Return to Answer

You can search the topics related to Chern-Laplacian on $(M,\omega)$ $$ \Delta^{C h} f=\Delta_d f+(d f, \theta)_\omega. $$ Actually it's just a Hodge- Laplacian plus a linear gradient term, here $\theta$ is the Lee form of $\omega$. For the first question, this requires the metric to be Gauduchon ($d^* \theta=0$), so we can integrate the gradient term by part to get that $$ \int_M \Delta^{Ch}f=0. $$ For the second question, for $$ \Delta^{Ch}u=f, $$ you can always find a solution when the metric is in the conformal class of the original metric, because in the conformal class there must be a Gauduchon metric, so if we consider $$ \left(\Delta^{C h}\right)^* g=\Delta_d g-(d g, \theta)_\eta $$ here $\eta$ is the new Gauduchon metric and $\theta$ is the Lee form of $\eta$. So $$ 0=\int_M u\left(\Delta^{C h}\right)^* u =\int_M\left(|\nabla u|^2-\frac{1}{2}\left(d u^2, \theta\right)\right) =\int_M|\nabla u|^2 , $$ which means kernel of $\left(\Delta^{C h}\right)^*$ is the constant set. Since the integral of $f$ is zero, it means that $f \in\left(\operatorname{ker}\left(\Delta^{C h}\right)^*\right)^{\perp}=\operatorname{imm} \Delta^{C h}$, which proves the existence of the solution.

And when the manifold is kahler, the gradient term will vanish, then the Chern connection is the same as Levi-Civita connection, you can refer to Daniele Angella's work on this topic, for example On Gauduchon connections with Kähler-like curvature

Besides, the prescribed Chern scalar curvature problem is related to Kazdan-Warner equation, but there are some open problems in this topic, you can read the following paper if you are interested On Chern-Yamabe problem

The prescribed Chern scalar curvature problem

The prescribed Gauduchon scalar curvature problem in almost Hermitian geometry

You can search the topics related to Chern-Laplacian on $(M,\omega)$ $$ \Delta^{C h} f=\Delta_d f+(d f, \theta)_\omega. $$ Actually it's just a Hodge- Laplacian plus a linear gradient term, here $\theta$ is the Lee form of $\omega$. For the first question, this requires the metric to be Gauduchon ($d^* \theta=0$), so we can integrate the gradient term by part to get that $$ \int_M \Delta^{Ch}f=0. $$ For the second question, for $$ \Delta^{Ch}u=f, $$ you can always find a solution when the metric is in the conformal class of the original metric, because in the conformal class there must be a Gauduchon metric, so if we consider $$ \left(\Delta^{C h}\right)^* g=\Delta_d g-(d g, \theta)_\eta $$ here $\eta$ is the new Gauduchon metric and $\theta$ is the Lee form of $\eta$. We will have $$ 0=\int_M u\left(\Delta^{C h}\right)^* u =\int_M\left(|\nabla u|^2-\frac{1}{2}\left(d u^2, \theta\right)\right) =\int_M|\nabla u|^2 , $$ which means kernel of $\left(\Delta^{C h}\right)^*$ is the constant set. Since the integral of $f$ is zero, it means that $f \in\left(\operatorname{ker}\left(\Delta^{C h}\right)^*\right)^{\perp}=\operatorname{imm} \Delta^{C h}$, which proves the existence of the solution.

And when the manifold is kahler, the gradient term will vanish, then the Chern connection is the same as Levi-Civita connection, you can refer to Daniele Angella's work on this topic, for example On Gauduchon connections with Kähler-like curvature

Besides, the prescribed Chern scalar curvature problem is related to Kazdan-Warner equation, but there are some open problems in this topic, you can read the following paper if you are interested On Chern-Yamabe problem

The prescribed Chern scalar curvature problem

The prescribed Gauduchon scalar curvature problem in almost Hermitian geometry

You can search the topics related to Chern-Laplacian on $(M,\omega)$ $$ \Delta^{C h} f=\Delta_d f+(d f, \theta)_\omega. $$ Actually it's just a Hodge- Laplacian plus a linear gradient term, here $\theta$ is the Lee form of $\omega$. For the first question, this requires the metric to be Gauduchon ($d^* \theta=0$), so we can integrate the gradient term by part to get that $$ \int_M \Delta^{Ch}f=0. $$ For the second question, for $$ \Delta^{Ch}u=f, $$ you can always find a solution when the metric is in the conformal class of the original metric, because in the conformal class there must be a Gauduchon metric, so if we consider $$ \left(\Delta^{C h}\right)^* g=\Delta_d g-(d g, \theta)_\eta $$ here $\eta$ is the new Gauduchon metric and $\theta$ is the Lee form of $\eta$. So $$ 0=\int_M u\left(\Delta^{C h}\right)^* u =\int_M\left(|\nabla u|^2-\frac{1}{2}\left(d u^2, \theta\right)\right) =\int_M|\nabla u|^2 , $$ which means kernel of $\left(\Delta^{C h}\right)^*$ is the constant set. Since the integral of $f$ is zero, it means that $f \in\left(\operatorname{ker}\left(\Delta^{C h}\right)^*\right)^{\perp}=\operatorname{imm} \Delta^{C h}$, which proves the existence of the solution.

