projplane.tex 36.1 KB
Newer Older
1
% Chapter 4, Topic _Linear Algebra_ Jim Hefferon
Jim Hefferon's avatar
Jim Hefferon committed
2
%  http://joshua.smcvt.edu/linearalgebra
3 4
%  2001-Jun-12
\topic{Projective Geometry}
Jim Hefferon's avatar
Jim Hefferon committed
5
\index{projective geometry|(}
6
There are geometries other than the familiar Euclidean one.
Jim Hefferon's avatar
Jim Hefferon committed
7 8
One such geometry arose when artists observed
that what a viewer sees is not necessarily what is there.
Jim Hefferon's avatar
Jim Hefferon committed
9
As an example, here is Leonardo da Vinci's 
Jim Hefferon's avatar
Jim Hefferon committed
10
\textit{The Last Supper}.\index{da Vinci, Leonardo}\index{Last Supper@\textit{Last Supper}}
Jim Hefferon's avatar
Jim Hefferon committed
11
% From http://upload.wikimedia.org/wikipedia/commons/7/77/DaVinci_LastSupper_high_res_2_nowatmrk.jpg
12
\begin{center} 
Jim Hefferon's avatar
Jim Hefferon committed
13
  \includegraphics[width=.6\textwidth]{LastSupper.jpg}
14
\end{center}
Jim Hefferon's avatar
Jim Hefferon committed
15 16
Look at where the ceiling meets the left and right walls.
In the room those lines are parallel but 
Jim Hefferon's avatar
Jim Hefferon committed
17
da~Vinci has painted lines that, if extended, would intersect.
Jim Hefferon's avatar
Jim Hefferon committed
18
The intersection is the 
19
\definend{vanishing point}.\index{vanishing point}\index{projection!central!vanishing point}
Jim Hefferon's avatar
Jim Hefferon committed
20
This aspect of perspective is familiar as an image of 
21 22
railroad tracks that appear to converge at the horizon.

Jim Hefferon's avatar
Jim Hefferon committed
23 24
Da~Vinci has adopted a model of how we see.
% of how
Jim Hefferon's avatar
Jim Hefferon committed
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's avatar
Jim Hefferon committed
29
as a point on the line where the wall meets the ceiling.
Jim Hefferon's avatar
Jim Hefferon committed
30
This first intersection point is what the person sees in that direction.
Jim Hefferon's avatar
Jim Hefferon committed
31 32
Overall what the person sees is the collection of three-dimensional
intersection points
Jim Hefferon's avatar
Jim Hefferon committed
33
projected to a common two dimensional image. 
34
\begin{center}
Jim Hefferon's avatar
Jim Hefferon committed
35
  \includegraphics[scale=.8]{ch4.5}
36
\end{center}
Jim Hefferon's avatar
Jim Hefferon committed
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's avatar
Jim Hefferon committed
43
from the viewer to~$C$ is not orthogonal to the image plane.
Jim Hefferon's avatar
Jim Hefferon committed
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's avatar
Jim Hefferon committed
50 51

