About the definition of coinduction of a module
$begingroup$
If $G$ is a group, $H$ a subgroup, and $N$ a left $mathbb{Z}[H]-$module, I've learned the following construction:
$$mathrm{coInd}_H^G(N) = mathrm{Hom}_{mathbb{Z}[H]}(mathbb{Z}[G], N)$$
where $mathbb{Z}[G]$ is given the structure of a left $mathbb{Z}[H]$-module setting
$$hcdot x = xh^{-1}$$
for every $xin G$ and $hin H$.
Then one makes $mathrm{coInd}_H^G(N)$ a left $mathbb{Z}[G]$-module declaring that for every $gin G$ and $varphiinmathrm{coInd}_H^G(N)$
$$(gcdotvarphi)(x) = varphi(g^{-1}x).$$
Is there some reason why we are taking all those inverses instead of just defining, in a way that looks more natural to me, the $H$-structure on $mathbb{Z}[G]$ setting $hcdot x = hx$ and then the $G-$structure on $mathrm{coInd}_H^G(N)$ setting $(gcdotvarphi)(x) = varphi(xg)$?
I think the two definitions are equivalent: if $varphiinmathrm{coInd}_H^G(N)$ one can define $psi:mathbb{Z}[G]to N$ such that $psi(x) = varphi(x^{-1})$ for every $xin G$. Now $psi$ is an element of the coinduced module constructed following the second definition, and the map sending every $varphi$ to the corresponding $psi$ gives an isomorphism. Am I wrong? If not, why is the first definition preferred over the second?
abstract-algebra modules representation-theory definition
edited Dec 14 '18 at 18:47
Arnaud D.
15.9k52443
asked Dec 14 '18 at 18:30
Francesco MiliziaFrancesco Milizia
564
$endgroup$
add a comment |
$begingroup$
If $G$ is a group, $H$ a subgroup, and $N$ a left $mathbb{Z}[H]-$module, I've learned the following construction:
$$mathrm{coInd}_H^G(N) = mathrm{Hom}_{mathbb{Z}[H]}(mathbb{Z}[G], N)$$
where $mathbb{Z}[G]$ is given the structure of a left $mathbb{Z}[H]$-module setting
$$hcdot x = xh^{-1}$$
for every $xin G$ and $hin H$.
Then one makes $mathrm{coInd}_H^G(N)$ a left $mathbb{Z}[G]$-module declaring that for every $gin G$ and $varphiinmathrm{coInd}_H^G(N)$
$$(gcdotvarphi)(x) = varphi(g^{-1}x).$$
Is there some reason why we are taking all those inverses instead of just defining, in a way that looks more natural to me, the $H$-structure on $mathbb{Z}[G]$ setting $hcdot x = hx$ and then the $G-$structure on $mathrm{coInd}_H^G(N)$ setting $(gcdotvarphi)(x) = varphi(xg)$?
I think the two definitions are equivalent: if $varphiinmathrm{coInd}_H^G(N)$ one can define $psi:mathbb{Z}[G]to N$ such that $psi(x) = varphi(x^{-1})$ for every $xin G$. Now $psi$ is an element of the coinduced module constructed following the second definition, and the map sending every $varphi$ to the corresponding $psi$ gives an isomorphism. Am I wrong? If not, why is the first definition preferred over the second?
abstract-algebra modules representation-theory definition
edited Dec 14 '18 at 18:47
Arnaud D.
15.9k52443
asked Dec 14 '18 at 18:30
Francesco MiliziaFrancesco Milizia
564
$endgroup$
$begingroup$
@SvanN Could you give a more detailed explanation? Which is the object that fails to be a left module?
$endgroup$
– Francesco Milizia
Dec 14 '18 at 20:40
$begingroup$
@SvanN I'm pretty sure that $mathrm{coInd}_H^G(N)$ as defined at the beginning of the question really is a left $G-$module.
$endgroup$
– Francesco Milizia
Dec 14 '18 at 21:01
$begingroup$
If $varphiin mathrm{coInd}_H^G(N)$, then $(g_1cdot(g_2cdotvarphi))(x) = (g_2cdotvarphi)(g_1^{-1}x) = varphi(g_2^{-1}g_1^{-1}x) = varphi((g_1g_2)^{-1}x) = ((g_1g_2)cdotvarphi)(x)$
$endgroup$
– Francesco Milizia
Dec 14 '18 at 21:10
$begingroup$
Whoops, my bad! Apologies
$endgroup$
– SvanN
Dec 14 '18 at 21:13
add a comment |
$begingroup$
If $G$ is a group, $H$ a subgroup, and $N$ a left $mathbb{Z}[H]-$module, I've learned the following construction:
$$mathrm{coInd}_H^G(N) = mathrm{Hom}_{mathbb{Z}[H]}(mathbb{Z}[G], N)$$
where $mathbb{Z}[G]$ is given the structure of a left $mathbb{Z}[H]$-module setting
$$hcdot x = xh^{-1}$$
for every $xin G$ and $hin H$.
