[Nakahara GTP] 11.5 Chern-Simons forms

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

11.5.1 Definition

Let $P_j(\mathcal{F})$ be an arbitrary $2j$-form characteristic class. Since $P_j(\mathcal{F})$ is closed, it can be written locally as an exact form by Poincare's lemma. Let us write $$\begin{align}P_j(\mathcal{F})=dQ_{2j-1}(\mathcal{A},\mathcal{F})\end{align}$$ where $Q_{2j-1}(\mathcal{A},\mathcal{F})\in \mathfrak{g}\otimes \Omega^{2j-1}(M)$. (Warning: This cannot be true globally. If $P_j=dQ_{2j-1}$ globally on a manifold $M$ without boundary, we would have $$\int_M P_{m/2}=\int_M dQ_{m-1}=\int_{\partial M}Q_{m-1}=0$$ where $m=\dim M$.) The $2j-1$ form $Q_{2j-1}(\mathcal{A},\mathcal{F})$ is called the Chern-Simons form of $P_j(\mathcal{F})$. From the proof of theorem 11.2(b), we find that $Q$ is given by the trangression of $P_j$, $$\begin{align} Q_{2j-1}(\mathcal{A},\mathcal{F})=TP_j(\mathcal{A},0)=j\int^1_0\tilde{P}_j(\mathcal{A},\mathcal{F}_t,\cdots,\mathcal{F}_t)dt\end{align}$$ where $\tilde{P}_j$ is the polarization of $P_j$, $\mathcal{F}=d\mathcal{A}+\mathcal{A}^2$ and we set $\mathcal{A}'=\mathcal{A}'=0$. Since $Q_{2j-1}$ depends on $\mathcal{F}$ and $\mathcal{A}$, we explicitly quote the $\mathcal{A}$-dependence. Of course, $\mathcal{A}'$ can be put equal to zero only on a local chart over which the bundle is trivial.

Suppose $M$ is an even-dimensional manifold ($\dim M=m=2l$) such that $\partial M\ne \varnothing$. Then it follows from Stokes' theorem that $$\begin{align}\int_MP_l(\mathcal{F})=\int_MdQ_{m-1}(\mathcal{A},\mathcal{F})=\int_{\partial M}Q_{m-1}(\mathcal{A},\mathcal{F}).\end{align}$$ The LHS takes its value in integers, and so does the RHS. Thus $Q_{m-1}$ is a characteristic class in its own right ant it describes the topology of the boundary $\partial M$.

11.5.2 The Chern-Simons form of the Chern character

As an example, let us work out the Chern-Simnons form of a Chern character $\mbox{ch}_j(\mathcal{F})$. The connection $\mathcal{A}_t$ which interpolates between $0$ and $\mathcal{A}$ is $$\begin{align}\mathcal{A}_t=t\mathcal{A}\end{align}$$ the corresponding curvature being $$\begin{align}\mathcal{F}_t=td\mathcal{A}+t^2\mathcal{A}^2=t\mathcal{F}+(t^2-t)\mathcal{A}^2.\end{align}$$ We find form (11.1.21) that $$\begin{align}Q_{2j-1}(\mathcal{A},\mathcal{F})=\frac{1}{(j-1)!}\left(\frac{i}{2\pi}\right)^j\int_0^1dt\ \mbox{str}(\mathcal{A},\mathcal{F}^{j-1}_t).\end{align}$$ For example, $$\begin{align}Q_1(\mathcal{A},\mathcal{F})=\frac{i}{2\pi}\int^1_0dt\ \mbox{tr}\mathcal{A}=\frac{i}{2\pi}\mbox{tr}\mathcal{A}\\ Q_3(\mathcal{A},\mathcal{F})=\left(\frac{i}{2\pi}\right)^2\int^1_0dt\ \mbox{str}(\mathcal{A},td\mathcal{A}+t^2\mathcal{A}^2)=\frac{1}{2}\left(\frac{i}{2\pi}\right)^2\mbox{tr}\left(\mathcal{A}d\mathcal{A}+\frac{2}{3}\mathcal{A}^3\right)\\ Q_5(\mathcal{A},\mathcal{F})=\frac{1}{2}\left(\frac{i}{2\pi}\right)^3\int^1_0dt\ \mbox{str}[\mathcal{A},(td\mathcal{A}+t^2\mathcal{A}^2)^2]=\frac{1}{6}\left(\frac{i}{2\pi}\right)^3\mbox{tr}\left[\mathcal{A}(d\mathcal{A})^2+\frac{3}{3}\mathcal{A}^3d\mathcal{A}+\frac{3}{5}\mathcal{A}^5\right]\end{align}$$

