An example of a Lie algebra with $[R,S]=R$
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What is an example of a complex linear Lie algebra $ L$ such that its radical $R$ is isomorphic to $mathbb C$ and $[R,S]=R$ where $S$ is the maximal semisimple subalgebra of $L$?
lie-algebras
asked Sep 1 at 7:50
Amrat A
1135
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What is an example of a complex linear Lie algebra $ L$ such that its radical $R$ is isomorphic to $mathbb C$ and $[R,S]=R$ where $S$ is the maximal semisimple subalgebra of $L$?
lie-algebras
asked Sep 1 at 7:50
Amrat A
1135
What have you tried? Also, are you sure that there always exists "the" (i.e. unique) maximal semisimple subalgebra?
– Torsten Schoeneberg
Sep 1 at 10:33
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
What is an example of a complex linear Lie algebra $ L$ such that its radical $R$ is isomorphic to $mathbb C$ and $[R,S]=R$ where $S$ is the maximal semisimple subalgebra of $L$?
lie-algebras
asked Sep 1 at 7:50
Amrat A
1135
What is an example of a complex linear Lie algebra $ L$ such that its radical $R$ is isomorphic to $mathbb C$ and $[R,S]=R$ where $S$ is the maximal semisimple subalgebra of $L$?
lie-algebras
lie-algebras
asked Sep 1 at 7:50
Amrat A
1135
asked Sep 1 at 7:50
Amrat A
1135
asked Sep 1 at 7:50
Amrat A
1135
asked Sep 1 at 7:50
Amrat A
1135
asked Sep 1 at 7:50
Amrat A
1135
1135
What have you tried? Also, are you sure that there always exists "the" (i.e. unique) maximal semisimple subalgebra?
– Torsten Schoeneberg
Sep 1 at 10:33
add a comment |Â
What have you tried? Also, are you sure that there always exists "the" (i.e. unique) maximal semisimple subalgebra?
– Torsten Schoeneberg
Sep 1 at 10:33
What have you tried? Also, are you sure that there always exists "the" (i.e. unique) maximal semisimple subalgebra?
– Torsten Schoeneberg
Sep 1 at 10:33
What have you tried? Also, are you sure that there always exists "the" (i.e. unique) maximal semisimple subalgebra?
– Torsten Schoeneberg
Sep 1 at 10:33
add a comment |Â
1 Answer
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1
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As I wrote in the comment, I am not sure if the question is well-defined as it is. However, I think the following observation should solve any specification of it.
If $L$ is a finite-dimensional Lie algebra over a field $k$ of characteristic $0$, and the radical $R$ of $L$ is one-dimensional, then $R$ is central, i.e. $[A, R] =0$ for any subalgebra $A subseteq L$.
Proof: Assume not, then since $R$ is an ideal and one-dimensional we have $[L, R] =R$. This means that the Lie algebra homomorphism $L rightarrow mathfrakgl(R)$, $x mapsto ad_R(x)$ is non-zero. Since $[R,R] =0$ because $R$ is one-dimensional, this homomorphism factors through a non-zero hence surjective homomorphism
$$L/R twoheadrightarrow mathfrakgl(R)$$
But the right hand side is just the one-dimensional abelian Lie algebra, whereas the left hand side is semisimple. This is impossible, as by factoring out the kernel it would induce an iso between a non-zero semisimple LA and an abelian LA.
answered Sep 2 at 19:42
Torsten Schoeneberg
2,7551732
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
As I wrote in the comment, I am not sure if the question is well-defined as it is. However, I think the following observation should solve any specification of it.
If $L$ is a finite-dimensional Lie algebra over a field $k$ of characteristic $0$, and the radical $R$ of $L$ is one-dimensional, then $R$ is central, i.e. $[A, R] =0$ for any subalgebra $A subseteq L$.