The operation of  
52
central projection preserves some geometric 
Jim Hefferon's avatar
Jim Hefferon committed
53
properties, for instance lines project to lines.
Jim Hefferon's avatar
Jim Hefferon committed
54 55
However, it fails to preserve some others.
One example is that equal 
Jim Hefferon's avatar
Jim Hefferon committed
56
length segments can project to segments of unequal length 
Jim Hefferon's avatar
Jim Hefferon committed
57
(above, $AB$ is longer than $BC$ because the
58
segment projected to $AB$ is closer to the viewer and 
Jim Hefferon's avatar
Jim Hefferon committed
59
closer things look bigger).
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's avatar
Jim Hefferon committed
67
We can think that each source point is pushed from the domain plane~$S$
Jim Hefferon's avatar
Jim Hefferon committed
68
outward to the image plane~$I$.
Jim Hefferon's avatar
Jim Hefferon committed
69
The second case of projection is that of the artist
Jim Hefferon's avatar
Jim Hefferon committed
70
pulling the source back to a canvas.
71 72 73
\begin{center}
  \includegraphics{ch4.7}
\end{center}
Jim Hefferon's avatar
Jim Hefferon committed
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. 
77
An example of this is when we use a pinhole to shine the 
Jim Hefferon's avatar
Jim Hefferon committed
78
image of a solar eclipse onto a paper.
79 80 81
\begin{center}
  \includegraphics{ch4.8}
\end{center}
Jim Hefferon's avatar
Jim Hefferon committed
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
87
$S$ to $I$.
Jim Hefferon's avatar
Jim Hefferon committed
88 89 90
We next look at three models of central projection, 
of increasing abstractness but
also of increasing uniformity.
Jim Hefferon's avatar
Jim Hefferon committed
91
The last model will bring out the linear algebra.
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's avatar
Jim Hefferon committed
96
Model this with parallel lines in a domain plane~$S$
97
and a projection via a $P$ to a codomain plane~$I$. 
Jim Hefferon's avatar
Jim Hefferon committed
98
(The gray lines shown are parallel to the $S$ plane and to the~$I$ plane.)
99 100 101
\begin{center}
  \includegraphics{ch4.9}
\end{center}
Jim Hefferon's avatar
Jim Hefferon committed
102
This single setting shows all three projection cases.
Jim Hefferon's avatar
Jim Hefferon committed
103
The first picture below shows $P$ acting as a movie projector by pushing
104
points from part of $S$ out to image points on the lower half of $I$.
Jim Hefferon's avatar
Jim Hefferon committed
105
The middle picture shows $P$ acting as the artist by 
106 107
pulling points from another part of $S$ back to  
image points in the middle of $I$.
Jim Hefferon's avatar
Jim Hefferon committed
108
In the third picture $P$ acts as the pinhole, projecting points from $S$
109
to the upper part of $I$.
Jim Hefferon's avatar
Jim Hefferon committed
110
This third picture is the trickiest\Dash the points that are
111
projected near to the vanishing point are the ones that are 
Jim Hefferon's avatar
Jim Hefferon committed
112
far out on the lower left of $S$. 
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's avatar
Jim Hefferon committed
122

Jim Hefferon's avatar
Jim Hefferon committed
123 124
There are two awkward things here.
First, neither of the two points in the domain
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's avatar
Jim Hefferon committed
128 129
(we say that these two are projected to infinity).
The second is that 
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's avatar
Jim Hefferon committed
134 135
(we say that the vanishing point is the image of a projection 
from infinity).
136 137 138 139
\begin{center}
  \includegraphics{ch4.13}
\end{center}

Jim Hefferon's avatar
Jim Hefferon committed
140
For a model that eliminates this awkwardness, 
Jim Hefferon's avatar
Jim Hefferon committed
141
cover the projector $P$ with a hemispheric dome.
Jim Hefferon's avatar
Jim Hefferon committed
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's avatar
Jim Hefferon committed
145
as with the movie projector.
Jim Hefferon's avatar
Jim Hefferon committed
146
It includes projecting things on the line further from $P$ than the dome,
Jim Hefferon's avatar
Jim Hefferon committed
147 148
as with the painter.
More subtly, it also includes things on the line that lie behind $P$,
Jim Hefferon's avatar
Jim Hefferon committed
149
as with the pinhole case.  
150 151 152
\begin{center} 
  \includegraphics{ch4.14}
\end{center}
Jim Hefferon's avatar
Jim Hefferon committed
153
More formally,
Jim Hefferon's avatar
Jim Hefferon committed
154
for any nonzero vector $\vec{v}\in\Re^3$, let the associated 
155
\definend{point $v$ in the projective plane}\index{point!in projective plane} 
Jim Hefferon's avatar
Jim Hefferon committed
156
be the set $\set{k\vec{v}\suchthat \text{$k\in\Re$ and $k\neq 0$}}$
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's avatar
Jim Hefferon committed
162
can be represented in any of these three ways.
163
\begin{equation*}
Jim Hefferon's avatar
Jim Hefferon committed
164
  \colvec[r]{1 \\ 2 \\ 3}
165 166 167
  \qquad
  \colvec{1/3 \\ 2/3 \\ 1}
  \qquad
Jim Hefferon's avatar
Jim Hefferon committed
168
  \colvec[r]{-2 \\ -4 \\ -6}
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's avatar
Jim Hefferon committed
173
for the point~$\ell$. 
174

