Musings

Let $\phi:\Sigma\to M$ be a map from the worldsheet, $\Sigma$ into a Riemannian Manifold, $(M,g)$. The NL$\sigma$M action is The Normal Coordinate Expansion is a particularly nice parametrization of fluctuations about the classical $\sigma$-model field $\phi$. I’ll use index-free notation, wherever possible. The Levi-Cevita connection, $\nabla$, on $M$ is torsion-free and metric-compatible, The Riemann curvature tensor is Consider a 1-parameter family of such $\sigma$-model maps, $\phi_t:\Sigma\to M$, $t\in[0,1]$ with $\phi_0=\phi$, our original $\sigma$-model map. We can equally-well think about this family as a map $\hat\phi:\Sigma\times[0,1]\to M$, with Given $\hat\phi$, we can pull back the connection, $\nabla$, on $M$ to a connection $\hat \nabla$ on $\Sigma\times [0,1]$. We don’t want to choose any old 1-parameter family, though. Let $\xi(x)$ be a section of the pullback tangent bundle, $\phi^*TM$. We wish to choose $\hat\phi$ so that we can extend $\xi(x)$ to $\xi(x,t)$ such that How do we achieve this? The idea is that, given a point $x\in\Sigma$, $\xi(x)$ gives a tangent vector to $M$ at the point $\phi(x)$. For this “initial condition”, we solve the geodesic equation and define the point $\phi_t(x)\in M$ to be $\gamma(t)$. This is guaranteed to be well-defined for small enough $t$. We can extend it to $t=1$ by considering $\xi$ to be sufficiently “small”. The extension of $\xi(x)$ to $\xi(x,t)$ is simply the one given by parallel-transporting $\xi$ along the curve $\gamma$, $$\hat\nabla_{\partial_t}\xi=0$$ or, with slight abuse of notation, (This is an abuse of notation because $\xi$ is not really a tensor on $M$. We typically have points $x,x'\in\Sigma$ with $\phi(x)=\phi(x')$ but $\xi(x)\neq\xi(x')$. This “mistake” will correct itself when we pull back to $\Sigma$.) Since $\Bigl[\tfrac{\partial}{\partial t},\partial_\mu\bigr]=0$ and the connection $\nabla$ is torsion-free, we can always exchange where or, with the same abuse of notation, Taking another covariant derivative with respect to $t$, we get the equation of geodesic deviation [1] $$\hat\nabla_{\partial_t}^2v =\hat\nabla_{\partial_t}\hat\nabla_{\partial_\mu}\xi =\hat R(\partial_t,\partial_\mu)\xi$$ or, in our abusive notation, Now say we wish to evaluate the $t$-dependence of the pull-back of some tensor $T$ on $M$, $$\phi_{t'}^* T = e^{t'\hat\nabla_{\partial_t}}(\hat\phi^*T)\vert_{t=0}$$ which we can, again, write as We are all set to apply this to the $\sigma$-model Lagrangian, We expand this using (12) and use (7),(10) and (11) to simplify the terms that result. Note that the $n^{th}$ order term in (12) is given by $\tfrac{ 1}{n}\nabla_\xi$ of the $(n-1)^{st}$ order term. The $0^{th}$ order term is $$\tfrac{1}{2}g(v,v)$$ At first order, we get $$g(v,\nabla_\xi v)= g(v,\nabla_v \xi)$$ Next comes $$\tfrac{1}{2}g(\nabla_\xi v,\nabla_v \xi)+\tfrac{1}{2}g(v,\nabla_\xi \nabla_v \xi)= \tfrac{1}{2}g(\nabla_v \xi,\nabla_v \xi) + \tfrac{1}{2}g(v,R(\xi,v) \xi)$$ At $3^{rd}$ order, we get $$\begin{split} \tfrac{1}{6}[2g(\nabla_v \xi,R(\xi,v )\xi)+g(\nabla_v \xi,R(\xi,v)\xi)& +g(v,(\nabla_\xi R)(\xi,v)\xi) +g(v,R(\xi,\nabla_v\xi)\xi)]\\ &=\tfrac{1}{6} g(v,(\nabla R)(\xi,\xi,v)\xi) +\tfrac{2}{3}g(\nabla_v\xi,R(\xi,v)\xi) \end{split}$$ where we used the symmetry of the Riemann tensor Finally, at $4^{th}$ order, we get $$\begin{split} \tfrac{1}{24}[g(\nabla_v \xi,(\nabla R)(\xi,\xi,v )\xi) &+g(v,(\nabla \nabla R)(\xi,\xi,\xi,v )\xi) +g(v,(\nabla R)(\xi,\xi,\nabla_v\xi)\xi) ]\\ &+\tfrac{1}{6}[g(R(\xi,v)\xi,R(\xi,v)\xi) +g(\nabla_v \xi,(\nabla R)(\xi,\xi,v)\xi) +g(\nabla_v \xi,R(\xi,\nabla_v \xi)\xi)]\\ =&\tfrac{1}{4} g(\nabla_v \xi,(\nabla R)(\xi,\xi,v)\xi) +\tfrac{1}{6}g(\nabla_v \xi,R(\xi,\nabla_v \xi)\xi)\\ &+\tfrac{1}{6}g(R(\xi,v)\xi,R(\xi,v)\xi) +\tfrac{1}{24}g(v,(\nabla \nabla R)(\xi,\xi,\xi,v )\xi) \end{split}$$ and so forth. Assembling all of these, we obtain Pulling back to $\Sigma$, we obtain the desired expansion of the $\sigma$-model lagrangian. In conventional notation, replace $v^i=\partial_\mu\phi^i$, $(\nabla_v\xi)^i=D_\mu \xi^i$ and $g(U,R(W,Z)V)=R_{ijkl}U^i V^j W^k Z^l$ to obtain the expressions found in Friedan [2] or Freedman et al [3].

Exercise 1: Compute the next term in the expansion, $$\begin{split} \tfrac{1}{12}g(\nabla_v\xi,(\nabla R)(\xi,\xi,\nabla_v\xi)\xi) +\tfrac{2}{15}g(R(\xi,v)\xi,R(\xi,\nabla_v\xi)\xi) +\tfrac{1}{15}g(\nabla_v\xi,(\nabla\nabla R)(\xi,\xi,\xi,v)\xi)\\ +\tfrac{7}{60}g(R(\xi,v)\xi,(\nabla R)(\xi,\xi,v)\xi) +\tfrac{1}{120}g(v,(\nabla\nabla\nabla R)(\xi,\xi,\xi,\xi,v)\xi) \end{split}$$

Exercise 2: Let $M$ be the $n$-sphere, $S^n$, with the round metric, $$ds^2= \frac{4 r^2(\mu) dx^i dx^i}{(1+|\mathbf{x}|^2)^2}$$ Show that the solution to the one-loop $\beta$-function equation is $$r^2(\mu) = r^2(\mu_0) + \frac{n-1}{4\pi}\log(\mu/\mu_0)$$