\magnification\magstep2 \def\Span#1{\langle#1\rangle} \def\implies{\mathrel{\Longrightarrow}} % \def\Glide{\mathop{\rm Glide}} % \def\Rot{\mathop{\rm Rot}} % \def\Transl{\mathop{\rm Transl}} % \def\Refl{\mathop{\rm Refl}} \def\SO{\mathop{\rm SO}} % \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 1996 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 (i)} What is meant by a motion of Euclidean $n$-space $\E^n$? State and prove a necessary and sufficient condition on an $n\times n$ matrix $A$ in order that the map defined by $$ {\bf x} \mapsto A {\bf x} + {\bf b}\quad\hbox{for ${\bf x}\in\E^n$} \eqno{(*)} $$ should be a motion of $\E^n$. \smallskip\itemitem{(ii)} Let $L_1$ and $L_2$ be the lines of $\E^2$ defined by the equations $$ L_1: (y=b)\quad\hbox{and}\quad L_2: (\cos\th)y=(\sin\th)x, $$ where $\th$ is a fixed angle between $0$ and $\pi$, and $\be\in\R$. Write down the reflections $R_1$, $R_2$ in $L_1$ and $L_2$ in the form $(*)$; proofs are not required. \smallskip\itemitem{(iii)} Calculate the composite motion $R_2\circ R_1$ in this form. \smallskip\itemitem{(iv)} Using the fact that both $R_1$ and $R_2$ fix the point $$ P=L_1\cap L_2=((\cot\th)\be,\be), $$ describe the effect of $R_2\circ R_1$ on $\E^2$ in geometric terms. \vskip.25in \hrule \medskip \hfill CONTINUED \dots \vfill\eject \itemitem{2.\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 the 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 \smallskip\itemitem{3.\quad (i)} Define a reflection of Euclidean $n$-space $\E^n$, and state without proof the theorem of the course expressing any motion of $\E^n$ as a composite of a number of reflections. Explain what it means in particular for a direct motion of $\R^3$ fixing the origin. Deduce that a $3\times3$ orthogonal matrix $A$ with $\det A=1$ has $+1$ as eigenvalue. (That is, any element $A\in\SO(3)$ fixes an axis.) \smallskip\itemitem{(ii)} Define the quaternion conjugate $q^*$ of a quaternion $q$, and explain how to find the inverse $q\1$ if $q\ne0$. Let $p,q$ be quaternions. Prove that $$ \hbox{$p$ is pure imaginary}\implies\hbox{$qpq\1$ is pure imaginary.} $$ Identify the space of pure imaginary quaternions $p=xi+yj+zk$ with Euclidean space $\E^3$ (with coordinates $x,y,z$). Assuming that $|q|^2=1$ for simplicity, prove that the map $r_q\colon\E^3\to\E^3$ given by $r_q(p)=qpq\1$ is an element of $\SO(3)$. \smallskip\itemitem{(iii)} Determine the rotation $r_q$ for the unit quaternion $$ q=c+si\quad\hbox{with}\quad c^2+s^2=1, $$ and express the result geometrically. \medskip \hrule \medskip \smallskip\item{4.} Let $S^2\subset\R^3$ be the sphere of radius $1$. Explain how to define the spherical distance $d(P,Q)$ between two points $P,Q\in S^2$. State without proof a version of the triangle inequality and explain briefly why it implies that the shortest path from $P$ to $Q$ is an arc of great circle. \smallskip\item{} Let $\De PQR$ be a triangle in spherical geometry, with right angle at $P$. State and prove from first principles a formula expressing the hypo\-tenuse $d(Q,R)$ in terms of the other two sides $d(P,Q)$ and $d(P,R)$. Verify that your formula approximates Pythagoras' theorem when the triangle is small compared with the radius of the sphere. [Hint: You may use the approximation $\cos a\approx 1-a^2+\cdots$ (higher order terms) for small $a$.] \vskip.25in \hrule \medskip \hfill CONTINUED \dots \vfill\eject \item{5.\enspace} Use your mathematical judgement to argue for the truth or otherwise of the following statements. \smallskip \itemitem{(a)} 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 taking $i\mapsto1,j\mapsto2$, for any pair $i,j$ of distinct elements of $\{1,2,\dots,n\}$, then $G=S_n$. \smallskip\itemitem{(b)} There exists a finite non-Abelian group of direct motions of the Euclidean plane $\E^2$. \smallskip\itemitem{(c)} There exists an affine linear map $\aff^1\to\aff^1$ of the affine line taking any set of 3 distinct points into any other. \smallskip\itemitem{(d)} Given 3 lines $L_1,L_2,L_3\subset\proj^3$, there is a unique line $M$ meeting all 3. \smallskip\itemitem{(e)} In hyperbolic geometry $\sH^2$, given a line $L$ and a point $P\notin L$, there is a unique line $M$ through $P$ perpendicular to $L$. \vskip.25in \hrule \medskip \hfill END \vfill\eject Answer to Q1. (i) A motion of $\E^n$ is a map $t\colon\E^n\to\E^n$ that preserves Euclidean distances, i.e., $d(t(P),t(Q))=d(P,Q)$ for all $P,Q$. \hfill[Definition, 3 marks] If $t$ is given by $x\mapsto Ax+b$ with $A$ and $n\times n$ matrix and $b$ a vector, $t$ is a motion if and only if $A$ is orthogonal. Indeed, the translation by $b$ does not affect distnaces, so that I need to prove that $x\mapsto Ax$ preserves distance from the origin, that is, the quadratic form $|x|^2={}^tx\cdot x$. But this holds if and only if it preserves the bilinear form ${}^txx$ on $\R^n$, that is, $$ {}^t(Ax)\cdot Ay ={}^tx({}^tA\cdot A)y={}^txy\quad\hbox{for all $x,y\in\R^n$,} $$ that is, ${}^tAA=1$. \hfill[Bookwork, 6 marks] (ii) The reflection in $L_1: (y=\be)$ is given by $x\mapsto x$, $y\mapsto2\be-y$, that is, $$ R_1\colon\left(\matrix{x\cr y}\right) \mapsto\left(\matrix{1&0\cr0&-1}\right)\left(\matrix{x\cr y}\right) +\left(\matrix{0\cr2\be}\right). $$ The reflection in $L_2:(\cos\th)y=(\sin\th)x$ is given by $$ R_2\colon\left(\matrix{x\cr y}\right)\mapsto \left(\matrix{\cos2\th&\sin2\th\cr\sin2\th&-\cos2\th}\right) \left(\matrix{x\cr y}\right). $$ (Several derivations are possible, for example, by computing the conjugate $\mathop{\rm Rotation(\th)\left(\matrix{1&0\cr0&-1}\right)} \mathop{\rm Rotation(-\th)}$, or by finding the fixed axis.) \hfill[4+4 marks] (iii) So $$ \eqalign{ R_2\circ R_1\colon\left(\matrix{x\cr y}\right)\mapsto {}&\left(\matrix{\cos2\th&\sin2\th\cr\sin2\th&-\cos2\th}\right) \left[\left(\matrix{1&0\cr0&-1}\right)\left(\matrix{x\cr y}\right) +\left(\matrix{0\cr2\be}\right) \right]\cr ={}&\left(\matrix{\cos2\th&-\sin2\th\cr\sin2\th&\cos2\th}\right) \left(\matrix{x\cr y}\right)+2\be \left(\matrix{\sin2\th\cr-\cos2\th}\right) } $$ \vfill\eject \hfill[4 marks] (iv) $R_2\circ R_1=\mathop{\rm Rotation(P,2\th)}$, where as given, $P=L_1\cap L_2=((\cot\th)\be,\be)$ (see Figure). \hfill[4 marks] \vskip3in \hfill[Standard type of question. $3+6+4+4+4+4=25$] \vfill\eject Answer to Q2. (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 Q3. (i) The reflection of $\E^n$ in a hyperplane $\Pi$ is the motion that fixes every $P\in\Pi$, and that takes a point $Q$ into $t(Q)$ constructed as the mirror image, that is, drop a perpendicular from $Q$ to $\Pi$ and set $t(Q)$ to be the point the same distance the other side. (In other words, $\Pi$ is the perpendicular bisector of $PQ$.) \proclaim Theorem. Any motion of $\E^n$ is a composite of at most $n+1$ reflections. Any motion fixing the first $k$ elements of a Euclidean frame of reference is a composite of at most $n+1-k$ reflections. In particular, if $n=3$ and $k=1$, any motion $t\colon\R^3\to\R^3$ is a composite of $\le3$ reflections. If $t$ is direct then the number of reflections is even, that is, $0$ (the identity) or 2. But the composite of 2 distinct reflections $\mathop{\rm Reflection(\Pi_1)}$, $\mathop{\rm Reflection(\Pi_2)}$ fixes the line $L=\Pi_1\cap\Pi_2$ (vector subspaces), as required. [Some students will probably do the last part ``Otherwise'' for fewer marks: There exists an orthonormal basis w.r.t.\ which $A$ is the block diagonal matrix, with each block either $1\times1$ block with entry $\pm1$, or the $2\times2$ rotation matrix $$ \left( \matrix { \cos\th&-\sin\th\cr \sin\th&\cos\th } \right) $$ In the $3\times3$ case there is at most one rotation block. If $\det A=1$ (that is, $A$ is direct) then either $A=1_3$ or $A$ has a block $+1$ and a rotation block.] \hfill[Statement of theorem. Deconstructing to $\R^3$ is unseen. 6+4 marks] (ii) If $q=a+bi+cj+dk$ then $q^*=a-bi-cj-dk$. Thus $q$ is real resp.\ pure imaginary if $q^*=q$ or $q^*=-q^*$. If $q\ne0$ then $$ qq^*=|q|^2=a^2+b^2+c^2+d^2 $$ is a positive real, so that $q\1=q^*/|q|^2$ is the inverse. Now for $a$ real, $qaq^*=aqq^*=a|q|^2$ is also real. If $p$ is pure imaginary then $p^*=-p$, so that $(qpq^*)^*=qp^*q^*=-qpq^*$, and $p$ is imaginary. Setting $r_q(p)=qpq\1$ then $r_q$ is a linear map $\R^3\to\R^3$, and $|r_q(p)|^2=|qpq\1|^2=(qpq^*)(qp^*q)/|q|^4=pp^*=|p^2|$, so that $r_q$ is a motion or orthogonal map. It is direct because $\det r_q$ is a continuous function from the nonzero quaternions to $\pm1$. \hfill[Bookwork. 7 marks] (d) Write $c=\cos\psi$, $s=\sin\psi$ and $q=c+si$. Then $q$ is a unit quaternion; $r_q(i)=qiq\1=i$ because $\R[i]=\C$ is commutative. And the action on $j,k$ is given by $$ \matrix { & r_q(j)=(c+si)j(c-si)=(c^2-s^2)j+2csk\cr \hbox{and}\quad & r_q(k)=(c+si)k(c-si)=-2csj+(c^2-s^2)k. } $$ This is a rotation of the $y,z$ plane by an angle $2\psi$. [Calculation similar to proof in course. 8 marks. $10+7+8=25$] \vfill\eject Answer to Q4. $S^2\subset\R^2$ is the sphere of radius $1$ centred at the origin. If $P=Q$ then define $d(P,Q)=0$; if $P=-Q$ (antipodal points) then define $d(P,Q)=\pi$. Otherwise $P,Q\in S^2$ are linearly independent points of $\R^3$ and span a \hbox{2-plane} which intersects $S^2$ in a great circle. Then define $d(P,Q)$ to be the length of the shorter arc from $P$ to $Q$. Since the arc length is $a=\angle POQ$, we get that $$ d(P,Q)=\arccos(P\cdot Q)\eqno{(*)} $$ (vector product). \hfill [Definition. 4 marks] The triangle inequality: $$ d(P,R)\le d(P,Q)+d(Q,R)\quad\hbox{for $P,Q,R\in S^2$,} $$ with equality if and only if $Q\in PR$, that is, $Q$ is contained in a shorter arc of greater circle from $P$ to $R$. \hfill [Memorable statement. 4 marks] If $C$ is any curve from $P$ to $Q$ then by definition of the Riemann integral, for any $\ep>0$, it can be approximated by a path $P=P_0,P_1,\dots,P_N=Q$ made up of arcs of great circles, so that $$ \hbox{length of $C$} \ge d(P_0,P_1)+d(P_1,P_2)+\cdots+d(P_{N-1},P_N)-\ep. $$ Omitting the intermediate steps decreases the right hand side, so by induction, length of $C\ge d(P,Q)$. Thus the arc of great circle is the shortest path. \hfill [Bookwork. Vague argument OK. 4 marks] Let $P=(1,0,0)$ be the North Pole. It is given that $\angle QPR=90^\circ$, so that I can assume that $Q$ is in the plane $y=0$, and $R$ in $z=0$. Suppose $d(P,Q)=a$ and $d(P,R)=b$. Then $Q=(\cos a,0,\sin a)$, $R=(\cos b,\sin b,0)$. Therefore $Q\cdot R=\cos a\cos b$, and formula $(*)$ gives $$ d(P,Q)=\arccos(\cos a\cos b). $$ \hfill [Easiest form of trig calculation used in notes. 8 marks] In other words, if $c$ is the hypothenuse then $\cos c=\cos a\cos b$. Now since $\cos a=1-a^2+\cdots$ (higher order terms), when $a,b,c$ are small, we get $(1-c^2)=(1-a^2)(1-b^2)$, that is, $c^2=a^2+b^2$ to good approximation. \hfill [Unseen. 5 marks] \hfill[$4+4+4+8+5=25$ \vfill\eject Answer to Q5. (a) True. By the given information, $G$ contains $(12)$, and $g=g_{ij}:i\mapsto1,j\mapsto2$, so by the conjugacy principle, it contains $g(12)g^{-1}=(ij)$. Obviously $S_n$ is generated by all the transpositions. \hfill[Tests conjugacy, 4.2--3 of book. 5 Marks] (b) False. A finite group $G\subset\mathop{\rm Eucl}(2)$ fixes a point $P_0$ (the centroid of any orbit $G\cdot P$); therefore $G$ has only reflections and rotations. If it's direct it only contains rotations around $P_0$, therefore is commutative. Thus the statement is false. \hfill[Tests various aspects of Euclidean groups. 5 Marks] (c) False. The ratio of distances is defined. The affine linear map $\fie\colon\aff^1\to\aff^1$ given by $x\mapsto ax+b$ is determined by its effect on $P=0$ and $Q=1$: because $\fie(0)=b$, and $\fie(1)=a+b$. Therefore, given another point $R=c$, the image $\fie(R)=ac+b$ is determined by $\fie(P)$ and $\fie(Q)$, so can't be changed. \hfill[Easy but unseen; like cross-ratio. 5 Marks] (d) False. There are infinitely many. For a counterexample, if $L_1$ and $L_2$ are disjoint, they span $\proj^3$ (by dimension of intersection), so every point of $L_3$ is on a line joining $L_1,L_2$, and this make infinitely many lines \hfill[Simple trap. 5 Marks] (e) True. Several arguments are possible, one by choosing $Q\in L$ to minimise distance $d(P,Q)$ (a continuous function on a compact set takes its inf), etc. Or consider a point $Q\in L$ moving from far over to the left (when the angle is very small) to far over to the right (when it is close to $\pi$. The angle is a continuous function so must take the value $\pi/2$ at some point. Or taking the reflection $t=\mathop{\rm Reflection}(L)$ and the line $Pt(P)$. Or use coordinate geometry in Lorentz space $\R^{-1,+2}$, with the Lorentz inner product $v\cdot_Lw$: suppose $L$ is given by $x_2=0$ and $P=(a,b_1,b_2)$ with $a^2=1+b_1^2+b_2^2$ ; then the point $$ Q=(c,d,0)={1\over a^2-b_1^2}(a,b_1,0)\in L. $$ Then $\{(c,d,0),(d,c,0),(0,0,1)\}$ is a new Lorentz basis in which $L:(x_2=0)$ and $M:(x_1=0)$. \hfill[The coordinate calc is hard unseen. 5 Marks] \hfill[$\le1$ mark for true/false. $5\times5=25$] \end