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Geometric quantization starts from a classical symplectic manifold $(P,\omega )$ and seeks a Hilbert space and quantum observables that reflect its geometry. It has two distinct stages: prequantization, which represents the full Poisson algebra on sections of a line bundle, and a choice of polarization in geometric quantization, which reduces the excessive phase-space dependence of those sections.

Use the conventions ${\iota }_{X_f}\omega =df$ and $\{f,g\}=\omega (X_f,X_g)$. A prequantum line bundle is a Hermitian line bundle $L\to P$ with unitary connection of curvatureExistence requiresfor every closed integral two-cycle $\Sigma$, equivalently integrality of the symplectic cohomology class. Necessity follows from the integral curvature periods of a unitary line bundle; conversely an integral class defines such a bundle, and one adjusts its connection to give the specified curvature. This is a restriction on the classical system, not a choice of local coordinates. For an exact cotangent symplectic form $\omega =-d\theta$, with $\theta =p_{j}d{q}_j$, the bundle can be trivial and a connection is $\nabla =d-i\theta /\hslash$. On a sphere with symplectic area $4\pi s$, the condition becomes $2s/\hslash \in \mathbb{Z}$. The sign of the curvature convention changes if one changes the Hamiltonian-vector-field convention or uses the dual bundle; the integrality condition is unaffected.

The Kostant-Souriau prequantum operator for a real observable isIts derivative part describes the classical Hamiltonian flow and the multiplication term corrects the bracket. Indeed,Since $X_f(g)=-\{f,g\}$, expanding the operator commutator givesWith the Hermitian bundle metric and Liouville measure, these operators are formally symmetric for real $f$ because Hamiltonian flows preserve symplectic volume. On noncompact spaces, domains and self-adjointness still require attention; the algebraic identity initially holds on an appropriate common smooth domain.

Prequantization alone is too large. Sections depend on $2n$ phase-space variables, and the canonical commutation relations are represented reducibly. A polarization is an involutive maximal isotropic distribution $\mathcal{F}\subset T_{\mathbb{C}}P$ of complex rank $n$. Polarized sections satisfyThe curvature vanishes on pairs of polarization vectors because $\omega {|}_{\mathcal{F}}=0$, and involutivity makes the equations locally compatible. A real polarization is tangent to a Lagrangian foliation. A complex polarization in a Kähler manifold chooses one complex tangent type. With the curvature sign used here, choose the holomorphic tangent distribution, giving antiholomorphic wavefunctions. The conjugate line-bundle and sign convention gives the more usual holomorphic description.

For $T^{\ast }Q$, the vertical real polarization is spanned locally by ${\partial }_{p_j}$. Since the canonical one-form has no vertical component, its polarized sections are functions of $q$ alone. In the above trivialization,One does not simply integrate these states over every momentum to retain the prequantum norm: they are constant along noncompact polarization leaves and that integral would diverge. The polarized inner product instead uses the configuration-space measure or an intrinsic half-density description. This recovers the Schrödinger realization.

Only observables whose Hamiltonian flows preserve $\mathcal{F}$ act directly on the polarized space. For example, the free-particle flow generated by $p^2/2$ sends vertical tangent vectors into directions with a configuration-space component, so its raw prequantum operator does not preserve the vertical polarization. To obtain the usual kinetic operator one needs an additional procedure, such as transporting the polarization and pairing back, rather than pretending every classical observable restricts automatically. The Blattner-Kostant-Sternberg pairing and related projection methods address this problem. The impossibility of quantizing all classical observables with every desired algebraic property is another reason to specify the admissible observable algebra.

A metaplectic correction tensors $L$ with a square root of the polarization's canonical determinant bundle when such a square root exists. These half-forms make the pairing more intrinsic and supply corrections to operator transport. They account for familiar Maslov and zero-point shifts; their existence is an extra topological condition. For a flat phase plane, take $z=q+ip$ and replace $\theta =pdq$ by the gauge-equivalent ${\theta }_s=(pdq-qdp)/2$. ThenThe antiholomorphic function therefore has Gaussian norm, the complex-conjugate Bargmann-Fock space. This explicit calculation fixes which polarization matches our curvature sign. Choosing the opposite tangent type with the same sign would give growing Gaussian sections, not the desired Hilbert space. On compact positive Kähler examples, the corresponding holomorphic description (or its conjugate in these conventions) gives finite-dimensional state spaces.