For example, $SU(2)$ gauge theory, $\mbox{ch}_2(\mathcal{F})=dQ_3(\mathcal{A},\mathcal{F})$ is $$\begin{align}\mbox{tr}[\epsilon^{\kappa\lambda\mu\nu}\mathcal{F}_{\kappa\lambda}\mathcal{F}_{\mu\nu}]=\partial_\kappa[2\epsilon^{\kappa\lambda\mu\nu}\mbox{tr}(\mathcal{A}_\lambda\partial_\mu\partial_\nu+\frac{2}{3}\mathcal{A}_\lambda\mathcal{A}_\mu\mathcal{A}_\nu)].\end{align}$$

11.5.3 Cartan's homotopy operator and applications

For later purpose, we define Cartan's homotopy formula following Zumino (1985) and Alvarez-Gaume and Ginsparg (1985). Let $$\begin{align}\mathcal{A}_t=\mathcal{A}_0+t(\mathcal{A}_1-\mathcal{A}_0)\quad \mathcal{F}_t=d\mathcal{A}_t+\mathcal{A}_t^2\end{align}$$ as before. Define an operator $l_t$ by $$\begin{align}l_t\mathcal{A}_t=0\quad l_t\mathcal{F}_t=\delta t(\mathcal{A}_1-\mathcal{A}_0).\end{align}$$ We require that $l_t$ be an anti-derivative (????), $$\begin{align}l_t(\eta_p\omega_q)=(l_t\eta_p)\omega_q+(-1)^p\eta_p(l_t\omega_q)\end{align}$$ for $\eta_p\in\Omega^p(M)$ and $\omega_q\in\Omega^q(M)$. We verify that $$(dl_t+l_td)\mathcal{A}_t)=l_t(\mathcal{F}_t-\mathcal{A}^2)=\delta t(\mathcal{A}_1-\mathcal{A}_0)=\delta t \frac{\partial \mathcal{A}_t}{\partial t}$$ and $$(dl_t+l_td)\mathcal{F}_t=d[\delta t(\mathcal{A}_1-\mathcal{A}_0)]+l_t[\mathcal{D}_t\mathcal{F}_t-\mathcal{A}_t\mathcal{F}_t+\mathcal{F}_t\mathcal{A}_t]\\ =\delta t[d(\mathcal{A}_1-\mathcal{A}_0)+\mathcal{A}_t(\mathcal{A}_1-\mathcal{A}_0)+(\mathcal{A}_1-\mathcal{A}_0)\mathcal{A}_t]\\ =\delta t\mathcal{D}_t(\mathcal{A}_1-\mathcal{A}_0)=\delta t\frac{\partial \mathcal{F}_t}{\partial t}$$ where we have used the Bianchi identity $\mathcal{D}_t\mathcal{F}_t=0$. This shows that for any polynomial $S(\mathcal{A},\mathcal{F})$ of $\mathcal{A}$ and $\mathcal{F}$, we obtain $$\begin{align}(dl_t+l_td)S(\mathcal{A}_t,\mathcal{F}_t)=\delta t \frac{\partial}{\partial t}S(\mathcal{A}_t,\mathcal{F}_t).\end{align}$$ On the RHS, $S$ should be a polynomial of $\mathcal{A}$ and $\mathcal{F}$ only and not of $d\mathcal{A}$ or $d\mathcal{F}$. (if $S$ does contain them, $d\mathcal{A}$ should be replaced by $\mathcal{F}-\mathcal{A}^2$ and $d\mathcal{F}$ by $\mathcal{D}\mathcal{F}-[\mathcal{A},\mathcal{F}]=-[\mathcal{A},\mathcal{F}]$.) Integrating (13) over $[0,1]$, we obtain Cartan's homotopy formula $$\begin{align}S(\mathcal{A}_1,\mathcal{F}_1)-S(\mathcal{A}_0,\mathcal{F}_0)=(dk_{01}+k_{01}d)S(\mathcal{A}_t,\mathcal{F}_t)\end{align}$$ where tha homotopy operator $k_{01}$ is defined by $$\begin{align}k_{01}S(\mathcal{A}_t,\mathcal{F}_t)\equiv \int^1_0\delta t l_t S(\mathcal{A}_t,\mathcal{F}_t).\end{align}$$ To operate $k_{01}$ on $S(\mathcal{A},\mathcal{F})$, we first replace $\mathcal{A}$ and $\mathcal{F}$ by $\mathcal{A}_t$ and $\mathcal{F}_t$, respectively, the operator $l_t$ on $S(\mathcal{A}_t,\mathcal{F}_t)$ and integrate over $t$.

