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Feb 12, 2015 at 16:45 history edited AAK CC BY-SA 3.0
added 2305 characters in body
Feb 12, 2015 at 16:41 comment added AAK @bananastack, in Thomason-Trobaugh this statement is Proposition 5.5.4.
Feb 12, 2015 at 16:38 history edited AAK CC BY-SA 3.0
added 2305 characters in body
Sep 7, 2014 at 15:38 history edited AAK CC BY-SA 3.0
added a reference
Sep 7, 2014 at 15:25 history edited AAK CC BY-SA 3.0
some clarification about the proof
Sep 7, 2014 at 10:57 comment added Urs Schreiber Thanks, Adeel, this is good. Have added these references here: ncatlab.org/nlab/show/algebraic+K-theory#Descent
Sep 7, 2014 at 0:31 comment added S. Carnahan I guess it's an intermediate step to the main result (Theorem 7.4). See Clark Barwick's answer: mathoverflow.net/questions/5580/…
Sep 7, 2014 at 0:12 comment added bananastack thanks @S.Carnahan: is that only apparently weaker than what Adeel claimed (by some manipulations of summands)? Or is it actually weaker?
Sep 7, 2014 at 0:01 comment added S. Carnahan @user125763 Lemma 5.5.1 in Thomason-Trobaugh says that for any perfect complex $F$ on $U$, there is a perfect complex $E$ on $X$ such that $F$ is isomorphic to a direct summand of $j^*E$ in $D^-(\mathcal{O}_U-Mod)$. This is, apparently, a result of Trobaugh's posthumous simulacrum appearing in Thomason's dream.
Sep 6, 2014 at 23:44 comment added bananastack Adeel, could you elaborate on your statement of extensions of perfect complexes? I mean, you seem to suggest that the obstruction to, say, a vector bundle extending from U to all of X lies in just $K_0(U)$? Is that true? I've never heard of this statement, so I'd like to know more.
Sep 6, 2014 at 22:14 history answered AAK CC BY-SA 3.0