11.3 Chern characters

11.3.1 Definitions

Among the characteristic classes, the Chern characters are of special importance due to their appearance in the Atiyah-Singer index theorem. The total Chern character is defined by

$$\mathrm{ch}(\mathcal{F})\equiv \mathrm{tr}\exp \left(\frac{\mathrm{i}\mathcal{F}}{2\pi }\right)=\sum _{j=1}\frac{1}{j!}\mathrm{tr}{\left(\frac{\mathrm{i}\mathcal{F}}{2\pi }\right)}^j.\text{\hspace{1em}(11.47)}$$

The $j$ th Chern character ${\mathrm{ch}}_j(\mathcal{F})$ is

$${\mathrm{ch}}_j(\mathcal{F})\equiv \frac{1}{j!}\mathrm{tr}{\left(\frac{\mathrm{i}\mathcal{F}}{2\pi }\right)}^j.\text{\hspace{1em}(11.48)}$$

If $2j>m=\dim M$ , ${\mathrm{ch}}_j(\mathcal{F})$ vanishes, hence $\mathrm{ch}(\mathcal{F})$ is a polynomial of finite order. Let us diagonalize $\mathcal{F}$ as

$$\frac{\mathrm{i}\mathfrak{F}}{2\pi }\to g^{-1}\left(\frac{\mathrm{i}\mathfrak{F}}{2\pi }\right)g=A\equiv \mathrm{diag}(x_1,\dots ,x_k)g\in \mathrm{GL}(k,\mathbb{C}).$$

The total Chern character is expressed as

$$\mathrm{tr}[\exp (A)]=\sum _{j=1}^k\exp \left(x_j\right).\text{\hspace{1em}(11.49)}$$

In terms of the elementary symmetric functions $S_r(x_j)$ , the total Chern character becomes

$$\begin{array}{rl}\sum _{j=1}^k\exp (x_j)& =\sum _{j=1}^k\left(1+x_j+\frac{1}{2!}{x}_j^2+\frac{1}{3!}{x}_j^3+\dots \right)\\ & =k+S_1\left(x_j\right)+\frac{1}{2!}\left[S_1{\left(x_j\right)}^2-2S_2\left(x_j\right)\right]+\dots .\end{array}\text{\hspace{1em}(11.50)}$$

Accordingly, each Chern character is expressed in terms of the Chern classes as

$${\mathrm{ch}}_0(\mathcal{F})=k\text{\hspace{1em}(11.51a)}$$

$${\mathrm{ch}}_1(\mathcal{F})=c_1(\mathcal{F})\text{\hspace{1em}(11.51b)}$$

$${\mathrm{ch}}_2(\mathcal{F})=\frac{1}{2}\left[c_1(\mathcal{F}{)}^2-2c_2(\mathcal{F})\right]\text{\hspace{1em}(11.51c)}$$

where $k$ is the fibre dimension of the bundle.

Example 11.2

Let $P$ be a U(1) bundle over $S^2$ . If ${\mathcal{A}}_{\mathrm{N}}$ and ${\mathcal{A}}_{\mathrm{S}}$ are the local connections on $U_{\mathrm{N}}$ and $U_{\mathrm{S}}$ defined in section 10.5, the field strength is given by ${\mathcal{F}}_i=\mathrm{d}{\mathcal{A}}_i$ ( $i=\mathrm{N},\mathrm{S}$ ). We have

$$\mathrm{ch}(\mathcal{F})=1+\frac{\mathrm{i}\mathcal{F}}{2\pi }\text{\hspace{1em}(11.52)}$$

where we have noted that ${\mathcal{F}}^n=0$ ( $n\ge 2$ ) on $S^2$ . This bundle describes the magnetic monopole. The magnetic charge $2g$ given by (10.94) is an integer expressed in terms of the Chern character as

$$N=\frac{\mathrm{i}}{2\pi }{\int }_{S^2}\mathcal{F}={\int }_{S^2}{\mathrm{ch}}_1(\mathcal{F}).\text{\hspace{1em}(11.53)}$$

