## How can I measure the Morse index in infinite dimensions?

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.

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 In your other question, you define the Morse index as the pair (V_-, V_+). Since it can happen that neither is unique, you presumably want to define this up to some notion of equivalence. Do you have a sense as to what that equivalence should be? (Or are you really only interested in the case where V_- finite dimensional?) – Andrew Stacey Dec 4 2009 at 9:45 Well, I would define it as the dimensions of the two spaces. But that's why I asked this question, to make sure that that definition makes sense. – Theo Johnson-Freyd Dec 4 2009 at 17:12 Dimension? One of them will be infinite so what do you mean by that? There are many better ways of doing it than by dimension which is rather a crude measure. What are your spaces, anyway? – Andrew Stacey Dec 4 2009 at 21:38

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 $V_-$.

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 This totally does the job I wanted, which was if some maximal (in the weak sense) subspace is finite-dimensional, than its dimension is the Morse index. Is the statement still true when V_- is infinite-dimensional, though? I had not intended to make it finite. – Theo Johnson-Freyd Dec 4 2009 at 7:49 If you can find any infinite $V_-$, then the Morse index is $\infty$", crudely. So the remaining question is whether there is a cardinal-valued Morse index. I don't know! My argument doesn't fully extend. If an unextendable negative subspace has cardinal dimension $\kappa$, then this argument only shows that another negative subspace has dimension at most $2^\kappa$. If $V$ has a countable basis, then you're set. Maybe the case of a countable topological basis is interesting, but it could then be a somewhat different question. – Greg Kuperberg Dec 4 2009 at 8:19 Reading between the lines, it looks as though the approach that Theo thought worked but failed was to find a complement of the negative subspace. I may be wrong, but this answer seems to be simply pointing out that if the negative subspace is finite dimensional then that complement always exists. – Andrew Stacey Dec 4 2009 at 11:45 I concede that I didn't read the other thread very carefully. :-) – Greg Kuperberg Dec 4 2009 at 14:43

I may well have missed the point of the question, if so I apologise. But in practical use, when you apply Morse theory in an infinite dimensional setting you almost always see an infinite index. Rather than look for a maximal negative definite subspace, the way around the problem is not to talk about the index of a single critical point, rather to talk of the relative index of two critical points. I explain how this goes below, but if you knew all this already and had some other application in mind, then leave a comment and I'll delete my answer. (Incidentally, the systematic application of Morse theory in infinite dimensions is due - to the best of my knowledge - to Andreas Floer and what I will briefly describe is usually called Floer theory or Floer homology.)

The reason to look at the relative index is that this gives the dimension of the space of gradient flow lines which run from one critical point to the other. These spaces are precisely what one must understand to build a complex and hence homology groups. In particular the spaces of dimension one give the boundary operator and the spaces of dimension two are used to check that d2=0. In fact, if you only care about the relative grading of the homology groups then the relative index is all that you really need.

So the next question is how to define the relative index. Here you can work backwards. It should be the dimension of the space of flow lines! Typically in Floer theory the gradient flow lines correspond to the solutions of some elliptic PDE. Considering flows between a fixed pair of critical points gives boundary conditions which make the problem Fredholm. The space of solutions then has a finite expected dimension which is taken as the definition of the relative index of the pair of critical points. To actually compute the index one can use the Atiyah-Singer index theorem or the spectral flow.

A classic example is the Chern-Simons functional on connections over a 3-manifold M. The critical points are flat connections and the gradient flow lines can be interpreted as instantons over the product of MxR. Saying the flow goes from one critical point to another says the instanton is asymptotic to one flat connection over M at minus infinity and the other flat connection at plus infinity. These boundary conditions make the instanton equation Fredholm and the space of solutions has an expected dimension which can be computed via the Atiyah-Singer index theorem. For more about this specific incarnation of Floer homology I recommend Donaldson's book "Floer homology groups in Yang-Mills theory".

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 Well, certainly don't delete your answer! Although the question I was asking was essentially about linear algebra, not applications :) – Theo Johnson-Freyd Dec 4 2009 at 17:09 I had a sneaking suspicion you were only focused on the linear algebra, but because you used the word "Morse" and just in case you had an application in mind I thought I should say something. – Joel Fine Dec 4 2009 at 17:57 It is true that modern interest in infinite-dimensional Morse theory is dominated by this relative case with bi-infinite index. In the Chern-Simons case there can't even be an absolute index because the Morse complex is only graded by a cyclic group. However, Morse's original application of Morse theory was to variations of curves, and Theo's question does fit that traditional setting. – Greg Kuperberg Dec 4 2009 at 22:13 @Greg Kuperberg, that's a very good point. – Joel Fine Dec 5 2009 at 11:22

It seems that simply one can't measure it. Here below is briefly described an example of a nondegenerate indefinite inner product space having no cardinal-valued Morse index.

Consider $(\ell^{2},<.,.>)$ as naturally embedded (via Riesz) into its (huge) algebraic dual, say $\mathcal{A}$, let $\mathcal{F}$ be the real vector space of all finitely supported functions from $\mathbb{R}$ to $\mathbb{\ell^{\textrm{2}}}$, and put $V$:= $\mathcal{A\times F}$. Next, write $\mathcal{A}$ as a direct sum $\mathcal{A}=\ell^2\oplus{E}$ (hence $\dim E=2^{c}$), let $\pi:\mathcal{A}$ $\to$ ${E}$ be the attached algebraic projection, and let $[.,.]$ be a scalar product on $E\times E$. If $u=$($\varphi,f)$, and $v=$($\psi,g)$ are in $V$, then define the bilinear symmetric pairing

$a(u,v)$:= - $[\pi \varphi,\pi \psi]$ + $\varphi(\sum_{t\in\mathbb{R}}g(t))+\psi(\sum_{t\in\mathbb{R}}f(t))$ + < $\sum_{t\in\mathbb{R}} f(t),\sum_{t\in\mathbb{R}}g(t))>$ - $\sum_{t\in\mathbb{R}}$ < $f(t),g(t)$> .

Define also the subspace $W$ of $V$ by $W$ := {${{ (\varphi,f)|\:\varphi=-\sum_{t\in\mathbb{R}}f(t)} }$}. Then is not hard to see that:

1) $W$ is negative definite (w.r.t. $a$), and $W^{\bot}$ = { 0 } , hence $a$ is non-degenerate.

2) [Using the C-B-S inequality and Riesz] Any maximal negative definite subspace $\mathcal{N}$ of $V$ containing $W$ is a linear subspace of $\ell^{2}\times\mathcal{F}$, hence $\dim$ $\mathcal{N}$ = $c$.

3) Any maximal negative definite subspace $\mathcal{M}$ of $V$ containing $E\times${ 0 } has $\dim\mathcal{M}$ $>$ $c$.

Consequently, $\mathcal{M}$ and $\mathcal{N}$ are not isomorphic as real vector spaces.

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