The KLR grading and the diagrammatic multiplication

Relation (1) has degree $0$ on both sides, and in (2) the identity $x_{a}e(\nu )=e(\nu )x_a$ has degree $2d_{{\nu }_a}$ while $x_a{x}_b=x_b{x}_a$ has degree $2d_{{\nu }_a}+2d_{{\nu }_b}$ on the $\nu$-component; relation (3) has degree $-b_{{\nu }_a{\nu }_{a+1}}=-b_{{\nu }_{a+1}{\nu }_a}$ on both sides by the symmetry of $b$ from [F1].

Relation (4) is homogeneous of degree $-2b_{{\nu }_a{\nu }_{a+1}}$: the left side ${\tau }_a^{2}e(\nu )$ has degree $2(-b_{{\nu }_a{\nu }_{a+1}})$, and the right side $Q_{{\nu }_a{\nu }_{a+1}}(x_a,x_{a+1})e(\nu )$ has degree $-2b_{{\nu }_a{\nu }_{a+1}}$ by [F2], since a monomial $u^r{v}^s$ of $Q_{ij}$ contributes $r(2d_i)+s(2d_j)$.

Relations (5) and (6) are homogeneous with their actual label-dependent degrees. In far crossing commutation, both products have degree $-b_{{\nu }_a{\nu }_{a+1}}-b_{{\nu }_b{\nu }_{b+1}}$ because the disjoint crossings interchange the same two color pairs. A far dot move has degree $2d_{{\nu }_b}-b_{{\nu }_a{\nu }_{a+1}}$ on both sides, as its untouched strand keeps the dot color. For adjacent unequal colors, dot sliding likewise preserves the degree of the crossing plus the transported dot. For equal colors $i$, the dot degree is $2d_i$ and the crossing degree is $-2d_i$, so both terms of ${\tau }_a{x}_a-x_{a+1}{\tau }_a=e(\nu )$ and ${\tau }_a{x}_{a+1}-x_a{\tau }_a=-e(\nu )$ have degree $0$, equal to the degree of the idempotent on the right. Thus every dot and far-crossing relation is homogeneous, without asserting that its common degree is always zero.

Relation (7), unequal outer colours: for colours $({\nu }_a,{\nu }_{a+1},{\nu }_{a+2})=(a^{\mathrm{\prime }},b^{\mathrm{\prime }},c^{\mathrm{\prime }})$ both braid words are sums of terms each moving the same three crossings over the same three pairs, so each has degree $-b_{a^{\mathrm{\prime }}{b}^{\mathrm{\prime }}}-b_{a^{\mathrm{\prime }}{c}^{\mathrm{\prime }}}-b_{b^{\mathrm{\prime }}{c}^{\mathrm{\prime }}}$, and the right side is $0$; on the equal-outer-colour case $(i,j,i)$ each braid word has the same degree $-2b_{ij}-2d_i$, while the correction numerator has degree $-2b_{ij}$ and its division by $x_{a+2}-x_a$, of degree $2d_i$, lowers the degree to $-2b_{ij}-2d_i$; for three equal colours $Q_{ii}=0$ and the ordinary nil-Hecke braid relation is homogeneous of degree $-6d_i$.

Use the free path algebra with vertices $\nu$, loops $x_{a}e(\nu )$ and arrows ${\tau }_{a}e(\nu ):\nu \to s_a\nu$, assigned the displayed degrees. Its vertex sum is the unit because $I$ is finite; decomposing the global generators by word idempotents identifies its presented quotient with $R_n(Q)$. Every defining component relation is homogeneous, so the two-sided ideal they generate is graded, and the quotient $R_n(Q)$ inherits a $\mathbb{Z}$-grading with $R_n(Q{)}_m={⨁}_{\nu ,\eta }e(\nu )R_n(Q{)}_{m}e(\eta )$; the weight refinement is inherited because each generator preserves the weight ${\sum }_a{\alpha }_{{\nu }_a}$ and each relation is weight-homogeneous, and $R(\beta )=e(\beta )R_{\mathrm{\mid }\beta \mathrm{\mid }}(Q)e(\beta )$ is a graded subalgebra for every $\beta \in Q_{+}$.