projplane.tex 36.1 KB
 Jim Hefferon committed Dec 05, 2011 1 % Chapter 4, Topic _Linear Algebra_ Jim Hefferon  Jim Hefferon committed Nov 15, 2013 2 % http://joshua.smcvt.edu/linearalgebra  Jim Hefferon committed Dec 05, 2011 3 4 % 2001-Jun-12 \topic{Projective Geometry}  Jim Hefferon committed Nov 29, 2013 5 \index{projective geometry|(}  Jim Hefferon committed Dec 05, 2011 6 There are geometries other than the familiar Euclidean one.  Jim Hefferon committed Nov 16, 2013 7 8 One such geometry arose when artists observed that what a viewer sees is not necessarily what is there.  Jim Hefferon committed Nov 15, 2013 9 As an example, here is Leonardo da Vinci's  Jim Hefferon committed Nov 16, 2013 10 \textit{The Last Supper}.\index{da Vinci, Leonardo}\index{Last Supper@\textit{Last Supper}}  Jim Hefferon committed Jan 16, 2012 11 % From http://upload.wikimedia.org/wikipedia/commons/7/77/DaVinci_LastSupper_high_res_2_nowatmrk.jpg  Jim Hefferon committed Dec 05, 2011 12 \begin{center}  Jim Hefferon committed Nov 15, 2013 13  \includegraphics[width=.6\textwidth]{LastSupper.jpg}  Jim Hefferon committed Dec 05, 2011 14 \end{center}  Jim Hefferon committed Jan 16, 2012 15 16 Look at where the ceiling meets the left and right walls. In the room those lines are parallel but  Jim Hefferon committed Nov 15, 2013 17 da~Vinci has painted lines that, if extended, would intersect.  Jim Hefferon committed Nov 16, 2013 18 The intersection is the  Jim Hefferon committed Feb 20, 2012 19 \definend{vanishing point}.\index{vanishing point}\index{projection!central!vanishing point}  Jim Hefferon committed Jan 16, 2012 20 This aspect of perspective is familiar as an image of  Jim Hefferon committed Dec 05, 2011 21 22 railroad tracks that appear to converge at the horizon.  Jim Hefferon committed Nov 16, 2013 23 24 Da~Vinci has adopted a model of how we see. % of how  Jim Hefferon committed Nov 15, 2013 25 26 27 28 % 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  Jim Hefferon committed Nov 16, 2013 29 as a point on the line where the wall meets the ceiling.  Jim Hefferon committed Nov 15, 2013 30 This first intersection point is what the person sees in that direction.  Jim Hefferon committed Nov 16, 2013 31 32 Overall what the person sees is the collection of three-dimensional intersection points  Jim Hefferon committed Nov 15, 2013 33 projected to a common two dimensional image.  Jim Hefferon committed Dec 05, 2011 34 \begin{center}  Jim Hefferon committed Nov 15, 2013 35  \includegraphics[scale=.8]{ch4.5}  Jim Hefferon committed Dec 05, 2011 36 \end{center}  Jim Hefferon committed Nov 15, 2013 37 38 39 40 41 42 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  Jim Hefferon committed Jan 16, 2012 43 from the viewer to~$C$ is not orthogonal to the image plane.  Jim Hefferon committed Nov 15, 2013 44 45 46 47 48 49 (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.)  Jim Hefferon committed Jan 16, 2012 50 51  The operation of  Jim Hefferon committed Dec 05, 2011 52 central projection preserves some geometric  Jim Hefferon committed Jan 16, 2012 53 properties, for instance lines project to lines.  Jim Hefferon committed Nov 15, 2013 54 55 However, it fails to preserve some others. One example is that equal  Jim Hefferon committed Jan 16, 2012 56 length segments can project to segments of unequal length  Jim Hefferon committed Nov 15, 2013 57 (above, $AB$ is longer than $BC$ because the  Jim Hefferon committed Dec 05, 2011 58 segment projected to $AB$ is closer to the viewer and  Jim Hefferon committed Jan 16, 2012 59 closer things look bigger).  Jim Hefferon committed Dec 05, 2011 60 61 62 63 64 65 66 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}  Jim Hefferon committed Nov 15, 2013 67 We can think that each source point is pushed from the domain plane~$S$  Jim Hefferon committed Nov 16, 2013 68 outward to the image plane~$I$.  Jim Hefferon committed Jan 16, 2012 69 The second case of projection is that of the artist  Jim Hefferon committed Nov 15, 2013 70 pulling the source back to a canvas.  Jim Hefferon committed Dec 05, 2011 71 72 73 \begin{center} \includegraphics{ch4.7} \end{center}  Jim Hefferon committed Nov 15, 2013 74 75 76 The two are different because first $S$ is in the middle and then $I$. One more configuration can happen, with $P$ in the middle.  Jim Hefferon committed Dec 05, 2011 77 An example of this is when we use a pinhole to shine the  Jim Hefferon committed Nov 15, 2013 78 image of a solar eclipse onto a paper.  