Mixing
Endomorphisms of irreducible submodules of simple algebras
up vote
1
down vote
favorite
I am quite new to Representation Theory and I am trying to wrap my head around some basics. I have some difficulties with the following setup:
Suppose I start with a group ring $R=K[G]$, where $K$ is a field of characteristic $0$ (not necessarily algebraically closed) and $G$ is a finite group (not necessarily abelian).
- I can decompose $R$ as a direct sum
$$ R = bigoplus e_{rho} R,$$
where each $e_{rho}$ is a central idempotent and for each $rho$, $e_{rho}R$ is a simple algebra. - Suppose I fix some $rho$ and take $A=e_{rho}R$. Then $A$ can be decomposed further as direct sum of simple left-modules
$$ A= bigopluslimits_{j=1}^f M_j, $$
- Consider now some $M_j$ as above and let $m in M_j$. Then, as $M_j subset A$, multiplication by $m$ seems to induce an $A$-endomorphism of $M_j$ and since $M_j$ is simple, this would imply by Schur's lemma that this map is an isomorphism.
Something seems off with the above reasoning, but I am unable to pinpoint where the error in the reasoning occurs. Any help or suggestions is greatly appreciated.
modules representation-theory
asked 2 days ago
mathfan230796
82
New contributor
mathfan230796 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
1
down vote
favorite
I am quite new to Representation Theory and I am trying to wrap my head around some basics. I have some difficulties with the following setup:
Suppose I start with a group ring $R=K[G]$, where $K$ is a field of characteristic $0$ (not necessarily algebraically closed) and $G$ is a finite group (not necessarily abelian).
- I can decompose $R$ as a direct sum
$$ R = bigoplus e_{rho} R,$$
where each $e_{rho}$ is a central idempotent and for each $rho$, $e_{rho}R$ is a simple algebra. - Suppose I fix some $rho$ and take $A=e_{rho}R$. Then $A$ can be decomposed further as direct sum of simple left-modules
$$ A= bigopluslimits_{j=1}^f M_j, $$
- Consider now some $M_j$ as above and let $m in M_j$. Then, as $M_j subset A$, multiplication by $m$ seems to induce an $A$-endomorphism of $M_j$ and since $M_j$ is simple, this would imply by Schur's lemma that this map is an isomorphism.
Something seems off with the above reasoning, but I am unable to pinpoint where the error in the reasoning occurs. Any help or suggestions is greatly appreciated.
modules representation-theory
asked 2 days ago
mathfan230796
82
New contributor
mathfan230796 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am quite new to Representation Theory and I am trying to wrap my head around some basics. I have some difficulties with the following setup:
Suppose I start with a group ring $R=K[G]$, where $K$ is a field of characteristic $0$ (not necessarily algebraically closed) and $G$ is a finite group (not necessarily abelian).
- I can decompose $R$ as a direct sum
$$ R = bigoplus e_{rho} R,$$
where each $e_{rho}$ is a central idempotent and for each $rho$, $e_{rho}R$ is a simple algebra. - Suppose I fix some $rho$ and take $A=e_{rho}R$. Then $A$ can be decomposed further as direct sum of simple left-modules
$$ A= bigopluslimits_{j=1}^f M_j, $$
- Consider now some $M_j$ as above and let $m in M_j$. Then, as $M_j subset A$, multiplication by $m$ seems to induce an $A$-endomorphism of $M_j$ and since $M_j$ is simple, this would imply by Schur's lemma that this map is an isomorphism.
Something seems off with the above reasoning, but I am unable to pinpoint where the error in the reasoning occurs. Any help or suggestions is greatly appreciated.
modules representation-theory
asked 2 days ago
mathfan230796
82
New contributor
mathfan230796 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I am quite new to Representation Theory and I am trying to wrap my head around some basics. I have some difficulties with the following setup:
Suppose I start with a group ring $R=K[G]$, where $K$ is a field of characteristic $0$ (not necessarily algebraically closed) and $G$ is a finite group (not necessarily abelian).
- I can decompose $R$ as a direct sum
$$ R = bigoplus e_{rho} R,$$
where each $e_{rho}$ is a central idempotent and for each $rho$, $e_{rho}R$ is a simple algebra. - Suppose I fix some $rho$ and take $A=e_{rho}R$. Then $A$ can be decomposed further as direct sum of simple left-modules
$$ A= bigopluslimits_{j=1}^f M_j, $$
- Consider now some $M_j$ as above and let $m in M_j$. Then, as $M_j subset A$, multiplication by $m$ seems to induce an $A$-endomorphism of $M_j$ and since $M_j$ is simple, this would imply by Schur's lemma that this map is an isomorphism.
