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?


1 Answer 1


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.