real analysis - How can I measure the Morse index in infinite dimensions?

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 $V$:= $mathcal{Atimes F}$. Next, write $mathcal{A}$
as a direct sum
$mathcal{A}=ell^2oplus{E}$ (hence
$dim E=2^{c}$),
let $pi:mathcal{A}$ $to$ ${E}$
be the attached algebraic projection, and let $[.,.]$
be a scalar product on $Etimes E$. If $u=$($varphi,f)$, and $v=$($psi,g)$
are in $V$, then define the bilinear symmetric pairing

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

Define also the subspace $W$ of $V$ by $W$ := {$ {{ (varphi,f)|:varphi=-sum_{tinmathbb{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 $ell^{2}timesmathcal{F}$, hence $dim$ $mathcal{N}$
= $c$.

3) Any maximal negative definite subspace $mathcal{M}$ of $V$ containing
$Etimes ${ 0 }$ $ has $dimmathcal{M}$ $>$ $c$.

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