3. Let 
be a symmetric Lorentzian inner product in
(i.e. it is non-degenerate and the corresponding quadratic form is of the type
+---, that is has exactly one positive
and 3 negative squares). A vector x is called a time-like vector if
.
Prove that for any two time-like vectors x,y we have the inverse
Cauchy-Schwarz inequality
Solution. Without loss of generality we can assume that the given inner product has the form
where 
, 
.
Let us consider the light cone
and the sets
of time-like and space-like vectors respectively.
Lemma. Any straight line passing through a time-like point intersects the light cone.
Proof. Let l be a straight line which contains a point 
.We need to
proof that l intersects C (i.e. has at least one common point with C)
Consider the hyperplane 
. Then l either is parallel to H
(i.e. 
is constant along l) or intersects H.
Note that all points of H are in 
except the origin which is in C.
Therefore if l intersects H, it should also intersect C. Now assume
that l is
parallel to H. Then l lies in the hyperplane 
. But the
intersection of
with this hyperplane is a 3-dimensional ball
. Clearly l can not be
confined in this
ball, so it should intersect its boundary which is in C.
Now let us consider 2 points 
and a straight line
.
The equality
considered as quadratic equation with respect to t, has a solution due to Lemma. Therefore the discriminant of this equation should be non-negative which exactly gives the desired inequality.
Remark. You might notice that we only used that 
, so the
inequality is true
for any time-like x and arbitrary y. But for any space-like or
isotropic y the inequality is
trivial because the right hand side is negative or 0.
8. Find a manifold M, 
, such that there exists no
Lorentzian metric on M.
Solution. We can assume without loss of generality that M is
connected.
Any Lorentzian metric on M in particular defines a division of all tangent
vectors into time-like, space-like and isotropic. Denote by 
the set
of all
time-like vectors. Each tangent space 
is in fact isomorphic to
with
the standard Lorentzian metric given by the formula
and then 
is separated by the hyperplane 
into 2 connected components
corresponding to 
and 
. It might happen that 
itself has
2 connected
components (as is the case e.g. for 
). Another possibility is that
is connected.
In this case moving along M and dragging with us a time-like vector, we
can return to
the another time-like vector over the same point 
but in a different
connected component of 
. Then the pairs
form a (connected) 2-fold covering manifold of M. To exclude this possibility let us take a manifold M which is simply-connected and therefore does not have any non-trivial coverings.
In particular let us consider the 4-dimensional
sphere 
. Over this sphere we can fix a choice of one of the two
connected components
of 
over each point 
continuously with respect to p.
Denote this
connected component by 
. It is a convex cone in 
. Locally in a
small neighbourhood
U of any given point we can choose
a continuous vector-field 
such that 
for all 
. Now let us take
a partition of unity 
subordinated to the covering of 
by
the described
neighbourhoods U. Then
is a continuous vector field on 
and 
for all 
because 
is convex. In particular 
for all 
. However such a
vector field does not
exist on 
(the ``haired sphere theorem") because the Euler
characteristics of 
does not vanish: 
. Therefore there is no Lorentzian
metric on 
.
Remark. It can be proved that in fact a Lorentzian metric on a
connected manifold
exists if and only if either M is not compact or M is compact and
.
9. ( A construction of the Lobachevsky plane). Consider the
group G of all
affine orientation preserving transformations of 
i.e. transformations
, 
, y>0. Write a left-invariant
Riemannian metric
on G, such that it coincides with the Euclidean metric 
at the
unit element
. (Here e corresponds to x=0, y=1).
( Answer: 
, or in other words
, 
.)
Solution. First note that for any Euclidean inner product in the Lie
algebra 
there exists unique left-invariant Riemannian metric on G which coincides
with the given
inner product on 
. So the required Riemannian metric on G is unique.
Clearly the group G is generated by translations 
and homotheties
(scalings) 
. The corresponding left multiplication
transformations on
G are (in coordinates x,y):
A Riemannian metric on G is left-invariant if and only if it is invariant
under both 
and 
for all 
and b>0.
The invariance under 
means that the components
(in coordinates 
) are independent on x. Therefore
these components
should be functions of y only.
Now let us use the notations: 
.
The invariance under the scalings 
means that
or
which is equivalent to
.
If the metric at e is 
, then we necessarily have
at all points.
Remark. By definition all the ``left translations" on G define
isometric
transformations of the Lobachevsky plane. In particular the translations
and homotheties 
are such isometries. But the left translations do not
exhaust
the group of all isometries which form a 3-dimensional Lie group. Still
missing are
``rotations" which are nontrivial transformations which leave just one
point fixed,
and also ``reflections" in a ``mirror" which can be an arbitrary geodesic.
10. In the previous exercise denote z=x+iy, 
, and show
that
any transformation 
with ad-bc=1 is an
isometry of G with the
metric 
.
( Hint: Observe that 
.)
Solution. Using the identity
we see that the group of the transformations 
with ad-bc=1 is generated by the transformations from Excercise 9 and by
the transformation 
. The last transformation preserves the
metric
as can be seen from the following calculation:
