Timeline for Is there any explicit result on the triangulated category of singularities of a curve?
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Apr 28, 2015 at 15:06 | comment | added | Zhaoting Wei | @AdeelKhan After some thinking I should say that I only know it in the smooth case also. Sorry for the misleading. Nevertheless if we can compute the Grothendieck group of $D^b_{coh}(X)$ for singular curves $X$, it will be solved. I guess $K_0(D^b_{coh}(X))$ is not far from $K_0(D^b_{coh}(\widetilde{X}))$ where $\widetilde{X}\to X$ is the resolution of singularities but I don't know how to prove it. | |
| Apr 28, 2015 at 14:52 | comment | added | AAK | Do you know a reference for that? I only knew that fact in the smooth case. | |
| Apr 28, 2015 at 13:55 | comment | added | Zhaoting Wei | @AdeelKhan Sure! Nevertheless for $X$ with arithmetic genus $\geq 1$, $D^b_{coh}(X)$ cannot have semiorthogonal decomposition but I wonder if $D_{sg}(X)$ still has. | |
| Apr 28, 2015 at 13:40 | comment | added | AAK | Just a comment: a semi-orthogonal decomposition of D^b_coh induces a semi-orthogonal decomposition of D_sg, according to Corollary 1.12 here. | |
| Apr 28, 2015 at 12:59 | history | asked | Zhaoting Wei | CC BY-SA 3.0 |