Real representation
$begingroup$
Define a representation $rho$ of a finite group $G$ over a $mathbb{C}$-vector space to be real if the space admits a basis for which matrix $rho(g)$ has real coefficients $forall g in G$.
I have to show that for ever $rho$ it is true that $rho otimes rho^*$ is always real ($rho^*$ is the dual representation).
I think I've got an answer but it's pretty ugly so i would like to know if there is a clever solution to this question.
group-theory finite-groups representation-theory
$endgroup$
|
show 1 more comment
$begingroup$
Define a representation $rho$ of a finite group $G$ over a $mathbb{C}$-vector space to be real if the space admits a basis for which matrix $rho(g)$ has real coefficients $forall g in G$.
I have to show that for ever $rho$ it is true that $rho otimes rho^*$ is always real ($rho^*$ is the dual representation).
I think I've got an answer but it's pretty ugly so i would like to know if there is a clever solution to this question.
group-theory finite-groups representation-theory
$endgroup$
$begingroup$
Maybe you can try something about the character since the character of $rho otimes rho^{ast}$ is real if "*" means the conjugate.
$endgroup$
– Aolong Li
Mar 23 '17 at 11:08
$begingroup$
The character should be real but I can't see how i can use this information to say that $rho$ is real
$endgroup$
– Tommaso Scognamiglio
Mar 23 '17 at 11:27
$begingroup$
Indeed, the character being real is not enough.
$endgroup$
– Tobias Kildetoft
Mar 23 '17 at 11:34
$begingroup$
@TommasoScognamiglio I thought you want to say $rho otimes rho^{ast}$ is real. I remember you can determine a representation by its character in the case of finite group. I will give more detail later after I find it in Serre's book.
$endgroup$
– Aolong Li
Mar 23 '17 at 11:36
2
$begingroup$
@AolongLi Yes, a complex representation is determined by its character. But the character being real does not imply that the representation is realizable over the reals.
$endgroup$
– Tobias Kildetoft
Mar 23 '17 at 11:43
|
show 1 more comment
$begingroup$
Define a representation $rho$ of a finite group $G$ over a $mathbb{C}$-vector space to be real if the space admits a basis for which matrix $rho(g)$ has real coefficients $forall g in G$.
I have to show that for ever $rho$ it is true that $rho otimes rho^*$ is always real ($rho^*$ is the dual representation).
I think I've got an answer but it's pretty ugly so i would like to know if there is a clever solution to this question.
group-theory finite-groups representation-theory
$endgroup$
Define a representation $rho$ of a finite group $G$ over a $mathbb{C}$-vector space to be real if the space admits a basis for which matrix $rho(g)$ has real coefficients $forall g in G$.
I have to show that for ever $rho$ it is true that $rho otimes rho^*$ is always real ($rho^*$ is the dual representation).
I think I've got an answer but it's pretty ugly so i would like to know if there is a clever solution to this question.
group-theory finite-groups representation-theory
group-theory finite-groups representation-theory
edited Mar 23 '17 at 11:26
Tommaso Scognamiglio
asked Mar 23 '17 at 8:19
Tommaso ScognamiglioTommaso Scognamiglio
492312
492312
$begingroup$
Maybe you can try something about the character since the character of $rho otimes rho^{ast}$ is real if "*" means the conjugate.
$endgroup$
– Aolong Li
Mar 23 '17 at 11:08
$begingroup$
The character should be real but I can't see how i can use this information to say that $rho$ is real
$endgroup$
– Tommaso Scognamiglio
Mar 23 '17 at 11:27
$begingroup$
Indeed, the character being real is not enough.
$endgroup$
– Tobias Kildetoft
Mar 23 '17 at 11:34
$begingroup$
@TommasoScognamiglio I thought you want to say $rho otimes rho^{ast}$ is real. I remember you can determine a representation by its character in the case of finite group. I will give more detail later after I find it in Serre's book.
