Action of $\text{SL}_2\mathbb{C}$ on $\mathbb{C}^3$ induces a 2:1 covering $\text{SL}_2\mathbb{C}\to \text{SO}_3\mathbb{C}$

Question

Exercise 7.17 in Fulton's Representation Theory reads,

Identify $\mathbb{C}^3$ with the space of traceless matrices in $M_2\mathbb{C}$ so that $g\in \text{SL}_2\mathbb{C}$ acts by $$A\mapsto gAg^{-1}$$ Show that this gives a 2:1 covering $$\text{SL}_2\mathbb{C}\to \text{SO}_3\mathbb{C}$$ which realizes the universal covering of $\text{SO}_3\mathbb{C}$.

It is not obvious to me what the identification should be, or how the 2:1 covering follows. Could anybody explain this problem?

Answer 1

$ \newcommand{\so}{\mathfrak{sl}} \newcommand{\gl}{\mathfrak{gl}} \newcommand{\bC}{\mathbb{C}} \DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\SL}{SL} \DeclareMathOperator{\SO}{SO} \DeclareMathOperator{\tr}{tr} \newcommand{\i}[1]{{#1}^{-1}} \newcommand{\c}[2]{{#2}{#1}\i{{#2}}} $Let $\so(2,\bC)$ denote the complex vector space of $2 \times 2$ complex traceless matrices.

1. Since similar matrices have the same trace, $$ \forall X \in \so(2,\bC), \; \forall A \in \GL(2,\bC), \; \c{X}{A} \in \so(2,\bC). $$ Hence, $(A,X) \mapsto \c{X}{A}$ defines a representation of $\GL(2,\bC)$ on the vector space $\so(2,\bC)$, which restricts to a representation $\pi_0 : \SL(2,\bC) \to \GL(\so(2,\bC))$ of $\SL(2,\bC)$ on $\so(2,\bC)$.

2. You should check that the vector space $\so(2,\bC)$ admits the basis of Pauli matrices $\{\sigma_1,\sigma_2,\sigma_2\}$, where $$ \sigma_1 = \begin{pmatrix}0&1\\1&0\end{pmatrix}, \quad \sigma_2 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad \sigma_3 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}. $$ Hence, one has an isomorphism $\gamma : \bC^3 \to \so(2,\bC)$ defined by $\gamma(x) := x_1 \sigma_1 + x_2 \sigma_2 + x_3 \sigma_3$, which in turn induces a representation $\pi : \SL(2,\bC) \to \GL(\bC^3) \cong \GL(3,\bC)$ of $\SL(2,\bC)$ on $\bC^3$ by $$ \forall A \in \SL(2,\bC), \; \pi(A) := \c{\pi_0(A)}{\gamma}. $$

Given all this, you must show that:

1. $\pi(\SL(2,\bC)) = \SO(3,\bC)$.

2. For any $T \in \SO(3,\bC)$, $\i{\pi}(T)$ has exactly two elements.

It is worth noting, in terms of wider context, that this construction is intimately linked to the intimately interrelated theories of Clifford algebras and of spin and spin$^\bC$ groups. In fact, the restriction of $\pi : \SL(2,\bC) \to \GL(3,\bC)$ to a representation of $\operatorname{SU}(2)$ on $\bC^3$ yields the accidental isomorphism $\operatorname{Spin}(3) \cong \operatorname{SU}(2)$, for, as you can check $\pi(\operatorname{SU}(2)) = \SO(3,\mathbb{R})$. Moreover, the isomorphism $\gamma : \bC^3 \to \so(2,\bC)$ yields the identification of $\bC^2$ as the unique (up to unitary equivalence) irreducible $\operatorname{Spin}(3)$-module.

Answer 2

$\DeclareMathOperator{\tr}{tr}$The map $(A,B) \mapsto \tr(AB)$ defines a symmetric, nondegenerate bilinear form on the space of traceless matrices in $M_2\mathbb{C}$. Clearly the action of $g$ preserves this form; hence, we get a map $SL_2\mathbb{C} \to SO_3\mathbb{C}$. To see that the map is 2:1, check that the kernel of this map is $\{1, -1\}$. Surjectivity is more difficult to check.