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Dec
17
revised cycle class as Chern class
added 238 characters in body
Dec
17
answered cycle class as Chern class
Dec
15
comment Topological/numerical constraints for the existence of more than one pencil
Yes, there is a topological characterization. One can use the so called the 1st characteristic variety (or jump locus) which is the set of rank one local systems with nontrivial $H^1$. This forms a variety and the positive irreducible components through $1$ are in bijection with pencils.
Dec
10
awarded  Enlightened
Dec
10
awarded  Nice Answer
Dec
4
answered Good lecture notes/books on Jacobian of hyperelliptic curve
Nov
19
revised Take contraction wrt a vector field twice and define kernel mod image. Does that give anything interesting?
added 86 characters in body
Nov
15
awarded  Necromancer
Nov
14
comment On a proposition in Hartshorne's paper “Ample vector bundles on curves”
OK, I misread it.
Nov
14
comment On a proposition in Hartshorne's paper “Ample vector bundles on curves”
What if you took $Y=A$ to be an elliptic curve, with $\phi$ the Frobenius? Isn't this a counterexample to (X)?
Nov
11
comment Optimal definition of “paving by affine spaces”?
To muddy the water further, Fulton, Intersection Theory, 1.9.1 uses an even broader definition: $X_i-X_{i-1}$ is a union of affine spaces of possibly varying dimensions. He refers to this as 'a scheme with a "cellular decomposition"'. He goes on to establish some desirable properties, such as the Chow group maps onto homology with a basis given by closures of the affine spaces. I suspect that this may be sufficient for most applications.
Nov
6
comment When did “Betti cohomology” come to be used the way it is today? (and how is it used)
I've only seen this usage among algebraic geometers. I assumed it was someone like Grothendieck who started this. I tend to understand it as with $\mathbb{Z}$ coefficients, unless that the author says otherwise. Finally, unless the field of definition lies in $\mathbb{R}$, there is no natural conjugation on $X(\mathbb{C})$. Of course, if you use cohomology with $\mathbb{C}$ as coefficients, then there is conjugation on that.
Nov
6
comment why are motives more serious than “naive” motives?
@birk my comment was partly a joke but not completely. If $R$ is an Artinian ring, then from the class of a module in $K_0(R) $ you can recover its length but you've lost everything else. In the same way passing to $K_0(Var)$ kills a lot; I don't see how you would recover the higher Chow group from its class.
Nov
5
comment why are motives more serious than “naive” motives?
For the same reason that Betti numbers are more "serious" than the Euler characteristic.
Nov
4
answered Which mapping class group representations come from algebraic geometry?
Oct
18
answered Take contraction wrt a vector field twice and define kernel mod image. Does that give anything interesting?
Oct
12
comment What is an infinite prime in algebraic topology?
If you take a $\mathbb{Q}$-algebra, for example a class in the Brauer group or a cohomology ring of a space, and tensor by $\mathbb{R}$ you actually loose information (e.g $Br(\mathbb{R})$ is much simpler than $Br(\mathbb{Q})$). Loss of information is not always a bad thing because the resulting objects may be easier to classify… but I guess you are after something else.
Oct
12
comment What is an infinite prime in algebraic topology?
I'm not a topologist, so this may be too naive, but we know from work of Quillen and Sullivan that rational homotopy theory is equivalent to the homotopy theory of a DGL or DGA over $\mathbb{Q}$. We could simply tensor this by $\mathbb{R}$ couldn't we? I know that the "real" in the paper "Real homotopy theory of theory of Kahler manifolds" by Deligne, Griffiths, Morgan, Sullivan refers to this process.
Oct
10
answered When is the Hodge diamond concentrated in $H^{n,n}$'s?
Oct
2
comment “Spreading out” locally free sheaves
The argument goes like this: If $M$ is a reflexive module over a $2$-dim regular local ring, then $depth(M)=2$. Now use Auslander-Buchsbaum-Serre to conclude that the projective dim $pd(M) = 2-depth(M) =0$.