# Maximal totally isotropic subspace in a space with a degenerate skew-symmetric bilinear form

Let $$V$$ be a real finite-dimensional vector space with a skew-symmetric bilinear form $$B \colon V \times V \to \mathbb{R}$$. In general, we assume that the form $$B$$ is degenerate. A subspace $$S \subset V$$ is called totally isotropic if $$S \subset S^\perp$$, where $$S^\perp$$ is the orthogonal complement of the subspace $$S$$ with respect to the form $$B$$. I need to prove the following statement: If $$\dim S = (1/2) ( \dim V + \dim V^\perp )$$, then $$S$$ is maximal totally isotropic.

As far as I understand, if $$S \subset V$$ is maximal totally isotropic, then it contains any totaly isotropic subspace. So, I begin my proof like this: Assume that there is a totally isotropic subspace $$S_1 \subset V$$ such that $$S_1 \not\subset S$$. This means that there is a vector $$x_0 \in S_1$$ such that $$x_0 \notin S$$. But I don’t understand what to do next? The only thing I was able to prove is that $$S = S^\perp$$.

I will be glad for any help.

• How did you get $S=S^{\perp}$? Dec 21, 2023 at 6:16
• From the equality $\dim S + \dim S^\perp = \dim V + \dim (V^\perp \cap S)$, it follows that $\dim S + \dim S^\perp \leq \dim V + \dim V^\perp$. This implies that $\dim S^\perp \leq (1/2)( \dim V + \dim V^\perp) = \dim S$. On the other hand, we have $S \subset S^\perp$ and therefore $\dim S^\perp \geq \dim S$. Since $S \subset S^\perp$, we obtain $S = S^\perp$. Dec 21, 2023 at 8:14
• But you only showed $\dim S^{\perp}\le \dim S$, which does not imply $S^{\perp}\subset S$, and so you may only conclude that $\dim S^{\perp}=\dim S$. By the way, I checked online and some sources say that maximal totally isotropic subspace means a totally isotropic subspace with maximal dimension, and so it is not immediately obvious that it has to contain very totally isotropic subspace Dec 21, 2023 at 8:38
• The inequality $\dim S^\perp \leq \dim S$ together with $\dim S^\perp \geq \dim S$ implies that $\dim S = \dim S^\perp$. Since $S \subset S^\perp$, doesn't this mean that $S = S^\perp$? Dec 21, 2023 at 8:54
• Ah, I see what you mean now, you’re right. But about the other point I made, you might wanna make sure what definition of maximal isotropic subspace you are using Dec 21, 2023 at 9:12

Let

$$$$\begin{array}{rcl}f&\colon &S \rightarrow {V}^{\ast }\\ &&x \mapsto B \left(x , \cdot \right) \end{array}$$$$

One has

$$$$\text{Ker} f = S \cap {V}^{\perp } \quad \Longrightarrow \quad \text{dim Ker} f \leqslant \text{dim} \left({V}^{\perp }\right)$$$$

and

$$$$\text{Image} \left(f\right) \subset {S'} \quad \Longrightarrow \quad \text{rank} \left(f\right) \leqslant \text{dim} \left(V\right)-\text{dim} \left(S\right)$$$$

where $${S'}$$ is the dual-orthogonal of $$S$$. By the rank-nullity theorem it follows that

$$$$\text{dim} \left(S\right) = \text{dim} \left(\text{Ker} f\right)+\text{rank} f \leqslant \text{dim} \left({V}^{\perp }\right)+\text{dim} \left(V\right)-\text{dim} \left(S\right)$$$$

Hence

$$$$\text{dim} \left(S\right) \leqslant \frac{1}{2} \left(\text{dim} {V}^{\perp }+\text{dim} V\right)$$$$

Clearly the equality case implies that $$S$$ is maximal.