Jim Hefferon's avatar
Jim Hefferon committed
175
This picture and definition
Jim Hefferon's avatar
Jim Hefferon committed
176 177
clarifies central projection
but there is still something ungainly about the dome 
Jim Hefferon's avatar
Jim Hefferon committed
178
model:~what happens when $P$ looks down?
Jim Hefferon's avatar
Jim Hefferon committed
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's avatar
Jim Hefferon committed
182
below.
Jim Hefferon's avatar
Jim Hefferon committed
183
Now the part of the line~$\ell$ that intersects the dome lies behind the page.  
Jim Hefferon's avatar
Jim Hefferon committed
184 185

That is, as
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's avatar
Jim Hefferon committed
188
(This brings out that the dome does not include the entire equator
189
or else
Jim Hefferon's avatar
Jim Hefferon committed
190
when the viewer is looking exactly along the equator then there would be 
Jim Hefferon's avatar
Jim Hefferon committed
191
two points in the line that are both on the dome. 
Jim Hefferon's avatar
Jim Hefferon committed
192
Instead we define the dome so that it includes the
Jim Hefferon's avatar
Jim Hefferon committed
193
points on the equator with a positive~$y$ coordinate, as well as the point
Jim Hefferon's avatar
Jim Hefferon committed
194
where $y=0$ and $x$ is positive.)
Jim Hefferon's avatar
Jim Hefferon committed
195
This discontinuity means that
196
we often have to treat equatorial points as a separate case.
Jim Hefferon's avatar
Jim Hefferon committed
197 198
So while the railroad track model of central projection
has three cases, the dome has two.
199

Jim Hefferon's avatar
Jim Hefferon committed
200
We can do better, we can reduce to a model having a single case.
201
Consider a sphere centered at the origin.
Jim Hefferon's avatar
Jim Hefferon committed
202 203
Any line through the origin intersects the sphere in two spots, said to be
antipodal.\index{antipodal points}
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's avatar
Jim Hefferon committed
207
Below, we show the two antipodal spots connected by a dashed line
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's avatar
Jim Hefferon committed
214
spots on the sphere is not as intuitive as the one-spot-per-point dome mode,
215
on the other hand
Jim Hefferon's avatar
Jim Hefferon committed
216
the awkwardness of the dome model is gone in that 
217
as a line of view slides from north to south, 
Jim Hefferon's avatar
Jim Hefferon committed
218
no sudden changes happen.
Jim Hefferon's avatar
Jim Hefferon committed
219
This central projection model is uniform.
220 221 222

So far we have described points in projective geometry.
What about lines?
Jim Hefferon's avatar
Jim Hefferon committed
223
What a viewer~$P$ at the origin sees as a line is shown below as 
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's avatar
Jim Hefferon committed
229 230
(We've included one of the projective points on this line 
to bring out a subtlety. 
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's avatar
Jim Hefferon committed
236
we can also describe a projective line with a triple of reals.
237 238 239
For instance, the members of this plane through the origin
in $\Re^3$
\begin{equation*}
Jim Hefferon's avatar
Jim Hefferon committed
240
  %\set{y\colvec[r]{1 \\ -1 \\ 0}+z\colvec[r]{1 \\ 0 \\ 1}\suchthat y,z\in\Re}=
241 242
  \set{\colvec{x \\ y \\ z}\suchthat x+y-z=0}
\end{equation*} 
Jim Hefferon's avatar
Jim Hefferon committed
243
project to a line that we can describe with
244
$\rowvec{1 &1 &-1}$
Jim Hefferon's avatar
Jim Hefferon committed
245
(using a row vector for this typographically distinguishes lines from points).
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's avatar
Jim Hefferon committed
250
to be the set $L=\set{k\vec{L}\suchthat \text{$k\in\Re$ and $k\neq 0$}}$.
251

