# The homotopy of universal Thom spectrum

Let $S^0_p$ be the $p$-adic sphere spectrum. Let $GL_1(S^0_p)$ be the set of unit componen of $\Omega^{\infty}S^0_p$. For any map $X \to BGL_1(S_p^0)$ we get a Thom spectrum call it $Mf$. Now consider the identity map on $GL_1(S^0_p)$, the Thom spectrum associated to that map is what I call the universal Thom spectra and denote it by $MGL_1(S^0_p)$.

My question is what are the homotopy groups of this Thom spectrum?

Assume that $R$ is a connective $E_\infty$ ring spectrum. Typically $GL_1(R)$ denotes the set of components in $\Omega^\infty R$ which span $GL_1(\pi_0 R)=\pi_0 R^\times$. I would call the unit component $SL_1(R)$. I will use this notation below.

In the formalism of Ando-Blumberg-Gepner-Hopkins-Rezk, the Thom spectrum functor takes based spaces over $BGL_1(R)$ to $R$-module spectra. Since the identity map $BGL_1(R)\rightarrow BGL_1(R)$ is terminal in the domain category its Thom spectrum $\mathcal{M}=MGL_1(R)$ receives a canonical map from any other such Thom spectrum. Note $MGL_1(R)$ is not always a terminal object in $R$-module spectra. $MSL_1(R)$ should satisfy a similar property among such Thom spectra which are $H\pi_0 R$-orientable.

Using the ABGHR formalism we can identify $$\mathcal{M}=\Sigma^\infty_+ EGL_1(S_p) \wedge^\mathbb{L}_{\Sigma^\infty_+ GL_1(S_p)} S_p.$$ This smash product only involves connective spectra, so $$\pi_0 \mathcal{M} \cong \pi_0(\Sigma^\infty_+ EGL_1(S_p)) \otimes_{\pi_0(\Sigma^\infty_+ GL_1(S_p))} \pi_0S_p\cong \mathbb{Z}\otimes_{\mathbb{Z}[\mathbb{Z}_p^\times]}\mathbb{Z}_p.$$ You can see this by examining the $\mathrm{Tor}$ spectral sequence, but there probably is a more elementary argument. As Neil Strickland pointed out, we can identify this torsion product (I would use the beginning of a free resolution of $\mathbb{Z}$ by free $\mathbb{Z}[\mathbb{Z}_p^\times]$-modules to calculate this): The torsion product is $$\mathbb{Z}_p/(gx-x)_{x\in \mathbb{Z}_p^\times,x\in \mathbb{Z}_p}.$$ If $p$ is odd this is 0 and if $p$ is 2 then this is $\mathbb{Z}/2$. This implies $MGL_1(S_p)\simeq *$ for $p$ odd.

Similarly $$\pi_0 MSL_1(S_p)\cong \pi_0 (\Sigma_+^\infty ESL_1 S_p)\otimes_{\pi_0 (\Sigma_+^\infty SL_1(S_p))} \pi_0(S_p)\cong \mathbb{Z}\otimes_\mathbb{Z}\mathbb{Z}_p\cong \mathbb{Z}_p.$$ If you take the subspace $G$ of $\Omega^\infty S_p$ spanned by the components $1+p\mathbb{Z}_p\subset \mathbb{Z}_p^\times$ (for $p=2$ this is $GL_1(S_2)$), then you get a Thom spectrum with $\pi_0 MG\cong \mathbb{Z}/p.$ When $p$ is $2$ the completion is unnecessary, $\pi_0 MGL_1(S)\cong \mathbb{Z}/2$.

As Neil Strickland pointed out, any $E_2$-ring spectrum, in particular any $E_\infty$ ring spectrum, $R$ with $\pi_0 R\cong \mathbb{Z}/p$ is an $H\mathbb{Z}/p$-algebra and hence a wedge of $H\mathbb{Z}/p$-modules. This implies the mod-$p$ homology is a sum of suspensions of copies of the dual Steenrod algebra. The generators of these copies of the dual Steenrod algebra are in 1-1 correspondence with generators of $\pi_*R$ as a graded $\mathbb{Z}/p$-vector space.

• In the original question $GL_1(S^0_p)$ was defined to be a connected component, so the tensor product should be over $\mathbb{Z}$ and the answer is $\mathbb{Z}_p$. If we instead take $GL_1(S^0_p)$ to be the union of invertible components, then we get what you wrote, but $\mathbb{Z}_p^\times$ is acting trivially on $\mathbb{Z}$ and by multiplication on $\mathbb{Z}_p$, so the tensor product is zero. – Neil Strickland Jun 7 '14 at 7:22
• We could instead use an intermediate space $G$ with $\pi_0(G)=1+p\mathbb{Z}_p$ and then we would get an $E_\infty$ Thom spectrum $\mathcal{M}$ with $\pi_0(\mathcal{M})=\mathbb{Z}/p$. Any such $E_\infty$ ring spectrum is known to be a wedge of suspensions of $H\mathbb{Z}/p$, so if you could calculate the homology, you would have a collapsing Adams spectral sequence for the homotopy. – Neil Strickland Jun 7 '14 at 7:27
• @Prasit: I did not fully understand your comment, but I will reiterate how my answer relates to your question. The case in your question is $MSL_1(S_p)$ whose $\pi_0$ is torsion-free. Since I suspected you meant to ask about $MGL_1(S_p)$ I included that case. That spectrum is universal in the sense that it is contractible and hence terminal. I am claiming that the homotopy groups of $MG$ are $\bZ/p$-modules and calculating them is equivalent to calculating the homology as as a comodule over the dual Steenrod algebra. I did not claim that the latter problem was easy; it is just algebraic. – Justin Noel Jun 8 '14 at 15:13