[Nakahara GTP] 11.1 Invariant polynomials and the Chern-Weil homomorphism
This article is one of Manifold, Differential Geometry, Fibre Bundle.
Characteristic classes are subsets of the cohomology classes of the base space and measure the non-triviality or twisting of a bundle. In this sense, they are obstructions which prevent a bundle from being a trivial bundle. Most of the characteristic classes are given by the de Rham cohomology classes.
Let $M$ be an $m$-dimensional manifold. An $r$-form $\omega\in \Omega^r(M)$ is closed if $d\omega=0$ and exact if $\omega=d\eta$ for some $\eta\in \Omega^{r-1}(M)$. The set of closed $r$-forms is denoted by $Z^r(M)$ and the set of exact $r$-forms by $B^r(M)$. Since $d^2=0$, it follows that $Z^r(M)\supset B^r(M)$. We define the $r$th de Rham cohomology group $H^r(M)$ by $$H^r(M)\equiv Z^r(M)/B^r(M).$$ In $H^r(M)$, two closed $r$-forms $\omega_1$ and $\omega_2$ are identified if $\omega_1-\omega_2=d\eta$ for some $\eta\in \Omega^{r-1}(M)$. Let $M$ be an $m$-dimensional manifold. The formal sum $$H^*(M)\equiv H^0(M)\oplus H^1(M)\oplus \cdots \oplus H^m(M)$$ is the cohomology ring with the product $\wedge:H^*(M)\times H^*(M)\rightarrow H^*(M)$ induced by $\wedge:H^p(M)\times H^q(M)\rightarrow H^{p+q}(M)$. Let $f:M\rightarrow N$ be a smooth map. The pullback $f^*:\Omega^r(N)\rightarrow \Omega^r(M)$ naturally induces a linear map $f^*:H^r(N)\rightarrow H^r(M)$ since $f^*$ commutes with the exterior derivative: $f^*d\omega=df^*\omega$. The pullback $f^*$ preserves the algebraic structure of the cohomology ring since $f^*(\omega\wedge\eta)=f^*\omega\wedge f^*\eta$.
11.1.1 Invariant polynomials
Let $M(k,\mathbb{C})$ be the set of complex $k\times k$ matrices. Let $S^r(M(k,\mathbb{C})$ denote the vector space of symmetric $r$-linear $\mathbb{C}$-valued functions on $M(k,\mathbb{C})$. In other words, a map $$\tilde{P}:\otimes^r M(k,\mathbb{C})\rightarrow \mathbb{C}$$ is an element of $S^r(M(k,\mathbb{C}))$ if it satisfies, in addition to linearity in each entry, the symmetry $$\begin{align}\tilde{P}(a_1,\cdots,a_i,\cdots,a_j,\cdots,a_r)=\tilde{P}(a_1,\cdots,a_j,\cdots,a_i,\cdots,a_r)\quad 1\le i,j\le r\end{align}$$ where $a_p\in GL(k,\mathbb{C})$. Let $$S^*(M(k,\mathbb{C}))\equiv \otimes^\infty_{r=0} S^r(M(k,\mathbb{C}))$$ denote the formal sum of symmetric multilinear $\mathbb{C}$-valued functions. We define a product of $\tilde{P}\in S^p(M(k,\mathbb{C}))$ and $\tilde{Q}\in S^q(M(k,\mathbb{C}))$ by $$\begin{align}\tilde{P}\tilde{Q}(X_1,\cdots,X_{p+q})\nonumber\\ =\frac{1}{(p+q)!}\sum_p \tilde{P}(X_{P(1)},\cdots,X_{P(p)})\tilde{Q}(X_{P(p+1)},\cdots,X_{P(p+1)},\cdots,X_{P(p+q)})\end{align}$$ where $P$ is the permutation of $(1,\cdots,p+q)$. $S^*(M(k,\mathbb{C}))$ is an algebra with this multiplication.
