we can vary both of the elements in any single row freely and independently, with-out destroying the 2-dependence. Show that the analogous property holds of the following 3-dependent block, even though all of its entries are distinct:
That is, show that the 4-by-4 matrix formed from the coefficients of the cubic poly-nomials with those triples of roots not only has determinant zero, but actually has rank only 2.
2.5 Proving the quadratic case
Now that we know precisely what the quadratic cases of the Projection and Witness Theorems say, we should discuss how to prove them.
2.5. PROVING THE QUADRATIC CASE 19 The Projection and Witness Theorems do not appear, as such, in standard texts on projective geometry, for the excellent reason that 2-dependence is an algebraic notion, not a geometric one. But those books do provide pieces which we can as-semble into a proof. Coxeter [12] shows that the six lines
{ai,bi} i∈[1..3]through a point O are the projections of the six vertices of a complete quadrilateral just when there exists an involution — that is, a self-inverse projectivity — of the pencil of lines through O that swaps ai with bi, for each i. Maxwell [33] shows that such an involution exists precisely when the three chords of the conic in Figure 2.1, that is, the lines hi := AiBifor i in [1. .3], are concurrent. And we already showed, in Ex-ercise 2.1-4, that the three chords(hi)are concurrent precisely when the six slopes have the algebraic property of 2-dependence. That chain of reasoning suffices to prove the Projection Theorem and the existence half of the Witness Theorem.
That assembled proof has a couple of unfortunate aspects: It doesn’t establish the uniqueness half of the Witness Theorem, and it relies on the concept of an in-volution — a concept that would be somewhat clumsy to generalize to the cubic case. Fortunately, it isn’t hard to prove the Projection Theorem and both halves of the Witness Theorem directly. Let’s do that, since proving the quadratic case directly will help to prepare us for proving the cubic case, in Chapter 9.
We begin with an overdue lemma about n-dependence. When we are testing an n-block of hyperplanes for n-dependence, it doesn’t matter what scale we use to measure the slopes of the hyperplanes in that pencil; that is, we can take any three distinct hyperplanes in that pencil — whichever ones we like — and assign, to them, the slopes 0, 1, and∞.
Lemma 2.5-1 The relation of n-dependence is projectively invariant, meaning the following. Let
{ui j}j∈[1..n] i∈[0..n] be any unordered n-block, and form a derived n-block
{vi j}j∈[1..n] i∈[0..n] by setting
vi j := aui j +b cui j +d,
where a, b, c, and d are fixed scalars with ad −bc nonzero. Then, the derived block is n-dependent just when the original block is.
Proof If we rewrite each entry ui j of the original block as a ratio u↑i j : u↓i j of ho-mogeneous coefficients, we can express the corresponding entryvi j of the derived block as the ratiovi j↑ :vi j↓, wherev↑i j :=au↑i j +bu↓i j andvi j↓ :=cu↑i j +du↓i j.
The original block is n-dependent just when the polynomials Ui(X):=(u↓i1X −u↑i1)· · ·(u↓inX −u↑in),
for i in [0 . . n], are linearly dependent. In a similar way, the derived block is n-dependent just when the polynomials
Vi(Y):=(vi1↓Y −vi1↑)· · ·(vin↓Y −v↑in)
20 CHAPTER 2. INTRODUCTION are linearly dependent. But we have
Vi(Y)=(a−cY)nUi
dY −b a−cY
,
and neither substituting the expression(dY −b)/(a−cY)for the variable X nor multiplying through by the fixed polynomial(a−cY)nhas any effect on the linear dependence. tu
Theorem 2.5-2 (Projection Theorem, quadratic case) Let O be any point in the plane, and let the six points{(Ai,Bi)}i∈[1..3]be the vertices of any complete quadri-lateral, none of whose four lines passes through O. The six lines{(ai,bi)}i∈[1..3]
joining O to the vertices of the quadrilateral then form a block that is 2-dependent and in which no line in any pair coincides with any line in any other pair.
Proof In order for a line in one pair to coincide with either of the two lines in either of the other two pairs, the center O of the projection would have to lie on one of the four lines of the quadrilateral, which is forbidden. So it suffices to show 2-dependence, which we shall do using analytic geometry, after choosing our co-ordinate system with some care.
The four points B1, B2, B3, and O are four points in the plane, with no three collinear. So we can choose our coordinate system so that those four points have the homogeneous coordinates
B1 =[1,0,0]
B2 =[0,1,0]
B3 =[0,0,1]
O =[1,1,1].
