\otimes \mathbf{1}) \Psi = \pm \Psi \ .$$ Due to the chirality (\[eq: chirality\_in\_10d\]) in 10 dimensions, Eq. (\[eq: chirality\_in\_4d\_Psi\]) implies $$\label{eq: chirality_in_6d_Psi}
{\left( \mathbf{1} \otimes \gamma_\chi \right)} \Psi = \pm \Psi \ ;$$ i.e., $\Psi$ is chiral also in six dimensions.

Let us next decompose Eq. (\[eq: 10d\_Dirac\]) into two terms as $$\Gamma^\mu {{\left[ A_\mu, \Psi \right]}}+ \Gamma^a {{\left[ A_a, \Psi \right]}}=0 \ .
\label{eq:ac-decompose}$$ Note that the first term and the second term have opposite chirality in 4-dimensions as well as in 6-dimensions, which implies that each term has to vanish separately. Thus, in order to obtain chiral fermions in four dimensions, we need to have Dirac zero modes in six dimensions; i.e., $$\Gamma^a {{\left[ A_a, \Psi \right]}}=0 \ .
\label{Dirac-eq-6d}$$

Let us now consider that $A_M$ is a classical solution with the quasi direct-product structure (\[eq: ansatz of A\]). Since $A_a = \mathbf{1} \otimes Y_a$, the general solution to (\[Dirac-eq-6d\]) can be obtained by decomposing $\Psi$ as $$\Psi = \psi^\mathrm{(4d)} \otimes \psi^\mathrm{(6d)} \ ,$$ where the 4-dimensional and 6-dimensional gamma matrices act only on $\psi^\mathrm{(4d)}$ and $\psi^\mathrm{(6d)}$, respectively. Thus, in order to satisfy (\[Dirac-eq-6d\]), we only need to require $$\label{eq: chirality_in_6d_psi}
\gamma^a {{\left[ Y_a, \psi_a^\mathrm{(6d)} \right]}}=0 \ 