Complexification of Clifford Algebra

Question

I'm studying the book "Spin Geometry" by Lawson and Michelson.
Let $Cl_{r,s}$ be the Clifford algebra of $\mathbb{R}^{r+s}$ with the quadratic form $q(x):=x_1^2+…+x_r^2-x_{r+1}^2-…-x_{r+s}^2$.
They define the complexification of $Cl_{r,s}$ as $Cl_{r,s}\otimes_{\mathbb{R}}\mathbb{C}$, and then they immediately say (with no proof) that it is obvious that $$Cl_{r,s}\otimes_{\mathbb{R}}\mathbb{C}\cong Cl(\mathbb{C}^{r+s},q\otimes \mathbb{C})$$
What do they mean by "$q\otimes \mathbb{C}$"? What does this quadratic form do and why does this isomorphism hold?

Answer 1

$ \newcommand\K{\mathbb K} \newcommand\Cl{\mathrm{Cl}} \newcommand\tensor\otimes \newcommand\C{\mathbb C} \newcommand\R{\mathbb R} $

We utilize the fact that $\C^n \cong \R^n\tensor_\R\C$. Define $q\tensor \C := q_\C : \C^n \to \C$ as the unique quadratic form such that $$ q_\C(v\tensor x) := q(v)x^2. $$ It follows easily that $$ q_\C(v\tensor 1 + v'\tensor i) = q(v) - q(v') + 2iB(v, v') $$ where $B(v, v') = \tfrac12(q(v + v') - q(v) - q(v'))$ is the bilinear form associated to $q$.

To show that $$ \Cl(\C^n, q_\C) \cong C := \Cl(\R^n, q)\tensor_\R\C \tag{$*$} $$ as complex algebras, it suffices to show that the RHS satisfies the universal property of complex Clifford algebras. We consider $\C^n \cong \R^n\tensor_\R\C$, and we let the inclusion $\C^n \to C$ be given by $v\tensor x \mapsto v\tensor x$ for $v \in \R^n$ and $x \in \C$ (so we are identifying $\R^n$ as a subset of $\Cl(\R^n, q)$. Then for any complex algebra $A$, let $f : \C^n \to A$ be a homomorphism such that $f(v\tensor x)^2 = q_\C(v\tensor x) = q(v)x^2$. Since $$ \Cl(\R^n, q)\tensor_\R\R \cong \Cl(\R^n, q) $$ as real algebras, the restriction of $f$ to $\R^n \to A$ as a real homomorphism and $A$ gives a unique real homomorphism $\phi : \Cl(\R^n, q) \to A$ such that $\phi(v) = f(v\tensor1)$ by the universal property of $\Cl(\R^n, q)$. Now define $\phi_\C : C \to A$ by $$ \phi_\C(X\tensor x) = \phi(X)x,\quad X \in \Cl(\R^n, q). $$ This is clearly a complex homomorphism. It immediately follows for $v \in \R^n$ that $$ \phi_\C(v\tensor x) = \phi(v)x = f(v\tensor1)x = f(v\tensor x). $$ $\phi_\C$ is automatically unique amongst maps $\phi' : C \to A$ such that $\phi'(v\tensor x) = f(v\tensor x)$ by virtue of being a homomorphism. This completes the proof of ($*$).