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Let $V$ be a real finite-dimensional vector space (I guess this forces $V$ to be $\mathbb{R}^n$). My intuition is that a vector $v\in V$ must be "glued" to the origin, since the origin is the only canonical thing that $V$ has (not even the basis is canonical, and I suspect the origin (aka. the zero vector) is the only vector that has the same representation under every basis).

But, in many contexts, vectors are not thought of as being "glued" to the origin, in particular when we think of them as "displacements": a "displacement vector" from $a$ to $b$ can be the same as a displacement vector from $c$ to $d$ under suitable conditions. (Intuitively, a "displacement vector" can be moved around without making it a different vector.)

In stark contrast, a "position" in Euclidean space cannot be moved around without making it a different position, since a position is only equal to itself and no other position.

So "displacements" and "positions" can be written as vectors, but clearly they don't behave the same.

In calculus texts, authors usually switch back and forth between vectors that are "glued" to the origin, and vectors that are "not glued" to the origin. But I find that this obscures the nature of what a vector is, and I'd like a rigorous distinction.

Spivak's A Comprehensive Introduction to Differential Geometry, Volume I (Chapter 3: The tangent bundle) suggests that the appropriate language is that of tangent bundles. Namely, at each point $x\in\mathbb{R}^n$ we have a copy of $\mathbb{R}^n$: its tangent space. So, a point together with its tangent space looks like $(x,\mathbb{R}^n)$. If we let $x$ vary over $\mathbb{R}^n$, I suppose the set of all tangent spaces would be the set $\{(x,\mathbb{R}^n) \ | \ x\in\mathbb{R}^n\}$, which looks suspiciously like $\mathbb{R}^n\times\mathbb{R}^n$, ie. $\mathbb{R}^{2n}$.

Now a "position vector" seems to be a vector in the original ${\mathbb R}^n$, and a "displacement vector" starting at $x \in {\mathbb R}^n$ seems to be a vector in the tangent space $(x, {\mathbb R}^n)$. Then a vector space always has an origin, and a vector is always "glued" to that origin. What allows us to "move vectors around" with impunity is that ${\mathbb R}^n$ is isomorphic to ${\mathbb R}^n$.

Moreover, a motivation for tangent spaces seems to be precisely to formalize the idea of "displacements" on a manifold. How this is related to the idea of affine spaces (which also seem to deal with "displacements"), I don't know.


  1. Is any part of the above discussion correct?
  2. When is a vector "glued" to the origin?
  3. What is a rigorous formulation of "position vectors" and "displacements vectors"? Does it use tangent bundles? Does it use affine spaces?
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    $\begingroup$ Allowing translations is precisely the difference between vector space and affine space. $\endgroup$ – Francesco Polizzi Aug 13 '17 at 18:04
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    $\begingroup$ A vector is just an equivalence class in the space, therefore any of its translations are equivalent to itself. Try to search for a definition of vectors using equivalent classes, This will answer your question $\endgroup$ – John D Aug 13 '17 at 18:10
  • $\begingroup$ I never think of a vector as being glued to the origin. A point's "position vector" is the displacement vector from the origin to the point. An element of $\mathbb R^n$ could be called either a "point" or a "vector", and which term is used suggests what you should visualize and what operations we are likely to perform. You can add two vectors but you would not add two points. (Btw, though it is not often discussed explicitly, I think that you can add a vector to a point to obtain a new point. And you can subtract a point A from a point B to get a vector -- the displacement vector from A to B.) $\endgroup$ – littleO Aug 13 '17 at 19:08
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  • $\begingroup$ Maybe this answer is relevant: math.stackexchange.com/a/2391075/259671 $\endgroup$ – Miguel Aug 14 '17 at 7:03
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I think it is helpful to think about these concepts on an arbitrary manifold. In that setting, there is no such thing as a position vector, because the manifold may not be a vector space; positions are specified instead by local coordinates.

