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Evolution of geometric quantities of a manifold

by Octavian on Sunday, June 14, 2009

Lemma

For a Riemannian manifold M with a time-dependent Riemannian metric g(t) which satisfies

\frac{\partial}{\partial t} g_{ij}(t) = h_{ij}(t)

then these will follow:

  1. Metric inverse
    \frac{\partial}{\partial t} g^{ij} = -h^{ij} = -g^{ik} g_{jl} h_{kl}
  2. Christoffel symbol
    \frac{\partial}{\partial t} \Gamma_{ij}^k = \frac{1}{2} g^{kl} \big( \nabla_i h_{jl} + \nabla_j h_{il} - \nabla_l h_{ij} \big)
  3. Riemann curvature tensor
    \frac{\partial}{\partial t} R_{ijk}^l = \frac{1}{2} g^{lp} \big( \nabla_i \nabla_j h_{kp} + \nabla_i \nabla_k h_{jp} - \nabla_i \nabla_p h_{jk} - \nabla_j \nabla_i h_{kp} - \nabla_j \nabla_k h_{ip} + \nabla_j \nabla_p h_{ik} \big)
  4. Ricci tensor
    \frac{\partial}{\partial t} R_{ij} = \frac{1}{2} g^{pq} \big( \nabla_q \nabla_i h_{jp} + \nabla_q \nabla_j h_{ip} - \nabla_q \nabla_p h_{ij} - \nabla_i \nabla_j h_{qp} \big)
  5. Scalar curvature
    \frac{\partial}{\partial t} R = -\Delta H + \nabla^p \nabla^q h_{pq} - h^{pq} R_{pq}, where H = g^{ij} h_{ij}
  6. Volume element
    \frac{\partial}{\partial t} d \mu = \frac{H}{2} d \mu
  7. Volume of manifold M
    \frac{\partial}{\partial t} \int_M d \mu = \int_M \frac{H}{2} d \mu
  8. Total scalar curvature on a closed manifold M
    \frac{\partial}{\partial t} \int_M R d \mu = \int_M (\frac{1}{2} RH - h^{ij} R_{ij}) d \mu

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From → Differential Geometry

4 Comments →
  1. juanmarqz permalink

    wow! let me tell you that those formulas are awesome… i wonder if you have one for the geodesic curvature, i mean, how it varies?
    thank u

    Reply
  2. Octavian permalink

    Dear Juanmarqz,

    I’m still studying about your question, because I’m still a beginner in this field. Especially because I only learn the bundle-like differential geometry, and am not familiar with the geometry in surfaces.
    Surely we can discuss it together ^^

    The first step to find the evolution of geodesic curvature is just finding the expression of geodesic curvature, related to metric tensor. Does the relation in this page help?

    Reply
  3. juanmarqz permalink

    My friend Octavian:
    We use Malyarenko notes to deduce a succinct formula for k_g (look at page 3), and from there, perhaps we can give the solution, Thanks

    Reply
  4. Octavian permalink

    I find a typo in page 3, it should be:
    “…Differentiating (2), we obtain
    t' = \frac{dx_u}{ds}u' + x_u u'' + \frac{dx_v}{ds}v'+ x_v v''“, not “…\frac{dx_u}{du}…”.

    The expression for \kappa_g in page 4 (the formula just above “Geodesics”) is too rigorous for us to variate it. I choose another way, it seems more efficient to variate \kappa_g from equation (1), but we must know how the normal and the tangent vectors evolve under metric evolution \frac{\partial g_{ij} (t)}{\partial t} = h_{ij} (t). It’s in progress, my bro…

    Reply

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