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Basically, I am trying to compute something with the Adams spectral sequence (as a toy example). The $E^2$ page reduced to computing $Ext^{s,t}_{\Gamma} (\mathbb{F}_2, \mathbb{F}_2)$, where $\Gamma = \mathbb{F}_2 x/ x^2$ and $deg(x) =1$. Keeping very close track of degree, I found that $Ext^{s,t}_{\Gamma} (\mathbb{F}_2, \mathbb{F}_2)$ is only nonzero in degree $s=t$. In this case, we have $\mathbb{F}_2 \{ \phi_t \} $ where $\phi_t \in Hom^t_{\mathbb{F}_2 x/x^2} (\Sigma^t \mathbb{F}_2 x/x^2, \mathbb{F}_2)$ sends $1$ to $1$ and $x$ to $0$. (the superscript $t$ denotes morphisms that lower degree by $t$.)

So I have come to the point that I need to understand the multiplicative structure of $Ext$. I know I am supposed to get that $\phi_1^2 = \phi_2$ (because I know what the spectral sequence is supposed to converge to--the 0th column should have a copy of $\mathbb{F}_2$ in each degree, but all of it generated by $\phi_1$ ). However, I am not really sure how to make sense of $\phi_1^2$. So my questions are

1) how to make sense of $\phi_1^2$? or have I confused things somewhere along the way

2) in order to work with the multiplicative structure in the second page of Adams spectral sequence, I need to understand the multiplicative structure of $Ext$. I would like to work directly with objects in $Ext$, but how does one multiply? and is it easier to do it this way or does one typically use the isomorphism between objects of $Ext$ and extensions, multiply the extensions (which is simple), and then translate back? But the isomorphism for higher Ext does not seem easy to work with.

Note: I specifically am trying to do with without the cobar complex.

Edit: I should add that the resolution I took was $\ldots \Sigma^2 \mathbb{F}_2 x/x^2 \xrightarrow{d} \Sigma \mathbb{F}_2 x/x^2 \xrightarrow{d} \mathbb{F}_2 x/x^2 \to \mathbb{F}_2$, where $d$ sends $\Sigma^{t} 1$ to $\Sigma^{t-1} x$ and everything else to $0$. I'm adding the $\Sigma$ to help keep track of degree.

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    $\begingroup$ You can actually pick any projective resolution, so what is probably optimal in this case is to take $... \rightarrow x^2\mathbb{F}_2[x]/x^4 \rightarrow x\mathbb{F}_2[x]/x^3 \rightarrow \mathbb{F}_2[x]/x^2 \rightarrow 0$. Then it boils down to calculating the endomorphism complex of this resolution. $\endgroup$ – Lev Soukhanov Jan 25 at 2:35
  • $\begingroup$ so your $\phi_t$ is a map which sends $x^t \in x^t\mathbb{F}_2[x] / x^{t+2}$ to $1$ in $\mathbb{F}_2$, and we need to somehow find its lift to the map of complexes, which shouldn't be too hard (it exists and unique up to homotopy due to rhs being projective resolution). I presume we can take $\phi_t$ to be the map which sends $\alpha \in x^k \mathbb{F}_2[x]/x^{k+2}$ to the $x^{-t}\alpha \in x^{k-t} \mathbb{F}_2[x]/x^{k-t+2}$ (so basically just shifts the complex by $t$ and maps the terms isomorphically where possible and to zero otherwise). $\endgroup$ – Lev Soukhanov Jan 25 at 2:41
  • $\begingroup$ That allows you to take compositions, and now the fact that $\phi_t \phi_s = \phi_{t+s}$ is clear. so if you want to compose exts, you need to resolve both modules and lift your maps to the maps of resolutions, and it is typically easy to do. (if my notation is unclear $x^k \mathbb{F}_2 / x^{k+2}$ is $\Sigma^k \mathbb{F}_2[x]/x^2$ in your notation) $\endgroup$ – Lev Soukhanov Jan 25 at 2:43
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So, maybe I should format this as an answer.

