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I'm studying curvilinear coordinate systems in $\mathbb{R^3}$, and after thinking about it for some days I still don't understand this.

When we have, say, cylindrical coordinates, we have a different vector basis for each point (a vector basis for the tangent space at each point), and those basis aren't all the same (if they were, we would have an affine coordinate system).

The position vector, a vector which takes the origin to any point in $\mathbb{R}^3$, can be expressed in cylindrical coordinates as $$\vec{r}=r\vec{e}_r+z\vec{e}_z$$

but, if the basis of $T_P\mathbb{R}^3$ for a specific point $P$ is only used for vectors "attatched" at $P$ or a neighbourhood of $P$, why can we express a vector from the origin to $P$ in coordinates of that basis?

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  • $\begingroup$ You already expressed the position vector in terms of $e_r$ and $e_z$, where $e_r$ is depending on the particular $(x,y,z)$ or corresponding $(r, \phi, z)$, but we are able to determine that $e_r$ for any possible position vector. So I do not understand your question. $\endgroup$ – mvw Aug 10 '18 at 12:14
  • $\begingroup$ @mvw my question is: how is it that a vector which is “not attached” to a point $P$ can be expressed as a linear combination of the basis vectors of $T_P\mathbb{R}^n$? $\endgroup$ – TeicDaun Aug 10 '18 at 14:25
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Say we have a point $P = (r, \theta, z)\in \Bbb R^3$, using cylindrical coordinates. This gives rise to a natural $\vec e_r$ which is of unit length and lies in the $xy$-plane and has angle $\theta$ to the $x$-axis, and a vector $\vec e_z$ which is just the unit vector along the $z$-axis. This way we have $$ P = r\vec e_r + z\vec e_z $$ A different point $Q$ would give a different $\vec e_r$ (and theoretiacally also a different $\vec e_z$, but by a quirk of the cylindrical coordinate system, that vector happens to be the same for any point). Thus $\vec e_r, \vec e_z\in \Bbb R^3$ aren't really well-defined by themselves, but only when given a specific point to work from.

At the same time, at this point $P$, we have the tangent space $T_P\Bbb R^3$. This is also a three-dimensional vector space, like $\Bbb R^3$, but it is not the same $\Bbb R^3$ which $P$ lies in. In this vector space we have a natural basis derived from the coordinate system around $P$, and that basis is sometimes called $\vec e_r, \vec e_\theta$ and $\vec e_z$ (the exact construction of these vectors uses the definition of $T_P\Bbb R^3$ extensively, so it will vary from book to book, and there are also different conventions as to what their lengths should be).

If we have a different point $Q\in \Bbb R^3$, then $T_Q\Bbb R^3$ is also a three-dimensional vector space, but it is different from $\Bbb R^3$ and it is different from $T_P\Bbb R^3$.

A vector ("point") in $\Bbb R^3$ has no natural translation into a vector in $T_P\Bbb R^3$, and neither does a vector in $T_Q\Bbb R^3$. However, both $T_P\Bbb R^3$ and $T_Q\Bbb R^3$ have their own naturally defined basis called $\vec e_r, \vec e_\theta$ and $\vec e_z$, but they only share their name because they are constructed the same way. A vector in $T_P\Bbb R^3$ is completely uncomparable to a vector in $T_Q\Bbb R^3$ (this is most easily seen by just looking at your definition of the tangent space; there is no natural way of making a vector of one space into a vector of the other (well, there is by translation, but that's not applicable to non-flat spaces and this a terrible habit to start)).

Along with the $\vec e_r, \vec e_z$ of $\Bbb R^3$, this is ripe for confusion. This is what I suspect has happened to you, and made you write this question. My best tip is to always be very aware of which vector space any vector is a part of when you write them down and manipulate them. If you do this by actually giving them different names (instead of calling all of them $\vec e_r$), then it's a little bit easier to keep things straight.

