# Clifford algebras as deformations of exterior algebras

$\def\Cl{\mathcal C\ell} \def\CL{\boldsymbol{\mathscr{C\kern-.1eml}}(\mathbb R)}$ 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 :-)

• There are lots of things happening in the neighborhood of the exterior algebra: it is the "ultimate degeneration", while arbitrarily near to it are all kinds of intermediate "strata" of various dimensions, corresponding to partially degenerate quadratic forms like $x_1^2$, $x_1^2+x_2^2$, $x_1^2+x_2^2+x_3^2$, etc. – მამუკა ჯიბლაძე May 6 '18 at 19:21
• It seems to me that it is possible to restrict to paths that are always nondegenerate away from 0; for example, fixed a signature $(p,d-p)$ one can consider the segment $t \mapsto \left(\begin{smallmatrix}t \mathbb I_p &0\\0&t \mathbb I_{d-p}\end{smallmatrix}\right)$. It is possible to simplify the problem in this special case? – Fosco May 6 '18 at 19:26
• One viewpoint you can take is to first consider the deformed algebra $T(V)[[\hbar]]/\langle vw + wv = 2\hbar q(v,w)\rangle$, and then set $\hbar = 1$. The first algebra is a deformation quantization of the Poisson algebra $\Lambda V\cong \mathrm{Sym} \Pi V$ with the Poisson bracket coming from the metric. – Bertram Arnold May 6 '18 at 21:05
• I can understand only part of your answer, alas! :-) Any reference or time for an extended explanation? Thanks! – Fosco May 6 '18 at 21:23
• There is a very easy way to make what you were trying to say precise. It does not involve paths in any "space of algebras". Rather, it involves paths in the space of associative products on a fixed vector space (which together make up an algebra), that is $(A,*_t)$. Setting $A=\Lambda^\bullet V$ and $*_0 = \wedge$, it is easy to find a formula for the product $*_t$ which reproduces the Clifford relation $v *_t v = t q(v)$. You'll find that it is even polynomial in $t$ and has the right $t=0$ limit. – Igor Khavkine May 6 '18 at 23:09

As Igor mentions in the comments, this is really a question about deformations of the multiplication map of the exterior algebra in the space of associative multiplications. Since this is a pretty general story, let me try to sketch how it works in this case.

Let's fix some field $k$ and a $k$-algebra $(A,\mu)$. We want to understand deformations of $(A,\mu)$, and to start only infinitesimal deformations. For this we fix an Artinian augmented $k$-algebra $R\to k$ and build a groupoid $\operatorname{Def}_R$: objects are flat $R$-algebras $B$ with an isomorphism $\rho: B\otimes_R k\cong A$, and morphisms from $(B,\rho)$ to $(B',\rho')$ are isomorphisms of $R$-algebras over $k$.

$R$ being Artinian just means that $R \cong k\oplus\mathfrak m$, with $\mathfrak m$ being nilpotent and finite-dimensional. In fact, one does not lose all that much by restricting to $R_n = k[\hbar]/(\hbar^{n+1})$. By flatness, any $R_n$-deformation is of the form $B\cong R_n\otimes_k A\cong A[\hbar]/(\hbar^{n+1})$, but with an interesting multiplication $\mu_{B}$. It is determined by $\mu_B(a,a') = \sum_{i=0}^n\hbar^i\mu_i(a,a')$ for some maps $\mu_i:A\otimes_k A\to A$. Similarly, any automorphism $\lambda$ of $B$ is determined by $\lambda(a) = \sum_{i=0}^n\hbar^n\lambda_i(a)$ for some maps $\lambda_i:A\to A$. Then the condition that $\mu_B$ is associative leads to the following constraints:

