# Contribution of the metric tensor towards the Riemann curvature tensor

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?

• 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 $x$. – Bence Racskó Dec 27 '17 at 21:07
• Or to put Uldreth's comment in different words, you need to know not just the metric but its second derivatives at that point. – Ben Crowell Dec 27 '17 at 21:13
• 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. – IanDsouza Dec 27 '17 at 21:43
• @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. – Ben Crowell Dec 27 '17 at 22:53
• @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 – IanDsouza Dec 27 '17 at 23:52

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 $g$.