Jim Hefferon committed Dec 05, 2011 79 80 81 \begin{center} \includegraphics{ch4.8} \end{center}  Jim Hefferon committed Jan 16, 2012 82 83 84 85 86  Although the three are not exactly the same, they are similar. We shall say that each is a central projection by $P$ of  Jim Hefferon committed Dec 05, 2011 87 $S$ to $I$.  Jim Hefferon committed Jan 16, 2012 88 89 90 We next look at three models of central projection, of increasing abstractness but also of increasing uniformity.  Jim Hefferon committed Nov 15, 2013 91 The last model will bring out the linear algebra.  Jim Hefferon committed Dec 05, 2011 92 93 94 95  %To illustrate some of the geometric effects of these projections, Consider again the effect of railroad tracks that appear to converge to a point.  Jim Hefferon committed Nov 15, 2013 96 Model this with parallel lines in a domain plane~$S$  Jim Hefferon committed Dec 05, 2011 97 and a projection via a $P$ to a codomain plane~$I$.  Jim Hefferon committed Nov 16, 2013 98 (The gray lines shown are parallel to the $S$ plane and to the~$I$ plane.)  Jim Hefferon committed Dec 05, 2011 99 100 101 \begin{center} \includegraphics{ch4.9} \end{center}  Jim Hefferon committed Nov 16, 2013 102 This single setting shows all three projection cases.  Jim Hefferon committed Jan 16, 2012 103 The first picture below shows $P$ acting as a movie projector by pushing  Jim Hefferon committed Dec 05, 2011 104 points from part of $S$ out to image points on the lower half of $I$.  Jim Hefferon committed Jan 16, 2012 105 The middle picture shows $P$ acting as the artist by  Jim Hefferon committed Dec 05, 2011 106 107 pulling points from another part of $S$ back to image points in the middle of $I$.  Jim Hefferon committed Jan 16, 2012 108 In the third picture $P$ acts as the pinhole, projecting points from $S$  Jim Hefferon committed Dec 05, 2011 109 to the upper part of $I$.  Jim Hefferon committed Jan 16, 2012 110 This third picture is the trickiest\Dash the points that are  Jim Hefferon committed Dec 05, 2011 111 projected near to the vanishing point are the ones that are  Jim Hefferon committed Nov 15, 2013 112 far out on the lower left of $S$.  Jim Hefferon committed Dec 05, 2011 113 114 115 116 117 118 119 120 121 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}  Jim Hefferon committed Jan 16, 2012 122   Jim Hefferon committed Nov 15, 2013 123 124 There are two awkward things here. First, neither of the two points in the domain  Jim Hefferon committed Dec 05, 2011 125 126 127 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  Jim Hefferon committed Nov 15, 2013 128 129 (we say that these two are projected to infinity). The second is that  Jim Hefferon committed Dec 05, 2011 130 131 132 133 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  Jim Hefferon committed Nov 15, 2013 134 135 (we say that the vanishing point is the image of a projection from infinity).  Jim Hefferon committed Dec 05, 2011 136 137 138 139 \begin{center} \includegraphics{ch4.13} \end{center}  Jim Hefferon committed Jan 16, 2012 140 For a model that eliminates this awkwardness,  Jim Hefferon committed Nov 15, 2013 141 cover the projector $P$ with a hemispheric dome.  Jim Hefferon committed Nov 16, 2013 142 143 144 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,  Jim Hefferon committed Nov 15, 2013 145 as with the movie projector.  Jim Hefferon committed Nov 16, 2013 146 It includes projecting things on the line further from $P$ than the dome,  Jim Hefferon committed Nov 15, 2013 147 148 as with the painter. More subtly, it also includes things on the line that lie behind $P$,  Jim Hefferon committed Nov 16, 2013 149 as with the pinhole case.  Jim Hefferon committed Dec 05, 2011 150 151 152 \begin{center} \includegraphics{ch4.14} \end{center}  Jim Hefferon committed Nov 15, 2013 153 More formally,  Jim Hefferon committed Nov 16, 2013 154 for any nonzero vector $\vec{v}\in\Re^3$, let the associated  Jim Hefferon committed Dec 05, 2011 155 \definend{point $v$ in the projective plane}\index{point!in projective plane}  Jim Hefferon committed Nov 16, 2013 156 be the set $\set{k\vec{v}\suchthat \text{$k\in\Re$and$k\neq 0$}}$  Jim Hefferon committed Dec 05, 2011 157 158 159 160 161 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  Jim Hefferon committed Nov 16, 2013 162 can be represented in any of these three ways.  