Symplectic version of "Gram-Schmidt" Let $w$ be a symplectic form on a vector space $V$ of dimension $2g$. Suppose we already have a free family $(a_1, \dots, a_g)$ such that $w(a_i, a_j) = 0$. I also have a family $(b_1, \dots, b_g)$ which verify that $(a_1, \dots, a_g, b_1, \dots, b_g)$ is a basis of $V$. I've read somewhere that there is a symplectic version of Gram-Schmidt to make the basis $(a_1, \dots, a_g, b_1, \dots, b_g)$ a symplectic one, but concretely, how does it work ?
Thank you !

Wikipedia cites Ana Cannas da Silva's Lectures on Symplectic Geometry for its assertion of this fact.  Prof. da Silva has kindly made a pdf copy of these lecture notes available from her staff page at ETH Zurich's web site.  The procedure you're looking for appears to be described in the proof of Theorem $1.1$ on pages $3–4$.
