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Averaging a Kähler form over translations with normalized Haar measure gives an invariant Kähler form in the same de Rham cohomology class. Positivity is preserved by averaging positive values on $(v,Jv)$, and the class is unchanged because translations are homotopic to the identity. Thus any Hodge metric on a complex torus can be replaced by an invariant one.

Every scalar holomorphic function on a compact connected complex manifold is constant by the maximum modulus principle. Apply this to each entry of the holomorphic Adjoint representation of a Lie group; its value is constantly the identity. Differentiation makes the complex Lie algebra abelian, hence connectedness makes the group abelian. The holomorphic Lie exponential map identifies it with ${\mathbb{C}}^n$ modulo a discrete subgroup. Compactness forces that subgroup to span the underlying real vector space, giving a full-rank Euclidean lattice and a complex torus.

A complex cylinder is a quotient $\mathbb{C}/\mathbb{Z}\tau$ by one nonzero complex translation. Unlike a compact complex torus, it is biholomorphic to ${\mathbb{C}}^{\ast }$ through a suitable exponential map.

For a complex torus $X=V/\Lambda$ of complex dimension $g$, constant forms give a natural isomorphism $H_{\bar{\partial }}^{p,q}(X)\cong {\bigwedge }^p{V}^{\ast }\otimes {\bigwedge }^q\overline{V^{\ast }}$. A flat Kähler metric and the Dolbeault Hodge decomposition prove that these constant forms are precisely the harmonic representatives. Thus $h^{p,q}=\left(\genfrac{}{}{0ex}{}{g}{p}\right)\left(\genfrac{}{}{0ex}{}{g}{q}\right)$ for $0\le p,q\le g$, and zero otherwise.

On a compact Kähler manifold $X$, if $P$ is a positive holomorphic line bundle, then $H^i(X,K_X\otimes P)=0$ for $i>0$. The canonical-bundle twist is part of the hypothesis and conclusion; positivity alone does not remove it on a general manifold. For a complex torus, the canonical bundle is trivial, so higher cohomology of a positive line bundle vanishes without an additional twist.

Let $X$ be a connected compact Riemann surface of genus $g$. Choose an oriented symplectic basis $a_1,\dots ,a_g,b_1,\dots ,b_g$ of $H_1(X,\mathbb{Z})$, with $a_i\cdot b_j={\delta }_{ij}$. For a closed complex differential form $\alpha$, write $A_i(\alpha )={\int }_{a_i}\alpha$ and $B_i(\alpha )={\int }_{b_i}\alpha$. The Riemann bilinear relations begin with the identityTo prove it, cut $X$ along the chosen cycles to obtain a fundamental polygon $P$. On this simply connected polygon, a closed differential form has a primitive $F$, with $dF=\alpha$. Stokes theorem gives ${\int }_P\alpha \wedge \beta ={\int }_{\partial P}F\beta$. The two copies of each edge have opposite orientations, and their values of $F$ differ by the period acquired in passing between them. Pairing the copies of the $a_i$ and $b_i$ edges gives, respectively, $-B_i(\alpha )A_i(\beta )$ and $A_i(\alpha )B_i(\beta )$. Summing proves the identity; the convention $a_i\cdot b_i=1$ fixes its sign.

For holomorphic one-forms $\omega ,\eta$, their wedge product is zero because locally both are multiples of $dz$. Thus the first holomorphic Riemann bilinear relations isFor a nonzero holomorphic one-form $\omega =f(z)dz$, the second relation follows by applying the bilinear identity to $\omega ,\overline{\omega }$:In particular, a holomorphic one-form with all $a$-periods zero must vanish. Since $\dim H^0(X,K_X)=g$, the map taking a holomorphic one-form to its $a$-periods is an isomorphism. We can therefore choose normalized holomorphic one-forms ${\omega }_1,\dots ,{\omega }_g$ with ${\int }_{a_i}{\omega }_j={\delta }_{ij}$. Put $B_{ij}={\int }_{b_i}{\omega }_j$. The first relation applied to ${\omega }_j,{\omega }_k$ gives $B_{jk}=B_{kj}$, and the second applied to ${\sum }_j{c}_j{\omega }_j$ gives $2{\overline{c}}^{,t}(\mathrm{Im}B)c>0$ for $c\ne 0$. ConsequentlyThese conclusions make the construction of the Jacobian variety work. Integration identifies the image of $H_1(X,\mathbb{Z})$ in $H^0(X,K_X{)}^{\ast }$ with the period lattice $\Lambda ={\mathbb{Z}}^g+B{\mathbb{Z}}^g$. The real-linear map $(m,n)\mapsto m+Bn$ is invertible: its imaginary part is $Yn$, and $Y$ is invertible. Hence $\Lambda$ is a discrete lattice of full real rank $2g$, andis a compact complex torus, rather than a non-Hausdorff quotient. Changing the symplectic basis changes the displayed coordinates but not this intrinsic quotient.

