A (partial) answer was given by Donu Arapura at Are most Kähler manifolds non-projective?

Let $C$ be a smooth projective curve of genus $g>0$, $\Gamma =\pi_1(C)$, and $\tilde{C}$ the universal cover. Choose an elliptic curve $E$ and a group homomorphism $h\colon \Gamma \to E$. Define an action of $\Gamma$ on $\tilde{C} \times E$ by $\gamma(x,y)=(\gamma x,y+h(\gamma))$, and let $S$ be the quotient.

$S$ is Kähler. If hh has infinite image, then $S$ is not algebraic.

Proof. $S$ is Kähler because $\tilde{C}\times E$ has an invariant Kähler metric. For the second statement, assume that $h$ has infinite image. Projection on the first factor gives a holomorphic map $f \colon S \to C$. The fibres of $f$ can be identified with $E$. Restricting a meromorphic function $F$ on $S$ to a fibre gives a meromorphic function on $E$ which is constant on the orbits $\{y+h(\gamma)\}$ and therefore constant. Therefore $F$ comes from $C$. This shows that transcendence degree of the field of meromorphic functions on $S$ is 1.

A (partial) answer was given by Donu Arapura at Are most Kähler manifolds non-projective?

Let $C$ be a smooth projective curve of genus $g>0$, $\Gamma =\pi_1(C)$, and $\tilde{C}$ the universal cover. Choose an elliptic curve $E$ and a group homomorphism $h\colon \Gamma \to E$. Define an action of $\Gamma$ on $\tilde{C} \times E$ by $\gamma(x,y)=(\gamma x,y+h(\gamma))$, and let $S$ be the quotient.

$S$ is Kähler. If hh has infinite image, then $S$ is not algebraic.

Proof. $S$ is Kähler because $\tilde{C}\times E$ has an invariant Kähler metric. For the second statement, assume that $h$ has infinite image. Projection on the first factor gives a holomorphic map $f \colon S \to C$. The fibres of $f$ can be identified with $E$. Restricting a meromorphic function $F$ on $S$ to a fibre gives a meromorphic function on $E$ which is constant on the orbits $\{y+h(\gamma)\}$ and therefore constant. Therefore $F$ comes from $C$. This shows that transcendence degree of the field of meromorphic functions on $S$ is 1.

A (particular) answer was given by Donu Arapura at Are most Kähler manifolds non-projective?

Let $C$ be a smooth projective curve of genus $g>0$, $\Gamma =\pi_1(C)$, and $\tilde{C}$ the universal cover. Choose an elliptic curve $E$ and a group homomorphism $h\colon \Gamma \to E$. Define an action of $\Gamma$ on $\tilde{C} \times E$ by $\gamma(x,y)=(\gamma x,y+h(\gamma))$, and let $S$ be the quotient.

$S$ is Kähler. If hh has infinite image, then $S$ is not algebraic.

Proof. $S$ is Kähler because $\tilde{C}\times E$ has an invariant Kähler metric. For the second statement, assume that $h$ has infinite image. Projection on the first factor gives a holomorphic map $f \colon S \to C$. The fibres of $f$ can be identified with $E$. Restricting a meromorphic function $F$ on $S$ to a fibre gives a meromorphic function on $E$ which is constant on the orbits $\{y+h(\gamma)\}$ and therefore constant. Therefore $F$ comes from $C$. This shows that transcendence degree of the field of meromorphic functions on $S$ is 1.

A (partial) answer was given by Donu Arapura at Are most Kähler manifolds non-projective?

Let $C$ be a smooth projective curve of genus $g>0$, $\Gamma =\pi_1(C)$, and $\tilde{C}$ the universal cover. Choose an elliptic curve $E$ and a group homomorphism $h\colon \Gamma \to E$. Define an action of $\Gamma$ on $\tilde{C} \times E$ by $\gamma(x,y)=(\gamma x,y+h(\gamma))$, and let $S$ be the quotient.

$S$ is Kähler. If hh has infinite image, then $S$ is not algebraic.

Proof. $S$ is Kähler because $\tilde{C}\times E$ has an invariant Kähler metric. For the second statement, assume that $h$ has infinite image. Projection on the first factor gives a holomorphic map $f \colon S \to C$. The fibres of $f$ can be identified with $E$. Restricting a meromorphic function $F$ on $S$ to a fibre gives a meromorphic function on $E$ which is constant on the orbits $\{y+h(\gamma)\}$ and therefore constant. Therefore $F$ comes from $C$. This shows that transcendence degree of the field of meromorphic functions on $S$ is 1.