As a part of my analysis course, one of the sections is "Metric Spaces". Pretty simple and intuitive definition given. But then, when discussing the Euclidean distance function, there was a small throwaway section about Norms. It said that the Euclidian distance function $d(x,y)$ is translation invariant, so therefore it is a Norm. However looking here, it seems that a Norm measures the length of a vector - not distance. So long story short, this triggered me to back-trace all these definitions until I had a coherent definition "stack" to work with, and no part of any of it was just "trust me bro, its maths magic".
I will first present the definition "stack" that I came up with so far, and after, I will present my concerns/questions about it, along with my current best guess.
Definition "stack"
1: Group
A group is a nonempty set $G$ equipped with some operation $(\cdot):G^2\to G$ that satisfies the group axioms. Additive groups are those who's operation is thought of as addition and denoted with $+$ symbol. i.e. $g + h$
2: (right) Group Action | References: 1
A (right) group action $a$ of group $G$ on set $X$, is a function $a:X \times G \to X$ that satisfies the (right) group action axioms. i.e. $a(x, g)$ Group actions can satisfy properties, such as being free and transitive.
3: Field | References 1
A field is some set $F$ equipped with addition $(+): F\times F\to F$ , and multiplication $(\cdot): F\times F\to F$, which obey the field axioms. i.e. $a \cdot (b+c)$
4: Vector Space | References: 1, 3
A vector space over a field $F$ is a nonempty set $V$ equipped with scalar multiplication $(\cdot \cdot):F \times V \to F$, and vector addition $(+):V^2\to V$, which obey the vector space axioms. Elements of $V$ are vectors. Elements of $F$ are scalars. Every vector space is also an additive group. i.e. $a \pmb v$, $\pmb u + \pmb v$
5: Basis | References: 4
Given a vector space $V$ over $F$, the subset $B \subset V$ is the basis of $V$ if every element of $V$ can be written as a unique, finite linear combination of elements of $B$. Every vector space has a basis (equivalent to choice axiom).
6: Dimension of Vector Space | References: 4, 6
The dimension of a vector space $V$ is the cardinality of it's basis vector.
7: Affine Space | References: 2, 4, 6
Given some vector space $\vec{A}$, an affine space $A$, is a nonempty set $A$ equipped with a free and transitive (right) group action $(+)$ of $\vec{A}$ on $A$. $\vec{A}$ is said to be the associated vector space of $A$. Elements of $A$ are called points. Elements of $\vec{A}$ are called translations. The dimension of the affine space $A$ is the dimension of it's associated vector space $\vec{A}$. The properties the $(+)$ are such that for every $P,Q \in A$, there exists exactly one $\pmb a \in \vec{A}$ such that $P + \pmb a = Q$. The translation $\pmb a$ is denoted as $\overrightarrow{PQ}$.
8: Metric Space
A metric space is the ordered pair $(M,d)$ where $M$ is a set and $d:M^2 \to \mathbb{R}$ is a metric on $M$, satisfying the metric axioms.
9: Normed Vector Space | References: 4, 8
Given a vector space $V$ over a subfield $F$ of the complex numbers $\mathbb{C}$, it becomes a Normed vector space when equipped with a function $\lvert \lvert \cdot \rvert \rvert: V \to \mathbb{R}$, called the Norm, such that the Norm has these properties. Normed spaces are also automatically equipped with a norm-induced metric of $d(x,y) = \lvert \lvert y-x \rvert \rvert$, therefore they are metric spaces.
10: Inner Product Space | References: 4, 9
$F$ is either $\mathbb{R}$ or $\mathbb{C}$. Given a vector space $V$ over $F$, it becomes an inner product space when equipped with a function $\langle \cdot , \cdot \rangle: V^2 \to F$, called the inner product, such that the inner product has these properties. Inner product spaces are also automatically equipped with the canonical Norm $\lvert \lvert x \rvert \rvert = \sqrt{ \langle x,x \rangle }$, therefore they are Normed vector spaces.
11: Euclidean Vector Space | References: 10, 6
Given an inner product space $\overrightarrow{E}$ over $\mathbb{R}$, it is a Euclidean vector space if it is finite-dimensional. The conventional notation for the inner product of Euclidean vector spaces is the dot notation: $\langle x,y \rangle = x \cdot y$, so the canonical Norm can be written as $\lvert \lvert x \rvert \rvert = \sqrt{ x \cdot x }$.
12: Euclidean Affine Space | 7, 8, 11
An affine space $E$ - who's associated vector space $\overrightarrow{E}$ is a Euclidean vector space, is called a Euclidean affine space. The metric on $E$ can be defined using the canonical Norm of inner product space $\overrightarrow{E}$, so that $d(P, Q) = \lvert \lvert \overrightarrow{PQ} \rvert \rvert$. With this, every Euclidean affine space is also a metric space.
Questions
Q: Are all of these definitions accurate? A: Probably correct. But maybe Euclidean vector spaces aren't necessarily over the Reals - any finitely dimensional inner product space will do. Hence the definition of Euclidean affine space really ought to specify that the associated Euclidean vector space is over the Reals - not complex.
Q: Is that really the only thing defining a Euclidean vector space? Are there any additional constraints? A: Yeah that's probably the only thing defining Euclidean vector spaces - no additional constraints.
Q: Is that the only thing defining Euclidean affine spaces? It says here "affine space over the reals", must the set $E$ have some connection with Reals somehow? A: Yeah that's probably the only thing defining Euclidean affine spaces - no additional constraints. "affine space over the reals" probably just means that the associated vector space $\overrightarrow{E}$ should be over Reals.
Q: Here, it defines the "Euclidean norm", to just be the canonical Norm of the inner product space $\overrightarrow{E}$. Are they the same thing then? In other places, "Euclidean norm" is expressed in terms of the dot product - in the space $\mathbb{R}^n$. A: Yeah the "Euclidean norm" just refers to the canonical Norm of inner product space $\overrightarrow{E}$, there isn't any additional constraints. It's just that every Hilbert space has an orthonormal basis hence $\overrightarrow{E}$ has one too. Given an orthonormal base $B$, you can use it to compute a coordinate vector $[v]_{B}$, for any $v \in \overrightarrow{E}$. Now $\overrightarrow{E}$'s inner product becomes equal to the dot product - i.e. $\langle v,u \rangle = [v]_{B}^T[u]_{B}$, therefore every possible Euclidean space is kinda equivalent to $\mathbb{R}^n$ equipped with the dot product. Hence when learning resources refer to the "Euclidean norm" in the context of $\mathbb{R}^n$ using the dot product, its because stuff you learn in this context is transferable to every other Euclidean affine space.
Are my guesses to my own questions accurate? Perhaps I am missing something?
Yes, your answers are accurate. You seem to have a pretty good understanding already.
A few comments:
Yes, specifically, they are isomorphic. However they are not canonically isomorphic, In other words, $\mathbb{R}^n$ has lots of isomorphisms to itself as a normed vector space.
Any vector space over the complex numbers is already a vector space over the reals. And any normed vector space over the complex numbers is still a normed vector space over the reals, because the norms are positive real numbers. In fact, any hermitian product is also a real inner product -- just take the real part. So a vector space, or an affine space, with an inner product, or not over the complex numbers isn't some totally different kind of thing as that kind of space over the reals. It's a real space, with some additional structure attached to it. But you can forget the complex structure and you still have a real space.