# A∞-algebras and group cohomology

Published: Thu 23 November 2006

In which the author, after a long session sweating blood with his advisor, manages to calculate the A-structures on the cohomology algebras $$H^*(C_4,\mathbb F_2)$$ and $$H^*(C_2\times C_2,\mathbb F_2)$$.

We will find the A-structures on the group cohomology ring by establishing an A-quasi-isomorphism to the endomorphism dg-algebra of a resolution of the base field. We'll write mi for operations on the group cohomology, and μi for operations on the endomorphism dg-algebra. The endomorphism dg-algebra has μ1=d and μ2=composition of maps, and all higher operations vanishing, in all our cases.

## Elementary abelian 2-group

Let's start with the easy case. Following to a certain the notation used in Dag Madsen's PhD thesis appendix (the Canonical Source of the A-structures of cyclic group cohomology algebras), and the recipe given in A-infinity algebras in representation theory, we may start by stating what we know as we start:

$$\Lambda = \mathbb F_2(C_2\times C_2) = \mathbb F_2[a,b]/(a^2,b^2)$$ our group algebra. We can resolve $$\mathbb F_2$$ using a neat canonical resolution, derived from the tensor products of the resolutions of the field with modules over the cyclic 2-group. This gives us the resolution
$$\begin{diagram} P : & \dots & \rTo & \Lambda^3 & \rTo & \Lambda^2 & \rTo & \Lambda^1 & \rTo & \mathbb F_2 & \rTo & 0 \end{diagram}$$
where each differential is given by a matrix D with Di,i=a and Di+1,i=b.

Recall that $$H^*(G,k)=H^*(\Hom_{kG}(P,P))$$. Thus $$\Hom_\Lambda(P,P)$$ is a dg-algebra, whose homology is precisely the group cohomology.

Now, by the minimality theorem (proven by Kadeishvili first, and reproven by a veritable host of mathematicians), there is a quasi-isomorphism of A-algebras $$H^*A \to A$$ that lifts the identity in homology. This we can use to figure out the A-structure for our cohomology ring: we know that the $$\Hom_\Lambda(P,P)$$ is an honest-to-glod dg-algebra, and thus has an A-structure with all higher multiplications (by which I mean 3-ary and higher) vanishing. We can also pick representatives for our cohomology ring elements as representatives of homotopy classes of chain maps $$P_{.}\to P_{.}$$. This gives us a quasi-isomorphism of dg-algebras $$H^*A\to A$$, lifting the identity, and which we can augment to an A-quasi-isomorphism.

Which is what we'll want to do now.

We'll (at this stage) use the fact that we know what $$H^*(C_2\times C_2,\mathbb F_2)$$ looks like: it's the algebra $$\mathbb F_2[x,y]$$. There are two 1-coclasses, both represented by a morphism $$\Lambda^2\to\mathbb F_2$$, namely one composing the projection onto the first factor with the augmentation map, and one composing the projection onto the second factor. We'll name the first of these x, and the second y, and note that they lift to chain maps $$P_{.}\to P_{.}$$ that shave off the first and last summand of $$\Lambda^i$$ respectively in each degree.

So for our quasi-isomorphism $$f_\dot$$, we now have $$f_1$$ sending each $$x^iy^j$$ to the chain map shaving off the first i and last j components. We also know $$m_1,m_2,\mu_1,\mu_2$$, namely 0, multiplication, differential and composition respectively. The first axiom we'll investigate for A-maps states that
$$f_1(m_1\otimes m_1)+f_1(m_2)+f_2(m_1)+\mu_1(f_2)+\mu_2(f_1\otimes f_1)=0$$
But now m1=0 and so this reduces to
$$f_1(m_2)+\mu_1(f_2)+\mu_2(f_1\otimes f_1)=0$$
so f2 is a map such that its differential is equal to the "commutator" of f1 and multiplication.

Now, pick some coclasses u,v. These will map to shaving maps as described above, and their product will map to a shaving map that does just the same as the composition of the individual shaving maps; so if u is xiyj, and v is xkyl, then f1(u) shaves off i components in the front and j components in the back, and f1(v) shaves off k in the front and l in the back. So, the composition of these two maps is the map that drops i+k components from the front, and j+l components from the back. On the other hand f1(uv)=f:sub:1(x:sup:i+kyj+l) is the map that drops i+k components from the front, and j+l components from the back.

