What's the correct notion of determinant of a bilinear pairing?

By a pairing on a vector space $V$, I mean a linear map $A : V \otimes V \to R$. If $V$ is $n$-dimensional ($n < \infty$), then I can define the determinant of $A$ by considering the canonical action of $A$ on $\bigwedge^n V \otimes \bigwedge^n V \to R$. In fact, I should define $\det A$ to be this map, which I will call $\bigwedge^n A$. Remember that $\bigwedge^n V$ is one-dimensional, and pairs canonically with the space of volume forms on $V$. So if $V$ has a volume form $\mathrm{vol}$, we can define $\det A$ as $\bigwedge^n A(\mathrm{vol}^{-1} \otimes \mathrm{vol}^{-1})$.

For comparison, if I had an operator $A: V\to V$, then I would look at the map $\bigwedge^n A: \bigwedge^n V \to \bigwedge^n V$. But $\bigwedge^n$ is one-dimensional, so $\bigwedge^n A$ consists of multiplication by a scalar, and this scalar is $\det A$. On the other hand, if you pick an identification $V \cong V^*$, then you can think of a pairing as an operator; the identification determines a non-zero volume form, and the notions of determinants are the same.

In any case, these definitions don't work for infinite-dimensional vector spaces, because there's no wedge-top''. I'd like a notion of "determinant of a pairing" like the zeta-function regularized determinant of an operator.

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Here's a way to test your definition. Let V be the space of smooth derichlet functions on [0,T]. I very firmly believe that the determinant of the pairing \int_0^T a f'(t) g'(t) dt (i.e. take derivatives and integrate, with a factor of a>0) should be T/a, where the "volume form" is "\int_0^T dx(t) dt". Via the usual pairing \int f(t) g(t) dt, my pairing is related to the map -a d^2/dt^2, which has zeta-function-regularized determinant 2T/\sqrt{a}. – Theo Johnson-Freyd Oct 8 '09 at 0:20

I'm struggling to understand your question. In the true wiki-spirit, let me try this publicly in the hope that you or someone else can clarify it for me.

You start with a finite dimensional definition. We have a linear map A: V⊗V → ℝ. From this you produce a map A: ⋀n V ⊗ ⋀n V → ℝ. I don't see this map. I presume that you are supposed to pair-up elements in the two factors and apply A, so you get A(vi ⊗ uj) for each i and j. That sort-of gives a matrix, not quite well-defined since I can swap rows and columns at the expense of a sign swap. Of course, taking the determinant of this matrix gives me something sensible. That means that I can define det(A) as a function to be det[A(vi ⊗ uj)]. Is this what you mean?

Let's see if that makes sense for your second paragraph where I do think I know what's going on. We start with A: V → V and for A: ⋀n V → ⋀n V. That I'm happy with. Then as that's one dimensional, I get a canonical scalar which I call det(A).

Now imagine we have an identification Ψ:V ≅ V*. Now you say that we can use this to convert an operator into a pairing, presumably via A ↦ (u⊗v ↦ Ψ(u)(Av). Ψ also gives us a volume form for ⋀n V, and I'm quite prepared to believe that the two notions coincide.

Now you want to do this in infinite dimensions.

Firstly, there is a notion of the top exterior power of an infinite dimensional vector space. There's absolutely no problem with that. However, if I'm correct in what you're trying to do about then that wouldn't help you since you need to take the determinant of an infinite matrix in order to define the determinant of a pairing. Also, it doesn't define the determinant of an arbitrary (even continuous) linear operator, so it wouldn't help with the other definition either.

Secondly, the "pairing" between pairings and operators in infinite dimensions isn't so tight as in finite dimensions. You talk blithely about picking an identification V ≅ V*. That suggests that you are thinking about Hilbert spaces. I'm not quite clear what a "smooth derichlet function" is (and I don't mean just the spelling of Dirichlet), so I can't be sure, but your example doesn't feel like a Hilbert space.

There is a notion of an operator with a determinant in infinite dimensions, but I'm not sure that this is going to help either.

A pairing is really a map V → V in disguise. Of course, there are spaces other than Hilbert spaces for which V is isomorphic to V, but generally one thinks about Hilbert spaces in this context. If not, and you only have a non-degenerate "standard" pairing on your space then you get an injection V → V* and there's no guarantee that the image of the map derived from the pairing will end up in V. To ensure that, you have to have strong conditions on the pairing, which probably preclude it from being an operator with a determinant.

However, you bring in zeta renormalisation, at which point I'm afraid I can't help. But I'm not sure that your question is too concerned with zeta renormalisation since that is, for you, a known quantity: you know what the zeta renormalisation of the determinant of your operator is, but you want to get it straight from a definition of the determinant of the pairing. Thus the point is the definition, not the renormalisation. At least, that's my reading of it.

