# Gradient and laplacian of a function defined on Riemannian manifold in local coordinates.

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?

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.

• Hi, thank you for confirming my calculations. I'm taking inspiration from exercise 8 of doCarmo's Riemannian geometry (he's using a geodesic frame, I'm just dropping it). The divergence of a vector field $X(p)$ si defined as the trace of the linear map $Y(p) \to \nabla_Y X(p)$. So I was trying to find a matrix representing such operator and compute the trace. Is this what I would need to do? Because I can't get very far at the moment with calculations (maybe it's a silly mistake). Commented Jul 16, 2020 at 15:50
• That is exactly what you'd need to do. When I was working out these expressions for the first time, I found it helpful to use colored pens to coordinate my calculations.
– Neal
Commented Jul 16, 2020 at 16:00
• I guess I'll ask a follow up question at some point to check my calculations then... Commented Jul 16, 2020 at 16:09
• One question though.. if the trace takes an operator and spits out a scalar (since it's sum of diagonal elements) how can $\Delta = div \; grad$ be well defined? Commented Jul 16, 2020 at 16:26
• $\Delta = \operatorname{div}\operatorname{grad}: f \mapsto \operatorname{tr} (v \mapsto \nabla_v (\operatorname{grad} f))$
– Neal
Commented Jul 16, 2020 at 16:32