Skip to main content
MathOverflow

Return to Answer

Corrected slight error: $f_iA$ should be $f_iAf_i$
Source Link
Jeremy Rickard
Jeremy Rickard
  • 35.2k
  • 2
  • 110
  • 151

Yes, $A^G$ must have finite global dimension, as it has the same kind of triangular form as $A$.

The Jacobson radical of $A$ is $$J(A)=\begin{pmatrix} 0 & M_{1,2} & \dots & M_{1,r} \\ 0 & 0 & \dots & M_{2,r} \\ \vdots & 0 & \ddots & \vdots \\ 0 & \dots & 0 & 0 \end{pmatrix},$$ and $$A/J(A)=A_1\times\dots\times A_r,$$ which can be identified with a subalgebra of $A$ in the obvious way.

The primitive central idempotents $e_1,\dots,e_s$ of $A/J(A)$ can be ordered so that $i<j\Rightarrow e_jJ(A)e_i=0$ (list the central idempotents of $A_1$ first, then those of $A_2$, etc.).

Since $G$ acts on $\{e_1,\dots,e_s\}$, preserving the property $e_jJ(A)e_i=0$, if $f_1,\dots,f_t$ are the $G$-orbit sums of the idempotents, then they can also be ordered so that $i<j\Rightarrow f_jJ(A)f_i=0$. Then taking $B_i=f_iA$$B_i=f_iAf_i$ and $N_{i,j}=f_iJ(A)f_j$, the triangular form of $A$ can be taken to be $$A=\begin{pmatrix} B_1 & N_{1,2} & \dots & N_{1,t} \\ 0 & B_2 & \dots & N_{2,t} \\ \vdots & 0 & \ddots & \vdots \\ 0 & \dots & 0 & B_t \end{pmatrix},$$ which is now stable under the action of $G$, so $$A^G=\begin{pmatrix} B_1^G & N_{1,2}^G & \dots & N_{1,t}^G \\ 0 & B_2^G & \dots & N_{2,t}^G \\ \vdots & 0 & \ddots & \vdots \\ 0 & \dots & 0 & B_t^G \end{pmatrix}.$$

Since $\operatorname{char}(k)$ does not divide $|G|$, and $B_i$ is semisimple for each $i$, $B_i^G$ is also semisimple, and so $A^G$ has finite global dimension, as sketched in the last paragraph of the question.

Yes, $A^G$ must have finite global dimension, as it has the same kind of triangular form as $A$.

The Jacobson radical of $A$ is $$J(A)=\begin{pmatrix} 0 & M_{1,2} & \dots & M_{1,r} \\ 0 & 0 & \dots & M_{2,r} \\ \vdots & 0 & \ddots & \vdots \\ 0 & \dots & 0 & 0 \end{pmatrix},$$ and $$A/J(A)=A_1\times\dots\times A_r,$$ which can be identified with a subalgebra of $A$ in the obvious way.

The primitive central idempotents $e_1,\dots,e_s$ of $A/J(A)$ can be ordered so that $i<j\Rightarrow e_jJ(A)e_i=0$ (list the central idempotents of $A_1$ first, then those of $A_2$, etc.).

Since $G$ acts on $\{e_1,\dots,e_s\}$, preserving the property $e_jJ(A)e_i=0$, if $f_1,\dots,f_t$ are the $G$-orbit sums of the idempotents, then they can also be ordered so that $i<j\Rightarrow f_jJ(A)f_i=0$. Then taking $B_i=f_iA$ and $N_{i,j}=f_iJ(A)f_j$, the triangular form of $A$ can be taken to be $$A=\begin{pmatrix} B_1 & N_{1,2} & \dots & N_{1,t} \\ 0 & B_2 & \dots & N_{2,t} \\ \vdots & 0 & \ddots & \vdots \\ 0 & \dots & 0 & B_t \end{pmatrix},$$ which is now stable under the action of $G$, so $$A^G=\begin{pmatrix} B_1^G & N_{1,2}^G & \dots & N_{1,t}^G \\ 0 & B_2^G & \dots & N_{2,t}^G \\ \vdots & 0 & \ddots & \vdots \\ 0 & \dots & 0 & B_t^G \end{pmatrix}.$$

Since $\operatorname{char}(k)$ does not divide $|G|$, and $B_i$ is semisimple for each $i$, $B_i^G$ is also semisimple, and so $A^G$ has finite global dimension, as sketched in the last paragraph of the question.