Then one makes $mathrm{coInd}_H^G(N)$ a left $mathbb{Z}[G]$-module declaring that for every $gin G$ and $varphiinmathrm{coInd}_H^G(N)$
$$(gcdotvarphi)(x) = varphi(g^{-1}x).$$
Is there some reason why we are taking all those inverses instead of just defining, in a way that looks more natural to me, the $H$-structure on $mathbb{Z}[G]$ setting $hcdot x = hx$ and then the $G-$structure on $mathrm{coInd}_H^G(N)$ setting $(gcdotvarphi)(x) = varphi(xg)$?
I think the two definitions are equivalent: if $varphiinmathrm{coInd}_H^G(N)$ one can define $psi:mathbb{Z}[G]to N$ such that $psi(x) = varphi(x^{-1})$ for every $xin G$. Now $psi$ is an element of the coinduced module constructed following the second definition, and the map sending every $varphi$ to the corresponding $psi$ gives an isomorphism. Am I wrong? If not, why is the first definition preferred over the second?
abstract-algebra modules representation-theory definition
edited Dec 14 '18 at 18:47
Arnaud D.
15.9k52443
asked Dec 14 '18 at 18:30
Francesco MiliziaFrancesco Milizia
564
$endgroup$
If $G$ is a group, $H$ a subgroup, and $N$ a left $mathbb{Z}[H]-$module, I've learned the following construction:
$$mathrm{coInd}_H^G(N) = mathrm{Hom}_{mathbb{Z}[H]}(mathbb{Z}[G], N)$$
where $mathbb{Z}[G]$ is given the structure of a left $mathbb{Z}[H]$-module setting
$$hcdot x = xh^{-1}$$
for every $xin G$ and $hin H$.
Then one makes $mathrm{coInd}_H^G(N)$ a left $mathbb{Z}[G]$-module declaring that for every $gin G$ and $varphiinmathrm{coInd}_H^G(N)$
$$(gcdotvarphi)(x) = varphi(g^{-1}x).$$
Is there some reason why we are taking all those inverses instead of just defining, in a way that looks more natural to me, the $H$-structure on $mathbb{Z}[G]$ setting $hcdot x = hx$ and then the $G-$structure on $mathrm{coInd}_H^G(N)$ setting $(gcdotvarphi)(x) = varphi(xg)$?
I think the two definitions are equivalent: if $varphiinmathrm{coInd}_H^G(N)$ one can define $psi:mathbb{Z}[G]to N$ such that $psi(x) = varphi(x^{-1})$ for every $xin G$. Now $psi$ is an element of the coinduced module constructed following the second definition, and the map sending every $varphi$ to the corresponding $psi$ gives an isomorphism. Am I wrong? If not, why is the first definition preferred over the second?
abstract-algebra modules representation-theory definition
abstract-algebra modules representation-theory definition
edited Dec 14 '18 at 18:47
Arnaud D.
15.9k52443
asked Dec 14 '18 at 18:30
Francesco MiliziaFrancesco Milizia
564
edited Dec 14 '18 at 18:47
Arnaud D.
15.9k52443
asked Dec 14 '18 at 18:30
Francesco MiliziaFrancesco Milizia
564
edited Dec 14 '18 at 18:47
Arnaud D.
15.9k52443
edited Dec 14 '18 at 18:47
Arnaud D.
15.9k52443
edited Dec 14 '18 at 18:47
Arnaud D.
15.9k52443
15.9k52443
asked Dec 14 '18 at 18:30
Francesco MiliziaFrancesco Milizia
564
asked Dec 14 '18 at 18:30
Francesco MiliziaFrancesco Milizia
564
asked Dec 14 '18 at 18:30
Francesco MiliziaFrancesco Milizia
564
564
$begingroup$
@SvanN Could you give a more detailed explanation? Which is the object that fails to be a left module?
$endgroup$
– Francesco Milizia
Dec 14 '18 at 20:40
$begingroup$
@SvanN I'm pretty sure that $mathrm{coInd}_H^G(N)$ as defined at the beginning of the question really is a left $G-$module.
$endgroup$
– Francesco Milizia
Dec 14 '18 at 21:01
$begingroup$
If $varphiin mathrm{coInd}_H^G(N)$, then $(g_1cdot(g_2cdotvarphi))(x) = (g_2cdotvarphi)(g_1^{-1}x) = varphi(g_2^{-1}g_1^{-1}x) = varphi((g_1g_2)^{-1}x) = ((g_1g_2)cdotvarphi)(x)$
$endgroup$
– Francesco Milizia
Dec 14 '18 at 21:10
$begingroup$
Whoops, my bad! Apologies
$endgroup$
– SvanN
Dec 14 '18 at 21:13
add a comment |
$begingroup$
@SvanN Could you give a more detailed explanation? Which is the object that fails to be a left module?
