## June 1, 2012

### Normal Coordinate Expansion

I’ve been spending several weeks at the Simons Center for Geometry and Physics. Towards the end of my stay, I got into a discussion with Tim Nguyen, about Ricci flow and nonlinear $\sigma$-models. He’d been reading Friedan’s PhD thesis, alongside Kevin Costello’s book. So I pointed him to some old notes of mine on the normal coordinate expansion, a key ingredient in renormalizing nonlinear $\sigma$-models, using the background-field method.

Then it occurred to me that the internet would be a much more useful place for those notes. So, since I have some time to kill, in JFK, here they are.

Let $\phi:\Sigma\to M$ be a map from the worldsheet, $\Sigma$ into a Riemannian Manifold, $(M,g)$. The NL$\sigma$M action is
(1)$S= \int_\Sigma \tfrac{1}{2}(\phi^*g)(\partial_\mu,\partial_\mu) d^2x$
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,
(2)$\begin{split} \nabla_X Y-\nabla_Y X-[X,Y]&\equiv T(X,Y)=0\\ X(g(Y,Z))&=g(\nabla_X Y,Z)+g(Y,\nabla_X Z) \end{split}$
The Riemann curvature tensor is
(3)$R(X,Y)=\nabla_X\nabla_Y-\nabla_Y \nabla_X-\nabla_{[X,Y]}$
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
(4)$\hat\phi(x,t)=\phi_t(x)$
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
(5)$\begin{split} \xi(x,t)&=\hat\phi_*\tfrac{\partial}{\partial t}\\ \hat\nabla_{\partial_t}\xi&=0 \end{split}$
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
(6)$\begin{gathered} \gamma:\, [0,1]\to M\\ \ddot{\gamma}^k + \Gamma_{ij}^k \dot{\gamma}^i\dot{\gamma}^k=0\\ \gamma(0)=\phi(x),\qquad \dot{\gamma}(0)=\xi(x) \end{gathered}$
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,
(7)$\nabla_\xi\xi=0$
(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
(8)$\hat\nabla_{\partial_t}v =\hat\nabla_{\partial_\mu}\xi$
where
(9)$v=\hat\phi_* \partial_\mu$
or, with the same abuse of notation,
(10)$\nabla_\xi v =\nabla_v \xi$
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,
(11)$\nabla_\xi\nabla_v \xi= R(\xi,v)\xi$
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
(12)$\phi^*(e^{t'\nabla_\xi} T) = \phi^*(T+t'\nabla_\xi T +\tfrac{t'^2}{2}\nabla^2_\xi T +\dots)$
We are all set to apply this to the $\sigma$-model Lagrangian,
(13)$\tfrac{1}{2}(\phi_t^*g)(\partial_\mu,\partial_\mu) = \phi_t^*(\tfrac{1}{2} g(v,v))$
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
(14)$g(U,R(V,W)Z)=g(W,R(Z,U)V)$
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
(15)$\begin{split} \phi_t^*\tfrac{1}{2}g(v,v)=\phi^*[\tfrac{1}{2}g(v,v)& +g(v,\nabla_v \xi) +\tfrac{1}{2}g(\nabla_v \xi,\nabla_v \xi) + \tfrac{1}{2}g(v,R(\xi,v) \xi)\\ &+\tfrac{1}{6} g(v,(\nabla R)(\xi,\xi,v)\xi) +\tfrac{2}{3}g(\nabla_v\xi,R(\xi,v)\xi)\\ &+\tfrac{1}{4} g(\nabla_v \xi,(\nabla R)(\xi,\xi,v)\xi) +\tfrac{1}{6}g(\nabla_v \xi,R(\xi,\nabla_v \xi)\xi)\\ & +\tfrac{1}{6}g(R(\xi,v)\xi,R(\xi,v)\xi) +\tfrac{1}{24}g(v,(\nabla \nabla R)(\xi,\xi,\xi,v )\xi)] \end{split}$
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 $r^2(\mu) = r^2(\mu_0) + \frac{n-1}{4\pi}\log(\mu/\mu_0)$
[1] S. Weinberg, Gravitation and Cosmology, (Wiley, 1972) p. 148.
[2] D. H. Friedan, “Nonlinear Models in $2+\epsilon$ Dimensions,” Ann. Phys. 163 (1985) 318.
[3] L. Alvarez-Gaume, D. Z. Freedman and S. Mukhi, “The Background Field Method and the Ultraviolet Structure of the Supersymmetric Nonlinear Sigma Model,” Ann. Phys. 134 (1981) 85.
Posted by distler at June 1, 2012 8:12 PM

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### Re: Normal Coordinate Expansion

Good notes in index free notation, thank you!

I was recently playing with “sigma”, the world function of deWitt, which is a bi-scalar. A covariant derivative of sigma transforms as a vector on one point and a scalar on the other. It appears that the expansion of the non-linear sigma model in xi, that you performed, coincides with that in the derivative of the world function. However, I’m having problems relating the two objects in a formal way. Perhaps you have notes on this too…

Posted by: OmarZ on June 27, 2012 12:12 PM | Permalink | Reply to this

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