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In an adapted smooth unitary frame with input-index connection convention, the Chern connection matrix has blocks $({\theta }_F,B;-{\overline{B}}^t,{\theta }_Q)$, where the second fundamental form of a holomorphic subbundle $B$ has type $(1,0)$. The Cartan curvature equation with input indices gives the displayed formula. If the ambient curvature is zero, ${\Theta }_F(X,\overline{X})=-B(X)B(X{)}^{†}$ is negative semidefinite. Dualizing the determinant line bundle therefore gives semipositive curvature.

On a compact Riemann surface, the degree of a holomorphic vector bundle is the degree of a line bundle of its determinant line bundle. A Hermitian metric on a holomorphic vector bundle and its Chern connection express this degree by the trace-curvature integral. The value is independent of the metric because the normalized trace curvature represents the First Chern class.

The top exterior power of a rank-$r$ vector bundle is its determinant line bundle. A short exact sequence of vector bundles gives a canonical isomorphism of the middle determinant with the tensor product of the other two. Locally, a subbundle frame followed by lifts of a quotient frame defines it; changing lifts only adds off-diagonal blocks. This is the determinant step in adjunction for a smooth submanifold.

For a rank-$k$ smooth vector bundle, take the exterior power ${\Lambda }^r{E}_p$ in each fibre. A local frame gives a trivialization with fibre ${\Lambda }^r{\mathbb{R}}^k$, and a transition matrix $A$ acts by ${\Lambda }^{r}A$. Functoriality preserves the cocycle identities, producing a smooth vector bundle of rank $\left(\genfrac{}{}{0ex}{}{k}{r}\right)$, or the zero bundle when $r>k$. Its top exterior power is the determinant line bundle.

The total space of a real vector bundle $E\to M$ has the exact sequence $0\to {\pi }^{\ast }E\to TE\to {\pi }^{\ast }TM\to 0$. Taking determinant line bundles gives the displayed identity. If $M$ is an orientable smooth manifold, the total space is orientable exactly when $\det E$ is trivial. Sufficiency follows by pullback; necessity follows by restricting the total-space determinant identity to the zero section. For the cotangent bundle, base and fibre coordinate determinants cancel, giving an orientation even when the base is not orientable.

Use homogeneous coordinates $X_0,\dots ,X_n$. Multiplication by these degree-one sections defines the sheaf morphismOn $U_i=D_{+}(X_i)$, write $t_j^{(i)}=X_j/X_i$, with $t_i^{(i)}=1$, and let $E_j^{(i)}$ be the standard vector in component $j$ using the local line bundle frame $X_i^{-1}$ of $\mathcal{O}(-1)$. Then $ϵ(E_j^{(i)})=t_j^{(i)}$, so $ϵ$ is surjective since $ϵ(E_i^{(i)})=1$. Its kernel is a locally free sheaf with basisIndeed any vector in the kernel is uniquely a sum of these vectors, by solving for its $i$-th component.

The Kähler differential sheaf ${\Omega }_{{\mathbf{P}}^n/k}^1$ has basis $d{t}_j^{(i)}$ on $U_i$, since this chart is a polynomial affine scheme. Define the local map $d{t}_j^{(i)}\mapsto F_j^{(i)}$. To check that these isomorphisms glue, on $U_i\cap U_{\ell }$ put $u=t_i^{(\ell )}$ and $v=t_j^{(\ell )}$. ThenThe twisting sheaf on projective space frame changes by $E_j^{(i)}=u^{-1}{E}_j^{(\ell )}$, henceFor $j=\ell$, use $t_{\ell }^{(\ell )}=1$, $d{t}_{\ell }^{(\ell )}=0$ and $F_{\ell }^{(\ell )}=0$; the same formula applies. Thus the differential and kernel frames have identical transition matrices. We obtain the cotangent form of the Euler sequence:This construction works over any field, including positive characteristic.

The Kähler differential sheaf has rank $n$. Taking determinant line bundles in this locally split short exact sequence of sheaves givesThe left side is the tensor product of $n+1$ copies of $\mathcal{O}(-1)$, and $\det (\mathcal{O})=\mathcal{O}$. ConsequentlyIt is the canonical bundle of projective space. The determinant identity can also be seen directly by taking the wedge of the kernel basis followed by a lift of the quotient basis; changing that lift by a kernel vector leaves the wedge unchanged.

For a nonzero holomorphic vector bundle $F$ on the curve, its degree of a holomorphic vector bundle and slope of a holomorphic vector bundle areWe obtain a bound independent of the particular holomorphic subbundle by fixing a Hermitian metric on a holomorphic vector bundle $E$ once and for all. Let $D_E$ be its Chern connection, $F_E=D_E^2$ its Chern curvature, and $\omega$ any positive area form on the compact Riemann surface. WriteHere $K$ is a smooth Hermitian endomorphism of $E$. By compactness, its largest eigenvalue is bounded above by some finite constant $C$ on the whole curve.