Example 11.9. Let us compute the Chern-Simons form of the Chern character using the homotopy formula. Let $S(\mathcal{A},\mathcal{F})=\mbox{ch}_{j+1}(\mathcal{F})$ and $\mathcal{A}_1=\mathcal{A}$, $\mathcal{A}_0=0$. Since $d\, \mbox{ch}_{j+1}(\mathcal{F})=0$, we have $$\mbox{ch}_{j+1}(\mathcal{F})=(dk_{01}+k_{01}d)\mbox{ch}_{j+1}(\mathcal{F}_t)=d[k_{01}\mbox{ch}_{j+1}(\mathcal{F}_t)].$$ Thus, $k_{01}\mbox{ch}_{j+1}(\mathcal{F})$ is identified with the Chern-Simons form $Q_{2j+1}(\mathcal{A},\mathcal{F})$. We find that $$\begin{align}k_{01}\mbox{ch}_{j+1}(\mathcal{F}_t)=\frac{1}{(j+1)!}k_{01}\ \mbox{tr}\left(\frac{i\mathcal{F}}{2\pi}\right)^{j+1}\nonumber\\ \frac{1}{(j+1)!}\left(\frac{i}{2\pi}\right)^{j+1}\int^1_0\delta tl_t\ \mbox{tr}(\mathcal{F}^{j+1}_t)\\ =\frac{1}{j!}\left( \frac{i}{2\pi}\right)^{j+1}\int^1_0\delta t \ \mbox{str}(\mathcal{A},\mathcal{F}^j_t)\end{align}$$ in agreement with (6).

