Clifford algebra $Cl_1 \cong \mathbb{C}$

Question

I'm reading Spin Geometry by Lawson and Michelsohn. In general Clifford algebras are defined by $Cl(V,q)=\mathcal{T}(V)/\mathfrak{I}(V)$, where $\mathcal{T}(V)$ is the tensor algebra of $V$ and $\mathfrak{I}(V)$ the ideal generated by the elements $v\otimes v+q(V)1$ ($q$ is a quadratic form). For $V=\mathbb{R}^{r+s}$ it follows that $q(x)=x_1^2+...+x_r^2-x_{r+1}^2-...-x_{r+s}^2$ and we write $Cl_{r,s}$.

In the book was mentioned that $Cl_{1,0} \cong \mathbb{C}$ and $Cl_{0,1} \cong \mathbb{R}\oplus\mathbb{R}$. How do I see that these Clifford algebras have dimension two?

Thanks for your help.

Answer 1

With basis $x$ (for $\mathbb{R}$ over $\mathbb{R}$) the tensor algebra is the polynomial algebra $\bigoplus_{n\geq 0}\mathbb{R}^{\otimes n}=\mathbb{R}[x]$, but $x^2=-1$ or $x^2=1$ in the two cases you list, so we get $Cl_{1,0}=\mathbb{R}[x]/(x^2+1)=\mathbb{C}$ or $Cl_{0,1}=\mathbb{R}[x]/(x^2-1)=\mathbb{R}^2$

Answer 2

When $\dim V=1$, we have $V=\mathrm{span}\{v\}$ for any nonzero $v$, and then the tensor algebra $T(V)$ is just the polynomial ring $\mathbb{R}[v]$. When quotienting, we get

$$ \frac{\mathbb{R}[v]}{(v^2+1)}\cong\mathbb{C}, \quad \frac{\mathbb{R}[v]}{(v^2-1)}\cong\mathbb{R}\oplus\mathbb{R}. $$

The first follows from the First Isomorphism Theorem applied to the $\mathbb{R}$-algebra homomorphism $\mathbb{R}[v]\to\mathbb{C}$ given by $f(v)\mapsto f(i)$. In other words, for any polynomial in $v$ we can (in the quotient) simplify any power of $v$ to be either a scalar or scalar multiple of $v$, so all elements are of the form $a+bv$, and $v^2=-1$.

The second follows from the Chinese Remainder Theorem, in other words

$$ \frac{\mathbb{R}[v]}{(v^2-1)}\cong\frac{\mathbb{R}[v]}{(v-1)}\oplus\frac{\mathbb{R}[v]}{(v+1)}\cong\mathbb{R}\oplus\mathbb{R}. $$

(The second isomorphism comes from another application of the 1st Iso Thm). Just as before, we can simplify any element of the quotient to be of the form $a+bv$, but now $v^2=+1$. In this case we can use a change of basis and rewrite this in terms of the orthogonal idempotents

$$e_1=\frac{1}{2}(1+v), \quad e_2=\frac{1}{2}(1-v).$$ That is, we can write all elements as $xe_1+ye_2$, and the relations $e_1e_2=e_2e_1=0$ and $e_1^2=e_1$, $e_2^2=e_2$ tell us that the multiplication rule for $xe_1+ye_2$ is the same as that for $(x,y)$ in $\mathbb{R}\oplus\mathbb{R}$, i.e. $e_1$ and $e_2$ correspond to the orthogonal idempotents $(1,0)$ and $(0,1)$ under an isomorphism.

(Also see: Peirce decomposition.)

In general, if $V$ is a vector space with basis $\{v_1,\cdots,v_n\}$, any element in the Clifford algebra $C\ell(V)$ (WRT some form) is a linear combination of 'noncommutative monomials' in $v_1,\cdots,v_n$, and any such monomial we can simplify to be a scalar multiple of some $v_{i_1}\cdots v_{i_k}$ for some indices with $1\le i_1<\cdots<e_k\le n$. Thus, $\dim C\ell(V)=2^n$; in fact there is a natural vector space isomorphism $C\ell(V)\cong\Lambda(V)$ where $\Lambda(V)$ is the exterior algebra (though, of course, this will not be an algebra isomorphism).