11.4 Pontrjagin and Euler classes

In the present section we will be concerned with the characteristic classes associated with a real vector bundle.

11.4.1 Pontrjagin classes

Let $E$ be a real vector bundle over an $m$ -dimensional manifold $M$ with ${\dim }_{\mathbb{R}}E=k$ . If $E$ is endowed with the fibre metric, we may introduce orthonormal frames at each fibre. The structure group may be reduced to $\mathrm{O}(k)$ from $\mathrm{GL}(k,\mathbb{R})$ . Since the generators of $\mathfrak{o}(k)$ are skew symmetric, the field strength $\mathcal{T}$ of $E$ is also skew symmetric. A skew-symmetric matrix $A$ is not diagonalizable by an element of a subgroup of $\mathrm{GL}(k,\mathbb{R})$ . It is, however, reducible to block diagonal form as

$$\begin{array}{rl}& A\to \left(\begin{array}{ccccc}0& {\lambda }_1& & & 0\\ -{\lambda }_1& 0& & & \\ & & 0& {\lambda }_2& \\ & & -{\lambda }_2& 0& \\ 0& & & & \ddots \end{array}\right)\\ & \to \left(\begin{array}{cccc}\mathrm{i}{\lambda }_1& & & \\ & -\mathrm{i}{\lambda }_1& 0& \\ & & \mathrm{i}{\lambda }_2& \\ & & & -\mathrm{i}{\lambda }_2\\ & 0& & \ddots \end{array}\right)\end{array}\text{\hspace{1em}(11.69)}$$

where the second diagonalization is achieved only by an element of $\mathrm{GL}(k,\mathbb{C})$ . If $k$ is odd, the last diagonal element is set to zero. For example, the generator of $\mathfrak{o}(3)=\mathfrak{so}(3)$ generating rotations around the $z$ -axis is

$$T_z=\left(\begin{array}{ccc}0& 1& 0\\ -1& 0& 0\\ 0& 0& 0\end{array}\right).$$

The total Pontrjagin class is defined by

$$p(\mathcal{F})\equiv \det \left(I+\frac{\mathcal{F}}{2\pi }\right).\text{\hspace{1em}(11.70)}$$

From the skew symmetry ${\mathcal{F}}^{\mathrm{t}}=-\mathcal{F}$ , it follows that

$$\det \left(I+\frac{\mathcal{F}}{2\pi }\right)=\det \left(I+\frac{{\mathcal{F}}^t}{2\pi }\right)=\det \left(I-\frac{\mathcal{F}}{2\pi }\right).$$

Therefore, $p(\mathcal{F})$ is an even function in $\mathcal{F}$ . The expansion of $p(\mathcal{F})$ is

$$p(\mathcal{F})=1+p_1(\mathcal{F})+p_2(\mathcal{F})+\dots \text{\hspace{1em}(11.71)}$$

where $p_j(\mathcal{F})$ is a polynomial of order $2j$ and is an element of $H^{4j}(M;\mathbb{R})$ . We note that $p_j(\mathcal{F})=0$ for either $2j>k=\dim E$ or $4j>\dim M$ .¹

Let us diagonalize $\mathcal{F}/2\pi$ as

$$\frac{\mathcal{F}}{2\pi }\to A\equiv \left(\begin{array}{ccccc}-\mathrm{i}{x}_1& & & & \\ & \mathrm{i}{x}_1& & 0& \\ & & -\mathrm{i}{x}_2& & \\ & 0& & \mathrm{i}{x}_2& \\ & & & & \ddots \end{array}\right)\text{\hspace{1em}(11.72)}$$

where $x_k\equiv -{\lambda }_k/2\pi$ , ${\lambda }_k$ being the eigenvalues of $\mathcal{F}$ . The sign has been chosen to simplify the Euler class defined here. The generating function of $p(\mathcal{F})$ is given by

$$p(\mathcal{F})=\det (I+A)=\prod _{i=1}^{[k/2]}\left(1+x_i^2\right)\text{\hspace{1em}(11.73)}$$

where

$$[k/2]=\to \{\begin{array}{ll}k/2& \text{i f}k\text{i s e v e n}\\ (k-1)/2& \text{i f}k\text{i s o d d}.\end{array}$$