Consider any pair of objects $A, B$ in an abelian category with, say, enough projective objects. Then, the extension group $Ext^i(A,B)$ is defined as $H^i(Hom(A^{\bullet}, B^{\bullet}))$, where $A^{\bullet}, B^{\bullet}$ are resolutions of objects $A, B$ respectively, and $A^{\bullet}$ needs to be projective, and no requirement on $B^{\bullet}$ but we will take it to be projective too to define composition.

Composition for Exts is then induced by composition $Hom(A^{\bullet}, B^{\bullet}) \otimes Hom(B^{\bullet}, C^{\bullet}) \rightarrow Hom(A^{\bullet}, C^{\bullet})$.

Let us denote $R = \mathbb{F}_2[x]/x^2$. Then, $\mathbb{F}_2$ admits the following resolution: $A^{\bullet} = ... \rightarrow \Sigma^2 R \rightarrow \Sigma R \rightarrow R \rightarrow 0$, the maps being multiplication by $x$.

The hom-complex $Hom(A^{\bullet},A^{\bullet})$ is then quasiisomorphic to $Hom(A^{\bullet}, \mathbb{F}_2)$ (it is always true for any pair of complexes such that left one is a complex of projective modules and the right one is switched to the quasiisomorphic one). Concretely, it means that any closed map $\phi: A^{\bullet} \rightarrow \mathbb{F}_2$ can be lifted to the map $\phi: A^{\bullet} \rightarrow A^{\bullet}$ uniquely up to homotopy. So, the map we lift is your $\phi_t$, and the lift is defined as $\phi_t: A^k \rightarrow A^{k-t} = Id$ for $k \geq t$ and $0$ otherwise.

It is now clear that $\phi_t \phi_s = \phi_{t+s}$

$\blacksquare$

the rule is basically "switch everything to resolutions and then map complexes instead of objects". it works very well if you have a category with enough projective or injective objects

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  • $\begingroup$ Thank you! In order to determine what the lift of $\phi_t$ is, I need to know what the quasi-isomorphism is, correct? Is it given by composing on the left with the projection $A^{\bullet} \to \mathbb{F}_2$? $\endgroup$ – Elise Jan 25 at 14:54
  • $\begingroup$ Yes, indeed (this is essentialy an unique choice because resolution is always quasiisomorphic to what it resolves, namely, its 0-th cohomology). You should also check another answer, it gives not only a recipe but puts into a broader context of the theory of quadratic algebras and Koszul duality (which I completely forgot about when writing this answer). $\endgroup$ – Lev Soukhanov Jan 26 at 9:53
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The composition product $Ext(N,P) \otimes Ext(M,N) \to Ext(M,P)$ can be computed as follows. Given cocycles $x : N_{s_1} \to \Sigma^{t_1} P$ and $y : M_{s_2} \to \Sigma^{t_2} N$, let $\{ y_s : M_{s+s_2} \to \Sigma^{t_2} N_s\}$ be a chain map lifting $y$. Then $xy$ is represented by the cocycle $x y_{s_1} : M_{s_1+s_2} \to \Sigma^{t_2} N_{s_1} \to \Sigma^{t_1+t_2} P $.

The point is that one needs (a finite bit of) the chain map lifting $y$, but only the cocycle $x$.

Saves effort.

(The $M_s$ and $N_s$ are the terms in projective resolutions of $M$ and $N$, of course.)

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In this case, you're dealing with a Koszul algebra ---this is precisely your observation that the spectral sequence is concentrated in the diagonal. Its Koszul dual is a polynomial algebra, and this gives you the multiplicative structure on $\mathsf{Ext}$: each diagonal entry is one dimensional with generator $x_s$, and you have that $x_sx_t=x_{t+s}$, so $\mathsf{Ext}$ is a polynonimal algebra generated by $x_1$.

This can be seen at the level of the (reduced) bar construction $B\Gamma$, whose dual computes your $E_2$-page. It has zero differential, and it is generated in each homological degree $s$ by $x_s = [x\vert\cdots\vert x]$ with $x$ appearing $s$ times. The product is dual to the coproduct of $B\Gamma$, which is simply $\Delta(x_s) = \sum x_i\otimes x_j$.

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