Unitary representations of finite groups over finite fields
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I would like to learn the basic theory of unitary representations of finite groups over finite fields.
Here, the unitary group $operatorname{GU}(n,mathbb{F}_{q^2})$ consists of all invertible transformations of $mathbb{F}_{q^2}^n$ that preserve the Hermitian form $langle x, y rangle = sum_{i in [n]} x_i y_i^q$, and "unitary representation" means a group homomorphism $rho colon G to operatorname{GU}(n,mathbb{F}_{q^2})$.
This is a special case of the usual notion of a representation $rho colon G to operatorname{GL}(n,mathbb{F}_{q^2})$.
Over the complex numbers, every representation $rho colon G to operatorname{GL}(n,mathbb{C})$ of a finite group $G$ is similar to a unitary representation $rho' colon G to operatorname{GU}(n,mathbb{C})$, in the sense that there is an invertible operator $M$ such that $rho'(g) = Mrho(g) M^{-1}$ for every $g in G$.
In this sense and others, the theory of unitary representations over $mathbb{C}$ is essentially the same as that of ordinary representations.
However, over finite fields the notions are distinct.
If $G$ is a finite group and $rho colon G to operatorname{GL}(n,mathbb{F}_{q^2})$ is a representation, there might not be an invertible operator $M$ such that $M rho(g) M^{-1} in operatorname{GU}(n,mathbb{F}_{q^2})$ for every $g in G$.
For example, $mathbb{Z}_5$ has a faithful 2-dimensional representation over $mathbb{F}_{3^2}$ that is not similar to any unitary representation, since 5 divides $|operatorname{GL}(2,mathbb{F}_{3^2})|$ but not $|operatorname{GU}(2,mathbb{F}_{3^2})|$.
Question:
Have unitary representations of finite groups over finite fields been systematically studied, and if so where can I learn the basics?
Here is one example of what I want to learn to do:
- Describe all the unitary representations of the dihedral group of order 8 when $q=11$.
At the moment I do not even know how to:
- Describe all the unitary representations of $mathbb{Z}_2 times mathbb{Z}_2$ when $q=3$.
Some other things I want to learn include:
Where Maschke's Theorem holds (i.e. $(|G|,q) = 1$ so that $mathbb{F}_{q^2}[G]$ is semisimple), does every unitary representation decompose as an orthogonal direct sum of irreducible unitary subrepresentations?
Again where Maschke's Theorem holds, is there any analogue of the Peter-Weyl Theorem to give the space $L^2(G)$ of functions $f colon G to mathbb{F}_{q^2}$ an orthogonal basis consisting of matrix elements for irreducible unitary representations?
What are some conditions for a modular representation to be similar to a unitary representation? (i.e. which subgroups of $operatorname{GL}(n,mathbb{F}_{q^2})$ are conjugate with subgroups of $operatorname{GU}(n,mathbb{F}_{q^2})$?)
Bonus for answers understandable to a humble analyst.
reference-request gr.group-theory rt.representation-theory finite-groups harmonic-analysis
New contributor
Joey Iverson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 3 more comments
$begingroup$
I would like to learn the basic theory of unitary representations of finite groups over finite fields.
Here, the unitary group $operatorname{GU}(n,mathbb{F}_{q^2})$ consists of all invertible transformations of $mathbb{F}_{q^2}^n$ that preserve the Hermitian form $langle x, y rangle = sum_{i in [n]} x_i y_i^q$, and "unitary representation" means a group homomorphism $rho colon G to operatorname{GU}(n,mathbb{F}_{q^2})$.
This is a special case of the usual notion of a representation $rho colon G to operatorname{GL}(n,mathbb{F}_{q^2})$.
Over the complex numbers, every representation $rho colon G to operatorname{GL}(n,mathbb{C})$ of a finite group $G$ is similar to a unitary representation $rho' colon G to operatorname{GU}(n,mathbb{C})$, in the sense that there is an invertible operator $M$ such that $rho'(g) = Mrho(g) M^{-1}$ for every $g in G$.
In this sense and others, the theory of unitary representations over $mathbb{C}$ is essentially the same as that of ordinary representations.
However, over finite fields the notions are distinct.
If $G$ is a finite group and $rho colon G to operatorname{GL}(n,mathbb{F}_{q^2})$ is a representation, there might not be an invertible operator $M$ such that $M rho(g) M^{-1} in operatorname{GU}(n,mathbb{F}_{q^2})$ for every $g in G$.
For example, $mathbb{Z}_5$ has a faithful 2-dimensional representation over $mathbb{F}_{3^2}$ that is not similar to any unitary representation, since 5 divides $|operatorname{GL}(2,mathbb{F}_{3^2})|$ but not $|operatorname{GU}(2,mathbb{F}_{3^2})|$.
Question:
Have unitary representations of finite groups over finite fields been systematically studied, and if so where can I learn the basics?
Here is one example of what I want to learn to do:
- Describe all the unitary representations of the dihedral group of order 8 when $q=11$.
