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Constructing the subsheaf. On an integral scheme, every nonempty open subset is an irreducible topological space and hence a connected space. A locally constant function from such an open subset to $F$ is constant. Thus the constant sheaf $\underline{F}$ has sections $F$ on every nonempty open subset, and the structure sheaf embeds in it by restriction to the generic point. Empty members of the given open cover may be discarded.

For every open subset $W\subseteq X$, defineThis is a subsheaf of the sheaf of modules $\underline{F}$: its condition is local, and compatible sections glue in $\underline{F}$, with the regularity conditions gluing in ${\mathcal{O}}_X$. On $U_i\cap U_j$, the hypothesis says that $f_j/f_i$ is a section of the sheaf of units of the structure sheaf. Therefore multiplication by it preserves ${\mathcal{O}}_X$, andIt follows that $\mathcal{L}{|}_{U_i}=f_i^{-1}{\mathcal{O}}_{U_i}$: the condition with index $i$ implies every other condition on an overlap, and the converse is part of the definition. Multiplication by $1/f_i$ is an isomorphism ${\mathcal{O}}_{U_i}\to \mathcal{L}{|}_{U_i}$, with inverse multiplication by $f_i$. Hence $\mathcal{L}$ is an invertible sheaf, or equivalently a line bundle, andThese local descriptions also give uniqueness as a subsheaf of $\underline{F}$. This construction is the divisor line bundle for the local rational equations $f_i$.

The tensor product. Refine the two open covers to $W_{ij}=U_i\cap V_j$. The ratiosare sections of the sheaf of units of the structure sheaf on the corresponding overlaps, so the products give a line bundle $\mathcal{M}\subseteq \underline{F}$ by the construction just proved. Multiplication in $F$ defines a sheaf morphismOn each $W_{ij}$ it sends the local generator $(1/f_i)\otimes (1/g_j)$ to the local generator $1/(f_i{g}_j)$. It is thus an isomorphism on an open cover, and hence an isomorphism globally. In particular,

The sign for projective twists. Write homogeneous coordinates $[x_0:\cdots :x_n]$ on projective space ${\mathbb{P}}_{\mathbb{Z}}^n$, and let $U_i=D_{+}(x_i)$ be its standard affine open cover. For any integer $m$, take the nonzero elements of the function fieldTheir ratios are $f_i/f_j=(x_j/x_i{)}^m$, which are sections of the sheaf of units of the structure sheaf on $U_i\cap U_j$, including when $m<0$. The twisting sheaf on projective space $\mathcal{O}(m)$ has local generator $x_i^m$ on $U_i$. Divide a rational section of $\mathcal{O}(m)$ by $x_0^m$ to embed it in the constant sheaf of the function field; the local generator becomes $(x_i/x_0{)}^m=1/f_i$. This identifies the constructed line bundle with $\mathcal{O}(m)$ and fixes the sign explicitly.

For $n\ge 1$, $\mathrm{Pic}({\mathbb{P}}_{\mathbb{Z}}^n)\cong \mathbb{Z}$, with generator $[\mathcal{O}(1)]$; the displayed data gives all its powers, with $\mathcal{O}(m)\otimes \mathcal{O}(r)\cong \mathcal{O}(m+r)$ by the product rule above. If $n=0$, projective space is $\mathrm{Spec}\mathbb{Z}$, its Picard group is zero, and the same formula gives only the trivial line bundle.

For a divisor on an algebraic curve $D$, let $L(D)=H^0(V,\mathcal{O}(D))$ and choose a basis of dimension $r+1$. Regarding the basis as sections of the divisor line bundle, evaluation gives ${\varphi }_D:V\to {\mathbb{P}}^r$. If the linear system of divisors has no base points, at least one section is nonzero at each point, and these homogeneous coordinates define a morphism. If the effective divisor has a fixed part $F$, first divide all sections by their common vanishing factor. The moving divisor $E=D-F$ is base-point-free, has the same section space, and gives the uniquely extended morphism on the smooth curve. Thus effectiveness alone does not make the raw evaluations everywhere nonzero, but cancellation supplies the usual map.

A morphism from a projective integral curve is a finite morphism onto its image precisely when it is nonconstant, equivalently $r\ge 1$: a positive-dimensional fiber of a curve would be the whole curve, while a nonconstant proper map with finite fibers is finite. It is an isomorphism onto its image precisely when its moving complete linear system is very ample. In section-space terms, putting $h=h^0(E)$, the criteria arewhere $Q=P$ is allowed. These express absence of base points, separation of distinct points, and separation of tangent directions. In the base-point-free convention $E=D$.