Jim Hefferon's avatar
Jim Hefferon committed
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 
254
\definend{incident} \Dash  the
Jim Hefferon's avatar
Jim Hefferon committed
255
point lies on the line, the line passes through the point \Dash  if and only
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's avatar
Jim Hefferon committed
266
of the form $k\cdot x+k\cdot y-k\cdot z=0$ for any nonzero~$k$.
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's avatar
Jim Hefferon committed
270
With this, we can do analytic projective geometry.
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's avatar
Jim Hefferon committed
274
meaning that for any projective point~$v$ incident with the line, any of $v$'s 
Jim Hefferon's avatar
Jim Hefferon committed
275
representative homogeneous coordinate vectors will 
Jim Hefferon's avatar
Jim Hefferon committed
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's avatar
Jim Hefferon committed
279
in projective geometry besides talking about the equation of a line, 
Jim Hefferon's avatar
Jim Hefferon committed
280 281
we also talk about the equation of a point. 
For the fixed point
282
\begin{equation*}
Jim Hefferon's avatar
Jim Hefferon committed
283
  v=\colvec[r]{1 \\ 2 \\ 3}
284
\end{equation*}
Jim Hefferon's avatar
Jim Hefferon committed
285 286
the property that characterizes 
lines incident on this point is that the components 
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's avatar
Jim Hefferon committed
291
This symmetry of the statements about lines and points is
Jim Hefferon's avatar
Jim Hefferon committed
292 293
the 
\definend{Duality Principle}\index{Duality Principle, of projective geometry} 
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's avatar
Jim Hefferon committed
297
in the projective plane two distinct lines determine one and only one point. 
Jim Hefferon's avatar
Jim Hefferon committed
298 299
Here is a picture showing two projective 
lines that cross in antipodal spots and thus 
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's avatar
Jim Hefferon committed
306
Contrast this with Euclidean geometry, where two unequal lines may
Jim Hefferon's avatar
Jim Hefferon committed
307
have a unique intersection or may be parallel.
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's avatar
Jim Hefferon committed
312 313
relationship between the two spaces:~we can view the 
projective plane as an extension of the Euclidean plane.
Jim Hefferon's avatar
Jim Hefferon committed
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  
317
antipodal spots from the sphere.
Jim Hefferon's avatar
Jim Hefferon committed
318
Conversely, we can view some points
Jim Hefferon's avatar
Jim Hefferon committed
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.)
322 323 324 325 326
\begin{center}
 \hfill
  \begin{tabular}{@{}c@{}}\includegraphics{ch4.18}\end{tabular}
 \hfill\llap{($**$)}
\end{center}
Jim Hefferon's avatar
Jim Hefferon committed
327
Thus we can think of projective space as consisting of the Euclidean plane 
328 329
with some extra points adjoined \Dash  
the Euclidean plane is embedded in the projective plane.
Jim Hefferon's avatar
Jim Hefferon committed
330
The extra points in projective space, the equatorial points,
Jim Hefferon's avatar
Jim Hefferon committed
331
are called \definend{ideal points}\index{ideal!point}%
332 333
\index{projective plane!ideal point} 
or \definend{points at infinity}\index{point!at infinity}
Jim Hefferon's avatar
Jim Hefferon committed
334
and the equator is called the 
Jim Hefferon's avatar
Jim Hefferon committed
335
\definend{ideal line}\index{ideal!line}%
336 337
\index{projective plane!ideal line} or 
\definend{line at infinity}\index{line at infinity}
Jim Hefferon's avatar
Jim Hefferon committed
338
(it is not a Euclidean line, it is a projective line). 
339

Jim Hefferon's avatar
Jim Hefferon committed
340 341 342
The advantage of this extension from the Euclidean plane 
to the projective plane
is that some of the nonuniformity 
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's avatar
Jim Hefferon committed
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). 
352

Jim Hefferon's avatar
Jim Hefferon committed
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.
357 358 359
Analytic projective geometry uses linear algebra.
For instance, for three points of the projective plane
$t$, $u$, and~$v$, 
Jim Hefferon's avatar
Jim Hefferon committed
360
setting up the equations for those points by fixing vectors representing each
Jim Hefferon's avatar
Jim Hefferon committed
361
shows that the three are collinear if and only if the resulting three-equation
362
system has infinitely many row vector solutions
Jim Hefferon's avatar
Jim Hefferon committed
363
representing their line.
Jim Hefferon's avatar
Jim Hefferon committed
364
That in turn holds if and only if this determinant is zero.
365
\begin{equation*}
Jim Hefferon's avatar
Jim Hefferon committed
366
  \begin{vmat}
367 368 369
    t_1  &u_1  &v_1  \\
    t_2  &u_2  &v_2  \\
    t_3  &u_3  &v_3  
Jim Hefferon's avatar
Jim Hefferon committed
370
  \end{vmat}
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's avatar
Jim Hefferon committed
374
Similarly, by duality, 
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.

