Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It is customary to use DeTurck's argument (or Hamilton's original one involving the Nash-Moser iteration) for proving local existence of the Ricci flow. I am wondering why one cannot use harmonic coordinates for this purpose, as can be done for the Einstein equations.

share|cite|improve this question
up vote 6 down vote accepted

There are at least two problems if you naively approach the problem from that point of view:

  1. How do you propose to time-evolve the harmonic coordinate system? For the Einstein's equation the coordinate system is more properly called "wave coordinates" as the coordinate functions solve the wave equation (compare to the Riemannian case when the coordinate functions solve the Laplace equation).
  2. Harmonic coordinate systems are not necessarily global. For hyperbolic equations like Einstein's equation, finite speed of propagation implies that you can cut up into small neighborhoods, and solve in the domain of dependence o that neighborhood in the (local) wave coordinate system. For parabolic equation one cannot use the same method.

That said, the DeTurck trick can be interpreted as the "harmonic coordinate" version of the proof. In the DeTurck trick you solve the modified Ricci flow with a vector field $X^a = g^{bc}\Gamma_{bc}^a$ where the $\Gamma$ are the analogues of the Christoffel symbol "relative to a fixed background metric". Now, the choice of harmonic coordinate system is one in which $0 = g^{bc}\Gamma_{bc}^a$ (the real Christoffel symbol relative to the coordinates now). So we can see DeTurck's trick as circumventing the difficulty in choosing a global, truly harmonic coordinate system, by compensating it with a time-dependent diffeomorphism that "gets rid of" that extra non-harmonicity.

The correct analogue for the DeTurck trick in Einstein's equations is not the simple o' wave coordinate system. Instead, it is what Choquet-Bruhat calls the "$\hat{e}$ wave gauge" and what is sometimes known as the "wave-map gauge"; for more details you should consult chapter VI.7 of her recent monograph General Relativity and the Einstein Equations.

share|cite|improve this answer
Thanks Willie. Just to add what I know, in his papers on prescribed Ricci curvature problem, DeTurck invented and applied this argument to several related problems, including the Einstein equations (even the inhomogeneous version). The argument seems simpler than the wave coordinate approach to the Cauchy problem of GR. I actually wondered why his argument did not get into GR textbooks. – timur Jul 24 '12 at 18:36
There's actually not much difference between the homogeneous and inhomogeneous version. The power of the DeTurck trick (and of the various wave gauges) is that the equations in that gauge becomes manifestly hyperbolic with a diagonal principal symbol, that is, the Ricci curvature term(s) can be replaced by (roughly speaking) $g^{ij}\partial^2_{ij}g_{kl}$. Now since Einstein's equation can always be written as $R_{ij} = T_{ij} - \frac12 T g_{ij}$ (moving the trace term to the other side) the principal part is purely determined by Ricci. – Willie Wong Jul 24 '12 at 19:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.