The short answer to my question may be a pointer to the right text. I will give all the background I know, and then ask my questions in list form.
Let A be an operator (on an infinite-dimensional vector space). You might as well assume that its spectrum is all real and positive. In fact, I only care when the spectrum is discrete and grows polynomially, but I hear that this stuff works more generally.
In general, A is not trace-class (the sum of the eigenvalues converges) or determinant-class (the product of the eigenvalues converges) — if the nth eigenvalue grows as np for some p>0, then it won't be. But there is a procedure to try to define a "trace" and "determinant" of A nevertheless.
Let us hope that for large enough s, the operators A-s (=exp(-s log A), and log A makes sense if the spectrum of A is positive) are trace-class. If so, then we can define ζA(s) = tr(A-s); it is analytic for Re(s) large enough. Let's hope that it has a single-valued meromorphic continuation and that this function (which I will also call ζA(s)) is smooth near s=0 and s=-1. All these hopes hold when the eigenvalues of A grow polynomially, whence ζA(s) can be compared to the Riemann zeta function.
Then we can immediately define the "regularized trace" TR A = ζA(-1) and the "regularized determinant" DET A = exp(-ζA'(-1)), where by ζA'(s) I mean the derivative of ζA(s) with respect to s. (If the eigenvalues λn are discrete, then ζA(s) = Σ λn-s, and so one would have TR A = Σ λn and DET A = Σ (log λn) λn-s |s=0, if they converged.) If A is trace- (determinant-) class, then TR A = tr A (DET A = det A).
So, here are my questions:
- Is it true that exp TR A = DET exp A?
- Let A(t) be a smooth family of operators (t is a real variable). Is it true that d/dt [ log DET A(t) ] = TR( A-1 dA/dt )? (I can prove this when A-1dA/dt is trace-class.)
- Is DET multiplicative, so that DET(AB) = DET A DET B? (I can prove this using 1. and 2., or using the part of 2. that I can prove if B is determinant-class.)
- Is TR cyclic, i.e. TR(AB) = TR(BA)?
- Is TR linear, i.e. TR(A + B) = TR A + TR B?
None of these are even obvious to me when A and B (or dA/dt) are simultaneously diagonalizable (except of course cyclicity), but of course in general they won't commute.