% Chapter 4, Topic _Linear Algebra_ Jim Hefferon % http://joshua.smcvt.edu/linearalgebra % 2001-Jun-12 \topic{Projective Geometry} \index{projective geometry|(} There are geometries other than the familiar Euclidean one. One such geometry arose when artists observed that what a viewer sees is not necessarily what is there. As an example, here is Leonardo da Vinci's \textit{The Last Supper}.\index{da Vinci, Leonardo}\index{Last Supper@\textit{Last Supper}} % From http://upload.wikimedia.org/wikipedia/commons/7/77/DaVinci_LastSupper_high_res_2_nowatmrk.jpg \begin{center} \includegraphics[width=.6\textwidth]{LastSupper.jpg} \end{center} Look at where the ceiling meets the left and right walls. In the room those lines are parallel but da~Vinci has painted lines that, if extended, would intersect. The intersection is the \definend{vanishing point}.\index{vanishing point}\index{projection!central!vanishing point} This aspect of perspective is familiar as an image of railroad tracks that appear to converge at the horizon. Da~Vinci has adopted a model of how we see. % of how % we project the three dimensional scene to a two dimensional image. Imagine a person viewing a room. From the person's eye, in every direction, carry a ray outward until it intersects something, such as a point on the line where the wall meets the ceiling. This first intersection point is what the person sees in that direction. Overall what the person sees is the collection of three-dimensional intersection points projected to a common two dimensional image. \begin{center} \includegraphics[scale=.8]{ch4.5} \end{center} This is a \emph{central projection}\index{projection!central}\index{central projection} from a single point. As the sketch shows, this projection is not orthogonal like the ones we have seen earlier because the line from the viewer to~$C$ is not orthogonal to the image plane. (This model is only an approximation\Dash it does not take into account such factors as that we have binocular vision or that our brain's processing greatly affects what we perceive. Nonetheless the model is interesting, both artistically and mathematically.) The operation of central projection preserves some geometric properties, for instance lines project to lines. However, it fails to preserve some others. One example is that equal length segments can project to segments of unequal length (above, $AB$ is longer than $BC$ because the segment projected to $AB$ is closer to the viewer and closer things look bigger). The study of the effects of central projections is projective geometry. There are three cases of central projection. The first is the projection done by a movie projector. \begin{center} \includegraphics{ch4.6} \end{center} We can think that each source point is pushed from the domain plane~$S$ outward to the image plane~$I$. The second case of projection is that of the artist pulling the source back to a canvas. \begin{center} \includegraphics{ch4.7} \end{center} The two are different because first $S$ is in the middle and then $I$. One more configuration can happen, with $P$ in the middle. An example of this is when we use a pinhole to shine the image of a solar eclipse onto a paper. \begin{center} \includegraphics{ch4.8} \end{center} Although the three are not exactly the same, they are similar. We shall say that each is a central projection by $P$ of $S$ to $I$. We next look at three models of central projection, of increasing abstractness but also of increasing uniformity. The last model will bring out the linear algebra. %To illustrate some of the geometric effects of these projections, Consider again the effect of railroad tracks that appear to converge to a point. Model this with parallel lines in a domain plane~$S$ and a projection via a $P$ to a codomain plane~$I$. (The gray lines shown are parallel to the $S$ plane and to the~$I$ plane.) \begin{center} \includegraphics{ch4.9} \end{center} This single setting shows all three projection cases. The first picture below shows $P$ acting as a movie projector by pushing points from part of $S$ out to image points on the lower half of $I$. The middle picture shows $P$ acting as the artist by pulling points from another part