Proof: Assume not, then since $R$ is an ideal and one-dimensional we have $[L, R] =R$. This means that the Lie algebra homomorphism $L rightarrow mathfrakgl(R)$, $x mapsto ad_R(x)$ is non-zero. Since $[R,R] =0$ because $R$ is one-dimensional, this homomorphism factors through a non-zero hence surjective homomorphism
$$L/R twoheadrightarrow mathfrakgl(R)$$
But the right hand side is just the one-dimensional abelian Lie algebra, whereas the left hand side is semisimple. This is impossible, as by factoring out the kernel it would induce an iso between a non-zero semisimple LA and an abelian LA.
answered Sep 2 at 19:42
Torsten Schoeneberg
2,7551732
add a comment |Â
up vote
1
down vote
accepted
As I wrote in the comment, I am not sure if the question is well-defined as it is. However, I think the following observation should solve any specification of it.
If $L$ is a finite-dimensional Lie algebra over a field $k$ of characteristic $0$, and the radical $R$ of $L$ is one-dimensional, then $R$ is central, i.e. $[A, R] =0$ for any subalgebra $A subseteq L$.
Proof: Assume not, then since $R$ is an ideal and one-dimensional we have $[L, R] =R$. This means that the Lie algebra homomorphism $L rightarrow mathfrakgl(R)$, $x mapsto ad_R(x)$ is non-zero. Since $[R,R] =0$ because $R$ is one-dimensional, this homomorphism factors through a non-zero hence surjective homomorphism
$$L/R twoheadrightarrow mathfrakgl(R)$$
But the right hand side is just the one-dimensional abelian Lie algebra, whereas the left hand side is semisimple. This is impossible, as by factoring out the kernel it would induce an iso between a non-zero semisimple LA and an abelian LA.
answered Sep 2 at 19:42
Torsten Schoeneberg
2,7551732
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
As I wrote in the comment, I am not sure if the question is well-defined as it is. However, I think the following observation should solve any specification of it.
If $L$ is a finite-dimensional Lie algebra over a field $k$ of characteristic $0$, and the radical $R$ of $L$ is one-dimensional, then $R$ is central, i.e. $[A, R] =0$ for any subalgebra $A subseteq L$.
Proof: Assume not, then since $R$ is an ideal and one-dimensional we have $[L, R] =R$. This means that the Lie algebra homomorphism $L rightarrow mathfrakgl(R)$, $x mapsto ad_R(x)$ is non-zero. Since $[R,R] =0$ because $R$ is one-dimensional, this homomorphism factors through a non-zero hence surjective homomorphism
$$L/R twoheadrightarrow mathfrakgl(R)$$
But the right hand side is just the one-dimensional abelian Lie algebra, whereas the left hand side is semisimple. This is impossible, as by factoring out the kernel it would induce an iso between a non-zero semisimple LA and an abelian LA.
answered Sep 2 at 19:42
Torsten Schoeneberg
2,7551732
As I wrote in the comment, I am not sure if the question is well-defined as it is. However, I think the following observation should solve any specification of it.
If $L$ is a finite-dimensional Lie algebra over a field $k$ of characteristic $0$, and the radical $R$ of $L$ is one-dimensional, then $R$ is central, i.e. $[A, R] =0$ for any subalgebra $A subseteq L$.
Proof: Assume not, then since $R$ is an ideal and one-dimensional we have $[L, R] =R$. This means that the Lie algebra homomorphism $L rightarrow mathfrakgl(R)$, $x mapsto ad_R(x)$ is non-zero. Since $[R,R] =0$ because $R$ is one-dimensional, this homomorphism factors through a non-zero hence surjective homomorphism
$$L/R twoheadrightarrow mathfrakgl(R)$$
But the right hand side is just the one-dimensional abelian Lie algebra, whereas the left hand side is semisimple. This is impossible, as by factoring out the kernel it would induce an iso between a non-zero semisimple LA and an abelian LA.
answered Sep 2 at 19:42
Torsten Schoeneberg
2,7551732
answered Sep 2 at 19:42
Torsten Schoeneberg
2,7551732
answered Sep 2 at 19:42
Torsten Schoeneberg
2,7551732
answered Sep 2 at 19:42
Torsten Schoeneberg
2,7551732
2,7551732
add a comment |Â
add a comment |Â
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What have you tried? Also, are you sure that there always exists "the" (i.e. unique) maximal semisimple subalgebra?
– Torsten Schoeneberg
Sep 1 at 10:33