Group cohomology of product with swapping (twisting) factors
$begingroup$
Let $M$ be a $G$-module, where $G = Bbb Z / 2 Bbb Z$. Define a $G$-module structure on $A = M oplus M$ by $g cdot (a,b) = (g cdot b, g cdot a)$.
What is the group cohomology $H^*(G, A)$ is terms of $H^*(G,M)$ ?
Thoughts:
I know that $H^n(G, -)$ commutes with products, but here the action on $A$ is not the diagonal action, so I'm not sure what to do here. Can we at least compute $H^1(G, A)$, provided that $H^1(G, M)=0$ ?
group-theory group-cohomology
asked Dec 24 '18 at 15:56
AlphonseAlphonse
2,225725
$endgroup$
add a comment |
$begingroup$
Let $M$ be a $G$-module, where $G = Bbb Z / 2 Bbb Z$. Define a $G$-module structure on $A = M oplus M$ by $g cdot (a,b) = (g cdot b, g cdot a)$.
What is the group cohomology $H^*(G, A)$ is terms of $H^*(G,M)$ ?
Thoughts:
I know that $H^n(G, -)$ commutes with products, but here the action on $A$ is not the diagonal action, so I'm not sure what to do here. Can we at least compute $H^1(G, A)$, provided that $H^1(G, M)=0$ ?
group-theory group-cohomology
asked Dec 24 '18 at 15:56
AlphonseAlphonse
2,225725
$endgroup$
1
$begingroup$
That does not look like a $G$-module structure to me.
$endgroup$
– Lord Shark the Unknown
Dec 24 '18 at 15:59
$begingroup$
@LordSharktheUnknown : thank you. Actually, I only need $G = Bbb Z / 2$. (It seems to work, then).
$endgroup$
– Alphonse
Dec 24 '18 at 16:05
$begingroup$
(My question comes from something like the $Bbb C/Bbb R$-Galois action on $Bbb C^{times} times Bbb C^{times}$.)
$endgroup$
– Alphonse
Dec 24 '18 at 16:14
add a comment |
$begingroup$
Let $M$ be a $G$-module, where $G = Bbb Z / 2 Bbb Z$. Define a $G$-module structure on $A = M oplus M$ by $g cdot (a,b) = (g cdot b, g cdot a)$.
What is the group cohomology $H^*(G, A)$ is terms of $H^*(G,M)$ ?
Thoughts:
I know that $H^n(G, -)$ commutes with products, but here the action on $A$ is not the diagonal action, so I'm not sure what to do here. Can we at least compute $H^1(G, A)$, provided that $H^1(G, M)=0$ ?
group-theory group-cohomology
asked Dec 24 '18 at 15:56
AlphonseAlphonse
2,225725
$endgroup$
Let $M$ be a $G$-module, where $G = Bbb Z / 2 Bbb Z$. Define a $G$-module structure on $A = M oplus M$ by $g cdot (a,b) = (g cdot b, g cdot a)$.
What is the group cohomology $H^*(G, A)$ is terms of $H^*(G,M)$ ?
Thoughts:
I know that $H^n(G, -)$ commutes with products, but here the action on $A$ is not the diagonal action, so I'm not sure what to do here. Can we at least compute $H^1(G, A)$, provided that $H^1(G, M)=0$ ?
group-theory group-cohomology
group-theory group-cohomology
asked Dec 24 '18 at 15:56
AlphonseAlphonse
2,225725
asked Dec 24 '18 at 15:56
AlphonseAlphonse
2,225725
edited Dec 24 '18 at 16:05
Alphonse
asked Dec 24 '18 at 15:56
AlphonseAlphonse
2,225725
asked Dec 24 '18 at 15:56
AlphonseAlphonse
2,225725
asked Dec 24 '18 at 15:56
AlphonseAlphonse
2,225725
2,225725
1
$begingroup$
That does not look like a $G$-module structure to me.
$endgroup$
– Lord Shark the Unknown
Dec 24 '18 at 15:59
$begingroup$
@LordSharktheUnknown : thank you. Actually, I only need $G = Bbb Z / 2$. (It seems to work, then).
$endgroup$
– Alphonse
Dec 24 '18 at 16:05
$begingroup$
(My question comes from something like the $Bbb C/Bbb R$-Galois action on $Bbb C^{times} times Bbb C^{times}$.)
$endgroup$
– Alphonse
Dec 24 '18 at 16:14
add a comment |
1
$begingroup$
That does not look like a $G$-module structure to me.
$endgroup$
– Lord Shark the Unknown
Dec 24 '18 at 15:59
$begingroup$
@LordSharktheUnknown : thank you. Actually, I only need $G = Bbb Z / 2$. (It seems to work, then).
