In order to get the bases of Clifford algebra, I want to understand the proof of the following proposition.
Proposition: Let $Cl(V)$ be the Clifford algebra of a $\mathbb{R}$-vector space $V$ with inner product $\langle\cdot, \cdot\rangle$. Let $V_1$ and $V_2$ be $\mathbb{R}$-vector spaces. Then, we get an isomorphism
$$Cl(V_1\oplus V_2)\cong Cl(V_1)\widehat{\otimes}Cl(V_2).$$
Here $\widehat{\otimes}$ is the tensor product with respect to the $\mathbb{Z}_2$-graded algebra.
Proof: We define
$$f:V_1\oplus V_2\to Cl(V_1)\widehat{\otimes}Cl(V_2),\ (v_1, v_2)\mapsto v_1\otimes 1 + 1\otimes v_2$$
which is a linear map between $\mathbb{R}$-vector spaces.
\begin{align*}
f(v_1,v_2)\cdot f(v_1,v_2) &=(v_1\otimes 1 + 1\otimes v_2)\widehat{\otimes}(v_1\otimes 1 + 1\otimes v_2)\\
&=(-1)^{deg(1)\cdot deg(v_2)}v_1^2\otimes 1 + (-1)^{deg(1)\cdot deg(1)}v_1\otimes v_2\\
&\quad\, +(-1)^{deg(v_2)\cdot deg(v_1)}v_1\otimes v_2 + (-1)^{deg(v_2)\cdot deg(1)}1\otimes v_2^2\\
&=-v_1^2\otimes 1 + v_1\otimes v_2 – v_1\otimes v_2 – 1\otimes v_2^2\\
&=-||(v_1, v_2)||^2\cdot (1\otimes 1)
\end{align*}
From this calculation, we induce a $\mathbb{R}$-linear map
$$F : Cl(V_1\oplus V_2)\to Cl(V_1)\widehat{\otimes}Cl(V_2)\ \text{ s.t. } F|_{V_1\oplus V_2} = f.$$
Finally, all we have to do is prove that $F$ is a bijection. However, I can not prove that $F$ is an injection. Please teach me.
Best Answer
To see that $F$ is injective, let $\{e_1, \dots, e_m\}$ and $\{e_{m+1}, \dots, e_{m+n}\}$ be bases of $V_1$ and $V_2$, then $\{e_J \mid J\subseteq \{1, \dots, m\}\}$ and $\{e_K \mid K \subseteq\{m+1, \dots, m+n\}\}$ are bases for $Cl(V_1)$ and $Cl(V_2)$ respectively, where $e_{\{a_1,\dots,a_l\}} = e_{a_1}\dots e_{a_l}$ if $a_1 < \dots < a_l$. These give rise to a basis of $Cl(V_1)\widehat{\otimes}Cl(V_2)$, namely $\{e_J\otimes e_K \mid J \subseteq \{1, \dots, m\}, K \subseteq \{m+1, \dots, m+n\}\}$. Likewise, $\{e_1, \dots, e_{m+n}\}$ is a basis for $V_1\oplus V_2$, so $\{e_I \mid I \subseteq \{1, \dots, m+n\}\}$ is a basis for $Cl(V_1\oplus V_2)$.
For a fixed $I = \{i_1, \dots, i_s\} \subseteq \{i_1, \dots, i_{m+n}\}$, suppose $i_r \leq m < i_{r+1}$, then
\begin{align*} F(e_I) &= F(e_{i_1}\dots e_{i_s})\\ &= F(e_{i_1})\dots F(e_{i_s})\\ &= f(e_{i_1})\dots f(e_{i_s})\\ &= (e_{i_1}\otimes 1)\dots(e_{i_r}\otimes 1)(1\otimes e_{i_{r+1}})\dots(1\otimes e_{i_s})\\ &= (e_{i_1}\dots e_{i_r})\otimes(e_{i_{r+1}}\dots e_{i_s})\\ &= e_J\otimes e_K \end{align*}
where $J = \{i_1,\dots, i_r\}$ and $K = I\setminus J = \{i_{r+1}, \dots, i_s\}$. Note that if $I'$ is another subset of $\{1, \dots, m+n\}$, then $F(e_{I'}) = e_{J'}\otimes e_{K'}$ with either $J' \neq J$, $K' \neq K$, or both. Therefore $\{F(e_I) \mid I \subseteq \{1,\dots, m+n\}\}$ is a linearly independent set, so $F$ is injective.
For completeness, $F$ is also surjective. To see this, note that $F$ is an algebra homomorphism, so the image of $F$ is a subalgebra of $Cl(V_1)\widehat{\otimes} Cl(V_2)$. Restricting $f$ to $V_1$, we see that there is unique map $Cl(V_1) \to Cl(V_1)\widehat{\otimes}\{1\}$ which extends $v \mapsto v\otimes 1$. By uniqueness, it must be the map given by $c \mapsto c\otimes 1$, so the image of $F$ contains $Cl(V_1)\widehat{\otimes}\{1\}$; likewise, it contains $\{1\}\widehat{\otimes}Cl(V_2)$. As the image of $F$ is a subalgebra of $Cl(V_1)\widehat{\otimes} Cl(V_2)$, it follows that $F$ is surjective.