From the proof of of proposition $3.26$ in Berline-Getzler-Vergne Heat Kernels and Dirac Operators:
We
simultaneously diagonalize the trace-class self-adjoint operators $(A_t)_{t>0}$
It is not clear to me why we can even do this. I know that for a compact operator $A$ on a Hilbert space we can write
$$A=\sum_{\lambda\in\sigma_p(A)}\lambda P_\lambda$$
where $P_\lambda$ is the projector onto the $\lambda$-eigenspace. But what about an uncountable family $(A_t)_{t>0}$?
I am pretty sure that the $A_t$ commute – in fact we should have $A_{t+h}=A_tA_h$. Can we assume that they all have the same eigenspaces (where the eigenvalue associated to an eigenspace depends on $t$) and is there a formula of the form
$$A_t=\sum_{i}\lambda_{i,t} P_i$$
where the $P_i$ are the projectors onto the aforementioned eigenspaces? Sorry, I haven't found any literature on this.
Best Answer
Write your total Hilbert space as $H=\bigoplus_{\lambda}{H_{\lambda}}$, where $H_{\lambda}$ is the eigenspace of $A_1$ for the eigenvalue $\lambda$ (it’s a “$L^2$-completed” orthonormal sum).
Since every $A_t$ commutes with $A_1$, every $H_{\lambda}$ is stable under every $A_t$.
If $\lambda \neq 0$, because $A_1$ is trace class, it is compact, thus $H_{\lambda}$ has finite dimension: so it has a basis simultaneously diagonalizing every $A_t$ (you can show, by induction on the dimension of the space, that every family $F$ of diagonalizable endomorphisms of a finite-dimensional vector space such that any two elements of $F$ commute is codiagonalizable).
If $\lambda =0$, then, for all $t >0$, the restriction of $A_t$ to $H_0$ is self-adjoint and nilpotent, hence zero. The conclusion follows.