# group cohomology

In this post we will talk about cohomology(from homological algebra), group cohomology and its applications.

Why should we study (co)homology? Of course there are historical motivations. Yet a modern point of view is to compensate for the non-exactness of a functor from a category(with good properties, for example, an abelian category) to another. So, one more fundamental question arises: why should we concern ourselves with exactness of a sequence? Perhaps the motivation for this comes from the fact that a splitting sequence of abelian groups implies that this sequence is exact. This is the most common necessary condition for a sequence to split.

Now let’s get down to some concrete things, we suppose that $\mathfrak{C}$ is an abelian category, one prototype example is the category of modules over a commutative ring with unit. We consider the associated ($\mathbb{Z}$-)graded category $\mathfrak{C}'$, in which objects are $(C_i,\delta_i)_{i\in\mathbb{Z}}$ where $\delta_i:C_i\rightarrow C_{i-1}$ is a morphism in $\mathfrak{C}$ such that $\delta_i\circ \delta_{i-1}=0$. So the first motivational question comes: why de we consider this kind of things? One motivation comes from differential manifolds. In the theory of differential manifolds, we have an important operator, the exterior differential operator, $d$, which acts on differential forms. Differential forms give a $\mathbb{Z}$(or $\mathbb{N}$)-graded real vector space, and $d\circ d=0$. This is an important property of $d$. And the category $\mathfrak{C}'$ can be seen as a generalization of this object.

Now a morphism from $(C_i,\delta_i),(D_i,\partial_i)$ is a sequence of maps $f_i:C_i\rightarrow D_i$ such that $\partial_i\circ f_i=f_{i-1}\circ \delta_i$. This definition is very reasonable.

For each such object $C=(C_i,\delta_i)$, we consider $H_n(C)=Ker(\delta_i)/Im(\delta_{i+1})$. This is of course a most fundamental concept in the theory of (co)homology. One straightforward motivation for these $H_n(C)$ is to compensate for the non-exactness of these maps $\delta_i$. If $f:C=(C_i,\delta_i)\rightarrow D=(D_i,\partial_i)$ is a morphism, then it is easy to verify that $f$ induces maps $H_n(f):H_n(C)\rightarrow H_n(D)$.

Now let’s consider another different question: given two maps $f,g:C=(C_i,\delta_i)\rightarrow D=(D_i,\partial_i)$, when will $H(f)=H(g)$? This is a rather interesting question. One sufficient condition comes from homotopy theory, which is, at first sight, completely different from homology theory. We say that two such maps $f,g$ are homotopy equivalent if there is a sequence of maps $h_n:C_n\rightarrow D_{n+1}$ such that $f_n-g_n=\partial_{n+1}\circ h_n+h_{n-1}\circ \delta_n$. It is very easy to show that if $f, g$ are homotopy equivalent, then they induce the same maps $H_n(f)=H_n(g),\forall n$. This is really a fundamental verification in homology theory. Recall the domains of $H_(f),H(g)$: they are $H_n(C)=Ker(\delta_n)/Im(\delta_{n+1})$. Thus for any $x\in H_n(C)$, we have that $f_n(x)-g_n(x)=\partial_{n+1}\circ h_n(x)+h_{n-1}\circ \delta_n(x)$. Note that $\delta_n(x)=0$, besides $\partial_{n+1}\circ h_n(x)\in Im(\partial_{n+1})$, thus $f_n(x)=g_n(x),\forall x\in H_n(C)$. It is also easy to show that homotopy equivalent is an equivalent relation. Note here why should we need the first term $\partial_{n+1}\circ h_n$? In fact these two terms can be seen from homotopy theory. That is what we are going to say in the next post.