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 *n*th eigenvalue grows as *n*^{p} 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 ζ

_{A}(

*s*) = tr(

*A*); it is analytic for Re(

^{-s}*s*) large enough. Let's hope that it has a single-valued meromorphic continuation and that this function (which I will also call ζ

_{A}(

*s*)) is smooth near

*s*=0 and

*s*=-1. All these hopes hold when the eigenvalues of

*A*grow polynomially, whence ζ

_{A}(

*s*) can be compared to the Riemann zeta function.

Then we can immediately define the "regularized trace" TR *A* = ζ_{A}(-1) and the "regularized determinant" DET *A* = exp(-ζ_{A}'(-1)), where by ζ_{A}'(*s*) I mean the derivative of ζ_{A}(*s*) with respect to *s*. (If the eigenvalues λ_{n} are discrete, then ζ_{A}(*s*) = Σ λ_{n}^{-s}, and so one would have TR *A* = Σ λ_{n} and DET *A* = Σ (log λ_{n}) λ_{n}^{-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/d*t*[ log DET*A*(_t_) ] = TR( A^{-1}d*A*/d*t*)? (I can prove this when A^{-1}d*A*/d*t*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 d*A*/d*t*) are simultaneously diagonalizable (except of course cyclicity), but of course in general they won't commute.