# Is sum of two symplectic subspaces symplectic?

Consider a symplectic vector space $$(\mathbb{R}^{2n}, \omega_0)$$ with standard symplectic form $$\omega_0$$ defined by: $$\omega_0(x,y) = xJ_0y^T$$ where $$J_0=\begin{pmatrix} 0 & I_n \\ -I_n & 0 \end{pmatrix}$$. Let $$W_1, W_2$$ be symplectic subspaces of $$\mathbb{R}^{2n}$$ such that $$W_1\cap W_2 = \{0\}$$. Is $$W_1 + W_2$$ also symplectic?

It seems to be true but neither can I give a proof nor a counter example.

• I presume you mean $W_{1} \cap W_{2} = \{ 0 \}$. Sep 18 '20 at 17:46
• Yes Im sorry thats a mistake Sep 18 '20 at 17:56

I got a counter example. Take $$W_1= span\{(1,0,0,0,0,0), (0,0,0,1,0,0)\}$$ and $$W_2= span\{(1,1,0,0,0,0), (0,0,0,1,0,1)\}$$. $$W_1,W_2$$ are symplectic because if $$e_1, e_2$$ denotes the basis elements of either $$W_1$$ or $$W_2$$, then $$\omega_0(e_1,e_2)= 1\neq 0$$. i.e., $$\omega_0$$ is non-degenerate on both $$W_1$$ and $$W_2$$. Clearly $$W_1 \cap W_2 = \{0\}.$$ But $$W_1 + W_2 = span \{(1,0,0,0,0,0), (0,0,0,1,0,0),(1,1,0,0,0,0), (0,0,0,1,0,1)\}$$ is not symplectic since $$(0,1,0,0,0,1) \in W_1 + W_2$$ and $$\omega_0((0,1,0,0,0,1), x)=0$$ $$\forall x\in W_1 +W_2$$.(i.e., $$\omega_0$$ is degenerate on $$W_1 + W_2$$.)
Furthermore, this will be true if $$W_1 \subset W_2^{\omega_0}$$ and $$W_2 \subset W_1^{\omega_0}$$. These conditions are sufficient but not neccessary.