Timeline for Semigroup cohomology of the semigroup of Hilbert polynomials
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Oct 8, 2017 at 19:41 | comment | added | Benjamin Steinberg | Ok. I haven't thought about hilbrrt polynomials in 20 years. | |
| Oct 8, 2017 at 18:38 | comment | added | მამუკა ჯიბლაძე | @BenjaminSteinberg In the linked text, everything is in terms of Hilbert functions. The value at 0 of the Hilbert function is 1 by definition, but its values only coincide with values of the Hilbert polynomial after certain $n_0$ (called Hilbert regularity). | |
| Oct 8, 2017 at 18:15 | comment | added | Benjamin Steinberg | I had thought from the article the OP linked that the value at 0 is 1. | |
| Oct 8, 2017 at 17:38 | comment | added | მამუკა ჯიბლაძე | It might be that minimal generators of the semigroup are Hilbert polynomials of irreducible varieties. Note however that the semigroup of varieties themselves (w. r. t. Cartesian product) is not cancellative: in an answer to a question here is mentioned an example by Shioda of pairwise nonisomorphic elliptic curves $E$, $E'$, $E''$ with $E\times E'\cong E\times E''$. | |
| Oct 8, 2017 at 17:11 | comment | added | მამუკა ჯიბლაძე | @BenjaminSteinberg Constant term of the Hilbert polynomial, although always an integer, is not 1 in general. For example, the Hilbert polynomial of a degree $d$ plane curve has constant term $-\frac{d(d-3)}2$. | |
| Oct 8, 2017 at 16:52 | comment | added | Benjamin Steinberg | If you can figure out the rank of that free abelian group then your cohomology is easy to get. | |
| Oct 8, 2017 at 16:52 | comment | added | Benjamin Steinberg | Your monoid consists of certain elements with consists ten 1 if I understood correctly. The constant term 1 polynomials form a free commutative monoid on the irreducible ones. And so this bigger monoid has a free abelian group of infinite rank as Grothendieck group. So your monoid also has a free abelian Grothendieck group. | |
| Oct 8, 2017 at 16:49 | comment | added | Benjamin Steinberg | Without a zero the cohomology is the same as the cohomology group of the Grothendieck group of this commutative cancellative monoid by Quillen's Thm A because the classifying space of the monoid and group are the same. | |
| Oct 8, 2017 at 15:47 | comment | added | Mingchen Xia | @მამუკაჯიბლაძე I do not include 0, so I can only reduce the problem a little bit by the universal coefficient theorem. | |
| Oct 8, 2017 at 15:43 | comment | added | Mingchen Xia | @BenjaminSteinberg You can find Macaulay's theorem here comp.uark.edu/~ashleykw/SCATCooper.pdf Page 6. I do not assume 0 is there, so the problem is more tricky | |
| Oct 8, 2017 at 15:41 | comment | added | Benjamin Steinberg | Yes. I assume that one would not want the zero for this reason. I suspect the group completion is free abelian but don't know enough about the semigroup in question. | |
| Oct 8, 2017 at 15:32 | comment | added | მამუკა ჯიბლაძე | @BenjaminSteinberg On the other hand, if zero is there then the answer will be zero - for example, since cohomology in question coincidies with the cohomology of the classifying space, which is contractible for semigroups with a zero. | |
| Oct 8, 2017 at 13:00 | comment | added | Benjamin Steinberg | If you don't include zero as a Hilbert polynomial you have a cancellative commutative semigroup and so its cohomology on a module with trivial actin is the same as the cohomology of its Grothendieck group (or group completion of group of fractions). | |
| Oct 8, 2017 at 11:49 | comment | added | Benjamin Steinberg | Do you count the zero polynomial? | |
| Oct 8, 2017 at 11:47 | comment | added | Benjamin Steinberg | What does Macaulay's theorem say? | |
| Oct 8, 2017 at 7:14 | history | asked | Mingchen Xia | CC BY-SA 3.0 |