$(V,rho)$ irreducible, im $rho subset mathfrak{sl}_n$, then $mathfrak{g}$ is semisimple
Let $mathfrak{g} subset mathfrak{gl}(V)$ for some finite
dimensional vector space $V$, over an algebraically closed field $mathbb{F}$ of
characateristic $0$.
Suppose that $rho: mathfrak{g} to mathfrak{gl}(V)$ and that
$V$ is irreducible as a $mathfrak{g}$ representation. Moreover, for any $x in
mathfrak{g}$, tr$(rho(x)) = 0$. Then $mathfrak{g}$ is semisimple.
My attempt:
I've used Lie's theorem and Lie's lemma to show that $rho(mathfrak{g})$ is semisimple;
the assumption of irreducibility allows us to conclude that the weight space of some
$lambda :$ rad$rho(mathfrak{g}) to mathbb{F}$ is $V$ itself.
From here, since tr$(rho(x)) = 0$, and since $h(v) = lambda(h)v$ for any $v in V$ and any $h in$ rad$rho(mathfrak{g})$, we have that rad$rho(mathfrak{g})$ must be zero, since char$mathbb{F} = 0$.
Is the fact the image is semisimple relevant? How I can go back to the original Lie algebra $mathfrak{g}$, or am I off track?
abstract-algebra lie-algebras
asked Dec 2 '18 at 18:36
Mariah
1,361518
add a comment |
Let $mathfrak{g} subset mathfrak{gl}(V)$ for some finite
dimensional vector space $V$, over an algebraically closed field $mathbb{F}$ of
characateristic $0$.
Suppose that $rho: mathfrak{g} to mathfrak{gl}(V)$ and that
$V$ is irreducible as a $mathfrak{g}$ representation. Moreover, for any $x in
mathfrak{g}$, tr$(rho(x)) = 0$. Then $mathfrak{g}$ is semisimple.
My attempt:
I've used Lie's theorem and Lie's lemma to show that $rho(mathfrak{g})$ is semisimple;
the assumption of irreducibility allows us to conclude that the weight space of some
$lambda :$ rad$rho(mathfrak{g}) to mathbb{F}$ is $V$ itself.
From here, since tr$(rho(x)) = 0$, and since $h(v) = lambda(h)v$ for any $v in V$ and any $h in$ rad$rho(mathfrak{g})$, we have that rad$rho(mathfrak{g})$ must be zero, since char$mathbb{F} = 0$.
Is the fact the image is semisimple relevant? How I can go back to the original Lie algebra $mathfrak{g}$, or am I off track?
abstract-algebra lie-algebras
asked Dec 2 '18 at 18:36
Mariah
1,361518
1
The assumptions are about the image... back to $mathfrak{g}$, you have the kernel $mathfrak{n}$ of $rho$, and so you know that $mathfrak{g}/mathfrak{n}$ is semisimple. Of course it tells you nothing about the structure of $mathfrak{n}$ itself; yet it at least says that the radical of $mathfrak{g}$ is contained in $mathfrak{n}$.
– YCor
Dec 2 '18 at 19:55
@YCor do you think there is a mistake in the question? In the question as it is posted it says to assume that $V$ is irreducible as a $mathfrak{g}$ representation and that tr$(rho(x)) = 0$ for any $x in mathfrak{g}$, Do you think it means to assume that any subspace which is stabilized under $mathfrak{g}$ itself is trivial or the whole space?
– Mariah
Dec 2 '18 at 20:00
But being stabilized by $mathfrak{g}$ or by $rho(mathfrak{g})$ means the same...
– YCor
Dec 2 '18 at 20:01
@YCor why? And do you have an idea about how to approach this question?
– Mariah
Dec 2 '18 at 20:06
add a comment |
Let $mathfrak{g} subset mathfrak{gl}(V)$ for some finite
dimensional vector space $V$, over an algebraically closed field $mathbb{F}$ of
characateristic $0$.
Suppose that $rho: mathfrak{g} to mathfrak{gl}(V)$ and that
$V$ is irreducible as a $mathfrak{g}$ representation. Moreover, for any $x in
mathfrak{g}$, tr$(rho(x)) = 0$. Then $mathfrak{g}$ is semisimple.