$endgroup$
– Francesco Milizia
Dec 14 '18 at 20:40
$begingroup$
@SvanN I'm pretty sure that $mathrm{coInd}_H^G(N)$ as defined at the beginning of the question really is a left $G-$module.
$endgroup$
– Francesco Milizia
Dec 14 '18 at 21:01
$begingroup$
If $varphiin mathrm{coInd}_H^G(N)$, then $(g_1cdot(g_2cdotvarphi))(x) = (g_2cdotvarphi)(g_1^{-1}x) = varphi(g_2^{-1}g_1^{-1}x) = varphi((g_1g_2)^{-1}x) = ((g_1g_2)cdotvarphi)(x)$
$endgroup$
– Francesco Milizia
Dec 14 '18 at 21:10
$begingroup$
Whoops, my bad! Apologies
$endgroup$
– SvanN
Dec 14 '18 at 21:13
$begingroup$
@SvanN Could you give a more detailed explanation? Which is the object that fails to be a left module?
$endgroup$
– Francesco Milizia
Dec 14 '18 at 20:40
$begingroup$
@SvanN Could you give a more detailed explanation? Which is the object that fails to be a left module?
$endgroup$
– Francesco Milizia
Dec 14 '18 at 20:40
$begingroup$
@SvanN I'm pretty sure that $mathrm{coInd}_H^G(N)$ as defined at the beginning of the question really is a left $G-$module.
$endgroup$
– Francesco Milizia
Dec 14 '18 at 21:01
$begingroup$
@SvanN I'm pretty sure that $mathrm{coInd}_H^G(N)$ as defined at the beginning of the question really is a left $G-$module.
$endgroup$
– Francesco Milizia
Dec 14 '18 at 21:01
$begingroup$
If $varphiin mathrm{coInd}_H^G(N)$, then $(g_1cdot(g_2cdotvarphi))(x) = (g_2cdotvarphi)(g_1^{-1}x) = varphi(g_2^{-1}g_1^{-1}x) = varphi((g_1g_2)^{-1}x) = ((g_1g_2)cdotvarphi)(x)$
$endgroup$
– Francesco Milizia
Dec 14 '18 at 21:10
$begingroup$
If $varphiin mathrm{coInd}_H^G(N)$, then $(g_1cdot(g_2cdotvarphi))(x) = (g_2cdotvarphi)(g_1^{-1}x) = varphi(g_2^{-1}g_1^{-1}x) = varphi((g_1g_2)^{-1}x) = ((g_1g_2)cdotvarphi)(x)$
$endgroup$
– Francesco Milizia
Dec 14 '18 at 21:10
$begingroup$
Whoops, my bad! Apologies
$endgroup$
– SvanN
Dec 14 '18 at 21:13
$begingroup$
Whoops, my bad! Apologies
$endgroup$
– SvanN
Dec 14 '18 at 21:13
add a comment |
1 Answer
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$begingroup$
I like your definition better, because it fits the general pattern "for an $Smathrm{-}Rmathrm{-}$bimodule $_SM_R$ and a left $S$-module $_SN$, $mathrm{Hom}_S(M, N)$ inherits left $R$-module structure from the right $R$-module structure of $M$".
I agree that the two definitions are equivalent. Perhaps the upshot of the first definition is that everything is defined in terms of left $H-$ and $G-$modules, without invoking right $G$-modules ($H$-modules, resp.).
But the fact that you can make this translation is not surprising, since the categories of left and right $G$-modules are equivalent for any group $G$: the equivalence is given precisely by this inverse of action operation, that is, a left $G$-module $_{mathbb{Z}[G]}M$ corresponds to the right $G$-module $M_{mathbb{Z}[G]},$ where the underlying Abelian group is the same, and the action is defined by $mcdot g:=g^{-1}cdot m$.
answered Dec 16 '18 at 5:29
Pavel ČoupekPavel Čoupek
4,45611126
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add a comment |
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$begingroup$
I like your definition better, because it fits the general pattern "for an $Smathrm{-}Rmathrm{-}$bimodule $_SM_R$ and a left $S$-module $_SN$, $mathrm{Hom}_S(M, N)$ inherits left $R$-module structure from the right $R$-module structure of $M$".
I agree that the two definitions are equivalent. Perhaps the upshot of the first definition is that everything is defined in terms of left $H-$ and $G-$modules, without invoking right $G$-modules ($H$-modules, resp.).
