I was trying to derive an expression of the gradient of a riemannian manifold.
Let $M$ be a Riemannian manifold of dimension $n$ and $f : M \to \mathbb{R}$ and let's define $grad f(p) : M \to\mathcal{X}(M)$ a vector field such that
$$
\langle grad \;f, v \rangle(p) = df_p(v)
$$
I want to derive an expression in terms of the metric for such gradient, here $d f_p$ represents the differential 1-form.
Here is my attempt to work out such formula. Let
$
B = \left\{ \frac{\partial}{\partial x^i}(p)\right\}_{i=1,\ldots,n}
$
be the basis of $T_p M$ in local coordinates, for each $p \in M$ therefore we want to represent $grad \; f(p)$ as
$$
grad \; f(p) = \sum_{i=1}^n a_i(p) \frac{\partial}{\partial x^i}(p),
$$
so the goal is to find the coefficents $a_i(p)$. The gradient map is linear by definition, so in order to be determined we need to apply it to the basis $B$. Doing so we get
$$
\langle grad \; f, \frac{\partial}{\partial x^j} \rangle(p) = \sum_{i=1}^n a_i(p) \langle \frac{\partial}{\partial x^i}, \frac{\partial}{\partial x^j} \rangle (p).
$$
By definition of the gradient operator we actually have for the lhs
$$
\langle grad \; f, \frac{\partial}{\partial x^j} \rangle = df_p \left( \frac{\partial}{\partial x^j} \right) = \frac{\partial f}{\partial x^j}(p),
$$
while by definition of Riemannian metric we have for the rhs
$$
\sum_{i=1}^n a_i(p) \langle \frac{\partial}{\partial x^i}, \frac{\partial}{\partial x^j} \rangle (p) = \sum_{i=1}^n a_i(p) g_{ij}(p)
$$
Therefore the gradient operator is fully determined if we solve the linear system
$$
\frac{\partial f}{\partial x^j}(p) = \sum_{i=1}^n a_i(p) g_{ij}(p) \;\;\; i = 1,\ldots,n\Leftrightarrow a_i(p) = \sum_{j=1}^n g^{ij}(p) \frac{\partial f}{\partial x^j}(p)
$$
where with $g^{ij}(p)$ I denote the element of the inverse of the metric tensor, which exists since it's SPD by definition. Therefore I endup with the expression
$$
grad \; f(p) = \sum_{i,j=1}^n g^{ij}(p) \frac{\partial f}{\partial x^j}(p) \frac{\partial}{\partial x^i}(p)
$$
Is this expression correct?
I was also trying to derive an expression for the laplacian but I don't know where to start at the moment, can you give me a clue maybe?
Best Answer
Yes, this expression is correct. Your argument is actually much more general: you have derived identification the metric provides between $T_p^*M$ and $T_pM$, via $\theta \mapsto \langle \theta, \cdot \rangle$, and then merely plugged in the coordinate expression for $df$.
One way to compute a coordinate expression for the Laplace operator $\Delta$ is to use the characterization $\Delta = \operatorname{div}\operatorname{grad}$. To follow this approach, I'd start by computing a coordinate expression for the divergence operator.