Jim Hefferon committed Dec 05, 2011 163 \begin{equation*}  Jim Hefferon committed Jan 16, 2012 164  \colvec[r]{1 \\ 2 \\ 3}  Jim Hefferon committed Dec 05, 2011 165 166 167  \qquad \colvec{1/3 \\ 2/3 \\ 1} \qquad  Jim Hefferon committed Jan 16, 2012 168  \colvec[r]{-2 \\ -4 \\ -6}  Jim Hefferon committed Dec 05, 2011 169 170 171 172 \end{equation*} Each of these is a \definend{homogeneous coordinate vector}\index{coordinates!homogeneous}% \index{homogeneous coordinate vector}\index{vector!homogeneous coordinate}  Jim Hefferon committed Nov 15, 2013 173 for the point~$\ell$.  Jim Hefferon committed Dec 05, 2011 174   Jim Hefferon committed Jan 16, 2012 175 This picture and definition  Jim Hefferon committed Nov 15, 2013 176 177 clarifies central projection but there is still something ungainly about the dome  Jim Hefferon committed Nov 15, 2013 178 model:~what happens when $P$ looks down?  Jim Hefferon committed Nov 16, 2013 179 180 181 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  Jim Hefferon committed Nov 15, 2013 182 below.  Jim Hefferon committed Nov 16, 2013 183 Now the part of the line~$\ell$ that intersects the dome lies behind the page.  Jim Hefferon committed Nov 15, 2013 184 185  That is, as  Jim Hefferon committed Dec 05, 2011 186 187 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.  Jim Hefferon committed Nov 15, 2013 188 (This brings out that the dome does not include the entire equator  Jim Hefferon committed Feb 20, 2012 189 or else  Jim Hefferon committed Nov 15, 2013 190 when the viewer is looking exactly along the equator then there would be  Jim Hefferon committed Nov 16, 2013 191 two points in the line that are both on the dome.  Jim Hefferon committed Nov 15, 2013 192 Instead we define the dome so that it includes the  Jim Hefferon committed Jan 16, 2012 193 points on the equator with a positive~$y$ coordinate, as well as the point  Jim Hefferon committed Nov 15, 2013 194 where $y=0$ and $x$ is positive.)  Jim Hefferon committed Jan 16, 2012 195 This discontinuity means that  Jim Hefferon committed Dec 05, 2011 196 we often have to treat equatorial points as a separate case.  Jim Hefferon committed Nov 15, 2013 197 198 So while the railroad track model of central projection has three cases, the dome has two.  Jim Hefferon committed Dec 05, 2011 199   Jim Hefferon committed Nov 15, 2013 200 We can do better, we can reduce to a model having a single case.  Jim Hefferon committed Dec 05, 2011 201 Consider a sphere centered at the origin.  Jim Hefferon committed Nov 15, 2013 202 203 Any line through the origin intersects the sphere in two spots, said to be antipodal.\index{antipodal points}  Jim Hefferon committed Dec 05, 2011 204 205 206 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.  Jim Hefferon committed Jan 16, 2012 207 Below, we show the two antipodal spots connected by a dashed line  Jim Hefferon committed Dec 05, 2011 208 209 210 211 212 213 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  Jim Hefferon committed Nov 15, 2013 214 spots on the sphere is not as intuitive as the one-spot-per-point dome mode,  Jim Hefferon committed Dec 05, 2011 215 on the other hand  Jim Hefferon committed Nov 16, 2013 216 the awkwardness of the dome model is gone in that  Jim Hefferon committed Dec 05, 2011 217 as a line of view slides from north to south,  Jim Hefferon committed Jan 16, 2012 218 no sudden changes happen.  Jim Hefferon committed Nov 16, 2013 219 This central projection model is uniform.  Jim Hefferon committed Dec 05, 2011 220 221 222  So far we have described points in projective geometry. What about lines?  Jim Hefferon committed Nov 16, 2013 223 What a viewer~$P$ at the origin sees as a line is shown below as  Jim Hefferon committed Dec 05, 2011 224 225 226 227 228 a great circle, the intersection of the model sphere with a plane through the origin. \begin{center} \includegraphics{ch4.16} \end{center}  Jim Hefferon committed Jan 16, 2012 229 230 (We've included one of the projective points on this line to bring out a subtlety.  Jim Hefferon committed Dec 05, 2011 231 232 233 234 235 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,  Jim Hefferon committed Nov 15, 2013 236 we can also describe a projective line with a triple of reals.  Jim Hefferon committed Dec 05, 2011 237 238 239 For instance, the members of this plane through the origin in $\Re^3$ \begin{equation*}  Jim Hefferon committed Jan 16, 2012 240  %\set{y\colvec[r]{1 \\ -1 \\ 0}+z\colvec[r]{1 \\ 0 \\ 1}\suchthat y,z\in\Re}=  Jim Hefferon committed Dec 05, 2011 241 242  \set{\colvec{x \\ y \\ z}\suchthat x+y-z=0} \end{equation*}  Jim Hefferon committed Nov 16, 2013 243 project to a line that we can describe with  Jim Hefferon committed Dec 05, 2011 244 $\rowvec{1 &1 &-1}$  Jim Hefferon committed Nov 16, 2013 245 (using a row vector for this typographically distinguishes lines from points).  