Projective spaces and B ezout’s theorem ﬁŒAÛ 0math.sjtu.edu.cn/faculty/mlai/teaching/summer2016/projectiveplane.… · (First half of 17th century) Fermat, Descartes,... analytic

Projective spaces and Bezout’s theorem �êAÛ{0

Projective spaces and Bezout’s theorem�êAÛ{0

Mijia Lai £5�\¤

[email protected]


Outline

1. History

2. Projective spaces

3. Conics and cubics

4. Bezout’s theorem and the resultant

5. Cayley-Bacharach theorem

6. Pappus’s theorem and Pascal’s theorem


Chronology of developments leading to algebraic geometry

(300 B.C.) Greeks: Euclid, Apollonius,... study of conics§�I�

(Middle Age) Arabs,... algebra, good notation of algebra,

(First half of 17th century) Fermat, Descartes,... analyticgeometry, )ÛAÛ

(Start of 19th century) Monge, Poncelet, Klein,... projectivegeometry, �KAÛ




















Main theme of algebraic geometry

(Second half of 19th century) Riemann, Plucker, Hilbert,Noether,... Algebraic geometry, �êAÛ

Study of the solutions of algebraic equations.

The study is naturally connected with commutative algebra,complex analysis, differential geometry, number theory, projectivegeometry, topology, etc.












Plane curves in Cartesian coordinates

Lines are of the form

ax + by + c = 0.

Ellipsex2

a2+

y 2

b2= 1.

Parabolay = ax2.

Hyperbolax2

a2− y 2

b2= 1.




ax + by + c = 0.

Ellipsex2

a2+

y 2

b2= 1.

Parabolay = ax2.

Hyperbolax2

a2− y 2

b2= 1.




ax + by + c = 0.

Ellipsex2

a2+

y 2

b2= 1.

Parabolay = ax2.

Hyperbolax2

a2− y 2

b2= 1.




ax + by + c = 0.

Ellipsex2

a2+

y 2

b2= 1.

Parabolay = ax2.

Hyperbolax2

a2− y 2

b2= 1.


Study from transformation-invariant point view

I One can choose coordinates freely, e.g., one makes a lineartransformation

x ′ = ax + by , y ′ = cx + dy .

Algebraic equation changes, but the shape does not change.

I Motivated by Perspective in drawing, which was introduced byItalian Renaissance painters and architects, one could alsofreely move the plane in R3 and consider projection of curveonto various planes. In doing so, one should add ’points atinfinity’ to a plane to make the projection a bijection. Thisextended plane is called the projective plane RP2.


Study from transformation-invariant point view

I One can choose coordinates freely, e.g., one makes a lineartransformation

x ′ = ax + by , y ′ = cx + dy .

Algebraic equation changes, but the shape does not change.

I Motivated by Perspective in drawing, which was introduced byItalian Renaissance painters and architects, one could alsofreely move the plane in R3 and consider projection of curveonto various planes. In doing so, one should add ’points atinfinity’ to a plane to make the projection a bijection. Thisextended plane is called the projective plane RP2.


Perspective



Projection between two planes

Light coming from O, there is no image for the line m ∈ P in Q,similarly, there is no pre-image for the line n ∈ Q in P. Thesolution is adding both planes a ’line at infinity’.


Projective spaces

A formal definition of RP2: R3 \ (0, 0, 0)/ ∼, where(x , y , z) ∼ (x ′, y ′, z ′) if there exists λ 6= 0 such that(x , y , z) = λ(x ′, y ′, z ′).It can be viewed as the set of all lines in R3 passing through origin.

We use the homogeneous coordinates [x : y : z ] to denote a pointin the projective plane, where x , y , z are not all zero.

Projective plane can be viewed as the union of plane z = 1 andlines that are parallel to z = 1 (’points at infinity’).


Plane curve↔ curve in the Projective plane

Any polynomial g(x , y) ∈ R[x , y ] of degree n can be homogenizedby zng( xz ,

yz ).

Examples: g(x , y) = 3x + 5y 2 − 2, after homogenization, we getf (x , y , z) = 3xz + 5y 2 − 2z2, a homogeneous polynomial in x , y , z .

f (x , y , z) = 0 ⊂ RP2 ⇔ g(x , y) = 0 ⊂ {z = 1}∪ ’points at infinity’.


