Mixing
cohomological dimension of groups vs cohomological dimension of subgroups
$begingroup$
Let $Gamma$ be a group and $Gamma^prime$ a subgroup. Then, $text{cd }Gamma^prime leq text{cd } Gamma$ because a projective resolution of $mathbb{Z}$ over $mathbb{Z}Gamma$ can also be regarded as a projective resolution of $mathbb{Z}$ over $mathbb{Z}Gamma^prime$.
Now, 'Cohomology of Groups' by Brown says that if $text{cd }Gamma < infty$ and $|Gamma : Gamma^prime |< infty$ (where that denotes the index), then the equality holds.
The proof goes as follows:
It can be shown that if $text{cd} Gamma= n$, then there is a free $mathbb{Z}Gamma$ module $F$ with $H^n(Gamma,F)neq 0$. Let $F^prime$ be a free $mathbb{Z}Gamma^prime$-module of the same rank. Suppose that $Fcong bigoplus_{I} mathbb{Z}Gamma$ and $F^primecong bigoplus_{I} mathbb{Z}Gamma^prime$.
Then, $text{Ind}_{Gamma^prime}^Gamma F^prime= mathbb{Z}Gamma otimes_{mathbb{Z}Gamma^prime} F^prime cong bigoplus_{I} mathbb{Z}Gamma otimes_{mathbb{Z}Gamma^prime} mathbb{Z}Gamma^prime cong F$, so by Shapiro's Lemma, $$H^n(Gamma^prime, F^prime)cong H^n(Gamma, F)neq 0.$$ Thus, $text{cd}Gamma^prime geq n$.
The problem is that I do not see where are we using the hypothesis that $|Gamma : Gamma^prime|<infty$. Can someone help me, please?
algebraic-topology homology-cohomology homological-algebra group-cohomology
asked Jan 6 at 14:26
KarenKaren
896
$endgroup$
add a comment |
$begingroup$
Let $Gamma$ be a group and $Gamma^prime$ a subgroup. Then, $text{cd }Gamma^prime leq text{cd } Gamma$ because a projective resolution of $mathbb{Z}$ over $mathbb{Z}Gamma$ can also be regarded as a projective resolution of $mathbb{Z}$ over $mathbb{Z}Gamma^prime$.
Now, 'Cohomology of Groups' by Brown says that if $text{cd }Gamma < infty$ and $|Gamma : Gamma^prime |< infty$ (where that denotes the index), then the equality holds.
The proof goes as follows:
It can be shown that if $text{cd} Gamma= n$, then there is a free $mathbb{Z}Gamma$ module $F$ with $H^n(Gamma,F)neq 0$. Let $F^prime$ be a free $mathbb{Z}Gamma^prime$-module of the same rank. Suppose that $Fcong bigoplus_{I} mathbb{Z}Gamma$ and $F^primecong bigoplus_{I} mathbb{Z}Gamma^prime$.
Then, $text{Ind}_{Gamma^prime}^Gamma F^prime= mathbb{Z}Gamma otimes_{mathbb{Z}Gamma^prime} F^prime cong bigoplus_{I} mathbb{Z}Gamma otimes_{mathbb{Z}Gamma^prime} mathbb{Z}Gamma^prime cong F$, so by Shapiro's Lemma, $$H^n(Gamma^prime, F^prime)cong H^n(Gamma, F)neq 0.$$ Thus, $text{cd}Gamma^prime geq n$.
The problem is that I do not see where are we using the hypothesis that $|Gamma : Gamma^prime|<infty$. Can someone help me, please?
algebraic-topology homology-cohomology homological-algebra group-cohomology
asked Jan 6 at 14:26
KarenKaren
896
$endgroup$
add a comment |
$begingroup$
Let $Gamma$ be a group and $Gamma^prime$ a subgroup. Then, $text{cd }Gamma^prime leq text{cd } Gamma$ because a projective resolution of $mathbb{Z}$ over $mathbb{Z}Gamma$ can also be regarded as a projective resolution of $mathbb{Z}$ over $mathbb{Z}Gamma^prime$.
Now, 'Cohomology of Groups' by Brown says that if $text{cd }Gamma < infty$ and $|Gamma : Gamma^prime |< infty$ (where that denotes the index), then the equality holds.
The proof goes as follows:
It can be shown that if $text{cd} Gamma= n$, then there is a free $mathbb{Z}Gamma$ module $F$ with $H^n(Gamma,F)neq 0$. Let $F^prime$ be a free $mathbb{Z}Gamma^prime$-module of the same rank. Suppose that $Fcong bigoplus_{I} mathbb{Z}Gamma$ and $F^primecong bigoplus_{I} mathbb{Z}Gamma^prime$.
