# NCG 2

I’ve decided that I must alternate between technical jargon and conceptual posts. If all my posts were like my last one, I wouldn’t make it to the first page of the article. So I think you know which category you belong to. NCG “odd number” will be the conceptual posts, and NCG “even number” will be more technical discussions.

Although Part 1 contained lots of interesting motivation to study NCG, I don’t really want to get into that. So we’ll just say, it occurs in physics a lot, and leave it at that. It turns out that when we say NCG we really mean noncommutative differential geometry as developed by A. Connes. This makes me happy, since I am a fan of differential geometry, and not so much of algebraic geometry (sorry to those aspiring algebraic geometers), and my big fear was that NCG was algebraic geometry but harder since things don’t commute.

I also need to apologize, because I assumed this was heading towards rings of functions which is why I brought up rings, but apparently we are headed down the noncommutative algebras path. So I’ll get on to exploring my first area of interest that has popped up.

In smooth manifolds we can give differential structure through the algebra of differential forms. In other words, given a smooth manifold M, and the exterior differential d, we have $(\Omega^*(M), d)$ that we call “the differential calculus over the algebra M.” Now I must admit that this language doesn’t really fit (since M is not an algebra), and I think what is meant is that where we have M we should really have $C^\infty (M)$ or the algebra of smooth functions on M. This is actually an algebra (the algebra of scalars), and we have the added bonus of being able to reconstruct M from its algebra of scalars.

So the moral of the story is, why do we need so much? Given an arbitrary algebra $\mathcal{A}$ and a differential (which we now need to define) satisfying $d^2=0$, we have $(\Omega^*(\mathcal{A}), d)$ or the differential calculus over the algebra $\mathcal{A}$.

So the generalization for the differential is not immediately obvious. Let $f\in\mathcal{A}$ such that the algebra has a representation on a Hilbert space $\mathcal{H}$. If we let F be an operator on the Hilbert space with spectral properties that make it look like the Dirac operator, then we define $df=i[F,f]$. This definition is really the product of an analogy with the Dirac operator and its commutation relations.

So this probably left everyone feeling extremely unsatisfied, but next time (4) I’ll build some examples as well as providing more solid grounds for interpreting spaces as their algebra of scalars.