Mixing
Confusion on order of group action on function
Clash Royale CLAN TAG#URR8PPP
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Let $Gamma$ be a group and $Gamma'leq Gamma$ be the subgroup. Let $Y$ be $Gamma'$ module. Define induction $Ind^Gamma_Gamma'Y=<f:Gammato YvertforallgammainGamma,gamma'inGamma',f(gamma'gamma)=gamma'f(gamma)>$ as abelian group generated by elements prescribed.
Then $Gamma$ action on $Ind^Gamma_Gamma'$ is defined by $gammainGamma, (gamma f)(x)=f(xgamma)$.
$textbfQ:$ When I try to verify the group action for $gamma_iinGamma$ $[(gamma_1gamma_2)f](x)=f(xgamma_1gamma_2)$ and $(gamma_1)[(gamma_2)f]=gamma_1(f(xgamma_2))=f(xgamma_2gamma_1)$, I cannot have $(gamma_1gamma_2)f=gamma_1(gamma_2(f))$. Have I done something wrong? Neukirch defined $Gamma$ action by $(gamma f)(x)=f(gamma x)$. Then I have no trouble to see $Gamma$ action here by $(gamma_1gamma_2f)(x)=f(gamma_1gamma_2x)=gamma_1(f(gamma_2x))=gamma_1((gamma_2)f)(x)$
Per Arnaud Mortier's comment, I think I figured out what is wrong in my mind setting. $gamma f(x)=f(xgamma)$ is setting left action of $f$ to right action on $x$. $(gammagamma' )f(x)=f(xcdot(gammagamma'))=f((xgamma')gamma)=gamma f(xgamma')=gamma(gamma' f)(x)$
For $gamma f(x)=f(gamma x)$(Neukirch's definition), I need $(gammagamma')f(x)=gamma(gamma' f)(x)=gamma' f(gamma x)=f(gamma'gamma x)$. I need to change left action to right action again here.
Ref. Lectures on Algebraic Geometry by Gunter Harder 2.2.4 Exercise 7.
abstract-algebra
asked Jul 26 at 23:17
user45765
2,1942718
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up vote
4
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Let $Gamma$ be a group and $Gamma'leq Gamma$ be the subgroup. Let $Y$ be $Gamma'$ module. Define induction $Ind^Gamma_Gamma'Y=<f:Gammato YvertforallgammainGamma,gamma'inGamma',f(gamma'gamma)=gamma'f(gamma)>$ as abelian group generated by elements prescribed.
Then $Gamma$ action on $Ind^Gamma_Gamma'$ is defined by $gammainGamma, (gamma f)(x)=f(xgamma)$.
$textbfQ:$ When I try to verify the group action for $gamma_iinGamma$ $[(gamma_1gamma_2)f](x)=f(xgamma_1gamma_2)$ and $(gamma_1)[(gamma_2)f]=gamma_1(f(xgamma_2))=f(xgamma_2gamma_1)$, I cannot have $(gamma_1gamma_2)f=gamma_1(gamma_2(f))$. Have I done something wrong? Neukirch defined $Gamma$ action by $(gamma f)(x)=f(gamma x)$. Then I have no trouble to see $Gamma$ action here by $(gamma_1gamma_2f)(x)=f(gamma_1gamma_2x)=gamma_1(f(gamma_2x))=gamma_1((gamma_2)f)(x)$
Per Arnaud Mortier's comment, I think I figured out what is wrong in my mind setting. $gamma f(x)=f(xgamma)$ is setting left action of $f$ to right action on $x$. $(gammagamma' )f(x)=f(xcdot(gammagamma'))=f((xgamma')gamma)=gamma f(xgamma')=gamma(gamma' f)(x)$
For $gamma f(x)=f(gamma x)$(Neukirch's definition), I need $(gammagamma')f(x)=gamma(gamma' f)(x)=gamma' f(gamma x)=f(gamma'gamma x)$. I need to change left action to right action again here.
