Mathematics 138
     Using SmProj.exe and LgProj.exe
To Analyze Projective Transformations
The WIndows programs SmProj.exe and LgProj.exe are on the course Web site.   They are the same, except that SmProj.exe handles data from the small projective senate, based on Z3, while LgProj.exe handles data from the large projective senate, based on Z5.  Luke Gustafson's PHP
software on the course Web site does many of the same calculations and handles projective senates based on larger finie fields.
For reference, here are the committees and coordinates for the small senate.
Amy	(0,0,1)
Bob      (1,0,1)
Cal	(2,0,1)
Dan	(0,1,1)
Eva	(1,1,1)	
Fay	(2,1,1)
Gus	(0,2,1)
Hal	(1,2,1)
Ian	(2,2,1)
Jim	(1,0,0)  (instructor at infinity with slope 0)
Ken	(1,1,0)  (instructor at infinity with slope 1)
Lou	(1,2,0)  (instructor at infinity with slope 2)
May	(0,1,0)  (instructor at infinity with slope infinity)

Committees:
athletics:  Amy, Bob, Cal, Jim        	(y = 0)			(0,1,0)
budget: Dan, Eva, Fay, Jim		(2y + z = 0)		(0,2,1)
compensation: Gus, Hal, Ian, Jim	(y + z = 0)		(0,1,1)
diversity: Amy, Dan, Gus, May		(x = 0)			(1,0,0)
education: Bob, Eva, Hal, May		(2x + z = 0)		(2,0,1)
football: Cal, Fay, Ian, May		(x + z = 0)		(1,0,1)
gifts    : Amy, Eva, Ian, Ken	           	(x + 2y = 0)		(1,2,0)
housing: Bob, Fay, Gus, Ken		(2x + y + z = 0)	  	(2,1,1)
insurance: Cal, Dan, Hal, Ken		(x + 2y + z = 0)		(1,2,1)
junior faculty: Amy, Fay, Hal, Lou            	(x + y = 0)		(1,1,0)
kitchen: Bob, Dan, Ian, Lou		(2x + 2y + z = 0)	(2,2,1)
library: Cal, Eva, Gus, Lou		(x + y + z = 0)		(1,1,1)
mystery: Jim, Ken, Lou, May (the committee at infinity) (z = 0) 	(0,0,1)
On the next page is the large projective senate.  