At the moment I do not even know how to:
- Describe all the unitary representations of $mathbb{Z}_2 times mathbb{Z}_2$ when $q=3$.
Some other things I want to learn include:
Where Maschke's Theorem holds (i.e. $(|G|,q) = 1$ so that $mathbb{F}_{q^2}[G]$ is semisimple), does every unitary representation decompose as an orthogonal direct sum of irreducible unitary subrepresentations?
Again where Maschke's Theorem holds, is there any analogue of the Peter-Weyl Theorem to give the space $L^2(G)$ of functions $f colon G to mathbb{F}_{q^2}$ an orthogonal basis consisting of matrix elements for irreducible unitary representations?
What are some conditions for a modular representation to be similar to a unitary representation? (i.e. which subgroups of $operatorname{GL}(n,mathbb{F}_{q^2})$ are conjugate with subgroups of $operatorname{GU}(n,mathbb{F}_{q^2})$?)
Bonus for answers understandable to a humble analyst.
reference-request gr.group-theory rt.representation-theory finite-groups harmonic-analysis
New contributor
Joey Iverson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Notice that finite-dimensional unitary representations are automatically semisimple; a minimal stable non-$0$ subspace is irreducible, and its orthogonal complement has smaller dimension. A similar argument shows that every semisimple unitary representation is an orthogonal direct sum of irreducibles.
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– LSpice
Apr 11 at 22:14
1
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@LSpice I see we need to be careful about what "irreducible" means for a unitary representation. Probably the right thing is to talk about unitary representations not just on $mathbb{F}_{q^2}^n$ but on general unitary spaces, i.e. vector spaces over $mathbb{F}_{q^2}$ equipped with nondegenerate sesquilinear forms. Then a "unitary subrepresentation" should mean an invariant subspace on which the form restricts nondegenerately (e.g. isotropic subspaces are out), and an "irreducible unitary representation" has no proper unitary subrepresentations.
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– Joey Iverson
Apr 12 at 23:39
1
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@LSpice I guess the orthogonal complement of a unitary subspace is again unitary, and that they make an algebraic direct sum together. Then your argument shows that every unitary representation is the orthogonal direct sum of irreducible unitary subrepresentations. Great!
$endgroup$
– Joey Iverson
Apr 12 at 23:41
1
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@LSpice One caveat is that an "irreducible unitary representation" is not necessarily irreducible as a $G$-module. For example, choose $G$ and $q$ so that $mathbb{F}_{q^2}[G]$ is not semisimple. The regular representation is unitary when we identify $mathbb{F}_{q^2}[G]$ with what I want to call $L^2(G)$. As a unitary representation it decomposes as a direct sum of irreducibles, but as a $G$-module it does not. Hence some of the irreducible unitary representations in the decomposition of $L^2(G)$ are not irreducible as $G$-modules.
$endgroup$
– Joey Iverson
Apr 12 at 23:42
1
$begingroup$
@LSpice If we use this definition of irreducibility, then the embedding of $V otimes V^*$ into $L^2(G)$ that you mentioned might not be injective. For instance, if $W subset V$ is an invariant subspace on which the form is degenerate, then we can choose $u in W setminus {0}$ and $v in (W cap W^perp) setminus {0}$, and the map you described sends $u otimes v^* mapsto 0$ (if I understand you correctly) since $langle rho(g) u, v rangle = 0$ for every $g in G$.
$endgroup$
– Joey Iverson
Apr 12 at 23:43
|
show 3 more comments
$begingroup$
I would like to learn the basic theory of unitary representations of finite groups over finite fields.
Here, the unitary group $operatorname{GU}(n,mathbb{F}_{q^2})$ consists of all invertible transformations of $mathbb{F}_{q^2}^n$ that preserve the Hermitian form $langle x, y rangle = sum_{i in [n]} x_i y_i^q$, and "unitary representation" means a group homomorphism $rho colon G to operatorname{GU}(n,mathbb{F}_{q^2})$.
This is a special case of the usual notion of a representation $rho colon G to operatorname{GL}(n,mathbb{F}_{q^2})$.
Over the complex numbers, every representation $rho colon G to operatorname{GL}(n,mathbb{C})$ of a finite group $G$ is similar to a unitary representation $rho' colon G to operatorname{GU}(n,mathbb{C})$, in the sense that there is an invertible operator $M$ such that $rho'(g) = Mrho(g) M^{-1}$ for every $g in G$.
In this sense and others, the theory of unitary representations over $mathbb{C}$ is essentially the same as that of ordinary representations.
However, over finite fields the notions are distinct.
If $G$ is a finite group and $rho colon G to operatorname{GL}(n,mathbb{F}_{q^2})$ is a representation, there might not be an invertible operator $M$ such that $M rho(g) M^{-1} in operatorname{GU}(n,mathbb{F}_{q^2})$ for every $g in G$.
For example, $mathbb{Z}_5$ has a faithful 2-dimensional representation over $mathbb{F}_{3^2}$ that is not similar to any unitary representation, since 5 divides $|operatorname{GL}(2,mathbb{F}_{3^2})|$ but not $|operatorname{GU}(2,mathbb{F}_{3^2})|$.
