Skip to main content
MathOverflow

Return to Answer

deleted 10 characters in body
Source Link
David Hill
David Hill
  • 1.5k
  • 8
  • 12

Then the answer is yes, finite dimensional irreducible $G_0$-modules are projective in the category of finite dimensional modules.

Note that $G_0=G_0'\oplus Z$, where $G_0'$ is semisimple and $Z$ is the center. Let $V$ be a finite dimensional $G_0$-module. Then, by Weyl's theorem, $V$ is completely irreducible as a $G_0'$-module. The elements of $Z$ act as scalars on each irreducible summand of $V$, by Schur's lemma. This means that the decomposition of $V$ into irreducible $G_0'$-modules is a decomposition as $G_0$-modules. It now follows that the category of finite dimensional $G_0$-modules is semi-simple, and irreducible=projective.

EDIT: This answer is based on the assumption that $G$ is a simple Lie superalgebra, so $G_0$ is reductive. At one time, there was a comment/answer by the OP indicating that this was the case of interest.

Then the answer is yes, finite dimensional irreducible $G_0$-modules are projective in the category of finite dimensional modules.

Note that $G_0=G_0'\oplus Z$, where $G_0'$ is semisimple and $Z$ is the center. Let $V$ be a finite dimensional $G_0$-module. Then, by Weyl's theorem, $V$ is completely irreducible as a $G_0'$-module. The elements of $Z$ act as scalars on each irreducible summand of $V$, by Schur's lemma. This means that the decomposition of $V$ into irreducible $G_0'$-modules is a decomposition as $G_0$-modules. It now follows that the category of finite dimensional $G_0$-modules is semi-simple, and irreducible=projective.

EDIT: This answer is based on the assumption that $G$ is a simple Lie superalgebra, so $G_0$ is reductive. At one time, there was a comment/answer by the OP indicating that this was the case of interest.

Then the answer is yes, finite dimensional irreducible $G_0$-modules are projective in the category of finite dimensional modules.

Note that $G_0=G_0'\oplus Z$, where $G_0'$ is semisimple and $Z$ is the center. Let $V$ be a finite dimensional $G_0$-module. Then, by Weyl's theorem, $V$ is completely irreducible as a $G_0'$-module. The elements of $Z$ act as scalars on each irreducible summand of $V$, by Schur's lemma. This means that the decomposition of $V$ into irreducible $G_0'$-modules is a decomposition as $G_0$-modules. It now follows that the category of finite dimensional $G_0$-modules is semi-simple, and irreducible=projective.

EDIT: This answer is based on the assumption that $G$ is a simple Lie superalgebra, so $G_0$ is reductive. At one time, there was a comment/answer indicating that this was the case of interest.

added 209 characters in body
Source Link
David Hill
David Hill
  • 1.5k
  • 8
  • 12

Then the answer is yes, finite dimensional irreducible $G_0$-modules are projective in the category of finite dimensional modules.

Note that $G_0=G_0'\oplus Z$, where $G_0'$ is semisimple and $Z$ is the center. Let $V$ be a finite dimensional $G_0$-module. Then, by Weyl's theorem, $V$ is completely irreducible as a $G_0'$-module. The elements of $Z$ act as scalars on each irreducible summand of $V$, by Schur's lemma. This means that the decomposition of $V$ into irreducible $G_0'$-modules is a decomposition as $G_0$-modules. It now follows that the category of finite dimensional $G_0$-modules is semi-simple, and irreducible=projective.

EDIT: This answer is based on the assumption that $G$ is a simple Lie superalgebra, so $G_0$ is reductive. At one time, there was a comment/answer by the OP indicating that this was the case of interest.

Then the answer is yes, finite dimensional irreducible $G_0$-modules are projective in the category of finite dimensional modules.

Note that $G_0=G_0'\oplus Z$, where $G_0'$ is semisimple and $Z$ is the center. Let $V$ be a finite dimensional $G_0$-module. Then, by Weyl's theorem, $V$ is completely irreducible as a $G_0'$-module. The elements of $Z$ act as scalars on each irreducible summand of $V$, by Schur's lemma. This means that the decomposition of $V$ into irreducible $G_0'$-modules is a decomposition as $G_0$-modules. It now follows that the category of finite dimensional $G_0$-modules is semi-simple, and irreducible=projective.

Then the answer is yes, finite dimensional irreducible $G_0$-modules are projective in the category of finite dimensional modules.

Note that $G_0=G_0'\oplus Z$, where $G_0'$ is semisimple and $Z$ is the center. Let $V$ be a finite dimensional $G_0$-module. Then, by Weyl's theorem, $V$ is completely irreducible as a $G_0'$-module. The elements of $Z$ act as scalars on each irreducible summand of $V$, by Schur's lemma. This means that the decomposition of $V$ into irreducible $G_0'$-modules is a decomposition as $G_0$-modules. It now follows that the category of finite dimensional $G_0$-modules is semi-simple, and irreducible=projective.

EDIT: This answer is based on the assumption that $G$ is a simple Lie superalgebra, so $G_0$ is reductive. At one time, there was a comment/answer by the OP indicating that this was the case of interest.

Source Link
David Hill
David Hill
  • 1.5k
  • 8
  • 12

Then the answer is yes, finite dimensional irreducible $G_0$-modules are projective in the category of finite dimensional modules.

Note that $G_0=G_0'\oplus Z$, where $G_0'$ is semisimple and $Z$ is the center. Let $V$ be a finite dimensional $G_0$-module. Then, by Weyl's theorem, $V$ is completely irreducible as a $G_0'$-module. The elements of $Z$ act as scalars on each irreducible summand of $V$, by Schur's lemma. This means that the decomposition of $V$ into irreducible $G_0'$-modules is a decomposition as $G_0$-modules. It now follows that the category of finite dimensional $G_0$-modules is semi-simple, and irreducible=projective.