Jim Hefferon committed Dec 05, 2011 246 247 248 249 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}  Jim Hefferon committed Nov 15, 2013 250 to be the set $L=\set{k\vec{L}\suchthat \text{$k\in\Re$and$k\neq 0$}}$.  Jim Hefferon committed Dec 05, 2011 251   Jim Hefferon committed Nov 16, 2013 252 253 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  Jim Hefferon committed Dec 05, 2011 254 \definend{incident} \Dash the  Jim Hefferon committed Jan 17, 2012 255 point lies on the line, the line passes through the point \Dash if and only  Jim Hefferon committed Dec 05, 2011 256 257 258 259 260 261 262 263 264 265 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  Jim Hefferon committed Nov 15, 2013 266 of the form $k\cdot x+k\cdot y-k\cdot z=0$ for any nonzero~$k$.  Jim Hefferon committed Dec 05, 2011 267 268 269 That is, the incidence formula is inherited from the three-space lines and planes of which $v$ and $L$ are projections.  Jim Hefferon committed Nov 16, 2013 270 With this, we can do analytic projective geometry.  Jim Hefferon committed Dec 05, 2011 271 272 273 For instance, the projective line $L=\rowvec{1 &1 &-1}$ has the equation $1v_1+1v_2-1v_3=0$,  Jim Hefferon committed Nov 16, 2013 274 meaning that for any projective point~$v$ incident with the line, any of $v$'s  Jim Hefferon committed Nov 15, 2013 275 representative homogeneous coordinate vectors will  Jim Hefferon committed Nov 16, 2013 276 277 278 satisfy the equation. This is true simply because those vectors lie on the three space plane. One difference from Euclidean analytic geometry is that  Jim Hefferon committed Nov 15, 2013 279 in projective geometry besides talking about the equation of a line,  Jim Hefferon committed Nov 16, 2013 280 281 we also talk about the equation of a point. For the fixed point  Jim Hefferon committed Dec 05, 2011 282 \begin{equation*}  Jim Hefferon committed Jan 16, 2012 283  v=\colvec[r]{1 \\ 2 \\ 3}  Jim Hefferon committed Dec 05, 2011 284 \end{equation*}  Jim Hefferon committed Nov 16, 2013 285 286 the property that characterizes lines incident on this point is that the components  Jim Hefferon committed Dec 05, 2011 287 288 289 290 of any representatives satisfy $1L_1+2L_2+3L_3=0$ and so this is the equation of $v$.  Jim Hefferon committed Nov 15, 2013 291 This symmetry of the statements about lines and points is  Jim Hefferon committed Nov 29, 2013 292 293 the \definend{Duality Principle}\index{Duality Principle, of projective geometry}  Jim Hefferon committed Dec 05, 2011 294 295 296 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,  Jim Hefferon committed Jan 16, 2012 297 in the projective plane two distinct lines determine one and only one point.  Jim Hefferon committed Nov 15, 2013 298 299 Here is a picture showing two projective lines that cross in antipodal spots and thus  Jim Hefferon committed Dec 05, 2011 300 301 302 303 304 305 cross at one projective point. \begin{center} \hfill \begin{tabular}{@{}c@{}}\includegraphics{ch4.17}\end{tabular} \hfill\llap{($*$)} \end{center}  Jim Hefferon committed Nov 15, 2013 306 Contrast this with Euclidean geometry, where two unequal lines may  Jim Hefferon committed Nov 16, 2013 307 have a unique intersection or may be parallel.  Jim Hefferon committed Dec 05, 2011 308 309 310 311 In this way, projective geometry is simpler, more uniform, than Euclidean geometry. That simplicity is relevant because there is a  Jim Hefferon committed Jan 16, 2012 312 313 relationship between the two spaces:~we can view the projective plane as an extension of the Euclidean plane.  Jim Hefferon committed Nov 16, 2013 314 315 316 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  Jim Hefferon committed Dec 05, 2011 317 antipodal spots from the sphere.  Jim Hefferon committed Nov 15, 2013 318 Conversely, we can view some points  Jim Hefferon committed Nov 16, 2013 319 320 321 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.)  Jim Hefferon committed Dec 05, 2011 322 323 324 325 326 \begin{center} \hfill \begin{tabular}{@{}c@{}}\includegraphics{ch4.18}\end{tabular} \hfill\llap{($**$)} \end{center}  Jim Hefferon committed Nov 16, 2013 327 Thus we can think of projective space as consisting of the Euclidean plane  Jim Hefferon committed Dec 05, 2011 328 329 with some extra points adjoined \Dash the Euclidean plane is embedded in the projective plane.  