2,516 reputation
12335
bio website math.sunysb.edu/~jstarr/…
location Stony Brook University, Stony Brook, NY USA
age
visits member for 5 years, 9 months
seen 6 mins ago

5h
answered Genus of a plane curve of the form $\prod_{i=1}^n (a_iX+b_iY+Z) = Z^n$
7h
comment Fibrations on blow-ups of $\mathbb{P}^2$
You are saying the same thing as what I said . . .
7h
comment Genus of a plane curve of the form $\prod_{i=1}^n (a_iX+b_iY+Z) = Z^n$
You said before that you know that there are singularities on the line $Z=0$, and now you are saying that there are not lines? Please read about the connection between the geometric genus and the singular points, for instance, in Example 3.9.2, p. 393 of Hartshorne's "Algebraic Geometry"
8h
comment Genus of a plane curve of the form $\prod_{i=1}^n (a_iX+b_iY+Z) = Z^n$
@AngeldelRio: If you can estimate from above the number of such intersection points on lines, then you can bound from below the geometric genus, assuming this curve is irreducible.
9h
comment Hilbert scheme of relative subschemes of lenght 2
Yes, you do have that. The standard obstruction theory for Hilbert schemes (as described in Artin's "Algebraization ... I" or in Chapter 1 of Koll'ar's "Rational curves ...") is actually an obstruction theory relative to a base $S$, at least as long as $\mathfrak{X}$ is flat over $S$ (otherwise there are some corrections). So the usual argument -- length $2$ schemes are LCI and the global obstruction is an $H^1$ that vanishes on zero dimensional schemes -- proves smoothness over $S$ of the relative Hilbert scheme.
10h
comment Genus of a plane curve of the form $\prod_{i=1}^n (a_iX+b_iY+Z) = Z^n$
If the projective lines given by $Z+\overline{a}_iX+\overline{b}_iY = 0$ are distinct, and if all pairwise intersections do not lie on the line $Z=0$, then it is straightforward to see that the curve is smooth.
10h
comment Genus of a plane curve of the form $\prod_{i=1}^n (a_iX+b_iY+Z) = Z^n$
Sorry: "pairwise linearly independent".
10h
comment Genus of a plane curve of the form $\prod_{i=1}^n (a_iX+b_iY+Z) = Z^n$
Even though the $a_i$ and $b_i$ are not integers, they are nonetheless algebraic integers. You can reduce your curve modulo a prime divisor $q$ of $n$. Then, if you check that the reductions $(\overline{a}_i,\overline{b}_i)$ are all nonzero and linearly independent, then the reduction curve is smooth so that abx's comment applies.
18h
comment Do modular forms show up in the cohomology of moduli spaces of unmarked curves?
@WillSawin: That is a good question. In characteristic $0$ I am certain that the pullback map on $H^q(-,\mathcal{O})$ from the coarse moduli space to the stack is an isomorphism of $k$-vector spaces. However, as you say, the formula from Hirzebruch-Riemann-Roch can be different. I know Toen found a modification for stacks. Probably the issue is that the pushforward map on K-theory works differently (since the invariant part of a finite dimensional representation of a finite group can have strictly smaller rank).
21h
comment Fibrations on blow-ups of $\mathbb{P}^2$
You know, Polizzi and I did give correct answers to one of your earlier questions, yet you did not accept either of our answers.
22h
answered Fibrations on blow-ups of $\mathbb{P}^2$
22h
comment Fibrations on blow-ups of $\mathbb{P}^2$
@S_Z_S: Your picture looks familiar, but didn't you have a different username before?
22h
comment Polynomials with Unique Critical Value
Probably you already noticed this, but the article of Aouira and Pfister does not include my example above. They are generalizing Saito's criterion in a different direction: trying to characterize when an isolated singularity has a non-nilpotent vector field, rather than trying to characterize when the critical locus is a single point.
23h
comment Polynomials with Unique Critical Value
Let me just spell that out, because I just confused myself about how this works! Let $k$ be algebraically closed of characteristic $2$. Let $a$, $b$ be elements in $k$ such that none of $a$, $b$, nor $a+b$ is $0$. Then the correct dehomogenized polynomial on $\mathbb{A}^2_k$ is $f(x,y) = xy(1-ax-by)(1+bx+ay)/(a+b)^2$. This is not weighted homogeneous (unless you allow both $x$ and $y$ to have weights $0$, in which case every polynomial is weighted homogeneous). Yet the critical locus is precisely $0$.
23h
comment Polynomials with Unique Critical Value
That is false in positive characteristic, cf. my answer to the following question: When is the kernel of the etale fundamental group in a fibration abelian?
1d
comment Calculations about the normal bundle of embedding of symmetric products
Every closed immersion of a local complete intersection scheme in a regular scheme is a regular embedding; I am certain this is somewhere in the appendix to Fulton's book. You can prove the formula for the Chern class by induction on $s$.
1d
comment Non-finitely generated, non-projective flat module, over a polynomial ring
Projective $R$-modules are not divisible. Every projective $R$-module is a direct summand of a free $R$-module, and every nonzero element in a free $R$-module obviously fails to be divisible. Thus, a divisible, flat $R$-module, e.g., the fraction field, is not projective.
2d
comment Do modular forms show up in the cohomology of moduli spaces of unmarked curves?
@WillSawin: "Can $\chi_a(\overline{\mathcal{M}}_g)$ possibly be estimated?" By Hirzebruch-Riemann-Roch, this equals the degree of the top graded piece of the Todd class (I am using the stack, so that everything is smooth). Of course the Todd class is a polynomial in the Chern classes. Bini studied the Chern classes, but I do not see how to use Bini's work to estimate the degree of the Todd class.
2d
comment Do modular forms show up in the cohomology of moduli spaces of unmarked curves?
One more consideration: any non-$(p,p)$ cohomology of the Satake compactification of $A_g$ will pullback to zero on $\overline{\mathcal{M}}_{1,n_1}\times \dots \times \overline{\mathcal{M}}_{1,n_k}$. This is because the Torelli map restricted to this locus factors through any of the many projections to the product $\overline{\mathcal{M}}_{1,1}\times \dots \overline{\mathcal{M}}_{1,1}$. As far as $\mathbb{Q}$-cohomology is concerned, this stack is essentially a product of copies of $\mathbb{P}^1$, thus has only $(p,p)$-cohomology.
Jul
26
comment Do modular forms show up in the cohomology of moduli spaces of unmarked curves?
In the "stable range", doesn't Mumford's conjecture / the Madsen-Weiss theorem tell us that the cohomology is all $(p,p)$?