First let's recall how to construct the finite dimensional irreducible representations of the Lorentz group. Say $J_i$ are the three rotation generators and $K_i$ are the three boost generators.
\begin{align*}
L_x = &\begin{pmatrix} 0&0&0&0 \\ 0&0&0&0 \\ 0&0&0&-1 \\ 0&0&1&0 \end{pmatrix}&
L_y = &\begin{pmatrix} 0&0&0&0 \\ 0&0&0&1 \\ 0&0&0&0 \\ 0&-1&0&0 \end{pmatrix}&
L_z = &\begin{pmatrix} 0&0&0&0 \\ 0&0&-1&0 \\ 0&1&0&0 \\ 0&0&0&0 \end{pmatrix}\\
K_x = &\begin{pmatrix} 0&1&0&0 \\ 1&0&0&0 \\ 0&0&0&0 \\ 0&0&0&0 \end{pmatrix}&
K_y = &\begin{pmatrix} 0&0&1&0 \\ 0&0&0&0 \\ 1&0&0&0 \\ 0&0&0&0 \end{pmatrix}&
K_z = &\begin{pmatrix} 0&0&0&1 \\ 0&0&0&0 \\ 0&0&0&0 \\ 1&0&0&0 \end{pmatrix}\\
\end{align*}
They satisfy
$$
[J_i, J_j] = \varepsilon_{ijk} J_k \hspace{1 cm} [K_i, K_j] = -\varepsilon_{ijk} J_k \hspace{1 cm} [J_i, K_j] = \varepsilon_{ijk}K_k.
$$
(Note that I am using the skew-adjoint convention for Lie algebra elements where I did not multiply by $i$.)
We then define
$$
A_i = \frac{1}{2} (J_i - i K_i) \hspace{2 cm} B_i = \frac{1}{2}(J_i + i K_i)
$$
which satisfy the commutation relations
$$
[A_i, A_j] = \varepsilon_{ijk} A_k \hspace{2cm} [B_i, B_j] = \varepsilon_{ijk} B_k \hspace{2cm} [A_i, B_j] = 0.
$$
Here is how you construct the representation of the Lorentz group: first, choose two non-negative half integers $j_1$ and $j_2$. These correspond to two spin $j$ representations of $\mathfrak{su}(2)$, which I will label
$$
\pi'_{j}.
$$
Recall that that
$$
\mathfrak{su}(2) = \mathrm{span}_\mathbb{R} \{ -\tfrac{i}{2} \sigma_x, -\tfrac{i}{2} \sigma_y, -\tfrac{i}{2} \sigma_z \}
$$
where
$$
[-\tfrac{i}{2} \sigma_i, -\tfrac{i}{2} \sigma_j] = -\tfrac{i}{2}\varepsilon_{ijk} \sigma_k.
$$
For this question, we only need to know the spin $1/2$ representation of $\mathfrak{su}(2)$, which is given by
$$
\pi'_{\tfrac{1}{2}}( -\tfrac{i}{2}\sigma_i) = -\tfrac{i}{2} \sigma_i.
$$
So okay, how do we construct the $(j_1, j_2)$ representation of the Lorentz group? Any Lie algebra element $X \in \mathfrak{so}(1,3)$ can written as a linear combination of $A_i$ and $B_i$:
$$
X = \sum_{i = 1}^3 (\alpha_i A_i + \beta_i B_i).
$$
(Note that we are actually dealing with the complexified version of the Lie algebra $\mathfrak{so}(1,3)$ because our definitions of $A_i$ and $B_i$ have factors of $i$, so $\alpha, \beta \in \mathbb{C}$.)
$A_i$ and $B_i$ form their own independent $\mathfrak{su}(2)$ algebras.
The Lie algebra representation $\pi'_{(j_1, j_2)}$ is then given by
\begin{align*}
\pi'_{(j_1, j_2)}(X) &= \pi'_{(j_1, j_2)}(\alpha_i A_i + \beta_j B_i) \\
&\equiv \pi'_{j_1}(\alpha_i A_i) \otimes \big( \pi'_{j_2}(\beta_j B_i) \big)^*
\end{align*}
where the star denotes complex conjugation.
