We know the greatest feature of Clifford algebra is coordinate-free. One can do vector operations without knowing the representation of vectors. And due to this very characteristic, Clifford or geometric algebra is believed to be a reinterpretation of differential geometry as suggested mainly by Hestenes and Doran.
But as far as I know, many manifold-related theorems depend on the topology of the manifold such as connectedness, compactness, boundaryless or not. I want to know how Clifford algebra behaves in different topologies?
A Clifford algebra is defined for a quadratic vector space - so it's a purely algebraic object not depending on a manifold. But if you have a spin manifold M, then, depending on the topological structure, it can admit several spin structures (they are classified by H^1(M;Z_2)), and each spin structure yields a spinor bundle, i.e., a bundle over M that looks *locally* as the spin representation constructed via the associated Clifford algebra, but globally not. For details, look at Th. Friedrich's book
"Dirac operators in Riemannian geometry" (AMS, GTM vol. 25).
On a given manifold M we may have several riemannian or pseudo-riemannian structures. From these we usually construct the bundle of Clifford algebras over M. This bundle can be trivial or not. On the other hand we may construct a bundle of Clifford algebras starting from a pseudo-Hermitian vector bundle over M, not necessarily the tangent bundle. Finally, we may consider vector-valued functions over M - this may lead to infinite-dimensional Clifford algebras - this happens in quantum field theory when we construct C* algebras of canonical anti-commutation relations (CAR). I think that Alain Connes was playing with something of this kind.
So, there are many different ways in which your question can be addressed. Probably the infinite-dimensional case is the most interesting one.
1. Clifford algebras are purly algebraic structures, so they do not depend on a manifold
2. On a maniold endowed with any scalar product (not necessarily riemannian) one can add a structure of Clifford bundle.
For both topics see my paper "mathematics for theoretical physics" (2014 edition) with some new theorems
For theuse of Clifford algebra in General Relativity and spinors see my paper "particles and fields" , with some useful computations on Cl algebras.
@Jean Claude Dutally is right, the existence of a Clifford bundle Cl \to M does not depend on the topology. The existence of spinors i.e. a the existence of a vector bundle S \to M together with a multiplication \mu Cl \tensor S \to S such that on each point x, the fibre S_x is isomorphic to the irreducible spinor representation of the Clifford algebra Cl_x does depend on the topology however. It is equivalent to the existence of a lift of the second Stiefel Whitney class w_2(M \in H^2(M, \Z/2\Z) ) to a class K in H^2(M, \Z). This is automatic if dim(M) = 4 by a theorem of Hirzebruch-Hopf.
Note this is less restrictive than the existence of a Spin structure which is a lift of the SO(n) Framebundle to a Spin(n) bundle.
Sideremark:
The Clifford bundle with its algebra bundle structure, is determined by the metric g on the manifold. Conversely the algebra structure determines the metric and therefore the Levi Civita connection. Even though Clifford bundles always exist they still require more structure than the exterior algebra (i.e. forms) and the exterior differential. The Clifford algebra approach promoted by the geometric algebra community, is therefore certainly not a replacement of modern (well 50 years old) coordinate and metric free differential geometry. Even when there is a metric involved, it tends to be a good idea to clearly distinguish between metric independent and metric dependent expressions. In my opinion avoiding exterior algebra and doing everything in a Clifford algebra is usually misguided because it becomes hard to see what does and what does not depend on the metric. For example the Maxwell equations can be written as one single coordinate independent equation using the geometric algebra formalism that writes the Faraday tensor and the current as an element of the Clifford algebra. The differential forms formulation of the Maxwell equation represents the Faraday tensor as a two form and the current as a three form and consists of two coordinate independent equations. However I believe two is actually better than one in this case because one equation is manifestly independent of a metric (dF = 0), and the other equation (d*F = J) manifestly does depend on the metric, but in a very restricted way through the Hodge star (and therefore only depends on the conformal class of the metric in dimension 4). When using spinors on manifolds however Clifford algebra bundles and spinor bundles as above are great and make it crystal clear how spinors and the metric are related!
- Badges
- Science topic
- Similar topics
- Mathematics
- Statistics