Let $V$ be a vector space over $\mathbb R$, and $a: V\otimes V\to \mathbb R$ a symmetric bilinear pairing. Recall that the *Morse index* of $a$ is the maximal dimension of any subspace $V\_- \subseteq V$ on which $a$ is negative-definite.


If $V$ is infinite-dimensional, it can be very hard to check every negative-definite subspace and compare their dimensions. Much easier is to exhibit a subspace $V\_- \subseteq V$ on which $a$ is negative-definite, which is not contained within any other negative subspace. But how do I know that the dimension of *this* negative subspace is the maximal dimension of *any* negative subspace? The approach I thought worked fails; see [this question](https://mathoverflow.net/questions/7709/splitting-a-space-into-positive-and-negative-parts/).



How can I measure the Morse index in infinite dimensions?
Let $V\_-$ be this subspace of $V$ defined in the weak sense that it is negative-definite and not extendible. There is a linear map $b:V \mapsto V\_-^\*$ given by the formula $b(v) = a(v,\cdot)$, where the right side is interpreted as a dual vector on $V\_-$. Since $V\_-$ is finite-dimensional by hypothesis, $\dim V\_- = \dim V\_-^\*$. Let $V\_-^\perp$ be the kernel of $b$. Now let $W$ be some other negative-definite subspace of $V$. I claim that $\dim W \le \dim V\_-$. Otherwise, $b$ has a kernel in $W$ because $W$ does not fit in the target of $b$. But if $w \in W \cap V\_-^\perp$ is non-zero, then $V\_-$ can be extended by $w$, contrary to the hypothesis.


Thus, any finite-dimensional negative-definite subspace which is not extendible has the maximum-dimension property. The proof has been arranged so that you never take the dual of $V$ itself, only of that of $V\_-$.


