[proofplan] We construct the Kähler form by descending the standard flat Kähler form on $\mathbb C^n$ to the quotient by the lattice. [Translation invariance](/theorems/4911) ensures that the form is well-defined on $X=\mathbb C^n/\Lambda$. Since the quotient map is a local biholomorphism, closedness, type, reality, and positivity may all be checked after pulling back to $\mathbb C^n$, where they are immediate from the explicit formula. [/proofplan]

[step:Put the standard flat form on $\mathbb C^n$] Let $z_j=x_j+i y_j$ be the standard complex coordinate functions on $\mathbb C^n$, where $x_j,y_j:\mathbb C^n\to\mathbb R$ are the real coordinate functions. Define the smooth complex-valued $2$-form \begin{align*} \omega_0=\frac{i}{2}\sum_{j=1}^n dz_j\wedge d\overline{z_j} \end{align*} on $\mathbb C^n$. Since $dz_j$ has type $(1,0)$ and $d\overline{z_j}$ has type $(0,1)$, the form $\omega_0$ has type $(1,1)$. Also, \begin{align*} \overline{\frac{i}{2}dz_j\wedge d\overline{z_j}}=-\frac{i}{2}d\overline{z_j}\wedge dz_j=\frac{i}{2}dz_j\wedge d\overline{z_j} \end{align*} so $\omega_0$ is real. Finally, using $z_j=x_j+i y_j$, we have \begin{align*} \frac{i}{2}dz_j\wedge d\overline{z_j}=dx_j\wedge dy_j, \end{align*} and therefore \begin{align*} \omega_0=\sum_{j=1}^n dx_j\wedge dy_j. \end{align*} Because each coefficient is constant and each coordinate $1$-form is closed, $d\omega_0=0$. [/step]

[step:Show that lattice translations preserve the flat form]For each $\lambda\in\Lambda$, define the translation map \begin{align*} \tau_\lambda:\mathbb C^n&\to\mathbb C^n, & z&\mapsto z+\lambda. \end{align*} For every coordinate function $z_j:\mathbb C^n\to\mathbb C$, the pullback satisfies \begin{align*} \tau_\lambda^*z_j=z_j+\lambda_j, \end{align*} where $\lambda_j\in\mathbb C$ is the $j$th coordinate of $\lambda$. Taking exterior derivatives gives \begin{align*} d(\tau_\lambda^*z_j)=dz_j, \end{align*} and similarly \begin{align*} d(\tau_\lambda^*\overline{z_j})=d\overline{z_j}. \end{align*} Hence \begin{align*} \tau_\lambda^*\omega_0 =\frac{i}{2}\sum_{j=1}^n dz_j\wedge d\overline{z_j} =\omega_0. \end{align*} Thus $\omega_0$ is invariant under every translation by an element of $\Lambda$.[/step]

[guided]The quotient $X=\mathbb C^n/\Lambda$ identifies points of $\mathbb C^n$ that differ by a lattice vector. Therefore a differential form on $\mathbb C^n$ can descend to the quotient only if it gives the same value at all points lying in the same $\Lambda$-orbit. This is exactly the invariance condition under the translations $\tau_\lambda$. Fix $\lambda\in\Lambda$. We write $\lambda=(\lambda_1,\dots,\lambda_n)$ with $\lambda_j\in\mathbb C$, and define \begin{align*} \tau_\lambda:\mathbb C^n&\to\mathbb C^n, & z&\mapsto z+\lambda. \end{align*} For the $j$th coordinate function $z_j:\mathbb C^n\to\mathbb C$, composition with $\tau_\lambda$ gives \begin{align*} \tau_\lambda^*z_j=z_j+\lambda_j. \end{align*} The constant term $\lambda_j$ disappears after applying the [exterior derivative](/theorems/1525), so \begin{align*} d(\tau_\lambda^*z_j)=d(z_j+\lambda_j)=dz_j. \end{align*} Complex conjugation gives the corresponding identity \begin{align*} d(\tau_\lambda^*\overline{z_j})=d\overline{z_j}. \end{align*} Substituting these identities into the explicit formula for $\omega_0$ yields \begin{align*} \tau_\lambda^*\omega_0 =\frac{i}{2}\sum_{j=1}^n d(\tau_\lambda^*z_j)\wedge d(\tau_\lambda^*\overline{z_j}) =\frac{i}{2}\sum_{j=1}^n dz_j\wedge d\overline{z_j} =\omega_0. \end{align*} Thus the standard flat form does not change under any lattice translation, which is the precise descent condition needed for a form on the quotient.[/guided]

[step:Descend the invariant form to the quotient torus] Let \begin{align*} \pi:\mathbb C^n&\to X, & z&\mapsto z+\Lambda \end{align*} be the quotient map. Since $\Lambda$ is a rank $2n$ lattice, its translation action on $\mathbb C^n$ is free, properly discontinuous, and holomorphic; hence $X$ is a complex manifold and $\pi$ is a local biholomorphism. We define a smooth $2$-form $\omega$ on $X$ by requiring \begin{align*} \pi^*\omega=\omega_0. \end{align*} This condition determines $\omega$ uniquely because $\pi$ is a surjective local diffeomorphism. It is also well-defined: if $z,z'\in\mathbb C^n$ satisfy $\pi(z)=\pi(z')$, then $z'=z+\lambda$ for some $\lambda\in\Lambda$, and the equality $\tau_\lambda^*\omega_0=\omega_0$ shows that the values prescribed from $z$ and $z'$ agree. Since $\pi$ is a local biholomorphism and $\pi^*\omega=\omega_0$ has type $(1,1)$ and is real, the descended form $\omega$ is a smooth real $(1,1)$-form on $X$. [/step]

[step:Verify closedness and positivity after pulling back to $\mathbb C^n$] Because pullback commutes with the exterior derivative, \begin{align*} \pi^*(d\omega)=d(\pi^*\omega)=d\omega_0=0. \end{align*} The map $\pi$ is a local diffeomorphism, so pullback by $\pi$ is pointwise injective on forms. Therefore $d\omega=0$. It remains to prove positivity. Let $x\in X$ and let $v\in T_xX$ be a nonzero tangent vector. Choose $z\in\mathbb C^n$ with $\pi(z)=x$. Since $d\pi_z:T_z\mathbb C^n\to T_xX$ is a real-linear isomorphism, there is a unique nonzero vector $u\in T_z\mathbb C^n$ such that $d\pi_z(u)=v$. Write \begin{align*} u=\sum_{j=1}^n a_j\partial_{x_j}\big|_z+\sum_{j=1}^n b_j\partial_{y_j}\big|_z \end{align*} with [real numbers](/page/Real%20Numbers) $a_j,b_j\in\mathbb R$. For the standard complex structure $J$ on $\mathbb C^n$, \begin{align*} Ju=\sum_{j=1}^n -b_j\partial_{x_j}\big|_z+\sum_{j=1}^n a_j\partial_{y_j}\big|_z. \end{align*} Using $\omega_0=\sum_{j=1}^n dx_j\wedge dy_j$, we obtain \begin{align*} \omega_0(u,Ju)=\sum_{j=1}^n (a_j^2+b_j^2)>0. \end{align*} Since $\pi$ is holomorphic, $d\pi_z(Ju)=Jv$. Therefore \begin{align*} \omega_x(v,Jv)=(\pi^*\omega)_z(u,Ju)=\omega_0(u,Ju)>0. \end{align*} Thus $\omega$ is positive. We have proved that $\omega$ is a closed positive real $(1,1)$-form on $X$, so $\omega$ is a Kähler form on the complex torus $X$. [/step]