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This question is actually a couple of interrelated questions all related to the "position vector" or "displacement vector" and its different roles in differential geometry and classical mechanics, but unfortunately I am not really sure how to disentangle them, so here it goes.

1) If I understand correctly, position vectors can only be written relative to some origin $O$, so that a position vector pointing to a point $P$ is really the vector $\vec{OP}$. My first problem is that it is unclear to me how to approach this mathematically rigorously. I can consider Physical space as an affine space, and consider the position vector $\vec{OP}$ as $\mathbf{r}_{op}= p - o$, the unique element that results from the substraction of $o$ and $p$, but this is unsatisfying because I can't really decide where to "put" the vector in physical space by this means. Specifically, if I choose two other points $p'$ and $o'$ that are the same distance apart I would get the same element from the affine space, so it seems to me that position vectors defined this way are not really "located" anywhere. Is this correct? Is there any way to "fix" this or a neater mathematical way to look at it?

2) In differential geometry and tensor calculus one often speaks of position vectors (for instance in wikipedia's article on curvilinear coordinates it is stated that coordinate frames are the derivatives of the "position vector"), but the fact that a manifold can have curvature really makes me uncomfortable and I don't see how the "straight" nature of position vectors can be reconciled with this. To exemplify: suppose one lived in the surface of a sphere, what would a position vector even mean in this context? Can the surface of a sphere be thought of as an affine space in some way? The alternative, I believe, is that one can only speak of position vectors in flat manifolds, and somehow position "ceases to be a vector" if space has curvature, but what about the "derivative of position vectors" as frame fields then?

I am sure that this questions are related in some way, anything that helps me to disentangle these things in my head would really be appreciated =)

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    $\begingroup$ I believe the Wikipedia article is referring only to curvilinear coordinates on an underlying flat space, e.g. polar coordinates in the usual Cartesian plane. If you start to work on a manifold with curvature, I don't think you can assign the usual vector space structure, and you should speak only of position coordinates, not position vectors. Then the coordinate frames are derivatives of the position coordinate--even in differential geometry, the derivative of a coordinate function still gives back a vector! $\endgroup$ Commented Mar 27, 2018 at 2:58
  • $\begingroup$ "Then the coordinate frames are derivatives of the position coordinate--even in differential geometry, the derivative of a coordinate function still gives back a vector!" Could you clarify this please? How does the derivative of a coordinate function give back a vector? For instance, is the derivative of $x^1 (p)$ with respect to $x^1(p)$ a vector in the tangent space $T_p$ ? I mean... I can kind of understand if you mean that the derivative operator itself is a vector, but I think the partial derivative of a scalar field should be a number. $\endgroup$
    – Ignacio
    Commented Mar 27, 2018 at 3:06
  • $\begingroup$ Sorry, I think I confused the issue. The partial derivative operators are the basis for the vector space, but if you have a parameterized path, the derivative of the position with respect to the parameter is a vector. Basically I was trying to get across that position is no longer a vector but the velocity still is. $\endgroup$ Commented Mar 27, 2018 at 16:59

1 Answer 1

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First, a few preliminaries regarding the rigorous definition of an affine space. If you are not interested in this, then you can skip to the bottom where I address your specific questions.

Given a set $S$ and a group $\big(G,\circ\big)$, we define a left group action $$\triangleright: G \times S \rightarrow S$$ $$(g,s) \mapsto g\triangleright s$$

such that

  1. The identity element $e\in G$ maps every element of $s$ to itself, and
  2. For all $g,h\in G$ and $s\in S$, we have that $g\triangleright (h \triangleright s) = (g\circ h)\triangleright s$

The stabilizer of a point $s\in S$ is the set of elements of $G$ which map $s$ to itself: $$\Sigma(s) := \big\{g \in G \ \big| \ g\triangleright s = s\big\}$$

  • A left action is called free if $\forall s\in S$: $\Sigma(s)=\{e\}$. In other words, the only element of $G$ which maps any $s\in S$ to itself is the identity element of $G$. Such an action has the property that if $g\triangleright s = h\triangleright s$, then we are immediately guaranteed that $g=h$.
  • A left action is called transitive if, for all $s,t\in S$, there exists some $g\in G$ such that $g \triangleright s = t$. That is, you can always get to any element of $S$ from any other by the action of some element of $G$.

Having made those definitions, we can define an affine space to be a triple $\big(A,V,\triangleright\big)$, where $A$ is a set, $V$ is a vector space, and $\triangleright$ is a free, transitive left action of $V$ on $A$, where we note that any vector space $V$ is, in particular, an abelian group with vector addition as the group operation. One typically calls elements of $A$ points and elements of $V$ displacement vectors.

Example: Let $A=\mathbb R^2$ as a set, and let $V = \mathbb R^2$ as a vector space. Given any point $p\equiv (a,b)\in A$ and any displacement vector $\vec v \equiv \langle x,y\rangle\in V$, define the left action

$$\vec v \triangleright p = \langle x,y\rangle \triangleright (a,b) := (a+x,b+y)$$

Unsurprisingly, any vector space is an affine space over itself, in analogy with this example.


Given a choice of origin point $O\in A$, any point $x\in A$ can be uniquely identified with a displacement vector $\vec v$ such that $\vec v \triangleright O = x$. Note that the existence of such a displacement vector is guaranteed because the left action is transitive, while its uniqueness is guaranteed because the left action is free.

In a slight abuse of notation, some people re-use the same symbole twice, and let $\vec x$ be the unique vector such that $\vec x \triangleright O = x$ (where the $x$ on the right hand side lives in $A$). For pedagogical purposes, it is probably better to denote such an object by $\vec v_x$ or something, but it is what it is.


