# Extending a basis to a symplectic basis

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!

• A bit late to see this, but it looks fine by my standards. – 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.