1. Prove that in 4-dimensional case the cyclic relation
is independent of the antisymmetry and symmetry relations
i.e. there exists a tensor of type (0,4) in 
such that it satisfies
(2) but not (1).
Solution. We will restrict ourselves to tensors at a point or,
equivalently, tensors
whose components are constants. In fixed coordinates such a tensor
is an arbitrary set of 256 numbers. Let us try to construct a tensor
such that the symmetry relations (2) are satisfied and
therefore (1) is not true. To see that this is possible it is sufficient to
see that
the symmetry group generated by the relations (2) applied to 
,
obviously leads to three non-intersecting sets of components, so extension of
to arbitrary
with the conditions (2) will not lead to a
contradiction.
2. Prove that in 2-dimensional case
where 
is the scalar curvature.
Solution. Both sides of the formula satisfy the symmetry relations
(2) above. Therefore
all components of each of them are expressed in terms of one (say
). It follows that
at each point they are proportional. To see the coincidence it is
sufficient to check that
their full contractions coincide i.e.
Since the left-hand side is S by definition of the scalar curvature, it
is sufficient to check
that
, which
is obvious because 
and
.
4. Find the mass of a black hole if its horizon sphere has the size
of an atom i.e.
the radius about 
cm.
Solution.
5. Prove that the Schwarzschild solution indeed has a singularity (which can not be corrected by changing coordinates).
Main ideas of the solution.
A. Let us say that a Lorentzian manifold is geodesically complete if every geodesic can be infinitely extended in both directions with respect to the natural parameter s (interval or proper time). It is physically natural to consider such a manifold as a space/time without singularities because any observer moving there in a spaceship can do it forever by the ship's internal clock.
Also if a manifold M can be included as an open set into a geodesically complete manifold, then we will consider M as having no true singularities, because the apparent singularities are caused by the fact that we consider only a part of a true space.
So a genuine singularity of M can be described as follows: M should have a geodesic that ends at a finite value of its natural parameter and can not be extended even in a larger Lorentzian manifold (in which M is isometrically imbedded).
B. How can we detect a genuine singularity of M? The easiest way is to
produce
a geodesic 
on M,
defined for 
, and a scalar field 
on M such that:
(i) 
is expressed locally as a non-singular function of 
and
their derivatives;
(ii) 
is unbounded as 
.
This would obviously imply that M has a genuine singularity (``at the end
of 
").
C. Now let us consider the Schwarzschild metric
How can we produce the geodesic 
and the scalar field 
for this
metric?
It is natural to take the geodesic 
which starts as a vertical
(radial) fall of the
test particle (or spaceship) from a distant point to the origin (r=0).
After crossing
the Schwarzschild radius 2m we should switch to other coordinates, e.g.
replace t,r
by 
as described in Sect. 19 of the Dirac's book. There r=0
corresponds to
and it is easy to see that our spaceship (moving along 
)
will reach
in a finite proper time.
D. Now let us try to produce an appropriate scalar field 
. It is
natural to
try to do this starting from the curvature tensor. Scalar curvature is a
natural scalar
field
but unfortunately it is identically 0 for the Schwarzschild metric because
even the Ricci
tensor vanishes identically.
But there are other scalar quantities which can be constructed
from the curvature tensor. Let us consider the curvature tensor
(with 2 upper suffixes). It is antisymmetric with respect to 
and
also with
respect to 
. Therefore for every point it defines a linear map on
antisymmetric tensors of type (2,0). (This map is in fact symmetric due to
the relation
.) Its eigenvalues or
coefficients of the characteristic
polynomial are scalar fields. For example the trace of this map is minus
the scalar curvature.
One of the scalar fields obtained in such a way is the ``Hilbert-Schmidt norm" of the curvature tensor which is
E. Let us calculate the scalar field 
from (4) for the Schwarzschild
metric (3)
starting at the points with r>2m (i.e. outside of the Schwarzschild
sphere). A standard
(though tedious) calculation shows that
Now note that 
is an analytic function of r for all r>0. It
follows that it is
given by the same formula even in the coordinates 
with
(See formula (19.8) in Dirac.)
It follows that we indeed meet a genuine singularity at r=0 (or 
).
7. Consider a satellite rotating around a black hole of the given
mass m (or given
Schwarzschild radius 2m) along a circular orbit of a radius R>2m
(outside the horizon sphere). Find
the period of the rotation by a distant observer clock and also by the clock in
the satellite. What happens with these periods as 
?
Solution. As usual denote 
,
and 
.
Then
and
For a given circular orbit we can choose coordinates so that
. Also 
.
It follows that 
. Symmetry implies that 
and 
are
constants. Denote
.
Equation (6) with 
takes the form
An easy calculation shows that
so we obtain
or
On the other hand (5) gives in our situation
Combining these 2 relations we find a and b:
It follows that by the satellite clock the period will be
Note that this period becomes 0 when R=3m which means that the
satellite should move
with the speed of light to remain on the circular orbit. Therefore the
circular orbits with
R<3m are impossible. (It follows that with the initial condition 
and any tangential initial velocity the spaceship will fall into the black
hole; to escape the
attraction of the black hole, it should move radially away from it.)
On the other hand by the distant observer clock the period will be
Note that the distant observer will not notice that something special
happens when
the radius becomes equal to 3m! And 
as a function of R
is exactly the same as in the classics (i.e. it obeys the 3rd Kepler law:
is
proportional to 
).
9. Let G be a compact Lie group with a biinvariant Riemannian
metric, X,Y,Z
left-invariant vector fields on G. Prove that 
and
.
Solution. The fact that one-parametric subgroups and their left
translations are geodesics
implies that 
for any left-invariant vector field Z on G.
Applying this to
Z=X+Y we obtain 
. On the other hand
due to the symmetry of the Levi-Civita connection. It follows that
.
Now
10. Let G be a compact Lie group with a biinvariant Riemannian
metric, X,Y,Z,W
left-invariant vector fields on G. Prove that 
and
.
Solution. Note first that 
, therefore
hence 
.
Now
