Mixing
For two invariant complements in the space of representation $T$. Prove that $T_{W_1}$ isomorphic to$...
$begingroup$
Let $W_{1}$ and $W_{2}$ be two invariant complements of the invariant space $U$ in the space of the representation $T$ prove that $T_{W_{1}}$ equivalent to $T_{W_{2}}$.
Invariant Complement Definition:
Two representations are equivalent iff:
Could anyone give me a hint for doing so please?
Thank you
representation-theory invariant-theory
edited Jan 23 at 6:51
hopefully
215114
asked Jan 22 at 21:43
IntuitionIntuition
1,089826
$endgroup$
add a comment |
$begingroup$
Let $W_{1}$ and $W_{2}$ be two invariant complements of the invariant space $U$ in the space of the representation $T$ prove that $T_{W_{1}}$ equivalent to $T_{W_{2}}$.
Invariant Complement Definition:
Two representations are equivalent iff:
Could anyone give me a hint for doing so please?
Thank you
representation-theory invariant-theory
edited Jan 23 at 6:51
hopefully
215114
asked Jan 22 at 21:43
IntuitionIntuition
1,089826
$endgroup$
$begingroup$
It would be easier to answer if you also give the definition for invariant complements and $T_{W_1}$.
$endgroup$
– Levent
Jan 22 at 22:01
$begingroup$
Okay sorry I will add the definitions @Levent
$endgroup$
– Intuition
Jan 22 at 23:21
$begingroup$
I have added the definition from the book the OP is using but it is pending peer reviewing @reuns
$endgroup$
– hopefully
Jan 23 at 6:12
add a comment |
$begingroup$
Let $W_{1}$ and $W_{2}$ be two invariant complements of the invariant space $U$ in the space of the representation $T$ prove that $T_{W_{1}}$ equivalent to $T_{W_{2}}$.
Invariant Complement Definition:
Two representations are equivalent iff:
Could anyone give me a hint for doing so please?
Thank you
representation-theory invariant-theory
edited Jan 23 at 6:51
hopefully
215114
asked Jan 22 at 21:43
IntuitionIntuition
1,089826
$endgroup$
Let $W_{1}$ and $W_{2}$ be two invariant complements of the invariant space $U$ in the space of the representation $T$ prove that $T_{W_{1}}$ equivalent to $T_{W_{2}}$.
Invariant Complement Definition:
Two representations are equivalent iff:
Could anyone give me a hint for doing so please?
Thank you
representation-theory invariant-theory
representation-theory invariant-theory
edited Jan 23 at 6:51
hopefully
215114
asked Jan 22 at 21:43
IntuitionIntuition
1,089826
edited Jan 23 at 6:51
hopefully
215114
asked Jan 22 at 21:43
IntuitionIntuition
1,089826
edited Jan 23 at 6:51
hopefully
215114
edited Jan 23 at 6:51
hopefully
215114
edited Jan 23 at 6:51
hopefully
215114
215114
asked Jan 22 at 21:43
IntuitionIntuition
1,089826
asked Jan 22 at 21:43
IntuitionIntuition
1,089826
asked Jan 22 at 21:43
IntuitionIntuition
1,089826
1,089826
$begingroup$
It would be easier to answer if you also give the definition for invariant complements and $T_{W_1}$.
$endgroup$
– Levent
Jan 22 at 22:01
$begingroup$
Okay sorry I will add the definitions @Levent
$endgroup$
– Intuition
Jan 22 at 23:21
$begingroup$
I have added the definition from the book the OP is using but it is pending peer reviewing @reuns
$endgroup$
– hopefully
Jan 23 at 6:12
add a comment |
$begingroup$
It would be easier to answer if you also give the definition for invariant complements and $T_{W_1}$.
$endgroup$
– Levent
Jan 22 at 22:01
$begingroup$
Okay sorry I will add the definitions @Levent
$endgroup$
– Intuition
Jan 22 at 23:21
$begingroup$
I have added the definition from the book the OP is using but it is pending peer reviewing @reuns
$endgroup$
– hopefully
Jan 23 at 6:12
$begingroup$
It would be easier to answer if you also give the definition for invariant complements and $T_{W_1}$.
$endgroup$
– Levent
Jan 22 at 22:01
$begingroup$
It would be easier to answer if you also give the definition for invariant complements and $T_{W_1}$.
