This is, I'm sure, an incredibly naive question, but: is there a simple explanation for why one should be interested in 1-cocycles?

Let me explain a bit. Given an action of a group $G$ on another group $A$ (the group structure of $A$ is respected by the action, in the sense that $\tau(ab)=(\tau a)(\tau b)$ for $a, b\in A$ and $\tau \in G$), a (1-)cocycle is a map $a: G\rightarrow A$ satisfying $a(\sigma\tau)=a(\sigma)\sigma(a(\tau))$. Now these objects are closely connected with group cohomology. My question is the following. Suppose that I didn't know the language of cohomology or homological algebra (which isn't much of a supposition). Is there a simple explanation for why these objects are interesting?

Note: I don't mean to seem dismissive of the larger edifice of group cohomology; even knowing nothing about it, it's clear to me that it is both interesting and incredibly powerful. But the cocycle condition is so simply stated, that it seems there should be a clear picture for what these objects are doing without delving into the more abstract machinery of group cohomology.

  • 5
    $\begingroup$ It's a great question. Not sure why you felt you needed to apologise so much for asking it. $\endgroup$ Mar 11, 2015 at 17:02

2 Answers 2


I believe there's a very simple motivation for the cocycle condition:

Suppose $f:X \to R$ is a function on a space $X$ with values in a ring $R$ (if you wish, $X$ is a manifold, and $R$ is $\mathbb{R}$ or $\mathbb{C}$, and the function $f$ is differentiable or even holomorphic). Suppose that $G$ is a group acting on $X$ (say on the left). Then for the function to be $G$-invariant is to say that $f(gx)=f(x)$ for all $x \in X$ and $g \in G$.

Often, this condition is too restrictive, i.e. there aren't enough functions satisfying this condition (as well as some other niceness conditions like holomorphic or whatever). Instead, you might want to condition the class of functions satisfying the following relation $$f(gx)=j(g,x) f(x),$$ where $j:G \times X \to R$ is a function. I'll give motivation for this type of relation below, but first I'd like to mention the consequences of such a relation.

Suppose there exists a nonzero $f$ satisfying this relation. Then for all $x \in X$ and $g_1,g_2 \in G$, we have $$f(g_1 g_2 x) = j(g_1 g_2, x) f(x).$$ On the other hand, we have

\begin{eqnarray*} f(g_1 g_2 x) &=& j(g_1,g_2 x) f(g_2 x) &=& j(g_1,g_2 x ) j(g_2,x) f(x) \end{eqnarray*}

Thus we find that $j(g_1 g_2, x) = j(g_1,g_2 x) j(g_2, x)$. If we let $\mathcal{O}^\times$ denote the group of nonzero functions on $X$ under multiplication (possibly only those satisfying some condition, like continuity or differentiability or holomorphicity), then we can think of $j$ as a map $G \to \mathcal{O}^\times$. We can think of $G$ as acting on $\mathcal{O}^\times$ on the right by precomposition, and then this condition on $j$ is precisely the cocycle condition for a map $G \to \mathcal{O}^\times$. In particular, it determines an element of $H^1(G,\mathcal{O}^\times)$.

So whence the condition $f(gx)=j(g,x) f(x)$? I'll start with some classical though possibly less conceptual motivation from special functions, then give the geometric interpretation in terms of line bundles.

Let $L$ be a lattice in the complex plane, i.e. the $\mathbb{Z}$-span of a $\mathbb{R}$-basis of $\mathbb{C}$. A theta function is a meromorphic function such that $\theta(z+\omega)=j(\omega,z)\theta(z)$ for all $z \in \mathbb{C}$, $\omega \in L$. More generally, if $X$ is a contractible Riemann surface, and $G$ is a group which acts on $X$ under sufficiently nice conditions, consider meromorphic functions $f$ on $X$ such that $f(gz)=j(g,z)f(z)$ for $z \in X$, $g \in G$, where $j: G \times X \to \mathbb{C}$ is holomorphic for fixed $g$. In the case of theta functions, $G$ is $L$, and $X$ is $\mathbb{C}$.

Another basic example is a modular form such as $G_{2k}(z)$, which satisfies $G_{2k}(g z) = (cz+d)^{2k} G_{2k}(z)$, where $g= \left(\begin{array}{cc} a & b \\ c & d \end{array}\right) \in G = SL_2(\mathbb{Z})$ acts as a fractional linear transformation. It follows automatically that something as simple as $(cz+d)^{2k}$ is a cocycle in group cohomology, since $G_{2k}$ is, for example, nonzero. In this case $X = \mathcal{H}$, the complex upper-half plane.

