In which the author, after a long session sweating blood with his
advisor, manages to calculate the A∞-structures on the
cohomology algebras [tex]H^*(C_4,\mathbb F_2)[/tex] and
[tex]H^*(C_2\times C_2,\mathbb F_2)[/tex].
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:
[tex]\Lambda = \mathbb F_2(C_2\times C_2) = \mathbb
F_2[a,b]/(a^2,b^2)[/tex] our group algebra. We can resolve
[tex]\mathbb F_2[/tex] 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
[tex]
\begin{diagram}
P : & \dots & \rTo & \Lambda^3 & \rTo & \Lambda^2 & \rTo &
\Lambda^1 & \rTo & \mathbb F_2 & \rTo & 0
\end{diagram}
[/tex]
where each differential is given by a matrix D with Di,i=a
and Di+1,i=b.
Recall that [tex]H^*(G,k)=H^*(\Hom_{kG}(P,P))[/tex]. Thus
[tex]\Hom_\Lambda(P,P)[/tex] 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 [tex]H^*A \to A[/tex] that
lifts the identity in homology. This we can use to figure out the
A∞-structure for our cohomology ring: we know that the
[tex]\Hom_\Lambda(P,P)[/tex] 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 [tex]P_{.}\to P_{.}[/tex]. This gives
us a quasi-isomorphism of dg-algebras [tex]H^*A\to A[/tex], 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
[tex]H^*(C_2\times C_2,\mathbb F_2)[/tex] looks like: it's the
algebra [tex]\mathbb F_2[x,y][/tex]. There are two 1-coclasses, both
represented by a morphism [tex]\Lambda^2\to\mathbb F_2[/tex], 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 [tex]P_{.}\to P_{.}[/tex] that shave off the first and last
summand of [tex]\Lambda^i[/tex] respectively in each degree.
So for our quasi-isomorphism [tex]f_\dot[/tex], we now have
[tex]f_1[/tex] sending each [tex]x^iy^j[/tex] to the chain map
shaving off the first i and last j components. We also know
[tex]m_1,m_2,\mu_1,\mu_2[/tex], namely 0, multiplication,
differential and composition respectively. The first axiom we'll
investigate for A∞-maps states that
[tex]
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[/tex]
But now m1=0 and so this reduces to
[tex]f_1(m_2)+\mu_1(f_2)+\mu_2(f_1\otimes f_1)=0[/tex]
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
[tex]\Lambda=\mathbb F_2C_4=\mathbb F_2[x]/(x^4)[/tex]
and the cohomology ring
[tex]\Gamma=\mathbb F_2[\xi,\eta]/(\xi^2)[/tex]
quasi-isomorphic to the endomorphism dg-algebra of a neat resolution
[tex]
P:&\dots&\Lambda&\rTo^{x}&\Lambda&\rTo^{x^3}&\Lambda&\rTo^{x}&\mathbb
F_2&\rTo&0[/tex]
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
[tex]\Hom_\Lambda(P,P)[/tex], 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
[tex]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}[/tex]
and
[tex]\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}[/tex]
and
[tex]\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}[/tex]
Working through the coherence axioms, the first we encounter is the
one defining [tex]f_2[/tex]. This is the one investigating the
relationship between
[tex]m_2(f_1\otimes f_1)[/tex] and [tex]f_1(m_2)[/tex]. So, we
pick [tex]u,v\in H^*(C_8,\mathbb F_2)[/tex], and investigate the
two expressions [tex]f_1(u)f_1(v)[/tex] and [tex]f_1(uv)[/tex].
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, [tex]\xi^2=0[/tex], so [tex]f_1(\xi^2)=0[/tex].
And [tex]f_1(\xi)f_1(\xi)[/tex] we can read off the following
diagram
[tex]\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}[/tex]
and thus we can conclude that [tex]f_1(\xi)f_1(\xi)=\cdot
x^2[/tex] 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
[tex]\Hom_\Lambda(P,P)[/tex] 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 [tex]\xi^2[/tex] under f1, and this particular
homotopy is what we'll want [tex]f_2(\xi,\xi)[/tex] 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
[tex]\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}[/tex]
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
[tex]\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}[/tex]
Thus [tex]f_2(\xi,\xi)=(x\; 0\; x\; 0)[1][/tex], 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
[tex]
\Phi_3=m_2(f_1\otimes f_2+f_2\otimes f_1)+f_2(1\otimes
m_2+m_2\otimes 1)[/tex]
(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
[tex]f_2(x,y)f_1(z)+f_2(x,yz)[/tex] for x,y odd
[tex]f_2(x,yz)+f_2(xy,z)[/tex] for x,z odd
[tex]f_1(x)f_2(y,z)+f_2(xy,z)[/tex] 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
[tex]\Phi_3(\xi,\xi,\xi)=f_1(\xi)f_2(\xi,\xi)+f_2(\xi,\xi)f_1(\xi)[/tex]
which is the sum of the maps given in the following two diagrams
[tex]\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}[/tex]
and
[tex]\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}[/tex]
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
[tex]\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)[/tex]
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
[tex]f_1(x)f_3(y,z,w)+f_3(xy,z,w)[/tex] for x even
[tex]f_3(xy,z,w)+f_3(x,yz,w)[/tex] for y even
[tex]f_3(x,yz,w)+f_3(x,y,zw)[/tex] for z even
[tex]f_3(x,y,z)f_1(w)+f_3(x,y,zw)[/tex] for w even
where the expressions vanish since all summands in each are the same
translations of [tex]f_3(\xi,\xi,\xi)[/tex].
And for [tex]\Phi_4(\xi,\xi,\xi,\xi)[/tex] we get
[tex]f_1(\xi)f_3(\xi,\xi,\xi)+f_2(\xi,\xi)f_2(\xi,\xi)+f_3(\xi,\xi,\xi)f_1(\xi)[/tex]
Again, we calculate each summand by using a corresponding diagram, and
get the three diagrams
[tex]\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}[/tex]
for f1f3 and
[tex]\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}[/tex]
for f3f1 and
[tex]\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}[/tex]
for f2f2
Thus we can read off that f2f2 vanishes, and that
the complete expression for [tex]\Phi_4(\xi,\xi,\xi,\xi)[/tex]
is the one we'd write down as
(1 1 1 1)[2]
which we recognize as [tex]f_1(\eta)[/tex], so we set
[tex]m_4(\xi,\xi,\xi,\xi)=\eta[/tex] and [tex]f_4=f_i=0[/tex].
And this concludes the calculation of the A∞-structure on
[tex]H^*(C_4,\mathbb F_2)[/tex], and also gives a rather clear hint
as to how to do it for [tex]H^*(C_{2^i},\mathbb F_2)[/tex] in
general.