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.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s