Classical differential operators with complex functions on Riemannian manifolds I am having some trouble understanding how to use the classical operators ($\nabla, \operatorname{div}, \Delta$) with complex functions on a Riemannian manifold $(M, g)$.
Consider the formula defining the gradient: $g(\nabla f,X) = \Bbb d f (X)$. If $f$ takes only real values everything is fine. If $f$ takes general complex values, then $\nabla f$ will be a vector field with complex components, and since $g$ is a section in a real vector bundle, what meaning does the above formula have?
I thought about complexifying the tangent bundle (by tensorizing with $\Bbb C$), but what should I do with $g$?
If I promote it to a Hermitian form $g_{\Bbb C} (U + \Bbb i V, X + \Bbb i Y) = g(U,V) + g(V,Y) + \Bbb i \big( g(V,X) - g(U,Y) \big)$ then the right-hand side of the definition of the gradient is $\Bbb C$-linear in both $f$ and $X$, while the right-hand side is sesqui-linear, so this would be a mistake. Is $\nabla$ conjugate-linear? This would shock me.
If, instead, I promote $g$ to a $\Bbb C$-bilinear form $g_{\Bbb C}$ then the definition above is non-contradictory, but if $X$ is a real vector field then $g_{\Bbb C} (\Bbb i X, \Bbb i X) \le 0$, which is a catastrophy.
Therefore, when I read things like $\int f \Delta h \Bbb \; d x = - \int g (\nabla f, \nabla h) \; \Bbb d x$, is this valid only for real-valued functions? Is the defintion of $\nabla$ changed for complex-valued functions? Who is $g$ in this formula?
 A: It seems natural to require that the operators $d$, $\nabla$, $\text{div}$, $\Delta$, etc., are complex linear and to work with a Hermitian metric $g$ on vector fields that's conjugate linear, say, in its second entry, as you describe. Then we have Hermitian $L^2$ inner products on the spaces of complex-valued functions: 
$$ \langle f, h \rangle_{L^2, \text{functions}} := \int f \bar{h} \, dx,$$
and vector fields, using the Hermitian pointwise metric $g$:
$$ \langle X, Y \rangle_{L^2, \text{vector fields}} := \int g(X,Y) \, dx,$$
and similarly for differential forms.
If we impose those conditions, then it seems we just have to be a little careful with adding "bars" to the formulas like the ones you mention. For example:
The gradient $\nabla$ satisfies
$$ g(\nabla f, X) = df( \bar{X}),$$
where $\bar{X}:=X_1 - i X_2$ if $X = X_1 + i X_2$ for real-valued $X_1$ and $X_2$.
(I admit, this still seems funny somehow; is this correct? Perhaps someone else can comment.)
Similarly, the Laplacian $\Delta = - \text{div} \, \nabla = d^\ast d$ satisfies
$$ \int f \Delta \bar{h} \, dx = \int g( \nabla f, \nabla h) \, dx = \int \hat{g} (df, dh) \, dx $$
where by $\hat{g}$ I mean the induced Hermitian metric on one-forms. Note that no "bar" appears when we rewrite the above as
$$ \langle f, \Delta h \rangle_{L^2, \text{functions}} = \langle \nabla f, \nabla h \rangle_{L^2, \text{vector fields}} = \langle df, d h \rangle_{L^2, \text{forms}}.$$