of $S$ back to image points in the middle of $I$. In the third picture $P$ acts as the pinhole, projecting points from $S$ to the upper part of $I$. This third picture is the trickiest\Dash the points that are projected near to the vanishing point are the ones that are far out on the lower left of $S$. Points in $S$ that are near to the vertical gray line are sent high up on $I$. \begin{center} \includegraphics{ch4.10} \hfil \includegraphics{ch4.11} \hfil \includegraphics{ch4.12} \end{center} There are two awkward things here. First, neither of the two points in the domain nearest to the vertical gray line (see below) has an image because a projection from those two is along the gray line that is parallel to the codomain plane (we say that these two are projected to infinity). The second is that the vanishing point in $I$ isn't the image of any point from $S$ because a projection to this point would be along the gray line that is parallel to the domain plane (we say that the vanishing point is the image of a projection from infinity). \begin{center} \includegraphics{ch4.13} \end{center} For a model that eliminates this awkwardness, cover the projector $P$ with a hemispheric dome. In any direction, defined by a line through the origin, project anything in that direction to the single spot on the dome where the line intersects. This includes projecting things on the line between $P$ and the dome, as with the movie projector. It includes projecting things on the line further from $P$ than the dome, as with the painter. More subtly, it also includes things on the line that lie behind $P$, as with the pinhole case. \begin{center} \includegraphics{ch4.14} \end{center} More formally, for any nonzero vector $\vec{v}\in\Re^3$, let the associated \definend{point $v$ in the projective plane}\index{point!in projective plane} be the set $\set{k\vec{v}\suchthat \text{$k\in\Re$and$k\neq 0$}}$ of nonzero vectors lying on the same line through the origin as $\vec{v}$. To describe a projective point we can give any representative member of the line, so that the projective point shown above can be represented in any of these three ways. \begin{equation*} \colvec[r]{1 \\ 2 \\ 3} \qquad \colvec{1/3 \\ 2/3 \\ 1} \qquad \colvec[r]{-2 \\ -4 \\ -6} \end{equation*} Each of these is a \definend{homogeneous coordinate vector}\index{coordinates!homogeneous}% \index{homogeneous coordinate vector}\index{vector!homogeneous coordinate} for the point~$\ell$. This picture and definition clarifies central projection but there is still something ungainly about the dome model:~what happens when $P$ looks down? Consider, in the sketch above, the part of $P$'s line of sight that comes up towards us, out of the page. Imagine that this part of the line falls, to the equator and below. Now the part of the line~$\ell$ that intersects the dome lies behind the page. That is, as the line of sight continues down past the equator, the projective point suddenly shifts from the front of the dome to the back of the dome. (This brings out that the dome does not include the entire equator or else when the viewer is looking exactly along the equator then there would be two points in the line that are both on the dome. Instead we define the dome so that it includes the points on the equator with a positive~$y$ coordinate, as well as the point where $y=0$ and $x$ is positive.) This discontinuity means that we often have to treat equatorial points as a separate case. So while the railroad track model of central projection has three cases, the dome has two. We can do better, we can reduce to a model having a single case. Consider a sphere centered at the origin. Any line through the origin intersects the sphere in two spots, said to be antipodal.\index{antipodal points} Because we associate each line through the origin with a point in the projective plane, we can draw such a point as a pair of antipodal spots on the sphere. Below, we show the two antipodal spots connected by a dashed line to emphasize that they are not two different points, the pair of spots together make one projective point. \begin{center} \includegraphics{ch4.15} \end{center} While drawing a point as a pair of antipodal spots on the sphere is not as intuitive as the one-spot-per-point dome mode, on the other hand the awkwardness of the dome model is gone in that as a line of view slides from north to south, no sudden changes happen. This central projection model is uniform. So far we have described points in projective geometry. What about lines? What a viewer~$P$ at the origin sees as a line is shown below as a great circle, the intersection of the model sphere with a plane through the origin. \begin{center} \includegraphics{ch4.16} \end{center} (We've included one of the projective points on this line to bring out a subtlety. Because two antipodal spots together make up a single projective point, the great circle's behind-the-paper part is the same set of projective points as its in-front-of-the-paper part.) Just as we did with each projective point, we can also describe a projective line with a triple of reals. For instance, the members of this plane through the origin in $\Re^3$ \begin{equation*} %\set{y\colvec[r]{1 \\ -1 \\ 0}+z\colvec[r]{1 \\ 0 \\ 1}\suchthat y,z\in\Re}= \set{\colvec{x \\ y \\ z}\suchthat x+y-z=0} \end{equation*} project to a line that we can describe with $\rowvec{1 &1 &-1}$ (using a row vector for this typographically distinguishes lines from points). In general, for any nonzero three-wide row vector $\smash{\vec{L}}$ we define the associated \definend{line in the projective plane},\index{projective plane!lines}% \index{line!in projective plane} to be the set $L=\set{k\vec{L}\suchthat \text{$k\in\Re$and$k\neq 0$}}$. The reason this description of a line as a triple is convenient is that in the projective plane a point $v$ and a line $L$ are \definend{incident} \Dash the point lies on the line, the line passes through the point \Dash if and only if a dot product of their representatives $v_1L_1+v_2L_2+v_3L_3$ is zero (\nearbyexercise{exer:IncidentIndReps} shows that this is independent of the choice of representatives $\smash{\vec{v}}$ and $\smash{\vec{L}}$). For instance, the projective point described above by the column vector with components $1$, $2$, and $3$ lies in the projective line described by $\rowvec{1 &1 &-1}$, simply because any vector in $\Re^3$ whose components are in ratio $1\mathbin :2\mathbin :3$ lies in the plane through the origin whose equation is of the form $k\cdot x+k\cdot y-k\cdot z=0$ for any nonzero~$k$. That is, the incidence formula is inherited from the three-space lines and planes of which $v$ and $L$ are projections. With this, we can do analytic projective geometry. For instance, the projective line $L=\rowvec{1 &1 &-1}$ has the equation $1v_1+1v_2-1v_3=0$, meaning that for any projective point~$v$ incident with the line, any of $v$'s representative homogeneous coordinate vectors will satisfy the equation. This is true simply because those vectors lie on the three space plane. One difference from Euclidean analytic geometry is that in projective geometry besides talking about the equation of a line, we also talk about the equation of a point. For the fixed point \begin{equation*} v=\colvec[r]{1 \\ 2 \\ 3} \end{equation*} the property that characterizes lines incident on this point is that the components of any representatives satisfy $1L_1+2L_2+3L_3=0$ and so this is the equation of $v$. This symmetry of the statements about lines and points is the \definend{Duality Principle}\index{Duality Principle, of projective geometry} of projective geometry:~in any true statement, interchanging point' with line' results in another true statement. For example, just as two distinct points determine one and only one line, in the projective plane two distinct lines determine one and only one point. Here is a picture showing two projective lines that cross in antipodal spots and thus cross at one projective point. \begin{center} \hfill \begin{tabular}{@{}c@{}}\includegraphics{ch4.17}\end{tabular} \hfill\llap{($*$)} \end{center} Contrast this with Euclidean geometry, where two unequal lines may have a unique intersection or may be parallel. In this way, projective geometry is simpler, more uniform, than Euclidean geometry. That simplicity is relevant because there is a relationship between the two spaces:~we can view the projective plane as an extension of the Euclidean plane. Draw the sphere model of the projective plane as the unit sphere in $\Re^3$. Take Euclidean $2$-space to be the plane $z=1$. As shown below, all of the points on the Euclidean plane are projections of antipodal spots from the sphere. Conversely, we can view some points in the projective plane as corresponding to points in Euclidean space. (Note that projective points on the equator don't correspond to points on the plane; instead we say these project out to infinity.) \begin{center} \hfill \begin{tabular}{@{}c@{}}\includegraphics{ch4.18}\end{tabular} \hfill\llap{($**$)} \end{center} Thus we can think of projective space as consisting of the Euclidean plane with some extra points adjoined \Dash the Euclidean plane is embedded in the projective plane. The extra points in projective space, the equatorial points, are called \definend{ideal points}\index{ideal!point}% \index{projective plane!ideal point} or \definend{points at infinity}\index{point!at infinity} and the equator is called the \definend{ideal line}\index{ideal!line}% \index{projective plane!ideal line} or \definend{line at infinity}\index{line at infinity} (it is not a Euclidean line, it is a projective line). The advantage of this extension from the Euclidean plane to the projective plane is that some of the nonuniformity of Euclidean geometry disappears. For instance, the projective lines shown above in~($*$) cross at antipodal spots, a single projective point, on the sphere's equator. If we put those lines into~($**$) then they correspond to Euclidean lines that are parallel. That is, in moving from the Euclidean plane to the projective plane, we move from having two cases, that distinct lines either intersect or are parallel, to having only one case, that distinct lines intersect (possibly at a point at infinity). A disadvantage of the projective plane is that we don't have the same familiarity with it as we have with the Euclidean plane. Doing analytic geometry in the projective plane helps because the equations lead us to the right conclusions. Analytic projective geometry uses linear algebra. For instance, for three points of the projective plane $t$, $u$, and~$v$, setting up the equations for those points by fixing vectors representing each shows that the three are collinear if and only if the resulting three-equation system has infinitely many row vector solutions representing their line. That in turn holds if and only if this determinant is zero. \begin{equation*} \begin{vmat} t_1 &u_1 &v_1 \\ t_2 &u_2 &v_2 \\ t_3 &u_3 &v_3 \end{vmat} \end{equation*} Thus, three points in the projective plane are collinear if and only if any three representative column vectors are linearly dependent. Similarly, by duality, three lines in the projective plane are incident on a single point if and only if any three row vectors representing them are linearly dependent. The following result is more evidence of the niceness of the geometry of the projective plane. These two triangles are \definend{in perspective}\index{perspective, triangles} from the point $O$ because their corresponding vertices are collinear. \begin{center} \includegraphics{ch4.19} \end{center} Consider the pairs of corresponding sides: the sides $T_1U_1$ and $T_2U_2$, the sides $T_1V_1$ and $T_2V_2$, and the sides $U_1V_1$ and $U_2V_2$. \definend{Desargue's Theorem}\index{Desargue's Theorem} is that when we extend the three pairs of corresponding sides, they intersect (shown here as the points $TU$, $TV$, and $UV$). What's more, those three intersection points are collinear. \begin{center} \includegraphics{ch4.23} \end{center} We will prove this using projective geometry. (We've drawn Euclidean figures because that is the more familiar image. To consider them as projective figures we can imagine that, although the line segments shown are parts of great circles and so are curved, the model has such a large radius compared to the size of the figures that the sides appear in our sketch to be straight.) For the proof we need a preliminary lemma \cite{Coxeter}:~if $W$, $X$, $Y$, $Z$ are four points in the projective plane, no three of which are collinear, then there are homogeneous coordinate vectors $\vec{w}$, $\vec{x}$, $\vec{y}$, and $\vec{z}$ for the projective points, and a basis $B$ for $\Re^3$, satisfying this. \begin{equation*} \rep{\vec{w}}{B}=\colvec[r]{1 \\ 0 \\ 0} \quad \rep{\vec{x}}{B}=\colvec[r]{0 \\ 1 \\ 0} \quad \rep{\vec{y}}{B}=\colvec[r]{0 \\ 0 \\ 1} \quad \rep{\vec{z}}{B}=\colvec[r]{1 \\ 1 \\ 1} \end{equation*} To prove the lemma, because $W$, $X$, and $Y$ are not on the same projective line, any homogeneous coordinate vectors $\vec{w}_0$, $\vec{x}_0$, and $\vec{y}_0$ do not line on the same plane through the origin in $\Re^3$ and so form a spanning set for $\Re^3$. Thus any homogeneous coordinate vector for $Z$ is a combination $\vec{z}_0=a\cdot\vec{w}_0+b\cdot\vec{x}_0+c\cdot\vec{y}_0$. Then let the basis be $B=\sequence{\vec{w},\vec{x},\vec{y}}$ and take $\vec{w}=a\cdot\vec{w}_0$, $\vec{x}=b\cdot\vec{x}_0$, $\vec{y}=c\cdot\vec{y}_0$, and $\vec{z}=\vec{z}_0$. To prove Desargue's Theorem use the lemma to fix homogeneous coordinate vectors and a basis. \begin{equation*} \rep{\vec{t}_1}{B}=\colvec[r]{1 \\ 0 \\ 0} \quad \rep{\vec{u}_1}{B}=\colvec[r]{0 \\ 1 \\ 0} \quad \rep{\vec{v}_1}{B}=\colvec[r]{0 \\ 0 \\ 1} \quad \rep{\vec{o}}{B}=\colvec[r]{1 \\ 1 \\ 1} \end{equation*} The projective point $T_2$ is incident on the projective line $OT_1$ so any homogeneous coordinate vector for $T_2$ lies in the plane through the origin in $\Re^3$ that is spanned by homogeneous coordinate vectors of $O$ and $T_1$: \begin{equation*} \rep{\vec{t}_2}{B}=a\colvec[r]{1 \\ 1 \\ 1} +b\colvec[r]{1 \\ 0 \\ 0} \end{equation*} for some scalars $a$ and $b$. Hence the homogeneous coordinate vectors of members $T_2$ of the line $OT_1$ are of the form on the left below. The forms for $U_2$ and $V_2$ are similar. \begin{equation*} \rep{\vec{t}_2}{B}=\colvec{t_2 \\ 1 \\ 1} \qquad \rep{\vec{u}_2}{B}=\colvec{1 \\ u_2 \\ 1} \qquad \rep{\vec{v}_2}{B}=\colvec{1 \\ 1 \\ v_2} \end{equation*} The projective line $T_1U_1$ is the projection of a plane through the origin in $\Re^3$. One way to get its equation is to note that any vector in it is linearly dependent on the vectors for $T_1$ and $U_1$ and so this determinant is zero. \begin{equation*} \begin{vmat} 1 &0 &x \\ 0 &1 &y \\ 0 &0 &z \end{vmat}=0 \qquad \Longrightarrow \qquad z=0 \end{equation*} The equation of the plane in $\Re^3$ whose image is the projective line $T_2U_2$ is this. \begin{equation*} \begin{vmat} t_2 &1 &x \\ 1 &u_2 &y \\ 1 &1 &z \end{vmat}=0 \qquad \Longrightarrow \qquad (1-u_2)\cdot x+(1-t_2)\cdot y+(t_2u_2-1)\cdot z=0 \end{equation*} Finding the intersection of the two is routine. \begin{equation*} T_1U_1\,\intersection\, T_2U_2 =\colvec{t_2-1 \\ 1-u_2 \\ 0} \end{equation*} (This is, of course, a homogeneous coordinate vector of a projective point.) The other two intersections are similar. \begin{equation*} T_1V_1\,\intersection\, T_2V_2 =\colvec{1-t_2 \\ 0 \\ v_2-1} \qquad U_1V_1\,\intersection\, U_2V_2 =\colvec{0 \\ u_2-1 \\ 1-v_2} \end{equation*} Finish the proof by noting that these projective points are on one projective line because the sum of the three homogeneous coordinate vectors is zero. Every projective theorem has a translation to a Euclidean version, although the Euclidean result may be messier to state and prove. Desargue's theorem illustrates this. In the translation to Euclidean space, we must treat separately the case where $O$ lies on the ideal line, for then the lines $T_1T_2$, $U_1U_2$, and $V_1V_2$ are parallel. The remark following the statement of Desargue's Theorem suggests thinking of the Euclidean pictures as figures from projective geometry for a sphere model with very large radius. That is, just as a small area of the world seems to people living there to be flat, the projective plane is locally Euclidean. We finish by pointing out one more thing about the projective plane. Although its local properties are familiar, the projective plane has a perhaps unfamiliar global property. The picture below shows a projective point. At that point we have drawn Cartesian axes, $xy$-axes. Of course, the axes appear in the picture at both antipodal spots, one in the northern hemisphere (that is, shown on the right) and the other in the south. Observe that in the northern hemisphere a person who puts their right hand on the sphere, palm down, with their thumb on the $y$~axis will have their fingers pointing along the $x$-axis in the positive direction. % But the antipodal axes give the opposite:~if a person puts their % right hand on the southern hemisphere spot on the sphere, palm on the % sphere's surface, % with their fingers pointing toward positive infinity on the $x$-axis, then % their thumb points on the $y$-axis toward negative infinity. % Instead, % to have their fingers point positively on the $x$-axis and their thumb % point positively on the $y$, a person must use their left hand. % Briefly, the projective plane is not orientable\Dash in this geometry, % left and right handedness are not fixed properties of figures. \begin{center} \includegraphics{ch4.24} \end{center} The sequence of pictures below show a trip around this space: the antipodal spots rotate around the sphere with the spot in the northern hemisphere moving up and over the north pole, ending on the far side of the sphere, and its companion coming to the front. (Be careful:~the trip shown is not halfway around the projective plane. It is a full circuit. The spots at either end of the dashed line are the same projective point. So by the third sphere below the trip has pretty much returned to the same projective point where we drew it starting above.) \begin{center} \begin{tabular}{@{}c@{}}\includegraphics{ch4.25}\end{tabular} \qquad\mbox{$\Longrightarrow$}\qquad \begin{tabular}{@{}c@{}}\includegraphics{ch4.26}\end{tabular} \qquad\mbox{$\Longrightarrow$}\qquad \begin{tabular}{@{}c@{}}\includegraphics{ch4.27}\end{tabular} \end{center} At the end of the circuit, the $x$~part of the $xy$-axes sticks out in the other direction. That is, for a person to put their thumb on the $y$-axis and have their fingers point positively on the $x$-axis, they must use their left hand. The projective plane is not orientable\Dash in this geometry, left and right handedness are not fixed properties of figures (said another way, we cannot describe a spiral as clockwise or counterclockwise). This exhibition of the existence of a non-orientable space raises the question of whether our universe orientable. Could an astronaut leave earth right-handed and return left-handed? \cite{Gardner} is a nontechnical reference. \cite{Clarke} is a classic science fiction story about orientation reversal. For an overview of projective geometry see \cite{CourantRobbins}. The approach we've taken here, the analytic approach, leads to quick theorems and % \Dash most importantly for us \Dash illustrates the power of linear algebra; see \cite{Hanes}, \cite{Ryan}, and \cite{Eggar}. But another approach, the synthetic approach of deriving the results from an axiom system, is both extraordinarily beautiful and is also the historical route of development. Two fine sources for this approach are \cite{Coxeter} or \cite{Seidenberg}. An easy and interesting application is in \cite{Davies}. \begin{exercises} \item What is the equation of this point? \begin{equation*} \colvec[r]{1 \\ 0 \\ 0} \end{equation*} \begin{answer} From the dot product \begin{equation*} 0=\colvec[r]{1 \\ 0 \\ 0}\dotprod\rowvec{L_1 &L_2 &L_3} =L_1 \end{equation*} we get that the equation is $L_1=0$. \end{answer} \item \begin{exparts} \partsitem Find the line incident on these points in the projective plane. \begin{equation*} \colvec[r]{1 \\ 2 \\ 3},\,\colvec[r]{4 \\ 5 \\ 6} \end{equation*} \partsitem Find the point incident on both of these projective lines. \begin{equation*} \rowvec{1 &2 &3},\,\rowvec{4 &5 &6} \end{equation*} \end{exparts} \begin{answer} \begin{exparts} \partsitem This determinant \begin{equation*} 0=\begin{vmat} 1 &4 &x \\ 2 &5 &y \\ 3 &6 &z \end{vmat} =-3x+6y-3z \end{equation*} shows that the line is $L=\rowvec{-3 &6 &-3}$. \partsitem $\colvec[r]{-3 \\ 6 \\ -3}$ \end{exparts} \end{answer} \item Find the formula for the line incident on two projective points. Find the formula for the point incident on two projective lines. \begin{answer} The line incident on \begin{equation*} u=\colvec{u_1 \\ u_2 \\ u_3} \qquad v=\colvec{v_1 \\ v_2 \\ v_3} \end{equation*} comes from this determinant equation. \begin{equation*} 0=\begin{vmat} u_1 &v_1 &x \\ u_2 &v_2 &y \\ u_3 &v_3 &z \end{vmat} =(u_2v_3-u_3v_2)\cdot x + (u_3v_1-u_1v_3)\cdot y + (u_1v_2-u_2v_1)\cdot z \end{equation*} The equation for the point incident on two lines is the same. \end{answer} \item \label{exer:IncidentIndReps} Prove that the definition of incidence is independent of the choice of the representatives of $p$ and $L$. That is, if $p_1$, $p_2$, $p_3$, and $q_1$, $q_2$, $q_3$ are two triples of homogeneous coordinates for $p$, and $L_1$, $L_2$, $L_3$, and $M_1$, $M_2$, $M_3$ are two triples of homogeneous coordinates for $L$, prove that $p_1L_1+p_2L_2+p_3L_3=0$ if and only if $q_1M_1+q_2M_2+q_3M_3=0$. \begin{answer} If $p_1$, $p_2$, $p_3$, and $q_1$, $q_2$, $q_3$ are two triples of homogeneous coordinates for $p$ then the two column vectors are in proportion, that is, lie on the same line through the origin. Similarly, the two row vectors are in proportion. \begin{equation*} k\cdot\colvec{p_1 \\ p_2 \\ p_3} =\colvec{q_1 \\ q_2 \\ q_3} \qquad m\cdot\rowvec{L_1 &L_2 &L_3} =\rowvec{M_1 &M_2 &M_3} \end{equation*} Then multiplying gives the answer $(km)\cdot (p_1L_1+p_2L_2+p_3L_3)=q_1M_1+q_2M_2+q_3M_3=0$. \end{answer} \item Give a drawing to show that central projection does not preserve circles, that a circle may project to an ellipse. Can a (non-circular) ellipse project to a circle? %Must it (in the sense that for any ellipse there is a projection %such that it would project to a circle)? \begin{answer} The picture of the solar eclipse \Dash unless the image plane is exactly perpendicular to the line from the sun through the pinhole \Dash shows the circle of the sun projecting to an image that is an ellipse. (Another example is that in many pictures in this Topic, we've shown the circle that is the sphere's equator as an ellipse, that is, a viewer of the drawing sees a circle as an ellipse.) The solar eclipse picture also shows the converse. If we picture the projection as going from left to right through the pinhole then the ellipse $I$ projects through $P$ to a circle~$S$. \end{answer} \item \label{exer:CorrProjPlaneEucPl} Give the formula for the correspondence between the non-equatorial part of the antipodal modal of the projective plane, and the plane $z=1$. \begin{answer} A spot on the unit sphere \begin{equation*} \colvec{p_1 \\ p_2 \\ p_3} \end{equation*} is non-equatorial if and only if $p_3\neq 0$. In that case it corresponds to this point on the $z=1$ plane \begin{equation*} \colvec{p_1/p_3 \\ p_2/p_3 \\ 1} \end{equation*} since that is intersection of the line containing the vector and the plane. \end{answer} % \item \label{exer:ELinesCorrPLines} % Consider the correspondence between the non-ideal points in % the projective plane and the Euclidean plane. % \begin{exparts} % \item Show that any Euclidean line corresponds to a projective line. % \item Show that parallel Euclidean lines correspond to projective % lines that meet at an ideal point. % \item Prove the converses of those two statements. % \end{exparts} % \item \label{exer:EuclidDesarg} % Give a statement of Desargue's Theorem for Euclidean geometry. % \item \label{exer:DesarAntipPict} % Draw a picture illustrating Desargue's Theorem on the antipodal model. \item (Pappus's Theorem) Assume that $T_0$, $U_0$, and $V_0$ are collinear and that $T_1$, $U_1$, and $V_1$ are collinear. Consider these three points: (i)~the intersection $V_2$ of the lines $T_0U_1$ and $T_1U_0$, (ii)~the intersection $U_2$ of the lines $T_0V_1$ and $T_1V_0$, and (iii)~the intersection $T_2$ of $U_0V_1$ and $U_1V_0$. \begin{exparts} \partsitem Draw a (Euclidean) picture. \partsitem Apply the lemma used in Desargue's Theorem to get simple homogeneous coordinate vectors for the $T$'s and $V_0$. \partsitem Find the resulting homogeneous coordinate vectors for $U$'s (these must each involve a parameter as, e.g., $U_0$ could be anywhere on the $T_0V_0$ line). \partsitem Find the resulting homogeneous coordinate vectors for $V_1$. (\textit{Hint:}~it involves two parameters.) \partsitem Find the resulting homogeneous coordinate vectors for $V_2$. (It also involves two parameters.) \partsitem Show that the product of the three parameters is $1$. \partsitem Verify that $V_2$ is on the $T_2U_2$ line. \end{exparts} \begin{answer} \begin{exparts} \partsitem Other pictures are possible, but this is one. \begin{center} \includegraphics{ch4.54} \end{center} The intersections $T_0U_1\,\intersection T_1U_0=V_2$, $T_0V_1\,\intersection T_1V_0=U_2$, and $U_0V_1\,\intersection U_1V_0=T_2$ are labeled so that on each line is a $T$, a $U$, and a $V$. \partsitem The lemma used in Desargue's Theorem gives a basis $B$ with respect to which the points have these homogeneous coordinate vectors. \begin{equation*} \rep{\vec{t}_0}{B}=\colvec[r]{1 \\ 0 \\ 0} \quad \rep{\vec{t}_1}{B}=\colvec[r]{0 \\ 1 \\ 0} \quad \rep{\vec{t}_2}{B}=\colvec[r]{0 \\ 0 \\ 1} \quad \rep{\vec{v}_0}{B}=\colvec[r]{1 \\ 1 \\ 1} \end{equation*} \partsitem First, any $U_0$ on $T_0V_0$ \begin{equation*} \rep{\vec{u}_0}{B}=a\colvec[r]{1 \\ 0 \\ 0} +b\colvec[r]{1 \\ 1 \\ 1} =\colvec{a+b \\ b \\ b} \end{equation*} has homogeneous coordinate vectors of this form \begin{equation*} \colvec{u_0 \\ 1 \\ 1} \end{equation*} ($u_0$ is a parameter; it depends on where on the $T_0V_0$ line the point $U_0$ is, but any point on that line has a homogeneous coordinate vector of this form for some $u_0\in\Re$). Similarly, $U_2$ is on $T_1V_0$ \begin{equation*} \rep{\vec{u}_2}{B}=c\colvec[r]{0 \\ 1 \\ 0} +d\colvec[r]{1 \\ 1 \\ 1} =\colvec{d \\ c+d \\ d} \end{equation*} and so has this homogeneous coordinate vector. \begin{equation*} \colvec{1 \\ u_2 \\ 1} \end{equation*} Also similarly, $U_1$ is incident on $T_2V_0$ \begin{equation*} \rep{\vec{u}_1}{B}=e\colvec[r]{0 \\ 0 \\ 1} +f\colvec[r]{1 \\ 1 \\ 1} =\colvec{f \\ f \\ e+f} \end{equation*} and has this homogeneous coordinate vector. \begin{equation*} \colvec{1 \\ 1 \\ u_1} \end{equation*} \partsitem Because $V_1$ is $T_0U_2\,\intersection\,U_0T_2$ we have this. \begin{equation*} g\colvec[r]{1 \\ 0 \\ 0}+h\colvec{1 \\ u_2 \\ 1} =i\colvec{u_0 \\ 1 \\ 1}+j\colvec[r]{0 \\ 0 \\ 1} \qquad\Longrightarrow\qquad \begin{aligned} g+h &= iu_0 \\ hu_2 &= i \\ h &= i+j \end{aligned} \end{equation*} Substituting $hu_2$ for $i$ in the first equation \begin{equation*} \colvec{hu_0u_2 \\ hu_2 \\ h} \end{equation*} shows that $V_1$ has this two-parameter homogeneous coordinate vector. \begin{equation*} \colvec{u_0u_2 \\ u_2 \\ 1} \end{equation*} \partsitem Since $V_2$ is the intersection $T_0U_1\,\intersection\,T_1U_0$ \begin{equation*} k\colvec[r]{1 \\ 0 \\ 0}+l\colvec{1 \\ 1 \\ u_1} =m\colvec[r]{0 \\ 1 \\ 0}+n\colvec{u_0 \\ 1 \\ 1} \qquad\Longrightarrow\qquad \begin{aligned} k+l &= nu_0 \\ l &= m+n \\ lu_1 &= n \end{aligned} \end{equation*} and substituting $lu_1$ for $n$ in the first equation \begin{equation*} \colvec{lu_0u_1 \\ l \\ lu_1} \end{equation*} gives that $V_2$ has this two-parameter homogeneous coordinate vector. \begin{equation*} \colvec{u_0u_1 \\ 1 \\ u_1} \end{equation*} \partsitem Because $V_1$ is on the $T_1U_1$ line its homogeneous coordinate vector has the form \begin{equation*} p\colvec[r]{0 \\ 1 \\ 0}+q\colvec{1 \\ 1 \\ u_1} =\colvec{q \\ p+q \\ qu_1} \tag*{($*$)}\end{equation*} but a previous part of this question established that $V_1$'s homogeneous coordinate vectors have the form \begin{equation*} \colvec{u_0u_2 \\ u_2 \\ 1} \end{equation*} and so this a homogeneous coordinate vector for $V_1$. \begin{equation*} \colvec{u_0u_1u_2 \\ u_1u_2 \\ u_1} \tag*{($**$)}\end{equation*} By ($*$) and ($**$), there is a relationship among the three parameters:~$u_0u_1u_2=1$. \partsitem The homogeneous coordinate vector of $V_2$ can be written in this way. \begin{equation*} \colvec{u_0u_1u_2 \\ u_2 \\ u_1u_2} =\colvec{1 \\ u_2 \\ u_1u_2} \end{equation*} Now, the $T_2U_2$ line consists of the points whose homogeneous coordinates have this form. \begin{equation*} r\colvec[r]{0 \\ 0 \\ 1}+s\colvec{1 \\ u_2 \\ 1} =\colvec{s \\ su_2 \\ r+s} \end{equation*} Taking $s=1$ and $r=u_1u_2-1$ shows that the homogeneous coordinate vectors of $V_2$ have this form. \end{exparts} %\textit{Author's note.} % Good thing that there are no more parts because we've used all of the % lower-case letters. \end{answer} \end{exercises} \index{projective geometry|)} \endinput