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-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 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*(0) and the "regularized determinant" DET *A* = exp(-ζ*A*'(0)), 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:


1. Is it true that exp TR *A* = DET exp *A*?
2. 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-1d*A*/d*t* is trace-class.)
3. 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.)
4. Is TR cyclic, i.e. TR(*AB*) = TR(*BA*)?
5. 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.



Zeta-function regularization of determinants and traces
I will answer some of my questions in the negative.


**3.**
First consider the case of rescaling an operator *A* by some (positive) number λ. Then ζλ*A*(s) = λ-sζ*A*(s), and so TR λ*A* = λ TR *A*. This is all well and good. How does the determinant behave? Define the "perceived dimension" DIM *A* to be logλ[ (DET λ*A*)/(DET *A*) ]. Then it's easy to see that DIM *A* = ζ*A*(0). What this means is that DET λ*A* = λζ*A*(0) DET *A*.


This is all well and good if the perceived dimension of a vector space does not depend on *A*. Unfortunately, it does. For example, the Hurwitz zeta functions ζ(s,μ) = Σ0∞(n+μ)-s (-μ not in **N**) naturally arise as the zeta functions of differential operators — e.g. as the operator *x*(d/d*x*) + μ on the space of (nice) functions on **R**. One can look up the values of this function, e.g. in Elizalde, et al. In particular, ζ(0,μ) = 1/2 - μ. Thus, let *A* and *B* be two such operators, with ζ*A* = ζ(s,α) and ζ*B* = ζ(s,β). For generic α and β, and provided *A* and *B* commute (e.g. for the suggested differential operators), then DET *AB* exists. But if DET were multiplicative, then:


DET(λ*AB*) = DET(λA) DET(*B*) = λ1/2 - α DET A DET B


but a similar calculation would yield λ1/2 - β DET A DET B.


This proves that DET is not multiplicative.


**1.**
My negative answer to 1. is not quite as satisfying, but it's OK. Consider an operator *A* (e.g. *x*(d/d*x*)+1) with eigenvalues 1,2,..., and so zeta function the Reimann function ζ(s). Then TR *A* = ζ(-1) = -1/12. On the other hand, exp *A* has eigenvalues e, e2, etc., and so zeta function ζexp *A*(s) = Σ e-ns = e-s/(1 - e-s) = 1/(es-1). This has a pole at s=0, and so DET exp *A* = lims→0 es/(es-1)2 = ∞. So question 1. is hopeless in the sense that *A* might be zeta-function regularizable but exp *A* not. I don't have a counterexample when all the zeta functions give finite values.


**5.**
As in my answer to 3. above, I will continue to consider the Hurwitz function ζ(*s,a*) = Σ*n*=0∞ (*n*+\_a\_)-*s*, which is the zeta function corresponding, for example, to the operator *x*(d/d*x*)+*a*, and we consider the case when *a* is not a nonpositive integer. One can look up various special values of (the analytic continuation) of the Hurwitz function, e.g. ζ(-*m,a*) = -*B**m*+1(*a*)/(*m*+1), where *Br* is the \_r\_th Bernoulli polynomial.


In particular,


TR(*x*(d/d*x*)+*a*) = -ζ(-1,*a*)/2 = -*a*2/2 + *a*/2 - 1/12


since, for example (from [Wikipedia](http://en.wikipedia.org/wiki/Bernoulli_polynomials)):


*B*2(*a*) = Σ*n*=02 1/(n+1) Σ*k*=0 *n* (-1)*k* {n \choose k} (*a*+\_k\_)2 = *a*2 - *a* + 1/6


Thus, consider the operator 2\_x\_(d/d*x*)+*a*+\_b\_. On the one hand:


TR(*x*(d/d*x*)+*a*) + TR(*x*(d/d*x*)+*b*) = -(*a*2+*b*2)/2 + (*a*+\_b\_)/2 - 1/6


On the other hand, TR is "linear" when it comes to multiplication by positive reals, and so:


TR(2\_x\_(d/d*x*)+*a*+\_b\_) = 2 TR(*x*(d/d*x*) + (*a*+\_b\_)/2) = -(*a*2+2\_ab\_+*b*2)/4 + (*a*+\_b\_)/2 - 1/6


In particular, we have TR(*x*(d/d*x*)+*a*) + TR(*x*(d/d*x*)+*b*) = TR( *x*(d/d*x*)+*a* + *x*(d/d*x*)+*b* ) if and only if *a*=\_b\_; otherwise 2\_ab\_ < *a*2+*b*2 is a strict inequality.


So the zeta-function regularized trace TR is not linear.


**0./2.**
My last comment is not so much to break number 2. above, but to suggest that it is limited in scope. In particular, for an operator *A* on an infinite-dimensional vector space, it is impossible for *A-s* to be trace-class for *s* in an open neighborhood of 0, and so if the zeta-function regularized DET makes sense, then det doesn't. (I.e. it's hopeless to say that det *A* = DET *A*.) Indeed, if the series converges for *s*=0, then it must be a finite sum.


Similarly, it is impossible for *A* to be trace class and also for *A-s* to be trace class for large *s*. If *A* is trace class, then its eigenvalues have finite sum, and in particular cluster near 0 (by the "divergence test" from freshman calculus). But then the eigenvalues of *A*-*s* tend to ∞ for positive *s*. I.e. it's hopeless to say that tr *A* = TR *A*.


My proof for 2. says the following. Suppose that d*A*/d*t* *A*-1 is trace class, and suppose that DET *A* makes sense as above. Then


d/d*t* [ DET *A* ] = (DET *A*)(tr d*A*/d*t* *A*-1)


I have no idea what happens, or even how to attack the problem, when d*A*/d*t* *A*-1 has a zeta-function-regularized trace.


