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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.

A Hodge metric is a Kähler metric whose Kähler form represents an integral cohomology class in real de Rham cohomology. Equivalently, its periods over integral two-cycles are integers. Some conventions require the class of $\omega /(2\pi )$ to be integral instead; this changes the normalization but not existence.

A geometric Kähler potential is locally a real smooth function $\phi$ with $\omega =i\partial \bar{\partial }\phi$ for the Kähler form. Its complex Hessian must be positive definite to define a Kähler metric. Conventions sometimes absorb a factor of $1/2$ into $\phi$. Potentials differing by the real part of a holomorphic function determine the same form. This topic concerns ordinary complex geometry; the existing undisambiguated Kähler potential entry concerns superfields.

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

Let ${\mathcal{A}}^{p,q}$ denote the sheaf of differential forms of type (p, q) printed in the PDF. Take the pointwise Hermitian inner product to be linear in its first argument, and use $\mathrm{vol}={\omega }^n/n!$. The bidegrees in the question specify the conjugate-linear Hodge star: it is uniquely characterized byThus $\ast (c\beta )=\bar{c}\ast \beta$, and $\ast$ maps bidegree $(p,q)$ to $(n-p,n-q)$. If $S$ is the complex-linear extension of the real Hodge star operator, then $\ast =S\circ \text{complex conjugation}$. The operator $S$ is real and $S^2=(-1{)}^{k(2n-k)}$ on degree $k=p+q$, soUsing the complex-linear star instead would give a different bidegree, $(n-q,n-p)$; keeping the convention explicit prevents that ambiguity.

The Hermitian Lefschetz operator is exterior multiplication by $\omega$. Define $\Lambda$ pointwise byIt lowers bidegree by $(1,1)$. In a unitary real coframe with $\omega ={\sum }_j{e}^j\wedge f^j$, it is $\Lambda ={\sum }_j{\iota }_{f_j}{\iota }_{e_j}$, the corresponding sum of interior product of a differential form operators. This gives its existence and identifies it as a smooth operator. On a compact manifold, integrate the pointwise equality against $\mathrm{vol}$ to obtainNo integration by parts is needed for this order-zero operator. Thus $\Lambda$ is both the pointwise and the global formal adjoint.

Write $[A,B]=AB-BA$, and let $k$ denote the degree of the input form. The given Lefschetz commutator is $[L,\Lambda ]=(k-n)I$. For $r=1$ the required formula is exactly this identity. If it holds for $r$, the commutator derivation identity givesThe second commutator acts on degree $k+2r$, so the coefficient on the right isThis proves the commutator formula for powers of the Lefschetz operator

For injectivity of powers of the Lefschetz operator, $r=0$ is immediate. Suppose $r\ge 1$ and $r+k\le n$. If $k=0$ or $1$, then $\Lambda \psi =0$, and $L^r\psi =0$ impliesThe scalar is nonzero, since $k+r\le n$ makes $k-n+r-1\le -1$. Repeating with the smaller power shows $\psi =0$. For $k\ge 2$, induct on $r+k$, treating all $r=0$ cases as already established. The same commutator calculation yields, with $c=r(k-n+r-1)\ne 0$,The induction hypothesis applies to $L^{r-1}$ on degree $k$, since $(r-1)+k<r+k$ and $(r-1)+k\le n$. Hence $\psi =L\alpha$ with $\alpha =c^{-1}\Lambda \psi$ of degree $k-2$. Moreover $L^{r+1}\alpha =L^r\psi =0$. Since $(r+1)+(k-2)=r+k-1$, induction also makes $L^{r+1}$ injective on that degree, so $\alpha =0$ and $\psi =0$. All operators preserve restriction to open sets; the argument applies on every open set. Therefore $L^r$ is an injective morphism of sheaves whenever $r+k\le n$.

