Zeta-function regularization of determinants and traces 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(0) and the "regularized determinant" DET A = exp(-ζA'(0)), 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.
 A: 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.
A: I don't have a specialized knowledge about regularizations, but as the first direction I think:
(1) yes, it's an identity if you look into definition of zeta-function (I thought of a different definition of zeta-regulaization)
(2) not sure, I'll check later. But it seems to be an identity as well.
(4, 5) TR only depens on eigenvalues. I don't remember what the status of eigen{AB} = eigen{BA} is, but I think its true under standard conditions on A and B -- if they are self-conjugate in a Hilbert space.
Let's say you try to compare regularized \det(1-AB) and \det(1-BA). I think the standard reasoning applies and says that those are equal; same might apply to (A+B).