Sometimes people forget to mention that you have to include the complex conjugation, but it won't work otherwise!
If $j_1 = 1/2$ and $j_2 = 1/2$, we have
\begin{equation*}
\pi_{\frac{1}{2}}'(A_i) = -\frac{i}{2}\sigma_i \otimes I \hspace{2cm} \big(\pi_{\frac{1}{2}}' (B_i)\big)^* = \frac{i}{2} I \otimes\sigma_i^*.
\end{equation*}
We can explicitly write out these tensor products in terms of a $2 \times 2 = 4$ dimensional basis. (Here I am using the so-called "Kronecker Product" to do this. That just a fancy name for multiplying all the elements of two $2\times 2$ cell-wise to get a $4 \times 4$ matrix.)
\begin{align*}
\pi_{(\frac{1}{2},\frac{1}{2})}'(A_x) &= -\frac{i}{2}\begin{pmatrix} 0&0&1&0 \\ 0&0&0&1 \\ 1&0&0&0 \\ 0&1&0&0 \end{pmatrix} & \big(\pi_{(\frac{1}{2},\frac{1}{2})}'(B_x)\big)^* &= \frac{i}{2}\begin{pmatrix} 0&1&0&0 \\ 1&0&0&0 \\ 0&0&0&1 \\ 0&0&1&0 \end{pmatrix} \\
\pi_{(\frac{1}{2},\frac{1}{2})}'(A_y) &= \frac{1}{2}\begin{pmatrix} 0&0&-1&0 \\ 0&0&0&-1 \\ 1&0&0&0 \\ 0&1&0&0 \end{pmatrix} & \big(\pi_{(\frac{1}{2},\frac{1}{2})}'(B_y)\big)^*&= \frac{1}{2}\begin{pmatrix} 0&-1&0&0 \\ 1&0&0&0 \\ 0&0&0&-1 \\ 0&0&1&0 \end{pmatrix} \\
\pi_{(\frac{1}{2},\frac{1}{2})}'(A_z) &= -\frac{i}{2}\begin{pmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&-1&0 \\ 0&0&0&-1 \end{pmatrix} & \big(\pi_{(\frac{1}{2},\frac{1}{2})}'(B_z)\big)^* &= \frac{i}{2}\begin{pmatrix} 1&0&0&0 \\ 0&-1&0&0 \\ 0&0&1&0 \\ 0&0&0&-1 \end{pmatrix}
\end{align*}
We can then write out the matrices of the rotations and boosts $J_i$ and $K_i$ using
$$
J_i = A_i + B_i \hspace{2cm} K_i = i(A_i - B_i).
$$
\begin{align*}
\pi'_{(\frac{1}{2},\frac{1}{2})}(J_x) &= \frac{i}{2}\begin{pmatrix} 0&1&-1&0 \\ 1&0&0&-1 \\ -1&0&0&1 \\ 0&-1&1&0 \end{pmatrix} & \pi'_{(\frac{1}{2},\frac{1}{2})}(K_x) &= \frac{1}{2}\begin{pmatrix} 0&1&1&0 \\ 1&0&0&1 \\ 1&0&0&1 \\ 0&1&1&0 \end{pmatrix} \\
\pi'_{(\frac{1}{2},\frac{1}{2})}(J_y) &= \frac{1}{2}\begin{pmatrix} 0&-1&-1&0 \\ 1&0&0&-1 \\ 1&0&0&-1 \\ 0&1&1&0 \end{pmatrix} & \pi'_{(\frac{1}{2},\frac{1}{2})}(K_y) &= \frac{i}{2}\begin{pmatrix} 0&1&-1&0 \\ -1&0&0&-1 \\ 1&0&0&1 \\ 0&1&-1&0 \end{pmatrix} \\ \pi'_{(\frac{1}{2},\frac{1}{2})}(J_z) &= \begin{pmatrix} 0&0&0&0 \\ 0&-i&0&0 \\ 0&0&i&0 \\ 0&0&0&0 \end{pmatrix} & \pi'_{(\frac{1}{2},\frac{1}{2})}(K_z) &= \begin{pmatrix} 1&0&0&0 \\ 0&0&0&0 \\ 0&0&0&0 \\ 0&0&0&-1 \end{pmatrix} \\
\end{align*}
These are strange matrices, although we can make them look much more suggestive in another basis. Define the matrix
\begin{equation*}
U = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & -i & 0\\ 0 & 1 & i & 0 \\ 1 & 0 & 0 &-1 \end{pmatrix}.
