Mathematics 152
Faculty Senate Affine Geometry

Undefined concepts:
An "instructor" (Euclid tried to define this as "that which has no part")
A "committee" (which, according to Euclid, "lies evenly with the members on
itself")
The relation of "serving:" an instructor may serve on a committee.

Axioms
A1. If A and B are distinct instructors, there is a unique committee on which
both serve.
A2. If A is an instructor who does not serve on committee c, there is a unique
committee d on which A serves that has no members in common with c.  (Committees c and d are said to be parallel.)
A3. There are at least two members on each committee; there are at least two
committees.
A4. (Desargues) Consider 6 distinct instructors, A, B, C, A', B', C', such that
a.  The committee on which A and B serve together is parallel to the committee
on which A' and B' serve together.
b.  The committee on which A and C serve together is parallel to the committee
on which A' and C' serve together.
c. The three committees determined by A and A', B and B', C and C' either are
all parallel or have precisely one instructor in common.
Then the committee on which B and C serve together is parallel to the committee on which B' and C' serve together.

A5. (Pappus) Suppose committees s and t have instructor O in common, with A, B,
and C also serving on s and D, E, F also serving on t.  If the committee on
which A and E both serve is parallel to the committee on which B and F both
serve, and the committee on which C and E both serve is parallel to the committee on which B and D both serve, then the committee on which A and D both
serve is parallel to the committee on which C and F both serve.

The smallest possible "tiny affine senate", with four instructors and six committees.
Instructors: Al, Bo, Cy, and Di
Committees:
athletics: Al, Bo
budget: Cy, Di
compensation: Al, Cy
diversity: Bo, Di
education: Al, Di
football: Bo, Cy

Theorems:
1. If a committee p has a member in common with one of two parallel committees q and r, it has a member in common with the other.

Proof: Assume the contrary, that p and q have A in common but p and r are parallel.
Then p and q would be distinct committees, both with A as a member and both parallel to r.  This contradicts axiom A2.
Therefore p and r are not parallel, so they intersect.

2. Given two committees r and s in an affine faculty senate, there is a one-to-one correspondence (a bijection, an invertible function) between the instructors serving on r and those serving on s.

Case 1: r and s are not parallel: they have member O in common.
Select member A of r and member A' of s, both different from O.
Let t = AA' be the committee on which both serve.
Choose any member B of r different from A and O
Let u be the committee parallel to t on which B serves.
Both s and t have A' as a member.
t is the unique committee through A' parallel to u, so s cannot be parallel to u.
So s and u have a member B' in common.
Define function f: r -->s so that B' = f(B) where B' is the common member of u and s
f is invertible: the inverse is B = g(B'), where B is constructed by choosing the committee through B' parallel to t and finding its common member with r.
So f establishes a one-to-one correspondence between r and s.
Case 2: r and s are parallel.
Choose instructors O and O' on r and s respectively and find the committee t on which O and O' both serve. Establish a bijection as in case 1 by constructing a committee parallel to t containing member B of r and finding what member it has in common with s, or vice versa.

Corollaries:
3. All committees have the same number n of instructors.
Proof:
Since f is invertible, r and s have the same number of members.  
4. Each instructor A serves on n+1 committees.
Proof: choose a committee r, with n members, on which A does not serve.
A serves with each member of r on a distinct committee.
A also serves on one committee that is parallel to r.
5. There are n+1 "pencils" of parallel committees, each containing n
committees, for a total of n(n+1) committees.
Proof: Start with the n+1 non-parallel committees on which A serves. Select one of these (call it s). Choose a committee t that intersects s, and construct one committee parallel or equal to s for each of the n members of t.
6. There are n2 instructors.
Proof: Select any instructor A.
This instructor serves on n+1 committees, each of which has n-1 members apart from A. Then there is A, for a total of (n+1)(n-1)+1 = n2 instructors.