Question:
Have unitary representations of finite groups over finite fields been systematically studied, and if so where can I learn the basics?
Here is one example of what I want to learn to do:
- Describe all the unitary representations of the dihedral group of order 8 when $q=11$.
At the moment I do not even know how to:
- Describe all the unitary representations of $mathbb{Z}_2 times mathbb{Z}_2$ when $q=3$.
Some other things I want to learn include:
Where Maschke's Theorem holds (i.e. $(|G|,q) = 1$ so that $mathbb{F}_{q^2}[G]$ is semisimple), does every unitary representation decompose as an orthogonal direct sum of irreducible unitary subrepresentations?
Again where Maschke's Theorem holds, is there any analogue of the Peter-Weyl Theorem to give the space $L^2(G)$ of functions $f colon G to mathbb{F}_{q^2}$ an orthogonal basis consisting of matrix elements for irreducible unitary representations?
What are some conditions for a modular representation to be similar to a unitary representation? (i.e. which subgroups of $operatorname{GL}(n,mathbb{F}_{q^2})$ are conjugate with subgroups of $operatorname{GU}(n,mathbb{F}_{q^2})$?)
Bonus for answers understandable to a humble analyst.
reference-request gr.group-theory rt.representation-theory finite-groups harmonic-analysis
New contributor
Joey Iverson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I would like to learn the basic theory of unitary representations of finite groups over finite fields.
Here, the unitary group $operatorname{GU}(n,mathbb{F}_{q^2})$ consists of all invertible transformations of $mathbb{F}_{q^2}^n$ that preserve the Hermitian form $langle x, y rangle = sum_{i in [n]} x_i y_i^q$, and "unitary representation" means a group homomorphism $rho colon G to operatorname{GU}(n,mathbb{F}_{q^2})$.
This is a special case of the usual notion of a representation $rho colon G to operatorname{GL}(n,mathbb{F}_{q^2})$.
Over the complex numbers, every representation $rho colon G to operatorname{GL}(n,mathbb{C})$ of a finite group $G$ is similar to a unitary representation $rho' colon G to operatorname{GU}(n,mathbb{C})$, in the sense that there is an invertible operator $M$ such that $rho'(g) = Mrho(g) M^{-1}$ for every $g in G$.
In this sense and others, the theory of unitary representations over $mathbb{C}$ is essentially the same as that of ordinary representations.
However, over finite fields the notions are distinct.
If $G$ is a finite group and $rho colon G to operatorname{GL}(n,mathbb{F}_{q^2})$ is a representation, there might not be an invertible operator $M$ such that $M rho(g) M^{-1} in operatorname{GU}(n,mathbb{F}_{q^2})$ for every $g in G$.
For example, $mathbb{Z}_5$ has a faithful 2-dimensional representation over $mathbb{F}_{3^2}$ that is not similar to any unitary representation, since 5 divides $|operatorname{GL}(2,mathbb{F}_{3^2})|$ but not $|operatorname{GU}(2,mathbb{F}_{3^2})|$.
Question:
Have unitary representations of finite groups over finite fields been systematically studied, and if so where can I learn the basics?
Here is one example of what I want to learn to do:
- Describe all the unitary representations of the dihedral group of order 8 when $q=11$.
At the moment I do not even know how to:
- Describe all the unitary representations of $mathbb{Z}_2 times mathbb{Z}_2$ when $q=3$.
Some other things I want to learn include:
Where Maschke's Theorem holds (i.e. $(|G|,q) = 1$ so that $mathbb{F}_{q^2}[G]$ is semisimple), does every unitary representation decompose as an orthogonal direct sum of irreducible unitary subrepresentations?
Again where Maschke's Theorem holds, is there any analogue of the Peter-Weyl Theorem to give the space $L^2(G)$ of functions $f colon G to mathbb{F}_{q^2}$ an orthogonal basis consisting of matrix elements for irreducible unitary representations?
What are some conditions for a modular representation to be similar to a unitary representation? (i.e. which subgroups of $operatorname{GL}(n,mathbb{F}_{q^2})$ are conjugate with subgroups of $operatorname{GU}(n,mathbb{F}_{q^2})$?)
Bonus for answers understandable to a humble analyst.
reference-request gr.group-theory rt.representation-theory finite-groups harmonic-analysis
reference-request gr.group-theory rt.representation-theory finite-groups harmonic-analysis
New contributor
Joey Iverson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Joey Iverson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 11 at 22:16
YCor
29.1k486141
29.1k486141
New contributor
Joey Iverson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Apr 11 at 22:08
Joey IversonJoey Iverson
462
462
New contributor
Joey Iverson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Joey Iverson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Joey Iverson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Notice that finite-dimensional unitary representations are automatically semisimple; a minimal stable non-$0$ subspace is irreducible, and its orthogonal complement has smaller dimension. A similar argument shows that every semisimple unitary representation is an orthogonal direct sum of irreducibles.