On a Kähler manifold, the Kähler identities, with the positive form convention used above, areTaking formal adjoints conjugates the scalar and reverses the order in the commutator. Since ${\Lambda }^{\ast }=L$, this givesTo derive the Kähler Laplacian identity, put $P=\partial$ and $B=\bar{\partial }$, so $P^2=B^2=0$ and $PB+BP=0$. Write $\{A,B\}=AB+BA$ for an anticommutator. The Kähler identities say $B^{\ast }=-i[\Lambda ,P]$ and $P^{\ast }=i[\Lambda ,B]$. Thereforebecause expansion leaves only terms containing $P^2$ or $B^2$. Furthermore,Expanding these expressions and using $PB=-BP$ makes them equal. Since $d=P+B$ and $d^{\ast }=P^{\ast }+B^{\ast }$, the mixed anticommutators already vanish, and consequentlyClosedness of the Kähler form implies $[L,P]=[L,B]=0$. Using the adjoint identity above,The equality of the three Laplacians therefore proves each Laplacian commutes with $L$.

For a compact Kähler manifold, the Dolbeault Hodge decomposition is the orthogonal decompositionHere harmonicity for the Dolbeault Laplacian and for the Hodge Laplacian agrees by the preceding identity. If $\bar{\partial }\alpha =0$ and $\alpha =h+\bar{\partial }u+{\bar{\partial }}^{\ast }v$, then $\bar{\partial }{\bar{\partial }}^{\ast }v=0$. Taking its inner product with $v$ gives $\Vert {\bar{\partial }}^{\ast }v{\Vert }^2=0$. Thus $\alpha$ is cohomologous to $h$. Conversely, a harmonic form is $\bar{\partial }$-closed, and a harmonic exact form $h=\bar{\partial }u$ satisfies $\Vert h{\Vert }^2=({\bar{\partial }}^{\ast }h,u)=0$. This proves existence and uniqueness of the harmonic representative and henceThese spaces are finite dimensional by the ellipticity of the Dolbeault Laplacian on the compact manifold.

Complex conjugation commutes with the real operator ${\Delta }_d$ and interchanges the bidegrees $(p,q)$ and $(q,p)$. It is therefore a conjugate-linear bijection of the harmonic spaces, proving Hodge symmetry. The real Hodge star operator commutes with ${\Delta }_d$, as does complex conjugation; hence their composition, our conjugate-linear Hodge star, sends harmonic $(p,q)$-forms bijectively to harmonic $(n-p,n-q)$-forms. Its square is the nonzero scalar established above. This proves Hodge duality, and gives

For the final hard Lefschetz isomorphism on Dolbeault cohomology, take $k=p+q\le n$ and $r=n-k$. This restriction is necessary to make the displayed power nonnegative. The Hermitian Lefschetz operator raises bidegree by $(1,1)$, and the Lefschetz operator preserves harmonic forms; thereforeis well defined. It is injective by injectivity of powers of the Lefschetz operator, since $r+k=n$. Hodge symmetry and Hodge duality give $h^{n-q,n-p}=h^{q,p}=h^{p,q}$, so it is a bijection between finite-dimensional spaces of equal dimension. Because $[L,\bar{\partial }]=0$, the map on harmonic representatives agrees with exterior multiplication by $[\omega {]}^{n-k}$ on Dolbeault cohomology. We concludeFor $p+q>n$, the corresponding valid statement is the inverse of the positive-power isomorphism from bidegree $(n-q,n-p)$ to $(p,q)$; a negative exterior-multiplication power is not defined.

Use the normalization in which a projective line has area $2\pi$. On the affine chart of Complex projective space where $Z_0\ne 0$, set $w_j=Z_j/Z_0$ and $S=1+{\sum }_j|w_j{|}^2$. Define the Fubini-Study form byOn another chart the corresponding potential differs by $\log |h{|}^2$ for a nowhere-zero holomorphic function $h$, whose $\partial \bar{\partial }$ is zero. Thus these local differential forms glue to a global form. It is real and closed. Its Hermitian coefficient matrix is positive definite, since for $v\ne 0$ the Cauchy-Schwarz inequality givesHence it is a Kähler form and in particular a symplectic form.

