Contents

# Contents

## Idea

The Thom space $Th(V)$ of a real vector bundle $V \to X$ over a topological space $X$ is the topological space obtained by first forming the disk bundle $D(V)$ of (unit) disks in the fibers of $V$ (with respect to a metric given by any choice of orthogonal structure) and then identifying to a point the boundaries of all the disks, i.e. forming the quotient topological space by the unit sphere bundle $S(V)$:

$Th(V) \coloneqq D(V)/S(V) \,.$

(N.B.: this is a quotient of the total spaces of the bundles taken in $Top$, not a bundle quotient in $Top_{/V}$.)

This is equivalently the mapping cone

$\array{ S(V) &\longrightarrow& * \\ {}^{\mathllap{p}}\downarrow &\swArrow& \downarrow \\ X & \longrightarrow & Th(V) }$

in Top of the sphere bundle of $V$. Therefore more generally, for $P \to X$ any n-sphere-fiber bundle over $X$ (spherical fibration), its Thom space is the the mapping cone

$\array{ P &\longrightarrow& * \\ {}^{\mathllap{p}}\downarrow &\swArrow& \downarrow \\ X & \longrightarrow & Th(P) }$

of the bundle projection.

For $X$ a compact topological space, $Th(V)$ is a model for the one-point compactification of the total space $V$.

The Thom space of the rank-$n$ universal vector bundle over the classifying space $B O(n)$ of the orthogonal group is usually denoted $M O(n)$. As $n$ ranges, these spaces form the Thom spectrum.

## Definition

###### Definition

Let $X$ be a topological space and let $V \to X$ be a vector bundle (topological vector bundle) over $X$ of rank $n$, which is associated to an O(n)-principal bundle. Equivalently this means that $V \to X$ is the pullback of the universal vector bundle $E_n \to B O(n)$ over the classifying space. Since $O(n)$ preserves the metric on $\mathbb{R}^n$, by definition, such $V$ inherits the structure of a metric space-fiber bundle. With respect to this structure:

1. the unit disk bundle $D(V) \to X$ is the subbundle of elements of norm $\leq 1$;

2. the unit sphere bundle $S(V)\to X$ is the subbundle of elements of norm $= 1$;

$S(V) \overset{i_V}{\hookrightarrow} D(V) \hookrightarrow V$;

3. the Thom space $Th(V)$ is the cofiber (formed in Top (prop.)) of $i_V$

$Th(V) \coloneqq cofib(i_V)$

canonically regarded as a pointed topological space.

$\array{ S(V) &\overset{i_V}{\longrightarrow}& D(V) \\ \downarrow &(po)& \downarrow \\ \ast &\longrightarrow& Th(V) } \,.$

If $V \to X$ is a general real vector bundle, then there exists an isomorphism to an $O(n)$-associated bundle and the Thom space of $V$ is, up to based homeomorphism, that of this orthogonal bundle.

###### Remark

If the rank of $V$ is positive, then $S(V)$ is non-empty and then the Thom space is the quotient topological space

$Th(V) \simeq D(V)/S(V) \,.$

However, in the degenerate case that the rank of $V$ vanishes, hence the case that $V = X\times \mathbb{R}^0 \simeq X$, then $D(V) \simeq V \simeq X$, but $S(V) = \emptyset$. Hence now the pushout defining the cofiber is

$\array{ \emptyset &\overset{i_V}{\longrightarrow}& X \\ \downarrow &(po)& \downarrow \\ \ast &\longrightarrow& Th(V) \simeq X_* } \,,$

which exhibits $Th(V)$ as the coproduct of $X$ with the point, hence as $X$ with a basepoint freely adjoined.

$Th(X \times \mathbb{R}^0) = Th(X) \simeq X_+ \,.$

## Properties

### Homotopy-theoretic nature

###### Proposition

Let $V \to X$ be a vector bundle over a CW-complex $X$. Then the Thom space $Th(V)$ (def. ) is equivalently the homotopy cofiber (def.) of the inclusion $S(V) \longrightarrow D(V)$ of the sphere bundle into the disk bundle.

