[Nakahara GTP] 11.2 Chern classes

 This article is one of Manifold, Differential Geometry, Fibre Bundle.

11.2.1 Definitions

Let $\pi:E\rightarrow M$ be a complex vector bundle whose fibre is $\mathbb{C}^k$. The structure group $G$ is a subgroup of $GL(k,\mathbb{C})$, and the gauge potential $\mathcal{A}$ and the field strength $\mathcal{F}$ take their values in $\mathfrak{g}$. Defined the total Chern class by $$\begin{align}c(\mathcal{F})\equiv \det \left( I+\frac{i\mathcal{F}}{2\pi}\right).\end{align}$$ Since $\mathcal{F}$ is a two-form, $c(\mathcal{F})$ is a direct sum of forms of even degrees, $$\begin{align}c(\mathcal{F})=1+c_1(\mathcal{F})+c_2(\mathcal{F})+\cdots\end{align}$$ where $c_j(\mathcal{F})\in \Omega^{2j}(M)$ is called the $j$th Chern class. In an $m$-dimensional manifold $M$, the Chern class $c_j(\mathcal{F})$ with $2j>m$ vanishes trivially. Irrespective of $\dim M$, the series terminates at $c_k(\mathcal{F})=\det (i\mathcal{F}/2\pi)$ and $c_j(\mathcal{F})=0$ for $j>k$. Since $c_j(\mathcal{F})$ is closed, it defines an element $[c_j(\mathcal{F})]$ of $H^{2j}(M)$.

Example 11.1. Let $F$ be a complex vector bundle with fibre $\mathbb{C}^2$ over $M$, where $G=SU(2)$ and $\dim M=4$. If we write the field $\mathcal{F}=\mathcal{F}^\alpha(\sigma_\alpha/2i)$, $\mathcal{F}^\alpha=\frac{1}{2}\mathcal{F}^\alpha_{\mu\nu}dx^\mu\wedge dx^\nu$, we have $$\begin{align}c(\mathcal{F})=\det \left( I+\frac{i}{2\pi} \mathcal{F}^\alpha(\sigma_\alpha/2i)\right)\nonumber\\ =\det \begin{pmatrix} 1+(i/2\pi)(\mathcal{F}^3/2i)& (i/2\pi)(\mathcal{F}^1-i\mathcal{F}^2)/2i\\ (i/2\pi)(\mathcal{F}^1+i\mathcal{F}^2)/2i& 1-(i/2\pi)(\mathcal{F}^3/2i)\end{pmatrix}\nonumber\\  =1+\frac{1}{4}\left( \frac{i}{2\pi}\right)^2 \left( \mathcal{F}^3\wedge\mathcal{F}^3+\mathcal{F}^1\wedge\mathcal{F}^1+\mathcal{F}^2\wedge\mathcal{F}^2\right).\end{align}$$ Individual Chern classes are $$\begin{align} c_0(\mathcal{F})=1\nonumber\\ c_1(\mathcal{F})=0\nonumber\\ c_2(\mathcal{F})=\left(\frac{i}{2\pi}\right)^2\sum\frac{\mathcal{F}^\alpha\wedge\mathcal{F}^\alpha}{4}=\det \left( \frac{i\mathcal{F}}{2\pi}\right).\end{align}$$ Higher Chern classes vanish identically.

For general fibre bundle, it is rather cumbersome to compute the Chern classes by expanding the determinant and it is desirable to find a formula which yields them more easily. This is done by diagonalizing the curvature form. The matrix form $\mathcal{F}$ is diagonalized by an appropriate matrix $g\in GL(k\mathcal{C})$ as $g^{-1}(i\mathcal{F}/2\pi)g=\mbox{diag}(x_1,\cdots,x_k)$, where $x_i$ is a two-form. This diagonal matrix will be denoted by $A$. For example, if $G=SU(k)$, the generators are chosen to be anti-Hermitian and a Hermitian matrix $i\mathcal{F}/2\pi$ can be diagonalized by $g\in SU(k)$. We have $$\begin{align}\det (1+A)=\det [\mbox{diag}(1+x_1,1+x_2,\cdots,1+x_k)=\prod^k_{j=1} (1+x_j)\nonumber\\ =1+(x_1+\cdots+x_k)+(x_1x_2+\cdots+x_{k-1}x_k)+\cdots+(x_1x_2+\cdots+x_k)\nonumber\\ =1+\mbox{tr}A+\frac{1}{2}\{(\mbox{tr}A)^2-\mbox{tr}A^2\}+\cdots+\det A.\end{align}$$ Observe that each term of (5) is an elemetary symmetric function of $\{x_j\}$, $$\begin{align}S_0(x_j)\equiv 1\nonumber\\ S_1(x_j)\equiv \sum^k_{j=1}x_j\nonumber\\ S_2(x_j)\equiv \sum_{i<j}x_ix_j\nonumber\\ \vdots\nonumber\\ S_k(x_j)\equiv x_1x_2\cdots x_k.\end{align}$$ Since $\det (I+A)$ is an invariant polynomial, we have $P(\mathcal{F})=P(g\mathcal{F}g^{-1})=P(2\pi A/i)$. Accordingly, we have, for general $\mathcal{F}$, $$\begin{align}c_0(\mathcal{F})=1\nonumber\\ c_1(\mathcal{F})=\mbox{tr}A=\mbox{tr}\left( g\frac{i\mathcal{F}}{2\pi}g^{-1}\right)=\frac{i}{2\pi}\mbox{tr}\mathcal{F}\nonumber\\ c_2(\mathcal{F})=\frac{1}{2}[(\mbox{tr}\mathcal{A})^2-\mbox{tr}\mathcal{A}^2]=\frac{1}{2}(i/2\pi)^2[\mbox{tr}\mathcal{F}\wedge\mbox{tr}\mathcal{F}-\mbox{tr}(\mathcal{F}\wedge\mathcal{F})]\nonumber\\ \vdots\nonumber\\ c_k(\mathcal{F})=\det A=(i/2\pi)^k\det \mathcal{F}.\end{align}$$