Although a characteristic class is gauge invariant, the Chern-Simons form need not be so. As an application of Cartan's homotopy formula, we compute the change in $Q_{2j+1}(\mathcal{A},\mathcal{F})$ under $\mathcal{A}\rightarrow \mathcal{A}^g=g^{-1}(\mathcal{A}+d)g$, $\mathcal{F}\rightarrow \mathcal{F}^g=g^{-1}\mathcal{F}g$. Consider the interpolating families $\mathcal{A}^g_t$ and $\mathcal{F}^g_t$ defined by $$\begin{align}\mathcal{A}^g_t\equiv tg^{-1}\mathcal{A}g+g^{-1}dg\\ \mathcal{F}^g_t\equiv d\mathcal{A}^g_t+(\mathcal{A}^g_t)^2=g^{-1}\mathcal{F}_tg\end{align}$$ where $\mathcal{F}_t\equiv t\mathcal{F}+(t^2-t)\mathcal{A}^2$. Note that $\mathcal{A}^g_0=g^{-1}dg$, $\mathcal{A}^g_1=\mathcal{A}^g$, $\mathcal{F}^g_0=0$ and $\mathcal{F}^g_1=\mathcal{F}^g$. Equation (15) yields $$\begin{align}Q_{2j+1}(\mathcal{A}^g,\mathcal{F}^g)-Q_{2j+1}(g^{-1}dg,0)=(dk_{01}+k_{01}d)Q_{2j+1}(\mathcal{A}^g_t,\mathcal{F}^g_t)$.\end{align}$$ For example, let $Q_{2j+1}$ be the Chern-Simons form of the Chern character $\mbox{ch}_{j+1}(\mathcal{F})$. Since $dQ_{2j+1}(\mathcal{A}^g_t,\mathcal{F}^g_t)=\mbox{ch}_{j+1}(\mathcal{F}^g_t)=\mbox{ch}_{j+1}(\mathcal{F}_t)$, we have $$\begin{align}k_{01}dQ_{2j+1}(\mathcal{A}^g_t,\mathcal{F}^g_t)=k_{01}\mbox{ch}_{j+1}(\mathcal{F}^g_t)\nonumber\\ = k_{01}\mbox{ch}_{j+1}(\mathcal{F}_t) = Q_{2j+1}(\mathcal{A},\mathcal{F})\end{align}$$ where the result of example 11.9 has been used to obtain the final equailty. Collection these results, we write (21) as $$\begin{align}Q_{2j+1}(\mathcal{A}^g,\mathcal{F}^g)-Q_{2j+1}(\mathcal{A},\mathcal{F})=Q_{2j+1}(g^{-1}dg,0)+d\alpha_{2j}\end{align}$$ where $\alpha_{2j}$ is a $2j$-form defined by $$\begin{align}\alpha_{2j}(\mathcal{A},\mathcal{F},v)\equiv k_{01}Q_{2j+1}(\mathcal{A}^g_t,\mathcal{F}^g_t)=k_{01}Q_{2j+1}(\mathcal{A}_t+v,\mathcal{F}_t)\end{align}$$ where $v\equiv dg\cdot g^{-1}$. (Note that $Q_{2j+1}(\mathcal{A},\mathcal{F})=Q_{2j+1}(g\mathcal{A}g^{-1}, g\mathcal{F}g^{-1})$.) The first term on the RHS of (23) is $$\begin{align}Q_{2j+1}(g^{-1}dg,0)=\frac{1}{j!}\left( \frac{i}{2\pi}\right)^{j+1}\int^1_0\delta t\, \mbox{tr}[g^{-1}dg\{(t^2-t)(g^{-1}dg)^2\}^j]\nonumber\\ =\frac{1}{j!}\left( \frac{i}{2\pi}\right)^{j+1}\mbox{tr}[(g^{-1}dg)^{2j+1}]\int^1_0\delta t(t^2-t)^j\nonumber\\ =(-1)^j\frac{j!}{(2j+1)!}\left( \frac{i}{2\pi}\right)^{j+1}\mbox{tr}[(g^{-1}dg)^{2j+1}]\end{align}$$ where we have noted that $\mathcal{F}_t=(t^2-t)(g^{-1}dg)^2$ and $$\int^1_0\delta t(t^2-t)^j=(-1)^jB(j+1,j+1)=(-1)^j\frac{(j!)^2}{(2j+1)!}$$ $B$ being the beta function. The $2j+1$ form $Q_{2j+1}(gdg,0)$ is closed and, hence, locally exact: $dQ_{2j+1}(g^{-1}dg,0)=\mbox{ch}_{j+1}(0)=0$.

