# An A∞-structure on the cohomology of D8

Published: Thu 30 November 2006

As a first unknown (kinda, sorta, it still falls under the category of path algebra quotients treated by Keller) A-calculation, I shall find the A-structure of $$H^*(D_8,\mathbb F_2)$$.

To do this, I fix the group algebra
$$\Lambda=\mathbb F_2[a,b]/(a^2,b^2,abab+baba)$$
and the cohomology ring
$$\Gamma=\mathbb F_2[x,y,z]/(xy)$$
with $$x,y\in\Gamma_1$$, $$z\in\Gamma_2$$
Furthermore, we pick a canonical nice resolution P, continuing the one I calculated previously. This has the i:th component Λi+1, and the differentials looking like
$$\begin{pmatrix} a&0&bab&0&0&0\\ 0&b&0&aba&0&0\\ 0&0&a&0&bab&0\\ 0&0&0&b&0&aba\\ 0&0&0&0&a&b \end{pmatrix}$$
for differentials starting in odd degree, and
$$\begin{pmatrix} a&0&bab&0&0\\ 0&b&0&aba&0\\ 0&0&a&0&bab\\ 0&0&0&b&aba \end{pmatrix}$$
for differentials starting in even degree. The first few you can see on the previous calculation, or if you don't want to bother, they are
$$\partial_0=\begin{pmatrix}a&b\end{pmatrix}$$
$$\partial_1=\begin{pmatrix}a&0&bab\\0&b&aba\end{pmatrix}$$
$$\partial_2=\begin{pmatrix}a&0&bab&0\\0&b&0&aba\\0&0&a&b\end{pmatrix}$$

Now, armed with this, we can get cracking. By lifting, we get canonical representating chain maps for x,y,z described, loosly, by the following:

x takes an element in $$\Lambda^{2k}$$, keeps the first, third, et.c. elements and throws out the even ordered elements; so $$x\cdot(a_1,a_2,a_3,a_4,a_5,a_6)=(a_1,0,a_3,0,a_5)$$
For an element in $$\Lambda^{2k+1}$$, the last element gets extra treatment, so
$$x\cdot(a_1,a_2,a_3,a_4,a_5,a_6,a_7)=(a_1,0,a_3,0,a_5,ab\cdot a_7)$$
For the lowest degrees, we also have
$$x_0 = \begin{pmatrix}1&0\end{pmatrix}$$
$$x_1 = \begin{pmatrix}1&0&0\\0&0&ab\end{pmatrix}$$
$$x_2 = \begin{pmatrix}1&0&0&0\\0&0&1&0\end{pmatrix}$$

This describes the image $$f_1(x)$$ of our quasi-isomorphism $$\Gamma\to\Hom_\Lambda(P,P)$$.

For $$f_1(y)$$, we get the interleaved effect: every second element, but now with even indices, and some extra treatments at the end. So an element in $$\Lambda^{2i}$$ will behave like
$$y\cdot(a_1,a_2,a_3,a_4,a_5,a_6)=(0,a_2,0,a_4,a_6)$$
and for an element in $$\Lambda^{2i+1}$$ we get
$$y\cdot(a_1,a_2,a_3,a_4,a_5,a_6,a_7)=(0,a_2,0,a_4,ba\cdot a_7,a_6)$$

The last generator, z, is rather boring. The corresponding chain map only lifts elements to other degrees: shaving off the first two components of whatever it is applied to.

We define $$f_1$$ on products of the generators by simply composing the corresponding chain maps as long as the product is defined. The interesting stuff, from an A point of view occurs when the product vanishes, thus for x,y in the first line. A calculation shows us that xy is the only interesting element of the ideal $$(xy)$$, since $$f_1(x)f_1(x)f_1(y)f_1(y)=0$$, and we define $$f_2(x^2,y)=f_1(x)f_2(x,y)$$ and $$f_2(x,y^2)=f_2(x,y)f_1(y)$$.

Thus, we'd be curious as to what happens with xy, and yx. Both products are zero in the cohomology ring; but the composition of the corresponding chain maps are not zero.

By calculation, we get $$f_1(x)f_1(y)$$ given in the first few step by the matrices
$$\begin{pmatrix}0&0&ba\end{pmatrix}$$
$$\begin{pmatrix}0&0&0&0\\0&0&0&ab\end{pmatrix}$$
and so on, with the lower right entry alternatingly ab and ba.

$$f_1(y)f_1(x)$$ is the same thing, but with ab and ba interchanged.

We want $$f_2(x,y)$$ and $$f_2(y,x)$$ to be homotopies between the 0 chain maps, and these two respectively. By juggling the relevant matrices in Magma for a while, I conclude that $$f_2(x,y)$$ has lower right entry a for all odd degree map components, and $$f_2(y,x)$$ has the entry above that b, same components. All even components vanish.

