I am struggling with the following exercise, which I have to do without Riemann-Roch:
Show that the dimensions of $H^0(\mathbb{P}^1, D)$ and $H^1(\mathbb{P}^1, D)$ depend only on the degree of D. Give formulas for these dimensions, and verify that the genus of $\mathbb{P}^1$ is $0$. (Hint: compute $H^0(\mathbb{P}^1, n\infty)$ and $H^1(\mathbb{P}^1, n\infty)$ using the standard open cover).
Definitions
- $D$ is a divisor on the projective line $\mathbb{P}^1$.
- $H^0(\mathbb{P}^1, D)$ and $H^1(\mathbb{P}^1, D)$ are the sheaf cohomologies for $\mathcal{O}_{\mathbb{P}^1}(D)$.
Attempt
I attempted the case for $H^0$ as follows.
First, the case given by the hint: if $D = n\infty$, then $\mathcal{O}_{\mathbb{P}^1}(D)(\mathbb{P}^1)$ consists of the elements $[(U, f)]$ in the function field such that $v_\infty(f) \ge -n$ and $v_P(f) \ge 0$ for all $P \ne \infty$. This latter condition implies that $f$ can be represented on a possibly smaller open $V$ by a polynomial in one variable (representing the coordinate of the affine line $\mathbb{A}^1 \subset \mathbb{P}^1$, by abuse of notation): $[(U, f)] = [(V, g)]$ where $g = \prod_{i=1}^r (x - a_i)^{e_i}$ with $e_i \ge 1$. Using a previous exercise, the former condition says that $-\deg(g) = v_\infty(f) \ge -n$ so $\deg(g) \le n = \deg(D)$. With this description, we see that $\mathcal{O}_{\mathbb{P}^1}(D)(\mathbb{P}^1)$ contains all polynomials of degree at most $n$, which is a space of dimension $n + 1 = \deg(D) + 1$ (one for each monomial) over $k$.
Now I'm not really sure about the general case. So I focus on another special case: $D = mP + n\infty$. By reasoning similar to the above, I get that $\mathcal{O}_{\mathbb{P}^1}(D)(\mathbb{P}^1)$ consists of all rational functions $g/h$ such that $\deg(g) \le \deg(h) + n$ and $h = (x - P)^s$ with $0 \le s \le m$. After looking at Basis for proper rational functions, it seems a $k$-basis for this space can be obtained as follows. First, by long division, we can separate out polynomials with no denominator from proper fractions. The former part will consist of polynomials of degree at most $n$, so it is of dimension $n + 1$. A basis for the latter part would be $\{ \frac{x^i}{(x-P)^s} \}_{1 \le s \le m}^{\ 0 \le i < s}$, which has cardinality $1 + 2 + \cdots + (m-1) = \frac{m(m-1)}{2}$. Combining both parts we get a dimension of $n + \frac{m(m-1)}{2} + 1$, but I don't see how to write this as an expression purely in terms of $\deg(D) = m + n$.
Similarly it seems I could generalize to a dimension of $\sum_P\frac{D(P)(D(P) - 1)}{2} + D(\infty) + 1$ for any effective divisor $D$, but again no clue on how to write this as just in terms of the degree of $D$, or how to generalize this to non-effective divisors.
Also no clue on the $H^1$ part.
The following is a fully elementary solution for $H^0(\mathbb{P}^1, D)$.
We have that $V(D) := H^0(\mathbb{P}^1, D) = \mathcal{O}_{\mathbb{P}^1}(D)(\mathbb{P}^1)$ consists of all rational functions $f$ satisfying $\mathrm{div}(f) + D \ge 0$. In particular, $v_\infty(f) \ge - D(\infty)$ so $\deg(f) \le D(\infty)$. Note that if $\deg(D) < 0$ we have $\deg(f) = D(\infty) < - \sum_{P \ne \infty} D(P)$, but this is a contradiction to the fact that $f$ is a polynomial multiple of $\prod_{P \ne \infty} (x - P)^{-D(P)}$ (because $v_P(f) \ge -D(P)$) and the fact that $\deg$ is a homomorphism. So, from now on we can assume $\deg(D) \ge 0$ and in particular $D(\infty) \ge -\sum_{P \ne \infty} D(P)$.
First we consider the case where $D$ is effective. Then $V(D)$ is the set of rational functions $f$ of the form $$\frac{g}{\prod_{P \ne \infty}(x - P)^{s_P}}$$ where $0 \le s_P \le D(P)$ and whose degree is at most $D(\infty)$. By partial fraction decomposition, we find that a basis for this space is $$ \{x^{D(\infty)}, \ldots, x, 1\} \cup \bigcup_{P \ne \infty} \left\{ \frac{1}{(x - P)^1}, \ldots, \frac{1}{(x - P)^{D(P)}} \right\} $$ so it has dimension $1 + D(\infty) + \sum_{P \ne \infty} D(P) = 1 + \deg(D)$.
To finish the proof here, instead of continuing with the elementary approach below, we could say that if $D$ is not effective then $V(D)$ and $V(D + \mathrm{div}(f))$ have the same dimension, by the theorem relating the divisor class group and the picard group, and $D + \mathrm{div}(f)$ is effective so this dimension is $1 + \deg(D + \mathrm{div}(f)) = 1 + \deg(D)$ since $\deg(\mathrm{div}(f)) = 0$ by the degree theorem.
If instead $D$ is effective except at $D(\infty) < 0$, by the assumption above there exists at least one $P \ne \infty$ such that $D(P) \ge |{D(\infty)}|$. Then $V$ is in bijection with its image under the injective $k$-linear map $f \mapsto (x - P)^{|D(\infty)|} f$ which is (by examining the description we've made above) $V\big(D - D(\infty)\infty + D(\infty)P\big)$, whose dimension we already worked out as $1 + \deg\big(D - D(\infty)\infty + D(\infty)P\big) = 1 + \deg(D)$.
Now suppose $D$ is not effective. Let $Z = \{P \in \mathbb{P}^1 \mid P \ne \infty,\ D(P) < 0\}$, and let $S$ be its complement in $\mathbb{P}^1 \setminus \infty$. Then $V(D)$ consists of rational functions of the form $$\frac{g \cdot \prod_{P \in Z} (x - P)^{|D(P)|}}{\prod_{P \in S}(x - P)^{s_P}}$$ where $0 \le s_P \le D(P)$ and whose degree is at most $D(\infty)$. Let $m = \sum_{P\in Z} |D(P)|$. Then $V(D)$ is the image of $V\big(D - m\infty - \sum_{P\in Z} D(P)P\big)$ under the injective $k$-linear map $$f \mapsto f \cdot \prod_{P \in Z}(x - P)^{|D(P)|} $$ The dimension of the latter, which is effective except possibly at $\infty$, is $1 + \deg\big(D - m\infty - \sum_{P\in Z} D(P)P \big) = 1 + \deg(D)$.
So in all cases we get a dimension of $\max \{0, 1 + \deg D\}$.