Representations of Nilpotent Lie Subalgebras

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Does every nilpotent lie subalgebra of gl(V) is representable by triangular matrices with 0 on the diagonal?




representation-theory lie-algebras




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  • My apologies for my erroneous earlier answer. Please see Andreas Cap's answer below.
    – peter a g
    Jul 28 at 14:47















up vote
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Does every nilpotent lie subalgebra of gl(V) is representable by triangular matrices with 0 on the diagonal?




representation-theory lie-algebras




share|cite|improve this question



















  • My apologies for my erroneous earlier answer. Please see Andreas Cap's answer below.
    – peter a g
    Jul 28 at 14:47













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Does every nilpotent lie subalgebra of gl(V) is representable by triangular matrices with 0 on the diagonal?




representation-theory lie-algebras




share|cite|improve this question











Does every nilpotent lie subalgebra of gl(V) is representable by triangular matrices with 0 on the diagonal?





representation-theory lie-algebras





share|cite|improve this question










share|cite|improve this question




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asked Jul 26 at 11:46









Or Kedar

1777




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  • My apologies for my erroneous earlier answer. Please see Andreas Cap's answer below.
    – peter a g
    Jul 28 at 14:47

















  • My apologies for my erroneous earlier answer. Please see Andreas Cap's answer below.
    – peter a g
    Jul 28 at 14:47
















My apologies for my erroneous earlier answer. Please see Andreas Cap's answer below.
– peter a g
Jul 28 at 14:47





My apologies for my erroneous earlier answer. Please see Andreas Cap's answer below.
– peter a g
Jul 28 at 14:47











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This is not true for rather simple reasons: Any one-dimensional subspace in $mathfrakgl(V)$ is an Abelian (and hence nilpotent) Lie subalgebra. This can be represented by upper triangular matrices with $0$ on the diagonal if and only if it is spanned by a nilpotent map.



Engel's theorem as mentioned in the comment by @peter_a_g is not a theorem about nipotent Lie subalgebras in $mathfrakgl(V)$ but about Lie subalgebras consisting of nilpotent endomorphisms of $V$. (It then implies that such a subalgebra is nilpotent, but the converse is not true.)






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  • I was tempted to leave my original stupid comment (albeit with a retraction) as a warning - but I did not. Thank you for the correction. BTW, unfortunately the at-sign-name in an answer does not seem to warn the designated person (i.e., does not mark 'recent messages' in red). This seems only to happen for comments. One should ask Stack Exchange why, though I bet the question has been asked before.
    – peter a g
    Jul 28 at 14:40











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1 Answer
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1 Answer
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active

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up vote
1
down vote













This is not true for rather simple reasons: Any one-dimensional subspace in $mathfrakgl(V)$ is an Abelian (and hence nilpotent) Lie subalgebra. This can be represented by upper triangular matrices with $0$ on the diagonal if and only if it is spanned by a nilpotent map.



Engel's theorem as mentioned in the comment by @peter_a_g is not a theorem about nipotent Lie subalgebras in $mathfrakgl(V)$ but about Lie subalgebras consisting of nilpotent endomorphisms of $V$. (It then implies that such a subalgebra is nilpotent, but the converse is not true.)






share|cite|improve this answer





















  • I was tempted to leave my original stupid comment (albeit with a retraction) as a warning - but I did not. Thank you for the correction. BTW, unfortunately the at-sign-name in an answer does not seem to warn the designated person (i.e., does not mark 'recent messages' in red). This seems only to happen for comments. One should ask Stack Exchange why, though I bet the question has been asked before.
    – peter a g
    Jul 28 at 14:40















up vote
1
down vote













This is not true for rather simple reasons: Any one-dimensional subspace in $mathfrakgl(V)$ is an Abelian (and hence nilpotent) Lie subalgebra. This can be represented by upper triangular matrices with $0$ on the diagonal if and only if it is spanned by a nilpotent map.



Engel's theorem as mentioned in the comment by @peter_a_g is not a theorem about nipotent Lie subalgebras in $mathfrakgl(V)$ but about Lie subalgebras consisting of nilpotent endomorphisms of $V$. (It then implies that such a subalgebra is nilpotent, but the converse is not true.)






share|cite|improve this answer





















  • I was tempted to leave my original stupid comment (albeit with a retraction) as a warning - but I did not. Thank you for the correction. BTW, unfortunately the at-sign-name in an answer does not seem to warn the designated person (i.e., does not mark 'recent messages' in red). This seems only to happen for comments. One should ask Stack Exchange why, though I bet the question has been asked before.
    – peter a g
    Jul 28 at 14:40













up vote
1
down vote










up vote
1
down vote









This is not true for rather simple reasons: Any one-dimensional subspace in $mathfrakgl(V)$ is an Abelian (and hence nilpotent) Lie subalgebra. This can be represented by upper triangular matrices with $0$ on the diagonal if and only if it is spanned by a nilpotent map.



Engel's theorem as mentioned in the comment by @peter_a_g is not a theorem about nipotent Lie subalgebras in $mathfrakgl(V)$ but about Lie subalgebras consisting of nilpotent endomorphisms of $V$. (It then implies that such a subalgebra is nilpotent, but the converse is not true.)






share|cite|improve this answer













This is not true for rather simple reasons: Any one-dimensional subspace in $mathfrakgl(V)$ is an Abelian (and hence nilpotent) Lie subalgebra. This can be represented by upper triangular matrices with $0$ on the diagonal if and only if it is spanned by a nilpotent map.



Engel's theorem as mentioned in the comment by @peter_a_g is not a theorem about nipotent Lie subalgebras in $mathfrakgl(V)$ but about Lie subalgebras consisting of nilpotent endomorphisms of $V$. (It then implies that such a subalgebra is nilpotent, but the converse is not true.)







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 27 at 8:54









Andreas Cap

9,599622




9,599622











  • I was tempted to leave my original stupid comment (albeit with a retraction) as a warning - but I did not. Thank you for the correction. BTW, unfortunately the at-sign-name in an answer does not seem to warn the designated person (i.e., does not mark 'recent messages' in red). This seems only to happen for comments. One should ask Stack Exchange why, though I bet the question has been asked before.
    – peter a g
    Jul 28 at 14:40

















  • I was tempted to leave my original stupid comment (albeit with a retraction) as a warning - but I did not. Thank you for the correction. BTW, unfortunately the at-sign-name in an answer does not seem to warn the designated person (i.e., does not mark 'recent messages' in red). This seems only to happen for comments. One should ask Stack Exchange why, though I bet the question has been asked before.
    – peter a g
    Jul 28 at 14:40
















I was tempted to leave my original stupid comment (albeit with a retraction) as a warning - but I did not. Thank you for the correction. BTW, unfortunately the at-sign-name in an answer does not seem to warn the designated person (i.e., does not mark 'recent messages' in red). This seems only to happen for comments. One should ask Stack Exchange why, though I bet the question has been asked before.
– peter a g
Jul 28 at 14:40





I was tempted to leave my original stupid comment (albeit with a retraction) as a warning - but I did not. Thank you for the correction. BTW, unfortunately the at-sign-name in an answer does not seem to warn the designated person (i.e., does not mark 'recent messages' in red). This seems only to happen for comments. One should ask Stack Exchange why, though I bet the question has been asked before.
– peter a g
Jul 28 at 14:40













 

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