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

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: $$$$0\neq \nabla_{a}g_{bc}=\tilde{\nabla}_{a}g_{bc}-g_{dc}C^{d}_{ba}-g_{bd}C^{d}_{ca}$$$$ 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?

• Are $\nabla$ and $\tilde\nabla$ arbitrary differential operators or are they covariant derivatives? Commented Jun 24, 2023 at 18:01
• At this point in the definition process, the uniqueness of the covariant derivative operator $\nabla$ has not been enforced. So both are arbitrary so long as they are 1) linear, 2) follow the leibnitz rule, 3) are commutative with contraction, and 4) agree when acting on scalar fields.
– B K
Commented Jun 24, 2023 at 21:00
• What are you trying to do? as in J.V. Gaiter's answer, the most natural choice is to leave the covariant derivative unitless. The partial derivatives $\partial_i$ can ocassionally carry units, though, if the coordinate patch does. I think it would be easier to give a satisfactory answer if you told us the context of your problem. Are you completely sure that you want to make your covariant derivative carry units? Commented Jun 26, 2023 at 5:11
• I am trying to find out what types of operators can be used in the context of relating two tangent spaces on a smooth manifold. In this case, I am following the prescription in General Relativity by Wald in section 3.1, pgs 31 - 34, without using metric compatibility to uniquely define each derivative operators. So now I can talk about the difference of operators, but I don't know which definitions are allowed to be used. I would like the (spacetime) covariant derivative to have units of $[L]^{-1}$ as they do in General Relativity when we use the convention of the metric tensor being unitless.
– B K
Commented Jun 26, 2023 at 6:08

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 $$T$$.

• So then in the opposite convention where the metric tensor is dimensionless and the differential operators have dimension, I am free to define units of the derivatives as I please. As long as I tack on the correct factor of proportionality.
– B K
Commented Jun 25, 2023 at 0:25
• There is no use of convention in my answer. In order to have a covariant derivative which has units you must define the exterior derivative to have units. Commented Jun 26, 2023 at 3:41
• For the case that the exterior derivative is defined to have units, is my statement correct?
– B K
Commented Jun 26, 2023 at 4:33

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.