4. Find all geodesics on the Lobachevsky plane.

Solution. We will present the Lobachevsky plane as (with the metric , z=x+iy).

First we will prove that the vertical ray is a geodesic for any fixed . To this end we will use the reflection which is an isometry. Consider a geodesic passing through a point of and having the vertical direction there. Then is a geodesic with the same initial conditions, hence which implies that .

Now let us use other isometries with , ad-bc=1. They should move to other geodesics. But in this way we can get all half-circles centered at y=0. It is easy to check that such a half-circle can be drawn through any chosen point in any chosen direction (except vertical ones). Therefore any geodesic is either a vertical ray or a half-circle centered at y=0.

Remark. (a) Note that the vertical rays can be obtained from the half-circles centered at y=0 in the limit as the radius goes to .

(b) Note that straight horizontal lines (with ) are not geodesics.

5. Let G be a Lie group with biinvariant Riemannian metric. Prove that geodesics passing through are exactly one-parametric subgroups, i.e. smooth maps such that and for all .

Solution. (i) Without loss of generality we can assume that G is connected (if not, we can switch to the connected component of e). We will denote left and right multiplication maps by as and respectively, so , . Both these maps are isometries.

(ii) We will use the following properties of the Lie group G with a biinvariant Riemannian metric.

A. G is a homogeneous space, i.e. for any there exists an isometry mapping x to y.

Indeed, it is sufficient to consider a left multiplication map by .

B. G is a symmetric space. This means that for any there exists an isometry such that and (i.e. induces the standard reflection of the tangent space at p).

Indeed, we can take . This map induces the standard reflection of , so is an isometry on . For arbitrary we use the identity to see that is also an isometry due to the biinvariance of the metric.

Now for any we can construct an isometry inducing the standard reflection of using the homogeneity of G (see A above): we can just define .

(iii) Note that homogeneity implies that G is complete, so all geodesics can be extended to all real values of their parameters (or, equivalently, have infinite length in both directions). Let be a geodesic such that . Then for all . Now let us fix t and consider the map . It is easy to see that it preserves the geodesic and in fact maps to . On the other hand from the explicit expression of above we find that it maps to . So we have

Taking u=0 we get . Then taking u=t we obtain . Arguing by induction we get for any integer and any . It follows that for any of the form where are arbitrary integers. Since t is arbitrary we can replace it by (with an arbitrary positive integer N) to obtain that for all rational . By continuity this is true then for all which proves that is a one-parametric subgroup.

(iv) Now we should prove that any one-parametric subgroup is geodesic. We can find a geodesic such that (as we proved it will be a one-parametric subgroup then) and have the same tangent vector when t=0. So it is sufficient to prove that a one-parametric subgroup with the given tangent vector at t=0 is unique.

(v) Let be a one-parametric subgroup with the given tangent vector at t=0:

We claim that it is possible to reconstruct by v. To this end consider the left-invariant vector field X on G such that . Then

for all . This means that is an integral curve of X, hence it is uniquely defined by v.

6. Describe all isometric transformations of the Lobachevsky plane (with the metric , z=x+iy) which leave a given point p fixed. They should include ``rotations" (which actually induce rotations in the tangent space ) and reflections with respect to a geodesic passing through p.

Solution. Because of homogeneity of it is sufficient to consider the case p=i. Then we can consider ``rotations"

which are isometries inducing the corresponding rotations in , and also a reflection which induces a reflection in . This transformations generate the whole group of orthogonal transformations of . Now note that any isometry with is uniquely defined by (which is a linear orthogonal transformation) because any geodesic passing through p with the tangent vector should be mapped by f to the geodesic passing through p with the tangent vector . Therefore we have one-one correspondence between the orthogonal transformations of and isometries of mapping p into itself.

7. Prove that the group of all isometries of is a 3-dimensional Lie group which is generated by the following transformations:

1) ``Translations": ;

2)``Homotheties" or ``Scalings": ;

3) ``Rotations around the point i":

4) Reflection .

Solution. The solution follows from the homogeneity of and the description of all isometries with a fixed point p given in the solution of the problem 6.

9. Calculate all the components of the curvature tensor for the paraboloid

(here ) with the metric which is induced by the standard Euclidean metric in .

Solution. It is most convenient to work in polar coordinates so that

Then dz=ardr, , and the Riemannian metric on is

We will take Then

Using the formula

we obtain

therefore

Now substituting this to the formula for the curvature tensor, we find

hence

In particular the sectional curvature (or Gauss curvature) is

(Here .)

Remark. Note that the coordinates are orthogonal but not orthonormal. Therefore to calculate in orthonormal coordinates we have to multiply it by a normalizing factor which can be easily obtained from the tensor transformation law. This exactly leads to the extra factor because the sectional curvature (or Gauss curvature) do not depend on coordinates and g=1 in orthonormal coordinates.

10. Calculate the curvature tensor on the standard sphere

where R>0.

( Hint: Use exercise 9 and symmetries.)

Solution. Note first that the sphere is a homogeneous space. The group of all orthogonal transformations of acts on transitively and all these transformations are isometries. Moreover they allow to move any orthonormal basis of a tangent space to any other orthonormal basis of any other tangent space. It follows that will be the same for any orthonormal basis X,Y in and any . Therefore it is sufficient to calculate it at the North Pole . Let us try to approximate the sphere near N by a paraboloid. To do this write the equation of the sphere near N as

where the term has vanishing derivatives of orders 0,1,2,3 at r=0. Therefore this term can be ignored in the calculation of the curvature tensor at N. Also the constant term R and the minus sign before do not change the curvature because replacing z by leads to an isometry. It follows that in the calculation of the sectional curvature (or Gauss curvature) at N we can replace by a paraboloid (near r=0) with . Using the calculations done in the solution of the problem 9, we see that the sectional curvature of the sphere at any point is equal to the one of the paraboloid at r=0 which is .

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