Jim Hefferon committed Nov 16, 2013 330 The extra points in projective space, the equatorial points,  Jim Hefferon committed Nov 27, 2013 331 are called \definend{ideal points}\index{ideal!point}%  Jim Hefferon committed Dec 05, 2011 332 333 \index{projective plane!ideal point} or \definend{points at infinity}\index{point!at infinity}  Jim Hefferon committed Nov 15, 2013 334 and the equator is called the  Jim Hefferon committed Nov 27, 2013 335 \definend{ideal line}\index{ideal!line}%  Jim Hefferon committed Dec 05, 2011 336 337 \index{projective plane!ideal line} or \definend{line at infinity}\index{line at infinity}  Jim Hefferon committed Jan 16, 2012 338 (it is not a Euclidean line, it is a projective line).  Jim Hefferon committed Dec 05, 2011 339   Jim Hefferon committed Nov 15, 2013 340 341 342 The advantage of this extension from the Euclidean plane to the projective plane is that some of the nonuniformity  Jim Hefferon committed Dec 05, 2011 343 344 345 346 347 348 349 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,  Jim Hefferon committed Nov 15, 2013 350 351 that distinct lines either intersect or are parallel, to having only one case, that distinct lines intersect (possibly at a point at infinity).  Jim Hefferon committed Dec 05, 2011 352   Jim Hefferon committed Nov 16, 2013 353 354 355 356 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.  Jim Hefferon committed Dec 05, 2011 357 358 359 Analytic projective geometry uses linear algebra. For instance, for three points of the projective plane $t$, $u$, and~$v$,  Jim Hefferon committed Nov 16, 2013 360 setting up the equations for those points by fixing vectors representing each  Jim Hefferon committed Nov 15, 2013 361 shows that the three are collinear if and only if the resulting three-equation  Jim Hefferon committed Dec 05, 2011 362 system has infinitely many row vector solutions  Jim Hefferon committed Nov 15, 2013 363 representing their line.  Jim Hefferon committed Nov 16, 2013 364 That in turn holds if and only if this determinant is zero.  Jim Hefferon committed Dec 05, 2011 365 \begin{equation*}  Jim Hefferon committed Jan 16, 2012 366  \begin{vmat}  Jim Hefferon committed Dec 05, 2011 367 368 369  t_1 &u_1 &v_1 \\ t_2 &u_2 &v_2 \\ t_3 &u_3 &v_3  Jim Hefferon committed Jan 16, 2012 370  \end{vmat}  Jim Hefferon committed Dec 05, 2011 371 372 373 \end{equation*} Thus, three points in the projective plane are collinear if and only if any three representative column vectors are linearly dependent.  Jim Hefferon committed Nov 15, 2013 374 Similarly, by duality,  Jim Hefferon committed Dec 05, 2011 375 376 377 378 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.  Jim Hefferon committed Feb 20, 2012 379 The following result is more evidence of the niceness  Jim Hefferon committed Nov 15, 2013 380 of the geometry of the projective plane.  Jim Hefferon committed Nov 16, 2013 381 These two triangles are  Jim Hefferon committed Nov 29, 2013 382 \definend{in perspective}\index{perspective, triangles} from the point $O$  Jim Hefferon committed Dec 05, 2011 383 384 385 386 387 388 389 390 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 $U_2V_2$.  Jim Hefferon committed Nov 15, 2013 391 \definend{Desargue's Theorem}\index{Desargue's Theorem}  Jim Hefferon committed Jan 16, 2012 392 is that when we extend the three pairs of corresponding  Jim Hefferon committed Nov 15, 2013 393 394 395 sides, they intersect (shown here as the points $TU$, $TV$, and $UV$). What's more, those three intersection points are collinear.  Jim Hefferon committed Dec 05, 2011 396 397 398 \begin{center} \includegraphics{ch4.23} \end{center}  Jim Hefferon committed Nov 15, 2013 399 400 401 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  Jim Hefferon committed Dec 05, 2011 402 403 404 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  Jim Hefferon committed Nov 15, 2013 405 figures that the sides appear in our sketch to be straight.)  Jim Hefferon committed Dec 05, 2011 406   Jim Hefferon committed Nov 15, 2013 407 408 409 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,  Jim Hefferon committed Dec 05, 2011 410 411 412 413 414 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*}  Jim Hefferon committed Jan 16, 2012 415  \rep{\vec{w}}{B}=\colvec[r]{1 \\ 0 \\ 0}  Jim Hefferon committed Dec 05, 2011 416  \quad  Jim Hefferon committed Jan 16, 2012 417  \rep{\vec{x}}{B}=\colvec[r]{0 \\ 1 \\ 0}  Jim Hefferon committed Dec 05, 2011 418  \quad  Jim Hefferon committed Jan 16, 2012 419  \rep{\vec{y}}{B}=\colvec[r]{0 \\ 0 \\ 1}  Jim Hefferon committed Dec 05, 2011 420  \quad  Jim Hefferon committed Jan 16, 2012 421  \rep{\vec{z}}{B}=\colvec[r]{1 \\ 1 \\ 1}  Jim Hefferon committed Dec 05, 2011 422 \end{equation*}  Jim Hefferon committed Nov 16, 2013 423 To prove the lemma,  Jim Hefferon committed Jan 16, 2012 424 because $W$, $X$, and $Y$ are not on the same projective line, any  Jim Hefferon committed Dec 05, 2011 425 homogeneous coordinate vectors  Jim Hefferon committed Jan 16, 2012 426 $\vec{w}_0$, $\vec{x}_0$, and $\vec{y}_0$ do not line on the same  Jim Hefferon committed Dec 05, 2011 427 428 plane through the origin in $\Re^3$ and so form a spanning set for $\Re^3$.  