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 $L(p_1 + p_2)$.

  • $\begingroup$ 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

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.)

  • $\begingroup$ 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
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    $\begingroup$ 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

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