# Christoffel symbols on a Lie group in Riemann normal coordinates

Consider a coordinate patch around the identity element in a Lie group given by the exponential mapping (Riemann normal coordinates). We have a Levi-Civita connection corresponding to the bi-invariant metric associated with the Killing form. What are the Christoffel symbols for this metric in the Riemann normal coordinates?

Here is a refinement of this question: The Christoffel symbols will have a Taylor expansion in the Riemann coordinates with the coefficients being some tensors constructed out of the Lie algebra structure constants. The first nonvanishing tensor is a 4-tensor constructed out of two factors of the structure constants. That one is easy to compute – this is related to the curvature 4-tensor at the identity. The question is – are the other tensors known in any closed form?

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Why is it interesting to construct the Christoffel symbol with respect to these co-ordinates? Isn't it more interesting to work with geometric invariants such as the curvature tensor? Also, it should be noted that you can do this only if the Lie group is semi-simple, and the metric will be Riemannian only if the group is compact. – Deane Yang Sep 17 '10 at 2:01
@Deane Yang, I suppose there could be a variety of reasons. Sometimes connections between fields are made via observations that fall out of what some may perceive to be uninteresting-looking computations. You don't have to look far to find them: Monsterous Moonshine, or the explicit computations of hyperbolic structures by people like Riley that inspired Thurston. By putting things into different notations and languages you give people more opportunities to spot connections. – Ryan Budney Sep 20 '10 at 2:43

This isn't a full answer but hopefully the beginning of one. I'm assuming it's a matrix group but it might work for all Lie groups anyway.

Let $G$ be a matrix group with Lie algebra $\mathfrak{g}$. Suppose $\gamma(t)$ is a geodesic in $G$ and $x(t)$ is the geodesic in the exponential coordinates, ie. $\gamma(t) = \exp(x(t))$. Then $x(t)$ satisfies an ODE $x''(t) + Q_{x(t)}(x'(t),x'(t)) = 0$, where, for given $x$, the $Q_x$ is some quadratic map $\mathfrak{g} \to \mathfrak{g}$. If we can find $Q$ then we just need to polarise it to a bilinear map $\mathfrak{g} \times \mathfrak{g} \to \mathfrak{g}$ and this bilinear map will contain all the Christoffel symbols.

There is a power series formula for the differential of the exponential map. See G. M. Tuynman, The Derivation of the Exponential Map of Matrices http://www.jstor.org/pss/2974511 - it implies:

$\frac{d}{d t}(\exp(x(t)) = \exp(x(t)) \left( \frac{1 - \exp(-\mathrm{ad}\ x(t))}{\mathrm{ad}\ x(t)}\right)x'(t)$

where $\mathrm{ad}\ x$ is the map which takes $y$ to $[x,y]$ and the fraction $\left( \frac{1 - \exp(-\mathrm{ad}\ x(t))}{\mathrm{ad}\ x(t)}\right)$ means $1 - \frac{1}{2!} \mathrm{ad}\ x(t) + \frac{1}{3!} \mathrm{ad}\ x(t)^2 - \frac{1}{4!} \mathrm{ad}\ x(t)^3 + \cdots,$ ie. it is the power series for the fraction as if $x(t)$ was a real number.

Because the covariant derivative is bi-invariant, the geodesic $\gamma$ must be of the form $\gamma(t) = \gamma(0) \exp(A t)$ for some $A \in \mathfrak{g}$. Differentiating $\exp(x(t)) = \gamma(0) \exp(A t):$

$\exp(x(t)) \left( \frac{1 - \exp(-\mathrm{ad}\ x(t))}{\mathrm{ad}\ x(t)}\right)x'(t) = \gamma(0) \exp(A t) A = \exp(x(t)) A.$

So $\left( \frac{1 - \exp(-\mathrm{ad}\ x(t))}{\mathrm{ad}\ x(t)}\right)x'(t) = A.$ Differentiating this,