379
The following result is more evidence of the niceness
Jim Hefferon's avatar
Jim Hefferon committed
380
of the geometry of the projective plane.
Jim Hefferon's avatar
Jim Hefferon committed
381
These two triangles are
Jim Hefferon's avatar
Jim Hefferon committed
382
\definend{in perspective}\index{perspective, triangles} from the point $O$ 
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's avatar
Jim Hefferon committed
391
\definend{Desargue's Theorem}\index{Desargue's Theorem}
Jim Hefferon's avatar
Jim Hefferon committed
392
is that when we extend the three pairs of corresponding 
Jim Hefferon's avatar
Jim Hefferon committed
393 394 395
sides, they intersect
(shown here as the points $TU$, $TV$, and $UV$).
What's more, those three intersection points are collinear.
396 397 398
\begin{center}
  \includegraphics{ch4.23}
\end{center}
Jim Hefferon's avatar
Jim Hefferon committed
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
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's avatar
Jim Hefferon committed
405
figures that the sides appear in our sketch to be straight.)
406

Jim Hefferon's avatar
Jim Hefferon committed
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,
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's avatar
Jim Hefferon committed
415
  \rep{\vec{w}}{B}=\colvec[r]{1 \\ 0 \\ 0}
416
  \quad
Jim Hefferon's avatar
Jim Hefferon committed
417
  \rep{\vec{x}}{B}=\colvec[r]{0 \\ 1 \\ 0}
418
  \quad
Jim Hefferon's avatar
Jim Hefferon committed
419
  \rep{\vec{y}}{B}=\colvec[r]{0 \\ 0 \\ 1}
420
  \quad
Jim Hefferon's avatar
Jim Hefferon committed
421
  \rep{\vec{z}}{B}=\colvec[r]{1 \\ 1 \\ 1}
422
\end{equation*}
Jim Hefferon's avatar
Jim Hefferon committed
423
To prove the lemma, 
Jim Hefferon's avatar
Jim Hefferon committed
424
because $W$, $X$, and $Y$ are not on the same projective line, any
425
homogeneous coordinate vectors 
Jim Hefferon's avatar
Jim Hefferon committed
426
$\vec{w}_0$, $\vec{x}_0$, and $\vec{y}_0$ do not line on the same
427 428
plane through the origin in $\Re^3$ and so form a 
spanning set for $\Re^3$.
Jim Hefferon's avatar
Jim Hefferon committed
429
Thus any homogeneous coordinate vector for $Z$ is a combination 
430
$\vec{z}_0=a\cdot\vec{w}_0+b\cdot\vec{x}_0+c\cdot\vec{y}_0$.
Jim Hefferon's avatar
Jim Hefferon committed
431
Then let the basis be $B=\sequence{\vec{w},\vec{x},\vec{y}}$ and take 
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's avatar
Jim Hefferon committed
435
and $\vec{z}=\vec{z}_0$.  
436