The Riemann bilinear relations also supply its polarization of a complex torus. Use the Hermitian form $H(z,w)=z^t{Y}^{-1}\overline{w}$, linear in the first argument, and set $E=-\mathrm{Im}H$. Symmetry of $B$ givesThus $E$ is an integral positive Riemann form on a complex torus. Its matrix on the displayed lattice basis is the unimodular symplectic matrix. The corresponding positive holomorphic line bundle, equivalently the one constructed from theta automorphy in Solution 4, gives a principal polarization of a complex torus. By the Kodaira embedding theorem this makes the complex torus projective, hence an abelian variety. The normalization $E=-\mathrm{Im}H$ here is important because $H$ was chosen linear in its first argument. If $g=0$, the spaces of holomorphic one-forms and periods are zero and the Jacobian variety is a point.

A complex analytic Lie group is a complex manifold whose multiplication and inversion are holomorphic maps between complex manifolds. Its identity tangent space is a complex Lie algebra, and the Adjoint representation of a Lie groupis holomorphic: differentiating the holomorphic conjugation map with respect to its second variable gives holomorphically varying matrix entries.

Every scalar holomorphic function on a compact connected complex manifold is constant. Indeed its modulus attains a maximum; the maximum modulus principle, applied in complex coordinate charts and along complex lines, makes it locally constant there, and connectedness propagates that constancy. Apply this to each entry of $\mathrm{Ad}$. Since ${\mathrm{Ad}}_e=I$, it follows that ${\mathrm{Ad}}_g=I$ for every $g$. Differentiation givesThus $\mathfrak{g}$ is abelian, and the connected group $G$ is abelian by the identity-neighbourhood argument in part (i). This proves that compact connected complex Lie groups are complex tori.

If $n={\dim }_{\mathbb{C}}G$, identify $\mathfrak{g}$ with ${\mathbb{C}}^n$. The Lie exponential map is holomorphic, additive, onto, and a local biholomorphism, by the holomorphic left-invariant differential equation and its identity derivative. Its kernel $B$ is a discrete subgroup of the additive space ${\mathbb{C}}^n$. The induced map yieldsas complex Lie groups, not merely as real groups. Compactness further forces $B$ to be a full-rank real lattice. If its real span $W$ were proper, projection would give a surjective continuous map ${\mathbb{C}}^n/B\to {\mathbb{R}}^{2n}/W$, contradicting compactness. The discrete-subgroup argument in part (i) therefore gives $B\cong {\mathbb{Z}}^{2n}$. The quotient is a complex torus; it need not split into one-dimensional complex tori.

State the characteristic convention explicitly: for real $\alpha ,\beta$, define the theta function with characteristics byBinary subscripts ${\vartheta }_{ab}$ often denote this function with $(\alpha ,\beta )=(a/2,b/2)$, $a,b\in \{0,1\}$. On compact subsets of the $(z,\tau )$ domain, a summand and any derivative are bounded by a polynomial in $|n|$ times $e^{-c{n}^2+C|n|}$. Normal convergence therefore gives an entire function of $z$, holomorphic dependence on $\tau$, and legitimate termwise differentiation. In particular the heat equation for theta functions with characteristics isCharacteristics are defined modulo integers with a phase: $\theta [\alpha +1,\beta ]=\theta [\alpha ,\beta ]$ and $\theta [\alpha ,\beta +1]=e^{2\pi i\alpha }\theta [\alpha ,\beta ]$. Shifting the summation index proves the spatial quasi-periodsThese multipliers describe a degree-one line bundle on the complex torus with periods $1,\tau$, rather than an ordinary elliptic function. The logarithmic derivative is unchanged under one and decreases by $2\pi i$ under tau. Pairing opposite sides of a fundamental cell therefore makes its contour integral $2\pi i$. The argument principle counts exactly one zero in a cell, with multiplicity.