$\frac{d}{dt}\left( \frac{1 - \exp(-\mathrm{ad}\ x(t))}{\mathrm{ad}\ x(t)}\right)x'(t) + \left( \frac{1 - \exp(-\mathrm{ad}\ x(t))}{\mathrm{ad}\ x(t)}\right)x''(t) = 0.$ So

$x''(t) + \left( \frac{1 - \exp(-\mathrm{ad}\ x(t))}{\mathrm{ad}\ x(t)}\right)^{-1} \frac{d}{dt}\left( \frac{1 - \exp(-\mathrm{ad}\ x(t))}{\mathrm{ad}\ x(t)}\right)x'(t)=0$, ie.

\begin{align}x''(t) &+ \left( 1 - \frac{1}{2!} \mathrm{ad}\ x(t) + \frac{1}{3!} \mathrm{ad}\ x(t)^2 - \frac{1}{4!} \mathrm{ad}\ x(t)^3 + \cdots \right)\\\\&\quad\cdot \left( - \frac{1}{2!} \mathrm{ad} x'(t) + \frac{1}{3!} (\mathrm{ad}\ x'(t) \mathrm{ad}\ x(t) + \mathrm{ad}\ x(t) \mathrm{ad}\ x'(t)) - \cdots \right) x'(t) = 0\\\\\end{align}

I'm not sure what to do now. We could compute a few terms of it to get some terms of $Q_x$, but it seems like there should be a nice formula, like the one above for the differential of $\exp$, or at least there should be a recursive formula for the coefficients of the series...? It's even possible that the formula simplifies WAY down, note eg that all the terms ending with $\mathrm{ad}(x'(t)) x'(t)$ die instantly.

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The Riemann normal coordinates on a group manifold are obtained by the application of the exponential map to its Lie algebra. This description (and notation) is based on the appendices of the following article by Marinov and terentyev.

http://physics.technion.ac.il/~site/wisw/memoriam/marinov/pub/Mar46.pdf

It is assumed that the group is semisimple.

I have written explicit expressions up to the third order. The Einstein summation convention is implemented.

The invariant metric on a semisimple Lie group manifold has the form:

$g_{ab}(\xi) = M_a^p(\xi) M_b^q(\xi) G_{pq}$

where: $G_{pq}$ is the Cartan-Killing form of the group Lie algebra.

The vielbeins $M_a^b(\xi)$ have the form:

$M_b^a(\xi) = \int_0^1 Ad(exp(\tau\xi))_b^a d\tau$

The Christoffel symbols are are given by:

$\Gamma_{ab}^c (\xi) = L_p^c (\xi)(\partial_a M_b^p +\partial_b M_a^p)$

where: $L_a^b(\xi)$ are the inverse vielbeins:

$L_a^b(\xi)M_b^c(\xi) = \delta_a^c$

given by:

$L_b^a(\xi) = \frac{\partial(log(exp(\eta)exp(\xi))^a)}{\partial \eta^b}|_{\eta=0}$

Using the Baker Campbell Hausdorff formula, we obtain the following expression for the inversse vielbeins

$L_b^a(\xi) = \delta_b^a +\frac{1}{2}C_{bc}^a\xi^c+\frac{1}{12}C_{bc}^pC_{pd}^a\xi^c\xi^d +\frac{1}{720}C_{bn}^m C_{mt}^r C_{rp}^q C_{qs}^a\xi^n \xi^t \xi^p \xi^s + ...$

Using the Hadamard formula, we obtain the following expression for the vielbeins

$M_b^a(\xi) = \delta_b^a -\frac{1}{2}C_{bc}^a\xi^c +\frac{1}{6}C_{bc}^pC_{pd}^a\xi^c\xi^d -\frac{1}{24}C_{bn}^m C_{mt}^r C_{rp}^a \xi^n \xi^t \xi^p +\frac{1}{120}C_{bn}^m C_{mt}^r C_{rp}^q C_{qs}^a\xi^n \xi^t \xi^p \xi^s + ...$

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