$endgroup$
– Aolong Li
Mar 23 '17 at 11:36
2
$begingroup$
@AolongLi Yes, a complex representation is determined by its character. But the character being real does not imply that the representation is realizable over the reals.
$endgroup$
– Tobias Kildetoft
Mar 23 '17 at 11:43
|
show 1 more comment
$begingroup$
Maybe you can try something about the character since the character of $rho otimes rho^{ast}$ is real if "*" means the conjugate.
$endgroup$
– Aolong Li
Mar 23 '17 at 11:08
$begingroup$
The character should be real but I can't see how i can use this information to say that $rho$ is real
$endgroup$
– Tommaso Scognamiglio
Mar 23 '17 at 11:27
$begingroup$
Indeed, the character being real is not enough.
$endgroup$
– Tobias Kildetoft
Mar 23 '17 at 11:34
$begingroup$
@TommasoScognamiglio I thought you want to say $rho otimes rho^{ast}$ is real. I remember you can determine a representation by its character in the case of finite group. I will give more detail later after I find it in Serre's book.
$endgroup$
– Aolong Li
Mar 23 '17 at 11:36
2
$begingroup$
@AolongLi Yes, a complex representation is determined by its character. But the character being real does not imply that the representation is realizable over the reals.
$endgroup$
– Tobias Kildetoft
Mar 23 '17 at 11:43
$begingroup$
Maybe you can try something about the character since the character of $rho otimes rho^{ast}$ is real if "*" means the conjugate.
$endgroup$
– Aolong Li
Mar 23 '17 at 11:08
$begingroup$
Maybe you can try something about the character since the character of $rho otimes rho^{ast}$ is real if "*" means the conjugate.
$endgroup$
– Aolong Li
Mar 23 '17 at 11:08
$begingroup$
The character should be real but I can't see how i can use this information to say that $rho$ is real
$endgroup$
– Tommaso Scognamiglio
Mar 23 '17 at 11:27
$begingroup$
The character should be real but I can't see how i can use this information to say that $rho$ is real
$endgroup$
– Tommaso Scognamiglio
Mar 23 '17 at 11:27
$begingroup$
Indeed, the character being real is not enough.
$endgroup$
– Tobias Kildetoft
Mar 23 '17 at 11:34
$begingroup$
Indeed, the character being real is not enough.
$endgroup$
– Tobias Kildetoft
Mar 23 '17 at 11:34
$begingroup$
@TommasoScognamiglio I thought you want to say $rho otimes rho^{ast}$ is real. I remember you can determine a representation by its character in the case of finite group. I will give more detail later after I find it in Serre's book.
$endgroup$
– Aolong Li
Mar 23 '17 at 11:36
$begingroup$
@TommasoScognamiglio I thought you want to say $rho otimes rho^{ast}$ is real. I remember you can determine a representation by its character in the case of finite group. I will give more detail later after I find it in Serre's book.
$endgroup$
– Aolong Li
Mar 23 '17 at 11:36
2
2
$begingroup$
@AolongLi Yes, a complex representation is determined by its character. But the character being real does not imply that the representation is realizable over the reals.
$endgroup$
– Tobias Kildetoft
Mar 23 '17 at 11:43
$begingroup$
@AolongLi Yes, a complex representation is determined by its character. But the character being real does not imply that the representation is realizable over the reals.
$endgroup$
– Tobias Kildetoft
Mar 23 '17 at 11:43
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Let $V$ be a complex vector space, say that a map $C:V to V$ is a real structure if it is a conjugate-linear involution, i.e. if $C^2=mathrm{id}_V$ and $C(v+lambda w)=C(v)+overline{lambda} C(w)$ for all $v,w in V, lambda in Bbb C$. In that case, we say that $C$ is a $G$-equivariant real structure if $C$ is $G$-equivariant.