Let $G$ be a matrix group and $\mathfrak{g}$ its Lie algebra. In practice, we take $G=GL(k,\mathbb{C})$, $U(k)$ or $SU(k)$. The Lie algebra $\mathfrak{g}$ is a subspace of $M(k,\mathbb{C})$(Just representation?) and we may consider the restrictions $S^r(\mathfrak{g})$ and $S^*(\mathfrak{g})$. $\tilde{P}\in S^r(\mathfrak{g})$ is said to be invariant if, for any $g\in G$ and $A_i\in \mathfrak{g}$. $\tilde{P}$ satisfies $$\begin{align}\tilde{P}(\mbox{Ad}_g A_1,\cdots,\mbox{Ad}_gA_r)=\tilde{P}(A_1,\cdots,A_r)\end{align}$$ where $\mbox{Ad}_gA_i=g^{-1}A_ig$. For example, $$\begin{align}\tilde{P}(A_1,A_2,\cdots,A_r)=\mbox{str}(A_1,A_2,\cdots,A_r)\equiv \frac{1}{r!}\sum_p\mbox{tr}(A_{P(1)},A_{P(2)},\cdots,A_{P(r)})\end{align}$$ is symmetric, $r$-linear and invariant, where 'str' stands for the symmetrized trace and is defined by the last equality. The set of $G$-invariant members of $S^r(\mathfrak{g})$ is denoted by $I^r(G)$. Note that $\mathfrak{g}_1=\mathfrak{g}_2$ does not necessarily imply $I^r(G_1)=I^r(G_2)$ (Lie group$\ne$Lie algebra). The product defined by (2) naturally induces a multiplication $$\begin{align}I^p(G)\otimes I^q(G)\rightarrow I^{p+q}(G).\end{align}$$ The sum $I^*(G)\equiv \bigotimes_{r\ge 0}I^r(G)$ is an algebra with this product.
Take $\tilde{P}\in I^r(G)$. The shorthand notation for the diagonal combination is $$\begin{align}P(A)\equiv \tilde{P}(A,A,\cdots,A)\quad A\in\mathfrak{g}.\end{align}$$ Clearly, $P$ is a polynomial of degree $r$ and called an invariant polynomial. $P$ is also $\mbox{Ad }G$-invariant, $$\begin{align}P(\mbox{Ad}_gA)=P(g^{-1}Ag)=P(A)\quad A\in\mathfrak{g},g\in G.\end{align}$$ For example, $\mbox{tr}(A^r)$ is an invariant polynomial obtained from (4). In general, an invariant polynomial may be written in terms of a sum of products of $P_r\equiv \mbox{tr}(A^r)$.
Conversely, any invariant polynomial $P$ defines an invariant and symmetric $r$-linear form $\tilde{P}$ by expanding $P(t_1A_1+\cdots+t_rA_r)$ as a polynomial in $t_i$. Then $1/r!$ times the coefficient of $t_1t_2\cdots t_r$ is invariant and symmetric by construction and is called the polarization of $P$. Take $P(A)\equiv \mbox{tr}(A^3)$, for example. Following the previous prescription, we expand $\mbox{tr}(t_1A_1+t_2A_2+t_3A_3)^3$ in powers of $t_1$, $t_2$, and $t_3$. The coefficient of $t_1t_2t_3$ is $$\mbox{tr}(A_1A_2A_3+A_1A_3A_2+A_2A_1A_3+A_2A_3A_1+A_3A_1A_2+A_3A_2A_1) = 3\mbox{tr}(A_1A_2A_3+A_2A_1A_3)$$ where the cyclicity of the trace has been used. The polarization is $$\tilde{P}(A_1,A_2,A_3)=\frac{1}{2}\mbox{tr}(A_1A_2A_3+A_2A_1A_3)=\mbox{str}(A_1,A_2,A_3).$$ (Very nice example!!!)
In the previous chapter, we introduced the local gauge potential $\mathcal{A}=\mathcal{A}_\mu dx^\mu$ and the field strength $\mathcal{F}=\frac{1}{2}\mathcal{F}_{\mu\nu}dx^\mu\wedge dx^\nu$ on a principal bundle. We have shown that these geometrical objects describe the associated vector bundles as well. Since the set of connections $\{\mathcal{A}_i\}$ describes the twisting of a fibre bundle, the non-triviality of a principal bundle is equally shared by its associated bundle. In fact, if (10.57) is employed as a definition of the local connection in a vector bundle, it can be defined even without reference to the principal bundle with which it is originally associated. Later, we encounter situations in which use of vector bundles is essential (the Whitney sum bundle, the splitting principle and so on).
