Cohomology ring with non-commutative coefficient ringIs the equivariant cohomology an equivariant cohomology?Finite generation of equivariant cohomology ringsEquivariant and basic cohomologyunits in non-commutative ring spectraWeights on equivariant cohomology?Krull dimension in equivariant cohomologyCohomology with coefficient in a Lie algebraCohomology of fiber bundles with non constant coefficientsIs there a geometric interpretation of the cohomology of an automorphism group acting on a universal deformation ring?Which rings are cohomology rings?
Cohomology ring with non-commutative coefficient ring
Is the equivariant cohomology an equivariant cohomology?Finite generation of equivariant cohomology ringsEquivariant and basic cohomologyunits in non-commutative ring spectraWeights on equivariant cohomology?Krull dimension in equivariant cohomologyCohomology with coefficient in a Lie algebraCohomology of fiber bundles with non constant coefficientsIs 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
asked Mar 23 at 3:25
KiuKiu
403212
$endgroup$
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
asked Mar 23 at 3:25
KiuKiu
403212
$endgroup$
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
asked Mar 23 at 3:25
KiuKiu
403212
$endgroup$
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
asked Mar 23 at 3:25
KiuKiu
403212
asked Mar 23 at 3:25
KiuKiu
403212
edited Apr 5 at 20:02
Kiu
asked Mar 23 at 3:25
KiuKiu
403212
asked Mar 23 at 3:25
KiuKiu
403212
asked Mar 23 at 3:25
KiuKiu
403212
403212
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
77.9k27320605
$endgroup$
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
77.9k27320605
$endgroup$
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
77.9k27320605
$endgroup$
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
77.9k27320605
$endgroup$
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
77.9k27320605
answered Mar 23 at 7:14
Qiaochu YuanQiaochu Yuan
77.9k27320605
answered Mar 23 at 7:14
Qiaochu YuanQiaochu Yuan
77.9k27320605
answered Mar 23 at 7:14
Qiaochu YuanQiaochu Yuan
77.9k27320605
77.9k27320605
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