Ref. Lectures on Algebraic Geometry by Gunter Harder 2.2.4 Exercise 7.
abstract-algebra
asked Jul 26 at 23:17
user45765
2,1942718
There is a notion of right action and left action, and the syntax should be adapted accordingly.
– Arnaud Mortier
Jul 26 at 23:24
@ArnaudMortier Yes. That is what I thought. My guess is $Hom(-,Y)$ is contravariant. That is why you want right action turned into left action.
– user45765
Jul 26 at 23:28
@ArnaudMortier Thanks for the hint. I think I figured it out.
– user45765
Jul 26 at 23:48
Dear @user45765, In the derivation you have added there are two errors that cancel one another to give you the desired result. Namely the second and fourth equalities are incorrect.
– Keenan Kidwell
Jul 27 at 3:26
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $Gamma$ be a group and $Gamma'leq Gamma$ be the subgroup. Let $Y$ be $Gamma'$ module. Define induction $Ind^Gamma_Gamma'Y=<f:Gammato YvertforallgammainGamma,gamma'inGamma',f(gamma'gamma)=gamma'f(gamma)>$ as abelian group generated by elements prescribed.
Then $Gamma$ action on $Ind^Gamma_Gamma'$ is defined by $gammainGamma, (gamma f)(x)=f(xgamma)$.
$textbfQ:$ When I try to verify the group action for $gamma_iinGamma$ $[(gamma_1gamma_2)f](x)=f(xgamma_1gamma_2)$ and $(gamma_1)[(gamma_2)f]=gamma_1(f(xgamma_2))=f(xgamma_2gamma_1)$, I cannot have $(gamma_1gamma_2)f=gamma_1(gamma_2(f))$. Have I done something wrong? Neukirch defined $Gamma$ action by $(gamma f)(x)=f(gamma x)$. Then I have no trouble to see $Gamma$ action here by $(gamma_1gamma_2f)(x)=f(gamma_1gamma_2x)=gamma_1(f(gamma_2x))=gamma_1((gamma_2)f)(x)$
Per Arnaud Mortier's comment, I think I figured out what is wrong in my mind setting. $gamma f(x)=f(xgamma)$ is setting left action of $f$ to right action on $x$. $(gammagamma' )f(x)=f(xcdot(gammagamma'))=f((xgamma')gamma)=gamma f(xgamma')=gamma(gamma' f)(x)$
For $gamma f(x)=f(gamma x)$(Neukirch's definition), I need $(gammagamma')f(x)=gamma(gamma' f)(x)=gamma' f(gamma x)=f(gamma'gamma x)$. I need to change left action to right action again here.
Ref. Lectures on Algebraic Geometry by Gunter Harder 2.2.4 Exercise 7.
abstract-algebra
asked Jul 26 at 23:17
user45765
2,1942718
Let $Gamma$ be a group and $Gamma'leq Gamma$ be the subgroup. Let $Y$ be $Gamma'$ module. Define induction $Ind^Gamma_Gamma'Y=<f:Gammato YvertforallgammainGamma,gamma'inGamma',f(gamma'gamma)=gamma'f(gamma)>$ as abelian group generated by elements prescribed.
Then $Gamma$ action on $Ind^Gamma_Gamma'$ is defined by $gammainGamma, (gamma f)(x)=f(xgamma)$.
$textbfQ:$ When I try to verify the group action for $gamma_iinGamma$ $[(gamma_1gamma_2)f](x)=f(xgamma_1gamma_2)$ and $(gamma_1)[(gamma_2)f]=gamma_1(f(xgamma_2))=f(xgamma_2gamma_1)$, I cannot have $(gamma_1gamma_2)f=gamma_1(gamma_2(f))$. Have I done something wrong? Neukirch defined $Gamma$ action by $(gamma f)(x)=f(gamma x)$. Then I have no trouble to see $Gamma$ action here by $(gamma_1gamma_2f)(x)=f(gamma_1gamma_2x)=gamma_1(f(gamma_2x))=gamma_1((gamma_2)f)(x)$
Per Arnaud Mortier's comment, I think I figured out what is wrong in my mind setting. $gamma f(x)=f(xgamma)$ is setting left action of $f$ to right action on $x$. $(gammagamma' )f(x)=f(xcdot(gammagamma'))=f((xgamma')gamma)=gamma f(xgamma')=gamma(gamma' f)(x)$
For $gamma f(x)=f(gamma x)$(Neukirch's definition), I need $(gammagamma')f(x)=gamma(gamma' f)(x)=gamma' f(gamma x)=f(gamma'gamma x)$. I need to change left action to right action again here.