Jim Hefferon's avatar
Jim Hefferon committed
437
To prove Desargue's Theorem use the lemma to fix homogeneous
438 439
coordinate vectors and a basis.
\begin{equation*}
Jim Hefferon's avatar
Jim Hefferon committed
440
  \rep{\vec{t}_1}{B}=\colvec[r]{1 \\ 0 \\ 0}
441
  \quad
Jim Hefferon's avatar
Jim Hefferon committed
442
  \rep{\vec{u}_1}{B}=\colvec[r]{0 \\ 1 \\ 0}
443
  \quad
Jim Hefferon's avatar
Jim Hefferon committed
444
  \rep{\vec{v}_1}{B}=\colvec[r]{0 \\ 0 \\ 1}
445
  \quad
Jim Hefferon's avatar
Jim Hefferon committed
446
  \rep{\vec{o}}{B}=\colvec[r]{1 \\ 1 \\ 1}    
447
\end{equation*}
Jim Hefferon's avatar
Jim Hefferon committed
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
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's avatar
Jim Hefferon committed
453 454
  \rep{\vec{t}_2}{B}=a\colvec[r]{1 \\ 1 \\ 1}
                          +b\colvec[r]{1 \\ 0 \\ 0}
455 456
\end{equation*}
for some scalars $a$ and $b$.
Jim Hefferon's avatar
Jim Hefferon committed
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. 
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's avatar
Jim Hefferon committed
467
The projective line $T_1U_1$ is the projection of a plane through the 
468
origin in $\Re^3$.
Jim Hefferon's avatar
Jim Hefferon committed
469
One way to get its equation is to note that any vector in it
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's avatar
Jim Hefferon committed
473
  \begin{vmat}
474 475 476
    1  &0  &x  \\
    0  &1  &y  \\
    0  &0  &z
Jim Hefferon's avatar
Jim Hefferon committed
477
  \end{vmat}=0
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's avatar
Jim Hefferon committed
486
  \begin{vmat}
487 488 489
    t_2  &1    &x  \\
    1    &u_2  &y  \\
    1    &1    &z
Jim Hefferon's avatar
Jim Hefferon committed
490
  \end{vmat}=0
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's avatar
Jim Hefferon committed
501
(This is, of course, a homogeneous coordinate vector of a projective point.)
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's avatar
Jim Hefferon committed
510
Finish the proof by noting that
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's avatar
Jim Hefferon committed
515
the Euclidean result may be messier to state and prove.
516
Desargue's theorem illustrates this.
Jim Hefferon's avatar
Jim Hefferon committed
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
519 520
$T_1T_2$, $U_1U_2$, and $V_1V_2$ are parallel.

Jim Hefferon's avatar
Jim Hefferon committed
521
The remark
522 523
following the statement of Desargue's Theorem suggests thinking
of the Euclidean pictures as figures from projective geometry 
Jim Hefferon's avatar
Jim Hefferon committed
524
for a sphere model with very large radius.
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.
528

Jim Hefferon's avatar
Jim Hefferon committed
529
We finish by pointing out one more thing about the projective plane. 
530
Although its local properties are familiar, the projective plane has
Jim Hefferon's avatar
Jim Hefferon committed
531
a perhaps unfamiliar global property.
532
The picture below shows a projective point.
533
At that point we have drawn Cartesian axes, $xy$-axes.
Jim Hefferon's avatar
Jim Hefferon committed
534
Of course, the axes appear in the picture at both antipodal spots, one in the
Jim Hefferon's avatar
Jim Hefferon committed
535
northern hemisphere (that is, shown on the right) 
536 537
and the other in the
south.
Jim Hefferon's avatar
Jim Hefferon committed
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 
542
along the $x$-axis in the 
Jim Hefferon's avatar
Jim Hefferon committed
543
positive direction.
Jim Hefferon's avatar
Jim Hefferon committed
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.
554 555 556
\begin{center}
  \includegraphics{ch4.24}
\end{center}
Jim Hefferon's avatar
Jim Hefferon committed
557 558
The sequence of pictures below 
show a trip around this space: 
Jim Hefferon's avatar
Jim Hefferon committed
559
the antipodal spots rotate around the sphere with the spot in the 
Jim Hefferon's avatar
Jim Hefferon committed
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 
565
projective point.
Jim Hefferon's avatar
Jim Hefferon committed
566
So by the third sphere below the trip has pretty much returned 
Jim Hefferon's avatar
Jim Hefferon committed
567
to the same projective point where we drew it starting above.) 
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's avatar
Jim Hefferon committed
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's avatar
Jim Hefferon committed
580
The projective plane is not orientable\Dash in this geometry,
Jim Hefferon's avatar
Jim Hefferon committed
581
left and right handedness are not fixed properties of figures
Jim Hefferon's avatar
Jim Hefferon committed
582 583
(said another way, 
we cannot describe a spiral as clockwise or
584 585 586
counterclockwise).

