TITLE: When is a vector "glued" to the origin?
QUESTION [22 upvotes]: 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.


Is any part of the above discussion correct?
When is a vector "glued" to the origin?
What is a rigorous formulation of "position vectors" and "displacements vectors"? Does it use tangent bundles? Does it use affine spaces?

REPLY [3 votes]: 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.