$endgroup$
– LSpice
Apr 11 at 22:14
1
$begingroup$
@LSpice I see we need to be careful about what "irreducible" means for a unitary representation. Probably the right thing is to talk about unitary representations not just on $mathbb{F}_{q^2}^n$ but on general unitary spaces, i.e. vector spaces over $mathbb{F}_{q^2}$ equipped with nondegenerate sesquilinear forms. Then a "unitary subrepresentation" should mean an invariant subspace on which the form restricts nondegenerately (e.g. isotropic subspaces are out), and an "irreducible unitary representation" has no proper unitary subrepresentations.
$endgroup$
– Joey Iverson
Apr 12 at 23:39
1
$begingroup$
@LSpice I guess the orthogonal complement of a unitary subspace is again unitary, and that they make an algebraic direct sum together. Then your argument shows that every unitary representation is the orthogonal direct sum of irreducible unitary subrepresentations. Great!
$endgroup$
– Joey Iverson
Apr 12 at 23:41
1
$begingroup$
@LSpice One caveat is that an "irreducible unitary representation" is not necessarily irreducible as a $G$-module. For example, choose $G$ and $q$ so that $mathbb{F}_{q^2}[G]$ is not semisimple. The regular representation is unitary when we identify $mathbb{F}_{q^2}[G]$ with what I want to call $L^2(G)$. As a unitary representation it decomposes as a direct sum of irreducibles, but as a $G$-module it does not. Hence some of the irreducible unitary representations in the decomposition of $L^2(G)$ are not irreducible as $G$-modules.
$endgroup$
– Joey Iverson
Apr 12 at 23:42
1
$begingroup$
@LSpice If we use this definition of irreducibility, then the embedding of $V otimes V^*$ into $L^2(G)$ that you mentioned might not be injective. For instance, if $W subset V$ is an invariant subspace on which the form is degenerate, then we can choose $u in W setminus {0}$ and $v in (W cap W^perp) setminus {0}$, and the map you described sends $u otimes v^* mapsto 0$ (if I understand you correctly) since $langle rho(g) u, v rangle = 0$ for every $g in G$.
$endgroup$
– Joey Iverson
Apr 12 at 23:43
|
show 3 more comments
$begingroup$
Notice that finite-dimensional unitary representations are automatically semisimple; a minimal stable non-$0$ subspace is irreducible, and its orthogonal complement has smaller dimension. A similar argument shows that every semisimple unitary representation is an orthogonal direct sum of irreducibles.
$endgroup$
– LSpice
Apr 11 at 22:14
1
$begingroup$
@LSpice I see we need to be careful about what "irreducible" means for a unitary representation. Probably the right thing is to talk about unitary representations not just on $mathbb{F}_{q^2}^n$ but on general unitary spaces, i.e. vector spaces over $mathbb{F}_{q^2}$ equipped with nondegenerate sesquilinear forms. Then a "unitary subrepresentation" should mean an invariant subspace on which the form restricts nondegenerately (e.g. isotropic subspaces are out), and an "irreducible unitary representation" has no proper unitary subrepresentations.
$endgroup$
– Joey Iverson
Apr 12 at 23:39
1
$begingroup$
@LSpice I guess the orthogonal complement of a unitary subspace is again unitary, and that they make an algebraic direct sum together. Then your argument shows that every unitary representation is the orthogonal direct sum of irreducible unitary subrepresentations. Great!
$endgroup$
– Joey Iverson
Apr 12 at 23:41
1
$begingroup$
@LSpice One caveat is that an "irreducible unitary representation" is not necessarily irreducible as a $G$-module. For example, choose $G$ and $q$ so that $mathbb{F}_{q^2}[G]$ is not semisimple. The regular representation is unitary when we identify $mathbb{F}_{q^2}[G]$ with what I want to call $L^2(G)$. As a unitary representation it decomposes as a direct sum of irreducibles, but as a $G$-module it does not. Hence some of the irreducible unitary representations in the decomposition of $L^2(G)$ are not irreducible as $G$-modules.
$endgroup$
– Joey Iverson
Apr 12 at 23:42
1
$begingroup$
@LSpice If we use this definition of irreducibility, then the embedding of $V otimes V^*$ into $L^2(G)$ that you mentioned might not be injective. For instance, if $W subset V$ is an invariant subspace on which the form is degenerate, then we can choose $u in W setminus {0}$ and $v in (W cap W^perp) setminus {0}$, and the map you described sends $u otimes v^* mapsto 0$ (if I understand you correctly) since $langle rho(g) u, v rangle = 0$ for every $g in G$.
$endgroup$
– Joey Iverson
Apr 12 at 23:43
$begingroup$
Notice that finite-dimensional unitary representations are automatically semisimple; a minimal stable non-$0$ subspace is irreducible, and its orthogonal complement has smaller dimension. A similar argument shows that every semisimple unitary representation is an orthogonal direct sum of irreducibles.
$endgroup$
– LSpice
Apr 11 at 22:14
$begingroup$
Notice that finite-dimensional unitary representations are automatically semisimple; a minimal stable non-$0$ subspace is irreducible, and its orthogonal complement has smaller dimension. A similar argument shows that every semisimple unitary representation is an orthogonal direct sum of irreducibles.