So they are the same. And thus we don't need to bother with any homotopies, or any higher order operations or higher order maps. We set f2 to be the zero map, and consider ourself finished and happy. The cohomology is a dg-algebra in its own right, and this is all there is to it in A-terms. And we're done.

This result implies, by the way, via a proposition from Keller, that the elementary abelian 2-groups have Koszul group algebras. An argument using restrictions of non-nilpotent coclasses to cyclic subgroups will tell you that these are the only finite groups that have Koszul group algebras.

With this example.

## Cyclic 4-group

This is the one canonical example known beforehand in group cohomology. It was calculated by Dag Madsen in his PhD-thesis, and cited ever since. I will perform the same calculation, but in a blinding detail you won't find in a thesis or a paper on the subject.

So, for starters, we find ourself with the group algebra
$$\Lambda=\mathbb F_2C_4=\mathbb F_2[x]/(x^4)$$
and the cohomology ring
$$\Gamma=\mathbb F_2[\xi,\eta]/(\xi^2)$$
quasi-isomorphic to the endomorphism dg-algebra of a neat resolution
$$P:&\dots&\Lambda&\rTo^{x}&\Lambda&\rTo^{x^3}&\Lambda&\rTo^{x}&\mathbb F_2&\rTo&0$$
where c in Γ represent the morphism m→cm.
Furthermore, as usual, the μi are all known, and m1=0, and m2 is the multiplication in Γ.

Once we've choosen representatives for the coclasses in $$\Hom_\Lambda(P,P)$$, we can lift this choice to an A-quasi-isomorphism. In this process, we'll find and define the relevant higher multiplications for Γ, thus finding the A-structure for that algebra.

Thus f1 sends
$$1\mapsto\begin{diagram} \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots\\ &&\dEq&&\dEq&&\dEq\\ \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots\\ \end{diagram}$$
and
$$\xi\mapsto\begin{diagram} \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots\\ &\rdTo^{x^2}&&\rdTo^{1}&&\rdTo^{x^2}&&\rdTo^{1}\\ \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots\\ \end{diagram}$$
and
$$\eta\mapsto\begin{diagram} \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots\\ &\rdTo(4,2)^{1}&&\rdTo(4,2)^{1}&&\rdTo(4,2)^{1}&&\\ \cdots&\rTo_x&\Lambda&\rTo_{x^3}&\Lambda&\rTo_x&\Lambda&\rTo_{x^3}&\cdots\\ \end{diagram}$$
Working through the coherence axioms, the first we encounter is the one defining $$f_2$$. This is the one investigating the relationship between
$$m_2(f_1\otimes f_1)$$ and $$f_1(m_2)$$. So, we pick $$u,v\in H^*(C_8,\mathbb F_2)$$, and investigate the two expressions $$f_1(u)f_1(v)$$ and $$f_1(uv)$$. Writing it down in detail tells us that the only point where a difference occurs is if both u and v are odd, and this case we can figure out from the case where u=v=ξ since η only translates chainmaps higher up in degree.
Now, $$\xi^2=0$$, so $$f_1(\xi^2)=0$$.
And $$f_1(\xi)f_1(\xi)$$ we can read off the following diagram
$$\begin{diagram} \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots\\ &\rdTo^{x^2}&&\rdTo^{1}&&\rdTo^{x^2}&&\rdTo^{1}\\ \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots\\ &\rdTo^{x^2}&&\rdTo^{1}&&\rdTo^{x^2}&&\rdTo^{1}\\ \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots\\ \end{diagram}$$
and thus we can conclude that $$f_1(\xi)f_1(\xi)=\cdot x^2$$ in degree 2. We shall be juggling maps a lot, so I will use the notation (x y z w)[i] for the map in $$\Hom_\Lambda(P,P)$$ that drops i degrees, and where the four last positions are multiplication by x, y, z and w respectively. So, with this notation, we have
f1(η)=(1 1 1 1)[2]
f1(ξ)=(1 x2 1 x2)[1]
f1(ξ)f:sub:1(ξ) = (x:sup:2 x2 x2 x2)[2]

Composing the lowest degree component of the map (x:sup:2 x2 x2 x2)[2] with the augmentation map, we see that in cohomology, it corresponds to the 0 element. So it is actually homotopic to the image of $$\xi^2$$ under f1, and this particular homotopy is what we'll want $$f_2(\xi,\xi)$$ to be.

