I am very sorry if this is a very trivial question or my formulation is inadequate, I am only an engineer, and I am not familiar with the topic. I am looking for the definition of the gradient of a function $f(x) \in \mathbb{R}^n$ for $x \in \mathbb{R}^n$ being on a $(n-1)$-dimensional manifold $M = \{x | m(x) = 0\}$ defined through a real valued and differentiable scalar function $m(x) = 0 \in \mathbb{R}^1$. The vector $f(x)$ is not necessarily tangent to $M$. After some reading I found the definition of an "intrinsic gradient" of a function $f(x)$ on $M$ through orthogonal projection \begin{equation} \nabla_M f(x) = J - \left(\frac{J\mu}{||\mu||^2}\right) \otimes \mu \end{equation} with \begin{equation} J = \nabla_x f(x) \in \mathbb{R}^{n \times n}\ , \quad \mu = \nabla_x m(x) \in \mathbb{R}^{n} \end{equation} and $\otimes$ being the dyadic product.
Is this $\nabla_M f(x)$ the quantity which tells me how $f(x)$ from point to point on the manifold changes? I can interpret this more or less by the perspective that the total change of $f(x)$ at $x$ is described by its normal gradient, but the change relevant to the points on the manifold is described by the part of the gradient tangential to the manifold. So the orthogonal projection filters the relevant part.
Question: Is this interpretation and definition correct? If not, could you recommend any literature for starting with this topic?
Actually, I have to understand the gradients (and divergence, for which I am also not sure of its definition) of functions defined on the manifold of rotational matrices $Q \in \mathbb{R}^{3 \times 3} $ in 3D, $M = \{Q | QQ^T=1_3 \wedge \det(Q)=1 \}$ with $1_3$ being the identity matrix in 3D. But before that I need to be sure that I understand first the easy stuff. Thanks!
EDIT 1: I just thought a little more and I dont know if the following definition of the divergence on a manifold is sensible and already defined. For $x \in \mathbb{R}^n$ I know that the divergence is defined as \begin{equation} {\rm div}{(f(x))} = (\nabla_x f(x)) \cdot 1_n = \sum_{\alpha=1}^n \frac{\partial f_\alpha}{\partial x_\alpha} = \sum_{\alpha=1}^n (\nabla_x f(x))\cdot(e_\alpha \otimes e_\alpha) \end{equation} with arbitrary orthonormal basis vectors $\{e_\alpha\}$. For a manifold $M = \{x|m(x)=0\}$ the matrix \begin{equation} 1_M = 1_n - \frac{\mu}{||\mu||}\otimes\frac{\mu}{||\mu||} , \quad \mu = \nabla_x m(x) \end{equation} represents the sum of the dyadic products of $(n-1)$ orthonormal tangential vectors $\tau_\alpha$ of the $(n-1)$-dimensional manifold at $x$, so \begin{equation} {\rm div}_M(f(x)) = (\nabla_M f(x)) \cdot 1_M = \sum_{\alpha=1}^{n-1} (\nabla_M f(x)) \cdot (\tau_\alpha\otimes\tau_\alpha) \end{equation} would give me the sum of the relevant changes of $f(x)$ in the different orthonormal tangential directions at $x$ on the manifold. This definition turns out (after some calculations) to be \begin{equation} {\rm div}_M(f(x)) = (\nabla_M f(x)) \cdot 1_M = {\rm div}(f(x)) - \frac{\mu}{||\mu||} \cdot \left( (\nabla_x f(x)) \frac{\mu}{||\mu||} \right) \end{equation} Is this the correct definition of the divergence on a manifold? This idea is for me geometrically intuitive but from a physical point of view I don't like it since singular points (wells and sinks) on the manifold for which the original divergence vanishes may not be identified as singular points on the manifold. I am aware that playing on a manifold of a bigger space inevitably will be blind to information orthogonal to it, but such a loss of information somehow troubles me.