$endgroup$
– Alphonse
Dec 24 '18 at 16:05
$begingroup$
(My question comes from something like the $Bbb C/Bbb R$-Galois action on $Bbb C^{times} times Bbb C^{times}$.)
$endgroup$
– Alphonse
Dec 24 '18 at 16:14
1
1
$begingroup$
That does not look like a $G$-module structure to me.
$endgroup$
– Lord Shark the Unknown
Dec 24 '18 at 15:59
$begingroup$
That does not look like a $G$-module structure to me.
$endgroup$
– Lord Shark the Unknown
Dec 24 '18 at 15:59
$begingroup$
@LordSharktheUnknown : thank you. Actually, I only need $G = Bbb Z / 2$. (It seems to work, then).
$endgroup$
– Alphonse
Dec 24 '18 at 16:05
$begingroup$
@LordSharktheUnknown : thank you. Actually, I only need $G = Bbb Z / 2$. (It seems to work, then).
$endgroup$
– Alphonse
Dec 24 '18 at 16:05
$begingroup$
(My question comes from something like the $Bbb C/Bbb R$-Galois action on $Bbb C^{times} times Bbb C^{times}$.)
$endgroup$
– Alphonse
Dec 24 '18 at 16:14
$begingroup$
(My question comes from something like the $Bbb C/Bbb R$-Galois action on $Bbb C^{times} times Bbb C^{times}$.)
$endgroup$
– Alphonse
Dec 24 '18 at 16:14
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For even $n>0$, $H^n$ is the kernel of $1-g$ factored by the image of $1+g$.
The kernel of $1-g$ is the set of solutions of $(a,b)=(gcdot b,gcdot a)$
which is equivalent to $b=gcdot a$. Thus the typical element of the
kernel is $(a,gcdot a)=(a,0)+gcdot(a,0)$ lying in the image of $1+g$.
Therefore $H^n(G,M)=0$ for even $n>0$.
A similar calculation gives $H^n(G,M)=0$ for odd $n$.
answered Dec 24 '18 at 17:56
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
add a comment |
$begingroup$
What you have written down is the $Bbb Z/2$-module $M otimes Bbb Z[Bbb Z/2]$.
The $G$-module $M otimes Bbb Z[G]$ is actually independent of the choice of action on $M$, up to isomorphism: the map $m otimes [g] mapsto g^-1m otimes [g]$ gives a $G$-equivariant isomorphism to $M otimes Bbb Z[G]$ with the trivial action on $M$. In particular we may as well assume the action is trivial.
If you choose a resolution $0to A_1 to A_2 to M to 0$ of $M$ by free abelian groups, this gives a short exact sequence $0 to A_1[Bbb Z/2] to A_2[Bbb Z/2] to M[Bbb Z/2] to 0$, so there is an associated long exact sequence in cohomology.
If I can convince you that the cohomology of $Bbb Z^infty[Bbb Z/2]$ is zero in degrees larger than $0$, and that $H^0(Bbb Z/2; A[Bbb Z/2]) cong A$ naturally in free abelian groups $A$, then the long exact sequence tells us that $H^*(Bbb Z/2; M[Bbb Z/2]) = 0$ in degrees larger than 0, and gives $H^0 cong M$.
But observe that $$H^*(Bbb Z/2; Bbb Z^infty[Bbb Z/2]) cong H^*(Bbb Z/2; Bbb Z[Bbb Z/2])^infty$$ and this may easily be seen to be zero at the level of the definition (in degrees larger than 0). $H^0$ is just the fixed point space, which should make the statements about it clear.
$endgroup$
$begingroup$
I remark that this same argument shows $M[G]$ has cohomology concentrated in degree 0 for any $G$ and any $G$-module $M$.
$endgroup$
– user98602
Dec 27 '18 at 16:33
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3051385%2fgroup-cohomology-of-product-with-swapping-twisting-factors%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For even $n>0$, $H^n$ is the kernel of $1-g$ factored by the image of $1+g$.
The kernel of $1-g$ is the set of solutions of $(a,b)=(gcdot b,gcdot a)$
which is equivalent to $b=gcdot a$. Thus the typical element of the
kernel is $(a,gcdot a)=(a,0)+gcdot(a,0)$ lying in the image of $1+g$.
Therefore $H^n(G,M)=0$ for even $n>0$.
A similar calculation gives $H^n(G,M)=0$ for odd $n$.
answered Dec 24 '18 at 17:56
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
add a comment |
$begingroup$
For even $n>0$, $H^n$ is the kernel of $1-g$ factored by the image of $1+g$.
The kernel of $1-g$ is the set of solutions of $(a,b)=(gcdot b,gcdot a)$
which is equivalent to $b=gcdot a$. Thus the typical element of the
kernel is $(a,gcdot a)=(a,0)+gcdot(a,0)$ lying in the image of $1+g$.
Therefore $H^n(G,M)=0$ for even $n>0$.