$endgroup$
– LSpice
Apr 11 at 22:14
1
1
$begingroup$
@LSpice I see we need to be careful about what "irreducible" means for a unitary representation. Probably the right thing is to talk about unitary representations not just on $mathbb{F}_{q^2}^n$ but on general unitary spaces, i.e. vector spaces over $mathbb{F}_{q^2}$ equipped with nondegenerate sesquilinear forms. Then a "unitary subrepresentation" should mean an invariant subspace on which the form restricts nondegenerately (e.g. isotropic subspaces are out), and an "irreducible unitary representation" has no proper unitary subrepresentations.
$endgroup$
– Joey Iverson
Apr 12 at 23:39
$begingroup$
@LSpice I see we need to be careful about what "irreducible" means for a unitary representation. Probably the right thing is to talk about unitary representations not just on $mathbb{F}_{q^2}^n$ but on general unitary spaces, i.e. vector spaces over $mathbb{F}_{q^2}$ equipped with nondegenerate sesquilinear forms. Then a "unitary subrepresentation" should mean an invariant subspace on which the form restricts nondegenerately (e.g. isotropic subspaces are out), and an "irreducible unitary representation" has no proper unitary subrepresentations.
$endgroup$
– Joey Iverson
Apr 12 at 23:39
1
1
$begingroup$
@LSpice I guess the orthogonal complement of a unitary subspace is again unitary, and that they make an algebraic direct sum together. Then your argument shows that every unitary representation is the orthogonal direct sum of irreducible unitary subrepresentations. Great!
$endgroup$
– Joey Iverson
Apr 12 at 23:41
$begingroup$
@LSpice I guess the orthogonal complement of a unitary subspace is again unitary, and that they make an algebraic direct sum together. Then your argument shows that every unitary representation is the orthogonal direct sum of irreducible unitary subrepresentations. Great!
$endgroup$
– Joey Iverson
Apr 12 at 23:41
1
1
$begingroup$
@LSpice One caveat is that an "irreducible unitary representation" is not necessarily irreducible as a $G$-module. For example, choose $G$ and $q$ so that $mathbb{F}_{q^2}[G]$ is not semisimple. The regular representation is unitary when we identify $mathbb{F}_{q^2}[G]$ with what I want to call $L^2(G)$. As a unitary representation it decomposes as a direct sum of irreducibles, but as a $G$-module it does not. Hence some of the irreducible unitary representations in the decomposition of $L^2(G)$ are not irreducible as $G$-modules.
$endgroup$
– Joey Iverson
Apr 12 at 23:42
$begingroup$
@LSpice One caveat is that an "irreducible unitary representation" is not necessarily irreducible as a $G$-module. For example, choose $G$ and $q$ so that $mathbb{F}_{q^2}[G]$ is not semisimple. The regular representation is unitary when we identify $mathbb{F}_{q^2}[G]$ with what I want to call $L^2(G)$. As a unitary representation it decomposes as a direct sum of irreducibles, but as a $G$-module it does not. Hence some of the irreducible unitary representations in the decomposition of $L^2(G)$ are not irreducible as $G$-modules.
$endgroup$
– Joey Iverson
Apr 12 at 23:42
1
1
$begingroup$
@LSpice If we use this definition of irreducibility, then the embedding of $V otimes V^*$ into $L^2(G)$ that you mentioned might not be injective. For instance, if $W subset V$ is an invariant subspace on which the form is degenerate, then we can choose $u in W setminus {0}$ and $v in (W cap W^perp) setminus {0}$, and the map you described sends $u otimes v^* mapsto 0$ (if I understand you correctly) since $langle rho(g) u, v rangle = 0$ for every $g in G$.
$endgroup$
– Joey Iverson
Apr 12 at 23:43
$begingroup$
@LSpice If we use this definition of irreducibility, then the embedding of $V otimes V^*$ into $L^2(G)$ that you mentioned might not be injective. For instance, if $W subset V$ is an invariant subspace on which the form is degenerate, then we can choose $u in W setminus {0}$ and $v in (W cap W^perp) setminus {0}$, and the map you described sends $u otimes v^* mapsto 0$ (if I understand you correctly) since $langle rho(g) u, v rangle = 0$ for every $g in G$.
$endgroup$
– Joey Iverson
Apr 12 at 23:43
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
We had to deal with this problem when classifying maximal subgroups of the finite classical groups, which is the aim of our book:
The Maximal Subgroups of the Low-Dimensional Finite Classical Groups, by
John N. Bray,
Derek F. Holt,
Colva M. Roney-Dougal.
The most difficult maximal subgroups to classify, are those in the so-called Aschbacher class ${mathscr S}$, consisting of absolutely irreducible subgroups that are almost simple mod scalars. Many of these arise as reductions of complex representations over finite fields. Tables of complex representations of groups that are close to simple are available up to dimension about $250$, but we needed to know which classical group the reduction lies in, which means identifying the fixed form.
We generally relied on Lemma 4.4.1 of the book, which says:
For a given absolutely irreducible representation over ${mathbb F}_{q^2}$ of a group $G$,
with Frobenius-Schur indicator $circ$, the image of $G$ under the representation consists of
isometries of a unitary form if and only if the action of the field automorphism
$sigma :x to x^q$ on the Brauer character is the same as complex conjugation.
In many cases, such as when $q$ is coprime to the group order, the Brauer character is just the ordinary complex character.
As an example, the reduction of the complex representation of degree $3$ of the $3$-fold cover $3.A_6$ of $A_6$ lies in ${rm PSL}(3,p)$ for primes $p equiv 1,4 bmod 15$, in ${rm PSU}(3,p)$ (as a subgroup of ${rm PSL}(3,p^2)$) when $p equiv 11,14 bmod 15$ (or when $p=5$), and in ${rm PSL}(3,p^2)$ without preserving a unitary form when $p equiv 2,3 bmod 5$.
$endgroup$
1
$begingroup$
Unless you think there's no risk of confusion, it may be worth it, for people like me who read "representation of a group over $mathbb F_{q^2}$" as "representation of (a group $G$ over $mathbb F_{q^2}$)", seeming to consider representations of algebraic groups, to re-phrase as "representation over $mathbb F_{q^2}$ of a group $G$", so that it's clear we're specifying the field of definition of the representation and not of the group (the latter meaningless anyway in the abstract setting considered).
$endgroup$
– LSpice
Apr 12 at 0:50
$begingroup$
OK, I have reworded it, but I was quoting the lemma directly from the book!
$endgroup$
– Derek Holt
Apr 12 at 2:03
add a comment |
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$begingroup$
We had to deal with this problem when classifying maximal subgroups of the finite classical groups, which is the aim of our book:
The Maximal Subgroups of the Low-Dimensional Finite Classical Groups, by
John N. Bray,
Derek F. Holt,
Colva M. Roney-Dougal.
The most difficult maximal subgroups to classify, are those in the so-called Aschbacher class ${mathscr S}$, consisting of absolutely irreducible subgroups that are almost simple mod scalars. Many of these arise as reductions of complex representations over finite fields. Tables of complex representations of groups that are close to simple are available up to dimension about $250$, but we needed to know which classical group the reduction lies in, which means identifying the fixed form.
We generally relied on Lemma 4.4.1 of the book, which says:
For a given absolutely irreducible representation over ${mathbb F}_{q^2}$ of a group $G$,
with Frobenius-Schur indicator $circ$, the image of $G$ under the representation consists of
isometries of a unitary form if and only if the action of the field automorphism
$sigma :x to x^q$ on the Brauer character is the same as complex conjugation.
In many cases, such as when $q$ is coprime to the group order, the Brauer character is just the ordinary complex character.
As an example, the reduction of the complex representation of degree $3$ of the $3$-fold cover $3.A_6$ of $A_6$ lies in ${rm PSL}(3,p)$ for primes $p equiv 1,4 bmod 15$, in ${rm PSU}(3,p)$ (as a subgroup of ${rm PSL}(3,p^2)$) when $p equiv 11,14 bmod 15$ (or when $p=5$), and in ${rm PSL}(3,p^2)$ without preserving a unitary form when $p equiv 2,3 bmod 5$.
$endgroup$
1
$begingroup$
Unless you think there's no risk of confusion, it may be worth it, for people like me who read "representation of a group over $mathbb F_{q^2}$" as "representation of (a group $G$ over $mathbb F_{q^2}$)", seeming to consider representations of algebraic groups, to re-phrase as "representation over $mathbb F_{q^2}$ of a group $G$", so that it's clear we're specifying the field of definition of the representation and not of the group (the latter meaningless anyway in the abstract setting considered).
$endgroup$
– LSpice
Apr 12 at 0:50
$begingroup$
OK, I have reworded it, but I was quoting the lemma directly from the book!
$endgroup$
– Derek Holt
Apr 12 at 2:03
add a comment |
$begingroup$
We had to deal with this problem when classifying maximal subgroups of the finite classical groups, which is the aim of our book:
The Maximal Subgroups of the Low-Dimensional Finite Classical Groups, by
John N. Bray,
Derek F. Holt,
Colva M. Roney-Dougal.
The most difficult maximal subgroups to classify, are those in the so-called Aschbacher class ${mathscr S}$, consisting of absolutely irreducible subgroups that are almost simple mod scalars. Many of these arise as reductions of complex representations over finite fields. Tables of complex representations of groups that are close to simple are available up to dimension about $250$, but we needed to know which classical group the reduction lies in, which means identifying the fixed form.
We generally relied on Lemma 4.4.1 of the book, which says:
For a given absolutely irreducible representation over ${mathbb F}_{q^2}$ of a group $G$,
with Frobenius-Schur indicator $circ$, the image of $G$ under the representation consists of
isometries of a unitary form if and only if the action of the field automorphism
$sigma :x to x^q$ on the Brauer character is the same as complex conjugation.
