2
$\begingroup$

Let $V$ be a symplectic vector space, i.e. a vector space with a non-degenerate alternating bilinear form, a so-called symplectic form. There is a theorem:

All symplectic forms on $V$ are ismorphic.

I have two questions about this:

1) Can somebody explain what it means precisely that two symplectic forms are isomorphic?

2) Could somebode give references on this statement with or without proof?

Thank you very much.

$\endgroup$
0
$\begingroup$

It can be proven that any finite-dimensional symplectic vector space of dimension $2n$ has a basis in which the symplectic form $\omega$ has the form:

$$\begin{pmatrix} 0 & I_n \\ -I_n & 0 \end{pmatrix}$$

This proves that two symplectic forms are isomorphic. A reference is Wikipedia.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Merci! And what does it mean if two such forms are isomorphic? $\endgroup$ – user586527 Aug 27 '18 at 16:20
  • $\begingroup$ It means that for any $\omega_1,\omega_2$ two symplectic forms, you can find an invertible linear transformation $\varphi$ such that for any $x,y \in V$ $\omega_1(x,y) = \omega_2(\varphi(x),\varphi(y))$. $\endgroup$ – mathcounterexamples.net Aug 27 '18 at 16:24
  • $\begingroup$ So, "Two symplectic vector-spaces of equal dimension are isomorphic" is right as well? $\endgroup$ – user586527 Aug 27 '18 at 16:46
  • $\begingroup$ Ate isomorphic as symplectic spaces. Yes. Any two real Vector spaces of same dimensions are isomorphic in general. $\endgroup$ – mathcounterexamples.net Aug 27 '18 at 16:49
  • 1
    $\begingroup$ What does ate mean? $\endgroup$ – user586527 Aug 27 '18 at 16:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy