External tensor product of irreducible representations is not irreducible?Algebraicity of holomorphic...
External tensor product of irreducible representations is not irreducible?
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I'm writing up some notes, and I realize I don't have a counterexample for something I suspect is false.
Dubious claim: If $(pi, V)$ and $(rho, W)$ are irreducible representations of two groups $G$ and $H$, respectively, then the "external" tensor product $pi boxtimes rho$ is an irreducible representation of $G times H$.
Of course this is true and well-known in the usual cases, e.g., when $G$ and $H$ are finite groups. The proof I know uses the converse to Schur's Lemma, or something similar.
Is there a nice counterexample for complex representations of some infinite groups? Published somewhere?
rt.representation-theory counterexamples
asked 2 days ago
MartyMarty
9,8382 gold badges36 silver badges74 bronze badges
$endgroup$
add a comment |
$begingroup$
I'm writing up some notes, and I realize I don't have a counterexample for something I suspect is false.
Dubious claim: If $(pi, V)$ and $(rho, W)$ are irreducible representations of two groups $G$ and $H$, respectively, then the "external" tensor product $pi boxtimes rho$ is an irreducible representation of $G times H$.
Of course this is true and well-known in the usual cases, e.g., when $G$ and $H$ are finite groups. The proof I know uses the converse to Schur's Lemma, or something similar.
Is there a nice counterexample for complex representations of some infinite groups? Published somewhere?
rt.representation-theory counterexamples
asked 2 days ago
MartyMarty
9,8382 gold badges36 silver badges74 bronze badges
$endgroup$
add a comment |
$begingroup$
I'm writing up some notes, and I realize I don't have a counterexample for something I suspect is false.
Dubious claim: If $(pi, V)$ and $(rho, W)$ are irreducible representations of two groups $G$ and $H$, respectively, then the "external" tensor product $pi boxtimes rho$ is an irreducible representation of $G times H$.
Of course this is true and well-known in the usual cases, e.g., when $G$ and $H$ are finite groups. The proof I know uses the converse to Schur's Lemma, or something similar.
Is there a nice counterexample for complex representations of some infinite groups? Published somewhere?
rt.representation-theory counterexamples
asked 2 days ago
MartyMarty
9,8382 gold badges36 silver badges74 bronze badges
$endgroup$
I'm writing up some notes, and I realize I don't have a counterexample for something I suspect is false.
Dubious claim: If $(pi, V)$ and $(rho, W)$ are irreducible representations of two groups $G$ and $H$, respectively, then the "external" tensor product $pi boxtimes rho$ is an irreducible representation of $G times H$.
Of course this is true and well-known in the usual cases, e.g., when $G$ and $H$ are finite groups. The proof I know uses the converse to Schur's Lemma, or something similar.
Is there a nice counterexample for complex representations of some infinite groups? Published somewhere?
rt.representation-theory counterexamples
rt.representation-theory counterexamples
asked 2 days ago
MartyMarty
9,8382 gold badges36 silver badges74 bronze badges
asked 2 days ago
MartyMarty
9,8382 gold badges36 silver badges74 bronze badges
asked 2 days ago
MartyMarty
9,8382 gold badges36 silver badges74 bronze badges
asked 2 days ago
MartyMarty
9,8382 gold badges36 silver badges74 bronze badges
asked 2 days ago
MartyMarty
9,8382 gold badges36 silver badges74 bronze badges
9,8382 gold badges36 silver badges74 bronze badges
add a comment |
add a comment |
1 Answer
1
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You can generate examples from standard counterexamples to (generalisations of) Schur's lemma.
Let E/F be a field extension. Let $G=H=E^times$, acting on the F-vector space E. Then the external tensor product is not irreducible, for example the kernel of the multiplication map $Eotimes_F Eto E$ is a submodule.
answered 2 days ago
Peter McNamaraPeter McNamara
5,82129 silver badges57 bronze badges
$endgroup$
$begingroup$
I think this works. So in the context of complex representations, let $G = H = {mathbb C}(T)^times$ acting on the complex vector space $V = {mathbb C}(T)$. It's kind of interesting to me to see if there's a countable-dimension example (over the complex numbers), but I won't move the goalposts here!
