Does Clifford algebra depend on the topology of manifold? 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 its very characteristc, Clifford or geometric algebra is believed to a reinterpretation of differential geometry suggested mainly by Hestenes and Doran. But as far as I know, many manifold-related theorem depends on the topology of the manifold such as connectedness, compactness, boundaryless or not. I want to know how Clifford algebra behave in different topologies?
 A: The Clifford algebra is defined for any vector space $V$ with quadratic form $q$ as $\operatorname{Cliff}(V;q) = T(V)/\langle x \otimes x + q(x) \rangle$ where $T(V) = \displaystyle \bigoplus_{n \ge 0} V^{\otimes n}$ is the tensor algebra on $V$. This makes sense in particular for the tangent space of any point on a manifold with the quadratic form coming from the metric, so to construct a Clifford bundle (a bundle over a pseudo-Riemannian manifold where each fiber is a Clifford algebra) we need only find a way to glue the fibers together to get the full vector bundle stucture.
The Clifford algebra on a vector space (with any choice of quadratic form) is canonically isomorphic as a vector space to the exterior algebra on that space (but not as an algebra). Extending this to vector bundles over a pseudo-Riemannian manifold, the Clifford bundle  can be topologized by defining this graded fiberwise vector space isomorphism $\operatorname{Cliff}(TM;g) \rightarrow \Lambda(TM)$ and pulling back the topology on $\Lambda(TM)$ (which you already know as it's a subbundle of the tensor bundle), which makes this into an isomorphism of vector bundles (but with different algebra structures). That is to say, excluding multiplication, the Clifford and exterior bundles are the same, and in particular they're the same topologically. There are more natural ways to topologize the Clifford bundle, e.g. via the associated bundle construction, but this is a quick and dirty way to get it done.
There are topological issues nearby though which can have a big effect on physics. If you want to construct an irreducible Clifford module (which is just a spinor bundle, something of clear importance for physics if e.g. you want to write down the Dirac equation), the base manifold (e.g. spacetime) needs to have a spin structure. Formally, this means that in signature $(m,n)$ there needs to be a lift of structure group from $\operatorname{SO}(m,n)$ to $\operatorname{Spin}(m,n)$ (its double cover). Such a lift exists if and only if the Stiefel–Whitney class $w_2 (TM) = 0$. Manifolds with such a lift are called "spin manifolds", and to even have global spinor fields and Dirac operators on them we need spacetime to be a spin manifold.