This exhibition of the existence of a 
Jim Hefferon's avatar
Jim Hefferon committed
587 588
non-orientable space raises the question of whether our universe
orientable.
Jim Hefferon's avatar
Jim Hefferon committed
589
Could 
590
an astronaut leave earth right-handed and return left-handed?
Jim Hefferon's avatar
Jim Hefferon committed
591
\cite{Gardner} is a nontechnical reference.
592 593
\cite{Clarke} is a classic science fiction story about 
orientation reversal.
594

Jim Hefferon's avatar
Jim Hefferon committed
595
For an overview of projective geometry see \cite{CourantRobbins}. 
596
The approach we've taken here, the analytic approach,
Jim Hefferon's avatar
Jim Hefferon committed
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}.
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's avatar
Jim Hefferon committed
606
An easy and interesting application is in \cite{Davies}.
607 608 609 610 611

\begin{exercises}
  \item 
    What is the equation of this point?
    \begin{equation*}
Jim Hefferon's avatar
Jim Hefferon committed
612
       \colvec[r]{1 \\ 0 \\ 0}
613 614 615 616
    \end{equation*}
    \begin{answer}
      From the dot product
      \begin{equation*}
Jim Hefferon's avatar
Jim Hefferon committed
617
        0=\colvec[r]{1 \\ 0 \\ 0}\dotprod\rowvec{L_1 &L_2 &L_3}
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's avatar
Jim Hefferon committed
627
           \colvec[r]{1 \\ 2 \\ 3},\,\colvec[r]{4 \\ 5 \\ 6}
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's avatar
Jim Hefferon committed
639
            0=\begin{vmat}
640 641 642
              1  &4  &x \\
              2  &5  &y \\
              3  &6  &z
Jim Hefferon's avatar
Jim Hefferon committed
643
            \end{vmat}
644 645 646
            =-3x+6y-3z
          \end{equation*}
          shows that the line is $L=\rowvec{-3 &6 &-3}$.
Jim Hefferon's avatar
Jim Hefferon committed
647
        \partsitem $\colvec[r]{-3 \\ 6 \\ -3}$
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's avatar
Jim Hefferon committed
660
      comes from this determinant equation.
661
      \begin{equation*}
Jim Hefferon's avatar
Jim Hefferon committed
662
        0=\begin{vmat}
663 664 665
          u_1  &v_1  &x  \\
          u_2  &v_2  &y  \\
          u_3  &v_3  &z
Jim Hefferon's avatar
Jim Hefferon committed
666
        \end{vmat}
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's avatar
Jim Hefferon committed
703
    %such that it would project to a circle)? 
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's avatar
Jim Hefferon committed
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.)
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's avatar
Jim Hefferon committed
792
            \rep{\vec{t}_0}{B}=\colvec[r]{1 \\ 0 \\ 0}
793
            \quad
Jim Hefferon's avatar
Jim Hefferon committed
794
            \rep{\vec{t}_1}{B}=\colvec[r]{0 \\ 1 \\ 0}
795
            \quad
Jim Hefferon's avatar
Jim Hefferon committed
796
            \rep{\vec{t}_2}{B}=\colvec[r]{0 \\ 0 \\ 1}
797
            \quad
Jim Hefferon's avatar
Jim Hefferon committed
798
            \rep{\vec{v}_0}{B}=\colvec[r]{1 \\ 1 \\ 1}
799 800 801 802
          \end{equation*}
        \partsitem
          First, any $U_0$ on $T_0V_0$
          \begin{equation*}
Jim Hefferon's avatar
Jim Hefferon committed
803 804
            \rep{\vec{u}_0}{B}=a\colvec[r]{1 \\ 0 \\ 0}
                               +b\colvec[r]{1 \\ 1 \\ 1}
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's avatar
Jim Hefferon committed
816 817
            \rep{\vec{u}_2}{B}=c\colvec[r]{0 \\ 1 \\ 0}
                                +d\colvec[r]{1 \\ 1 \\ 1}
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's avatar
Jim Hefferon committed
826 827
            \rep{\vec{u}_1}{B}=e\colvec[r]{0 \\ 0 \\ 1}
                                +f\colvec[r]{1 \\ 1 \\ 1}
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's avatar
Jim Hefferon committed
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}
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's avatar
Jim Hefferon committed
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}
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's avatar
Jim Hefferon committed
881
             p\colvec[r]{0 \\ 1 \\ 0}+q\colvec{1 \\ 1 \\ u_1}
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's avatar
Jim Hefferon committed
905
             r\colvec[r]{0 \\ 0 \\ 1}+s\colvec{1 \\ u_2 \\ 1}
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's avatar
Jim Hefferon committed
916
\index{projective geometry|)}
917
\endinput