Question

Let \(C\) be a nonsingular quartic, \(P_{1}, P_{2}, P_{3} \in C\). Let \(D=P_{1}+P_{2}+P_{3} .\) Let \(L\) and \(L^{\prime}\) be lines such that \(L . \cdot C=P_{1}+P_{2}+P_{4}+P_{5}, L^{\prime} \cdot C=P_{1}+P_{3}+P_{6}+P_{7} .\) Suppose these seven points are distinct. Show that \(D\) is not linearly equivalent to any other effective divisor. (Hint: Apply the residue theorem to the conic \(L L^{\prime} .\) ) Investigate in a similar way other divisors of small degree on quartics with various types of multiple points.

Short answer

In this exercise, we have shown that the divisor \(D = P_1 + P_2 + P_3\) is not linearly equivalent to any other effective divisor. We achieved this by expressing the intersection divisors for the lines \(L\) and \(L'\) and their conic, and then applying the residue theorem to analyze the possible linear equivalence of divisor D to other effective divisors. By showing that D cannot be linearly equivalent to any divisor formed by removing any set of two distinct points from the intersection divisor of \(L L'\) and \(C\), we concluded that D is not linearly equivalent to any other effective divisor.

Step-by-step solution

Calculate the intersection divisors for L and L'

First, let's write out the intersection divisors for \(L\) and \(L'\) with the curve \(C\). We are given that the lines intersect C in the following points: (For line L) : \(P_1+P_2+P_4+P_5\) (For line L'): \(P_1+P_3+P_6+P_7\)

Calculate the intersection divisor for the conic LL'

The intersection divisor for the conic \(L L'\) with the curve \(C\) can be written as the sum of the intersection divisors of both lines. Therefore, the intersection divisor for the conic LL' with C is: \(P_1+P_2+P_4+P_5 + P_1+P_3+P_6+P_7\) We can simplify this as: \(2P_1+P_2+P_3+P_4+P_5+P_6+P_7\)

Apply the residue theorem to the divisor of LL'

The residue theorem can be applied to the intersection divisor of the conic \(L L'\) and \(C\) obtained in step 2. The Residue Theorem states: "The sum of the residues of a meromorphic function over all its poles on a curve is always zero" We are given that \(C\) is a nonsingular quartic curve, and therefore has genus 3. If we consider the divisor of LL', it is of degree 8. Now let's analyze the possible linear equivalence of our divisor D to other effective divisors.

Check for linear equivalence with divisor D

Our divisor \(D = P_1 + P_2 + P_3\) is a degree 3 divisor. Note that if D were linearly equivalent to a divisor created by removing one point from the intersection divisor of \(L L'\) with \(C\), it would be a divisor of degree 7, which is not possible due to the degree of D. Now consider the intersection divisor of the conic \(L L'\) and \(C\) in step 2. Suppose our divisor, \(D\), is linearly equivalent to a divisor \(D'\) obtained by removing two distinct points from the intersection divisor of \(L L'\) with \(C\). Suppose we remove \(P_4\) and \(P_5\). Then, we have: \(D' = 2P_1+P_2+P_3+P_6+P_7\) We also have the relation for functional equivalence, that is: \(D \sim D' \implies D - D' \sim 0\) This implies that there exists a function \(f\) on the curve \(C\) with divisor \((f) = D - D'\). Hence: \((f) = (P_1 + P_2 + P_3) - (2P_1 + P_2 + P_3 + P_6 + P_7)\) Simplifying, we have: \((f) = -P_1 - P_6 - P_7\) Since degree of \((f)\) must be non-negative, we have a contradiction. Thus, the divisor \(D\) cannot be linearly equivalent to any other divisor formed by removing any set of two distinct points from the intersection divisor of \(L L'\) and \(C\). Therefore, the divisor \(D = P_1 + P_2 + P_3\) is not linearly equivalent to any other effective divisor, as required.

Key concepts

Quartic Curve

A quartic curve is an algebraic curve of the fourth degree, which means it is defined by a polynomial equation where the highest degree of any term is four. Quartic curves can take on various shapes and possess different properties depending on the coefficients of the terms in the equation. In the context of the given problem, we're working with a nonsingular quartic. This means that the curve has no self-intersections or cusps, which implies that it is a smooth curve.

Such curves are of significant interest in algebraic geometry because they can provide insights into more complex structures. Their nonsingular nature is crucial when calculating intersections with lines, as it ensures that the intersections will be well-defined points. In our situation, the curve intersects with two lines, resulting in several distinct points, denoted as P₁, P₂, and so forth. The significance of these points is further explored when discussing divisors and their linear equivalence on the curve.

Linear Equivalence of Divisors

The concept of divisors plays a fundamental role in algebraic geometry. A divisor is a formal sum of points on a curve, with possibly integer coefficients assigned to each point. In our case, the divisor D is made up of three points on the curve, P₁ + P₂ + P₃.

Understanding Linear Equivalence

Two divisors are said to be linearly equivalent, denoted as D ~ D', if there is a non-zero rational function f on the curve such that the divisor of f (noted as (f)) is the difference of the two divisors, D - D'. This is an equivalence relation that groups divisors into classes where each class can be thought of as representing a sort of 'shape' on the curve.

In simple terms, linear equivalence is a way to classify divisors that behave 'similarly' by considering the function that 'transforms' one into the other. In the solved problem, we found that the divisor D could not be linearly equivalent to another effective divisor by analyzing its possible transformations using the function f. This is a crucial discovery, as it signifies that D occupies a unique position on the quartic curve.

Residue Theorem

The residue theorem is a powerful tool in complex analysis, and it extends to algebraic geometry through the context of Riemann surfaces, which can be viewed as complex algebraic curves. In essence, it states that for a meromorphic function (one that is holomorphic except for isolated poles) on a complex curve, the sum of the residues at all its poles is zero.

Application in Algebraic Curves

When we apply the residue theorem in the context of algebraic curves, we're essentially using it to add up the contributions from the poles of a meromorphic function on the curve. For the problem given, the theorem helps us deduce certain constraints about the possible divisors on a curve.

By employing the residue theorem to the conic formed by the lines L and L', we are seeking to understand the nature of the function f that could potentially demonstrate linear equivalence between the original divisor D and another one. However, the contradiction that arises as per the theorem's application indicates that no such function can exist, reinforcing that D is in its own distinct linear equivalence class on the nonsingular quartic curve we are examining.