Thom-Splitting: Application of steenrod squares-1

Introduction

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In this blog, we explore how Steenrod squares and the Steenrod algebra can be used to study problems in homotopy theory. Our main objective is to understand the Thom splitting, which states that the Thom spectrum $MO$ is weakly homotopy equivalent to the wedge sum
$$ \bigvee_i \Sigma^i H\mathbb{Z}_2. $$
Here, $\mathbb{Z}_2$ refers to the field with two elements, not the 2-adic integers. Our goal is to construct a map $$ f: MO \to \bigvee_i \Sigma^i H\mathbb{Z}_2, $$ and then show that this map is a weak homotopy equivalence.

Outline of the Argument

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We will proceed in the following steps:

• First, compute $H^{\ast}(BO)$, where all (co)homology is taken with mod 2 coefficients.
• Dualize the cohomology ring to obtain a graded $Z_2$-algebra structure on $H_{\ast}(BO)$, which is free.
• Use the Thom isomorphism to identify $H_{\ast}(MO) \cong H_{\ast}(BO)$ as graded $\mathbb{Z}_2$-algebras.
• Show that $H_{\ast}(MO)$ is a cofree graded comodule over the dual Steenrod algebra $\mathcal{A}^{\ast}$.
• Dualizing again, we deduce that $H^{\ast}(MO)$ is a free graded module over the Steenrod algebra $\mathcal{A}$.

Constructing the Map

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From this structure, it follows that the graded homotopy group $$ [MO, H\mathbb{Z}_2] $$ is a free graded module over $$ [H\mathbb{Z}_2, H\mathbb{Z}_2]. $$ Let ${\alpha_i}$ be a basis for this module. Then we obtain a canonical map of spectra: $$ MO \to \prod_i \Sigma^{|\alpha_i|} H\mathbb{Z}_2 \to \bigvee_i \Sigma^{|\alpha_i|} H\mathbb{Z}_2, $$ where the second map is the projection from the product to the wedge.

To show that this map is a weak homotopy equivalence, consider its homotopy cofiber. Since the map induces an isomorphism on mod-2 homology, the homology of the cofiber is trivial. By the mod-2 Hurewicz theorem, this implies the homotopy groups of the cofiber are also trivial. Hence, the map is a weak equivalence.

Applications

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The Thom splitting enables a straightforward computation of the ring $\pi_{\ast}(MO)$. Furthermore, by the Thom–Pontryagin construction, we know that $\pi_{\ast}(MO)$ is isomorphic to the ring of unoriented cobordism classes. Thus, the splitting allows us to explicitly compute this important cobordism ring.

What is $MO$

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Simply $MO$ is the Thom spectra corresponding to $BO$.

Little bit about Thom Spectra

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Let, $V \to X$ be a vector bundle over $X$(say it is a CW complex) and $S^V$ be the corresponding sphere bundle with the infinite section $s:X \to S^V$. Now we define Thom space, $$Th(V):= \text{Cofiber}({s})$$

Example: The Thom space of the trivial bundle $X \times R^n \to X$ is $\Sigma^n X_{+}$.

We call a vector bundle orientable if there exist $\alpha \in H^{\dim V}(Th(V))$ such that it pulled back to a generator of $S^{\dim V}$ under the following pullback square.

If $V \to X$ and $W \to Y$ are two vector bundles we define $V \boxtimes W$ to be the the product-vector bundle on $X \times Y$. Now, $$Th(V \boxtimes W)= Th(V)\wedge Th(W)$$

With this we are ready to define $MO$. Recall that $BO(n)$ is the classifying space of all $n$-ranked real vector bundle. Let, $\gamma_n:EO(n)\to BO(n)$ be the uviversal tautological vector bundle. Then we define, $MO_n := Th(\gamma_n)$. Now note that, we have the following pull back diagram of vector bundles,

This will give us a map from $Th(\gamma_n\oplus \varepsilon) \to Th(\gamma_{n+1})$, here $\varepsilon$ is the trivial line bundle. This is basically the structure map of the $MO$ spectra.