Lemma If $V$ is a complex representation of a group $G$, then there is a real representation (by which I mean a representation on a real vector space) $W$ of $G$ such that $Bbb C otimes_{Bbb R} W cong V$ (this is equivalent to being a real representation as defined in the question) iff $V$ admits a $G$-equivariant real structure.
Proof from a sophisticated viewpoint, this is an instance of Galois descent, but let's give an elementary argument. If $Bbb C otimes_{Bbb R} W cong V$, then $C(z otimes w)=overline{z} otimes w$ defines a $G$-equivariant real structure.
Suppose that $C$ is a $G$-equivariant real structure on $V$, then $W:= {v in V mid C(v)=v}$ is a $G$-stable $Bbb R$ subspace of $V$ and one gets a $Bbb C$-linear $G$-equivariant map $varphi:Bbb C otimes_{Bbb R} W to V, zotimes w to zw$.
To see that this is an isomorphism, note that $V=Woplus iW$, as we can write $v=frac{v+C(v)}{2}+ifrac{v-C(v)}{2i}$ with $frac{v+C(v)}{2},frac{v-C(v)}{2i} in W$ and clearly $W cap iW = {0}$. One also has $Bbb C= Bbb R oplus i Bbb R$, so $Bbb C otimes_{Bbb R} W cong W oplus iW$ and the isomorphism is compatible with this. (Note:This can be extended to an equivalence of suitable categories)
To apply this to the problem, let $G$ be a finite group and let $rho:G to mathrm{GL}(V)$ be a finite-dimensional complex representation, then choose a $G$-invariant scalar product $langle -,- rangle$ on $V$, this defines a $G$-equivariant conjugate-linear isomorphism $psi:V to V^*$ given by $v mapsto langle v,-rangle$ (assuming the convention that the inner product is linear in the second argument)
Now consider the map $C:V otimes_{Bbb C} V^*to V otimes_{Bbb C}V^*, votimes xi mapsto psi^{-1}(xi) otimes psi(v)$, $C^2= mathrm{id}$ and $C$ is conjugate-linear and $G$-equivariant, so by the lemma, $rho otimes rho^*$ is real.
$endgroup$
1
$begingroup$
Great answer (+1), but it is perhaps helpful if you edit to clarify that when you say 'a real representation $W$' you mean an actual really real representation, that is, a representation on a real vector space $W$, and not a 'real' representation as defined in the original post.
$endgroup$
– Vincent
Oct 31 '18 at 21:59
add a comment |
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$begingroup$
Let $V$ be a complex vector space, say that a map $C:V to V$ is a real structure if it is a conjugate-linear involution, i.e. if $C^2=mathrm{id}_V$ and $C(v+lambda w)=C(v)+overline{lambda} C(w)$ for all $v,w in V, lambda in Bbb C$. In that case, we say that $C$ is a $G$-equivariant real structure if $C$ is $G$-equivariant.
Lemma If $V$ is a complex representation of a group $G$, then there is a real representation (by which I mean a representation on a real vector space) $W$ of $G$ such that $Bbb C otimes_{Bbb R} W cong V$ (this is equivalent to being a real representation as defined in the question) iff $V$ admits a $G$-equivariant real structure.
Proof from a sophisticated viewpoint, this is an instance of Galois descent, but let's give an elementary argument. If $Bbb C otimes_{Bbb R} W cong V$, then $C(z otimes w)=overline{z} otimes w$ defines a $G$-equivariant real structure.
Suppose that $C$ is a $G$-equivariant real structure on $V$, then $W:= {v in V mid C(v)=v}$ is a $G$-stable $Bbb R$ subspace of $V$ and one gets a $Bbb C$-linear $G$-equivariant map $varphi:Bbb C otimes_{Bbb R} W to V, zotimes w to zw$.