That makes three of the four lines of the quadrilateral have simple coefficients. Re-call that, by our convention, the point with homogeneous coordinates [w,x,y] lies on the line with homogeneous coefficientshc,d,eijust when cw+d x+ey=0.
So the line q = A1B2B3 of the quadrilateral has the homogeneous coefficients q = h1,0,0i, while we similarly have r = B1A2B3 = h0,1,0iand s = B1B2A3 = h0,0,1i. The fourth line a = A1A2A3of the quadrilateral, however, has homoge-neous coefficients about which we know very little; let’s say that a = hf,g,hi.
It is easy to calculate the homogeneous coordinates of the points(Ai)in terms of the homogeneous coefficients f , g, and h of the line a; we have
A1 =[0,h,−g]
A2 =[−h,0, f ] A3 =[g,−f,0].
2.5. PROVING THE QUADRATIC CASE 21 We can then calculate the homogeneous coefficients of the six lines that connect the center point O to the six vertices of the quadrilateral:
a1= h−g−h,g,hi b1 = h0,1,−1i a2= hf,−f −h,hi b2 = h−1,0,1i a3= hf,g,−f −gi b3 = h1,−1,0i.
Our next task is to choose a definition of slope, in the pencil of lines through the center point O; by Lemma 2.5-1, it doesn’t matter which definition we choose.
Let’s adopt, as our measure of slope, the first homogeneous coefficient divided by the last — so the three distinct lines b1, b2, and b3 have the slopes 0,−1, and∞. (The middle coefficient doesn’t appear explicitly in this recipe, but it is not being ignored; we apply this recipe only to lines through the point O, whose three ho-mogeneous coefficients sum to zero.)
We then test for 2-dependence by plugging those slopes into the 3-by-3 deter-minant in Equation 2.4-1, getting
0 g+h −h
−f f −h h f −f −g 0
,
and simple algebra shows that this determinant is identically zero. tu
Theorem 2.5-3 (Witness Theorem, quadratic case) Let O be any point in the plane and let {(ai,bi)}i∈[1..3] be any ordered, 2-dependent block of lines through O in which no line in any pair coincides with any line in any other pair. Then, there exist complete quadrilaterals{(Ai,Bi)}i∈[1..3], each of whose six vertices lies on the corresponding line and none of whose four lines passes through O. Further-more, any two such quadrilaterals are related by a projective transformation of the plane that fixes O and every line through O.
Proof Since no line in any pair coincides with any line in any other pair, we de-duce from Exercise 2.4-3 that the condition of 2-dependence determines each line uniquely, given the other five lines. In particular, the line a3is so determined.
To construct a witnessing quadrilateral, we proceed as in Exercise 2.1-2. We choose a point Bion the line bi, for each i in [1. .3], being careful only that the three points B1, B2, and B3are not collinear and that none of them coincides with O. We then construct the points A1 :=a1∩B2B3and A2 :=a2∩B1B3, neither of which can possibly coincide with O. Since the lines a1 and a2 are distinct, we deduce that the line a := A1A2 does not pass through O. We construct the point A3 := a∩ B1B2and the line a03 := O A3. The six points{(A1,B1), (A2,B2), (A3,B3)}
form a complete quadrilateral, none of whose four lines passes through O. Hence, from the Projection Theorem, it follows that the ordered block of lines{(a1,b1), (a2,b2), (a30,b3)}is 2-dependent. Since the relation of 2-dependence determines
22 CHAPTER 2. INTRODUCTION
o
H1 H2
H3
h A1
A2
A3
B1
B2
B3
k
Figure 2.5: A 2-block of points along a line o whose 2-dependence is demonstrated by the collinearity of the three points(Hi), each of which is the intersection of a pair of tangents of a conic curve k, of which the line o is also a tangent.
the line a3uniquely, given the other five lines, the lines a3 and a03 must coincide, so the quadrilateral that we have constructed is a witness.
The only freedom of choice that we had, in carrying out this construction, was to select the point Bion the line bi, for i in [1. .3]. It follows that every witnessing quadrilateral can be produced by that construction, provided that the points(Bi)are chosen correctly. If{(A0i,Bi0)}i∈[1..3] and{(A00i,Bi00)}i∈[1..3] are any two witnessing quadrilaterals, the unique projective transformation of the plane that maps the four points(O,B10,B20,B30)to(O,B100,B200,B300)fixes O and every line through O and also carries the first witnessing quadrilateral to the second. tu