The Small Affine Faculty Senate

Instructors:
Amy,Bob,Cal,Dan,Eva,Fay,Gus,Hal,Ian	

Committees:
athletics:  Amy, Bob, Cal
budget: Dan, Eva, Fay
compensation: Gus, Hal, Ian
diversity: Amy, Dan, Gus
education: Bob, Eva, Hal
football: Cal, Fay, Ian
gifts: Amy, Eva, Ian
housing: Bob, Fay, Gus
insurance: Cal, Dan, Hal
junior faculty: Amy, Fay, Hal
kitchen: Bob, Dan, Ian
library: Cal, Eva, Gus

The Medium Affine Faculty Senate

Instructors:
Adam,Beth,Carl,Dave,Emma,Fred,Greg,Herb,
Irma,Jane,Kate,Lynn,Mike,Neil,Owen,Phil

Committees:
athletics:  Adam, Beth, Carl, Dave
budget: Emma, Fred, Greg, Herb
compensation: Irma, Jane, Kate, Lynn
diversity: Mike, Neil, Owen, Phil
education: Adam, Emma, Irma, Mike
football: Beth, Fred, Jane, Neil
gifts:  Carl, Greg, Kate, Owen
housing: Dave, Herb, Lynn, Phil
insurance: Adam, Fred, Kate, Phil
junior faculty: Beth, Emma, Lynn, Owen
kitchen: Carl, Herb, Irma, Neil
library: Dave, Greg, Jane, Mike
medical: Adam, Herb, Jane, Owen
night school: Beth, Greg, Irma, Phil
overseers: Carl, Fred, Lynn, Mike
planning: Dave, Emma, Kate, Neil
quality: Adam, Greg, Lynn, Neil
recruiting: Beth, Herb, Kate, Mike
steering: Carl, Emma, Jane, Phil
tenure: Dave, Fred, Irma, Owen

The Large Affine Faculty Senate

Instructors:
Alice,Betty,Chris,Danny,Emily,Frank,Gavin,Helen,Isaac,
James,Kevin,Larry,Maria,Nancy,Oscar,Patty,Quint,
Ralph,Sally,Terry,Uriah,Viola,Wally,Xenia,Yoric

Committees:
athletics:  Alice, Betty, Chris, Danny, Emily
budget: Frank, Gavin, Helen, Isaac, James
compensation: Kevin, Larry, Maria, Nancy, Oscar
diversity: Patty, Quint, Ralph, Sally, Terry
education: Uriah, Viola, Wally, Xenia, Yoric
football: Alice, Frank, Kevin, Patty, Uriah
gifts: Betty, Gavin, Larry, Quint, Viola  
housing: Chris, Helen, Maria, Ralph, Wally
insurance: Danny, Isaac, Nancy, Sally, Xenia
junior faculty: Emily, James, Oscar, Terry, Yoric
kitchen: Alice, James, Nancy, Ralph, Viola
library: Betty, Frank, Oscar, Sally, Wally
medical: Chris, Gavin, Kevin, Terry, Xenia
night school: Danny, Helen, Larry, Patty, Yoric
overseers: Emily, Isaac, Maria, Quint, Uriah
planning: Alice, Helen, Oscar, Quint, Xenia
quality: Betty, Isaac, Kevin, Ralph, Yoric
recruiting: Chris, James, Larry, Sally, Uriah
steering: Danny, Frank, Maria, Terry, Viola
tenure: Emily, Gavin, Nancy, Patty, Wally
union: Alice, Isaac, Larry, Terry, Wally
vacation: Betty, James, Maria, Patty, Xenia
workplace: Chris, Frank, Nancy, Quint, Yoric
xenophobia: Danny, Gavin, Oscar, Ralph, Uriah
yearbook: Emily, Helen, Kevin, Sally, Viola
zoology: Alice, Gavin, Maria, Sally, Yoric
1derful: Betty, Helen, Nancy, Terry, Uriah
2torial: Chris, Isaac, Oscar, Patty, Viola
3quarter: Danny, James, Kevin, Quint, Wally
4casting: Emily, Frank, Larry, Ralph, Xenia

 

Definition of addition on the members of a committee.
Diagram on the Web: "adding instructors" up through step 4.
To add members A and C, proceed as follows:
Select a committee s, and select an instructor O as the "additive identity."
Select an arbitrary instructor B not serving on s. 
Let t be the committee parallel to s on which B serves. 
Let u be the committee on which O and B serve. 
Let v be the committee parallel or equal to u on which A serves.
Let D be the instructor who serves on both t and v. 
Let w be the committee on which B and C serve. 
Let x be the committee parallel or equal to w on which D serves.
Then A + C is defined to be the instructor who serves on both s and x.