Plane curve↔ curve in the Projective plane

Any polynomial g(x , y) ∈ R[x , y ] of degree n can be homogenizedby zng( xz ,

yz ).

Examples: g(x , y) = 3x + 5y 2 − 2, after homogenization, we getf (x , y , z) = 3xz + 5y 2 − 2z2, a homogeneous polynomial in x , y , z .

f (x , y , z) = 0 ⊂ RP2 ⇔ g(x , y) = 0 ⊂ {z = 1}∪ ’points at infinity’.


Within this identification, we could study zeros of homogeneouspolynomials in RP2. They are usual plane curves with ’points atinfinity’ added.

Degree one (line): ax + by + cz = 0.

Degree two (quadratic curve):ax2 + by 2 + cz2 + dxy + exz + fyz = 0.

Degree three (cubic curve): ax3 + by 3 + · · ·+ jxyz︸︷︷︸10 terms

= 0.


Within this identification, we could study zeros of homogeneouspolynomials in RP2. They are usual plane curves with ’points atinfinity’ added.

Degree one (line): ax + by + cz = 0.

Degree two (quadratic curve):ax2 + by 2 + cz2 + dxy + exz + fyz = 0.

Degree three (cubic curve): ax3 + by 3 + · · ·+ jxyz︸︷︷︸10 terms

= 0.


Classification of quadratic curves in RP2

x ′ = ax + by + cz , y ′ = dx + ey + fz , z ′ = gx + hy + iz

is called a linear transformation from RP2 → RP2.

Theorem

Any curve of degree 2 in RP2 can be transformed into one the followingtypes:

I x2 = 0, a double line;

I x2 + y 2 = 0, a point;

I x2 − y 2 = 0, two lines,

I x2 + y 2 + z2 = 0, the empty set;

I x2 + y 2 − z2 = 0, the unit circle (conic).


Classification of quadratic curves in RP2

x ′ = ax + by + cz , y ′ = dx + ey + fz , z ′ = gx + hy + iz

is called a linear transformation from RP2 → RP2.

Theorem

Any curve of degree 2 in RP2 can be transformed into one the followingtypes:

I x2 = 0, a double line;

I x2 + y 2 = 0, a point;

I x2 − y 2 = 0, two lines,

I x2 + y 2 + z2 = 0, the empty set;

I x2 + y 2 − z2 = 0, the unit circle (conic).


Conics

Ellipses, parabolas, hyperbolas are all same as circles. (They allarise as conic sections, i.e., intersection of a plane with a cone invarious position, thus from projective point view they are the same)

Conic sections


Cubics

A cubic curve is irreducible if it cannot factor out degree 1 or 2factors.

Theorem

Any irreducible cubic curve can be transformed into

y 2 = x3 + ax2 + bx + c .

In complex projective plane, such a curve looks like a torus, so it iscalled an elliptic curve.


Cubics


Theorem


y 2 = x3 + ax2 + bx + c .



Cubics


Theorem


y 2 = x3 + ax2 + bx + c .



Common zeros: solutions of algebraic systems

Examples: 3x + 2y − 1 = 0 and 3x + 2y − 5 = 0By homogenization, they become 3x + 2y − z = 0 and3x + 2y − 5z = 0, by eliminating z , we get 3x + 2y = 0, and z = 0,thus they intersect at [−2, 3, 0] ∈ RP2. This is a ’point at infinity’.So in projective plane, any two lines intersect at exactly one point.

Example: x2 − y 2 = 1, x − y = 0By homogenization, they become x2 − y 2 = z2 and x − y = 0, itfollows that z = 0, and x = y , thus they intersect at [1 : 1 : 0], butthe intersection has a multiplicity 2.


Bezout’s theorem

Theorem (Bezout)

Let f = 0 and g = 0 be two homogeneous polynomials in x , y , zvariables of degree n and m respectively, assume they don’t haveany common component, then they intersect at most m · n pointsin RP2. Assumption as above, f and g intersect at exactly m · npoints, counting multiplicities, in the complex projective plane.