My attempt:
I've used Lie's theorem and Lie's lemma to show that $rho(mathfrak{g})$ is semisimple;
the assumption of irreducibility allows us to conclude that the weight space of some
$lambda :$ rad$rho(mathfrak{g}) to mathbb{F}$ is $V$ itself.
From here, since tr$(rho(x)) = 0$, and since $h(v) = lambda(h)v$ for any $v in V$ and any $h in$ rad$rho(mathfrak{g})$, we have that rad$rho(mathfrak{g})$ must be zero, since char$mathbb{F} = 0$.
Is the fact the image is semisimple relevant? How I can go back to the original Lie algebra $mathfrak{g}$, or am I off track?
abstract-algebra lie-algebras
asked Dec 2 '18 at 18:36
Mariah
1,361518
Let $mathfrak{g} subset mathfrak{gl}(V)$ for some finite
dimensional vector space $V$, over an algebraically closed field $mathbb{F}$ of
characateristic $0$.
Suppose that $rho: mathfrak{g} to mathfrak{gl}(V)$ and that
$V$ is irreducible as a $mathfrak{g}$ representation. Moreover, for any $x in
mathfrak{g}$, tr$(rho(x)) = 0$. Then $mathfrak{g}$ is semisimple.
My attempt:
I've used Lie's theorem and Lie's lemma to show that $rho(mathfrak{g})$ is semisimple;
the assumption of irreducibility allows us to conclude that the weight space of some
$lambda :$ rad$rho(mathfrak{g}) to mathbb{F}$ is $V$ itself.
From here, since tr$(rho(x)) = 0$, and since $h(v) = lambda(h)v$ for any $v in V$ and any $h in$ rad$rho(mathfrak{g})$, we have that rad$rho(mathfrak{g})$ must be zero, since char$mathbb{F} = 0$.
Is the fact the image is semisimple relevant? How I can go back to the original Lie algebra $mathfrak{g}$, or am I off track?
abstract-algebra lie-algebras
abstract-algebra lie-algebras
asked Dec 2 '18 at 18:36
Mariah
1,361518
asked Dec 2 '18 at 18:36
Mariah
1,361518
edited Dec 2 '18 at 23:04
asked Dec 2 '18 at 18:36
Mariah
1,361518
asked Dec 2 '18 at 18:36
Mariah
1,361518
asked Dec 2 '18 at 18:36
Mariah
1,361518
1,361518
1
The assumptions are about the image... back to $mathfrak{g}$, you have the kernel $mathfrak{n}$ of $rho$, and so you know that $mathfrak{g}/mathfrak{n}$ is semisimple. Of course it tells you nothing about the structure of $mathfrak{n}$ itself; yet it at least says that the radical of $mathfrak{g}$ is contained in $mathfrak{n}$.
– YCor
Dec 2 '18 at 19:55
@YCor do you think there is a mistake in the question? In the question as it is posted it says to assume that $V$ is irreducible as a $mathfrak{g}$ representation and that tr$(rho(x)) = 0$ for any $x in mathfrak{g}$, Do you think it means to assume that any subspace which is stabilized under $mathfrak{g}$ itself is trivial or the whole space?
– Mariah
Dec 2 '18 at 20:00
But being stabilized by $mathfrak{g}$ or by $rho(mathfrak{g})$ means the same...
– YCor
Dec 2 '18 at 20:01
@YCor why? And do you have an idea about how to approach this question?
– Mariah
Dec 2 '18 at 20:06
add a comment |
1
The assumptions are about the image... back to $mathfrak{g}$, you have the kernel $mathfrak{n}$ of $rho$, and so you know that $mathfrak{g}/mathfrak{n}$ is semisimple. Of course it tells you nothing about the structure of $mathfrak{n}$ itself; yet it at least says that the radical of $mathfrak{g}$ is contained in $mathfrak{n}$.
– YCor
Dec 2 '18 at 19:55
@YCor do you think there is a mistake in the question? In the question as it is posted it says to assume that $V$ is irreducible as a $mathfrak{g}$ representation and that tr$(rho(x)) = 0$ for any $x in mathfrak{g}$, Do you think it means to assume that any subspace which is stabilized under $mathfrak{g}$ itself is trivial or the whole space?