\end{equation*}
Amazingly,
\begin{equation*}
U^{-1} \big( \pi'_{(\frac{1}{2},\frac{1}{2})}(L_i) \big) U = L_i \hspace{1cm}U^{-1} \big( \pi'_{(\frac{1}{2},\frac{1}{2})}(K_i) \big) U = K_i.
\end{equation*}
Therefore, the $(\tfrac{1}{2}, \tfrac{1}{2})$ representation is equivalent to the regular "vector" representation of $SO^+(1,3)$. However, these "vectors" live in $\mathbb{C}^4$, not $\mathbb{R}^4$, which people usually don't mention.
It's pretty annoying that P&S just give you
$$S^{\mu \nu} = \frac{i}{4} [\gamma^{\mu},\gamma^{\nu}]$$
from thin air, here is a way to derive it similar to Bjorken-Drell's derivation (who start from the Dirac equation) but from the Clifford algebra directly, assuming that products of the gamma matrices form a basis. Given a Clifford algebra of $\gamma^{\mu}$'s satisfying
\begin{align}
\{ \gamma^{\mu} , \gamma^{\mu} \} = 2 \eta^{\mu \nu} I
\end{align}
we note that for an invertible transformation $S$ we have
\begin{align}
2 \eta^{\mu \nu} I &= 2 \eta^{\mu \nu} S^{-1} S \\
&= S^{-1}(2 \eta^{\mu \nu}) S \\
&= S^{-1}\{ \gamma^{\mu} , \gamma^{\mu} \} S \\
&= \{ S^{-1} \gamma^{\mu} S, S^{-1}\gamma^{\mu} S \} \\
&= \{ \gamma'^{\mu} , \gamma'^{\mu} \}
\end{align}
showing us that the Clifford algebra of matrices
$$\gamma'^{\mu} = S^{-1} \gamma^{\mu} S$$
also satisfies the Clifford algebra, hence any set of matrices satisfying the Clifford algebra can be obtained from a given set $\gamma^{\mu}$ using a non-singular transformation $S$. Since the anti-commutation relations involve the metric $\eta_{\mu \nu}$, and we know the metric is left invariant under Lorentz transformations
$$\eta^{\mu \nu} = \Lambda^{\mu} \, _{\rho} \Lambda^{\nu} \, _{\sigma} \eta^{\rho \sigma} $$
this immediately implies
\begin{align}
2 \eta^{\mu \nu} I &= \Lambda^{\mu} \, _{\rho} \Lambda^{\nu} \, _{\sigma} 2 \eta^{\rho \sigma} I \\
&= \Lambda^{\mu} \, _{\rho} \Lambda^{\nu} \, _{\sigma} \{ \gamma^{\rho} , \gamma^{\sigma} \} \\
&= \{ \Lambda^{\mu} \, _{\rho} \gamma^{\rho} , \Lambda^{\nu} \, _{\sigma} \gamma^{\sigma} \} \\
&= \{ \gamma'^{\mu} , \gamma'^{\mu} \}
\end{align}
which shows that the Lorentz transformation of a gamma matrix also satisfies the Clifford algebra, and so is itself a gamma matrix, and hence can be expressed in terms of some non-singular transformation $S$
\begin{align}
\gamma'^{\mu} &= \Lambda^{\mu} \, _{\nu} \gamma^{\nu} \\
&= S^{-1} \gamma^{\mu} S
\end{align}
where $S$ is to be determined. Since the operators $S$ represent performing a Lorentz transformation on $\gamma^{\mu}$, and Lorentz transformations on fields expand as $I - \frac{i}{2}\omega_{\mu \nu} M^{\mu \nu}$, we expand $\Lambda$ and $S$ as
