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I was wondering about something concerning tilting objects... Suppose we are given a split algebraic torus $G=\mathbb{G}^n_m$, that is a linearly reductive group with no semisimple part, and let $\mathrm{Repr}(G)$ be the category of finite dimensional representations of $G$. Since $G$ is linearly reductive this category is semisimple. Now consider the derived category $D^b(\mathrm{Repr}(G))$. Does this category have a tilting object?

I mean if there is an object $\mathcal{T}$ that generates $D^b(\mathrm{Repr}(G))$ then just take the simple representations occuring in $\mathcal{T}$, that are finitely many and sum them up to get a tilting object. But then these finitely many simple representations would generate $D^b(\mathrm{Repr}(G))$ too... but since $\mathbb{G}^n_m$ has infinitely many simple representations there is one not occuring in that sum. Hence it cannot be that $D^b(\mathrm{Repr}(G))$ admits a tilting object. Is this argument ok?

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The argument is OK. – Sasha Jan 28 '14 at 11:50
It might help to remind people how "tilting object" is defined in this setting. Also, I don't see the point of "linearly" here, since you look only at tori. I guess a similar argument could be mafde for any nontrivial reductive group, at least in characteristic 0. Having infinitely many isomorphism classes of simple modules in a semisimple category seems to be the main ingredient ruling out a tilting object in the derived category. – Jim Humphreys Jan 28 '14 at 14:15

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