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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.

Filippo
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  • Surely this $(A_t)$ has important properties -- is it a semigroup? For in that case it is not just "a uncountable family" of operators. If it is the heat semigroup, then $A_t=e^{t\Delta}$, so diagonalizing $\Delta$ you simultaneously diagonalize all $A_t$-s. – Giuseppe Negro Apr 26 '23 at 10:00
  • @GiuseppeNegro Yes, $(A_t)$ is a semi-group (prop. 2.17). However, AFAIU the book simultaneously diagonalizes the semi-group in order to establish that we can diagonalize $\Delta$, not the other way around. – Filippo Apr 26 '23 at 10:44
  • Oh really? Surprising to me. The fact that you badly cited the book didn't help. You should write the authors before the title! :-) (A link, if available, would also help) – Giuseppe Negro Apr 26 '23 at 10:45
  • Thanks a lot for listening to my suggestion concerning the citation. – Giuseppe Negro Apr 26 '23 at 14:36
  • @GiuseppeNegro Thank you for the suggestion, I agree that it is a good habit to provide a link to the references. – Filippo Apr 26 '23 at 14:48

1 Answers1

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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.

Aphelli
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  • Thank you for your answer. I have a doubt though: AFAUI you showed that we can essentially WLOG consider the finite-dimensional case. But even in the finite-dimensional case, it is not clear to me that we can simulanously diagonalize an uncountable family of commuting self-adjoint operators (in fact this really surprises me, as my references only consider finite families of operators). What does the induction step look like? – Filippo Apr 25 '23 at 13:21
  • If every operator is scalar, we’re done. Otherwise, we can write $V=V_1 \oplus V_2$ with nonzero $V_i$, such that $V_1$ is an eigenspace for one operator in the family, and $V_2$ is the sum of its remaining eigenspaces. Then each $V_i$ is stable under every operator in the family, so you can co-digonalize the operators on $V_1$ and $V_2$ separately and consider the reunion of the bases. – Aphelli Apr 25 '23 at 18:29
  • I understand, thank you for the comment! The only part I don't understand is how you conclude that$$\forall t: A_t|_{H_0}=0.$$I assume that by nilpotent you mean$$\forall t:\forall v\in H_0:(A_t\circ A_t)v=0$$ – Filippo Apr 26 '23 at 14:04
  • ...but all I can show is that $(A_1\circ A_t)v=0$ for $v\in H_0$. Could you please elaborate? – Filippo Apr 26 '23 at 14:05
  • Let $v \in H_0$ and $t >0$. There is some $N>0$ such that $Nt>1$. Then $A_t^N(v)=A_{Nt}(v)=A_{Nt-1}(A_1(v))=0$. So $A_t^N$ vanishes on $H_0$. But a self-adjoint nilpotent operator is zero. – Aphelli Apr 26 '23 at 18:49
  • Thank you very much for the explanation! – Filippo Apr 26 '23 at 19:01
  • Is "codiagonalize" synonymous to "simultaneously diagonalize"? – Filippo Apr 26 '23 at 19:05
  • Yes, sorry if that was unclear. – Aphelli Apr 26 '23 at 21:31
  • It surprises me that noone upvoted your answer. It seems correct to me. Accepted and upvoted. – Filippo Apr 28 '23 at 16:20