Mixing
For $U$ a $G-$invariant subspace of $V$ a $G-$representation, show that $T_g(U)=U$
$begingroup$
Prove that if the subspace $U$ of the space of the representation $T:Gto GL(V)$ of $G$ is invariant, then $T(g)U = U$ for all $g in G.$
Could any one give me a hint how to solve this question ?
representation-theory invariant-theory
edited Jan 22 at 21:00
Antonios-Alexandros Robotis
10.5k41741
asked Jan 22 at 20:48
IntuitionIntuition
1,089826
$endgroup$
add a comment |
$begingroup$
Prove that if the subspace $U$ of the space of the representation $T:Gto GL(V)$ of $G$ is invariant, then $T(g)U = U$ for all $g in G.$
Could any one give me a hint how to solve this question ?
representation-theory invariant-theory
edited Jan 22 at 21:00
Antonios-Alexandros Robotis
10.5k41741
asked Jan 22 at 20:48
IntuitionIntuition
1,089826
$endgroup$
add a comment |
$begingroup$
Prove that if the subspace $U$ of the space of the representation $T:Gto GL(V)$ of $G$ is invariant, then $T(g)U = U$ for all $g in G.$
Could any one give me a hint how to solve this question ?
representation-theory invariant-theory
edited Jan 22 at 21:00
Antonios-Alexandros Robotis
10.5k41741
asked Jan 22 at 20:48
IntuitionIntuition
1,089826
$endgroup$
Prove that if the subspace $U$ of the space of the representation $T:Gto GL(V)$ of $G$ is invariant, then $T(g)U = U$ for all $g in G.$
Could any one give me a hint how to solve this question ?
representation-theory invariant-theory
representation-theory invariant-theory
edited Jan 22 at 21:00
Antonios-Alexandros Robotis
10.5k41741
asked Jan 22 at 20:48
IntuitionIntuition
1,089826
edited Jan 22 at 21:00
Antonios-Alexandros Robotis
10.5k41741
asked Jan 22 at 20:48
IntuitionIntuition
1,089826
edited Jan 22 at 21:00
Antonios-Alexandros Robotis
10.5k41741
edited Jan 22 at 21:00
Antonios-Alexandros Robotis
10.5k41741
edited Jan 22 at 21:00
Antonios-Alexandros Robotis
10.5k41741
10.5k41741
asked Jan 22 at 20:48
IntuitionIntuition
1,089826
asked Jan 22 at 20:48
IntuitionIntuition
1,089826
asked Jan 22 at 20:48
IntuitionIntuition
1,089826
1,089826
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Well, I'm assuming $V$ is a $mathbb{C}-$vector space and $T:Gto GL(V)$ is your representation. The fact that $U$ is $G-$invariant means that for each element $gin G$ $T_g(U)subseteq U$. However, we also know that $T_g$ is a nonsingular linear transformation, so $dim T_g(U)=dim U$ so, $T_g(U)=U$.
answered Jan 22 at 20:58
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Well, I'm assuming $V$ is a $mathbb{C}-$vector space and $T:Gto GL(V)$ is your representation. The fact that $U$ is $G-$invariant means that for each element $gin G$ $T_g(U)subseteq U$. However, we also know that $T_g$ is a nonsingular linear transformation, so $dim T_g(U)=dim U$ so, $T_g(U)=U$.
answered Jan 22 at 20:58
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
$endgroup$
add a comment |
$begingroup$
Well, I'm assuming $V$ is a $mathbb{C}-$vector space and $T:Gto GL(V)$ is your representation. The fact that $U$ is $G-$invariant means that for each element $gin G$ $T_g(U)subseteq U$. However, we also know that $T_g$ is a nonsingular linear transformation, so $dim T_g(U)=dim U$ so, $T_g(U)=U$.
answered Jan 22 at 20:58
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
$endgroup$
add a comment |
$begingroup$
Well, I'm assuming $V$ is a $mathbb{C}-$vector space and $T:Gto GL(V)$ is your representation. The fact that $U$ is $G-$invariant means that for each element $gin G$ $T_g(U)subseteq U$. However, we also know that $T_g$ is a nonsingular linear transformation, so $dim T_g(U)=dim U$ so, $T_g(U)=U$.
answered Jan 22 at 20:58
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
$endgroup$
Well, I'm assuming $V$ is a $mathbb{C}-$vector space and $T:Gto GL(V)$ is your representation. The fact that $U$ is $G-$invariant means that for each element $gin G$ $T_g(U)subseteq U$. However, we also know that $T_g$ is a nonsingular linear transformation, so $dim T_g(U)=dim U$ so, $T_g(U)=U$.
answered Jan 22 at 20:58
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
answered Jan 22 at 20:58
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
answered Jan 22 at 20:58
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
answered Jan 22 at 20:58
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
10.5k41741
add a comment |
add a comment |
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