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Write ${\Omega }_C^1={\Omega }_{C/k}^1$. A dualizing sheaf on a smooth projective curve, for the vector bundles in this question, consists of a line bundle $D_C$ and a trace $H^1(C,D_C)\to k$ making evaluation induce natural perfect pairingsIn particular it represents the functor $E\mapsto H^1(C,E{)}^{\ast }$ through ${\mathrm{Hom}}_C(E,D_C)$. We construct it by reducing to the proved residue duality on the projective line.

Choose a separating nonconstant rational function $t$ on $C$. Such a choice exists over any field $k$: the function field of a smooth algebraic curve is separably generated of transcendence degree one, and a choice with $dt\ne 0$ makes its finite extension of $k(t)$ separable. The map $(t:1)$ extends across poles using $(1:t^{-1})$ in the discrete valuation ring at each point. It gives a finite generically separable morphismIndeed it is nonconstant, hence has finite fibres, and is proper; a proper quasi-finite morphism is finite. This finite map is flat: locally its direct-image module is finite and torsion-free over a discrete valuation ring, and hence free. The same argument works componentwise if needed.

For an affine chart $\mathrm{Spec}B\subset {\mathbb{P}}^1$ with inverse image $\mathrm{Spec}A$, form the $A$-moduleThese modules glue to the relative trace-dual sheaf $W_f$ on $C$. There are two algebraic identities used here. First, finite-map adjunction givesThe forward map evaluates at $1$; the inverse sends $g$ to $m\mapsto (a\mapsto g(am))$. Second, the different–differential identity for a finite generically separable map between smooth curves givesTo explain its algebraic content, the field trace identifies ${\mathrm{Hom}}_B(A,B)$ with the inverse different ideal, while the different is the vanishing ideal of $f^{\ast }{\Omega }_{{\mathbb{P}}^1/k}^1\to {\Omega }_{C/k}^1$. These identities include wild ramification; one must not replace the different by just ramification index minus one. This is the identity allowed to be quoted in the question; the finite-map description is recorded in Stacks Project, finite morphisms, and its differential form in Stacks Project, Section 53.12.

Set $D_C=W_f\otimes f^{\ast }{\Omega }_{{\mathbb{P}}^1/k}^1$. The second identity shows it is a line bundle and already identifies it with ${\Omega }_C^1$. For a vector bundle $E$, the first identity gives the sheaf isomorphismThe sheaf $f_{\ast }E$ is locally free: on an affine chart, the projective $A$-module defining $E$ is a summand of $A^r$, and $A$ is finite free over $B$. Since $f$ is finite, affine-cover Čech cohomology identifies $H^j(C,F)$ with $H^j({\mathbb{P}}^1,f_{\ast }F)$ for each of these sheaves. Apply the already established duality on ${\mathbb{P}}^1$ to $f_{\ast }E$. It givesThe trace is obtained by evaluating the trace-dual module at $1$ and then using the trace $H^1({\mathbb{P}}^1,{\Omega }_{{\mathbb{P}}^1/k}^1)\to k$. The adjunction formulas show the displayed isomorphisms are induced by evaluation against this trace and are natural in $E$. They thus establish the requested dualizing property, not merely the existence of an invertible differential sheaf. Uniqueness follows from the Yoneda lemma applied to the representing functor on vector bundles.

Use the finite map $f:C\to {\mathbb{P}}^1$ from the preceding solution. The inverse images of the two standard affine charts and of their intersection are affine, because $f$ is finite. The vanishing of quasi-coherent cohomology on an affine scheme therefore lets their two-term Čech cochain complex compute $H^i(C,E)$. It has terms only in degrees zero and one, provingThis actually applies to every quasi-coherent sheaf on $C$.

For a smooth projective curve with $H^0(C,{\mathcal{O}}_C)=k$, define its genus of a smooth projective curve byThe preceding Serre duality also gives $g={\dim }_k{H}^0(C,{\Omega }_C^1)$. The condition on constants holds in particular for a geometrically connected smooth projective curve. It makes $g$ equal to the arithmetic genus $1-\chi ({\mathcal{O}}_C)$. We state the theorem with this usual convention and then give the formula valid without that condition.

For a divisor on an algebraic curve $D$ and a canonical divisor $K$ with $\mathcal{O}(K)\cong {\Omega }_C^1$, the Riemann-Roch theorem isThe degree over an arbitrary field is weighted by residue degrees: ${\mathrm{deg}}_{k}D={\sum }_p{n}_p[k(p):k]$ when $D={\sum }_p{n}_{p}p$. Equivalently, for a line bundle $L$,We now prove it via Riemann-Roch via elementary modifications.

For any closed point $p$ and any line bundle $L$, the exact sequencehas a skyscraper sheaf as its quotient, with $k$-dimension $[k(p):k]$. This follows locally from a uniformizer of the discrete valuation ring at $p$: a local line-bundle generator modulo that uniformizer spans one copy of $k(p)$. The quotient has zero higher cohomology, as it is the direct image from a finite affine scheme. The long exact sequence in sheaf cohomology and the proved higher vanishing giveEvery line bundle has a nonzero rational section of a line bundle, whose local orders give a divisor on an algebraic curve $D$ with $L\cong \mathcal{O}(D)$. Iterating the displayed identity, adding or subtracting closed points with their multiplicities, givesFor the constants convention above, $\chi ({\mathcal{O}}_C)=1-g$. The dualizing sheaf on a smooth projective curve gives $h^1(\mathcal{O}(D))=h^0(\mathcal{O}(K-D))$, which converts this Euler-characteristic identity into the stated Riemann-Roch theorem. This proves the formula over arbitrary $k$, without assuming the closed points are $k$-rational.

For vector bundles, the corresponding formula isTo see it, saturate a rational rank-one subspace to a line subbundle and induct on rank. Both Euler characteristic of a coherent sheaf and the degree of the determinant are additive in the resulting short exact sequence. The line-bundle formula just proved starts the induction.

Taking $L={\Omega }_C^1$ and using duality gives $h^0({\Omega }_C^1)=g$, $h^1({\Omega }_C^1)=1$, and hence ${\mathrm{deg}}_k{\Omega }_C^1=2g-2$. In particular a line bundle of degree greater than $2g-2$ has zero first cohomology: its dual twisted by ${\Omega }_C^1$ has negative degree, and a negative-degree line bundle has no nonzero section, since any section would have an effective divisor of zeros.

If the term “curve over $k$” is used without requiring $H^0({\mathcal{O}}_C)=k$, the unconditional formula we proved isWriting $g=h^1({\mathcal{O}}_C)$ then replaces $1-g$ by $h^0({\mathcal{O}}_C)-g$. For example, ${\mathbb{P}}_{k^{\prime }}^1$ viewed as a $k$-curve for a finite separable extension $k^{\prime }/k$ has $h^0({\mathcal{O}}_C)=[k^{\prime }:k]$, so a literal constant term $1$ would be incorrect. Alternatively the arithmetic genus $p_a=1-\chi ({\mathcal{O}}_C)$ gives the formula with constant $1-p_a$. The constants hypothesis for the usual genus is recorded in Stacks Project, genus of a curve.

Here is the quadratic-form criterion for an Anosov flow. Assume that the generator $F$ is nowhere zero and form the transverse vector bundleSince $D{\varphi }_{t}F(x)=F({\varphi }_{t}x)$, the derivative induces a linear cocycle $P_t:{\mathcal{E}}_x\to {\mathcal{E}}_{{\varphi }_{t}x}$. Suppose there is a continuous nondegenerate indefinite quadratic form $Q_x$ on each fibre, with constant signature of a quadratic form, such that $t\mapsto Q_{{\varphi }_{t}x}(P_t\xi )$ is differentiable andis continuous and positive definite on every transverse fibre. Then ${\varphi }_t$ is an Anosov flow. An equivalent way to specify $Q$ is a quadratic form on $TN$ whose radical is exactly $\mathbb{R}F$ and whose induced form and derivative satisfy these conditions on the quotient.

On the closed manifold, compactness makes the positivity uniform: for some $c>0$, $\mathcal{D}{Q}_x(\xi )\ge c\Vert \xi {\Vert }^2$. Thus the positive and negative cones are strictly separated by the linearized flow, yielding uniform unstable and stable transverse subbundles and their invariant lifts. The positive index is the unstable dimension and the negative index is the stable dimension. Strict positivity must be imposed transversely; imposing it on the flow direction would be impossible because that direction is carried into itself.

