Algebraic Geometry Jottings 5

Prev TOC Next

We’ve been looking at Kendig‘s two definitions of intersection multiplicity; now let’s look at Fulton‘s.

Fulton characterizes the multiplicity I(EF) with seven properties (§3.3). (Fulton calls it the intersection number. Also, he writes I(P,EF) for the intersection number at P. I’ll usually assume P is the origin O, and omit writing it.) The last three properties stand out:

  1.  I(EF) ≥ m(E)m(F), with equality if and only if E and F have no tangent lines in common. Here m(E) and m(F) are the multiplicities of E and F (explained below). As Fulton notes, all we really need is a much weaker property: I(xy)=1. That is, the lines x=0 and y=0 (the axes) intersect with multiplicity 1 at the origin.
  2. I(EFG) = I(EF) + I(EG).
  3. I(EF) = I(E∩(F+AE)) for any polynomial A.

The multiplicity of a curve at the origin O is the order of its polynomial, i.e., the degree of its lowest order term(s). Take our favorite E and F:

E(x,y)=(x^2+y^2)^2+3x^2y-y^3=0
F(x,y)=(x^2+y^2)^3-4x^2y^2=0

The multiplicity (at O) of E is 3, and of F is 4. We know from the pictures in the first post that E has three distinct branches at O, and F has four. Intuitively, property (5) says that every intersection of a branch of E with a branch of F contributes 1 to the intersection number when the branches intersect transversally, and more than one if they are tangent.

Property (6) seems obvious once you notice that FG=0 defines the union of the curves F=0 and G=0. Of course, F and G might have a common factor—a common component of the two curves. Property (6) then says that the component counts double when intersecting E with FG.

Fulton calls property (7) “the least intuitive”. Let’s see if the “level curve” idea sheds any light. Recall how this works: we parametrize a branch of E as (xE(t), yE(t)), passing through O when t=0;  xE(t) and yE(t) are power series in t. We plug (xE(t), yE(t)) into the polynomial F(x,y), obtaining a function F(t). The order of F(t) gives the multiplicity of the branch-curve intersection.

Now, E is the level curve E(x,y)=0, and (xE(t), yE(t)) moves along a branch of it, so E(t)=E(xE(t), yE(t)) is identically zero. So

(F+AE)(t) = F(t)+A(t)E(t) = F(t)

So the order is the same for both F and F+AE, all branch-curve intersections have the same multiplicity, and therefore the curve-curve multiplicity is the same.

As Fulton notes, the seven properties imply an algorithm for computing intersection numbers. Fulton illustrates it for our favorite curves E and F (see p.40), incorporating some shortcuts. Let’s see how it goes. He uses some auxiliary curves:

E=(x^2+y^2)^2+3x^2y-y^3
F=(x^2+y^2)^3-4x^2y^2
F'=F-(x^2+y^2)E=yG, where
G=(x^2+y^2)(y^2-3x^2)-4x^2y
G'=G+3E=yH, where
H=5x^2-3y^2+4y^3+4x^2y

Here’s the strategy: use property (7) to whittle down the higher-order part. For E, this is (x2+y2)2, and for F it is (x2+y2)3, so forming F′ = F–(x2+y2)E cancels it out. We’re lucky that F′ factors as yG. Note the total degrees of F, F′, and G: 6, 5, 4.

We can’t use (7) to reduce the total degree of G, so instead we target the degree of the x-part—what we get by striking out all terms involving y, i.e., G(x,0). This is –3x4. So we form G′ = G+3E (because the x-part of E is x4). Now we are lucky that G′ factors as yH.

The auxiliary curves G and H look like this:

Auxiliary Curves G and H

Recall that F, the four-petal rose, has two branches tangent to the x-axis at O. G has only one, and H has none. Note that we remove a factor of y passing from F′ to G, and from G′ to H.

The full computation of I(EF) looks like this:

I(EF) = I(EF′) = I(EyG),  by (7)
I(EyG) = I(Ey) + I(EG),  by (6)
I(EG) = I(EG′) = I(EyH),  by (7)
I(EyH) = I(Ey) + I(EH),  by (6)
I(EH) = m(E) m(H) = 3×2 = 6,  by (5)
I(Ey) = I(x4y),  by (7)
I(x4y) = m(x4) m(y) = 4×1,  by (5)
I(EF) = I(Ey) + I(Ey) + I(EH) = 4+4+6 = 14

—the same result we’ve seen before.

I’ll look at Fulton’s other definition in Post 7.

Prev TOC Next

Leave a comment

Filed under Algebraic Geometry

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.