– Mariah
Dec 2 '18 at 20:00
But being stabilized by $mathfrak{g}$ or by $rho(mathfrak{g})$ means the same...
– YCor
Dec 2 '18 at 20:01
@YCor why? And do you have an idea about how to approach this question?
– Mariah
Dec 2 '18 at 20:06
1
1
The assumptions are about the image... back to $mathfrak{g}$, you have the kernel $mathfrak{n}$ of $rho$, and so you know that $mathfrak{g}/mathfrak{n}$ is semisimple. Of course it tells you nothing about the structure of $mathfrak{n}$ itself; yet it at least says that the radical of $mathfrak{g}$ is contained in $mathfrak{n}$.
– YCor
Dec 2 '18 at 19:55
The assumptions are about the image... back to $mathfrak{g}$, you have the kernel $mathfrak{n}$ of $rho$, and so you know that $mathfrak{g}/mathfrak{n}$ is semisimple. Of course it tells you nothing about the structure of $mathfrak{n}$ itself; yet it at least says that the radical of $mathfrak{g}$ is contained in $mathfrak{n}$.
– YCor
Dec 2 '18 at 19:55
@YCor do you think there is a mistake in the question? In the question as it is posted it says to assume that $V$ is irreducible as a $mathfrak{g}$ representation and that tr$(rho(x)) = 0$ for any $x in mathfrak{g}$, Do you think it means to assume that any subspace which is stabilized under $mathfrak{g}$ itself is trivial or the whole space?
– Mariah
Dec 2 '18 at 20:00
@YCor do you think there is a mistake in the question? In the question as it is posted it says to assume that $V$ is irreducible as a $mathfrak{g}$ representation and that tr$(rho(x)) = 0$ for any $x in mathfrak{g}$, Do you think it means to assume that any subspace which is stabilized under $mathfrak{g}$ itself is trivial or the whole space?
– Mariah
Dec 2 '18 at 20:00
But being stabilized by $mathfrak{g}$ or by $rho(mathfrak{g})$ means the same...
– YCor
Dec 2 '18 at 20:01
But being stabilized by $mathfrak{g}$ or by $rho(mathfrak{g})$ means the same...
– YCor
Dec 2 '18 at 20:01
@YCor why? And do you have an idea about how to approach this question?
– Mariah
Dec 2 '18 at 20:06
@YCor why? And do you have an idea about how to approach this question?
– Mariah
Dec 2 '18 at 20:06
add a comment |
1 Answer
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Counterexample: $mathfrak{g}:=mathfrak{gl}_2(Bbb C)$ which has a one-dimensional centre $mathfrak{z}$, and let $rho$ be the compositition
$$rho: mathfrak{gl}_2(Bbb C) twoheadrightarrow mathfrak{gl}_2(Bbb C) / mathfrak{z} simeq mathfrak{sl}_2(Bbb C) subset mathfrak{gl}_2(Bbb C).$$
Maybe what is meant in the question is that the given inclusion $mathfrak{g} subsetmathfrak{gl}(V)$ itself is irreducible when viewed as representation? Then you're done if you managed to show that $rho(mathfrak g)$ is semisimple.
answered Dec 2 '18 at 23:28
Torsten Schoeneberg
3,8512833
Maybe! But can you explain why $rho (mathfrak{g})$ is semisimple implies the original is as well in that case? I'm not sure I see it:)
– Mariah
Dec 2 '18 at 23:31
Can you explain what you mean by "given inclusion 𝔤⊂𝔤𝔩(V) itself is irreducible when viewed as representation? Then you're done if you managed to show that ρ(𝔤) is semisimple"?
– Mariah
Dec 3 '18 at 11:54
It's hard to explain if you don't see this immediately. According to the question, $mathfrak{g} subset mathfrak{gl}(V)$. That is what I mean by "given inclusion". Then if one views this inclusion as a map from $mathfrak{g}$ to $mathfrak{gl}(V)$ (sending an element $x$ to itself), it is an injective homomorphism of Lie algebras. Call it $rho$. You get $mathfrak{g} simeq rho(mathfrak{g})$. (I am not sure though if this is indeed how the question is meant. Because if it was, one could just say that the traces of elements of $mathfrak{g}$ seen as elements of $mathfrak{gl}(V)$ are $0$.)