The characteristic function is a translate of the unshifted theta function:At $z=(1+\tau )/2$, the unshifted series cancels in pairs $n,-n-1$. Hence the zeros of theta functions with characteristics occur exactly and simply atReindexing also gives $\theta [\alpha ,\beta ](-z)=\theta [-\alpha ,-\beta ](z)$. For half-integer characteristics the parity is $e^{4\pi i\alpha \beta }$, so only $(1/2,1/2)$ is odd and vanishes at zero; the other three theta constants are nonzero.

The modular transformations of theta characteristics are obtained from Gaussian Poisson summation and elementary phase rearrangement. With the holomorphic branch of $\sqrt{-i\tau }$ positive at $\tau =i$,For the first formula, apply Poisson summation to $F(t)=e^{\pi i\tau (t+\alpha {)}^2+2\pi i(t+\alpha )(z+\beta )}$. Completing the square in its Fourier transform givesRearranging ${\sum }_{n}F(n)={\sum }_m\hat{F}(m)$ proves the stated inversion law. For translation by one, use $e^{\pi i{n}^2}=e^{\pi in}$ in the defining series; the remaining characteristic phase is precisely the displayed factor.

Write $Q=e^{\pi i\tau }$, so the modular-form variable of Question 1 is $q=Q^2$, and put ${\Theta }_2=\theta [1/2,0](0,\tau )$, ${\Theta }_3=\theta [0,0](0,\tau )$ and ${\Theta }_4=\theta [0,1/2](0,\tau )$. Inversion interchanges ${\Theta }_2,{\Theta }_4$ and fixes ${\Theta }_3$, each with factor $\sqrt{-i\tau }$. Translation interchanges ${\Theta }_3,{\Theta }_4$ and multiplies ${\Theta }_2$ by $e^{\pi i/4}$. The Jacobi triple product givesIt also gives the conventional odd theta function ${\theta }_1=-\theta [1/2,1/2]$ its product $2Q^{1/4}\sin (\pi z){\prod }_{m\ge 1}(1-Q^{2m})(1-2Q^{2m}\cos (2\pi z)+Q^{4m})$. Differentiate at zero and multiply the three even products to obtain the Jacobi derivative formulaIn the constant-product simplification, ${\prod }_m(1+Q^{2m})(1-Q^{4m-2})=1$, by separating even and odd positive exponents. The products prove nonvanishing of all three even constants as well.

The Jacobi abstruse identity can be proved directly, without assuming it from the products. Splitting pairs of summation indices according to whether their sum and difference are even or odd givesFor example, $u=(n+m)/2,v=(n-m)/2$ are either both integers or both half-integers, and $n^2+m^2=2u^2+2v^2$. The alternating signs give the second identity; shifting both original indices by one half gives the third. Squaring the first two and subtracting then proves ${\Theta }_3^4={\Theta }_2^4+{\Theta }_4^4$.

These identities connect theta theory to modular forms and elliptic curves. The fourth powers are weight-two forms on the principal level-two congruence subgroup. The modular lambda function $\lambda ={\Theta }_2^4/{\Theta }_3^4$ obeys $\lambda (-1/\tau )=1-\lambda (\tau )$ and $\lambda (\tau +1)=\lambda (\tau )/(\lambda (\tau )-1)$; the characteristic permutation kernel is $\Gamma (2)$. It parametrizes an elliptic curve with its ordered two-torsion in Legendre form $y^2=x(x-1)(x-\lambda )$. The corresponding Klein j-invariant is $256(1-\lambda +{\lambda }^2{)}^3/[{\lambda }^2(1-\lambda {)}^2]$. Finally the same products giveIndeed ${\Theta }_2{\Theta }_3{\Theta }_4=2Q^{1/4}{\prod }_m(1-Q^{2m}{)}^3$, whose eighth power divided by $256$ is $q{\prod }_m(1-q^m{)}^{24}$. This recovers the normalized modular discriminant of Question 1 and displays the bridge between Gaussian theta sums and the lattice discriminant of Question 2.

