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.

## 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…