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This is a theorem mentioned in Hansjorg Geiges' Symplectic Geometry book while proving Darboux's theorem for local realisation of a contact form on a manifold $M^{2n+1}$ (see $\S2.5.1$ p.67) I haven't seen this theorem in any textbook or on the net. Could anyone state it for me?

Karthik C
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1 Answers1

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Perhaps it's referring to the result that for any nonsingular skew-symmetric form on a finite-dimensional vector space $V$ over any field, there is a basis $e_1, ..., e_n, f_1, ..., f_n$ of $V$ such that $\langle e_i, f_i \rangle = 1$, $\langle f_i, e_i \rangle = -1$, and all other pairings between basis vectors are 0.

Ted
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  • So to prove this, I start with a symplectic basis for the kernel of $\alpha$ at $p$ (we can find one) and then if we find $z$ such that $d\alpha(z,x)=0$ for all $x$ in the kernel then we are sure $i_zd\alpha\equiv0$, for anyway $d\alpha(z,z)=0$ and $z$ together with $\ker\alpha$ span the vector space. So here $z$ lies in a 1-d space and we can scale to ensure $\alpha(z)=1$. Is this a legitimate way to go about the proof? – Karthik C Mar 20 '13 at 08:27
  • Can you explain your notation more? What is $i_z$? What is a symplectic basis? I would have assumed a symplectic basis was the type of basis described in the theorem I quoted above, but if you're trying to prove this theorem, then you can't start with such a basis... – Ted Mar 20 '13 at 15:59
  • Oh sorry $i_z\omega$ for a 2-form $\omega$ is defined as $i_z\omega(x)=\omega(z,x)$ and yes indeed, a symplectic basis is one you described in your answer. – Karthik C Mar 24 '13 at 10:53
  • @Host-website-on-iPage You can find a proof in the wonderful linear algebra book of Kostrikin and Manin, in section 3.3 combined with section 2.10. – Alex Shpilkin Mar 24 '18 at 03:52