For a nonconstant holomorphic map $f:X\to Y$ between compact connected Riemann surfaces, its degree of a holomorphic map is the number of preimages of a regular value. More generally it is the constant sum ${\sum }_{p\in f^{-1}(q)}{e}_p$, where $e_p$ is the local multiplicity. The Riemann-Hurwitz formula is

Negation is well-defined on the complex torus because $-\Lambda =\Lambda$. It is holomorphic with holomorphic inverse itself, so $\psi$ is biholomorphic. Its fixed points satisfy $2z\in \Lambda$. If $\Lambda =\mathbb{Z}{\omega }_1+\mathbb{Z}{\omega }_2$, they are the four distinct classes

Away from these points, choose a small coordinate disk $U$ disjoint from $\psi (U)$. The quotient map is a homeomorphism from $U$ onto $\pi (U)$, so transfer its complex coordinate to that open set of $S^{\prime }$. The transition between two such lifted charts is induced either by identity or by $\psi$, hence is holomorphic. These charts give the required Riemann-surface structure, and $\pi$ is locally a holomorphic bijection there.

The quotient has generic degree two. At a fixed point, a local coordinate centred there changes under $\psi$ as $u\mapsto -u$. The assumed holomorphic quotient map is invariant under this operation, so its local expansion is even and has multiplicity at least two. Since the total degree is two, it is exactly two. There are no other ramification points. The domain torus has genus one, so Riemann-Hurwitz formula gives

Let $X$ be a connected compact Riemann surface of geometric genus $g$, equivalently a smooth projective curve over $\mathbb{C}$, and put $V=H^0(X,K_X)$. Choose a symplectic basis $a_1,\dots ,a_g,b_1,\dots ,b_g$ of $H_1(X,\mathbb{Z})$ with $a_i\cdot b_j={\delta }_{ij}$. For a closed differential form $\alpha$, write $A_i(\alpha )={\int }_{a_i}\alpha$ and $B_i(\alpha )={\int }_{b_i}\alpha$. The underlying bilinear identity isIt holds for any two smooth closed one-forms. In particular, the Riemann bilinear relations for holomorphic differential forms arefor $\omega ,\eta \in V$ and $\omega \ne 0$ in the inequality. The inequality fixes the orientation and the sign convention.

To prove the identity, cut $X$ along the symplectic basis to a fundamental polygon $P$. On $P$, a closed one-form $\alpha$ has a primitive $F$. The Stokes theorem givesOpposite copies of each cut have opposite orientations, while their values of $F$ differ by the corresponding period of $\alpha$. Pairing those edges therefore leaves $A_i(\alpha ){\int }_{b_i}\beta -B_i(\alpha ){\int }_{a_i}\beta$ for the $i$th handle. Summing proves the identity. The same calculation can be made with a small disk about the polygon's vertices removed; its extra boundary terms tend to zero, so the vertices cause no additional term.

For holomorphic differential forms $\omega ,\eta$, their wedge product of differential forms vanishes: locally both are multiples of $dz$. This proves the first relation. If $\omega =f(z)dz$ and $z=x+iy$, thenA nonzero holomorphic differential form is nonzero on an open set, so the integral is strictly positive. Applying the identity to $\omega ,\overline{\omega }$ proves the second relation.

These relations first prove that the map $V\to {\mathbb{C}}^g$ taking $a$-periods is injective: if every $A_i(\omega )$ vanishes, the positive integral just computed would vanish. Since $\dim V=g$ by the Riemann-Roch theorem, it is an isomorphism. Consequently there is a unique basis ${\omega }_1,\dots ,{\omega }_g$ with ${\int }_{a_j}{\omega }_i={\delta }_{ij}$. Set ${\tau }_{ij}={\int }_{b_j}{\omega }_i$. The first relation gives ${\tau }_{ij}={\tau }_{ji}$, and the second gives, for any nonzero column $c$,Thus the normalized period matrix of a complex torus is symmetric and has positive-definite imaginary part:

The integration map sends a cycle $\gamma$ to the functional $\omega \mapsto {\int }_{\gamma }\omega$ in $V^{\ast }$. In the normalized basis its image is the period latticeIndeed, the real-linear map $(x,y)\mapsto x+\tau y$ from ${\mathbb{R}}^{2g}$ to ${\mathbb{C}}^g$ is an isomorphism: its imaginary part is $Yy$, and $Y$ is invertible. Its image of ${\mathbb{Z}}^{2g}$ is therefore discrete and has compact quotient. This proves thatis a well-defined $g$-dimensional complex torus, rather than merely a quotient by an arbitrary subgroup.

