Timeline for Which sets of lattice points have rational generating functions?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 17, 2011 at 18:39 | vote | accept | Alex Fink | ||
| May 17, 2011 at 15:54 | answer | added | David E Speyer | timeline score: 12 | |
| May 16, 2011 at 16:21 | comment | added | Alex Fink | Here's an approach similar to Gerry's I'd been trying. Take a linear map $\mathbb N^d\to\mathbb N$ with finite fibers, so that the gf for the size of the fibers is the image of $f$ under $t_i\mapsto t^{a_i}$, with all $a_i>0$. The size of the fibers is integral and bounded by a polynomial, so by the one-variable theory it has to be a quasi-polynomial. It seems like that should impose useful conditions on $f$... | |
| May 16, 2011 at 16:19 | history | edited | Alex Fink | CC BY-SA 3.0 |
give the gf a name
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| May 16, 2011 at 16:05 | comment | added | Alex Fink | Qiaochu: yes, I meant to allow for the affine subsemigroup to be 0, or in general to be generated by any finite subset of $\mathbb N^d$. I've edited the question to say the same thing more polyhedrally. | |
| May 16, 2011 at 16:05 | history | edited | Alex Fink | CC BY-SA 3.0 |
add an equivalent polyhedral condition to the conjectured answer
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| May 15, 2011 at 11:18 | comment | added | Gerry Myerson | I wonder if this works. Multiply through by the denominator of the rational function; the vanishing of most of the terms in the product gives you a recurrence relation, in many variables, satisfied by the coefficients. Then maybe it's possible to eliminate all but one variable, and apply the well-known theory of one-variable recurrences. | |
| May 15, 2011 at 5:36 | comment | added | Torsten Ekedahl | The affine semigroup could consist of $0$ only which allows for $P$ to be finite. | |
| May 15, 2011 at 5:12 | comment | added | Qiaochu Yuan | Can you clarify what a finitely-generated module over an affine sub-semigroup is? Any definition of this I can think of would require that $P$ be infinite. | |
| May 15, 2011 at 4:22 | history | asked | Alex Fink | CC BY-SA 3.0 |