It is well-known, that if $A = \mathrm{id} + S$ is a bounded operator on a separable Hilbert Space $H$ with $S$ trace-class, then there is a well-defined notion of determinant, e.g. in terms of the singular value decomposition of $S$.


Now let $H$ and $K$ be two separable, infinite-dimensional Hilbert Spaces. To define the determinant for operators between $H$ and $K$, one could now try to make the following definition.


**Def.** If for a bounded linear operator $A: H \longrightarrow K$, there exists a unitary transformation $U: H \longrightarrow K$ such that $A-U \in N(H, K)$, the space of nuclear operators from $H$ to $K$, define
$$ |\det(A)| = |\det(U^{-1}A)|.$$
Note that $U^{-1}A - \mathrm{id}$ is trace-class by the assumption on $A$ so that this makes sense.


*Question:* Is this well-defined, i.e. independent of the choice of $U$. A quick calculation shows that this boils down to the question: 


**If $V^{-1}U-\mathrm{id}$ or equivalently $U-V$ is trace-class, is then necessarily $|\det(V^{-1}U)| = 1$ ?**


*Remark*: Notice that without some kind of orientation, you can obviously only define the modulus of the determinant. But is there a notion of orientability on infinite-dimensional Hilbert spaces?



Infinite Determinant between different Hilbert Spaces
It seems this implies that $V^{-1}U$ can be diagonalised isn't it ?


Indeed: $V^{-1}U$ is a unitary, let just call it $W$. If $1-W$ is trace class it is in particular compact, moreover it is normal (as $W$ is unitary) hence it can be diagonalised in some Hilbert basis which obviously also diagonalize $W$.


The deteminant is then a convergent product of eigen values of $W$, which are all complex number of module $1$ so this conclude the proof.


