Non-abelian cohomologies
$begingroup$
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 $mathbb{R}$). Is there a notion of equivariant cohomology for $(X, G)$ with coefficients in a non-commutative algebra $A$?
at.algebraic-topology cohomology
edited 15 hours ago
Ali Taghavi
8052085
asked yesterday
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 $mathbb{R}$). Is there a notion of equivariant cohomology for $(X, G)$ with coefficients in a non-commutative algebra $A$?
at.algebraic-topology cohomology
edited 15 hours ago
Ali Taghavi
8052085
asked yesterday
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
yesterday
$begingroup$
Thanks for the comment.
$endgroup$
– Kiu
yesterday
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
yesterday
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 $mathbb{R}$). Is there a notion of equivariant cohomology for $(X, G)$ with coefficients in a non-commutative algebra $A$?
at.algebraic-topology cohomology
edited 15 hours ago
Ali Taghavi
8052085
asked yesterday
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 $mathbb{R}$). 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 15 hours ago
Ali Taghavi
8052085
asked yesterday
KiuKiu
393112
edited 15 hours ago
Ali Taghavi
8052085
asked yesterday
KiuKiu
393112
edited 15 hours ago
Ali Taghavi
8052085
edited 15 hours ago
Ali Taghavi
8052085
edited 15 hours ago
Ali Taghavi
8052085
8052085
asked yesterday
KiuKiu
393112
asked yesterday
KiuKiu
393112
asked yesterday
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
yesterday
$begingroup$
Thanks for the comment.
$endgroup$
– Kiu
yesterday
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
yesterday
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
yesterday
$begingroup$
Thanks for the comment.
$endgroup$
– Kiu
yesterday
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
yesterday
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
yesterday
$begingroup$
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
yesterday
$begingroup$
Thanks for the comment.
$endgroup$
– Kiu
yesterday
$begingroup$
Thanks for the comment.
$endgroup$
– Kiu
yesterday
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
yesterday
$begingroup$
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
yesterday
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 yesterday
Qiaochu YuanQiaochu Yuan
77.5k27317603
$endgroup$
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 yesterday
Qiaochu YuanQiaochu Yuan
77.5k27317603
$endgroup$
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 yesterday
Qiaochu YuanQiaochu Yuan
77.5k27317603
$endgroup$
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 yesterday
Qiaochu YuanQiaochu Yuan
77.5k27317603
$endgroup$
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 yesterday
Qiaochu YuanQiaochu Yuan
77.5k27317603
answered yesterday
Qiaochu YuanQiaochu Yuan
77.5k27317603
answered yesterday
Qiaochu YuanQiaochu Yuan
77.5k27317603
answered yesterday
Qiaochu YuanQiaochu Yuan
77.5k27317603
77.5k27317603
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
yesterday
$begingroup$
Thanks for the comment.
$endgroup$
– Kiu
yesterday
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
yesterday