Thus, for the first nonzero component, we have
$$\begin{pmatrix}0&0&0\\0&0&a\end{pmatrix}$$
for x,y and
$$\begin{pmatrix}0&0&b\\0&0&0\end{pmatrix}$$
for y,x.
Now, if we look at Φ3, most of the possibilities vanish because of the way we defined f2 for the non xy, yx cases. Thus, the only interesting entries remaining are xyx and yxy. These are, respectively,
$$\Phi_3(x,y,x)=f_1(x)f_2(y,x)+f_2(x,y)f_1(x)$$
$$\Phi_3(y,x,y)=f_1(y)f_2(x,y)+f_2(y,x)f_1(y)$$
and calculation of these expressions is a matter of composing the chain maps we have a tentative grasp of already. This gives us the case of xyx alternatingly between b in lower right corner for all even degrees and a just to the left of it for odd degrees, thus starting with the matrices
$$\begin{pmatrix}0&0&b\end{pmatrix}$$
$$\begin{pmatrix}0&0&0\\0&a&0\end{pmatrix}$$
and for the case yxy, we have a similar pattern, but with a in the lower right for odd degrees and b above it for even degrees, giving the first two maps as
$$\begin{pmatrix}0&0&a\end{pmatrix}$$
$$\begin{pmatrix}0&0&b\\0&0&0\end{pmatrix}$$

So, we can immediately conclude that our m3 is going to be the zero element of Γ, since none of these chain maps are homotopic to any non-trivial coclass representatives. Thus we'll need to find homotopies from these to the zero chain maps for our values of f3.

A homotopy suitable for f3(x,y,x) would be h with
$$h_0=\begin{pmatrix}0&0\end{pmatrix}$$
$$h_1=\begin{pmatrix}0&0&0\\0&0&1\end{pmatrix}$$
$$h_2=\begin{pmatrix}0&0&0&0\\0&0&0&1\\0&0&0&0\end{pmatrix}$$
$$h_3=\begin{pmatrix}0&0&0&0&0\\0&0&0&1&0\\0&0&0&0&0\\0&0&0&0&1\end{pmatrix}$$
$$h_4=\begin{pmatrix}0&0&0&0&0&0\\0&0&0&1&0&0\\0&0&0&0&0&0\\0&0&0&0&0&1\\0&0&0&0&0&0\end{pmatrix}$$
It goes on, but I have not yet discerned a pattern clear enough to describe a generic element of this chain map. I probably should, at some point.
For f3(y,x,y), we would need a homotopy for the other sequence, and calculations lead me to put down h with
$$h_0=\begin{pmatrix}0&0\end{pmatrix}$$
$$h_1=\begin{pmatrix}0&0&1\\0&0&0\end{pmatrix}$$
$$h_2=\begin{pmatrix}0&0&1&0\\0&0&0&0\\0&0&0&0\end{pmatrix}$$
$$h_3=\begin{pmatrix}0&0&1&0&0\\0&0&0&0&0\\0&0&0&0&1\\0&0&0&0&0\end{pmatrix}$$
$$h_4=\begin{pmatrix}0&0&1&0&0&0\\0&0&0&0&0&0\\0&0&0&0&1&0\\0&0&0&0&0&0\\0&0&0&0&0&0\end{pmatrix}$$
and here we can discern a pattern to the matrices. They will have a sequence of 1 starting out at the third column, and going down skipping every second place.

Armed with these calculations, we may set out to calculate for our pleasure Φ4. Due to the heuristic we use in defining the fi for things "lifted" from the generators we've defined above, we shall discard anything except for xyxy and yxyx from study; all other cases will just give the relations we use in calculating f2 or f3 from the cases we've calculated.

Thus, we'll be interested in
$$\Phi_4(x,y,x,y)=f_1(x)f_3(y,x,y)+f_2(x,y)f_2(x,y)+f_3(x,y,x)f_1(y)$$
and
$$\Phi_4(y,x,y,x)=f_1(y)f_3(x,y,x)+f_2(y,x)f_2(y,x)+f_3(y,x,y)f_1(x)$$
and thus we can, by using the already calculated matrices and a CAS (I use Magma right now) just calculate the matrices. In both these cases we get the same thing: the matrices
$$\begin{pmatrix}0&0&1\end{pmatrix}$$
$$\begin{pmatrix}0&0&1&0\\0&0&0&1\end{pmatrix}$$
$$\begin{pmatrix}0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\end{pmatrix}$$
and so on; which are precisely the matrices we chose for our $$f_1(z)$$.

Thus, we'll set m4(x,y,x,y)=m:sub:4(y,x,y,x)=z and f4=0, and stop calculating right here.