Let $P(M,\mathbb{C})$ be a principal bundle. We extend the domain of invariant polynomials form $\mathfrak{g}$ to $\mathfrak{g}$-valued $p$-forms on $M$. For $A_i\eta_i$ ($A_i\in\mathfrak{g}$, $\eta\in\Omega^{p_i}(M);\ 1\le i\le r$), we define $$\begin{align}\tilde{P}(A_1\eta_1,\cdots,A_r\eta_r)\equiv \eta_1\wedge\cdots\wedge\eta_r\tilde{P}(A_1,\cdots,A_r).\end{align}$$ For example, corresponding to (4), we have $$\mbox{str}(A_1\eta_1,\cdots,A_r\eta_r)=\eta_1\wedge\cdots\wedge\eta_r\mbox{str}(A_1,\cdots,A_r).$$ The diagonal combination is $$\begin{align}P(A\eta)\equiv \eta\wedge\cdots\wedge\eta P(A).\end{align}$$ THe action $\tilde{P}$ or $P$ on general elements is given by the $r$-linearity. In particular, we are interested in the invariant polynomial of the form $P(\mathcal{F}$ in the following. The importance of invariant polynomials resides in the following fundamental theorem.
Theorem 11.1. (Chern-Weil theorem) Let $P$ be an invariant polynomial. Then $P(\mathcal{F})$ satisfies
(a) $dP(\mathcal{F})=0$. (Closed)
(b) Let $\mathcal{F}$ and $\mathcal{F}'$ be curvature two-forms corresponding to different connections $\mathcal{A}$ and $\mathcal{A}'$. Then the difference $P(\mathcal{F}')-P(\mathcal{F})$ is exact.
Proof. (a) It is sufficient to prove that $dP(\mathcal{F})=0$ for an invariant polynomial $P_r(\mathcal{F})$ which is homogeneous of degree $r$, since any invariant polynomial can be decomposed into homogeneous polynomials. First consider the identity, $$\tilde{P}_r(g^{-1}_tX_1g_t,\cdots,g^{-1}_tX_rg_t)=\tilde{P}_r(X_1,\cdots,X_r)$$ where $g_t\equiv \exp (tX)$ and $X,X_i\in \mathfrak{g}$. By putting $t=0$ after differentiation with respect to $t$, obtain $$\begin{align}\sum^r_{i=1}\tilde{P}_r(X_1,\cdots,[X_i,X],\cdots,X_r)=0.\end{align}$$
Next, let $A$ be a $\mathfrak{g}$-valued $p$-form and $\Omega_i$ be a $\mathfrak{g}$-form ($1\le i\le r$). Without loss of generality, we may take $A=X\eta$ and $\Omega_i=X_i\eta_i$ where $X,X_i\in\mathfrak{g}$ and $\eta$ ($\eta_i$) is a $p$-form ($p_i$-form). Define $$\begin{align}[\Omega_i,A]\equiv \eta_i\wedge\eta[X_i,X]=X_iX(\eta_i\wedge\eta)-(-1)^{pp_i}XX_i(\eta\wedge\eta_i).\end{align}$$ Let us note that $$\tilde{P}_r(\Omega_1,\cdots,[\Omega_i,A],\cdots,\Omega_r)\\ =\eta_1\wedge\cdots\wedge\eta_i\wedge\eta\wedge\cdots\eta_r\tilde{P}_r(X_1,\cdots,X_iX,\cdots,X_r)\\ -(-1)^{p\cdot p_i}\eta_1\wedge\cdots\wedge\eta\wedge\eta_i\wedge\cdots\wedge\eta_r\tilde{P}_r(X_1,\cdots,XX_i,\cdots,X_r)\\ =\eta\wedge\eta_1\wedge\cdots\wedge\eta_r(-1)^{p(p_1+\cdots+p_i)}\times \tilde{P}_r(X_1,\cdots,[X_i,X],\cdots,X_r).$$ From this and (10), we find $$\begin{align}\sum^r_{i=1}(-1)^{p(p_1+\cdots+p_i)}\tilde{P}_r(\Omega_1,\cdots,[\Omega_i,A],\cdots,\Omega_r)=0.\end{align}$$
