[proofplan] We verify the defining properties of a Kähler form on the product complex manifold. The projection pullbacks are real forms of type $(1,1)$ because the projections intertwine the product complex structure with the complex structures on the factors. Closedness follows from the naturality of the [exterior derivative](/theorems/1525) and the closedness of $\omega_X$ and $\omega_Y$. Finally, positivity is checked on a tangent vector $(v,w)$, where the product form splits into the sum of the two positive factor contributions. [/proofplan]

[step:Check that the projection pullbacks are real forms of type $(1,1)$]Fix a point $(x,y)\in X\times Y$. Under the standard identification \begin{align*} T_{(x,y)}(X\times Y)=T_xX\oplus T_yY, \end{align*} the differentials of the projections are the linear maps \begin{align*} d(\pi_X)_{(x,y)}:T_xX\oplus T_yY&\to T_xX, & d(\pi_X)_{(x,y)}(v,w)&=v, \end{align*} and \begin{align*} d(\pi_Y)_{(x,y)}:T_xX\oplus T_yY&\to T_yY, & d(\pi_Y)_{(x,y)}(v,w)&=w. \end{align*} Thus $\pi_X$ and $\pi_Y$ are holomorphic with respect to the product complex structure, since \begin{align*} d(\pi_X)_{(x,y)}(J_{X\times Y}(v,w))=J_Xv \end{align*} and \begin{align*} d(\pi_Y)_{(x,y)}(J_{X\times Y}(v,w))=J_Yw. \end{align*} Let $\xi=(v_1,w_1)$ and $\eta=(v_2,w_2)$ be tangent vectors in $T_{(x,y)}(X\times Y)$. Since $\omega_X$ has type $(1,1)$, the pullback satisfies \begin{align*} (\pi_X^*\omega_X)_{(x,y)}(J_{X\times Y}\xi,J_{X\times Y}\eta)=\omega_{X,x}(J_Xv_1,J_Xv_2). \end{align*} By the type $(1,1)$ property of $\omega_X$, this equals \begin{align*} \omega_{X,x}(v_1,v_2)=(\pi_X^*\omega_X)_{(x,y)}(\xi,\eta). \end{align*} The same argument gives \begin{align*} (\pi_Y^*\omega_Y)_{(x,y)}(J_{X\times Y}\xi,J_{X\times Y}\eta)=(\pi_Y^*\omega_Y)_{(x,y)}(\xi,\eta). \end{align*} Therefore both pullbacks have type $(1,1)$, and so their sum $\omega$ has type $(1,1)$. Since $\omega_X$ and $\omega_Y$ are real forms and pullback preserves real-valuedness of forms, $\omega$ is real.[/step]

