Mixing
Definition of the $0$-coboundary in group cohomology
$begingroup$
I'm trying to learn, what is group cohomology. Since I'm not a matematician, the general definition is too abstract to me (at least for the time being), and requires too much category theory and homological algebra. First I'm trying to catch the "technical" definition presented by Wikipedia. This definition starts with the definition of the modules of cochains for $n=0,1,...$, and the coboundary maps from the $n$-th module to the $n+1$-th as follows:
"For $nge0$, let $C^n(G,M)$ be the group of all functions from $G^n$ to $M$. This is an abelian group; its elements are called the (inhomogeneous) $n$-cochains. The coboundary homomorphisms
$begin{cases} d^{n+1} : C^n (G,M) to C^{n+1}(G,M)\ left(d^{n+1}varphiright) (g_1, ldots, g_{n+1}) = g_1 varphi(g_2,dots, g_{n+1}) + sum_{i=1}^n (-1)^i varphi left (g_1,ldots, g_{i-1}, g_i g_{i+1}, ldots, g_{n+1} right ) (-1)^{n+1}varphi(g_1,ldots, g_n) end{cases}$
"
This is OK for $n>0$, but it has no sense when $n=0$. What is the definition of the $d^1$ coboundary map?
group-cohomology
asked Jan 28 at 7:14
mmamma
622413
$endgroup$
add a comment |
$begingroup$
I'm trying to learn, what is group cohomology. Since I'm not a matematician, the general definition is too abstract to me (at least for the time being), and requires too much category theory and homological algebra. First I'm trying to catch the "technical" definition presented by Wikipedia. This definition starts with the definition of the modules of cochains for $n=0,1,...$, and the coboundary maps from the $n$-th module to the $n+1$-th as follows:
"For $nge0$, let $C^n(G,M)$ be the group of all functions from $G^n$ to $M$. This is an abelian group; its elements are called the (inhomogeneous) $n$-cochains. The coboundary homomorphisms
$begin{cases} d^{n+1} : C^n (G,M) to C^{n+1}(G,M)\ left(d^{n+1}varphiright) (g_1, ldots, g_{n+1}) = g_1 varphi(g_2,dots, g_{n+1}) + sum_{i=1}^n (-1)^i varphi left (g_1,ldots, g_{i-1}, g_i g_{i+1}, ldots, g_{n+1} right ) (-1)^{n+1}varphi(g_1,ldots, g_n) end{cases}$
"
This is OK for $n>0$, but it has no sense when $n=0$. What is the definition of the $d^1$ coboundary map?
group-cohomology
asked Jan 28 at 7:14
mmamma
622413
$endgroup$
add a comment |
$begingroup$
I'm trying to learn, what is group cohomology. Since I'm not a matematician, the general definition is too abstract to me (at least for the time being), and requires too much category theory and homological algebra. First I'm trying to catch the "technical" definition presented by Wikipedia. This definition starts with the definition of the modules of cochains for $n=0,1,...$, and the coboundary maps from the $n$-th module to the $n+1$-th as follows:
"For $nge0$, let $C^n(G,M)$ be the group of all functions from $G^n$ to $M$. This is an abelian group; its elements are called the (inhomogeneous) $n$-cochains. The coboundary homomorphisms
$begin{cases} d^{n+1} : C^n (G,M) to C^{n+1}(G,M)\ left(d^{n+1}varphiright) (g_1, ldots, g_{n+1}) = g_1 varphi(g_2,dots, g_{n+1}) + sum_{i=1}^n (-1)^i varphi left (g_1,ldots, g_{i-1}, g_i g_{i+1}, ldots, g_{n+1} right ) (-1)^{n+1}varphi(g_1,ldots, g_n) end{cases}$
"
This is OK for $n>0$, but it has no sense when $n=0$. What is the definition of the $d^1$ coboundary map?
group-cohomology
asked Jan 28 at 7:14
mmamma
622413
$endgroup$
I'm trying to learn, what is group cohomology. Since I'm not a matematician, the general definition is too abstract to me (at least for the time being), and requires too much category theory and homological algebra. First I'm trying to catch the "technical" definition presented by Wikipedia. This definition starts with the definition of the modules of cochains for $n=0,1,...$, and the coboundary maps from the $n$-th module to the $n+1$-th as follows:
"For $nge0$, let $C^n(G,M)$ be the group of all functions from $G^n$ to $M$. This is an abelian group; its elements are called the (inhomogeneous) $n$-cochains. The coboundary homomorphisms
$begin{cases} d^{n+1} : C^n (G,M) to C^{n+1}(G,M)\ left(d^{n+1}varphiright) (g_1, ldots, g_{n+1}) = g_1 varphi(g_2,dots, g_{n+1}) + sum_{i=1}^n (-1)^i varphi left (g_1,ldots, g_{i-1}, g_i g_{i+1}, ldots, g_{n+1} right ) (-1)^{n+1}varphi(g_1,ldots, g_n) end{cases}$
"
This is OK for $n>0$, but it has no sense when $n=0$. What is the definition of the $d^1$ coboundary map?
group-cohomology
group-cohomology
asked Jan 28 at 7:14
mmamma
622413
asked Jan 28 at 7:14
mmamma
622413
asked Jan 28 at 7:14
mmamma
622413
asked Jan 28 at 7:14
mmamma
622413
asked Jan 28 at 7:14
mmamma
622413
622413
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For $n=0$, $G^n$ has only one element, so a function $G^nto M$ is essentially an element $min M$ (the image of the element of $G^n$).
