# The cell structure of Thom spectra

I would like to understand the cell structure of integrally oriented Thom spectra. A Thom spectrum over a space $X$ is something you can build from a stable spherical bundle, which is classified by a map $\varphi: X \to BGL_1 \mathbb{S}$. When $R$ is an $A_\infty$ ring spectrum, it's possible to deloop $GL_1 R$ once to get $BGL_1 R$, and the spherical bundle $\varphi$ is said to be $R$-orientable when the map $$X \xrightarrow{\varphi} BGL_1 \mathbb{S} \xrightarrow{BGL_1 \eta} BGL_1 R$$ is null. The point of an orientation (i.e., a choice of lift of $\varphi$ to the fiber of $BGL_1 \eta$) is that it buys you a Thom isomorphism $$R \wedge T(\varphi) = R \wedge \Sigma^\infty_+ X.$$

$\newcommand{\Z}{\mathbb{Z}}\renewcommand{\S}{\mathbb{S}}\renewcommand{\phi}{\varphi}\newcommand{\sm}{\wedge}\newcommand{\Susp}{\Sigma}\newcommand{\Loops}{\Omega}$I would like to specify that my bundle is $H\Z$-oriented --- I think that'll be important for a couple reasons below. What it means homotopically for $\phi$ to be $H\Z$-orientable is evident by looking at the homotopy groups of $BGL_1 R$, which are given by the following formula: $$\pi_n GL_1 R = \begin{cases}(\pi_0 R)^\times & \hbox{when n = 0,} \\ \pi_n R & \hbox{otherwise.}\end{cases}$$ In the case of $BGL_1 \S$, these groups are complicated because $\pi_n \S$ is complicated, but at least we know $\pi_1 BGL_1 \S = \Z/2$. The spectrum $H\Z$ is easier, where we have the complete calculation $BGL_1 H\Z = \Susp H\Z/2$, and the map $BGL_1 \eta$ is an isomorphism in degree $1$. Altogether this means...

1. ... that an $H\Z$-oriented spherical fibration has a unique lift to the fiber, and
2. ... that the fiber has known homotopy type as well. It is given as a connected cover $$\operatorname{fib}(BGL_1 \S \to BGL_1 H\Z) = \Loops^\infty \S^1[2, \infty).$$

Now with all the pieces in place, let me ask the baby version of my question first: suppose that $X$ takes the form $X = S^n$, $n \ge 2$. Then the spherical bundle $\phi$ is actually selecting an element $\omega \in \pi_{n-1} \S$ in the stable homotopy groups of spheres.

Can the Thom spectrum $T(\phi)$ be identified with $M_0(\omega)$, the cone on $\omega$ with bottom cell in dimension $0$?

This is true in the paltry number of cases that I know classically: the projective spaces $\mathbb{R}\mathrm{P}^1$ and $\mathbb{C}\mathrm{P}^1$ can be identified with the spheres $S^1$ and $S^2$ respectively, and they carry reduced tautological line bundles whose Thom spectra can be identified with $\Susp^{-1} \Susp^\infty \mathbb{R}\mathrm{P}^2$ and $\Susp^{-2} \Susp^\infty \mathbb{C}\mathrm{P}^2$, which themselves are the Moore spectra $M_0(2\iota)$ and $M_0(\eta)$. (Of course, $\mathbb{R}\mathrm{P}^1$ doesn't carry an $H\mathbb{Z}$-orientable bundle, but that's not super important to me that this example is captured by an answer --- it's just motivation.) The proofs of these that I know (from Atiyah's Thom Complexes) don't go in a way that lend themselves to generalization along the lines of the highlighted question, though.

Now the more serious question: I'm much more interested to know what happens in the case where $X$ is a cell complex with more than one cell. Toward that end:

What information about the attaching maps of the cells of $\Susp^\infty X$ and about the action of the map $\phi$ is necessary to describe the attaching maps of the cells of $T(\phi)$?

Here's a guess at a way of phrasing a response to this question: one possible way of encoding the attaching maps of $\Susp^\infty_+ X$ is to claim perfect knowledge of the differentials in the Atiyah-Hirzebruch spectral sequence $$H_*(X; \pi_* \S) \Rightarrow \pi_* \Susp^\infty_+ X.$$ So my question could be interpreted as: what do I need to know about the map $\phi$ to use it to twist the differentials in that Atiyah-Hirzebruch spectral sequence to study the spectral sequence $$H_*(X; \pi_* \S) \cong H_*(T(\phi); \pi_* \S) \Rightarrow \pi_* T(\phi)$$ instead? (Note: the isomorphism in this last line is where I finally use the $H\Z$-orientable hypothesis. I think it's required to begin comparing cell structures using these methods.)

