Determining maximal Cohen-Macaulay modules over an invariant ring
$begingroup$
Suppose that $G$ is a finite small (i.e. reflection-free) subgroup of $text{GL}(n,mathbb{C})$ acting on $S := mathbb{C}[x_1, dots, x_n]$. Set $R := S^G$. By 5.20 Corollary of this, the maximal Cohen Macaulay modules of $R$ are in bijection with the irreps of $G$ and are given by $M_j := (S otimes_mathbb{C} V_j)^G$, where $V_j$ is the $j$th irrep. If $V_j$ is one dimensional with corresponding character $chi_j$, then $M_j$ is isomorphic to
begin{align*}
{ f in S mid g cdot f = chi_j(g)^{-1} f text{ for all } g in G }.
end{align*}
Is there a similar description of the maximal Cohen-Macaulay module $M_j$ when $V_j$ has dimension 2 or greater? Preferably one which only relies on knowing the character, and not requiring one to construct the corresponding representation explicitly.
abstract-algebra ring-theory representation-theory invariant-theory cohen-macaulay
asked Jan 15 at 15:58
lokodizlokodiz
1,4341819
$endgroup$
add a comment |
$begingroup$
Suppose that $G$ is a finite small (i.e. reflection-free) subgroup of $text{GL}(n,mathbb{C})$ acting on $S := mathbb{C}[x_1, dots, x_n]$. Set $R := S^G$. By 5.20 Corollary of this, the maximal Cohen Macaulay modules of $R$ are in bijection with the irreps of $G$ and are given by $M_j := (S otimes_mathbb{C} V_j)^G$, where $V_j$ is the $j$th irrep. If $V_j$ is one dimensional with corresponding character $chi_j$, then $M_j$ is isomorphic to
begin{align*}
{ f in S mid g cdot f = chi_j(g)^{-1} f text{ for all } g in G }.
end{align*}
Is there a similar description of the maximal Cohen-Macaulay module $M_j$ when $V_j$ has dimension 2 or greater? Preferably one which only relies on knowing the character, and not requiring one to construct the corresponding representation explicitly.
abstract-algebra ring-theory representation-theory invariant-theory cohen-macaulay
asked Jan 15 at 15:58
lokodizlokodiz
1,4341819
$endgroup$
$begingroup$
Just to fix notation: small here is a subgroup of $mathrm{GL}(V)$ is small if it does not contain any element with codimension one fixed space.
$endgroup$
– Stephen
Jan 16 at 18:36
$begingroup$
Why do you expect something similar to the linear case for $(S otimes V_j)^G$? You could, of course, use the idempotent corresponding to the character of $V_j$ to write this without mentioning the representation at all: you are looking at occurences of $V_j^*$ in $S$.
$endgroup$
– Stephen
Jan 16 at 18:38
1
$begingroup$
I suppose my confusion stems from the fact that I'm not sure how I'm supposed to view $(S otimes V_j)^G$ as an $R$-submodule of $S$ (with reference to 5.20 Corollary again). Is there some obvious way to view $(S otimes V_j)^G$ inside of $S$ that I'm missing? Possibly using the idempotent, as you mentioned.
$endgroup$
– lokodiz
Jan 16 at 19:13
add a comment |
$begingroup$
Suppose that $G$ is a finite small (i.e. reflection-free) subgroup of $text{GL}(n,mathbb{C})$ acting on $S := mathbb{C}[x_1, dots, x_n]$. Set $R := S^G$. By 5.20 Corollary of this, the maximal Cohen Macaulay modules of $R$ are in bijection with the irreps of $G$ and are given by $M_j := (S otimes_mathbb{C} V_j)^G$, where $V_j$ is the $j$th irrep. If $V_j$ is one dimensional with corresponding character $chi_j$, then $M_j$ is isomorphic to
begin{align*}
{ f in S mid g cdot f = chi_j(g)^{-1} f text{ for all } g in G }.
end{align*}
Is there a similar description of the maximal Cohen-Macaulay module $M_j$ when $V_j$ has dimension 2 or greater? Preferably one which only relies on knowing the character, and not requiring one to construct the corresponding representation explicitly.
abstract-algebra ring-theory representation-theory invariant-theory cohen-macaulay
asked Jan 15 at 15:58
lokodizlokodiz
1,4341819
$endgroup$
Suppose that $G$ is a finite small (i.e. reflection-free) subgroup of $text{GL}(n,mathbb{C})$ acting on $S := mathbb{C}[x_1, dots, x_n]$. Set $R := S^G$. By 5.20 Corollary of this, the maximal Cohen Macaulay modules of $R$ are in bijection with the irreps of $G$ and are given by $M_j := (S otimes_mathbb{C} V_j)^G$, where $V_j$ is the $j$th irrep. If $V_j$ is one dimensional with corresponding character $chi_j$, then $M_j$ is isomorphic to
begin{align*}
{ f in S mid g cdot f = chi_j(g)^{-1} f text{ for all } g in G }.
end{align*}
Is there a similar description of the maximal Cohen-Macaulay module $M_j$ when $V_j$ has dimension 2 or greater? Preferably one which only relies on knowing the character, and not requiring one to construct the corresponding representation explicitly.
abstract-algebra ring-theory representation-theory invariant-theory cohen-macaulay
abstract-algebra ring-theory representation-theory invariant-theory cohen-macaulay
asked Jan 15 at 15:58
lokodizlokodiz
1,4341819
asked Jan 15 at 15:58
lokodizlokodiz
1,4341819
edited Jan 16 at 19:10
lokodiz
asked Jan 15 at 15:58
lokodizlokodiz
1,4341819
asked Jan 15 at 15:58
lokodizlokodiz
1,4341819
asked Jan 15 at 15:58
lokodizlokodiz
1,4341819
1,4341819
$begingroup$
Just to fix notation: small here is a subgroup of $mathrm{GL}(V)$ is small if it does not contain any element with codimension one fixed space.
