Let $C$ be a smooth projective curve of genus $g$ over $k$. Then $\dim C = 1$ and $\omega_C$ is a line bundle of degree $2g - 2$ (by Riemann-Roch).

For the structure sheaf $\mathcal{O}_C$, Serre duality gives:

and

Since $h^0(C, \mathcal{O}_C) = 1$ (constant functions) and $h^1(C, \mathcal{O}_C) = g$ (by definition of genus), we get $h^0(C, \omega_C) = g$, which is the classical fact that the space of holomorphic differentials on a Riemann surface has dimension equal to the genus.

Let $C$ be a smooth projective curve of genus $g$, and let $\mathcal{L}$ be a line bundle on $C$. Then $\mathcal{L}^\vee = \mathcal{L}^{-1}$ and Serre duality states:

This is often written as:

Combined with Riemann-Roch, which states $h^0(C, \mathcal{L}) - h^1(C, \mathcal{L}) = \deg(\mathcal{L}) + 1 - g$, this gives the complete picture of cohomology dimensions for line bundles on curves.

For a curve $C$ of genus $g$, we have $\omega_C \otimes \omega_C^{-1} \cong \mathcal{O}_C$. Applying Serre duality to $\omega_C$:

Both sides have dimension $g$. The dual pairing

is perfect and plays a central role in the theory of Jacobians and theta functions.

Let $X = \mathbb{P}^n_k$. The canonical sheaf is $\omega_{\mathbb{P}^n} = \mathcal{O}_{\mathbb{P}^n}(-n-1)$.

For the twisted structure sheaf $\mathcal{O}_{\mathbb{P}^n}(m)$, we have:

and

Serre duality gives:

For example, when $i = 0$ and $m \geq 0$:

The left side has dimension $\binom{m+n}{n}$ (homogeneous polynomials of degree $m$), while the right side vanishes since $-m-n-1 < -n-1$. This matches the fact that $H^n(\mathbb{P}^n, \mathcal{O}_{\mathbb{P}^n}(d)) = 0$ for $d > -n-1$.

Consider $H^n(\mathbb{P}^n, \omega_{\mathbb{P}^n}) = H^n(\mathbb{P}^n, \mathcal{O}_{\mathbb{P}^n}(-n-1))$. By Serre duality with $i = n$ and $\mathcal{F} = \omega_{\mathbb{P}^n}$:

The right side is one-dimensional (spanned by constant functions), so $H^n(\mathbb{P}^n, \omega_{\mathbb{P}^n}) \cong k$, confirming that the trace map is an isomorphism.

More generally, for any $m$:

which is $k$ if $m = -n-1$, and zero otherwise.

Let $S$ be a smooth projective surface over $k$. The canonical sheaf $\omega_S$ is a line bundle, often denoted $K_S$ when viewed as a divisor class.

For a line bundle $\mathcal{L}$ on $S$, Serre duality gives:

In particular:

• $H^0(S, \mathcal{L})^\vee \cong H^2(S, \omega_S \otimes \mathcal{L}^{-1})$
• $H^1(S, \mathcal{L})^\vee \cong H^1(S, \omega_S \otimes \mathcal{L}^{-1})$ (self-dual middle cohomology)
• $H^2(S, \mathcal{L})^\vee \cong H^0(S, \omega_S \otimes \mathcal{L}^{-1})$

For $\mathcal{L} = \mathcal{O}_S$, we get:

the geometric genus of the surface.

Let $Q \subset \mathbb{P}^3$ be a smooth quadric surface. Then $Q \cong \mathbb{P}^1 \times \mathbb{P}^1$ and we can compute the canonical sheaf using the adjunction formula:

Actually, $\mathcal{O}_Q(-2) \cong \mathcal{O}_{\mathbb{P}^1 \times \mathbb{P}^1}(-1, -1)$ under the isomorphism $Q \cong \mathbb{P}^1 \times \mathbb{P}^1$.

For $\mathcal{L} = \mathcal{O}_Q(a, b)$, Serre duality gives:

For example:

• $H^0(Q, \mathcal{O}_Q(1, 1))$ has dimension 4 (the linear forms restricted to $Q$)
• $H^2(Q, \mathcal{O}_Q(1, 1))^\vee \cong H^0(Q, \mathcal{O}_Q(-3, -3)) = 0$, which matches

This can be verified using the Künneth formula and the cohomology of $\mathbb{P}^1$.

A K3 surface is a smooth projective surface $S$ with $\omega_S \cong \mathcal{O}_S$ and $H^1(S, \mathcal{O}_S) = 0$.

Serre duality for $\mathcal{O}_S$ gives:

Both are one-dimensional. The pairing

is perfect, with both sides spanned by the constant function 1 and a generator of $H^2(S, \mathcal{O}_S)$ respectively.

For the middle cohomology:

Since this vanishes for K3 surfaces, the self-duality is vacuous. However, for a general surface, this gives a symmetric bilinear form on $H^1(S, \mathcal{O}_S)$.

Let $X$ be a smooth projective variety of dimension $n$. The arithmetic genus is defined by:

where $\chi(\mathcal{O}_X) = \sum_{i=0}^n (-1)^i h^i(X, \mathcal{O}_X)$ is the Euler characteristic.

By Serre duality, $h^i(X, \mathcal{O}_X) = h^{n-i}(X, \omega_X)$. Therefore:

Reindexing with $j = n - i$:

This shows that $\chi(\mathcal{O}_X)$ and $\chi(\omega_X)$ are related by a sign depending on dimension.

