An A∞-structure on the cohomology of D8

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 [tex]H^*(D_8,\mathbb F_2)[/tex].

To do this, I fix the group algebra

[tex]\Lambda=\mathbb F_2[a,b]/(a^2,b^2,abab+baba)[/tex]
and the cohomology ring
[tex]\Gamma=\mathbb F_2[x,y,z]/(xy)[/tex]
with [tex]x,y\in\Gamma_1[/tex], [tex]z\in\Gamma_2[/tex]

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

[tex]\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}[/tex]
for differentials starting in odd degree, and
[tex]\begin{pmatrix}
a&0&bab&0&0\\
0&b&0&aba&0\\
0&0&a&0&bab\\
0&0&0&b&aba
\end{pmatrix}[/tex]
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
[tex]\partial_0=\begin{pmatrix}a&b\end{pmatrix}[/tex]
[tex]\partial_1=\begin{pmatrix}a&0&bab\\0&b&aba\end{pmatrix}[/tex]
[tex]\partial_2=\begin{pmatrix}a&0&bab&0\\0&b&0&aba\\0&0&a&b\end{pmatrix}[/tex]

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 [tex]\Lambda^{2k}[/tex], keeps the first, third, et.c. elements and throws out the even ordered elements; so [tex]x\cdot(a_1,a_2,a_3,a_4,a_5,a_6)=(a_1,0,a_3,0,a_5)[/tex]

For an element in [tex]\Lambda^{2k+1}[/tex], the last element gets
extra treatment, so
[tex]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)[/tex]
For the lowest degrees, we also have
[tex]x_0 = \begin{pmatrix}1&0\end{pmatrix}[/tex]
[tex]x_1 = \begin{pmatrix}1&0&0\\0&0&ab\end{pmatrix}[/tex]
[tex]x_2 = \begin{pmatrix}1&0&0&0\\0&0&1&0\end{pmatrix}[/tex]

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

For [tex]f_1(y)[/tex], we get the interleaved effect: every second element, but now with even indices, and some extra treatments at the end. So an element in [tex]\Lambda^{2i}[/tex] will behave like

[tex]y\cdot(a_1,a_2,a_3,a_4,a_5,a_6)=(0,a_2,0,a_4,a_6)[/tex]

and for an element in [tex]\Lambda^{2i+1}[/tex] we get

[tex]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)[/tex]

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 [tex]f_1[/tex] 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 [tex](xy)[/tex], since [tex]f_1(x)f_1(x)f_1(y)f_1(y)=0[/tex], and we define [tex]f_2(x^2,y)=f_1(x)f_2(x,y)[/tex] and [tex]f_2(x,y^2)=f_2(x,y)f_1(y)[/tex].

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 [tex]f_1(x)f_1(y)[/tex] given in the first few step by the matrices

[tex]\begin{pmatrix}0&0&ba\end{pmatrix}[/tex]
[tex]\begin{pmatrix}0&0&0&0\\0&0&0&ab\end{pmatrix}[/tex]
and so on, with the lower right entry alternatingly ab and ba.

[tex]f_1(y)f_1(x)[/tex] is the same thing, but with ab and ba interchanged.

We want [tex]f_2(x,y)[/tex] and [tex]f_2(y,x)[/tex] 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 [tex]f_2(x,y)[/tex] has lower right entry a for all odd degree map components, and [tex]f_2(y,x)[/tex] has the entry above that b, same components. All even components vanish.

Thus, for the first nonzero component, we have

[tex]\begin{pmatrix}0&0&0\\0&0&a\end{pmatrix}[/tex]
for x,y and
[tex]\begin{pmatrix}0&0&b\\0&0&0\end{pmatrix}[/tex]
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,

[tex]\Phi_3(x,y,x)=f_1(x)f_2(y,x)+f_2(x,y)f_1(x)[/tex]
[tex]\Phi_3(y,x,y)=f_1(y)f_2(x,y)+f_2(y,x)f_1(y)[/tex]
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
[tex]\begin{pmatrix}0&0&b\end{pmatrix}[/tex]
[tex]\begin{pmatrix}0&0&0\\0&a&0\end{pmatrix}[/tex]
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
[tex]\begin{pmatrix}0&0&a\end{pmatrix}[/tex]
[tex]\begin{pmatrix}0&0&b\\0&0&0\end{pmatrix}[/tex]

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

[tex]h_0=\begin{pmatrix}0&0\end{pmatrix}[/tex]
[tex]h_1=\begin{pmatrix}0&0&0\\0&0&1\end{pmatrix}[/tex]
[tex]h_2=\begin{pmatrix}0&0&0&0\\0&0&0&1\\0&0&0&0\end{pmatrix}[/tex]
[tex]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}[/tex]
[tex]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}[/tex]
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

[tex]h_0=\begin{pmatrix}0&0\end{pmatrix}[/tex]
[tex]h_1=\begin{pmatrix}0&0&1\\0&0&0\end{pmatrix}[/tex]
[tex]h_2=\begin{pmatrix}0&0&1&0\\0&0&0&0\\0&0&0&0\end{pmatrix}[/tex]
[tex]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}[/tex]
[tex]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}[/tex]
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 f_i 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

[tex]\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)[/tex]

and
[tex]\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)[/tex]
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
[tex]\begin{pmatrix}0&0&1\end{pmatrix}[/tex]
[tex]\begin{pmatrix}0&0&1&0\\0&0&0&1\end{pmatrix}[/tex]
[tex]\begin{pmatrix}0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\end{pmatrix}[/tex]
and so on; which are precisely the matrices we chose for our
[tex]f_1(z)[/tex].

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.

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