Theorem 7: 
The definition of A + C is independent of the choice of B.
Consider an alternative auxiliary instructor B'

Easy case: B' serves on the same committee t as O and B
Diagram on the Web: "Adding instructors" steps 5 and 6
Construct parallels to committee s, t that includes B and t' that includes B'.
Since OB and OB' are the same committee, the same committee u on which A serves is parallel to both.  This intersects t at D and t' at D'.
Construct x parallel to BC through D, intersecting s at E
and x' joining D' to E
Now apply A4(Desargues) to the triangles BB'C and DD'E
BD(committee t), B'D'(committee t') and CE (committee s) are all parallel.
Also BB' || DD' and BC || DE
By A4, then B'C || D'E.
When constructing the sum A + C using B' , we need a committee parallel to B'C that includes D'. The committee x' is the unique one with this property, so the sum is E.

Harder case: B' does not serve on the same committee as O and B
This needs a preliminary step.
Diagram on the Web: "Adding instructors (hard case)":

Construct parallels to committee s, t that includes B and t' that includes B'.
Construct AD parallel to OB with instructor D on committee t
Construct AD' parallel to OB' with instructor D' on committee t'
Now apply A4 to triangles OBB' and ADD' to conclude that BB' and DD'
The remainder of the proof is the same as above.

Theorem 8: A + O = A for any A in r.
Diagram on the Web: "Additive identity"

Proof:
Call the committee on which O and A serve committee s.
Choose an instructor B, and construct committee t parallel to s that includes T
Construct AD parallel to OB
Now the committee that includes B and the second operand C+O is OB.
The parallel committee through D is AD, which intersects s in instructor A
A + O and A are the same instructor.

Theorem 9. For any instructor A on committee r, there is an instructor -A such that
 A + (-A)  = O.
Diagram on the Web : "Additive Inverse":

Proof:
As usual, select committee t through B parallel to s, u through A parallel to OB, and let D be the member that t and u have in common.  Now construct x through B parallel to OD.  This is not parallel to t (or s) so it has a member -A in common with s.
Add A + -A using B as the auxiliary member; the sum is O.
So -A is the additive inverse of A.
Theorem 10. Addition is associative:  (A+C) + E = A + (C+E)
Diagram on the Web "Addition is associative":

Proof (this one is really incomprehensible without a diagram)
The secret is to choose three members of committee t that will suffice to carry out all four of the additions that need to be done
Left-hand side
Add A + C using auxiliary members B (and D) on committee t.
Construct OB' parallel to BC, with B' a member of t.
Now add E to the sum A + C, choosing auxiliary member B' so that point D is again used. Call the result G.
Right-hand side:
Add C + E using B'; the second auxiliary member is B. Call the sum F.
Now B'E, BF, and DG are all parallel.
Finally, add A + (C+E) using B; the second auxiliary member is D. This requires constructing a committee parallel to BF that includes D.  But this must be DG.

It now follows from general principles about groups that the instructors on a committee form a group under addition.  This means that O is a unique two-sided identity and that -A is a unique two-sided inverse. 

Theorem 11. (the converse of axiom A4)
Consider 6 distinct instructors, A, B, C, A', B', C', such that
a.  The committee on which A and B serve together is parallel to the committee
on which A' and B' serve together.
b.  The committee on which A and C serve together is parallel to the committee
on which A' and C' serve together.
c.  The committee on which B and C serve together is parallel to the committee
on which B' and C' serve together.
A, B, and C do not serve on a single committee, nor do A', B' and C'
Then the three committees on which A and A', B and B', and C and C' serve
together either are all parallel or all have a single member in common.