The relations also provide its canonical principal polarization of a complex torus. With the convention that a Hermitian form is conjugate-linear in its first argument, put $H(z,w)={\overline{z}}^{,t}{Y}^{-1}w$ and $E=\mathrm{Im}H$. Symmetry of $\tau$ givesIn particular $E$ is integral and unimodular on $\Lambda$, and it is compatible with the complex structure. These are exactly the positivity and integrality conditions for a Riemann form on a complex torus; they make the complex torus a Jacobian variety with a principal polarization of a complex torus. If instead one uses a first-argument-linear Hermitian form, the same alternating form is $-\mathrm{Im}H$. Changing the symplectic basis changes the coordinates but not this polarized quotient. For $g=0$, $V=0$ and the Jacobian variety is a point.

Fix $p_0\in X$ on a connected compact Riemann surface, and construct its Jacobian variety by periods as in the first solution. For a divisor on an algebraic curve $E={\sum }_p{n}_{p}p$ of degree zero, defineChanging the paths adds a period lattice element; changing $p_0$ adds a common integral multiplied by $\sum n_p=0$. The Abel theorem for divisors saysEquivalently, effective divisors on an algebraic curve of the same degree have the same Abelian sum exactly when they are equivalent under linear equivalence of divisors.

We need a meromorphic version of the Riemann bilinear relations. Let $\eta$ be a differential of the third kind, with residues $n_p$ at its simple poles, and let $\omega$ be a holomorphic differential form. Choose representatives of the cycles and paths avoiding the poles in the same cut polygon, and let $F(p)={\int }_{p_0}^p\omega$ there. ThenHere the equality is an equality of the chosen complex numbers, before taking any quotient. To prove it, remove small disks about the poles. Since $d(F\eta )=\omega \wedge \eta =0$ on the remaining polygon, its boundary integral is zero. Pairing opposite outer edges gives the left side, just as in the first solution. The clockwise boundary about $p$ contributes $-2\pi i{n}_{p}F(p)$ by the residue theorem. Moving these terms to the other side proves the formula and its sign.

Such an $\eta$ exists whenever $\sum n_p=0$. Here is the required existence argument. Let $S$ be the reduced support of $E$, and considerLocally the last map takes $cdz/z$ to $c$. By Serre duality, the obstruction in $H^1(K_X)\cong \mathbb{C}$ is dual to restriction of constant functions, so it is the sum of the residues. Thus the kernel of this sum is exactly the image of global meromorphic differentials. Alternatively, the Riemann-Roch theorem gives $h^0(K_X(S))-h^0(K_X)=|S|-1$, and the residue theorem identifies that image with the same codimension-one subspace. Subtracting a linear combination of the normalized holomorphic differential forms makes $A_i(\eta )=0$ for every $i$.

Suppose now that $\mathrm{AJ}(E)=0$. Let $u_i={\sum }_p{n}_p{\int }_{p_0}^p{\omega }_i$ with the paths used above. There are integer columns $m,n$ such that $u=m+\tau n$. The normalized meromorphic bilinear formula gives $B_i(\eta )=2\pi i{u}_i$. Thereforehas $a$-periods $-2\pi i{n}_i$ and $b$-periods $2\pi i{m}_i$. Its integrals around the poles are also in $2\pi i\mathbb{Z}$. These cycles generate the homology of the punctured surface, sois a single-valued, nowhere-zero holomorphic function away from the support of $E$. Near $p$, write ${\eta }^{\prime }=n_{p}dz/z+h(z)dz$. Then $f=z^{n_p}\exp (H(z))$ times a nonzero constant, with $H^{\prime }=h$. It extends meromorphically with order exactly $n_p$. Hence $\mathrm{div}(f)=E$.

Conversely, if $E=\mathrm{div}(f)$, take $\eta =df/f$. It has residues $n_p$ and every period lies in $2\pi i\mathbb{Z}$: continuation of a local logarithm of $f$ changes it by an integral multiple of $2\pi i$. Write $A_i(\eta )=2\pi i{r}_i$ and $B_i(\eta )=2\pi i{s}_i$. Applying the meromorphic bilinear formula to ${\omega }_i$ givesThus $u\in \Lambda$ and $\mathrm{AJ}(E)=0$, completing both directions without assuming the conclusion as a property of the Jacobian variety.