Now if $min M$ is such an element, or correespondingly $f: *mapsto m$ is the map $G^0to M$, you have $d^1f $ which is a map $G^1=Gto M$, and the formula gives $d^1f (g) = gcdot f() + displaystylesum_{i=1}^0$something $+ (-1)^{-1}f()= gcdot m - m$.
Indeed a sum going from $1$ to $0$ of anything is $0$, and for the first and last terms, you remove on element of the list $(g_1,...,g_{n+1})$, which, for $n=0$, has only one element, so you get a list with $0$ elements : great : $f$ can take those as arguments; ans it returns $m$.
Any map of the form $gmapsto gcdot m- m$ is called a principal derivation $Gto M$, and the kernel of $d^2 : C^1(G,M)to C^2(G,M)$ consists of derivations, that is, maps $d: Gto M$ such that $d(gh) = d(g) + gd(h)$. So $H^1(G,M)$ is the set of derivations modulo principal derivations.
answered Jan 28 at 9:10
MaxMax
15.7k11143
$endgroup$
$begingroup$
The name principal derivation is new for me. Is it a synonym for principal crossed homomorphism?
$endgroup$
– mma
Jan 29 at 4:37
$begingroup$
Yes, according to this : encyclopediaofmath.org/index.php/Crossed_homomorphism
$endgroup$
– Max
Jan 29 at 8:51
$begingroup$
Thanks! Meanwhile I found that in Weibel's An Introduction to Homological algebra it is explicitely formulated (definition 6.4.1)
$endgroup$
– mma
Jan 30 at 5:45
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%2f3090575%2fdefinition-of-the-0-coboundary-in-group-cohomology%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
$begingroup$
For $n=0$, $G^n$ has only one element, so a function $G^nto M$ is essentially an element $min M$ (the image of the element of $G^n$).
Now if $min M$ is such an element, or correespondingly $f: *mapsto m$ is the map $G^0to M$, you have $d^1f $ which is a map $G^1=Gto M$, and the formula gives $d^1f (g) = gcdot f() + displaystylesum_{i=1}^0$something $+ (-1)^{-1}f()= gcdot m - m$.
Indeed a sum going from $1$ to $0$ of anything is $0$, and for the first and last terms, you remove on element of the list $(g_1,...,g_{n+1})$, which, for $n=0$, has only one element, so you get a list with $0$ elements : great : $f$ can take those as arguments; ans it returns $m$.
Any map of the form $gmapsto gcdot m- m$ is called a principal derivation $Gto M$, and the kernel of $d^2 : C^1(G,M)to C^2(G,M)$ consists of derivations, that is, maps $d: Gto M$ such that $d(gh) = d(g) + gd(h)$. So $H^1(G,M)$ is the set of derivations modulo principal derivations.
answered Jan 28 at 9:10
MaxMax
15.7k11143
$endgroup$
$begingroup$
The name principal derivation is new for me. Is it a synonym for principal crossed homomorphism?
$endgroup$
– mma
Jan 29 at 4:37
$begingroup$
Yes, according to this : encyclopediaofmath.org/index.php/Crossed_homomorphism
$endgroup$
– Max
Jan 29 at 8:51
$begingroup$
Thanks! Meanwhile I found that in Weibel's An Introduction to Homological algebra it is explicitely formulated (definition 6.4.1)
$endgroup$
– mma
Jan 30 at 5:45
add a comment |
$begingroup$
For $n=0$, $G^n$ has only one element, so a function $G^nto M$ is essentially an element $min M$ (the image of the element of $G^n$).
Now if $min M$ is such an element, or correespondingly $f: *mapsto m$ is the map $G^0to M$, you have $d^1f $ which is a map $G^1=Gto M$, and the formula gives $d^1f (g) = gcdot f() + displaystylesum_{i=1}^0$something $+ (-1)^{-1}f()= gcdot m - m$.
Indeed a sum going from $1$ to $0$ of anything is $0$, and for the first and last terms, you remove on element of the list $(g_1,...,g_{n+1})$, which, for $n=0$, has only one element, so you get a list with $0$ elements : great : $f$ can take those as arguments; ans it returns $m$.
Any map of the form $gmapsto gcdot m- m$ is called a principal derivation $Gto M$, and the kernel of $d^2 : C^1(G,M)to C^2(G,M)$ consists of derivations, that is, maps $d: Gto M$ such that $d(gh) = d(g) + gd(h)$. So $H^1(G,M)$ is the set of derivations modulo principal derivations.
answered Jan 28 at 9:10
MaxMax
15.7k11143
$endgroup$
$begingroup$
The name principal derivation is new for me. Is it a synonym for principal crossed homomorphism?
