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 ζλA(s) = λ-sζA(s), and so TR λA = λ TR A. This is all well and good. How does the determinant behave? Define the "perceived dimension" DIM A to be logλ[ (DET λA)/(DET A) ]. Then it's easy to see that DIM A = ζA(0). What this means is that DET λA = λζA(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/dx) + 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 ζA = ζ(s,α) and ζB = 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(λAB) = 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/dx)+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) = Σn=0∞ (n+_a_)-s, which is the zeta function corresponding, for example, to the operator x(d/dx)+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) = -Bm+1(a)/(m+1), where Br is the _r_th Bernoulli polynomial.
In particular,
TR(x(d/dx)+a) = -ζ(-1,a)/2 = -a2/2 + a/2 - 1/12
since, for example (from Wikipedia):
B2(a) = Σn=02 1/(n+1) Σk=0 n (-1)k {n \choose k} (a+_k_)2 = a2 - a + 1/6
Thus, consider the operator 2_x_(d/dx)+a+_b_. On the one hand:
TR(x(d/dx)+a) + TR(x(d/dx)+b) = -(a2+b2)/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/dx)+a+_b_) = 2 TR(x(d/dx) + (a+_b_)/2) = -(a2+2_ab_+b2)/4 + (a+_b_)/2 - 1/6
In particular, we have TR(x(d/dx)+a) + TR(x(d/dx)+b) = TR( x(d/dx)+a + x(d/dx)+b ) if and only if a=_b_; otherwise 2_ab_ < a2+b2 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 dA/dt A-1 is trace class, and suppose that DET A makes sense as above. Then
d/dt [ DET A ] = (DET A)(tr dA/dt A-1)
I have no idea what happens, or even how to attack the problem, when dA/dt A-1 has a zeta-function-regularized trace.