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I've never quite 'got' Massey products - this question, I guess, is to work out a small example that might shed some light for me.

So following Wikipedia, let $\Gamma$ be a differential graded algebra with differential $d$, let $\bar{u} = (-1)^{|u|+1}u$ and let $[u]$ denote the cohomology class of an element $u \in \Gamma$. Then suppose that $uv = 0$ and $vw=0$. Then there are $s,t$ such that $d(s)=\bar{u}v$ and $d(t)=\bar{v}w$, and then the cycle $\bar{s}w+\bar{u}t$ represents an element of the Massey product $\langle u,v,w\rangle$.

The particular example I will be interested in is applying this to the cobar complex in the Adams spectral sequence for the sphere spectrum (at $p=2$). This has $E_2$ term $\text{Ext}_{\mathcal{A}}^s(\mathbb{F}_2,\Sigma^t \mathbb{F}_2)$. We have the following theorem

The 2-line $\text{Ext}^2_{\mathcal{A}}(\mathbb{F}_2,\Sigma^\ast \mathbb{F}_2)$ is the graded vector space generated by $h_ih_j$ subject to relations $h_ih_j = h_jh_i$ and $h_ih_{i+1}=0$.

The notes of Paul Goerss then point out that this is the defining relation for a Massey product and that it is a 'fun exercise' to show that $\langle h_0,h_1h_0 \rangle = h_1^2$ and $\langle h_1,h_0,h_1 \rangle = h_0h_2$

I'm interested in doing these calculations, but I can't seem to work out how! Firstly, I presume the first is a typo and it should be $\langle h_0,h_1,h_0 \rangle $? Secondly I feel like, no matter what classes I chose for $s$ and $t$ that the answer should involve $h_1$, since it is taking the place of $u$ and $w$ in the above.

If anyone can shed any light on how to do these calculations, it would be much appreciated.

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Surely the first is a typo and should be as you say it is. Have you tried working with cocycles in a minimal resolution as opposed to the bar complex? that will probably be a lot easier. You will only need the minimal resolution in a small range. Maybe try looking at what the massey products could be in $Ext_{A(1)}(\mathbb{F}_2,\mathbb{F}_2)$ where $A(1)$ is the subalgebra of the steenrod algebra generated by $sq^1$ and $sq^2$. –  Sean Tilson Aug 5 '12 at 22:22

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