Gradient on the hyperbolic plane

Question

Consider the hyperbolic plane \begin{align} \mathbb{H}^2=\{(x,y)\in \mathbb{R}_+\times \mathbb{R}\} \end{align} with some metric $g$. Is there something like the Voss-Weyl formula for the gradient? I want to use this formula to compute $b\cdot \nabla$ for some vector field $b$ on $\mathbb{H}^2$ in terms of the gradient $\nabla_g$with respect to the given metric $g$. So I'm not looking for $\nabla X$ but only the "gradient vector" representation. I would be grateful for any hints

Answer 1

The key point is that your metric $g$ sets up a non-degenerate pairing between the two vector fields. This can be used to define a gradient in the following way: The gradient $\nabla_g u$ of a smooth function $u: \mathbb{H}^2 \to \mathbb{R}$ is defined as the unique vector field satisfying $$ \langle \nabla_g u, V \rangle_g=du(V) $$ for all vector fields $V \in \mathfrak{X}(\mathbb{H^2})$ and $du$ is the differential; a $1$-form in this case. However, if we examine this in coordinates, this means that $$ \nabla_g u^T G V=\nabla u^T V $$ where $G >0$ is a symmetrix, $2 \times 2$ positive definite matrix representing the metric tensor and $du(V)=\nabla u^T V$. Note we did another remarkable thing here: We identified the differential with a vector. In the euclidean case, this is pretty much clear and well-known.

Since this identity has to hold for any $V \in \mathfrak{X}(\mathbb{H}^2)$, the problem boilds down to a matrix-vector equation. Next, we can take the transpose and just invert $G$ to arrive at the formula $$ \nabla_g u=G^{-1}\nabla u $$ You now have a matrix-vector type formula for $\nabla_g u$. You can choose any type of metric you like, i.e. $$ G=\frac{1}{y^2}\begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix} $$ and try this formula out.