$endgroup$
– Marty
yesterday
2
$begingroup$
There won't be an example of countable dimension over C. Over any field, an irrep always has endomorphism ring a division algebra. There are no division algebras of countable dimension over C. Thus for countable dimension, you have Schur's Lemma so your proof of irreducibility of the tensor product should go through.
$endgroup$
– Peter McNamara
yesterday
$begingroup$
Well, I know the first part well. But I use the converse of Schur's Lemma to prove irreducibility of the tensor product. One can prove $End_{G times H}(pi boxtimes rho$ is ${mathbb C}$ using Schur's Lemma... but then what? Or am I missing something easy here?
$endgroup$
– Marty
yesterday
1
$begingroup$
If your representations (one is enough) are finite-dimensional over a ground field it works in the same conditions as for finite groups (free is the field is algebraically closed, needs extra assumptions else). Anyway, as PeterMcNamara's answer tells, it is much more a question of field theory rather than a one of representation theory.
$endgroup$
– Aurélien Djament
yesterday
$begingroup$
@Marty, you can use Jacobson density theorem on V and W to prove Votimes W is irreducible (this approach also has the benefit of working in categories which are not semisimple). Write to me if you want more details.
$endgroup$
– Peter McNamara
yesterday
|
show 1 more comment
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
You can generate examples from standard counterexamples to (generalisations of) Schur's lemma.
Let E/F be a field extension. Let $G=H=E^times$, acting on the F-vector space E. Then the external tensor product is not irreducible, for example the kernel of the multiplication map $Eotimes_F Eto E$ is a submodule.
answered 2 days ago
Peter McNamaraPeter McNamara
5,82129 silver badges57 bronze badges
$endgroup$
$begingroup$
I think this works. So in the context of complex representations, let $G = H = {mathbb C}(T)^times$ acting on the complex vector space $V = {mathbb C}(T)$. It's kind of interesting to me to see if there's a countable-dimension example (over the complex numbers), but I won't move the goalposts here!
$endgroup$
– Marty
yesterday
2
$begingroup$
There won't be an example of countable dimension over C. Over any field, an irrep always has endomorphism ring a division algebra. There are no division algebras of countable dimension over C. Thus for countable dimension, you have Schur's Lemma so your proof of irreducibility of the tensor product should go through.
$endgroup$
– Peter McNamara
yesterday
$begingroup$
Well, I know the first part well. But I use the converse of Schur's Lemma to prove irreducibility of the tensor product. One can prove $End_{G times H}(pi boxtimes rho$ is ${mathbb C}$ using Schur's Lemma... but then what? Or am I missing something easy here?
$endgroup$
– Marty
yesterday
1
$begingroup$
If your representations (one is enough) are finite-dimensional over a ground field it works in the same conditions as for finite groups (free is the field is algebraically closed, needs extra assumptions else). Anyway, as PeterMcNamara's answer tells, it is much more a question of field theory rather than a one of representation theory.
$endgroup$
– Aurélien Djament
yesterday
$begingroup$
@Marty, you can use Jacobson density theorem on V and W to prove Votimes W is irreducible (this approach also has the benefit of working in categories which are not semisimple). Write to me if you want more details.
$endgroup$
– Peter McNamara
yesterday
|
show 1 more comment
$begingroup$
You can generate examples from standard counterexamples to (generalisations of) Schur's lemma.
Let E/F be a field extension. Let $G=H=E^times$, acting on the F-vector space E. Then the external tensor product is not irreducible, for example the kernel of the multiplication map $Eotimes_F Eto E$ is a submodule.
answered 2 days ago
Peter McNamaraPeter McNamara
5,82129 silver badges57 bronze badges
$endgroup$
$begingroup$
I think this works. So in the context of complex representations, let $G = H = {mathbb C}(T)^times$ acting on the complex vector space $V = {mathbb C}(T)$. It's kind of interesting to me to see if there's a countable-dimension example (over the complex numbers), but I won't move the goalposts here!
$endgroup$
– Marty
yesterday
2
$begingroup$
There won't be an example of countable dimension over C. Over any field, an irrep always has endomorphism ring a division algebra. There are no division algebras of countable dimension over C. Thus for countable dimension, you have Schur's Lemma so your proof of irreducibility of the tensor product should go through.