In (11.73) only even powers appear, reflecting the skew symmetry. Each Pontrjagin class is computed from (11.73) as

$$p_j(\mathcal{F})=\sum _{i_1<i_2<\cdots <i_j}^{[k/2]}{x}_{i_1}^2{x}_{i_2}^2\dots x_{i_j}^2.\text{\hspace{1em}(11.74)}$$

To write $p_j(\mathcal{F})$ in terms of the curvature two-form $\mathcal{F}/2\pi$ , we first note that

$$\mathrm{tr}{\left(\frac{\mathcal{F}}{2\pi }\right)}^{2j}=\mathrm{tr}{A}^{2j}=2(-1{)}^j\sum _{i=1}^{[k/2]}{x}_i^{2j}.$$

It then follows that

$$p_1(\mathcal{F})=\sum _i{x}_i^2=-\frac{1}{2}{\left(\frac{1}{2\pi }\right)}^2\mathrm{tr}{\mathcal{F}}^2\text{\hspace{1em}(11.75a)}$$

$$\begin{array}{rl}{p}_2(\mathcal{F})& =\sum _{i<j}{x}_i^2{x}_j^2=\frac{1}{2}\left[{\left(\sum _i{x}_i^2\right)}^2-\sum _i{x}_i^4\right]\\ & =\frac{1}{8}{\left(\frac{1}{2\pi }\right)}^4\left[{\left(\mathrm{tr}{\mathcal{F}}^2\right)}^2-2\mathrm{tr}{\mathcal{F}}^4\right]\end{array}\text{\hspace{1em}(11.75b)}$$

$$\begin{array}{rl}{p}_3(\mathcal{F})& =\sum _{i<j<k}{x}_i^2{x}_j^2{x}_k^2\\ & =\frac{1}{48}{\left(\frac{1}{2\pi }\right)}^6\left[-{\left(\mathrm{tr}{\mathcal{F}}^2\right)}^3+6\mathrm{tr}{\mathcal{F}}^2\mathrm{tr}{\mathcal{F}}^4-8\mathrm{tr}{\mathcal{F}}^6\right]\end{array}\text{\hspace{1em}(11.75c)}$$

$$\begin{array}{rl}& p_4(\mathcal{F})=\sum _{i<j<k<l}{x}_i^2{x}_j^2{x}_k^2{x}_l^2\\ & =\frac{1}{384}{\left(\frac{1}{2\pi }\right)}^8[(\mathrm{tr}{\mathcal{F}}^2{)}^4-12(\mathrm{tr}{\mathcal{F}}^2{)}^2\mathrm{tr}{\mathcal{F}}^4+32\mathrm{tr}{\mathcal{F}}^2\mathrm{tr}{\mathcal{F}}^6\\ & +12{\left(\mathrm{tr}{\mathcal{F}}^4\right)}^2-48\mathrm{tr}{\mathcal{F}}^8]\end{array}\text{\hspace{1em}(11.75d)}$$

中

$$p_{[k/2]}(\mathcal{F})=x_1^2{x}_2^2\dots x_{[k/2]}^2={\left(\frac{1}{2\pi }\right)}^k\det \mathcal{F}.\text{\hspace{1em}(11.75e)}$$