– Torsten Schoeneberg
Dec 3 '18 at 23:02
Ah ok, I didn't realise you meant that $rho$ was the inclusion map itself. Then yes of course in that case we'd be done. I need to ask around see what is actually meant here..
– Mariah
Dec 3 '18 at 23:10
add a comment |
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Counterexample: $mathfrak{g}:=mathfrak{gl}_2(Bbb C)$ which has a one-dimensional centre $mathfrak{z}$, and let $rho$ be the compositition
$$rho: mathfrak{gl}_2(Bbb C) twoheadrightarrow mathfrak{gl}_2(Bbb C) / mathfrak{z} simeq mathfrak{sl}_2(Bbb C) subset mathfrak{gl}_2(Bbb C).$$
Maybe what is meant in the question is that the given inclusion $mathfrak{g} subsetmathfrak{gl}(V)$ itself is irreducible when viewed as representation? Then you're done if you managed to show that $rho(mathfrak g)$ is semisimple.
answered Dec 2 '18 at 23:28
Torsten Schoeneberg
3,8512833
Maybe! But can you explain why $rho (mathfrak{g})$ is semisimple implies the original is as well in that case? I'm not sure I see it:)
– Mariah
Dec 2 '18 at 23:31
Can you explain what you mean by "given inclusion 𝔤⊂𝔤𝔩(V) itself is irreducible when viewed as representation? Then you're done if you managed to show that ρ(𝔤) is semisimple"?
– Mariah
Dec 3 '18 at 11:54
It's hard to explain if you don't see this immediately. According to the question, $mathfrak{g} subset mathfrak{gl}(V)$. That is what I mean by "given inclusion". Then if one views this inclusion as a map from $mathfrak{g}$ to $mathfrak{gl}(V)$ (sending an element $x$ to itself), it is an injective homomorphism of Lie algebras. Call it $rho$. You get $mathfrak{g} simeq rho(mathfrak{g})$. (I am not sure though if this is indeed how the question is meant. Because if it was, one could just say that the traces of elements of $mathfrak{g}$ seen as elements of $mathfrak{gl}(V)$ are $0$.)
– Torsten Schoeneberg
Dec 3 '18 at 23:02
Ah ok, I didn't realise you meant that $rho$ was the inclusion map itself. Then yes of course in that case we'd be done. I need to ask around see what is actually meant here..
– Mariah
Dec 3 '18 at 23:10
add a comment |
Counterexample: $mathfrak{g}:=mathfrak{gl}_2(Bbb C)$ which has a one-dimensional centre $mathfrak{z}$, and let $rho$ be the compositition
$$rho: mathfrak{gl}_2(Bbb C) twoheadrightarrow mathfrak{gl}_2(Bbb C) / mathfrak{z} simeq mathfrak{sl}_2(Bbb C) subset mathfrak{gl}_2(Bbb C).$$
Maybe what is meant in the question is that the given inclusion $mathfrak{g} subsetmathfrak{gl}(V)$ itself is irreducible when viewed as representation? Then you're done if you managed to show that $rho(mathfrak g)$ is semisimple.
answered Dec 2 '18 at 23:28
Torsten Schoeneberg
3,8512833
Maybe! But can you explain why $rho (mathfrak{g})$ is semisimple implies the original is as well in that case? I'm not sure I see it:)
– Mariah
Dec 2 '18 at 23:31
Can you explain what you mean by "given inclusion 𝔤⊂𝔤𝔩(V) itself is irreducible when viewed as representation? Then you're done if you managed to show that ρ(𝔤) is semisimple"?
– Mariah
Dec 3 '18 at 11:54
It's hard to explain if you don't see this immediately. According to the question, $mathfrak{g} subset mathfrak{gl}(V)$. That is what I mean by "given inclusion". Then if one views this inclusion as a map from $mathfrak{g}$ to $mathfrak{gl}(V)$ (sending an element $x$ to itself), it is an injective homomorphism of Lie algebras. Call it $rho$. You get $mathfrak{g} simeq rho(mathfrak{g})$. (I am not sure though if this is indeed how the question is meant. Because if it was, one could just say that the traces of elements of $mathfrak{g}$ seen as elements of $mathfrak{gl}(V)$ are $0$.)