A real rank-$d$ vector bundle $p:E\to M$ is a continuous family of $d$-dimensional real vector spaces, locally trivial by homeomorphisms $p^{-1}(U)\cong U\times {\mathbb{R}}^d$ which are linear on every fiber. Transition matrices are continuous maps into ${\mathrm{GL}}_d(\mathbb{R})$. The Grassmannian ${\mathrm{Gr}}_k({\mathbb{R}}^n)$ is the space of $k$-dimensional linear subspaces; its topology can be defined by identifying each plane $W$ with its orthogonal projection matrix $P_W$. Local graph charts identify nearby planes with graphs of maps $W_0\to W_0^{\perp }$.

The tautological bundle is $E_{\mathrm{taut}}=\{(W,v):v\in W\}\to {\mathrm{Gr}}_k({\mathbb{R}}^n)$. In graph charts a basis of $W_0$ gives a continuously varying basis of each graph, proving local triviality. The universal quotient bundle has fiber $Q_W={\mathbb{R}}^n/W$. The map $[v]\mapsto (I-P_W)v$ identifies it continuously with the orthogonal-complement bundle $W^{\perp }$, whose graph charts similarly supply local frames. Thus $Q_{\mathrm{taut}}$ is a rank-$(n-k)$ quotient vector bundle.

For the given $E\to M$, at each $m$ choose a basis of $E_m$ and global sections of a vector bundle realizing its basis vectors, using the evaluation-surjectivity assumption. Those $d$ sections remain independent on a neighbourhood of $m$, since the corresponding determinant in a local trivialization is nonzero at $m$ and continuous. Compactness gives a finite cover by such neighbourhoods. Collect all the selected sections as $s_1,\dots ,s_N$; they span every fiber. They define a fiberwise surjective vector bundle morphismIts kernels have constant dimension $N-d$ and form the kernel bundle of a surjective vector bundle morphism. More concretely, in a local frame of $E$ the map has a continuous full-row-rank matrix $A(m)$ and its kernel projection isThis varies continuously and is unchanged by an invertible change of target frame. The Grassmannian as projection matrices therefore makesa continuous map. The fiber quotient map induced by $F_m$ is an isomorphism ${\mathbb{R}}^N/\ker F_m\to E_m$. The same local matrix formula gives continuously varying right inverses, so these fiber isomorphisms give a vector bundle isomorphism. This proves the classification of vector bundles by a universal quotient bundle:The displayed equality in the classification statement is understood as bundle isomorphism, not literal equality of total spaces.

Finally, the universal quotient bundle need not have a nowhere-zero section. Take ${\mathrm{Gr}}_1({\mathbb{R}}^2)={\mathbb{RP}}^1$. Its quotient identifies with $W^{\perp }$, a Möbius line bundle. Write $W_{\theta }=\mathrm{span}(\cos \theta ,\sin \theta )$, with $\theta$ modulo $\pi$. A section of the complement line has the form $a(\theta )(-\sin \theta ,\cos \theta )$, and being single-valued requires $a(\theta +\pi )=-a(\theta )$. If $a$ is nonzero anywhere, the intermediate value theorem forces a zero between that point and its translate by $\pi$; if it is nowhere nonzero the conclusion is immediate. Hence this rank-one quotient bundle has no nowhere-zero section.

For a rank-$r$ real vector bundle with a fiber metric, the Thom isomorphism theorem says that if the bundle is oriented over the coefficient ring $R$, there is a Thom class $u\in H^r(D(E),S(E);R)$ restricting to the chosen generator on every fiber pair, andis an isomorphism. For $R={\mathbb{F}}_2$, every real vector bundle has the required orientation. The disk bundle retracts to $B$, so apply the pair long exact sequence and the Thom isomorphism to get the Gysin sequence of a sphere bundleThe first map is multiplication by the Euler class of a vector bundle, the pullback to the zero section of the absolute image of $u$. The connecting map followed by inverse Thom isomorphism gives ${\pi }_{!}$.

Apply this over ${\mathbb{F}}_2$ to the tautological real line bundle $\gamma$ on ${\mathbb{RP}}^n$. Its unit sphere bundle is $S^n$, with projection sending a unit vector to its line. Put $a=e_2(\gamma )=w_1(\gamma )\in H^1({\mathbb{RP}}^n;{\mathbb{F}}_2)$, the mod-two Euler class of a real line bundle. For $n\ge 1$ the degree-zero pullback $H^0({\mathbb{RP}}^n)\to H^0(S^n)$ is an isomorphism. Exactness therefore makes the subsequent transfer zero, and multiplication by $a$ from $H^0$ into $H^1$ injective. In intermediate degrees the sphere cohomology vanishes, making each successive multiplication by $a$ an isomorphism. At degree $n$, the map from $H^{n-1}$ into $H^n$ is injective; the transfer $H^n(S^n)\to H^n({\mathbb{RP}}^n)$ is surjective because $H^{n+1}({\mathbb{RP}}^n)=0$. Since $H^n(S^n)={\mathbb{F}}_2$, that top group is one-dimensional too, and $a^n$ is its nonzero generator. The same endpoint reasoning covers $n=1$. This gives the Gysin proof of mod-two projective-space cohomology and the mod-two cohomology ring of real projective space:For $n=0$ this says simply $H^{\ast }({\mathbb{RP}}^0)={\mathbb{F}}_2$.

For a self-map of ${\mathbb{RP}}^4$, the induced map has $f^{\ast }a=\lambda a$ with $\lambda \in {\mathbb{F}}_2$, so $f^{\ast }{a}^j={\lambda }^j{a}^j$. The mod-two alternating trace isfor either value of $\lambda$. The alternating trace on homology equals that on the finite cellular chain complex; consequently this is the reduction modulo two of the integral Lefschetz number. That integer is nonzero, and the Lefschetz fixed-point theorem forces a fixed point. Thus every self-map of the real projective four-space has a fixed point, a case of the Lefschetz fixed-point property of even-dimensional real projective space.

For the odd-dimensional example, let $J(x_1,x_2,x_3,x_4)=(-x_2,x_1,-x_4,x_3)$, so $J^2=-I$. Its projectivization isIt is well-defined and continuous on ${\mathbb{RP}}^3$. A fixed projective point would have $Jv=tv$ for a nonzero real vector and real $t$, forcing $-v=J^{2}v=t^{2}v$ and therefore $t^2=-1$, impossible. This is a fixed-point-free complex-structure map on odd-dimensional real projective space.

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.

Work with integral cohomology. For a frame $v=(v_1,\dots ,v_k)$ in the complex Stiefel manifold, defineThese fibres form the complement bundle on a complex Stiefel manifold, a complex vector subbundle of the trivial rank-$n$ bundle. Indeedis a smooth orthogonal projector of constant complex rank $n-k$; locally projecting suitable fixed basis vectors and applying Gram-Schmidt process provides bundle frames. The underlying real vector bundle has rank $r=2(n-k)$ and is oriented by its complex structure: a complex change of frame has real determinant equal to the positive number $|{\det }_{\mathbb{C}}A{|}^2$.

A point of its sphere bundle is $(v_1,\dots ,v_k,w)$ with $\Vert w\Vert =1$ and $w$ orthogonal to the frame. Appending $w$ gives a diffeomorphismwhose inverse forgets the last vector while retaining it as the unit vector in the complement. Thus the asserted homotopy equivalence is actually an identification of sphere bundles, for $0\le k<n$.

Start the induction at $V_0({\mathbb{C}}^n)=\{\mathrm{point}\}$, with cohomology the exterior algebra on no generators. SupposeThis is a free abelian group in every degree, and for $k\ge 1$ its smallest positive degree is $2n-2k+1=r+1$. Hence $H^r(V_k;\mathbb{Z})=0$. This vanishing holds at $k=0$ as well, since the base is a point and $r=2n>0$. The Euler class $e(E)$ belongs to this zero group and is therefore zero.

For this oriented rank-$r$ vector bundle, the Gysin sequence of a sphere bundle has segmentBecause $e(E)=0$, this becomesIn particular $H^{\ast }(V_{k+1};\mathbb{Z})$ is torsion-free. At $j=r-1$, choose a class $a\in H^{r-1}(V_{k+1};\mathbb{Z})$ with ${\pi }_{!}a=1$. Its restriction to the sphere fibre is an integral generator. The projection formula for sphere bundle integration, with the conventional fibre-integration sign, isIt can be obtained from the relative cohomology of the disk/sphere pair and the Thom isomorphism theorem. Consequently every cohomology class of $V_{k+1}$ is uniquely expressible as ${\pi }^{\ast }c+{\pi }^{\ast }b⌣a$: subtract the second term using its fibre integral, then use exactness to identify the remaining term; injectivity of ${\pi }^{\ast }$ and the projection formula prove uniqueness. This establishes an additive splitting by the old monomials and the old monomials multiplied by $a$.

