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)?



Representations of Pin vs. Representations of Clifford
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 $A$."


