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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...

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  • $\begingroup$ I am guessing from context that when you say that $A$ is a "finite dimensional $k$-algebra", what you mean is that $A$ is finite dimensional as a $k$-vector space, not that $A$ has finite Krull dimension. $\endgroup$
    – Jason Starr
    Commented Jan 19, 2014 at 13:32

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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.

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  • $\begingroup$ Shucks, you beat me to it. $\endgroup$
    – Jason Starr
    Commented Jan 19, 2014 at 15:07
  • $\begingroup$ @JasonStarr : And I gave you an hour's head start, too! :-) $\endgroup$
    – Jeremy Rickard
    Commented Jan 19, 2014 at 15:11
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    $\begingroup$ @abx: I've edited my answer to add an example where $\operatorname{char}(k)$ is prime to $|G|$. $\endgroup$
    – Jeremy Rickard
    Commented Jan 19, 2014 at 17:37
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    $\begingroup$ @JasonStarr: Yes, that's true. A Maschke-like proof shows that for any (right) $A^G$-module $M$, the natural $A^G$-module map $M\to M\otimes_{A^G}A$ is a split embedding. If $A$ is semisimple, then $M\otimes_{A^G}A$ (and hence $M$) is projective. $\endgroup$
    – Jeremy Rickard
    Commented Jan 20, 2014 at 10:16
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    $\begingroup$ And there's a paper of George Bergman (Groups acting on hereditary rings. Proc. London Math. Soc. (3) 23 (1971), 70-82) that proves that if $A$ is hereditary and $|G|$ is invertible in $A$, then $A^G$ is also hereditary. $\endgroup$
    – Jeremy Rickard
    Commented Jan 20, 2014 at 10:31

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