First let me review the definition of first non-commutative cohomology. Let $G$ be a group and $A$ a left $G$-group, i.e. for any $\sigma, \tau\in G$ and $a, b\in A$, one has $\sigma(\tau(a))=(\sigma\tau)(a), \sigma(ab)=\sigma(a)\sigma(b)$.

A 1-cocycle from $G$ to $A$ is a function $f: G\to A$ such that $f(\sigma\tau)=f(\sigma)\sigma(f(\tau))$ for any $\sigma, \tau\in G$. Two 1-cocycles $f, g$ are equivalent if there exists $c\in A$ such that $f(\sigma)=c^{-1}g(\sigma)\sigma(c)$ for all $\sigma\in G$, this is easily verified to be an equivalence relation. Then the first non-commutative cohomology $H^1(A,G)$ is defined to be the set of 1-cocycle modulo equivalence relation. This is a pointed set with base point given by the trivial cocycle.

My question is: both relations $f(\sigma\tau)=f(\sigma)\sigma(f(\tau))$ and $f(\sigma)=c^{-1}g(\sigma)\sigma(c)$ seem to be unnatural. Does there exists a good motivation for these relations?

  • $\begingroup$ Well... they are the same as for commutative cohomology! $\endgroup$ Jan 17, 2013 at 21:45
  • $\begingroup$ @MarianoSuárez-Alvarez: I guess one can motivate this by working out some examples. There are many in descent theory, e.g. Brauer-Severi varieties. But I would really like to see if there is a motivation come from 'first principles', whatever that means :) $\endgroup$
    – minimax
    Jan 17, 2013 at 22:05
  • $\begingroup$ I doubt there is a nice motivation for non-abelian cohomology (you can take the high road and follow Giraud aand so on...) except for the fact that it is what works. The cocycle condition is precisely what you find when you try to describe many objects (forms over subfields, fiber bundles, &c) $\endgroup$ Jan 17, 2013 at 22:09

2 Answers 2


Here is one application that might provide some motivation. Given a finite group $G,$ one might consider a representations of $G.$ These are finite dimensional $\mathbb{F}$-vector spaces $V$ on which $G$ acts as linear transformations. But what is a linear transformation?

You may be thinking it's a matrix. But you'd be wrong.

A technical point that you may have glossed over when first learning linear algebra is that matrices are representations of linear transformations with respect to a chosen basis. Said in another way, given a basis $\beta$ of $V$ one obtains a homomorphism from $[\cdot]_{\beta}: End(V) \rightarrow M_n(\mathbb{F})$ from the ring of linear transformations of $V$ to $n$ by $n$ matrices over $F.$ And given a different basis $\beta'$ one may obtain a different homomorphism $[\cdot]_{\beta'}.$ Luckily, these homomorphisms are related by conjugation by the change of basis matrix $M_{\beta, \beta'}.$ Now one might like to think of a representation of $G$ as a homomorphism from $G$ to $GL_n(\mathbb{F}),$ but in order to associate such a map to a representation one must first choose a basis and in general there is no best choice. To avoid such a choice one instead associates to the representation the set of all homomorphisms from $G$ to $GL_n(\mathbb{F})$ obtained upon choosing a basis for $V.$ This set of homomorphisms is exactly a $GL_n(\mathbb{F})$ conjugacy orbit in $Hom(G,GL_n(\mathbb{F}))$ i.e. an element of $H^1(G,GL_n(\mathbb{F}))$ where $G$ acts trivially on $Gl_n(\mathbb{F}).$ On the other hand, to any element of $c \in H^1(G,GL_n(\mathbb{F}))$ we may associate the representation given by any lift of $c$ to $C^1(G,GL_n(\mathbb{F})).$ One checks that these maps are inverses and thus we see there is a bijection

$$H^1(G,GL_n(\mathbb{F})) \cong \{\text{n-dimensional Linear Representations of G over } \mathbb{F}\}.$$

