Homological algebra using nonabelian groups
$begingroup$
Can homological algebra be done with nonabelian groups? In particular, can homology or cohomology be defined on chain complexes of nonabelian groups? I know that Abelian categories are the choice settings for homological algebra, but the notions of kernel and cokernel (which seem to be all that is necessary to define homology) seem to make sense for nonabelian groups as well, if we define $operatorname{coker}(f : G to H)$ to be the quotient of $H$ by the normal subgroup generated by $operatorname{im} f$.
For example, given a sequence of nonabelian groups
$$
dotsb to C_3 xrightarrow{partial_3} C_2 xrightarrow{partial_2} C_1 xrightarrow{partial_1} C_0 to 0
$$
with $partial_{n} circ partial_{n+1} = 0$, is it useful to define the homology groups $H_n(C_bullet) = ker partial_n/N$, where $N$ is the normal subgroup generated by $operatorname{im} partial_{n+1}$? By useful I mean whether it respects homological algebra, assembles into useful long exact sequences, all the usual stuff.
algebraic-topology homology-cohomology homological-algebra abelian-categories
asked Dec 6 '18 at 3:36
Herng YiHerng Yi
1,4641023
$endgroup$
add a comment |
$begingroup$
Can homological algebra be done with nonabelian groups? In particular, can homology or cohomology be defined on chain complexes of nonabelian groups? I know that Abelian categories are the choice settings for homological algebra, but the notions of kernel and cokernel (which seem to be all that is necessary to define homology) seem to make sense for nonabelian groups as well, if we define $operatorname{coker}(f : G to H)$ to be the quotient of $H$ by the normal subgroup generated by $operatorname{im} f$.
For example, given a sequence of nonabelian groups
$$
dotsb to C_3 xrightarrow{partial_3} C_2 xrightarrow{partial_2} C_1 xrightarrow{partial_1} C_0 to 0
$$
with $partial_{n} circ partial_{n+1} = 0$, is it useful to define the homology groups $H_n(C_bullet) = ker partial_n/N$, where $N$ is the normal subgroup generated by $operatorname{im} partial_{n+1}$? By useful I mean whether it respects homological algebra, assembles into useful long exact sequences, all the usual stuff.
algebraic-topology homology-cohomology homological-algebra abelian-categories
asked Dec 6 '18 at 3:36
Herng YiHerng Yi
1,4641023
$endgroup$
$begingroup$
Is the image of a group homomorphism between two non-abelian groups always a normal subgroup? I'm pretty sure that the fact that nonabelian groups aren't modules is the reason why they are not of interest to mathematicians. One thing that I'm thinking of is the long exact sequence of fibrations. I don't know anything about this but from the wiki page, there doesn't seem to be any requirement that any of the homotopy groups are necessarily abelian.
$endgroup$
– Yunus Syed
Dec 6 '18 at 3:52
$begingroup$
@YunusSyed, $operatorname{im}(f : G to H)$ is not necessarily a normal subgroup of $H$, which is why I define $operatorname{coker} f$ to be $H$ quotiented over the normal subgroup generated by $operatorname{im} f$. To define homology or cohomology, all one seems to need is a notion of kernel and cokernel (like in an Abelian category). But is this useful? Thank you for the connection to homotopy groups though, that's an intriguing possibility that I had not considered.
$endgroup$
– Herng Yi
Dec 6 '18 at 4:03
$begingroup$
@Hemg Yi: If $f:Gto H$ is a crossed module, that image is a normal subgroup in $H$.
$endgroup$
– Tim Porter
Dec 9 '18 at 10:24
add a comment |
$begingroup$
Can homological algebra be done with nonabelian groups? In particular, can homology or cohomology be defined on chain complexes of nonabelian groups? I know that Abelian categories are the choice settings for homological algebra, but the notions of kernel and cokernel (which seem to be all that is necessary to define homology) seem to make sense for nonabelian groups as well, if we define $operatorname{coker}(f : G to H)$ to be the quotient of $H$ by the normal subgroup generated by $operatorname{im} f$.
For example, given a sequence of nonabelian groups
$$
dotsb to C_3 xrightarrow{partial_3} C_2 xrightarrow{partial_2} C_1 xrightarrow{partial_1} C_0 to 0
$$
with $partial_{n} circ partial_{n+1} = 0$, is it useful to define the homology groups $H_n(C_bullet) = ker partial_n/N$, where $N$ is the normal subgroup generated by $operatorname{im} partial_{n+1}$? By useful I mean whether it respects homological algebra, assembles into useful long exact sequences, all the usual stuff.
algebraic-topology homology-cohomology homological-algebra abelian-categories
asked Dec 6 '18 at 3:36
Herng YiHerng Yi
1,4641023
$endgroup$
Can homological algebra be done with nonabelian groups? In particular, can homology or cohomology be defined on chain complexes of nonabelian groups? I know that Abelian categories are the choice settings for homological algebra, but the notions of kernel and cokernel (which seem to be all that is necessary to define homology) seem to make sense for nonabelian groups as well, if we define $operatorname{coker}(f : G to H)$ to be the quotient of $H$ by the normal subgroup generated by $operatorname{im} f$.
