Timeline for Comparing self-equivalences of a triangulated category and automorphisms of its Grothendieck group
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 21, 2017 at 20:05 | vote | accept | მამუკა ჯიბლაძე | ||
| Oct 20, 2017 at 15:53 | comment | added | Will Sawin | @მამუკაჯიბლაძე No, because $Hom(\mathcal O(-1),\mathcal O(1))\neq Hom(\mathcal O(1), \mathcal O(-1))$. | |
| Oct 20, 2017 at 15:45 | answer | added | Jeremy Rickard | timeline score: 5 | |
| Oct 20, 2017 at 12:25 | comment | added | მამუკა ჯიბლაძე | @pbelmans Is it obvious that automorphisms of P$^1$ induce identity on $K_0$? Also, cannot interchanging $O(1)$ and $O(-1)$ be realized by an autoequivalence? | |
| Oct 20, 2017 at 11:55 | comment | added | pbelmans | @მამუკა The auto-equivalence group of $\mathbb{P}^1$ consists of standard auto-equivalences: tensoring with a line bundle, pullback along an automorphism, or shifting. Automorphisms induce the identity on the Grothendieck group, shifting is multiplication by $-1$. So that's why I ignored these. | |
| Oct 20, 2017 at 11:43 | comment | added | მამუკა ჯიბლაძე | @JasonStarr well this would be even "less surjective". | |
| Oct 20, 2017 at 11:43 | comment | added | მამუკა ჯიბლაძე | @pbelmans but why is any self-equivalence given by tensoring by a line bundle? | |
| Oct 20, 2017 at 11:19 | comment | added | Jason Starr | Maybe the OP should ask about self-equivalences that preserve a dualizing object. | |
| Oct 20, 2017 at 10:52 | comment | added | pbelmans | I might be overlooking something, but isn't the derived category of $\mathbb{P}^1$ already an example where it is not surjective? Tensoring by a line bundle will not give you the automorphism which permutes the copies of $\mathbb{Z}$ in $\mathrm{K}_0(\mathbb{P}^1)$. | |
| Oct 20, 2017 at 10:33 | history | asked | მამუკა ჯიბლაძე | CC BY-SA 3.0 |