the localization of the zero modes, we need to consider the right-handed and left-handed components separately, which can be extracted as $$\varphi_{\mathrm{R} \alpha} = 
\frac{1+ \gamma_\chi}{2} \varphi_\alpha \ , \quad
\varphi_{\mathrm{L} \alpha} = 
\frac{1- \gamma_\chi}{2} \varphi_\alpha \ .
\label{eq: unitary_transf_psi}$$

Furthermore, we should take into account that the classical solution of the form (\[eq: form\_Ya\]) has symmetry under $$\label{eq: UEX}
Y_a  \mapsto  g \, Y_a \, g^\dagger \ ,$$ where the $g \in \mathrm{SU}(N_{\rm Y})$ is block-diagonal $$g =
\begin{pmatrix}
g^{(1)} & 0 \\
0 & g^{(2)}
\end{pmatrix}$$ with $g^{(1)} \in \mathrm{SU}(N_{\rm Y}^{(1)})$ and $g^{(2)} \in \mathrm{SU}(N_{\rm Y}^{(2)})$, which preserves the structure . Under this transformation, the wave functions of the corresponding Dirac operator transform as $$\label{eq: varphi_LR_prime}
\varphi_\alpha \mapsto 
g^{(1)} \varphi_\alpha g^{(2)\dagger}  \ .
$$ We choose $g^{(1)}$ and $g^{(2)}$ in in such a way that $\varphi_{\mathrm{R}1}$ is diagonal with positive elements in the descending order. This is nothing but the singular value decomposition of $\varphi_{\mathrm{R}1}$, where the diagonal elements correspond to the singular values.

In what follows, we change the index $a$ of $Y_a$ from $a=4,\ldots,9$ to $a=1,\ldots,6$. From the picture of the intersecting D-branes, we expect the emergence of Dirac zero modes when the two branes intersect 