For example, given a sequence of nonabelian groups
$$
dotsb to C_3 xrightarrow{partial_3} C_2 xrightarrow{partial_2} C_1 xrightarrow{partial_1} C_0 to 0
$$
with $partial_{n} circ partial_{n+1} = 0$, is it useful to define the homology groups $H_n(C_bullet) = ker partial_n/N$, where $N$ is the normal subgroup generated by $operatorname{im} partial_{n+1}$? By useful I mean whether it respects homological algebra, assembles into useful long exact sequences, all the usual stuff.
algebraic-topology homology-cohomology homological-algebra abelian-categories
algebraic-topology homology-cohomology homological-algebra abelian-categories
asked Dec 6 '18 at 3:36
Herng YiHerng Yi
1,4641023
asked Dec 6 '18 at 3:36
Herng YiHerng Yi
1,4641023
edited Dec 6 '18 at 7:52
Herng Yi
asked Dec 6 '18 at 3:36
Herng YiHerng Yi
1,4641023
asked Dec 6 '18 at 3:36
Herng YiHerng Yi
1,4641023
asked Dec 6 '18 at 3:36
Herng YiHerng Yi
1,4641023
1,4641023
$begingroup$
Is the image of a group homomorphism between two non-abelian groups always a normal subgroup? I'm pretty sure that the fact that nonabelian groups aren't modules is the reason why they are not of interest to mathematicians. One thing that I'm thinking of is the long exact sequence of fibrations. I don't know anything about this but from the wiki page, there doesn't seem to be any requirement that any of the homotopy groups are necessarily abelian.
$endgroup$
– Yunus Syed
Dec 6 '18 at 3:52
$begingroup$
@YunusSyed, $operatorname{im}(f : G to H)$ is not necessarily a normal subgroup of $H$, which is why I define $operatorname{coker} f$ to be $H$ quotiented over the normal subgroup generated by $operatorname{im} f$. To define homology or cohomology, all one seems to need is a notion of kernel and cokernel (like in an Abelian category). But is this useful? Thank you for the connection to homotopy groups though, that's an intriguing possibility that I had not considered.
$endgroup$
– Herng Yi
Dec 6 '18 at 4:03
$begingroup$
@Hemg Yi: If $f:Gto H$ is a crossed module, that image is a normal subgroup in $H$.
$endgroup$
– Tim Porter
Dec 9 '18 at 10:24
add a comment |
$begingroup$
Is the image of a group homomorphism between two non-abelian groups always a normal subgroup? I'm pretty sure that the fact that nonabelian groups aren't modules is the reason why they are not of interest to mathematicians. One thing that I'm thinking of is the long exact sequence of fibrations. I don't know anything about this but from the wiki page, there doesn't seem to be any requirement that any of the homotopy groups are necessarily abelian.
$endgroup$
– Yunus Syed
Dec 6 '18 at 3:52
$begingroup$
@YunusSyed, $operatorname{im}(f : G to H)$ is not necessarily a normal subgroup of $H$, which is why I define $operatorname{coker} f$ to be $H$ quotiented over the normal subgroup generated by $operatorname{im} f$. To define homology or cohomology, all one seems to need is a notion of kernel and cokernel (like in an Abelian category). But is this useful? Thank you for the connection to homotopy groups though, that's an intriguing possibility that I had not considered.
$endgroup$
– Herng Yi
Dec 6 '18 at 4:03
$begingroup$
@Hemg Yi: If $f:Gto H$ is a crossed module, that image is a normal subgroup in $H$.
$endgroup$
– Tim Porter
Dec 9 '18 at 10:24
$begingroup$
Is the image of a group homomorphism between two non-abelian groups always a normal subgroup? I'm pretty sure that the fact that nonabelian groups aren't modules is the reason why they are not of interest to mathematicians. One thing that I'm thinking of is the long exact sequence of fibrations. I don't know anything about this but from the wiki page, there doesn't seem to be any requirement that any of the homotopy groups are necessarily abelian.
$endgroup$
– Yunus Syed
Dec 6 '18 at 3:52
$begingroup$
Is the image of a group homomorphism between two non-abelian groups always a normal subgroup? I'm pretty sure that the fact that nonabelian groups aren't modules is the reason why they are not of interest to mathematicians. One thing that I'm thinking of is the long exact sequence of fibrations. I don't know anything about this but from the wiki page, there doesn't seem to be any requirement that any of the homotopy groups are necessarily abelian.
$endgroup$
– Yunus Syed
Dec 6 '18 at 3:52
$begingroup$
@YunusSyed, $operatorname{im}(f : G to H)$ is not necessarily a normal subgroup of $H$, which is why I define $operatorname{coker} f$ to be $H$ quotiented over the normal subgroup generated by $operatorname{im} f$. To define homology or cohomology, all one seems to need is a notion of kernel and cokernel (like in an Abelian category). But is this useful? Thank you for the connection to homotopy groups though, that's an intriguing possibility that I had not considered.
$endgroup$
– Herng Yi
Dec 6 '18 at 4:03
$begingroup$
@YunusSyed, $operatorname{im}(f : G to H)$ is not necessarily a normal subgroup of $H$, which is why I define $operatorname{coker} f$ to be $H$ quotiented over the normal subgroup generated by $operatorname{im} f$. To define homology or cohomology, all one seems to need is a notion of kernel and cokernel (like in an Abelian category). But is this useful? Thank you for the connection to homotopy groups though, that's an intriguing possibility that I had not considered.
$endgroup$
– Herng Yi
Dec 6 '18 at 4:03
$begingroup$
@Hemg Yi: If $f:Gto H$ is a crossed module, that image is a normal subgroup in $H$.
$endgroup$
– Tim Porter
Dec 9 '18 at 10:24
$begingroup$
@Hemg Yi: If $f:Gto H$ is a crossed module, that image is a normal subgroup in $H$.
$endgroup$
– Tim Porter
Dec 9 '18 at 10:24
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
A good setting for generalizing homological algebra is that of homological categories. There is a book on the subject due to Borceux and Bourn references at the linked article, as well as a very approachable shorter book by Bourn alone. The definition is noticeably more complex than that of an abelian category or a topos, but quite natural with sufficient explanation.
