$\DeclareMathOperator\Cl{Cl}\DeclareMathOperator\CCl{\mathbb Cl}\DeclareMathOperator\SO{SO}$Let $M$ be an oriented closed Riemannian manifold of dimension $n$ and $\CCl(M)= \Cl(M)\otimes \mathbb{C}$ be the complexified Clifford algebra bundle over $M$. Here $\Cl(M)$ is the real Clifford algebra bundle of $M$ defined by: $\Cl(M)= P_{\SO}(M)\times_{O(n)} Cl(\mathbb{R}^{n})$, where $P_{\SO}(M)$ is the $\SO(n)$ bundle over $M$ and $\Cl(\mathbb{R}^{n})$ is the real Clifford algebra over $\mathbb{R}^{n}$ and the action of $SO(n)$ on $Cl(\mathbb{R}^{n})$ is the natural action given by the homomorphism $SO(n)\rightarrow Aut(Cl(\mathbb{R}^{n}))$ . Let $\Gamma (M)$ denotes the set of all continuous sections of the complexified Clifford algebra bundle $\CCl(M)$. Indeed $\Gamma (M)$ is a unital associative algebra over $\mathbb{C}$. Now it is well known that complex Clifford algebras are rigid (rigid means there are no nontrivial formal deformations). Can anyone give me a clue how to compute Hochschild cohomology groups of the algebra $\Gamma (M)$ or more precisely is it rigid? Any suitable referrence regarding this will also be helpful.

$\begingroup$ So you are saying that a rigid bundle admits no formal deformations? To say it is "equivalent to a null deformation" is a long winded way of saying the same and one can say, obfuscatory. $\endgroup$– Mozibur UllahSep 22 at 3:41

$\begingroup$ Or rather defined as such ... $\endgroup$– Mozibur UllahSep 22 at 3:57
1 Answer
Let's try this: assume that the complex vector bundle you start with has even fiber dimension such that the corresponding Clifford algebra bundle is a bundle of complex matrix algebras. Suppose further that we are in a Spin$^c$ situation, i.e. the bundle of Clifford algebras is the bundle of the endomorphisms of a chosen Spin bundle. In this case, the algebra of sections of the Clifford bundle are Morita equivalent to the algebra of complexvalued smooth functions on the base. In fact, this is an equivalent statement to the existence of a Spin structure, a classical result from noncommutative geometry.
Now: Morita equivalent unital algebras have isomorphic Hochschild cohomologies, this settles your first question since the Hochschild cohomology of the smooth functions is known to be the complex sections of the Grassmann algebra bundle of $TM$. That is the classical HochschildKostantRosenberg theorem.
But it also settles your initial question since every formal deformation of the functions induces a unique formal deformation of the endomorphisms up to equivalence of deformations. This is one of the nice aspects of Morita theory: you deform the bimodule, i.e. the sections of the Spin bundle, as a right module over the functions, in a uniqueuptoequivalence way since it is finitely generated projective (SerreSwan). Then this induces a deformation of the endomorphism in a uniqueuptoequivalence way such that they act as right module endomorphisms. Together we have a deformed bimodule structure. You can find this in papers of Henrique Bursztyn and myself with many details and various additional aspects.
In the end you get a onetoone correspondence of the equivalence classes of deformations of the functions with the deformations of the sections of the endomorphism bundle aka Clifford bundle. In particular, thanks to Kontsevich's existence theorem, any (formal) Poisson structure gives such a deformation into a star product. We thus have zillions of formal deformations, many inequivalent ones. In particular, it is not at all rigid, as soon as the dimension of your manifold is at least $2$ (to have nontrivial Poisson structures).
In odd dimensions or without the assumption of Spin$^c$, this argument will not work that easily. Hmm....

$\begingroup$ Thanks a lot. I'll try to understand it thoroughly. $\endgroup$ Sep 23 at 1:57