Let $F\subseteq E$ be a holomorphic subbundle of positive rank $r$, with the induced metric. Use the smooth orthogonal splitting of a vector subbundle $E=F\oplus F^{\perp }$. In this splitting the connection has block formwhere $D_F$ is the projected Chern connection and $B$ is the second fundamental form of a holomorphic subbundle. The latter has type $(1,0)$: the $(0,1)$ part of $D_E$ preserves the holomorphic subbundle. The adjoint $B^{†}$ includes conjugation of the form factor and has type $(0,1)$. Squaring this block matrix givesThis derives the curvature formula for a holomorphic subbundle in the column-vector convention used here. Locally write $B=bdz$. ThenThe degree of a holomorphic vector bundle is the trace-curvature integral, since the trace is the curvature of the determinant line bundle. HenceThe last integrand is a nonnegative real two-form and is independent of the local unitary frame and coordinate. Since $K\le C{I}_E$, the compressed trace is at most $rC$ pointwise. ThereforeThis is the subbundle slope bound from ambient Chern curvature. The right side depends only on $E$ and the initially chosen metrics, and is finite. It proves the requested uniform upper bound for every positive-rank subbundle, without assuming any stability property of $E$ or using a bound that depends on $F$.

If a vector bundle $E\to M$ has rank $k$, form the exterior power of a vector bundle fibrewise: its fibre at $p$ is ${\Lambda }^r{E}_p$. A vector bundle trivialization over $U$ induces one with fibre ${\Lambda }^r{\mathbb{R}}^k$. Transition matrices $A_{\alpha \beta }(p)$ induce ${\Lambda }^r{A}_{\alpha \beta }(p)$, which are smooth and obey the cocycle law because exterior powers preserve composition. They glue to a smooth bundle of rank $\left(\genfrac{}{}{0ex}{}{k}{r}\right)$; for $r>k$ it is the zero bundle. At $r=k$ it is the determinant line bundle.

A real line bundle is trivial precisely when it has a nowhere-zero smooth section: a section $s$ gives the bundle isomorphism $(p,t)\mapsto ts(p)$ from $M\times \mathbb{R}$. Apply this to ${\Lambda }^n{T}^{\ast }M$. Suppose first that an oriented atlas is given. Its local coordinate formsare positive multiples of one another on every overlap. Choose a subordinate smooth partition of unity ${\rho }_{\alpha }\ge 0$. Extend ${\rho }_{\alpha }{\nu }_{\alpha }$ by zero outside its chart and set $\nu ={\sum }_{\alpha }{\rho }_{\alpha }{\nu }_{\alpha }$. Local finiteness makes this smooth. At any point at least one weight is positive, and all summands are nonnegative multiples of the same positive top form. Therefore $\nu$ never vanishes, proving triviality of the line bundle.

Conversely let $\nu$ be a nowhere-zero section of ${\Lambda }^n{T}^{\ast }M$. In each sufficiently small connected chart write $\nu =a_{\alpha }d{x}_{\alpha }^1\wedge \cdots \wedge d{x}_{\alpha }^n$. Its coefficient has constant nonzero sign. If necessary reverse the first coordinate, making $a_{\alpha }>0$. On overlaps the equality of the two expressions givesso each transition determinant is positive. These charts cover $M$, provingFor dimension zero both assertions hold with the empty exterior product equal to one.

For the orientability of a vector bundle total space, write $\pi :E\to M$, with rank $r$ and orientable base. The vertical tangent vectors at $v\in E_p$ identify with $E_p$ by differentiating translation in that fibre. The differential of $\pi$ therefore gives the exact sequence of vector bundles on $E$Taking determinant line bundles givesTo see the determinant identification directly, wedge a vertical frame with any lifts of a base frame; altering a lift by a vertical vector does not change that top wedge. Since $M$ is orientable, $\det TM$ is trivial. Consequently $\det TE\cong {\pi }^{\ast }\det E$. If $\det E$ is trivial, so is its pullback, making the total space orientable. Conversely, if $E$ is orientable as a manifold, its tangent determinant is trivial; restricting this isomorphism to the zero section gives triviality of $\det E$. The tangent and cotangent determinant lines are dual, so either tests orientability. We obtain

For a cotangent bundle the base need not be orientable. If base coordinates change from $x$ to $y=y(x)$ with Jacobian $J$, fibre coordinates of the same covector change from $\xi$ to $\eta =J^{-T}\xi$. The full coordinate change on $T^{\ast }M$ has block JacobianThus every induced cotangent coordinate change preserves orientation, even when the base change reverses it. These charts form an oriented atlas on the total space, proving every cotangent bundle is orientable. Equivalently the canonical symplectic form of the cotangent bundle has a nowhere-zero top exterior power, as in cotangent bundle orientation.

