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For a given element $f\in\pi_n^s$ by Freudenthal theorem, it is known that $f$ pulls back to an element of $\pi_{2n+1}S^{n+1}$. At a prime $p$, whether or not if $f$ pulls back further depends on the iterated application of the James fibration or the $EHP$-sequence. What I would like to know is that if, at the prime $2$ or any other prime (at least small primes such as $3$ or $5$), there is any result which upon reasonable homological conditions on $f$, a lower bound on $l$ so that $f$ pulls back to $\pi_{2n+1-l}S^{n+1-l}$ is given?!?

For instance, for $f$ being one of the Hopf invariant one elements, it is known that $f$ does not pull back further. This could be seen only by applying the EHP sequence once. The possible, and related, homological conditions are that (1) $f$ is detected by a primary operation (2) $f$ is detected in the $1$-line of the ASS.

So, to make some elaboration, we may ask what is the answer in the case of Kervaire invariant one elements or the element detected by $\pi_*J$ and some answer in this case is known after Mahowald (or maybe a complete answer, I am not sure!).

Here, by homological condition, I mean something in terms of the ASS or a generalised Adams spectral sequence based some homology theory $E$. I must say that I would be surprise if something like this exists, but I might be wrong!

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    $\begingroup$ It might be somehow different from what you are asking, but if you know about $f$ in Adams $E_2$, then you can look at the $\Lambda$-algebra, which is a DGA $\Lambda$ with $H(\Lambda)$ computing Adams $E_2$. The advantage of $\Lambda$-algebra is that it comes with a natural filtration $\Lambda(n)$, and each $\Lambda(n)$ computes $E_2$ of unstable Adams for $S^n$. The Curtis algorithm can tell you how elements in $H(\Lambda(n))$ change when $n$ grows, thus tell you in Adams $E_2$, which unstable Adams $f$ can come from. $\endgroup$
    – Mingcong Zeng
    Dec 17, 2017 at 1:21
  • $\begingroup$ @MingcongZeng Actually, this is sort of tools that I had in mind, thanks for pointing at this. $\endgroup$
    – user51223
    Dec 17, 2017 at 8:16

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