Real representation












6












$begingroup$


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.










share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















6












$begingroup$


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.










share|cite|improve this question











$endgroup$












  • $begingroup$
    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














6












6








6


0



$begingroup$


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.










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















2












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 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













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 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


















2












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 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
















2












2








2





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








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
















  • 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




















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