This is intended to be the subject of a short postgraduate talk in UCC. At times there will be little attempt at rigour — mostly I am just concerned with ideas, motivation and giving a flavour of the philosophy. Also it is fully possible that I have got it completely wrong in my interpretation!


It is a theme in mathematics that geometry and algebra are dual:

\text{Geometry }\leftrightarrow \text{ Algebra}

Arguably this theme began when Descartes began to answer questions about synthetic geometry using the (largely) algebraic methods of coordinate geometry. Since then this duality has been extended and refined to consider:

\text{Spaces }\leftrightarrow \text{ Algebra of Functions on the Space}

Here a space is a set of points with some additional structure, and the idea is that for a given space, there will be a canonical algebra of functions on the space. For example, given a compact, Hausdorff topological space X, the canonical algebra of functions is C(X) — the continuous functions on X.

In this talk we will be mostly concerned with what properties of the space we can recover from the algebra of functions on it — in the above example we could in fact recover everything from the canonical algebra of functions. In this talk mostly we will not necessarily be talking about the canonical algebra of functions.


Let X=\{a_1,a_2,a_3\} be a set of three distinct points in space. Now consider F(X) the space of complex valued functions X\rightarrow \mathbb{C}. Now to define f\in F(X), all we must do is choose three complex numbers:

f(a_i)=\lambda_i; for i=1,2,3.

Hence every f\in F(X) may be uniquely written in the form:


That is \{\mathbf{1}_{\{a_i\}}\} is a basis of F(X) so \text{dim } F(X)=3.

Now I would argue that the only feature of this space is that \text{card}(X)=3. So for a finite set X such as this one, with no additional structure at all, the algebra of functions F(X) can tell us everything about X.

Group Multiplication

Suppose we are told that G is a finite set but that there is a binary relation \star\subset G\times G such that \star satisfies the group axioms. We know from the last section that F(G) can recover the order of the group:

|G|=\text{dim }F(G)=:n,

\Rightarrow G=\{g_1,g_2,\dots,g_n\}.

Can F(G) do more? Can F(G) recover the group law? Well if we make the assumption that the functions f\in F(G) know what to do with g_i\star g_j for all i,j=1,\dots,n then yes we can.

Choose k\in F(G) to be an injective function; that is k can distinguish between points:

k(g_i)=k(g_j)\Rightarrow g_i=g_j.

Now evaluate each of the elements of G to generate n distinct complex numbers:

k(g_i)=\lambda_i; i=1,\dots,n.

Now evaluate the n\times n products:

k(g_i\star g_j); i,j=1,\dots,n.

Now if

k(g_i\star g_j)=\lambda_k=k(g_k),

the injectivity of k forces g_i\star g_j=g_k. Hence we can construct the multiplication tables.

In particular, if

k(g_e\star g_i)=\lambda_i; for all i=1,\dots,n,

then g_e=e — the identity on G.

What about Infinite Cardinality

Let V be a finite dimensional real vector space of dimension n. Now the cardinality of V is \aleph_0, but an algebra of functions (careful!) on V can give us the dimension of V.

Consider L(V) — the space of all linear maps on V. Let \{e_1,\dots,e_n\} be a basis for V and suppose T is a linear map on V:

T(\mathbf{u}+\lambda\mathbf{v})=T\mathbf{u}+\lambda T\mathbf{v}.

Let \mathbf{v}\in V such that




=\sum_{i=1}^n\alpha_i Te_i.

Hence T is defined by what it does to the basis vectors. Say


so we need n real numbers for each of the n basis vectors to define a linear map. Therefore L(V) is of dimension n\times n (duh!). So if \text{dim }L(V)=n\times n, then \text{dim }V=n.


Consider the interval X=[0,1]. In the norm topology it is connected which means that we cannot represent X as a union of disjoint subsets. Consider the continuous functions on XC(X). Now call p\in C(X) a projection if p^2=p. That is

[p(x)]^2=p(x); for all x\in X,

\Leftrightarrow p(x)=0 or 1.

Now this means that p either takes the value 0 or the value 1.

Suppose p is a non-zero projection and set

A=\{x\in X:p(x)=1\}.

Now p=\mathbf{1}_{A}. Now it should be clear that either A=\emptyset or X; otherwise p is not continuous as it will have jump discontinuities on the boundary of A.

Hence the only projections on the connected set X are the trivial projections 0 and \mathbf{1}_X.

