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 $\alpha_n$?

  • $\begingroup$ I think there is a kind of answer in page 107-114 of Ravenel's orange book, or something which you can extract your answer from it. There is also Paul Turner's Manchester thesis which some work on T(n) spectra and mod $2$ orientations is done, but I am not sure where to find a pdf file! $\endgroup$
    – user51223
    Dec 2, 2018 at 5:48
  • $\begingroup$ And if I remember correctly, $T(i)$ spectra interpolate between $X(n)$ and $X(n+1)$, so aren't they just the $F(n)$ spectra defined in Revanel's Orange book? $\endgroup$
    – user51223
    Dec 2, 2018 at 5:54
  • $\begingroup$ @user51223 Thanks. I know of that part of Ravenel's book, but I don't know how to use it to answer my question. Where can I find a copy of his thesis? It seems really hard to find a copy online. The spectra T(i) do not interpolate between X(n) and X(n+1); instead, they split off of X(p^i)_p much in the same way as BP splits off of MU_p. $\endgroup$
    – skd
    Dec 2, 2018 at 15:22
  • $\begingroup$ @user51223 I think I have a copy of Paul Turner's thesis in my office, which I can check tomorrow, but I am fairly sure that it does not contain the material that you mention. Perhaps you are misremembering where you saw it? $\endgroup$ Dec 2, 2018 at 22:25
  • $\begingroup$ @ Neil Strickland Yes, it is quite possible that I have been misremembering this. At the time of writing the above comment, I thought that it is in his thesis where he talks about HZ/2 or MO orientations and formal groups laws related to it! Unfortunately, I did not make a copy of his thesis! $\endgroup$
    – user51223
    Dec 3, 2018 at 7:08

1 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.

  • $\begingroup$ Thanks. I believe that what you wrote actually hints that the answer is probably positive. Namely, looping the map $\Omega^2 S^3 \to B\mathrm{GL}_1(\mathbb{S})$ gives the map $\Omega^3 S^3 \to \mathrm{GL}_1(\mathbb{S}) \subseteq \text{colim }\Omega^n S^n$. This factors through the map $\Omega^3 S^3 \to \Omega^3 S^3$, which is multiplication by (1-p) on homotopy. In particular, taking $\pi_{2p-3}$ gives a map $\pi_{2p}(S^3) \to \pi_{2p-3}(\mathbb{S})$ which sends $\alpha_1'$ to $\Sigma^\infty((1-p)\alpha_1')$. This map is clearly not the inclusion, but it just multiplies by the unit (1-p). $\endgroup$
    – skd
    Dec 3, 2018 at 14:52
  • $\begingroup$ well, I thought it is a decomposability argument and hence gives a negative answer! $\endgroup$
    – user51223
    Dec 3, 2018 at 16:39
  • $\begingroup$ Sorry, I'm not sure I understand why $\alpha_1$ would be a decomposable element. $\endgroup$
    – skd
    Dec 3, 2018 at 16:43
  • $\begingroup$ You're right! At the beginning, I was taking a map $S^3\to BSG$ and factorise $\alpha$ through it. I now think that the above arguments prove what you were after. Bearing in mind that $S^1\to BSG$ corresponds to $1-p\in{_p\pi_{2p-3}^{st}}\simeq\mathbb{Z}/p$ so the above composition indeed would be equal to $\alpha_1$ as you say and the orientation follows! I was just not sure about the identification of $T(1)$ as a Thom spectrum as I have assumed you are using as you say it has $\alpha_1=0$ which I was not sure about its meaning! $\endgroup$
    – user51223
    Dec 4, 2018 at 5:42
  • $\begingroup$ Thanks again. Note that $T(1)$ is defined to be the free $\mathbf{E}_1$-ring with $\alpha_1 = 0$, so it is defined as the Thom spectrum of that map. I guess the second part of my question is still unanswered. Do you have any thoughts on that? $\endgroup$
    – skd
    Dec 4, 2018 at 5:47

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