Invariant subspace of a representation of a Lie algebra
0
$begingroup$
This may be silly. Let $mathfrak{g}$ be an Lie algebra, with $V$ a nonzero representation.
Denote the invariant subspace $V^{mathfrak{g}}={vin V|x.v=0,forall xin mathfrak{g}}$.
Question:Is $V^{mathfrak{g}}$ a subrepresentation of $V$? It seems to satisfy all axioms to become a subrepresentation. But I just wonder why the textbook is vague about it (namely, Lorentz's book) and calling it a subspace.
representation-theory lie-algebras
asked Dec 6 '18 at 3:47
Weimin JiangWeimin Jiang
11
$endgroup$
add a comment |
0
$begingroup$
This may be silly. Let $mathfrak{g}$ be an Lie algebra, with $V$ a nonzero representation.
Denote the invariant subspace $V^{mathfrak{g}}={vin V|x.v=0,forall xin mathfrak{g}}$.
Question:Is $V^{mathfrak{g}}$ a subrepresentation of $V$? It seems to satisfy all axioms to become a subrepresentation. But I just wonder why the textbook is vague about it (namely, Lorentz's book) and calling it a subspace.
representation-theory lie-algebras
asked Dec 6 '18 at 3:47
Weimin JiangWeimin Jiang
11
$endgroup$
$begingroup$
The textbook says that it is a subspace, which is true. But it is also invariant, yes. Did you check any another book so far?
$endgroup$
– Dietrich Burde
Dec 6 '18 at 13:40
$begingroup$
@DietrichBurde Yes, I checked Hall's Book(GTM 222), which also just mentions it as a subspace. I asked my professor about it, he/she was also vague about it, say that $V^{mathfrak{g}}$ already "use up" all the $mathfrak{g}$ actions, which makes me confused. That's why I ask here. Thanks for the response.
$endgroup$
– Weimin Jiang
Dec 6 '18 at 17:04
$begingroup$
Anything that satisfies the axioms of X is an X. But we don't write out all the things something is every time we talk about it.
$endgroup$
– Artimis Fowl
Dec 7 '18 at 8:11
$begingroup$
It is a subrepresentation but not interesting in the sense that $ mathfrak{g} $ acts trivially on $ V^{mathfrak{g}} $ and this is what your professor perhaps meant to say - "$ V^{mathfrak{g}} $ uses up all the $ mathfrak{g} $-actions". So it is as good as being a subspace only.
$endgroup$
– hellHound
Dec 10 '18 at 19:50
add a comment |
0
0
0
$begingroup$
This may be silly. Let $mathfrak{g}$ be an Lie algebra, with $V$ a nonzero representation.
Denote the invariant subspace $V^{mathfrak{g}}={vin V|x.v=0,forall xin mathfrak{g}}$.
Question:Is $V^{mathfrak{g}}$ a subrepresentation of $V$? It seems to satisfy all axioms to become a subrepresentation. But I just wonder why the textbook is vague about it (namely, Lorentz's book) and calling it a subspace.
representation-theory lie-algebras
asked Dec 6 '18 at 3:47
Weimin JiangWeimin Jiang
11
$endgroup$
This may be silly. Let $mathfrak{g}$ be an Lie algebra, with $V$ a nonzero representation.
Denote the invariant subspace $V^{mathfrak{g}}={vin V|x.v=0,forall xin mathfrak{g}}$.
Question:Is $V^{mathfrak{g}}$ a subrepresentation of $V$? It seems to satisfy all axioms to become a subrepresentation. But I just wonder why the textbook is vague about it (namely, Lorentz's book) and calling it a subspace.
representation-theory lie-algebras
representation-theory lie-algebras
asked Dec 6 '18 at 3:47
Weimin JiangWeimin Jiang
11
asked Dec 6 '18 at 3:47
Weimin JiangWeimin Jiang
11
asked Dec 6 '18 at 3:47
Weimin JiangWeimin Jiang
11
asked Dec 6 '18 at 3:47
Weimin JiangWeimin Jiang
11
asked Dec 6 '18 at 3:47
Weimin JiangWeimin Jiang
11
11
$begingroup$
The textbook says that it is a subspace, which is true. But it is also invariant, yes. Did you check any another book so far?
$endgroup$
– Dietrich Burde
Dec 6 '18 at 13:40
$begingroup$
@DietrichBurde Yes, I checked Hall's Book(GTM 222), which also just mentions it as a subspace. I asked my professor about it, he/she was also vague about it, say that $V^{mathfrak{g}}$ already "use up" all the $mathfrak{g}$ actions, which makes me confused. That's why I ask here. Thanks for the response.
$endgroup$
– Weimin Jiang
Dec 6 '18 at 17:04
$begingroup$
Anything that satisfies the axioms of X is an X. But we don't write out all the things something is every time we talk about it.
$endgroup$
– Artimis Fowl
Dec 7 '18 at 8:11
$begingroup$
It is a subrepresentation but not interesting in the sense that $ mathfrak{g} $ acts trivially on $ V^{mathfrak{g}} $ and this is what your professor perhaps meant to say - "$ V^{mathfrak{g}} $ uses up all the $ mathfrak{g} $-actions". So it is as good as being a subspace only.
