Is there some physical intuition behind Clifford Algebras?

Question

The mathematically rigorous definition of a Clifford Algebra is as follows:

Let $V$ be a vector space over a field $\mathbb{K}$ and let $Q : V\to \mathbb{K}$ be a quadratic form on $V$. A Clifford algebra $C\ell(V,Q)$ is a unital associative algebra over $\mathbb{K}$ together with a linear map $i: V\to C\ell(V,Q)$ satisfying $i(v)^2=Q(v)1$ for all $v\in V$ defined by the following universal property: given any associative algebra $A$ over $\mathbb{K}$ and any linear map $j: V\to A$ such that $j(v)^2=Q(v)1_A$ for all $v\in V$ there is a unique algebra homomorphism $f : C\ell(V,Q)\to A$ such that $f\circ i = j$.

As a definition this is fine. Then we see how to construct these objects and prove results and so on.

My point is that Clifford Algebras seems to be quite important in Physics. Just for the sake of an example, they appear on the definition of spinors. But I believe I have seem them being refered in other contexts.

What I want to ask is: regarding the importance of Clifford Algebras in Physics, is there some nice intuition behind them? Although I'm able to understand mathematically this definition and use it I can't have any intuition about it. Is there any physical intuition about Clifford Algebras?

Answer 1

It's often nice to think of natural numbers as a subset of the integers. And to think of the integers as a subset of the rationals. The rationals as a subset of the algebraic. The algebraic as a subset of the reals.

And some people even like to think of the reals as a subset of the complex.

But there are lots of things going on there since those numbers are all used as objects and as operations. But keep in mind that it's nice to have a playground where all your toys are already there.

Now let's bring in the physics. In physics you don't need to use $\mathbb R^3$ technically you could use anything bijective with that as a model. And same for spacetime, you don't have to use $\mathbb R^4$ because anything bijective could work as well. As a vector space you could map over the vector addition and the scaling by scalars. But if you want to respect a metric that wouldn't be enough.

When you have a metric you can also talk about hyperplanes, and the vectors orthogonal to them. You might want to talk about the plane spanned by two vectors. You might want to talk about the 3d subspace spanned by three vectors. You might want to talk about the reflection across a hyperplane orthogonal to a vector.

So lots of objects and lots of operations. And just like you liked having a larger algebra that contained your smaller algebras, so you might might want an algebra everything can live inside.

A mathematician will say something isn't a vector if you can't add it. And even if you can add then, they would say it's just a group if you can't also scale it by a scalar. But the vectors are just a subset of the full multivector algebra. And they aren't even a subalgebra. So they aren't really a finished system. The only way you could they are finished is if you pretend you can't multiply them.

By that standard the imaginary numbers are just a vector space because you can pretend not to be able to multiply them.

But in physics we have things like spin and angular momentum and magnetic fields that truly require planes to describe them. And in relativity you have to admit planes as new objects because in 4d a 2d plane isn't a hyperplane so you can't just point at some vectors and give them a different name.

So we need scalars and we need vectors and we need planes too. That's life, and geometry.

If we introduce all these objects we can then talk about their relationships. But each one is sometimes an object and sometimes an operator. Just like matrices. A square matrix can multiply another square matrix treating the other like an object when that matrix could itself be an operator.

That's what an algebra is fundamentally.

So you want to have your objects and your operations. In fact most objects could just be transformations of reference objects. And the quadratic form is just trying to specify the metric to be respected by the transformations, the operations. Which end up giving you the full algebra.

So the set of products of vectors is natural as a set of operations. The full algebra isn't exactly needed, but it can be useful as a basis.

Answer 2

I like the subject of the question, which I've often pondered myself.

Timaeus's answer is very instructive.

My own, much shorter but possibly even more intuitive answer (betraying perhaps my lack of in-depth expertise, of course) to user1620696's question, is that Clifford algebras simply encode many more aspects of (i.e. ways of describing) the properties of linear spaces - in particular the combination of rotations with standard vectors (displacements).

My reference to rotations is probably equivalent to what Timaeus discusses as planes and hyperplanes.

Comments and corrections appreciated.