Jim Hefferon committed Jan 16, 2012 429 Thus any homogeneous coordinate vector for $Z$ is a combination  Jim Hefferon committed Dec 05, 2011 430 $\vec{z}_0=a\cdot\vec{w}_0+b\cdot\vec{x}_0+c\cdot\vec{y}_0$.  Jim Hefferon committed Nov 15, 2013 431 Then let the basis be $B=\sequence{\vec{w},\vec{x},\vec{y}}$ and take  Jim Hefferon committed Dec 05, 2011 432 433 434 $\vec{w}=a\cdot\vec{w}_0$, $\vec{x}=b\cdot\vec{x}_0$, $\vec{y}=c\cdot\vec{y}_0$,  Jim Hefferon committed Nov 15, 2013 435 and $\vec{z}=\vec{z}_0$.  Jim Hefferon committed Dec 05, 2011 436   Jim Hefferon committed Nov 15, 2013 437 To prove Desargue's Theorem use the lemma to fix homogeneous  Jim Hefferon committed Dec 05, 2011 438 439 coordinate vectors and a basis. \begin{equation*}  Jim Hefferon committed Jan 16, 2012 440  \rep{\vec{t}_1}{B}=\colvec[r]{1 \\ 0 \\ 0}  Jim Hefferon committed Dec 05, 2011 441  \quad  Jim Hefferon committed Jan 16, 2012 442  \rep{\vec{u}_1}{B}=\colvec[r]{0 \\ 1 \\ 0}  Jim Hefferon committed Dec 05, 2011 443  \quad  Jim Hefferon committed Jan 16, 2012 444  \rep{\vec{v}_1}{B}=\colvec[r]{0 \\ 0 \\ 1}  Jim Hefferon committed Dec 05, 2011 445  \quad  Jim Hefferon committed Jan 16, 2012 446  \rep{\vec{o}}{B}=\colvec[r]{1 \\ 1 \\ 1}  Jim Hefferon committed Dec 05, 2011 447 \end{equation*}  Jim Hefferon committed Nov 16, 2013 448 449 The projective point $T_2$ is incident on the projective line $OT_1$ so any homogeneous coordinate vector for $T_2$ lies in the  Jim Hefferon committed Dec 05, 2011 450 451 452 plane through the origin in $\Re^3$ that is spanned by homogeneous coordinate vectors of $O$ and $T_1$: \begin{equation*}  Jim Hefferon committed Jan 16, 2012 453 454  \rep{\vec{t}_2}{B}=a\colvec[r]{1 \\ 1 \\ 1} +b\colvec[r]{1 \\ 0 \\ 0}  Jim Hefferon committed Dec 05, 2011 455 456 \end{equation*} for some scalars $a$ and $b$.  Jim Hefferon committed Nov 16, 2013 457 458 459 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.  Jim Hefferon committed Dec 05, 2011 460 461 462 463 464 465 466 \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*}  Jim Hefferon committed Nov 15, 2013 467 The projective line $T_1U_1$ is the projection of a plane through the  Jim Hefferon committed Dec 05, 2011 468 origin in $\Re^3$.  Jim Hefferon committed Jan 16, 2012 469 One way to get its equation is to note that any vector in it  Jim Hefferon committed Dec 05, 2011 470 471 472 is linearly dependent on the vectors for $T_1$ and $U_1$ and so this determinant is zero. \begin{equation*}  Jim Hefferon committed Jan 16, 2012 473  \begin{vmat}  Jim Hefferon committed Dec 05, 2011 474 475 476  1 &0 &x \\ 0 &1 &y \\ 0 &0 &z  Jim Hefferon committed Jan 16, 2012 477  \end{vmat}=0  Jim Hefferon committed Dec 05, 2011 478 479 480 481 482 483 484 485  \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*}  Jim Hefferon committed Jan 16, 2012 486  \begin{vmat}  Jim Hefferon committed Dec 05, 2011 487 488 489  t_2 &1 &x \\ 1 &u_2 &y \\ 1 &1 &z  Jim Hefferon committed Jan 16, 2012 490  \end{vmat}=0  Jim Hefferon committed Dec 05, 2011 491 492 493 494 495 496 497 498 499 500  \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*}  Jim Hefferon committed Nov 15, 2013 501 (This is, of course, a homogeneous coordinate vector of a projective point.)  Jim Hefferon committed Dec 05, 2011 502 503 504 505 506 507 508 509 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*}  Jim Hefferon committed Jan 16, 2012 510 Finish the proof by noting that  Jim Hefferon committed Dec 05, 2011 511 512 513 514 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  Jim Hefferon committed Jan 16, 2012 515 the Euclidean result may be messier to state and prove.  Jim Hefferon committed Dec 05, 2011 516 Desargue's theorem illustrates this.  Jim Hefferon committed Jan 16, 2012 517 518 In the translation to Euclidean space, we must treat separately the case where $O$ lies on the ideal line, for then the lines  Jim Hefferon committed Dec 05, 2011 519 520 $T_1T_2$, $U_1U_2$, and $V_1V_2$ are parallel.  