bio | website | math.purdue.edu/~dvb |
---|---|---|
location | ||
age | ||
visits | member for | 4 years, 9 months |
seen | 6 hours ago | |
stats | profile views | 15,918 |
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 $pd(M) = 2-depth(M) =0$. |
Sep 30 |
awarded | Explainer |
Sep 25 |
comment |
Holomorphic vector bundles over $\mathbb{CP}^1$ and elliptic curves
It wasn't clear if you already know the answer: they are sums of line bundles. This is contained what I believe is Grothendieck's first paper in algebraic geometry. If you don't read French you can find a proof in a book by Okonek, Spindler & Schneider, on vector bundles on projective spaces |
Sep 24 |
awarded | Enlightened |
Sep 24 |
awarded | Nice Answer |
Sep 12 |
comment |
Simply-connected 4-manifolds can be blown up and down to complex projective planes. How about non-simply-connected ones?
But the fundamental group would have to be free for that to work, wouldn't it? |
Sep 12 |
comment |
Simply-connected 4-manifolds can be blown up and down to complex projective planes. How about non-simply-connected ones?
Given that blow up/downs won't change the fundamental group, I'm not sure what kind of answer you're hoping for. |