Let $P$ be an SU(2) bundle over $S^4$ . The total Chern class of $P$ is given by (11.27). The total Chern character is

$$\mathrm{ch}(\mathcal{F})=2+\mathrm{tr}\left(\frac{\mathrm{i}\mathcal{F}}{2\pi }\right)+\frac{1}{2}\mathrm{tr}{\left(\frac{\mathrm{i}\mathcal{F}}{2\pi }\right)}^2.\text{\hspace{1em}(11.54)}$$

$\mathrm{Ch}(\mathcal{F})$ terminates at ${\mathrm{ch}}_2(\mathcal{F})$ since ${\mathcal{F}}^n=0$ for $n\ge 3$ . Moreover, $\mathrm{tr}\mathcal{F}=0$ for $G=\mathrm{SU}(2)$ , $n\ge 2$ . As we found in section 10.5, the instanton number is given by

$$\frac{1}{2}{\int }_{S^4}\mathrm{tr}{\left(\frac{\mathrm{i}\mathcal{F}}{2\pi }\right)}^2={\int }_{S^4}{\mathrm{ch}}_2(\mathcal{F}).\text{\hspace{1em}(11.55)}$$

In both cases, ${\mathrm{ch}}_j$ measures how the bundle is twisted when local pieces are patched together.

Example 11.3

Let $P$ be a U(1) bundle over a $2m$ -dimensional manifold $M$ . The $m$ th Chern character is

$$\begin{array}{rl}\frac{1}{m!}\mathrm{tr}{\left(\frac{\mathrm{i}\mathfrak{F}}{2\pi }\right)}^m& =\frac{1}{m!}{\left(\frac{\mathrm{i}}{2\pi }\right)}^m{\left[\frac{1}{2}{\mathfrak{F}}_{\mu v}\mathrm{d}{x}^{\mu }\wedge \mathrm{d}{x}^v\right]}^m\\ & =\frac{1}{m!}{\left(\frac{\mathrm{i}}{4\pi }\right)}^m{\mathcal{F}}_{{\mu }_1{v}_1}\dots {\mathcal{F}}_{{\mu }_m{v}_m}\mathrm{d}{x}^{{\mu }_1}\wedge \mathrm{d}{x}^{v_1}\wedge \cdots \wedge \mathrm{d}{x}^{{\mu }_m}\wedge \mathrm{d}{x}^{v_m}\\ & ={\left(\frac{\mathrm{i}}{4\pi }\right)}^m{ϵ}^{{\mu }_1{v}_1\dots {\mu }_m{v}_m}{\mathcal{F}}_{{\mu }_1{v}_1}\dots {\mathcal{F}}_{{\mu }_m{v}_m}\mathrm{d}{x}^1\wedge \cdots \wedge \mathrm{d}{x}^{2m}\end{array}$$

which describes the U(1) anomaly in $2m$ -dimensional space, see chapter 13.

Example 11.4

Let $L$ be a complex line bundle. It then follows that

$$\mathrm{ch}(L)=\mathrm{tr}\exp \left(\frac{\mathrm{i}\mathcal{F}}{2\pi }\right)={\mathrm{e}}^x=1+xx\equiv \frac{\mathrm{i}\mathcal{F}}{2\pi }.\text{\hspace{1em}(11.56)}$$

For example, let $L\stackrel{\pi }{\to }\mathbb{C}{P}^1$ be the canonical line bundle over $\mathbb{C}{P}^1=S^2$ . The Fubini-Study metric yields the curvature

$$\mathcal{F}=-\partial \bar{\partial }\ln (1+|z{|}^2)=-\frac{\mathrm{d}z\wedge \mathrm{d}\bar{z}}{(1+z\bar{z}{)}^2}\text{\hspace{1em}(11.57)}$$

see example 8.8. In real coordinates $z=x+\mathrm{i}y=r\exp (\mathrm{i}\theta )$ , we have

$$\mathcal{F}=2\mathrm{i}\frac{\mathrm{d}x\wedge \mathrm{d}y}{(1+x^2+y^2{)}^2}=2\mathrm{i}\frac{r\mathrm{d}r\wedge \mathrm{d}\theta }{(1+r^2{)}^2}.\text{\hspace{1em}(11.58)}$$

