For projective line $\mathbb{P}^1$:

| $d$ | $h^0(\mathbb{P}^1, \mathcal{O}(d))$ | $h^1(\mathbb{P}^1, \mathcal{O}(d))$ | |-----|--------------------------------------|--------------------------------------| | $\cdots$ | $0$ | $\cdots$ | | $-4$ | $0$ | $3$ | | $-3$ | $0$ | $2$ | | $-2$ | $0$ | $1$ | | $-1$ | $0$ | $0$ | | $0$ | $1$ | $0$ | | $1$ | $2$ | $0$ | | $2$ | $3$ | $0$ | | $3$ | $4$ | $0$ | | $4$ | $5$ | $0$ | | $\cdots$ | $\cdots$ | $0$ |

Note the symmetry: $h^1(\mathbb{P}^1, \mathcal{O}(-d-2)) = h^0(\mathbb{P}^1, \mathcal{O}(d))$ for all $d$.

For projective plane $\mathbb{P}^2$:

| $d$ | $h^0(\mathbb{P}^2, \mathcal{O}(d))$ | $h^1(\mathbb{P}^2, \mathcal{O}(d))$ | $h^2(\mathbb{P}^2, \mathcal{O}(d))$ | |-----|--------------------------------------|--------------------------------------|--------------------------------------| | $\cdots$ | $0$ | $0$ | $\cdots$ | | $-5$ | $0$ | $0$ | $6$ | | $-4$ | $0$ | $0$ | $3$ | | $-3$ | $0$ | $0$ | $1$ | | $-2$ | $0$ | $0$ | $0$ | | $-1$ | $0$ | $0$ | $0$ | | $0$ | $1$ | $0$ | $0$ | | $1$ | $3$ | $0$ | $0$ | | $2$ | $6$ | $0$ | $0$ | | $3$ | $10$ | $0$ | $0$ | | $4$ | $15$ | $0$ | $0$ | | $\cdots$ | $\cdots$ | $0$ | $0$ |

The middle cohomology $H^1$ vanishes identically. The symmetry is: $h^2(\mathbb{P}^2, \mathcal{O}(-d-3)) = h^0(\mathbb{P}^2, \mathcal{O}(d))$.

For projective 3-space $\mathbb{P}^3$:

| $d$ | $h^0$ | $h^1$ | $h^2$ | $h^3$ | |-----|-------|-------|-------|-------| | $-6$ | $0$ | $0$ | $0$ | $10$ | | $-5$ | $0$ | $0$ | $0$ | $4$ | | $-4$ | $0$ | $0$ | $0$ | $1$ | | $-3$ | $0$ | $0$ | $0$ | $0$ | | $-2$ | $0$ | $0$ | $0$ | $0$ | | $-1$ | $0$ | $0$ | $0$ | $0$ | | $0$ | $1$ | $0$ | $0$ | $0$ | | $1$ | $4$ | $0$ | $0$ | $0$ | | $2$ | $10$ | $0$ | $0$ | $0$ | | $3$ | $20$ | $0$ | $0$ | $0$ | | $4$ | $35$ | $0$ | $0$ | $0$ |

Both $H^1$ and $H^2$ vanish identically, illustrating the gap phenomenon.

Global sections of $\mathcal{O}(d)$ correspond to homogeneous polynomials of degree $d$ in $n+1$ variables $X_0, \ldots, X_n$. The dimension is: $h^0(\mathbb{P}^n, \mathcal{O}(d)) = \binom{n+d}{n} = \binom{n+d}{d}$

Explicit examples:

• $n=2, d=2$: $h^0 = \binom{4}{2} = 6$. Basis: $\{X_0^2, X_0X_1, X_0X_2, X_1^2, X_1X_2, X_2^2\}$.
• $n=3, d=1$: $h^0 = \binom{4}{1} = 4$. Basis: $\{X_0, X_1, X_2, X_3\}$.
• $n=1, d=3$: $h^0 = \binom{4}{1} = 4$. Basis: $\{X_0^3, X_0^2X_1, X_0X_1^2, X_1^3\}$.

For $d < 0$, there are no nonzero global sections since a homogeneous polynomial of negative degree in positive degree variables is trivial.

For $d < 0$, a global section of $\mathcal{O}(d)$ would be given by a collection of Laurent polynomials on each affine chart $U_i = D_+(X_i)$ that glue compatibly. On $U_i$, sections are of the form $f/X_i^d$ where $f$ is homogeneous of degree $d$.

For $d < 0$, such a section must have poles along the hyperplane $\{X_i = 0\}$ in each chart. However, since $\mathbb{P}^n$ is covered by the $U_i$ and the complement of $U_i$ is the hyperplane $\{X_i = 0\}$, a global section cannot have poles anywhere.

Therefore, $H^0(\mathbb{P}^n, \mathcal{O}(d)) = 0$ for $d < 0$.

By Serre duality: $H^n(\mathbb{P}^n, \mathcal{O}(d)) \cong H^0(\mathbb{P}^n, \mathcal{O}(-d-n-1))^\vee$

For $d \leq -n-1$, we have $-d-n-1 \geq 0$, so: $h^n(\mathbb{P}^n, \mathcal{O}(d)) = h^0(\mathbb{P}^n, \mathcal{O}(-d-n-1)) = \binom{n+(-d-n-1)}{n} = \binom{-d-1}{n}$

Concrete cases:

• $n=1, d=-3$: $h^1(\mathbb{P}^1, \mathcal{O}(-3)) = \binom{3-1}{1} = \binom{2}{1} = 2$.
• $n=2, d=-4$: $h^2(\mathbb{P}^2, \mathcal{O}(-4)) = \binom{4-1}{2} = \binom{3}{2} = 3$.
• $n=3, d=-5$: $h^3(\mathbb{P}^3, \mathcal{O}(-5)) = \binom{5-1}{3} = \binom{4}{3} = 4$.

Consider $\mathbb{P}^3$ and compute $H^1(\mathbb{P}^3, \mathcal{O}(d))$ for various $d$:

For any $d$:

• If $d \geq 0$: By the main theorem, $H^1(\mathbb{P}^3, \mathcal{O}(d)) = 0$.
• If $d < 0$: By Serre duality, $H^1(\mathbb{P}^3, \mathcal{O}(d)) \cong H^2(\mathbb{P}^3, \mathcal{O}(-d-4))^\vee$.
  • For $-d-4 \geq 0$, i.e., $d \leq -4$, we would need $H^2(\mathbb{P}^3, \mathcal{O}(-d-4)) \neq 0$.
  • But $H^2(\mathbb{P}^3, \mathcal{O}(m)) = 0$ for all $m$ by the main theorem.

Therefore, $H^1(\mathbb{P}^3, \mathcal{O}(d)) = 0$ for all $d \in \mathbb{Z}$.

Similarly, $H^2(\mathbb{P}^3, \mathcal{O}(d)) = 0$ for all $d$. The only nonzero cohomology occurs at $i=0$ (for $d \geq 0$) and $i=3$ (for $d \leq -4$).

Consider the standard cover $U_0 = D_+(X_0)$ and $U_1 = D_+(X_1)$ of $\mathbb{P}^1$.

The Čech complex is: $\Gamma(U_0, \mathcal{O}(-2)) \oplus \Gamma(U_1, \mathcal{O}(-2)) \xrightarrow{\delta} \Gamma(U_0 \cap U_1, \mathcal{O}(-2))$

We have:

• $\Gamma(U_0, \mathcal{O}(-2)) = k[X_1/X_0] \cdot X_0^2$ (Laurent monomials $X_1^j/X_0^{j+2}$)
• $\Gamma(U_1, \mathcal{O}(-2)) = k[X_0/X_1] \cdot X_1^2$ (Laurent monomials $X_0^i/X_1^{i+2}$)
• $\Gamma(U_0 \cap U_1, \mathcal{O}(-2)) = k[X_0/X_1, X_1/X_0]$ (all Laurent monomials)

The coboundary map $\delta(s_0, s_1) = s_0|_{U_0 \cap U_1} - s_1|_{U_0 \cap U_1}$.

A 1-cocycle is an element $\phi \in \Gamma(U_0 \cap U_1, \mathcal{O}(-2))$ such that $\phi$ is not in the image of $\delta$.

The image of $\delta$ consists of Laurent polynomials that can be written as a difference of a polynomial in $X_1/X_0$ and a polynomial in $X_0/X_1$ (after multiplying by $X_0^2$ or $X_1^2$).

Consider $\phi = 1/(X_0 X_1)$. In the $U_0$ chart, this is $1/(X_0^2 \cdot (X_1/X_0)) = 1/X_0^2 \cdot X_0/X_1$, which has a pole at $X_0 = 0$. Similarly, it has a pole at $X_1 = 0$ in the $U_1$ chart. Therefore, $\phi$ cannot be written as $\delta(s_0, s_1)$.

Thus $H^1(\mathbb{P}^1, \mathcal{O}(-2)) \cong k$ is generated by the class of $1/(X_0 X_1)$.

For $\mathbb{P}^2$ with cover $U_0, U_1, U_2$, the Čech complex for $\mathcal{O}(d)$ is: $C^0 \xrightarrow{\delta^0} C^1 \xrightarrow{\delta^1} C^2$ where:

• $C^0 = \Gamma(U_0, \mathcal{O}(d)) \oplus \Gamma(U_1, \mathcal{O}(d)) \oplus \Gamma(U_2, \mathcal{O}(d))$
• $C^1 = \Gamma(U_0 \cap U_1, \mathcal{O}(d)) \oplus \Gamma(U_0 \cap U_2, \mathcal{O}(d)) \oplus \Gamma(U_1 \cap U_2, \mathcal{O}(d))$
• $C^2 = \Gamma(U_0 \cap U_1 \cap U_2, \mathcal{O}(d))$

For $d = -3$, we compute $H^2(\mathbb{P}^2, \mathcal{O}(-3))$:

• $C^2 = k[X_0/X_1, X_0/X_2, X_1/X_0, X_1/X_2, X_2/X_0, X_2/X_1]$ (Laurent polynomials)
• Elements of degree $-3$ in $C^2$ that are not in $\text{im}(\delta^1)$ form $H^2$.

The element $1/(X_0 X_1 X_2)$ has poles everywhere and cannot be a coboundary. It generates $H^2(\mathbb{P}^2, \mathcal{O}(-3)) \cong k$.

The canonical sheaf of $\mathbb{P}^1$ is $\omega_{\mathbb{P}^1} = \mathcal{O}(-2)$. Serre duality states: $H^1(\mathbb{P}^1, \mathcal{O}(d)) \cong H^0(\mathbb{P}^1, \mathcal{O}(-d-2))^\vee$

Verifying from the table:

• $h^1(\mathbb{P}^1, \mathcal{O}(-3)) = 2 = h^0(\mathbb{P}^1, \mathcal{O}(1))$
• $h^1(\mathbb{P}^1, \mathcal{O}(-2)) = 1 = h^0(\mathbb{P}^1, \mathcal{O}(0))$
• $h^1(\mathbb{P}^1, \mathcal{O}(-1)) = 0 = h^0(\mathbb{P}^1, \mathcal{O}(-1))$
• $h^1(\mathbb{P}^1, \mathcal{O}(0)) = 0 = h^0(\mathbb{P}^1, \mathcal{O}(-2))$
• $h^1(\mathbb{P}^1, \mathcal{O}(1)) = 0 = h^0(\mathbb{P}^1, \mathcal{O}(-3))$

The duality is perfect.