The holomorphic normal bundle is the quotient $N_V=T^{1,0}M{|}_V/T^{1,0}V$, of rank $r$. The tangent sequence is a short exact sequence of holomorphic vector bundles:Taking determinant line bundles gives $\det (T^{1,0}M){|}_V\cong \det (T^{1,0}V)\otimes \det N_V$. Dualizing and rearranging proves adjunction for a smooth submanifold:The determinant identity follows locally by adjoining lifts of a quotient frame to a subbundle frame; changing the lifts adds only off-diagonal blocks, so does not change the determinant.

For a smooth divisor, choose reduced local defining functions $f_i$, with $f_i=u_{ij}{f}_j$ and $u_{ij}$ holomorphic and nowhere zero. The holomorphic line bundle associated to a divisor $[V]={\mathcal{O}}_M(V)$ has local frames $e_i=f_i^{-1}$, with $e_j=u_{ij}{e}_i$. The products $f_i{e}_i$ define its canonical section vanishing on $V$. Along $V$,Thus $v\mapsto d{f}_i(v)e_i$ patches to a map $T^{1,0}M{|}_V\to [V]{|}_V$, whose kernel is $T^{1,0}V$ and which is surjective because each reduced defining function has nonzero differential there. Hence $N_V\cong [V]{|}_V$. This is the normal bundle of a smooth analytic hypersurface. The divisor construction uses a closed hypersurface: an arbitrary nonclosed embedded hypersurface need not be a divisor. For example, a punctured line in ${\mathbb{C}}^2$ cannot be the support of a divisor, since its missing limit point would also lie in every local holomorphic zero set containing that line.

Let $s$ be the section of $E$ defining $V$. The intended hypothesis is that it is a regular zero locus: $ds$ has rank $r$ on $V$. In a local trivialization, differentiation of the component functions definesThis is intrinsic: differentiating a change-of-frame matrix introduces terms multiplied by $s$, which vanish on $V$. Its kernel is $T^{1,0}V$, so it induces an isomorphism of holomorphic normal bundles. ConsequentlyThis is the normal bundle of a regular zero locus. Here $\det E=L_1\otimes \cdots \otimes L_r$.

Regularity is essential if “vanishing” is read only as equality of underlying sets. On ${\mathbb{P}}^2$, the section $Z_0^2$ of $\mathcal{O}(2)$ vanishes set-theoretically on a smooth line $V$. Its derivative vanishes on $V$, andThus smoothness of the underlying zero set alone does not establish the claimed normal-bundle isomorphism. The preceding proof applies when the equations define the reduced smooth submanifold, equivalently are transverse to zero.

For Complex projective space, put $\mathcal{O}(a)=[aH]$. At a line $\ell \subset {\mathbb{C}}^{n+1}$, the tangent space is $\mathrm{Hom}(\ell ,{\mathbb{C}}^{n+1}/\ell )$. Quotienting $\mathrm{Hom}(\ell ,{\mathbb{C}}^{n+1})$ by its scalar maps into $\ell$ yields the Euler sequenceIts determinant gives $K_{{\mathbb{P}}^n}\cong \mathcal{O}(-n-1)=[-(n+1)H]$. For a smooth complete intersection defined regularly by the homogeneous equations, apply the normal-bundle result to $E={⨁}_i\mathcal{O}(d_i)$:This is the canonical bundle of a regular projective complete intersection. The same regularity qualification is needed here: the squared-line example would otherwise predict ${\mathcal{O}}_{{\mathbb{P}}^1}(-1)$ for its canonical bundle, whereas the line has $K_V={\mathcal{O}}_{{\mathbb{P}}^1}(-2)$.

A Spin-c structure on an oriented rank-$n$ real vector bundle is a lift of its oriented frame bundle to ${\mathrm{Spin}}^c(n)=(\mathrm{Spin}(n)\times U(1))/\{(1,1),(-1,-1)\}$. Its determinant line bundle is associated with $[s,z]\mapsto z^2$; its First Chern class reduces modulo two to the second Stiefel–Whitney class. When such structures exist, their isomorphism classes form an affine space over $H^2$; tensoring by a complex line bundle of first Chern class $h$ changes the determinant class by $2h$. On a closed oriented three-manifold they always exist. On a four-manifold, the determinant class is characteristic: $⟨c_1(\mathfrak{s}),v⟩\equiv v\cdot v(\mathrm{mod}2)$.