From $\mathrm{ch}(\mathcal{F})=1+\mathrm{tr}(\mathrm{i}\mathcal{F}/2\pi )$ , we have

$${\mathrm{ch}}_1(\mathcal{F})=-\frac{1}{\pi }\frac{r\mathrm{d}r\wedge \mathrm{d}\theta }{(1+r^2{)}^2}.\text{\hspace{1em}(11.59)}$$

${\mathrm{Ch}}_1(L)$ , the integral of ${\mathrm{ch}}_1(\mathcal{F})$ over $S^2$ is an integer,

$${\mathrm{Ch}}_1(L)=-\frac{1}{\pi }\int \frac{r\mathrm{d}r\mathrm{d}\theta }{(1+r^2{)}^2}=-{\int }_1^{\infty }{t}^{-2}\mathrm{d}t=-1.\text{\hspace{1em}(11.60)}$$

11.3.2 Properties of the Chern characters

Theorem 11.5

(a) (Naturality) Let $E\stackrel{\pi }{\to }M$ be a vector bundle with $F={\mathbb{C}}^k$ . Let $f:N\to M$ be a smooth map. Then

$$\mathrm{ch}\left(f^{\ast }E\right)=f^{\ast }\mathrm{ch}(E).\text{\hspace{1em}(11.61)}$$

(b) Let $E$ and $F$ be vector bundles over a manifold $M$ . The Chern characters of $E\otimes F$ and $E\oplus F$ are given by

$$\mathrm{ch}(E\otimes F)=\mathrm{ch}(E)\wedge \mathrm{ch}(F)\text{\hspace{1em}(11.62a)}$$

$$\mathrm{ch}(E\oplus F)=\mathrm{ch}(E)\oplus \mathrm{ch}(F).\text{\hspace{1em}(11.62b)}$$

Proof

(a) follows from theorem 11.2(a).

(b) These results are immediate from the definition of the ch-polynomial. Let

$$\mathrm{ch}(A)=\sum \frac{1}{j!}\mathrm{tr}{\left(\frac{\mathrm{i}A}{2\pi }\right)}^j$$

be a polynomial of a matrix $A$ . Suppose $A$ is a tensor product of $B$ and $C$ , $A=B\otimes C=B\otimes I+I\otimes C$ (note that ${\mathcal{F}}_{E\otimes F}={\mathcal{F}}_E\otimes I+I\otimes {\mathcal{F}}_F$ ). Then we find that

$$\begin{array}{rl}\mathrm{ch}(B\otimes C)& =\sum _j\frac{1}{j!}{\left(\frac{\mathrm{i}}{2\pi }\right)}^j\mathrm{tr}(B\otimes I+I\otimes C{)}^j\\ & =\sum _j\frac{1}{j!}{\left(\frac{\mathrm{i}}{2\pi }\right)}^j\sum _{m=1}^j\left(\genfrac{}{}{0ex}{}{j}{m}\right)\mathrm{tr}(B^m)\mathrm{tr}(C^{j-m})\\ & =\sum _m\frac{1}{m!}\mathrm{tr}{\left(\frac{\mathrm{i}B}{2\pi }\right)}^m\sum _n\frac{1}{n!}\mathrm{tr}{\left(\frac{\mathrm{i}C}{2\pi }\right)}^n=\mathrm{ch}(B)\mathrm{ch}(C).\end{array}$$

Equation (11.62a) is proved if $B$ is replaced by ${\mathcal{F}}_E$ and $C$ by ${\mathcal{F}}_F$ .

If $A$ is block diagonal,

$$A=\left(\begin{array}{cc}B& 0\\ 0& C\end{array}\right)=B\oplus C$$

we have

$$\begin{array}{rl}\mathrm{ch}(B\oplus C)& =\sum \frac{1}{j!}{\left(\frac{\mathrm{i}}{2\pi }\right)}^j\mathrm{tr}(B\oplus C{)}^j\\ & =\sum \frac{1}{j!}{\left(\frac{1}{2\pi }\right)}^j[\mathrm{tr}(B^j)+\mathrm{tr}(C^j)]=\mathrm{ch}(B)+\mathrm{ch}(C).\end{array}$$

This relation remains true when $A,B$ and $C$ are replaced by ${\mathcal{F}}_{E\oplus F},{\mathcal{F}}_E$ and ${\mathcal{F}}_F$ respectively.