To establish the ring, $|a|=r-1$ is odd, so graded commutativity of the cup product gives $a^2=-a^2$, and hence $2a^2=0$. The torsion-freeness just proved now gives $a^2=0$. Graded commutativity also gives the required anticommutation with each old odd generator, and the old relations persist under the injective ring map ${\pi }^{\ast }$. The additive splitting proves that there are no additional relations. This is exactly Gysin ring splitting for an odd-dimensional sphere bundle, and supplies a new exterior generator of degreeInduction therefore proves the full ring calculationOver $\mathbb{Z}$, graded commutativity by itself only kills twice an odd square; torsion-freeness is the essential extra step in this integral cohomology of a complex Stiefel manifold proof.

Finally an ordered orthonormal $n$-frame is a unitary matrix, so $V_n({\mathbb{C}}^n)=U(n)$. Each exterior generator may be included once or omitted from a basis monomial. The rank of degree $j$ is therefore the number of subsets of the generator degrees $1,3,\dots ,2n-1$ summing to $j$. Multiplication of their two-term generating factors givesThe highest exponent is $1+3+\cdots +(2n-1)=n^2$, agreeing with the real dimension of the unitary group, and the total rank is $2^n$.

Work with real vector bundles; for complex bundles replace the inner products below by positive-definite Hermitian forms. A smooth rank-$r$ vector bundle is a smooth total space $E$ with a smooth projection $\pi :E\to M$, a real vector space structure on each fiber $E_p$, and an open cover with vector bundle trivializationsEach ${\Phi }_i$ is a diffeomorphism over $U_i$ and is linear on fibers. The overlap maps have the form $(p,v)\mapsto (p,A_{ji}(p)v)$, where $A_{ji}:U_i\cap U_j\to G{L}_r(\mathbb{R})$ is smooth. The rank of a vector bundle may be specified separately on each connected component.

A smooth fiber metric is a positive-definite symmetric bilinear form $h_p$ on each fiber, with smoothly varying coefficients in every local frame. Equivalently it is a smooth section of ${\mathrm{Sym}}^2{E}^{\ast }$ which is positive definite at every point. Each trivialization supplies a local fiber metric $h_i$ by pulling back the Euclidean inner product.

A smooth manifold is Hausdorff and second-countable, hence paracompact. Choose a smooth, locally finite partition of unity $\{{\varphi }_i\}$ subordinate to a trivializing cover, with ${\varphi }_i\ge 0$, $\mathrm{supp}{\varphi }_i\subset U_i$, and ${\sum }_i{\varphi }_i=1$. DefineExtend each weighted term by zero outside its trivializing set; its support lies inside that set, so the extension is smooth. Local finiteness makes the sum smooth. At a point $p$ and for $0\ne v\in E_p$, every nonzero term ${\varphi }_i(p)h_i(v,v)$ is positive and at least one such term occurs. Thus $h_p(v,v)>0$. This proves that every smooth vector bundle admits a smooth fiber metric.

Let $r=\mathrm{rank}F$ and $s=\mathrm{rank}E$. Around any $p$, choose a local frame $f_1,\dots ,f_r$ of the vector subbundle $F$. Choose additional smooth sections $e_{r+1},\dots ,e_s$ of $E$ which complement these vectors at $p$. They can be obtained by constant-coordinate sections in a vector bundle trivialization. The determinant of the resulting frame is nonzero at $p$, hence on a sufficiently small neighborhood. Thusis an adapted local frame of $E$ near $p$.

On the fiberwise quotient, the classes $[e_{r+1}],\dots ,[e_s]$ form a basis. Give the quotient vector bundle its local coordinates byAn overlap between adapted frames has a smooth transition function of a vector bundle represented by an invertible block upper triangular matrix,The zero lower-left block expresses that both sets of first $r$ frame vectors span $F$. Its lower-right block $D$ is smooth and invertible. Passing to quotient coordinates removes the upper block, leaving precisely the transition $D$. These lower-right blocks satisfy the cocycle identity because the original transition matrices do. Consequently they define a compatible smooth bundle atlas of rank $s-r$. This proves $E/F$ is a smooth vector bundle, and its fiberwise quotient map $q:E\to E/F$ is a smooth vector bundle morphism. The construction is independent of the adapted frames: any two choices have the same type of compatible transition. The usual quotient topology agrees with this atlas, since in adapted coordinates $q$ is an open linear projection on each fiber.

