\magnification\magstep2
\def\Span#1{\langle#1\rangle}
\def\implies{\mathrel{\Longrightarrow}}
 \def\Aff{\mathop{\rm Aff}}
 \def\Eucl{\mathop{\rm Eucl}}
 \def\Glide{\mathop{\rm Glide}}
 \def\Refl{\mathop{\rm Refl}}
 \def\Rot{\mathop{\rm Rot}}
 \def\SO{\mathop{\rm SO}}
 \def\O{\mathop{\rm O}}
 \def\Transl{\mathop{\rm Transl}}
 \def\cosh{\mathop{\rm cosh}}
 \def\sinh{\mathop{\rm sinh}}
 \def\arccosh{\mathop{\rm arccosh}}
% \def\Cup{\bigcup}
\def\1{^{-1}}
\def\al{\alpha}
\def\be{\beta}
\def\ep{\varepsilon}
\def\la{\lambda}
\def\th{\theta}
\def\fie{\varphi}
\def\De{\Delta}
\def\La{\Lambda}
\def\Si{\Sigma}
\def\R{{\bf R}}
\def\C{{\bf C}}
\def\E{{\bf E}}
\def\proj{{\bf P}}
\def\aff{{\bf A}}
\def\sH{{\cal H}}

\parindent0cm

MA2430

THE UNIVERSITY OF WARWICK

SECOND YEAR EXAMINATION: JUNE 1998

GEOMETRY

Time allowed: 2 hours

\medskip
\hrule\vskip.6mm\hrule
\medskip

{\it Read carefully the instructions on the answer book and make sure that the
particulars required are entered on it.}

\medskip
\hrule
\medskip

ANSWER 4 QUESTIONS

If you have answered more than the required 4 questions, you will only be
given credit for your 4 best answers.

\medskip
\hrule\vskip.6mm\hrule
\medskip

\parindent7mm

\itemitem{1.\quad (a)} Define a motion of the Euclidean plane. Describe in
words and figures the different types of motion, and explain without
proof the theorem classifying motions into types.

\itemitem{(b)} In the figure below, $PAQB$ and $DPCQ$ are two rectangles
having a common diagonal, and the median lines $L$, $M$ and vectors $\bf
a$, $\bf b$ are as indicated.

\vskip2.25in

\itemitem{} Set $g_1=\Glide(L,{\bf a})$ and $g_2=\Glide(M,{\bf b})$.
By considering its effect on the points $P$ and $A$ in the diagram,
determine the composite $g_2\circ g_1$ of the two glide reflections.

\vskip.25in
\hrule
\medskip \hfill CONTINUED \dots
\vfill\eject

\itemitem{2.\quad (a)} The model of the non-Euclidean hyperbolic plane
used in the course is the upper sheet $\sH^2$ of the hyperbola
$x^2+y^2-t^2=-1$ in Lorentz space $\R^{2,1}$. Straight lines of $\sH^2$
are the ``great hyper\-bolas'' defined as the intersection of $\sH^2$ with
planes of $\R^{2,1}$. Define the hyperbolic distance $d(P,Q)$ between two
points of $\sH^2$. Explain how the Lorentz group acts as a group of
motions (isometries) on $\sH^2$; how transitive is this action on points
and lines of $\sH^2$? (No proofs are required.)

\itemitem{(b)} Let $\triangle PQR$ be a hyperbolic triangle with right
angle at $P$ and $d(P,Q)=s$, $d(P,R)=t$. State and prove from first
principles a formula expressing the length of the hypotenuse $d(Q,R)$ in
terms of the lengths $s$ and $t$ of the other two sides. [Hint: First use
the transitivity described in (a) to put the triangle in a simple form.]

\itemitem{(c)} Use your formula and the approximation $\cosh s\approx
1+s^2/2+s^4/4!$ to verify that the hypotenuse is greater than the
Euclidean value $\sqrt{s^2+t^2}$, and approximates it when the distances
$s$ and $t$ are small.

\vskip.25in
\hrule
\vskip.25in

\itemitem{3.\quad (a)} What is an affine transformation $t\colon
\aff^2\to\aff^2$ of the affine plane $\aff^2$? Explain without
detailed proof how $t$ can be represented in matrix terms.

\itemitem{(b)} Write down if possible an affine transformation $t$ which
takes
 $$
 (0,0)\mapsto(2,0),\quad (1,0)\mapsto(0,1) \quad\hbox{and}\quad
 (1,1)\mapsto(a,b),
 $$
where $a,b$ are two real numbers with $a+2b\ne2$. What is the reason for
forbidding $a+2b=2$?

\itemitem{(c)} Explain briefly why angles and distances are not in
general defined in the affine geometry of $\aff^2$. What properties of
a triple of points can be defined?

\vskip.25in
\hrule
\medskip \hfill CONTINUED \dots
\vfill\eject

\itemitem{4.\quad (a)} Define the notions of affine linear subspace and
linear span in affine space $\aff^n_\R$. State and prove the theorem on
dimension of intersection of affine linear subspaces $E,F\subset\aff^n_\R$
(results on vector spaces may be used freely). Explain briefly why this
result is more convenient to state in projective space $\proj^n_\R$.

\smallskip\itemitem{(b)} Consider $\triangle PQR$ and $\triangle P'Q'R'$
in projective 3-space $\proj^3$, where
 $$
\matrix{P=(0,0,0,1),&Q=(1,0,0,1),&R=(0,1,0,1),\cr\cr
P'=(0,0,1,1),&Q'=(1,0,1,1),&R'=(0,1,1,1).}
 $$
By writing out the lines $PP'$, $QQ'$, $RR'$, show that $\triangle PQR$ and
$\triangle P'Q'R'$ are in perspectivity. [Hint: you may want to take coordinates $x,y,z,t$ and to visualise
$PQRP'Q'R'$ as a triangular prism in the affine piece $t=1$.] Now determine the
points of intersection of the corresponding sides of the triangles, for example:
 $$
\matrix
{QR:(x+y=t,z=0),\quad Q'R':(x+y=t,z=t),\cr\cr
\qquad\implies QR\cap Q'R'=(1:-1:0:0).}
 $$

\smallskip\itemitem{(c)} State and prove Desargues' theorem for two
triangles $\triangle PQR$ and $\triangle P'Q'R'$ spanning distinct planes
in $\proj^3_\R$ and in perspectivity from a point $O$. Verify it in the
example of (b).

\vskip.25in
\hrule
\medskip \hfill CONTINUED \dots
\vfill\eject

\item{5.\enspace} Give a brief mathematical argument for or against each
of the following five statements, which may be true, false, or
ambiguously formulated.

\smallskip\itemitem{(a)} Two distinct reflections of Euclidean space
$\R^n$ commute if and only if they generate a group of order 4.

\smallskip\itemitem{(b)} There are motions of the sphere $S^2$ having
exactly $0$, $2$ or $\infty$ fixed points.

\smallskip\itemitem{(c)} Let $S_n$ be the symmetric group on
$\{1,2,\dots,n\}$ and $G\subset S_n$ a subgroup; if $G$ contains the
transposition $(12)$ and an element of order $n$, then $G=S_n$.

\smallskip\itemitem{(d)} The group $\SO(2)$ of rotations of the Euclidean
plane is a normal subgroup of the Euclidean group $\Eucl(2)$.

\smallskip\itemitem{(e)} The theorems of affine geometry are also valid
in Euclidean space.

\smallskip\item{} [{\it This is not a multiple choice question}, and no
credit will be given for ``true'' or ``false'' without justification.
Credit will be given for the quality of your argument, irrespective of
whether you conclude the statement is ``true'' or ``false''. Feel free to
add your own interpretation of any statement that seems ambiguous.]

\vskip.25in
\hrule
\medskip

\hfill END

\vfill\eject

Answer to Q1.

(a) $\E^2=\R^2$ with the usual Euclidean metric. A Euclidean motion is an
isometry $t\colon\E^2\to\E^2$, that is, a bijection that preserves
distances.

\hfill[Definition, 4 marks]

\proclaim Theorem. Classification of motions into 4 types:
\item{1.} Translation $\Transl(b)$ in vector $b$
\item{2.} Rotation $\Rot(P,\theta)$ through $\theta$ about centre $P$
\item{3.} Reflection $\Refl(L)$ in line $L$
\item{4.} Glide (or glide-reflection) $\Glide(L,a)$. This is the
reflection in a line $L$ composed with a translation in a vector $a$
along $L$.

\hfill[Bookwork, 9 marks]

Following the indication, in the figure, $g_1(P)=Q$ and $g_2(Q)=P$.
Therefore $g_2\circ g_1$ fixes $P$. Both $g_1$ and $g_2$ are opposite
motions, so we expect $g_2\circ g_1$ to be direct. Thus it should be a
rotation about centre $P$. (We also guess that the angle of rotation
should be $2\theta$.)

Now $g_1$ takes $A$ to a point $A'=g_1(A)$ along the extension of $BQ$,
with $\angle CQA'=\theta$. In turn $g_2$ takes $C$ to a point $C'$ on the
extension of $DP$, so that $\angle APC'=\theta$, and takes $A'$ to $A''$
so that $\angle C'PA''=\theta$, therefore $\angle APA''=2\theta$.
Therefore $g_2\circ g_1=\Rot(P,2\theta)$.

\hfill[Problem 6+6 marks]

\vskip1cm

\hfill[Standard type of question. $4+9+6+6=25$]

\vfill\eject

Answer to Q2.

(a) Write $({\bf x}\cdot_L{\bf y})$ for the Lorentz inner product. Then
for $P={\bf x}$, $Q={\bf y}$, the hyperbolic distance $d(P,Q)$ is defined
by $\cosh d(P,Q)=({\bf x}\cdot_L{\bf y})$. (Not required:
$\hbox{r.h.s.}>0$, so $\arccosh$ is defined.)

The Lorentz group $\SO^+(2,1)$ is the orthogonal group of the Lorentz
inner product preserving the components of the light cone, that is, the
group of $3\times3$ matrixes preserving $\cdot_L$. It acts on $\R^{2,1}$
preserving everything, therefore acts on $\sH^2$ preserving distances,
that is, as motions.

$\SO^+(2,1)$ acts transitively on points of $\sH^2$, on tangent
directions at points, and at orthogonal frames at a point.

\hfill[Definitions 8 marks]

(b) By the transitivity just discussed, given a right-angled triangle
$\triangle PQR$, we can choose coordinates so that $P=(0,0,1)$ and the
sides $PQ$ and $PR$ are the positive $x$ and $y$ axes. Then $Q=(\sinh
s,0,\cosh s)$ and $R=(0,\sinh t,\cosh t)$, with $s,t>0$. Hence
 $$
\cosh d(Q,R)=\cosh s\cosh t.
 $$
This is the required formula.

\hfill[Simple case of ``main formula'', 12 marks]

(c) It is enough to prove: $\cosh s\cosh t>\cosh(\sqrt{s^2+t^2})$, since
$\cosh$ is increasing. Using the given approximation, the left hand side
is
 $$
(1+{s^2\over2}+{s^4\over24})(1+{t^2\over2}+{t^4\over24})
=1+{s^2+t^2\over2}+{s^4\over24}+{t^4\over24}+{s^2t^2\over4}+\cdots
 $$
and the right hand side
 $$
1+{s^2+t^2\over2}+{(s^2+t^2)^2\over24}
=1+{s^2+t^2\over2}+{s^4\over24}+{t^4\over24}+{2s^2t^2\over24}.
 $$
We win because the coeff of $s^2t^2$ is 1/4 in the first case, and
1/12 in the second.