In many cases, such as when $q$ is coprime to the group order, the Brauer character is just the ordinary complex character.
As an example, the reduction of the complex representation of degree $3$ of the $3$-fold cover $3.A_6$ of $A_6$ lies in ${rm PSL}(3,p)$ for primes $p equiv 1,4 bmod 15$, in ${rm PSU}(3,p)$ (as a subgroup of ${rm PSL}(3,p^2)$) when $p equiv 11,14 bmod 15$ (or when $p=5$), and in ${rm PSL}(3,p^2)$ without preserving a unitary form when $p equiv 2,3 bmod 5$.
$endgroup$
1
$begingroup$
Unless you think there's no risk of confusion, it may be worth it, for people like me who read "representation of a group over $mathbb F_{q^2}$" as "representation of (a group $G$ over $mathbb F_{q^2}$)", seeming to consider representations of algebraic groups, to re-phrase as "representation over $mathbb F_{q^2}$ of a group $G$", so that it's clear we're specifying the field of definition of the representation and not of the group (the latter meaningless anyway in the abstract setting considered).
$endgroup$
– LSpice
Apr 12 at 0:50
$begingroup$
OK, I have reworded it, but I was quoting the lemma directly from the book!
$endgroup$
– Derek Holt
Apr 12 at 2:03
add a comment |
$begingroup$
We had to deal with this problem when classifying maximal subgroups of the finite classical groups, which is the aim of our book:
The Maximal Subgroups of the Low-Dimensional Finite Classical Groups, by
John N. Bray,
Derek F. Holt,
Colva M. Roney-Dougal.
The most difficult maximal subgroups to classify, are those in the so-called Aschbacher class ${mathscr S}$, consisting of absolutely irreducible subgroups that are almost simple mod scalars. Many of these arise as reductions of complex representations over finite fields. Tables of complex representations of groups that are close to simple are available up to dimension about $250$, but we needed to know which classical group the reduction lies in, which means identifying the fixed form.
We generally relied on Lemma 4.4.1 of the book, which says:
For a given absolutely irreducible representation over ${mathbb F}_{q^2}$ of a group $G$,
with Frobenius-Schur indicator $circ$, the image of $G$ under the representation consists of
isometries of a unitary form if and only if the action of the field automorphism
$sigma :x to x^q$ on the Brauer character is the same as complex conjugation.
In many cases, such as when $q$ is coprime to the group order, the Brauer character is just the ordinary complex character.
As an example, the reduction of the complex representation of degree $3$ of the $3$-fold cover $3.A_6$ of $A_6$ lies in ${rm PSL}(3,p)$ for primes $p equiv 1,4 bmod 15$, in ${rm PSU}(3,p)$ (as a subgroup of ${rm PSL}(3,p^2)$) when $p equiv 11,14 bmod 15$ (or when $p=5$), and in ${rm PSL}(3,p^2)$ without preserving a unitary form when $p equiv 2,3 bmod 5$.
$endgroup$
We had to deal with this problem when classifying maximal subgroups of the finite classical groups, which is the aim of our book:
The Maximal Subgroups of the Low-Dimensional Finite Classical Groups, by
John N. Bray,
Derek F. Holt,
Colva M. Roney-Dougal.
The most difficult maximal subgroups to classify, are those in the so-called Aschbacher class ${mathscr S}$, consisting of absolutely irreducible subgroups that are almost simple mod scalars. Many of these arise as reductions of complex representations over finite fields. Tables of complex representations of groups that are close to simple are available up to dimension about $250$, but we needed to know which classical group the reduction lies in, which means identifying the fixed form.
We generally relied on Lemma 4.4.1 of the book, which says:
For a given absolutely irreducible representation over ${mathbb F}_{q^2}$ of a group $G$,
with Frobenius-Schur indicator $circ$, the image of $G$ under the representation consists of
isometries of a unitary form if and only if the action of the field automorphism
$sigma :x to x^q$ on the Brauer character is the same as complex conjugation.
In many cases, such as when $q$ is coprime to the group order, the Brauer character is just the ordinary complex character.
As an example, the reduction of the complex representation of degree $3$ of the $3$-fold cover $3.A_6$ of $A_6$ lies in ${rm PSL}(3,p)$ for primes $p equiv 1,4 bmod 15$, in ${rm PSU}(3,p)$ (as a subgroup of ${rm PSL}(3,p^2)$) when $p equiv 11,14 bmod 15$ (or when $p=5$), and in ${rm PSL}(3,p^2)$ without preserving a unitary form when $p equiv 2,3 bmod 5$.
edited Apr 12 at 2:03
answered Apr 11 at 23:36
Derek HoltDerek Holt
27.6k464113
27.6k464113
1
$begingroup$
Unless you think there's no risk of confusion, it may be worth it, for people like me who read "representation of a group over $mathbb F_{q^2}$" as "representation of (a group $G$ over $mathbb F_{q^2}$)", seeming to consider representations of algebraic groups, to re-phrase as "representation over $mathbb F_{q^2}$ of a group $G$", so that it's clear we're specifying the field of definition of the representation and not of the group (the latter meaningless anyway in the abstract setting considered).
