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:



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> How to prove that $det(F\_{g,h} \cdot F\_{h,k}) \cong det(F\_{g,h}) \otimes det(F\_{h,k})$?
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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.



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> How to prove that if $S,T$ are two Fredholm operators with $S-T$ being trace-class then $detS \cong detT$ are *canonically* isomorphic?
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I would be very grateful for any help.



Determinant line of Fredholm operators and composition of morphisms
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.