The canonical sheaf of $\mathbb{P}^2$ is $\omega_{\mathbb{P}^2} = \mathcal{O}(-3)$. Serre duality states: $H^2(\mathbb{P}^2, \mathcal{O}(d)) \cong H^0(\mathbb{P}^2, \mathcal{O}(-d-3))^\vee$

Verifying:

• $h^2(\mathbb{P}^2, \mathcal{O}(-5)) = 6 = h^0(\mathbb{P}^2, \mathcal{O}(2))$
• $h^2(\mathbb{P}^2, \mathcal{O}(-4)) = 3 = h^0(\mathbb{P}^2, \mathcal{O}(1))$
• $h^2(\mathbb{P}^2, \mathcal{O}(-3)) = 1 = h^0(\mathbb{P}^2, \mathcal{O}(0))$
• $h^2(\mathbb{P}^2, \mathcal{O}(-2)) = 0 = h^0(\mathbb{P}^2, \mathcal{O}(-1))$

The middle cohomology satisfies: $H^1(\mathbb{P}^2, \mathcal{O}(d)) \cong H^1(\mathbb{P}^2, \mathcal{O}(-d-3))^\vee$ Since both sides vanish for all $d$, the duality is trivially satisfied.

The Hilbert polynomial of $\mathbb{P}^n$ is defined as: $P(d) = \chi(\mathcal{O}_{\mathbb{P}^n}(d)) = \sum_{i=0}^n (-1)^i h^i(\mathbb{P}^n, \mathcal{O}(d))$

For $d \geq 0$: $P(d) = h^0(\mathbb{P}^n, \mathcal{O}(d)) = \binom{n+d}{n}$

This is a polynomial in $d$ of degree $n$ with leading coefficient $1/n!$.

For $\mathbb{P}^1$: $P(d) = d + 1$ For $\mathbb{P}^2$: $P(d) = \frac{1}{2}(d+1)(d+2) = \frac{d^2 + 3d + 2}{2}$ For $\mathbb{P}^3$: $P(d) = \frac{1}{6}(d+1)(d+2)(d+3)$

For negative $d$:

• If $-n \leq d < 0$: $\chi(\mathcal{O}(d)) = 0$ (all terms vanish)
• If $d < -n$: $\chi(\mathcal{O}(d)) = (-1)^n \binom{-d-1}{n}$

For $\mathbb{P}^1$ and $d = -2$: $\chi(\mathcal{O}(-2)) = h^0 - h^1 = 0 - 1 = -1 = (-1)^1 \binom{2-1}{1} = -1$

The Hilbert series of $S = k[X_0, \ldots, X_n]$ is: $H(t) = \sum_{d=0}^\infty h^0(\mathbb{P}^n, \mathcal{O}(d)) t^d = \sum_{d=0}^\infty \binom{n+d}{n} t^d = \frac{1}{(1-t)^{n+1}}$

For $\mathbb{P}^1$: $H(t) = \frac{1}{(1-t)^2} = 1 + 2t + 3t^2 + 4t^3 + \cdots$ For $\mathbb{P}^2$: $H(t) = \frac{1}{(1-t)^3} = 1 + 3t + 6t^2 + 10t^3 + \cdots$

The coefficients match the dimensions from the tables.

Taking the long exact sequence in cohomology and using $H^i(\mathbb{P}^n, \mathcal{O}(1)) = 0$ for $i > 0$:

For $\mathbb{P}^2$, the Euler sequence gives: $0 \to H^0(\mathbb{P}^2, \mathcal{O}) \to H^0(\mathbb{P}^2, \mathcal{O}(1))^{\oplus 3} \to H^0(\mathbb{P}^2, T_{\mathbb{P}^2}) \to$ $\to H^1(\mathbb{P}^2, \mathcal{O}) \to H^1(\mathbb{P}^2, \mathcal{O}(1))^{\oplus 3} \to H^1(\mathbb{P}^2, T_{\mathbb{P}^2}) \to \cdots$

