Non-abelian cohomologies The Next CEO of Stack OverflowAre the strata of Nakajima quiver varieties simply-connected? Do they have odd cohomology?Equivariant homology of $Omega X$/-space (references needed)?Finite generation of equivariant cohomology ringsIs there a high-concept explanation of the dual Steenrod algebra as the automorphism group scheme of the formal additive group?Equivariant and basic cohomologyIs there any “deep” relation between the localization theorem of equivariant cohomology and the localization theorem of equivariant K-theoryWeights on equivariant cohomology?Krull dimension in equivariant cohomologyIs there a geometric interpretation of the cohomology of an automorphism group acting on a universal deformation ring?Which rings are cohomology rings?
Non-abelian cohomologies
The Next CEO of Stack OverflowAre the strata of Nakajima quiver varieties simply-connected? Do they have odd cohomology?Equivariant homology of $Omega X$/-space (references needed)?Finite generation of equivariant cohomology ringsIs there a high-concept explanation of the dual Steenrod algebra as the automorphism group scheme of the formal additive group?Equivariant and basic cohomologyIs there any “deep” relation between the localization theorem of equivariant cohomology and the localization theorem of equivariant K-theoryWeights on equivariant cohomology?Krull dimension in equivariant cohomologyIs there a geometric interpretation of the cohomology of an automorphism group acting on a universal deformation ring?Which rings are cohomology rings?
$begingroup$
Let A be a non-commutative algebra and let X be some geometric space (such as a topological space or an algebraic variety or scheme). Is there a notion of cohomology ring of X with coefficients in A? What is the correct set up to consider cohomologies with non-commutative coefficients?
If a topological group $G$ acts on a space $X$ one can construct its equivariant cohomology ring $H^*_G(X)$ (say with coefficients in $mathbbR$). Is there a notion of equivariant cohomology for $(X, G)$ with coefficients in a non-commutative algebra $A$?
at.algebraic-topology cohomology
edited Mar 24 at 11:07
Ali Taghavi
13352085
asked Mar 23 at 3:25
KiuKiu
393112
$endgroup$
add a comment |
$begingroup$
Let A be a non-commutative algebra and let X be some geometric space (such as a topological space or an algebraic variety or scheme). Is there a notion of cohomology ring of X with coefficients in A? What is the correct set up to consider cohomologies with non-commutative coefficients?
If a topological group $G$ acts on a space $X$ one can construct its equivariant cohomology ring $H^*_G(X)$ (say with coefficients in $mathbbR$). Is there a notion of equivariant cohomology for $(X, G)$ with coefficients in a non-commutative algebra $A$?
at.algebraic-topology cohomology
edited Mar 24 at 11:07
Ali Taghavi
13352085
asked Mar 23 at 3:25
KiuKiu
393112
$endgroup$
1
$begingroup$
Although it's not what you're asking for, the Galois cohomology sets $operatorname H^1(E/F, mathbb G(E))$ with coefficients in the $E$-rational points of an algebraic group $mathbb G$ make sense even if $mathbb G$ is non-Abelian.
$endgroup$
– LSpice
Mar 23 at 3:48
$begingroup$
Thanks for the comment.
$endgroup$
– Kiu
Mar 23 at 4:10
1
$begingroup$
I think a related question to ask is that are there examples of cohomology theories such that the non-commutativity of the coefficient ring helps to detect something more that the case when the coefficient ring is commutative?
$endgroup$
– user51223
Mar 24 at 2:25
add a comment |
$begingroup$
Let A be a non-commutative algebra and let X be some geometric space (such as a topological space or an algebraic variety or scheme). Is there a notion of cohomology ring of X with coefficients in A? What is the correct set up to consider cohomologies with non-commutative coefficients?
If a topological group $G$ acts on a space $X$ one can construct its equivariant cohomology ring $H^*_G(X)$ (say with coefficients in $mathbbR$). Is there a notion of equivariant cohomology for $(X, G)$ with coefficients in a non-commutative algebra $A$?
at.algebraic-topology cohomology
edited Mar 24 at 11:07
Ali Taghavi
13352085
asked Mar 23 at 3:25
KiuKiu
393112
$endgroup$
Let A be a non-commutative algebra and let X be some geometric space (such as a topological space or an algebraic variety or scheme). Is there a notion of cohomology ring of X with coefficients in A? What is the correct set up to consider cohomologies with non-commutative coefficients?