A connection on a vector bundle is an $\mathbb{R}$-linear mapwhich satisfies $\nabla (fs)=df\otimes s+f\nabla s$ for every smooth function $f$ and section $s$. Here ${\Omega }^1(M;E)=\Gamma (T^{\ast }M\otimes E)$. Equivalently, writing ${\nabla }_{X}s=(\nabla s)(X)$, it is linear over smooth functions in $X$, linear over $\mathbb{R}$ in $s$, and obeysChoose a trivializing open cover and a local frame $e_{i1},\dots ,e_{ir}$ on each $U_i$. Differentiating the coefficient functions gives a local connection on a vector bundle:No compatibility of these local connections is assumed. Choose a smooth locally finite partition of unity $\{{\varphi }_i\}$ subordinate to the cover and defineextending each weighted term by zero outside $U_i$. Its support lies inside $U_i$, so each extension is smooth, and local finiteness makes the whole sum smooth. The operation is linear over smooth functions in $X$. For the Leibniz rule,using ${\sum }_i{\varphi }_i=1$. This proves every smooth vector bundle admits a connection by the construction of a vector bundle connection by a partition of unity.

Throughout the algebraic questions, $C$ is a smooth projective connected algebraic curve over the algebraically closed field $k$. This is the curve convention needed for the Jacobian variety; for an arbitrary singular or nonproper curve the assertions need not hold. Choose $P_0\in C(k)$ and write $\pi :C\times T\to T$.

The degree-$d$ Picard functor of a curve iswith pullback as its action on morphisms. Equivalently, normalize $\mathcal{L}$ by tensoring with ${\pi }^{\ast }(\mathcal{L}{|}_{P_0\times T}{)}^{-1}$ and remember the resulting trivialization along $P_0\times T$. Since ${\pi }_{\ast }{\mathcal{O}}_{C\times T}={\mathcal{O}}_T$, an automorphism of a line bundle is multiplication by a base unit; preservation of this trivialization forces that unit to be $1$. Thus normalized line bundles have effective descent and the displayed functor is already the appropriate sheaf, rather than just a list of classes over fields.

Here is a construction of its representing scheme. The symmetric product of a curve $C^{(r)}$ represents relative effective degree-$r$ Cartier divisors. It is projective, and its completed local ring at $\sum m_i{Q}_i$ is a power-series ring in the elementary symmetric coefficients of $m_i$ local parameters at each $Q_i$. Thus it is smooth of dimension $r$, in every characteristic. PutThis is open by the semicontinuity theorem for coherent cohomology. It is nonempty: choose $g$ points successively imposing independent vanishing conditions on $H^0(C,K_C)$, so that $H^0(C,K_C-D)=0$, and use Serre duality. The Riemann-Roch theorem gives $h^0({\mathcal{O}}_C(D))=1$ on $U$.

For any fixed line bundle $B$ of degree $g-d$, use a copy $U_B$ of $U$ with the universal normalized line bundle $\mathcal{O}(D)\otimes B^{-1}$. This represents exactly the open subfunctor on which $H^1({\mathcal{L}}_t\otimes B)=0$. Indeed, cohomology and base change for line bundles on a curve makes ${\pi }_{\ast }(\mathcal{L}\otimes B)$ a rank-one vector bundle. The evaluation morphismhas a unique relative zero Cartier divisor of degree $g$ and recovers the point of $U_B$. No choice of a global generator on $T$ is needed. Conversely that zero Cartier divisor recovers $\mathcal{L}$ modulo a bundle from $T$.

These charts cover: given any degree-$d$ line bundle $L$, choose a nonspecial effective degree-$g$ Cartier divisor $D$ and take $B=\mathcal{O}(D)\otimes L^{-1}$. On the overlap of $U_B$ and $U_{B^{\prime }}$, the unique effective representative of $\mathcal{O}(D)\otimes B^{-1}\otimes B^{\prime }$ is constructed by the same rank-one evaluation map. It therefore gives a regular transition morphism; uniqueness gives both the inverse and the cocycle identity. Gluing gives a scheme $J^d$ representing ${\mathcal{P}}^d$, with a universal normalized line bundle.

