Define $\phi(n) = 4n - 2$. Is there a proof that on the second list of unstable Adams spectral sequence (for all spheres) there are no elements in squares $(n, m)$ such that $m < \phi(n)$. The problem is that there are plenty of elements on the first list (in lambda-algebra) in this region $m < \phi(n)$.
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1$\begingroup$ This is hard to follow. What is the convention for indexing that you are using? By "no elements", do you mean in the $E_2$-term, or the $E_\infty$-term? $\endgroup$– Charles RezkCommented Feb 18, 2017 at 15:48
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$\begingroup$ You are right, let me explain in details. First, I am interested in case $p=3$. Of-course $E_1$-term is $\Lambda$-algebra itself. Any admissible monom $x\in \Lambda$ has the length (the number of generators in $x$) and the degree (the sum of degrees of generators in $x$, where degree of $\lambda_i$ is $4i-1$ and degree of $\mu_i$ is $4i$). Then by the square $(n,m)$ in $E_1$ I mean the vector space (over $\mathbb{Z}_3$) spanned by all admissible monoms of length $n$ and degree $m$. The question is there a proof that each square $(n,m)$ on $E_2$ is zero vector space if $m < \phi(n)$. $\endgroup$– SamarkandCommented Feb 18, 2017 at 16:36
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