For a slightly less trivial example of an affine space, consider the following set:

$$ A := \big\{ (a,b,c)\in \mathbb R^3 \ \big| \ z = 5\big\}$$

$A$ is not a vector subspace of $\mathbb R^3$ because it does not include the origin. However, if we equip it with the vector space $V = \mathbb R^2$ and the left action $$\langle x,y\rangle \triangleright (a,b,c) := (a+x,b+y,c)$$

then it does indeed become an affine space.



Now, on to your questions.

Specifically, if I choose two other points p′ and o′ that are the same distance apart I would get the same element from the affine space, so it seems to me that position vectors defined this way are not really "located" anywhere. Is this correct? Is there any way to "fix" this or a neater mathematical way to look at it?

Yes, that is correct. There's nothing to fix, in the sense that this is a feature of affine spaces rather than a bug.

The trouble with attaching vectors to points is that each point then has its own copy of the vector space, and there is no immediately obvious way to compare or combine vectors which are attached to different points.

Put a different way, an affine space comes equipped with a single, universal vector space, rather than a bunch of individual vector spaces which would need to connect together using a quite a bit of extra structure (e.g. an affine connection).


[...] the fact that a manifold can have curvature really makes me uncomfortable and I don't see how the "straight" nature of position vectors can be reconciled with this.

You are right to be uncomfortable. Indeed, spaces which possess intrinsic curvature are disqualified from being affine spaces because, for example, transporting a vector around a closed loop generically results in a different vector from the one you started with - this stands in stark contrast to the fairly obvious $$\vec x + \vec a + \vec b + (-\vec a) + (-\vec b) = \vec x$$ which we would have if we were working in an affine space.

I believe, is that one can only speak of position vectors in flat manifolds, and somehow position "ceases to be a vector" if space has curvature[...]

Yes, that is accurate.

[...]but what about the "derivative of position vectors" as frame fields then?

Flat space is not the same as "Cartesian coordinates." Obviously the Euclidean plane $\mathbb E^2$ is flat in the sense that it has no intrinsic curvature, but one can still describe it using polar coordinates, which have a natural basis which varies over $\mathbb E^2$.



EDIT: I noticed that while I tried to address the questions in the body of your post, I didn't actually address the question in the title.

Given an affine space $\big(A,V,\triangleright\big)$, an affine frame is a choice of origin point $O\in A$ and a basis $\{\vec v_i\}$ of $V$. Given such a frame, the position vector field is a vector-valued function $$\vec R:A \rightarrow V$$ $$a \mapsto \vec R(a)$$

where $\vec R(a)$ is the unique vector such that $\vec R(a) \triangleright O = a$. It is a field in the sense that is a function defined at every point on the underlying set $A$.

Given a curve $\gamma : \mathbb R \rightarrow A$ (which one might suggestively call a trajectory), we can define the position vector along the curve

$$\vec R_\gamma : \mathbb R \rightarrow V$$ $$ t \mapsto \vec R \big(\gamma(t)\big)$$

This object is fundamentally what we're talking about when we refer to the time-dependent position vector of a particle.

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  • $\begingroup$ So, if I understand correctly a "position vector field" is not really a field in the sense that things are fields in diff geometry, it is not really a section of the tangent bundle, where one chooses one tangent vector per tangent space, it really is the bijection between the affine space and the vector space that you can construct after choosing a specific point on the affine space. The "trajectory" and the "time dependent position vector" that you wrote about are also related via this bijection (is it really just a bijection? I feel like it should be something more)(…) $\endgroup$
    – Ignacio
    Commented Mar 27, 2018 at 7:28
  • $\begingroup$ I assume that you can do a similar thing locally in manifolds, if you have a trajectory of the form $\gamma : \mathbb{R} \rightarrow M$ you can select coordinate charts along this trajectory and for each of these coordinate charts you can define $q_{\gamma}$, coordinates along the trajectory in an analogous way. If I understood correctly then those coordinates are not necessarily vectorial in nature, they belong to an open subset of $\mathbb{R^d}$.(...) $\endgroup$
    – Ignacio
    Commented Mar 27, 2018 at 7:28
  • $\begingroup$ I feel, however, that choosing coordinates and defining coordinates along a trajectory and choosing an origin and defining a position vector along a trajectory are not totally analogous, you can also define coordinates in an affine space “on top of” your definition of an “affine frame”. I think I may need to think about this a bit more and digest it, I am missing the exact connection between the position vector and coordinates in an affine space in order to totally understand this.(...) $\endgroup$
    – Ignacio
    Commented Mar 27, 2018 at 7:29
  • $\begingroup$ Right now while I’m writing this it ocurrs to me that position vectors feel like some sort of unnecessary baggage that we have in euclidean space but that we get rid of in manifolds, which is kinda dumb, but maybe the point is that it is useful to “vectorize” position when we can.As a side note, your answer is amazing, and I really appreciate your notation, I could tell you made quite an effort to use an unambiguous notation and it really aids understanding, I believe I didn’t fully understand what an affine space was before reading your definition. $\endgroup$
    – Ignacio
    Commented Mar 27, 2018 at 7:29
  • $\begingroup$ @Ignacio Yes, everything you say looks right. It happens often in Hamiltonian dynamics that one says that a point $x$ in phase space $X$ has coordinates $(q,p)$, but then starts talking about derivatives of $q$ and $p$ as though that makes any sense. What we really differentiate are the maps $Q:X \rightarrow \mathbb R$ and $P:X\rightarrow\mathbb R$ where $Q\big((q,p)\big) = q$ and $P\big((q,p)\big)=p$, and their corresponding restrictions to a curve. $\endgroup$
    – J. Murray
    Commented Mar 27, 2018 at 12:03

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