• $\mu_0 = \mu$ by checking compatibility with $\rho$. From now on, I'll write $\cdot = \mu$. Similarly, $\lambda_0 = \operatorname{id}$.
• $\mu_1(f,g)h + \mu_1(fg,h) = f\mu_1(g,h) + \mu_1(f,gh)$, and $\lambda$ is a morphism from $\mu_1$ to $\mu_2$ iff $\mu_1(f,g) - \mu_2(f,g) = \lambda_1(fg) - f\lambda_1(g) - \lambda_1(f)g$. The first condition is that $\mu_1$ is a closed $1$-cocycle in the Hochschild chain complex $HH^n(A) = \operatorname{Hom}(A^{\otimes(n+1)},A)$ with differential $$\mathrm d\mu(a_0,\dots,a_{n+1}) = a_0\mu(a_1,\dots,a_n) + \sum_{i=0}^{n-1} (-1)^{i+1}\mu(a_0,\dots,a_ia_{i+1},\dots,a_n) + (-1)^{n+1}\mu(a_0,\dots,a_{n-1})a_n\ .$$ The multiplications of two such cocycles are isomorphic iff the cocycles are cohomologous. In this case the isomorphisms are a torsor over the set of closed $0$-cycles. All in all, the category $\operatorname{Def}_{k[x]/x^2}$ has objects in bijection with $H^1(HH(A))$, and each object has automorphism group $H^0(HH(A))$.
• $f\mu_2(g,h) - \mu_2(fg,h) + \mu_2(f,gh) - \mu_2(f,g)h = \mu_1(f,\mu_1(g,h)) - \mu_1(\mu_1(f,g),h)$. Thus given a first-order deformation $\mu_1$, one may form the element $\mu_1(-,\mu_1(-,-)) - \mu_1(\mu_1(-,-),-)$, which one can check to be a closed $2$-cocycle. The existence of $\mu_2$ is equivalent to the corresponding cohomology class vanishing. Thus for every first-order deformation there is an obstruction lying in $H^2(HH(A))$ which vanishes iff this first-order deformation extends to a second-order deformation. In this case the extensions are a torsor over $H^1(HH(A))$, and similarly for the automorphisms.
• This story continues: If we have found $\mu_1,\dots,\mu_n$, the existence of $\mu_{n+1}$ is equivalent to some obstruction class in $H^2(HH(A))$ vanishing, in which case the posible extensions are a torsor over $H^1(HH(A))$. Note however that the obstruction depends on all previous choices.

In fact, essentially all of these formal deformation problems arise in the following way (at least if $k$ is characteristic $0$): One has a dgla $(\mathfrak g,[-,-])$ and a natural isomorphism $\operatorname{Def}_{k\oplus\mathfrak m} \cong \{x\in \mathfrak m\otimes \mathfrak g^1\mid \mathrm dx + \frac{1}{2}[x,x] = 0\}//\mathfrak m\otimes \mathfrak g^0$, where the nilpotent Lie algebra $\mathfrak m\otimes \mathfrak g^0$ acts on this set of "Maurer-Cartan elements" by the Lie bracket and one takes the action groupoid of the action of the corresponding nilpotent Lie group. In our case the dgla is $HH^*(A)$ equipped with the Gerstenhaber bracket which takes two cocycles and takes the sum over all ways to plug one into the other (with some signs).

In general, the deformation functor only depends on the dgla up to quasiisomorphism. This has one important consequence: If $\mathfrak g$ is formal, i.e. quasiisomorphic to its cohomology, then a Maurer-Cartan element of $\mathfrak g$ is "the same" as an element $x\in H^1(\mathfrak g)$ satisfying $[x,x] = 0$, i.e. a first-order deformation which extends to a second-order deformation. This then gives rise to a Maurer-Cartan element for any $R$, i.e. a lift to arbitrarily high order in $\hbar$, without having to check whether any more obstructions vanish! One can then put all of these lifts together to obtain an associative multiplication $\mu_\hbar = \mu + \cdots$ on $A[[\hbar]]$.

For a differential geometer, the most interesting case is that $A = C^\infty(M)$. In this case one should restrict to the subcomplex of the Hochschild complex of maps $C^\infty(M)^{\otimes(n+1)}\to C^\infty(M)$ which are differential operators in each variable. The Hochschild-Kostant-Rosenberg theorem computes the homology of this complex as polyvector fields $\oplus_{i\ge 1} \Gamma(M,\Lambda^i TM)$. The bracket on the homology groups is the Schouten bracket (extension of the Lie bracket as a biderivation). Amazingly, as proven by Kontsevich this dgla is actually formal! By the previous discussion this means that a first-order deformation, i.e. a biderivation $\pi\in\Gamma(M,\Lambda^2 TM)$, extends to all orders iff it extends to a second-order deformation, i.e. iff $[\pi,\pi] = 0$, iff $f\otimes g\mapsto \{f,g\} = \pi(\mathrm df,\mathrm dg)$ is a Poisson structure on $M$. This is called deformation quantization since quantization should associate to each function $f$ on $M$ an operator $\widehat f$ on some Hilbert space such that $[\widehat f,\widehat g] = \hbar \widehat{\{f,g\}} + O(\hbar^2)$. Deformation quantization does away with the Hilbert space and formally expands $\widehat f \cdot \widehat g = \widehat{f\cdot g} + \sum_{i\ge 1}\hbar^i\widehat{\mu_i(f,g)}$.

What does any of this have to do with your question about Clifford algebras? One can think of the graded commutative algebra $\Lambda^*V$ as functions on the supermanifold $\Pi V^\vee$ ("$V$ dual in odd degree"). The metric is a graded antisymmetric bilinear form on $V^\vee$ and hence induces a symplectic, and by extension Poisson, structure on this supermanifold, which by the previous discussion has an essentially unique deformation quantization. In fact, one can give an explicit formula for it by putting the right "fermion signs" into the Moyal product. This is a very special deformation quantization since only finitely many $\mu_i$ are nonzero. Thus for any value of $\hbar$ the series $\sum_{i\ge 0}\hbar^i \mu_i$ converges, and one may set $\hbar = 1$ to recover the Clifford algebra.

