Equivalent definition of Spin group in terms of automorphisms

Question

Let $\mathrm{Cl}(\mathbb{R}^n)$ denote the (real) Clifford algebra on $\mathbb{R}^n$ with respect to the Euclidean inner product. Let $\mathrm{Cl}^0({\mathbb{R}^n})$ denote the even part of $\mathrm{Cl}(\mathbb{R}^n)$, i.e. the span of elements of the form $e_{i_1}\dots e_{i_{2k}}$ for some $k$, where $e_1,\dots,e_n$ is the standard basis of $\mathbb{R}^n$.

The group $\mathrm{Pin}(n)$ is defined to be the multiplicative subgroup of $\mathrm{Cl}(\mathbb{R}^n)^\times$ generated by the unit vectors of $\mathbb{R}^n$, i.e. $\mathrm{Pin}(n)$ consists of products of unit vectors.

One then defines $\mathrm{Spin}(n)$ as:

$$\mathrm{Spin}(n)=\mathrm{Pin}(n)\cap\mathrm{Cl}^0({\mathbb{R}^n})$$

Now $\mathrm{Spin}(n)$ has the property that it acts as automorphisms on $\mathrm{Cl}(\mathbb{R}^n)$ via conjugation, in such a way that it acts on the the copy of $\mathbb{R}^n\subseteq\mathrm{Cl}(\mathbb{R}^n)$ via $\mathrm{SO}(n)$.

Question: Can one define $\mathrm{Spin}(n)$ directly using this property (possibly with additional constraints)?

Answer 1

$ \newcommand\R{\mathbb R} \newcommand\Cl{\mathrm{Cl}} \newcommand\tr{\mathrm{tr}} \newcommand\form[1]{\langle#1\rangle} \newcommand\Orthog{\mathrm O} \newcommand\Pin{\mathrm{Pin}} \newcommand\Spin{\mathrm{Spin}} $Yes.

Define the Lipschitz group $$ \Gamma(n) = \{x \in \Cl(\R^n)^\times \;:\; \forall v\in\R^n.\:\hat xvx^{-1} \in \R^n\} $$ where $\hat x$ is the main involution of $x$, i.e. the unique extension of $v \mapsto -v$ in $\R^n$ to an automorphism of $\Cl(\R^n)$. Then $$ \Pin(n) = \{x \in \Gamma(n) \;:\; \tilde xx = 1\},\quad \Spin(n) = \Pin(n)\cap\Cl^0(\R^n) $$ where $\tilde x$ is the main anti-involution, i.e. the unique extension of the identity on $\R^n$ to an anti-automorphism of $\Cl(\R^n)$.

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We see immediately that for any $v \in \R^n$ and $x \in \Gamma(n)$ $$ (\hat xvx^{-1})^2 = \widehat{(x\hat v\hat x^{-1})}\hat xvx^{-1} = xv^2x^{-1} = v^2 $$ because $x\hat v\hat x^{-1} = -xv\hat x^{-1}$ is a vector. So the action of $\Gamma(n)$ on $\R^n$ belongs to $\Orthog(n)$. Because $\hat vwv^{-1} = -vwv^{-1}$ is easily seen to be a reflection along $v$, by the Cartan-Dieudonne Theorem we have both that $\Gamma(n) \to \Orthog(n)$ is surjective and that every element of $\Gamma(n)$ is a product of at most $n$ vectors.

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This story is fairly easy to generalize to any vector space over any field with characteristic $\ne2$ with any quadratic form.