Since $H^1(\mathbb{P}^2, \mathcal{O}) = 0$ and $H^1(\mathbb{P}^2, \mathcal{O}(1)) = 0$: $H^1(\mathbb{P}^2, T_{\mathbb{P}^2}) = 0$

Computing dimensions:

• $h^0(\mathbb{P}^2, \mathcal{O}) = 1$
• $h^0(\mathbb{P}^2, \mathcal{O}(1)) = 3$
• Thus $h^0(\mathbb{P}^2, T_{\mathbb{P}^2}) = 3 \cdot 3 - 1 = 8$

These are the infinitesimal automorphisms of $\mathbb{P}^2$, corresponding to the $8$-dimensional group $\text{PGL}_3(k)$.

Twisting the Euler sequence by $\mathcal{O}(d)$ gives: $0 \to \mathcal{O}(d) \to \mathcal{O}(d+1)^{\oplus (n+1)} \to T_{\mathbb{P}^n}(d) \to 0$

For $\mathbb{P}^1$ and $d = -1$: $0 \to \mathcal{O}(-1) \to \mathcal{O}^{\oplus 2} \to T_{\mathbb{P}^1}(-1) \to 0$

Taking cohomology: $0 \to H^0(\mathcal{O}(-1)) \to H^0(\mathcal{O})^{\oplus 2} \to H^0(T_{\mathbb{P}^1}(-1)) \to H^1(\mathcal{O}(-1)) \to H^1(\mathcal{O})^{\oplus 2}$

Since $h^0(\mathcal{O}(-1)) = 0$, $h^0(\mathcal{O}) = 1$, and $h^1(\mathcal{O}(-1)) = 0$, $h^1(\mathcal{O}) = 0$: $0 \to k^{\oplus 2} \to H^0(T_{\mathbb{P}^1}(-1)) \to 0$

Thus $h^0(T_{\mathbb{P}^1}(-1)) = 2$.

For the next cohomology: $H^1(\mathcal{O}(-1)) \to H^1(\mathcal{O})^{\oplus 2} \to H^1(T_{\mathbb{P}^1}(-1)) \to H^2(\mathcal{O}(-1))$

All terms are zero, so $H^1(T_{\mathbb{P}^1}(-1)) = 0$.

A key consequence is that for any $d > 0$ and $0 < i < n$: $H^i(\mathbb{P}^n, \mathcal{O}(d)) = 0$

This is a special case of Serre's vanishing theorem. It implies that for sufficiently positive twists, coherent sheaves on projective varieties have vanishing higher cohomology.

For example, on $\mathbb{P}^2$:

• $H^1(\mathbb{P}^2, \mathcal{O}(d)) = 0$ for all $d \geq 0$
• More generally, if $\mathcal{F}$ is coherent on $\mathbb{P}^2$, then $H^1(\mathbb{P}^2, \mathcal{F}(d)) = 0$ for $d \gg 0$

Consider the ideal sheaf sequence of a point $p \in \mathbb{P}^2$: $0 \to \mathcal{I}_p \to \mathcal{O}_{\mathbb{P}^2} \to \mathcal{O}_p \to 0$

Twisting by $\mathcal{O}(d)$ and taking cohomology: $0 \to H^0(\mathcal{I}_p(d)) \to H^0(\mathcal{O}(d)) \to H^0(\mathcal{O}_p(d)) \to H^1(\mathcal{I}_p(d)) \to H^1(\mathcal{O}(d)) \to \cdots$

For $d \geq 1$:

• $H^0(\mathcal{O}(d)) = k[X_0, X_1, X_2]_d$ has dimension $\binom{d+2}{2}$
• $H^0(\mathcal{O}_p(d)) = k$ (evaluating at $p$)
• $H^1(\mathcal{O}(d)) = 0$