I apologise that this isn't anyway remotely close to an answer, but it's a bit long for a comment and might help someone else give you an actual answer. I thought it worth sharing my attempts to understand the question, though. If not, I'll happily delete it.

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Yes, I think that your clarification at the beginning is what I had in mind. I tend to work in indices --- actually in Penrose's diagrams --- and then what I meant is clear: a pairing is an object A_{ij} with two lowered indices; take its nth tensor power A_{i_1j_i}\dots A_{i_nj_n}, antisymmetrize the i_ks (i.e.\ project to the antisymmetric part) and antisymmetrize the j_ks. This defines what I claim is the correct notion of det A as a symmetric bilinear pairing on \bigwedge^n V. – Theo Johnson-Freyd Oct 8 '09 at 20:13
As to your later points, I'm fully aware that V and V* are not isomorphic --- even in Hilbert space, unless by V* you mean bounded linear functionals. And anyway the example I care about is not bounded. Can you please point me to the right description of "wedge-top" of an infinite-dimensional vector space? (Oh, and sorry about the misspelling. The case I care about is smooth functions on [0,T] that vanish at the end-points. The operator -d^2/dt^2 takes such functions to functions that need not vanish, but Fourier still gives the eigenvalues; this is what's done in the physics literature.) – Theo Johnson-Freyd Oct 8 '09 at 20:20

I apologize to Andrew Stacey in advance since what I was going to say is basically what he's saying.

Consider a finite-dimensional space. There's a notion of determinant of an operator (operator is defined as map $V \otimes V^* \to \mathbb R$). This is a simple notion, just a product of eigenvalues, or a trace on a one-dimensional space of $V^{top} \to V^{top}$.

Now you can find an isomorphism $V^* \to V$ for any non-degenerate scalar form $(\dot, \dot)$ — so there's a way to say that for pairing and scalar form there's a determinant that is a number. But still let's not think of it as a determinant of a pairing — it's really a determinant of an operator.

The situation is basically the same with infinite-dimensional spaces. Select your favorite regularization scheme — a way to compute products of infinite series of numbers — (e.g. zeta-regularization) and define for an operator $X$ determinant $\det X$ as product of its eigenvalues, if it converges, or leave it undefined it still doesn't. For a pairing $A$ you can do that provided there's a scalar form fixed, that is whenever you have a Hilbert space.

You can think about wedge-top form here, but then you have to define wedge-top for an inifite-dimensional space. Sure, it can be done formally and you'll be able to define the action of operator on it by referring to the above paragraph for the meaning of det, but I don't know of any sufficiently different way to start from wedge-top and get det from there.

I'll look up the references. Note that the philosophy of physics is slightly different. They don't need to compute dets of arbitrary operators, rather they have some specific operators for which they know the answer must be finite and the fact that our initial computation gives infinity is because it just an approximation to the right calculation and a stupid approximation at that.

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I'm intrigued by the regularisation. It's something I don't know much about. Do you have a good reference where I can learn it? What's det(2I) for example? Take any space you like, providing it is infinite dimensional. – Loop Space Oct 8 '09 at 20:20
This much I know, but I think you said it more clearly. What I'm getting at is: there's a good notion of determinant of a pairing that is not real-number valued but rather is valued in the space of bilinear maps on wedge-top (see above). Since wedge-top is one-dimensional, there is a one-dimensional space of these. A choice of inner product picks out a distinguished element of wedge-top (inverse to the volume form), and evaluating my bilinear form on this distinguished element gives your (more standard) determinant. I'm hoping for a version of this story for infinite-dimensional spaces. – Theo Johnson-Freyd Oct 8 '09 at 20:25
@Andrew: I don't have any good references, but apparently these regularizations are standard in the physics world (zeta-regularization was spearheaded by Hawking). I learned what I know about it from Takhtajan's book on Quantum Mechanics, and from random googling. I hope that my linked-to question will elicit some good references. Anyway, I don't know what det(2I) is: no power of 2I is trace-class. In the case of Dirichlet functions on [0,T], I believe the answer should be 1/2, but not for any mathematical reason, and the best answer I can get zeta to give is 1/\sqrt{2}. – Theo Johnson-Freyd Oct 8 '09 at 20:31
My question about det(2I) was aimed at Ilya, hence putting it in a comment here, since Ilya said "For an operator X you can define det(X) as -- suitably regularized (for example, zeta-regularized) -- product of its eigenvalues." As this is completely unknown to me, I was trying to understand it via an example, 2I being the simplest I could think of that would also be illustrative. – Loop Space Oct 9 '09 at 7:09
I'm sorry for the misunderstanding. What I meant was that you define det to be some number if your scheme if the scheme gives a finite answer, so, no, I don't know any scheme to make det(2*I) finite. – Ilya Nikokoshev Oct 11 '09 at 21:15