In which cases can a metric be defined on a vector space such that there exists an isomorphism to $\mathbb R^n$? I've just learned about vector spaces and I'm curious about for which vector spaces can a metric be defined such that one could create an isomorphism to Euclidean space.
I'm also interested if, in order to define any geometry, it requires the definition of a metric alone, or there have to be other properties.
For example, according to my understanding, Euclidean spaces have a "colloquial" geometric interpretation because they are topological spaces with a metric structure.
My initial hunch was that not all vector spaces can be used to construct a geometry, and for a geometric interpretation to exist the space must have some type of metric structure, or some type of property that formalizes distances.
I would like to know if this thinking is correct or wrong and some guidance on what I should learn/read next to delve into this topic.
I'm a Math undergraduate learning linear algebra right now after having learned some group theory, but I haven't learned about topology yet. I might not have the proper terminology, I would just like to learn more about this.
 A: One thing I think isn't emphasised enough at early undergraduate level -- maybe it can't usefully be emphasised at that stage? -- is that natural objects like $\mathbb{R}$ are many things at once. What I mean by that is the following: $\mathbb{R}$ is (1) just a set, (2) a topological space, (3) an abelian group, (4) a ring, (5) a vector space over $\mathbb{Q}$, (6) a well-ordered set...
Which should we think of it as? Well, informally, we tend to think of it as all of these things at once, because some of them are useful for algebra, some are useful for analysis, some are useful for geometry, etc. But they're all subtly different things.
(For example, as a set, $\mathbb{R}$ is isomorphic to $\mathbb{R}\setminus \{\pi\}$, but that makes no sense as abelian groups; however, as abelian groups, $\mathbb{R}$ is isomorphic to $\mathbb{C}$, but that's not true as rings; and you can definitely construct rings that are isomorphic to $\mathbb{R}$ as rings, but not as topological spaces...)
Now, part of what you're asking is "when is a vector space $V$ over a field $F$ isomorphic-as-a-vector-space to $\mathbb{R}^n$ over $\mathbb{R}$?". This bit actually isn't too hard: if you go back and unpack the definition of "isomorphism (of vector spaces)", you'll see that $F$ has to be equal to $\mathbb{R}$ (so if $F = \mathbb{Q}$ or $\mathbb{C}(t)$ or $\mathbb{F}_7$, then $V$ is never isomorphic to $\mathbb{R}^n$); and one of the main goals of a first course in linear algebra is to prove that, when $F = \mathbb{R}$,
$$V \cong \mathbb{R}^n \Leftrightarrow \dim_\mathbb{R} V = n.$$
Let's suppose that $f: V\to \mathbb{R}^n$ is an isomorphism of vector spaces, and suppose we have some favourite metric on $\mathbb{R}^n$ already - say, $d_{\mathbb{R}^n}(x,y)$ returns the distance between two points $x,y\in\mathbb{R}^n$.  Let's now look at the question in your post title: when can we define a metric on $V$ that means that $V$ is isomorphic to $\mathbb{R}^n$ in some metric space kind of way too? Well, this is actually quite easy too: if $u,v\in V$, then $f(u),f(v)\in\mathbb{R}^n$, so we can just define $d_V(u,v) = d_{\mathbb{R}^n}(f(u),f(v))$. Then $f$ is both an isomorphism of vector spaces and of metric spaces.
So far, so good, but I haven't actually said anything about "geometry" or "Euclidean space" yet. What is our favourite metric on $\mathbb{R}^n$? Well, it's this one - essentially, the distance between two points in space is given by Pythagoras's theorem relative to the axes. Once you've got that, you can define straight lines (e.g. the line between $x$ and $y$ is $\{x + t(y-x) : 0\leq t\leq 1\}$), circles (e.g. as the set of all points a fixed radius from a fixed centre), dot products (from the polarisation identity), angles (e.g. using the cosine rule), etc etc, and you can go ahead and do all of Euclidean geometry from there.
I said above that the distance was defined relative to the axes. Axes are given by a choice of basis. If you change the basis, do you get the same notion of distance? Not necessarily: if you change the basis orthonormally, yes, but it's also possible for basis $B_2$ to look non-orthonormal from the perspective of basis $B_1$ and vice-versa (think: coordinate axes on the plane that aren't at right angles to each other), and this will change the notion of distance. Linear algebra has formalisms for dealing with this too.
Finally, I might as well ramble off topic a little and mention -- since you sometimes mention "geometry" without qualifying it with "Euclidean" -- that there are some natural non-Euclidean geometries too. Spherical geometry is really important if you happen to live on a big sphere, and hyperbolic geometry is important in relativity (Escher liked it too). These aren't vector spaces, I suppose, but both of them can be built from $\mathbb{R}$ (just using slightly different notions of space and distance), and they have some strange properties (e.g. the angles of a triangle don't add up to $180^\circ$ any more!). You might be interested in nonarchimedean geometries, such as those built off the $p$-adic numbers, which also have strange properties (e.g. circles have multiple centres!).
If you're interested in this kind of stuff, I very highly recommend Keith Carne's notes on geometry and groups.
A: There is an old adage in geometry that any set equipped with a topology shall be regarded as a space.
The usefulness of metric (or normed) spaces, and how they are used in analysis; metrics on manifolds and their use in differential geometry; in a sense, the very existence of differential geometry... all these facts contribute to substantiate the claim that whatever you "geometry" is about, it is at least a statement about a topological space.
The main problem with this very general point, that seemingly puts general topology above all other possible kinds of "geometries" one might study, is that it is too general: in order to be able to encompass all possible instances of the notion of space, the bare notion of topology became so general to bring basically no information about the nuts and bolts of a specific question about a specific class of spaces. This, I argue, is what Andrew D. Hwang suggests when they say that "modern mathematics moves the other way around from experience-based spatial intuition to mathematical foundation". We have different classes of mathematical languages to speak about different classes of problems and solve them: metric spaces, Riemannian geometry, functional analysis, algebraic geometry... All these notions of geometry are tied together by the overarching notion of topological space, but knowing only general topology gives you little information about the specifics of your problem in (say) functional analysis; or to put it differently, all these areas make substantial use of general topology, in very different ways, to tackle very different problems.
For example, in any given concrete setting you might possibly think of (apart from the task of disproving a conjecture by building on purpose a pathological object) your spaces do not carry mere topologies, but topologies with additional properties: you wan to sudy Hausdorff topologies, or metrizable topologies, or manifolds, or compact manifolds of some sort, or Banach spaces (=complete normed spaces), and then down in the hole of Hilbert spaces, Sobolev spaces,... you name it, the zoology of names mathematicians have come up with in the last century is basically infinite (and somewhat dense in the surroundings of Poland...).
Now, all this begs an important question:

