# How to generate a Lagrangian subspace of a symplectic space $\mathbb{R}^{4}$?

I am studying symplectic geometry. My question is motivated by the following question in a textbook: "Is it possible for a $2$-dim subspace of a $4$-dim symplectic vector space to be neither symplectic nor Lagrangian? If so, find necessary conditions for this. If not, generalise, state and prove the corresponding result".

Let us take $\mathbb{R}^{4}$ with the standard symplectic form $\omega$. So if $v = (v_{1},v_{2},v_{3},v_{4}) \in \mathbb{R}^{4}$ and $w = (w_{1}, w_{2}, w_{3}, w_{4}) \in \mathbb{R}^{4}$, then we have

$$\omega(v, w) = v_{1}w_{2} - v_{2}w_{1} + v_{3}w_{4} - v_{4}w_{3}$$

I want to generate a Lagrangian subspace $W$ of $\mathbb{R}^{4}$ (i.e. such that $W = W^{\perp}$) and also a symplectic subspace $V$ of $\mathbb{R}^{4}$ (i.e. $V \cap V^{\perp} = \{0\}$).

I choose $X = \mbox{span} \{(1,0,0,0) , (0,0,1,0) \}$, so an $v_{1} - v_{3}$ plane. I try to find an $\omega$-orthogonal subspace to $X$ in two ways.

First, I take an arbitrary vector $v \in X$ and express it as $v = (\lambda_{1}, 0, \lambda_{2}, 0)$ for some $\lambda_{1}, \lambda_{2} \in \mathbb{R}$. By definition of symplectic orthogonality, I find $w$ such that $\omega(v, w) = 0$. This yields $\lambda_{1}w_{2} + \lambda_{2}w_{4}=0$ and hence $w = (w_{1}, -\lambda_{2}, w_{3}, \lambda_{1})$ where $w_{1}$ and $w_{3}$ are free. So $X^{\perp} = \mbox{span} \{ w \}$ and $X \cap X^{\perp} = \emptyset$, so $X$ is neither symplectic nor Lagrangian, answering the earlier question.

Now for the second approach. Take $\lambda_{2} = 0$. This yields $w_{2} = 0$, $w_{1}, w_{3}, w_{4}$ free. Then take $\lambda_{1} = 0$. This yields $w_{4}=0$ and $w_{1}, w_{2}, w_{3}$ free. Combining, we see that $w_{2} = w_{4} = 0$ and we have $w = (w_{1},0,w_{3},0)$. Thus we see that $X= X^{\perp}$ and $X$ is Lagrangian!

So taking two approaches, I get two different answers.

Where am I going wrong?

(1) Two linear subspaces never have empty intersection, since they have at least in common the origin. (2) You wrote that $X^{\perp} = \mathrm{span}\{w\}$, which suggests that it is one-dimensional, but you clearly stated before that it is generated by the two free variables $w_1$ and $w_3$. (3) The equation $w_2 \lambda_1 + w_4 \lambda_2 = 0$ usually not only has $(w_2, w_4) = (-\lambda_2, \lambda_1)$ as a solution, but also any multiple of this solution. (4) The crucial one. If you pick $w \in X^{\perp}$, then $\omega(v,w) = 0$ for all $v \in X$, that is for any choice of $\lambda_1$ and $\lambda_2$. This forces $w = (w_1, 0, w_3, 0)$ (with $w_1, w_3$ free), so $X^{\perp} = X$ and the space is Lagrangian.
When the subspace has dimension 2, a more straightforward approach is possible. Choose a basis $\{e_1, e_2 \}$ of $X$ and note that $\omega(e_1, e_1) = 0 = \omega(e_2, e_2)$ and that $\alpha := \omega(e_1, e_2) = - \omega(e_2, e_1)$. So the (non)degeneracy of $\omega$ on $X$ is completely determined by whether $\alpha = 0$ or not. If it is 0, $X$ is isotropic (which is equivalent to Lagrangian if and only if the ambient space has dimension 4); Otherwise, it is symplectic.
Hence, there are only two possibilities for a two-dimensional subspace. Through this approach, it is easy to see that your explicit choice of $X$ is Lagrangian.