$endgroup$
– mma
Jan 29 at 4:37
$begingroup$
Yes, according to this : encyclopediaofmath.org/index.php/Crossed_homomorphism
$endgroup$
– Max
Jan 29 at 8:51
$begingroup$
Thanks! Meanwhile I found that in Weibel's An Introduction to Homological algebra it is explicitely formulated (definition 6.4.1)
$endgroup$
– mma
Jan 30 at 5:45
add a comment |
$begingroup$
For $n=0$, $G^n$ has only one element, so a function $G^nto M$ is essentially an element $min M$ (the image of the element of $G^n$).
Now if $min M$ is such an element, or correespondingly $f: *mapsto m$ is the map $G^0to M$, you have $d^1f $ which is a map $G^1=Gto M$, and the formula gives $d^1f (g) = gcdot f() + displaystylesum_{i=1}^0$something $+ (-1)^{-1}f()= gcdot m - m$.
Indeed a sum going from $1$ to $0$ of anything is $0$, and for the first and last terms, you remove on element of the list $(g_1,...,g_{n+1})$, which, for $n=0$, has only one element, so you get a list with $0$ elements : great : $f$ can take those as arguments; ans it returns $m$.
Any map of the form $gmapsto gcdot m- m$ is called a principal derivation $Gto M$, and the kernel of $d^2 : C^1(G,M)to C^2(G,M)$ consists of derivations, that is, maps $d: Gto M$ such that $d(gh) = d(g) + gd(h)$. So $H^1(G,M)$ is the set of derivations modulo principal derivations.
answered Jan 28 at 9:10
MaxMax
15.7k11143
$endgroup$
For $n=0$, $G^n$ has only one element, so a function $G^nto M$ is essentially an element $min M$ (the image of the element of $G^n$).
Now if $min M$ is such an element, or correespondingly $f: *mapsto m$ is the map $G^0to M$, you have $d^1f $ which is a map $G^1=Gto M$, and the formula gives $d^1f (g) = gcdot f() + displaystylesum_{i=1}^0$something $+ (-1)^{-1}f()= gcdot m - m$.
Indeed a sum going from $1$ to $0$ of anything is $0$, and for the first and last terms, you remove on element of the list $(g_1,...,g_{n+1})$, which, for $n=0$, has only one element, so you get a list with $0$ elements : great : $f$ can take those as arguments; ans it returns $m$.
Any map of the form $gmapsto gcdot m- m$ is called a principal derivation $Gto M$, and the kernel of $d^2 : C^1(G,M)to C^2(G,M)$ consists of derivations, that is, maps $d: Gto M$ such that $d(gh) = d(g) + gd(h)$. So $H^1(G,M)$ is the set of derivations modulo principal derivations.
answered Jan 28 at 9:10
MaxMax
15.7k11143
answered Jan 28 at 9:10
MaxMax
15.7k11143
answered Jan 28 at 9:10
MaxMax
15.7k11143
answered Jan 28 at 9:10
MaxMax
15.7k11143
15.7k11143
$begingroup$
The name principal derivation is new for me. Is it a synonym for principal crossed homomorphism?
$endgroup$
– mma
Jan 29 at 4:37
$begingroup$
Yes, according to this : encyclopediaofmath.org/index.php/Crossed_homomorphism
$endgroup$
– Max
Jan 29 at 8:51
$begingroup$
Thanks! Meanwhile I found that in Weibel's An Introduction to Homological algebra it is explicitely formulated (definition 6.4.1)
$endgroup$
– mma
Jan 30 at 5:45
add a comment |
$begingroup$
The name principal derivation is new for me. Is it a synonym for principal crossed homomorphism?
$endgroup$
– mma
Jan 29 at 4:37
$begingroup$
Yes, according to this : encyclopediaofmath.org/index.php/Crossed_homomorphism
$endgroup$
– Max
Jan 29 at 8:51
$begingroup$
Thanks! Meanwhile I found that in Weibel's An Introduction to Homological algebra it is explicitely formulated (definition 6.4.1)
$endgroup$
– mma
Jan 30 at 5:45
$begingroup$
The name principal derivation is new for me. Is it a synonym for principal crossed homomorphism?
$endgroup$
– mma
Jan 29 at 4:37
$begingroup$
The name principal derivation is new for me. Is it a synonym for principal crossed homomorphism?
$endgroup$
– mma
Jan 29 at 4:37
$begingroup$
Yes, according to this : encyclopediaofmath.org/index.php/Crossed_homomorphism
$endgroup$
– Max
Jan 29 at 8:51
$begingroup$
Yes, according to this : encyclopediaofmath.org/index.php/Crossed_homomorphism
$endgroup$
– Max
Jan 29 at 8:51
$begingroup$
Thanks! Meanwhile I found that in Weibel's An Introduction to Homological algebra it is explicitely formulated (definition 6.4.1)
$endgroup$
– mma
Jan 30 at 5:45
$begingroup$
Thanks! Meanwhile I found that in Weibel's An Introduction to Homological algebra it is explicitely formulated (definition 6.4.1)
$endgroup$
– mma
Jan 30 at 5:45
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%2f3090575%2fdefinition-of-the-0-coboundary-in-group-cohomology%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
