Are vectors equivalence classes over ordered pair of points?

Question

Reading Vector Calculus and Linear Algebra by Hubbard and Hubbard I got to know about points and vectors (about points for the first time actually) where they define the point essentially as an n-tuple in $\mathbb{R^n}$ and vectors as increments in $\mathbb{R^n}$

and since I recently learned how to construct a set of one kind of object from another by using equivalence classes that got me thinking.

Can we say that vectors are equivalence classes of ordered pair of points where $$\left(a,b\right) \sim \left(c,d\right) \iff a - b = c - d $$

It seems that I can define almost all of the usual vector properties or axioms here. Also, is the space of new equivalence classes isomorphic to $\mathbb{R^n}$ ?

in this light can I say that now I can define a vector in (note that earlier vector was not a member of $\mathbb{R^n}$) $\ \mathbb{R^n}$ (since the isomorphic spaces have the same structure, again I am new to this so I don't know properly). it's an increment with all the usual vector properties such that there exists another space isomorphic to $\mathbb{R^n}$ whose tuple's equivalence classes are the vectors of $\mathbb{R^n}$ and that's how I say that a tuple in $\mathbb{R^n}$ can be seen both as a vector and a point.

I know that making a theory of your own or asking about your opinion is not an up to the mark post But I think it goes in the right direction and do touch the spirit of the relation between point and vector.

Answer 1

There are both philosophical/physical and mathematical questions at hand. I'll answer the mathematical questions first.

Consider an arbitrary vector space $V$ over underlying field $F$ (in OP's case, $V = \mathbb{R}^n$ and $F = \mathbb{R}$). Consider the equivalence relation $\sim \subseteq (V^2)^2$ defined by $(a, b) \sim (c, d) \iff a - b = c - d$.

Let's note that $\sim$ is the kernel pair of the map $f(a, b) = a - b$, which is a linear map. Therefore, we see that $(V^2 / \sim) = V^2 / \ker (f)$ using the standard definition of quotient spaces. The obvious vector space structure OP alludes to on $V^2 / \sim$ is simply the quotient space structure on $V^2 / \ker(f)$.

Note further that $f : V^2 \to V$ is a surjective function, and therefore, $V^2 / \ker(f)$ is isomorphic to $V$.

Note that it makes no sense to "define" a vector this way, since in order to get the vector space structure on $V^2 / \sim$, we already need to have a vector space structure on $V$. But it does provide another way to think about vectors, and that's always useful.

Philosophically, a vector is simply an element of a vector space, and that's all there is to it. If you want to come up with physical intuition for what a vector looks like in space, this isn't a bad description for Euclidean space. But as you have probably learned by now, the pseudo-Riemannian manifold of spacetime that is the universe is very far from Euclidean space (thanks, Einstein). Thus, we need better notions of vectors. In particular, the relevant physical notion is that of a vector bundle - in particular, the tangent bundle - which provides the notion of a vector at a point.

In Euclidean space, there's a natural way to take a vector located at point $a$ and transport the vector to one located at point $b$. In particular, we can transport a vector along any path from point $a$ to point $b$, and it doesn't matter which path we pick. So the notion of a "vector at a point" is largely redundant, since there's a canonical way to transport vectors from one point to another.

By contrast, for a curved manifold like a sphere, we still have a way to take a tangent vector to the sphere at point $a$ and transport it along a path to point $b$. But depending on which path we pick from point $a$ to point $b$, we might get a different result.

So when discussing tangent vectors on curved spaces, it's a very bad intuition to view tangent vectors as equivalence classes of pairs of points.