There is a homomorphism from the group of (isomorphism classes of) self-equivalences of a triangulated category to the automorphism group of its Grothendieck group. Is this homomorphism surjective? If not, is there a calculable obstruction? Are there some natural classes of triangulated categories for which it is surjective? In any case, what would be a good reference for this question?
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asked Oct 20, 2017 at 10:33
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5$\begingroup$ 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 $\mathrm{K}_0(\mathbb{P}^1)$. $\endgroup$– pbelmansCommented Oct 20, 2017 at 10:52
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$\begingroup$ Maybe the OP should ask about self-equivalences that preserve a dualizing object. $\endgroup$– Jason StarrCommented Oct 20, 2017 at 11:19
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$\begingroup$ @pbelmans but why is any self-equivalence given by tensoring by a line bundle? $\endgroup$– მამუკა ჯიბლაძეCommented Oct 20, 2017 at 11:43
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$\begingroup$ @JasonStarr well this would be even "less surjective". $\endgroup$– მამუკა ჯიბლაძეCommented Oct 20, 2017 at 11:43
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1$\begingroup$ @მამუკაჯიბლაძე No, because $Hom(\mathcal O(-1),\mathcal O(1))\neq Hom(\mathcal O(1), \mathcal O(-1))$. $\endgroup$– Will SawinCommented Oct 20, 2017 at 15:53
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1 Answer
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One obstruction is that if all Hom-sets are finite dimensional vector spaces, and for all objects $X$ and $Y$, $\text{Hom}(X,Y[i])=0$ for all but finitely many $i$, then any self-equivalence must preserve the bilinear form $$\langle [X],[Y]\rangle =\sum_i(-1)^i\dim\text{Hom}(X,Y[i])$$ on the Grothendieck group.
answered Oct 20, 2017 at 15:45