More to the point, the concept subsumes groups, as well as other nice algebraic categories which have a zero object that are "groupish" enough, such as rings without unit and Lie algebras, and more exotically, the opposite of the category of pointed sets. However, such categories as unital rings and monoids (even commutative monoids) are not homological, essentially because they lack a sufficiently strong notion of kernel. Most of the standard homological algebra results hold in a homological category, including the snake lemma and the resulting long exact sequence in homology.
answered Dec 6 '18 at 7:10
Kevin CarlsonKevin Carlson
32.7k23271
$endgroup$
add a comment |
$begingroup$
There's no reason to expect this to be useful, and as far as I know it's not. An abstract conceptual way to think about where homological algebra comes from is the Dold-Kan correspondence, which says that nonnegatively graded chain complexes of abelian groups (for simplicity) are equivalent to simplicial abelian groups, and this equivalence sends the homology groups of a chain complex to the simplicial homotopy groups of the corresponding simplicial abelian group. Why you would care about simplicial abelian groups is a long story, but the short version is that they have the same homotopy theory as topological abelian groups.
The Dold-Kan correspondence is valid more generally with abelian groups replaced by an abelian category, but is not at all valid for groups: simplicial groups (which hav the same homotopy theory as topological groups, which are very interesting!) are much more complicated than chain complexes of groups.
answered Dec 6 '18 at 5:15
Qiaochu YuanQiaochu Yuan
278k32584920
$endgroup$
$begingroup$
Of course, one can often make sense of first cohomology with coefficients in a nonabelian group with has good properties, but this is not at all what OP was asking about.
$endgroup$
– Mike Miller
Dec 6 '18 at 5:36
$begingroup$
There is a Dold-Kan correspondence for groups, and generally objects in any semi-abelian category, although certainly it doesn't apply directly to chain complexes as in the abelian case.
$endgroup$
– Kevin Carlson
Dec 6 '18 at 7:13
$begingroup$
@MikeMiller, could you give an example where the first cohomology with coefficients in a nonabelian group is interesting?
$endgroup$
– Herng Yi
Dec 6 '18 at 7:54
1
$begingroup$
There is good reason to suppose that your comments are useful as homotopical algebra generalises homological algebra in many of the ways that the original questioner was asking for and simplicial homotopy theory is a key factor in many settings for homotopical algebra.
$endgroup$
– Tim Porter
Dec 9 '18 at 10:27
add a comment |
$begingroup$
Taking up the point made by Qiaochu Yuan, chain complexes, at least the positive ones of abelian groups correspond to simplicial abelian groups. There is a well defined homotopy theory of simplicial groups and this has all that you might want (and more!). The Moore complex of a simplicial group (see the nLab article on this) is a chain complex of groups (which need not be abelian). It has the nice property that the image of the boundary map is always a normal subgroup so you do not need to take normal closures or things like that. Its Homology groups are the same as the homotopy groups of the simplicial group. There are long exact sequences etc as you would expect.
There is a Dold-Kan theorem in this case as well, but the objects corresponding to chain complexes encode a lot more of the structure. You can get a feeling for what this extra structure is in the nLab entry on hypercrossed complexes and in the paper: P. Carrasco and A. M. Cegarra, Group-theoretic Algebraic Models for Homotopy Types, J. Pure Appl. Alg., 75, (1991), 195 – 235.
Finally in reply to Hemg Yi's comment, there is a lot of work on non-abelian extensions of groups and it is exactly the cohomology with non-abelian coefficients that play the key role. a fairly simple article on this is by Ronnie Brown and myself, On the Schreier theory of non - abelian extensions: generalisations
and computations, Proc. Royal Irish Acad. 96, (1996) 213 - 227, which may give you a flavour.
answered Dec 7 '18 at 16:10
Tim PorterTim Porter
75648
$endgroup$
add a comment |
$begingroup$
To give further background to the question here is an extract from google scholar with a search on "Frohlich homological"' (A. Frohlich was a well known worker in algebraic number theory):
Non‐Abelian Homological Algebra I. Derived Functors and Satellites,
A Fröhlich - Proceedings of the London Mathematical Society, 1961 - Wiley Online Library
The purpose of this forthcoming series of papers is the development of a homology (or
cohomology) theory based on structures with non-commutative addition. The module over a
ring, which was hitherto the basic concept for homological algebra, will here be replaced by …
A search on "Lue homological" also gives relevant articles. One of the points that comes out from his 1971, 1981 articles, derived from Frohlich's work, is that his form of the construction of the cohomology is by classes of representative algebraic objects: for example, in the case of groups these objects are commonly called crossed $n$-fold extensions.
The more usual approach is in terms of classes of cocycles. To my mind, a difficulty here is: what do you do with a cocycle when you have got it? By contrast, all sorts of things can be done with or asked about algebraic objects.
This is relevant to the algebraic objects used in the book Nonabelian Algebraic Topology, where, following J.H.C. Whitehead, his crossed modules, and what are now called crossed complexes, and other structures, are used to model some homotopy types. One aspect of the approach
is that the complications of, say, a 3-cocycle definition, are, following work of J. Huebschmann, put into the differentials of a "standard free crossed resolution", so that the 3-cocycle becomes a morphism of crossed complexes; there are many homotopical methods for constructing such morphisms.
To give further background, note that in a letter dated 02/05/1983 Alexander Grothendieck wrote:
Don’t be surprised by my supposed efficiency in digging out the right
kind of notions – I have just been following, rather let myself be pulled
ahead, by that very strong thread (roughly: understand non commutative
cohomology of topoi!) which I kept trying to sell for about ten or twenty
years now, without anyone ready to ``buy” it, namely to do the work. So
finally I got mad and decided to work out at least an outline by myself.
These were his ideas in "Pursuing Stacks".
So I think there is still a lot to take in, assess, and evaluate!
answered Dec 8 '18 at 17:09
Ronnie BrownRonnie Brown
11.9k12938
$endgroup$
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%2f3028000%2fhomological-algebra-using-nonabelian-groups%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A good setting for generalizing homological algebra is that of homological categories. There is a book on the subject due to Borceux and Bourn references at the linked article, as well as a very approachable shorter book by Bourn alone. The definition is noticeably more complex than that of an abelian category or a topos, but quite natural with sufficient explanation.