11.2.2 Properties of Chern classes

We will deal with several vector bundels in the following. We often denote the Chern class of a vector bundle $E$ by $c(E)$. If the specification of the curvature is required, we write $c(\mathcal{F}_E)$.

Thoerem 11.3. Let $\pi:E\rightarrow M$ be a vector bundle with $G=GL(k,\mathbb{C})$ and $F=\mathbb{C}^k$.

(a) (Naturality) Let $f:N\rightarrow M$ be a smooth map. Then $$\begin{align}c(f^*E)=f^*c(E).\end{align}$$

(b) Let $\pi':F\rightarrow M$ be another vector bundle with $F=\mathbb{C}^l$ and $G=GL(l,\mathbb{C})$. The total Chern class of a Whiney sum bundle $E\oplus F$ is $$\begin{align} c(E\oplus F)=c(E)\wedge c(F).\end{align}$$

Proof. (a) The natruality follows directly from theorem 11.2(a). Since the curvature of $f^*E$ is $\mathcal{F}_{f^*E}=f^*\mathcal{F}_E$, the total Chern class of $f^*E$ is $$c(f^*E)=\det \left( I+\frac{i}{2\pi}\mathcal{F}_{f^*E}\right)=\det\left( I+\frac{i}{2\pi}f^*\mathcal{F}_E\right)\\ =f^*\det \left( I+\frac{i}{2\pi}\mathcal{F}_E\right) =f^*c(E).$$

(b) Let us consider the Chern polynomial of a matrix $$A=\begin{pmatrix}B&0\\0&C\end{pmatrix}.$$ (Note that the curvature of a Whitney sum bundle is block diagonal: $\mathcal{F}_{E\oplus F}=\mbox{diag}(\mathcal{F}_E,\mathcal{F}_F)$.) We find that $$\det \left( I+\frac{iA}{2\pi}\right) =\det \begin{pmatrix} I+\frac{iB}{2\pi}& 0\\ 0& I+\frac{iC}{2\pi}\end{pmatrix}\\ =\det \left( I+\frac{iB}{2\pi}\right) \det\left( I+\frac{iC}{2\pi}\right)=c(B)c(C).$$ This relation remains true when $B$ and $C$ are replaced by $\mathcal{F}_E$ and $\mathcal{F}_F$, namely $$c(\mathcal{F}_{E\oplus F})=c(\mathcal{F}_E)\wedge c(\mathcal{F}_F).$$

11.2.3 Splitting principle

Let $E$ be a Whitney sum of $n$ complex line bundles, $$\begin{align}E=L_1\oplus L_2\oplus \cdots\oplus L_n.\end{align}$$ From (9), we have $$\begin{align}c(E)=c(L_1)c(L_2)\cdots c(L_n)\end{align}$$ where the product is the exterior product of differential forms. Since $c_r(L)=0$ for $r\ge 2$, we write $$\begin{align}c(L_i)=1+c_1(L_i)\equiv 1+x_i.\end{align}$$ Then (11) becomes $$\begin{align}c(E)=\prod^n_{i=1}(1+x_i).\end{align}$$ Comparing this with (5), we find that the Chern class of an $n$-dimensional vector bundle $E$ is identical with that of the Whitney sum of $n$ complex line bundles. Although $E$ is not a Whitney sum of complex line bundles in general, as far as the Chern classes are concerned, we may pretend that this is the case. This is called the splitting principle and we accept this fact without proof.

Intuitively speaking, if the curvature $\mathcal{F}$ is diagonalized, the complex vector space on which $g$ acts splits into $k$ independent pices: $\mathbb{C}^k\rightarrow \mathbb{C}\oplus \cdots\oplus \mathbb{C}$. An eigenvalue $x_i$ is a curvature in each complex line bundle. Since diagonalizable matrices are dense in $M(n,\mathbb{C})$, any matrix may be approximated by a diagonal one as closely as we wish. Hence, the splitting priciple applies to any matrix.

