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.

  • $\begingroup$ How did you get $S=S^{\perp}$? $\endgroup$
    – Divide1918
    Dec 21, 2023 at 6:16
  • $\begingroup$ 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 $S = S^\perp$. $\endgroup$ Dec 21, 2023 at 8:14
  • $\begingroup$ 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 $\endgroup$
    – Divide1918
    Dec 21, 2023 at 8:38
  • $\begingroup$ 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 $S = S^\perp$? $\endgroup$ Dec 21, 2023 at 8:54
  • 2
    $\begingroup$ 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 $\endgroup$
    – Divide1918
    Dec 21, 2023 at 9:12

1 Answer 1



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

One has

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


\begin{equation}\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)\end{equation}

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

\begin{equation}\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)\end{equation}


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

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


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .