What is the horizontal space of trivial hermitian line bundle? Suppose $L=M\times\Bbb C$ is a trivial holomorphic line bundle on a complex manifold $M$, and suppose there is a Hermitian fibre metric $h$ on $L$. 

Question: What is the horizontal space of $T_{(p,z)}L$ with respect to the canonical connection on $L$.

I am a bit mixed up with all the definitions there are. Here the canonical connection is given by a connection 1-form, but we need to translate this into a splitting of the tangent space into horizontal and vertical space. There is clearly an obvious splitting of $T_{(p,z)}L$ and I am wondering if this is just that:

We have $T_{(p,z)}L=T_{(p,z)}(M\times\Bbb C)=T_pM\oplus\Bbb C$. Is that the desired splitting?

 A: Let $z$ be the standard vertical coordinate on $L$ so that we can write $h = \rho\ dz \otimes \overline{dz}$ for some real function $\rho$ on $M$. The connection induced by $h$ satisfies $\nabla h = 0$, so plugging in the standard basis we get $$d \rho =\nabla (h(\partial_z, \bar\partial_z)) =h(\nabla \partial_z,\bar\partial_z) + h(\partial_z, \nabla\bar\partial_z).$$
From what I can gather, the compatibility condition with the complex structure is that the $(0,1)$-component of $\nabla$ is just $\bar \partial$. This means that the second term above vanishes, and the first is just $\rho \omega$ for $\omega$ the single 1-form associated to the holomorphic frame $\{\partial_z\}$ for $L$. Thus we can write $\omega = d \log \rho$, and the general expression for the connection is just $$\nabla s = d s + s\ d \log \rho.$$
Thus in order for a section $s$ to be $\nabla$-parallel we require $ds = -s\ d \log \rho$, so the horizontal subspace of $T_{(p,z)} L \simeq T_p M \oplus \Bbb C$ is $$\{(v,w) \in T_pM \oplus \Bbb C : w = -z\ d(\log \rho)(v)\}.$$
