# Thom spectra, tmf, and Weierstrass curve Hopf Algebroid

Let $$X(4)$$ be the Thom spectrum associated to $$\Omega SU(4) \to \Omega SU \simeq BU$$. Since $$X(4)$$ is a homotopy commutative ring spectrum, for any spectrum Y we can construct a resolution $$Y \wedge X(4) \to Y \wedge X(4) \wedge X(4) \to Y \wedge X(4) \wedge X(4) \wedge X(4) \to \dots$$ of Y. (Sorry but I don't know how to type the cosimplicial diagram. This is just construction of the ANSS with respect to $$X(4)$$.) In Rezk's note on tmf, he claims that the resolution for $$Y = tmf$$ is the same as the cobar complex obtained from the Weierstrass Hopf algebroid $$(\mathbb{Z}[a_1, a_2, a_3, a_4, a_6], \mathbb{Z}[a_1, a_2, a_3, a_4, a_6][r, s, t])$$ (Thm 14.5). I want to know a proof of this theorem because this construction is used in Bauer's paper, but I haven't find a written proof of it. I think Hopkins-Mahowald report also states similar result (Cor 2.4), but I don't see how to conclude this from the theorem above. Does anybody know how to compute this, at least? Following is my scatterer thoughts:

1. Since $$Y \wedge X(4)$$ has a complex orientation of degree 4 (green book, 6.5.3), we know that $$\pi_*(tmf \wedge X(4) \wedge X(4)) = \pi_* (tmf \wedge X(4))[r, s, t]$$. So we have to compute $$\pi_* (tmf \wedge X(4))$$ first. Can we compute this from $$\pi_* (tmf \wedge MU)$$? Mathew's paper calculate the latter so I am wondering if I can use this.

2. As Hopkins-Mahowald says, homotopy classes $$a_1 \dots a_6$$ come from $$\pi_*(X(4))$$. However I don't know the computation of $$\pi_*(X(4))$$ or image of Hurewicz homomorphsim. Also, what are so special about $$tmf$$ or $$eo_2$$ so that these elements forms a homotopy group? Do we have a explanation of this computation? Why $$X(4)$$ is used here?