What is a topological space, exactly?

As a working category theorist, my response is something like: I do not quite know, and I don't have to bother too much (in fact, the very notion of topological space looks a bit obsolete from the point of view of category theory; but this is a different, interesting story).
This does not mean I do not know the definition of a topological space, or that I do not recognise the notion of topology as one of the deepest achievements of mathematics in the twentieth century. It is rather an invitation to take the tenet above ("a space is just a shorthand for a topological space") and approach with a clear mind the fact of life that very undesirable things happen when you push this seemingly innocuous idea to its natural limit.
Category theory can -or so I believe- substantiate a little bit more formally this take.
As you probably know, it is a banality of mathematical practice that all objects of a given "kind" organise themselves in a category. Topological spaces make no exception, because there is a category where

*

*Objects are topological spaces, i.e. pairs $(X, O_X)$ where $O_X$ is a topology on a set $X$;

*morphisms are continuous functions.

There is no surprise in this definition: the identity function $1 : X \to X$ is continuous if you take the same topology on $X$, and the composition of two continuous functions is continuous. Being able to build a category-like structure from any given interesting mathematical object is, as I said, a banality of mathematical life.
A likewise elementary example of a functor now is the following correspondence $U$:

on objects, forget that a set $X$ has a topology $O_X$, on morphisms forget that a function $f : X \to Y$ is continuous: this yields "the mere set $X$" and "the mere function $f$ in $\sf Set$" from a continuous map $f : (X,O_X)\to (Y,O_Y)$.

Again, this construction is near to a tautology, it's an easy example of a functor $U : {\sf Top} \to {\sf Set}$, because (evidently) $U$ sends the identity continuous function of $(X,O_X)$ to the identity function of the set $X$, and it preserves composition: $U(g\circ f)=U(g)\circ U(f)$.
What is way more interesting now is that the comment you've been given (that there is no geometry in a "space" equipped with a trivial choice of topology/metric) can be substantiated as follows:

There is a functor ${\sf Set} \to {\sf Top}$ going in the opposite direction of $U$ that regards a set as a degenerate kind of topological space. Actually there are two such functors, and they are both related to $U$ in some way, because I can think of at least two perfectly legitimate topologies on any given set $A$:


*

*the discrete topology, where every subset is open; this defines a topological space $A^\delta$.

*the indiscrete topology, where $\varnothing, A$ are the only two open subsets of $A$; this defines a topological space $A^\iota$.

In both cases (I will leave this to you as a riddle to solve with the bare definition of continuity), any function $A\to B$ gives rise to a continuous function $f^\delta : A^\delta \to B^\delta$, and $f^\iota : A^\iota \to B^\iota$.
This is one way to explain why exactly these choices of topologies are stupid ones: a topology on a set $X$ is an additional piece of structure I have put on the set. This additional piece of information, when done coherently for all sets $X,Y,\dots$ selects only some functions $f : X \to Y$ between two sets, the continuous ones, i.e those that preserve the structure. If no function is left out in this filtering process, because the topologies were too simple, I know nothing more about $X,Y$ than their bare structure as sets. So I know very little: I can basically just count how big $X$ and $Y$ are... not exactly a statement about a "geometric" property of $X$ and $Y$, right?
A: You can not necessarily define a natural topology on a vector space, so often there is no good geometric interpretation.
However, sometimes the field $K$ has a natural topology (for example $K$ is the rational numbers, the real numbers, the complex numbers, a finite field or the $p$-adic numbers). These fields $K$ are called topological fields. For a topological field $K$, we automatically get a topology on each finite-dimensional vector space $K^n$ as well.
If we identify $\mathbb{C}$ with $\mathbb{R}^2$ by sending $a+bi$ to $(a,b)$, then the topologies on $\mathbb{C}$ and $\mathbb{R}^2$ as above agree with each other. So we can say that $\mathbb{C}$ and $\mathbb{R}^2$ are homeomorphic (which means: isomorphic as topological spaces). Further, we also get that $\mathbb{C}^n \cong \mathbb{R}^{2n}$.
This is the only case where some $K^n$ is homeomorphic to some $\mathbb{R}^m$. The underlying reason is that there are only two topological fields that are both connected and locally compact, namely $\mathbb{R}$ and $\mathbb{C}$. So if $K^n \cong \mathbb{R}^m$, then $K^n$ has to be connected and locally compact, so $K$ is connected and locally compact as well, but then $K = \mathbb{R}$ or $K=\mathbb{C}$.