$endgroup$
– Peter McNamara
yesterday
$begingroup$
Well, I know the first part well. But I use the converse of Schur's Lemma to prove irreducibility of the tensor product. One can prove $End_{G times H}(pi boxtimes rho$ is ${mathbb C}$ using Schur's Lemma... but then what? Or am I missing something easy here?
$endgroup$
– Marty
yesterday
1
$begingroup$
If your representations (one is enough) are finite-dimensional over a ground field it works in the same conditions as for finite groups (free is the field is algebraically closed, needs extra assumptions else). Anyway, as PeterMcNamara's answer tells, it is much more a question of field theory rather than a one of representation theory.
$endgroup$
– Aurélien Djament
yesterday
$begingroup$
@Marty, you can use Jacobson density theorem on V and W to prove Votimes W is irreducible (this approach also has the benefit of working in categories which are not semisimple). Write to me if you want more details.
$endgroup$
– Peter McNamara
yesterday
|
show 1 more comment
$begingroup$
You can generate examples from standard counterexamples to (generalisations of) Schur's lemma.
Let E/F be a field extension. Let $G=H=E^times$, acting on the F-vector space E. Then the external tensor product is not irreducible, for example the kernel of the multiplication map $Eotimes_F Eto E$ is a submodule.
answered 2 days ago
Peter McNamaraPeter McNamara
5,82129 silver badges57 bronze badges
$endgroup$
You can generate examples from standard counterexamples to (generalisations of) Schur's lemma.
Let E/F be a field extension. Let $G=H=E^times$, acting on the F-vector space E. Then the external tensor product is not irreducible, for example the kernel of the multiplication map $Eotimes_F Eto E$ is a submodule.
answered 2 days ago
Peter McNamaraPeter McNamara
5,82129 silver badges57 bronze badges
answered 2 days ago
Peter McNamaraPeter McNamara
5,82129 silver badges57 bronze badges
answered 2 days ago
Peter McNamaraPeter McNamara
5,82129 silver badges57 bronze badges
answered 2 days ago
Peter McNamaraPeter McNamara
5,82129 silver badges57 bronze badges
5,82129 silver badges57 bronze badges
$begingroup$
I think this works. So in the context of complex representations, let $G = H = {mathbb C}(T)^times$ acting on the complex vector space $V = {mathbb C}(T)$. It's kind of interesting to me to see if there's a countable-dimension example (over the complex numbers), but I won't move the goalposts here!
$endgroup$
– Marty
yesterday
2
$begingroup$
There won't be an example of countable dimension over C. Over any field, an irrep always has endomorphism ring a division algebra. There are no division algebras of countable dimension over C. Thus for countable dimension, you have Schur's Lemma so your proof of irreducibility of the tensor product should go through.
$endgroup$
– Peter McNamara
yesterday
$begingroup$
Well, I know the first part well. But I use the converse of Schur's Lemma to prove irreducibility of the tensor product. One can prove $End_{G times H}(pi boxtimes rho$ is ${mathbb C}$ using Schur's Lemma... but then what? Or am I missing something easy here?
$endgroup$
– Marty
yesterday
1
$begingroup$
If your representations (one is enough) are finite-dimensional over a ground field it works in the same conditions as for finite groups (free is the field is algebraically closed, needs extra assumptions else). Anyway, as PeterMcNamara's answer tells, it is much more a question of field theory rather than a one of representation theory.
$endgroup$
– Aurélien Djament
yesterday
$begingroup$
@Marty, you can use Jacobson density theorem on V and W to prove Votimes W is irreducible (this approach also has the benefit of working in categories which are not semisimple). Write to me if you want more details.
$endgroup$
– Peter McNamara
yesterday
|
show 1 more comment
$begingroup$
I think this works. So in the context of complex representations, let $G = H = {mathbb C}(T)^times$ acting on the complex vector space $V = {mathbb C}(T)$. It's kind of interesting to me to see if there's a countable-dimension example (over the complex numbers), but I won't move the goalposts here!
