link between symplectic characterisation and isotropic or symplectic subspaces I've stumbled on this exercise
I have a symplectic capacity $c$ and $O$ an open set of $\mathbb{R}^{2n}$ and $W$ a $2n-2$ dimensional subspace $W$ of $\mathbb{R}^{2n}$. I have to show that $c(O+W) = \infty$ if $W^\perp$ is isotropic otherwise, $c(O+W) \in (0;\infty)$.
I see no link in my course between a capacity and isotropic or symplectic subspaces. Any hint to get me started?
 A: That $W^\perp$ is isotropic implies $W^\perp \subset W$ (that is, $W$ is coisotropic). Let $e\in W$. Then since $W^\perp\subset W$, there is a symplectic basis $\alpha =\{e_1, \cdots, e_n, f_1, \cdots, f_n\}$ so that 
$$W = \operatorname{span}\{ e_1, \cdots, e_n, f_3, \cdots, f_n\}.$$
With respect to the basis $\alpha$ the symplectic form is written 
$$\omega = \sum_i e^i \wedge f^i.$$
Now consider the "ball" 
$$B_R = \{ x_1e_1 + \cdots +x_n e_n + y_1 f_1 + +\cdots y_n f_n: \sum_i x_i^2 + y_i^2 <R\}.$$
Then for any $r>0$, 
$$\Phi_r (x_1, \cdots, x_n , y_1, \cdots , y_n) = \left( rx_1, rx_2, x_3, \cdots, x_n, \frac{1}{r} y_1, \frac{1}{r} y_2, y_3, \cdots, y_n\right)$$
is a symplectomorphism and for all $R>0$, there $r'$ large so that $\Phi_r (B_R) \subset O+W$. This implies $c(O+W) \ge c(B_R)$ for all $R$ and thus is infinite. 
For the second part, I assume that $O$ is bounded (or the statement is false). If $W$ is not coisotropic, then $W^\perp$ is not in $W$. Since $\dim W + \dim W^\perp = 2n$, we have 
$$ W \cap W^\perp = \{0\}.$$
(The case that $W\cap W^\perp$ is one dimensional is impossible). Thus there are symplectic basis $\alpha$ so that 
$$W = \{e_2, \cdots, e_n, f_2, \cdots, f_n\}.$$
Thus $O +W$, under a symplectomorphism, is a subset of 
$$B^2_R \times\mathbb R^{2n-2}\subset \mathbb R^2 \times \mathbb R^{2n-2}$$
and hence $c(O+W) \le c(B_R^2 \times \mathbb R^{2n-2}) = \pi R^2$. It is clear that the capacity is nonzero, $c\in (0,\infty)$. 
