Quotient of annihilator of simple n-dim $mathfraksl_2$-module is $M_2(mathbbC)$?

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Suppose $V^(n)$ is the simple $mathfrakg=mathfraksl_2$-module of dimension $n$, and let $I_n$ be the annihilator, an ideal of the universal enveloping algebra $U(mathfrakg)$. Why is $U(I_n)=U(mathfrakg)/I_n$ isomorphic to the matrix ring $M_n(mathbbC)$? I can identify $U(I_n)$ as a subalgebra of $M_n(mathbbC)$, and then $V^(n)$ is naturally a simple $n$-dimensional $U(I_n)$-module.

The justification I read is that by Artin-Wedderburn, the only simple finite dimensional complex algebra with a simple $n$-dimensional module is $M_n(mathbbC)$, so $M_n(mathbbC)$ is a quotient of $U(I_n)$. I think the result then follows just by counting the dimensions as complex vector spaces, to see they're equal. But how does the observation from Artin-Wedderburn imply $M_n(mathbbC)$ is a quotient? I don't see any sort of obvious surjective map or anything.

modules representation-theory lie-algebras

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edited Jul 24 at 5:38

asked Jul 24 at 3:21

Hailie Mathieson

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Suppose $V^(n)$ is the simple $mathfrakg=mathfraksl_2$-module of dimension $n$, and let $I_n$ be the annihilator, an ideal of the universal enveloping algebra $U(mathfrakg)$. Why is $U(I_n)=U(mathfrakg)/I_n$ isomorphic to the matrix ring $M_n(mathbbC)$? I can identify $U(I_n)$ as a subalgebra of $M_n(mathbbC)$, and then $V^(n)$ is naturally a simple $n$-dimensional $U(I_n)$-module.

The justification I read is that by Artin-Wedderburn, the only simple finite dimensional complex algebra with a simple $n$-dimensional module is $M_n(mathbbC)$, so $M_n(mathbbC)$ is a quotient of $U(I_n)$. I think the result then follows just by counting the dimensions as complex vector spaces, to see they're equal. But how does the observation from Artin-Wedderburn imply $M_n(mathbbC)$ is a quotient? I don't see any sort of obvious surjective map or anything.

modules representation-theory lie-algebras

share|cite|improve this question

edited Jul 24 at 5:38

asked Jul 24 at 3:21

Hailie Mathieson

731518

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Suppose $V^(n)$ is the simple $mathfrakg=mathfraksl_2$-module of dimension $n$, and let $I_n$ be the annihilator, an ideal of the universal enveloping algebra $U(mathfrakg)$. Why is $U(I_n)=U(mathfrakg)/I_n$ isomorphic to the matrix ring $M_n(mathbbC)$? I can identify $U(I_n)$ as a subalgebra of $M_n(mathbbC)$, and then $V^(n)$ is naturally a simple $n$-dimensional $U(I_n)$-module.

The justification I read is that by Artin-Wedderburn, the only simple finite dimensional complex algebra with a simple $n$-dimensional module is $M_n(mathbbC)$, so $M_n(mathbbC)$ is a quotient of $U(I_n)$. I think the result then follows just by counting the dimensions as complex vector spaces, to see they're equal. But how does the observation from Artin-Wedderburn imply $M_n(mathbbC)$ is a quotient? I don't see any sort of obvious surjective map or anything.

modules representation-theory lie-algebras

share|cite|improve this question

edited Jul 24 at 5:38

asked Jul 24 at 3:21

Hailie Mathieson

731518

Suppose $V^(n)$ is the simple $mathfrakg=mathfraksl_2$-module of dimension $n$, and let $I_n$ be the annihilator, an ideal of the universal enveloping algebra $U(mathfrakg)$. Why is $U(I_n)=U(mathfrakg)/I_n$ isomorphic to the matrix ring $M_n(mathbbC)$? I can identify $U(I_n)$ as a subalgebra of $M_n(mathbbC)$, and then $V^(n)$ is naturally a simple $n$-dimensional $U(I_n)$-module.

