# Coordinates of the tensor of the second fundamental form for a submanifold

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

For local choice of unit normal vector field $\nu$, the second fundamental form of $M$, a $(0,2)-$tensor is given by

$$A(V,W) = - \langle S_V W, \nu \rangle_e = \langle W, \overline{\nabla}_V \nu \rangle_e$$

or in coordinates $A = (h_{ij})$ by

$$h_{ij} = - \left\langle \frac{\partial^2 F}{\partial x_jx_i},\nu \right\rangle_e = \left\langle \frac{\partial F}{\partial x_i},\frac{\partial \nu}{\partial x_j} \right\rangle_e.$$

A relevant information on the lecture notes 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 call the image $F(M)$ a hypersurface. We will often identify points on M with their image under the immersion, if there is no risk of confusion. Let $x = (x_1, \cdots , x_n)$ be a local coordinate system on M[...] We denote by

$$g_{ij} = \left\langle \frac{\partial F}{\partial x_i}, \frac{\partial F}{\partial x_j} \right\rangle$$

My doubt is why the expression for $h_{ij}$ is that from above?

$\textbf{My attempt in order to understand the coordinates$h_{ij}$:}$

Firstly, I think that $h_{ij} = A \left( \frac{\partial F}{\partial x_i}, \frac{\partial F}{\partial x_j} \right)$, then

$h_{ij} = - \left\langle S_{\frac{\partial F}{\partial x_i}} \frac{\partial F}{\partial x_j}, \nu \right\rangle_e = - \left\langle \left( \overline{\nabla}_{\frac{\partial F}{\partial x_i}} \frac{\partial F}{\partial x_j} \right)^{\perp}, \nu \right\rangle_e$

and

$h_{ij} = \left\langle \frac{\partial F}{\partial x_j}, \overline{\nabla}_{\frac{\partial F}{\partial x_i}} \nu \right\rangle_e$

Secondly, I think that if $U_{\alpha} \subset \mathbb{R}^n$ are open sets and $\phi_{\alpha}: U_{\alpha} \longrightarrow M$ are charts of the manifold $M$, then $F \circ \phi_{\alpha}: U_{\alpha} \longrightarrow F(M) \subset \mathbb{R}^{n+1}$ are charts of the submanifold $F(M)$, therefore the $\frac{\partial}{\partial x_i} = d (F \circ \phi_{\alpha}) (e_i)$ for $i = 1, \cdots, n$, where $e_i$ is an element of canonic basis of $\mathbb{R}^n$, but I don't sure about this because in the literature of Riemmanian Geometry $\frac{\partial}{\partial x_i}$ denote an element of coordinate basis of $T_pM$.

Your attempt is mostly correct with a remark in order. In differential geometry the notation plays a crucial role, and the devil is in the details. What is happening here is a lot of tacit or implicit identifications, that are mentioned but not really reflected in the notation.

First of all, there is an identification along the charts $\phi_a$, taking into account that they are diffemorphisms. Thus, the tangential vector fields on $M$ are generated by $$\frac{\partial F}{\partial x^i} := dF(\frac{\partial}{\partial x^i})$$

Secondly, there is an identification along the pullback $F^*$, and the fields $\frac{\partial F}{\partial x^i}$ are thought as the same as the fields from the coordinate frame $\frac{\partial}{\partial x^i}$.

For instance, the first fundamental form, aka the metric $g$ on $M$ is given on page 3 of your notes as

$$g_{i j} := \langle \frac{\partial F}{\partial x^i}, \frac{\partial F}{\partial x^j} \rangle$$

that in the pullback can be written as familiar as $g_{i j} = g(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j})$, which is actually a definition of the intrinsic metric as the pullback of the ambient metric.