###### Proof

The Thom space is defined as the ordinary cofiber of $S(V)\to D(V)$. Under the given assumption, this inclusion is a relative cell complex inclusion, hence a cofibration in the classical model structure on topological spaces (thm.). Therefore in this case the ordinary cofiber represents the homotopy cofiber (def.).

The equivalence to the following alternative model for this homotopy cofiber is relevant when discussing Thom isomorphisms and orientation in generalized cohomology:

###### Proposition

Let $V \to X$ be a vector bundle over a CW-complex $X$. Write $V \setminus X$ for the complement of its 0-section. Then the Thom space $Th(V)$ (def. ) is homotopy equivalent to the mapping cone of the inclusion $(V \setminus X) \hookrightarrow V$ (hence to the pair $(V,V \setminus X)$ in the language of generalized (Eilenberg-Steenrod) cohomology).

###### Proof

The mapping cone of any map out of a CW-complex represents the homotopy cofiber of that map (exmpl.). Moreover, transformation by (weak) homotopy equivalences between morphisms induces a (weak) homotopy equivalence on their homotopy fibers (prop.). But we have such a weak homotopy equivalence, given by contracting away the fibers of the vector bundle:

$\array{ V\setminus X &\longrightarrow& V \\ {}^{\mathllap{\in W_{cl}}}\downarrow && \downarrow^{\mathrlap{\in W_{cl}}} \\ S(V) &\hookrightarrow& D(V) } \,.$

### Behaviour under direct sum of vector bundles

###### Proposition

Let $V_1,V_2 \to X$ be two real vector bundles. Then the Thom space (def. ) of the direct sum of vector bundles $V_1 \oplus V_2 \to X$ is expressed in terms of the Thom space of the pullback bundles $V_2|_{D(V_1)}$ and $V_2|_{S(V_1)}$ of $V_2$ to the disk/sphere bundle of $V_1$ as

$Th(V_1 \oplus V_2) \simeq Th(V_2|_{D(V_1)})/Th(V_2|_{S(V_1)}) \,.$
###### Proof

Notice that

1. $D(V_1 \oplus V_2) \simeq D(V_2|_{Int D(V_1)}) \cup S(V_1)$;

2. $S(V_1 \oplus V_2) \simeq S(V_2|_{Int D(V_1)}) \cup Int D(V_2|_{S(V_1)})$.

(Since a point at radial distance $r$ in $V_1 \oplus V_2$ is a point at radius $r_1 \leq r$ in $V_2$ and a point at radius $\sqrt{r^2 - r_1^2}$ in $V_1$.)

###### Proposition

For $V$ a vector bundle then the Thom space (def. ) of $\mathbb{R}^n \oplus V$, the direct sum of vector bundles with the trivial rank $n$ vector bundle, is homeomorphic to the smash product of the Thom space of $V$ with the $n$-sphere (the $n$-fold reduced suspension).

$Th(\mathbb{R}^n \oplus V) \simeq S^n \wedge Th(V) = \Sigma^n Th(V) \,.$
###### Proof

Apply prop. with $V_1 = \mathbb{R}^n$ and $V_2 = V$. Since $V_1$ is a trivial bundle, then

$V_2|_{D(V_1)} \simeq V_2\times D^n$

(as a bundle over $X\times D^n$) and similarly

$V_2|_{S(V_1)} \simeq V_2\times S^n \,.$
###### Remark

Prop. implies that for every vector bundle $V$ the sequence of spaces $Th(\mathbb{R}^n \oplus V)$ forms a suspension spectrum: this is the Thom spectrum of $V$.

###### Example

By prop. and remark the Thom space (def. ) of a trivial vector bundle of rank $n$ is the $n$-fold suspension of the base space

\begin{aligned} Th(X \times \mathbb{R}^n) & \simeq S^n \wedge Th(X\times \mathbb{R}^0) \\ & \simeq S^n \wedge (X_+) \end{aligned} \,.

