Unstable Greek letter elements

Question

A theorem of Hopkins and Mahowald states that the Thom spectrum of the map $\Omega^2 S^3 \to B\mathrm{GL}_1(\mathbb{S}_{(p)})$ classifying the element $p$ is exactly $\mathrm{H}\mathbf{F}_p$. Let $T(1)$ denote the free $\mathbf{E}_1$-ring with $\alpha_1 = 0$ (so that at $p=2$, it is the 2-localization of the $\mathbf{E}_2$(!)-ring spectrum $X(2)$). The following question stemmed from attempting to understand whether there is a $p$-local orientation $T(1) \to \mathrm{H}\mathbf{F}_p$ which is a map of Thom spectra, i.e., if there is an orientation which comes from Thom-ifying a map of spaces $\Omega S^{2p-1} \to \Omega^2 S^3$ over $B\mathrm{GL}_1(\mathbb{S}_{(p)})$.

The element $\alpha_1\in \pi_{2p-3}(\mathbb{S})$ desuspends to an element $\alpha_1'$ of the unstable homotopy group $\pi_{2p}(S^3)$. The unstable element $\alpha_1'$ gets us a map $S^{2p-2} \to \Omega^2 S^3$, and hence a map $\Omega S^{2p-1} \to \Omega^2 S^3$. Does this Thom-ify to an orientation $T(1) \to \mathrm{H}\mathbf{F}_p$? This would follow, I think, if we knew that the map $\Omega^2 S^3 \to B\mathrm{GL}_1(\mathbb{S}_{(p)})$ is an isomorphism on $\pi_{2p-2}$, but I don't know if this is true.

A related question is the following. The map $\Omega S^{2p-1} \to \Omega^2 S^3$ is in turn is adjoint to a map $\Sigma \Omega S^{2p-1} \to \Omega S^3$. The James splitting tells us that the source splits as $\bigvee_{n\geq 0} S^{2n(p-1)+1}$, so we get maps $S^{2n(p-1)+1} \to \Omega S^3$, which are exactly elements of $\pi_{2n(p-1)+2}(S^3)$. Are these elements (nonzero multiples of) the desuspensions of the other $\alpha$-family elements $\alpha_n$?

Answer 1

This is a long comment about the ``existence of orientation'' part of your question, not a precise answer!

It seems to me the answer is negative (although I was trying to prove it is positive!). To see this, let's note that in your question, $GL_1(\mathbb{S})$ is the same as $SG$ (sometimes also denoted $F$) which the space of stable maps $S^0\to S^0$ of degree $1$ and the monoid strcuture is the multiplicative one arising from composition of stable maps. As a space $SG$ is homotopy equivalent to $Q_0S^0$ where $QS^0=\mathrm{colimit}\ \Omega^iS^i$ and $Q_iS^0$ denotes the path component corresponding to $i\in\pi_0QS^0\simeq\pi_0^{st}\simeq\mathbb{Z}$.

Now, for $\alpha_1\in {_p\pi_{2p-3}^{st}}$ viewed as a mapping $$S^{2p-3}\to Q_0S^0\simeq Q_1S^0=SG=\Omega(BSG)$$ and its adjoint $S^{2p-2}\to BSG$ the infinite loop structure on $BSG$, allows to extend this to a loop map $\Omega S^{2p-1}\to BSG$ and it seems that you want $T(1)$ to be the Thom spectrum of this map, right?

Let's take the element $S^1\to BSG$ whose Thom spectrum is $H\mathbb{F}_p$. If my understanding is correct, then in order to show that an orientation as in your question exists, it would be enough to show that the composition $$S^{2p-2}\stackrel{\alpha'}{\to} \Omega S^3\to BSG\ (*)$$ is the same as $\alpha$.

It think there could be different ways to get a contradiction to the existence of such a decomposition. For instance, it shows that $\alpha_1$ is a decomposable element in the ring $\pi_*^{st}$ which I presume it is not!

Moreover, after using loop structure of $BSG$ to extend $(*)$ to a composition of loop maps as $$\Omega S^{2p-1}\stackrel{\alpha'}{\to} \Omega^2S^3\to BSG$$ then upon Thomifying one gets $T(1)\to H\mathbb{F}_p$ as desired. Since you know about the pull back of $\alpha$ to an element of $\pi_{2p}S^3$ then it would be enough to show that looping $\Omega^2S^3\to BSG$ one gets the inclusion $\Omega^3S^3\to SG=Q_1S^0$. Note that from the beginning, since $S^{2p-3}$ is path connected for $p>1$ then we could work with $\Omega^3_0S^3$. Now, I am not sure about the last claim that the loop maps is the inclusion! Perhaps, using multiplicative structure on $SG$ would give the desired contradiction!

These might lead to an answer, I guess!

ADDED The $\alpha_n$ elements live in ${_2\pi_*}J$ which coincide with the image of the $J$ homomorphism $SO\to SG$ is $p>2$. Since you are already using James-splitting, then one way to think about this is to consider the adjoint mapping $f:S^{2n(p-1)+1}\to\Omega S^3$ and compose it with some suitable James-Hopf map, $j_t:\Omega S^3\to \Omega S^{2t+1}$ and see if the composition is nontrivial or not. The image of $J$ is known, and if the composition $j_t\circ f$ is nontrivial then there might be a chance that this gives you the $\alpha_n$ elements. On the other hand, since somehow these elements arise from odd-primary Hopd invariant one problem then by analogy first look at the prime $p=2$ and see starting from $\eta$, $\nu$, or $\sigma$, you can get a family whose all elements are nontrivial, and if so then your guess might be correct.