Ref. Lectures on Algebraic Geometry by Gunter Harder 2.2.4 Exercise 7.
abstract-algebra
asked Jul 26 at 23:17
user45765
2,1942718
edited Jul 26 at 23:47
asked Jul 26 at 23:17
user45765
2,1942718
asked Jul 26 at 23:17
user45765
2,1942718
asked Jul 26 at 23:17
user45765
2,1942718
2,1942718
There is a notion of right action and left action, and the syntax should be adapted accordingly.
– Arnaud Mortier
Jul 26 at 23:24
@ArnaudMortier Yes. That is what I thought. My guess is $Hom(-,Y)$ is contravariant. That is why you want right action turned into left action.
– user45765
Jul 26 at 23:28
@ArnaudMortier Thanks for the hint. I think I figured it out.
– user45765
Jul 26 at 23:48
Dear @user45765, In the derivation you have added there are two errors that cancel one another to give you the desired result. Namely the second and fourth equalities are incorrect.
– Keenan Kidwell
Jul 27 at 3:26
add a comment |Â
There is a notion of right action and left action, and the syntax should be adapted accordingly.
– Arnaud Mortier
Jul 26 at 23:24
@ArnaudMortier Yes. That is what I thought. My guess is $Hom(-,Y)$ is contravariant. That is why you want right action turned into left action.
– user45765
Jul 26 at 23:28
@ArnaudMortier Thanks for the hint. I think I figured it out.
– user45765
Jul 26 at 23:48
Dear @user45765, In the derivation you have added there are two errors that cancel one another to give you the desired result. Namely the second and fourth equalities are incorrect.
– Keenan Kidwell
Jul 27 at 3:26
There is a notion of right action and left action, and the syntax should be adapted accordingly.
– Arnaud Mortier
Jul 26 at 23:24
There is a notion of right action and left action, and the syntax should be adapted accordingly.
– Arnaud Mortier
Jul 26 at 23:24
@ArnaudMortier Yes. That is what I thought. My guess is $Hom(-,Y)$ is contravariant. That is why you want right action turned into left action.
– user45765
Jul 26 at 23:28
@ArnaudMortier Yes. That is what I thought. My guess is $Hom(-,Y)$ is contravariant. That is why you want right action turned into left action.
– user45765
Jul 26 at 23:28
@ArnaudMortier Thanks for the hint. I think I figured it out.
– user45765
Jul 26 at 23:48
@ArnaudMortier Thanks for the hint. I think I figured it out.
– user45765
Jul 26 at 23:48
Dear @user45765, In the derivation you have added there are two errors that cancel one another to give you the desired result. Namely the second and fourth equalities are incorrect.
– Keenan Kidwell
Jul 27 at 3:26
Dear @user45765, In the derivation you have added there are two errors that cancel one another to give you the desired result. Namely the second and fourth equalities are incorrect.