Now suppose the genus is two. The Riemann-Roch theorem gives $\mathrm{deg}K=2$ and $h^0(K)=2$. There cannot be a nonconstant function with only a single simple pole on a positive-genus smooth projective curve: it would give a degree-one map to ${\mathbb{P}}^1$, an isomorphism. Hence $h^0(P)=1$ for every point $P$. Applying Riemann-Roch theorem to $K-P$ gives $h^0(K-P)=h^0(P)=1$, so the canonical linear system has no base point. The resulting map ${\varphi }_K:V\to {\mathbb{P}}^1$ has degree $\mathrm{deg}K=2$.

Set $D=K+P_1+P_2$, of degree four. The Riemann-Roch theorem gives $h^0(D)=3$ and $h^0(D-P)=2$ for every $P$, since the residual divisors have negative degree. Thus ${\varphi }_D$ is a base-point-free map to ${\mathbb{P}}^2$ with nondegenerate image $C$. Degree of the pullback of a general line givesThe image cannot be a line because the three sections are independent. The only alternatives are a birational quartic or a double cover of a conic. In the latter case identify the smooth conic with ${\mathbb{P}}^1$. Its hyperplane bundle has degree two, so $D\sim 2A$, where $A$ is a degree-two divisor with $h^0(A)\ge 2$. Again Riemann-Roch theorem implies $h^0(K-A)=h^0(A)-1\ge 1$. Since $K-A$ has degree zero, it must be linearly equivalent to zero; hence $A\sim K$. This would force $P_1+P_2\sim K$, contrary to the hypothesis. Therefore the image is a plane quartic and the map is birational. A smooth plane quartic has genus $(4-1)(4-2)/2=3$, while its normalization here has genus two. Consequently $\boxed{{\varphi }_D\text{ is birational onto a singular plane quartic}}$.

The canonical bundle is $K_M={\bigwedge }^m(T^{1,0}M{)}^{\ast }$, the holomorphic line bundle of holomorphic top forms on a complex $m$-manifold. For an embedded complex submanifold $V$ of dimension $n$, the holomorphic normal bundle is the quotientTaking determinants and duals gives adjunction for a smooth submanifold:For a closed smooth hypersurface, $[V]={\mathcal{O}}_M(V)$ is its divisor line bundle, equipped with a holomorphic section whose zero divisor is $V$ with multiplicity one. If $f_{\alpha }$ locally defines $V$, local meromorphic generators $1/f_{\alpha }$ transform by the holomorphic units $f_{\alpha }/f_{\beta }$. The stated normal-bundle relation isEquivalently the conormal is ${\mathcal{O}}_M(-V){|}_V$. The closed-divisor interpretation is needed to define $[V]$; an arbitrary nonclosed local submanifold does not define that global divisor bundle.

For a holomorphic subbundle $F\subset E$, define $D_{F}s=\pi D_{E}s$ on sections of $F$, with $\pi$ acting on the bundle factor of a bundle-valued form. It obeys the Leibniz rule. Since $F$ is holomorphic, $D_E^{0,1}s={\bar{\partial }}_{E}s$ lies in $F$, so $D_F^{0,1}={\bar{\partial }}_F$. For $s,t\in F$, their inner products with the normal components of $D_{E}s,D_{E}t$ vanish; hence projecting preserves metric compatibility. The uniqueness proved for the Chern connection identifiesThis is the projected Chern connection.

Choose the adapted smooth unitary frame and retain Question 2's input-index convention. The connection matrix is skew-Hermitian. Its upper-left block is ${\theta }_1$ for $D_F$; writing the lower-left block as $-A$ forces the upper-right block to be ${\overline{A}}^t$:Holomorphicity of $F$ makes the normal part of $D_E^{0,1}s$ vanish, so $B={\overline{A}}^t$ has type $(1,0)$, and $A$ has type $(0,1)$. The matrix $B$ represents the second fundamental form of a holomorphic subbundle in this input-index convention. These types would appear in the opposite blocks with coefficient-column indices; they must not be swapped while retaining the printed minus-sign Cartan equation.

Take the upper-left block of the Cartan curvature equation with input indices. It isTherefore the curvature formula for a holomorphic subbundle isThe sign and the order of the factors both follow from the displayed block multiplication.