Finding a suitable polarization is therefore a substantial part of the construction. A globally smooth real Lagrangian foliation may fail to exist; for example a regular real polarization on $S^2$ would require a line field that its tangent topology forbids. Singular foliations and noncompact leaves complicate both states and inner products. Although the restricted prequantum connection is locally flat on a Lagrangian leaf, a global nonzero parallel section requires trivial holonomy. In an integrable system this selects the Bohr-Sommerfeld quantization leaves; including the half-form correction gives the familiar action conditionin the usual Maslov-index convention. Complex polarizations require compatible global complex geometry and positivity. Different choices can lead to quite different realizations, and a natural unitary comparison is not automatic. Thus integral curvature provides prequantization; polarization and its global compatibility determine the actual quantum state space.

Use the covariant exterior derivative on the endomorphism bundle and remember that $F$ has degree two. The Bianchi identity is $d_{A}F=dF+A\wedge F-F\wedge A=0$. The local formula from part (b) givesThe two lines cancel, establishingBecause this is an endomorphism-valued equation and the operators transform covariantly under a change of local frame, this local calculation proves the global identity.

For a line bundle over $k=\mathbb{R}$ or $\mathbb{C}$, every fibre endomorphism is multiplication by a scalar. Thus $\mathrm{End}E$ is canonically the trivial scalar bundle, even when $E$ itself is nontrivial. In a local frame $A$ is a scalar one-form, so $A\wedge A=0$ and $F=dA$ locally. Also $A\wedge F=F\wedge A$, since their degrees are one and two. The Bianchi identity therefore reduces to the ordinary closedness conditionThis does not assert that the local scalar one-forms $A$ combine into a global one-form.

For two connections on a vector bundle $\nabla ,{\nabla }^{\prime }$, the difference is $C^{\infty }(M)$-linear in the section: the two $df\otimes s$ terms cancel. It is consequently a global endomorphism-valued one-form $a$. In rank one, $a$ is a global ordinary scalar one-form. Locally $A^{\prime }=A+a$, and the scalar cross terms and $a\wedge a$ vanish, givingThis is a global equality, so the two closed differential forms differ by an exact differential form. It proves the connection-independent curvature class of a line bundle:

Use the quotient convention for projective space. To a scheme $T$, associate the set of pairswhere $\mathcal{L}$ is a line bundle, modulo isomorphisms $\mathcal{L}\to {\mathcal{L}}^{\prime }$ carrying $q$ to $q^{\prime }$. Pullback defines the contravariant functor. On $T=\mathrm{Spec}A$, this means rank-one projective module quotients of $A^{n+1}$; the quotient need not be free. Thus merely listing generating tuples modulo multiplication by a unit would omit points over rings with nontrivial Picard group.

This functor is represented by ${\mathbb{P}}_{\mathbb{Z}}^n$. For each $i$, the locus where $q(e_i)$ generates $\mathcal{L}$ is open. These loci cover $T$, since a surjection onto a rank-one free stalk must have some coefficient a unit in its local ring. On this locus, trivialize by $q(e_i)$ and writeThese functions specify a map to the standard chart $U_i=\mathrm{Spec}\mathbb{Z}[t_j:j\ne i]$. Their transition rules are exactly those of projective space, so the chart maps glue. Conversely the universal quotient on projective space pulls back to the given pair, and the construction is natural and unique. This proves that projective space represents invertible quotients.

Write its universal exact sequence asA first-order change of the quotient, with target trivialized locally, is a map $\delta q:V\otimes \mathcal{O}\to \mathcal{O}(1)$. Changing the target trivialization by $1+\epsilon h$ changes $\delta q$ by $hq$. Restriction to $\mathcal{K}$ removes this ambiguity, and every local map $\mathcal{K}\to \mathcal{O}(1)$ extends because the displayed sequence splits locally. Consequently the tangent sheaf of a scheme isThis identification can also be seen on $U_i$: the first-order change of $t_j$ is $\delta q(e_j)-t_j\delta q(e_i)$ after setting $q(e_i)=1$. Thus these maps are precisely the coordinate tangent directions.

Apply $\mathcal{H}om(-,\mathcal{O}(1))$ to the universal sequence. Since its terms are locally free and it splits locally, this gives the Euler sequenceIt remains exact after any base change. Taking its dual of a sheaf over a field $k$ givesFor $n=1$, the first term has rank one; taking determinants yieldsExplicitly, for $u=t^{-1}$ on the overlap of the two affine charts, $du=-t^{-2}dt$, the same transition as $\mathcal{O}(-2)$ after changing one local frame's sign.

Cover the projective line by $U_0=\mathrm{Spec}k[t]$ and $U_1=\mathrm{Spec}k[t^{-1}]$. On their overlap the coordinate is invertible. Choose frames $e_0,e_1$ of the twisting sheaf on projective space $\mathcal{O}(n)$ with $e_1=t^n{e}_0$. The vanishing of quasi-coherent cohomology on an affine scheme applies to both charts and their intersection, so the Čech cochain complexcomputes the sheaf cohomology. Its kernel identifies with $k[t]\cap t^{n}k[t^{-1}]$, and its cokernel isFor $n\ge 0$, the intersection has basis $1,t,\dots ,t^n$; for $n<0$ it is zero. In the quotient, powers $t^j$ with $j\ge 0$ or $j\le n$ disappear. The surviving basis for $n\le -2$ is $t^{n+1},\dots ,t^{-1}$; for $n\ge -1$ none survives. HenceThis is the Čech cohomology of twists on the projective line.

By the preceding question, $\omega ={\Omega }_{{\mathbb{P}}^1/k}^1=\mathcal{O}(-2)$. In the $dt$ frame its degree-one Čech cohomology is one-dimensional, represented by $t^{-1}dt$. DefineThis is well defined: differentials regular on $U_0$ have nonnegative powers, and those regular on $U_1$ have powers at most $-2$ when expressed using $dt$. Evaluation and the cup product give natural pairingsFor $E=\mathcal{O}(n)$ with $n\ge 0$, the basis element $t^a$ in $H^0(E)$ pairs with $t^{-a-1}dt$ in $H^1(E^{\vee }\otimes \omega )$ and with no other element of this dual Laurent basis. The pairing matrix is a permutation matrix, hence invertible. For $n\le -2$ the same computation exchanges degrees zero and one. For $n=-1$ both sides vanish. Thus the pairings are perfect for every line bundle on ${\mathbb{P}}^1$.

Here we used the classification of line bundles on the projective line: they are $\mathcal{O}(a)$ for integers $a$. One direct justification is that every line bundle has a nonzero rational section and hence is associated with a divisor on an algebraic curve. The finite closed point defined by an irreducible polynomial $p(t)$ is linearly equivalent to $(\mathrm{deg}p)\infty$, using the principal divisor of $p(t)$. Thus every divisor is equivalent to its degree times $\infty$, proving the classification over arbitrary $k$.

To extend the pairing to every locally free coherent sheaf $E$, induct on rank, without assuming the splitting theorem of the next question. A one-dimensional subspace of the generic fibre, saturated in $E$, gives an exact sequencewith $L$ a line bundle and $Q$ a lower-rank vector bundle. This saturated line subbundle on a smooth curve exists because its stalk and its torsion-free quotient are free over the discrete valuation rings of the curve. The dual sequence, tensored with $\omega$, is exact as well, since the original sequence splits locally.