\hfill[Tricky, 5 marks]

\hfill[Easiest possible question on $\sH^2$. $8+12+5=20$]

\vfill\eject

The left hand side is
 $$
\sum {s^{2i}\over2i!}\times \sum {t^{2j}\over2j!} = \sum_{i,j}
{s^{2i}t^{2j}\over(2i!2j!)}
 $$
The right hand side is
 $$
\sum {(s^2+t^2)^i\over2i!}
 $$
Lhs $>$ rhs follows by simple manipulations from
$\left(\matrix{2d\cr 2i}\right)>\left(\matrix{d\cr i}\right)$.

\vfill\eject


Answer to Q3.

An affine transformation $t\colon\aff^2\to\aff^2$ is a bijective map with
the affine linear property $t(\la x+(1-\la)y)=\la t(x)+(1-\la)t(y)$ for
all $x,y\in\aff^2$.

$t$ can be written in matrix terms as $t(x)=Ax+b$ where $A$ is a
nonsingular $2\times2$ matrix and $b$ a vector.

\hfill[Definition 6 marks]

The translation vector is $(2,0)$. After translating back by $(2,0)$, the
matrix comes from $(1,0)\mapsto (-2,1)$ and $(0,1)\mapsto(a,b-1)$, so that
finally
 $$
 t\left(\matrix{x\cr y}\right)=
 \left(\matrix{-2&a\cr 1&b-1}\right)\left(\matrix{x\cr y}\right)
 +\left(\matrix{2\cr0}\right).
 $$
If $a+2b=2$ then the matrix is singular, so the whole plane is mapped to
the line $x+2y=1$, so that $t$ is not an affine transformation.

\hfill[Problem 8 marks]

Geometric properties are the properties invariant under the allowed
transformations. In affine geometry, you can change nonzero distances
arbitrarily, e.g., by $(x,y)\mapsto(\la x,\mu y)$. Similarly, you can
change angles arbitrarily, e.g., by a ``shear''
$(x,y)\mapsto (x+\la y,y)$. Thus distances and angles are not affine
properties.

Given 3 points $P,Q,R$ in $\aff^2$, whether or not their are collinear is
preserved by affine transformations, so is a property defined in affine
geometry. If $P,Q,R$ are collinear, the ratio of distances $d(P,Q):d(P,R)$
is preserved by affine transformations, so likewise, is a property defined
in affine geometry.

\hfill[Official course ideology, 11 marks]

\vskip1cm

\hfill[Standard type of question 6+8+11=25]

\vfill\eject

Answer to Q4.

(a) $\aff^n=\R^n$. An affine linear subspace is of the form $u_0+V$ where
$u_0\in\R^n$ and $V\subset\R^n$ is a vector subspace. Given
a nonempty set
$\Si\subset\aff^n$, its span $\Span{\Si}$ is the smallest affine linear subspace
containing $\Si$. In other words, if $u_0\in\Si$ is any point,
 $$
\Span{\Si}=u_0+V,\quad\hbox{where}\quad V=\Span{\Si-u_0}.
 $$

The theorem says that for $E,F$ affine linear subspaces,
 $$
\dim E+\dim F=\dim(E\cap F)+\dim\Span{E,F},
 $$
provided $E\cap F\ne\emptyset$. For the proof, just pick $u_0\in E\cap F$, so
that $E=u_0+U$, $F=u_0+V$ for some vector subspaces $U,V\subset\R^n$. Then
 $$
E\cap F=u_0+(U\cap V),\quad\Span{E,F}=u_0+(U+V),
 $$
and the formula therefore follows from that for vector spaces.

The same formula holds for all $E,F$ without exception in $\proj^n$, provided we
write $\dim\emptyset=-1$. The lecturer seemed to think this was really good.