More to the point, the concept subsumes groups, as well as other nice algebraic categories which have a zero object that are "groupish" enough, such as rings without unit and Lie algebras, and more exotically, the opposite of the category of pointed sets. However, such categories as unital rings and monoids (even commutative monoids) are not homological, essentially because they lack a sufficiently strong notion of kernel. Most of the standard homological algebra results hold in a homological category, including the snake lemma and the resulting long exact sequence in homology.
answered Dec 6 '18 at 7:10
Kevin CarlsonKevin Carlson
32.7k23271
$endgroup$
add a comment |
$begingroup$
A good setting for generalizing homological algebra is that of homological categories. There is a book on the subject due to Borceux and Bourn references at the linked article, as well as a very approachable shorter book by Bourn alone. The definition is noticeably more complex than that of an abelian category or a topos, but quite natural with sufficient explanation.
More to the point, the concept subsumes groups, as well as other nice algebraic categories which have a zero object that are "groupish" enough, such as rings without unit and Lie algebras, and more exotically, the opposite of the category of pointed sets. However, such categories as unital rings and monoids (even commutative monoids) are not homological, essentially because they lack a sufficiently strong notion of kernel. Most of the standard homological algebra results hold in a homological category, including the snake lemma and the resulting long exact sequence in homology.
answered Dec 6 '18 at 7:10
Kevin CarlsonKevin Carlson
32.7k23271
$endgroup$
add a comment |
$begingroup$
A good setting for generalizing homological algebra is that of homological categories. There is a book on the subject due to Borceux and Bourn references at the linked article, as well as a very approachable shorter book by Bourn alone. The definition is noticeably more complex than that of an abelian category or a topos, but quite natural with sufficient explanation.
More to the point, the concept subsumes groups, as well as other nice algebraic categories which have a zero object that are "groupish" enough, such as rings without unit and Lie algebras, and more exotically, the opposite of the category of pointed sets. However, such categories as unital rings and monoids (even commutative monoids) are not homological, essentially because they lack a sufficiently strong notion of kernel. Most of the standard homological algebra results hold in a homological category, including the snake lemma and the resulting long exact sequence in homology.
answered Dec 6 '18 at 7:10
Kevin CarlsonKevin Carlson
32.7k23271
$endgroup$
A good setting for generalizing homological algebra is that of homological categories. There is a book on the subject due to Borceux and Bourn references at the linked article, as well as a very approachable shorter book by Bourn alone. The definition is noticeably more complex than that of an abelian category or a topos, but quite natural with sufficient explanation.
More to the point, the concept subsumes groups, as well as other nice algebraic categories which have a zero object that are "groupish" enough, such as rings without unit and Lie algebras, and more exotically, the opposite of the category of pointed sets. However, such categories as unital rings and monoids (even commutative monoids) are not homological, essentially because they lack a sufficiently strong notion of kernel. Most of the standard homological algebra results hold in a homological category, including the snake lemma and the resulting long exact sequence in homology.
answered Dec 6 '18 at 7:10
Kevin CarlsonKevin Carlson
32.7k23271
edited Dec 6 '18 at 7:22
answered Dec 6 '18 at 7:10
Kevin CarlsonKevin Carlson
32.7k23271
answered Dec 6 '18 at 7:10
Kevin CarlsonKevin Carlson
32.7k23271
answered Dec 6 '18 at 7:10
Kevin CarlsonKevin Carlson
32.7k23271
32.7k23271
add a comment |
add a comment |
$begingroup$
There's no reason to expect this to be useful, and as far as I know it's not. An abstract conceptual way to think about where homological algebra comes from is the Dold-Kan correspondence, which says that nonnegatively graded chain complexes of abelian groups (for simplicity) are equivalent to simplicial abelian groups, and this equivalence sends the homology groups of a chain complex to the simplicial homotopy groups of the corresponding simplicial abelian group. Why you would care about simplicial abelian groups is a long story, but the short version is that they have the same homotopy theory as topological abelian groups.
The Dold-Kan correspondence is valid more generally with abelian groups replaced by an abelian category, but is not at all valid for groups: simplicial groups (which hav the same homotopy theory as topological groups, which are very interesting!) are much more complicated than chain complexes of groups.
answered Dec 6 '18 at 5:15
Qiaochu YuanQiaochu Yuan
278k32584920
$endgroup$
$begingroup$
Of course, one can often make sense of first cohomology with coefficients in a nonabelian group with has good properties, but this is not at all what OP was asking about.
$endgroup$
– Mike Miller
Dec 6 '18 at 5:36
$begingroup$
There is a Dold-Kan correspondence for groups, and generally objects in any semi-abelian category, although certainly it doesn't apply directly to chain complexes as in the abelian case.
$endgroup$
– Kevin Carlson
Dec 6 '18 at 7:13
$begingroup$
@MikeMiller, could you give an example where the first cohomology with coefficients in a nonabelian group is interesting?
$endgroup$
– Herng Yi
Dec 6 '18 at 7:54
1
$begingroup$
There is good reason to suppose that your comments are useful as homotopical algebra generalises homological algebra in many of the ways that the original questioner was asking for and simplicial homotopy theory is a key factor in many settings for homotopical algebra.
$endgroup$
– Tim Porter
Dec 9 '18 at 10:27
add a comment |
$begingroup$
There's no reason to expect this to be useful, and as far as I know it's not. An abstract conceptual way to think about where homological algebra comes from is the Dold-Kan correspondence, which says that nonnegatively graded chain complexes of abelian groups (for simplicity) are equivalent to simplicial abelian groups, and this equivalence sends the homology groups of a chain complex to the simplicial homotopy groups of the corresponding simplicial abelian group. Why you would care about simplicial abelian groups is a long story, but the short version is that they have the same homotopy theory as topological abelian groups.
The Dold-Kan correspondence is valid more generally with abelian groups replaced by an abelian category, but is not at all valid for groups: simplicial groups (which hav the same homotopy theory as topological groups, which are very interesting!) are much more complicated than chain complexes of groups.
answered Dec 6 '18 at 5:15
Qiaochu YuanQiaochu Yuan
278k32584920
$endgroup$
$begingroup$
Of course, one can often make sense of first cohomology with coefficients in a nonabelian group with has good properties, but this is not at all what OP was asking about.
$endgroup$
– Mike Miller
Dec 6 '18 at 5:36
$begingroup$
There is a Dold-Kan correspondence for groups, and generally objects in any semi-abelian category, although certainly it doesn't apply directly to chain complexes as in the abelian case.
$endgroup$
– Kevin Carlson
Dec 6 '18 at 7:13
$begingroup$
@MikeMiller, could you give an example where the first cohomology with coefficients in a nonabelian group is interesting?
$endgroup$
– Herng Yi
Dec 6 '18 at 7:54
1
$begingroup$
There is good reason to suppose that your comments are useful as homotopical algebra generalises homological algebra in many of the ways that the original questioner was asking for and simplicial homotopy theory is a key factor in many settings for homotopical algebra.
$endgroup$
– Tim Porter
Dec 9 '18 at 10:27
add a comment |
$begingroup$
There's no reason to expect this to be useful, and as far as I know it's not. An abstract conceptual way to think about where homological algebra comes from is the Dold-Kan correspondence, which says that nonnegatively graded chain complexes of abelian groups (for simplicity) are equivalent to simplicial abelian groups, and this equivalence sends the homology groups of a chain complex to the simplicial homotopy groups of the corresponding simplicial abelian group. Why you would care about simplicial abelian groups is a long story, but the short version is that they have the same homotopy theory as topological abelian groups.
The Dold-Kan correspondence is valid more generally with abelian groups replaced by an abelian category, but is not at all valid for groups: simplicial groups (which hav the same homotopy theory as topological groups, which are very interesting!) are much more complicated than chain complexes of groups.
answered Dec 6 '18 at 5:15
Qiaochu YuanQiaochu Yuan
278k32584920
$endgroup$
There's no reason to expect this to be useful, and as far as I know it's not. An abstract conceptual way to think about where homological algebra comes from is the Dold-Kan correspondence, which says that nonnegatively graded chain complexes of abelian groups (for simplicity) are equivalent to simplicial abelian groups, and this equivalence sends the homology groups of a chain complex to the simplicial homotopy groups of the corresponding simplicial abelian group. Why you would care about simplicial abelian groups is a long story, but the short version is that they have the same homotopy theory as topological abelian groups.
The Dold-Kan correspondence is valid more generally with abelian groups replaced by an abelian category, but is not at all valid for groups: simplicial groups (which hav the same homotopy theory as topological groups, which are very interesting!) are much more complicated than chain complexes of groups.
answered Dec 6 '18 at 5:15
Qiaochu YuanQiaochu Yuan
278k32584920
answered Dec 6 '18 at 5:15
Qiaochu YuanQiaochu Yuan
278k32584920
answered Dec 6 '18 at 5:15
Qiaochu YuanQiaochu Yuan
278k32584920
answered Dec 6 '18 at 5:15
Qiaochu YuanQiaochu Yuan
278k32584920
278k32584920
$begingroup$
Of course, one can often make sense of first cohomology with coefficients in a nonabelian group with has good properties, but this is not at all what OP was asking about.
$endgroup$
– Mike Miller
Dec 6 '18 at 5:36
$begingroup$
There is a Dold-Kan correspondence for groups, and generally objects in any semi-abelian category, although certainly it doesn't apply directly to chain complexes as in the abelian case.
$endgroup$
– Kevin Carlson
Dec 6 '18 at 7:13
$begingroup$
@MikeMiller, could you give an example where the first cohomology with coefficients in a nonabelian group is interesting?
$endgroup$
– Herng Yi
Dec 6 '18 at 7:54
1
$begingroup$
There is good reason to suppose that your comments are useful as homotopical algebra generalises homological algebra in many of the ways that the original questioner was asking for and simplicial homotopy theory is a key factor in many settings for homotopical algebra.
$endgroup$
– Tim Porter
Dec 9 '18 at 10:27
add a comment |
$begingroup$
Of course, one can often make sense of first cohomology with coefficients in a nonabelian group with has good properties, but this is not at all what OP was asking about.
$endgroup$
– Mike Miller
Dec 6 '18 at 5:36
$begingroup$
There is a Dold-Kan correspondence for groups, and generally objects in any semi-abelian category, although certainly it doesn't apply directly to chain complexes as in the abelian case.
$endgroup$
– Kevin Carlson
Dec 6 '18 at 7:13
$begingroup$
@MikeMiller, could you give an example where the first cohomology with coefficients in a nonabelian group is interesting?
$endgroup$
– Herng Yi
Dec 6 '18 at 7:54
1
$begingroup$
There is good reason to suppose that your comments are useful as homotopical algebra generalises homological algebra in many of the ways that the original questioner was asking for and simplicial homotopy theory is a key factor in many settings for homotopical algebra.
$endgroup$
– Tim Porter
Dec 9 '18 at 10:27
$begingroup$
Of course, one can often make sense of first cohomology with coefficients in a nonabelian group with has good properties, but this is not at all what OP was asking about.
$endgroup$
– Mike Miller
Dec 6 '18 at 5:36
$begingroup$
Of course, one can often make sense of first cohomology with coefficients in a nonabelian group with has good properties, but this is not at all what OP was asking about.
$endgroup$
– Mike Miller
Dec 6 '18 at 5:36
$begingroup$
There is a Dold-Kan correspondence for groups, and generally objects in any semi-abelian category, although certainly it doesn't apply directly to chain complexes as in the abelian case.
$endgroup$
– Kevin Carlson
Dec 6 '18 at 7:13
$begingroup$
There is a Dold-Kan correspondence for groups, and generally objects in any semi-abelian category, although certainly it doesn't apply directly to chain complexes as in the abelian case.
$endgroup$
– Kevin Carlson
Dec 6 '18 at 7:13
$begingroup$
@MikeMiller, could you give an example where the first cohomology with coefficients in a nonabelian group is interesting?
$endgroup$
– Herng Yi
Dec 6 '18 at 7:54
$begingroup$
@MikeMiller, could you give an example where the first cohomology with coefficients in a nonabelian group is interesting?
$endgroup$
– Herng Yi
Dec 6 '18 at 7:54
1
1
$begingroup$
There is good reason to suppose that your comments are useful as homotopical algebra generalises homological algebra in many of the ways that the original questioner was asking for and simplicial homotopy theory is a key factor in many settings for homotopical algebra.
$endgroup$
– Tim Porter
Dec 9 '18 at 10:27
$begingroup$
There is good reason to suppose that your comments are useful as homotopical algebra generalises homological algebra in many of the ways that the original questioner was asking for and simplicial homotopy theory is a key factor in many settings for homotopical algebra.
$endgroup$
– Tim Porter
Dec 9 '18 at 10:27
add a comment |
$begingroup$
Taking up the point made by Qiaochu Yuan, chain complexes, at least the positive ones of abelian groups correspond to simplicial abelian groups. There is a well defined homotopy theory of simplicial groups and this has all that you might want (and more!). The Moore complex of a simplicial group (see the nLab article on this) is a chain complex of groups (which need not be abelian). It has the nice property that the image of the boundary map is always a normal subgroup so you do not need to take normal closures or things like that. Its Homology groups are the same as the homotopy groups of the simplicial group. There are long exact sequences etc as you would expect.
There is a Dold-Kan theorem in this case as well, but the objects corresponding to chain complexes encode a lot more of the structure. You can get a feeling for what this extra structure is in the nLab entry on hypercrossed complexes and in the paper: P. Carrasco and A. M. Cegarra, Group-theoretic Algebraic Models for Homotopy Types, J. Pure Appl. Alg., 75, (1991), 195 – 235.
Finally in reply to Hemg Yi's comment, there is a lot of work on non-abelian extensions of groups and it is exactly the cohomology with non-abelian coefficients that play the key role. a fairly simple article on this is by Ronnie Brown and myself, On the Schreier theory of non - abelian extensions: generalisations
and computations, Proc. Royal Irish Acad. 96, (1996) 213 - 227, which may give you a flavour.
answered Dec 7 '18 at 16:10
Tim PorterTim Porter
75648
$endgroup$
add a comment |
$begingroup$
Taking up the point made by Qiaochu Yuan, chain complexes, at least the positive ones of abelian groups correspond to simplicial abelian groups. There is a well defined homotopy theory of simplicial groups and this has all that you might want (and more!). The Moore complex of a simplicial group (see the nLab article on this) is a chain complex of groups (which need not be abelian). It has the nice property that the image of the boundary map is always a normal subgroup so you do not need to take normal closures or things like that. Its Homology groups are the same as the homotopy groups of the simplicial group. There are long exact sequences etc as you would expect.
There is a Dold-Kan theorem in this case as well, but the objects corresponding to chain complexes encode a lot more of the structure. You can get a feeling for what this extra structure is in the nLab entry on hypercrossed complexes and in the paper: P. Carrasco and A. M. Cegarra, Group-theoretic Algebraic Models for Homotopy Types, J. Pure Appl. Alg., 75, (1991), 195 – 235.
Finally in reply to Hemg Yi's comment, there is a lot of work on non-abelian extensions of groups and it is exactly the cohomology with non-abelian coefficients that play the key role. a fairly simple article on this is by Ronnie Brown and myself, On the Schreier theory of non - abelian extensions: generalisations
and computations, Proc. Royal Irish Acad. 96, (1996) 213 - 227, which may give you a flavour.
answered Dec 7 '18 at 16:10
Tim PorterTim Porter
75648
$endgroup$
add a comment |
$begingroup$
Taking up the point made by Qiaochu Yuan, chain complexes, at least the positive ones of abelian groups correspond to simplicial abelian groups. There is a well defined homotopy theory of simplicial groups and this has all that you might want (and more!). The Moore complex of a simplicial group (see the nLab article on this) is a chain complex of groups (which need not be abelian). It has the nice property that the image of the boundary map is always a normal subgroup so you do not need to take normal closures or things like that. Its Homology groups are the same as the homotopy groups of the simplicial group. There are long exact sequences etc as you would expect.
There is a Dold-Kan theorem in this case as well, but the objects corresponding to chain complexes encode a lot more of the structure. You can get a feeling for what this extra structure is in the nLab entry on hypercrossed complexes and in the paper: P. Carrasco and A. M. Cegarra, Group-theoretic Algebraic Models for Homotopy Types, J. Pure Appl. Alg., 75, (1991), 195 – 235.
Finally in reply to Hemg Yi's comment, there is a lot of work on non-abelian extensions of groups and it is exactly the cohomology with non-abelian coefficients that play the key role. a fairly simple article on this is by Ronnie Brown and myself, On the Schreier theory of non - abelian extensions: generalisations
and computations, Proc. Royal Irish Acad. 96, (1996) 213 - 227, which may give you a flavour.
answered Dec 7 '18 at 16:10
Tim PorterTim Porter
75648
$endgroup$
Taking up the point made by Qiaochu Yuan, chain complexes, at least the positive ones of abelian groups correspond to simplicial abelian groups. There is a well defined homotopy theory of simplicial groups and this has all that you might want (and more!). The Moore complex of a simplicial group (see the nLab article on this) is a chain complex of groups (which need not be abelian). It has the nice property that the image of the boundary map is always a normal subgroup so you do not need to take normal closures or things like that. Its Homology groups are the same as the homotopy groups of the simplicial group. There are long exact sequences etc as you would expect.
There is a Dold-Kan theorem in this case as well, but the objects corresponding to chain complexes encode a lot more of the structure. You can get a feeling for what this extra structure is in the nLab entry on hypercrossed complexes and in the paper: P. Carrasco and A. M. Cegarra, Group-theoretic Algebraic Models for Homotopy Types, J. Pure Appl. Alg., 75, (1991), 195 – 235.
Finally in reply to Hemg Yi's comment, there is a lot of work on non-abelian extensions of groups and it is exactly the cohomology with non-abelian coefficients that play the key role. a fairly simple article on this is by Ronnie Brown and myself, On the Schreier theory of non - abelian extensions: generalisations
and computations, Proc. Royal Irish Acad. 96, (1996) 213 - 227, which may give you a flavour.
answered Dec 7 '18 at 16:10
Tim PorterTim Porter
75648
edited Dec 8 '18 at 6:52
answered Dec 7 '18 at 16:10
Tim PorterTim Porter
75648
answered Dec 7 '18 at 16:10
Tim PorterTim Porter
75648
answered Dec 7 '18 at 16:10
Tim PorterTim Porter
75648
75648
add a comment |
add a comment |
$begingroup$
To give further background to the question here is an extract from google scholar with a search on "Frohlich homological"' (A. Frohlich was a well known worker in algebraic number theory):
Non‐Abelian Homological Algebra I. Derived Functors and Satellites,
A Fröhlich - Proceedings of the London Mathematical Society, 1961 - Wiley Online Library
The purpose of this forthcoming series of papers is the development of a homology (or
cohomology) theory based on structures with non-commutative addition. The module over a
ring, which was hitherto the basic concept for homological algebra, will here be replaced by …
A search on "Lue homological" also gives relevant articles. One of the points that comes out from his 1971, 1981 articles, derived from Frohlich's work, is that his form of the construction of the cohomology is by classes of representative algebraic objects: for example, in the case of groups these objects are commonly called crossed $n$-fold extensions.
The more usual approach is in terms of classes of cocycles. To my mind, a difficulty here is: what do you do with a cocycle when you have got it? By contrast, all sorts of things can be done with or asked about algebraic objects.
This is relevant to the algebraic objects used in the book Nonabelian Algebraic Topology, where, following J.H.C. Whitehead, his crossed modules, and what are now called crossed complexes, and other structures, are used to model some homotopy types. One aspect of the approach
is that the complications of, say, a 3-cocycle definition, are, following work of J. Huebschmann, put into the differentials of a "standard free crossed resolution", so that the 3-cocycle becomes a morphism of crossed complexes; there are many homotopical methods for constructing such morphisms.
To give further background, note that in a letter dated 02/05/1983 Alexander Grothendieck wrote:
Don’t be surprised by my supposed efficiency in digging out the right
kind of notions – I have just been following, rather let myself be pulled
ahead, by that very strong thread (roughly: understand non commutative
cohomology of topoi!) which I kept trying to sell for about ten or twenty
years now, without anyone ready to ``buy” it, namely to do the work. So
finally I got mad and decided to work out at least an outline by myself.
These were his ideas in "Pursuing Stacks".
So I think there is still a lot to take in, assess, and evaluate!
answered Dec 8 '18 at 17:09
Ronnie BrownRonnie Brown
11.9k12938
$endgroup$
add a comment |
$begingroup$
To give further background to the question here is an extract from google scholar with a search on "Frohlich homological"' (A. Frohlich was a well known worker in algebraic number theory):
Non‐Abelian Homological Algebra I. Derived Functors and Satellites,
A Fröhlich - Proceedings of the London Mathematical Society, 1961 - Wiley Online Library
The purpose of this forthcoming series of papers is the development of a homology (or
cohomology) theory based on structures with non-commutative addition. The module over a
ring, which was hitherto the basic concept for homological algebra, will here be replaced by …
A search on "Lue homological" also gives relevant articles. One of the points that comes out from his 1971, 1981 articles, derived from Frohlich's work, is that his form of the construction of the cohomology is by classes of representative algebraic objects: for example, in the case of groups these objects are commonly called crossed $n$-fold extensions.
The more usual approach is in terms of classes of cocycles. To my mind, a difficulty here is: what do you do with a cocycle when you have got it? By contrast, all sorts of things can be done with or asked about algebraic objects.
This is relevant to the algebraic objects used in the book Nonabelian Algebraic Topology, where, following J.H.C. Whitehead, his crossed modules, and what are now called crossed complexes, and other structures, are used to model some homotopy types. One aspect of the approach
is that the complications of, say, a 3-cocycle definition, are, following work of J. Huebschmann, put into the differentials of a "standard free crossed resolution", so that the 3-cocycle becomes a morphism of crossed complexes; there are many homotopical methods for constructing such morphisms.
To give further background, note that in a letter dated 02/05/1983 Alexander Grothendieck wrote:
Don’t be surprised by my supposed efficiency in digging out the right
kind of notions – I have just been following, rather let myself be pulled
ahead, by that very strong thread (roughly: understand non commutative
cohomology of topoi!) which I kept trying to sell for about ten or twenty
years now, without anyone ready to ``buy” it, namely to do the work. So
finally I got mad and decided to work out at least an outline by myself.
These were his ideas in "Pursuing Stacks".
