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\begin{document}

\centerline {\bf Riemannian Geometry and General Relativity}
\centerline{\bf (Differential Geometry-2, MTH 3412)}
\centerline{\bf Home Assignement 1. Solutions of selected problems.}
\centerline{\bf Professor M.Shubin}
\centerline{\bf Winter 1997}
%\medskip\noindent
%{\bf Textbooks:}
%\smallskip\noindent
%1) {\it Riemannian Geometry}, by Manfredo Perdig\~ao do Carmo.
%Birkh\"auser, Boston, 1993.
%\smallskip\noindent
%2) {\it General Theory of Relativity}, by P.A.M.Dirac. Princeton
%University Press, 1996.

\medskip\noindent
{\bf 3.} Let $\langle,\rangle$ be a symmetric Lorentzian inner product in
$\RR^4$
(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 {\it time-like} vector if
$\langle x,x\rangle>0$.
Prove that for any two time-like vectors $x,y$  we have the inverse
Cauchy-Schwarz inequality
$\langle x,y\rangle^2\ge \langle x,x\rangle\langle y,y\rangle\;.$

\medskip
{\bf Solution.} Without loss of generality we can assume that the given
inner product
has the form
$$\langle x,y\rangle=x^0y^0-x^1y^1-x^2y^2-x^3y^3,$$
where $x=(x^0,x^1,x^2,x^3)$, $y=(y^0,y^1,y^2,y^3)$.

Let us consider the light cone
$$C=\{x|\,\langle x,x\rangle=0\}$$
and the sets
$$C^+=\{x|\,\langle x,x\rangle>0\},\quad C^-=\{x|\,\langle x,x\rangle<0\}$$
of time-like and space-like vectors respectively.
\medskip

{\bf Lemma.} {\it Any straight line passing through  a  time-like point
intersects
the light cone.}

\medskip
{\bf Proof.}  Let $l$ be a  straight line  which contains a point  $x\in
C^+$.We need to
proof that $l$ intersects $C$ (i.e. has at least one common point with $C$)

Consider the hyperplane $H=\{y|\,y^0=0\}$. Then $l$ either is parallel to $H$
(i.e. $x^0$ is constant along $l$) or intersects $H$.
Note that all points of $H$ are in $C^-$ 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 $\{y|\,y^0=x^0\}$. But the
intersection of
$C^+$ with this hyperplane  is a 3-dimensional ball
$\{y|\,y^0=x^0,\;(y^1)^2+(y^2)^2+(y^3)^2<(x^0)^2\}$. Clearly $l$ can not be
confined in this
ball, so it should intersect its boundary which is in $C$. 
% $\ecarre$

\medskip
Now let us consider 2 points $x,y\in C^+$ and a straight line
$l_{x,y}=\{x+ty|\,t\in\RR\}$.
The equality
$$0=\langle x+ty,x+ty\rangle=\langle x,x\rangle+2t\langle x,y\rangle+t^2\langle y,y\rangle$$
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.

\medskip
{\bf Remark.} You might notice that we only used that $x\in C^+$, 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 $0$.

\medskip\noindent
{\bf 8.} Find a manifold $M$, $\dim_{\RR} M=4$, such that there exists no
Lorentzian metric on $M$.

\medskip
{\bf 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 $C^+$ the set
of all
time-like vectors. Each tangent space $T_pM$ is in fact isomorphic to
$\RR^4$ with
the standard Lorentzian metric  given by the formula
$$\langle v,w\rangle=v^0w^0-v^1w^1-v^2w^2-v^3w^3,$$
and then  $C^+_p=C^+\cap T_pM=\{v|\,\langle v,v\rangle>0\}$
is separated by the hyperplane $H=\{v|\,v^0=0\}$ into 2 connected components
corresponding to $v^0>0$ and $v_0<0$. It might happen that $C^+$ itself has
2 connected
components (as is the case e.g. for $\RR^4$). Another possibility is that
$C^+$ 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 $p\in M$ but in a different
connected component of $C^+_p$. Then the pairs
$$\{p, \hbox{connected component of}\ T_pM\}$$
 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 $S^4$. Over this sphere we can fix a choice of one of the two
connected components
of $C^+_p$ over each point $p\in S^4$ continuously with respect to $p$.
Denote this
connected component by $D^+_p$. It is a convex cone in $T_pM$. Locally in a
small neighbourhood
$U$ of any given point  we can choose
a continuous vector-field $v_U$ such that $v_U(p)\in D^+_p$ for all $p\in
U$. Now let us take
a partition of unity $\{\phi_U\}$ subordinated to the covering  of $S^4$ by
the described
neighbourhoods $U$. Then
$$v=\sum_U\phi_Uv_U$$
is a continuous vector field on $S^4$ and $v(p)\in D^+_p$ for all $p\in
S^4$ because $D_p^+$
is convex. In particular $v(p)\not=0$ for all $p\in S^4$. However such a
vector field does not
exist on $S^4$ (the ``haired sphere theorem") because the Euler
characteristics of $S^4$
does not vanish: $\chi(S^4)=2\not=0$. Therefore there is no Lorentzian
metric on $S^4$. 
% $\ecarre$

\medskip
{\bf 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
$\chi(M)=0$.

