Homogeneous degree check for the corrected KLR braid relation

The common degree of the two words. The three crossings of ${\psi }_r{\psi }_{r+1}{\psi }_{r}e(i)$ cross the colour pairs $(i_r,i_{r+1})$, $(i_r,i_{r+2})$, $(i_{r+1},i_{r+2})$ of the successive permuted sequences, so its degree is $d(i)$; the other braid word has the same multiset of crossed pairs, hence the same degree $d(i)$.

The four cases of the correction. Substituting the cyclic Cartan entries: (a) if $i_{r+2}=i_r\to i_{r+1}$ then $d(i)=-(-1)-2-(-1)=0$ and the correction is $+1$; (b) if $i_{r+2}=i_r\leftarrow i_{r+1}$ then $d(i)=0$ and the correction is $-1$; (c) if $e=2$ and $i_{r+2}=i_r\rightleftarrows i_{r+1}$, then $a_{i_r{i}_{r+1}}=a_{i_{r+1}{i}_{r+2}}=-2$ and $a_{i_r{i}_{r+2}}=a_{00}=2$, so $d(i)=2+2-2=2$, matching the degree $2$ of $y_r+y_{r+2}-2y_{r+1}$; (d) in all other cases the correction is $0$. This is the source's case list, and the nonzero correction requires equal outer colours, while zero corrections remain homogeneous without asserting that their degree is zero.

Conclusion. In each nonzero case the correction term has degree $d(i)$, equal to the common degree of the two braid words computed in step 1.1; in the remaining cases the correction vanishes, so the identity is homogeneous at degree $d(i)$. Hence the single-edge corrected braid relations have degree $0$ and the $e=2$ double-edge corrected braid relation has degree $2$, both matching the two sides. The displayed source word order is ${\psi }_r{\psi }_{r+1}{\psi }_r-{\psi }_{r+1}{\psi }_r{\psi }_{r+1}$, which equals the native opposite-order $\tau$-difference under $\psi =-\tau$. No choice is used.