Something seems off with the above reasoning, but I am unable to pinpoint where the error in the reasoning occurs. Any help or suggestions is greatly appreciated.
modules representation-theory
modules representation-theory
asked 2 days ago
mathfan230796
82
New contributor
mathfan230796 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
mathfan230796
82
New contributor
mathfan230796 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
mathfan230796
82
New contributor
mathfan230796 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
mathfan230796
82
asked 2 days ago
mathfan230796
82
82
New contributor
mathfan230796 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
mathfan230796 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
mathfan230796 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Multiplication by an element of $A$ is typically not an endomorphism of an $A$-module, if $A$ is not commutative. Indeed, suppose $M$ is a left $A$-module and $ain A$. For $f(x)=ax$ to be an endomorphism $Mto M$, we would need to have $f(bx)=bf(x)$ for all $bin A$, which means $a(bx)=b(ax)$. There is no reason for this to be true if $abneq ba$.
So, in your setup, left multiplication by $m$ is typically not an endomorphism of $M_j$.
On the other hand, though, right-multiplication by $m$ actually is an endomorphism of $M_j$ (it maps $M_j$ to itself since for any $xin M_j$, $xmin M_j$ because $min M_j$). Your argument is then (almost) correct and says that right multiplication by $m$ is an isomorphism on $M_j$ unless it is $0$ (don't forget the latter possibility!).
This might seem surprising but you can verify it quite explicitly. We can identify $A$ with a matrix ring $M_f(D)$ over some division ring $D$, with $M_j$ being the matrices whose only nonzero column is the $j$th column. You can then verify that if $x,min M_j$ then $xm=xd$ where $din D$ is the $(j,j)$ entry of the matrix $m$ (none of the other entries matter since $x$ has only one nonzero column). Since $D$ is a division ring, $d$ is either $0$ or a unit, and so $xmapsto xd$ is either $0$ or a bijection on $M_j$.
answered 2 days ago
Eric Wofsey
175k12202326
thank you very much for this! :)
– mathfan230796
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Multiplication by an element of $A$ is typically not an endomorphism of an $A$-module, if $A$ is not commutative. Indeed, suppose $M$ is a left $A$-module and $ain A$. For $f(x)=ax$ to be an endomorphism $Mto M$, we would need to have $f(bx)=bf(x)$ for all $bin A$, which means $a(bx)=b(ax)$. There is no reason for this to be true if $abneq ba$.
So, in your setup, left multiplication by $m$ is typically not an endomorphism of $M_j$.
On the other hand, though, right-multiplication by $m$ actually is an endomorphism of $M_j$ (it maps $M_j$ to itself since for any $xin M_j$, $xmin M_j$ because $min M_j$). Your argument is then (almost) correct and says that right multiplication by $m$ is an isomorphism on $M_j$ unless it is $0$ (don't forget the latter possibility!).
This might seem surprising but you can verify it quite explicitly. We can identify $A$ with a matrix ring $M_f(D)$ over some division ring $D$, with $M_j$ being the matrices whose only nonzero column is the $j$th column. You can then verify that if $x,min M_j$ then $xm=xd$ where $din D$ is the $(j,j)$ entry of the matrix $m$ (none of the other entries matter since $x$ has only one nonzero column). Since $D$ is a division ring, $d$ is either $0$ or a unit, and so $xmapsto xd$ is either $0$ or a bijection on $M_j$.
answered 2 days ago
Eric Wofsey
175k12202326
thank you very much for this! :)
– mathfan230796
2 days ago
add a comment |
up vote
2
down vote
accepted
Multiplication by an element of $A$ is typically not an endomorphism of an $A$-module, if $A$ is not commutative. Indeed, suppose $M$ is a left $A$-module and $ain A$. For $f(x)=ax$ to be an endomorphism $Mto M$, we would need to have $f(bx)=bf(x)$ for all $bin A$, which means $a(bx)=b(ax)$. There is no reason for this to be true if $abneq ba$.
So, in your setup, left multiplication by $m$ is typically not an endomorphism of $M_j$.
On the other hand, though, right-multiplication by $m$ actually is an endomorphism of $M_j$ (it maps $M_j$ to itself since for any $xin M_j$, $xmin M_j$ because $min M_j$). Your argument is then (almost) correct and says that right multiplication by $m$ is an isomorphism on $M_j$ unless it is $0$ (don't forget the latter possibility!).
