TITLE: Map Laplacian in terms of covariant derivatives
QUESTION [5 upvotes]: I stumbled upon the following definition:

Let 

$\mathcal{M}$ be a manifold,
$g_{ij}$, $h_{ij}$ be two Riemannian metrics on $\mathcal{M}$,
$\psi : \mathcal{M} \to \mathcal{M}$ be a twice differentiable map,
$C_1$, $C_2$ be two charts on $\mathcal{M}$,
$\Psi := C_2 \circ \psi \circ C_1^{-1}$.

The map Laplacian $\Delta_{g,h}$ is defined by $$ (C_2 \circ
 (\Delta_{g,h} \psi) \circ C_1^{-1})^q := g^{ij}\left( \partial_i
 \partial_j \Psi^q 
 -  \Gamma(g)^k_{ij} \, \partial_k \Psi^q
 + (\Gamma(h)^q_{mn} \circ \Psi) \, \partial_i \Psi^m \, \partial_j \Psi^n \right) . $$

Basically, my question is how to make sense of the above beast, but since this is probably too broad, I would like to narrow it down to: can the above formula be expressed in terms of covariant derivatives? 

What I got so far:
I recognise the term $\partial_i  \partial_j \Psi^q -  \Gamma(g)^k_{ij}\, \partial_k \Psi^q = \nabla_i \nabla_j \Psi^q$ as the second covariant derivative of $\Psi^q$ such that $g^{ij} \nabla_i \nabla_j \Psi^q = \nabla^i \nabla_j \Psi^q$ can be interpreted as the Laplacian of $\Psi^q$. Since $\Psi^q$ depends on the chart $C_2$, I also expect there to be some term to correct for that, which is probably the last one, but I cannot make that last statement any more precise. 

Update: Another question would be whether it is possible to give a physical intuition for this operator. For example, if $\mathcal{M} = \mathbb{R}^3$, the map Laplacian becomes the vector Laplacian appearing in the Navier-Stokes equation where it describes the friction in the fluid (diffusion of momentum). This intuition does not generalise to the situation at hand, however, because $\psi$ is a map from the manifold onto itself, not from the tangent space (which is $T\mathcal{M} = \mathbb{R}^3 = \mathcal{M}$ for $\mathcal{M} = \mathbb{R}^3$) onto itself.

REPLY [3 votes]: The map Laplacian has a very simple coordinate free expression - it is (as you'd expect for something called the Laplacian) the trace of the second derivative:
$$ \Delta_{g,h} \psi = {\rm tr}_g \nabla D \psi.$$
The complexity here is hidden in that $\nabla$: we need a covariant derivative that can act on $D \psi$. We can interpret $D \psi$ as a section of the bundle $T^* M \otimes \psi^* TM$, which can be naturally equipped with the tensor product connection formed from $\nabla^g$ and $\psi^* \nabla^h$. The Christoffel symbols of this connection have components coming from both factors ($\Gamma(g)$ directly and $\Gamma(h)$ via $D\psi$), and thus you get the two correction terms in the coordinate formula.