$endgroup$
– Levent
Jan 22 at 22:01
$begingroup$
Okay sorry I will add the definitions @Levent
$endgroup$
– Intuition
Jan 22 at 23:21
$begingroup$
Okay sorry I will add the definitions @Levent
$endgroup$
– Intuition
Jan 22 at 23:21
$begingroup$
I have added the definition from the book the OP is using but it is pending peer reviewing @reuns
$endgroup$
– hopefully
Jan 23 at 6:12
$begingroup$
I have added the definition from the book the OP is using but it is pending peer reviewing @reuns
$endgroup$
– hopefully
Jan 23 at 6:12
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $V=Uoplus W$. Define the quotient module $V/U$ by $gcdot (v+U)=gcdot v+U$ for $vin V$ and $gin G$. Can you show that this indeed defines a module? Then use the canonical quotient map $q:V=Uoplus Wrightarrow V/U$ and restrict it to $W$. Then $qmid_W$ is injective since $Wcap U={0}$ and surjective by the fact that $V=U+W$. Also observe that $q(gcdot w)=gcdot w+U =gcdot(w+U)=gcdot q(w)$ so $q$ is a $G$-module morphism. Hence $q$ is an isomorphism between $G$-modules. As a result any $G$-invariant complement of $U$ in $V$ is isomorphic to the module $V/U$ which implies that they are all in fact isomorphic as $G$-modules.
Edit : To clarify, I would like to explain a bit more. For a submodule $U$ of $V$, you can define the quotient module $V/U$ as a I explained above. Now, if $W$ is also a submodule of $V$ satisfying $V=Uoplus W$, the quotient map $q:Vrightarrow V/U$ will be injective when restricted to $W$ (if $q(w)=0+U$, then $w+U=0+U$ so $win U$, and since $Ucap W={0}$ we get $w=0$. Hence $qmid_W$ is injective). Moreover, it will be surjective since for every element $v+U$ in $V/U$ we can write $v=w+u$ for some $win W$ and $uin U$ (this is true because $V=U+W$) and then $v+U=w+u+U=w+U=q(w)$. Lastly, it is a $G$-module morphism as I have shown above. Then we deduce that $qmid_W$ is in fact an isomorphism of two $G$-modules, $W$ and $V/U$. Now this is true for every submodule $W$ of $V$ satisfying $V=Uoplus W$. Hence if both $W_1$ and $W_2$ satisfy this property, we conclude that they are both isomorphic to $V/U$ and thus, they are in fact isomorphic to each other.
answered Jan 23 at 9:27
LeventLevent
2,729925
$endgroup$
2
$begingroup$
@Intuition I'd say at first $W $ (and your $W_1,W_2$) doesn't exist. You only have a representation $G,V$ and a sub-representation $G,U$. Show that $q : V to V/U$ is well-defined and commutes with the action of $G$, obtaining a representation $G,V/U$. Then, assume such $W$ exists (a complementary subspace which is a subrepresentation of $G,V$) and conclude
$endgroup$
– reuns
Jan 23 at 20:38
1
$begingroup$
@Intuition You need to make things precise. It is always the case that for some subspace $W$ we have $V = U oplus W$. But for most such $W$ you won't have $gcdot w in W$ for every $w in W,g in G$. So in the context of $G$-modules/representations you can't identify $V/U$ with $W$ as you did in other contexts. The point of the decomposition in irreducible representations is to find if such a $W$ with $gcdot w in W$ exists.
$endgroup$
– reuns
Jan 23 at 21:18
1
$begingroup$
@hopefully I edited.
$endgroup$
– Levent
Jan 24 at 0:59
1
$begingroup$
@Idonotknow yes, I think it is a trivial step since $q$ is the canonical quotient of vector spaces, i.e. already a linear map.
$endgroup$
– Levent
Jan 25 at 13:51
1
$begingroup$
@Intuition We are trying to show that two $G$-modules are isomorphic, not two vector spaces. Hence we need to find a $G$-module morphism that has an inverse. If you want to prove that two $G$-modules are isomorphic, you cannot just write a vector space isomorphism, you also need to show that the map respects $G$-action.
