Let $X$ be a genus $g$ Riemann surface, $p\in X$. I’m trying to find an accessible proof of the following fact: there is a sequence of integers
$$0 < n_1 < n_2 < \ldots < n_g \leq 2g-1$$
such that there is no meromorphic function on $X$ with a single order $n_j$ pole at $p$.
I’ve been told this is related to “Weierstrass gaps” or the “Weierstrass theorem” (I don’t know anything about these, as I’m currently taking an Introduction to Riemann surfaces course for the first time). I found a Wikipedia article that discusses the problem, but the proof they give is just a sketch and there’s already some steps I don’t understand. For instance, they say that “from the Riemann–Roch theorem the dimension of the spaces $L(kP)$ eventually increments by exactly $1$ as we move to the right”, but I couldn’t deduce this by myself.
Could someone refer me to a proof suitable for a student (with as few prerequisites as possible)?
The fundamental observation is that there exists an $f\in \mathcal{M}(X)$ with exactly an order $n$ pole at $P$ if and only if $L(nP)/L((n-1)P) \ne 0.$ In that case, there exists $f\in L(nP)\setminus L((n-1)P)$ and that function does the trick: because $f\in L(nP)\setminus L((n-1)P)$, it has an order $n$ pole at $P$, but not a pole of order $\le n-1$, else it would lie in $L((n-1)P)$.
Anyway, with this out of the way this turns into a problem of understanding $\ell(nP) := \dim L(nP)$ for varying $n\ge 0$. $\ell(0) = 1$, because this just consists of the constant functions. Riemann-Roch says $$ \ell(D)- \ell(K-D) = \deg D + 1 - g. $$ We know $\deg(K) = 2g-2$ so that when $\deg(D) \ge 2g-1$, $\ell(K-D) = 0$ : this follows from the fact that $K-D$ is of negative degree. Consequently, we have complete understanding of $\ell(D)$ for $\deg D \ge 2g-1$. Indeed, by Riemann-Roch it equals $\deg D + 1 -g$. So, for $n\ge 2g-1$, $\ell(nP) = n + 1 -g$. So, the question you are asking (and the Weierstrass Gap problem) has to do with understanding the sequence of integers $\ell(nP)$ for the "blind spot" of Riemann-Roch. Namely, for $1\le n\le 2g-2$. The next ingredient we need is:
The first equality is by definition. The inequality is proved as follows: if $f,g\in L(nP)$ then expanding using Laurent series at $P$, we can find $c\in \Bbb{C}^*$ such that $f-cg$ has an order $\le n-1$ pole at $P$. Consequently, $f-cg\in L((n-1)P)$. This proves the statement.
So, we understand our sequence of numbers well now. For $n\ge 0$, $\ell(nP)$ is a monotone increasing sequence of positive integers, which increases by at most $1$ at each step. Moreover, $\ell((2g-1)P) =g.$ Therefore, there must be a total of $g-1$ integers in the range $1\le n \le 2g-1$ for which $\ell(nP) - \ell((n-1)P) = 1$. There must then also be $g$ integers in $1\le n \le 2g-1$ such that $\ell(nP) = \ell((n-1)P)$. Putting everything together, we have proven that there are integers $$ 1\le n_1<\cdots<n_g\le 2g-1 $$ such that there is no $f\in \mathcal{M}(X)$ with a pole of order exactly $n_i$ at $P$.