$endgroup$
– Marty
yesterday
2
$begingroup$
There won't be an example of countable dimension over C. Over any field, an irrep always has endomorphism ring a division algebra. There are no division algebras of countable dimension over C. Thus for countable dimension, you have Schur's Lemma so your proof of irreducibility of the tensor product should go through.
$endgroup$
– Peter McNamara
yesterday
$begingroup$
Well, I know the first part well. But I use the converse of Schur's Lemma to prove irreducibility of the tensor product. One can prove $End_{G times H}(pi boxtimes rho$ is ${mathbb C}$ using Schur's Lemma... but then what? Or am I missing something easy here?
$endgroup$
– Marty
yesterday
1
$begingroup$
If your representations (one is enough) are finite-dimensional over a ground field it works in the same conditions as for finite groups (free is the field is algebraically closed, needs extra assumptions else). Anyway, as PeterMcNamara's answer tells, it is much more a question of field theory rather than a one of representation theory.
$endgroup$
– Aurélien Djament
yesterday
$begingroup$
@Marty, you can use Jacobson density theorem on V and W to prove Votimes W is irreducible (this approach also has the benefit of working in categories which are not semisimple). Write to me if you want more details.
$endgroup$
– Peter McNamara
yesterday
$begingroup$
I think this works. So in the context of complex representations, let $G = H = {mathbb C}(T)^times$ acting on the complex vector space $V = {mathbb C}(T)$. It's kind of interesting to me to see if there's a countable-dimension example (over the complex numbers), but I won't move the goalposts here!
$endgroup$
– Marty
yesterday
$begingroup$
I think this works. So in the context of complex representations, let $G = H = {mathbb C}(T)^times$ acting on the complex vector space $V = {mathbb C}(T)$. It's kind of interesting to me to see if there's a countable-dimension example (over the complex numbers), but I won't move the goalposts here!
$endgroup$
– Marty
yesterday
2
2
$begingroup$
There won't be an example of countable dimension over C. Over any field, an irrep always has endomorphism ring a division algebra. There are no division algebras of countable dimension over C. Thus for countable dimension, you have Schur's Lemma so your proof of irreducibility of the tensor product should go through.
$endgroup$
– Peter McNamara
yesterday
$begingroup$
There won't be an example of countable dimension over C. Over any field, an irrep always has endomorphism ring a division algebra. There are no division algebras of countable dimension over C. Thus for countable dimension, you have Schur's Lemma so your proof of irreducibility of the tensor product should go through.
$endgroup$
– Peter McNamara
yesterday
$begingroup$
Well, I know the first part well. But I use the converse of Schur's Lemma to prove irreducibility of the tensor product. One can prove $End_{G times H}(pi boxtimes rho$ is ${mathbb C}$ using Schur's Lemma... but then what? Or am I missing something easy here?
$endgroup$
– Marty
yesterday
$begingroup$
Well, I know the first part well. But I use the converse of Schur's Lemma to prove irreducibility of the tensor product. One can prove $End_{G times H}(pi boxtimes rho$ is ${mathbb C}$ using Schur's Lemma... but then what? Or am I missing something easy here?
$endgroup$
– Marty
yesterday
1
1
$begingroup$
If your representations (one is enough) are finite-dimensional over a ground field it works in the same conditions as for finite groups (free is the field is algebraically closed, needs extra assumptions else). Anyway, as PeterMcNamara's answer tells, it is much more a question of field theory rather than a one of representation theory.
$endgroup$
– Aurélien Djament
yesterday
$begingroup$
If your representations (one is enough) are finite-dimensional over a ground field it works in the same conditions as for finite groups (free is the field is algebraically closed, needs extra assumptions else). Anyway, as PeterMcNamara's answer tells, it is much more a question of field theory rather than a one of representation theory.
$endgroup$
– Aurélien Djament
yesterday
$begingroup$
@Marty, you can use Jacobson density theorem on V and W to prove Votimes W is irreducible (this approach also has the benefit of working in categories which are not semisimple). Write to me if you want more details.
$endgroup$
– Peter McNamara
yesterday
$begingroup$
@Marty, you can use Jacobson density theorem on V and W to prove Votimes W is irreducible (this approach also has the benefit of working in categories which are not semisimple). Write to me if you want more details.
$endgroup$
– Peter McNamara
yesterday
|
show 1 more comment
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