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Could someone please tell me if there is any difference between the concepts "linear series" and "linear systems" on algebraic curves?

Also, for smooth plane curves of degree $n$, what is the main difference between linear system of "curves of degree $d$" in the case $d\leq n-3$, and the case $d>n-3$?

user108555
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  • In your second sentence, $n$ is the degree of the curves, and $d$ is degree of the linear series, or... what? – Brenin Nov 24 '13 at 22:37
  • Thanks for asking. In this case, $d$ is the degree of the curves that form the linear system. That is the curves cutting out the curve of degree $n$. – user108555 Nov 25 '13 at 00:46
  • So if you speak about curves of degree $d$, the linear series is on $\mathbb P^2$ (not on the curve). I still do not understand how this is related to a curve of degree $n$. Is it a curve that you fix in advance? what do you mean by cut out, precisely? – Brenin Nov 25 '13 at 01:19
  • A curve $\mathcal{C}$, of degree $n$, is fixed in advance. Now consider the vector space of curves of degree $d$. Intersecting each curve of degree $d$ with $\mathcal{C}$ will produce a positive divisor of degree $nd$ (Bezout's thm). For instance, if $d=n-3$ the divisor cut out on $\mathcal{C}$ will have degree n(n-3) and it could be a canonical divisor. – user108555 Nov 25 '13 at 09:25
  • My first question: how should we call this set of equivalent divisors? Linear system or linear series? Regardless of what it is called, is there any remarkable difference (as linear ser/sys) from the cases $d \leq n−3$ to the cases $d>n−3$? – user108555 Nov 25 '13 at 09:26
  • "Linear system" and "linear series" are, to the best of my knowledge, completely interchangeable. For the second question, what kind of "remarkable difference" do you have in mind? –  Nov 25 '13 at 09:46
  • The obvious property is that if $d>n-3$ then all these divisors on $\mathcal C$ will be nonspecial, i.e. they have $h^1=0$. – Brenin Nov 25 '13 at 10:48

1 Answers1

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First question:

The terms linear system and linear series are completely interchangeable, it's just a matter of taste. Here's the definition of a linear system of divisors:

Def: A divisor $D$ is linearly equivalent to $D'$ if there exist a globally defined rational function $f:C\to k$ such that $D+(f) = D'$.

Def: A divisor $D$ is effective if the order of every point is non-negative.

Def: Given a divisor $D$ on a curve $C$, the complete linear system (or complete linear series) $|D|$ associated to $D$ is the set of all effective divisors on $C$ which are linearly equivalent to $D$.

Def: A linear system (or linear series) is a linear subspace of a complete linear system.

Second question:

Recall that the Riemann-Roch theorem for smooth curves states that $$ h^0(D) - h^0(K-D) = \deg(D) - g + 1, $$ where:

  • $g$ is the genus of the curve $C$ which, by the genus-degree formula for smooth plane curves, is given by $$ g = \frac{(n-2)(n-1)}{2} $$

  • $h^0(D)$ denotes the dimension of the linear series $|D|$ as a vector space over the ground field $k$

  • $K$ is any canonical divisor of $C$ and $h^0(K-D)$ denotes the dimension of the linear series $|K-D|$ as a vector space over the ground field $k$

Further, recall that as soon as $\deg(E)<0$ we have $h^0(E) = 0$, i.e. the linear series $|E|$ consists of $E$ only.

Now, since the degree of a canonical divisor $K$ is given by (to see this just plug $D=0$ in the Riemann-Roch formula above) $$ \deg(K) = 2g-2 = n\cdot(n-3), $$ we deduce that, if $D$ is the divisor of degree $n\cdot d$ consisting of the points of intersection between $C$ and another plane curve of degree $d$, we have $$ d > n-3 \implies \deg(D)>\deg(K) \implies h^0(K-D) = 0. $$ Therefore in the case $d > n-3$ the dimension of the linear series $|D|$ can be easily computed using the Riemann-Roch formula: in this case indeed we have $$ h^0(D) = n\cdot d - g + 1 = \frac{n\cdot d - (n-2)(n-1) + 2}{2} = \frac{n\cdot (2d -n+ 3)}{2}. $$ On the other hand, if $d \leq n-3$ the dimension of $|D|$ is harder to compute, because of the tricky term $h^0(K-D)$ appearing in the Riemann-Roch formula.

Boris Bukh
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Abramo
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  • Thanks for the comment. Any clue about the second question? – user108555 Nov 26 '13 at 13:38
  • @user108555: Please let me know if everything is clear and if you have further questions – Abramo Nov 26 '13 at 18:12
  • In your second efinition you must replace "...is the set of all divisors on $C$ which are linearly equivalent to $D$" by "...is the set of all effective divisors on $C$ which are linearly equivalent to $D$" – Georges Elencwajg Feb 03 '19 at 19:37