11.2.4 Universal bundles and classifying spaces

Let $\pi:E\rightarrow M$ be a vector bundle with fibre $\mathbb{C}^k$. It is known that we can always find a bundle $\pi':\bar{E}\rightarrow M$ such that $$\begin{align}E\oplus \bar{E}\simeq M\times \mathbb{C}^n\end{align}$$ for some $n\ge k$. The fibre $F_p$ of $E$ at $p\in M$ is a $k$-plane lying in $\mathbb{C}^n$. Let $G_{k,n}(\mathbb{C})$ be the Grassmann manifold defined in example 8.4.

Example 8.4. Let $M_{k,n}(\mathbb{C})$ be the set of $k\times n$ matrices of rank $k$ ($k\le n$). Take $A,B\in M_{k,n}(\mathbb{C})$ and define an equivalence relation by $A\sim B$ if there exists $g\in GL(k,\mathbb{C})$ such that $B=gA$.

The manifold $G_{k,n}(\mathbb{C}$ is the set of $k$-plane in $\mathbb{C}^n$. Similarly to the canonical line bundle, we define the canonical $k$-plane bundle $L_{k,n}(\mathbb{C})$ over $G_{k,n}(\mathbb{C})$ with the fibre $\mathbb{C}^k$. Consider a map $f:M\rightarrow G_{k,n}(\mathbb{C})$ which maps a point $p$ to the $k$-plane $F_p$ in $\mathbb{C}^n$.

Theorem 11.4. Let $M$ be a manifold with $\dim M=m$ and let $\pi:E\rightarrow M$ be a complex vector bundle with the fibre $\mathbb{C}^k$. Then there exists a natural number $N$ such that for $n>N$, 

(a) there exists a map $f: M\rightarrow G_{k,n}(\mathbb{C})$ such that $$\begin{align}E\simeq f^*L_{k,n}(\mathbb{C})\end{align}$$

(b) $f^*L_{k,n}(\mathbb{C})\simeq g^*L_{k,n}(\mathbb{C})$ if and only if $f,g:M\rightarrow G_{k,n}(\mathbb{C})$ are homotopic.

The proof is found in Chern (1979). For example, if $\pi:E\rightarrow M$ is a complex line bundle, then there exists a bundle $\pi':\bar{E}\rightarrow M$ such that $E\oplus \bar{E}\simeq M\times \mathbb{C}^n$ and a map $f:M\rightarrow G_{1,n}(\mathbb{C})\simeq \mathbb{C}P^{n-1}$ such that $E=f^*L$, $L$ being the canonical line bundle over $\mathbb{C}P^{n-1}$. Moreover, if $f\sim g$, then $f^*L$ is equivalent to $g^*L$. Theorem 11.4 shows that the classification of vector bundle reduces to that of the homotopy classes of the map $M\rightarrow G_{k,n}(\mathbb{C})$.

It is convenient to define the classifying space $G_k(\mathbb{C})$. Regarding a $k$-plane in $\mathbb{C}^n$ as that in $\mathbb{C}^{n+1}$, we have natural inclusions. $$\begin{align} G_{k,k}(\mathbb{C})\hookrightarrow G_{k,k+1}(\mathbb{C})\hookrightarrow \cdots\hookrightarrow G_k(\mathbb{C})\end{align}$$ where $$\begin{align} G_k(\mathbb{C})\equiv \bigcup^\infty_{n=k}G_{k,n}(\mathbb{C}).\end{align}$$ Correspondingly, we have the universal bundle $L_k\rightarrow G_k(\mathbb{C})$ whose fibre is $\mathbb{C}^k$. For any complex vector bundle $\pi:E\rightarrow M$ with fibre $\mathbb{C}^k$, there exists a map $f:M\rightarrow G_k(\mathbb{C})$ such that $E=f^*L_k(\mathbb{C})$.

Let $\pi:E\rightarrow M$ be a vector bundle. A characteristic class $\chi$ is defined as a map $\chi:E\rightarrow \chi(E)\in H^*(M)$ such that $$\begin{align}\chi(f^*E)=f^*\chi (E)\quad \mbox{(naturality)}\\ \chi(E)=\chi(E')\quad \mbox{if }E\mbox{ is equivalent to }E'.\end{align}$$ The map $f^*$ on the LHS of (18) is a pullback of the bundle while $f^*$ on the RHS is that of the cohomology class. Since the homotopy class $[f]$ of $f:M\rightarrow G_k(\mathbb{C})$ uniquely defines the pullback $$\begin{align}f^*H^*(G_k)\rightarrow H^*(M)\end{align}$$ an element $\chi(E)=f^*\chi (G_k)$ proves to be useful in classifying complex vector bundles over $M$ with $\dim E=k$. For each choice of $\chi(G_k)$, there exists a characteristic class in $E$.

The Chern class $c(E)$ is also defined axiomatically by

(i) $c(f^*E)=f^*c(E)$ (naturality)

(ii) $c(E)=c_0(E)\oplus c_1(E)\oplus \cdots \oplus c_k(E)$ ($c_i(E)\in H^{2i}(M);\ c_i(E)=0\ i>k$

(iii) $c(E\oplus F)=c(E)c(F)$ (Whitney sum)

(iv) $c(L)=1+x$ (normalization)

$L$ being the canonical line bundle over $\mathbb{C}P^n$. It can be shown that these axioms uniquely define the Chern calss as (1).