# Are Thom spectra MU, MSO and K-theory spectra KU, KO modules over some truncations of the sphere spectrum?

The Thom spectrum MO is a module over the ring spectrum π≤0S=HZ, where S is the sphere spectrum. In particular, MO is equivalent to the Eilenberg-MacLane spectrum Hπ*(MO). On the other hand, MU and MSO are not modules over π≤0S, because they have nontrivial k-invariants.

I wonder if the above result about MO can be generalized to other Thom spectra like MU and MSO by considering higher truncations π≤kS of the sphere spectrum. Such a result would be interesting because it is related to the question of representing Thom spectra as (weak) simplicial objects in the k-category of symmetric monoidal (k+1)-groups (k-groupoids with invertible objects), in the sense that interpreting symmetric monoidal (k+1)-groups as stable homotopy k-types and taking the homotopy colimit should give back the Thom spectrum under consideration. Such models are interesting because they are more strict then (say) Segal's Γ-spaces. A negative answer would necessarily preclude the existence of such models because a stable homotopy k-type is a module over π≤kS, and the homotopy colimit of a simplicial diagram of modules over π≤kS in the category of spectra is a again a module over π≤kS.

Is MU or MSO a module over the ring spectrum π≤kS for some k>0? Same question for KU and KO.

(Incidentally, the above result for MO implies that MO can be represented as a simplicial abelian group. I wonder if there is a geometric model for MO as a simplicial abelian group along the lines of Galatius-Madsen-Tillmann-Weiss theorem, i.e., n-simplices should be related to unoriented n-manifolds. Such a model would necessarily make explicit use of properties of unoriented manifolds as opposed to oriented or stably complex manifolds.)

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I would be very surprised if this worked for $MU$. – Dylan Wilson Mar 12 '13 at 23:42

For any $k$ and any $0<n<\infty$, $K(n)\wedge \pi_{\leq k}S=0$. Indeed, this is true for any spectrum with finitely many homotopy groups, since $K(n)\wedge H\mathbb{Z}=0$ and any such spectrum has a finite filtration into Eilenberg-MacLane spectra. If a spectrum $M$ admits a $\pi_{\leq k}S$-module structure, then $M$ is a retract of $\pi_{\leq k}S\wedge M$, and so $K(n)\wedge M$ is a retract of $K(n)\wedge\pi_{\leq k}S\wedge M=0$ and hence $K(n)\wedge M=0$. But this is not true for $n=1$ and all of your examples (for $MSO$, you must work at an odd prime).