The construction is of finite type, not an infinite union of indispensable charts. Choose $m\ge 2g-1$ and a fixed line bundle $F$ of degree $m-d$. Every $L\otimes F$ has a nonzero section by the Riemann-Roch theorem, so the universal effective Cartier divisor gives a surjection $C^{(m)}\to J^d$. The inverse images of the charts cover the projective, hence quasi-compact, symmetric product of a curve $C^{(m)}$. A finite subcover gives finitely many charts for $J^d$. This surjection also makes $J^d$ irreducible. The charts show that it is smooth of dimension $g$.

For properness, apply the valuative criterion for properness. Let $R$ be a discrete valuation $k$-algebra with fraction field $K$, and let $L_K$ be a degree-$d$ line bundle on $C_K$. A rational section represents it by a Cartier divisor. Its closure in the regular surface $C_R$ is again a Cartier divisor, since regular local rings are factorial. Thus $L_K$ extends to a line bundle on $C_R$, whose fibre degree remains $d$ by constancy of line bundle degree in a family. Two extensions with the same generic class differ by a vertical Cartier divisor. The special fibre is irreducible, so that Cartier divisor is an integer multiple of the entire fibre, the principal Cartier divisor of a power of a uniformizer. The extensions therefore give the same normalized Picard class. This proves existence and uniqueness in the valuative criterion, hence separatedness and properness of the finite-type scheme $J^d$.

Consequently ${\mathrm{Pic}}_C^d={\mathrm{Jac}}_C^d$ is a smooth proper $k$-variety of dimension $g$. Tensoring by ${\mathcal{O}}_C(-d{P}_0)$ identifies it with $J^0$. On $J^0$ the tensor product and inverse of line bundles induce the algebraic group operations; it is the Jacobian variety, an abelian variety.

Write $X={\mathrm{Pic}}^{g-1}(C)$. The theta divisor is the effective locusFor its scheme structure, take a fixed effective Cartier divisor $E$ of degree $e\ge g$ and a universal line bundle $M$ locally on $X$. Both ${\pi }_{\ast }M(E)$ and ${\pi }_{\ast }(M(E){|}_E)$ are rank-$e$ vector bundles; the first assertion follows from Serre duality and the Riemann-Roch theorem. Define $\Theta$ by the determinant of their evaluation map. Its kernel is $H^0(C,M)$, so this is precisely the required locus. Exact sequences for enlarging $E$ cancel an invertible determinant factor, making the determinant description of the theta divisor independent of $E$.

This is a genuine irreducible reduced Cartier divisor. Choose $g-1$ points imposing independent conditions on $H^0(C,K_C)$. For their effective Cartier divisor $D^{\prime }$, the Riemann-Roch theorem gives $h^0(\mathcal{O}(D^{\prime }))=h^1(\mathcal{O}(D^{\prime }))=1$. These points form a nonempty open set in the irreducible symmetric product of a curve $C^{(g-1)}$; the Abel map of an algebraic curve is generically one-to-one there, so its image has dimension $g-1$. At such a class $M$, the first-order change of the evaluation determinant is the cup-product functionalUnder Serre duality its dual is multiplication $H^0(M)\otimes H^0(K_C\otimes M^{-1})\to H^0(K_C)$. The product of two nonzero sections on an integral algebraic curve is nonzero. Thus the functional is nonzero and the determinant has multiplicity one generically. Its irreducible support is consequently the reduced theta divisor.