– Torsten Schoeneberg
Dec 3 '18 at 23:02
Ah ok, I didn't realise you meant that $rho$ was the inclusion map itself. Then yes of course in that case we'd be done. I need to ask around see what is actually meant here..
– Mariah
Dec 3 '18 at 23:10
add a comment |
Counterexample: $mathfrak{g}:=mathfrak{gl}_2(Bbb C)$ which has a one-dimensional centre $mathfrak{z}$, and let $rho$ be the compositition
$$rho: mathfrak{gl}_2(Bbb C) twoheadrightarrow mathfrak{gl}_2(Bbb C) / mathfrak{z} simeq mathfrak{sl}_2(Bbb C) subset mathfrak{gl}_2(Bbb C).$$
Maybe what is meant in the question is that the given inclusion $mathfrak{g} subsetmathfrak{gl}(V)$ itself is irreducible when viewed as representation? Then you're done if you managed to show that $rho(mathfrak g)$ is semisimple.
answered Dec 2 '18 at 23:28
Torsten Schoeneberg
3,8512833
Counterexample: $mathfrak{g}:=mathfrak{gl}_2(Bbb C)$ which has a one-dimensional centre $mathfrak{z}$, and let $rho$ be the compositition
$$rho: mathfrak{gl}_2(Bbb C) twoheadrightarrow mathfrak{gl}_2(Bbb C) / mathfrak{z} simeq mathfrak{sl}_2(Bbb C) subset mathfrak{gl}_2(Bbb C).$$
Maybe what is meant in the question is that the given inclusion $mathfrak{g} subsetmathfrak{gl}(V)$ itself is irreducible when viewed as representation? Then you're done if you managed to show that $rho(mathfrak g)$ is semisimple.
answered Dec 2 '18 at 23:28
Torsten Schoeneberg
3,8512833
answered Dec 2 '18 at 23:28
Torsten Schoeneberg
3,8512833
answered Dec 2 '18 at 23:28
Torsten Schoeneberg
3,8512833
answered Dec 2 '18 at 23:28
Torsten Schoeneberg
3,8512833
3,8512833
Maybe! But can you explain why $rho (mathfrak{g})$ is semisimple implies the original is as well in that case? I'm not sure I see it:)
– Mariah
Dec 2 '18 at 23:31
Can you explain what you mean by "given inclusion 𝔤⊂𝔤𝔩(V) itself is irreducible when viewed as representation? Then you're done if you managed to show that ρ(𝔤) is semisimple"?
– Mariah
Dec 3 '18 at 11:54
It's hard to explain if you don't see this immediately. According to the question, $mathfrak{g} subset mathfrak{gl}(V)$. That is what I mean by "given inclusion". Then if one views this inclusion as a map from $mathfrak{g}$ to $mathfrak{gl}(V)$ (sending an element $x$ to itself), it is an injective homomorphism of Lie algebras. Call it $rho$. You get $mathfrak{g} simeq rho(mathfrak{g})$. (I am not sure though if this is indeed how the question is meant. Because if it was, one could just say that the traces of elements of $mathfrak{g}$ seen as elements of $mathfrak{gl}(V)$ are $0$.)
– Torsten Schoeneberg
Dec 3 '18 at 23:02
Ah ok, I didn't realise you meant that $rho$ was the inclusion map itself. Then yes of course in that case we'd be done. I need to ask around see what is actually meant here..
– Mariah
Dec 3 '18 at 23:10
add a comment |
Maybe! But can you explain why $rho (mathfrak{g})$ is semisimple implies the original is as well in that case? I'm not sure I see it:)
– Mariah
Dec 2 '18 at 23:31
Can you explain what you mean by "given inclusion 𝔤⊂𝔤𝔩(V) itself is irreducible when viewed as representation? Then you're done if you managed to show that ρ(𝔤) is semisimple"?
– Mariah
Dec 3 '18 at 11:54
It's hard to explain if you don't see this immediately. According to the question, $mathfrak{g} subset mathfrak{gl}(V)$. That is what I mean by "given inclusion". Then if one views this inclusion as a map from $mathfrak{g}$ to $mathfrak{gl}(V)$ (sending an element $x$ to itself), it is an injective homomorphism of Lie algebras. Call it $rho$. You get $mathfrak{g} simeq rho(mathfrak{g})$. (I am not sure though if this is indeed how the question is meant. Because if it was, one could just say that the traces of elements of $mathfrak{g}$ seen as elements of $mathfrak{gl}(V)$ are $0$.)