$endgroup$
– Levent
Jan 25 at 13:52
|
show 9 more comments
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1 Answer
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1 Answer
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$begingroup$
Let $V=Uoplus W$. Define the quotient module $V/U$ by $gcdot (v+U)=gcdot v+U$ for $vin V$ and $gin G$. Can you show that this indeed defines a module? Then use the canonical quotient map $q:V=Uoplus Wrightarrow V/U$ and restrict it to $W$. Then $qmid_W$ is injective since $Wcap U={0}$ and surjective by the fact that $V=U+W$. Also observe that $q(gcdot w)=gcdot w+U =gcdot(w+U)=gcdot q(w)$ so $q$ is a $G$-module morphism. Hence $q$ is an isomorphism between $G$-modules. As a result any $G$-invariant complement of $U$ in $V$ is isomorphic to the module $V/U$ which implies that they are all in fact isomorphic as $G$-modules.
Edit : To clarify, I would like to explain a bit more. For a submodule $U$ of $V$, you can define the quotient module $V/U$ as a I explained above. Now, if $W$ is also a submodule of $V$ satisfying $V=Uoplus W$, the quotient map $q:Vrightarrow V/U$ will be injective when restricted to $W$ (if $q(w)=0+U$, then $w+U=0+U$ so $win U$, and since $Ucap W={0}$ we get $w=0$. Hence $qmid_W$ is injective). Moreover, it will be surjective since for every element $v+U$ in $V/U$ we can write $v=w+u$ for some $win W$ and $uin U$ (this is true because $V=U+W$) and then $v+U=w+u+U=w+U=q(w)$. Lastly, it is a $G$-module morphism as I have shown above. Then we deduce that $qmid_W$ is in fact an isomorphism of two $G$-modules, $W$ and $V/U$. Now this is true for every submodule $W$ of $V$ satisfying $V=Uoplus W$. Hence if both $W_1$ and $W_2$ satisfy this property, we conclude that they are both isomorphic to $V/U$ and thus, they are in fact isomorphic to each other.
answered Jan 23 at 9:27
LeventLevent
2,729925
$endgroup$
2
$begingroup$
@Intuition I'd say at first $W $ (and your $W_1,W_2$) doesn't exist. You only have a representation $G,V$ and a sub-representation $G,U$. Show that $q : V to V/U$ is well-defined and commutes with the action of $G$, obtaining a representation $G,V/U$. Then, assume such $W$ exists (a complementary subspace which is a subrepresentation of $G,V$) and conclude
$endgroup$
– reuns
Jan 23 at 20:38
1
$begingroup$
@Intuition You need to make things precise. It is always the case that for some subspace $W$ we have $V = U oplus W$. But for most such $W$ you won't have $gcdot w in W$ for every $w in W,g in G$. So in the context of $G$-modules/representations you can't identify $V/U$ with $W$ as you did in other contexts. The point of the decomposition in irreducible representations is to find if such a $W$ with $gcdot w in W$ exists.
$endgroup$
– reuns
Jan 23 at 21:18
1
$begingroup$
@hopefully I edited.
$endgroup$
– Levent
Jan 24 at 0:59
1
$begingroup$
@Idonotknow yes, I think it is a trivial step since $q$ is the canonical quotient of vector spaces, i.e. already a linear map.
$endgroup$
– Levent
Jan 25 at 13:51
1
$begingroup$
@Intuition We are trying to show that two $G$-modules are isomorphic, not two vector spaces. Hence we need to find a $G$-module morphism that has an inverse. If you want to prove that two $G$-modules are isomorphic, you cannot just write a vector space isomorphism, you also need to show that the map respects $G$-action.
$endgroup$
– Levent
Jan 25 at 13:52
|
show 9 more comments
$begingroup$
Let $V=Uoplus W$. Define the quotient module $V/U$ by $gcdot (v+U)=gcdot v+U$ for $vin V$ and $gin G$. Can you show that this indeed defines a module? Then use the canonical quotient map $q:V=Uoplus Wrightarrow V/U$ and restrict it to $W$. Then $qmid_W$ is injective since $Wcap U={0}$ and surjective by the fact that $V=U+W$. Also observe that $q(gcdot w)=gcdot w+U =gcdot(w+U)=gcdot q(w)$ so $q$ is a $G$-module morphism. Hence $q$ is an isomorphism between $G$-modules. As a result any $G$-invariant complement of $U$ in $V$ is isomorphic to the module $V/U$ which implies that they are all in fact isomorphic as $G$-modules.