$endgroup$
– Stephen
Jan 16 at 18:36
$begingroup$
Why do you expect something similar to the linear case for $(S otimes V_j)^G$? You could, of course, use the idempotent corresponding to the character of $V_j$ to write this without mentioning the representation at all: you are looking at occurences of $V_j^*$ in $S$.
$endgroup$
– Stephen
Jan 16 at 18:38
1
$begingroup$
I suppose my confusion stems from the fact that I'm not sure how I'm supposed to view $(S otimes V_j)^G$ as an $R$-submodule of $S$ (with reference to 5.20 Corollary again). Is there some obvious way to view $(S otimes V_j)^G$ inside of $S$ that I'm missing? Possibly using the idempotent, as you mentioned.
$endgroup$
– lokodiz
Jan 16 at 19:13
add a comment |
$begingroup$
Just to fix notation: small here is a subgroup of $mathrm{GL}(V)$ is small if it does not contain any element with codimension one fixed space.
$endgroup$
– Stephen
Jan 16 at 18:36
$begingroup$
Why do you expect something similar to the linear case for $(S otimes V_j)^G$? You could, of course, use the idempotent corresponding to the character of $V_j$ to write this without mentioning the representation at all: you are looking at occurences of $V_j^*$ in $S$.
$endgroup$
– Stephen
Jan 16 at 18:38
1
$begingroup$
I suppose my confusion stems from the fact that I'm not sure how I'm supposed to view $(S otimes V_j)^G$ as an $R$-submodule of $S$ (with reference to 5.20 Corollary again). Is there some obvious way to view $(S otimes V_j)^G$ inside of $S$ that I'm missing? Possibly using the idempotent, as you mentioned.
$endgroup$
– lokodiz
Jan 16 at 19:13
$begingroup$
Just to fix notation: small here is a subgroup of $mathrm{GL}(V)$ is small if it does not contain any element with codimension one fixed space.
$endgroup$
– Stephen
Jan 16 at 18:36
$begingroup$
Just to fix notation: small here is a subgroup of $mathrm{GL}(V)$ is small if it does not contain any element with codimension one fixed space.
$endgroup$
– Stephen
Jan 16 at 18:36
$begingroup$
Why do you expect something similar to the linear case for $(S otimes V_j)^G$? You could, of course, use the idempotent corresponding to the character of $V_j$ to write this without mentioning the representation at all: you are looking at occurences of $V_j^*$ in $S$.
$endgroup$
– Stephen
Jan 16 at 18:38
$begingroup$
Why do you expect something similar to the linear case for $(S otimes V_j)^G$? You could, of course, use the idempotent corresponding to the character of $V_j$ to write this without mentioning the representation at all: you are looking at occurences of $V_j^*$ in $S$.
$endgroup$
– Stephen
Jan 16 at 18:38
1
1
$begingroup$
I suppose my confusion stems from the fact that I'm not sure how I'm supposed to view $(S otimes V_j)^G$ as an $R$-submodule of $S$ (with reference to 5.20 Corollary again). Is there some obvious way to view $(S otimes V_j)^G$ inside of $S$ that I'm missing? Possibly using the idempotent, as you mentioned.
$endgroup$
– lokodiz
Jan 16 at 19:13
$begingroup$
I suppose my confusion stems from the fact that I'm not sure how I'm supposed to view $(S otimes V_j)^G$ as an $R$-submodule of $S$ (with reference to 5.20 Corollary again). Is there some obvious way to view $(S otimes V_j)^G$ inside of $S$ that I'm missing? Possibly using the idempotent, as you mentioned.
$endgroup$
– lokodiz
Jan 16 at 19:13
add a comment |
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$begingroup$
Just to fix notation: small here is a subgroup of $mathrm{GL}(V)$ is small if it does not contain any element with codimension one fixed space.
$endgroup$
– Stephen
Jan 16 at 18:36
$begingroup$
Why do you expect something similar to the linear case for $(S otimes V_j)^G$? You could, of course, use the idempotent corresponding to the character of $V_j$ to write this without mentioning the representation at all: you are looking at occurences of $V_j^*$ in $S$.
$endgroup$
– Stephen
Jan 16 at 18:38
1
$begingroup$
I suppose my confusion stems from the fact that I'm not sure how I'm supposed to view $(S otimes V_j)^G$ as an $R$-submodule of $S$ (with reference to 5.20 Corollary again). Is there some obvious way to view $(S otimes V_j)^G$ inside of $S$ that I'm missing? Possibly using the idempotent, as you mentioned.
$endgroup$
– lokodiz
Jan 16 at 19:13