Question

Let \(D\) be a divisor, and let \(V\) be a subspace of \(L(D)\) (as a vector space). The set of effective divisors \(\\{\operatorname{div}(f)+D \mid f \in V, f \neq 0\\}\) is called a linear series. If \(f_{1}, \ldots, f_{r+1}\) is a basis for \(V\), then the correspondence \(\operatorname{div}\left(\sum \lambda_{i} f_{i}\right)+D \mapsto\left(\lambda_{1}, \ldots, \lambda_{r+1}\right)\) sets up a one-to-one correspondence between the linear series and \(\mathbb{P}^{r}\). If \(\operatorname{deg}(D)=\) \(n\), the series is often called a \(g_{n}^{r}\). The series is called complete if \(V=L(D)\), i.e., every effective divisor linearly equivalent to \(D\) appears. (a) Show that, with \(C, E\) as in Section 1, the series \(\\{\operatorname{div}(G)-E \mid G\) is an adjoint of degree \(n\) not containing \(C\\}\) is complete. (b) Assume that there is no \(P\) in \(X\) such that \(\operatorname{div}(f)+D \geq P\) for all nonzero \(f\) in \(V\). (This can always be achieved by replacing \(D\) by a divisor \(\left.D^{\prime} \leq D .\right)\) For each \(P \in X\), let \(H_{P}=\\{f \in V \mid \operatorname{div}(f)+D \geq P\) or \(f=\) \(0\\}\), a hyperplane in \(V\). Show that the mapping \(P \mapsto H_{P}\) is a morphism \(\varphi_{V}\) from \(X\) to the projective space \(\mathbb{P}^{*}(V)\) of hyperplanes in \(V .\) (c) A hyperplane \(M\) in \(\mathbb{P}^{*}(V)\) corresponds to a line \(m\) in \(V\). Show that \(\varphi_{V}^{-1}(M)\) is the divisor \(\operatorname{div}(f)+D\), where \(f\) spans the line \(m\). Show that \(\varphi_{V}(X)\) is not contained in any hyperplane of \(\mathbb{P}^{*}(V)\). (d) Conversely, if \(\varphi: X \rightarrow \mathbb{P}^{r}\) is any morphism whose image is not contained in any hyperplane, show that the divisors \(\varphi^{-1}(M)\) form a linear system on \(X .\) (Hint: If \(D=\) \(\varphi^{-1}\left(M_{0}\right)\), then \(\left.\varphi^{-1}(M)=\operatorname{div}\left(M / M_{0}\right)+D .\right)\) (e) If \(V=L(D)\) and \(\operatorname{deg}(D) \geq 2 g+1\), show that \(\varphi_{V}\) is one-to-one. (Hint: See Corollary 3 .) Linear systems are used to map curves to and embed curves in projective spaces.

Short answer

Question: Prove that the given linear series is complete and describe its properties, including the morphism φ_V and its inverse, and the correspondence between morphism and linear systems. Answer: To prove that the given linear series is complete, we need to show that for every effective divisor D' linearly equivalent to D, there exists a function G in V such that div(G) - E = D'. We can define a morphism φ_V: X → P*(V) by P ↦ H_P where H_P is a hyperplane in V. The inverse of the morphism φ_V is given by φ_V^(-1)(M) = div(f) + D. The image of φ_V is not contained in any hyperplane of P*(V), and given a morphism φ: X → P^r whose image is not contained in any hyperplane, the divisors φ^(-1)(M) form a linear system. Finally, φ_V is a one-to-one morphism when deg(D) ≥ 2g + 1.

Step-by-step solution

(a) Proving the completeness of the linear series

To prove that the given series is complete, we need to show that for every effective divisor D' linearly equivalent to D, there exists a function G in V such that div(G) - E = D'. Since D' is linearly equivalent to D, there exists a rational function f on X such that div(f) = D' - D. Let G = fC, where C is the function defining the canonical divisor. Then div(G) = D' - D + E. Adding E to both sides, we have div(G) + E = D'. Thus, the series is complete.

(b) Defining the morphism φ_V

For each P in X, let H_P = {f ∈ V | div(f) + D ≥ P or f = 0}, which is a hyperplane in V. We can define a mapping φ_V: X → P*(V) by P ↦ H_P. To show that this is a morphism, we need to check that the map is continuous. Since the coefficients of the functions f in V are in the projective space, for any open subset U of P*(V), the set φ_V^(-1)(U) is also open in X. This proves that the mapping is continuous, and hence, φ_V is a morphism.

(c) Properties of the morphism φ_V and its inverse

Let M be a hyperplane in P*(V) and let m be the corresponding line in V. If we take the inverse of the morphism φ_V, we can show that φ_V^(-1)(M) = div(f) + D for some f ∈ V that spans m. Since M corresponds to the hyperplane H_P, we have φ_V^(-1)(M) = P such that div(f) + D ≥ P for some f ∈ V. Thus, φ_V^(-1)(M) = div(f) + D. Now, we need to show that the image of φ_V is not contained in any hyperplane of P*(V). If the image of φ_V was contained in a hyperplane, there would exist a nonzero linear combination of functions in V such that this linear combination vanishes identically on X. But this would imply that X is contained in a lower-dimensional space, which is a contradiction. Hence, φ_V(X) is not contained in any hyperplane of P*(V).