The reader should verify that

$$p(E\oplus F)=p(E)\wedge p(F).\text{\hspace{1em}(11.76)}$$

It is easy to guess that the Pontrjagin classes are written in terms of Chern classes. Since Chern classes are defined only for complex vector bundles, we must complexify the fibre of $E$ so that complex numbers make sense. The resulting vector bundle is denoted by $E^{\mathbb{C}}$ . Let $A$ be a skew-symmetric real matrix. We find that

$$\begin{array}{rl}\det (I+\mathrm{i}A)& =\det \left(\begin{array}{ccccc}1+x_1& & & 0& \\ & 1-x_1& & & \\ & & 1+x_2& & \\ & 0& & 1-x_2& \\ & & & & \ddots \end{array}\right)\\ & =\prod _{i=1}^{[k/2]}(1-x_i^2)=1-p_1(A)+p_2(A)-\dots \end{array}$$

from which it follows that

$$p_j(E)=(-1{)}^j{c}_{2j}\left(E^{\complement }\right).\text{\hspace{1em}(11.77)}$$

Example 11.5

Let $M$ be a four-dimensional Riemannian manifold. When the orthonormal frame $\{{\hat{e}}_{\alpha }\}$ is employed, the structure group of the tangent bundle $TM$ may be reduced to $\mathrm{O}(4)$ . Let $\mathcal{R}=\frac{1}{2}{\mathcal{R}}_{\alpha \beta }{\theta }^{\alpha }\wedge {\theta }^{\beta }$ be the curvature two-form $(\mathcal{R}$ should not be confused with the scalar curvature). For the tangent bundle, it is common to write $p(M)$ instead of $p(\mathcal{R})$ . We have

$$\det \left(I+\frac{\mathcal{R}}{2\pi }\right)=1-\frac{1}{8{\pi }^2}\mathrm{tr}{\mathcal{R}}^2+\frac{1}{128{\pi }^4}[(\mathrm{tr}{\mathcal{R}}^2{)}^2-2\mathrm{tr}{\mathcal{R}}^4].\text{\hspace{1em}(11.78)}$$

Each Pontrjagin class is given by

$$p_0(M)=1\text{\hspace{1em}(11.79a)}$$

$$p_1(M)=-\frac{1}{8{\pi }^2}\mathrm{tr}{\mathcal{R}}^2=-\frac{1}{8{\pi }^2}{\mathcal{R}}_{\alpha \beta }{\mathcal{R}}_{\beta \alpha }\text{\hspace{1em}(11.79b)}$$

$$p_2(M)=\frac{1}{128{\pi }^4}\left[{\left(\mathrm{tr}{\mathcal{R}}^2\right)}^2-2\mathrm{tr}{\mathcal{R}}^4\right]={\left(\frac{1}{2\pi }\right)}^4\det \mathcal{R}.\text{\hspace{1em}(11.79c)}$$

Although $p_2(M)$ vanishes as a differential form, we need it in the next subsection to compute the Euler class.

11.4.2 Euler classes

Let $M$ be a 2l-dimensional orientable Riemannian manifold and let $TM$ be the tangent bundle of $M$ . We denote the curvature by $\mathcal{R}$ . It is always possible to reduce the structure group of $TM$ down to $\mathrm{SO}(2l)$ by employing an orthonormal frame. The Euler class $e$ of $M$ is defined by the square root of the 4l-form $p_l$ ,

$$e(A)e(A)=p_l(A).\text{\hspace{1em}(11.80)}$$

Both sides should be understood as functions of a $2l\times 2l$ matrix $A$ and not of the curvature $\mathcal{R}$ , since $p_1(\mathcal{R})$ vanishes identically. However, $e(M)\equiv e(\mathcal{R})$ thus defined is a $2l$ -form and, indeed, gives a volume element of $M$ . If $M$ is an odd-dimensional manifold we define $e(M)=0$ , see later.

