Michi’s blog » A∞-algebras and group cohomology

## A∞-algebras and group cohomology

• November 23rd, 2006
• 3:48 pm

In which the author, after a long session sweating blood with his advisor, manages to calculate the A-structures on the cohomology algebras and .

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:

our group algebra. We can resolve 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

where each differential is given by a matrix D with Di,i=a and Di+1,i=b.

Recall that . Thus 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 that lifts the identity in homology. This we can use to figure out the A-structure for our cohomology ring: we know that the 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 . This gives us a quasi-isomorphism of dg-algebras , 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 looks like: it’s the algebra . There are two 1-coclasses, both represented by a morphism , 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 that shave off the first and last summand of respectively in each degree.

So for our quasi-isomorphism , we now have sending each to the chain map shaving off the first i and last j components. We also know , namely 0, multiplication, differential and composition respectively. The first axiom we’ll investigate for A-maps states that

But now m1=0 and so this reduces to

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)=f1(xi+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

and the cohomology ring

quasi-isomorphic to the endomorphism dg-algebra of a neat resolution

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 , 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

and

and

Working through the coherence axioms, the first we encounter is the one defining . This is the one investigating the relationship between
and . So, we pick , and investigate the two expressions and . 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, , so .
And we can read off the following diagram

and thus we can conclude that 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 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)f1(ξ) = (x2 x2 x2 x2)[2]

Composing the lowest degree component of the map (x2 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 under f1, and this particular homotopy is what we’ll want to be.

So we’ll want a homotopy h, defined by that
dh+hd = (x2 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

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

Thus , 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

(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
for x,y odd
for x,z odd
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

which is the sum of the maps given in the following two diagrams

and

By reading off the diagrams, we note that this sum is the chain map
(x3 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 (x3 x3 x3 x3)[2] is a map h with
dh+hd = (x3 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

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
for x even
for y even
for z even
for w even
where the expressions vanish since all summands in each are the same translations of .

And for we get

Again, we calculate each summand by using a corresponding diagram, and get the three diagrams

for f1f3 and

for f3f1 and

for f2f2

Thus we can read off that f2f2 vanishes, and that the complete expression for is the one we’d write down as
(1 1 1 1)[2]
which we recognize as , so we set and .

And this concludes the calculation of the A-structure on , and also gives a rather clear hint as to how to do it for in general.

### 3 People had this to say...

It’s perhaps not as much “fun” as computing by hand, but you could also appeal to a result of Keller’s, also proved by me with some coauthors, that in the Ext algebra for an algebra A, the higher multiplications on tensor products of Ext^1 reflect relations in the original algebra A. See my paper “A-infinity structure on Ext-algebras”, joint with D.-M. Lu, Q.-S. Wu and J. J. Zhang, available from my web page and also from the ArXiv.

(Caveat: in our paper, we require the algebra A to be graded connected, so it wouldn’t apply directly to a group algebra. However, in the cases you consider here, you can put a grading on kG to make it graded connected, and then appeal to the theorem. Keller’s result might not require this grading.)

• Michi
• April 9th, 2007
• 20:22

John: It doesn’t really show here – but most of my study of A structures is done with both the Lu-Palmieri-Wu-Zhang papers in addition to the Keller papers. I’m interested in trying to go at it as it were from the other direction – not reading off the A-structures from the group algebra, but rather establishing, in the first line as an exercise for my own sake, the connection from the Ext-algebra with A-structure how to nail down the group algebra.

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