Yes, $A^G$ must have finite global dimension, as it has the same kind of triangular form as $A$.

The Jacobson radical of $A$ is $$J(A)=\begin{pmatrix} 0 & M_{1,2} & \dots & M_{1,r} \\ 0 & 0 & \dots & M_{2,r} \\ \vdots & 0 & \ddots & \vdots \\ 0 & \dots & 0 & 0 \end{pmatrix},$$ and $$A/J(A)=A_1\times\dots\times A_r,$$ which can be identified with a subalgebra of $A$ in the obvious way.

The primitive central idempotents $e_1,\dots,e_s$ of $A/J(A)$ can be ordered so that $i<j\Rightarrow e_jJ(A)e_i=0$ (list the central idempotents of $A_1$ first, then those of $A_2$, etc.).

Since $G$ acts on $\{e_1,\dots,e_s\}$, preserving the property $e_jJ(A)e_i=0$, if $f_1,\dots,f_t$ are the $G$-orbit sums of the idempotents, then they can also be ordered so that $i<j\Rightarrow f_jJ(A)f_i=0$. Then taking $B_i=f_iAf_i$ and $N_{i,j}=f_iJ(A)f_j$, the triangular form of $A$ can be taken to be $$A=\begin{pmatrix} B_1 & N_{1,2} & \dots & N_{1,t} \\ 0 & B_2 & \dots & N_{2,t} \\ \vdots & 0 & \ddots & \vdots \\ 0 & \dots & 0 & B_t \end{pmatrix},$$ which is now stable under the action of $G$, so $$A^G=\begin{pmatrix} B_1^G & N_{1,2}^G & \dots & N_{1,t}^G \\ 0 & B_2^G & \dots & N_{2,t}^G \\ \vdots & 0 & \ddots & \vdots \\ 0 & \dots & 0 & B_t^G \end{pmatrix}.$$

Since $\operatorname{char}(k)$ does not divide $|G|$, and $B_i$ is semisimple for each $i$, $B_i^G$ is also semisimple, and so $A^G$ has finite global dimension, as sketched in the last paragraph of the question.

Source Link
Jeremy Rickard
Jeremy Rickard
  • 35.2k
  • 2
  • 110
  • 151

Yes, $A^G$ must have finite global dimension, as it has the same kind of triangular form as $A$.

The Jacobson radical of $A$ is $$J(A)=\begin{pmatrix} 0 & M_{1,2} & \dots & M_{1,r} \\ 0 & 0 & \dots & M_{2,r} \\ \vdots & 0 & \ddots & \vdots \\ 0 & \dots & 0 & 0 \end{pmatrix},$$ and $$A/J(A)=A_1\times\dots\times A_r,$$ which can be identified with a subalgebra of $A$ in the obvious way.

The primitive central idempotents $e_1,\dots,e_s$ of $A/J(A)$ can be ordered so that $i<j\Rightarrow e_jJ(A)e_i=0$ (list the central idempotents of $A_1$ first, then those of $A_2$, etc.).

Since $G$ acts on $\{e_1,\dots,e_s\}$, preserving the property $e_jJ(A)e_i=0$, if $f_1,\dots,f_t$ are the $G$-orbit sums of the idempotents, then they can also be ordered so that $i<j\Rightarrow f_jJ(A)f_i=0$. Then taking $B_i=f_iA$ and $N_{i,j}=f_iJ(A)f_j$, the triangular form of $A$ can be taken to be $$A=\begin{pmatrix} B_1 & N_{1,2} & \dots & N_{1,t} \\ 0 & B_2 & \dots & N_{2,t} \\ \vdots & 0 & \ddots & \vdots \\ 0 & \dots & 0 & B_t \end{pmatrix},$$ which is now stable under the action of $G$, so $$A^G=\begin{pmatrix} B_1^G & N_{1,2}^G & \dots & N_{1,t}^G \\ 0 & B_2^G & \dots & N_{2,t}^G \\ \vdots & 0 & \ddots & \vdots \\ 0 & \dots & 0 & B_t^G \end{pmatrix}.$$

Since $\operatorname{char}(k)$ does not divide $|G|$, and $B_i$ is semisimple for each $i$, $B_i^G$ is also semisimple, and so $A^G$ has finite global dimension, as sketched in the last paragraph of the question.