One can broaden the category of representations of $G$ by considering semilinear representations. These are finite dimensional vector spaces $V$ together with an action of $G$ on $V$ and $\mathbb{F},$ such that

  1. $g(cv)=g(c)g(v)$ for all $c \in \mathbb{F}$ and $v \in V$
  2. $g(v_1 + v_2)=g(v_1) + g(v_2)$ for all $v_1,v_2 \in V.$
  3. $G$ acts on $\mathbb{F}$ as field automorphisms

One can then tell a similar story as in the case of linear representations. Given a semilinear representation $V$ of $G$ one obtains for every basis $\beta$ of $V$ a map $f_{\beta}:G \rightarrow GL_n(\mathbb{F})$ which sends $g$ to $[g]_{\beta}.$ One checks that such a map satisfies $f_{\beta}(g_1g_2) = f_{\beta}(g_1)g_1(f_\beta(g_2))$ and for a second basis $\beta'$ of $V$ the two maps $f_{\beta}$ and $f_{\beta'}$ are related by $M_{\beta,\beta'}^{-1}f_{\beta}(g)g(M_{\beta,\beta'}) = f_{\beta'}.$ Associating to $V$ the set of all such maps we obtain an element of $H^1(G,GL_n(\mathbb{F}))$ where the action of $G$ on $GL_n(\mathbb{F})$ is induced from the action of $G$ on $\mathbb{F}.$ Calling the action of $G$ on $\mathbb{F}$ $\Sigma.$ It follows by an identical argument as above

$$H^1(G,GL_n(\mathbb{F})) \cong \{\text{n-dimensional Semi-Linear Representations of G over } \mathbb{F} \text{ where G acts on } \mathbb{F} \text{ by } \Sigma\}.$$


It is a well-known fact in algebraic topology that for an abelian group $A$ (or more generally, any sheaf of abelian groups $A$) and any topological space $X$, the sheaf cohomology group $H^1 (X, A)$ is in bijection with the set of $A$-torsors (a.k.a. principal $A$-bundles) on $X$. Roughly speaking, a local trivialisation for an $A$-torsor gives a Čech $1$-cocycle for $A$, and the cohomology class of this cochain depends only on the isomorphism class of the $A$-torsor and not the choice of trivialisation. As I understand it, the definition of $H^1$ for non-abelian $A$ is designed so that this theorem remains true.

So how do we bring this idea to group cohomology? Well, for any discrete group $G$ and any $G$-module $A$, the group cohomology $H^* (G, A)$ is naturally isomorphic to topos cohomology $H^* (\mathbf{B} G, A)$, where $\mathbf{B} G$ is the category of all left $G$-sets. The definition of $A$-torsor makes sense in any topos and for any internal group $A$, and in the case of $\mathbf{B} G$ it comes down to this: an $A$-torsor is an inhabited left $G$-set $X$ equipped with a $G$-equivariant right $A$-action such that the map $X \times A \to X \times X$ sending $(x, a)$ to $(x, x \odot a)$ is a bijection.

In particular, for any choice of $x$, the map $a \to x \odot a$ is a bijection of sets $A \to X$. But this is not a $G$-equivariant bijection: after all, $g \cdot (x, a) = (g \cdot x, g \cdot a)$, and that gets mapped to $(g \cdot x, (g \cdot x) \odot (g \cdot a))$, which by $G$-equivariance is equal to $g \cdot (x, x \odot a) = (g \cdot x, g \cdot (x \odot a))$; so $$(g \cdot x) \odot (g \cdot a) = g \cdot (x \odot a)$$ but it is not necessarily true that $x \odot (g \cdot a) \stackrel{?}{=} g \cdot (x \odot a)$. Regardless, we have a set-theoretic bijection, so for each $g$ in $G$ there exists a unique element $f(g)$ of $A$ such that $$g \cdot x = x \odot f(g)$$ and given another $h$ in $G$, we have \begin{align} x \odot f (h g) & = h g \cdot x \\ & = h \cdot (g \cdot x) \\ & = h \cdot (x \odot f(g)) \\ & = (h \cdot x) \odot (h \cdot f(g)) \\ & = (x \odot f(h)) \odot (h \cdot f(g)) \\ & = x \odot (f(h) (h \cdot f(g))) \end{align} which by uniqueness implies $$f (h g) = f(h) (h \cdot f(g))$$ which lo and behold is the condition defining a $1$-cocycle for $A$. (You could do all this for a left $A$-action on $X$, but that gives a different equation.)

