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.

  • 2
    $\begingroup$ Have you asked Borcherds? $\endgroup$
    – KConrad
    Oct 29, 2014 at 0:24
  • $\begingroup$ I second @KConrad, since you can walk down the hall and knock on his door --- my memory is that he is often in, and usually happy to talk to graduate students. And he certainly is sensitive to other conventions issues related to Clifford algebras in his Lie theory classes. If you get an answer, do be sure to post it here, of course. $\endgroup$ Oct 29, 2014 at 0:47
  • 1
    $\begingroup$ It is almost totally irrelevant whether you take tangent or cotangent bundles; all the theory is for Riemann manifolds, where both are naturally isomorphic. $\endgroup$ Oct 29, 2014 at 20:39
  • 1
    $\begingroup$ More interesting: on Clifford algebras there are two natural anti-involutions $\iota^{\pm}$, related via negation on the odd part, giving Clifford norms $u\mapsto u\iota^{\pm}(u)=\iota^{\pm}(u)u$ on the group GPin of units whose conjugation preserves $V$. As algebraic groups, their kernels Pin$^{\pm}$ are central extensions of O($q$)=O($-q$) by $\mu_2$ for which the fiber over $r_v$ for non-isotropic $v$ consists of points with respective orders 4 and 2, so these are non-isomorphic as central extensions. Atiyah-Bott-Shapiro use $\iota^{-}$, Chevalley & Serre use $\iota^+$. Et tu, Qiaochu? $\endgroup$
    – user27920
    Oct 30, 2014 at 6:11
  • $\begingroup$ @Johannes: I had in mind the possibility of wanting some kind of functoriality with respect to morphisms. For functoriality it seems to matter which one you pick. $\endgroup$ Oct 30, 2014 at 6:44

2 Answers 2


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.

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 $gh$.)

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

  • 11
    $\begingroup$ Even the last technical comment by itself deserves a +1 - this subtlety is surprising (to me). $\endgroup$ Oct 30, 2014 at 0:26
  • 6
    $\begingroup$ @Peter: these kinds of higher coherence issues can usually be interpreted topologically by passing through the homotopy hypothesis. In the topological setting, André is pointing out that for a map $f : B \mathbb{Z}_2 \to X$ to be nullhomotopic it does not suffice for $f$ to induce the zero map on $\pi_1$ (since e.g. $B \mathbb{Z}_2$ has interesting higher cohomology and $X$ could be an Eilenberg-MacLane space). The existence of $\alpha$ is a statement about the $1$-skeleton of $B \mathbb{Z}_2$ and the further coherence condition is a statement about the $2$-skeleton. $\endgroup$ Oct 30, 2014 at 6:49
  • $\begingroup$ That also is very enlightening, thanks! $\endgroup$ Oct 30, 2014 at 12:03
  • 9
    $\begingroup$ @PeterSamuelson, just don't get mixed up and assign the last technical comment a -1. $\endgroup$
    – LSpice
    Jun 2, 2017 at 17:28

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 $\epsilon =1$.

  • 4
    $\begingroup$ I always take $\epsilon =-1$; because this turns the Dirac operator into a formally self-adjoint one and the Laplacian into a nonnegative operator. Everything else becomes extremely unnatural when doing analysis of elliptic operators. $\endgroup$ Oct 29, 2014 at 20:23
  • 16
    $\begingroup$ Then you are a geometer. $\endgroup$ Oct 29, 2014 at 20:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.