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When doing differential geometry, physicists often use $$\mathrm{d}\vec{r} = \mathrm{d}x^i\space\vec{e}_i$$ for many different things. For instance, they define the holonomic basis $\{\vec{e}^{\space\prime}_a\}$ relative to a coordinate system $\{x'^a\}$ by imposing $$\mathrm{d}\vec{r} = \mathrm{d}x'^{a}\space\vec{e}^{\space\prime}_a \implies \vec{e}^{\space\prime}_a=\frac{\partial\vec{r}}{\partial x'^a}$$ and they compute the quadratic form of the metric $\mathrm{d}s^2$ as $\mathrm{d}\vec{r}\cdot\mathrm{d}\vec{r}$.

Computing the differential of a vector field ($\vec{r}=x^i\vec{e}_i$, in this case) feels strange, as in differential geometry differentials are usually considered to be alternating $k$-forms, so it would only make sense to talk about the differential of a scalar field (aka its exterior derivative).

Not only that, the "true" definitions of holonomic bases and $\mathrm{d}s^2$ don't use this $\mathrm{d}\vec{r}$ at all.

EDIT: in fact, taking the derivative of $\vec{r}$, or any other vector field, is something we are not allowed to do in a general differentiable manifold without a connection, so we obviously wouldn't define a holonomic basis like that. A holonomic basis would basically be the basis formed by the tangent vectors $\partial/\partial x'^a$.

After thinking about it, I thought the differential of a vector field might just be $$\mathrm{d}\vec{\varphi} = (\nabla_i\varphi^j)\space\vec{e}_j\otimes\mathrm{d}x^i,$$ so maybe $\mathrm{d}\vec{r} = \mathrm{d}x^i\space\vec{e}_i$ means $\mathrm{d}\vec{r} = \mathrm{d}x^i\otimes\vec{e}_i$? How is $\mathrm{d}\vec{r}$ rigorously defined, otherwise?

TeicDaun
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  • I think phisicist use the differenctial of a vector as "a very very tiny change of the vector". And express it in some geometric basis system. All of it, most of times, in Euclidean space. Mathematcis go further and define abstract conceps that for some criteria match what phisicist need. So you can define $d \vec{r}$ more genereically and use some partcular case for "real" world. – Ripi2 Apr 09 '21 at 23:05
  • Yes @Ripi2, but we not only think about it like that conceptually, we also do computations and derive results from it, so there must be a way to connect the physicist picture and the mathematician picture. That's where my question was going. – TeicDaun Apr 10 '21 at 10:38

1 Answers1

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You're correct. it's a tensor product. Abstractly, on a vector space $V$, $d\vec{r} \in V\otimes V^*$. Any element of this tensor space defines a map $V \rightarrow V$. For example, if $e\otimes \omega \in V\otimes V^*$, then it defines a map $$ v \mapsto \langle \omega,v\rangle e. $$ In particular, given any frame of vector fields, $(\vec{e}_1, \dots, \vec{e}_n)$ with the dual basis of $1$-form $(\omega^1, \dots, \omega^n)$, the definition of $d\vec{r}$ is $$ d\vec{r} = \vec{e}_i\otimes \omega^i $$ This definition is invariant under change of basis. In particular, the map associated with it is simply the identity map, $$ \langle d\vec{r}, v\rangle = \langle \vec{e}_i\otimes \omega^i, v^je_j\rangle = \vec{e}_iv^j\langle \omega^i,e_j\rangle = v^ie_i = v. $$

If you have a coordinate system $(x^1, \dots, x^n)$, then you can set $(\vec{e}_1, \dots, \vec{e}_n)$ equal to the coordinate vector fields and $(\omega^1, \dots, \omega^n) = (dx^1, \dots, dx^n)$.

If you have another coordinate system $(y^1, \dots, y^n)$ with corresponding coordinate vector fields $(\vec{f}_1, \dots, \vec{f}_n)$ and dual frame $(dy^1, \dots, dy^n)$, then $$ d\vec{r} = \vec{e}_i\otimes dx^i = \vec{f}_i\otimes dy^i. $$ From here, I think it should be straightforward to derive all of the formulas you've written down.

Deane
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  • Can we do this in any differentiable manifold, or does something here assume a Euclidean manifold? – TeicDaun Apr 09 '21 at 22:51
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    Everything I wrote actually works on a manifold. The difference arises only when you try to differentiate the equations I wrote down. On $\mathbb{R}^n$, $d\vec{r}$ really is a vector of scalar $1$-forms, so you can take the exterior derivative again. On a manifold, the tensor $I = e_i\otimes \omega^i$ is still the identity map on $T_*M$, but you can differentiate it only if you have a connection. If $M$ is a Riemannian manifold, then you can do this using the Levi-Civita connection. – Deane Apr 09 '21 at 23:02
  • So, on a differentiable manifold, you have $\vec{\partial}_i\otimes\mathrm{d}x^i$, but you cannot say it is the differential of $\vec{r} = x^i\vec{\partial}_i$ unless you have a connection (such as the Levi-Civita connection, in a Riemannian manifold) and its covariant derivative actually equals the identity (1,1)-tensor. Did I understand correctly? – TeicDaun Apr 10 '21 at 10:35
  • As a note, I think it's worth pointing out that the space $V\otimes V^*$ is isomorphic to $Hom_{\mathbb{R}}(V,V)$ (the collection of all linear maps from $V$ to itself. Remember, the differential of a map between manifolds is nothing more than a family of maps between their tangent spaces – Nick Castillo Apr 10 '21 at 14:36
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    For an arbitrary manifold, you can let $\vec{r}$ be the identity map from the manifold to itself. Its differential $d\vec{r}$ is then the identity map on each tangent space. That all can be done without any connection. However, if you want to differentiate $d\vec{r}$, which is a tensor in $T_M\otimes T^M$, in a coordinate-free way, then a connection is needed. – Deane Apr 11 '21 at 22:20