Next, let's take a closer look at $h_{i j} = A(\frac{\partial F}{\partial x^i}, \frac{\partial F}{\partial x^j})$, following the notation and definition, adopted in the notes (and the question):

\begin{align} h_{i j} & = A(\frac{\partial F}{\partial x^i}, \frac{\partial F}{\partial x^j}) = - \langle \big( S_{\frac{\partial F}{\partial x^i}} \frac{\partial F}{\partial x^j} \big)^{\perp}, \nu \rangle_{e} = - \langle S_{\frac{\partial F}{\partial x^i}} \frac{\partial F}{\partial x^j} , \nu \rangle_{e} \\ & = - \langle \overline{\nabla}_{\frac{\partial F}{\partial x^i}} \frac{\partial F}{\partial x^j} , \nu \rangle_{e} = - \langle \overline{\nabla}_{dF(\frac{\partial}{\partial x^i})} \frac{\partial F}{\partial x^j} , \nu \rangle_{e} \equiv - \langle \overline{\nabla}_{\frac{\partial}{\partial x^i}} \frac{\partial F}{\partial x^j}, \nu \rangle_{e} \\ & = - \langle \frac{\partial^2 F}{\partial x^i \partial x^j}, \nu \rangle_{e} \end{align}

where the last line follows from the definition of $\overline{\nabla}$:

$$\overline{\nabla}_{\frac{\partial}{\partial x^i}} Y = \big( \frac{\partial Y^{1}}{\partial x^i}, \dots, \frac{\partial Y^{n+1}}{\partial x^i} \big)$$

and

$$\frac{\partial F}{\partial x^i} = \begin{pmatrix} \frac{\partial F^{1}}{\partial x^i} \\ \dots \\ \frac{\partial F^{n+1}}{\partial x^i} \end{pmatrix}$$

and you need to spot the cheating with pullbacks :)

• Let $g_M$ the metric on $M$ and $g_{F(M)}$ the metric on $F(M)$. I see that $g_{F(M)} = F^*g_M$, because $(g_{F(M)})_{ij} = \langle \frac{\partial F}{\partial x_i} , \frac{\partial F}{\partial x_j} \rangle = \langle dF(\frac{\partial }{\partial x_i}), dF(\frac{\partial }{\partial x_j}) \rangle = F^*g_{F(M)}(\frac{\partial }{\partial x_i},\frac{\partial }{\partial x_j}) = (g_{M})_{ij}$, but I can't see how this helps me. Commented May 26, 2018 at 15:31
• @George sure, it takes a little more effort to grasp. I've added a few lines o calculation that may help you better. Commented May 27, 2018 at 1:08
• I had developed about this computation for $h_{ij}$, but I'm stuck in understand why the penultimate equality is true, because, at first, it doesn't make sense for me $\overline{\nabla}_{\frac{\partial}{\partial x^i}} \frac{\partial F}{\partial x^j}$, because $\frac{\partial}{\partial x^i}$ it's a vector field on $T_pM$, $\frac{\partial F}{\partial x^j}$ it's a vector field on $T_{F(p)}F(M)$ and $\overline{\nabla}$ takes two vector fields on $T_{F(p)}F(M)$ and return a vector field on $T_{F(p)}F(M)$, but I will try spot the cheating with pullbacks. Thanks! :) Commented May 27, 2018 at 13:35
• just a doubt: is this the definition of pull back of a connection duetosymmetry.com/notes/notes-on-the-pullback-connection/… ? I'm asking this because I never studied pull back of connections in my self-study on Riemannian Geometry. Commented May 27, 2018 at 14:13
• @George 1) Yes, that is the one, mentioned in your nice link. 2) You have spotted the cheating: indeed the connections around the $\equiv$ are different, but 3) it is easier to think in terms of identifications (this is why I started my answer with them), than in terms of pullbacks (because you will need to use a more involved notation, similar to what is used in the link). 4) The most thorough treatment of pullbacks I found in a wonderful book of B.Andrews and C.Hopper "The Ricci Flow in Riemannian Geometry" that is freely available here. Commented May 28, 2018 at 8:56