Resultant

Given two polynomials f (x) = anxn + · · ·+ a0 andg(x) = bmxm + · · ·+ b0, the resultant R(f , g) is defined to be

R(f , g) = amn bnm

∏(αi − βj),

where αi are roots of f (x), and βj are those of g(x). (Byfundamental theorem of algebra, αi , βj exist as complex numbers)

Proposition

R(f , g) = 0 if and only if f and g have a common root.


In terms of coefficients, we have

R(f , g) =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

an an−1 a0

an an−1 a0

· · · · · ·an an−1 a0

bm b0

bm b0

· · · · · ·bm b0

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣


Determine a quadratic curve

Any two points determine a line uniquely. How many pointsdetermine a quadratic curve uniquely?

Heuristics: suppose the curve is ax2 + bxy + cy 2 + dx + ey + f = 0,then plugging given points, we will get a system of linear equationson a, b, c , d , e, f - the six unknowns. Multiplying by a non-zeroconstant results in the same curve, so in general if given fivepoints, we get five equations for six unknowns, from linear algebra,we know the solution set is at least one-dimensional.


Determine a quadratic curve

Any two points determine a line uniquely. How many pointsdetermine a quadratic curve uniquely?

Heuristics: suppose the curve is ax2 + bxy + cy 2 + dx + ey + f = 0,then plugging given points, we will get a system of linear equationson a, b, c , d , e, f - the six unknowns. Multiplying by a non-zeroconstant results in the same curve, so in general if given fivepoints, we get five equations for six unknowns, from linear algebra,we know the solution set is at least one-dimensional.


Uniqueness

Uniqueness of such curve relies on Bezout’s theorem (2× 2 < 5).Namely, the given five points should be in the ’general position’.

Theorem

Any five points in RP2, no three of which are collinear, lie onexactly one conic.

Theorem

Any five points in RP2, no four of which are collinear, lie onexactly one quadratic curve.


Uniqueness


Theorem


Theorem



Uniqueness


Theorem


Theorem



Cubic curve

For a cubic curve, the same linear algebra reason suggests that anynine points can determine a cubic curve. What about theuniqueness? The issue is subtle as 9 = 3× 3.

Theorem (Cayley-Bacharach)

Let P(x , y) = 0 and Q(x , y) = 0 be two cubic curves that intersect (overC) in precisely nine distinct points p1, · · · , p9. Let R(x , y) be a cubicpolynomial that vanishes on eight of these points (say p1, · · · , p8). ThenR is a linear combination of P and Q, in particular, R also vanishes onthe ninth point p9.


Cubic curve

For a cubic curve, the same linear algebra reason suggests that anynine points can determine a cubic curve. What about theuniqueness? The issue is subtle as 9 = 3× 3.

Theorem (Cayley-Bacharach)

Let P(x , y) = 0 and Q(x , y) = 0 be two cubic curves that intersect (overC) in precisely nine distinct points p1, · · · , p9. Let R(x , y) be a cubicpolynomial that vanishes on eight of these points (say p1, · · · , p8). ThenR is a linear combination of P and Q, in particular, R also vanishes onthe ninth point p9.


Pappus’ theorem

Theorem (Pappus)

Let l , l ′ be two distinct lines, let A1,A2,A3 be three distinct points on lnot on l ′, and let B1,B2,B3 be three distinct points on l ′ not on l.Suppose that for ij = 12, 23, 31, the line AiBj and AjBi intersect at Cij .Then the three points C12, C23 and C31 are collinear.


Pascal’s theorem

Theorem (Pascal)

Let A1,A2,A3,B1,B2,B3 be distinct points on a conic σ. Suppose thatfor ij = 12, 23, 31, the lines AiBj and AjBi intersect at Cij . Then thepoints C12, C23 and C31 are collinear.


References and further reading topics

References:

1. Robert Bix, Conics and Cubics £�g�Úng�¤§

2. Frances Kirwan, Complex Algebraic Curves £E�ê�¤§

3. Terence Tao, Pappus’s theorem and elliptic curves, His blog:What’s new?

I Figure out the ’multiplicity’ mentioned in this lecture.

I Why certain cubic curves are called Elliptic curves?

I Resultant

Documents

Projective spaces and B ezout’s theorem ﬁŒAÛ 0math.sjtu.edu.cn/faculty/mlai/teaching/summer2016/projectiveplane.… · (First half of 17th century) Fermat, Descartes,... analytic