Timeline for Cohomology and deformations of moduli of vector bundles
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 14, 2016 at 5:35 | comment | added | naf | If $g=r=2$ the moduli space is affine. For general $g$ and $r=2$ there is a relatively explicit description of the moduli space (if $C$ is hyperelliptic) due to Desale and Ramanan from which you can probably check whether the vanishing holds. | |
| Jan 13, 2016 at 17:25 | comment | added | Peter Dalakov | I think the (Kodaira-Spencer) map $H^1(T_C)\to H^1(T_X)$ is only known to be injective, see Hitchin's "Flat connections and geometric quantization", section 5. There are isoms $H^1(T_C)\to H^0(Sym^2 T_X)$ though, ibid. | |
| Jan 13, 2016 at 17:23 | comment | added | Jason Starr | I made an arithmetic mistake in computing the codimension above. It should be $(r-1)(g-1)-1$. So if $(r-1)(g-1) - 1 \geq 4$, i.e., $(r-1)(g-1) \geq 5$, we might naively expect the corresponding cohomology with supports in the boundary to vanish in the range necessary to give vanishing of $h^2(X,\mathcal{O}_X)$. | |
| Jan 13, 2016 at 17:05 | comment | added | Jason Starr | @OlivierBenoist: You are certainly correct about the complement of a point in $\mathbb{P}^2$. A naive estimate suggests that the strictly semistable locus in the moduli space has codimension, approximately, $(r-2)(g-1)$ (something seems wrong when $r$ equals $2$ ...). If the compact moduli space were smooth, removing a subset of very high codimension would not affect $h^2(X,\mathcal{O}_X)$. | |
| Jan 13, 2016 at 17:02 | comment | added | Olivier Benoist | @IMeasy Yes: I agree. I am not claiming that your statement that $H^1(X,\mathcal{O}_X)=0$ is false (I do not know that), but that the argument you give does not suffice to prove it (as far as I understand). | |
| Jan 13, 2016 at 16:41 | comment | added | IMeasy | sorry Olivier, I might not have been clear. I am only talking about the case when $X$ is the moduli space. | |
| Jan 13, 2016 at 16:03 | comment | added | Olivier Benoist | It is not clear to me why $H^1(X,\mathcal{O}_X)=0$. I do not think that your argument suffices: if $X$ is $\mathbb{P}^2$ with one point removed, $Pic(X)=\mathbb{Z}$, but $H^1(X,\mathcal{O}_X)$ is nonzero (and in fact infinite-dimensional). | |
| Jan 13, 2016 at 16:02 | comment | added | IMeasy | sorry, I meant: Thank you, Jason. Yes in fact that's what I knew. In the case of stable bundles with trivial determinant the variety is still smooth and unirational, but just quasi-projective... | |
| Jan 13, 2016 at 15:55 | comment | added | IMeasy | Yes in fact that's what I knew | |
| Jan 13, 2016 at 14:49 | comment | added | Jason Starr | Certainly there exist smooth varieties such that $H^2(X,\mathcal{O}_X)$ is nonozero. However, the coarse moduli space of stable vector bundles is smooth and unirational, so it might well be that $h^2(X,\mathcal{O}_X)$ is zero. When the degree of the (fixed) determinant is prime to the rank, then $X$ is projective, smooth and unirational, hence $h^2(X,\mathcal{O}_X)$ is zero if the characteristic is $0$ (and probably there is an argument valid in all characteristics). | |
| Jan 13, 2016 at 14:16 | history | asked | IMeasy | CC BY-SA 3.0 |