Now for some geometric motivation. I'll stick to the case of elliptic curves, though I might rewrite this and make it more general. We define an elliptic curve to be $E=\mathbb{C}/L$ for a two-dimensional lattice $L$. Note that the first homology group of this elliptic curve is isomorphic to $L$ precisely because it is a quotient of the universal cover $\mathbb{C}$ by $L$. We will see that a theta function is a section of a line bundle on an elliptic curve. Since any line bundle can be lifted to $\mathbb{C}$, the universal cover, and any line bundle over a contractible space is trivial, the line bundle is a quotient of the trivial line bundle over $\mathbb{C}$. We can define a function $j(\omega,z):L \times \mathbb{C} \to \mathbb{C} \setminus \{0\}$. Then we identify $(z,w) \in \mathbb{C}^2$ (i.e. the line bundle over $\mathbb{C}$) with $(z+\omega,j(\omega,z)w)$. For this equivalence relation to give a well-defined bundle over $\mathbb{C}/L$, we need the following: Suppose $\omega_1,\omega_2 \in L$. Then $(z,w)$ is identified with $(z+\omega_1+\omega_2,j(\omega_1+\omega_2,z)w$. But $(z,w)$ is identified with $(z+\omega_1,j(\omega_1,z)w)$, which is identified with $(z+\omega_1+\omega_2,j(\omega_2,z+\omega_1)j(\omega_1,z)w)$. In other words, this forces $j(\omega_1+\omega_2,z) = j(\omega_2,z+\omega_1)j(\omega_1,z)$. This means that, if we view $j$ as a function from $L$ to the set of non-vanishing holomorphic functions $\mathbb{C} \to \mathbb{C}$, with (right) L-action on this set defined by $(\omega f)(z) \mapsto f(z+\omega)$, then $j$ is in fact a $1$-cocyle in the language of group cohomology. Thus $H^1(L,\mathcal{O}(\mathbb{C}))$, where $\mathcal{O}(\mathbb{C})$ denotes the (additive) $L$-module of holomorphic functions on $\mathbb{C}$, classifies line bundles over $\mathbb{C}/L$. What's more is that this set is also classified by the sheaf cohomology $H^1(E,\mathcal{O}(E)^{\times})$ (where $\mathcal{O}(E)$ is the sheaf of holomorphic functions on $E$, and the $\times$ indicates the group of units of the ring of holomorphic functions). That is, we can compute the sheaf cohomology of a space by considering the group cohomology of the action of the homology group on the universal cover! In addition, the $0$th group cohomology (this time of the meromorphic functions, not just the holomorphic ones) is the invariant elements under $L$, i.e. the elliptic functions, and similarly the $0$th sheaf cohomology is the global sections, again the elliptic functions.

Once one constructs a line bundle on $E$ as a quotient of a line bundle on $\mathbb{C}$ via the cocycle $j$, a section of that line bundle corresponds to a function $f$ on the complex plane satisfying the cocycle condition. Similarly, modular forms are sections of line bundles on modular curves.

  • 6
    $\begingroup$ A very nice clarification. It is maybe not unnatural to wonder, by the way, if all cocycles of the action of the group $SL(2, \mathbb{Z})$ on the $\mathcal{M}(\mathcal{H})$ are of the form $(cz + d)^k$ for some $k$? $\endgroup$ Feb 5, 2014 at 9:29
  • $\begingroup$ Sorry for the late reply - this is lovely! $\endgroup$ Mar 28, 2014 at 21:19
  • $\begingroup$ @streetcar277: Yes. This is because, as David explains in his answer, the cocycles in question correspond to complex line bundles on $\mathcal{H}$, which are $\operatorname{SL}_2(\mathbb R)$-equivariant. It turns out that each such is uniquely determined by its first Chern class which is an integer and, actually, corresponds precisely to the $k$ in the automorphy factor you write down. $\endgroup$ Oct 31, 2020 at 14:40
  • $\begingroup$ Equivalently, the isotropy group of the fibre is the unitary group $U(1)$, hence a line bundle gives rise to a character on $U(1)$, and those are in bijection with integers. Here is a very nice reference for all of this, which even classifies all equivariant vector bundles: projecteuclid.org/download/pdf_1/euclid.ijm/1258138184 $\endgroup$ Oct 31, 2020 at 14:42
  • $\begingroup$ An afterthought: I should have written "Yes, up to a coboundary". The point is that isom. classes of line bundles correspond to cohomology classes of cocycles, so there is some ambiguity in choosing the cocycle corresponding to a fixed line bundle. In the modular forms literature, one usually imposes a further condition on the cocycles by demanding that their absolute value be of the form $|cz+d|^k$; it is then equal to $(cz+d)^k$ times a function $\operatorname{SL_2}(\mathbb Z) \rightarrow S^1$ with certain properties, called a "multiplier system". See Rankin's book on modular forms, Ch. 3. $\endgroup$ Nov 3, 2020 at 16:36

Here's on simple interpretation of $H^1(G, A)$ that you can check directly.

If $G$ acts on $A$, then we may construct a semidirect product $G \ltimes A$. Then $H^1(G,A)$ corresponds to the conjugacy classes of complements to $A$ in the semidirect product. (In general, $H^1(G,A)$ only a pointed set - a set with a distinguished element (the identically 1 function); it is not a group unless $A$ is abelian.)

Proof: Any complement must be of the form $\{(g, a(g)) : g \in G\}$ for some $a : G \to A$. Check that the condition for such a function $a$ to be a complement is precisely the cocycle condition, and that if you change a cocycle by a coboundary, that corresponds to conjugation. The "special" cocycle that is identically 1 corresponds to the original complement $G$.

  • 1
    $\begingroup$ Sorry, what do you mean by "complements?" $\endgroup$ Feb 24, 2013 at 2:52
  • 1
    $\begingroup$ en.wikipedia.org/wiki/Complement_%28group_theory%29 $\endgroup$
    – Ted
    Feb 24, 2013 at 3:16
  • 2
    $\begingroup$ As a minor side: the coboundaries seem not under considerations by OP, who wants but a motivation of cocycles, not including coboundaries, while your interpretation even clarifies the coboundaries. In any case, thanks for the answer. $\endgroup$
    – awllower
    Apr 3, 2013 at 6:17
  • $\begingroup$ @Ted, How is $H^1(G,G)$ defined? Can you kindly explain. If $A$ is not abelian, then i do not know how this is defined. Thanks. $\endgroup$
    – user114539
    Feb 26, 2016 at 10:10
  • $\begingroup$ @user114539 Here's an online source: groupprops.subwiki.org/wiki/… It's basically the same definition as in the abelian case, except that you can't prove that the cohomology set is a group anymore. You can find a lot more details and discussion in Serre's book "Galois cohomology", Chapter 5. $\endgroup$
    – Ted
    Mar 2, 2016 at 7:19

You must log in to answer this question.

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