Homological algebra using nonabelian groups
up vote
6
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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 4 hours ago
Herng Yi
1,4391023
add a comment |
up vote
6
down vote
favorite
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 4 hours ago
Herng Yi
1,4391023
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.
– Yunus Syed
4 hours ago
@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.
– Herng Yi
3 hours ago
add a comment |
up vote
6
down vote
favorite
up vote
6
down vote
favorite
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 4 hours ago
Herng Yi
1,4391023
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 4 hours ago
Herng Yi
1,4391023
asked 4 hours ago
Herng Yi
1,4391023
edited 6 mins ago
asked 4 hours ago
Herng Yi
1,4391023
asked 4 hours ago
Herng Yi
1,4391023
asked 4 hours ago
Herng Yi
1,4391023
1,4391023
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.
– Yunus Syed
4 hours ago
@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.
– Herng Yi
3 hours ago
add a comment |
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.
– Yunus Syed
4 hours ago
@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.
– Herng Yi
3 hours ago
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.
– Yunus Syed
4 hours ago
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.
– Yunus Syed
4 hours ago
@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.
– Herng Yi
3 hours ago
@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.
– Herng Yi
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
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 2 hours ago
Qiaochu Yuan
275k32578914
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.
– Mike Miller
2 hours ago
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.
– Kevin Carlson
45 mins ago
@MikeMiller, could you give an example where the first cohomology with coefficients in a nonabelian group is interesting?
– Herng Yi
4 mins ago
add a comment |
up vote
1
down vote
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 48 mins ago
Kevin Carlson
32.1k23270
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
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 2 hours ago
Qiaochu Yuan
275k32578914
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.
– Mike Miller
2 hours ago
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.
– Kevin Carlson
45 mins ago
@MikeMiller, could you give an example where the first cohomology with coefficients in a nonabelian group is interesting?
– Herng Yi
4 mins ago
add a comment |
up vote
3
down vote
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 2 hours ago
Qiaochu Yuan
275k32578914
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.
– Mike Miller
2 hours ago
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.
– Kevin Carlson
45 mins ago
@MikeMiller, could you give an example where the first cohomology with coefficients in a nonabelian group is interesting?
– Herng Yi
4 mins ago
add a comment |
up vote
3
down vote
up vote
3
down vote
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 2 hours ago
Qiaochu Yuan
275k32578914
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 2 hours ago
Qiaochu Yuan
275k32578914
answered 2 hours ago
Qiaochu Yuan
275k32578914
answered 2 hours ago
Qiaochu Yuan
275k32578914
answered 2 hours ago
Qiaochu Yuan
275k32578914
275k32578914
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.
– Mike Miller
2 hours ago
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.
– Kevin Carlson
45 mins ago
@MikeMiller, could you give an example where the first cohomology with coefficients in a nonabelian group is interesting?
– Herng Yi
4 mins ago
add a comment |
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.
– Mike Miller
2 hours ago
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.
– Kevin Carlson
45 mins ago
@MikeMiller, could you give an example where the first cohomology with coefficients in a nonabelian group is interesting?
– Herng Yi
4 mins ago
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.
– Mike Miller
2 hours ago
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.
– Mike Miller
2 hours ago
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.
– Kevin Carlson
45 mins ago
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.
– Kevin Carlson
45 mins ago
@MikeMiller, could you give an example where the first cohomology with coefficients in a nonabelian group is interesting?
– Herng Yi
4 mins ago
@MikeMiller, could you give an example where the first cohomology with coefficients in a nonabelian group is interesting?
– Herng Yi
4 mins ago
add a comment |
up vote
1
down vote
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 48 mins ago
Kevin Carlson
32.1k23270
add a comment |
up vote
1
down vote
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 48 mins ago
Kevin Carlson
32.1k23270
add a comment |
up vote
1
down vote
up vote
1
down vote
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 48 mins ago
Kevin Carlson
32.1k23270
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 48 mins ago
Kevin Carlson
32.1k23270
edited 36 mins ago
answered 48 mins ago
Kevin Carlson
32.1k23270
answered 48 mins ago
Kevin Carlson
32.1k23270
answered 48 mins ago
Kevin Carlson
32.1k23270
32.1k23270
add a comment |
add a comment |
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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.
– Yunus Syed
4 hours ago
@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.
– Herng Yi
3 hours ago