Now let $M$ be a complex torus and $V$ a closed smooth hypersurface. The torus has a translation-invariant flat Kähler metric, and $T^{1,0}M$ and $K_M$ are holomorphically trivial. Give $E=T^{1,0}M{|}_V$ its flat metric and apply the preceding formula to $F=T^{1,0}V$. Since ${\Theta }_E=0$,For a $(1,0)$ tangent vector $X$, the latter coefficient is $\mathrm{tr}(B(X)B(X{)}^{†})\ge 0$, so $i{\Theta }_{K_V}$ is a semipositive real $(1,1)$ form. Thus $K_V$ is a semipositive holomorphic line bundle. This proves the semipositive canonical bundle of a submanifold of a complex torus assertion directly from curvature. By adjunction for a smooth submanifold, $K_V\cong [V]{|}_V$.

Restrict a positive metric on $L$ to $V$. The tensor-product metric onhas curvature $i{\Theta }_{L{|}_V}+ai{\Theta }_{K_V}$, which is positive for every nonnegative integer $a$. HenceThe integer condition is implicit in a tensor power. No assertion of strict positivity of $[V]{|}_V$ is required.

For vanishing, write $E_a=L\otimes [V{]}^{\otimes a}$. The allowed Kodaira vanishing theorem says that on a compact Kähler manifold $X$, $H^i(X,K_X\otimes P)=0$ for $i>0$ when $P$ is a positive holomorphic line bundle. Since $K_M$ is trivial, its application on the torus gives $H^i(M,E_0)=H^i(M,L)=0$ for $i>0$.

For every $a\ge 1$, the divisor section gives the exact sequencewhere $i:V\hookrightarrow M$. Crucially,and $P_{a-1}$ is positive by the proved restriction statement. The induced Kähler metric makes $V$ compact Kähler, so Kodaira vanishing theorem on $V$ gives $H^i(V,E_a{|}_V)=0$ for all $i>0$. In the long exact sequence in sheaf cohomology, the segmentthen has zero outside terms by induction. ThereforeThis proves vanishing for divisor twists on a complex torus using restriction positivity and the divisor exact sequence, rather than presuming ambient positivity of every divisor twist.

Take the hyperplane divisor $D=(t_0=0)$. On $U_i=D_{+}(t_i)$ its local equation is $g_i=t_0/t_i$, a non-zero-divisor; on $U_0$ it is the unit one. Thus it is an effective Cartier divisor. On overlaps, $g_i/g_j=t_j/t_i$ is a unit.

The divisor line bundle ${\mathcal{O}}_X(D)$ is locally generated by $1/g_i=t_i/t_0$. Map this generator to the local generator $t_i$ of the twisting sheaf on projective space ${\mathcal{O}}_X(1)$. On overlaps both generators transform by $t_i/t_j$, so the maps glue and giveFor $n=0$ the hyperplane is empty and this is the zero Cartier divisor; the conclusion still holds.

Let $K$ be the function field of the integral scheme $X$ and choose a nonzero rational section of a line bundle $s\in {\mathcal{L}}_{\eta }$. Choose an open trivializing cover $U_i$ with generators $e_i$ for the invertible sheaf $\mathcal{L}$. On each nonempty $U_i$, write $s=a_i{e}_i$ with $a_i\in K^{\ast }$. If $e_i=u_{ij}{e}_j$ on an overlap, then $a_j=a_i{u}_{ij}$, so $a_i/a_j=u_{ij}^{-1}$ is a regular unit. Consequently the local rational functions $a_i$ define a Cartier divisor $D$.

The divisor line bundle ${\mathcal{O}}_X(D)$ restricts to $a_i^{-1}{\mathcal{O}}_{U_i}$. Define its local isomorphism from $\mathcal{L}$ by $e_i\mapsto a_i^{-1}$. On an overlap,so the maps glue. HenceThe rational section maps to the rational function $1$, which also fixes the sign convention for $D$. Choosing a different nonzero rational section changes $D$ by a principal Cartier divisor, so its linear equivalence of Cartier divisors class is unchanged. No projectivity or Noetherian hypothesis is required.

A Cartier divisor is semiample if some positive multiple is a basepoint-free divisor, equivalently its divisor line bundle has a globally generated positive power. Such a power defines a Kodaira map $f$ with $\mathcal{O}(mD)=f^{\ast }\mathcal{O}(1)$. Semiampleness implies nefness but need not imply ampleness: a fibre divisor of a morphism to a curve is a basic example.

The tangent bundle of a complex torus is holomorphically trivial with a flat Hermitian metric. Apply the curvature formula for a holomorphic subbundle to $TV\subset TM{|}_V$: its determinant dual $K_V$ has semipositive curvature. For a closed smooth hypersurface, adjunction for a smooth submanifold identifies $K_V$ with the restriction of its divisor line bundle. Thus a positive ambient line bundle remains positive on the hypersurface after tensoring by any nonnegative power of the restricted divisor bundle.