I was wondering if there is any proof available for Morse-Bott lemma in the infinite dimensional case. In the finite dimensional case we have the following:

Suppose $f\colon M \to \mathbb R$ is a smooth function on $M$ where $M^n$ is a finite dimensional manifold. Let $N^k$ be a non-degenerate critical submanifold of $M$. Then for every point $x$ in $N$, there exist an open set containing $U$ and a diffeomorphism $\phi\colon U\to \mathbb R^n$ such that:

i)- $\phi(x)=0$.

ii)- $\phi(U\cap N)\subset \{0\}\times \mathbb R^k$, in other words, $\phi$ is a regular chart for $N$.

iii)- $f\circ\phi^{-1}(y)= -y_1^2-\dots--y_{\lambda}^2+y_{\lambda+1}^2+\dots +y_{n-k}^2+f(x)$,

where $\lambda$ is the index of $f$ on $N$.

The infinite dimensional analog would be:

If $\mathcal N$ is a non-degenerate critical submanifold for $f\colon \mathcal M \to \mathbb R$ where $\mathcal M$ is a Hilbert manifold. Prove for every point $x$ in $\mathcal N$, there is a regular chart $(U,\phi)$ for $\mathcal N$ containing $x$ into $T_x\mathcal M$ such that:

$f\circ\phi^{-1}(y)=-\Vert Py\Vert^2+\Vert (I-P)(y)\Vert^2+f(x)$, where $P\colon N_x\mathcal N \to E_{-}$ is the orthogonal projection from the normal space of $\mathcal N$ at $x$ to the negative space of Hessian at $x$.

I tried to see if one can extend the proof for the finite dimensional case to the infinite dimensional case, but there is an argument in the proof about "somehow smoothly diagonalizing" a family of symmetric matrices, which I don't know whether such a procedure is possible for self-adjoint operators over a Hilbert space or not. We may assume Hessian of each critical point can be represented by a compact self-adjoint operator, but I am not still sure that we can do such a procedure. The lemma is important in the sense that it implies $E_{-}N$ is a smooth vector bundle over N and when we pass a critical level the disk bundle of this vector bundle attaches to the previous level along its boundary (the unit sphere bundle). You can find a proof for the finite dimensional case in:




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