Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $(V,\omega)$ be a symplectic vector space of dimension $2n$. How can I show that for an isotropic subspace $S \subset V$, there exists a symplectic basis $(A_i,B_i)$ for $V$ such that $S= $ span$(A_1,...,A_k)$ for some $k$.

It is related to this thread. I understand the symplectic case, but I cannot apply it to isotropic case.

share|cite|improve this question
up vote 5 down vote accepted

Choose any basis $A_1,\dots, A_k$ of $S$. By non-degeneracy of $\omega$ there exist $C_i\in V$, $i=1\dots k$, s.t. $\omega(A_i,C_j)=\delta_{ij}$. You can the find $B_i$'s of the form $B_i=C_i+\sum_{j<i}a_{ij} A_k$ ($a_{ij}$'s are numbers) so that $\omega(B_i,B_j)=0$ ($\omega(A_i,B_j)=\delta_{ij}$ is automatic): find $a_{ij}$'s first for $i=1$, then $i=2$ etc. Now $\operatorname{span}(A_1,\dots,A_k,B_1,\dots,B_k)$ is a symplectic subspace, so use the answer you linked to (to complete $A_1,\dots,A_k,B_1,\dots,B_k$ to a symplectic basis of $V$).

share|cite|improve this answer
I don't see where you used $S$ is isotropic. – user20353 Dec 30 '11 at 23:34
@user20353: to get $\omega(A_i,A_j)=0$, which is then also used to get $\omega(A_i,B_j)=\delta_{ij}$ (see the def. of $B_i$'s) – user8268 Dec 31 '11 at 7:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.