So we'll want a homotopy h, defined by that
dh+hd = (x:sup:2 x2 x2 x2)[2]
so we can immediately conclude that h needs to be of degree 1, since composition with d will add another degree step. So our h will look something like
$$\begin{diagram} \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots\\ &\rdTo^{h}&&\rdTo^{h}&&\rdTo^{h}&&\rdTo^{h}\\ \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots \end{diagram}$$
and we can check what happens as we chase through the diagram. If we start at an odd-indexed position, we'll get the two components
hd(o) = h(x o) = x h(o)
dh(o) = x3 h(o)
for o of odd degree. If we instead start in an even degree, we get
hd(e) = h(x3e) = x3h(e)
dh(e) = x h(e)
where e is an element of even degree. By close inspection in the diagram, we note that the expressions involving x3 all involve h applied on elements of odd degree, and so we'll set those to vanish, and fill in the needed values by letting h(e)=x. Thus we get the chain map
h = (x 0 x 0) [1]
or in diagrammatic form
$$\begin{diagram} \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots\\ &\rdTo^{0}&&\rdTo^{x}&&\rdTo^{0}&&\rdTo^{x}\\ \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots \end{diagram}$$

Thus $$f_2(\xi,\xi)=(x\; 0\; x\; 0)[1]$$, and f2 on odd and odd elements are all translates of this, and all other parameters to f2 give us a zero map. This brings us to a point where we can start investigating m3.

We can extract f1m3 and m1f3 from the 3rd A-morphism axiom, and put the rest into a map of its own. This will end up to be something supposed to be homotopic to the image of m3, and so we can define m3 and the homotopy once we have them.

The rest of the axiom is
$$\Phi_3=m_2(f_1\otimes f_2+f_2\otimes f_1)+f_2(1\otimes m_2+m_2\otimes 1)$$
(where I am using the fact that we're in characteristic 2 extensively)
Applying this on elements x,y,z gives us a possibility to distinguish between cases.

If only one element is of odd degree, then every f2 occurring will have at least one even argument, and so will vanish.

If two elements are of odd degree, we get the expressions
$$f_2(x,y)f_1(z)+f_2(x,yz)$$ for x,y odd
$$f_2(x,yz)+f_2(xy,z)$$ for x,z odd
$$f_1(x)f_2(y,z)+f_2(xy,z)$$ for y,z odd.
Each of these vanish if we take the behaviour of f2 for higher odd coclasses into account: these are just translates of the behaviour defined in f2(ξ,ξ), and so should vanish, since we defined f2 on pairs of odd classes to just be translates of the value on ξ and ξ.

Remains the case with all three elements of odd degree. Again, higher odd elements behave by translating the behaviour of ξ, and so it is enough to study the behaviour on ξ, ξ, ξ.