For completeness, the effective degree-$d$ version lives on the symmetric product of a curve $X^{(d)}=X^d/{\mathfrak{S}}_d$. The sum of integrals on $X^d$ is invariant under the finite symmetric group and descends to the Abelian sum map $u_d$. The quotient is locally described by elementary symmetric coordinates, so repeated points are included. The proof above identifies its fibre over $u_d(D)$ withExplicitly, a nonzero global section gives its effective zero divisor, and two sections give the same divisor precisely when their quotient is constant. These maps are holomorphic in local symmetric coordinates. More intrinsically, a family of effective divisors on an algebraic curve in this fixed class determines a line of sections of ${\mathcal{O}}_X(D)$, after locally trivializing any line bundle pulled back from the parameter space. Conversely such a line of sections gives its family of zero divisors on an algebraic curve. These operations are inverse and unaffected by that trivialization, proving the identification with the projective space in families as well as on points.

Finally this description identifies the analytic Jacobian variety with ${\mathrm{Pic}}^0(X)$. Every degree-zero line bundle has a meromorphic section: after twisting by a sufficiently large effective divisor, the Riemann-Roch theorem supplies a nonzero global section. Its divisor represents the original line bundle. The correspondence to the period quotient is well-defined and injective by the theorem. It is surjective because one may choose $g$ distinct points with independent evaluations of holomorphic differential forms; the derivative of the sum of their Abel integrals is then an isomorphism. Hence the subgroup generated by point differences contains an open subset of the connected complex torus, and must be the whole group. Local integration coordinates, or the exponential sequence for ${\mathcal{O}}_X$, give the same holomorphic identification. This justifies the subsequent use of degree-$d$ line bundles as points of a translate of the Jacobian variety.

Choose a projective embedding of the abelian surface $X$. By Bertini's theorem, a general hyperplane section $C$ is a smooth projective curve that is connected. It is ample, so $C^2>0$. Choose $p\in C$ and translate the inclusion so it sends $p$ to the identity of $X$. The universal property of the Jacobian variety then gives a homomorphism of abelian varietieswhose composition with the based Abel map of an algebraic curve is this translated inclusion.

The image is a closed connected algebraic subgroup, since $J(C)$ is projective, and it contains a translate of $C$. If the image had dimension one, that irreducible curve would be the image subgroup itself. A distinct coset is disjoint from it and has the same numerical divisor class: translation acts trivially on cohomology, being homotopic to the identity on the complex torus. Its self-intersection would therefore be zero, contradicting $C^2>0$. The image has dimension two and is all of $X$. Thus $\Phi$ is surjective, as expressed by the principle that an ample curve on an abelian surface generates the surface.

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.

The Appell–Humbert theorem identifies the positive Hermitian form $H_1$ with a holomorphic line bundle whose First Chern class is represented, in our linear-first convention, byIts integral over the square fundamental domain is $1$, so $\mathrm{deg}L=1$. In particular the oriented Chern pairing is $-E_1(1,i)=1$; using $E_1(1,i)$ without adjusting the Hermitian convention would give the wrong sign.

The complex torus has a nowhere-vanishing differential $dz$, so its canonical divisor is trivial and its genus is one. A nonzero section of a line bundle of negative degree would have an effective divisor of negative degree, which is impossible. Thus $H^0(X,L^{-1})=0$. The Riemann-Roch theorem and Serre duality giveTherefore $\boxed{h^0(X,L)=1}$. The choice of semicharacter of a complex lattice changes the degree-zero component of the line bundle, but not this calculation.

A holomorphic group endomorphism of the complex torus lifts to multiplication by a scalar $c\in \mathbb{C}$, as in Question 1. It descends exactly when $c\Lambda \subseteq \Lambda$. Applying this condition to $1$ gives $c\in \mathbb{Z}[i]$. Conversely any Gaussian integer $c$ preserves $\mathbb{Z}[i]$, since the latter is a ring. Addition of endomorphisms is addition of scalars, and composition is multiplication of scalars. Thus, as rings,The wording “endomorphism algebra” here denotes this integral endomorphism ring of an elliptic curve. If it instead meant the rational algebra ${\mathrm{End}}^0(X)=\mathrm{End}(X){\otimes }_{\mathbb{Z}}\mathbb{Q}$, that algebra would be $\mathbb{Q}(i)$.