Next, consider the derivative, $$\begin{align}d\tilde{P}_r(\Omega_1,\cdots,\Omega_r)=d(\eta_1\wedge\cdots\wedge\eta_r)\tilde{P}_r(X_1,\cdots,X_r)\nonumber\\ =\sum^r_{i=1}(-1)^{(p_1+\cdots+p_{i-1})}(\eta_1\wedge\cdots\wedge d\eta_i\wedge\cdots\wedge\eta_r)\times \tilde{P}_r(X_1,\cdots,X_i,\cdots,X_r)\nonumber\\ =\sum^r_{i=1}(-1)^{(p_1+\cdots+p_{i-1})}\tilde{P}_r(\Omega_1,\cdots,d\Omega_i,\cdots,\Omega_r).\end{align}$$
Let $A=\mathcal{A}$ and $\Omega_i=\mathcal{F}$ in (12) and (13) for which $p=1$ and $p_i=2$. By adding $0$ of the form (12) to (13) we have $$\begin{align}d\tilde{P}_r(\mathcal{F},\cdots,\mathcal{F}) \sum^r_{i=1}[\tilde{P}_r(\mathcal{F},\cdots,d\mathcal{F},\cdots,\mathcal{F})+\tilde{P}_r(\mathcal{F},\cdots,[\mathcal{A},\mathcal{F}],\cdots,\mathcal{F})] =\sum^r_{i=1}\tilde{P}_r(\mathcal{F},\cdots,\mathcal{D}\mathcal{F},\cdots,\mathcal{F})=0\end{align}$$ since $\mathcal{D}\mathcal{F}=d\mathcal{F}+[\mathcal{A},\mathcal{F}]=0$ (the Bianchi identity). We have proved $$dP_r(\mathcal{F})=d\tilde{P}_r(\mathcal{F},\cdots,\mathcal{F})=0.$$
(b) Let $\mathcal{A}$ and $\mathcal{A}'$ be two connections on $E$ and let $\mathcal{F}$ and $\mathcal{F}'$ be the respective field strengths. Define an interpolating gauge potential $\mathcal{A}_t$, by $$\begin{align}\mathcal{A}_t\equiv \mathcal{A}+t\theta\quad \theta\equiv (\mathcal{A}'-\mathcal{A})\quad 0\le t\le 1\end{align}$$ so that $\mathcal{A}_0=\mathcal{A}$ and $\mathcal{A}_1=\mathcal{A}'$. The corresponding field strength is $$\begin{align}\mathcal{F}_t\equiv d\mathcal{A}_t+\mathcal{A}_t\wedge\mathcal{A}_t=\mathcal{F}+t\mathcal{D}\theta+t^2\theta^2\end{align}$$ where $\mathcal{D}\theta=d\theta+[\mathcal{A},\theta]=d\theta+\mathcal{A}\wedge\theta+\theta\wedge\mathcal{A}$ ($\theta^2=\theta\wedge\theta$). We first note that $$\begin{align}P_r(\mathcal{F}')-P_r(\mathcal{F})=P_r(\mathcal{F}_1)-P_r(\mathcal{F}_0)\nonumber\\ =\int_0^1dt\frac{d}{dt}P_r(\mathcal{F}_r)=r\int^1_0dt\tilde{P}_r\left( \frac{d}{dt}\mathcal{F}_t,\mathcal{F}_t,\cdots,\mathcal{F}_t\right).\end{align}$$ (Use symmetric property) From (16), we find that $$\begin{align}\frac{d}{dt}P_r(\mathcal{F}_t)=r\tilde{P}_r(\mathcal{D}\theta+2t\theta^2,\mathcal{F}_t,\cdots,\mathcal{F}_t)\nonumber\\ =r\tilde{P}_r(\mathcal{D}\theta,\mathcal{F}_t,\cdots,\mathcal{F}_t)+2rt\tilde{P}_r(\theta^2,\mathcal{F}_t,\cdots,\mathcal{F}_t).\end{align}$$
