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)?

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.

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)?

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.

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)?