Proof:
Case 1: Suppose that two of the committees (say AA' and BB') are parallel.
Diagram on the Web: "Desargues, parallel case"
By A2, there is a unique committee (call it s) that includes C and that is parallel to AA' and BB'.
Choose a committee that includes A' and is parallel to AC.  This intersects committee s. Call the member that they have in common C''.
Now we have three parallel committees, AA', BB', and CC'', and AB||A'B', AC||A'C''.
So by axiom A4, BC || B'C''.
Now there is a unique committee through A' parallel to AC, and there is a unique committee through B' parallel to BC. These committees intersect in C''. But by hypothesis they intersect also in C'.  Therefore C' = C'' and CC' is parallel to AA' and BB'.

Case 2: Suppose that two of the committees (say AA' and BB') intersect at a point O.
Diagram on the Web: "Desargues, concurrent case"
By A1, there is a unique committee (call it s) that includes C and that passes through O.
Choose a committee that includes A' and is parallel to AC.  This intersects committee s. Call the member that they have in common C''.
Now we have three concurrent committees, AA', BB', and CC'', and AB||A'B', AC||A'C''.
So by axiom A4, BC || B'C''.
Again, there is a unique committee through A' parallel to AC, and there is a unique committee through B' parallel to BC. These committees intersect in C''. But by hypothesis they intersect also in C'.  Therefore C' = C'' and CC' is concurrent with AA' and BB'.

Theorem 12. Addition is commutative: A+C = C+A
Proof: 
Diagram on the Web: Addition is commutative.
Add A+C using auxiliary members B and D; call the result E.
Next start to add C+A using auxiliary member B.  This leads to a second auxiliary member D' with CD' parallel to OB.
Now look at triangles ADE and D'CB
AD || D'C (both were constructed parallel to OB)
AE || D'B (the first is s; the second is t)
DE || CB (to add A+C, DE was constructed this way)
Therefore by the theorem just proved, AD', DC, and EB are either parallel or concurrent
Finally, look at triangles ADB and D'CE.
AD', DC, and BE are either parallel or concurrent (as was just shown)
AD || D'C  (as was previously stated)
DB || CE   (the first is t; the second is s)
So by Axiom A4, AB || D'E
To finish adding C+A, we need a committee parallel to AB with D' as a member.
D'E is (the only) such committee; so C + A = E.
Definition of multiplication on the members of a committee:
To multiply A and C, proceed as follows:
Select a committee s, a member O as the "additive identity" and a different member I as the "multiplicative identity."
Select an auxiliary instructor B not serving on s. 
Let v be the committee on which A serves that is parallel or equal to the
committee t on which I and B serve. 
Let D be the member of v who also belongs to the committee u on which O and B
serve. 
Let x be the committee on which D serves that is parallel or equal to the
committee w on which B and C serve.
Then AC is defined to be the instructor who serves on both s and x.

Theorem 13. The definition of AC is independent of the choice of B.
Call the alternative auxiliary instructor B'
As with addition, there are two cases.
Case 1: I, B, and B' are collinear.
Diagram on the Web: "Multiplication is commutative":

Proof:
Construct committees that intersect s at O, t including B and t' including B'
Construct ADD' parallel to IBB' with instructor D on committee t and instructor D' on committee t'.
Construct x parallel to BC through D, intersecting s at E.
Construct x' with D' and E as members
Now apply A4 to BB'C and DD'E
BD (committee t), B'D' (committee t') and CE (committee s) are concurrent at O
Also BB' || DD' and BC || DE
By A4, then B'C || D'E
When constructing the product AC using B', we need a committee parallel to B'C that includes D'. The unique committee with this property is x', so the product is E.

Case 2: B' is not collinear with I and B.
Diagram on the Web: "Multiplication is commutative: general case":

First construct DD' as above.
Then apply A4 to triangles IBB' and ADD' to conclude that BB' || DD'
The remainder of the proof is the same as above. 

Theorem 14. AI = A.
Consider the case C = I (to prove AI = A)
Diagram on the Web "Multiplicative identity":
Proof:
The committee on which C(=I) serves that is parallel to AD is IB
d B are the same, so x = BC and the product is C.
 
Theorem 15  AO = OA = O.
Case 1: C = O (to prove that AO = O)

The committee w has O as a member, and the parallel (equal in this case) committee that includes D is also w, which has only O in common with s.