So I think there is still a lot to take in, assess, and evaluate!
answered Dec 8 '18 at 17:09
Ronnie BrownRonnie Brown
11.9k12938
$endgroup$
add a comment |
$begingroup$
To give further background to the question here is an extract from google scholar with a search on "Frohlich homological"' (A. Frohlich was a well known worker in algebraic number theory):
Non‐Abelian Homological Algebra I. Derived Functors and Satellites,
A Fröhlich - Proceedings of the London Mathematical Society, 1961 - Wiley Online Library
The purpose of this forthcoming series of papers is the development of a homology (or
cohomology) theory based on structures with non-commutative addition. The module over a
ring, which was hitherto the basic concept for homological algebra, will here be replaced by …
A search on "Lue homological" also gives relevant articles. One of the points that comes out from his 1971, 1981 articles, derived from Frohlich's work, is that his form of the construction of the cohomology is by classes of representative algebraic objects: for example, in the case of groups these objects are commonly called crossed $n$-fold extensions.
The more usual approach is in terms of classes of cocycles. To my mind, a difficulty here is: what do you do with a cocycle when you have got it? By contrast, all sorts of things can be done with or asked about algebraic objects.
This is relevant to the algebraic objects used in the book Nonabelian Algebraic Topology, where, following J.H.C. Whitehead, his crossed modules, and what are now called crossed complexes, and other structures, are used to model some homotopy types. One aspect of the approach
is that the complications of, say, a 3-cocycle definition, are, following work of J. Huebschmann, put into the differentials of a "standard free crossed resolution", so that the 3-cocycle becomes a morphism of crossed complexes; there are many homotopical methods for constructing such morphisms.
To give further background, note that in a letter dated 02/05/1983 Alexander Grothendieck wrote:
Don’t be surprised by my supposed efficiency in digging out the right
kind of notions – I have just been following, rather let myself be pulled
ahead, by that very strong thread (roughly: understand non commutative
cohomology of topoi!) which I kept trying to sell for about ten or twenty
years now, without anyone ready to ``buy” it, namely to do the work. So
finally I got mad and decided to work out at least an outline by myself.
These were his ideas in "Pursuing Stacks".
So I think there is still a lot to take in, assess, and evaluate!
answered Dec 8 '18 at 17:09
Ronnie BrownRonnie Brown
11.9k12938
$endgroup$
To give further background to the question here is an extract from google scholar with a search on "Frohlich homological"' (A. Frohlich was a well known worker in algebraic number theory):
Non‐Abelian Homological Algebra I. Derived Functors and Satellites,
A Fröhlich - Proceedings of the London Mathematical Society, 1961 - Wiley Online Library
The purpose of this forthcoming series of papers is the development of a homology (or
cohomology) theory based on structures with non-commutative addition. The module over a
ring, which was hitherto the basic concept for homological algebra, will here be replaced by …
A search on "Lue homological" also gives relevant articles. One of the points that comes out from his 1971, 1981 articles, derived from Frohlich's work, is that his form of the construction of the cohomology is by classes of representative algebraic objects: for example, in the case of groups these objects are commonly called crossed $n$-fold extensions.
The more usual approach is in terms of classes of cocycles. To my mind, a difficulty here is: what do you do with a cocycle when you have got it? By contrast, all sorts of things can be done with or asked about algebraic objects.
This is relevant to the algebraic objects used in the book Nonabelian Algebraic Topology, where, following J.H.C. Whitehead, his crossed modules, and what are now called crossed complexes, and other structures, are used to model some homotopy types. One aspect of the approach
is that the complications of, say, a 3-cocycle definition, are, following work of J. Huebschmann, put into the differentials of a "standard free crossed resolution", so that the 3-cocycle becomes a morphism of crossed complexes; there are many homotopical methods for constructing such morphisms.
To give further background, note that in a letter dated 02/05/1983 Alexander Grothendieck wrote:
Don’t be surprised by my supposed efficiency in digging out the right
kind of notions – I have just been following, rather let myself be pulled
ahead, by that very strong thread (roughly: understand non commutative
cohomology of topoi!) which I kept trying to sell for about ten or twenty
years now, without anyone ready to ``buy” it, namely to do the work. So
finally I got mad and decided to work out at least an outline by myself.
These were his ideas in "Pursuing Stacks".
So I think there is still a lot to take in, assess, and evaluate!
answered Dec 8 '18 at 17:09
Ronnie BrownRonnie Brown
11.9k12938
edited Dec 9 '18 at 10:04
answered Dec 8 '18 at 17:09
Ronnie BrownRonnie Brown
11.9k12938
answered Dec 8 '18 at 17:09
Ronnie BrownRonnie Brown
11.9k12938
answered Dec 8 '18 at 17:09
Ronnie BrownRonnie Brown
11.9k12938
11.9k12938
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%2f3028000%2fhomological-algebra-using-nonabelian-groups%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
$begingroup$
Is the image of a group homomorphism between two non-abelian groups always a normal subgroup? I'm pretty sure that the fact that nonabelian groups aren't modules is the reason why they are not of interest to mathematicians. One thing that I'm thinking of is the long exact sequence of fibrations. I don't know anything about this but from the wiki page, there doesn't seem to be any requirement that any of the homotopy groups are necessarily abelian.
$endgroup$
– Yunus Syed
Dec 6 '18 at 3:52
$begingroup$
@YunusSyed, $operatorname{im}(f : G to H)$ is not necessarily a normal subgroup of $H$, which is why I define $operatorname{coker} f$ to be $H$ quotiented over the normal subgroup generated by $operatorname{im} f$. To define homology or cohomology, all one seems to need is a notion of kernel and cokernel (like in an Abelian category). But is this useful? Thank you for the connection to homotopy groups though, that's an intriguing possibility that I had not considered.
$endgroup$
– Herng Yi
Dec 6 '18 at 4:03
$begingroup$
@Hemg Yi: If $f:Gto H$ is a crossed module, that image is a normal subgroup in $H$.
$endgroup$
– Tim Porter
Dec 9 '18 at 10:24