The only reason we can identify positions with vectors in Euclidean space and move them around is that (1) Euclidean space is a vector space, and (2) the space carries an affine connection, which supplies a notion of parallel transport. You can't even "move around" tangent vectors on an arbitrary manifold unless you have a connection (though every Riemannian manifold comes with a natural one, the Levi-Civita connection); and in addition, as Ted says in his answer, you can't imagine those tangent vector glued "wherever you like" unless the tangent bundle is trivializable. The tangent bundle on a vector space is always trivializable.


A very useful example to keep in mind is polar coordinates in the Euclidean plane. Here a position is specified by $(r,\theta)$ or, especially in physics and engineering, a "position vector" $r\mathbf{\hat{r}}$. The former is the manifold point of view; the latter exploits the fact that we are laying curvilinear coordinates on what, at bottom, is a vector space. But you quickly learn this thing called a position vector is not really a position vector, because the vector $\mathbf{\hat{r}}$ changes from point to point. It doesn't live in the underlying space; rather, it lives in the tangent bundle, and the tangent spaces change from point to point. We are only able to think of it as a bona fide position vector because we imagine uprooting it from its tangent space and gluing it instead to the tangent space at the origin. But, to repeat, in a general manifold, the underlying space isn't necessarily a vector space and doesn't necessarily have a zero vector.

In general, to specify a position we just gives its local coordinates, and that is that: we don't need a vector to specify a position. You might wonder, then, where velocity vectors come from, because we're so used to getting velocity vectors by differentiating position vectors. But velocity vectors in arbitrary manifolds don't come from differentiating position vectors. They come from partial differentiation in directions specified by the local coordinates. More precisely, given local coordinates $(x_1, x_2)$, we represent a velocity vector as the operator $a\frac{\partial}{\partial x_1}+b\frac{\partial}{\partial x_2}$. (This is part of the reason, I think, many people struggle with the abstract definition of a tangent vector. For years they have been allowed to think of a velocity vector as the derivative of a position vector.)

You can see how the two points of view diverge when you ask a physicist or engineer to tell undergraduates how to represent velocity or acceleration vectors in planar polar coordinates. Most likely, they will draw a nice picture showing why the (time) derivative (along some curve) of $\mathbf{\hat{r}}$ is $\dot{\theta}\mathbf{\hat{\theta}}$. But the picture only works if you can parallel transport the tangent basis vectors $\mathbf{\hat{r}}$ and $\mathbf{\hat{\theta}}$ at a later point back along the curve to the tangent space at the initial point -- and it is exactly this seemingly innocent feature of the picture that requires an affine connection in general. What the calculation is really doing is exploiting the standard Euclidean connection to specify the covariant derivative / Christoffel symbols of planar polar coordinates.

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    $\begingroup$ I agree, but this raises a fundamental pedagogical concern: should we tell the students of basic geometry that all their intuitions are wrong and they will understand everything in a couple of years when they reach differential geometry? $\endgroup$ – Miguel Aug 14 '17 at 7:08
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    $\begingroup$ @Miguel: Well, I wish some proviso to that effect would be included in elementary courses: "warning: the notion of a 'position vector' doesn't make sense in general curved spaces." In my own case, I never understood what mechanics books (e.g. Kleppner and Kolenkow) were doing when they were differentiating position and velocity vectors in polar coordinates until I understood, several years later, they were using the standard affine connection to do the covariant derivative. A footnote to this effect would have spared me much misery, rote memorization, and confusion. $\endgroup$ – symplectomorphic Aug 14 '17 at 14:59
  • $\begingroup$ I try to do it, in the same way I try to emphasize general metrics, rather than Euclidean inner product, but most students prefer memorization... and confusion :) $\endgroup$ – Miguel Aug 14 '17 at 15:48
  • $\begingroup$ I think that just as you can take the difference of two points in $\mathbb R^n$ to obtain a vector, on a smooth manifold you can take the difference of two very closely neighboring points to obtain a tangent vector. I think this allows an intuitive understanding of the velocity vector for a path on a manifold. The velocity vector is the difference of two closely neighboring points on the path, divided by the time taken to travel from the first point to the second point. This is quite similar to our standard understanding of velocity vectors for paths in $\mathbb R^n$. $\endgroup$ – littleO Aug 14 '17 at 19:03
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    $\begingroup$ @littleO: I agree that is a nice heuristic but still I think it is one that can lead to confusion: it essentially glosses over the fundamental change in perspective required when we switch from a linear to a nonlinear object. Your story is right infinitesimally. But the Lie group isn't the Lie algebra, e.g. And in any case, your story doesn't involve position vectors (a canonical vector pointing from the origin to a point), which was my key point: in general there is no talk of position vectors, and that is no problem, because there is no need for them. $\endgroup$ – symplectomorphic Aug 14 '17 at 19:27
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Your discussion is correct, but somewhat special to $\mathbb{R}^n$. If you try to do the same thing on a sphere, you run into problems. You still have a tangent bundle with an $\mathbb{R}^2$ at each point, but you can't "move vectors around" in the same way that you can in $\mathbb{R}^n$, because any continuous vector field on a sphere must be 0 somewhere. So if you start with a nonzero vector at some point on the sphere, you can't continuously assign a corresponding nonzero vector at all points of the sphere.

