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Jul 20, 2021 at 4:03 comment added Li Yutong Thank you! It makes sense to me when $D$ is Cariter.
Jul 20, 2021 at 3:21 comment added Sándor Kovács If $D$ is a Cartier divisor, then $Sing(X)\cap D\subseteq Sing(D)$, so it is at least of codimension 2 in $D$, in which case that equality is true if you restrict to strict resolutions (i.e., only blow up centers inside $Sing(X,D)$).
Jul 20, 2021 at 3:13 comment added Sándor Kovács You are right, this needs more love, but I think it's still possible. Cheers,
Jul 19, 2021 at 16:01 history edited Daniele Tampieri CC BY-SA 4.0
Defined mathematics operators `\Supp` and `\Exc`.
S Jul 19, 2021 at 15:44 history suggested Evans Gambit CC BY-SA 4.0
fixed grammer, suggestion for the title
Jul 19, 2021 at 14:40 review Suggested edits
S Jul 19, 2021 at 15:44
Jul 19, 2021 at 14:11 history asked Li Yutong CC BY-SA 4.0