• Eric, the answer to your questions is "yes" (and if you use the cellular filtration, you can get a spectral sequence that works even in the non-$HZ$-orientable case, but it starts at $E_1$). What source for a definition of Thom spectra is comfortable? Jul 30, 2013 at 17:34
• I'm happy with your favorite definition of a Thom spectrum or with the definition that makes computing the bottom pieces of this spectral sequence clearest in toy examples. I'll probably be happier still, though, if we can avoid phrases like "$\infty$-colimit of an $X$-indexed diagram". Jul 30, 2013 at 18:13
• Some short discussion of this question and Tyler's answer in homotopy theory chat: chat.stackexchange.com/transcript/message/10575569#10575569 Aug 1, 2013 at 2:09
• Turns out Mosher did some calculations along these lines in the 60s: sciencedirect.com/science/article/pii/0040938368900268 . May 19, 2014 at 16:43
• The earliest result I know in this direction is Milnor's Lemma 1 (for which he says he'll give the easy proof in a later paper) for the case X = S^n in "On the Whitehead homomorphism J". Bull. Amer. Math. Soc. 64 1958 79–82. Feb 27, 2015 at 19:45

EDIT: Let's try and turn this disaster into something with a moderate amount of workability by making some simplifying assumptions. I wouldn't recommend reading the previous version.

First, let's assume $X$ is a finite cell complex, coming from a filtration by subcomplexes $X^{(d)}$. (We'll allow ourselves the possibility of $X^{(d)}$ being formed from $X^{(d-1)}$ by attaching multiple cells of any necessary dimension.) The Thom spectrum functor of an infinite complex is (at the least, equivalent to) the colimit of the Thom spectra on its finite subcomplexes, and since our questions are about cellular filtrations we're not losing important details by restricting to finite objects.

Let's also assume for simplification that we have a topological group $G$ that acts by based maps on $S^n$, and a map $\alpha: X \to BG$ classifying a principal $G$-bundle $P \to X$. From this, we get a Thom space $X^\alpha = S^n \wedge_G P_+$. Let's assume for simplicity that we're talking about the associated virtual bundle of degree zero so that the Thom spectrum is the suspension spectrum $\Sigma^{-n} \Sigma^\infty X^\alpha$.

For each cell $D^k \to X$, choose a lift $D^k \to P$, which extends uniquely to a $G$-map $G \times D^k \to P$. These give $P$ a $G$-equivariant cell structure. If $P^{(d)}$ is the preimage of $X^{(d)}$, then $P^{(d)}/P^{(d-1)}$ is a wedge of copies of $G_+ \wedge S^{j_i}$. Every new cell of $P^{(d+1)}$ is attached via an equivariant boundary map $G \times S^{k-1} \to P^{(d)}$. On Thom spaces we get a filtration by $S^n \wedge_G P^{(d)}_+$, with the quotients in the filtration being wedges of $S^{n+k}$. In particular, the attaching maps between different layers are determined by $G$-equivariant maps $G_+ \wedge S^{k-1} \to \bigvee G_+ \wedge S^{j_i}$; taking the Thom space turns this into maps $S^{n+k-1} \to \bigvee S^{n+j_i}$.

Now take this whole picture and take suspension spectra.

We have a suspension spectrum $Y = \Sigma^\infty P_+$, which has an action of the group algebra $\mathbb{S}[G] = \Sigma^\infty G_+$. Taking suspension spectra preserves colimits, so that we can translate space-level identifications. We find that the suspension spectrum of $X$ is $\mathbb{S} \wedge_{\mathbb{S}[G]} Y$ where $G$ acts trivially on $\mathbb{S}$, while the Thom spectrum is $S^0 \wedge_{\mathbb{S}[G]} Y$. Here we view $S^0 = \Sigma^{-n} \Sigma^\infty S^n$ as being equivalent to $\mathbb{S}$, but inheriting its appropriate $G$-action.

We also have a filtration of $Y$ by submodules $Y^d = \Sigma^\infty_+ P^{(d)}_+$. The layers $Y^d / Y^{d-1}$ are wedges $\bigvee \mathbb{S}[G] \wedge S^{j_i}$. Let's use this filtration to build the "resolution" $$Y^0 \leftarrow \Sigma^{-1} Y^1/Y^0 \leftarrow \Sigma^{-2} Y^2/Y^0 \leftarrow \cdots.$$ On homotopy groups, this gives a chain complex $C$ with $C_d = \pi_* (\Sigma^{-d} Y^d / Y^{d-1})$ of (graded) free modules over $\pi_* \mathbb{S}[G]$. It has the following properties.