\begin{align}
\Lambda^{\mu} \, _{\nu} &= \delta ^{\mu} \, _{\nu} + \omega^{\mu} \, _{\nu} \\
S &= I - \frac{i}{2} \omega_{\mu \nu} \Sigma^{\mu \nu}
\end{align}
where $\Sigma^{\mu \nu}$ must be anti-symmetric and constructed from a basis of gamma matrices, hence from
\begin{align}
\gamma'^a &= \Lambda^a \, _{\mu} \gamma^{\mu} \\
&= (\delta^a \, _{\mu} + \omega^a \, _{\mu})\gamma^{\mu} \\
&= \gamma^a + \omega^a \, _{\mu} \gamma^{\mu} \\
&= \gamma^a + \omega_{b \mu} \eta^{a b} \gamma^{\mu} \\
&= \gamma^a + \omega_{b \mu} \eta^{a [b} \gamma^{\mu]} \\
&= \gamma^a + \frac{1}{2} \omega_{b \mu} (\eta^{a b} \gamma^{\mu} - \eta^{a \mu} \gamma^b) \\
&= \gamma^a + \frac{1}{2} \omega_{\nu} (\eta^{a \mu} \gamma^{\nu} - \eta^{a \nu} \gamma^{\mu}) \\
&= S^{-1} \gamma^a S \\
&= (I - \frac{i}{2} \omega_{\mu \nu} \Sigma^{\mu \nu}) \gamma^a (I + \frac{i}{2} \omega_{\mu \nu} \Sigma^{\mu \nu}) \\
&= \gamma^a - \frac{i}{2} \omega_{\mu \nu} [\gamma^a, \Sigma^{\mu \nu}]
\end{align}
we have the relation (which can be interpreted as saying that $\gamma^a$ transforms as a vector under spinor representations of Lorentz transformations, as e.g. in Tong's QFT notes)
\begin{align}
i (\eta^{a \mu} \gamma^{\nu} - \eta^{a \nu} \gamma^{\mu}) = [\gamma^a, \Sigma^{\mu \nu}]
\end{align}
and we know $\Sigma^{\mu \nu}$, since it is anti-symmetric, must involve a product's of $\gamma$ matrices (because of the 16-dimensional basis formed from Clifford algebra elements), only two by the left-hand side, and from
\begin{align}
\gamma^{\mu} \gamma^{\nu} &= - \gamma^{\nu} \gamma^{\mu} , \ \ \ \mu \neq \nu, \\
\gamma^{\mu} \gamma^{\mu} &= \gamma^{\nu} \gamma^{\mu} , \ \ \ \mu = \nu,
\end{align}
we expect that
\begin{align}
\Sigma^{\mu \nu} &= c [\gamma^{\mu},\gamma^{\nu}] \\
&= c (\gamma^{\mu} \gamma^{\nu} - \gamma^{\nu} \gamma^{\mu}) \\
&= 2 c ( \gamma^{\mu} \gamma^{\nu} - \eta^{\mu \nu})
\end{align}
for some $c$ which we constrain by the (vector) relation above
\begin{align}
i (\eta^{a \mu} \gamma^{\nu} - \eta^{a \nu} \gamma^{\mu}) &= [\gamma^a, \Sigma^{\mu \nu}] \\
&= c [\gamma^a, 2( \gamma^{\mu} \gamma^{\nu} - \eta^{\mu \nu})] \\
&= 2 c [\gamma^a, \gamma^{\mu} \gamma^{\nu}] \\
&= 2 c ( \gamma^{\mu} [\gamma^a,\gamma^{\nu}] + [\gamma^a, \gamma^{\mu}] \gamma^{\nu}) \\
&= 2 c [ \gamma^{\mu} 2( \gamma^a \gamma^{\nu} - \eta^{a \nu}) + 2( \gamma^a \gamma^{\mu} - \eta^{a \mu}) \gamma^{\nu}] \\
&= 4 c [ \gamma^{\mu} ( \gamma^a \gamma^{\nu} - \eta^{a \nu}) + ( \gamma^{\mu} \gamma^{a} + 2 \eta^{a \mu} - \eta^{a \mu}) \gamma^{\nu}] \\
&= 4 c (\eta^{a \mu} \gamma^{\nu} - \eta^{a \nu} \gamma^{\mu}).
