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Quotient of a quotient.
The 2019 Stack Overflow Developer Survey Results Are InThe universal enveloping algebra of a loop algebra as a quotient of the free associative algebra.Irreducible representation and reductive Lie algebraProjective objects in BGG category $mathcalO$ are projective $U(mathfrakg)$-modules?Why is $[mathfrakq, mathfrakg]$ the smallest ideal $mathfraka$ of $mathfrakg$ such that $mathfrakg/mathfraka$ is reductive.Whitehead's lemma (Lie algebras) for reductive Lie algebras.Highest weight module of direct sum of two Lie algebrasVerma module analogue for non-semisimple Lie algebrasSemisimple Lie algebra representation to reductive lie algebraQuotient of annihilator of simple n-dim $mathfraksl_2$-module is $M_2(mathbbC)$?About nondegeneration of Killing form
$begingroup$
Let $mathfrakg$ be a complex reductive Lie algebra. Let $M,N,L$ be finite dimensional $mathfrakg$-modules.
Suppose $M$ is a quotient of $N$ and $N$ is a quotient of $L$.
My question: Is $M$ a quotient of $L$?
modules lie-algebras
asked Mar 23 at 2:01
James CheungJames Cheung
1236
$endgroup$
add a comment |
$begingroup$
Let $mathfrakg$ be a complex reductive Lie algebra. Let $M,N,L$ be finite dimensional $mathfrakg$-modules.
Suppose $M$ is a quotient of $N$ and $N$ is a quotient of $L$.
My question: Is $M$ a quotient of $L$?
modules lie-algebras
asked Mar 23 at 2:01
James CheungJames Cheung
1236
$endgroup$
$begingroup$
Where presumably "being a quotient" means "being isomorphic to a quotient"? (Otherwise it's almost never true.)
$endgroup$
– Torsten Schoeneberg
Mar 23 at 10:43
$begingroup$
Yes. That is what I meant. Thank you for pointing out this issue.
$endgroup$
– James Cheung
Mar 26 at 8:37
add a comment |
$begingroup$
Let $mathfrakg$ be a complex reductive Lie algebra. Let $M,N,L$ be finite dimensional $mathfrakg$-modules.
Suppose $M$ is a quotient of $N$ and $N$ is a quotient of $L$.
My question: Is $M$ a quotient of $L$?
modules lie-algebras
asked Mar 23 at 2:01
James CheungJames Cheung
1236
$endgroup$
Let $mathfrakg$ be a complex reductive Lie algebra. Let $M,N,L$ be finite dimensional $mathfrakg$-modules.
Suppose $M$ is a quotient of $N$ and $N$ is a quotient of $L$.
My question: Is $M$ a quotient of $L$?
modules lie-algebras
modules lie-algebras
asked Mar 23 at 2:01
James CheungJames Cheung
1236
asked Mar 23 at 2:01
James CheungJames Cheung
1236
asked Mar 23 at 2:01
James CheungJames Cheung
1236
asked Mar 23 at 2:01
James CheungJames Cheung
1236
asked Mar 23 at 2:01
James CheungJames Cheung
1236
1236
$begingroup$
Where presumably "being a quotient" means "being isomorphic to a quotient"? (Otherwise it's almost never true.)
$endgroup$
– Torsten Schoeneberg
Mar 23 at 10:43
$begingroup$
Yes. That is what I meant. Thank you for pointing out this issue.
$endgroup$
– James Cheung
Mar 26 at 8:37
add a comment |
$begingroup$
Where presumably "being a quotient" means "being isomorphic to a quotient"? (Otherwise it's almost never true.)
$endgroup$
– Torsten Schoeneberg
Mar 23 at 10:43
$begingroup$
Yes. That is what I meant. Thank you for pointing out this issue.
$endgroup$
– James Cheung
Mar 26 at 8:37
$begingroup$
Where presumably "being a quotient" means "being isomorphic to a quotient"? (Otherwise it's almost never true.)
$endgroup$
– Torsten Schoeneberg
Mar 23 at 10:43
$begingroup$
Where presumably "being a quotient" means "being isomorphic to a quotient"? (Otherwise it's almost never true.)
$endgroup$
– Torsten Schoeneberg
Mar 23 at 10:43
$begingroup$
Yes. That is what I meant. Thank you for pointing out this issue.
$endgroup$
– James Cheung
Mar 26 at 8:37
$begingroup$
Yes. That is what I meant. Thank you for pointing out this issue.
$endgroup$
– James Cheung
Mar 26 at 8:37
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $q: L twoheadrightarrow M$ is the canonical quotient map for $M$, and $p: M twoheadrightarrow N$ is the canonical quotient map, then $pq: L to N$ is surjective, and thus $N cong L / mathrmker(pq)$.
