---- $X$ is a [[topological space]]. > [!theorem] Theorem. ([[The Thom isomorphism theorem]]) > Let $R$ be a (say, [[commutative ring|commutative]]) [[ring]]. Let $E \xrightarrow{\pi}X$ be a $d$-dimensional [[vector bundle]], and $\{ \varepsilon_{x} \}_{x \in X}$ be an $R$-[[orientation of a vector bundle|orientation]] of $E$. Put $E^{\sharp}:=E - s_{0}(X)$. > > Then: > > 1. $H^{i}(E, E^{\sharp};R)=0$ for $i<d$; > 2. There is a unique class $u_{E} \in H^{d}(E, E^{\sharp};R)$ which restricts[^1] to $\varepsilon_{x}$ on each fiber. This is known as the **Thom class** of $(E, \{ \varepsilon_{x} \})$. > 3. The map $\Phi$ given by the [[relative cup product|composition]] $H^{i}(X; R) \xrightarrow[\simeq]{\pi^{*}}H^{i}(E;R) \xrightarrow{- \smile u_{E}} H^{i+d}(E, E^{\sharp}; R)$ > is an [[isomorphism]]. > > > [!note] Remark. > > Though the proof does construct $u_{E}$, what matters far more is that it exists and is unqiue. > ^note > > [!definition] Definition. (Euler class) > We define the **Euler class** $e(E) \in H^{d}(X; R)$ of the [[vector bundle]] $E$ to be the image of $u_{E}$ under the composition $H^{d}(E, E^{\sharp}; R) \xrightarrow{q^{*}} H^{d}(E, R) \xrightarrow[\simeq]{s_{0}^{*}} H^{d}(X; R).$ This is an example of a **characteristic class**, which is a cohomology class related to an oriented vector bundle that behaves nicely under pullback. Indeed, given a [[vector bundle]] $E \xrightarrow{\pi}X$ and a map $f:Y \to X$, we can form a [[pullback of a vector bundle|pullback]] > > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRADMA9AKgAIBREAF9S6TLnyEUARnJVajFm0Eix2PASKzp8+s1aIQADWGiQGdZKJkd1PUsMBNYfJhQA5vCKh2AJwgAtkiyIDgQSABMdooGIAA6cWhYpj7+QYgAzNRhkdH6bOwg1Ax0AEYwDAAK4hpSIL5Y7gAWOCkcaUhkoeGZeQ4cPLwJSUUgJeVVNVaGDc2tqu2Bndk9IfaxCU10OMDsQi5CQA > \begin{tikzcd} > f^* E \arrow[d, "f^* \pi"'] \arrow[r, "\hat{f}"] & E \arrow[d, "\pi"] \\ > Y \arrow[r, "f"'] & X > \end{tikzcd} > \end{document} > ``` > This pullback respects the Euler class[^2]: $e\big( f^{*}(E) \big)=f^{*}e(E) \in H^{d}(Y ; R).$ > > > [!basicexample] > Since [[vector subbundle|subbundles]] are [[pullback of a vector bundle|pullbacks]], this implies that the Euler class of a subbundle is gotten merely by restricting the Euler class. ^definition > [!basicproperties] Properties of $u_{E}$. > - $u_{E} \smile u_{E}=\pi^{*}\big( e(E)\big) \smile u _E \in H^{2d}(E, E^{\sharp};R)$ ^properties > [!basicproperties] Properties of $e(E)$. > - (Normalization) [[Euler class obstructs existence of nonvanishing section]]. This is one of the characteristic class axioms; another is the 'commuting with pullbacks' property above. > - [[the Euler class of an odd-rank oriented vector bundle is torsion]] ^properties > [!proof]- Proof. ([[The Thom isomorphism theorem]]) > (Suppressing $R$ from the notation for readability.) > > **The trivialized case.** > > We start with the case where $E$ is a trivial vector bundle, then patch the result up for a general vector bundle. > > > > So suppose $E = X \times \mathbb{R}^{d}$. Recall $H^{*}(\mathbb{R}^{d}, \mathbb{R}^{d}-\{ 0 \})=\begin{cases} > R & *=d \\ > 0 & * \neq d. > \end{cases}$ > In particular, the $R$-[[module|modules]] are [[free module|free]] and a ([[relative singular homology|relative]] version of) [[Kunneth Theorem for singular cohomology|Kunneth's theorem]][^3] tells [[cross product on cohomology|the map]] $\times: H^{*}(X) \otimes_{R} H^{*}(\mathbb{R}^{d}, \mathbb{R}^{d}-\{ 0 \}) \xrightarrow{\cong}H^{*}(\overbrace{ X \times \mathbb{R}^{d} }^{ E }, \overbrace{ X \times (\mathbb{R}^{d}-\{ 0 \}) }^{ E^{\sharp} })$ is an [[isomorphism]] of [[graded ring|graded rings]], where in grading $i$ there's an isomorphism of $R$-[[module|modules]] $\bigoplus_{k+\ell=i}H^{k}(X) \otimes_{R} H^{\ell}(\mathbb{R}^{d}, \mathbb{R}^{d}-\{ 0 \}) \xrightarrow{\cong} H^{i}(E, E^{\sharp}).$ > > The claim of the Thom isomorphism theorem now follows immediately: > 1. If $i < d$, then $\ell \leq i <d$ and $H^{\ell}(\mathbb{R}^{d}, \mathbb{R}^{d}-\{ 0 \})=H^{\ell}(\mathbb{S}^{d})=0$ in each summand on the LHS. Hence the RHS is zero: $H^{i}(E, E^{\sharp})=0$. > 2. The only nonvanishing summand for $H^{d}(E, E^{\sharp})$ is when $k=0$, $\ell=d$. So in this case $H^{0}(X) \otimes_{R} H^{d}(\mathbb{R}^{d}, \mathbb{R}^{d}-\{ 0 \}) \cong H^{d}(E, E^{\sharp}).$ > This defines and determines the Thom class $u_{E} \in H^{d}(E, E^{\sharp})$ to be $u_{E}=$ ? > 3. He didn't say anything here? > > > > **The general case.** See handwritten notes. ---- #### [^1]: By 'restricts', one means the following. For $x \in X$, we know the [[topological pair|map of pairs]] $(E_{x}, E_{x}^{\sharp}) \hookrightarrow (E, E^{\sharp})$ [[relative singular homology|induces]] a morphism $H^{d}(E, E^{\sharp}; R) \to H^{d}(E_{x}, E_{x}^{\sharp};R)$ on [[singular cohomology|cohomology]]. By 'restricts' we mean that this morphism sends $u_{E}$ to $\varepsilon_{x}$, for each $x \in X$. [^2]: Indeed, since we have a fiberwise [[isomorphism]] $(f^{*}E)_{y} \cong E_{f(y)}$, an $R$-[[orientation of a vector bundle|orientation]] for $E$ induces one for $f^{*}E$, and we know $f^{*}(u_{E})=u_{f^{*}E}$ by uniqueness of the Thom class. [^3]: Actually, this isn't quite the relative Kunneth we showed. It is really regular Kunneth plus an application of the [[five lemma]]. But don't worry too much about that. ----- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```