If a topological group $G$ acts on a space $X$ one can construct its equivariant cohomology ring $H^*_G(X)$ (say with coefficients in $mathbbR$). Is there a notion of equivariant cohomology for $(X, G)$ with coefficients in a non-commutative algebra $A$?
at.algebraic-topology cohomology
at.algebraic-topology cohomology
edited Mar 24 at 11:07
Ali Taghavi
13352085
asked Mar 23 at 3:25
KiuKiu
393112
edited Mar 24 at 11:07
Ali Taghavi
13352085
asked Mar 23 at 3:25
KiuKiu
393112
edited Mar 24 at 11:07
Ali Taghavi
13352085
edited Mar 24 at 11:07
Ali Taghavi
13352085
edited Mar 24 at 11:07
Ali Taghavi
13352085
13352085
asked Mar 23 at 3:25
KiuKiu
393112
asked Mar 23 at 3:25
KiuKiu
393112
asked Mar 23 at 3:25
KiuKiu
393112
393112
1
$begingroup$
Although it's not what you're asking for, the Galois cohomology sets $operatorname H^1(E/F, mathbb G(E))$ with coefficients in the $E$-rational points of an algebraic group $mathbb G$ make sense even if $mathbb G$ is non-Abelian.
$endgroup$
– LSpice
Mar 23 at 3:48
$begingroup$
Thanks for the comment.
$endgroup$
– Kiu
Mar 23 at 4:10
1
$begingroup$
I think a related question to ask is that are there examples of cohomology theories such that the non-commutativity of the coefficient ring helps to detect something more that the case when the coefficient ring is commutative?
$endgroup$
– user51223
Mar 24 at 2:25
add a comment |
1
$begingroup$
Although it's not what you're asking for, the Galois cohomology sets $operatorname H^1(E/F, mathbb G(E))$ with coefficients in the $E$-rational points of an algebraic group $mathbb G$ make sense even if $mathbb G$ is non-Abelian.
$endgroup$
– LSpice
Mar 23 at 3:48
$begingroup$
Thanks for the comment.
$endgroup$
– Kiu
Mar 23 at 4:10
1
$begingroup$
I think a related question to ask is that are there examples of cohomology theories such that the non-commutativity of the coefficient ring helps to detect something more that the case when the coefficient ring is commutative?
$endgroup$
– user51223
Mar 24 at 2:25
1
1
$begingroup$
Although it's not what you're asking for, the Galois cohomology sets $operatorname H^1(E/F, mathbb G(E))$ with coefficients in the $E$-rational points of an algebraic group $mathbb G$ make sense even if $mathbb G$ is non-Abelian.
$endgroup$
– LSpice
Mar 23 at 3:48
$begingroup$
Although it's not what you're asking for, the Galois cohomology sets $operatorname H^1(E/F, mathbb G(E))$ with coefficients in the $E$-rational points of an algebraic group $mathbb G$ make sense even if $mathbb G$ is non-Abelian.
$endgroup$
– LSpice
Mar 23 at 3:48
$begingroup$
Thanks for the comment.
$endgroup$
– Kiu
Mar 23 at 4:10
$begingroup$
Thanks for the comment.
$endgroup$
– Kiu
Mar 23 at 4:10
1
1
$begingroup$
I think a related question to ask is that are there examples of cohomology theories such that the non-commutativity of the coefficient ring helps to detect something more that the case when the coefficient ring is commutative?
$endgroup$
– user51223
Mar 24 at 2:25
$begingroup$
I think a related question to ask is that are there examples of cohomology theories such that the non-commutativity of the coefficient ring helps to detect something more that the case when the coefficient ring is commutative?
$endgroup$
– user51223
Mar 24 at 2:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Is there a notion of cohomology ring of X with coefficients in A?
Yes, and nothing new is needed. The underlying additive group of $A$ is abelian so you take cohomology with coefficients in that abelian group; then the multiplication on $A$ is a bilinear map $A times A to A$ which induces a map
$$H^n(X, A) times H^m(X, A) to H^n+m(X, A)$$
in the usual way. That is, the construction is exactly the same as for cohomology with coefficients in a commutative ring; commutativity of the multiplication is not actually used in the construction. So this isn't actually "nonabelian cohomology" in the sense that term is usually meant.
answered Mar 23 at 7:14
Qiaochu YuanQiaochu Yuan
77.6k27318604
$endgroup$
add a comment |
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$begingroup$
Is there a notion of cohomology ring of X with coefficients in A?
Yes, and nothing new is needed. The underlying additive group of $A$ is abelian so you take cohomology with coefficients in that abelian group; then the multiplication on $A$ is a bilinear map $A times A to A$ which induces a map
$$H^n(X, A) times H^m(X, A) to H^n+m(X, A)$$
in the usual way. That is, the construction is exactly the same as for cohomology with coefficients in a commutative ring; commutativity of the multiplication is not actually used in the construction. So this isn't actually "nonabelian cohomology" in the sense that term is usually meant.
answered Mar 23 at 7:14
Qiaochu YuanQiaochu Yuan
77.6k27318604
$endgroup$
add a comment |
$begingroup$
Is there a notion of cohomology ring of X with coefficients in A?
Yes, and nothing new is needed. The underlying additive group of $A$ is abelian so you take cohomology with coefficients in that abelian group; then the multiplication on $A$ is a bilinear map $A times A to A$ which induces a map
$$H^n(X, A) times H^m(X, A) to H^n+m(X, A)$$
in the usual way. That is, the construction is exactly the same as for cohomology with coefficients in a commutative ring; commutativity of the multiplication is not actually used in the construction. So this isn't actually "nonabelian cohomology" in the sense that term is usually meant.
answered Mar 23 at 7:14
Qiaochu YuanQiaochu Yuan
77.6k27318604
$endgroup$
add a comment |
$begingroup$
Is there a notion of cohomology ring of X with coefficients in A?
Yes, and nothing new is needed. The underlying additive group of $A$ is abelian so you take cohomology with coefficients in that abelian group; then the multiplication on $A$ is a bilinear map $A times A to A$ which induces a map
$$H^n(X, A) times H^m(X, A) to H^n+m(X, A)$$
in the usual way. That is, the construction is exactly the same as for cohomology with coefficients in a commutative ring; commutativity of the multiplication is not actually used in the construction. So this isn't actually "nonabelian cohomology" in the sense that term is usually meant.
answered Mar 23 at 7:14
Qiaochu YuanQiaochu Yuan
77.6k27318604
$endgroup$
Is there a notion of cohomology ring of X with coefficients in A?
Yes, and nothing new is needed. The underlying additive group of $A$ is abelian so you take cohomology with coefficients in that abelian group; then the multiplication on $A$ is a bilinear map $A times A to A$ which induces a map
$$H^n(X, A) times H^m(X, A) to H^n+m(X, A)$$
in the usual way. That is, the construction is exactly the same as for cohomology with coefficients in a commutative ring; commutativity of the multiplication is not actually used in the construction. So this isn't actually "nonabelian cohomology" in the sense that term is usually meant.
answered Mar 23 at 7:14
Qiaochu YuanQiaochu Yuan
77.6k27318604
answered Mar 23 at 7:14
Qiaochu YuanQiaochu Yuan
77.6k27318604
answered Mar 23 at 7:14
Qiaochu YuanQiaochu Yuan
77.6k27318604
answered Mar 23 at 7:14
Qiaochu YuanQiaochu Yuan
77.6k27318604
77.6k27318604
add a comment |
add a comment |
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$begingroup$
Although it's not what you're asking for, the Galois cohomology sets $operatorname H^1(E/F, mathbb G(E))$ with coefficients in the $E$-rational points of an algebraic group $mathbb G$ make sense even if $mathbb G$ is non-Abelian.
$endgroup$
– LSpice
Mar 23 at 3:48
$begingroup$
Thanks for the comment.
$endgroup$
– Kiu
Mar 23 at 4:10
1
$begingroup$
I think a related question to ask is that are there examples of cohomology theories such that the non-commutativity of the coefficient ring helps to detect something more that the case when the coefficient ring is commutative?
$endgroup$
– user51223
Mar 24 at 2:25