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A nondegenerate symmetric form with a half-dimensional isotropic subspace has zero signature

Statement

Let $V$ be a finite-dimensional real vector space with a symmetric bilinear form $B$ (Bilinear forms, and symmetric, skew-symmetric, and alternating bilinear forms), and let $W \subseteq V$ be totally isotropic: $B (w, w^{'}) = 0$ for all $w, w^{'} \in W$ . If $B$ is nondegenerate (The matrix, left and right radicals, rank, and nondegeneracy of a bilinear form on a finite-dimensional space) and $2 \dim W = \dim V$ , then $sign (B) = b_{+} - b_{-} = 0$ , with $b_{+} = b_{-} = \frac{1}{2} \dim V$ ; equivalently, in a suitable basis $B$ is an orthogonal direct sum of hyperbolic planes.

Facts & Assumptions

Given: A finite-dimensional real vector space $V$ with a symmetric bilinear form $B$ , and a totally isotropic subspace $W \subseteq V$ with $2 \dim W = \dim V$ and $B$ nondegenerate.

[F1]

If a basis diagonalizes $B$ with $p$ positive, $q$ negative and $r$ zero diagonal entries, the inertia is $(p, q, r)$ , the rank is $p + q$ , the signature is $p - q$ , and Sylvester's law makes the triple independent of the diagonalizing basis (Positive and negative definiteness, the inertia $(p, q, r)$ , rank $p + q$ , and signature $p - q$ of a real symmetric bilinear or quadratic form).

[F2]

Every symmetric bilinear form on a finite-dimensional real vector space is congruent to exactly one normal form $diag (I_{p}, - I_{q}, 0_{r})$ with $p + q + r = \dim V$ (Sylvester's law of inertia: every real symmetric form is congruent to $diag (I_{p}, - I_{q}, 0_{r})$ , and $(p, q, r)$ is unique).

[F3]

In a basis $B$ with coordinate columns $x = [u]_{B}$ and $y = [v]_{B}$ one has $B (u, v) = x^{T} [B]_{B} y$ ; the left and right radicals are the sets of vectors pairing to zero with everything, and $B$ is nondegenerate exactly when both radicals vanish (The matrix, left and right radicals, rank, and nondegeneracy of a bilinear form on a finite-dimensional space).

[F4]

A symmetric bilinear form satisfies $B (u, v) = B (v, u)$ for all $u, v$ (Bilinear forms, and symmetric, skew-symmetric, and alternating bilinear forms).

[F5]

On $R^{N}$ the Euclidean inner product is positive definite: $⟨ x, x ⟩ = \sum_{k < N} x_{k}^{2} \geq 0$ , and $⟨ x, x ⟩ = 0$ holds only for $x = 0$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F6]

For a linear map $T : U \to U^{'}$ with $U$ finite-dimensional, $\dim U = \dim \ker T + \dim im T$ (Rank-nullity: $\dim_{F} V = nullity T + rank T$ ).

[F7]

A subspace of a finite-dimensional space is finite-dimensional and has dimension at most that of the ambient space (If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ ).

Proof

Proof technique: direct; diagonal normal form and the positivity of sums of squares.

1.1givenF1

If $\dim V = 0$ then $V = 0$ and $W = 0$ , the inertia is $(0, 0, 0)$ , so $sign (B) = 0$ and $b_{+} = b_{-} = 0 = \frac{1}{2} \dim V$ .

1.2givenF2F3

Suppose $\dim V = n$ . By [F2], fix a basis $e_{1}, \dots, e_{n}$ of $V$ in which $B$ has matrix $diag (I_{p}, - I_{q}, 0_{r})$ with $p + q + r = n$ ; thus $B (e_{i}, e_{j}) = ε_{i} δ_{i j}$ with $ε_{i} = 1$ for $i \leq p$ , $ε_{i} = - 1$ for $p < i \leq p + q$ , and $ε_{i} = 0$ otherwise. No $ε_{i}$ is zero: if $ε_{i} = 0$ , then $B (e_{i}, e_{j}) = 0$ for every $j$ , so $B (e_{i}, -)$ vanishes on the basis and hence on $V$ , putting the nonzero vector $e_{i}$ in the left radical, contrary to nondegeneracy. Hence $r = 0$ and $p + q = n$ .

