Good afternoon!

I tried to understand the following fact about symplectic linear algebra. Given a Lagrangian $L$ subspace of a symplectic vector space $(V,\Omega)$, one can extend each basis of $L$ to a symplectic basis. I tried to do the proof by myself and by use of "Ana Cannas da Silva-Lectures on Symplectic Geometry" but I am still not sure whether it is ok. What do you think?

The necessary condition that $\Omega(e_i, e_j)=0 ~\forall i,j=1,...,n$ is fullfilled bacause $L$ is a Lagrangian, meaning that $L = L^{\Omega} = \{v \in V: \Omega(v,l)=0~ \forall l \in L\}$.

Now we have to find $n$ elements ($L$ Lagrangian, i.e. $\dim(V) = \frac{1}{2}\dim(L)$) $f_1, ..., f_n \in L^{\Omega}=L$ such that $\Omega(f_i, e_i) = 1$, $\Omega(f_i, e_j) = 0$ and $\Omega(f_i, f_j)=0$ for all $i \neq j = 1,...,n$.

Let $\{e_1, ..., e_n\}$ be such a basis.

(1) Define the set $W := span(e_2, e_3,...,e_n) \subset L$. Since $\Omega$ is nondegenerate we can always find an element $\tilde{f}_1 \in W^{\Omega}=\{v \in V: \Omega(v,w)=0~ \forall w \in W\}$ with $\Omega(e_1, \tilde{f}_1) \neq 0$. Take $f_1=\frac{\tilde{f}_1}{\Omega(e_1, \tilde{f}_1)}$. Then $\Omega(f_1, e_1) = 1$. Furthermore $\Omega(f_1,e_i)=0$ because $f_1 \in W^{\Omega}$. Note that $V_1 := \text{span}(e_1, f_1) \subset W^{\Omega}$ and with some effort one can show that $V = V_1 \bigoplus V_1^{\Omega}$. If $V_1^{\Omega} =\emptyset$, we are ready.

(2) $e_2 \in V_1^{\Omega}$. Analogue to above because of the nondegeneracy of $\Omega$ and $e_2 \neq 0$ there exists an element $\tilde{f_2} \in V_1^{\Omega}$ with $\Omega(e_2, \tilde{f_2}) \neq 0$. Take $f_2=\frac{\tilde{f}_2}{\Omega(e_1, \tilde{f}_2)}$. Then $\Omega(e_2, f_2) = 1$ and $\Omega(e_1, f_2) = 0 = \Omega(f_1, f_2)$. Again one can show that $V = (V_1 \bigoplus V_2) \bigoplus (V_1 \bigoplus V_2)^{\Omega}$. If $(V_1 \bigoplus V_2)^{\Omega} =\emptyset$, we are ready.

(3) Now because L was Lagrangian and a subset of a finite vector space, this procedure ends.

Thanks to all who have looked at this!

  • $\begingroup$ A bit late to see this, but it looks fine by my standards. $\endgroup$ – anakhro Jul 6 '17 at 18:58

It is not quite clear why there should be $f_1 \in W_1^\Omega$ such that $\Omega (e_1, f_1) \neq 0$. You must use the hypothesis of $Y$ being Lagrangian.

Suppose $\dim V = 2n$. Let $W_1 = \text{span}\{e_2,\ldots,e_n\}$. Consider the symplectic orthogonal $W_1^\Omega$. Notice that $W_1$ is isotropic
then $W_1 \subset W_1^\Omega$. Since $\Omega (e_1, e_j) = 0$ for every $j = 2,\ldots,n$, $e_1 \in W_1^\Omega - W_1$.

Now $\dim W_1 = n-1$ which implies that $\dim W_1^\Omega = n+1$. Hence there is $f_1 \in W_1^{\Omega}-Y$. We want to make sure that $\Omega (e_1,f_1) \neq 0$. Suppose that is not the case, then $\Omega (f_1, e_j) = 0$, for each $j = 1,\ldots, n$ which implies that $f_1 \in Y^\Omega = Y$ ($Y$ is Lagrangian), a contradiction. Without loss we may choose $f_1$ such that $\Omega (e_1, f_1) = 1$.

Next we consider $Z = \text{span}\{e_1,f_1\}$, it follows that $Z$ is symplectic, that is, $Z \cap Z^\Omega = \{0\}$. From this we have $V = Z \oplus Z^\Omega$ where $\dim Z^\Omega = 2n - 2$. Since $W_1 \subseteq Z^\Omega$ and $\dim W_1 = n - 1$, $W_1$ is a Lagrangian subspace of $W_2^\Omega$. We apply the exact same argument above for with $W_2 = \{e_3, \ldots, e_n\}$ and so on.


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