$\def\Cl{\mathcal C\ell}
\def\CL{\boldsymbol{\mathscr{C\kern-.1eml}}(\mathbb R)}$
I'm not an expert in neither of the fields I'm touching, so don't be too rude with me :-) here's my question.

A well known definition of Clifford algebras is the following: 

> 
> Fix a quadratic form $q$ on a vector space $V$[¹] and consider the quotient of the tensor algebra $T(V)$ under the ideal $\mathfrak i\_q = \langle v\otimes v - q(v){\bf 1}\mid v\in V\rangle$; 
> 
> 
> *Remark*: as such quotient enjoys a suitable universal property, it is easy to see that $(V,q)\mapsto \Cl(V,q)$ determines a functor from the category of quadratic spaces to the category of algebras.
> 
> 
> *Remark:* It's easy to see that if $q=0$ the Clifford algebra $\Cl(V,0)$ coincides with the exterior algebra $\bigwedge^\*\! V$ (this is clear, $\mathfrak i\_0=\langle v\otimes v\mid v\in V\rangle$!).
> 
> 
> 

Now.

I would like to know to which extent it is possible to regard Clifford algebras as "deformations of exterior algebras": I believe that fixing a suitable real number $0<\epsilon \ll 1$, and a map $t\mapsto q\_t$ defined for $t\in]-\epsilon,\epsilon[$ with codomain the space of quadratic forms, a useful interpretation of this construction regards $\Cl(V,q\_\epsilon)$ as a suitably small deformation of $\bigwedge^\*\! V$.[²] 

I can try to be more precise, but this pushes my moderate knowledge of differential geometry to its limit. I'm pretty sure there is a topology on the *don't say moduli, don't say moduli*--"space" of Clifford algebras $\CL$ so that I can differentiate those curves $\gamma : I \to \CL$ for which $\gamma(0)=\bigwedge^\*\!V$. 

> 
> * If this is true, what is the linear term of a "series expansion"
>  $$\textstyle
> \gamma(t) \approx \bigwedge^\*\!V + t(\text{higher order terms})?
> $$
> * Is there any hope to prove that the functor $\Cl$ is "analytic" in the sense that there are suitable vector-space valued coefficients $a\_n(V,q)$ for which
>  $$\textstyle
> \Cl(V,q) \cong \bigoplus\_{n\ge 0} a\_n(V,q)\otimes V^{\otimes n}
> $$
>  or even less prudently: $\Cl(V,q) \cong \bigoplus\_{n\ge 0} a\_n(V,q)\otimes (\bigwedge^\*V)^{\otimes n}$ (I'd call these "semi-analytic", or something)?
> * Regarding quadratic forms as symmetric matrices, one can wonder what are the properties of Clifford algebras induced by the fact that $\|q\|\ll 1$. For example, an exterior algebra $\Cl(V,0)$ decomposes as a direct sum of homogeneous components
>  $$\textstyle
> \Cl(V,0) \cong \bigoplus\_{p=0}^{\dim V} \bigwedge^p\!V
> $$
>  is there a similar decomposition of $\Cl(V,q\_\epsilon)$ for a sufficiently small $\epsilon$[³]?
> 
> 
> 

I can't help but admit I wasn't able to come up with a good idea for this question. :-) especially because well, I'm not a geometer.

---

[¹] I will assume that vector spaces are finite dimensional, over a field of characteristic zero, sorry if you like arithmetic geometry! Even better, if you want, vector spaces are real.

[²] I guess it is possible to say simply that I'm given a "Clifford bundle" $p : E \to \mathbb R$, and the preimage of a small neighbourhood of zero is a neighbourhood of $(V, 0)$ -the zero quadratic space over the typical fiber of $p$.

[³] I must admit this might be a trivial question: I remember that $\Cl(V,q)$ has an even-odd decomposition in a natural way, but I can't find a reference for a complete decomposition in irreducibles. And, well, I don't have a book on this topic :-)