Now consider X=[0,1]\cup[2,3]. Now X is certainly disconnected but \mathbf{1}_{[0,1]} and \mathbf{1}_{[2,3]} are continuous non-trivial projections.

In general (!), if C(X) contains non-trivial projections, then X is disconnected.


To consider compactness, equip a set Y with a metric and the induced topology. Suppose X is a compact subset of Y such that X is compact. For the purposes of this piece, we will suppose that X compact just means that X is closed and bounded (X will look like \mathbb{R}^n.)

Now let C(X) be the set of continuous functions on X.

Extreme Value Theorem

A continuous function on a compact set A attains it’s absolute maximum and minimum on A.

In our example, every continuous function on X attains its maximum on X so in this case there exists a positive M_f\in\mathbb{R} such that:

|f(x)|\leq M_f; for all x\in X    (*).

Similarly, if X is open or unbounded (i.e. not compact), then we can see that a condition such as (*) need not necessarily hold.

For example, X=(0,1) is bounded but not closed. The continuous function x\mapsto 1/x is unbounded.

Also,  X=[0,\infty) is closed but not bounded. The identity function on Xx\mapsto x is unbounded.

So if we are given a space C(X) then we know that X cannot be compact if C(X) contains unbounded functions.


A space X is Hausdorff if any two points x,y\in X there exists neighbourhoods of x and yU and V say, such that U and V are disjoint.

With some extra assumptions on X (namely that it is normal and locally compact), consider C(X). If for all f\in C(X), there exists a pair x\neq y such that f(x)=f(y) then X cannot be Hausdorff. Why?

Urysohn’s Lemma

If A,B are disjoint closed sets in a normal space X, then there exists a continuous function f:X\rightarrow\mathbb{R} such that f_{|A\cup B}=\mathbf{1}_A.

Under the assumptions, Urysohn’s Lemma guarantees that there is a function f\in C(X) such that f(x)=1 and f(y)=0. That A and B exist are by local compactness.

So we pick this function f and say well f(x)=1f(y)=0. Contradiction. That is x,y must be in the same neighbourhood and Hausdorffness is destroyed.

Non-Commutative Spaces

Note that in all these examples (except when we looked at L(V)), the  algebra of functions had a common structure:

  1. Vector Space — for any complex valued function we can define point-wise the functions f+g and \lambda f for \lambda \in \mathbb{C}.
  2. Normed Space — there are various norms we could put on the algebra of functions; supremum norm, one norm, two norm, etc.
  3. Inner Product Space — again, for example, \langle f,g\rangle=\int f(x)\overline{g(x)}\,dx. We can go further if we pick the correct algebra of functions — we can pick the algebra of functions to be complete, so that we have a Hilbert Space (a complete inner product space).
  4. Associative Algebra — we can define pointwise a product on the algebra of functions.
  5. *-Algebra — the algebra of functions takes on an involution {}^*, namely the conjugation:


Any algebra which has these five features (with a few conditions on how the norm interacts with the product and the involution), is known as a C*-algebra and by and large, the canonical algebra of functions on a space will have this structure.

Because of the commutativity of \mathbb{C}, the algebra of functions on X is commutative:

f(x)g(x)=g(x)f(x); for all x\in X.

There is a beautiful theorem of Gelfand that states that every abelian C*-algebra is isomorphic to an algebra of functions on a space.

But what about non-commutative C*-algebras? How about:

\text{Commutative Space }\leftrightarrow \text{ Commutative Algebra of Functions}

\text{Non-Commutative Space }\leftrightarrow \text{ Non-Commutative Algebra of Functions}.

Ordinarily we look at a space X, and then look at the induced algebra of functions F(X). Our work above has shown that often it is equally valid to look at a C*-algebra, say F, and look at the induced space X(F). Why not do this for a non-commutative C*-algebra F?

We could then say things about X(F) based on F. We have to realise that X(F) is not going to look like our intuitive idea of space as a set of points. This is hard to imagine so don’t bother — looking at F itself should be equally valid. For example, two definitions we could decide to use could be:

  1. X(F) is connected if F contains non-trivial projections
  2. X(F) is compact if F contains unbounded elements

At this point I admit that this about the height of my understanding. I really should check this out properly, but I think Heisenberg’s uncertainty principle and a whole pile of modern quantum theory works on non-commutative spaces so in essence they are saying the very geometry of this universe is non-commutative!

For a proper and accurate account see