But the fact that you can make this translation is not surprising, since the categories of left and right $G$-modules are equivalent for any group $G$: the equivalence is given precisely by this inverse of action operation, that is, a left $G$-module $_{mathbb{Z}[G]}M$ corresponds to the right $G$-module $M_{mathbb{Z}[G]},$ where the underlying Abelian group is the same, and the action is defined by $mcdot g:=g^{-1}cdot m$.
answered Dec 16 '18 at 5:29
Pavel ČoupekPavel Čoupek
4,45611126
$endgroup$
add a comment |
$begingroup$
I like your definition better, because it fits the general pattern "for an $Smathrm{-}Rmathrm{-}$bimodule $_SM_R$ and a left $S$-module $_SN$, $mathrm{Hom}_S(M, N)$ inherits left $R$-module structure from the right $R$-module structure of $M$".
I agree that the two definitions are equivalent. Perhaps the upshot of the first definition is that everything is defined in terms of left $H-$ and $G-$modules, without invoking right $G$-modules ($H$-modules, resp.).
But the fact that you can make this translation is not surprising, since the categories of left and right $G$-modules are equivalent for any group $G$: the equivalence is given precisely by this inverse of action operation, that is, a left $G$-module $_{mathbb{Z}[G]}M$ corresponds to the right $G$-module $M_{mathbb{Z}[G]},$ where the underlying Abelian group is the same, and the action is defined by $mcdot g:=g^{-1}cdot m$.
answered Dec 16 '18 at 5:29
Pavel ČoupekPavel Čoupek
4,45611126
$endgroup$
add a comment |
$begingroup$
I like your definition better, because it fits the general pattern "for an $Smathrm{-}Rmathrm{-}$bimodule $_SM_R$ and a left $S$-module $_SN$, $mathrm{Hom}_S(M, N)$ inherits left $R$-module structure from the right $R$-module structure of $M$".
I agree that the two definitions are equivalent. Perhaps the upshot of the first definition is that everything is defined in terms of left $H-$ and $G-$modules, without invoking right $G$-modules ($H$-modules, resp.).
But the fact that you can make this translation is not surprising, since the categories of left and right $G$-modules are equivalent for any group $G$: the equivalence is given precisely by this inverse of action operation, that is, a left $G$-module $_{mathbb{Z}[G]}M$ corresponds to the right $G$-module $M_{mathbb{Z}[G]},$ where the underlying Abelian group is the same, and the action is defined by $mcdot g:=g^{-1}cdot m$.
answered Dec 16 '18 at 5:29
Pavel ČoupekPavel Čoupek
4,45611126
$endgroup$
I like your definition better, because it fits the general pattern "for an $Smathrm{-}Rmathrm{-}$bimodule $_SM_R$ and a left $S$-module $_SN$, $mathrm{Hom}_S(M, N)$ inherits left $R$-module structure from the right $R$-module structure of $M$".
I agree that the two definitions are equivalent. Perhaps the upshot of the first definition is that everything is defined in terms of left $H-$ and $G-$modules, without invoking right $G$-modules ($H$-modules, resp.).
But the fact that you can make this translation is not surprising, since the categories of left and right $G$-modules are equivalent for any group $G$: the equivalence is given precisely by this inverse of action operation, that is, a left $G$-module $_{mathbb{Z}[G]}M$ corresponds to the right $G$-module $M_{mathbb{Z}[G]},$ where the underlying Abelian group is the same, and the action is defined by $mcdot g:=g^{-1}cdot m$.
answered Dec 16 '18 at 5:29
Pavel ČoupekPavel Čoupek
4,45611126
edited Dec 16 '18 at 5:35
answered Dec 16 '18 at 5:29
Pavel ČoupekPavel Čoupek
4,45611126
answered Dec 16 '18 at 5:29
Pavel ČoupekPavel Čoupek
4,45611126
answered Dec 16 '18 at 5:29
Pavel ČoupekPavel Čoupek
4,45611126
4,45611126
add a comment |
add a comment |
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$begingroup$
@SvanN Could you give a more detailed explanation? Which is the object that fails to be a left module?
$endgroup$
– Francesco Milizia
Dec 14 '18 at 20:40
$begingroup$
@SvanN I'm pretty sure that $mathrm{coInd}_H^G(N)$ as defined at the beginning of the question really is a left $G-$module.
$endgroup$
– Francesco Milizia
Dec 14 '18 at 21:01
$begingroup$
If $varphiin mathrm{coInd}_H^G(N)$, then $(g_1cdot(g_2cdotvarphi))(x) = (g_2cdotvarphi)(g_1^{-1}x) = varphi(g_2^{-1}g_1^{-1}x) = varphi((g_1g_2)^{-1}x) = ((g_1g_2)cdotvarphi)(x)$
$endgroup$
– Francesco Milizia
Dec 14 '18 at 21:10
$begingroup$
Whoops, my bad! Apologies
$endgroup$
– SvanN
Dec 14 '18 at 21:13