More generally, a manifold in which it's possible to "move vectors around" from one tangent space to another continuously is called a parallelizable manifold (search that term for a more precise definition).

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    $\begingroup$ Wait: you don't need a manifold to be parallelizable to "move tangent vectors around," right? You just need a notion of parallel transport -- that is, an affine connection. Or am I missing something? $\endgroup$ – symplectomorphic Aug 13 '17 at 19:36
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    $\begingroup$ @symplectomorphic If you have a connection, you can move vectors along any path. If you want the outcome to only depend on the initial and final point but not the path in between, you need more structure. Or you might ignore the parallel aspect of transport and just look for a way to identify all the tangent spaces. This is trivializability of the tangent bundle and seems to be intended in the answer (which I only understood after following the link). $\endgroup$ – Joonas Ilmavirta Aug 13 '17 at 20:46
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    $\begingroup$ @Joonas Illmavirta: that is helpful, thanks. An affine connection allows us to find a "corresponding vector" (in Ted's language) at any other particular position; trivializability allows us to find a "corresponding vector" at all other positions simultaneously, which is definitely what we have in mind when we think of a Euclidean vector as being glued "wherever we like." $\endgroup$ – symplectomorphic Aug 13 '17 at 20:56
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It is easy to prove that an isomorphism from one vector space $V$ to another, $W,$ must map the origin of $V$ to the origin of $W$. But for any non-origin vector $v\in V$ and any non-origin vector in $w\in W,$ there is some isomorphism that maps $v$ to $w.$ "Isomorphism" in this case means a mapping that is

  • linear, and
  • one-to-one, and
  • onto.

One could say that a position vector is a point in an affine space and a displacement vector, which takes some position vectors to others, is a point in a vector space. In an affine space there is no privileged point which is the "origin"; in a vector space there is one. The "privileged" displacement is the zero displacement; there is no privileged position.

One can look at it like this: Suppose John Doe and Richard Roe disagree on which location in physical space should be considered the origin. They both attempt to compute a linear combination of several points. Let's say those several points are in Bennington, Vermont and Pyongyang, North Korea and the location of the International Space Station at noon today (Greenwich time). They disagree on the results because they disagree on where the origin is. Except that if the sum of the coefficients is $1,$ then the will agree despite their differing choice of origin. A linear combination in which the sum of the coefficients is $1$ is called an affine combination. So one could say that a vector space is a set of points equipped with an operation of linear combination satisfying the algebraic laws of linear combinations, and an affine space is a set of points equipped with an operation of affine combination satisfying the algebraic laws of affine combinations. And an affine space becomes a vector space as soon as you choose one of its points to be the origin.

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(This is more of a long comment than an answer) Vectors represent different things, as you say, and some are natural to fix at the origin, like positions, and some can move around freely, like translations. Some are even glued to other points, like vector fields.

Vectors may be used to represent each of these things, and what type of quantity a vector represents is what decides if it is glued to the origin or not. So I suggest that you at all times are conscious about what the vector you are working with represents (and remember that there can be more than one representation at once, like a position being translated).

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