Note also that $$\mathcal{D}\mathcal{F}_t=d\mathcal{F}_t+[\mathcal{A},\mathcal{F}_t]=-[\mathcal{A}_t,\mathcal{F}_t]=t[\mathcal{F}_t,\theta]$$ where use has been made of the Bianchi identity $\mathcal{D}_t\mathcal{F}_t=d\mathcal{F}_t+[\mathcal{A}_t,\mathcal{F}_t]=0$. ($\mathcal{D}$ is the covariant derivative with respect to $\mathcal{A}$ while $\mathcal{D}_t$ is that with respect to $\mathcal{A}_t$.) It then follows that $$\begin{align}d[\tilde{P}_r(\theta,\mathcal{F}_t,\cdots,\mathcal{F}_t)]=\tilde{P}_r(d\theta,\mathcal{F}_t,\cdots,\mathcal{F}_t)-(r-1)\tilde{P}_r(\theta, d\mathcal{F}_t,\cdots,\mathcal{F}_t)\nonumber\\ =\tilde{P}_r(\mathcal{D}\theta,\mathcal{F}_t,\cdots,\mathcal{F}_t)-(r-1)\tilde{P}_r(\theta,\mathcal{D}\mathcal{F}_t,\cdots,\mathcal{F}_t)\nonumber\\ =\tilde{P}_r(\mathcal{D}\theta,\mathcal{F}_t,\cdots,\mathcal{F}_t)-(r-1)t\tilde{P}_r(\theta,[\mathcal{F}_t,\theta],\mathcal{F}_t,\cdots,\mathcal{F}_t)\end{align}$$ where we have added a $0$ of the form (12) to change $d$ to $\mathcal{D}$. If we take $\Omega_1=A=\theta,\Omega_2=\cdots=\Omega_m=\mathcal{F}_t$ in (12), we have $$2\tilde{P}_r(\theta^2,\mathcal{F}_t,\cdots,\mathcal{F}_t)+(r-1)\tilde{P}_r(\theta,[\mathcal{F}_t,\theta],\mathcal{F}_t,\cdots,\mathcal{F}_t)=0.$$
From (18), (19) and the previous identity, we obtain $$\frac{d}{dt}P_r(\mathcal{F}_t)=rd[\tilde{P}_r(\theta,\mathcal{F}_t,\cdots,\mathcal{F}_t)].$$ We finally find that $$\begin{align}P_r(\mathcal{F}')-P_r(\mathcal{F})=d\left[ r\int^1_0\tilde{P}_r(\mathcal{A}'-\mathcal{A},\mathcal{F}_t,\cdots,\mathcal{F}_t)dt\right].\end{align}$$ It shows that $P_r(\mathcal{F}')$ differes from $P_r(\mathcal{F})$ by an exact form. $\blacksquare$
We define the trangression $TP_r(\mathcal{A}',\mathcal{A})$ of $P_r$ by $$\begin{align}TP_r(\mathcal{A}',\mathcal{A})\equiv r\int^1_0dt\tilde{P}_r(\mathcal{A}'-\mathcal{A},\mathcal{F}_t,\cdots,\mathcal{F}_t)\end{align}$$ where $\tilde{P}_r$ is the polarization of $P_r$(Sum over permutation). Transgressions will play an important role when we discuss Chern-Simons forms in section 11.5. Let $\dim M=m$. Since $P_m(\mathcal{F}')$ differs from $P_m(\mathcal{F})$ by an exact form, their integrals over a manifold $M$ without a boundary should be the same: $$\begin{align} \int_MP_m(\mathcal{F}')-\int_MP_m(\mathcal{F})=\int_MdTP_m(\mathcal{A}',\mathcal{A})=\int_{\partial M}P_m(\mathcal{A}',\mathcal{A})=0.\end{align}$$
As has been proved, an invariant polynomial is closed and, in general, non-trivial. Accordingly, it defines a cohomology class of $M$. Theorem 11.1(b) ensures that this cohomology class is independent of the gauge potential chosen. The cohomology class thus defined is called the characteristic class. The characteristic class defined by an invariant polynomial $P$ is denoted by $\chi_E(P)$ where $E$ is a fibre bundle on which connections and curvatures ae defined. [Remark: Since a principal bundel and its associated bundles share the same gauge potential and field strengths, the Chern-Weil theorem applies equally to both bundles. Accordingly, $E$ can be either a principal bundle or a vector bundle.]