$endgroup$
– LSpice
Apr 12 at 0:50
$begingroup$
OK, I have reworded it, but I was quoting the lemma directly from the book!
$endgroup$
– Derek Holt
Apr 12 at 2:03
add a comment |
1
$begingroup$
Unless you think there's no risk of confusion, it may be worth it, for people like me who read "representation of a group over $mathbb F_{q^2}$" as "representation of (a group $G$ over $mathbb F_{q^2}$)", seeming to consider representations of algebraic groups, to re-phrase as "representation over $mathbb F_{q^2}$ of a group $G$", so that it's clear we're specifying the field of definition of the representation and not of the group (the latter meaningless anyway in the abstract setting considered).
$endgroup$
– LSpice
Apr 12 at 0:50
$begingroup$
OK, I have reworded it, but I was quoting the lemma directly from the book!
$endgroup$
– Derek Holt
Apr 12 at 2:03
1
1
$begingroup$
Unless you think there's no risk of confusion, it may be worth it, for people like me who read "representation of a group over $mathbb F_{q^2}$" as "representation of (a group $G$ over $mathbb F_{q^2}$)", seeming to consider representations of algebraic groups, to re-phrase as "representation over $mathbb F_{q^2}$ of a group $G$", so that it's clear we're specifying the field of definition of the representation and not of the group (the latter meaningless anyway in the abstract setting considered).
$endgroup$
– LSpice
Apr 12 at 0:50
$begingroup$
Unless you think there's no risk of confusion, it may be worth it, for people like me who read "representation of a group over $mathbb F_{q^2}$" as "representation of (a group $G$ over $mathbb F_{q^2}$)", seeming to consider representations of algebraic groups, to re-phrase as "representation over $mathbb F_{q^2}$ of a group $G$", so that it's clear we're specifying the field of definition of the representation and not of the group (the latter meaningless anyway in the abstract setting considered).
$endgroup$
– LSpice
Apr 12 at 0:50
$begingroup$
OK, I have reworded it, but I was quoting the lemma directly from the book!
$endgroup$
– Derek Holt
Apr 12 at 2:03
$begingroup$
OK, I have reworded it, but I was quoting the lemma directly from the book!
$endgroup$
– Derek Holt
Apr 12 at 2:03
add a comment |
Joey Iverson is a new contributor. Be nice, and check out our Code of Conduct.
Joey Iverson is a new contributor. Be nice, and check out our Code of Conduct.
Joey Iverson is a new contributor. Be nice, and check out our Code of Conduct.
Joey Iverson is a new contributor. Be nice, and check out our Code of Conduct.
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Notice that finite-dimensional unitary representations are automatically semisimple; a minimal stable non-$0$ subspace is irreducible, and its orthogonal complement has smaller dimension. A similar argument shows that every semisimple unitary representation is an orthogonal direct sum of irreducibles.
$endgroup$
– LSpice
Apr 11 at 22:14
1
$begingroup$
@LSpice I see we need to be careful about what "irreducible" means for a unitary representation. Probably the right thing is to talk about unitary representations not just on $mathbb{F}_{q^2}^n$ but on general unitary spaces, i.e. vector spaces over $mathbb{F}_{q^2}$ equipped with nondegenerate sesquilinear forms. Then a "unitary subrepresentation" should mean an invariant subspace on which the form restricts nondegenerately (e.g. isotropic subspaces are out), and an "irreducible unitary representation" has no proper unitary subrepresentations.
$endgroup$
– Joey Iverson
Apr 12 at 23:39
1
$begingroup$
@LSpice I guess the orthogonal complement of a unitary subspace is again unitary, and that they make an algebraic direct sum together. Then your argument shows that every unitary representation is the orthogonal direct sum of irreducible unitary subrepresentations. Great!
$endgroup$
– Joey Iverson
Apr 12 at 23:41
1
$begingroup$
@LSpice One caveat is that an "irreducible unitary representation" is not necessarily irreducible as a $G$-module. For example, choose $G$ and $q$ so that $mathbb{F}_{q^2}[G]$ is not semisimple. The regular representation is unitary when we identify $mathbb{F}_{q^2}[G]$ with what I want to call $L^2(G)$. As a unitary representation it decomposes as a direct sum of irreducibles, but as a $G$-module it does not. Hence some of the irreducible unitary representations in the decomposition of $L^2(G)$ are not irreducible as $G$-modules.
$endgroup$
– Joey Iverson
Apr 12 at 23:42
1
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@LSpice If we use this definition of irreducibility, then the embedding of $V otimes V^*$ into $L^2(G)$ that you mentioned might not be injective. For instance, if $W subset V$ is an invariant subspace on which the form is degenerate, then we can choose $u in W setminus {0}$ and $v in (W cap W^perp) setminus {0}$, and the map you described sends $u otimes v^* mapsto 0$ (if I understand you correctly) since $langle rho(g) u, v rangle = 0$ for every $g in G$.
$endgroup$
– Joey Iverson
Apr 12 at 23:43