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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?

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You made several mistakes in the first approach. I mention them from last to first, as the first mistake is the crucial one, but the others are also to be mentioned.

(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.

Note that your second approach is another way to express (4).


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.

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  • $\begingroup$ Thank you for the nice explanation @Jordan Payette $\endgroup$ – Alex Feb 24 '17 at 15:22

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