\end{align}
This gives the result $c = i/4$. The generator of Lorentz transformations of gamma matrices is
\begin{align}
\Sigma^{\mu \nu} &= \dfrac{i}{4} [\gamma^{\mu},\gamma^{\nu}] \\
&= \dfrac{i}{2}(\gamma^{\mu} \gamma^{\nu} - \eta^{\mu \nu}) \ \ \text{i.e.} \\
S &= I - \frac{i}{2} \omega_{\mu \nu} (\frac{i}{4} [\gamma^{\mu},\gamma^{\nu}]) \\
&= I + \dfrac{1}{8} \omega_{\mu \nu} [\gamma^{\mu},\gamma^{\nu}].
\end{align}
Using the fact that the gamma matrices transform as a vector under the spinor representation of an infinitesimal Lorentz transformation,
\begin{align}
[\Sigma^{\mu \nu}, \gamma^{\rho}] = i (\gamma^{\mu} \eta^{\nu \rho} - \gamma^{\nu} \eta^{\mu \rho})
\end{align}
we can show the spinor representation of a Lorentz transformation satisfies the Lorentz algebra commutation relations, since for $\rho \neq \sigma$
\begin{align}
[\Sigma^{\mu \nu},\Sigma^{\rho \sigma}] &= \frac{i}{2}[\Sigma^{\mu \nu},\gamma^{\rho} \gamma^{\sigma}] \\
&= \frac{i}{2}([\Sigma^{\mu \nu},\gamma^{\rho} ] \gamma^{\sigma} + \gamma^{\rho} [\Sigma^{\mu \nu}, \gamma^{\sigma}]) \\
&= \frac{i}{2}\{ i (\gamma^{\mu} \eta^{\nu \rho} - \gamma^{\nu} \eta^{\mu \rho}) \gamma^{\sigma} + \gamma^{\rho} i (\gamma^{\mu} \eta^{\nu \sigma} - \gamma^{\nu} \eta^{\mu \sigma}) \} \\
&= - \frac{1}{2}\{ \gamma^{\mu} \eta^{\nu \rho} \gamma^{\sigma} - \gamma^{\nu} \eta^{\mu \rho} \gamma^{\sigma} + \gamma^{\rho} \gamma^{\mu} \eta^{\nu \sigma} - \gamma^{\rho} \gamma^{\nu} \eta^{\mu \sigma} \} \\
&= \frac{i}{2}\{ \eta^{\nu \rho} (2 \Sigma^{\mu \sigma} + \eta^{\mu \sigma}) - \eta^{\mu \rho} (2 \Sigma^{\nu \sigma} - \eta^{\nu \sigma}) + (2 \Sigma^{\rho \mu} - \eta^{\rho \mu}) \eta^{\nu \sigma} - (2 \Sigma^{\rho \nu}) - \eta^{\rho \nu}) \eta^{\mu \sigma} \} \\
&= i ( \eta^{\nu \rho} \Sigma^{\mu \sigma} - \eta^{\mu \rho} \Sigma^{\nu \sigma} + \Sigma^{\rho \mu} \eta^{\nu \sigma} - \Sigma^{\rho \nu} \eta^{\mu \sigma} ).
