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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(-1) and the "regularized determinant" DET A = exp(-ζA'(-1)), 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:

  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.

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2 Answers 2

up vote 2 down vote accepted

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.

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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).

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Great, although any chance you can either spell out the reasoning or point me to a reference? For example, here's what I know about (1). Let A have eigenvalues A_n. Then DET exp A is: compute \sum A_n \exp(-s A_n), extend, and evaluate at s=0. exp TR A is: compute \exp( \sum A_n^{-s}), extend, and evaluate at s=-1. Is there a change of variables to get from one of these functions to the other? Otherwise, I don't know why their extensions should match up. But perhaps I'm being dumb. –  Theo Johnson-Freyd Oct 8 '09 at 20:38
    
I'll return to this on the weekend and see if I'm right :). –  Ilya Nikokoshev Oct 8 '09 at 22:01
    
Ok, it's certainly my mistake --- I basically thought det exp A is defined as exp tr A which would make (1) trivial. But for the reference, I think books.google.com/… might help. –  Ilya Nikokoshev Oct 11 '09 at 21:12

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