spaces. Recall that $T\in{\operatorname{\mathsf{mod}}}\Lambda$ is a *tilting $\Lambda$-module* if it satisfies the following two conditions:

-   Rigidity condition: ${\operatorname{Ext}\nolimits}^i_\Lambda(T,T)=0$ for all $i>0$,

-   Generation condition: ${\mathsf{D}}^{{\operatorname{b}\nolimits}}({\operatorname{\mathsf{mod}}}A)=\operatorname{\mathsf{thick}}T$, where $\operatorname{\mathsf{thick}}T$ is the smallest triangulated subcategory of ${\mathsf{D}}^{{\operatorname{b}\nolimits}}({\operatorname{\mathsf{mod}}}A)$ which is closed under direct summands and contains $T$.

The existence of such a tilting bundle gives rise to a derived equivalence between $\Lambda$ and ${\operatorname{End}\nolimits}_\Lambda(T)$. We will give a simple proof of the following result in the language of orders:

Let $\Lambda$ be a GL order on ${\mathbb{P}}^d$.

-   There exists a tilting bundle $T$ in ${\operatorname{\mathsf{mod}}}\Lambda$.

-   We have a triangle equivalence ${\mathsf{D}}^{{\operatorname{b}\nolimits}}({\operatorname{\mathsf{mod}}}\Lambda)\simeq{\mathsf{D}}^{{\operatorname{b}\nolimits}}({\operatorname{\mathsf{mod}}}{\operatorname{End}\nolimits}_\Lambda(T))$.

-   $\Lambda$ has global dimension $d$.

In fact, we will explicitly construct a tilting bundle $T$. Crucially, our proof is geometric for it uses the theorem of Beilinson [@Be] regarding the existence of a tilting bundle on ${\mathbb{P}}^d$.

A similar construction of tilting bundles in a more general setup will be discussed in a joint work [@ILO] with Oppermann. [**Acknowledgements.**]{} The authors thank Kenneth Chan for valuable discussion leading to this work. They thank Colin Ingalls, Gustavo Jasso and Steffen Oppermann for stimulating discussions.

Tilting bundles on orders {#Tilting}
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Let ${\mathbb{P}}^d$ be a projective $d$-space and fix $n\geq 1$ 