The two long exact sequences in sheaf cohomology, after dualizing one, are compatible with the evaluation pairings. In Čech cohomology, this compatibility follows by lifting a section locally: the differences of its lifts represent its connecting class, and contracting those differences with a dual cocycle gives the transpose connecting map, with the usual sign. Thus the connecting maps are adjoint; changing one vertical map's sign gives a commuting diagram. Its rows areThe maps for $L$ and $Q$ are isomorphisms by the line-bundle calculation and induction. Exactness then makes the maps for $E$ isomorphisms, by the Five lemma (or by comparing the relevant kernels and cokernels). All groups are finite-dimensional because the curve is projective and the sheaves coherent. Therefore residue duality on the projective line provesThese isomorphisms are natural in $E$. Higher sheaf cohomology vanishes by the same two-chart complex.

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 $k$ is algebraically closed and $C$ is a connected smooth projective curve. For a scheme $T$ over $k$, let $\pi :C_T=C{\times }_{k}T\to T$. The degree-$d$ Picard functor of a curve assignsOne can equivalently take the associated faithfully flat sheaf. It is essential to divide out by line bundles from the base: a parameter space should remember the family on $C$, not an independent base line bundle.

Choose $p_0\in C(k)$. A class can be normalized as $L\otimes {\pi }^{\ast }(L{|}_{p_0\times T}{)}^{-1}$, with its canonical trivialization along $p_0\times T$. Since ${\pi }_{\ast }{\mathcal{O}}_{C_T}={\mathcal{O}}_T$, an automorphism of a line bundle is multiplication by a base unit, and preserving this normalization forces that unit to be $1$. Descent of line bundles is effective, so normalized bundles satisfy the faithfully flat sheaf condition with no automorphism ambiguity. Thus this normalized description works on families, including nonreduced bases, rather than only on $k$-points.

We construct a representing scheme by Picard charts from nonspecial divisors. Fix $N\ge 2g$ and put $r=N-g$. The symmetric product of a curve $C^{(g)}$ represents relative effective Cartier divisors on a curve of degree $g$ and has a universal divisor $\Delta$. This uses the scheme symmetric product, not the naive set of unordered $T$-points. Locally on a smooth curve, a divisor of length $m$ is described by a monic polynomial in a smooth coordinate; its coefficients are elementary symmetric functions, so this construction includes coincident points in any characteristic. These local descriptions give the usual effective-divisor interpretation of the symmetric product.

Let $W\subset C^{(g)}$ be the open set where $H^1(C,\mathcal{O}(E))=0$. It is open by semicontinuity theorem for coherent cohomology. The Riemann-Roch theorem gives $h^0(\mathcal{O}(E))=1$ there, so $E$ is the unique effective divisor in its line-bundle class. For every fixed effective divisor $B$ of degree $r$, take a copy $W_B$ of $W$, with familyof degree $N$. Normalize it along $p_0$ as above.

This chart represents the open subfunctor defined byIndeed, for a family satisfying that condition, cohomology and base change for line bundles on a curve makes $V={\pi }_{\ast }L(-B)$ a line bundle on $T$, with formation commuting with base change. The rank is one by Riemann-Roch theorem. The evaluation mapis a nonzero section on each fibre, up to the base twist. Its zero scheme is a relative effective Cartier divisor on a curve $E$ of degree $g$. To check the family assertion, source and target are flat over $T$ and the fibre maps are injections of line bundles on an integral curve, so their cokernel is flat over $T$. Their zeros are Cartier, and the zero scheme is proper with finite fibres of length $g$, hence finite flat of degree $g$. It therefore defines a unique map $T\to W_B$. Conversely this construction recovers the family $\mathcal{O}(B+E)$ modulo a base line bundle.

The base-change step controls infinitesimal families as well. Locally, a finite free complex computes the two cohomology groups. Vanishing of its degree-one cokernel on fibres makes the last differential surjective by the Nakayama lemma. Splitting that differential leaves a locally free degree-zero kernel of rank one and commutes with base change. This is the standard finite-complex form of the permitted cohomology semicontinuity and base-change results.

