vector space – Some Mathematical Farts

A $k$ -algebra $A$ where $k$ is a field, is a ring with the added structure of being a vector space over $k$ (with respect to the same addition) where scalar multiplication behaves nicely with the ring multiplication. $\mu(ab) = (\mu a)b = a(\mu b)$ for $\mu \in k$ and $a, b \in A$ . The atypical example is a matrix algebra of dimension $n$ with matrix elements in $k$ denoted $M_{n}(k)$ . There is a notion of algebra morphism which is not too hard to figure out with the above or a quick google, so we have a notion of isomorphism.

Matrix algebras are very special, they have vectors in which they can act on. For an $n$ -dimensional matrix algebra these objects are the tuples $(x_1, \cdots, x_n)$ where $x_i \in k$ . As a collection, they are denoted $k^n$ . These form a vector space and as algebras $M_{n}(k) \cong \text{Aut}(k^n)$ . Where the algebra operation in $M_n(k)$ is matrix multiplication and $\text{Aut}(k^n)$ is the algebra of vector space automorphisms $k^n \longrightarrow k^n$ with the algebra operation being composition.

The definition of an $A$ -module can be found elsewhere but to motivate its origins, it is precisely the articulation of assigning some ‘vectors’ in some vector space $V$ to a general $k$ -algebra $A$ which $A$ can act on $V$ similarly to matrices on tuples. If you struggle to remember the definition of an $A$ -module just think like this and you should be able to reconstruct it.

Modules can even be defined for a ring $R$ , with the relaxation of the structure of a vector space. And so we can assign ‘vectors’ to rings by looking at $R$ -modules. So we can ask the question, what are $\mathbb{Z}$ -modules? These are in fact abelian groups (why?) and so the collection of possible ‘vectors’ for $\mathbb{Z}$ can be thought of as just abelian groups.

I think that’s pretty nice.

A normed space is a vector space $V$ over a field $\mathbb{F}$ equipped with a so called ‘norm’ which is a function $\lVert . \rVert : V \longrightarrow \mathbb{R}$ that satisfies a list of axioms. The intuition behind such a function is to give each elements in $V$ a ‘size’. This is a strong piece of information and allows one to vision $V$ as a metric space and hence induce a topology onto $V$ . I am going to assume you’ve seen these concepts before and have knowledge of continuous functions in particular their topological definition and how they act on compact sets.

Looking at the sequence normed space $\longrightarrow$ metric space $\longrightarrow$ topology one can ask the very natural questions: what conditions on $V$ and $\lVert . \rVert$ give the same topology? What normed spaces don’t have the same topology? etc etc etc. The list goes on.

As the name of this post suggests we are going to show that when our underlying vector space $V$ is of a fixed finite dimension (we are going to assume $V$ is over $\mathbb{R}$ or $\mathbb{C}$ ) then there is only one topology that is induced. In some sense this means we know what notion of neighborhoods to expect if we’re in the finite world playing over the most ‘natural’ field. There is only one such notion.

In a more precise fashion what we are going to show is that any two norms on a finite dimensional space $V$ over $\mathbb{R}$ or $\mathbb{C}$ determine the same topology.

For convenience we will let $\mathbb{F}$ be our background field which will either be the real numbers or the complex numbers. Suppose $V$ has dimension $n$ with basis $e_1,...,e_n$ . This means every element in $V$ has a unique representation of the form $\sum_{i=1}^{n}\lambda_i e_i$ where $\lambda_i \in \mathbb{F}$ for all $i$ . To get our result we shall define the ‘special’ Euclidean norm on $V$ as $\lVert x\rVert_{E} := (\sum_{i=1}^{n}|\lambda_i|^2)^{\frac{1}{2}}$ . One can check this is a norm. We also let $\lVert.\rVert : V \longrightarrow \mathbb{R}$ be an arbitrary norm on $V$ .

How do we show these norms induce the same topology? We find constants $0 \neq M, m \in \mathbb{R}$ such that $m \lVert x \rVert_{E} \leq \lVert x \rVert \leq M \lVert x\rVert_{E}$ for all $x \in V$ . The reason why this is sufficient is because in terms of the induced topologies an open ball in one topology will contain an open ball of the other and vice versa. If you’re not convinced take a moment to try to see why(It’s very instructive in getting familiar with the theory).

Let us start by determining $M$ . Let $x \in V$ by using the unique representation in terms of our basis and the axioms of a norm we can estimate the norm like so $\lVert x \rVert = \lVert \sum_{i=1}^{n} \lambda_i e_i \rVert \leq \sum_{i=1}^{n} \lVert \lambda_i e_i \rVert = \sum_{i=1}^{n} |\lambda_i| \lVert x\rVert$ . Using Cauchy-Schwartz we can estimate even further and deduce $\lVert x \rVert \leq (\sum_{i=1}^{n}|\lambda_i|^2)^\frac{1}{2} (\sum_{i=1}^{n}| \lVert e_i \rVert^2)^\frac{1}{2}$ hence we have $\lVert x \rVert \leq M \lVert x \rVert_{E}$ where $M = (\sum_{i=1}^{n} \lVert e_i \rVert^2)^\frac{1}{2} > 0$ .