General construction: In general if we have a map from a spectra $f:X\to BO$ then define $Th(f)_n = Th(f_n^{\ast}\gamma_n)$. For example :

• The map $BSO \to BO$ gives rise to sthe spectrum $MSO$ whose homotopy ring is the un-orented cobordsim ring.
• The map from $BU \to BO$ gives rise to $MU$ whose homotopy ring is the oriented cobordism ring.
• The map from $\ast \to BO$ gives rise to the sphere spectra whose homotopy ring gives rise to framed cobordism ring.

Towards the main proof

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Recall that, $H_{\ast}(E;R) = [S,E\wedge HR]$ and $H^{\ast}(E;R)= [E, HR]$. Where $E$ is a spectra, $HR$ is the eilenberg maclane spectra and $S$ is the sphere spectra. Re call that $HR$ is a ring spectrum, if $E$ is a ring spectrum then $H_{\ast}(E;R)$ can be given a ring structure using the composition of following maps:

$$f\wedge g : S \to E \wedge HR \wedge E \wedge HR \simeq E \wedge E \wedge HR \wedge HR \to E \wedge HR$$ last map is smash of the multiplication maps of the ring spectrum.

At the begining of the blog we have talked about this ring and the cohomology ring.

Step 1: $H_{\ast}(MO)$

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For the rest of the blog we will use $F$ for $Z/2Z$.

Result 1: As rings $H^{\ast}(BO(n);F)\simeq F[w_1,\cdots, w_n]$ where $w_i \in H^n(BO(n);F)$ is the whitney class of $BO(n)$.

To prove this consider the map $f : BO(1)^n \to BO(n)$ under $\gamma_1 \cdots \times \gamma_1 \to \gamma_n$. Note that $f^{\ast}$ is map from $H^{\ast}(BO(n))\to F[X]^{\otimes n} = F[x_1,\cdots, x_n]$

By the naturality of w.s-classes we get that $f^{\ast}(w(\gamma_n)) = w(\gamma_1 \times \cdots \gamma_1)$. Note that, $w(\gamma_1\times \cdots \gamma_1)= \sum s_i(x_1,\cdots,x_n)$ here $s_i$ means $i$-th symmetric polynomial. Thus the map $f^{\ast}$ is injective. Thus $H^{\ast}(BO(n))$ is isomorphic to the subring generated by the symmetric polynomial generated by $x_i$ which are precisely the whitney classes so the theorem is proved.

Now we will dualize the above ap $f^{\ast}$ to get a map from $H_{\ast}(BO(1)^n) \to H_{\ast}(BO(n))$. Let, $y_i$ be the dual if $x^i$ then the dualize map is

$$F[y_1,\cdots]^{\otimes n} \to H_{\ast}(BO(n))$$

surjective where we have quoteinted by the symmetric polynomial. Now colimit commutes with homology (as it does with homotopy groups). Thus by taking colimit we get, $$H_{\ast}(BO) = \lim_n F[y_{i_1}\cdots y_{i_n}: i_1\leq \cdots \leq i_n] = F[y_1,y_2,\cdots]$$

Thom isomorphism: If $V \to X$ is a vector bundle of rank $n$ the following two spectra are weak equivalent in the category of specta $$HF \wedge Th(V) \equiv HF \wedge \Sigma^n X_{+} $$

Here we are not bothered about orientation. So what we get using Thom isomorphism is the following :

$$[S, HF \wedge MO(n)] = [S, HF \wedge \Sigma^n BO(n)_{+}]$$ Taking the colimit over $n$ we get,

$$H_{\ast}(MO;F) \simeq F[\tilde{y_i}: i \in \mathbb{N}]$$ here $\tilde{y}_i$ corresponds to the image of $y_i$ under the Thom isomorphism. Dualizing the above thing we get $$H^{\ast}(MO;F) = F[p_1,\cdots]$$ Here $p_i$ are duals of $\tilde{y}_i$.

Step 2: $H^{\ast}(MO;F)$ as $\mathcal{A}$-module

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For this step we need to understand what $\tilde{y}_i$ are.