$(V,rho)$ irreducible, im $rho subset mathfrak{sl}_n$, then $mathfrak{g}$ is semisimple












2












$begingroup$



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







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    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 $mathfrak{n}$.
    $endgroup$
    – YCor
    Dec 2 '18 at 19:55










  • $begingroup$
    @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?
    $endgroup$
    – Mariah
    Dec 2 '18 at 20:00












  • $begingroup$
    But being stabilized by $mathfrak{g}$ or by $rho(mathfrak{g})$ means the same...
    $endgroup$
    – YCor
    Dec 2 '18 at 20:01










  • $begingroup$
    @YCor why? And do you have an idea about how to approach this question?
    $endgroup$
    – Mariah
    Dec 2 '18 at 20:06
















2












$begingroup$



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







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    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 $mathfrak{n}$.
    $endgroup$
    – YCor
    Dec 2 '18 at 19:55










  • $begingroup$
    @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?
    $endgroup$
    – Mariah
    Dec 2 '18 at 20:00












  • $begingroup$
    But being stabilized by $mathfrak{g}$ or by $rho(mathfrak{g})$ means the same...
    $endgroup$
    – YCor
    Dec 2 '18 at 20:01










  • $begingroup$
    @YCor why? And do you have an idea about how to approach this question?
    $endgroup$
    – Mariah
    Dec 2 '18 at 20:06














2












2








2


1



$begingroup$



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







share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 23:04







Mariah

















asked Dec 2 '18 at 18:36









MariahMariah

1,3831518




1,3831518








  • 1




    $begingroup$
    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 $mathfrak{n}$.
    $endgroup$
    – YCor
    Dec 2 '18 at 19:55










  • $begingroup$
    @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?
    $endgroup$
    – Mariah
    Dec 2 '18 at 20:00












  • $begingroup$
    But being stabilized by $mathfrak{g}$ or by $rho(mathfrak{g})$ means the same...
    $endgroup$
    – YCor
    Dec 2 '18 at 20:01










  • $begingroup$
    @YCor why? And do you have an idea about how to approach this question?
    $endgroup$
    – Mariah
    Dec 2 '18 at 20:06














  • 1




    $begingroup$
    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 $mathfrak{n}$.
    $endgroup$
    – YCor
    Dec 2 '18 at 19:55










  • $begingroup$
    @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?
    $endgroup$
    – Mariah
    Dec 2 '18 at 20:00












  • $begingroup$
    But being stabilized by $mathfrak{g}$ or by $rho(mathfrak{g})$ means the same...
    $endgroup$
    – YCor
    Dec 2 '18 at 20:01










  • $begingroup$
    @YCor why? And do you have an idea about how to approach this question?
    $endgroup$
    – Mariah
    Dec 2 '18 at 20:06








1




1




$begingroup$
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 $mathfrak{n}$.
$endgroup$
– YCor
Dec 2 '18 at 19:55




$begingroup$
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 $mathfrak{n}$.
$endgroup$
– YCor
Dec 2 '18 at 19:55












$begingroup$
@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?
$endgroup$
– Mariah
Dec 2 '18 at 20:00






$begingroup$
@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?
$endgroup$
– Mariah
Dec 2 '18 at 20:00














$begingroup$
But being stabilized by $mathfrak{g}$ or by $rho(mathfrak{g})$ means the same...
$endgroup$
– YCor
Dec 2 '18 at 20:01




$begingroup$
But being stabilized by $mathfrak{g}$ or by $rho(mathfrak{g})$ means the same...
$endgroup$
– YCor
Dec 2 '18 at 20:01












$begingroup$
@YCor why? And do you have an idea about how to approach this question?
$endgroup$
– Mariah
Dec 2 '18 at 20:06




$begingroup$
@YCor why? And do you have an idea about how to approach this question?
$endgroup$
– Mariah
Dec 2 '18 at 20:06










1 Answer
1






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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:)
    $endgroup$
    – Mariah
    Dec 2 '18 at 23:31












  • $begingroup$
    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"?
    $endgroup$
    – Mariah
    Dec 3 '18 at 11:54










  • $begingroup$
    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 $0$.)
    $endgroup$
    – Torsten Schoeneberg
    Dec 3 '18 at 23:02










  • $begingroup$
    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..
    $endgroup$
    – Mariah
    Dec 3 '18 at 23:10











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023019%2fv-rho-irreducible-im-rho-subset-mathfraksl-n-then-mathfrakg-i%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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:)
    $endgroup$
    – Mariah
    Dec 2 '18 at 23:31












  • $begingroup$
    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"?
    $endgroup$
    – Mariah
    Dec 3 '18 at 11:54










  • $begingroup$
    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 $0$.)
    $endgroup$
    – Torsten Schoeneberg
    Dec 3 '18 at 23:02










  • $begingroup$
    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..
    $endgroup$
    – Mariah
    Dec 3 '18 at 23:10
















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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:)
    $endgroup$
    – Mariah
    Dec 2 '18 at 23:31












  • $begingroup$
    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"?
    $endgroup$
    – Mariah
    Dec 3 '18 at 11:54










  • $begingroup$
    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 $0$.)
    $endgroup$
    – Torsten Schoeneberg
    Dec 3 '18 at 23:02










  • $begingroup$
    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..
    $endgroup$
    – Mariah
    Dec 3 '18 at 23:10














1












1








1





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 2 '18 at 23:28









Torsten SchoenebergTorsten Schoeneberg

3,9112833




3,9112833












  • $begingroup$
    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:)
    $endgroup$
    – Mariah
    Dec 2 '18 at 23:31












  • $begingroup$
    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"?
    $endgroup$
    – Mariah
    Dec 3 '18 at 11:54










  • $begingroup$
    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 $0$.)
    $endgroup$
    – Torsten Schoeneberg
    Dec 3 '18 at 23:02










  • $begingroup$
    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..
    $endgroup$
    – Mariah
    Dec 3 '18 at 23:10


















  • $begingroup$
    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:)
    $endgroup$
    – Mariah
    Dec 2 '18 at 23:31












  • $begingroup$
    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"?
    $endgroup$
    – Mariah
    Dec 3 '18 at 11:54










  • $begingroup$
    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 $0$.)
    $endgroup$
    – Torsten Schoeneberg
    Dec 3 '18 at 23:02










  • $begingroup$
    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..
    $endgroup$
    – Mariah
    Dec 3 '18 at 23:10
















$begingroup$
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:)
$endgroup$
– Mariah
Dec 2 '18 at 23:31






$begingroup$
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:)
$endgroup$
– Mariah
Dec 2 '18 at 23:31














$begingroup$
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"?
$endgroup$
– Mariah
Dec 3 '18 at 11:54




$begingroup$
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"?
$endgroup$
– Mariah
Dec 3 '18 at 11:54












$begingroup$
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 $0$.)
$endgroup$
– Torsten Schoeneberg
Dec 3 '18 at 23:02




$begingroup$
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 $0$.)
$endgroup$
– Torsten Schoeneberg
Dec 3 '18 at 23:02












$begingroup$
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..
$endgroup$
– Mariah
Dec 3 '18 at 23:10




$begingroup$
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..
$endgroup$
– Mariah
Dec 3 '18 at 23:10


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023019%2fv-rho-irreducible-im-rho-subset-mathfraksl-n-then-mathfrakg-i%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Probability when a professor distributes a quiz and homework assignment to a class of n students.

Image
0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
Read more

Aardman Animations

Image
Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
Read more

Are they similar matrix

Image
0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
Read more