Can the fix point set of a nontrivial irreducible complex representation of a finite odd order group be non...
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I'm trying to show that if $G$ is a finite odd order group then, all of its nontrivial complex representations are of complex type (i.e., it is not realisable over the reals).
(I have answered it here: If $G$ is a finite non-trivial group of odd order, it has an irreducible representation not realisable over the reals.)
Let $V$ be such a representation, and let $V^G:={vin V: gv=v, forall g in G}$ be its fix point set. Using some theorems in Bröcker's book about Representations of Compact Lie Groups, I can solve this problem if I show that $V^G=0$. Is that true?
finite-groups representation-theory
asked Jan 14 at 0:08
Andre GomesAndre Gomes
920516
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add a comment |
$begingroup$
I'm trying to show that if $G$ is a finite odd order group then, all of its nontrivial complex representations are of complex type (i.e., it is not realisable over the reals).
(I have answered it here: If $G$ is a finite non-trivial group of odd order, it has an irreducible representation not realisable over the reals.)
Let $V$ be such a representation, and let $V^G:={vin V: gv=v, forall g in G}$ be its fix point set. Using some theorems in Bröcker's book about Representations of Compact Lie Groups, I can solve this problem if I show that $V^G=0$. Is that true?
finite-groups representation-theory
asked Jan 14 at 0:08
Andre GomesAndre Gomes
920516
$endgroup$
1
$begingroup$
Well, what about the trivial representation?
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– Matt Samuel
Jan 14 at 0:11
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It should be a non-trivial representation*, I'll change @MattSamuel
$endgroup$
– Andre Gomes
Jan 14 at 0:12
$begingroup$
I have answered the first question here: math.stackexchange.com/questions/1033844/…
$endgroup$
– Andre Gomes
Jan 14 at 17:24
add a comment |
$begingroup$
I'm trying to show that if $G$ is a finite odd order group then, all of its nontrivial complex representations are of complex type (i.e., it is not realisable over the reals).
(I have answered it here: If $G$ is a finite non-trivial group of odd order, it has an irreducible representation not realisable over the reals.)
Let $V$ be such a representation, and let $V^G:={vin V: gv=v, forall g in G}$ be its fix point set. Using some theorems in Bröcker's book about Representations of Compact Lie Groups, I can solve this problem if I show that $V^G=0$. Is that true?
finite-groups representation-theory
asked Jan 14 at 0:08
Andre GomesAndre Gomes
920516
$endgroup$
I'm trying to show that if $G$ is a finite odd order group then, all of its nontrivial complex representations are of complex type (i.e., it is not realisable over the reals).
(I have answered it here: If $G$ is a finite non-trivial group of odd order, it has an irreducible representation not realisable over the reals.)
Let $V$ be such a representation, and let $V^G:={vin V: gv=v, forall g in G}$ be its fix point set. Using some theorems in Bröcker's book about Representations of Compact Lie Groups, I can solve this problem if I show that $V^G=0$. Is that true?
finite-groups representation-theory
finite-groups representation-theory
asked Jan 14 at 0:08
Andre GomesAndre Gomes
920516
asked Jan 14 at 0:08
Andre GomesAndre Gomes
920516
edited Jan 14 at 17:25
Andre Gomes
asked Jan 14 at 0:08
Andre GomesAndre Gomes
920516
asked Jan 14 at 0:08
Andre GomesAndre Gomes
920516
asked Jan 14 at 0:08
Andre GomesAndre Gomes
920516
920516
1
$begingroup$
Well, what about the trivial representation?
$endgroup$
– Matt Samuel
Jan 14 at 0:11
$begingroup$
It should be a non-trivial representation*, I'll change @MattSamuel
$endgroup$
– Andre Gomes
Jan 14 at 0:12
$begingroup$
I have answered the first question here: math.stackexchange.com/questions/1033844/…
$endgroup$
– Andre Gomes
Jan 14 at 17:24
add a comment |
1
$begingroup$
Well, what about the trivial representation?
$endgroup$
– Matt Samuel
Jan 14 at 0:11
$begingroup$
It should be a non-trivial representation*, I'll change @MattSamuel
$endgroup$
– Andre Gomes
Jan 14 at 0:12
$begingroup$
I have answered the first question here: math.stackexchange.com/questions/1033844/…
$endgroup$
– Andre Gomes
Jan 14 at 17:24
1
1
$begingroup$
Well, what about the trivial representation?
$endgroup$
– Matt Samuel
Jan 14 at 0:11
$begingroup$
Well, what about the trivial representation?
$endgroup$
– Matt Samuel
Jan 14 at 0:11
$begingroup$
It should be a non-trivial representation*, I'll change @MattSamuel
$endgroup$
– Andre Gomes
Jan 14 at 0:12
$begingroup$
It should be a non-trivial representation*, I'll change @MattSamuel
$endgroup$
– Andre Gomes
Jan 14 at 0:12
$begingroup$
I have answered the first question here: math.stackexchange.com/questions/1033844/…
$endgroup$
– Andre Gomes
Jan 14 at 17:24
$begingroup$
I have answered the first question here: math.stackexchange.com/questions/1033844/…
$endgroup$
– Andre Gomes
Jan 14 at 17:24
add a comment |
1 Answer
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Any nonzero vector $vin V^G$ spans a $1$-dimensional trivial sub-representation. If $V$ is irreducible of dimension $>1$, then it follows that $V^G=0$. If $dim V=1$, then $V^Gneq 0$ implies $V$ is the trivial representation.
answered Jan 14 at 16:30
David HillDavid Hill
9,6561619
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Any nonzero vector $vin V^G$ spans a $1$-dimensional trivial sub-representation. If $V$ is irreducible of dimension $>1$, then it follows that $V^G=0$. If $dim V=1$, then $V^Gneq 0$ implies $V$ is the trivial representation.
answered Jan 14 at 16:30
David HillDavid Hill
9,6561619
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add a comment |
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Any nonzero vector $vin V^G$ spans a $1$-dimensional trivial sub-representation. If $V$ is irreducible of dimension $>1$, then it follows that $V^G=0$. If $dim V=1$, then $V^Gneq 0$ implies $V$ is the trivial representation.
answered Jan 14 at 16:30
David HillDavid Hill
9,6561619
$endgroup$
add a comment |
$begingroup$
Any nonzero vector $vin V^G$ spans a $1$-dimensional trivial sub-representation. If $V$ is irreducible of dimension $>1$, then it follows that $V^G=0$. If $dim V=1$, then $V^Gneq 0$ implies $V$ is the trivial representation.
answered Jan 14 at 16:30
David HillDavid Hill
9,6561619
$endgroup$
Any nonzero vector $vin V^G$ spans a $1$-dimensional trivial sub-representation. If $V$ is irreducible of dimension $>1$, then it follows that $V^G=0$. If $dim V=1$, then $V^Gneq 0$ implies $V$ is the trivial representation.
answered Jan 14 at 16:30
David HillDavid Hill
9,6561619
answered Jan 14 at 16:30
David HillDavid Hill
9,6561619
answered Jan 14 at 16:30
David HillDavid Hill
9,6561619
answered Jan 14 at 16:30
David HillDavid Hill
9,6561619
9,6561619
add a comment |
add a comment |
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1
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Well, what about the trivial representation?
$endgroup$
– Matt Samuel
Jan 14 at 0:11
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It should be a non-trivial representation*, I'll change @MattSamuel
$endgroup$
– Andre Gomes
Jan 14 at 0:12
$begingroup$
I have answered the first question here: math.stackexchange.com/questions/1033844/…
$endgroup$
– Andre Gomes
Jan 14 at 17:24