Therefore: $h^0(\mathcal{I}_p(d)) = \binom{d+2}{2} - 1$ $H^1(\mathcal{I}_p(d)) = 0$

For $d = 0$:

• $h^0(\mathcal{I}_p) = 0$ (no global functions vanishing at a point)
• From $0 \to 0 \to k \xrightarrow{\sim} k \to H^1(\mathcal{I}_p) \to 0$, we get $H^1(\mathcal{I}_p) = 0$

For $d = -1$:

• $h^0(\mathcal{I}_p(-1)) = h^0(\mathcal{O}(-1)) = 0$
• The sequence $0 \to 0 \to 0 \to H^1(\mathcal{I}_p(-1)) \to 0$ gives $H^1(\mathcal{I}_p(-1)) = 0$

For a curve $C \subset \mathbb{P}^2$ of degree $d$ given by a homogeneous polynomial, the arithmetic genus is: $p_a(C) = \frac{(d-1)(d-2)}{2}$

This can be computed using the exact sequence: $0 \to \mathcal{O}_{\mathbb{P}^2}(-d) \to \mathcal{O}_{\mathbb{P}^2} \to \mathcal{O}_C \to 0$

The arithmetic genus is $p_a(C) = 1 - \chi(\mathcal{O}_C)$. From the long exact sequence: $\chi(\mathcal{O}_C) = \chi(\mathcal{O}_{\mathbb{P}^2}) - \chi(\mathcal{O}_{\mathbb{P}^2}(-d))$

We have:

• $\chi(\mathcal{O}_{\mathbb{P}^2}) = 1$
• $\chi(\mathcal{O}_{\mathbb{P}^2}(-d)) = h^0(\mathcal{O}(-d)) - h^1(\mathcal{O}(-d)) + h^2(\mathcal{O}(-d))$

For $d \geq 1$:

• $h^0(\mathcal{O}(-d)) = 0$
• $h^1(\mathcal{O}(-d)) = 0$
• $h^2(\mathcal{O}(-d)) = \binom{d-1}{2} = \frac{d(d-1)}{2}$ (by Serre duality with $\mathcal{O}(d-3)$)

Wait, we need to be careful. For $d \geq 3$: $h^2(\mathcal{O}_{\mathbb{P}^2}(-d)) = h^0(\mathcal{O}_{\mathbb{P}^2}(d-3)) = \binom{d-3+2}{2} = \binom{d-1}{2}$

Thus: $\chi(\mathcal{O}_C) = 1 - \left(0 - 0 - \binom{d-1}{2}\right) = 1 + \binom{d-1}{2} = 1 + \frac{(d-1)(d-2)}{2}$

Therefore: $p_a(C) = 1 - \chi(\mathcal{O}_C) = 1 - 1 - \frac{(d-1)(d-2)}{2} = -\frac{(d-1)(d-2)}{2}$

Actually, there's a sign issue. Let me recalculate. For $d \geq 3$, $-d \leq -3$, so: $h^2(\mathcal{O}_{\mathbb{P}^2}(-d)) = h^0(\mathcal{O}(d-3))$

For $d = 3$: $h^2(\mathcal{O}(-3)) = h^0(\mathcal{O}(0)) = 1$ For $d = 4$: $h^2(\mathcal{O}(-4)) = h^0(\mathcal{O}(1)) = 3$

So $\chi(\mathcal{O}(-d)) = -h^2(\mathcal{O}(-d))$ for $d \geq 3$.

Thus: $\chi(\mathcal{O}_C) = 1 - \left(-\binom{d-1}{2}\right) = 1 + \binom{d-1}{2}$

And: $p_a(C) = 1 - \chi(\mathcal{O}_C) = -\binom{d-1}{2} = \frac{(d-1)(d-2)}{2}$

For a line ($d=1$): $p_a = 0$. For a conic ($d=2$): $p_a = 0$. For a cubic ($d=3$): $p_a = 1$.