Jim Hefferon committed Nov 16, 2013 521 The remark  Jim Hefferon committed Dec 05, 2011 522 523 following the statement of Desargue's Theorem suggests thinking of the Euclidean pictures as figures from projective geometry  Jim Hefferon committed Nov 16, 2013 524 for a sphere model with very large radius.  Jim Hefferon committed Feb 20, 2012 525 526 527 That is, just as a small area of the world seems to people living there to be flat, the projective plane is locally Euclidean.  Jim Hefferon committed Dec 05, 2011 528   Jim Hefferon committed Nov 16, 2013 529 We finish by pointing out one more thing about the projective plane.  Jim Hefferon committed Feb 20, 2012 530 Although its local properties are familiar, the projective plane has  Jim Hefferon committed Nov 16, 2013 531 a perhaps unfamiliar global property.  Jim Hefferon committed Dec 05, 2011 532 The picture below shows a projective point.  Jim Hefferon committed Feb 20, 2012 533 At that point we have drawn Cartesian axes, $xy$-axes.  Jim Hefferon committed Nov 16, 2013 534 Of course, the axes appear in the picture at both antipodal spots, one in the  Jim Hefferon committed Nov 15, 2013 535 northern hemisphere (that is, shown on the right)  Jim Hefferon committed Feb 20, 2012 536 537 and the other in the south.  Jim Hefferon committed Nov 16, 2013 538 539 540 541 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  Jim Hefferon committed Feb 20, 2012 542 along the $x$-axis in the  Jim Hefferon committed Nov 16, 2013 543 positive direction.  Jim Hefferon committed Nov 15, 2013 544 545 546 547 548 549 550 551 552 553 % 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.  Jim Hefferon committed Dec 05, 2011 554 555 556 \begin{center} \includegraphics{ch4.24} \end{center}  Jim Hefferon committed Nov 15, 2013 557 558 The sequence of pictures below show a trip around this space:  Jim Hefferon committed Nov 16, 2013 559 the antipodal spots rotate around the sphere with the spot in the  Jim Hefferon committed Nov 15, 2013 560 561 562 563 564 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  Jim Hefferon committed Dec 05, 2011 565 projective point.  Jim Hefferon committed Nov 15, 2013 566 So by the third sphere below the trip has pretty much returned  Jim Hefferon committed Nov 16, 2013 567 to the same projective point where we drew it starting above.)  Jim Hefferon committed Dec 05, 2011 568 569 570 571 572 573 574 575 576 \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.  Jim Hefferon committed Nov 15, 2013 577 578 579 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.  Jim Hefferon committed Nov 16, 2013 580 The projective plane is not orientable\Dash in this geometry,  Jim Hefferon committed Nov 15, 2013 581 left and right handedness are not fixed properties of figures  Jim Hefferon committed Nov 16, 2013 582 583 (said another way, we cannot describe a spiral as clockwise or  Jim Hefferon committed Dec 05, 2011 584 585 586 counterclockwise). This exhibition of the existence of a  Jim Hefferon committed Nov 15, 2013 587 588 non-orientable space raises the question of whether our universe orientable.  Jim Hefferon committed Nov 16, 2013 589 Could  Jim Hefferon committed Feb 20, 2012 590 an astronaut leave earth right-handed and return left-handed?  Jim Hefferon committed Jan 16, 2012 591 \cite{Gardner} is a nontechnical reference.  Jim Hefferon committed Feb 20, 2012 592 593 \cite{Clarke} is a classic science fiction story about orientation reversal.  Jim Hefferon committed Dec 05, 2011 594   Jim Hefferon committed Nov 15, 2013 595 For an overview of projective geometry see \cite{CourantRobbins}.  Jim Hefferon committed Dec 05, 2011 596 The approach we've taken here, the analytic approach,  Jim Hefferon committed Nov 16, 2013 597 598 599 600 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}.  Jim Hefferon committed Dec 05, 2011 601 602 603 604 605 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}.  Jim Hefferon committed Nov 15, 2013 606 An easy and interesting application is in \cite{Davies}.  Jim Hefferon committed Dec 05, 2011 607 608 609 610 611  \begin{exercises} \item What is the equation of this point? \begin{equation*}  Jim Hefferon committed Jan 16, 2012 612  \colvec[r]{1 \\ 0 \\ 0}  Jim Hefferon committed Dec 05, 2011 613 614 615 616  \end{equation*} \begin{answer} From the dot product \begin{equation*}  Jim Hefferon committed Jan 16, 2012 617  0=\colvec[r]{1 \\ 0 \\ 0}\dotprod\rowvec{L_1 &L_2 &L_3}  Jim Hefferon committed Dec 05, 2011 618 619 620 621 622 623 624 625 626  =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*}  Jim Hefferon committed Jan 16, 2012 627  \colvec[r]{1 \\ 2 \\ 3},\,\colvec[r]{4 \\ 5 \\ 6}  Jim Hefferon committed Dec 05, 2011 628 629 630 631 632 633 634 635 636 637 638  \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*}  Jim Hefferon committed Jan 16, 2012 639  0=\begin{vmat}  Jim Hefferon committed Dec 05, 2011 640 641 642  1 &4 &x \\ 2 &5 &y \\ 3 &6 &z  Jim Hefferon committed Jan 16, 2012 643  \end{vmat}  Jim Hefferon committed Dec 05, 2011 644 645 646  =-3x+6y-3z \end{equation*} shows that the line is $L=\rowvec{-3 &6 &-3}$.  