# Vector differential rigourous definition

I don't understand how vector differential $$d\vec{r}$$ is defined rigorously. For a differential $$dx$$, it can be defined rigorously through differential geometry, where $$dx$$ is interpreted as a covector. However this view doesn't seem to work for $$d\vec{r}$$. For example, in Cartesian coordinates $$d\vec{r} = dx\space \vec{e_x} + dy\space \vec{e_y}$$. I can understand $$dx + dy$$ as a one form, but I don't know how to interpret $$dx\space \vec{e_x} + dy\space \vec{e_y}$$.

I'm coming from curvilinear coordinates wikipedia article, where $$d\vec{r}$$ is used extensively to get result that helps prove vector operators in non-Cartesian coordinate system.

It seems like you know some differential geometry, so I would recommend you read Loomis and Sternberg's book Advanced Calculus. In appendix II of chapter $$11$$, they introduce the notion of a vector valued differential form. The definition is as follows:

Let $$M$$ be a differentiable manifold, and $$E$$ a Banach space over $$\Bbb{R}$$. An $$E$$-valued exterior differential form on $$M$$, of degree $$p$$, is a function $$\omega$$ which assigns to each $$x \in M$$, a function \begin{align} \omega(x): \underbrace{(T_xM) \times \dots \times (T_xM)}_{p \text{ times}} \to E, \end{align} whereby $$\omega(x)$$ is (continuous) multilinear and alternating.

Typically when we discuss differential forms, we take $$E= \Bbb{R}$$. Hence, we get the notion of a 1-form like $$dx,dy,dz$$. If instead we only assume that $$E$$ is a finite-dimensional real vector space, then by choosing a basis $$\{e_1, \dots, e_n\}$$ for $$E$$, we can uniquely write every $$E$$-valued exterior differential form $$\omega$$ as $$$$\omega = \sum_{i=1}^n \omega_ie_i$$$$ where $$\omega_i$$ are $$\Bbb{R}$$-valued exterior differential forms (i.e the usual kind which you might be used to). In such a case, we might write the form $$\omega$$ in vector notation as $$(\omega_1, \dots, \omega_n)$$ (again, refer to the book for a slightly more detailed explanation).

A concrete example can be when $$E = \Bbb{R}^2$$, and we choose the standard basis $$e_1,e_2$$. Then, once we have the real-valued 1-forms $$dx$$ and $$dy$$, we can define the vector-differential form $$d\boldsymbol{r}$$ as $$$$d\boldsymbol{r} := dx \cdot e_1 + dy \cdot e_2 = (dx,dy)$$$$ where the symbols have the meaning as explained above.

• Thank you for your answer. Can you explain how the result in curvilinear coordinates wiki article can be derived? In that article the difference is the basis is not global. E.g. In polar coordinate system, the basis at each point is different. But nevertheless the result still looks like it is using $d\vec{r}$ as a differential form. Here is the link: en.wikipedia.org/wiki/… – Rui Liu Jun 28 '19 at 18:16
• @RuiLiu unfortunately, I'm just starting to learn differential geomtery, and I find the classical notations/language used in that article extremely extremely confusing (it even says on the wikipedia page that the article has several issues) For instance, a lot of terms in the article are not even properly defined. Also, I don't have a firm enough understanding of differential geometry to make the "translation" between the vague classical manner of speaking and the more modern/abstract terminology, so I'm afraid I can't help you beyond this. – peek-a-boo Jun 28 '19 at 20:48
• Thank you anyway! – Rui Liu Jun 29 '19 at 10:10
• Hi @peek-a-boo. Regarding the wikipedia page on curvilinear coordinates, I think I now get a rather satisfactory understanding. If you're interested, I have posted my answer here math.stackexchange.com/a/3278237/459936 – Rui Liu Jun 29 '19 at 22:58

I think I now understand it a bit more. To answer this question, I don't $$d\mathbf{r}$$ is a differential form.