The usual basis (in the “physics convention”) of $(1, 3)$ is given by six matrices $J^i$, $K^i$ with \begin{align} [J^i, J^j] &= iε^{ijk} J^k \\ [J^i, K^j] &= iε^{ijk} K^k \\ [K^i, K^j] &= -iε^{ijk} J^k \end{align} The elements $Λ ∈ \mathrm{SO}(1, 3)^+$ are then given by \begin{equation} Λ = \exp\!\big(-i(\vec{α} ⋅ \vec{J} + \vec{η} ⋅ \vec{K})\big) \end{equation}
We can go to the complexification $(1, 3)_ℂ$ and choose a new basis $T_{\mathrm{L}/\mathrm{R}}^i = \frac{1}{2} (J^i ± iK^i)$ such that \begin{align} [T_{\mathrm{L}}^i, T_{\mathrm{L}}^j] &= iε^{ijk} T_{\mathrm{L}}^k \\ [T_{\mathrm{R}}^i, T_{\mathrm{R}}^j] &= iε^{ijk} T_{\mathrm{R}}^k \\ [T_{\mathrm{L}}^i, T_{\mathrm{R}}^j] &= 0 \end{align} This demonstrates that $(1, 3)_ℂ ≅ (2)_ℂ ⊕ (2)_ℂ$.
Now, my question is that since $(1, 3)_ℂ$ decomposes into a direct sum, I would expect the corresponding group to be a direct product: \begin{equation} \exp\!\big((1, 3)_ℂ\big) ≅ \exp\!\big((2)_ℂ ⊕ (2)_ℂ\big) = \mathrm{SU}(2)_ℂ × \mathrm{SU}(2)_ℂ \end{equation} So, in my mind, the elements of this group should be block-diagonal matrices in some form. However, using the Baker–Campbell–Hausdorff formula, it can be shown that \begin{equation} Λ = \exp\!\big(-i(\vec{α} ⋅ \vec{J} + \vec{η} ⋅ \vec{K})\big) = \exp\!\big(-i(\vec{α} - i\vec{η}) ⋅ \vec{T}_{\mathrm{L}})\big) \exp\!\big(-i(\vec{α} + i\vec{η}) ⋅ \vec{T}_{\mathrm{R}})\big) \end{equation} and while there is a product of two matrices on the right, these matrices are multiplied together instead of being put into a block matrix. How can these two ideas be reconciled?