As for $\alpha_{2j}$ we have, for example, $$\begin{align}\alpha_2=\frac{1}{2}\left(\frac{i}{2\pi}\right)^2\int^1_0l_t\, \mbox{tr}[(\mathcal{A}_t+v)\mathcal{F}_t-\frac{1}{3}(\mathcal{A}_t+v)^3]\nonumber\\ =\frac{1}{2}\left(\frac{i}{2\pi}\right)^2\int^1_0\delta t\, \mbox{tr}(-t\mathcal{A}^2-v\mathcal{A})\nonumber\\ =-\frac{1}{2}\left(\frac{i}{2\pi}\right)^2\mbox{tr}(v\mathcal{A})\end{align}$$ where we have noted that $$\mbox{tr}\mathcal{A}^2=dx^\mu\wedge dx^\nu \mbox{tr}(\mathcal{A}_\mu\mathcal{A}_\nu)=-dx^\nu\wedge dx^\mu\mbox{tr}(\mathcal{A}_\nu\mathcal{A}_\mu)=0.$$

Example 11.10. In three-dimensional spacetime, a gauge theory may have a gauge-invariant mass term given by the Chern-Simons three form. Since the Chern-Simons form changes by a locally exact form under a gauge transformation, the action remains invariant. We ristrict ourselves to the $U(1)$ gauge theory for simplicity. Consider the Lagrangian (we put $\mathcal{A}=iA$, $\mathcal{F}=iF$) $$\begin{align}\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{1}{4}m\epsilon^{\lambda\mu\nu}F_{\lambda\mu}A_\nu\end{align}$$ where $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$. Note that the second term in the Chern-Simons form of the second Chern character $F^2$ of the $U(1)$ bundle. The field equation is $$\begin{align}\partial_\mu F^{\mu\nu}+m\star  F^\nu=0\end{align}$$ where $$\star F^\mu=\frac{1}{2}\epsilon^{\mu\kappa\lambda}F_{\kappa\lambda}\quad F^{\mu\nu}=\epsilon^{\mu\nu\lambda}\star F_\lambda.$$ The Bianchi identity $$\begin{align}\partial_\mu \star F^\mu=0\end{align}$$ follows from (28) as a consequence of the skew symmetric of $F^{\mu\nu}$. It is easy to verify that the field equation is invariant under a gauge transformation, $$\begin{align}A_\mu\rightarrow A_\mu+\partial_\mu\theta\end{align}$$ while the Lagrangian changes by a total derivative, $$\begin{align}\mathcal{L}\rightarrow -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+\frac{1}{4}m\epsilon^{\lambda\mu\nu}F_{\lambda\mu}(A_\nu+\partial_v\theta)=\mathcal{L}+\frac{1}{2}m\partial_\nu(\star F^\nu\theta).\end{align}$$ Equation (11.106b) shows taht the last term on the RHS is identified with $$Q_3(A^\theta,F^\theta)-Q_3(A,F)\sim (A+d\theta)dA-AdA\sim d(\theta dA).$$ If we assume that $F$ falls off at large spacetime distances, this term does not contribute to the action: $$\begin{align}\int d^3x\mathcal{L}\rightarrow \int d^3x\mathcal{L}+\frac{m}{2}\int d^3x\partial_\nu (\star F^\nu\theta)=\int d^3x \mathcal{L}.\end{align}$$

Let us show that (27) describes a massive field. We first wirte (28) as $$\epsilon^{\mu\nu\alpha}\partial_\mu \star F_\alpha=-m\star F^\nu.$$ Multiplying $\epsilon_{\kappa \lambda\nu}$ on both sides, we have $$\partial_\lambda\star F_\kappa-\partial_\kappa \star F_\lambda=-mF_{\kappa\lambda}.$$ Taking the $\partial^\lambda$-derivative and using (29), we find that $$\begin{align}(\partial^\lambda\partial_\lambda+m^2)\star F_\kappa=0\end{align}$$ which shows that $\star F_\kappa$ is a massive vector field of mass $m$.