The classical Riemann-Roch theorem for a line bundle $\mathcal{L}$ on a curve $C$ of genus $g$ states:

Using Serre duality, $h^1(C, \mathcal{L}) = h^0(C, \omega_C \otimes \mathcal{L}^{-1})$, so we can rewrite this as:

For $\mathcal{L} = \mathcal{O}_C$, this gives $1 - h^0(C, \omega_C) = 1 - g$, hence $h^0(C, \omega_C) = g$.

For $\mathcal{L} = \omega_C$, we get $h^0(C, \omega_C) - h^0(C, \mathcal{O}_C) = (2g - 2) + 1 - g = g - 1$, which gives $h^0(C, \omega_C) = g$ again (since $h^0(C, \mathcal{O}_C) = 1$).

Serre duality thus provides an independent proof of key special cases of Riemann-Roch and relates the dimensions of cohomology groups for dual bundles.

For a smooth projective variety $X$ over $\mathbb{C}$ of dimension $n$, the Hodge numbers $h^{p,q}(X) = h^q(X, \Omega_X^p)$ satisfy:

(complex conjugation symmetry) and

(Serre duality).

For a surface ($n = 2$), this gives a diamond pattern:

• $h^{0,0} = 1$
• $h^{1,0} = h^{0,1}$ and $h^{2,0} = h^{0,2}$ (complex conjugation)
• $h^{0,0} = h^{2,2} = 1$ and $h^{1,0} = h^{1,2}$, $h^{0,1} = h^{2,1}$ (Serre duality)
• $h^{1,1}$ is self-dual

These symmetries severely constrain the possible Hodge diamonds and are fundamental to the classification of surfaces.

Let $f: X \to Y$ be a smooth projective morphism of relative dimension $n$ between smooth varieties. The relative canonical sheaf is $\omega_{X/Y} = \omega_X \otimes f^* \omega_Y^{-1}$.

Grothendieck duality for $f$ gives, for a coherent sheaf $\mathcal{F}$ on $X$:

as sheaves on $Y$ (where the dual is taken sheaf-theoretically).

For example, if $X = C \times Y$ where $C$ is a curve of genus $g$ and $f: C \times Y \to Y$ is projection, then for $\mathcal{F} = \mathcal{O}_C \boxtimes \mathcal{E}$ (where $\mathcal{E}$ is a locally free sheaf on $Y$):

Relative duality then states $R^1 f_* (\mathcal{O}_C \boxtimes \mathcal{E}) \cong (R^0 f_*(\mathcal{O}_C \boxtimes (\mathcal{E}^\vee \otimes \omega_{C/Y})))^\vee$.

Let $X$ be a smooth projective variety of dimension $n$, and let $Y \subset X$ be a smooth closed subvariety of codimension $c$. Consider the ideal sheaf $\mathcal{I}_Y$.

We have an exact sequence:

The dual sequence is:

Since $\mathcal{O}_X^\vee \cong \mathcal{O}_X$, we have $\mathcal{I}_Y^\vee \not\cong \mathcal{I}_Y$ in general.

For Serre duality applied to $\mathcal{I}_Y$:

This is useful for computing cohomology of ideal sheaves. For instance, if $Y$ is a point on a curve $C$, then $\mathcal{I}_Y = \mathcal{O}_C(-Y)$ (a line bundle), and Serre duality gives the classical result for line bundles.

Let $X$ be a smooth projective variety of dimension $n$, and let $p \in X$ be a point. Consider the skyscraper sheaf $k_p$ at $p$ (i.e., $k_p$ has stalk $k$ at $p$ and zero elsewhere).

The dual sheaf $k_p^\vee = \mathcal{H}om(k_p, \mathcal{O}_X)$ is zero since there are no non-zero morphisms from a skyscraper at $p$ to $\mathcal{O}_X$ (as $\mathcal{O}_X$ is torsion-free).

Therefore, $k_p^\vee \otimes \omega_X = 0$, and Serre duality gives:

But we know that $H^0(X, k_p) = k$ (the global sections are just the value at $p$), and all other cohomology groups vanish. This seems contradictory, but the resolution is that Serre duality as stated applies most cleanly to locally free sheaves or more generally to sheaves with nice duality properties.

For skyscraper sheaves, one must use the more sophisticated formulation in terms of $\mathcal{E}xt$ sheaves and local duality.

Kodaira vanishing states that for a smooth projective variety $X$ over $\mathbb{C}$ and an ample line bundle $\mathcal{L}$:

Combined with Serre duality for $\mathcal{F} = \mathcal{L}^{-1}$:

we conclude that when $\mathcal{L}$ is ample:

This means that for an anti-ample line bundle $\mathcal{L}^{-1}$, all cohomology vanishes except possibly in the top degree. This is a powerful consequence of combining Kodaira vanishing with Serre duality.

Let $X = \mathbb{P}^n$ and let $\mathcal{F}$ be a coherent sheaf. We say $\mathcal{F}$ is $m$-regular if:

By Serre duality:

So $\mathcal{F}$ is $m$-regular if and only if:

Equivalently, setting $j = n - i$:

This shows that $m$-regularity of $\mathcal{F}$ is closely related to vanishing properties of $\mathcal{F}^\vee$ twisted by appropriate line bundles, via Serre duality.