Is the geometric representation of $\mathbb{R}^n$ in any sense canonical? This is somewhat of a soft question, but it's been bothering me for a while so I really need to ask it. When we study vector spaces in an abstract setting, we're introduced to a coordinate-free way to think about the objects involved, and learn to dissociate them from the geometric visualization of arrows or points living in a system of perpendicular cartesian grid. Metric notions are also generalized, so that we speak about norms instead of lengths, bilinear forms instead of dot products. Then we learn that no computational power is lost in this generalization, since all finite-dimensional inner product spaces are isomorphic to $\mathbb{R}^n$ equipped with the dot product, allowing us to 'get a grip' on an arbitrary abstract space.
So far so good. But now comes my problem: if all of this is true, then why do we still represent  $\mathbb{R}^n$ geometrically? No textbook or lecture I've ever come across ever distinguishes between $\mathbb{R}^n$  and geometric Euclidean space; in fact, $\mathbb{R}^n$  is often $\it{called}$ Euclidean space. Why is this the case? Is the geometric representation somehow privileged? Why not identify $\mathbb{R}^n$ with the space of non-constant polynomials in n powers of $x$ with real coefficients? Is the space of geometric vectors equipped with the physical angle-measuring dot product 'special'?
 A: 
No textbook or lecture I've ever come across ever distinguishes between $\Bbb R^n$ and geometric Euclidean space

This is because no textbook or lecture you've ever come across that mentioned $\Bbb R^n$ as Euclidean space was about actual geometry. Instead they were about algebra or analysis or analytic geometry, where the differences between actual Euclidean space and $\Bbb R^n$ were not important, and therefore were swept under the rug.
But there are differences. Two in particular:

*

*$\Bbb R^n$ has a distinguished point, $\mathbf 0$. $n$-dimensional Euclidean space does not. However, if you select an arbitrary point to act as the origin, you can use geometric constructions to define a vector space structure and an inner product on Euclidean space.

*$\Bbb R^n$ has a distinguished orthonormal basis. Even given an origin, Euclidean space does not. This is the difference between $\Bbb R^n$ and an arbitrary $n$-dimensional inner product space.



if all of this is true, then why do we still represent $\Bbb R^n$ geometrically?

We don't, particularly. I think this is more often said to represent Euclidean space algebraically/analytically. That is, to point out that $\Bbb R^n$ offers a model of the Euclidean axioms.

Why is this the case? Is the geometric representation somehow privileged? Why not identify $\Bbb R^n$ with the space of non-constant polynomials in n powers of x with real coefficients? Is the space of geometric vectors equipped with the physical angle-measuring dot product 'special'?

Consider teaching a class to fresh students of analytic geometry or linear algebra, and describing $\Bbb R^3$ as being like "the space of second-degree polynomials with real coefficients". How many of your students would find such a description to be helpful? Zero, of course. They will have no concept of such a thing. It would make no sense to them at all.
But describe it as Euclidean geometric space, and they will know what you are talking about. They've had bits of geometry taught to them from the beginning of their schooling. They know what points and lines are, and triangles and circles. Whatever their schooling, they will at least know what you are referring to.
There are two reasons for likening $\Bbb R^n$ to Euclidean space. In analytic geometry, it is done to apply algebra and analytic concepts to geometry, providing new ways to handle geometric ideas. In linear algebra, it may be done to go in the reverse direction: to allow the student to see vector spaces as an extension of geometry, something that they are already familiar with. This allows them to apply their geometric intuition in understanding these new spaces.
