What are the "correct" conventions for defining Clifford algebras?

Question

I have three related questions about conventions for defining Clifford algebras.

1) Let $(V, q)$ be a quadratic vector space. Should the Clifford algebra $\text{Cliff}(V, q)$ have defining relations $v^2 = q(v)$ or $v^2 = -q(v)$?

2) Should $\text{Cliff}(n)$ denote the Clifford algebra generated by $n$ anticommuting square roots of $1$ or by $n$ anticommuting square roots of $-1$? That is, after you pick an answer to 1), should $\text{Cliff}(n)$ be $\text{Cliff}(\mathbb{R}^n, \| \cdot \|)$ or $\text{Cliff}(\mathbb{R}^n, - \| \cdot \|)$? More generally, after you pick an answer to 1), should $\text{Cliff}(p, q)$ be the Clifford algebra associated to the quadratic form of signature $(p, q)$ or of signature $(q, p)$?

3) Let $(X, g)$ be a Riemannian manifold with Riemannian metric $g$. After you pick an answer to 1), should the bundle of Clifford algebras $\text{Cliff}(X)$ associated to $X$ be given fiberwise by $\text{Cliff}(T_x(X), \pm g_x)$ or by $\text{Cliff}(T_x^{\ast}(X), \pm g_x^{\ast})$?

For 1), on the one hand, $v^2 = q(v)$ seems very natural, especially if you think of the Clifford algebra functor as a version of the universal enveloping algebra functor, and it is used in Atiyah-Bott-Shapiro. On the other hand, Lawson-Michelson and Berline-Getzler-Vergne use $v^2 = -q(v)$, I think because they want $\text{Cliff}(\mathbb{R}^n, \| \cdot \|)$ to be the Clifford algebra generated by $n$ anticommuting square roots of $-1$. This is, for example, the correct Clifford algebra to write down if you want to write down a square root of the negative of the Laplacian (which is positive definite).

For 2), this choice affects the correct statement of the relationship between $\text{Cliff}(n)$-modules and real $K$-theory, but there is something very confusing going on here, namely that with either convention, $\text{Cliff}(n)$-modules are related to both $KO^n$ and $KO^{-n}$; see Andre Henriques' MO question on this subject.

For 3), whatever the answer to 1) or 2) I think everyone agrees that $\text{Cliff}(X)$ should be given fiberwise by $n$ anticommuting square roots of $-1$, where $n = \dim X$, so once you fix an answer to 1) that fixes the signs. The choice of sign affects the correct statement of the Thom isomorphism in K-theory.

Lawson-Michelson use the tangent bundle but Berline-Getzler-Vergne use the cotangent bundle. The tangent bundle seems natural if you want to think of Clifford multiplication as a deformation of a covariant derivative, and the cotangent bundle seems natural if you want to think of the Clifford bundle as a deformation of exterior forms. I'm not sure how important this choice is.

Anyway, I just want to know whether there are good justifications to sticking to one particular set of conventions so I can pick a consistent one for myself; reconciling the conventions of other authors is exhausting, especially because I haven't decided what conventions I want to use.

Answer 1

This is not really an answer, but rather a meta-answer as to why there exist many conventions in the first place.

The symmetric monoidal category $\mathit{sVect}$ of super-vector spaces has a non-trivial involution $J$. The symmetric monoidal functor $J:\mathit{sVect}\to \mathit{sVect}$ is the identity at the level of objects and at the level of morphisms. But the coherence $J(V \otimes W) \xrightarrow{\cong} J(V) \otimes J(W)$ is non-trivial. It is given by $-1$ on $V_{odd} \otimes W_{odd}$ and $+1$ on the rest.

The image of $\mathit{Cliff}(V,q)$ under $J$ is $\mathit{Cliff}(V,-q)$. So anything that you do with one convention can equally well be done with the other convention.

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Over the complex numbers, $J$ is equivalent to the identity functor. The symmetric monoidal natural transformation $J\Rightarrow Id$ that exhibits the equivalence acts as $i$ on the odd part and as $1$ on the even part of any super-vector space.

Over the reals, $J$ is not equivalent to the identity functor, as can be seen from the fact that $\mathit{Cliff}(\mathbb R,|\cdot|^2)\not\simeq\mathit{Cliff}(\mathbb R,-|\cdot|^2)$.

One last technical comment: Over $\mathbb C$, the action of $\mathbb Z/2$ on $\mathit{sVect}$ defined by $J$ is still non-trivial, despite the fact that $J$ is trivial. A trivialization of the action isn't just an equivalence $\alpha:J\cong Id$. For such an equivalence to trivialize the action, it would need to satisfy the further coherence $\alpha\circ \alpha = 1$, which isn't satisfied by any choice of $\alpha$. (To trivialize the action of a group $G$, one needs to trivialize the actions of each $g\in G$ in such a way that the trivializations of $g,h\in G$ compose to the trivialization of $gh$.)

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Now, as far as practical things are concerned, I would recommend minimizing the number of minus signs that you end up writing down.

Answer 2

To my mind, it has to do with p.d.e.'s and to your background, geometry or physics. Let me explain. $\newcommand{\pa}{\partial}$

Fix $\epsilon=\pm 1$. Define the Clifford algebra using the $\epsilon$ rule

$$ c_jc_k+c_kc_j=2\epsilon\delta_{jk} $$

and we form the Dirac operator

$$ D=\sum_k c_k\pa_{x^k} $$

Its square is

$$D^2=\epsilon\sum_k \pa_{x^k}^2. $$

Dirac wanted this square to be a Laplacian. If you are a geometer, then your Laplacian is

$$-\sum_k\pa_{x^k}^2, $$

so you would prefer $\epsilon =-1$. If you are a physicist, then your Laplacian is

$$\sum_k\pa_{x^k}^2 $$

and you would prefer $\epsilon =1$.