Let $M$ be a Riemannian manifold of dimension $n$ which is isometrically embedded in some Euclidean space $\mathbb{R}^N$ (say by some embedding $\phi:M\to\mathbb{R}^N$). Suppose we have a smooth function $f:M\to\mathbb{R}$ and a fixed vector $w\in\mathbb{R}^N$.
I define a vector field $X:M\to TM$ by
$$X(m)=f(m)P_mw,$$
where $P_m:T_{\phi(m)}\mathbb{R}^N\to T_mM$ is defined by projecting from $T_{\phi(m)}\mathbb{R}^N$ to $\phi_*T_{\phi(m)}M$ and then seeing the result as an element of $T_mM$.
I would like to compute the divergence
$$\operatorname{div}(X)=\,???$$
but I am quite stuck. My differential geometry is a bit rusty, the extrinsic definition of the vector field stumps me quite a bit, and I was hoping someone here would have a good idea about how to proceed.
What I would expect to come out of this intuitively is something of the form
$$\operatorname{div}(X) = g(\nabla f,P_mw) + f(x)(\text{some kind of curvature term}).$$
The first term seems to come out quite naturally from the product rule, while the remaining term involves differentiating the embedding and I would expect it to give me indications of how the manifold bends in a direction or the other.
I am trying to work in special charts (eg normal coordinates around a point and possibly using an adapted orthonormal frame centered at the point) but with no success for now. I will update my question if I make significant progress.
Best Answer
Let $P \colon \iota^*(T\Bbb R^N) \to TM$ be the orthogonal projection, where $\iota\colon (M,g) \to (\Bbb R^N, \langle \cdot, \cdot \rangle)$ is an isometric embedding. We use $\nabla$ for the Levi-Civita connection of $M$ and $\overline{\nabla}$ for that of $\Bbb R^N$. Let $I\!I$ be the vector valued second fundamental form, defined by $I\! I(u,v) = \overline{\nabla}_uv - \nabla_u v \DeclareMathOperator{\trace}{trace}$. Let $H$ be the mean curvature vector defined by $H = \trace (I\! I)$. Let $\{e_1,\ldots,e_n\}$ be a local orthonormal frame on $M$. We have $H = \sum_{j=1}^n I\! I(e_j,e_j)$ by definition. Then $\DeclareMathOperator{\Div}{div} \newcommand{\nablab}{\overline{\nabla}}$ \begin{align} \Div(Pw) &= \trace(\nabla (Pw)) & \text{by definition}\\ &= \sum_{j=1}^n g(\nabla_{e_j}(Pw),e_j) & \text{by definition}\\ &= \sum_{j=1}^n e_j\,g(Pw,e_j) - g(Pw,\nabla_{e_j}e_j) & \text{since } \nabla \text{ is metric for } g\\ &= \sum_{j=1}^n e_j\,\langle Pw,e_j\rangle - \langle Pw, \nabla_{e_j}e_j \rangle & \text{by } g=\iota^*\langle\cdot,\cdot\rangle\\ &= \sum_{j=1}^n e_j\, \langle w,e_j \rangle - \langle w, \nabla_{e_j}e_j \rangle & \text{by orthogonality properties}\\ &= \sum_{j=1}^n \langle \nablab_{e_j}w,e_j\rangle + \langle w, \nablab_{e_j}e_j\rangle - \langle w, \nabla_{e_j}e_j \rangle & \text{since } \nablab \text{ is metric for } \langle \cdot,\cdot \rangle \\ &= \sum_{j=1}^n \langle w, \nablab_{e_j}e_j - \nabla_{e_j}e_j\rangle & \text{since } \nablab w = 0\\ &= \sum_{j=1}^n \langle w, I\!I(e_j,e_j)\rangle & \text{by definition} \\ &= \langle w, \sum_{j=1}^n I\!I(e_j,e_j)\rangle & \text{by linearity}\\ &= \langle w, H \rangle & \text{by definition} \end{align} (note that for these computations to be completely rigorous, one needs to extend $\{e_1,\ldots,e_n\}$ in a neighbourhood in $\Bbb R^N$ of their domain, and then evaluate on $M$.) Now, notice that $$ \Div(X) = \Div(fPw) = \trace(\nabla (f Pw)) = \trace (df \otimes Pw + f\nabla (Pw)) = df(Pw) + f \Div(Pw). $$ It finally follows that $$ \Div(X) = \langle \nabla f, w \rangle + f \langle w,H\rangle, $$ where we have used that $df(Pw) = g(\nabla f, Pw) = \langle \nabla f,w \rangle$. Note that the curvature term is of extrinsic nature (it involves the mean curvature of the embedding $\iota\colon M \to \Bbb R^N$), and not intrinsic.
This formula is still valid for any isometric embedding $M \hookrightarrow N$ if $w$ is asked to be parallel in $N$ (or at least along $M$). If $w$ is not parallel, there is some "horizontal ambient divergence" $\sum_{j=1}^n\langle \nablab_{e_j}w,e_j\rangle$ involved.