And when the manifold is kahler, the gradient term will vanish, then the Chern connection is the same as Levi-Civita connection, you can refer to Daniele Angella's work on this topic, for example On Gauduchon connections with Kähler-like curvature

On Chern-Yamabe problem

Besides, the prescribed Chern scalar curvature problem is related to Kazdan-Warner equation, but there is some open problems in this topic, you can read the following paper if you are interested

The prescribed Chern scalar curvature problem

The prescribed Gauduchon scalar curvature problem in almost Hermitian geometry

You can search the topics related to Chern-Laplacian on $(M,\omega)$ $$ \Delta^{C h} f=\Delta_d f+(d f, \theta)_\omega. $$ Actually it's just a Hodge- Laplacian plus a linear gradient term, here $\theta$ is the Lee form of $\omega$. For the first question, this requires the metric to be Gauduchon ($d^* \theta=0$), so we can integrate the gradient term by part to get that $$ \int_M \Delta^{Ch}f=0. $$ For the second question, for $$ \Delta^{Ch}u=f, $$ you can always find a solution when the metric is in the conformal class of the original metric, because in the conformal class there must be a Gauduchon metric, so if we consider $$ \left(\Delta^{C h}\right)^* g=\Delta_d g-(d g, \theta)_\eta $$ here $\eta$ is the new Gauduchon metric and $\theta$ is the Lee form of $\eta$. So $$ 0=\int_M u\left(\Delta^{C h}\right)^* u =\int_M\left(|\nabla u|^2-\frac{1}{2}\left(d u^2, \theta\right)\right) =\int_M|\nabla u|^2 , $$ which means kernel of $\left(\Delta^{C h}\right)^*$ is the constant set. Since the integral of $f$ is zero, it means that $f \in\left(\operatorname{ker}\left(\Delta^{C h}\right)^*\right)^{\perp}=\operatorname{imm} \Delta^{C h}$, which proves the existence of the solution.

And when the manifold is kahler, the gradient term will vanish, then the Chern connection is the same as Levi-Civita connection, you can refer to Daniele Angella's work on this topic, for example On Gauduchon connections with Kähler-like curvature

Besides, the prescribed Chern scalar curvature problem is related to Kazdan-Warner equation, but there are some open problems in this topic, you can read the following paper if you are interested On Chern-Yamabe problem

The prescribed Chern scalar curvature problem

The prescribed Gauduchon scalar curvature problem in almost Hermitian geometry

You can search the topics related to Chern-Laplacian on $(M,\omega)$ $$ \Delta^{C h} f=\Delta_d f+(d f, \theta)_\omega. $$ Actually it's just a Hodge- Laplacian plus a linear gradient term, here $\theta$ is the Lee form of $\omega$. For the first question, this requires the metric to be Gauduchon ($d^* \theta=0$), so we can integrate the gradient term by part to get that $$ \int_M \Delta^{Ch}f=0. $$ For the second question, for $$ \Delta^{Ch}u=f, $$ you can always find a solution when the metric is in the conformal class of the original metric, because in the conformal class there must be a Gauduchon metric, so if we consider $$ \left(\Delta^{C h}\right)^* g=\Delta_d g-(d g, \theta)_\eta $$ here $\eta$ is the new Gauduchon metric and $\theta$ is the Lee form of $\eta$. So $$ 0=\int_M u\left(\Delta^{C h}\right)^* u =\int_M\left(|\nabla u|^2-\frac{1}{2}\left(d u^2, \theta\right)\right) =\int_M|\nabla u|^2 , $$ which means kernel of $\left(\Delta^{C h}\right)^*$ is the constant set. Since the integral of $f$ is zero, it means that $f \in\left(\operatorname{ker}\left(\Delta^{C h}\right)^*\right)^{\perp}=\operatorname{imm} \Delta^{C h}$, which proves the existence of the solution.

You can search the topics related to Chern-Laplacian on $(M,\omega)$ $$ \Delta^{C h} f=\Delta_d f+(d f, \theta)_\omega. $$ Actually it's just a Hodge- Laplacian plus a linear gradient term, here $\theta$ is the Lee form of $\omega$. For the first question, this requires the metric to be Gauduchon ($d^* \theta=0$), so we can integrate the gradient term by part to get that $$ \int_M \Delta^{Ch}f=0. $$ For the second question, for $$ \Delta^{Ch}u=f, $$ you can always find a solution when the metric is in the conformal class of the original metric, because in the conformal class there must be a Gauduchon metric, so if we consider $$ \left(\Delta^{C h}\right)^* g=\Delta_d g-(d g, \theta)_\eta $$ here $\eta$ is the new Gauduchon metric and $\theta$ is the Lee form of $\eta$. So $$ 0=\int_M u\left(\Delta^{C h}\right)^* u =\int_M\left(|\nabla u|^2-\frac{1}{2}\left(d u^2, \theta\right)\right) =\int_M|\nabla u|^2 , $$ which means kernel of $\left(\Delta^{C h}\right)^*$ is the constant set. Since the integral of $f$ is zero, it means that $f \in\left(\operatorname{ker}\left(\Delta^{C h}\right)^*\right)^{\perp}=\operatorname{imm} \Delta^{C h}$, which proves the existence of the solution.

And when the manifold is kahler, the gradient term will vanish, then the Chern connection is the same as Levi-Civita connection, you can refer to Daniele Angella's work on this topic, for example On Gauduchon connections with Kähler-like curvature

On Chern-Yamabe problem

Besides, the prescribed Chern scalar curvature problem is related to Kazdan-Warner equation, but there is some open problems in this topic, you can read the following paper if you are interested

The prescribed Chern scalar curvature problem

The prescribed Gauduchon scalar curvature problem in almost Hermitian geometry