What if we choose a different element of $X$, say $x'$? Again, whatever $x'$ is, there is a unique element $c$ of $A$ such that $$x' = x \odot c$$ and so if $f'$ is the $1$-cocycle associated with $x'$, then $$g \cdot x' = x' \odot f'(g) = x \odot (c f'(g))$$ and $$g \cdot x' = g \cdot (x \odot c) = (g \cdot x) \odot (g \cdot c) = x \odot (f(g) (g \cdot c))$$ and therefore $$f'(g) = c^{-1} f(g) (g \cdot c)$$ which is precisely the definition of equivalence of $1$-cochains.

OK, you say, so every $A$-torsor gives rise to a cohomology class of $1$-cocycles, and obviously this construction depends only on the isomorphism class of the $A$-torsor. What about the other way around? Let $f : G \to A$ be a $1$-cocycle and let $X = A$ with the following $G$-action: $$g \cdot x = x f(g)$$ This is a $G$-action, because $f(1_G) = 1_A$ and $$h g \cdot x = x f(h g) = x f(h) (h \cdot f(g)) = (h \cdot x) (h \cdot f(g)) = h \cdot (x f(g)) = h \cdot (g \cdot x)$$ and obviously $A$ acts on itself $G$-equivariantly because $A$ is a $G$-group. Observe also that $X$ has a canonical element, namely $1_A$, and it is easy to see that the $1$-cocycle that it defines is exactly $f$. On the other hand, if $f$ is the $1$-cocycle corresponding to an $A$-torsor $\tilde{X}$ and an element $\tilde{x}$, then there is a unique $(G, A)$-equivariant isomorphism $X \to \tilde{X}$ sending $1_A$ to $\tilde{x}$. Thus the two constructions are mutually inverse up to isomorphism, and $H^1(G, A)$ really does classify $A$-torsors up to isomorphism.

  • $\begingroup$ I really like this answer, very cleanly motivated. Thanks! $\endgroup$
    – minimax
    Jan 20, 2013 at 3:14
  • 1
    $\begingroup$ I don't see how this is motivation :-) It's like saying that the proof that dimension of a vector space is well-defined is motivation for the definition of dimension! this is just «the fact that it is what works» spelled out in one particular situation where it works. $\endgroup$ Feb 4, 2013 at 18:12
  • $\begingroup$ I quite like the idea that $\check{H}^1$ classifies torsors, so it's motivation enough for me. The fact that it fits into a long exact sequence then becomes a theorem, rather than a definition as it is in the derived functor framework... $\endgroup$
    – Zhen Lin
    Feb 4, 2013 at 18:56
  • $\begingroup$ Sure, that is nice. But that is not really a motivation, in that is not a priori in any sense. You set out to classify torsors and, as you do it, you find the cocycle equations. Another example, maybe more obvious: that the Poincaré-Birkhoff-Witt theorem is a direct consequence of the Jacobi identity for Lie algebras does not turn it into a motivation for considering algerbras which satisfy the identity. $\endgroup$ Feb 4, 2013 at 22:47
  • $\begingroup$ I really don't follow what you're saying. The question, as I read it, asks why $H^1 (G, -)$ is defined as it is, not why we should consider $H^1 (G, -)$ at all. I admit I don't have any good answer for that question. $\endgroup$
    – Zhen Lin
    Feb 4, 2013 at 22:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .