# The tangential gradient of a function $f$ defined on a manifold

I'm self study Riemannian Geometry to be able to understand this lecture notes about Mean Curvature Flow. The first chapter is a review of Riemannian Geometry and I'm stuck in the following part:

"Using the isomorphism induced by the metric $$g$$ we can regard $$\nabla f$$ also as element of the tangent space, in this case it is called the gradient of $$f$$. The gradient of $$f$$ can be identified with a vector in $$R^{n+1}$$ via the differential $$dF$$; such a vector is called the tangential gradient of $$f$$ and is denoted by $$\nabla^M f$$, given in coordinates by"

$$\nabla^M f = \nabla^i f \frac{\partial F}{\partial x_i} = g^{ij} \frac{\partial f}{\partial x_j} \frac{\partial F}{\partial x_i}$$

The word ”tangential” comes from the equivalent definition of $$\nabla^M f$$ in case $$f$$ is a function defined on the ambient space $$R^{n+1}$$. It can be checked that $$\nabla^M f$$ is the projection of the standard Euclidean gradient $$DF$$ onto the tangent space of $$M$$, that is,

$$\nabla^M f = Df - \langle Df, \nu \rangle_e \nu,$$

where $$\nu$$ is a local choice of unit normal to $$M$$.

A relevant consideration done in the beginning of the chapter is

We restrict ourselves to manifolds of codimension 1 in an Euclidean ambient space, i.e. we consider a n-dimensional smooth manifold $$M$$, without boundary, either closed or complete and non-compact and an immersion (or embedding)

$$F: M \longrightarrow \mathbb{R}^{n+1}$$

[...]We denote by $$g_{ij} := \langle \frac{\partial F}{\partial x_i}. \frac{\partial F}{\partial x_j} \rangle_e$$ the induced metric on $$M$$, where $$\langle \cdot, \cdot \rangle_e$$ is the Euclidean scalar product on $$\mathbb{R}^{n+1}$$.

My doubt is what I need to check exactly? I need to show that $$g^{ij} \frac{\partial f}{\partial x_j} \frac{\partial F}{\partial x_i} = Df - \langle Df, \nu \rangle_e$$ or I need prove that the $$\textit{gradient}$$ of $$f$$ (which is different of the definition of $$\textit{tangential gradient}$$ of $$f$$) is equal to $$Df - \langle Df, \nu \rangle_e$$ identifying $$\textit{gradient}$$ of $$f$$ with the $$\textit{tangential gradient}$$ of $$f$$? In the first case, I think the argument is just as it is in this topic, but I don't know how to prooced if the case is the second.

$$\textbf{P.S.:}$$

Other consideration that maybe be relevant: my background are Linear Algebra, Metric Spaces, Analysis on $$\mathbb{R^n}$$ and Differential Geometry.

The only thing you need to "check" (essentially) is that $$D \widetilde{f} \circ DF=df,$$ which is trivial by the chain rule, where I am denoting by $\widetilde{f}$ the function as defined on $\mathbb{R}^{n+1}$ (so that $\widetilde{f} \circ F=f$). $^{(1)}$

Now the rest is linear algebra (which may also be considered something to "check"), since given $X_p \in T_pM$, \begin{align*} df(X_p)=D\widetilde{f}( DF(X_p))&=\langle \nabla \widetilde{f}, DF(X_p\rangle) \\ &=\langle (\nabla \widetilde{f})^{\perp}+(\nabla \widetilde{f})^{\top}, DF(X_p)\rangle \\ &=\langle (\nabla \widetilde{f})^{\top}, DF( X_p) \rangle \\ &=\langle DF^{-1}(\nabla \widetilde{f}^{\top}), X_p \rangle, \end{align*} so the gradient of $f$ at some $p \in M$ is just the tangential gradient vector of $\widetilde{f}$ (under the identification of $DF$ $^{(2)}$). I am assuming you are giving $M$ the pull-back metric, of course.

$^{(1)}$Just an observation: There may be some identifications here, depending on definitions. If you take $Df$ to mean the usual definition as given in a standard course in analysis, then you must identify $T_p\mathbb{R}^n$ and $\mathbb{R}^n$ in the usual way. This is also the difference between $df: T_pM \to \mathbb{R}$ or $df: T_pM \to T_p\mathbb{R}$, which you allude to in the comments.

$^{(2)}$This is the identification he mentions at

The gradient of $f$ can be identified with a vector in $R^{n+1}$ via the differential $dF$.

• I understand the second part assuming the first, but the first part isn't clear for me yet. First, $f$ is a map $f: M \longrightarrow \mathbb{R}$, then $df_p: T_pM \longrightarrow T_{f(p)} \mathbb{R}$, right? But $F: M \longrightarrow \mathbb{R}^{n+1}$, then $DF: T_pM \longrightarrow T_{F(p)} F(M) \subset \mathbb{R}^{n+1}$ and, consequently, DF|_{T_pM} \neq df, right? I saw that you're brazilian, I'm brazilian too. I'm studying Riemannian geometry by this notes: mat.ufmg.br/~rodney/notas_de_aula/geometria_riemanniana.pdf written in portuguese. – George May 23 '18 at 20:53
• Is your definition of Xf the same as defined in these lecture notes on page $15$? – George May 23 '18 at 20:54
• @George Good to see a fellow brazilian! : ) I will put some more details in the answer. OBS: I thought "F" just referred to the extension of $f$, so there are some clashing notations here. I'll fix them up. – Aloizio Macedo May 23 '18 at 21:18
• @S.Maths it's $\nabla^M f = Df - \langle Df, \nu \rangle_e \nu$. – George Jul 19 at 22:41
• @S.Maths, $\nabla^M f$ is the tangential component of $Df \in \mathbb{R}^{n+1}$, the usual derivative in an Euclidean space. Are you refirring $\nabla f$ as $\nabla^M f$? – George Jul 23 at 19:23