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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 σ\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 σ\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 ϕ:ΣM\phi:\Sigma\to M be a map from the worldsheet, Σ\Sigma into a Riemannian Manifold, (M,g)(M,g). The NLσ\sigmaM action is
(1)S= Σ12(ϕ *g)( μ, μ)d 2xS= \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 σ\sigma-model field ϕ\phi. I’ll use index-free notation, wherever possible. The Levi-Cevita connection, \nabla, on MM is torsion-free and metric-compatible,
(2) XY YX[X,Y] T(X,Y)=0 X(g(Y,Z)) =g( XY,Z)+g(Y, XZ) \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)= X Y Y X [X,Y] R(X,Y)=\nabla_X\nabla_Y-\nabla_Y \nabla_X-\nabla_{[X,Y]}
Consider a 1-parameter family of such σ\sigma-model maps, ϕ t:ΣM\phi_t:\Sigma\to M, t[0,1]t\in[0,1] with ϕ 0=ϕ\phi_0=\phi, our original σ\sigma-model map. We can equally-well think about this family as a map ϕ^:Σ×[0,1]M\hat\phi:\Sigma\times[0,1]\to M, with
(4)ϕ^(x,t)=ϕ t(x) \hat\phi(x,t)=\phi_t(x)
Given ϕ^\hat\phi, we can pull back the connection, \nabla, on MM to a connection ^\hat \nabla on Σ×[0,1]\Sigma\times [0,1]. We don’t want to choose any old 1-parameter family, though. Let ξ(x)\xi(x) be a section of the pullback tangent bundle, ϕ *TM\phi^*TM. We wish to choose ϕ^\hat\phi so that we can extend ξ(x)\xi(x) to ξ(x,t)\xi(x,t) such that
(5)ξ(x,t) =ϕ^ *t ^ tξ =0 \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 xΣx\in\Sigma, ξ(x)\xi(x) gives a tangent vector to MM at the point ϕ(x)\phi(x). For this “initial condition”, we solve the geodesic equation
(6)γ:[0,1]M γ¨ k+Γ ij kγ˙ iγ˙ k=0 γ(0)=ϕ(x),γ˙(0)=ξ(x) \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 ϕ t(x)M\phi_t(x)\in M to be γ(t)\gamma(t). This is guaranteed to be well-defined for small enough tt. We can extend it to t=1t=1 by considering ξ\xi to be sufficiently “small”. The extension of ξ(x)\xi(x) to ξ(x,t)\xi(x,t) is simply the one given by parallel-transporting ξ\xi along the curve γ\gamma, ^ tξ=0 \hat\nabla_{\partial_t}\xi=0 or, with slight abuse of notation,
(7) ξξ=0\nabla_\xi\xi=0
(This is an abuse of notation because ξ\xi is not really a tensor on MM. We typically have points x,xΣx,x'\in\Sigma with ϕ(x)=ϕ(x)\phi(x)=\phi(x') but ξ(x)ξ(x)\xi(x)\neq\xi(x'). This “mistake” will correct itself when we pull back to Σ\Sigma.) Since [t, μ]=0\Bigl[\tfrac{\partial}{\partial t},\partial_\mu\bigr]=0 and the connection \nabla is torsion-free, we can always exchange
(8)^ tv=^ μξ \hat\nabla_{\partial_t}v =\hat\nabla_{\partial_\mu}\xi
where
(9)v=ϕ^ * μ v=\hat\phi_* \partial_\mu
or, with the same abuse of notation,