Then, $text{Ind}_{Gamma^prime}^Gamma F^prime= mathbb{Z}Gamma otimes_{mathbb{Z}Gamma^prime} F^prime cong bigoplus_{I} mathbb{Z}Gamma otimes_{mathbb{Z}Gamma^prime} mathbb{Z}Gamma^prime cong F$, so by Shapiro's Lemma, $$H^n(Gamma^prime, F^prime)cong H^n(Gamma, F)neq 0.$$ Thus, $text{cd}Gamma^prime geq n$.
The problem is that I do not see where are we using the hypothesis that $|Gamma : Gamma^prime|<infty$. Can someone help me, please?
algebraic-topology homology-cohomology homological-algebra group-cohomology
asked Jan 6 at 14:26
KarenKaren
896
$endgroup$
Let $Gamma$ be a group and $Gamma^prime$ a subgroup. Then, $text{cd }Gamma^prime leq text{cd } Gamma$ because a projective resolution of $mathbb{Z}$ over $mathbb{Z}Gamma$ can also be regarded as a projective resolution of $mathbb{Z}$ over $mathbb{Z}Gamma^prime$.
Now, 'Cohomology of Groups' by Brown says that if $text{cd }Gamma < infty$ and $|Gamma : Gamma^prime |< infty$ (where that denotes the index), then the equality holds.
The proof goes as follows:
It can be shown that if $text{cd} Gamma= n$, then there is a free $mathbb{Z}Gamma$ module $F$ with $H^n(Gamma,F)neq 0$. Let $F^prime$ be a free $mathbb{Z}Gamma^prime$-module of the same rank. Suppose that $Fcong bigoplus_{I} mathbb{Z}Gamma$ and $F^primecong bigoplus_{I} mathbb{Z}Gamma^prime$.
Then, $text{Ind}_{Gamma^prime}^Gamma F^prime= mathbb{Z}Gamma otimes_{mathbb{Z}Gamma^prime} F^prime cong bigoplus_{I} mathbb{Z}Gamma otimes_{mathbb{Z}Gamma^prime} mathbb{Z}Gamma^prime cong F$, so by Shapiro's Lemma, $$H^n(Gamma^prime, F^prime)cong H^n(Gamma, F)neq 0.$$ Thus, $text{cd}Gamma^prime geq n$.
The problem is that I do not see where are we using the hypothesis that $|Gamma : Gamma^prime|<infty$. Can someone help me, please?
algebraic-topology homology-cohomology homological-algebra group-cohomology
algebraic-topology homology-cohomology homological-algebra group-cohomology
asked Jan 6 at 14:26
KarenKaren
896
asked Jan 6 at 14:26
KarenKaren
896
asked Jan 6 at 14:26
KarenKaren
896
asked Jan 6 at 14:26
KarenKaren
896
asked Jan 6 at 14:26
KarenKaren
896
896
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
See here for Shapiro's lemma : it holds for induction for homology.
For cohomology, it only holds with coinduction; but if $[Gamma :Gamma'] < infty$ then induction and coinduction coincide, so Shapiro's lemma holds in cohomology with the induced module.
answered Jan 6 at 16:58
MaxMax
13.7k11142
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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%2f3063920%2fcohomological-dimension-of-groups-vs-cohomological-dimension-of-subgroups%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
See here for Shapiro's lemma : it holds for induction for homology.
For cohomology, it only holds with coinduction; but if $[Gamma :Gamma'] < infty$ then induction and coinduction coincide, so Shapiro's lemma holds in cohomology with the induced module.
answered Jan 6 at 16:58
MaxMax
13.7k11142
$endgroup$
add a comment |
$begingroup$
See here for Shapiro's lemma : it holds for induction for homology.
For cohomology, it only holds with coinduction; but if $[Gamma :Gamma'] < infty$ then induction and coinduction coincide, so Shapiro's lemma holds in cohomology with the induced module.
answered Jan 6 at 16:58
MaxMax
13.7k11142
$endgroup$
add a comment |
$begingroup$
See here for Shapiro's lemma : it holds for induction for homology.
For cohomology, it only holds with coinduction; but if $[Gamma :Gamma'] < infty$ then induction and coinduction coincide, so Shapiro's lemma holds in cohomology with the induced module.
answered Jan 6 at 16:58
MaxMax
13.7k11142
$endgroup$
See here for Shapiro's lemma : it holds for induction for homology.
For cohomology, it only holds with coinduction; but if $[Gamma :Gamma'] < infty$ then induction and coinduction coincide, so Shapiro's lemma holds in cohomology with the induced module.
answered Jan 6 at 16:58
MaxMax
13.7k11142
answered Jan 6 at 16:58
MaxMax
13.7k11142
answered Jan 6 at 16:58
MaxMax
13.7k11142
answered Jan 6 at 16:58
MaxMax
13.7k11142
13.7k11142
add a comment |
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%2f3063920%2fcohomological-dimension-of-groups-vs-cohomological-dimension-of-subgroups%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