That was easy enough, right? If you’ve been doing mathematics long enough you notice the majority of mathematical proofs have an easy direction and a hard direction. That was the easy direction. Determining $m$ is more difficult, it isn’t insanely hard, and relies on the Heine-Borel Theorem and some properties of continuous functions.

Let us first define a function $f_{\lVert . \rVert} : \mathbb{F}^n \longrightarrow \mathbb{R}$ by $f(\lambda_1,...,\lambda_n) := \lVert \sum_{i=1}^{n} \lambda_i e_i \rVert$ . I mentioned continuity and so it’s not surprising that $f_{\lVert . \rVert}$ is continuous with respect to the standard topology on $\mathbb{F}^n$ .

To show $f_{\lVert . \rVert}$ is continuous we exploit the triangle inequality of the norm $\lVert . \rVert$ . Let $\epsilon > 0$ and pick $\delta = \epsilon$ and $(\alpha_1,...,\alpha_n), (\beta_1,...,\beta_n) \in \mathbb{F}^n$ be such that $(\sum_{i =1}^n |\alpha_i - \beta_i|^2)^{1/2} < \delta$ and let $H := \max_{1 \leq i \leq n} (\lVert e_i \rVert )$ . We estimate $| \; \lVert \sum_{i=1}^n \alpha_i e_i \rVert - \lVert \sum_{i=1}^n \beta_i e_i \rVert \; | \leq \lVert \sum_{i=1}^n (\alpha_i - \beta_i) e_i \rVert \leq \sum_{i=1}^n |\alpha_i - \beta_i | \lVert e_i \rVert$ $\leq H \sum_{i=1}^n |\alpha_i - \beta_i |$ $\leq n^{\frac{1}{2}} H(\sum_{i =1}^n |\alpha_i - \beta_i|^2)^{1/2} \leq n^{\frac{1}{2}} H\epsilon$ with the second to last inequality coming from Cauchy-Schwartz. Hence $f_{\lVert . \rVert }$ is continuous.

We also define the set $U := \{\lambda_1,...,\lambda_n) \mathbb{F}^n : \sum_{i=1}^{n} |\lambda_i|^2 = 1 \}$ which is the unit-circle in $\mathbb{F}^n$ . It follows that $U$ is compact with respect to the standard norm on $\mathbb{F}^n$ . To show this we use the the Heine-Borel Theorem which in its statement means it is necessary and sufficient to show that $U$ is closed and bounded as we are working over $\mathbb{F}^n$ .

The fact $U$ is bounded follows straight from the definition so we only need to show that $U$ is closed. This is the part where $f_{\lVert .\rVert}$ comes in as one can check that $U = f_{\lVert . \rVert_{E} }^{-1}(\{1\})$ where $f_{\lVert . \rVert_{E} }^{-1}$ is the pre-image of $f_{\lVert . \rVert_{E} }$ . As $\{1\}$ is closed in $\mathbb{R}$ and $f_{\lVert . \rVert_{E} }$ is continuous we have that $U$ is closed and hence it is compact by the Heine-Borel Theorem.

We’ve done the set-up so now let us get onto proving our actual result. To do this we shall restrict our function $f_{\lVert . \rVert }$ onto the set $U$ and ask ourself what the value $m := \inf_{x \in U} (f_{\lVert . \rVert }(x) )$ is? As $f_{\lVert .\rVert }$ is continuous and $U$ is compact we can use a property of continuous functions to deduce that there exists an $\alpha \in A$ such that $m = f_{\lVert . \rVert }(\alpha)$ . Suppose $\alpha = (\alpha_1,..,\alpha_n)$ and that $m =0$ then $f(\alpha_1,...,\alpha_n) = \lVert \sum_{i=1}^{n} \alpha_i e_i \rVert = 0$ and so $\sum_{i=1}^{n} \alpha_i e_i = 0$ by properties of the norm and by the linear independence of the basis we must have $\alpha_i = 0$ for all $i \in \{1,...,n\}$ . This means $\alpha = 0$ which is a contradiction as $0 \notin U$ . Bringing this all together $m > 0$ and by the definition of $m$ we have shown $\lVert x \rVert \geq m > 0$ when $\lVert x \rVert_{E} = 1$ .

All $x \in V$ can be written in the form $x = \lVert x \rVert_{E}y$ for some $y \in V$ such that $\lVert y \rVert_{E} = 1$ or equivalently where $y = \sum_{i=1}^{n}\lambda_i e_i$ where $(\lambda_1,...,\lambda_n) \in U$ we can say $\lVert x \rVert = \lVert x \rVert_{E} \lVert y \rVert > m\lVert x \rVert_{E}$ where $m \neq 0$ .

What we’ve shown is that every possible norm on a vector space of a specified dimension over $\mathbb{R}$ or $\mathbb{C}$ induces the same topology, in other words, as the title suggests there is one topology to rule all normed spaces of a fixed dimension… over $\mathbb{R}$ or $\mathbb{C}$ .

Tag: vector space

Algebras and why modules are a thing

It’s all local for topological vector spaces

One Topology to rule all normed spaces of a fixed dimension