Alice	(0,0,1)
Betty	(1,0,1)
Chris	(2,0,1)
Danny	(3,0,1)
Emily	(4,0,1)
Frank	(0,1,1)
Gavin	(1,1,1)
Helen	(2,1,1)
Isaac	(3,1,1)
James	(4,1,1)
Kevin	(0,2,1)
Larry	(1,2,1)
Maria	(2,2,1)
Nancy	(3,2,1)
Oscar	(4,2,1)
Patty	(0,3,1)
Quint	(1,3,1)
Ralph	(2,3,1)
Sally	(3,3,1)
Terry	(4,3,1)
Uriah	(0,4,1)
Viola	(1,4,1)
Wally	(2,4,1)
Xenia	(3,4,1)
Yoric	(4,4,1)
Zelda	(1,0,0)   (member at infinity with slope 0)
1nita	(1,1,0)   (member at infinity with slope 1)
2pelo	(1,2,0)   (member at infinity with slope 2)
3some	(1,3,0)   (member at infinity with slope 3)
4tune	(1,4,0)   (member at infinity with slope 4)
5larm	(0,1,0)   (member at infinity with slope infinity)
Committees:
athletics:  Alice, Betty, Chris, Danny, Emily, Zelda		 (0,1,0)
budget: Frank, Gavin, Helen, Isaac, James, Zelda 		 (0,4,1)
compensation: Kevin, Larry, Maria, Nancy, Oscar, Zelda	(0,2,1)
diversity: Patty, Quint, Ralph, Sally, Terry, Zelda		(0,3,1)
education: Uriah, Viola, Wally, Xenia, Yoric, Zelda		(0,1,1)
football: Alice, Frank, Kevin, Patty, Uriah, 5larm		(1,0,0)
gifts: Betty, Gavin, Larry, Quint, Viola, 5larm  		(4,0,1)
housing: Chris, Helen, Maria, Ralph, Wally, 5larm		(2,0,1)
insurance: Danny, Isaac, Nancy, Sally, Xenia, 5larm               	(3,0,1)
junior: Emily, James, Oscar, Terry, Yoric, 5larm		(1,0,1)
kitchen: Alice, James, Nancy, Ralph, Viola, 4tune		(1,1,0)
library: Betty, Frank, Oscar, Sally, Wally, 4tune		(4,4,1)
medical: Chris, Gavin, Kevin, Terry, Xenia, 4tune		(2,2,1)
night: Danny, Helen, Larry, Patty, Yoric, 4tune		(3,3,1)
overseers: Emily, Isaac, Maria, Quint, Uriah, 4tune		(1,1,1)
planning: Alice, Helen, Oscar, Quint, Xenia, 3some                	(1,3,0)
quality: Betty, Isaac, Kevin, Ralph, Yoric, 3some		(4,2,1)
recruiting: Chris, James, Larry, Sally, Uriah, 3some		(2,1,1)
steering: Danny, Frank, Maria, Terry, Viola, 3some		(3,4,1)
tenure: Emily, Gavin, Nancy, Patty, Wally, 3some		(1,3,1)
union: Alice, Isaac, Larry, Terry, Wally, 2pelo		(1,2,0)
vacation: Betty, James, Maria, Patty, Xenia, 2pelo		(4,3,1)
workplace: Chris, Frank, Nancy, Quint, Yoric, 2pelo		(2,4,1)
xenophobia: Danny, Gavin, Oscar, Ralph, Uriah, 2pelo	(3,1,1)
yearbook: Emily, Helen, Kevin, Sally, Viola, 2pelo		(1,2,1)
zoology: Alice, Gavin, Maria, Sally, Yoric. 1nita		(1,4,0)
1derful: Betty, Helen, Nancy, Terry, Uriah. 1nita		(4,1,1)
2torial: Chris, Isaac, Oscar, Patty, Viola. 1nita		(2,3,1)
3quarter: Danny, James, Kevin, Quint, Wally. 1nita		(3,2,1)
4casting: Emily, Frank, Larry, Ralph, Xenia. 1nita		(1,4,1)
5roptics: Zelda, 1nita, 2pelo,3some, 4tune, 5larm		(0,0,1)

You can calculate the committee on which two instructors serve.  This is done with cross products:

The arithmetic is done modulo 5 for the large senate, so the answer is converted to
and then tosince all multiples of a vector represent the same committee.
Similarly, 
The answer is converted toand then to

You can also easily determine all the instructors on a committee.  The overseers committee has equation x + y + z (1,1,1) so all the instructors on it have coordinates that sum to 0 modulo 5.
Dually, Gavin has coordinates (1,1,1) and so he serves on all committees for which the coordinates (a,b,c) (meaning ax + by +cz = 0) satisfy a + b + c.
Another way of looking at this (and the way it is implemented) is that instructor v serves on committee x if the inner product <x, v> = 0.

You can create any possible projective transformation of the senate by building a nonsingular 3x3 matrix.  There are always four different matrices (conveniently, one each with determinant 1, 2, 3, and 4) that represent the same projective transformation.
The three combo boxes under "Transformation Matrix" let you select the three columns of the matrix.  This makes it easy to specify the image of (1,0,0), (0,1,0), and (0,0,1).  Because there are four possible columns that give the same image for one of thse vectors, you can also specify the image of (1,1,1) in accordance with the Fundamental Theorem.

When you build a matrix, the contents of the drop-down lists are automatically adjusted so that you cannot create a singular matrix.  For the first column, this eliminates just (0,0,0), leaving 124 choices.  For the second, all multiples of the first column are eliminated, leaving 125-5 = 120 choices.  For the third column, all 25 linear combinations of the first two columns are eliminated, leaving 125-25 = 100 choices.   There are four matrices for each transformation, but this means that there are 124 * 120 * 100 / 4 = 372000 possible transfomations to study.