Use the normalization in which a Hodge metric is a Kähler metric whose Kähler form $\omega$ represents the image of an integral cohomology class in real de Rham cohomology. Equivalently, all its periods on integral two-cycles are integers. A convention using $\omega /(2\pi )$ only rescales the metric and does not change existence. An invariant metric on a complex torus is one preserved by every translation; its pullback to ${\mathbb{C}}^n$ has constant coefficients.

Let $T={\mathbb{C}}^n/\Lambda$ carry a Hodge metric. Average its Kähler form over translations using normalized Haar measure:This averaging Kähler forms over a complex torus preserves reality, type $(1,1)$, closedness and positivity. Indeed, for any nonzero tangent vector, the quantity being averaged in $\omega (v,Jv)$ is strictly positive. Every translation is homotopic to the identity, since a path from $0$ to $a$ supplies such a homotopy. Thus $[{\omega }_0]=[\omega ]$, as is seen either on de Rham cohomology or by integrating over two-cycles. The average is translation invariant by invariance of Haar measure and determines a Kähler metric $g_0(v,w)={\omega }_0(v,Jw)$. Its class is still integral. The reverse implication is immediate. Therefore a Hodge metric exists exactly when an invariant one does.

Here is the Riemann bilinear criterion for a period matrix, with the signs kept explicit. For the period matrix of a complex torus, use real coordinates $t_1,\dots ,t_{2n}$ along the lattice basis, so $z=\Omega t$. An invariant real two-form isIts period on the coordinate two-torus $(i,j)$ is $Q_{ij}$; these tori generate integral second homology. Thus integrality of the class is exactly $Q\in M_{2n}(\mathbb{Z})$. Positivity of a Kähler form makes $Q$ nonsingular. Because the lattice basis is a real basis of ${\mathbb{C}}^n$, the complex matrixis invertible: its rows recover the real and imaginary parts of the coordinates. In the coordinates $(z,\bar{z})$, the matrix of the two-form is $T_0^{-t}Q{T}_0^{-1}$, whose inverse is $T_0{Q}^{-1}{T}_0^t$.

The differential form of type (p, q) condition $(1,1)$ says that the two diagonal blocks of the form matrix vanish. Since the form is nonsingular, this is equivalent to the diagonal blocks of its inverse vanishing. Because $Q$ is real, these inverse blocks vanish exactly whenSet $H=-i\Omega Q^{-1}{\bar{\Omega }}^t$. Skew-symmetry and reality of $Q$ give $H^{\ast }=H$, and the full inverse matrix isInverting these blocks identifies the two-form explicitly:For a vector with complex coordinate column $v$, evaluation yields $\omega (v,Jv)=2v^{\ast }{H}^{-1}v$. Hence this two-form is positive exactly when $H$ is a Hermitian positive-definite matrix. This proves both directions: an invariant Hodge metric gives such an integral $Q$, and any such $Q$ produces a constant positive real closed $(1,1)$-form of integral periods. In particular,As a sign check, for $\Omega =(I,iI)$ and $Q=\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right)$ the last matrix is $2I$, and the associated form is $(i/2)\sum d{z}_a\wedge d{\bar{z}}_a$.

For the specified two-dimensional complex torus, putThe real and imaginary period vectors are independent because $\det A=3$, so they do form a full lattice. Suppose a polarization matrix $Q$ existed. Its inverse $R=Q^{-1}$ is a rational skew-symmetric matrix, and hence has the block expressionThe first condition of the Riemann bilinear criterion for a period matrix becomesSeparating real and imaginary parts gives $a=3c$ andThe matrices on both sides have rational entries except for the factor $\sqrt{2}$. Irrationality of $\sqrt{2}$ therefore implies $B=B^t$ and $BJ+JB=0$. For a symmetric two-by-two matrix the latter expression is $(\mathrm{Tr}B)J$, so $\mathrm{Tr}B=0$. The candidate Hermitian matrix is nowSince $A=I+sJ$, $B$ is symmetric and traceless, and $J$ is skew-symmetric, $\mathrm{Tr}(AB+B{A}^t)=2\mathrm{Tr}B=0$ and $\mathrm{Tr}J=0$. Thus $\mathrm{Tr}H=0$. A Hermitian positive-definite matrix has strictly positive eigenvalues, hence strictly positive trace. This contradiction proves