The following symbols used in this paper, page 16.
Let $\Sigma \subset M$ be a hypersurface.
So I have
$$ A(\nabla\varphi,\nabla\psi) = A_{ij}\nabla^i \varphi \nabla^j \psi \tag{1}$$
where $A_{ij}$ is the second fundamental form of $\Sigma$ and $\varphi \in C^\infty(\Sigma)$ is some real function.
Next I have
$$H \left< \nabla\varphi , \nabla \psi \right> = H\ \nabla^i \varphi \nabla_i \psi \tag{2}$$
where $H = g^{ij}A_{ij}$ is the mean curvature which is a trace of the second fundamental form $A_{ij}$.
Questions:
- If I replace $\psi$ with $\phi$, then I can write (1) as
$$A(\nabla\varphi,\nabla\varphi) = A|\nabla\varphi|^2 \tag{3}$$
and (2) as
$$ H\ \nabla^i \varphi \nabla_i \varphi = H |\nabla\varphi|^2 \tag{4}$$
I am pretty sure that (4) is right, but how about (3) is it right or wrong?
- What is the difference between brackets $(\cdot , \cdot)$ and $\left< \cdot , \cdot \right>$?
Additional information:
$|A|^2 = A^{ij}A_{ij} \tag{5}$
where $A$ is the shape operator of $\Sigma$.
Thank you.
In this context, $\langle \cdot, \cdot\rangle$ is the metric on $\Sigma$ and there is no such thing as $(\cdot, \cdot)$.
The second fundamental form is defined as $$ A : T_x\Sigma \times T_x\Sigma \to \mathbb R, \ \ \ A(X, Y) = \langle \nabla^M_X Y, v\rangle,$$ here $v$ is the unit normal vector.
Thus there is no such thing as $(\cdot, \cdot)$, and $A|\nabla \varphi|^2$ does not make sense too.
On the other hand, $H \langle \nabla\varphi, \nabla \varphi\rangle$ does make sense: $H$ is the mean curvature, which is just a function. Thus $H \langle \nabla\varphi, \nabla \varphi\rangle$ is the multiplication between $H$ and the metric $\langle\cdot, \cdot\rangle$.