To get a better feel of the Riemann curvature tensor and sectional curvature:
- Work through one of the definitions of the Riemann curvature tensor and sectional curvature with a $2$-dimensional sphere of radius $r$.
- Define the hyperbolic plane as the space-like "unit sphere" of $3$-dimensional Minkowski space, defined using an inner product with signature $(-,+,+)$. Work out the sectional and Riemann curvature of that
- Repeat #1 and #2 for the $n$-dimensional sphere and hyperbolic space, as well as flat space
Sectional curvature determines Riemann curvature:
That the sectional curvature uniquely determines the Riemann curvature is a consequence of the following:
- The Riemann curvature tensor is a quadratic form on the vector space of $\Lambda^2T_xM$
- The sectional curvature function corresponds to evaluating the Riemann curvature tensor (as a quadratic form) on decomposable elements of $\Lambda^2T_xM$
- There is a basis of $\Lambda^2T_xM$ consisting only of decomposable elements
Added in response to Anirbit's comment
Perhaps you shouldn't try to compute the curvature too soon. First, make sure you understand the Riemannian metric of the unit sphere and hyperbolic space inside out. There are many ways to do this. But the most concrete way I know is to use stereographic projection of the sphere onto a hyperplane orthogonal to the last co-ordinate axis. Either the hyperplane through the origin or the one through the south pole works fine. This gives you a very nice set of co-ordinates on the whole sphere minus one point. Work out the Riemannian metric and the Christoffel symbols. Also, work out formulas for an orthonormal frame of vector fields and the corresponding dual frame of 1-forms. Figure out the covariant derivatives of these vector fields and the corresponding dual connection 1-forms.
After you do this, do everything again with hyperbolic space, which is the hypersurface
$-x_0^2 + x_1^2 + \cdots + x_n^2 = -1$ with $x_0 > 0$
in Minkowski space with the Riemannian metric induced by the flat Minkowski metric. You can do stereographic projection just like for the sphere but onto the unit $n$-disk given by
$x_1^2 + \cdots + x_n^2 = 1$ and $x_0 = 0$,
where the formula for the hyperbolic metric looks just like the spherical metric in stereographic co-ordinates but with a sign change in appropriate places. This is the standard conformal model of hyperbolic space.
After you understand this inside out, you can use these pictures to figure out why the $n$-sphere and its metric is given by $O(n+1)/O(n)$ and hyperbolic space by $O(n,1)/O(n)$ and why the metrics you've computed above correspond to the natural invariant metric on these homogeneous spaces. You can then check that the formulas for invariant metrics on homogeneous spaces give you the same answers as above.
Use references only for the general formulas for the metric, connection (including Christoffel symbols), and curvature. I recommend that you try to work out these examples by hand yourself instead of trying to follow someone else's calculations. If possible, however, do it with another student who is also trying to learn this at the same time.
If, however, you want to peek at a reference for hints, I recommend the book by Gallot, Hulin, and Lafontaine. I suspect that the book by Thurston is good too (I studied his notes when I was a student). For invariant Riemannian metrics on a homogeneous space, I recommend the book by Cheeger and Ebin (available cheap from AMS! When I was a student, I had to pay a hundred dollars for this little book but it was well worth it).
But mostly when I was learning this stuff, I did and redid the same calculations many times on my own. I was never able to learn much more than a bare outline of the ideas from either books or lectures. Just try to get a rough idea of what's going on from the books, but do the details yourself.
Here's an answer to an analogous question, not Bill's original question, but also a question about how to specify (in a simple way) a non-positively curved metric on a compact Riemann surface $C$ of genus $g>1$, in this instance, one that has been specified as an algebraic curve somewhere (as opposed to being given it as a surface in $3$-space).
The construction is easy: If the curve has been specified as an algebraic curve, then, more-or-less by algorithmic means, one can write down a basis for the holomorphic differentials on $C$ (which is a complex vector space of dimension $g$). Now select two of these differentials, say, $\omega$ and $\eta$, that have no common zeroes on $C$. (Again, this can be tested algebraically). Now consider the metric $g = \omega\circ\bar\omega + \eta\circ\bar\eta$. This $g$ will have non-positive curvature. In fact, the curvature will vanish at only a finite number of points and will otherwise be strictly negative. (Of course, you can add more terms. If you take a basis $\omega_1,\ldots,\omega_g$ of the holomorphic differentials on $C$, then the metric $g = \omega_1\circ\bar{\omega_1} +\cdots + \omega_g\circ\bar{\omega_g}$ will have strictly negative curvature except when $C$ is hyperelliptic, in which case, the curvature will vanish at the Weierstrass points of $C$.)
For example, if you take a hyperelliptic curve, say $y^2 = (x-\lambda_1)(x-\lambda_2)\cdots(x-\lambda_{2g+2})$ (with the $\lambda_i$ being distinct and, say, nonzero), then a basis for the holomorphic differentials will be given by $\omega_i = x^{i-1}dx/y$ for $i = 1,\ldots, g$. Moreover, $\omega_1$ and $\omega_g$ (for example) have no common zeros. Thus, the smooth metric $g = (1 + |x|^{2(g-1)})|dx|^2/|y|^2$ has negative curvature on this curve except at a finite number of points.
Best Answer
I'll briefly spell out what others have pointed to concerning geodesics on surfaces of revolution (or more generally, surfaces with a 1-parameter group of symmetries), because it's nice and not as widely understood as it should be.
Geodesics on surfaces of revolution conserve angular momentum about the central axis, so the geodesic flow splits into 2-dimensional surfaces having constant energy (~length) and angular momentum (The more general principle is that the inner product of the tangent to a geodesic with any infinitesimal isometry of a Riemannian manifold is constant). The surfaces are generically toruses. The shadow of these toruses on the surface of revolution is an annulus, a component of a set of $r \ge r_0$, where on each point with $r > r_0$ there are two vectors having the given angular momentum, but they merge at the boundary, both becoming tangent to the boundary of the annulus. If you sketch the picture, you will see the torus. The geodesics correspond to the physical phenomenon of the pattern of string or thread mechnically but passively wound around a cylinder. As string builds up in the middle, geodesics start to oscillate back and forth in a sinusoidal pattern, further amplifying the bulge in the middle.
To find the geodesic from point x to point y, you need to know which angular momentum will take you from x to y. For any two meridian circles and any choice of angular momentum, the geodesics of given angular momentum map one circle to the other by a rotation. Both the angle of rotation of the map and the length of the particular family of geodesics traversing the annulus is given by an integral over an interval cutting across the annulus, since the the slope of the vector field at all intervening points is known. I have an aversion to actual symbolic computation so I won't give you example formulas, but I believe this should meet your criterion for explicitness.
But to take a step back: this question, asking for an explicit formula, has an unstated (and probably unintended) connotation that is worth examining: this use of language implicitly suggests that non-symbolic forms are less worthy. I don't know the background motivation for the question, but an alternative question for some purposes would be to give example of surfaces where you can exhibit the distance function. Communication of mathematics is biassed toward symbolic forms. However, for many people and many purposes, some kind of graphical representation of the distance function, and/or diagrams or explanations of why it is what it is as well as a striaghtforward method for computing it, would often be better than a symbolic answer.
The geodesic flow of course is an ordinary differential equation. It is a vector field on the 3-manifold of unit-length tangent vectors to the surface, defined by very easy equations: the vectors are tangent to the surface, and their derivative (= the 2nd derivative of a geodesic arc) is normal to the surface. The solutions may not always have a nice symbolic form, but they always have a nice and easy-to-compute geometric form. Finding the distance involves the implicit function theorem, but this is easy and intuitive. One could, for instance, easily draw a parametric surface that is the graph of distance as a function of position directly from solutions to the ODE (which no doubt sometimes even have reasonable symbolic representations). Both the ODE for the geodesic flow and the inverse function to give distance as a function of position are easy to compute numerically, and easy to understand qualitatively.