Theorem 11.2. Let $P$ be an invariant polynomial in $I^*(G)$ and $E$ be a fibre bundle over $M$ with structure group $G$.
(a) The map $$\begin{align}\chi_E:I^*(G)\rightarrow H^*(M)\end{align}$$ defined by $P\rightarrow \chi_E(P)$ is a homomorphism (Weil homomorphism).
(b) Let $f:N\rightarrow M$ be a differentiable map. For the pullback bundle $f^*E$ of $E$, we have the so-called naturality $$\begin{align}\chi_{f^*E}=f^*\chi_E.\end{align}$$
Proof. (a) Take $P_r\in I^r(G)$ and $P_s\in I^s(G)$. If we write $\mathcal{F}=\mathcal{F}^\alpha T_\alpha$, we have $$(P_rP_s)(\mathcal{F})\\ =\mathcal{F}^{\alpha_1}\wedge\cdots\wedge\mathcal{F}^{\alpha_r}\wedge\mathcal{F}^{\beta_1}\wedge\cdots\wedge\mathcal{F}^{\beta_s}\times\frac{1}{(r+s)!}\tilde{P}_r(T_{\alpha_1},\cdots,T_{\alpha_r})\tilde{P}_s(T_{\beta_1},\cdots,T_{\beta_s})\\ =P_r(\mathcal{F})\wedge P_s(\mathcal{F}).$$ Then (a) follows since $P_r(\mathcal{F}),P_s(\mathcal{F})\in H^*(M)$.
(b) Let $\mathcal{A}$ be a gauge potential of $E$ and $\mathcal{F}=d\mathcal{A}+\mathcal{A}\wedge\mathcal{A}$. It is easy to verify that the pullback $f^*\mathcal{A}$ is a connection in $f^*E$. In fact, let $\mathcal{A}_i$ and $\mathcal{A}_j$ be local connections in overlapping charts $U_i$ and $U_j$ of $M$. If $t_{ij}$ is a transition function on $U_i\cap U_j$, the transition function on $f^*E$ is given by $f^*t_{ij}=t_{ij}\circ f$. The pullback $f^*\mathcal{A}_i$ and $f^*\mathcal{A}_j$ are related as $$f^*\mathcal{A}_j=f^*(t^{-1}_{ij}\mathcal{A}_it_{ij}+t^{-1}_{ij}dt_{ij})=(f^*t^{-1}_{ij})(f^*\mathcal{A}_i)(f^*\mathcal{A}_i)(f^*t_{ij})+(f^*t^{-1}_{ij})(df^*t_{ij}).$$ This shows that $f^*\mathcal{A}$ is, indeed, a local connection on $f^*E$ (local connection is natural). The corresponding field strength on $f^*E$ is $$d(f^*\mathcal{A}_i)+f^*\mathcal{A}_i\wedge f^*\mathcal{A}_i=f^*[d\mathcal{A}_i+\mathcal{A}_i\wedge\mathcal{A}_i]=f^*\mathcal{F}_i.$$ (field strensth is natural) Hence, $f^*P(\mathcal{F}_i)=P(f^*\mathcal{F}_i)$, that is $f^*\chi_E(P)=\chi_{f^*E}(P)$. $\blacksquare$
Corollary 11.1. Characteristic classes of a trivial bundle are trivial.
Proof. Let $\pi:E\rightarrow M$ be a trivial bundle. Since $E$ is trivial, there exists a map $f:M\rightarrow \{p\}$ such that $E=f^*E_0$ where $E_0\rightarrow \{p\}$ is a bundle over a point $p$. All the de Rham cohomology groups of a point are trivial and so are the characteristic classes. Theorem 11.2(b) ensures that the characteristic classes $\chi_E$ ($=f^*\chi_{E_0}$) of $E$ are also trivial. $\blacksquare$