A similar calculation gives $H^n(G,M)=0$ for odd $n$.
answered Dec 24 '18 at 17:56
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
add a comment |
$begingroup$
For even $n>0$, $H^n$ is the kernel of $1-g$ factored by the image of $1+g$.
The kernel of $1-g$ is the set of solutions of $(a,b)=(gcdot b,gcdot a)$
which is equivalent to $b=gcdot a$. Thus the typical element of the
kernel is $(a,gcdot a)=(a,0)+gcdot(a,0)$ lying in the image of $1+g$.
Therefore $H^n(G,M)=0$ for even $n>0$.
A similar calculation gives $H^n(G,M)=0$ for odd $n$.
answered Dec 24 '18 at 17:56
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
For even $n>0$, $H^n$ is the kernel of $1-g$ factored by the image of $1+g$.
The kernel of $1-g$ is the set of solutions of $(a,b)=(gcdot b,gcdot a)$
which is equivalent to $b=gcdot a$. Thus the typical element of the
kernel is $(a,gcdot a)=(a,0)+gcdot(a,0)$ lying in the image of $1+g$.
Therefore $H^n(G,M)=0$ for even $n>0$.
A similar calculation gives $H^n(G,M)=0$ for odd $n$.
answered Dec 24 '18 at 17:56
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered Dec 24 '18 at 17:56
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered Dec 24 '18 at 17:56
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered Dec 24 '18 at 17:56
Lord Shark the UnknownLord Shark the Unknown
109k1163136
109k1163136
add a comment |
add a comment |
$begingroup$
What you have written down is the $Bbb Z/2$-module $M otimes Bbb Z[Bbb Z/2]$.
The $G$-module $M otimes Bbb Z[G]$ is actually independent of the choice of action on $M$, up to isomorphism: the map $m otimes [g] mapsto g^-1m otimes [g]$ gives a $G$-equivariant isomorphism to $M otimes Bbb Z[G]$ with the trivial action on $M$. In particular we may as well assume the action is trivial.
If you choose a resolution $0to A_1 to A_2 to M to 0$ of $M$ by free abelian groups, this gives a short exact sequence $0 to A_1[Bbb Z/2] to A_2[Bbb Z/2] to M[Bbb Z/2] to 0$, so there is an associated long exact sequence in cohomology.
If I can convince you that the cohomology of $Bbb Z^infty[Bbb Z/2]$ is zero in degrees larger than $0$, and that $H^0(Bbb Z/2; A[Bbb Z/2]) cong A$ naturally in free abelian groups $A$, then the long exact sequence tells us that $H^*(Bbb Z/2; M[Bbb Z/2]) = 0$ in degrees larger than 0, and gives $H^0 cong M$.
But observe that $$H^*(Bbb Z/2; Bbb Z^infty[Bbb Z/2]) cong H^*(Bbb Z/2; Bbb Z[Bbb Z/2])^infty$$ and this may easily be seen to be zero at the level of the definition (in degrees larger than 0). $H^0$ is just the fixed point space, which should make the statements about it clear.
$endgroup$
$begingroup$
I remark that this same argument shows $M[G]$ has cohomology concentrated in degree 0 for any $G$ and any $G$-module $M$.
$endgroup$
– user98602
Dec 27 '18 at 16:33
add a comment |
$begingroup$
What you have written down is the $Bbb Z/2$-module $M otimes Bbb Z[Bbb Z/2]$.
The $G$-module $M otimes Bbb Z[G]$ is actually independent of the choice of action on $M$, up to isomorphism: the map $m otimes [g] mapsto g^-1m otimes [g]$ gives a $G$-equivariant isomorphism to $M otimes Bbb Z[G]$ with the trivial action on $M$. In particular we may as well assume the action is trivial.
If you choose a resolution $0to A_1 to A_2 to M to 0$ of $M$ by free abelian groups, this gives a short exact sequence $0 to A_1[Bbb Z/2] to A_2[Bbb Z/2] to M[Bbb Z/2] to 0$, so there is an associated long exact sequence in cohomology.
If I can convince you that the cohomology of $Bbb Z^infty[Bbb Z/2]$ is zero in degrees larger than $0$, and that $H^0(Bbb Z/2; A[Bbb Z/2]) cong A$ naturally in free abelian groups $A$, then the long exact sequence tells us that $H^*(Bbb Z/2; M[Bbb Z/2]) = 0$ in degrees larger than 0, and gives $H^0 cong M$.
But observe that $$H^*(Bbb Z/2; Bbb Z^infty[Bbb Z/2]) cong H^*(Bbb Z/2; Bbb Z[Bbb Z/2])^infty$$ and this may easily be seen to be zero at the level of the definition (in degrees larger than 0). $H^0$ is just the fixed point space, which should make the statements about it clear.