\hfill[Bookwork. 11 marks]

(b) $PP':(x=y=0)$, $QQ':(x=t,y=0)$ and $RR':(x=0,y=t)$ intersect in $O=(0010)$.
(By the hint, the sides of the prism are vertical and parallel so meet at
$(0010)$.)

Now $PQ:(y=z=0)$, $P'Q':(y=0,z=t)$ intersect in $C=(1000)$,

$PR:(x=z=0)$, $P'R':(x=0,z=t)$ intersect in $B=(0100)$,

and $QR\cap Q'R'$ intersect in $A=(1:-1:0:0)$ is given.

\hfill[Easier than lectured material. 7 marks]

(c) \proclaim Desargues' theorem. If $\triangle PQR$ and $\triangle P'Q'R'$ are
in perspectivity from a point $O$ then the 3 points of intersection
 $$
A=QR\cap Q'R',\quad B=PR\cap P'R',\quad C=PQ\cap P'Q'
 $$
of the corresponding sides are collinear.

Proof. If the two triangles span distinct planes $\Pi=\Span{\triangle
PQR}\ne\Pi'=\Span{\triangle P'Q'R'}$ then this is obvious: $OPQP'Q'$ is a plane
so $PQ$ and $P'Q'$ intersect in a point $C$, etc., and so $C$ is
contained in lines in the planes $\Pi$, $\Pi'$, therefore is contained in the
line $L=\Pi\cap\Pi'$. Similarly for $B,C$. Q.E.D.

The above 3 points $(1:-1:0:0),(0100),(1000)$ are obviously on $z=t=0$.

\hfill[The easy case of Desargues theorem. 3 marks. $11+7+7=25$]

\vfill\eject

Answer to Q5. 

(a) True. Two possible arguments: by abstract groups $r_1^2=r_2^2$ and
$r_1r_2=r_2r_1$ gives the multiplication table of a 4-group. Or two
distinct reflection commute if and only if their defining hyperplanes are
perpendicular, and then in suitable coordinates the group is
$x,y\mapsto\pm x,\pm y$.

\hfill[Easy start, but groups is the last chapter of notes. 5 Marks]

(b) True. A rotation about an axis has 2 fixed points. A reflection in a
plane has a whole equator of fixed points. The antipodal inversion has no
fixed points.

\hfill[Tests $\O(3)$ as group of motions of $S^2$. 5 Marks]

(c) False, or True depending on interpretation. False: if $n=2k$ with $k$
odd then an element $g=\hbox{(2-cycle)($k$-cycle)}$ has order $n$.
Obviously $G=\Span{g,(12)}$ is not transitive on $\{1,2,\dots,n\}$.

True: if you interpret element $g$ of order $n$ to mean $n$-cycle,
then $G=\Span{g,(12)}=S_n$.

Partial proof: Messing around $g$ and $(12)$, you eventually get
$g_{ij}:i\mapsto1,j\mapsto2$ for each $i,j$ so by the conjugacy principle,
it contains $g_{ij}(12)g_{ij}^{-1}=(ij)$. Obviously $S_n$ is generated by
all the transpositions.

\hfill[Tests conjugacy, 5.4--5 of book. 5 Marks]

(d) False. $\SO(2)$ means rotation about the origin $0$. If a Euclidean
motion $g$ takes $0\mapsto P=g(0)$ then $g\SO(2)g\1$ is the group of
rotations around $P$, so $g\SO(2)g\1\ne\SO(2)$, and it's not normal.

\hfill[Conjugacy was stressed in 5.4 of book. 5 Marks]

(e) To argue for true: The Erlangen program says that theorems of affine
geometry are statements invariant under the affine group. Therefore these
statements are invariant under the subgroup $\Eucl(n)$, so Euclidean
theorems.

To argue for false: in affine geometry, it is a theorem that the affine
group is 2-transitive, but that is false for the Euclidean group.

\hfill[Test Erlangen program and common sense. 5 Marks]

\hfill[No credit for true/false. $5\times5=25$]

\end