Let $\mathbb{M} \subset \mathbb{R}^3$ be a $C^2$ manifold with normal vector ${\bf n}$. Then, for any scalar-valued function $f$, we can write its surface gradient as: \begin{align} \nabla_{\mathbb{M}} f = \left(I-{\bf n}{\bf n}^T\right)\nabla f, \end{align} where $\nabla$ is the $\mathbb{R}^3$ gradient (treated here as a column vector). Clearly, $P = \left(I-{\bf n}{\bf n}^T\right)$ is a projection operator onto the tangent plane.
Let ${\bf u} = [u,v,w]^T$ be a vector field in $\mathbb{R}^3$ such that ${\bf u} \cdot {\bf n} = 0$, in other words, ${\bf u}$ is tangent to $\mathbb{M}$, but has 3 components. It is common in mechanics (and other applications) to define the gradient of this vector field as \begin{align} \nabla {\bf u} = \begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} & \frac{\partial u}{\partial z} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} & \frac{\partial v}{\partial z} \\ \frac{\partial w}{\partial x} & \frac{\partial w}{\partial y} & \frac{\partial w}{\partial z} \end{bmatrix}. \end{align} My question is, what is the surface gradient of ${\bf u}$ if ${\bf u} \cdot {\bf n} = 0$? Is it simply the following? \begin{align} \nabla_{\mathbb{M}} {\bf u} = P \nabla {\bf u}. \end{align} If not, what is the correct expression?
I'd appreciate an answer in terms of these $\mathbb{R}^3$ operators rather than intrinsic derivatives. Also, a symbolic expression for $\nabla_{\mathbb{M}} {\bf u}$ would be greatly appreciated.