(d) Morphism and linear system correspondence

Given a morphism φ: X → P^r whose image is not contained in any hyperplane, let D = φ^(-1)(M₀). Then φ^(-1)(M) = div(M / M₀) + D. We need to show that the divisors φ^(-1)(M) form a linear system. Since φ is a morphism between projective spaces, the divisors φ^(-1)(M) correspond to the set {div(f) + D | f ∈ V}, which is indeed a linear system on X.

(e) One-to-one property of the morphism φ_V

Let V = L(D) and suppose deg(D) ≥ 2g + 1. To show that φ_V is one-to-one, we need to prove that for any two distinct points P and Q in X, there exists a function f ∈ V such that div(f) + D ≥ P but div(f) + D < Q. If div(f) ≥ P, then the order of f at P is non-negative. Since the degree of D is greater than or equal to 2g + 1 and the degree of div(f) is g, we can apply the Corollary 3, and there exists a function g ∈ V such that the order of g at Q is positive. Then, the function f = h - g, where h ∈ V is a function that vanishes at Q (hence div(h) ≥ Q), has the desired property. Thus, φ_V is a one-to-one morphism when deg(D) ≥ 2g + 1.

Key concepts

Divisor Class Group

The concept of a divisor class group is fundamental in algebraic geometry, playing a crucial role in understanding the structure of algebraic curves and surfaces. It is formed by taking divisors on a variety, which are formal sums of subvarieties (like points on curves, curves on surfaces, etc.), and factoring out linear equivalence.

Two divisors are linearly equivalent if their difference is the divisor of some rational function. The divisor class group, denoted often as Cl(X) for a variety X, measures the failure of unique factorization in the coordinate ring of X and reflects the geometry of the variety itself. It is particularly interesting because it captures both topological and arithmetic properties of the space.

In the context of the exercise, effective divisors that are linearly equivalent to a given divisor D form a complete linear series, which can be seen as an element of the divisor class group. Understanding this group helps students tackle problems related to mapping algebraic varieties into projective spaces and studying their geometric properties.

Projective Space

Projective space, expressed as \(\mathbb{P}^r\), where \((r\) stands for the dimension, is a key concept in geometry and provides an extension of Euclidean space where the concept of infinity is treated in a seamless manner. Points in projective space can represent lines through the origin in \((r+1\))-dimensional space, thus unifying the treatment of points at infinity with those in affine space.

In algebraic geometry, projective space is where many curves and varieties live. When we map an algebraic curve into a projective space via a linear series, we're essentially embedding its properties into a more 'global' framework, where the behavior at infinity is well-defined.

In our exercise, the relationship between the linear series and projective space is highlighted. The basis functions \((f_i\)) of the subspace \((V\)) map to coordinates in \((\mathbb{P}^r\)), suggesting that every element in the linear series can be viewed as a point in projective space, a crucial insight for understanding morphisms between varieties.

Morphism

A morphism in algebraic geometry is essentially a function between two algebraic varieties that preserves their structure. Think of it like a bridge that respects the properties and features of the territories it's connecting.

Morphisms play an indispensible role in understanding how different geometric objects are related. For instance, in our problem, the mapping \((P \mapsto H_P\)) assigns to each point \((P\)) in the variety \((X\)) a hyperplane \((H_P\)) in the linear series subspace \((V\)). This morphism, \((\varphi_V\)), essentially provides a way to 'project' the variety into a new space, revealing information about the variety's structure.

By proving this mapping is a morphism, we are asserting the kind of continuity and compatibility that is central for the morphisms to faithfully translate geometric and algebraic information from one context to another, an insight that's pivotal for a student's grasp of algebraic geometry concepts.

Effective Divisor

An effective divisor is a formal sum of points (in the case of curves) or other subvarieties, with non-negative coefficients. You can think of an effective divisor as marking specific locations on a curve or surface with certain 'multiplicities' indicating how many times each is counted.

For instance, if we say \((\operatorname{div}(f)+D\)) where \((f\)) is a non-zero element of \((V\)), this represents an effective divisor that tells us the zeros of \((f\)) but adjusted by the divisor \((D\)). This concept is integral when dealing with linear series, as they consist of effective divisors linearly equivalent to a given divisor \((D\)).

Understanding effective divisors, therefore, is crucial for students tackling problems involving the mapping of varieties since they reflect the intersection of a variety with other geometrical entities, thereby providing insights into the variety's structure. When analyzing the completeness of a linear series or morphisms between spaces, effective divisors are the key characters that students need to appreciate.