– Torsten Schoeneberg
Dec 3 '18 at 23:02
Ah ok, I didn't realise you meant that $rho$ was the inclusion map itself. Then yes of course in that case we'd be done. I need to ask around see what is actually meant here..
– Mariah
Dec 3 '18 at 23:10
Maybe! But can you explain why $rho (mathfrak{g})$ is semisimple implies the original is as well in that case? I'm not sure I see it:)
– Mariah
Dec 2 '18 at 23:31
Maybe! But can you explain why $rho (mathfrak{g})$ is semisimple implies the original is as well in that case? I'm not sure I see it:)
– Mariah
Dec 2 '18 at 23:31
Can you explain what you mean by "given inclusion 𝔤⊂𝔤𝔩(V) itself is irreducible when viewed as representation? Then you're done if you managed to show that ρ(𝔤) is semisimple"?
– Mariah
Dec 3 '18 at 11:54
Can you explain what you mean by "given inclusion 𝔤⊂𝔤𝔩(V) itself is irreducible when viewed as representation? Then you're done if you managed to show that ρ(𝔤) is semisimple"?
– Mariah
Dec 3 '18 at 11:54
It's hard to explain if you don't see this immediately. According to the question, $mathfrak{g} subset mathfrak{gl}(V)$. That is what I mean by "given inclusion". Then if one views this inclusion as a map from $mathfrak{g}$ to $mathfrak{gl}(V)$ (sending an element $x$ to itself), it is an injective homomorphism of Lie algebras. Call it $rho$. You get $mathfrak{g} simeq rho(mathfrak{g})$. (I am not sure though if this is indeed how the question is meant. Because if it was, one could just say that the traces of elements of $mathfrak{g}$ seen as elements of $mathfrak{gl}(V)$ are $0$.)
– Torsten Schoeneberg
Dec 3 '18 at 23:02
It's hard to explain if you don't see this immediately. According to the question, $mathfrak{g} subset mathfrak{gl}(V)$. That is what I mean by "given inclusion". Then if one views this inclusion as a map from $mathfrak{g}$ to $mathfrak{gl}(V)$ (sending an element $x$ to itself), it is an injective homomorphism of Lie algebras. Call it $rho$. You get $mathfrak{g} simeq rho(mathfrak{g})$. (I am not sure though if this is indeed how the question is meant. Because if it was, one could just say that the traces of elements of $mathfrak{g}$ seen as elements of $mathfrak{gl}(V)$ are $0$.)
– Torsten Schoeneberg
Dec 3 '18 at 23:02
Ah ok, I didn't realise you meant that $rho$ was the inclusion map itself. Then yes of course in that case we'd be done. I need to ask around see what is actually meant here..
– Mariah
Dec 3 '18 at 23:10
Ah ok, I didn't realise you meant that $rho$ was the inclusion map itself. Then yes of course in that case we'd be done. I need to ask around see what is actually meant here..
– Mariah
Dec 3 '18 at 23:10
add a comment |
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The assumptions are about the image... back to $mathfrak{g}$, you have the kernel $mathfrak{n}$ of $rho$, and so you know that $mathfrak{g}/mathfrak{n}$ is semisimple. Of course it tells you nothing about the structure of $mathfrak{n}$ itself; yet it at least says that the radical of $mathfrak{g}$ is contained in $mathfrak{n}$.
– YCor
Dec 2 '18 at 19:55
@YCor do you think there is a mistake in the question? In the question as it is posted it says to assume that $V$ is irreducible as a $mathfrak{g}$ representation and that tr$(rho(x)) = 0$ for any $x in mathfrak{g}$, Do you think it means to assume that any subspace which is stabilized under $mathfrak{g}$ itself is trivial or the whole space?
– Mariah
Dec 2 '18 at 20:00
But being stabilized by $mathfrak{g}$ or by $rho(mathfrak{g})$ means the same...
– YCor
Dec 2 '18 at 20:01
@YCor why? And do you have an idea about how to approach this question?
– Mariah
Dec 2 '18 at 20:06