Zeta-function regularization of determinants and traces

Question

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 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/dt [ log DET A(t) ] = TR( A^-1 dA/dt )? (I can prove this when A^-1dA/dt 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 dA/dt) are simultaneously diagonalizable (except of course cyclicity), but of course in general they won't commute.

Answer 1

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 ζ_λA(s) = λ^-sζ_A(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 log_λ[ (DET λA)/(DET A) ]. Then it's easy to see that DIM A = ζ_A(0). What this means is that DET λA = λ^ζ_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/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 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(λAB) = DET(λA) DET(B) = λ^{1/2 - α} DET A DET B

but a similar calculation would yield λ^{1/2 - 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 ζ_{exp A}(s) = Σ e^-ns = e^-s/(1 - e^-s) = 1/(e^s-1). This has a pole at s=0, and so DET exp A = lim_s→0 e^s/(e^s-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/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) = -B_m+1(a)/(m+1), where B_r 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) = Σ_n=0² 1/(n+1) Σ_k=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.

Answer 2

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).