How do units 'change' when we move to the language of differential forms?

Question

Consider a 2D Euclidean vector as we're taught in first year: $p = x \>\hat{i} + y \> \hat{j} $ where x and y are in meters. If one goes looking for the units of a unit vector they will be told that since they are vectors divided by their magnitudes, they are unit less. Thus the vector, and the vectors components, have the same units.

In the language of differential forms, our (co)vector would be: $p = x \> dx + y \> dy$ however our vector now has units of square length. Similarly its dual vector in the coordinate basis, would be $p = x \> \partial_x + y \> \partial_y$ which is unit less.

My question is, should it be the vector or its components which have the correct units? When we say a force is a Newton, is this a demand on the vector, or upon the vector components?

Answer 1

You are perfectly free to assign units of length to $dx$ and units of inverse length to $\partial_x$.

This doesn't change any of the conclusion of dimensional analysis, as we can see from several examples. First, this assignment gives additional units of $(\text{length})^{n-m}$ to rank $(m, n)$ tensors, i.e. those with $n$ covariant indices and $m$ contravariant indices. But since we always equate tensors of the same rank in covariant equations, these units just cancel out. We're left with just the units of the tensor components, which are by definition the same as before.

Tensor operations don't mess this up. For example, we can always contract a covariant and contravariant index, turning a rank $(m, n)$ tensor to rank $(m-1, n-1)$, but this doesn't change the units. Also, we can integrate differential forms, but this works out perfectly as well. For example, in the usual notation, $$W = \int \mathbf{F} \cdot d \mathbf{x}$$ which tells us that $W$ has the dimensions of force times length. But in differential form notation, $$W = \int_\gamma f, \quad f = F_i \ dx_i.$$ The components of the force differential form $f$ have dimensions of force, while $dx_i$ has units of length, so $W$ has units of force times length again.