Why is the first nontrivial $p$-local stable stem cyclic?

Question

Let $\pi_\ast^{(p)}$ be the ring of $p$-local stable homotopy groups of spheres. This is a nonnegatively graded ring, with $\mathbb Z_{(p)}$ in degree $0$. The first nonvanishing positive degree element is in degree $2p-3$, and $\pi_{2p-3}^{(p)} = \mathbb Z / p$ is a cyclic group.

Question: Why is the first positive degree group cyclic?

Remarks:

• Obviously an answer is: “Well, you just compute it, and this is the result you get.” I’m basically wondering if there is something more enlightening to say.

• For example, when $p = 2$, $\pi_1^{(2)}$ is generated by the Hopf fibration $\eta$. So I’m asking why there aren’t any more Hopf fibrations. (I’m not actually asking why the Hopf fibration becomes torsion after stabilization, although that’s also an interesting question, and maybe an answer here will also shed light on that question.)

• I suppose another answer would be “well, if the first two nontrivial stable stems happened to be in the same degree, wouldn’t that be an odd coincidence?” I agree with the suggestion here, and maybe one way of answering my question would be to flesh this out a bit, and give some heuristic from which this would follow as a probabilistic statement.

• I think I actually have a better sense for why the first nonvanishing degree is in degree $2p-3 = 2(p-1)-1$, having something to do with the dual Steenrod algebra having its first generator in that same degree. But these sorts of considerations don’t seem to explain the cyclicity.

Answer 1

Let me expand on my comment under the question.

By Nakaoka's theorem, the cohomology of the infinite symmetric group $\Sigma_\infty$ satisfies cohomological stability. In particular, up to degree (more than) $2p-2$, the map $\Sigma_p\to\Sigma_\infty$ is a $p$-local cohomology isomorphism. The cohomology $H^*(\Sigma_p,\mathbb{Z}_{(p)})$ is given by the invariants of the cohomology of a Sylow $p$-subgroup (which is cyclic of order $p$) under the action of the normaliser. So it's $\mathbb{Z}_{(p)}[x]/(px)$ with $|x|=2p-2$. Using universal coefficients, the homology $H_*(\Sigma_p,\mathbb{Z}_{(p)})$ is zero in degrees less than $2p-3$, and $\mathbb{Z}/p$ in degree $2p-3$. So the same is true of $H_*(\Sigma_\infty,\mathbb{Z}_{(p)})$.

Now the stable stems are the homotopy groups of $QS^0$. Localising at $p$, and using the Hurewicz theorem, the first non-zero homotopy group is the same as the first non-zero homology group of the identity component. But by the Barratt-Priddy-Quillen theorem, the identity component of $QS^0$ is equivalent to the Quillen plus construction $(B\Sigma_\infty)^+$, so its homology is the same as the homology of $B\Sigma_\infty$. It follows that the first non-zero stable homotopy group at $p$ occurs in degree $2p-3$ and is isomorphic to $\mathbb{Z}/p$.

Answer 2

Maybe one aspect that could illuminate this fact, outside of raw computation, is that these classes are in the image of the J-homomorphism. The J-homomorphism is surjective in a range (higher chromatic classes don't start until dimension $2p^2-p-2$) and by Adams's work, its image is finite cyclic.

Answer 3

Here is an approach, which I think goes back to Serre's thesis. It is a mere computation, but it doesn't have Eilenberg-MacLane spaces or the Steenrod algebra in it. We just keep using the Serre spectral sequence of $\Omega X\to \ast\to X$ and the fact that it is a spectral sequence of rings.

Let $n>1$ be odd.

Step 1: From knowing the cohomology of $S^n$, you work out that the cohomology ring of $\Omega S^n$ looks like a polynomial ring in one degree $n-1$ generator as long as degree $<p(n-1)$.

Step 2: From that, you work out that the cohomology ring of $\Omega^2 S^n$ looks like cohomology of $S^{n-2}$ in degrees $<p(n-1)-1$. In particular it looks like $S^{n-2}$ long enough so that Step 3 is just like Step 1:

Step 3: The cohomology ring of $\Omega^3 S^n$ looks like a polynomial ring in one degree $n-3$ generator until you hit degree $p(n-3)$.

Step 4 is like Step 2: the cohomology ring of $\Omega^4 S^n$ looks like cohomology of $S^{n-4}$ in degrees $<p(n-3)-1$.

Carry on like this until you have a description, in low degrees, of the cohomology of $\Omega^{n-1}S^n$. The details of what goes on at the edge of the range do not matter except for the last couple of steps. $H^\ast\Omega^{n-3}S^n$ looks like $S^3$ up to $3p-3$, I think. So $H^\ast\Omega^{n-2}S^n$ is polynomial on $x\in H^2$ until you get to $H^{2p}$, which is infinite cyclic but generated by $x^p/p$. So $H^\ast\Omega^{n-1}S^n$ looks like $S^1$ until you see a group of order $p$ at $H^{2p-3}$. It follows that the ($p$-local) cohomology of the universal cover of $\Omega^{n-1}S^n$ begins with $H^{2p-3}$ of order $p$, so that the homotopy of the same space begins with $\pi_{2p-4}$ of order $p$, so that in the homotopy of $S^n$ the first thing after $\pi_n$ is $\pi_{n+2p-3}$ of order $p$.

Note that this is an unstable result: for all odd $n>1$, the first $p$-torsion in $\pi_\ast S^n$ is cyclic of order $p$ in $\pi_{n+2p-3}$.

Answer 4

Re your side question: here's Fred Cohen's favorite proof that the Hopf map becomes torsion after stabilization. Consider the diagram

$$\require{AMScd}\begin{CD} S(\mathbb{C}^2) @>>f> S(\mathbb{C}^2) \\ @VV\eta{}V @VV\eta{}V\\ \mathbb{C}P^1 @>>g> \mathbb{C}P^1 \end{CD}$$ in which $f(z_1,z_2) = (z_1^k,z_2^k)$ and $g(z_1/z_2) = (z_1/z_2)^k$. Then $\deg(f) = k^2$ and $\deg(g) = k$, so we have $\eta \circ k^2 = k \circ \eta$. After suspension, the maps commute, so we can rewrite this as $(k^2-k)\eta = 0$. Thus $\eta$ must be annihilated by $\gcd_k(k^2-k) = 2$.

(See also Henriques' answer re $\pi_3^s$: third stable homotopy of spheres via geometry