Note that this did not use any structure of $mathfrakg$ or the finite-dimensionality of the modules $L$, $M$, and $N$; it follows only from the fact that if $varphi: A twoheadrightarrow B$ is a surjective morphism of "modules," then $B cong A / mathrmker(varphi)$. This is true for abelian groups, and a "module" of any kind should be an abelian group with extra structure. In fact, this fact is true* in any abelian category, which are the most general categories of modules.
*: modulowink wink replacing surjectivity with the categorical notion of epimorphism.
answered Mar 23 at 2:31
Joshua MundingerJoshua Mundinger
2,9171028
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
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oldest
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active
oldest
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active
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votes
$begingroup$
If $q: L twoheadrightarrow M$ is the canonical quotient map for $M$, and $p: M twoheadrightarrow N$ is the canonical quotient map, then $pq: L to N$ is surjective, and thus $N cong L / mathrmker(pq)$.
Note that this did not use any structure of $mathfrakg$ or the finite-dimensionality of the modules $L$, $M$, and $N$; it follows only from the fact that if $varphi: A twoheadrightarrow B$ is a surjective morphism of "modules," then $B cong A / mathrmker(varphi)$. This is true for abelian groups, and a "module" of any kind should be an abelian group with extra structure. In fact, this fact is true* in any abelian category, which are the most general categories of modules.
*: modulowink wink replacing surjectivity with the categorical notion of epimorphism.
answered Mar 23 at 2:31
Joshua MundingerJoshua Mundinger
2,9171028
$endgroup$
add a comment |
$begingroup$
If $q: L twoheadrightarrow M$ is the canonical quotient map for $M$, and $p: M twoheadrightarrow N$ is the canonical quotient map, then $pq: L to N$ is surjective, and thus $N cong L / mathrmker(pq)$.
Note that this did not use any structure of $mathfrakg$ or the finite-dimensionality of the modules $L$, $M$, and $N$; it follows only from the fact that if $varphi: A twoheadrightarrow B$ is a surjective morphism of "modules," then $B cong A / mathrmker(varphi)$. This is true for abelian groups, and a "module" of any kind should be an abelian group with extra structure. In fact, this fact is true* in any abelian category, which are the most general categories of modules.
*: modulowink wink replacing surjectivity with the categorical notion of epimorphism.
answered Mar 23 at 2:31
Joshua MundingerJoshua Mundinger
2,9171028
$endgroup$
add a comment |
$begingroup$
If $q: L twoheadrightarrow M$ is the canonical quotient map for $M$, and $p: M twoheadrightarrow N$ is the canonical quotient map, then $pq: L to N$ is surjective, and thus $N cong L / mathrmker(pq)$.
Note that this did not use any structure of $mathfrakg$ or the finite-dimensionality of the modules $L$, $M$, and $N$; it follows only from the fact that if $varphi: A twoheadrightarrow B$ is a surjective morphism of "modules," then $B cong A / mathrmker(varphi)$. This is true for abelian groups, and a "module" of any kind should be an abelian group with extra structure. In fact, this fact is true* in any abelian category, which are the most general categories of modules.
*: modulowink wink replacing surjectivity with the categorical notion of epimorphism.
answered Mar 23 at 2:31
Joshua MundingerJoshua Mundinger
2,9171028
$endgroup$
If $q: L twoheadrightarrow M$ is the canonical quotient map for $M$, and $p: M twoheadrightarrow N$ is the canonical quotient map, then $pq: L to N$ is surjective, and thus $N cong L / mathrmker(pq)$.
Note that this did not use any structure of $mathfrakg$ or the finite-dimensionality of the modules $L$, $M$, and $N$; it follows only from the fact that if $varphi: A twoheadrightarrow B$ is a surjective morphism of "modules," then $B cong A / mathrmker(varphi)$. This is true for abelian groups, and a "module" of any kind should be an abelian group with extra structure. In fact, this fact is true* in any abelian category, which are the most general categories of modules.
*: modulowink wink replacing surjectivity with the categorical notion of epimorphism.
answered Mar 23 at 2:31
Joshua MundingerJoshua Mundinger
2,9171028
edited Mar 23 at 3:25
answered Mar 23 at 2:31
Joshua MundingerJoshua Mundinger
2,9171028
answered Mar 23 at 2:31
Joshua MundingerJoshua Mundinger
2,9171028
answered Mar 23 at 2:31
Joshua MundingerJoshua Mundinger
2,9171028
2,9171028
add a comment |
add a comment |
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$begingroup$
Where presumably "being a quotient" means "being isomorphic to a quotient"? (Otherwise it's almost never true.)
$endgroup$
– Torsten Schoeneberg
Mar 23 at 10:43
$begingroup$
Yes. That is what I meant. Thank you for pointing out this issue.
$endgroup$
– James Cheung
Mar 26 at 8:37