To see that this is an isomorphism, note that $V=Woplus iW$, as we can write $v=frac{v+C(v)}{2}+ifrac{v-C(v)}{2i}$ with $frac{v+C(v)}{2},frac{v-C(v)}{2i} in W$ and clearly $W cap iW = {0}$. One also has $Bbb C= Bbb R oplus i Bbb R$, so $Bbb C otimes_{Bbb R} W cong W oplus iW$ and the isomorphism is compatible with this. (Note:This can be extended to an equivalence of suitable categories)
To apply this to the problem, let $G$ be a finite group and let $rho:G to mathrm{GL}(V)$ be a finite-dimensional complex representation, then choose a $G$-invariant scalar product $langle -,- rangle$ on $V$, this defines a $G$-equivariant conjugate-linear isomorphism $psi:V to V^*$ given by $v mapsto langle v,-rangle$ (assuming the convention that the inner product is linear in the second argument)
Now consider the map $C:V otimes_{Bbb C} V^*to V otimes_{Bbb C}V^*, votimes xi mapsto psi^{-1}(xi) otimes psi(v)$, $C^2= mathrm{id}$ and $C$ is conjugate-linear and $G$-equivariant, so by the lemma, $rho otimes rho^*$ is real.
$endgroup$
1
$begingroup$
Great answer (+1), but it is perhaps helpful if you edit to clarify that when you say 'a real representation $W$' you mean an actual really real representation, that is, a representation on a real vector space $W$, and not a 'real' representation as defined in the original post.
$endgroup$
– Vincent
Oct 31 '18 at 21:59
add a comment |
$begingroup$
Let $V$ be a complex vector space, say that a map $C:V to V$ is a real structure if it is a conjugate-linear involution, i.e. if $C^2=mathrm{id}_V$ and $C(v+lambda w)=C(v)+overline{lambda} C(w)$ for all $v,w in V, lambda in Bbb C$. In that case, we say that $C$ is a $G$-equivariant real structure if $C$ is $G$-equivariant.
Lemma If $V$ is a complex representation of a group $G$, then there is a real representation (by which I mean a representation on a real vector space) $W$ of $G$ such that $Bbb C otimes_{Bbb R} W cong V$ (this is equivalent to being a real representation as defined in the question) iff $V$ admits a $G$-equivariant real structure.
Proof from a sophisticated viewpoint, this is an instance of Galois descent, but let's give an elementary argument. If $Bbb C otimes_{Bbb R} W cong V$, then $C(z otimes w)=overline{z} otimes w$ defines a $G$-equivariant real structure.
Suppose that $C$ is a $G$-equivariant real structure on $V$, then $W:= {v in V mid C(v)=v}$ is a $G$-stable $Bbb R$ subspace of $V$ and one gets a $Bbb C$-linear $G$-equivariant map $varphi:Bbb C otimes_{Bbb R} W to V, zotimes w to zw$.
To see that this is an isomorphism, note that $V=Woplus iW$, as we can write $v=frac{v+C(v)}{2}+ifrac{v-C(v)}{2i}$ with $frac{v+C(v)}{2},frac{v-C(v)}{2i} in W$ and clearly $W cap iW = {0}$. One also has $Bbb C= Bbb R oplus i Bbb R$, so $Bbb C otimes_{Bbb R} W cong W oplus iW$ and the isomorphism is compatible with this. (Note:This can be extended to an equivalence of suitable categories)
To apply this to the problem, let $G$ be a finite group and let $rho:G to mathrm{GL}(V)$ be a finite-dimensional complex representation, then choose a $G$-invariant scalar product $langle -,- rangle$ on $V$, this defines a $G$-equivariant conjugate-linear isomorphism $psi:V to V^*$ given by $v mapsto langle v,-rangle$ (assuming the convention that the inner product is linear in the second argument)
Now consider the map $C:V otimes_{Bbb C} V^*to V otimes_{Bbb C}V^*, votimes xi mapsto psi^{-1}(xi) otimes psi(v)$, $C^2= mathrm{id}$ and $C$ is conjugate-linear and $G$-equivariant, so by the lemma, $rho otimes rho^*$ is real.
