Determinant line of Fredholm operators and composition of morphisms

Question

Let $P$ be a polarization of a Hilbert space $\mathcal{H}$, i.e. a bounded idempotent: consider a group $G=GL_{res}(\mathcal{H}):=\{g \in GL(\mathcal{H}): [g,P] \in HS\}$ (where $HS$ is the set of all Hilbert Schmidt operators). For $g \in G$ let $P_g:=gPg^{-1}$ and define $F_{g,h}=P_gP_h-P_h+I$. One can check that $F_{g,h}$ are Fredholm: therefore one can define the so called determinant line $det(F_{g,h})$ to be $\Lambda^{top}(ker(F_{g,h})) \otimes \Lambda^{top}(coker(F_{g,h}))^*$. Using this data one can construct a (small) category where (the set of) objects is $G$ and for $g,h \in G$ we put $Hom(g,h)=det(F_{g,h})$. However it is not immediate that this constitutes a category: in order to do so one have to prove some identifications:

How to prove that $det(F_{g,h} \cdot F_{h,k}) \cong det(F_{g,h}) \otimes det(F_{h,k})$?

Of course both spaces are in fact one dimensional therefore there is an abstract isomorphism but I suspect that this isomorphism should be somehow canonical and natural. I would guess that in order to construct such an isomorphism maybe one has to use the fact that $Index(F_{g,h}F_{h,k})=Index(F_{g,h})+Index(F_{h,k})$.

Once we have this identification one can show that in fact $F_{g,h}F_{h,k}-F_{g,k}$ is a trace-class operator.

How to prove that if $S,T$ are two Fredholm operators with $S-T$ being trace-class then $detS \cong detT$ are canonically isomorphic?

I would be very grateful for any help.

Answer 1

As mentioned in the comments the first part of the question is in Abbonandolo and Majers "Infinite dimensional Grassmannians". My Hilbert spaces are real and separable and infinite dimensional.

I want to state that I basically know next to nothing about trace class operators, so take the rest with a grain of salt.

To every Fredholm operator one can associate its determinant line. An important fact is that this defines an actual line bundle over the space of Fredholm operators $\mathrm{det}\rightarrow \Phi(\mathbb{H})$. This is the unique (up to isomorphism) line bundle over $\Phi(\mathbb{H})$ that is non-trivial over each component of $\Phi(\mathbb{H})$. (This uses the fact that all components are homotopy equivalent and that $\pi_1(\Phi(\mathbb H))\cong\mathbb{Z}/2)$.

A continuous family of Fredholm operators is a continuous map $f:X\rightarrow \Phi(\mathbb H)$. One can use this map to pull back the determinant line bundle over $X$. Let's assume that $X$ is connected. Then the bundle $f^*\det$ is a trivial bundle over $X$ if and only if $f_*:\pi_1(X)\rightarrow \pi_1(\Phi(\mathbb H))$ is the trivial homomorphism.

Let $S$ be a Fredholm operator. Let $X$ be the space of Fredholm operators T such that $S-T$ is trace class, and $f:X\rightarrow \Phi(\mathbb H)$ the inclusion. I think that $X$ is an affine space, hence contractible, from which it follows that $f_*$ is trivial. This means that the determinant line bundle is trivial over $X$. You can use the triviality of the determinant line bundle to give a canonical isomorphism between two different operators of trace class.