When you click "Show Effects," the matrix is applied to each instructor, its transpose is applied to each committee, and the results are displayed.  If an eigenvector is found, it is followed by a string like *2*, where the number between the stars is the eigenvalue. Since a matrix A and its transpose have the same characteristic equation, you will see the same eigenvectors for instructors and committees.  The example above has no eigenvectors, something that could not happen in real projective geometry since any cubic over the reals has at least one real root.

Now it is easy to investigate all the possibilities where the matrix A has eigenvalues.
Case 1.  The characteristic equation has only one root.  We can always multiply the matrix by a constant, not changing the projective transformation that it represents, so that the eigenvalue is 1.
But there may be 1, 2, or 3 eigenvectors.  If there is only one eigenvector, the root may be either a single root or a triple root.
Case 1a. Triple root, but 1 eigenvector only.  There is one fixed committee and one fixed instructor who serves on that committee.  The easiest way to obtain this case is to put 1s on the main diagonal.  Then, either immediately above or below the diagonal, put any two nonzero entries.  You can also put a nonzero entry in the corner next to these.  This will make the matrix look like .
The fixed instructor is clearly (1,0,0) - Zelda and the fixed committee is (0,0,1) - 5roptics, on which Zelda serves.  Make the 5roptics committee be the committee at infinity and remove it. Now you have a transformation of the affine plane that maps each original point of the affine plane into another point, but that has no fixed points and no fixed lines.  You can think of this as a Galilean
transformation - a shear followed by a translation, but the fixed line of the shear is horizontal rather than vertical, and the translation must have a nonzero vertical component.

Case 1b. There are two eigenvectors.  These span a two-dimebsional subspace of R3 in which every vector is fixed, so the projective transformation has one committee that is fixed pointwise. One member of this committee (5larm) is fixed "linewise."  This is an elation.  If the fixed committee is (0,0,1) then the resulting transformation of the affine plane is just a translation, but if a committee other than the fixed committee is removed, the resulting affine tranformation is more complicated.

Case 1c: There are three linearly independent eigenvectors, so the transformation is the identity.
Case 1d: The characteristic equation has only one root in the field. There must be one eigenvector, but in this case the fixed instructor does not serve on the fixed committee.  Removing the fixed committee leaves an affine plane where the transformation has one fixed point but no fixed lines (a rotation, for example)

Case 2. There are two distinct eigenvalues.  We can multiply A by a constant to make either of them 1.  There are two cases, depending on whether of not the repeated eigenvalue has only a single associated eigenvector or a whole plane of eigenvectors.
Case 2a. The repeated eigenvalue has only a single eigenvector.  There are two fixed instructors and two fixed committees.  Here is a simple example, shown with the small senate so that you can see both of the fixed instructors.

Jim serves on the mystery committee, and Amy serves on the athletics committee.

Case 2b: The repeated eigenvalue has a two-dimensional subspace of eigenvectors.  In projective terms this is a committee that is fixed pointwise. There is one other fixed instructor not on the fixed committee, who is fixed "linewise."  If the square of both eigenvalues is the same, then the square of the matrix is the identity, and the transformation is a harmonic homology.

If the eigenvalues do not have the same square (this could not happen with the small senate) then the transformation is a homology but not a harmonic homology.  The instructors and committees that are not fixed are not simply interchanged in pairs.

Case 3: The matrix A has three distinct nonzero eigenvalues (again, this cannot happen with the small senate).  Now there are three eigenvectors.  Each pair spans a plane that is fixed, but not pointwise.  So there are three fixed committees also.

Now try choosing any of the 372000 transformations.  Apart from the specific fixed instructors and fixed committees involved, each should behave like one of the cases that has been described.  By removing one of the fixed committees, it is always possible to leave an affine plane in which the effect of the transformation is easy to describe.  However, if you remove a committee that is not fixed, the resulting transformation of the remaining instructors is not an affine collineation.  Instructors get "sent to infinity."  Committees that used to intersect now become parallel if their former point of intersection got "moved to infinity."  This is the sort of projective transformation that is of interest to artists, where parallel lines in one plane (railroad tracks on the ground) become converted to intersecting lines in another plane (the picture).