These chart subfunctors cover all degree-$N$ families. For a geometric-fibre line bundle $L$ of degree $N$, the preceding solution gives $H^1(L)=0$ and $h^0(L)=N+1-g=r+1$. Choose $r$ points successively so that each imposes one independent condition on the current section space. At each stage that space is nonzero; a nonzero section has finitely many zeros, so a point outside its zero set can be chosen, also outside the finitely many previously chosen points. For their sum $B$,and Riemann-Roch theorem forces $h^1(L(-B))=0$. The points may be chosen in $C(k)$ even when the fibre field extends $k$, since $C(k)$ is infinite and a nonzero section on the base-changed curve has only finitely many zeros. Thus fixed divisors $B$ defined over $k$ suffice. Upper semicontinuity gives an open neighbourhood on the base for each successful $B$.

It remains to glue the charts explicitly. On $W_B$, the locusis open. The line bundle inside the braces has degree $g$; its unique section up to scale has an effective zero divisor $E^{\prime }$. Base change and the evaluation construction make $E\mapsto E^{\prime }$ a morphism $W_{B{B}^{\prime }}\to W_{B^{\prime }}$. It satisfies $\mathcal{O}(B+E)\cong \mathcal{O}(B^{\prime }+E^{\prime })$. Reversing $B,B^{\prime }$ gives its inverse. On triple overlaps the transition maps compose correctly because the residual effective divisor is unique. Hence the copies $W_B$ glue along these open isomorphisms to a scheme $P^N$.

The normalized bundles $L_B$ glue as well: on overlaps their normalized isomorphisms exist and are unique, so they satisfy the cocycle condition. Let ${\mathcal{P}}_N$ denote the resulting universal normalized line bundle on $C\times P^N$. For any $T$, the open cover where the conditions for some $B$ hold gives compatible maps $T\to W_B$, and thus a unique map $T\to P^N$. Pulling back ${\mathcal{P}}_N$ reverses this procedure. We have established a natural bijectionThis proves representability of the full functor, not just a parametrization of individual line bundles. The nonspecial-divisor construction is also recorded in Stacks Project, Picard scheme of a curve.

Finally tensoring by ${\mathcal{O}}_C((N-d)p_0)$ is a natural isomorphism ${\mathrm{Pic}}_C^d\to {\mathrm{Pic}}_C^N$ for every integer $d$. Transporting the universal bundle back constructs its degree-$d$ representative $P^d$, usually denoted ${\mathrm{Pic}}_{C/k}^d$. Thus every degree component is represented. These identifications depend on the chosen point and twist; intrinsically the degree-zero component is a group scheme and each degree component is its torsor.

An isogeny of elliptic curves $\varphi :E_1\to E_2$ over a field $K$ is a nonconstant morphism preserving the identity and addition. It is surjective and finite: its image is a closed connected subgroup, and a nonconstant morphism of smooth projective curves is finite. The degree of an isogeny is the degree of its induced extension of function fields. A separable isogeny has degree equal to the number of geometric kernel points; in general the degree is the length of the finite kernel scheme, incorporating inseparability. The degree of an isogeny multiplies under composition, and degree one means an isomorphism. In characteristic zero all elliptic isogenies are separable. In positive characteristic the Frobenius isogeny illustrates inseparability, while an elliptic isogeny of degree prime to the characteristic is separable.

The homomorphism group of elliptic curves includes the zero homomorphism, the constant identity map, and every nonzero member is an elliptic isogeny. The operations are pointwise addition and negation; for endomorphisms, composition also gives the endomorphism ring of an elliptic curve. We put $q(0)=0$ and $q(\varphi )=\mathrm{deg}\varphi$ otherwise. We now prove the positive definiteness of degree on elliptic-curve homomorphisms in full.

Let $E=E_2$, and choose a Weierstrass model. Its coordinate $x$ has a double pole at $O$, and the generic fibre of $x$ consists of $P$ and $-P$. On $E\times E$ the rational function $x(P)-x(Q)$ therefore has Weil divisorwhere $\Delta =\{P=Q\}$, ${\Delta }_{-}=\{P=-Q\}$, $V=\{O\}\times E$, and $H=E\times \{O\}$. The generic zeros on the two diagonals have multiplicity one, and the generic poles on $V,H$ have multiplicity two; their intersections have codimension two and add no Weil divisor components. This argument also uses the general Weierstrass involution in characteristic $2$, where $-P$ need not mean changing just the sign of $y$.