\medskip\noindent
{\bf 9.} ({\it A construction of  the Lobachevsky plane}). Consider the
group $G$ of all
affine orientation preserving transformations of $\RR$ i.e. transformations
$g_{x,y}:t\mapsto yt+x$, $x\in\RR$, $y>0$. Write a left-invariant
Riemannian metric
on $G$, such that it coincides with the Euclidean metric $dx^2+dy^2$ at the
unit element
$e\in G$. (Here $e$ corresponds to $x=0, y=1$).

\noindent
({\it Answer:} $ds^2={dx^2+dy^2\over y^2}$, or in other words
$g_{12}=0$, $g_{11}=g_{22}=1/y^2$.)

\medskip
{\bf Solution.} First note that for any Euclidean inner product in the Lie
algebra $T_eG$
there exists unique left-invariant Riemannian metric on $G$ which coincides
with the given
inner product on $T_eG$. So the required Riemannian metric on $G$ is unique.

Clearly the group $G$ is generated by translations $\Phi_a=g_{a,1}:t\mapsto
t+a$ and homotheties
(scalings) $H_b=g_{0,b}:t\mapsto bt$. The corresponding left multiplication
transformations on
$G$ are (in coordinates $x,y$):
$$\phi_a:(x,y)\mapsto (x+a,y),\qquad h_b:(x,y)\mapsto (bx,by)\,. $$
A Riemannian metric on $G$ is left-invariant if and only if it is invariant
under both $\phi_a$
and $h_b$ for all $a\in\RR$ and $b>0$.

The invariance under $\phi_a$ means that the components
$g_{\mu\nu}$ (in coordinates $(x,y)$) are independent on $x$. Therefore
these components
should be functions of $y$ only.

Now let us use the notations: $x^1=x,\ x^2=y$.
The invariance under the scalings $h_b$ means that
$$g_{\mu\nu}(by)d(bx)^\mu d(bx)^\nu=g_{\mu\nu}(y)dx^\mu dx^\nu\,,$$
or
$g_{\mu\nu}(by)=b^{-2}g_{\mu\nu}(y)$ which is equivalent to
$g_{\mu\nu}(y)=y^{-2}g_{\mu\nu}(1)$.
If the metric at $e$ is $dx^2+dy^2$, then we necessarily have
$ds^2=y^{-2}(dx^2+dy^2)$
at all points.
%  $\ecarre$

\medskip
{\bf Remark.} By definition all the ``left translations" on $G$ define
isometric
transformations of the Lobachevsky plane. In particular the translations
$\phi_a$
and homotheties $h_b$ 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.


\medskip\noindent
{\bf 10.} In the previous exercise denote $z=x+iy$, $i=\sqrt{-1}$, and show
that
any transformation $z\mapsto {az+b\over cz+d}$ with $ad-bc=1$ is an
isometry of $G$ with the
metric $ds^2$.

\noindent
({\it Hint}: Observe that $ds^2=-{4dzd\bar z\over (z-\bar z)^2}$.)

\medskip
{\bf Solution.} Using the identity
$${az+b\over cz+d}={a\over c}\Biggl(1-{ad-bc\over a(cz+d)}\Biggr)=
\Biggl({a\over c}-{1\over c(cz+d)}\Biggr)\,,$$
we see that the group of the transformations $z\mapsto {az+b\over cz+d}$
with $ad-bc=1$ is generated by the transformations from Excercise 9 and by
the transformation $z\mapsto -1/z$. The last transformation preserves the
metric
as can be seen from the following calculation:
$${d\Bigl(-{1\over z}\Bigr)d\Bigl(-{1\over \bz}\Bigr)\over \Bigl(-{1\over
z}+{1\over \bz}\Bigr)^2}
={\Bigl({dz\over z^2}\Bigr)\Bigl({d\bz\over \bz^2}\Bigr)\over
\Bigl({z-\bz\over z\bz}\Bigr)^2}
={dzd\bz\over (z-\bz)^2}\;.\qquad $$

% $\ecarre$

\end{document}