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Jul 21, 2023 at 0:08 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Mar 23, 2023 at 0:07 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Feb 20, 2023 at 23:26 answer added Tomo timeline score: 1
Dec 5, 2022 at 0:41 comment added Tomo Thank you @D.-C.Cisinski, that completely clarifies the confusion of my second paragraph.
Dec 4, 2022 at 11:00 comment added D.-C. Cisinski In fact, one can also see this in a more elementary way. See Section B.2 in this paper: cambridge.org/core/journals/compositio-mathematica/article/…
Dec 4, 2022 at 10:39 comment added D.-C. Cisinski Hence, at the end of the day, $D_{cons}(X,\mathbf{Q}_\ell)$ has the same objects as $D_{cons}(X,\mathbf{Z}_\ell)$ and the mapping spaces of $D_{cons}(X,\mathbf{Q}_\ell)$ are those of $D_{cons}(X,\mathbf{Z}_\ell)$ tensored by $\mathbf{Q}$.
Dec 4, 2022 at 10:34 comment added D.-C. Cisinski For references: Theorem 2.35 in Benjamin Antieau, David Gepner, Jeremiah Heller arxiv.org/abs/1610.07207 for the vanishing of $K_{-1}$. For the interpretation of $K_{-1}=0$ in terms of Verdier quotients, this is in the paper of Marco Schlichting Negative K-theory of derived categories (end of Section 1).
Dec 4, 2022 at 10:26 comment added D.-C. Cisinski In fact, since $D_{cons}(X,\mathbf{Z}_\ell)$ has a bounded $t$-structure, we have a vanishing $K$-group in degree $-1$: $K_{-1}(D_{cons}(X,\mathbf{Z}_\ell))=0$. That means precisely that all the Verdier quotients of $D_{cons}(X,\mathbf{Z}_\ell)$ already are idempotent complete.
Dec 4, 2022 at 7:17 history edited Tomo CC BY-SA 4.0
Second paragraph had some nonsense
Dec 4, 2022 at 2:32 history asked Tomo CC BY-SA 4.0