Inverting the parallel transport via path-ordered exponential

Question

In Riemann geometry one can formally solve the parallel transport equation

$$ \dot{v}^\mu + \Gamma^\mu_{\rho\sigma} \, u^\rho \, v^\sigma = 0 $$

of a vector $v$ along a curve with unit tangent vector $u^\mu = \dot{x}^\mu$ using the path-ordered exponential

$$ P^\mu_\nu(s,0) = \left( \text{P exp} - \int_0^s ds \, \Gamma \, u \right)^\mu_\nu $$

$$ v^\mu(s) = P^\mu_\nu(s,0) \, v^\nu(0) $$

Suppose we have

$$ \langle v(s), w(s) \rangle = \langle P(s,0) \, v(0), P(s,0) \, w(0) \rangle $$

with

$$ \langle v, w \rangle = g_{\mu\nu} \, v^\mu \, w^\nu $$

Question: can one show that the path-ordered exponentials cancel?

$$ \langle v(s), w(s) \rangle = \langle v(0), w(0) \rangle $$

Can you give some reference to this path ordered exp? In any case the integral in the formula for P should be an exponent — magma, Commented Sep 2, 2020 at 2:57
Please refer to Carroll‘s lecture notes, parallel propagator, eq. (3.45) ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll3.html — TomS, Commented Sep 2, 2020 at 5:10
The answer is rather simple in SU(N) gauge theories b/c one sees explicitly that the path ordered exponential $P$ is a SU(N) matrix, and that reversing the path order sends each factor to its own inverse; one can use the algebraic properties of the su(n) generators. For the connection coefficients $\Gamma$ I do not see this algebraic property, so the inversion stays on the formal level. — TomS, Commented Sep 2, 2020 at 5:19
Ok, this is my guess: reversing the path order sends the tangent vector $u \to -u$, that means each factor $(1 + \Gamma u) ds \to (1 - \Gamma u) ds$ which should proof that $P$ is an SO(3,1) matrix, and that for any path $C$ we have $P^{-1}[C] = P[-C]$ as expected. Hope this is sufficient. — TomS, Commented Sep 2, 2020 at 6:04

TomS · Accepted Answer · 2020-09-02 11:09:09Z

1

The answer ist „yes“.

Lecture Notes on General Relativity

Sean M. Carroll, Institute for Theoretical Physics

Chapter 3. Curvature

answered Sep 2, 2020 at 11:09

TomS

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