\end{align}
This method generalizes from $SO(3,1)$ to $SO(N)$, see e.g. Kaku QFT Sec. 2.6, and the underlying reason for doing any of this in the first place is that one seeks to find projective representations which arise due to the non-simple-connectedness of these orthogonal groups. Regarding your question about arbitrary metrics $g_{\mu \nu}$, this method applies to, and arises due to the non-simple-connectedness of, special orthogonal groups, you can't generalize to arbitrary metrics, this is a problem which can be circumvented in supergravity and superstring theory using veilbein's.
References:
- Bjorken, J.D. and Drell, S.D., 1964. Relativistic quantum mechanics; Ch. 2.
- Kaku, M., 1993. Quantum field theory: a modern introduction. Oxford Univ. Press; Sec. 2.6.
- Tong, Quantum Field Theory Notes http://www.damtp.cam.ac.uk/user/tong/qft.html.
- http://www.damtp.cam.ac.uk/user/examples/D18S.pdf
- Does $GL(N,\mathbb{R})$ own spinor representation? Which group is its covering group? (Kaku's QFT textbook)
Best Answer
Here is a mathematical derivation. We use the sign convention $(+,-,-,-)$ for the Minkowski metric $\eta_{\mu\nu}$.
I) First recall the fact that
This follows partly because:
There is a bijective isometry from the Minkowski space $(\mathbb{R}^{1,3},||\cdot||^2)$ to the space of $2\times2 $ Hermitian matrices $(u(2),\det(\cdot))$, $$\mathbb{R}^{1,3} ~\cong ~ u(2) ~:=~\{\sigma\in {\rm Mat}_{2\times 2}(\mathbb{C}) \mid \sigma^{\dagger}=\sigma \} ~=~ {\rm span}_{\mathbb{R}} \{\sigma_{\mu} \mid \mu=0,1,2,3\}, $$ $$\mathbb{R}^{1,3}~\ni~\tilde{x}~=~(x^0,x^1,x^2,x^3) \quad\mapsto \quad\sigma~=~x^{\mu}\sigma_{\mu}~\in~ u(2), $$ $$ ||\tilde{x}||^2 ~=~x^{\mu} \eta_{\mu\nu}x^{\nu} ~=~\det(\sigma), \qquad \sigma_{0}~:=~{\bf 1}_{2 \times 2}.\tag{1}$$
There is a group action $\rho: SL(2,\mathbb{C})\times u(2) \to u(2)$ given by $$g\quad \mapsto\quad\rho(g)\sigma~:= ~g\sigma g^{\dagger}, \qquad g\in SL(2,\mathbb{C}),\qquad\sigma\in u(2), \tag{2}$$ which is length preserving, i.e. $g$ is a pseudo-orthogonal (or Lorentz) transformation. In other words, there is a Lie group homomorphism
$$\rho: SL(2,\mathbb{C}) \quad\to\quad O(u(2),\mathbb{R})~\cong~ O(1,3;\mathbb{R}) .\tag{3}$$
Since $\rho$ is a continuous map and $SL(2,\mathbb{C})$ is a connected set, the image $\rho(SL(2,\mathbb{C}))$ must again be a connected set. In fact, one may show so there is a surjective Lie group homomorphism$^1$
$$\rho: SL(2,\mathbb{C}) \quad\to\quad SO^+(u(2),\mathbb{R})~\cong~ SO^+(1,3;\mathbb{R}) , $$ $$\rho(\pm {\bf 1}_{2 \times 2})~=~{\bf 1}_{u(2)}.\tag{4}$$