1.3givenF3F4algebra

Let $w = \sum_{i = 1}^{n} a_{i} e_{i} \in W$ . Since $W$ is totally isotropic, $0 = B (w, w) = \sum_{i \leq p} a_{i}^{2} - \sum_{i > p} a_{i}^{2}$ by the diagonal matrix formula [F3]; equivalently $\sum_{i \leq p} a_{i}^{2} = \sum_{i > p} a_{i}^{2}$ .

2.1step 1.3F3F5

The coordinate projection $π_{+} : W \to V_{+} : = span (e_{1}, \dots, e_{p})$ , $π_{+} (w) = \sum_{i \leq p} a_{i} e_{i}$ , is injective: if $π_{+} (w) = 0$ , then $\sum_{i \leq p} a_{i}^{2} = 0$ , so by step 1.3 also $\sum_{i > p} a_{i}^{2} = 0$ , and by the positive definiteness [F5] applied to the coordinate vector of $π_{-} (w) = \sum_{i > p} a_{i} e_{i}$ we get $π_{-} (w) = 0$ ; hence $w = 0$ .

3.1step 2.1F6F7given

By [F6] applied to the injective linear map $π_{+}$ , whose image lies in the $p$ -dimensional space $V_{+}$ with basis $e_{1}, \dots, e_{p}$ , we get $\dim W = \dim im π_{+} \leq \dim V_{+} = p$ by [F7].

4.1step 1.3step 2.1step 3.1F5F6F7given

The projection $π_{-} : W \to span (e_{p + 1}, \dots, e_{p + q})$ onto the negative coordinates is injective by the same argument: if $π_{-} (w) = 0$ , then $\sum_{i > p} a_{i}^{2} = 0$ , so by step 1.3 also $\sum_{i \leq p} a_{i}^{2} = 0$ , so $π_{+} (w) = 0$ and step 2.1 gives $w = 0$ . Applying [F6] and [F7] as in step 3.1 to the subspace $span (e_{p + 1}, \dots, e_{p + q})$ of dimension $q$ gives $\dim W \leq q$ .

5.1step 3.1step 4.1F1given

Since $2 \dim W = n = p + q$ and $\dim W \leq \min (p, q)$ , both $p$ and $q$ equal $\dim W = \frac{1}{2} \dim V$ . By [F1] the inertia is $(m, m, 0)$ with $m = \dim W$ , so $b_{+} = b_{-} = m = \frac{1}{2} \dim V$ and $sign (B) = p - q = 0$ .

6.1step 1.2step 5.1F3F4algebra∎

Equivalently, the basis can be chosen hyperbolic: with $m = \dim W$ and $u_{i} : = e_{i} + e_{m + i}$ , $g_{i} : = \frac{1}{2} (e_{i} - e_{m + i})$ for $1 \leq i \leq m$ , bilinearity and [F4] give $B (u_{i}, u_{i}) = 1 - 1 = 0$ , $B (g_{i}, g_{i}) = \frac{1}{4} (1 - 1) = 0$ and $B (u_{i}, g_{i}) = \frac{1}{2} (1 + 1) = 1$ , while orthogonality of the diagonal basis makes the planes $span (u_{i}, g_{i})$ pairwise orthogonal with $V = ⨁_{i = 1}^{m} span (u_{i}, g_{i})$ . Thus $B$ is an orthogonal direct sum of $m$ hyperbolic planes.

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Sources

Daniel S. Freed, Bordism: Old and New (lecture notes, UT Austin, Fall 2012) (standard reference, not scraped)
John Milnor and James Stasheff, Characteristic Classes (re-typeset scan; original pagination) (standard reference, not scraped)