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    $\begingroup$ Thank you very much for your answer. By the way, if $\vec{r}=r\vec{e}_r + z\vec{e}_z$ lives in $\mathbb{R}^3$ and not in $T_P\mathbb{R}^3$, does its derivative with respect to time (maybe I'm thinking a bit too much as a physicist right now) also live in $\mathbb{R}^3$? Or does it live in $T_P\mathbb{R}^3$? $\endgroup$ – TeicDaun Aug 10 '18 at 14:09
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    $\begingroup$ @PabloT. Given a function $f:\Bbb R\to\Bbb R^3$, $f'(t_0)$ lives inside $T_{f(t_0)}\Bbb R^3$. For instance, if $f(0)=P$, we have $f'(0)\in T_P\Bbb R^3$. Under reasonable niceness assumptions, of course (specifically that $f$ is differentiable). In fact, one common definition of $T_P\Bbb R^3$ is as equivalence classes of (differentiable) curves going through $P$ where two curves are equivalent iff they have the same derivative at $P$. $\endgroup$ – Arthur Aug 10 '18 at 14:28
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    $\begingroup$ (cont.) Thus, in a sense, the elements of $T_P\Bbb R$ can be seen as the set of all possible derivatives a curve through $P$ could have. With a suitable vector space structure, of course. Under this definition, $\vec e_r$ is most naturally defined (in my opinion) as the equivalence class of / derivative of the curve $t\mapsto (t,\theta, z)$ at $t=r$, and similarly for the other two basis vectors. $\endgroup$ – Arthur Aug 10 '18 at 14:35
  • $\begingroup$ Oh, I feel like my differential geometry notes make a ton of sense, now. Thank you very much. One last question: is the space where $P$ lives an affine space, or just a "vector space of points"? $\endgroup$ – TeicDaun Aug 10 '18 at 15:39
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So we have a surface $S$ (in your example, a cylinder) and a set of tangent spaces $\{T_P|P\in S\}$. It is best to think of each tangent space in the set of tangent spaces as being a completely separate instance of $\mathbb{R}^2$.

If you think of $S$ as being embedded in $\mathbb{R}^3$ then it is tempting to think of the tangent spaces as being planes in $\mathbb{R}^3$, but this is misleading. If they were really planes in $\mathbb{R}^3$ then the tangent spaces of two nearby points would overlap. In your example, the tangent spaces at $P$ and $Q$ would intersect along a line in $\mathbb{R}^3$ or maybe overlap completely if $P$ and $Q$ are on the same generator of the cylinder. In fact there is no intersection of $T_P$ and $T_Q$ if $P \ne Q$; $T_P$ and $T_Q$ are completely separate vector spaces.

So any basis vectors you choose in $T_P$ are completely separate from the basis vectors that you might choose in some other tangent space $T_Q$ - and also completely separate from any basis vectors in the "ambient" space $\mathbb{R}^3$.

This is still true even when $S$ is a plane. Although it is tempting to think of the tangent space to a point $P$ in a plane $S$ as consisting of vectors in $S$ with their "root" at $P$, this is doubly misleading picture because you are then tempted to add or compare tangent vectors at different points.

It is possible to define relationships between the vectors in $T_P$ and the vectors in tangent spaces of points "near to" $P$, but this requires an additional structure called a "connection".

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  • $\begingroup$ Aren't the tangent spaces versions of $\mathbb{R}^3$ and not $\mathbb{R}^2$? I mean, we don't really have one cylinder embedded in $\mathbb{R}^3$ because we have 3 free coordinates $(r,\theta,z)$, and for every specific $r$ they belong to a different cylinder. $\endgroup$ – TeicDaun Aug 10 '18 at 14:18
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    $\begingroup$ @Pablo_T. I misread your question and though you were asking about tangent spaces to a specific cylinder i.e. a single value of $r$. If you are looking at tangent spaces across the whole of $\mathbb{R}^3$ then, yes, each tangent space is a copy of $\mathbb{R}^3$. $\endgroup$ – gandalf61 Aug 10 '18 at 14:26

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