The format of this post will be a little unorganized. I start off with the definition of group cohomology without giving much motivation. Then, I properly define a bar resolution for groups in complete detail. With this setup, I properly motivate group cohomology, work out some extremely interesting examples to see why one should care about this particular free resolution. I’ve also decided to make another post later on a generalization of the bar resolution to the case of monads and a further generalization to algebras. I am currently taking a course on Homotopy Type Theory and I’ve found it blissfully interesting so there might be a few posts in the future on that too. I’ve only recently noticed that I’ve never actually posted anything on either algebraic topology or homotopy theory, which is actually my main interest so I might update that soon enough.
A representation of a group over a field
is just a
module. It is a common philosophy that one can study the structure of a group by studying its representations. In some sense, group cohomology relates to topology but I’ll get to this later on in the post.
Consider as a
where
acts trivially and let
be a representation,i.e
module. Extend this to a
. The group cohomology
with coefficients in
is defined by
. One can immediately see that
. A map
in
satisfies
. But since the action on
is trivial,
. This implies that
. Sending
to any element of
corresponds exactly to the fixed points
. So,
corresponds to the fixed points
.
Bar resolution
So, let’s describe these things in terms of the bar resolution. The calculation of the functor entails finding a projective resolution $latex F_{\bullet} \mapsto X \to 0$ and applying the contravariant
functor. Taking ‘homology’ yields
. The bar resolution actually gives such a resolution, in fact, a free resolution.
Let be the free
on
, that is, the set of symbols
for
and
where all those expressions are simply formal symbols for the generated free module. We have the following free resolution, whose maps will be described soon.
[Insert Image]
where
the augmentation map is given by
on the basis elements.
is given by:
where the maps
is defined on the basis of the free module which extends to the entire group.
,
for
. Keep note of the dimension above. Finally,
.
There is topological motivation for this seemingly bizarre construction.Let us say that the aim is to somehow construct a simplicial object from . For every ordered
tuple, one can insert a
simplex and
can act on these faces by diagonal action:
If any element of the tuple is , the simplex degenerates into lower dimension, for example from the differential maps we saw above,
if
. The differnetial maps match with the this interpretation of degeneracy and face maps of the simiplicial object.You can alternatively also define a normalized version of the bar resolution where the maps
are replaced by tuples where the element are non-identity. It is easy to check that the sequence is a chain complex though this involves some annoying calculations. Also, the sequence is exact as we’ll show in the following lemma.
Lemma 1:
The sequence above is a chain complex that is for
and
. It is also a
-free resolution i.e it is exact.
Proof of Lemma 1:
Step 1:Proving that it is a chain complex
Pick a basis element .
.
I encourage the reader to work out the case for . Now, for
, we can do a little trick to simplify the calculations. Define a new set of
modules
by
for
with the
action given by a diagonal action on the tuples as follows
.
I will omit the details but essentially, what happens is that where we have a bijection of those tuples. So, we just hacked our module
to be a free
module on one lower dimension. Verify that it is indeed a bijection and you get an isomorphism
for
given by
.
Isomorphism follows trivially from the fact that left multiplication by is a bijection on a group. There may be other ways to map the basis elements too.
Define by
.
We have that by the usual calculation one encounters in chain complexes. This will obviously be useful soon. We shall prove that
from which it follows that
iff
since
is an isomorphism.
Now apply to this expression and check that it is equal to
. It is a slightly annoying calculation but much less painful than dealing with the original equation. Now, it remains to deal with the case for
(we’ve already done j=1), which I’ve already left as an exercise to the reader. Anyways, moving onto the next step
Step 2:Complex is chain-homotopic to the identity
We define the chain homotopy for
(taking the -1 case as
)on the basis elements by
, simply adding the element
to the symbol. We simply use the isomorphism
and transfer the entire problem to the
‘s to get
so we get that
. Now I leave it to the reader to verify through simple calculations that
for
. Lower cases can be dealt with separately.
Group cohomology again and a few applciations
Ok, now that we’ve constructed this module resolution. By definition,
Now, it’s great that we’ve got a free resolution but obviously the hope is that one gets some meaningful results with the bar resolution. Let’s observe a few key ideas.
on the level of sets. Basically, assigning the values at the basis elements gives all maps from
to
by extending linearly.
Motivation
Let’s look at some low-dimensional cases to see what exactly is going on.
I describe in the first section of the post.
Just like as described in the first section, even , the collection of maps
since specifying the values on the basis elements determines the entire homomorphism. Under this representation, the boundary maps
can be written more explicitly as:
We can call the elements of
and the elements of
as
, the cycles and boundaries respectively or cocycles and coboundaries, if one wishes to be more precise. It is a chain complex and it is exact though I won’t bother much with the details. The upshot is that know we have the following description of the cohomology group
.
Great! Now we have just one final simple thing to put in place before we look at some applications.
Theorem 1
Suppose is a short exact sequence of
modules, then there is right-derived long exact sequence:
I covered the proof of this theorem for the general case of derived functors in my other post.
I guess, as a simple consequence of the proof of dimension shifting I completed in the other post(or you could prove easily on your own), we have the following theorem which is useful in the context of group cohomology.