$endgroup$
1
$begingroup$
Great answer (+1), but it is perhaps helpful if you edit to clarify that when you say 'a real representation $W$' you mean an actual really real representation, that is, a representation on a real vector space $W$, and not a 'real' representation as defined in the original post.
$endgroup$
– Vincent
Oct 31 '18 at 21:59
add a comment |
$begingroup$
Let $V$ be a complex vector space, say that a map $C:V to V$ is a real structure if it is a conjugate-linear involution, i.e. if $C^2=mathrm{id}_V$ and $C(v+lambda w)=C(v)+overline{lambda} C(w)$ for all $v,w in V, lambda in Bbb C$. In that case, we say that $C$ is a $G$-equivariant real structure if $C$ is $G$-equivariant.
Lemma If $V$ is a complex representation of a group $G$, then there is a real representation (by which I mean a representation on a real vector space) $W$ of $G$ such that $Bbb C otimes_{Bbb R} W cong V$ (this is equivalent to being a real representation as defined in the question) iff $V$ admits a $G$-equivariant real structure.
Proof from a sophisticated viewpoint, this is an instance of Galois descent, but let's give an elementary argument. If $Bbb C otimes_{Bbb R} W cong V$, then $C(z otimes w)=overline{z} otimes w$ defines a $G$-equivariant real structure.
Suppose that $C$ is a $G$-equivariant real structure on $V$, then $W:= {v in V mid C(v)=v}$ is a $G$-stable $Bbb R$ subspace of $V$ and one gets a $Bbb C$-linear $G$-equivariant map $varphi:Bbb C otimes_{Bbb R} W to V, zotimes w to zw$.
To see that this is an isomorphism, note that $V=Woplus iW$, as we can write $v=frac{v+C(v)}{2}+ifrac{v-C(v)}{2i}$ with $frac{v+C(v)}{2},frac{v-C(v)}{2i} in W$ and clearly $W cap iW = {0}$. One also has $Bbb C= Bbb R oplus i Bbb R$, so $Bbb C otimes_{Bbb R} W cong W oplus iW$ and the isomorphism is compatible with this. (Note:This can be extended to an equivalence of suitable categories)
To apply this to the problem, let $G$ be a finite group and let $rho:G to mathrm{GL}(V)$ be a finite-dimensional complex representation, then choose a $G$-invariant scalar product $langle -,- rangle$ on $V$, this defines a $G$-equivariant conjugate-linear isomorphism $psi:V to V^*$ given by $v mapsto langle v,-rangle$ (assuming the convention that the inner product is linear in the second argument)
Now consider the map $C:V otimes_{Bbb C} V^*to V otimes_{Bbb C}V^*, votimes xi mapsto psi^{-1}(xi) otimes psi(v)$, $C^2= mathrm{id}$ and $C$ is conjugate-linear and $G$-equivariant, so by the lemma, $rho otimes rho^*$ is real.
$endgroup$
Let $V$ be a complex vector space, say that a map $C:V to V$ is a real structure if it is a conjugate-linear involution, i.e. if $C^2=mathrm{id}_V$ and $C(v+lambda w)=C(v)+overline{lambda} C(w)$ for all $v,w in V, lambda in Bbb C$. In that case, we say that $C$ is a $G$-equivariant real structure if $C$ is $G$-equivariant.
Lemma If $V$ is a complex representation of a group $G$, then there is a real representation (by which I mean a representation on a real vector space) $W$ of $G$ such that $Bbb C otimes_{Bbb R} W cong V$ (this is equivalent to being a real representation as defined in the question) iff $V$ admits a $G$-equivariant real structure.
Proof from a sophisticated viewpoint, this is an instance of Galois descent, but let's give an elementary argument. If $Bbb C otimes_{Bbb R} W cong V$, then $C(z otimes w)=overline{z} otimes w$ defines a $G$-equivariant real structure.
