I was wondering about something concerning tilting objects... Suppose we are give 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 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 finitly 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 can not be that $D^b(\mathrm{Repr}(G))$ admits a tilting object. Is this argument ok???

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?