On a complex projective line with $w=x+iy$ this is $2(1+|w{|}^2{)}^{-2}dx\wedge dy$, whose total area is $2\pi$. Thus $[{\omega }_{\mathrm{FS}}/(2\pi )]$ is the positive generator, equivalently $c_1(\mathcal{O}(1))$. Another common normalization uses half this form and gives line area $\pi$; the scale must be carried consistently into symplectic reduction.

Write $L\alpha =\omega \wedge \alpha$, the Lefschetz operator of a Kähler manifold. The metric and volume form define the $L^2$ inner product on smooth complex forms, and $\Lambda =L^{\ast }$ is its formal adjoint. Similarly, ${\bar{\partial }}^{\ast }$ is the formal adjoint of the Dolbeault operator, characterized by $⟨\bar{\partial }\beta ,\alpha ⟩=⟨\beta ,{\bar{\partial }}^{\ast }\alpha ⟩$. The Dolbeault Laplacian isOn the compact manifold without boundary, integration by parts givesIf the Laplacian vanishes, both terms are zero. Conversely, if both operators annihilate $\alpha$, the defining formula annihilates it. Thus harmonicity is equivalent to being both $\bar{\partial }$-closed and ${\bar{\partial }}^{\ast }$-closed.

Because $\omega$ is closed and has type $(1,1)$, $\bar{\partial }\omega =0$ and $[L,\bar{\partial }]=0$. The supplied identity from the Kähler identities gives $[L,{\bar{\partial }}^{\ast }]=-i\partial$. With ordinary commutators for the even-degree operator $L$,The last identity follows from $d^2=0$. This also proves that the Lefschetz operator preserves harmonic forms.

For the cohomology map one can work directly with forms: $\bar{\partial }({\omega }^k\wedge \alpha )={\omega }^k\wedge \bar{\partial }\alpha$, since $\omega$ has even degree. It takes closed forms to closed forms and exact forms to exact forms. Therefore the $k$th power of $L$ inducesThe bidegree is $(p+k,q+k)$, including the zero groups outside the dimension range. No isomorphism claim is needed here.

Finally put $\theta =-i\partial f$, so $\bar{\partial }\theta =i\partial \bar{\partial }f={\omega }^{\prime }-\omega$. LetBoth Kähler forms are $\bar{\partial }$-closed, so $\bar{\partial }{T}_k=0$. For a closed representative $\alpha$,This primitive has type $(p+k,q+k-1)$; the degree-one sign produces no additional term because $T_k\wedge \alpha$ is closed. HenceThis is the dependence of Lefschetz maps on the Dolbeault class.

A positive holomorphic line bundle $F$ admits a Hermitian metric whose Chern connection curvature satisfies that $i{F}_F$ is a positive real (1, 1)-form. In a local holomorphic local frame with squared length $e^{-\phi }$, the local formula for the Chern connection on a line bundle gives $F_F=\partial \bar{\partial }\phi$, so positivity means $i\partial \bar{\partial }\phi$ is positive definite. Its closedness makes $\omega =i{F}_F$ a Kähler form; use this form to define the operators below.

Let $n={\dim }_{\mathbb{C}}X>0$, choose a Hermitian metric on $E$, and equip $G_m=E\otimes F^{-m}$ with the tensor-product metric. The curvature of a tensor product connection givesOn $G_m$-valued zero-forms, the Lefschetz commutator is $[L,\Lambda ]=-n\mathrm{id}$. Thus the Bochner-Kodaira-Nakano identity givesThis last operator is a fixed smooth self-adjoint bundle endomorphism. Compactness supplies a finite $C$ such that $⟨R_{E}v,v⟩\ge -C\Vert v{\Vert }^2$ at every point. For a holomorphic section $s$ of $G_m$, $D^{\prime \prime }s=0$ and $D^{\prime \prime \ast }s=0$ by degree, so integrating the identity givesChoose an integer $m_0$ with $m_{0}n>C$. Then $\boxed{H^0(X,E\otimes F^{-m})=0(m\ge m_0)}$. The threshold depends on the fixed bundle $E$, as the curvature bound makes explicit. Positive complex dimension is necessary: on a zero-dimensional manifold positivity is vacuous and a nonzero fibre has nonzero sections for every twist.