Now let $L={\mathcal{O}}_C(D)$ have degree $g$. The Riemann-Roch theorem gives $h^0(L)\ge 1$, while evaluation at $P$ givesIf $h^0(L)\ge 2$, this kernel is nonzero for every $P$, so the entire image of ${\varphi }_D$ lies in $\Theta$. If $h^0(L)=1$, write $s$ for a nonzero section. The kernel is nonzero exactly at the zeros of $s$, so the image is not contained in $\Theta$. Moreover $h^1(L)=0$, and the family $L(-P)$ is computed by the rank-one evaluation map $H^0(L)\otimes {\mathcal{O}}_C\to L$. Its determinant is $s$ itself. This proves the scheme-theoretic restriction of the theta divisor to an Abel curve:Here the intersection is regarded as a Cartier divisor on $C$. Since $D$ was specified as a divisor class, it means its unique effective representative, with all multiplicities. The map ${\varphi }_D$ is an embedding: distinct points with the same image would give a degree-one map $C\to {\mathbb{P}}^1$; its cotangent map is evaluation of canonical differentials, which is surjective because $K_C$ is base-point-free. To see the latter, $h^0({\mathcal{O}}_C(P))=1$ for genus at least one, and the Riemann-Roch theorem gives $h^0(K_C-P)=g-1$. Thus the pullback also describes the intersection on the embedded curve.

It remains to count sections on $X$. Let $s_{\Theta }$ be the section defining the theta divisor, and take any $s\in H^0(X,{\mathcal{O}}_X(\Theta ))$. The quotient $r=s/s_{\Theta }$ is a rational function. Choose a fixed degree-$2g-1$ line bundle $T$ and the morphismIt is surjective: for every $M\in X$, the degree-$g$ bundle $T\otimes M^{-1}$ has a nonzero section by the Riemann-Roch theorem. On a general coordinate curve, put $D=T-{\sum }_{j\ne i}{P}_j$. The $g-1$ general points impose independent conditions on the $g$-dimensional space $H^0(C,T)$, leaving $h^0(\mathcal{O}(D))=1$. This independence follows inductively by choosing each new point away from the common zero set of the remaining nonzero sections. The restriction already proved identifies ${\mathcal{O}}_X(\Theta )$ on that coordinate curve with ${\mathcal{O}}_C(D)$, whose section space is one-dimensional. Hence $r\circ \beta$ is constant along every general coordinate curve.

A rational function on a product of integral curves which is constant in each factor is constant: on a common dense open set, successively varying one coordinate identifies its value with the value at a fixed general tuple. Therefore $r\circ \beta \in k$, and surjectivity of $\beta$ gives $r\in k$. Every $s$ is a scalar multiple of $s_{\Theta }$, proving

Put $n=3g-3$. To specify the Abel map of an algebraic curve into $X={\mathrm{Pic}}^{g-1}(C)$, useThe canonical-class twist is necessary: an untwisted sum has degree $3g-3$, rather than $g-1$. Let $L={\mathcal{O}}_C(2K_C)$. The Riemann-Roch theorem and Serre duality give $h^0(L)=n$ and $h^1(L)=0$.

On $C\times C^n$, write ${\Gamma }_i$ for the graph of the $i$th coordinate, $\mathcal{D}={\sum }_i{\Gamma }_i$, and $\pi$ for projection to $C^n$. The locally free sheaf $E_L={\pi }_{\ast }(L{|}_{\mathcal{D}})$ has rank $n$; at repeated points it records values and jets. Evaluation gives a morphism of rank-$n$ vector bundlesIts kernel at a tuple is $H^0(C,2K_C-\sum z_j)$. By the Riemann-Roch theorem, this space has the same dimension as $H^0(C,\sum z_j-K_C)$, since both line bundles have degree $g-1$. Thus its determinant vanishes exactly along ${\alpha }^{\ast }\Theta$. This is also an equality of Cartier divisors with multiplicities: the evaluation complex computes the sheaf cohomology of $L(-\mathcal{D})$; Serre duality identifies its determinant equation, up to an invertible factor, with the defining theta evaluation equation for $K_C\otimes L^{-1}(\mathcal{D})$.

To calculate the determinant line, set ${\mathcal{D}}_i={\Gamma }_1+\cdots +{\Gamma }_i$. Restriction to the extra graph givesProjection is exact on these finite relative divisors. On ${\Gamma }_i\cong C^n$, the first term is ${\mathrm{pr}}_i^{\ast }L\otimes \mathcal{O}(-{\sum }_{j<i}{\Delta }_{ij})$, where ${\Delta }_{ij}=\{z_i=z_j\}$. Taking determinants inductively yields the evaluation determinant on a product of curves:The determinant of the fixed space $H^0(C,L)$ contributes only a constant line. ThereforeThe diagonal subtraction is essential: ordinary evaluation determinants vanish when columns coincide, whereas restriction to a repeated-point Cartier divisor uses jets. The restriction exact sequences account for those factors in every characteristic.