Edit : To clarify, I would like to explain a bit more. For a submodule $U$ of $V$, you can define the quotient module $V/U$ as a I explained above. Now, if $W$ is also a submodule of $V$ satisfying $V=Uoplus W$, the quotient map $q:Vrightarrow V/U$ will be injective when restricted to $W$ (if $q(w)=0+U$, then $w+U=0+U$ so $win U$, and since $Ucap W={0}$ we get $w=0$. Hence $qmid_W$ is injective). Moreover, it will be surjective since for every element $v+U$ in $V/U$ we can write $v=w+u$ for some $win W$ and $uin U$ (this is true because $V=U+W$) and then $v+U=w+u+U=w+U=q(w)$. Lastly, it is a $G$-module morphism as I have shown above. Then we deduce that $qmid_W$ is in fact an isomorphism of two $G$-modules, $W$ and $V/U$. Now this is true for every submodule $W$ of $V$ satisfying $V=Uoplus W$. Hence if both $W_1$ and $W_2$ satisfy this property, we conclude that they are both isomorphic to $V/U$ and thus, they are in fact isomorphic to each other.
answered Jan 23 at 9:27
LeventLevent
2,729925
$endgroup$
2
$begingroup$
@Intuition I'd say at first $W $ (and your $W_1,W_2$) doesn't exist. You only have a representation $G,V$ and a sub-representation $G,U$. Show that $q : V to V/U$ is well-defined and commutes with the action of $G$, obtaining a representation $G,V/U$. Then, assume such $W$ exists (a complementary subspace which is a subrepresentation of $G,V$) and conclude
$endgroup$
– reuns
Jan 23 at 20:38
1
$begingroup$
@Intuition You need to make things precise. It is always the case that for some subspace $W$ we have $V = U oplus W$. But for most such $W$ you won't have $gcdot w in W$ for every $w in W,g in G$. So in the context of $G$-modules/representations you can't identify $V/U$ with $W$ as you did in other contexts. The point of the decomposition in irreducible representations is to find if such a $W$ with $gcdot w in W$ exists.
$endgroup$
– reuns
Jan 23 at 21:18
1
$begingroup$
@hopefully I edited.
$endgroup$
– Levent
Jan 24 at 0:59
1
$begingroup$
@Idonotknow yes, I think it is a trivial step since $q$ is the canonical quotient of vector spaces, i.e. already a linear map.
$endgroup$
– Levent
Jan 25 at 13:51
1
$begingroup$
@Intuition We are trying to show that two $G$-modules are isomorphic, not two vector spaces. Hence we need to find a $G$-module morphism that has an inverse. If you want to prove that two $G$-modules are isomorphic, you cannot just write a vector space isomorphism, you also need to show that the map respects $G$-action.
$endgroup$
– Levent
Jan 25 at 13:52
|
show 9 more comments
$begingroup$
Let $V=Uoplus W$. Define the quotient module $V/U$ by $gcdot (v+U)=gcdot v+U$ for $vin V$ and $gin G$. Can you show that this indeed defines a module? Then use the canonical quotient map $q:V=Uoplus Wrightarrow V/U$ and restrict it to $W$. Then $qmid_W$ is injective since $Wcap U={0}$ and surjective by the fact that $V=U+W$. Also observe that $q(gcdot w)=gcdot w+U =gcdot(w+U)=gcdot q(w)$ so $q$ is a $G$-module morphism. Hence $q$ is an isomorphism between $G$-modules. As a result any $G$-invariant complement of $U$ in $V$ is isomorphic to the module $V/U$ which implies that they are all in fact isomorphic as $G$-modules.
Edit : To clarify, I would like to explain a bit more. For a submodule $U$ of $V$, you can define the quotient module $V/U$ as a I explained above. Now, if $W$ is also a submodule of $V$ satisfying $V=Uoplus W$, the quotient map $q:Vrightarrow V/U$ will be injective when restricted to $W$ (if $q(w)=0+U$, then $w+U=0+U$ so $win U$, and since $Ucap W={0}$ we get $w=0$. Hence $qmid_W$ is injective). Moreover, it will be surjective since for every element $v+U$ in $V/U$ we can write $v=w+u$ for some $win W$ and $uin U$ (this is true because $V=U+W$) and then $v+U=w+u+U=w+U=q(w)$. Lastly, it is a $G$-module morphism as I have shown above. Then we deduce that $qmid_W$ is in fact an isomorphism of two $G$-modules, $W$ and $V/U$. Now this is true for every submodule $W$ of $V$ satisfying $V=Uoplus W$. Hence if both $W_1$ and $W_2$ satisfy this property, we conclude that they are both isomorphic to $V/U$ and thus, they are in fact isomorphic to each other.