In this case, we get
$$\Phi_3(\xi,\xi,\xi)=f_1(\xi)f_2(\xi,\xi)+f_2(\xi,\xi)f_1(\xi)$$
which is the sum of the maps given in the following two diagrams
$$\begin{diagram} \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots\\ &\rdTo^{0}&&\rdTo^{x}&&\rdTo^{0}&&\rdTo^{x}\\ \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots\\ &\rdTo^{x^2}&&\rdTo^{1}&&\rdTo^{x^2}&&\rdTo^{1}\\ \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots\\ \end{diagram}$$
and
$$\begin{diagram} \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots\\ &\rdTo^{x^2}&&\rdTo^{1}&&\rdTo^{x^2}&&\rdTo^{1}\\ \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots\\ &\rdTo^{0}&&\rdTo^{x}&&\rdTo^{0}&&\rdTo^{x}\\ \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots\\ \end{diagram}$$
By reading off the diagrams, we note that this sum is the chain map
(x:sup:3 x3 x3 x3)[2]
and we can further note that this corresponds to the cocycle 0 of degree 2 (since the augmentation composed with x3 vanishes), so we put m3=0, to correspond to what this is homotopic to. And f3(ξ,ξ,ξ) needs to be mapped precisely to this homotopy.
So, a homotopy h between 0 and (x:sup:3 x3 x3 x3)[2] is a map h with
dh+hd = (x:sup:3 x3 x3 x3)[2]
From the same considerations as above, we can conclude that h will have the two components
hd(o) = h(x o) = x h(o)
dh(o) = x3 h(o)
for o of odd degree. If we instead start in an even degree, we get
hd(e) = h(x3e) = x3h(e)
dh(e) = x h(e)
where e is an element of even degree.
Now, this time, since we want the x3 to occur, we can pick h to be 0 on even degree components and the identity on odd degree components, giving us
h = (0 1 0 1)[1]
and this being the image of f3(ξ,ξ,ξ).
Going ever onwards, the next axiom we check is the one, that when we leave m1f4 and f1m4 out of the mix, we'll get
$$\Phi_4=m_2(f_1\otimes f_3+f_2\otimes f_2+f_3\otimes f_1) + f_2(1\otimes m_3+m_3\otimes 1) \\ + f_3(1\otimes1\otimes m_2+1\otimes m_2\otimes1+m_2\otimes1\otimes1)$$
and again, we can start looking at cases based on number of odd arguments.

For only one odd argument, all the f2 and f3 will vanish.

For two odd arguments, the same thing will happen.

For three odd arguments, we get the expressions
$$f_1(x)f_3(y,z,w)+f_3(xy,z,w)$$ for x even
$$f_3(xy,z,w)+f_3(x,yz,w)$$ for y even
$$f_3(x,yz,w)+f_3(x,y,zw)$$ for z even
$$f_3(x,y,z)f_1(w)+f_3(x,y,zw)$$ for w even
where the expressions vanish since all summands in each are the same translations of $$f_3(\xi,\xi,\xi)$$.
And for $$\Phi_4(\xi,\xi,\xi,\xi)$$ we get
$$f_1(\xi)f_3(\xi,\xi,\xi)+f_2(\xi,\xi)f_2(\xi,\xi)+f_3(\xi,\xi,\xi)f_1(\xi)$$
Again, we calculate each summand by using a corresponding diagram, and get the three diagrams
$$\begin{diagram} \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots\\ &\rdTo^{1}&&\rdTo^{0}&&\rdTo^{1}&&\rdTo^{0}\\ \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots\\ &\rdTo^{x^2}&&\rdTo^{1}&&\rdTo^{x^2}&&\rdTo^{1}\\ \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots\\ \end{diagram}$$
for f1f3 and
$$\begin{diagram} \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots\\ &\rdTo^{x^2}&&\rdTo^{1}&&\rdTo^{x^2}&&\rdTo^{1}\\ \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots\\ &\rdTo^{1}&&\rdTo^{0}&&\rdTo^{1}&&\rdTo^{0}\\ \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots\\ \end{diagram}$$
for f3f1 and
$$\begin{diagram} \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots\\ &\rdTo^{0}&&\rdTo^{x}&&\rdTo^{0}&&\rdTo^{x}\\ \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots\\ &\rdTo^{0}&&\rdTo^{x}&&\rdTo^{0}&&\rdTo^{x}\\ \cdots&\rTo^x&\Lambda&\rTo^{x^3}&\Lambda&\rTo^x&\Lambda&\rTo^{x^3}&\cdots\\ \end{diagram}$$
for f2f2
Thus we can read off that f2f2 vanishes, and that the complete expression for $$\Phi_4(\xi,\xi,\xi,\xi)$$ is the one we'd write down as
(1 1 1 1)[2]
which we recognize as $$f_1(\eta)$$, so we set $$m_4(\xi,\xi,\xi,\xi)=\eta$$ and $$f_4=f_i=0$$.

And this concludes the calculation of the A-structure on $$H^*(C_4,\mathbb F_2)$$, and also gives a rather clear hint as to how to do it for $$H^*(C_{2^i},\mathbb F_2)$$ in general.