This might seem surprising but you can verify it quite explicitly. We can identify $A$ with a matrix ring $M_f(D)$ over some division ring $D$, with $M_j$ being the matrices whose only nonzero column is the $j$th column. You can then verify that if $x,min M_j$ then $xm=xd$ where $din D$ is the $(j,j)$ entry of the matrix $m$ (none of the other entries matter since $x$ has only one nonzero column). Since $D$ is a division ring, $d$ is either $0$ or a unit, and so $xmapsto xd$ is either $0$ or a bijection on $M_j$.
answered 2 days ago
Eric Wofsey
175k12202326
thank you very much for this! :)
– mathfan230796
2 days ago
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Multiplication by an element of $A$ is typically not an endomorphism of an $A$-module, if $A$ is not commutative. Indeed, suppose $M$ is a left $A$-module and $ain A$. For $f(x)=ax$ to be an endomorphism $Mto M$, we would need to have $f(bx)=bf(x)$ for all $bin A$, which means $a(bx)=b(ax)$. There is no reason for this to be true if $abneq ba$.
So, in your setup, left multiplication by $m$ is typically not an endomorphism of $M_j$.
On the other hand, though, right-multiplication by $m$ actually is an endomorphism of $M_j$ (it maps $M_j$ to itself since for any $xin M_j$, $xmin M_j$ because $min M_j$). Your argument is then (almost) correct and says that right multiplication by $m$ is an isomorphism on $M_j$ unless it is $0$ (don't forget the latter possibility!).
This might seem surprising but you can verify it quite explicitly. We can identify $A$ with a matrix ring $M_f(D)$ over some division ring $D$, with $M_j$ being the matrices whose only nonzero column is the $j$th column. You can then verify that if $x,min M_j$ then $xm=xd$ where $din D$ is the $(j,j)$ entry of the matrix $m$ (none of the other entries matter since $x$ has only one nonzero column). Since $D$ is a division ring, $d$ is either $0$ or a unit, and so $xmapsto xd$ is either $0$ or a bijection on $M_j$.
answered 2 days ago
Eric Wofsey
175k12202326
Multiplication by an element of $A$ is typically not an endomorphism of an $A$-module, if $A$ is not commutative. Indeed, suppose $M$ is a left $A$-module and $ain A$. For $f(x)=ax$ to be an endomorphism $Mto M$, we would need to have $f(bx)=bf(x)$ for all $bin A$, which means $a(bx)=b(ax)$. There is no reason for this to be true if $abneq ba$.
So, in your setup, left multiplication by $m$ is typically not an endomorphism of $M_j$.
On the other hand, though, right-multiplication by $m$ actually is an endomorphism of $M_j$ (it maps $M_j$ to itself since for any $xin M_j$, $xmin M_j$ because $min M_j$). Your argument is then (almost) correct and says that right multiplication by $m$ is an isomorphism on $M_j$ unless it is $0$ (don't forget the latter possibility!).
This might seem surprising but you can verify it quite explicitly. We can identify $A$ with a matrix ring $M_f(D)$ over some division ring $D$, with $M_j$ being the matrices whose only nonzero column is the $j$th column. You can then verify that if $x,min M_j$ then $xm=xd$ where $din D$ is the $(j,j)$ entry of the matrix $m$ (none of the other entries matter since $x$ has only one nonzero column). Since $D$ is a division ring, $d$ is either $0$ or a unit, and so $xmapsto xd$ is either $0$ or a bijection on $M_j$.
answered 2 days ago
Eric Wofsey
175k12202326
edited 2 days ago
answered 2 days ago
Eric Wofsey
175k12202326
answered 2 days ago
Eric Wofsey
175k12202326
answered 2 days ago
Eric Wofsey
175k12202326
175k12202326
thank you very much for this! :)
– mathfan230796
2 days ago
add a comment |
thank you very much for this! :)
– mathfan230796
2 days ago
thank you very much for this! :)
– mathfan230796
2 days ago
thank you very much for this! :)
– mathfan230796
2 days ago
add a comment |
mathfan230796 is a new contributor. Be nice, and check out our Code of Conduct.
mathfan230796 is a new contributor. Be nice, and check out our Code of Conduct.
mathfan230796 is a new contributor. Be nice, and check out our Code of Conduct.
mathfan230796 is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005491%2fendomorphisms-of-irreducible-submodules-of-simple-algebras%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