Let us see how the splitting principle works in this case. Let $L_j$ ( $1\le j\le k$ ) be complex line bundles. From (11.62b) we have, for $E=L_1\oplus L_2\oplus \cdots \oplus L_k$ ,

$$\mathrm{ch}(E)=\mathrm{ch}\left(L_1\right)\oplus \mathrm{ch}\left(L_2\right)\oplus \cdots \oplus \mathrm{ch}\left(L_k\right).\text{\hspace{1em}(11.63)}$$

Since $\mathrm{ch}(L_i)=\exp (x_i)$ , we find

$$\mathrm{ch}(E)=\prod _{j=1}^k\exp \left(x_j\right)\text{\hspace{1em}(11.64)}$$

which is simply (11.50). Hence, the Chern character of a general vector bundle $E$ is given by that of a Whitney sum of $k$ complex line bundles. The characteristic classes themselves cannot differentiate between two vector bundles of the same base space and the same fibre dimension. What is important is their integral over the base space.

11.3.3 Todd classes

Another useful characteristic class associated with a complex vector bundle is the Todd class defined by

$$\mathrm{Td}(\mathcal{F})=\prod _j\frac{x_j}{1-{\mathrm{e}}^{-x_j}}\text{\hspace{1em}(11.65)}$$

where the splitting principle is understood. If expanded in powers of $x_j$ , $\mathrm{Td}(\mathcal{F})$ becomes

$$\begin{array}{rl}\mathrm{Td}\left(\mathcal{F}\right)& =\prod _j\left(1+\frac{1}{2}{x}_j+\sum _{k\ge 1}(-1{)}^{k-1}\frac{B_k}{(2k)!}{x}_j^{2k}\right)\\ & =1+\frac{1}{2}\sum _j{x}_j+\frac{1}{12}\sum _j{x}_j^2+\frac{1}{4}\sum _{j<k}{x}_j{x}_k+\dots \\ & =1+\frac{1}{2}{c}_1(\mathcal{F})+\frac{1}{12}\left[c_1(\mathcal{F}{)}^2+c_2(\mathcal{F})\right]+\dots \end{array}\text{\hspace{1em}(11.66)}$$

where the $B_k$ are the Bernoulli numbers

$$B_1=\frac{1}{6}{B}_2=\frac{1}{30}{B}_3=\frac{1}{42}{B}_4=\frac{1}{30}{B}_5=\frac{5}{66}\dots .$$

The first few terms of (11.66) are:

$${\mathrm{Td}}_0(\mathcal{F})=1\text{\hspace{1em}(11.67a)}$$

$${\mathrm{Td}}_1(\mathcal{F})=\frac{1}{2}{c}_1\text{\hspace{1em}(11.67b)}$$

$${\mathrm{Td}}_2(\mathcal{F})=\frac{1}{12}\left(c_1^2+c_2\right)\text{\hspace{1em}(11.67c)}$$

$${\mathrm{Td}}_3(\mathcal{F})=\frac{1}{24}{c}_1{c}_2\text{\hspace{1em}(11.67d)}$$

$${\mathrm{Td}}_4(\mathcal{F})=\frac{1}{720}\left(-c_1^4+4c_1^2{c}_2+3c_2^2+c_1{c}_3-c_4\right)\text{\hspace{1em}(11.67e)}$$

$${\mathrm{Td}}_5(\mathcal{F})=\frac{1}{1440}\left(-c_1^3{c}_2+3c_1{c}_2^2+c_1^2{c}_3-c_1{c}_4\right)\text{\hspace{1em}(11.67f)}$$

where $c_i$ stands for $c_i(\mathfrak{F})$

Exercise 11.2

Let $E$ and $F$ be complex vector bundles over $M$ . Show that

$$\mathrm{Td}(E\oplus F)=\mathrm{Td}(E)\wedge \mathrm{Td}(F).\text{\hspace{1em}(11.68)}$$

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