(10) ξv= vξ\nabla_\xi v =\nabla_v \xi
Taking another covariant derivative with respect to tt, we get the equation of geodesic deviation [1] ^ t 2v=^ t^ μξ=R^( t, μ)ξ \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) ξ vξ=R(ξ,v)ξ\nabla_\xi\nabla_v \xi= R(\xi,v)\xi
Now say we wish to evaluate the tt-dependence of the pull-back of some tensor TT on MM, ϕ t *T=e t^ t(ϕ^ *T)| t=0 \phi_{t'}^* T = e^{t'\hat\nabla_{\partial_t}}(\hat\phi^*T)\vert_{t=0} which we can, again, write as
(12)ϕ *(e t ξT)=ϕ *(T+t ξT+t 22 ξ 2T+)\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 σ\sigma-model Lagrangian,
(13)12(ϕ t *g)( μ, μ)=ϕ t *(12g(v,v)) \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 thn^{th} order term in (12) is given by 1n ξ\tfrac{ 1}{n}\nabla_\xi of the (n1) st(n-1)^{st} order term. The 0 th0^{th} order term is 12g(v,v) \tfrac{1}{2}g(v,v) At first order, we get g(v, ξv)=g(v, vξ) g(v,\nabla_\xi v)= g(v,\nabla_v \xi) Next comes 12g( ξv, vξ)+12g(v, ξ vξ)=12g( vξ, vξ)+12g(v,R(ξ,v)ξ) \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 rd3^{rd} order, we get 16[2g( vξ,R(ξ,v)ξ)+g( vξ,R(ξ,v)ξ) +g(v,( ξR)(ξ,v)ξ)+g(v,R(ξ, vξ)ξ)] =16g(v,(R)(ξ,ξ,v)ξ)+23g( vξ,R(ξ,v)ξ) \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) g(U,R(V,W)Z)=g(W,R(Z,U)V)
Finally, at 4 th4^{th} order, we get 124[g( vξ,(R)(ξ,ξ,v)ξ) +g(v,(R)(ξ,ξ,ξ,v)ξ)+g(v,(R)(ξ,ξ, vξ)ξ)] +16[g(R(ξ,v)ξ,R(ξ,v)ξ)+g( vξ,(R)(ξ,ξ,v)ξ)+g( vξ,R(ξ, vξ)ξ)] = 14g( vξ,(R)(ξ,ξ,v)ξ)+16g( vξ,R(ξ, vξ)ξ) +16g(R(ξ,v)ξ,R(ξ,v)ξ)+124g(v,(R)(ξ,ξ,ξ,v)ξ) \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)ϕ t *12g(v,v)=ϕ *[12g(v,v) +g(v, vξ)+12g( vξ, vξ)+12g(v,R(ξ,v)ξ) +16g(v,(R)(ξ,ξ,v)ξ)+23g( vξ,R(ξ,v)ξ) +14g( vξ,(R)(ξ,ξ,v)ξ)+16g( vξ,R(ξ, vξ)ξ) +16g(R(ξ,v)ξ,R(ξ,v)ξ)+124g(v,(R)(ξ,ξ,ξ,v)ξ)] \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 Σ\Sigma, we obtain the desired expansion of the σ\sigma-model lagrangian. In conventional notation, replace v i= μϕ iv^i=\partial_\mu\phi^i, ( vξ) i=D μξ i(\nabla_v\xi)^i=D_\mu \xi^i and g(U,R(W,Z)V)=R ijklU iV jW kZ lg(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, 112g( vξ,(R)(ξ,ξ, vξ)ξ)+215g(R(ξ,v)ξ,R(ξ, vξ)ξ)+115g( vξ,(R)(ξ,ξ,ξ,v)ξ) +760g(R(ξ,v)ξ,(R)(ξ,ξ,v)ξ)+1120g(v,(R)(ξ,ξ,ξ,ξ,v)ξ) \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 MM be the nn-sphere, S nS^n, with the round metric, ds 2=4r 2(μ)dx idx i(1+|x| 2) 2 ds^2= \frac{4 r^2(\mu) dx^i dx^i}{(1+|\mathbf{x}|^2)^2} Show that the solution to the one-loop β\beta-function equation is r 2(μ)=r 2(μ 0)+n14πlog(μ/μ 0) 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+ϵ2+\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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