Martin's Blog

Ordinary multiple points

Posted by martin on Monday, 10 May 2010 at 21:41

Singular points in a curve are places where curve fails to be smooth: intuitively, multiple points of the curve pile up on top of each other. In this post I will describe a simple invariant of curve singularities, the multiplicity, which essentially counts how many points are piled up there. In the simplest case of an ordinary multiple point, I describe how to use the previous post’s algorithm to compute a power series for each branch of the curve near the singularity.

Curve y^2 - x^3 - x^2 = 0
$y^2 - x^3 - x^2 = 0$

Defining multiplicity

In the case of $y^2 - x^3 - x^2$, intuition suggests that the origin lies on this curve “twice”: there are two “branches” passing through the origin, and if we could look at each branch separately, the origin should be a smooth point of that branch.

One way of justifying this is to say that, for $x$ and $y$ small, the dominant terms of the equation are the quadratic ones: $y^2 - x^2$. This factorises as $(y-x)(y+x)$, and the lines $y-x = 0$, $y+x = 0$ are the tangents to the two branches of $C$.

In general, we define the multiplicity $r$ of a curve $C : f(x, y) = 0$ at $(0, 0)$ to be the lowest total degree of a non-zero term in $f$ (by “total degree” I mean the sum of the $x$ and $y$ degrees).

The curve $C$ is approximated near the origin by the curve $C'$ whose equation $g(x, y) = 0$ is obtained by picking all the total-degree $r$ terms of $f$ and throwing away the rest.

This gives a homogeneous polynomial of degree $r$, which (over an algebraically closed field) splits as a product of $r$ linear factors - in other words the curve $g(x, y) = 0$ is just a union of lines through the origin. To find the factors, set $y = 1$ and decomposing the resulting one-variable polynomial $g(x, 1)$. (This one-variable polynomial might have degree less than $r$, but in this case the missing factors of $g(x, y)$ are factors of $y$).

Curve y^2 - x^3 = 0
$y^2 - x^3 = 0$

The tangent lines to $C$ at $(0, 0)$ are the lines defined by the linear factors of $g(x, y)$. We see that there are at most $r$ tangent lines. There may be fewer tangent lines if some are repeated - for example $y^2 - x^3 = 0$ which has multiplicity 2, with the line $y = 0$ appearing as a tangent line twice.

A singular point with no tangent lines repeated - i.e. it has multiplicity $r$ and $r$ distinct tangent lines is called an ordinary multiple point.

Recall that a point on $C$ is non-singular iff at least one of $\partial f/\partial x$, $\partial f/\partial y$ is non-zero. In other words, the origin is non-singular iff $f$ has a non-zero degree 1 term. Hence a point on $C$ is non-singular if and only if its multiplicity is exactly 1.

Furthermore, at a non-singular point, the tangent line defined above is the same as the usual tangent line.

Intersection multiplicities

Here is another, more geometric, way of defining the multiplicity of a point on a plane curve.

We first define the intersection multiplicity between a curve and a line.

Suppose that $L$ is the line $y = ax$ and $C$ the curve $f(x, y) = 0$, passing through the origin. Substitute $y = ax$ in $f(x, y) = 0$ to get the one-variable polynomial $f(x, ax) = 0$. Then the intersection multiplicity between $L$ and $C$ at $(0, 0)$ is the multiplicity of $0$ as a root of this polynomial.

If $L$ is the $y$-axis, we need to substitute $x = 0$ in $f(x, y) = 0$ instead.

We see that the intersection multiplicity is a positive integer, at most the degree of $f$.

Furthermore, the intersection multiplicity between $L$ and $C$ at $(0, 0)$ is always at least $r$, the multiplicity of $(0, 0)$ on $C$, because only terms of degree at least $r$ can appear in $f(x, ax)$.

The intersection multiplicity is greater than $r$ if and only if the total-degree $r$ terms of $f(x, ax)$ vanish: in other words iff $y = ax$ is a linear factor of the polynomial obtained by picking out the total degree $r$ terms of $f(x, y)$. (This works also for $x = 0$.)

We conclude that: the multiplicity of $P$ on $C$ is the minimum (over all lines $L$ through $P$) of the intersection multiplicities of $L$ with $C$ at $P$; the intersection multiplicity is equal to the multiplicity of $P$ on $C$ for all but finitely many lines $L$; and the exceptional lines are precisely the tangents to $C$ at $P$.

In the simplest case, a tangent line has intersection multiplicity with $C$ at $P$ one greater than the multiplicity of $C$ in $P$. This may fail to hold even at an ordinary multiple point, indeed even at a non-singular point, such as the origin on the curve $y = x^3$. Conversely, a point may fail to be ordinary while all its tangents still have intersection multiplicity one greater than the multiplicity of the point: for example at the origin of $y^2 = x^3$, the only tangent is the $x$-axis with intersection multiplicity 3.

Solving for power series at an ordinary multiple point

In the previous post I gave an algorithm for constructing a power series solution to a polynomial $f(x, y) = 0$ at a non-singular point where the curve does not have a vertical tangent. Today I shall consider an ordinary point of multiplicity $r$ which does not have a vertical tangent. For simplicity suppose that the chosen point is the origin.

Obviously we cannot get a unique power series expansion for $y$, because there should be one for each branch. So we begin by picking a branch, with tangent $y = a_1 x$.

A power series expansion along this branch will have the form $y = a_1 x + \ldots$. We can remove a factor of $x$ and write $y = x\tilde{y}$; then we are looking for a power series $\tilde{y} = a_1 + \ldots$ satisfying $f(x, \tilde{y}x) = 0$ (this replacing of $y$ by $x\tilde{y}$ is an example of the important process of blowing up).

The polynomial $f(x, \tilde{y}x)$ has a factor of $x^r$; say $f(x, x\tilde{y}) = x^r \tilde{f}(x, \tilde{y})$. The $x^r$ is irrelevant to finding a power series expansion for $\tilde{y}$, so we focus on $\tilde{f}$.

Let $\tilde{g}(\tilde{y}) = \tilde{f}(0, \tilde{y})$. This polynomial is formed from the total-degree r terms of $f(x, y)$, and its roots are the gradients of the tangents to $C$.

Since our point is an ordinary multiple point, these roots are distinct. So we can apply the algorithm of the previous post to find a unique power series $\tilde{y} = a_1 + \ldots$ satisfying $\tilde{f}(x, \tilde{y}) = 0$.

For example, let $f(x, y) = y^2 - x^2 - x^3$, and choose the branch with gradient +1. We get $\tilde{f}(x, \tilde{y}) = \tilde{y}^2 - x - 1$, and the power series for this branch of the original curve is $y = x + \frac{1}{2}x^2 - \frac{1}{8}x^3 + \frac{1}{16}x^4 - \frac{5}{128}x^5 + \ldots $

Tags ,

Trackbacks

Use the following link to trackback from your own site:
http://www.martinorr.name/blog/trackbacks?article_id=171

Comments

No comments.

Comments are disabled

Archives

Syndicate