# Unstable Greek letter elements

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 $$\alpha_n$$?

• 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! Dec 2, 2018 at 5:48
• 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? Dec 2, 2018 at 5:54
• @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.
– skd
Dec 2, 2018 at 15:22
• @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? Dec 2, 2018 at 22:25
• @ 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! Dec 3, 2018 at 7:08

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!

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.

• 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).
– skd
Dec 3, 2018 at 14:52
• well, I thought it is a decomposability argument and hence gives a negative answer! Dec 3, 2018 at 16:39
• Sorry, I'm not sure I understand why $\alpha_1$ would be a decomposable element.
– skd
Dec 3, 2018 at 16:43
• 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! Dec 4, 2018 at 5:42
• 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?
– skd
Dec 4, 2018 at 5:47