1
$\begingroup$

Hi I'm reviewing Riemann Surfaces and this is a question from an old text of exam. We did until Riemann-Roch Therorem. I can't understand which space $L$ is, therefore the real meaning of this answer. I hope there's someone who can explain what's the text asking; I'd really appreciate it. Thank you.

Let $X = \mathbb{P}^1(\mathbb{C})$, $\quad $ $p_1 = [0,1] = 0 \quad p_2 = [1,0] = \infty$

(1) Compute the dimension of $L(p_1 + p_2)$.

(2) Write a base for $L(p_1 + p_2)$.

$\endgroup$
  • $\begingroup$ Where are you getting your notation from? If it is from notes then you'll have to ask your teacher, unless that's a standard notation. Otherwise find out what text books is your course based on. $\endgroup$ – DonAntonio Feb 21 '17 at 21:57
4
$\begingroup$

The vector subspace $L(p_1+p_2)\subset Rat (\mathbb P^1)=\mathbb C(z)$ consists of those rational functions which have at worst simple poles at $p_1$ and $p_2$, so that $$L(p_1+p_2)=\mathbb C\oplus \mathbb Cz\oplus \mathbb C \frac 1z$$ Thus $L(p_1+p_2)$ is a complex vector space of dimension $3$ with basis $\{1,z,\frac 1z\}$.
(You definitely don't need Riemann-Roch, a difficult theorem, to solve this question.)

$\endgroup$
  • $\begingroup$ Thank you very much for your kind answer, I thought of $L(D)$ as the space of meromorphic functions with poles limited by a divisor $D$. By the way, do you know where I can find something about the space you mentioned ? (In order to solve exercise like this, because I didn't quite understand from where you take $1,z $ and $\frac{1}{z}$. $\endgroup$ – lor123 Feb 21 '17 at 22:27
  • 1
    $\begingroup$ Dear lor123, I'm glad I could help you. An excellent reference is Miranda's Algebraic Curves and Riemann Surfaces. As to your exercise, the rational functions having only a pole of order at most $1$ in $\mathbb C$ are of the form $\frac {P(z)}{z}$ where $P(z)$ is a polynomial. And if you want moreover this rational function to have at most a pole of order $1$ at infinity then $P(z)$ has to be of degree at most $2$, so that finally your rational function must have the form $\frac {a+bz+cz^2}{z}$ . $\endgroup$ – Georges Elencwajg Feb 21 '17 at 23:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.