The justification I read is that by Artin-Wedderburn, the only simple finite dimensional complex algebra with a simple $n$-dimensional module is $M_n(mathbbC)$, so $M_n(mathbbC)$ is a quotient of $U(I_n)$. I think the result then follows just by counting the dimensions as complex vector spaces, to see they're equal. But how does the observation from Artin-Wedderburn imply $M_n(mathbbC)$ is a quotient? I don't see any sort of obvious surjective map or anything.

modules representation-theory lie-algebras

share|cite|improve this question

edited Jul 24 at 5:38

asked Jul 24 at 3:21

Hailie Mathieson

731518

share|cite|improve this question

share|cite|improve this question

edited Jul 24 at 5:38

edited Jul 24 at 5:38

edited Jul 24 at 5:38

asked Jul 24 at 3:21

Hailie Mathieson

731518

asked Jul 24 at 3:21

Hailie Mathieson

731518

asked Jul 24 at 3:21

Hailie Mathieson

731518

731518

add a comment | 

add a comment | 

1 Answer
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Saying the $mathfrakg$-module $V=V^(n)$ is simple is equivalent to saying that the associated representation
$$rho:U(mathfrakg)tomathrmEnd_mathbbC(V)$$
is surjective. Now, $kerrho=I_n$, so by the fundamental homomorphism theorem
$$
U(mathfrakg)/I_ncongmathrmEnd_mathbbC(V)cong M_n(mathbbC).
$$

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answered Jul 25 at 19:37

David Hill

7,9311618

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1 Answer
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1 Answer
1

active

oldest

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active

oldest

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up vote
1
down vote



accepted

Saying the $mathfrakg$-module $V=V^(n)$ is simple is equivalent to saying that the associated representation
$$rho:U(mathfrakg)tomathrmEnd_mathbbC(V)$$
is surjective. Now, $kerrho=I_n$, so by the fundamental homomorphism theorem
$$
U(mathfrakg)/I_ncongmathrmEnd_mathbbC(V)cong M_n(mathbbC).
$$

share|cite|improve this answer

answered Jul 25 at 19:37

David Hill

7,9311618

add a comment | 

up vote
1
down vote



accepted

Saying the $mathfrakg$-module $V=V^(n)$ is simple is equivalent to saying that the associated representation
$$rho:U(mathfrakg)tomathrmEnd_mathbbC(V)$$
is surjective. Now, $kerrho=I_n$, so by the fundamental homomorphism theorem
$$
U(mathfrakg)/I_ncongmathrmEnd_mathbbC(V)cong M_n(mathbbC).
$$

share|cite|improve this answer

answered Jul 25 at 19:37

David Hill

7,9311618

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Saying the $mathfrakg$-module $V=V^(n)$ is simple is equivalent to saying that the associated representation
$$rho:U(mathfrakg)tomathrmEnd_mathbbC(V)$$
is surjective. Now, $kerrho=I_n$, so by the fundamental homomorphism theorem
$$
U(mathfrakg)/I_ncongmathrmEnd_mathbbC(V)cong M_n(mathbbC).
$$

share|cite|improve this answer

answered Jul 25 at 19:37

David Hill

7,9311618

Saying the $mathfrakg$-module $V=V^(n)$ is simple is equivalent to saying that the associated representation
$$rho:U(mathfrakg)tomathrmEnd_mathbbC(V)$$
is surjective. Now, $kerrho=I_n$, so by the fundamental homomorphism theorem
$$
U(mathfrakg)/I_ncongmathrmEnd_mathbbC(V)cong M_n(mathbbC).
$$

share|cite|improve this answer

answered Jul 25 at 19:37

David Hill

7,9311618

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 19:37

David Hill

7,9311618

answered Jul 25 at 19:37

David Hill

7,9311618

answered Jul 25 at 19:37

David Hill

7,9311618

7,9311618

add a comment | 

add a comment | 

 

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