at 23:38
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Just a comment: given a nondegenerate quadratic form $q$, the group that preserves $q$ depends on the index of $q$. We obtain the orthogonal group when $q$ is positive definite, but when $q$ is not positive definite (but still nondegenerate) we can have other things. For example, when the index of $q$ is 1 (here the index of $q$ is the dimension of the "largest" subspace where $q$ is negative definite; I don't know if this definition is standard), we obtain the Lorentz group: en.wikipedia.org/wiki/Lorentz_group . –  Ronaldo Apr 10 '11 at 23:54
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as for the geometric algebra here is a free copy archive.org/details/geometricalgebra033556mbp –  yoyo Apr 11 '11 at 1:42
    
Thanks a lot to both, very helpful; unfortunately I don't have the status necessary to give you points. –  gary Apr 14 '11 at 1:04
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1 Answer

up vote 1 down vote accepted

Over the reals the intersection of the orthogonal and symplectic groups in even dimension is isomorphic to the unitary group in the half dimension.

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

This is the 2 out of 3 property expressing the compatibility the symplectic structure with the symmetric bilinear form of the orthogonal group.

The orthogonal group over $Z_2$ is a subgroup of the symplectic group because a symmetric bilinear form is also alternating (since $-1 = +1$).

The full symplectic group $Sp(2n, Z_2)$ can be realized from the action of the $2^n$ dimensional Clifford group on the bits of the binary representation of the basis vectors (up to a phase) as explained in Daniel Gottesmann's paper.

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Thanks, David, that's what I was looking for. –  gary Jun 22 '11 at 16:05
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