Timeline for Zeta function of $\Delta[\text{det},m]$
Current License: CC BY-SA 3.0
11 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Mar 5, 2017 at 16:31 | history | edited | Alexey Milovanov | CC BY-SA 3.0 |
added 211 characters in body
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| Mar 5, 2017 at 16:24 | history | undeleted | Alexey Milovanov | ||
| Mar 5, 2017 at 12:16 | history | deleted | Alexey Milovanov | via Vote | |
| Mar 5, 2017 at 10:52 | comment | added | Gerry Myerson | Also posted to (and answered at) m.se, math.stackexchange.com/questions/2171324/is-delta-det-m-smooth without notice to either site. | |
| Mar 5, 2017 at 10:32 | comment | added | Dan Petersen | Since you said you already have a proof assuming $\Delta[det,m]$ smooth, I haven't tried to work out what precise bounds you get in this way. But the key is that the degrees of these polynomials are bounded by the sum of the Betti numbers of $\Delta[det,m]$, and these are bounded by the degrees of the defining equations. One reference is a paper by Katz, "Sums of Betti numbers in arbitrary characteristic". | |
| Mar 5, 2017 at 10:30 | comment | added | Dan Petersen | oh, I see. Then your proof in the case $\Delta[det, m]$ smooth should work just as well in the general case. The only part of the Weil conjecture that fails for singular $X$ is the Riemann hypothesis (i.e. the estimate for the absolute values of the roots of the factors $P_i(T)$ of the zeta function). It is still true for $X$ a singular compact variety that the zeta function is rational and the factors in the factorization have degrees given by the Betti numbers. (If $X$ is not compact then you need to work with the compactly supported Betti numbers.) | |
| Mar 5, 2017 at 10:18 | comment | added | Alexey Milovanov | @DanPetersen some polynomial in $m$. | |
| Mar 5, 2017 at 10:15 | comment | added | Dan Petersen | what is poly(m)? | |
| Mar 5, 2017 at 8:41 | history | edited | Alexey Milovanov |
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| Mar 4, 2017 at 22:06 | history | asked | Alexey Milovanov | CC BY-SA 3.0 |