The Kähler form and the Hermitian metric give the $L^2$ inner product on vector-bundle-valued differential forms, using volume ${\omega }^n/n!$. Define the formal adjoint ${\bar{\partial }}_E^{\ast }$ and the elliptic, self-adjoint, nonnegative Dolbeault LaplacianIts harmonic space isThe equality follows from $⟨{\Delta }_E^{\prime \prime }\alpha ,\alpha ⟩=\Vert {\bar{\partial }}_E\alpha {\Vert }^2+\Vert {\bar{\partial }}_E^{\ast }\alpha {\Vert }^2$. On compact $X$, the bundle-valued Dolbeault Hodge decomposition states that this space is finite dimensional and thatThe sum is orthogonal for the $L^2$ inner product and all summands here consist of smooth forms. Every Dolbeault cohomology class has a unique harmonic representative, giving $H_{\bar{\partial }}^{p,q}(X,E)\cong {\mathcal{H}}^{p,q}(X,E)$ and, by the Dolbeault theorem, the corresponding sheaf cohomology isomorphism. This is a decomposition for ${\bar{\partial }}_E$, whose square is zero; it does not require the full Chern connection to be flat.

Choose a positive Hermitian metric on $L$. Its curvature form $\omega =i{F}_h$ is a Kähler form on the compact complex manifold of dimension one. Part (c) gives a threshold $k_0$ with $H^1(X,\mathcal{O}(L^k))=0$ for every $k\ge k_0$. We use this to prescribe a finite principal part of a meromorphic section.

Fix a smooth cutoff $\chi$ supported inside the given coordinate chart and equal to one near $x_0$. Let $P(z)={\sum }_{j=-r}^{-1}{a}_j{z}^j$. On the punctured manifold define $t_k=\chi P(z){\zeta }^k$ inside the chart, extended by zero outside. Its Dolbeault operatoris smooth globally: it vanishes near the pole and near the boundary of the chart. It is ${\bar{\partial }}_{L^k}$-closed, either by the square-zero identity or because there are no $(0,2)$-forms in complex dimension one. The vanishing of $H^1(X,\mathcal{O}(L^k))$, together with the Dolbeault theorem, therefore gives a global smooth section $u_k$ with ${\bar{\partial }}_{L^k}{u}_k={\eta }_k$.

The section $s_k=t_k-u_k$ is holomorphic on $X\setminus \{x_0\}$. Near $x_0$, ${\eta }_k=0$, so $u_k=f_k(z){\zeta }^k$ with $f_k$ holomorphic through zero. Its Taylor series then giveswhere the second series converges near zero and its coefficients are those of $-f_k$. The threshold is determined by the fixed bundle $L$ and $X$, and is independent of the prescribed coefficients. In fact the same threshold works for every finite pole order $r$, since the Dolbeault cohomology obstruction is the same $H^1(X,L^k)$ for every such cutoff construction.

For a potential already smooth on all of $\mathbb{C}$, the associated form is a Kähler form exactly when the displayed two positivity conditions hold. The first gives a positive metric coefficient off the origin; the second is positivity of $f_{z\bar{z}}(0)$. The associated real metric is $2f_{z\bar{z}}(d{x}^2+d{y}^2)$. Positivity away from the origin alone permits a degenerate metric at the origin.

A complex torus with period matrix of a complex torus $\Omega$ admits a Hodge metric precisely when there is a nonsingular integral skew-symmetric matrix $Q$ satisfying the displayed conditions. The entries of $Q$ are the periods of the invariant Kähler form over lattice two-tori. The first condition imposes type $(1,1)$, and the second imposes positivity; the factor $-i$ corresponds to the convention $\omega (v,Jv)>0$.

For a positive Hermitian holomorphic line bundle on compact $X$, using its curvature as the Kähler form, a linear lower bound in $k$ for the full Dolbeault Laplacian on positive antiholomorphic degrees rules out harmonic forms for sufficiently large $k$. The Dolbeault theorem and Dolbeault Hodge decomposition then imply $H^{q>0}(X,L^k)=0$.