 Solution

First use the actual PDF boost bracket $[K_i,K_j]=-i{ϵ}_{ijk}{J}_k$; replacing the final $J_k$ by $K_k$, as the TeX transcription does, would be a different and incorrect algebra. Expanding the three brackets yieldsThusThe two sets have SU(2)-type brackets. Their compact real form exponentiates to $SU(2)\times SU(2)$, whereas their full complexification exponentiates to $SL(2,\mathbb{C})\times SL(2,\mathbb{C})$. This does not turn the physical real Lorentz group into a compact product. With Hermitian $J,K$, $A^{†}=B$, and the two factors are tied by the Lorentz reality condition; the connected physical spin cover is $SL(2,\mathbb{C})$.

For the right-handed spinor representation, the supplied sigma formula gives ${\bar{\sigma }}^{ij}={ϵ}_{ijk}{\sigma }_k/2$ and ${\bar{\sigma }}^{0i}=i{\sigma }_i/2$. ThereforeIt is the $(0,1/2)$ representation in this definition of $A,B$. These are finite-dimensional representation matrices; the boosts are not Hermitian for a positive-definite spinor inner product, as expected for a noncompact group. They are distinct from the operator-component commutator coefficients discussed in Question 1.

For the group map, encode a real Minkowski vector by the Hermitian matrixThe trace identity $\frac{1}{2}\mathrm{Tr}({\bar{\sigma }}^{\mu }{\sigma }_{\nu })={\delta }^{\mu }{}_{\nu }$ recovers its contravariant components. For $N\in SL(2,\mathbb{C})$, $X^{\prime }=NX{N}^{†}$ is Hermitian and has the same determinant. Expanding $X^{\prime }$ in that basis givesThe coefficients are real because the trace of a product of Hermitian matrices is real. Equality of determinants for all $x$ proves ${\Lambda }^T\eta \Lambda =\eta$. The action for $N_1{N}_2$ is the composition of the two actions, proving the homomorphism law.

Since $SL(2,\mathbb{C})$ is connected and the identity maps to the identity, its image has determinant $+1$ and preserves time orientation. More directly, positive-definite matrices representing future timelike vectors remain positive definite under $X\mapsto NX{N}^{†}$. The kernel is $\{I,-I\}$: if every Hermitian $X$ is fixed, $X=I$ first makes $N$ unitary, and then commutation with all Hermitian matrices makes it a scalar with determinant one.

Conversely, $SU(2)$ gives every spatial rotation, and positive Hermitian determinant-one matrices $\exp (\mathbf{b}\cdot \mathbf{\sigma }/2)$ give every pure boost. Rotation-boost decomposition yields every proper orthochronous Lorentz transformation. Thus the formula gives the Lorentz spinor double coverIt is a map into $SO(1,3)$ as asked, with precisely its identity component as image; it does not produce disconnected time-reversing transformations.