Case 2: O = A (to prove that OC = O)
Proof:
The second auxiliary member is O, and the any committee that includes O of course has only instructor O is common with s.

Theorem 16: If A is different from O, there is an inverse member V such that AV = I.
Diagram on the Web: Multiplicative inverse:

Proof:
After choosing B and constructing D, find the committee w that is parallel to DI and has B as a member.  Let V be the member that w and s have in common.
Use B as auxiliary instructor and construct D.
The committee x that includes D and is parallel to BV is DI.
Theorem 17. Multiplication is associative: (AC)E = A(CE)
Diagram on the Web: "Multiplication is associative":
Proof:
As with the corresponding proof for addition, the secret is to choose three members of committee t that will suffice to carry out all four multiplications.
Left-hand side:
Mutiply AC using auxiliary members B (and D) on committee t.
Construct IB' parallel to BC, with B' a member of t.
Now multiply the product AC by E, choosing auxiliary member B' so that D is again used. Call the result G
Right-hand side:
Multiply CE using B'. The second auxiliary member is B.  Call the result F.
Now B'E, BF, and DG are all parallel
Finally, multiply AF using auxiliary member B. The second auxiliary member is D.
The final step requires constructing a committee parallel to BF that includes D. But 
this must be DG, so A(CE) also equals G.

Theorem 18. Multiplication is commutative
Diagram on the Web "Multiplication is commutative":

Proof: 
Multiply AC using auxiliary members B and D. Call the product R.
This means that AD || IB and DR || BC
Multiply CA using auxiliary member B
Start by constructing D' so that CD' || IB  
Now B, D, D' are concurrent, and so are A, C, R.
Also BC || DR and AD || CD'
By Axiom A5, then, AB || D'R
To complete the multiplication CA, construct a committee parallel to AB that includes D'.  The only such committee is D'R, and so the product is CA = R.

Theorem 19. Multiplication is distributive with respect to addition: (A+C)E = AE + CE.
This is proved on p. 67 of Bennett, Affine and Projective Geometry, though there is an error in the accompanying diagram. There is a diagram on the Web: "Distributive Law."  There are 15 points on the diagram!  It is an interesting proof, but you are not responsible for it.
Theorem 20. Any committee with n members in an affine faculty senate is isomorphic to
the finite field with n elements.
Proof: We have verified all the axioms for a field!

Theorem 21. For an affine faculty senate with three non-collinear instructors O, Ix, and Iy chosen, the instructors are in one-to-one correspondence with the elements of a two-dimensional vector space over a finite field.
Proof: Here is how to associate a coordinate pair with every instructor in an affine faculty senate:
1. Choose an instructor O as the "origin"
2. Choose committee sx on which O and Ix serve and committee sy on which O and Iy serve.  On both committees O is the additive identity. On sx instructor Ix is the multiplicative identity, and on sy instructor Iy is the multiplicative identity.

3. Construct the committee parallel to sx through Iy and the committee through sy parallel to Ix.  Find their intersection.  This is the instructor II that has coordinates (1,1).
4. Find the committee y=x that includes O and the instructor II. With O as additive identity and II as multiplicative identity, identify each other instructor on this committee with some element z
of the finite field with n elements. This will involve an arbitrary choice in the case of the medium senate (n=4), since either of the two remaining instructors could be associated with the element [x] and the other with [x+1].  Every instructor on this committee will have two equal coordinates (z,z). 
5 For each instructor Z on the committee y=x, with coordinates (z,z), construct the committee parallel to sx through Z and thereby determine the instructor on Iy with coordinates (0,z). Likewise construct the committee through Z parallel to sy to determine the instructor on Ix with coordinates (z,0)   Now you have associated an element of the finite field Fn with n elements to each member of sx and of sy.
6. For an arbitrary instructor A, identify the committee tx parallel or equal to sx
on which A serves and the committee ty parallel or equal to sy on which A
serves.  The y-coordinate of A, Ay, is the element of Fn associated with the
instructor that tx and sy have in common. The x-coordinate of A, Ax, is the element
of Fn associated with the instructor that ty and sx have in common. So each instructor A has been identified with a coordinate pair (Ax,Ay).