answered Jan 23 at 9:27
LeventLevent
2,729925
$endgroup$
Let $V=Uoplus W$. Define the quotient module $V/U$ by $gcdot (v+U)=gcdot v+U$ for $vin V$ and $gin G$. Can you show that this indeed defines a module? Then use the canonical quotient map $q:V=Uoplus Wrightarrow V/U$ and restrict it to $W$. Then $qmid_W$ is injective since $Wcap U={0}$ and surjective by the fact that $V=U+W$. Also observe that $q(gcdot w)=gcdot w+U =gcdot(w+U)=gcdot q(w)$ so $q$ is a $G$-module morphism. Hence $q$ is an isomorphism between $G$-modules. As a result any $G$-invariant complement of $U$ in $V$ is isomorphic to the module $V/U$ which implies that they are all in fact isomorphic as $G$-modules.
Edit : To clarify, I would like to explain a bit more. For a submodule $U$ of $V$, you can define the quotient module $V/U$ as a I explained above. Now, if $W$ is also a submodule of $V$ satisfying $V=Uoplus W$, the quotient map $q:Vrightarrow V/U$ will be injective when restricted to $W$ (if $q(w)=0+U$, then $w+U=0+U$ so $win U$, and since $Ucap W={0}$ we get $w=0$. Hence $qmid_W$ is injective). Moreover, it will be surjective since for every element $v+U$ in $V/U$ we can write $v=w+u$ for some $win W$ and $uin U$ (this is true because $V=U+W$) and then $v+U=w+u+U=w+U=q(w)$. Lastly, it is a $G$-module morphism as I have shown above. Then we deduce that $qmid_W$ is in fact an isomorphism of two $G$-modules, $W$ and $V/U$. Now this is true for every submodule $W$ of $V$ satisfying $V=Uoplus W$. Hence if both $W_1$ and $W_2$ satisfy this property, we conclude that they are both isomorphic to $V/U$ and thus, they are in fact isomorphic to each other.
answered Jan 23 at 9:27
LeventLevent
2,729925
edited Jan 24 at 0:59
answered Jan 23 at 9:27
LeventLevent
2,729925
answered Jan 23 at 9:27
LeventLevent
2,729925
answered Jan 23 at 9:27
LeventLevent
2,729925
2,729925
2
$begingroup$
@Intuition I'd say at first $W $ (and your $W_1,W_2$) doesn't exist. You only have a representation $G,V$ and a sub-representation $G,U$. Show that $q : V to V/U$ is well-defined and commutes with the action of $G$, obtaining a representation $G,V/U$. Then, assume such $W$ exists (a complementary subspace which is a subrepresentation of $G,V$) and conclude
$endgroup$
– reuns
Jan 23 at 20:38
1
$begingroup$
@Intuition You need to make things precise. It is always the case that for some subspace $W$ we have $V = U oplus W$. But for most such $W$ you won't have $gcdot w in W$ for every $w in W,g in G$. So in the context of $G$-modules/representations you can't identify $V/U$ with $W$ as you did in other contexts. The point of the decomposition in irreducible representations is to find if such a $W$ with $gcdot w in W$ exists.
$endgroup$
– reuns
Jan 23 at 21:18
1
$begingroup$
@hopefully I edited.
$endgroup$
– Levent
Jan 24 at 0:59
1
$begingroup$
@Idonotknow yes, I think it is a trivial step since $q$ is the canonical quotient of vector spaces, i.e. already a linear map.
$endgroup$
– Levent
Jan 25 at 13:51
1
$begingroup$
@Intuition We are trying to show that two $G$-modules are isomorphic, not two vector spaces. Hence we need to find a $G$-module morphism that has an inverse. If you want to prove that two $G$-modules are isomorphic, you cannot just write a vector space isomorphism, you also need to show that the map respects $G$-action.
$endgroup$
– Levent
Jan 25 at 13:52
|
show 9 more comments
2
$begingroup$
@Intuition I'd say at first $W $ (and your $W_1,W_2$) doesn't exist. You only have a representation $G,V$ and a sub-representation $G,U$. Show that $q : V to V/U$ is well-defined and commutes with the action of $G$, obtaining a representation $G,V/U$. Then, assume such $W$ exists (a complementary subspace which is a subrepresentation of $G,V$) and conclude
$endgroup$
– reuns
Jan 23 at 20:38
1
$begingroup$
@Intuition You need to make things precise. It is always the case that for some subspace $W$ we have $V = U oplus W$. But for most such $W$ you won't have $gcdot w in W$ for every $w in W,g in G$. So in the context of $G$-modules/representations you can't identify $V/U$ with $W$ as you did in other contexts. The point of the decomposition in irreducible representations is to find if such a $W$ with $gcdot w in W$ exists.
