This question is related to this MO question and this MSE question.

Let $E$ be a hermitian holomorphic vector bundle over a hermitian manifold $X$. The bundle $\bigwedge^{\bullet,\bullet}X\otimes E$ has an induced hermitian metric. As $E$ is holomorphic, there is a Dolbeault operator $\bar{\partial}_E$ which acts on

$$\Omega^{\bullet,\bullet}(X, E) = \Gamma\left(X, \bigwedge\nolimits^{\!\bullet,\bullet}X\otimes E\right),$$

and it has adjoint $\bar{\partial}_E^*$. Both $\bar{\partial}_E$ and $\bar{\partial}_E^*$ are first order differential operators and

$$\Delta_{\bar{\partial}_E} = \bar{\partial}_E\bar{\partial}_E^* + \bar{\partial}_E^*\bar{\partial}_E$$

is a second order differential operator. Furthermore, $2\Delta_{\bar{\partial}_E}$ is a generalised Laplacian; that is,

$$\sigma_2(2\Delta_{\bar{\partial}_E})(\xi) = - \|\xi\|^2\operatorname{id}_{E_x}$$

where $x = \pi(\xi)$ and $\pi : TX \to X$ is the natural projection map.

Proposition 9.1.27(Nicolaesu's Lectures on the Geometry of Manifolds)Let $L$ be a generalised Laplacian on $F$. There is a unique metric connection $\nabla$ on $F$ and $\mathfrak{R} \in \operatorname{End}(F)$ such that

$$L = \nabla^*\nabla + \mathfrak{R}.$$

I want to know what the connection is for the generalised Laplacian $2\Delta_{\bar{\partial}_E}$ under the assumption that $X$ is Kähler. If I'm not mistaken, the following result is what I need.

Proposition 3.67(Berline, Getzler, & Vergne's Heat Kernels and Dirac Operators)Let $\mathcal{W}$ be a holomorphic vector bundle with hermitian metric on a Kähler manifold $X$. The tensor product of the Levi-Civita connection with the canonical connection of $\mathcal{W}$ is a Clifford connection on the Clifford module $\bigwedge(T^{0,1}X)^*\otimes\mathcal{W}$, with associated Dirac operator $\sqrt{2}(\bar{\partial}_E + \bar{\partial}_E^*)$.

First of all, $\sqrt{2}(\bar{\partial}_E+\bar{\partial}_E^*)$ is a Dirac operator for $2\Delta_{\bar{\partial}_E}$, i.e

$$\left(\sqrt{2}(\bar{\partial}_E+\bar{\partial}_E^*)\right)^2 = 2\Delta_{\bar{\partial}_E}.$$

If $\mathcal{W} = \bigwedge(T^{1,0}X)^*\otimes E$ where the first factor has the metric induced by the metric on $X$ and $E$ is equipped with a metric such that the tensor product metric agrees with the metric on $\mathcal{W}$. Then

$$\bigwedge(T^{0,1}X)^*\otimes\bigwedge(T^{1,0}X)^*\otimes E \cong \bigwedge\nolimits^{\!\bullet, \bullet}X\otimes E.$$

The connection on $\bigwedge(T^{0,1}X)^*$ is the Levi-Civita connection, and the connection on $\bigwedge(T^{1,0}X)^*$ is the Chern connection, but as the manifold is Kähler, the Chern connection coincides with the Levi-Civita connection. Therefore on

$$\left(\bigwedge(T^{0,1}X)^*\otimes\bigwedge(T^{1,0}X)^*\right)\otimes E \cong\bigwedge\nolimits^{\!\bullet, \bullet}X\otimes E$$

the connection is given by the Levi-Civita connection on the first factor and the Chern connection on the second factor.

- Is this the metric connection from
Proposition 9.1.27for the generalised Laplacian $2\Delta_{\bar{\partial}_E}$?- If so, is it also the metric connection for the generalised Laplacian $2\Delta_{\partial_E}$?
- Can any of this be generalised to the case where $X$ is just a hermitian manifold?

I'm still getting used to the whole language involved in this setup, so if it seems that I've missed something, please let me know.

**Added later:** Thanks to the two answers, several of my questions have been answered. If $X$ is a Kähler manifold, then on $\Omega^{p,q}(X, E)$ we have $\Delta_{\bar{\partial}_E} = \nabla^*\nabla + \mathfrak{R}_1$ and $\Delta_{\partial_E} = \nabla^*\nabla + \mathfrak{R}_2$ where, in both cases, $\nabla$ is the connection induced by the Chern connection. Combining these two decompositions, we have $\Delta_{\bar{\partial}_E} = \Delta_{\partial_E} + (\mathfrak{R}_1 - \mathfrak{R}_2)$. As $\Delta_{\bar{\partial}_E} = \Delta_{\partial_E} + [iF_{\nabla}, \Lambda]$ we see that $\mathfrak{R}_1 - \mathfrak{R}_2 = [iF_{\nabla}, \Lambda]$.

In the case that $X$ is a hermitian manifold, I am struggling to see how to generalise the above consideration. I know that $$\Delta_{\bar{\partial}_E} = \Delta_{\partial_E} + [iF_{\nabla}, \Lambda] + [\bar{\partial}_E, [L, \Lambda_{\bar{\partial}\omega}]] - [\partial_E, [L, \Lambda_{\partial\omega}]].$$ Combining the decompositions $\Delta_{\bar{\partial}_E} = \nabla_1^*\nabla_1 + \mathfrak{R}_1$ and $\Delta_{\partial_E} = \nabla_2^*\nabla_2 + \mathfrak{R}_2$, we have $$\Delta_{\bar{\partial}_E} = \Delta_{\partial_E} + (\nabla_1^*\nabla_1 - \nabla_2^*\nabla_2) + (\mathfrak{R}_1 - \mathfrak{R}_2).$$ Therefore $$(\nabla_1^*\nabla_1 - \nabla_2^*\nabla_2) + (\mathfrak{R}_1 - \mathfrak{R}_2) = [iF_{\nabla}, \Lambda] + [\bar{\partial}_E, [L, \Lambda_{\bar{\partial}\omega}]] - [\partial_E, [L, \Lambda_{\partial\omega}]].$$ What I can't deduce is the correspondence between the terms. On the right hand side, the first term is order zero, whilst the other two terms are order one. On the left hand side, the first bracket is at most order two, but I'm guessing they have the same principal part so it is actually order one. Is $\nabla_1 \neq \nabla_2$ or are they equal so that only $\mathfrak{R}_1 - \mathfrak{R}_2$ represents the right hand side. Surely this can't be the case as $\mathfrak{R}_1$ and $\mathfrak{R}_2$ are order zero, aren't they?