$endgroup$
$begingroup$
I remark that this same argument shows $M[G]$ has cohomology concentrated in degree 0 for any $G$ and any $G$-module $M$.
$endgroup$
– user98602
Dec 27 '18 at 16:33
add a comment |
$begingroup$
What you have written down is the $Bbb Z/2$-module $M otimes Bbb Z[Bbb Z/2]$.
The $G$-module $M otimes Bbb Z[G]$ is actually independent of the choice of action on $M$, up to isomorphism: the map $m otimes [g] mapsto g^-1m otimes [g]$ gives a $G$-equivariant isomorphism to $M otimes Bbb Z[G]$ with the trivial action on $M$. In particular we may as well assume the action is trivial.
If you choose a resolution $0to A_1 to A_2 to M to 0$ of $M$ by free abelian groups, this gives a short exact sequence $0 to A_1[Bbb Z/2] to A_2[Bbb Z/2] to M[Bbb Z/2] to 0$, so there is an associated long exact sequence in cohomology.
If I can convince you that the cohomology of $Bbb Z^infty[Bbb Z/2]$ is zero in degrees larger than $0$, and that $H^0(Bbb Z/2; A[Bbb Z/2]) cong A$ naturally in free abelian groups $A$, then the long exact sequence tells us that $H^*(Bbb Z/2; M[Bbb Z/2]) = 0$ in degrees larger than 0, and gives $H^0 cong M$.
But observe that $$H^*(Bbb Z/2; Bbb Z^infty[Bbb Z/2]) cong H^*(Bbb Z/2; Bbb Z[Bbb Z/2])^infty$$ and this may easily be seen to be zero at the level of the definition (in degrees larger than 0). $H^0$ is just the fixed point space, which should make the statements about it clear.
$endgroup$
What you have written down is the $Bbb Z/2$-module $M otimes Bbb Z[Bbb Z/2]$.
The $G$-module $M otimes Bbb Z[G]$ is actually independent of the choice of action on $M$, up to isomorphism: the map $m otimes [g] mapsto g^-1m otimes [g]$ gives a $G$-equivariant isomorphism to $M otimes Bbb Z[G]$ with the trivial action on $M$. In particular we may as well assume the action is trivial.
If you choose a resolution $0to A_1 to A_2 to M to 0$ of $M$ by free abelian groups, this gives a short exact sequence $0 to A_1[Bbb Z/2] to A_2[Bbb Z/2] to M[Bbb Z/2] to 0$, so there is an associated long exact sequence in cohomology.
If I can convince you that the cohomology of $Bbb Z^infty[Bbb Z/2]$ is zero in degrees larger than $0$, and that $H^0(Bbb Z/2; A[Bbb Z/2]) cong A$ naturally in free abelian groups $A$, then the long exact sequence tells us that $H^*(Bbb Z/2; M[Bbb Z/2]) = 0$ in degrees larger than 0, and gives $H^0 cong M$.
But observe that $$H^*(Bbb Z/2; Bbb Z^infty[Bbb Z/2]) cong H^*(Bbb Z/2; Bbb Z[Bbb Z/2])^infty$$ and this may easily be seen to be zero at the level of the definition (in degrees larger than 0). $H^0$ is just the fixed point space, which should make the statements about it clear.
answered Dec 24 '18 at 17:39
user98602
$begingroup$
I remark that this same argument shows $M[G]$ has cohomology concentrated in degree 0 for any $G$ and any $G$-module $M$.
$endgroup$
– user98602
Dec 27 '18 at 16:33
add a comment |
$begingroup$
I remark that this same argument shows $M[G]$ has cohomology concentrated in degree 0 for any $G$ and any $G$-module $M$.
$endgroup$
– user98602
Dec 27 '18 at 16:33
$begingroup$
I remark that this same argument shows $M[G]$ has cohomology concentrated in degree 0 for any $G$ and any $G$-module $M$.
$endgroup$
– user98602
Dec 27 '18 at 16:33
$begingroup$
I remark that this same argument shows $M[G]$ has cohomology concentrated in degree 0 for any $G$ and any $G$-module $M$.
$endgroup$
– user98602
Dec 27 '18 at 16:33
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3051385%2fgroup-cohomology-of-product-with-swapping-twisting-factors%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
That does not look like a $G$-module structure to me.
$endgroup$
– Lord Shark the Unknown
Dec 24 '18 at 15:59
$begingroup$
@LordSharktheUnknown : thank you. Actually, I only need $G = Bbb Z / 2$. (It seems to work, then).
$endgroup$
– Alphonse
Dec 24 '18 at 16:05
$begingroup$
(My question comes from something like the $Bbb C/Bbb R$-Galois action on $Bbb C^{times} times Bbb C^{times}$.)
$endgroup$
– Alphonse
Dec 24 '18 at 16:14