Representations of Pin vs. Representations of Clifford

Question

This may be total nonsense. But I need to know the answer quickly and I am too tired to think about it thoroughly. Let $k$ be a positive integer. Roe's "Elliptic Operators" claims that there is a 1-to-1 correspondence between:

• representations of the Clifford algebra $\operatorname{Cl}\mathbb R^k$ of the vector space $\mathbb R^k$ with the standard inner product;

• representations of the Pin group of this vector space (i. e., of the subgroup of the multiplicative group of $\operatorname{Cl}\mathbb R^k$ generated by vectors from $\mathbb R^k$) on which the element $-1$ of the Pin group acts as $-\operatorname{id}$;

• representations of the subgroup $\left\lbrace \pm e_1^{i_1}e_2^{i_2}...e_k^{i_k} \mid 0\leq i_1,i_2,...,i_k\leq 1 \right\rbrace$ of the Pin group (where $\left(e_1,e_2,...,e_n\right)$ is the standard orthogonal basis of $\mathbb R^k$) on which the group element $-1$ acts as $-\operatorname{id}$.

I do see how representations restrict from the above to the below, and also how there is a 1-to-1 correspondence between the first and the third kind of representations. But is it really that obvious that there are no "strange" representations of the second kind? I mean, why is a representation of the Pin group uniquely given by how it behaves on $-1$, $e_1$, $e_2$, ..., $e_k$ ?

Any help welcome, I'd already be glad to know whether it's really that obvious or not.

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EDIT: This seems to have caused some confusion. Here is the core of the question:

Assume that we have a representation $\rho$ of the Pin group $\operatorname{Pin}\mathbb R^k$ such that $\rho\left(-1\right)=-\operatorname{id}$. This, in particular, means an action of each unit vector. By linearity, we can extend this to an action of every vector. Is this always a representation (i.e., does the sum of two vectors always act as the sum of their respective actions)?

Answer 1

I'm not sure whether a representation of an algebra $A$ means a representation of the unit group of $A$, or an $A$-module. With the second interpretation, the statement is false.

Let's take $k=2$ and use the negative definite inner product. (This example will occur inside any larger example.) So the Clifford algebra is generated by $e_1$ and $e_2$, subject to $e_1^2=e_2^2=1$ and $e_1 e_2 = - e_2 e_1$. Let $S$ be the $2$-dimensional representation $$\rho_S(e_1) = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \quad \rho_S(e_2) = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.$$ Let $V=S^{\otimes 3}$. I claim that $V$ is a $\operatorname{Pin}$ representation where $-1$ acts by $- \operatorname{Id}$, but $V$ is not a module for the Clifford algebra.

For all $\theta$, the vector $v(\theta) := \cos \theta e_1 + \sin \theta e_2$ is in the Pin group. Clearly, $$\rho_S( v(\theta)) = \begin{pmatrix} \cos \theta & \sin \theta \\ \sin \theta & - \cos \theta \end{pmatrix}.$$ Then $\rho_V(v(\theta))$ is an $8 \times 8$ matrix I don't care to write down, whose entries are degree $3$ polynomials in $\sin \theta$ and $\cos \theta$. The point is, $$ \rho_V( v(\theta) ) \neq \cos \theta \rho_V(e_1) + \sin \theta \rho_V(e_2).$$ So $V$ is not an $A$-module. It is easy to build similar examples for the other signatures.

I'm not sure what happens if we read "representation of $A$" as "representation of the unit group of $A$."

Answer 2

Thanks to everyone who posted here. It is not "obvious" to me what I was thinking of here, and I'm embarrassed that this argument has stood unchanged in the book since the first edition in 1988 or so. I appreciate your pointing the issue out. There aren't any plans currently for a new edition of the book - I just dragged the old TeX files out and found that I can no longer compile them - but if one does come to be I will ensure that appropriate corrections are incorporated.

Answer 3

David Speyer gave a very nice counterexample, so I'd like to follow up with some holistic reasons for why we should not expect such a bijection:

1.Clifford algebras are basically matrix algebras with some extra noise (see the wikipedia page), and matrix algebras have very poor representation theory. In particular, all (finite dimensional) representations of a Clifford algebra are direct sums drawn from a finite set of irreducibles, while you can take any representation of Pin where -1 acts as -Id, and take a tensor product with a representation of Pin where -1 acts as Id (i.e., any representation of the orthogonal group) to get something new. David used the particular example of the tensor square of a spinor rep. These group representations are therefore parametrized by something about as big as the monoid semiring $\mathbb{N}[\Lambda^+]$ on the dominant integral weights.

2.If you want a representation of Pin to come from a Clifford algebra, you need the restriction to a maximal torus to have a very specific set of characters. Otherwise, linearity gets violated. You can in fact easily classify such characters, since Clifford algebras have very few irreducible representations. The characters span a finite dimensional vector space, unlike the characters of suitable representations of Pin.

Answer 4

For the equivalence between (3) and (2), let $(\rho,V)$ denote a representation of type (3). Therefore, we can define the action of the generator $a = \sum \alpha_i e_i \in \mathbb{R}^n$ of $Pin$ as follows: $\tilde{\rho}(a) = \sum \alpha_i \rho(e_i)$. To do this, we used that $\rho(-1) = -id$, because then there is no need to insert $\rho(-1)$ if $\alpha_i < 0$. Checking that these satisfy the relations of the generators of $Pin$ is then not hard (note that the standard inner product on $\mathbb{R}^n$ makes $\rho(e_i)$ and $\rho(e_j)$ on $V$ commute and use the relations in the group in (3)). It is clear that restriction $\tilde{\rho}$ to the group in (3) gives back $\rho$ again.