# Orientation and sign convention in 2D electrodynamics using differential forms

I've been following this paper for a treatment of electrodynamics using differential forms. In particular, they demonstrate that Maxwell's equations expressed using differential forms are form-invariant with regards to dimensionality. They then develop 2D electrodynamics with $$\mathcal{E}$$, a 1-form representing the electric field intensity and $$\mathcal{D}$$, being a 1-form representing the electric flux density, related by the constitutive relation $$\mathcal{D} = \varepsilon {\star}\mathcal{E}$$, where $$\star$$ represents the 2D Hodge star operator, which notably behaves as $${\star}dx = dy, {\star}dy = -dx$$. This is in contrast to the 3D case, where $$\mathcal{E}$$ is a 1-form and $$\mathcal{D}$$ is a 2-form.

In 3.5, the paper notes that $$\mathcal{D}$$ is actually a twisted 1-form. This is due to the fact that, while both $$\mathcal{E}$$ and $$\mathcal{D}$$ can be represented by their equipotential lines, and integrals count the number of such lines pierced along the path, a path integral of $$\mathcal{E}$$ represents potential change along the path, while a path integral of $$\mathcal{D}$$ represents the amount of flux flowing perpendicularly across the path.

My confusion arises from not being able to make sense of a sign convention for a very simple example of this scenario. Consider $$\mathcal{E} = dx + dy$$, giving $$\mathcal{D} = -\varepsilon dx + \varepsilon dy$$. Let $$\mathcal{C}$$ be a curve along the $$x$$-axis parameterized by $$\mathbf{r}(t) = t\widehat{\mathbf{i}}$$, $$0 \leq t \leq 1$$. Then we have $$\int_{\mathcal{C}} \mathcal{E} = \int_0^1 \mathcal{E}(\mathbf{r}'(t)) dt = \int_0^1 \mathcal{E}(\widehat{\mathbf{i}}) dt = \int_0^1 dt = 1$$

while $$\int_{\mathcal{C}} \mathcal{D} = \int_0^1 \mathcal{D}(\mathbf{r}'(t)) dt = \int_0^1 \mathcal{D}(\widehat{\mathbf{i}}) dt = \int_0^1 - \varepsilon dt = - \varepsilon$$ However, it seems from visualizing the situation that the potential change along $$\mathcal{C}$$ and the flux across $$\mathcal{C}$$ should both be positive (electric field lines are going up and to the right).

I expect that the confusion arises since, in 3D, we cannot simply re-use the same manifold on which we integrated $$\mathcal{E}$$ to integrate $$\mathcal{D}$$, since the former must be a curve, while the latter must be a surface.

It seems that, perhaps, we must choose the opposite orientation for $$\mathcal{C}$$ when integrating $$\mathcal{D}$$ instead of $$\mathcal{E}$$. It seems then that, in 2D, there may be a difference between a "path-like" curve and a "surface-like" curve, with regards to orientation. Is there a way to make sense of this?