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Let $\dim(V)=6$. Show that $Sp(V,\Omega)$ acts transitively on $V^*-\{0\}$, where $\Omega$ here is a symplectic 2 form on $V$. ($V^*$ here is algebraic dual of $V$)

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Does $V^*$ denote the algebraic dual? What is the action? People unfamiliar with the question will have a hard time guessing. –  rschwieb Jan 24 '13 at 15:33
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This $\Omega$ is not present in the body of the text. –  Berci Jan 24 '13 at 15:56
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@rschwieb: all of the stuff mentioned in the question is standard and natural. Still, since $\Omega$ is non-degenerate, we can translate all questions about forms to equivalent questions about vectors by $v \mapsto \Omega(v, \cdot)$ and its inverse, so it's a bit puzzling that the author insists on working with $V^*$. –  Marek Jan 24 '13 at 16:44
    
i edited it again –  hassan joulani Jan 24 '13 at 16:44
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@Berci: $\Omega$ is implicit in the definition of $\operatorname{Sp}(V) = \operatorname{Sp}(V,\Omega)$ (the linear maps preserving the symplectic form). –  Martin Jan 24 '13 at 16:49

1 Answer 1

Hint. Show first that any non-zero vector is part of a symplectic basis.

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