$$\newcommand{\MU}{\mathrm{MU}} \newcommand{\SU}{\mathrm{SU}} \newcommand{\tmf}{\mathrm{tmf}} \newcommand{\ko}{\mathrm{ko}} \newcommand{\BGL}{\mathrm{BGL}} \newcommand{\ku}{\mathrm{ku}} \newcommand{\GL}{\mathrm{GL}} \newcommand{\RP}{\mathbf{R}P}$$ Suppose $$R$$ is a homotopy commutative ring such that $$X(n)_\ast(R)$$ is concentrated in even degrees for some $$n\geq 0$$. Then we can form the graded stack $$M_R$$ associated to the graded Hopf algebroid $$(A, \Gamma):= (X(n)_\ast(R), X(n)_\ast(X(n) \otimes R))$$. This will be isomorphic to the graded stack associated to the graded Hopf algebroid $$(A', \Gamma'):= (\MU_\ast(R), \MU_\ast(\MU \otimes R))$$. To see this, note that there is an isomorphism of algebras $$\MU_\ast(R) \cong X(n)_\ast(R)[x_{n+1}, x_{n+2}, \cdots],$$ where $$|x_i| = 2i$$. Similarly, $$\MU_\ast(\MU \otimes R) \cong X(n)_\ast(X(n) \otimes R)[x_{n+1}, x_{n+2}, \cdots][t_{n+1}, t_{n+2}, \cdots].$$ where $$|t_i| = 2i$$. (In fact, these can be lifted to equivalences at the level of spectra $$\MU\otimes R \simeq \MU\otimes_{X(n)} (X(n) \otimes R) \simeq X(n) \otimes R \otimes \Omega(\SU/\SU(n))_+;$$ but this is at the expense of multiplicativity: $$\Omega(\SU/\SU(n))$$ isn't an $$\mathbf{E}_2$$-space.) So $$(A', \Gamma')$$ is isomorphic to $$(A[x_{n+1}, \cdots], \Gamma[x_{n+1}, \cdots, t_{n+1}, \cdots])$$. One can now check from the Hopf algebroid structure that they must present the same stack, namely $$M_R$$.
Now, let us try to understand $$X(4) \otimes \tmf$$. As a warmup, let us first understand $$X(2) \otimes \ko$$ (this is in Section 3.2 of "From spectra to stacks", see the comments). Observe that $$X(2)$$ is the Thom spectrum of the map $$\Omega S^3 \to \BGL_1(S)$$ which detects $$\eta\in \pi_1(S)$$ on the bottom cell of the source. There is an EHP sequence $$S^2 \to \Omega S^3 \to \Omega S^5$$, using which one can show that $$X(2) \simeq C\eta \otimes S/\!/\nu$$, where $$S/\!/\nu$$ is the Thom spectrum of the map $$\Omega S^5 \to \BGL_1(S)$$ which detects $$\nu\in \pi_3(S)$$ on the bottom cell of the source. Therefore, $$\ko \otimes X(2) \simeq \ko \otimes C\eta \otimes S/\!/\nu$$. We also know that $$\ko \otimes C\eta\simeq \ku$$ (the Wood equivalence), so that $$\ko \otimes X(2) \simeq \ku \otimes S/\!/\nu$$. Since $$\nu = 0$$ in $$\ku$$, we have $$\ku \otimes S/\!/\nu \simeq \ku[\Omega S^5]$$, so we conclude that $$\ko \otimes X(2) \simeq \ku[\Omega S^5]$$. At the level of homotopy, this is $$\mathbf{Z}[\beta, x_4]$$ with $$|\beta|=2$$ and $$|x_4|=4$$.
(Remark: the above argument only shows that the equivalence $$\ko \otimes X(2) \simeq \ku \otimes S/\!/\nu$$ is true additively. Indeed, we used the Wood equivalence $$\ko \otimes C\eta\simeq \ku$$, which is a priori just an equivalence of spectra. To show that $$\ko \otimes X(2) \simeq \ku \otimes S/\!/\nu$$ as ring spectra, we need to equip $$\ko/\eta$$ with an $$\mathbf{E}_\infty$$-ring structure and show that the Wood equivalence is one of $$\mathbf{E}_\infty$$-rings. There are a few ways of doing this. For example, it is a consequence of the fact that the map $$\eta: S^1 \to \GL_1(\ko)$$ detecting $$\eta$$ is a map of infinite loop spaces; in fact, this map factors through an infinite loop map $$\RP^\infty \to \GL_1(\ko)$$. It corresponds to a map $$\Sigma \mathbf{Z}/2 \to \mathrm{gl}_1(\ko)$$, which is the inclusion of a summand. But this is getting to be a bit of a digression.)
Let us now return to $$X(4) \otimes \tmf$$. (I don't know the details of the original argument due to Hopkins and Mahowald, so let me explain a possibly different approach. Most of the details of this story, such as the spectra $$T(2)$$ and $$B$$ below, are in my paper https://sanathdevalapurkar.github.io/files/thom.pdf. ) For simplicity, let me localize at $$2$$. Then $$X(4)$$ splits as a wedge of copies of an $$\mathbf{E}_2$$-ring called $$T(2)$$. In fact, $$T(2)$$ is the Thom spectrum of the bundle over $$\Omega \mathrm{Sp}(2)$$ given by the $$\mathbf{E}_2$$-map $$\Omega \mathrm{Sp}(2) \to \Omega \SU(4) \to \Omega\SU \simeq \mathrm{BU}.$$ Since $$\SU(4)/\mathrm{Sp}(2) \simeq S^5$$, we have that $$X(4)$$ is the Thom spectrum of a map $$\Omega S^5 \to \BGL_1(T(2))$$. This map is $$2$$-locally null, so $$X(4) \simeq T(2)[\Omega S^5]$$, i.e., $$T(2)[a_2]$$ with $$|a_2|=4$$.
So we only need to understand $$\tmf \otimes T(2)$$. There is a map $$T(2) \to \mathbf{F}_2$$ which is injective on mod $$2$$ homology, and its image is $$\mathrm{H}_\ast(T(2); \mathbf{F}_2) \cong \mathbf{F}_2[\zeta_1^2, \zeta_2^2]$$. There is an $$\mathbf{E}_1$$-ring $$B$$ such that $$T(2) \simeq B \otimes DA_1$$, and which has a map $$B \to T(2)$$. The composite $$B \to T(2) \to \mathbf{F}_2$$ on mod $$2$$ homology is injective, and its image is $$\mathrm{H}_\ast(B; \mathbf{F}_2) \cong \mathbf{F}_2[\zeta_1^8, \zeta_2^4]$$. Then, $$\tmf \otimes T(2) \simeq \tmf \otimes DA_1 \otimes B \simeq \mathrm{BP}\langle 2\rangle \otimes B,$$ whose homotopy groups are $$\pi_\ast(\mathrm{BP}\langle 2\rangle)[x_8, y_{12}] \cong \mathbf{Z}_{(2)}[v_1, v_2, x_8, y_{12}]$$. Here, $$|v_1| = 2$$, $$|v_2| = 6$$, $$|x_8|=8$$, and $$|y_{12}|=12$$. (Note that we used the Wood-type equivalence $$\tmf \otimes DA_1 \simeq \mathrm{BP}\langle 2\rangle$$; just as with the Wood equivalence $$\ku \otimes C\eta \simeq \ku$$, this is a priori just an equivalence of spectra. But I think one can actually equip $$\tmf \otimes DA_1$$ with at least an $$\mathbf{E}_2$$-ring structure such that the Wood-type equivalence is one of $$\mathbf{E}_2$$-rings.) Therefore, $$\pi_\ast(\tmf_{(2)} \otimes X(4)) \simeq \pi_\ast(\tmf_{(2)} \otimes T(2)[\Omega S^5]) \cong \pi_\ast(\mathrm{BP}\langle 2\rangle)[a_2, x_8, y_{12}] \cong \mathbf{Z}_{(2)}[v_1, a_2, v_2, x_8, y_{12}].$$ This is exactly the desired calculation (recall that $$|a_2| = 4$$), at least $$2$$-locally. The same sort of argument works $$3$$-locally.