Theorem 2(Dimension shifting):
If is a short exact sequence of
modules such that
is trivial, then we have the following natural(wrt to chian complex morphisms) isomorphisms
for
.
The result essentially allows us to compute homology by computing homology of with respect to some other representation in a lower dimension. Now, let us move onto some examples.
The Extension Problem
This is probably most the classical application of group cohomology which I am sure the reader is already quite familiar with so I shall discuss it only little detail.
A classical problem in group theory was to find all possible extensions of a group by a normal subgroup. If we have two groups , it is natural to ask what groups
are extensions of
by
, that is, fit into the exact sequence
An obvious example is a the direct product . One could also add the extra condition that it be a split exact sequence which is equivalent to a semidirect-product
. Two extensions are said to be equivalent if the following diagrams commute:
[Insert Image]
Let’s think of a few extensions of groups. For the groups , we have the trivial extension
and the well known extension
where
acts by inversion on
.This is the dihedral group
of order 6.(Sorry if I have messed up the order of placement of the normal subgroup, I never remember how it works. I think I may have even messed up the semidirect product symbol too. Who cares, I always hated that symbol anyways and abhor the idea of trying to fix it).
We’re going to prove that the elements of the cohomology group are in one-to-one correspondence to extensions up to equivalence. We begin by first defining a section on the level of sets as follows: If there is an extension of the form
[Insert Image]
We would like to define an action on
from the exact sequence. Since the sequence is exact, there exists some
such that $latex \pi(e_{g})=g$. A section is basically a choice of such
‘s. We’ll explain the representative notation soon.
acts on
by conjugation
. Note that
and also that conjugation by this element
since
is abelian. So, the action is independent of the choice of
.
Let the section be . Each
represents an equivalence class
. Both
and
are in the same equivalence class
for
, so by exactness, there is a unique
satisfying
Define a function by
, the value determined above. Note that every element in
can be written uniquely as
by exactness. Let’s see how multiplication works in
:
. Since
,
. Since
is abelian
where
is the action of
defined above.
Associativity of multiplication in (I leave it to the reader to check this) gives the equality:
which is exactly the condition for
to be in
. Finally, let
be another section of the extension and let
be its factor set. We’ll show next that
and
differ by a coboundary, an element of
which implies that every extension has a well-defined cohomology class in
.
Obviously, lie in the same coset
since they both map to
under
. This implies that there is a unique
such that
. Define
by
.
Also,
Equating the two expressions yields,
giving the needed result.
Now, we have to prove the opposite assertion , that, an element of yields a unique(upto equivalence) extension of
by
. I encourage the reader to find the remaining details in a textbook like Dummit and Foote or Aluffi(I am nor sure if he covers it though).
The element in
corresponds to a split extension because then the section
is a genuine homomorphism and so
.
Schur-Zassenhaus Theorem
Consider the exact sequence of groups:
[Insert Image]
If then the sequence is split, equivalently,
If the order of , a normal subgroup of
, is coprime to the order of the index of
in
, then there is a semidirect product such that
for some complement
.
We’ll first deal with the abelian case and that’ll turn out to be useful in the general proof. Towards, that we have the following Lemma.
Lemma 2:
Let be a finite group of order
and let
be an abelian group with
, then
are
modules, that is, they are annihilated on multiplication by
.
Proof of Lemma 2:
Let be the bar resolution of
and let
be the endomorphism of
given by multiplication by
on
and multiplication by
for
for
. We show that
is nullhomotopic. Consider the chain homotopy
given by the formula on the basis elements
I leave it as an exercise to the reader to show that it is nullhomotopic, a trivial calculation.
Corollary 2.1
Let be a finite group of order
and let
be either a vector space over
or a
module then
for
.
Proof of Corollary 2.1
Multiplication by is an isomorphism on
in either case. Applying
to the bar resolution
, consider a class
.
which is
only if
is
. So,
.
This resolves the Schur-Zassenhaus theorem for the abelian case below.
Corollary 2.2(Abelian Case)
If is an abelian group, then
so every extension of
by
is split.
With this, we reduce the proof of the general case to this specific case through some techniques. Before that though, there’s just a random group theory fact we’re going to use.
Lemma 3:
Let be a group and let
by a Sylow-p subgroup. Then,
is a subgroup of
.
Proof of Lemma 3:
Proof of Schur Zassenhaus Theorem
We proceed by induction on , the order of
, the first element in the sequence. Let
be a prime that divides $n=ord(A)$ and consider
, a Sylow-p subgroup of
. We know that
by standard group theory via. the class equation, for example. Consider
, the normalizer of
in
. Using Frattinis’s argument,
.
So, the index divides
and
is a subgroup of
by Lemma 3. So,
. So, we get an extension
by the Second Isomorphism Theorem and using Frattini’s argument. If
, using the induction hypothesis, this sequence splits so there is a subgroup of
which is isomorphic to
. On the other hand, if
, we get that
and we get the extension
by the Third Isomorphism theorem. This splits by the induction hypothesis. Let
be the subgroup of
isomorphic to
. Pulling back, let
where
is the natural quotient map
. From this, we get the extension,
. Using Corollary 2.2(since
is abelian), there is subgroup of
isomorphic to
so there is a subgroup of
isomorphic to
which, as established, is isomorphic to
. Q.E.D