global dimension of invariant algebra Suppose we are given a finite dimensional $k$-algebra $A$ with an action of a finite group $G$. Suppose $\mathrm{gldim}(A)$ is finite. What is the relation between $\mathrm{gldim}(A)$ and $\mathrm{gldim}(A^G)$ ? Is $\mathrm{gldim}(A^G)$ finite too ?
Such algebras occur for example when on considers a smooth projective variety over $k$ with an action of a finite group $G$. If $X$ has a $G$-equivariant tilting bundle $\mathcal{T}$ then $\mathrm{End}_G(\mathcal{T})=\mathrm{End}(\mathcal{T})^G$ would be such an $k$-algebra as above...
 A: If $k$ has prime characteristic dividing $|G|$ then there are natural examples where $\operatorname{gldim}(A^G)=\infty$.
For example, let $A=\operatorname{End}_k(kG)$ be the ring of linear endomorphisms of the regular $kG$-module, which is a matrix ring over $k$ and so has global dimension zero, and let $G$ act by conjugation. Then $A^G=\operatorname{End}_{kG}(kG)\cong kG$, which has infinite global dimension.
And here's an example where $\operatorname{char}(k)$ doesn't divide $|G|$.
Let $k$ be any field with characteristic different from two. Let $A$ be the path algebra of the quiver with two vertices and one arrow in each direction between the two vertices (call the arrows $a$ and $b$), modulo the relation $ab=0$. So $A$ is five-dimensional, with basis $\{e,f,a,b,ba\}$, where $e$ and $f$ are orthogonal idempotents, $ea=a=af$, $fb=b=be$.
The global dimension of $A$ is two. 
Let $G$ be a cyclic group of order two, where a generator acts on $A$ by fixing $e$ and $f$, and multiplying $a$ and $b$ by $-1$.
Then $A^G$ has basis $\{e,f,ba\}$, and is isomorphic to $k[x]/(x^2)\times k$ (where $x=ba$), which has infinite global dimension.