Example 11.6

Let $M=S^2$ and consider the tangent bundle $T{S}^2$ . From example 7.14, we find the curvature two-form,

$${\mathcal{R}}_{\theta \varphi }=-{\mathcal{R}}_{\varphi \theta }={\sin }^2\theta \frac{\mathrm{d}\theta \wedge \mathrm{d}\varphi }{\sin \theta }=\sin \theta \mathrm{d}\theta \wedge \mathrm{d}\varphi$$

where we have noted that $g_{\theta \theta }={\sin }^2\theta$ . Although $p_1(S^2)=0$ as a differential form, we compute it to find the Euler form. We have

$$\begin{array}{rl}{p}_1(S^2)& =-\frac{1}{8{\pi }^2}\mathrm{tr}{\mathcal{R}}^2=-\frac{1}{8{\pi }^2}[{\mathcal{R}}_{\theta \varphi }{\mathcal{R}}_{\varphi \theta }+{\mathcal{R}}_{\varphi \theta }{\mathcal{R}}_{\theta \varphi }]\\ & ={\left(\frac{1}{2\pi }\sin \theta d\theta \wedge d\varphi \right)}^2\end{array}$$

from which we read off

$$e\left(S^2\right)=\frac{1}{2\pi }\sin \theta d\theta \wedge d\varphi .\text{\hspace{1em}(11.81)}$$

It is interesting to note that

$${\int }_{S^2}e\left(S^2\right)=\frac{1}{2\pi }{\int }_0^{2\pi }\mathrm{d}\varphi {\int }_0^{\pi }\mathrm{d}\theta \sin \theta =2\text{\hspace{1em}(11.82)}$$

which is the Euler characteristic of $S^2$ , see section 2.4. This is not just a coincidence. Let us take another convincing example, a torus $T^2$ . Since $T^2$ admits a flat connection, the curvature vanishes identically. It then follows that $e(T^2)\equiv 0$ and $\chi (T^2)=0$ . These are special cases of the Gauss-Bonnet theorem,

$${\int }_{M}e(M)=\chi (M)\text{\hspace{1em}(11.83)}$$

for a compact orientable manifold $M$ . If $M$ is odd dimensional both $e$ and $\chi$ vanish, see (6.39).

In general, the determinant of a $2l\times 2l$ skew-symmetric matrix $A$ is a square of a polynomial called the Pfaffian $\mathrm{Pf}(A)$ ,²

$$\det A=\mathrm{Pf}(A{)}^2.\text{\hspace{1em}(11.84)}$$

We show that the Pfaffian is given by

$$\mathrm{Pf}(A)=\frac{(-1{)}^l}{2^{l}l!}\sum _P\mathrm{sgn}(P)A_{P(1)P(2)}{A}_{P(3)P(4)}\dots A_{P(2l-1)P(2l)}\text{\hspace{1em}(11.85)}$$

where the phase has been chosen for later convenience. We first note that a skew-symmetric matrix $A$ can be block diagonalized by an element of $\mathrm{O}(2l)$ as

$$S^{\mathrm{t}}AS=\Lambda =\left(\begin{array}{cccccc}0& {\lambda }_1& & & & \\ -{\lambda }_1& 0& & & 0& \\ & & 0& {\lambda }_2& & \\ & & -{\lambda }_2& 0& & \\ & & & & \ddots & \\ & 0& & & 0& {\lambda }_l\\ & & & & -{\lambda }_l& 0\end{array}\right).\text{\hspace{1em}(11.86)}$$

It is easy to see that

$$\det A=\det \Lambda =\prod _{i=1}^l{\lambda }_i^2.$$

To compute $\mathrm{Pf}(\Lambda )$ , we note that the non-vanishing terms in (11.85) are of the form $A_{12}{A}_{34}\dots A_{2l-1,2l}$ . Moreover, there are $2^l$ ways of changing the suffices as $A_{ij}\to A_{ji}$ , such as

$$A_{12}{A}_{34}\dots A_{2l-1,2l}\to A_{21}{A}_{34}\dots A_{2l-1,2l}$$

and $l!$ permutations of the pairs of indices, for example,

$$A_{12}{A}_{34}\dots A_{2l-1,2l}\to A_{34}{A}_{12}\dots A_{2l-1,2l}.$$

Hence, we have

$$\mathrm{Pf}(\Lambda )=(-1{)}^l{A}_{12}{A}_{34}\dots A_{2l-1,2l}=(-1{)}^l\prod _{i=1}^l{\lambda }_i.$$