$endgroup$
– hellHound
Dec 10 '18 at 19:50
add a comment |
$begingroup$
The textbook says that it is a subspace, which is true. But it is also invariant, yes. Did you check any another book so far?
$endgroup$
– Dietrich Burde
Dec 6 '18 at 13:40
$begingroup$
@DietrichBurde Yes, I checked Hall's Book(GTM 222), which also just mentions it as a subspace. I asked my professor about it, he/she was also vague about it, say that $V^{mathfrak{g}}$ already "use up" all the $mathfrak{g}$ actions, which makes me confused. That's why I ask here. Thanks for the response.
$endgroup$
– Weimin Jiang
Dec 6 '18 at 17:04
$begingroup$
Anything that satisfies the axioms of X is an X. But we don't write out all the things something is every time we talk about it.
$endgroup$
– Artimis Fowl
Dec 7 '18 at 8:11
$begingroup$
It is a subrepresentation but not interesting in the sense that $ mathfrak{g} $ acts trivially on $ V^{mathfrak{g}} $ and this is what your professor perhaps meant to say - "$ V^{mathfrak{g}} $ uses up all the $ mathfrak{g} $-actions". So it is as good as being a subspace only.
$endgroup$
– hellHound
Dec 10 '18 at 19:50
$begingroup$
The textbook says that it is a subspace, which is true. But it is also invariant, yes. Did you check any another book so far?
$endgroup$
– Dietrich Burde
Dec 6 '18 at 13:40
$begingroup$
The textbook says that it is a subspace, which is true. But it is also invariant, yes. Did you check any another book so far?
$endgroup$
– Dietrich Burde
Dec 6 '18 at 13:40
$begingroup$
@DietrichBurde Yes, I checked Hall's Book(GTM 222), which also just mentions it as a subspace. I asked my professor about it, he/she was also vague about it, say that $V^{mathfrak{g}}$ already "use up" all the $mathfrak{g}$ actions, which makes me confused. That's why I ask here. Thanks for the response.
$endgroup$
– Weimin Jiang
Dec 6 '18 at 17:04
$begingroup$
@DietrichBurde Yes, I checked Hall's Book(GTM 222), which also just mentions it as a subspace. I asked my professor about it, he/she was also vague about it, say that $V^{mathfrak{g}}$ already "use up" all the $mathfrak{g}$ actions, which makes me confused. That's why I ask here. Thanks for the response.
$endgroup$
– Weimin Jiang
Dec 6 '18 at 17:04
$begingroup$
Anything that satisfies the axioms of X is an X. But we don't write out all the things something is every time we talk about it.
$endgroup$
– Artimis Fowl
Dec 7 '18 at 8:11
$begingroup$
Anything that satisfies the axioms of X is an X. But we don't write out all the things something is every time we talk about it.
$endgroup$
– Artimis Fowl
Dec 7 '18 at 8:11
$begingroup$
It is a subrepresentation but not interesting in the sense that $ mathfrak{g} $ acts trivially on $ V^{mathfrak{g}} $ and this is what your professor perhaps meant to say - "$ V^{mathfrak{g}} $ uses up all the $ mathfrak{g} $-actions". So it is as good as being a subspace only.
$endgroup$
– hellHound
Dec 10 '18 at 19:50
$begingroup$
It is a subrepresentation but not interesting in the sense that $ mathfrak{g} $ acts trivially on $ V^{mathfrak{g}} $ and this is what your professor perhaps meant to say - "$ V^{mathfrak{g}} $ uses up all the $ mathfrak{g} $-actions". So it is as good as being a subspace only.
$endgroup$
– hellHound
Dec 10 '18 at 19:50
add a comment |
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$begingroup$
The textbook says that it is a subspace, which is true. But it is also invariant, yes. Did you check any another book so far?
$endgroup$
– Dietrich Burde
Dec 6 '18 at 13:40
$begingroup$
@DietrichBurde Yes, I checked Hall's Book(GTM 222), which also just mentions it as a subspace. I asked my professor about it, he/she was also vague about it, say that $V^{mathfrak{g}}$ already "use up" all the $mathfrak{g}$ actions, which makes me confused. That's why I ask here. Thanks for the response.
$endgroup$
– Weimin Jiang
Dec 6 '18 at 17:04
$begingroup$
Anything that satisfies the axioms of X is an X. But we don't write out all the things something is every time we talk about it.
$endgroup$
– Artimis Fowl
Dec 7 '18 at 8:11
$begingroup$
It is a subrepresentation but not interesting in the sense that $ mathfrak{g} $ acts trivially on $ V^{mathfrak{g}} $ and this is what your professor perhaps meant to say - "$ V^{mathfrak{g}} $ uses up all the $ mathfrak{g} $-actions". So it is as good as being a subspace only.
$endgroup$
– hellHound
Dec 10 '18 at 19:50