Symplectic version of "Gram-Schmidt"

Question

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 !

lonza leggiera · Accepted Answer · 2022-02-08T01:00:36.133

7

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 $3–4$.

edited Feb 08 '22 at 01:00

answered Feb 08 '22 at 00:30

lonza leggiera

31,984

1

Indeed. Also, regarding google-able keywords, "geometric algebra" (e.g., E. Artin's old book by that name) is a somewhat larger body of info that discusses such things. – paul garrett Feb 08 '22 at 00:47
Thank you very much to both of you ! – Mathuser Feb 08 '22 at 11:15

Symplectic version of "Gram-Schmidt"

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