Jim Hefferon committed Jan 16, 2012 647  \partsitem $\colvec[r]{-3 \\ 6 \\ -3}$  Jim Hefferon committed Dec 05, 2011 648 649 650 651 652 653 654 655 656 657 658 659  \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*}  Jim Hefferon committed Jan 16, 2012 660  comes from this determinant equation.  Jim Hefferon committed Dec 05, 2011 661  \begin{equation*}  Jim Hefferon committed Jan 16, 2012 662  0=\begin{vmat}  Jim Hefferon committed Dec 05, 2011 663 664 665  u_1 &v_1 &x \\ u_2 &v_2 &y \\ u_3 &v_3 &z  Jim Hefferon committed Jan 16, 2012 666  \end{vmat}  Jim Hefferon committed Dec 05, 2011 667 668 669 670 671 672 673 674 675 676 677 678 679 680 681 682 683 684 685 686 687 688 689 690 691 692 693 694 695 696 697 698 699 700 701 702  =(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  Jim Hefferon committed Jan 16, 2012 703  %such that it would project to a circle)?  Jim Hefferon committed Dec 05, 2011 704 705 706 707 708 709  \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  Jim Hefferon committed Jan 16, 2012 710 711  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.)  Jim Hefferon committed Dec 05, 2011 712 713 714 715 716 717 718 719 720 721 722 723 724 725 726 727 728 729 730 731 732 733 734 735 736 737 738 739 740 741 742 743 744 745 746 747 748 749 750 751 752 753 754 755 756 757 758 759 760 761 762 763 764 765 766 767 768 769 770 771 772 773 774 775 776 777 778 779 780 781 782 783 784 785 786 787 788 789 790 791  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*}  Jim Hefferon committed Jan 16, 2012 792  \rep{\vec{t}_0}{B}=\colvec[r]{1 \\ 0 \\ 0}  Jim Hefferon committed Dec 05, 2011 793  \quad  Jim Hefferon committed Jan 16, 2012 794  \rep{\vec{t}_1}{B}=\colvec[r]{0 \\ 1 \\ 0}  Jim Hefferon committed Dec 05, 2011 795  \quad  Jim Hefferon committed Jan 16, 2012 796  \rep{\vec{t}_2}{B}=\colvec[r]{0 \\ 0 \\ 1}  Jim Hefferon committed Dec 05, 2011 797  \quad  Jim Hefferon committed Jan 16, 2012 798  \rep{\vec{v}_0}{B}=\colvec[r]{1 \\ 1 \\ 1}  Jim Hefferon committed Dec 05, 2011 799 800 801 802  \end{equation*} \partsitem First, any $U_0$ on $T_0V_0$ \begin{equation*}  Jim Hefferon committed Jan 16, 2012 803 804  \rep{\vec{u}_0}{B}=a\colvec[r]{1 \\ 0 \\ 0} +b\colvec[r]{1 \\ 1 \\ 1}  Jim Hefferon committed Dec 05, 2011 805 806 807 808 809 810 811 812 813 814 815  =\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*}  Jim Hefferon committed Jan 16, 2012 816 817  \rep{\vec{u}_2}{B}=c\colvec[r]{0 \\ 1 \\ 0} +d\colvec[r]{1 \\ 1 \\ 1}  Jim Hefferon committed Dec 05, 2011 818 819 820 821 822 823 824 825  =\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*}  Jim Hefferon committed Jan 16, 2012 826 827  \rep{\vec{u}_1}{B}=e\colvec[r]{0 \\ 0 \\ 1} +f\colvec[r]{1 \\ 1 \\ 1}  Jim Hefferon committed Dec 05, 2011 828 829 830 831 832 833 834 835 836  =\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*}  Jim Hefferon committed Jan 16, 2012 837 838  g\colvec[r]{1 \\ 0 \\ 0}+h\colvec{1 \\ u_2 \\ 1} =i\colvec{u_0 \\ 1 \\ 1}+j\colvec[r]{0 \\ 0 \\ 1}  Jim Hefferon committed Dec 05, 2011 839 840 841 842 843 844 845 846 847 848 849 850 851 852 853 854 855 856 857 858  \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*}  Jim Hefferon committed Jan 16, 2012 859 860  k\colvec[r]{1 \\ 0 \\ 0}+l\colvec{1 \\ 1 \\ u_1} =m\colvec[r]{0 \\ 1 \\ 0}+n\colvec{u_0 \\ 1 \\ 1}  Jim Hefferon committed Dec 05, 2011 861 862 863 864 865 866 867 868 869 870 871 872 873 874 875 876 877 878 879 880  \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*}  Jim Hefferon committed Jan 16, 2012 881  p\colvec[r]{0 \\ 1 \\ 0}+q\colvec{1 \\ 1 \\ u_1}  Jim Hefferon committed Dec 05, 2011 882 883 884 885 886 887 888 889 890 891 892 893 894 895 896 897 898 899 900 901 902 903 904  =\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*}  Jim Hefferon committed Jan 16, 2012 905  r\colvec[r]{0 \\ 0 \\ 1}+s\colvec{1 \\ u_2 \\ 1}  Jim Hefferon committed Dec 05, 2011 906 907 908 909 910 911 912 913 914 915  =\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}  Jim Hefferon committed Nov 29, 2013 916 \index{projective geometry|)}  Jim Hefferon committed Dec 05, 2011 917 \endinput