The Lie group $SL(2,\mathbb{C})=\pm e^{sl(2,\mathbb{C})}$ has Lie algebra $$ sl(2,\mathbb{C}) ~=~ \{\tau\in{\rm Mat}_{2\times 2}(\mathbb{C}) \mid {\rm tr}(\tau)~=~0 \} ~=~{\rm span}_{\mathbb{C}} \{\sigma_{i} \mid i=1,2,3\}.\tag{5}$$
The Lie group homomorphism $\rho: SL(2,\mathbb{C}) \to O(u(2),\mathbb{R})$ induces a Lie algebra homomorphism $$\rho: sl(2,\mathbb{C})\to o(u(2),\mathbb{R})\tag{6}$$ given by $$ \rho(\tau)\sigma ~=~ \tau \sigma +\sigma \tau^{\dagger}, \qquad \tau\in sl(2,\mathbb{C}),\qquad\sigma\in u(2), $$ $$ \rho(\tau) ~=~ L_{\tau} +R_{\tau^{\dagger}},\tag{7}$$ where we have defined left and right multiplication of $2\times 2$ matrices $$L_{\sigma}(\tau)~:=~\sigma \tau~=:~ R_{\tau}(\sigma), \qquad \sigma,\tau ~\in~ {\rm Mat}_{2\times 2}(\mathbb{C}).\tag{8}$$
II) Note that the Lorentz Lie algebra $so(1,3;\mathbb{R}) \cong sl(2,\mathbb{C})$ does not$^2$ contain two perpendicular copies of, say, the real Lie algebra $su(2)$ or $sl(2,\mathbb{R})$. For comparison and completeness, let us mention that for other signatures in $4$ dimensions, one has
$$SO(4;\mathbb{R})~\cong~[SU(2)\times SU(2)]/\mathbb{Z}_2, \qquad\text{(compact form)}\tag{9}$$
$$SO^+(2,2;\mathbb{R})~\cong~[SL(2,\mathbb{R})\times SL(2,\mathbb{R})]/\mathbb{Z}_2.\qquad\text{(split form)}\tag{10}$$
The compact form (9) has a nice proof using quaternions
$$(\mathbb{R}^4,||\cdot||^2) ~\cong~ (\mathbb{H},|\cdot|^2)\quad\text{and}\quad SU(2)~\cong~ U(1,\mathbb{H}),\tag{11}$$
see also this Math.SE post and this Phys.SE post. The split form (10) uses a bijective isometry
$$(\mathbb{R}^{2,2},||\cdot||^2) ~\cong~({\rm Mat}_{2\times 2}(\mathbb{R}),\det(\cdot)).\tag{12}$$
To decompose Minkowski space into left- and right-handed Weyl spinor representations, one must go to the complexification, i.e. one must use the fact that
Note that Refs. 1-2 do not discuss complexification$^2$. One can more or less repeat the construction from section I with the real numbers $\mathbb{R}$ replaced by complex numbers $\mathbb{C}$, however with some important caveats.
There is a bijective isometry from the complexified Minkowski space $(\mathbb{C}^{1,3},||\cdot||^2)$ to the space of $2\times2 $ matrices $({\rm Mat}_{2\times 2}(\mathbb{C}),\det(\cdot))$, $$\mathbb{C}^{1,3} ~\cong ~ {\rm Mat}_{2\times 2}(\mathbb{C}) ~=~ {\rm span}_{\mathbb{C}} \{\sigma_{\mu} \mid \mu=0,1,2,3\}, $$ $$ M(1,3;\mathbb{C})~\ni~\tilde{x}~=~(x^0,x^1,x^2,x^3) \quad\mapsto \quad\sigma~=~x^{\mu}\sigma_{\mu}~\in~ {\rm Mat}_{2\times 2}(\mathbb{C}) , $$ $$ ||\tilde{x}||^2 ~=~x^{\mu} \eta_{\mu\nu}x^{\nu} ~=~\det(\sigma).\tag{13}$$ Note that forms are taken to be bilinear rather than sesquilinear.
