Can the units of an arbitrary differential operator be completely arbitrary on a smooth manifold?

Question

On a smooth manifold, given any two derivative operators $\tilde{\nabla}_{a}$ and $\nabla_{a}$, there exists a connection $C^{c}_{ab}$ such that, acting on a metric tensor $g_{bc}$, we have: \begin{equation} 0\neq \nabla_{a}g_{bc}=\tilde{\nabla}_{a}g_{bc}-g_{dc}C^{d}_{ba}-g_{bd}C^{d}_{ca} \end{equation} Are we allowed to choose any differential operator of any units given that we assume the derivative has a constant of proportionality such that the units match up?

For example, can I choose $\tilde{\nabla}_{a}$ to be $\tilde{\partial}_a$ with units $[Length]^{-1}$, and $\nabla_{a}$ to be $\partial_{a}$ with units $[Mass]^{-1}$ as long as $\partial_{a}g_{bc}=Q_{abc}=\kappa N_{abc}$ where $\kappa$ is a constant of proportionality with the units $\frac{[Mass]}{[Length]}$ so long as $g_{bc}$ is a function of both mass and length?

Answer 1

Any covariant derivative must be dimensionless by the following argument involving the Leibniz rule. Let $\nabla$ be a covariant derivative on $M$, $f$ a smooth dimensionless function on $M$ and $T$ a section of a tensor bundle of arbitrary units. One has $\nabla (fT) =f(\nabla T) + T\otimes df$ by the Leibniz rule. Since the local expression for $df$ is $\frac{\partial f}{\partial x^i} dx^i$, $df$ is necessarily dimensionless. For the right hand side of Leibniz rule equation to make sense, the units of $f\nabla T$ must be the same as those of $T\otimes df$ which are just those of $T$.

Answer 2

An abstract manifold, like an abstract vector space, has no concept of units. The physical universe itself has no units associated to it. Another way to say this is that the laws of geometry and physics remain the same, no matter what units are used to write them down.

Units appear only when you introduce measuring instruments. In a vector space, this is a choice of an inner or dot product. With that, you can now define the length of a vector and the distance between two points. For a Riemannian manifold, points, tangent vectors, 1-forms per se have no units. If we view the norm of a tangent vector as length, then $g(v,v)$ has units of length squared. So you can view the metric, as a function of tangent vectors, as having units of length squared. Now you can figure out units of other tensors by checking how they rescale if the metric is rescaled. The dual metric, which is usually written $g(\theta,\theta)$, there $\theta$ is a cotangent vector, has units of length$^{-2}$. Since any covariant derivative $\nabla_XY$ is invariant under scaling of the metric, it is unitless. This includes the Levi-Civita connection. What’s interesting is that Riemann curvature tensor has units of length${}^{2}$ and the Ricci tensor is unitless. Their norms, however, have units of length${}^{-2}$.

Keeping track of units is in fact quite useful in not just physics but also in pure math. I’ve never seen any systematic discussion of this, but everyone working in PDEs and geometric analysis uses it.