– Keenan Kidwell
Jul 27 at 3:26
add a comment |Â
1 Answer
1
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votes
up vote
1
down vote
accepted
The correct computation to verify that $(gamma_1gamma_2)f=gamma_1(gamma_2f)$ goes as follows. Given $xinGamma$, we have
$$((gamma_1gamma_2)f)(x)=f(x(gamma_1gamma_2))=f((xgamma_1)gamma_2)
=(gamma_2 f)(xgamma_1)=(gamma_1(gamma_2f))(x)text.$$
Note also that in your definition of the induced $Gamma$-module, you do not need to include "abelian group generated by [the] elements prescribed," because the indicated functions already form a group under pointwise addition (using the operation on $Y$) of functions.
answered Jul 27 at 3:30
Keenan Kidwell
18.9k13069
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The correct computation to verify that $(gamma_1gamma_2)f=gamma_1(gamma_2f)$ goes as follows. Given $xinGamma$, we have
$$((gamma_1gamma_2)f)(x)=f(x(gamma_1gamma_2))=f((xgamma_1)gamma_2)
=(gamma_2 f)(xgamma_1)=(gamma_1(gamma_2f))(x)text.$$
Note also that in your definition of the induced $Gamma$-module, you do not need to include "abelian group generated by [the] elements prescribed," because the indicated functions already form a group under pointwise addition (using the operation on $Y$) of functions.
answered Jul 27 at 3:30
Keenan Kidwell
18.9k13069
add a comment |Â
up vote
1
down vote
accepted
The correct computation to verify that $(gamma_1gamma_2)f=gamma_1(gamma_2f)$ goes as follows. Given $xinGamma$, we have
$$((gamma_1gamma_2)f)(x)=f(x(gamma_1gamma_2))=f((xgamma_1)gamma_2)
=(gamma_2 f)(xgamma_1)=(gamma_1(gamma_2f))(x)text.$$
Note also that in your definition of the induced $Gamma$-module, you do not need to include "abelian group generated by [the] elements prescribed," because the indicated functions already form a group under pointwise addition (using the operation on $Y$) of functions.
answered Jul 27 at 3:30
Keenan Kidwell
18.9k13069
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The correct computation to verify that $(gamma_1gamma_2)f=gamma_1(gamma_2f)$ goes as follows. Given $xinGamma$, we have
$$((gamma_1gamma_2)f)(x)=f(x(gamma_1gamma_2))=f((xgamma_1)gamma_2)
=(gamma_2 f)(xgamma_1)=(gamma_1(gamma_2f))(x)text.$$
Note also that in your definition of the induced $Gamma$-module, you do not need to include "abelian group generated by [the] elements prescribed," because the indicated functions already form a group under pointwise addition (using the operation on $Y$) of functions.
answered Jul 27 at 3:30
Keenan Kidwell
18.9k13069
The correct computation to verify that $(gamma_1gamma_2)f=gamma_1(gamma_2f)$ goes as follows. Given $xinGamma$, we have
$$((gamma_1gamma_2)f)(x)=f(x(gamma_1gamma_2))=f((xgamma_1)gamma_2)
=(gamma_2 f)(xgamma_1)=(gamma_1(gamma_2f))(x)text.$$
Note also that in your definition of the induced $Gamma$-module, you do not need to include "abelian group generated by [the] elements prescribed," because the indicated functions already form a group under pointwise addition (using the operation on $Y$) of functions.
answered Jul 27 at 3:30
Keenan Kidwell
18.9k13069
answered Jul 27 at 3:30
Keenan Kidwell
18.9k13069
answered Jul 27 at 3:30
Keenan Kidwell
18.9k13069
answered Jul 27 at 3:30
Keenan Kidwell
18.9k13069
18.9k13069
add a comment |Â
add a comment |Â
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There is a notion of right action and left action, and the syntax should be adapted accordingly.
– Arnaud Mortier
Jul 26 at 23:24
@ArnaudMortier Yes. That is what I thought. My guess is $Hom(-,Y)$ is contravariant. That is why you want right action turned into left action.
– user45765
Jul 26 at 23:28
@ArnaudMortier Thanks for the hint. I think I figured it out.
– user45765
Jul 26 at 23:48
Dear @user45765, In the derivation you have added there are two errors that cancel one another to give you the desired result. Namely the second and fourth equalities are incorrect.
– Keenan Kidwell
Jul 27 at 3:26