Let $V$ be a vector space over $\mathbb R$. A *symmetric bilinear pairing* on $V$ is a linear map $a: V\otimes V \to \mathbb R$. Because $\mathbb R$ is characteristic not-two, I will freely confuse symmetric bilinear pairings with quadratic forms; if $v\in V$, I will write $av^2$ for $a(v\otimes v)$; and $av_1v_2$ for $a(v_1\otimes v_2)$. The pairing $a$ is *positive (negative) definite* on $V$ if $av^2$ is strictly positive (negative) whenever $v\neq 0$. The pairing $a$ is *nondegenerate* if the corresponding map $a: V\to V^\*: v \mapsto a(v\otimes-)$ has trivial kernel.

Two subspaces $V_1,V_2 \subseteq V$ are *orthogonal* if $av_1v_2 = 0$ for any $v_1\in V_1$ and any $v_2\in V_2$. If $V_1\subseteq V$, its *orthogonal complemenet* is the maximal subspace $V_2\subseteq V$ so that $V_1$ and $V_2$ are orthogonal (it always exists, and may intersect $V_1$). A subspace $V_+\leq V$ is *positive (negative)* if $a|\_{V_+ \otimes V_+}$ is positive- (negative-) definite. A subspace is *maximally positive (negative)* if it is positive (negative) and not contained in any other positive (negative) subspace. Maximally positive (negative) subspaces exist by Zorn's lemma.

Let $a$ be a nondegenerate symmetric bilinear pairing on $V$. If $V$ is finite-dimensional, then the following are true (e.g. by running Gram-Schmidt):

- Let $V_+$ and $V_-$ be maximally positive and negative subspaces. Then $V = V_+ + V_-$. Since $V_+ \cap V_- = 0$, this presents $V$ as a direct sum.
- Let $V_+$ be a maximally positive subspace. Then its orthogonal complement is maximal negative.

An important corrolarry of 1. is that any two maximally positive (negative) subspaces have the same dimension. Then we can define the *signature* of the pairing $a$ as the pair $(\dim V_+,\dim V_-)$. When $a$ is not nondegenerate, the signature is actually a triple; the third term is the dimension of the kernel of the map $a: V\to V^\*$. ($a$ induces a nondegenerate symmetric pairing on $V / \ker a$.)

**My question is:** are statements 1., 2. true when $V$ is infinite-dimensional? If not, what (topologico-analytical, say) conditions on $V$ assure that they are? (Or at least that 1. is; I don't really care about 2.)