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Studying hyperplane arrangements in Stanley's Enumerative Combinatorics, the finite field method is explained and (for completeness here) relies on the following theorem:

Let $\mathcal{A}$ be an arrangement in $\mathbb{Q}^n$, and suppose $L(\mathcal{A})\simeq L(\mathcal{A}_q)$ for some prime power $q$. Then $$\chi_\mathcal{A}(q)=\#\left(\mathbb{F}^n_q-\bigcup_{H\in\mathcal{A}_q}H\right).$$

Using this theorem is straightforward enough but there's a small detail I can't really grasp: in all examples Stanley applies the theorem to get some expression $\chi_\mathcal{A}(q)$ and then directly makes a conclusion of $\chi_\mathcal{A}(x)$ for general $x$. For me, $q$ is fixed although (slightly) arbitrary so $\chi_\mathcal{A}(q)$ is, you know, a value that happens to be a polynomial in $q$. It feels like drawing the conclusion that $f(x)=x^2+2x-1$ from knowing that $f(3)=14$. Is it because there's infinitely many such $q$'s? Something else?

I'm rather certain this has some combinatorics-flavored answer where the variable is there to more or less just keep different numbers apart (and evaluating functions at a specific point is more of a nice feature than the point of the procedure), but it's hard to be certain of your own thoughts with such things.

Not a Salmon Fish
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Anna Lindeberg
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  • I should learn this! It seems to be that your question is related to this one, where the root system stuff simply says a lot more about the inherent symmetries of the hyperplane arrangement. Again assuming that your question is about an arrangement of hyperplanes. But, this is mostly speculation by me. – Jyrki Lahtonen Jan 05 '23 at 03:38
  • Anyway (more speculation), it seems likely that we, indeed, get the same polynomial for most choices of prime $p$ (and $q$ a power of such a prime). In vaguely similar problems in number theory there are often only finitely many primes, where something exceptional happens. Comparing with the other question, I guess that this polynomial simply "counts" the numbers of hyperplanes of various dimensions gotten as intersections of the system. Of course, it is a live possibility that I'm way off base :-) – Jyrki Lahtonen Jan 05 '23 at 03:44
  • Another possibility is that if we know, from elsewhere in the theory, that the coefficients of $\chi$ are integers of bounded size, then some large enough input will uniquely identify it. For example, if we know that $\chi(x)$ is a quadratic polynomial with integer coefficients of absolute value at most $3$, and that $\chi(1000)=1001999$, then we can conclude that $\chi(x)=x^2+2x-1$. – Jyrki Lahtonen Jan 05 '23 at 03:50
  • It's indeed about hyperplane arrangements! Stanley just shortens it to "arrangements" only. I think you're onto something with the boundedness of the coefficients – although I'm uncertain of exactly how. Per definition $\chi_{\mathcal{A}}(x)=\sum_{t\in L(\mathcal{A})}\mu(0,t)x^{dim(t)}$ (which is a handful) i.e. each coefficient is a finite sum of integers, since $\mu$ is integer-valued. I'll try to follow this trail somehow... – Anna Lindeberg Jan 05 '23 at 12:24

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