Suppose that $C$ is a $G$-equivariant real structure on $V$, then $W:= {v in V mid C(v)=v}$ is a $G$-stable $Bbb R$ subspace of $V$ and one gets a $Bbb C$-linear $G$-equivariant map $varphi:Bbb C otimes_{Bbb R} W to V, zotimes w to zw$.
To see that this is an isomorphism, note that $V=Woplus iW$, as we can write $v=frac{v+C(v)}{2}+ifrac{v-C(v)}{2i}$ with $frac{v+C(v)}{2},frac{v-C(v)}{2i} in W$ and clearly $W cap iW = {0}$. One also has $Bbb C= Bbb R oplus i Bbb R$, so $Bbb C otimes_{Bbb R} W cong W oplus iW$ and the isomorphism is compatible with this. (Note:This can be extended to an equivalence of suitable categories)
To apply this to the problem, let $G$ be a finite group and let $rho:G to mathrm{GL}(V)$ be a finite-dimensional complex representation, then choose a $G$-invariant scalar product $langle -,- rangle$ on $V$, this defines a $G$-equivariant conjugate-linear isomorphism $psi:V to V^*$ given by $v mapsto langle v,-rangle$ (assuming the convention that the inner product is linear in the second argument)
Now consider the map $C:V otimes_{Bbb C} V^*to V otimes_{Bbb C}V^*, votimes xi mapsto psi^{-1}(xi) otimes psi(v)$, $C^2= mathrm{id}$ and $C$ is conjugate-linear and $G$-equivariant, so by the lemma, $rho otimes rho^*$ is real.
edited Nov 30 '18 at 22:46
answered Oct 31 '18 at 21:01
MatheinBoulomenosMatheinBoulomenos
8,0351936
8,0351936
1
$begingroup$
Great answer (+1), but it is perhaps helpful if you edit to clarify that when you say 'a real representation $W$' you mean an actual really real representation, that is, a representation on a real vector space $W$, and not a 'real' representation as defined in the original post.
$endgroup$
– Vincent
Oct 31 '18 at 21:59
add a comment |
1
$begingroup$
Great answer (+1), but it is perhaps helpful if you edit to clarify that when you say 'a real representation $W$' you mean an actual really real representation, that is, a representation on a real vector space $W$, and not a 'real' representation as defined in the original post.
$endgroup$
– Vincent
Oct 31 '18 at 21:59
1
1
$begingroup$
Great answer (+1), but it is perhaps helpful if you edit to clarify that when you say 'a real representation $W$' you mean an actual really real representation, that is, a representation on a real vector space $W$, and not a 'real' representation as defined in the original post.
$endgroup$
– Vincent
Oct 31 '18 at 21:59
$begingroup$
Great answer (+1), but it is perhaps helpful if you edit to clarify that when you say 'a real representation $W$' you mean an actual really real representation, that is, a representation on a real vector space $W$, and not a 'real' representation as defined in the original post.
$endgroup$
– Vincent
Oct 31 '18 at 21:59
add a comment |
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$begingroup$
Maybe you can try something about the character since the character of $rho otimes rho^{ast}$ is real if "*" means the conjugate.
$endgroup$
– Aolong Li
Mar 23 '17 at 11:08
$begingroup$
The character should be real but I can't see how i can use this information to say that $rho$ is real
$endgroup$
– Tommaso Scognamiglio
Mar 23 '17 at 11:27
$begingroup$
Indeed, the character being real is not enough.
$endgroup$
– Tobias Kildetoft
Mar 23 '17 at 11:34
$begingroup$
@TommasoScognamiglio I thought you want to say $rho otimes rho^{ast}$ is real. I remember you can determine a representation by its character in the case of finite group. I will give more detail later after I find it in Serre's book.
$endgroup$
– Aolong Li
Mar 23 '17 at 11:36
2
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@AolongLi Yes, a complex representation is determined by its character. But the character being real does not imply that the representation is realizable over the reals.
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– Tobias Kildetoft
Mar 23 '17 at 11:43