For the sum and difference maps $a(P,Q)=P+Q$ and $s(P,Q)=P-Q$, the divisors $a^{\ast }(O)$ and $s^{\ast }(O)$ are ${\Delta }_{-}$ and $\Delta$. Consequently, for the degree-one line bundle $L={\mathcal{O}}_E(O)$,Pull back along $P\mapsto (\varphi (P),\psi (P))$ and take degrees of line bundles on $E_1$. A nonconstant map pulls back $L$ to degree equal to its degree, while a constant map pulls it back to a trivial bundle of degree zero. We obtain the degree parallelogram lawUsing the line-bundle identity is important: it still restricts correctly when $\varphi =\psi$ or $\varphi =-\psi$, when restriction of the displayed rational function would be identically zero.

Since $q(-\varphi )=q(\varphi )$, applying the identity to $n\varphi$ and $\varphi$ gives the second-difference recursionStarting from $q(0)=0$ and $q(\varphi )$ yields $q(n\varphi )=n^{2}q(\varphi )$ for all integers $n$. DefineIt is symmetric and odd in each variable. Applying the parallelogram identity to the four terms in its first expression givesInterchanging $u,w$ and using oddness also gives $B(u+w,v)-B(u-w,v)=2B(w,v)$. Adding proves $B(u+w,v)=B(u,v)+B(w,v)$; symmetry proves additivity in the other variable. Thus $q$ is a quadratic form with associated rational bilinear form $B$.

Every nonzero homomorphism has positive degree, so $q(\varphi )>0$ for $\varphi \ne 0$. The scaling identity also shows that the homomorphism group is a torsion-free group. Hence $q$ extends to its rational vector space. On the span of any finitely many rationally independent homomorphisms, the matrix of $B$ is rational. Positivity on nonzero rational vectors implies nonnegativity on real vectors by density. If the real matrix had a nontrivial kernel, its rational linear equations would have a nonzero rational solution, contradicting positivity. Thus its real extension is a positive-definite quadratic form. This establishes the requested positivity as well as the quadratic identities.

An important companion to degree is the dual isogeny. Under $E\cong {\mathrm{Pic}}^0(E)$ via $P\mapsto [P-O]$, the pullback ${\varphi }^{\ast }:{\mathrm{Pic}}^0(E_2)\to {\mathrm{Pic}}^0(E_1)$ defines $\hat{\varphi }:E_2\to E_1$. Divisor pushforward corresponds to $\varphi$, and pushforward after pullback multiplies divisor classes by $n=\mathrm{deg}\varphi$. Therefore $\varphi \hat{\varphi }=[n]$. Composing shows that $\hat{\varphi }\varphi -[n]$ takes values in the finite kernel of $\varphi$; a morphism from a connected smooth curve into this finite scheme must be constant, and its identity value is zero. ThusThe last equality follows from $\mathrm{deg}[n]=n^2$, already proved by applying $q(m\varphi )=m^{2}q(\varphi )$ to the identity map. The dual isogeny is unique, since the difference of two choices composed with $\varphi$ would vanish and $\varphi$ is surjective. It follows that being isogenous is an equivalence relation: duality supplies symmetry, and composition supplies transitivity. For a finite subgroup of geometric points of order prime to the characteristic, the quotient curve and its quotient map give an elliptic isogeny with that kernel; in general one uses finite subgroup schemes. This is the geometric source of explicit descent maps, including the two-isogeny formula in Question 4.

Finally elliptic isogenies preserve the rational rank of an abelian group over a number field. Their dual compositions are multiplication by $n$, so after tensoring the rational point groups with $\mathbb{Q}$ the elliptic isogeny becomes invertible. For a single curve, its endomorphism algebra in characteristic zero is $\mathbb{Q}$ or an imaginary quadratic field, giving ordinary integer multiplication or complex multiplication. In positive characteristic a supersingular curve can instead have a quaternionic endomorphism algebra. The positive degree form controls the sizes of these homomorphisms, while duality makes their arithmetic and geometric properties accessible.