Dimension of totally isotropic subspaces for a given quadratic form

Question

Let $X$ be a real linear space with the quadratic form given in a basis $(e_1,...,e_r,e_{r+1},...,e_{r+s} )$ by

$$Q(x)=\sum_{i=1}^r x_i^2-\sum_{i=r+1}^{r+s} x_i^2$$ for $x=\sum_{i=1}^{r+s}x_i e_i.$

A linear subspace $V\subset X$ is called totally isotropic if $Q|_V=0$. I know that all maximal (in the sense of inclusion) totally isotropic subspace of $X$ have the same dimension.

Assume for example that $r\ge s$. How to prove the maximal isotropic subspaces for the $Q$ have dimension $r$?

Thanks.

Answer 1

Let $V$ be a maximal totally isotropic subspace. Let $W$ be a subspace of dimension $\max\{r,s\}$ on which $Q$ is positive or negative definite. (For instance, if $\max\{r,s\}=r$, let $W$ be the span of $e_1,\ldots,e_r$, and if $\max\{r,s\}=s$, let $W$ be the span of $e_{r+1},\ldots, e_s$.) Then \begin{equation*} \dim(V\cap W) = \dim(V) + \dim(W) - \dim(V+W) \end{equation*} is a basic result from linear algebra. Since $V$ is totally isotropic, $V\cap W=0$, so we get \begin{align*} \dim(V) &= \dim(V+W)-\dim(W) \\ &\leq \dim(X)-\max\{r,s\} \\ &= r + s - \max\{r,s\} \\ &= \min\{r,s\}. \end{align*} On the other hand, certainly we have $\dim(V)\geq\min\{r,s\}$, because $V$ is a maximal totally isotropic subspace and we know that there exists a totally isotropic subspace with dimension $\min\{r,s\}$.

Therefore $\dim(V)=\min\{r,s\}$.

EDIT: @JanVysoky is correct, I was too glib in claiming the inequality $\dim(V)\geq\min\{r,s\}$. Let $B$ be the symmetric bilinear form on $X$ corresponding to the quadratic form $Q$. Consider the subspace $V^\perp$ defined by \begin{equation*} V^\perp = \{v^\perp\in X:B(v,v^\perp)=0\text{ for all } v\in V\}. \end{equation*} The theory of bilinear forms tells us that, since $Q$ and hence $B$ is non-degenerate, $\dim(V^\perp) = \dim(X)-\dim(V)$. Let $W$ be a totally isotropic subspace of dimension $\min\{r,s\}$. (For instance, if $\min\{r,s\}=s$, let $W$ be the span of $e_1+e_{r+s},e_2+e_{r+s-1},\ldots,e_s+e_{r+1}$ as @Bach said in the comments.) The maximality of $V$ implies $V^\perp\cap W=0$. Then \begin{align*} 0 &= \dim(V^\perp\cap W) \\ &= \dim(V^\perp) + \dim(W) - \dim(V^\perp + W) \\ &\geq (\dim(X) - \dim(V)) + \dim(W) - \dim(X) \\ &= \min\{r,s\} - \dim(V) \end{align*} Therefore $\dim(V)\geq\min\{r,s\}$.

Answer 2

Recall that symmetric bilinear forms over $\Bbb R$ or over any real closed field are parametrized by their signature and their rank. In particular non-degenerate forms are parametrized by their signature. The common dimension of every maximal isotropic subspace is called the Witt index of your quadratic form $Q$, and usually denoted by $\nu(Q)$. In fact

Claim For a non-degenerate quadratic form $Q$ over $\Bbb R$, $\nu(Q)=\frac 1 2(n-|s(Q)|)$ where $s(Q)$ is the signature of the associated bilinear form of $Q$.

Answer 3

Here is my (I believe corrected) proof of the frakbak's statement $\dim(V) \geq \min \{r,s\}$:

First, I will need a simple (almost obvious) lemma:

Lemma: Let $W \subseteq X$ be any subspace where $X$ is positive semi-definite. Let $Q$ has a signature $(r,s)$. Then there holds an inequality $s \leq codim(W)$.

Similarly, if $W \subseteq X$ is any subspace where $X$ is negative semi-definite, then $r \leq codim(W)$.

proof of Lemma: The number $s$ can be defined as a supremum over the set of dimensions of all negative definite subspaces of $X$. Let $N \subseteq X$ be any negative definite subspace and assume that $Q$ is positive semi-definite on $W$.

Then certainly $W \cap N = \{0\}$, and the quotient map $q: X \rightarrow X/W$ maps $N$ bijectively onto a subspace $q(N) \subseteq X/W$. This shows that $\dim(N) = \dim(q(N)) \leq \dim(X/W) = codim(W)$. Taking the supremum over all such $N$ the gives $s \leq codim(W)$.

The proof for negative semi-definite subspace $W$ is analogous.

Now to the actual proof: For a contradiction, assume that $\dim(V) < \min\{r,s\}$.

We claim that necessarily, there is then a non-zero and isotropic vector $x \in V^{\perp} \setminus V$, where $V^{\perp}$ is the orthogonal complement induced by the symmetric bilinear form $B$ corresponding to $Q$.

Indeed, suppose that this is not true. As $V$ is isotropic, we have $V \subseteq V^{\perp}$ and one can write $V^{\perp} = V \oplus N$ for some complementary subspace $N$. Observe that $N$ is non-trivial as $\dim(V^{\perp}) > \max \{r,s\} \geq \min \{r,s\} > \dim(V)$.

Let $v \in V$ and $0 \neq x \in N$ be arbitrary. Then $v + x \in V^{\perp} \setminus V$ is (by assumption) not isotropic:

$0 \neq B(v+x,v+x) = B(x,x)$.

But this means that $N$ contains no non-zero isotropic vectors and thus $N$ is positive or negative definite with respect to $Q$. Consequently the entire subspace $V^{\perp}$ is positive semi-definite or negative semi-definite. Using the Lemma above, this shows that at least one of the inequalities $s \leq codim(V^{\perp}) = \dim(V)$ and $r \leq codim(V^{\perp}) = dim(V)$ must be true.

But this would imply $\dim(V) \geq \min\{r,s\}$, which cannot happen.

Whence at least one of the non-zero vectors in $V^{\perp} \setminus V$ must be isotropic, say $x$. Setting $V' = V \oplus \mathbb{R} \{x\}$ would give us an isotropic subspace of $X$ properly containing $V$, which is a contradiction with the maximality of $V$.

This shows that our original assumption $\dim(V) < \min\{r,s\}$ must be wrong. Q.E.D.