0
$\begingroup$

If I specify a metric tensor in terms of a choice of bases in the co-tangent space at a given point in a manifold, how much information does the metric tell me about the curvature tensor at that point? Can I calculate the curvature tensor entirely from the metric tensor?

$\endgroup$
  • 5
    $\begingroup$ At a single point? None at all. You need to know the metric in the "second order infinitesimal neighborhood" of a given point $x$ to be able to tell the curvature at $x$. $\endgroup$ – Bence Racskó Dec 27 '17 at 21:07
  • 1
    $\begingroup$ Or to put Uldreth's comment in different words, you need to know not just the metric but its second derivatives at that point. $\endgroup$ – Ben Crowell Dec 27 '17 at 21:13
  • $\begingroup$ I can still specify it as $g _{\mu\nu} (x)$ at a point in the manifold. The explicit $x$ dependence should contain information about the second order derivatives. $\endgroup$ – IanDsouza Dec 27 '17 at 21:43
  • $\begingroup$ @IanDsouza: If you're specifying it as a function of x, then you're not specifying it at a point, you're specifying it at all points. $\endgroup$ – Ben Crowell Dec 27 '17 at 22:53
  • $\begingroup$ @BenCrowell I guess expressing $g\mu\nu(x)$ might make sense at given point if you're considering infinitesimal neighborhoods (as you pointed out the second derivative might be important here). I'm a little weary of using the same function $g\mu\nu(x)$ over the entire manifold because $g\mu\nu(x)$ belongs to the (0,2) tensor space defined at some point, say $p_1$. At any other point, wouldn't you have to define a separate (0,2) tensor space at say point $p_2$ which won't be the same tensor space defined at $p_1$ though there might pull back maps that might take you from one space to the other $\endgroup$ – IanDsouza Dec 27 '17 at 23:52
0
$\begingroup$

Yes you can, but you need to know the metric up to the second derivative, so the metric in a neighb. of the manifold. You can do that for the Levi-Civita connection, since it is completely defined by the metric $g$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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