## Representations of Pin vs. Representations of Clifford

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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 $\mathrm{Cl}\mathbb R^k$ 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 $\mathrm{Cl}\mathbb R^k$ generated by vectors from $\mathbb R^k$) on which the element $-1$ of the Pin group acts as $-\mathrm{id}$;

• 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 $-\mathrm{id}$.

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.

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 $\mathrm{Pin}\mathbb R^k$ such that $\rho\left(-1\right)=-\mathrm{id}$. 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)?

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For 3 -> 1, extend the induced map R^k --> End(V) using the universal property of Cl(R^k). – Michael May 10 2010 at 23:55
If I understand correctly, the question basically boils down to whether every rep of Pin is the restriction of a rep of Cl? – Eric O. Korman May 11 2010 at 1:21
Every representation of Pin which has -1 acting as -id. (The others are clearly taboo.) – darij grinberg May 11 2010 at 6:44
I remember being really confused by this when I read that stuff awhile back, and I wound up turning to Spin Geometry by Lawson and Michelsohn in order to understand the construction of the spinor bundle.  I guess I'm relieved I didn't dwell on it too much!  Still, the only part of this discussion that is needed for the rest of the book is the existence and uniqueness of the spin representation in even dimensions, and uniqueness even survives the mistake.  I think I'll discuss this with Prof. Roe when I see him again next week. – Paul Siegel May 11 2010 at 23:52
Actually, Roe uses this to characterize the irreducible representations of the Clifford algebra. In my eyes this is quite an overkill, since it is easy to see that the representation map from the Clifford algebra to the endomorphism ring of the half-spin representation (or, in the odd case, to the sum of the endomorphism rings of the two half-spin representations) is bijective. – darij grinberg May 12 2010 at 7:13

## 5 Answers

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 $\mathrm{Pin}$ representation where $-1$ acts by $- \mathrm{Id}$, 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 $A$."

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1
Representations of the unit group $\Gamma$ are representations of Pin with some rubbish attached. In general, you have a central extension $0 \to K^\times \to \Gamma \to O(V) \to 1$, and Pin is the subgroup of $\Gamma$ with spinor norm 1. In the case above, a representation of Pin becomes a representation of $\Gamma$ once you choose a commuting action of the multiplicative group of positive reals. – S. Carnahan♦ May 11 2010 at 13:23
Thanks, David! I assumed that "representation of an algebra" is generally understood as "representation of the algebra" and not "representation of the unit group", because otherwise it would be a bit pointless to talk about representations of path algebras of acyclic quivers, and similar nilpotent stuff... I'm still wondering: does this work if the Clifford algebra is over a real vector space with a real, positive-definite inner product? – darij grinberg May 11 2010 at 14:24

1
Correction: My previous comment holds when $\Gamma$ is the Clifford group, which is not in general the group of units. Invertible elements need to satisfy a twisted conjugation condition to lie in the Clifford group, and this condition is typically nontrivial. Since the Clifford algebra is basically a matrix algebra, its group of units looks a lot like a general linear group (which has a lot more representations than a matrix algebra). – S. Carnahan♦ May 11 2010 at 17:12

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I think this argument works in the case that the inner product is definite (which is the case you are considering anyways). In this case, the Pin group is generated by the set of all normalized non-zero vectors. Therefore, by the universal property of Cl, a rep of Pin gives a unique rep of Cl and restricting this rep gives you back the rep of Pin you started with since they agree on unit vectors. Thus every rep of Pin is the restriction of a rep of Cl. If you know where all the $e_i$'s get mapped, you then get a unique rep of Cl which gives a unique rep of Pin.

If the inner product is not definite you will still get a rep of Cl but I can't see how you know that it will agree with the original rep of Pin.

EDIT: I have assumed that the map we get from the set of all unit vectors (by restricting the map from Pin), is the restriction of a linear map from $R^n$. This doesn't seem to be obviously the case, but is equivalent to the statement that every rep of Pin is the restriction of a rep of Cl.

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Sorry, but could you be more precise about "by the universal property of Cl"? – darij grinberg May 11 2010 at 6:43
en.wikipedia.org/wiki/… – Orbicular May 11 2010 at 6:57
So...? Why do we have a linear map from V to the endomorphism space to begin with? How do we know that the normalization of $v+w$ acts as a scalar multiple of the sum of the actions of $v$ and $w$? – darij grinberg May 11 2010 at 7:04
This has nothing to do with the signature of the inner product. – José Figueroa-O'Farrill May 11 2010 at 9:18

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.

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My difficulty is proving that (2) -> (3) -> (2) gives the same representation that we started with - not (3) -> (2) -> (3). – darij grinberg May 11 2010 at 10:53

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.

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Your reason 1 was exactly the reason why I had doubts about the statement. Otherwise I would have accepted Roe's "The proof is obvious"... – darij grinberg May 11 2010 at 15:43
It would have been nice if you had included some indication to that effect in the statement of the question.  It looks like many people were thrown off by the expectation that the statement was true. – S. Carnahan♦ May 11 2010 at 16:27
You are right: Reading my first post again, I don't see an indication that I disbelieved the assertion. I didn't expect I could ever fail to be negative enough... – darij grinberg May 11 2010 at 17:11

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.

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I thought that one of the biggest advantages of TeX over proprietary typesetting systems was that it would be forever compilable (i.e. backward compatible). Is it not true??? – Victor Protsak Jul 9 2010 at 22:44
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