Therefore a general Thom space may be thought of as a “twisted reduced suspension”, with twist encoded by a vector bundle (or rather by its underlying spherical fibration). See at Thom spectrum – For infinity-module bundles for more on this.

Correspondingly the Thom isomorphism for a given Thom space is a twisted version of the suspension isomorphism.

###### Proposition

For $V_1 \to X_1$ and $V_2 \to X_2$ to vector bundles, let $V_1 \boxtimes V_2 \to X_1 \times X_2$ be the direct sum of vector bundles of their pullbacks to $X_1 \times X_2$. The corresponding Thom space is the smash product of the individual Thom spaces:

$Th(V_1 \boxtimes V_2) \simeq Th(V_1) \wedge Th(V_2) \,.$
###### Remark

Prop. induces on the Thom spectra of remark the structure of ring spectra.

### CW-structure

If the base space of the vector bundle carries the structure of a CW-complex, then its Thom space (def. ) canonically inherits the structure of a CW-complex, too:

###### Lemma

Let $V \to X$ be a vector bundle of rank $n \geq 1$. over a CW-complex $X$.

Then $Th(V)$ has the structure of a CW-complex with

1. $S(E)/S(E)$ the only 0-cell

2. precisely one $(n+k)$-cell $D^{k+n}\to Th(V)$ for each $k$-cell $D^k \to X$ of $X$, given as the pullback

$\array{ D^{k+n} &\longrightarrow& D(V) &\longrightarrow& D(V)/S(V) = Th(V) \\ \downarrow &(pb)& \downarrow \\ D^k &\longrightarrow& X } \,.$

(e.g. Cruz 04, lemma 6)

In particular, $Th(V)$ has a single $n$-cell and an $(n+1)$-cell for each 1-cell of $X$. There are no cells in $Th(C)$ between dimension $0$ and $n$. The cellular boundary of an $(n+1)$-cell is 0 if $V$ is orientable over the corresponding 1-cell of $X$, and it is twice the $n$-cell in the opposite case. Thus $H^n(Th(V);\mathbb{Z})$ is $\mathbb{Z}$ if $V$ is orientable and $0$ if $V$ is non-orientable. In the orientable case a generator of $H^n(Th(V);\mathbb{Z})$ restricts to a generator of $H^n(S^n;\mathbb{Z})$ in the “fiber” $S^n$ of $Th(V)$ over the 0-cell of $X$, hence the same is true for all the “fibers” $S^n$ and so one has a Thom class.

### Cohomology

###### Remark

Given a vector bundle $V \to X$ of rank $n$, then the reduced ordinary cohomology of its Thom space $Th(V)$ (def. ) vanishes in degrees $\lt n$:

$\tilde H^{\bullet \lt n}(Th(V)) \simeq H^{\bullet \lt n}(D(V), S(V)) \simeq 0 \,.$
###### Proof

Consider the long exact sequence of relative cohomology (here)

$\cdots \to H^{\bullet-1}(D(V)) \overset{i^\ast}{\longrightarrow} H^{\bullet-1}(S(V)) \longrightarrow H^\bullet(D(V), S(V)) \longrightarrow H^{\bullet}(D(V)) \overset{i^\ast}{\longrightarrow} H^{\bullet}(S(V)) \to \cdots \,.$

Since the cohomology in degree $k$ only depends on the $k$-skeleton, and since for $k \lt n$ the $k$-skeleton of $S(V)$ equals that of $X$, and since $D(V)$ is even homotopy equivalent to $X$, the morhism $i^\ast$ is an isomorphism in degrees lower than $n$. Hence by exactness of the sequence it follows that $H^{\bullet \lt n}(D(V),S(V)) = 0$.

## Examples

### Thom space of universal line bundle

Due to:

Further discussion: