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This question is rather vague. Are there any natural situations which involve Laurent polynomials of the form $$\sum q^{a_i}\in\mathbb{Z}[q,q^{-1}]$$ where the $a_i$'s are either Euler characteristics of some spaces (possibly all subspaces of one fixed space), or more generally, indices of some elliptic operators? I've stumbled across such a beast, but am unsure how to interpret it. I was thinking at first that it was an element of $K_{S^1} (pt)$ or something, but in that case the exponents are telling us about which $S^1$ representations show up in the appropriate bundles, not the indices of the operators, right? Is there some relation with the index? (Please tell me if I'm talking nonsense! I don't really know this K-theory stuff).

Maybe the answer I'm looking for doesn't involve K-theory, anyway. Does anyone have any ideas? I'd love to hear about any and everything!

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2 Answers 2

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  1. Knot polynomials like the Jones polynomial
  2. Perturbative expansions of Feynman Integrals
  3. Heat kernel asymptotics, and other universal polynomials in characteristic classes.
  4. Generating functions associated to combinatorial problems.
  5. Poincare Polynomials of Topological Spaces.
  6. Hilbert Polynomials.
  7. Certain families of orthogonal polynomials, generally associated to representation theory.
  8. The A-polynomial

I am sure there are more.

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  • $\begingroup$ Ah many of these are quite related to the things I've been looking at! But can you explain some of them to me? For example, as far as I was aware, the exponents in the Poincaré polynomial just correspond to the gradings on your space (or vector space), and it's the coefficients which resemble Euler characteristics. Unfortunately, in my case the Euler characteristics (or indices of operators) show up in the exponents, not the coefficients. Is there some natural operation that would be responsible for this? (my same complaint holds or 1, as far as I know.) How about 2? and 8? Thanks a lot! $\endgroup$
    – Sam Lewallen
    Commented Mar 29, 2010 at 4:28
  • $\begingroup$ If you were looking at the Poincare Polynomial of a filtered Floer chain complex, then the index of the operators could appear in the exponents because of the Maslov index, and whatever you used to get your filtration. In infinitely many dimensions, the Maslov index is computed via the Atiyah-Singer index theorem, and it can involve the kinds of shifting you are thinking of. Also the terms can show up just from needing to adjust the boundary operator so it respects the filtration. It show up in 2, from standard formulas for doing asymptotic expansion using steepest descent. $\endgroup$
    – Charlie Frohman
    Commented Mar 29, 2010 at 10:08
  • $\begingroup$ The A-polynomial is in some sense dual to the Jones polynomial. You can find a primitive version of this duality in Frohman, Gelca, Lofaro in the Transactions of the AMS. However, Thang Le and Stavros developed these ideas more clearly, and many of Stavros great recent conjectures flow from this viewpoint. It would be cool to Categorify the A-polynomial, so that the homology theory was "Dual" to the Colored Jones Khovanov homology. $\endgroup$
    – Charlie Frohman
    Commented Mar 29, 2010 at 10:12
  • $\begingroup$ One more thing. In the computation of the original Floer homology of Seifert Fibered spaces you got components of representations, like what you are seeing. Kirk and Klassen worked those out. When you perturb a component to make it transverse, it delivers its Euler characteristic of critical points. $\endgroup$
    – Charlie Frohman
    Commented Mar 29, 2010 at 11:24
  • $\begingroup$ hmm, wasn't there something longer here before? I was about to comment. It was super useful! I'm still digesting all the things you've said... anyway, thanks very very much, and I'll reply soon! $\endgroup$
    – Sam Lewallen
    Commented Mar 30, 2010 at 17:25
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I met this guy in quantum cohomology. More precisely, in examples of quantum coefficient ring. So if you tag the this problem with "symplectic geometry" or "symplectic topology" I think many people there will have better understanding.

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  • $\begingroup$ hmm thanks! Can you be more specific? $\endgroup$
    – Sam Lewallen
    Commented Mar 30, 2010 at 17:27

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