Mixing
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
asked Jul 26 at 11:46
Or Kedar
1777
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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
asked Jul 26 at 11:46
Or Kedar
1777
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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up vote
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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
asked Jul 26 at 11:46
Or Kedar
1777
Does every nilpotent lie subalgebra of gl(V) is representable by triangular matrices with 0 on the diagonal?
representation-theory lie-algebras
asked Jul 26 at 11:46
Or Kedar
1777
asked Jul 26 at 11:46
Or Kedar
1777
asked Jul 26 at 11:46
Or Kedar
1777
asked Jul 26 at 11:46
Or Kedar
1777
1777
My apologies for my erroneous earlier answer. Please see Andreas Cap's answer below.
– peter a g
Jul 28 at 14:47
add a comment |Â
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
add a comment |Â
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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.)
answered Jul 27 at 8:54
Andreas Cap
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
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
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.)
answered Jul 27 at 8:54
Andreas Cap
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
add a comment |Â
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.)
answered Jul 27 at 8:54
Andreas Cap
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
add a comment |Â
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.)
answered Jul 27 at 8:54
Andreas Cap
9,599622
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.)
answered Jul 27 at 8:54
Andreas Cap
9,599622
answered Jul 27 at 8:54
Andreas Cap
9,599622
answered Jul 27 at 8:54
Andreas Cap
9,599622
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
add a comment |Â
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
add a comment |Â
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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