Consider a linear subspace Y of a symplectic vector space (V,w). Its symplectic orthogonal is defined by
$$Y^O=\{v\in V|w(v,u)=0~ \forall ~u \in Y\}$$
Now, apparently, we must have
$$(Y^O)^O=Y$$
for any Y, but this confuses me a bit. Here is what I'm thinking:
Let's say V is for simplicity 6 dimensional and has a basis of unit vectors $e_1,e_2,e_3,f_1,f_2,f_3$ such that $$w(e_i,f_j)=\delta_{i,j} ~~,~~ w(e_i,e_j)=w(f_i,f_j)=0$$
Then Y might be e.g. a subspace with basis $e_1,e_2,f_1$. Following the definition above, only $e_3,f_3$ would have vanishing symplectic product with all three of these vectors, so that $e_3,f_3$ are a basis of $Y^O$. But then by the same logic $(Y^O)^O$ would have a basis $e_1,e_2,f_1,f_2$, which is more than the original 3 vectors and cannot be equivalent to Y. So it seems $Y\neq (Y^O)^O$ after all.
Is there some mistake in the above logic? How should the argument properly go?
The problem with your example is that you've missed another independent element of $Y^O$: $e_2$ is also orthogonal to $Y$, so that $e_2, e_3, f_3$ is a basis of $(Y^O)^O$.
I think the statement only holds in the finite-dimensional case, and if we assume that $V$ is finite-dimensional then we can use a dimension-counting argument to prove it. Suppose that $V$ is $k$-dimensional and $Y$ is $n$-dimensional, and choose a basis $b_1, \ldots, b_n$ of Y. Use this to define a map $\phi : V \to F^n$ (where $F$ is the field over which $V$ is defined) by $$ \phi(v) = (w(b_1, v), \ldots, w(b_n, v)) $$ By the non-degeneracy of $w$, this map is surjective. Its kernel is therefore $k-n$ dimensional, and is of course equal to $Y^O$. Now define a similar map $\phi^O$ using a basis for $Y^O$, and note that $Y \subset \ker(\phi^O)$, and $\dim(Y) = \dim(\ker(\phi^O))$. Therefore $Y = \ker(\phi^O) = (Y^O)^O$.
There might be a cleaner argument...