Given a surface $\mathcal S$, I want to compute the Laplace-Beltrami operator of a tangent vector field/function $\mathbf v: \mathcal S \to T\mathcal S$. Definitions/what I know:
There exists a parametrization $\mathbf {X}(x^1,x^2)$ of the surface $\mathcal S$ (with local coordinates $x^1, x^2$), which defines two tangent vectors to each point on the surface $\mathbf e_i = \frac{\partial \mathbf X}{\partial x^I} = \partial_i \mathbf X$ (basis vectors, orthogonal but not necessarily orthonormal).
The covariant derivative of a vector $\mathbf v$ with contravariant components $v^i$ is $\nabla_i v^j = \mathbf e^j \cdot \partial_i \mathbf v = \partial_i v^j + \Gamma_{ik}^j v^k$, where $\Gamma_{ik}^j$ are the Christoffel symbols.
The gradient $\nabla_{\mathcal S}$ of a scalar function $u: \mathcal M\to \mathbb R$ is given by $\nabla_{\mathcal S} u = \mathbf e^i \partial_i u$, and the divergence $\nabla_{\mathcal S}\cdot$ of a vector function $\mathbf v$ is given by $\nabla_{\mathcal S} \mathbf v = \nabla_i v^i$.
The Laplace-Beltrami operator is defined for a scalar function $u$ by $\Delta_{\mathcal S} u = \nabla_{\mathcal S} \cdot \nabla_{\mathcal S}u = g^{ij} \nabla_i \nabla_j u$, where $g^{ij}$ is the inverse of the metric $g_{ij}$.
My questions:
Is the Laplace-Beltrami operator of a vector function $\mathbf v$ given by $\Delta_{\mathcal S} \mathbf v = \nabla_{\mathcal S} \cdot \nabla_{\mathcal S} \mathbf v$?
How does that look (contravariant) componentwise? Is it $[\Delta_{\mathcal S} \mathbf v]^k = g^{ij} \nabla_i \nabla_j v^k$?
I found in a reference a different expression: $\Delta_{\mathcal S} \mathbf v = \mathbf e_i \nabla_j\nabla^j v^i - 2 \mathbf n C_j^i \nabla_i v^j - \mathbf e_k v^i C_i^j C_j^k$, where $\mathbf n$ is the normal to the surface and $C_{ij} = -\mathbf n \cdot \partial_i\partial_j\mathbf X$ is the curvature tensor. This is obtained by "several times applying the Gauss-Weingarten relations": $\partial_i\mathbf e_j = -C_{ij} \mathbf n + \Gamma_{ij}^k \mathbf e_k$ and $\partial_i \mathbf n = C_i^j \mathbf e_j$. Is this maybe the correct expression instead of the one in question 1?