$endgroup$
– reuns
Jan 23 at 21:18
1
$begingroup$
@hopefully I edited.
$endgroup$
– Levent
Jan 24 at 0:59
1
$begingroup$
@Idonotknow yes, I think it is a trivial step since $q$ is the canonical quotient of vector spaces, i.e. already a linear map.
$endgroup$
– Levent
Jan 25 at 13:51
1
$begingroup$
@Intuition We are trying to show that two $G$-modules are isomorphic, not two vector spaces. Hence we need to find a $G$-module morphism that has an inverse. If you want to prove that two $G$-modules are isomorphic, you cannot just write a vector space isomorphism, you also need to show that the map respects $G$-action.
$endgroup$
– Levent
Jan 25 at 13:52
2
2
$begingroup$
@Intuition I'd say at first $W $ (and your $W_1,W_2$) doesn't exist. You only have a representation $G,V$ and a sub-representation $G,U$. Show that $q : V to V/U$ is well-defined and commutes with the action of $G$, obtaining a representation $G,V/U$. Then, assume such $W$ exists (a complementary subspace which is a subrepresentation of $G,V$) and conclude
$endgroup$
– reuns
Jan 23 at 20:38
$begingroup$
@Intuition I'd say at first $W $ (and your $W_1,W_2$) doesn't exist. You only have a representation $G,V$ and a sub-representation $G,U$. Show that $q : V to V/U$ is well-defined and commutes with the action of $G$, obtaining a representation $G,V/U$. Then, assume such $W$ exists (a complementary subspace which is a subrepresentation of $G,V$) and conclude
$endgroup$
– reuns
Jan 23 at 20:38
1
1
$begingroup$
@Intuition You need to make things precise. It is always the case that for some subspace $W$ we have $V = U oplus W$. But for most such $W$ you won't have $gcdot w in W$ for every $w in W,g in G$. So in the context of $G$-modules/representations you can't identify $V/U$ with $W$ as you did in other contexts. The point of the decomposition in irreducible representations is to find if such a $W$ with $gcdot w in W$ exists.
$endgroup$
– reuns
Jan 23 at 21:18
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@Intuition You need to make things precise. It is always the case that for some subspace $W$ we have $V = U oplus W$. But for most such $W$ you won't have $gcdot w in W$ for every $w in W,g in G$. So in the context of $G$-modules/representations you can't identify $V/U$ with $W$ as you did in other contexts. The point of the decomposition in irreducible representations is to find if such a $W$ with $gcdot w in W$ exists.
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– reuns
Jan 23 at 21:18
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@hopefully I edited.
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– Levent
Jan 24 at 0:59
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@hopefully I edited.
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– Levent
Jan 24 at 0:59
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@Idonotknow yes, I think it is a trivial step since $q$ is the canonical quotient of vector spaces, i.e. already a linear map.
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– Levent
Jan 25 at 13:51
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@Idonotknow yes, I think it is a trivial step since $q$ is the canonical quotient of vector spaces, i.e. already a linear map.
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– Levent
Jan 25 at 13:51
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@Intuition We are trying to show that two $G$-modules are isomorphic, not two vector spaces. Hence we need to find a $G$-module morphism that has an inverse. If you want to prove that two $G$-modules are isomorphic, you cannot just write a vector space isomorphism, you also need to show that the map respects $G$-action.
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– Levent
Jan 25 at 13:52
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@Intuition We are trying to show that two $G$-modules are isomorphic, not two vector spaces. Hence we need to find a $G$-module morphism that has an inverse. If you want to prove that two $G$-modules are isomorphic, you cannot just write a vector space isomorphism, you also need to show that the map respects $G$-action.
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– Levent
Jan 25 at 13:52
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It would be easier to answer if you also give the definition for invariant complements and $T_{W_1}$.
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– Levent
Jan 22 at 22:01
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Okay sorry I will add the definitions @Levent
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– Intuition
Jan 22 at 23:21
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I have added the definition from the book the OP is using but it is pending peer reviewing @reuns
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– hopefully
Jan 23 at 6:12