Talk:Symplectic vector space

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I redirected this page to symplectic manifold (which symplectic topology and symplectic geometry also go to) and then upon further reflection I reverted the page to the way it was. Shouldn't this page discuss a symplectic vector space and not a symplectic manifold (as it does now). Briefly a symplectic vector space is a vector space with a nondegenerate, alternating form. The tangent space to every symplectic manifold is a symplectic vector space. Actually, the example given on this page really is just a vector space. -- Fropuff 07:03, 2004 Feb 24 (UTC)

The article is not complete but is probably not a stub anymore. Should that line be deleted? Sympleko 10:57, 13 Apr 2005 (UTC)

[edit] symplectic matrix

I'm completely befuddled by this recent edit:

(Note that this is not the same thing as a symplectic matrix, which is a different concept discussed below).

which was used to replace

An ordered basis can always be found to express this matrix as a symplectic matrix.

The new sentance seems to be trying to say that something isn't something, but I'm not sure what the two somethings are. The old sentance seemed quite clear to me: a skew symmetric matrix can always be made symplectic by change of coords ... right? why remove such a sentance ? Confused ... linas 03:43, 11 May 2005 (UTC)

Symplectic matrices are the coordinate representations of symplectic transformations on a symplectic vector space. They presevere the non-degenerate skew-symmetric matrix which is the coordinate representation of the symplectic form. Symplectic matrices are only defined with respect to said non-degenerate skew-symmetric matrix. To say that the symplectic form has a coordinate representation as a symplectic matrix doesn't really make any sense: with respect to which matrix? -- Fropuff 05:03, 2005 May 11 (UTC)

[edit] Why only define a symplectic space over the real numbers?

Shouldn't this article be far more general? It is completely possible to define symplectic spaces over arbitrary fields. It might be possible though, that those spaces only make since over fiels with characeristic unequal to 2. See Quadratic_form. —Preceding unsigned comment added by 169.229.55.42 (talk • contribs) 01:26, 26 January 2006

In short, yes. The real case is the easiest one to handle, and is the one that is revelant to symplectic manifolds. In any case, I believe everything that is stated in the article works over arbitrary fields with char ≠ 2. I need to look up how the characteristic 2 case is handled. I suspect one just uses nondegenerate alternating forms rather than skew-symmetric ones, but there may be additional subtleties. -- Fropuff 02:08, 26 January 2006 (UTC)

[edit] Equal or just Isomorphic?

Shouldn't a Lagrangian subspace W of a symplectic vector space be merely isomorphic to W-perp, not equal? Sdenton-at-math.ucdavis 22:52, 16 February 2007 (UTC)

No. If that were the case any subspace whose dimension was half that of V would be Lagrangian, which is not a very useful notion. -- Fropuff 00:15, 17 February 2007 (UTC)