Thus, we conclude that a block diagonal matrix $\Lambda$ satisfies

$$\det \Lambda =\mathrm{Pf}(\Lambda {)}^2.$$

To show that (11.84) is true for any skew-symmetric matrices (not necessarily block diagonal) we use the following lemma, ³

$$\mathrm{Pf}\left(X^{\mathrm{t}}AX\right)=\mathrm{Pf}(A)\det X.\text{\hspace{1em}(11.87)}$$

If $S^{\mathrm{t}}AS=\Lambda$ for $S\in \mathrm{O}(2l)$ , we have $A=S\Lambda S^{\mathrm{t}}$ , hence

$$\mathrm{Pf}\left(S\Lambda S^{\mathrm{t}}\right)=\mathrm{Pf}(\Lambda )\det S=(-1{)}^l\prod _{i=1}^l{\lambda }_i\det S.$$

We finally find $\det A=\mathrm{Pf}(A{)}^2$ for a skew-symmetric matrix $A$ .

Note that $\mathrm{Pf}(A)$ is $\mathrm{SO}(2l)$ invariant but changes sign under an improper rotation $S$ (det $S=-1$ ) of $\mathrm{O}(2l)$ .

Exercise 11.3

Show that the determinant of an odd-dimensional skew-symmetric matrix vanishes. This is why we put $e(M)=0$ for an odd-dimensional manifold.

The Euler class is defined in terms of the curvature $\mathcal{R}$ as

$$\begin{array}{rl}e(M)& =\mathrm{Pf}(\mathcal{R}/2\pi )\\ & =\frac{(-1{)}^l}{(4\pi {)}^{l}l!}\sum _P\mathrm{sgn}(P){\mathcal{R}}_{P(1)P(2)}\dots {\mathcal{R}}_{P(2l-1)P(2l)}.\end{array}\text{\hspace{1em}(11.88)}$$

The generating function is obtained by taking $x_j=-{\lambda }_i/2\pi$

$$e(x)=x_1{x}_2\dots x_l=\prod _{i=1}^l{x}_i.\text{\hspace{1em}(11.89)}$$

The phase $(-1{)}^l$ has been chosen to simplify the RHS.

Example 11.7

Let $M$ be a four-dimensional orientable manifold. The structure group of $TM$ is SO(4), see example 11.5. The Euler class is obtained from (11.88) as

$$e(M)=\frac{1}{2(4\pi {)}^2}{ϵ}^{ijkl}{\mathcal{R}}_{ij}\wedge {\mathcal{R}}_{kl}.\text{\hspace{1em}(11.90)}$$

This is in agreement with the result of example 11.5. The relevant Pontrjagin class is

$$p_2(M)=\frac{1}{128{\pi }^4}[(\mathrm{tr}{\mathcal{R}}^2{)}^2-2\mathrm{tr}{\mathcal{R}}^4]=x_1^2{x}_2^2.$$

Since $e(M)=x_1{x}_2$ , we have $p_2(M)=e(M)\wedge e(M)$ . This is written as a matrix identity,

$$\frac{1}{128{\pi }^4}[(\mathrm{tr}{A}^2{)}^2-2\mathrm{tr}{A}^4]={\left(\frac{1}{2(4\pi {)}^4}{ϵ}^{ijkl}{A}_{ij}{A}_{kl}\right)}^2.$$

11.4.3 Hirzebruch $L$ -polynomial and $\hat{A}$ -genus

The Hirzebruch $L$ -polynomial is defined by

$$\begin{array}{rl}L(x)& =\prod _{j=1}^k\frac{x_j}{\tanh x_j}\\ & =\prod _{j=1}^k\left(1+\sum _{n\ge 1}(-1{)}^{n-1}\frac{2^{2n}}{(2n)!}{B}_n{x}_j^{2n}\right)\end{array}\text{\hspace{1em}(11.91)}$$

