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.
Best Answer
Suppose by contradiction it is. Write it as $G/K$ where $G$ is the identity component of the isometry group and $K$ is compact, and $G$ acts faithfully on $G/K$. Since $G$ is connected, maximal compact subgroups are connected. Since $G/K$ is not contractible, $K$ is not maximal compact and it follows that maximal compact subgroups have codimension $1$. Since in noncompact simple Lie groups, maximal compact subgroups have codimension $\ge 2$, it follows that $G$ has no noncompact simple factor.
Suppose by contradiction that $G$ is not solvable. Then it has a simple compact connected subgroup $S$. Let $K'$ be a maximal compact subgroup. Then $K'\cap S$ has codimension $\le 1$ in $S$. Since a simple compact group has no subgroup of codimension $1$, it follows that $S\subset K'$. In turn, since $K$ has codimension $1$ in $K'$, by the same argument we deduce that $S\subset K$. Hence $S$ is contained in the intersection of all conjugates of $K$, which is trivial. Contradiction.
So $G$ is a connected solvable Lie group. Let $M$ be a closed connected normal subgroup of codimension $1$. If $MK=G$ then $M$ has smaller dimension and acts transitively, so we can argue by induction. Hence, assuming that $G$ has minimal dimension, we have $MK\neq G$, so $K\subset M$. Since in an abelian connected Lie group the intersection of all codimension 1 closed connected subgroups is trivial, we deduce that $K\subset \overline{[G,G]}$. The latter is nilpotent, and hence $K$ is a central torus in $\overline{[G,G]}$. Since the action of $G$ on the maximal torus of $\overline{[G,G]}$ is trivial, we deduce that $K$ is central, hence trivial. So $G$ is a 2-dimensional Lie group, and in particular is orientable.
(We have proved that if a noncompact connected surface can be endowed with a homogeneous Riemannian metric, then it is diffeomorphic to a Lie group, and hence to the plane, the cylinder. Of course various approaches to this result exist and some have already been mentioned in the comments.)