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:

*

*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.


*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?
 A: $\newcommand{\MU}{\mathrm{MU}} \newcommand{\SU}{\mathrm{SU}} \newcommand{\tmf}{\mathrm{tmf}} \newcommand{\ko}{\mathrm{ko}} \newcommand{\BGL}{\mathrm{BGL}} \newcommand{\ku}{\mathrm{ku}} \newcommand{\GL}{\mathrm{GL}} \newcommand{\RP}{\mathbf{R}P}$
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.