• It gives a spectral sequence with $E_2$-term $H_{**}(C)$ converging to $\pi_* Y$.

• Smashing over $\mathbb{S}[G]$ with $\mathbb{S}$ tensors this chain complex over $\pi_* \mathbb{S}[G]$ with $\pi_* \mathbb{S}$. This chain complex comes from the cellular filtration of $\Sigma^\infty X_+$, and its homology is the $E_2$-term of the cellular spectral sequence for the homotopy of $\Sigma^\infty X_+$. (Standard techniques allow us to say there's a spectral sequence starting with $Tor^{\pi_* \mathbb{S}[G]}_{***}(\pi_* \mathbb{S}, H_{**}(C))$ and converging to the $E_2$-term.)

• Smashing over $\mathbb{S}[G]$ with $S^0$ tensors this chain complex over $\pi_* \mathbb{S}[G]$ with $\pi_* S^0$ (in this case, $G$ acts nontrivially!). This chain complex comes from the cellular filtration of the Thom spectrum, and its homology is the $E_2$-term of the cellular spectral sequence for the homotopy of $\Sigma^\infty X_+$. (Again, there's a spectral sequence starting with $Tor^{\pi_* \mathbb{S}[G]}_{***}(\pi_* S^0, H_{**}(C))$ and converging to the $E_2$-term.)

The upshot of this is that, in terms of attaching maps, both the suspension spectrum of $X$ and the Thom spectrum of this degree-zero virtual bundle have cellular filtrations with the same cells. The relation between the adjacent attaching maps is this: they have common lifts from $\pi_* \mathbb{S}$ to $\pi_* \mathbb{S}[G]$, but one specialization has $G$ act trivially and one has $G$ act via its action on spheres.

Examples help.

If $G = O(1)$, then $\pi_* \mathbb{S}[G]$ is literally the group algebra over $\pi_* \mathbb{S}$ on $\mathbb{Z}/2$. The principal bundle over your space is a chain complex of free modules over this group algebra (and if $X = \mathbb{RP}^\infty$, it is a resolution of $\pi_* \mathbb{S}$). The two specializations make the generator ${-1}$ acts trivially and nontrivially respectively.

If $G = U(1)$, then $\pi_* \mathbb{S}[G]$ is even more interesting: it is a graded-commutative ring $\pi_* \mathbb{S}[d] / (d^2 = \eta d)$, where $|d| = 1$. The element $d$ comes from the canonical element in $\pi_1 U(1)$. The principal bundle again gives us a chain complex of free modules over this (and if $X = \mathbb{CP}^\infty$, it is the resolution of $\pi_* \mathbb{S}$ by free modules of rank $1$ given by alternately multiplying by $d$ and $d + \eta$). The two specializations make the generator act by $0$ and by $\eta$ respectively.

If $X = S^k$, then we can get into gory detail. We can use the cell decomposition with one zero-cell and one $k$-cell; the principal bundle is then formed by attaching $G \times D^k$ to $G$ along an equivariant map $G \times S^{k-1} \to G$ (so we need to be careful when $k=1$); without loss of generality we can make a choice sending the basepoint to the identity. This is literally an element in $\alpha \in \pi_{k-1}(G) = \pi_k BG$. Take suspension spectra and unwind: our chain complex is a two-term complex $\pi_* \mathbb{S}[G] \leftarrow \Sigma^{k-1} \pi_* \mathbb{S}[G]$. The map if $k \gt 1$ is given by (the image of) our element $\alpha$, while if $k = 1$ it is multiplication by the difference $1 - \alpha$. The two specializations make $\alpha$ go to zero (recovering the suspension of $S^k$) and to the appropriate image of $\pi_{k-1} G \to \pi_{k-1} Map(S^n,S^n) \to \pi_{k-1} \Omega^n S^n$ (recovering the cone on $\alpha$ or $1-\alpha$).

Secondary maps, like the $d_2$ in the spectral sequences I mentioned, are harder; suddenly you are trying to relate attaching maps for cells of two different objects. There may be some last bit of voodoo possible by taking these, which are basically Toda brackets, and trying to lift these to something equivariant. Seems like an interesting problem.

• Hooray! By refreshing this every few hours, I made it finish! Sep 28, 2013 at 4:46