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This question goes in the bucket of "this must be well known, but I don't see it and am not sure where to look it up."

Given two Laurent power series $A(t)=\sum_{k>N}a_kt^k$ and $B(t)=\sum_{k>M}b_kt^k$ for $a_k,b_k\in F$ a field, we say that these are expansions at 0 and $\infty$ of a rational function $f$ if in the field $F((t))$, we have that $f(t)=A(t)$ and $f(t^{-1})=B(t)$.

Now, think about $\,a_k$ and $\,b_k$ as formal variables:

Are there any relations between $a_k$ and $b_k$, or any other way of expressing algebraically that $A(t) = B(t^{-1})$?

It seems to me that there can be no relations in the strong sense that I can fix any finite number of $a_*$ and $b_*$ and complete to get such compatible expansions, but maybe there's some more subtle relation I've missed.

EDIT: After feeding the answer below, I see what I should have thought of before: this is actually a topological property. We can endow the ring generated by $a_k$ and $b_k$ with a topology such that the ideal generated by the $m\times m$ minors of the Hankel matrix for each $m$ give a basis of neighborhoods of $0$. A map of this ring to the base field $F$ (with the discrete topology) is continuous if and only if one of these neighborhoods is killed, and we have a rational function that we have the two expansions of.

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    $\begingroup$ [deleted an earlier comment] Just to make sure I understand, you are interested in something like the following situation: $f(t) = (2+t^2) / (1-t)$, so we can form a Laurent expansion about $t=0$ to get $(2+t^2)(1+t+t^2+\dots)=$ whatever, then $f(t^{-1})= (2t^2+1) / (t^2-t)$ whose Laurent expansion about $t=0$ is $-t^{-1}(2t^2+1)(1+t+t^2+\dots)=$ whatever. Is that the setup you're describing? $\endgroup$
    – Yemon Choi
    Jul 10 '18 at 2:20
  • $\begingroup$ I think any relations might eventually be more about rational functions than anything else. One possibility is to look for roots of the denominator, which can appear in the coefficients as the geometric coefficients. In other words, look for partial fraction decompositions in the power series. $\endgroup$
    – user44191
    Jul 10 '18 at 3:10
  • $\begingroup$ Note that a global field is dense in any finite product of its local fields. Take, for example, the global field $F_q(T)$ of $P_{F_q}^1$ and its local fields at $T = 0$ and $T = \infty$. $\endgroup$
    – Marty
    Jul 10 '18 at 4:59
  • $\begingroup$ Perhaps Lemma (1.3.4) in Kapranov's The elliptic curve in S-duality theory, arXiv:math.AG/0001005 is relevant. $\endgroup$
    – Balazs
    Jul 10 '18 at 5:21
  • $\begingroup$ @Balazs that lemma essentially says that $A(t)-B(t^{-1})$ must be of the form $\sum_{n\in \mathbb Z} \left(P_1(n)a_1^n+\cdots P_k(n)a_k^n\right)t^n$. Now remains the problem of recognizing sequences of coefficients which are of this form. There is no single identity that will do the job, and my answer shows that you can get away with asking every large enough minor of the Hankel matrix to vanish. $\endgroup$ Jul 10 '18 at 7:45
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One common criterion that is used to algebraically encode whether a power series $\sum_{n\geq 0} a_nx^n$ represents a rational function is that the infinite Hankel matrix $(a_{i+j-1})$ be of finite rank. If this is the kind of criterion you are looking for then for your pair of series $A(t)$ and $B(t)$ satisfy $A(t)=B(t^{-1})$ if and only if the (bi infinite) Hankel matrix associated to $$C(t)=A(t)-B(t^{-1})$$ is of finite rank. That is, express $A(t)-B(t^{-1})$ as a power series $\sum_{n\in \mathbb Z}c_n x^n$ and then check that the bi infinite matrix $(c_{i+j-1})$ has finite rank. This is the same as saying that certain determinants with entries of the form $a_k-b_{-k}$ vanish.

More explicitly this is encoding the fact that if, for example, $A(t)$ looks like $$\sum_{n\geq 0}\left(P_1(n)\alpha_{1}^n+\cdots P_k(n)\alpha_k^n\right)t^n$$ then $B(t)$ looks like $$-\sum_{n\geq 0}\left(P_1(-n)\alpha_{1}^{-n}+\cdots P_k(-n)\alpha_k^{-n}\right)t^n.$$ In other words as formal power series we have a "vanishing relation" $$\sum_{n\in \mathbb Z} P(n)\alpha^nz^n=0$$ This last relation is a nonrigorous way of expressing what I said above, but I like thinking about it as a relation in the same sense as in Brion's formula from polyhedral generating functions/equivariant localization (and it can be made rigorous using the same methods).

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  • $\begingroup$ Also notice that this answer confirms your observation about fixing a finite number of $A$ and $B$ coefficients and then completing the rest. For example I can fix arbitrary values for $P(1),P(2),...,P(k)$ as well as $P(0),P(-1),...,P(-r)$ and then find a polynomial $P$ that takes these values. This is because you need infinitely many relations (aka arbitrarily large determinants to vanish). $\endgroup$ Jul 10 '18 at 3:33
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    $\begingroup$ Is there a specific meaning for the rank? From what I can see, it seems to be the degree of the denominator "away from 0 and infinity", tentatively. $\endgroup$
    – user44191
    Jul 10 '18 at 3:38
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    $\begingroup$ This result goes back at least to Pólya in 1928--29, as discussed on page 524 of Enumerative Combinatorics, vol. 1, second ed. $\endgroup$ May 27 '20 at 3:58

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