[guided]We first make explicit what the product complex structure does. At a point $(x,y)\in X\times Y$, a tangent vector is a pair $(v,w)$ with $v\in T_xX$ and $w\in T_yY$. The product complex structure is the [linear map](/page/Linear%20Map) \begin{align*} J_{X\times Y,(x,y)}:T_xX\oplus T_yY&\to T_xX\oplus T_yY, & (v,w)&\mapsto (J_{X,x}v,J_{Y,y}w). \end{align*} The projection maps forget one component, so their differentials are \begin{align*} d(\pi_X)_{(x,y)}(v,w)=v \end{align*} and \begin{align*} d(\pi_Y)_{(x,y)}(v,w)=w. \end{align*} These formulas show that each projection commutes with the relevant complex structures: \begin{align*} d(\pi_X)_{(x,y)}(J_{X\times Y}(v,w))=d(\pi_X)_{(x,y)}(J_Xv,J_Yw)=J_Xv \end{align*} and \begin{align*} d(\pi_Y)_{(x,y)}(J_{X\times Y}(v,w))=d(\pi_Y)_{(x,y)}(J_Xv,J_Yw)=J_Yw. \end{align*} Now let $\xi=(v_1,w_1)$ and $\eta=(v_2,w_2)$ be tangent vectors at $(x,y)$. To check that $\pi_X^*\omega_X$ has type $(1,1)$, we use the defining formula for pullback of a $2$-form: \begin{align*} (\pi_X^*\omega_X)_{(x,y)}(\xi,\eta)=\omega_{X,x}(d(\pi_X)_{(x,y)}\xi,d(\pi_X)_{(x,y)}\eta)=\omega_{X,x}(v_1,v_2). \end{align*} Applying this formula after inserting the product complex structure gives \begin{align*} (\pi_X^*\omega_X)_{(x,y)}(J_{X\times Y}\xi,J_{X\times Y}\eta)=\omega_{X,x}(J_Xv_1,J_Xv_2). \end{align*} Because $\omega_X$ is a Kähler form, it is a real form of type $(1,1)$ on $X$, hence \begin{align*} \omega_{X,x}(J_Xv_1,J_Xv_2)=\omega_{X,x}(v_1,v_2). \end{align*} Combining the last two displays proves \begin{align*} (\pi_X^*\omega_X)_{(x,y)}(J_{X\times Y}\xi,J_{X\times Y}\eta)=(\pi_X^*\omega_X)_{(x,y)}(\xi,\eta). \end{align*} The proof for $\pi_Y^*\omega_Y$ is the same with the $Y$-component: \begin{align*} (\pi_Y^*\omega_Y)_{(x,y)}(J_{X\times Y}\xi,J_{X\times Y}\eta)=\omega_{Y,y}(J_Yw_1,J_Yw_2)=\omega_{Y,y}(w_1,w_2)=(\pi_Y^*\omega_Y)_{(x,y)}(\xi,\eta). \end{align*} Thus both summands are of type $(1,1)$, and their sum is also of type $(1,1)$. Since the values of $\omega_X$ and $\omega_Y$ are real and pullback only evaluates these real forms on pushed-forward tangent vectors, both pullbacks are real forms. Therefore $\omega=\pi_X^*\omega_X+\pi_Y^*\omega_Y$ is a real form of type $(1,1)$.[/guided]

[step:Use naturality of the exterior derivative to prove closedness] The exterior derivative commutes with pullback by smooth maps. Since $\pi_X$ and $\pi_Y$ are smooth and $\omega_X$ and $\omega_Y$ are Kähler forms, we have \begin{align*} d\omega=d(\pi_X^*\omega_X+\pi_Y^*\omega_Y). \end{align*} By linearity of $d$, \begin{align*} d\omega=d(\pi_X^*\omega_X)+d(\pi_Y^*\omega_Y). \end{align*} By naturality of $d$ under pullback, \begin{align*} d\omega=\pi_X^*(d\omega_X)+\pi_Y^*(d\omega_Y). \end{align*} Because $\omega_X$ and $\omega_Y$ are closed, \begin{align*} d\omega_X=0 \end{align*} and \begin{align*} d\omega_Y=0. \end{align*} Therefore \begin{align*} d\omega=0. \end{align*} [/step]

[step:Compute the positivity of the product form componentwise] Let $(x,y)\in X\times Y$, and let $\xi=(v,w)\in T_xX\oplus T_yY$ be nonzero. Then \begin{align*} J_{X\times Y}\xi=(J_Xv,J_Yw). \end{align*} Using the defining formula for pullback, \begin{align*} \omega_{(x,y)}(\xi,J_{X\times Y}\xi)=\omega_{X,x}(v,J_Xv)+\omega_{Y,y}(w,J_Yw). \end{align*} Since $\omega_X$ and $\omega_Y$ are positive Kähler forms, \begin{align*} \omega_{X,x}(v,J_Xv)\ge 0 \end{align*} with equality if and only if $v=0$, and \begin{align*} \omega_{Y,y}(w,J_Yw)\ge 0 \end{align*} with equality if and only if $w=0$. Because $\xi=(v,w)$ is nonzero, at least one of $v$ and $w$ is nonzero. Hence at least one summand is strictly positive and the other is nonnegative, so \begin{align*} \omega_{(x,y)}(\xi,J_{X\times Y}\xi)>0. \end{align*} Thus $\omega$ is positive with respect to the product complex structure. [/step]

[step:Conclude that the product form is Kähler] We have shown that $\omega$ is real, has type $(1,1)$, is closed, and is positive on every nonzero tangent vector of $X\times Y$ with respect to the product complex structure. These are precisely the defining properties of a Kähler form. Therefore \begin{align*} \omega=\pi_X^*\omega_X+\pi_Y^*\omega_Y \end{align*} is a Kähler form on $X\times Y$. [/step]