where the $B_n$ are Bernoulli numbers, see (11.66). The function $L(x)$ is even in $x_j$ and can be written in terms of the Pontrjagin classes,

$$L(\mathfrak{F})=1+\frac{1}{3}{p}_1+\frac{1}{45}(-p_1^2+7p_2)+\frac{1}{945}(2p_1^3-13p_1{p}_2+62p_3)+\dots \text{\hspace{1em}(11.92)}$$

where $p_j$ stands for $p_j(\mathcal{F})$ . From the splitting principle, we find that

$$L(E\oplus F)=L(E)\wedge L(F).\text{\hspace{1em}(11.93)}$$

The $\hat{A}$ ( $A$ -roof) genus $\hat{A}(\mathcal{F})$ is defined by

$$\begin{array}{rl}\hat{A}(\mathcal{F})& =\prod _{j=1}^k\frac{x_j/2}{\sinh (x_j/2)}\\ & =\prod _{j=1}^k\left(1+\sum _{n\ge 1}(-1{)}^n\frac{\left(2^{2n}-2\right)}{(2n)!}{B}_n{x}_j^{2n}\right).\end{array}\text{\hspace{1em}(11.94)}$$

This is an even function of $x_j$ and can be expanded in $p_j$ . $\hat{A}$ is also called the Dirac genus by physicists. It satisfies

$$\hat{A}(E\oplus F)=\hat{A}(E)\wedge \hat{A}(F).\text{\hspace{1em}(11.95)}$$

$\hat{A}$ is written in terms of the Pontrjagin classes as

$$\begin{array}{rl}& \hat{A}(\mathcal{F})=1-\frac{1}{24}{p}_1+\frac{1}{5760}(7p_1^2-4p_2)\\ & +\frac{1}{967680}(-31p_1^3+44p_1{p}_2-16p_3)+\dots .\end{array}\text{\hspace{1em}(11.96)}$$

Example 11.8

Let $M$ be a compact connected and orientable four-dimensional manifold. Let us consider the symmetric bilinear form $\sigma :H^2(M;\mathbb{R})\times H^2(M;\mathbb{R})\to \mathbb{R}$ defined by

$$\sigma ([\alpha ],[\beta ])={\int }_M\alpha \wedge \beta .\text{\hspace{1em}(11.97)}$$

$\sigma$ is a $b^2\times b^2$ symmetric matrix where $b^2=\dim H^2(M;\mathbb{R})$ is the Betti number. Clearly $\sigma$ is non-degenerate since $\sigma ([\alpha ],[\beta ])=0$ for any $[\alpha ]\in H^2(M;\mathbb{R})$ implies $[\beta ]=0$ . Let $p(q)$ be the number of positive (negative) eigenvalues of $\sigma$ . The Hirzebruch signature of $M$ is

$$\tau (M)\equiv p-q.\text{\hspace{1em}(11.98)}$$

According to the Hirzebruch signature theorem (see section 12.5), this number is also given in terms of the $L$ -polynomial as

$$\tau (M)={\int }_M{L}_1(M)=\frac{1}{3}{\int }_M{p}_1(M).\text{\hspace{1em}(11.99)}$$

Endnotes

来源与版本

正文：英文 · OCR 机器稿 · 待校对

核对状态：OCR 机器稿 · 待校对

原书 PDF

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Footnotes

1. Although $p_m(\mathcal{F})=0$ , $p_m(B)$ need not vanish for a matrix $B$ . $p_m$ will be used to define the Euler class later. ↩

2. See proposition 1.3. The definition here differs in phase from that in section 1.5. It turns out to be convenient to choose the present phase convention in the definition of the Euler class. ↩

3. Since $\det (X^{\mathrm{t}}AX)=(\det X{)}^2\det A$ , we have $\mathrm{Pf}(X^{\mathrm{t}}AX)=\pm \mathrm{Pf}(A)\det X$ . Here the plus sign should be chosen since $\mathrm{Pf}(I^{\mathrm{t}}AI)=\mathrm{Pf}(A)$ . ↩