There is a surjective Lie group homomorphism$^3$
$$\rho: SL(2,\mathbb{C}) \times SL(2,\mathbb{C}) \quad\to\quad SO({\rm Mat}_{2\times 2}(\mathbb{C}),\mathbb{C})~\cong~ SO(1,3;\mathbb{C})\tag{14}$$ given by $$(g_L, g_R)\quad \mapsto\quad\rho(g_L, g_R)\sigma~:= ~g_L\sigma g^{\dagger}_R, $$ $$ g_L, g_R\in SL(2,\mathbb{C}),\qquad\sigma~\in~ {\rm Mat}_{2\times 2}(\mathbb{C}).\tag{15} $$
The Lie group $SL(2,\mathbb{C})\times SL(2,\mathbb{C})$ has Lie algebra $sl(2,\mathbb{C})\oplus sl(2,\mathbb{C})$.
The Lie group homomorphism
$$\rho: SL(2,\mathbb{C})\times SL(2,\mathbb{C}) \quad\to\quad SO({\rm Mat}_{2\times 2}(\mathbb{C}),\mathbb{C})\tag{16}$$ induces a Lie algebra homomorphism $$\rho: sl(2,\mathbb{C})\oplus sl(2,\mathbb{C})\quad\to\quad so({\rm Mat}_{2\times 2}(\mathbb{C}),\mathbb{C})\tag{17}$$ given by $$ \rho(\tau_L\oplus\tau_R)\sigma ~=~ \tau_L \sigma +\sigma \tau^{\dagger}_R, \qquad \tau_L,\tau_R\in sl(2,\mathbb{C}),\qquad \sigma\in {\rm Mat}_{2\times 2}(\mathbb{C}), $$ $$ \rho(\tau_L\oplus\tau_R) ~=~ L_{\tau_L} +R_{\tau^{\dagger}_R}.\tag{18}$$
The left action (acting from left on a two-dimensional complex column vector) yields by definition the (left-handed Weyl) spinor representation $(\frac{1}{2},0)$, while the right action (acting from right on a two-dimensional complex row vector) yields by definition the right-handed Weyl/complex conjugate spinor representation $(0,\frac{1}{2})$. The above shows that
References:
Anthony Zee, Quantum Field Theory in a Nutshell, 1st edition, 2003.
Anthony Zee, Quantum Field Theory in a Nutshell, 2nd edition, 2010.
$^1$ It is easy to check that it is not possible to describe discrete Lorentz transformations, such as, e.g. parity $P$, time-reversal $T$, or $PT$ with a group element $g\in GL(2,\mathbb{C})$ and formula (2).
$^2$ For a laugh, check out the (in several ways) wrong second sentence on p.113 in Ref. 1: "The mathematically sophisticated say that the algebra $SO(3,1)$ is isomorphic to $SU(2)\otimes SU(2)$." The corrected statement would e.g. be "The mathematically sophisticated say that the group $SO(3,1;\mathbb{C})$ is locally isomorphic to $SL(2,\mathbb{C})\times SL(2,\mathbb{C})$." Nevertheless, let me rush to add that Zee's book is overall a very nice book. In Ref. 2, the above sentence is removed, and a subsection called "More on $SO(4)$, $SO(3,1)$, and $SO(2,2)$" is added on page 531-532.
$^3$ It is not possible to mimic an improper Lorentz transformations $\Lambda\in O(1,3;\mathbb{C})$ [i.e. with negative determinant $\det (\Lambda)=-1$] with the help of two matrices $g_L, g_R\in GL(2,\mathbb{C})$ in formula (15); such as, e.g., the spatial parity transformation $$P:~~(x^0,x^1,x^2,x^3) ~\mapsto~ (x^0,-x^1,-x^2,-x^3).\tag{19}$$ Similarly, the Weyl spinor representations are representations of (the double cover of) $SO(1,3;\mathbb{C})$ but not of (the double cover of) $O(1,3;\mathbb{C})$. E.g. the spatial parity transformation (19) intertwine between left-handed and right-handed Weyl spinor representations.