• This is a fascinating answer and it will take me "several years of Sundays" to understand it :-) – Fosco May 9 '18 at 20:03
• Thanks! I tried to be detailed, but here is the short version of the story: Whenever you have a deformation problem, you should first try to understand the corresponding formal deformation problem, which is governed by some dgla. If this dgla is formal, a deformation to all orders is uniquely determined by its first-order deformation, which has to be unobstructed. So a deformation of $\Lambda^*V$ "is" a Poisson structure on $\Pi V^\vee$, of which $\operatorname{Sym}^2(V)$ is a special example. The other answer gives the explicit formula for the multiplication in terms of the Poisson bracket. – Bertram Arnold May 10 '18 at 10:27

In addition to the answer of Bertram Arnold, let me point out that there is a very explicit formula for "fermionic" Weyl-Moyal product. Let us assume that your vector space $V$ (or module) is defined over a ring containing the rationals. This is sort of crucial in the following and in prime characteristic the story is slightly different.

Part of the sport is now to work without using a basis in order to have things applicable in a ring-theoretic framework as well. If you do not care about this, then many of the computations simplify drastically.

Consider the Grassmann algebra $\Lambda^\bullet(V)$ and define the following operator \begin{equation} P_q\colon \Lambda^\bullet(V) \otimes \Lambda^\bullet(V) \longrightarrow \Lambda^{\bullet-1}(V) \otimes \Lambda^{\bullet-1}(V) \end{equation} explicitly on factorizing tensor \begin{equation} P_q(v_1 \wedge \cdots \wedge v_k \otimes w_1 \wedge \cdots \wedge w_l) = \sum_{i=1}^k \sum_{j=1}^l (-1)^{j+k-i-1} q(v_i, w_j) v_1 \wedge \stackrel{i}{\cdots} \wedge v_k \otimes w_1 \wedge \stackrel{j}{\cdots} \wedge w_l \end{equation} This turns now out to be a biderivation (with signs), i.e. one has some straightforward Leibniz rules which, on the other hand, determine $P_q$ completely. Next, one needs the operators $P_{12} = P_q \otimes 1$, $P_{23} = 1 \otimes P_q$ and $P_{13} = (1 \otimes \tau) \circ (P \otimes 1) \circ (1 \otimes \tau)$ on the triple tensor product where $\tau$ is the usual flip of tensor factors. Here you have to be careful to obey Koszul's rule for tensor products of graded things correctly...

If you denote the Grassmann algebra product now by $\mu$ then the above Leibniz rules can be summarized as \begin{equation} P_q \circ (\mu \otimes 1) = (\mu \otimes 1) \circ (P_{23} + P_{13}) \quad \textrm{and} \quad P_q \circ (1 \otimes \mu) = (1\otimes \mu) \circ (P_{12} + P_{13}) \end{equation} A last explicit computation now shows that the three operators $P_{12}$, $P_{13}$ and $P_{23}$ commute pairwise.

From this it is a general argument (and quite illustrative to prove) that the formula \begin{equation} a \star_q b = \mu \circ \exp(\hbar P_q) (a \otimes b) \end{equation} defines an associative product as a formal power series in $\hbar$: this goes back at least to Gerstenhaber himself in the famous papers on deformation theory of associative algebras (I think in the III part...?).

A final observation now shows that you can safely set $\hbar = 1$ since the operator $P_q$ decreases the Grassmann degree in both tensor factors by one and hence the exponential series is just a finite sum.

By the very construction, the product $\star_q$ is written as a series in $q$.

OK, now I constructed you a funny associative (unital) algebra, where is the Clifford algebra? First, a really simple computation shows that \begin{equation} v \star_q w + w \star_q v = 2q(v, w) 1 \end{equation} for $v, w \in V$. Hence one obtains a unique algebra homomorphism from the Clifford algebra to this deformed Grassmann algebra, inducing this map on generators. Denote this map by $\sigma$ (for symbol calculus). Conversely, define a quantization map from the Grassmann algebra to the Clifford algebra explicitly by \begin{equation} Q(v_1 \wedge \cdots \wedge v_k) = \frac{1}{k!} \sum_{\pi \in S_k} (-1)^\pi v_{\pi(1)} \bullet \cdots \bullet v_{\pi(k)} \end{equation} where you treat the right hand side as products in the Clifford algebra. It requires now a bit of a computation (but just computation) to show that $Q$ and $\sigma$ are inverse to each other. In particular, both of them are algebra morphisms.

• Thank you, unfortunately I can accept only one answer. – Fosco May 9 '18 at 20:03