Polarities 
A polarity of a projective plane is a bijection between the points(instructors) and lines(committees) of the plane.  The simplest polarity to describe is one that associates the instructor with coordinates (a, b, c) and the committee whose equation is ax + by + cz = 0.  There is nothing really special about this; it depends entirely on how we assigned coordinates.
Clicking the "Analyze Polarity" button will display this polarity.  If the instructor serves on the associated committee, the instructor and committee are both called "lightlike" or "self-conjugate" and are marked with asterisks.  For the small senate there are 4 lightlike instructors; for the large senate there are 6. 
When an instructor and committee are associated by a polarity, the instructor is called the "pole" of the committee and the committee is called the "polar line" of the instructor.  For shorthand, we can call the instructor and committee the "dual" of each other.

Doing the same thing in R3 would have been less interesting, because if you plug (a,b,c) into the equation ax + by + cz = 0 you get a2 + b2 + c2 = 0, which has no nonzero solutions.  It is a nice feature of the finite senates that choosing the simplest imaginable polarity leads to something that is much more similar to hyperbolic geometry than to elliptic geometry.
Now take the lightlike committees and find all their points of intersection.  These are called "spacelike instructors."  There are 4*3/2 = 6 of them for the small senate and 6*5/2 = 15 of them for the large senate.
All instructors that are not lightlike or spacelike are called timelike.  There are 13-4-6 = 3 of them for the small senate and 31-6-15 = 10 of them for the large senate.

Dually, take the lightlike instructors and find all the committees joining them.  These are called "spacelike committees."  There are 4*3/2 = 6 of them for the small senate and 6*5/2 = 15 of them for the large senate.
All committees that are not lightlike or spacelike are called timelike.  There are 13-4-6 = 3 of them for the small senate and 31-6-15 = 10 of them for the large senate.

Here are the results for the small senate.

Notice that the timelike instructors are dual to the timelike committees and the spacelike instructors are dual to the spacelike committees.

Now think of the lightlike instructors as defining a set of points that we will call a "conic." Why? Because the equation for a lightlike instructor is can be written as
z2 = 2(x2 + y2) (mod 3, of course) for the small senate
z2 = 4(x2 + y2) (mod 5, of course) for the large senate
and this sort of equation describes a cone in R3.
When we drop a "line at infinity" and graph the projective plane in an affine plane, we see the intersection of the cone with the plane: a circle, ellipse, parabola, or hyperbola.

Every lightlike instructor is a point on this "conic" and every lightlike committee includes one point of the "conic".  We can think of a lightlike committee as "tangent to the conic," while a spacelike committee, with two instructors on the conic, is a "secant line."  A timelike committee does not intersect the conic at all.

Orthogonal transformations
As in R3, define an orthogonal matrix A by the property that AtA = I.
This means that if v and w are columns, then <v,v> = <w,w> = 1 and <v,w> = 0.

To create an orthogonal matrix, check the "Orthogonal only" box.  The combo boxes will adjust themselves so that you can only choose orthogonal unit vectors.

When you apply an orthogonal transformation to the senate, the lightlike instructors are mapped to lightlike instructors, spacelike to spacelike, and timelike to timelike. Dually, the lightlike committees are mapped to lightlike committees, spacelike to spacelike, and timelike to timelike.

This opens up a possibility for new geometries.  We can, for example, keep only the timelike instructors and only the spacelike committees, and any orthogonal transformation will transform them among themselves.

More general polarities:
If we choose any nonsingular symmetric matrix B, we can define a polarity as follows:
Given an instructor v, Bv gives the coordinates for its polar line
Given a committee x, B-1x gives the coordinates for its pole.
Viewed in terms of instructors and committees, such a polarity works just like the original one. It has the crucial property that if y is on the polar line, x , of x, then x is on the polar line h of y.
Proof: x = Bx, so <y, Bx> = 0.
Since B is symmetric, <y, Bx> = <By, x> so <h,x> = 0.
When choosing B, each combo box sets both a row and a column.  You must heed warnings about singular matrices.

For a general polarity, an orthogonal matrix A must satisfy AtBA = B
Then an instructor v is lightlike if he serves on his dual committee, i.e.
<v, Bv> = 0.
If orthogonal transformation A maps v to Av,
<Av, BAv> = <v, AtBAv> = <v, Bv> = 0.
so lightlike instructors are mapped to lightlike instructors and everything else falls into place.