# Why are the divided difference operators of the nil Hecke ring only of degree 1?

In the paper "A Diagrammatic Approach to the Categorification of Quantum Groups I" (arXiv, journal, MSN), Khovanov and Lauda put a strand with a single dot in degree 2, and put the crossing operator in degree 1, and set these to be the generators of a graded ring.

However, when they define an action of their (graded) diagram ring on a given (graded) polynomial ring, applying the crossing operator a homogeneous polynomial with integral coefficients in this graded ring yields a polynomial in odd degree, which shouldn't be possible (as attaching dots to strands represents raising the degree by 1, and thus all polynomials sit in even degree). Is there something I'm missing here?

• "as attaching dots to strands represents raising the degree by 1, and thus all polynomials sit in even degree" Do you mean attaching dots raises the degree by 2? – Sean Clark Nov 9 '17 at 20:38

It is not true that the crossing is necessarily in degree $1$. More precisely, if $\delta_{k,\mathbf i}$ is the crossing between strands $k$ and $k+1$ for the sequence $\mathbf i=(i_1,\ldots, i_m)$, then $$\deg(\delta_{k,\mathbf i})=-i_k\cdot i_{k+1}=\begin{cases}-2&\text{ if }i_k=i_{k+1},\\ 1 &\text{ if } i_k\text{ is connected to }i_{k+1}\text{ in the graph }\Gamma,\\ 0&\text{ otherwise.}\end{cases}.$$
In particular, in the case $i_k=i_{k+1}$ (such as in $\S$ 2.2 (3), which given your question title may be the case you are principally interested in), $\delta_{k,\mathbf i}$ indeed acts as a divided difference operator, hence lowers the degree of the polynomial by $2$, hence indeed has degree $-2$.
If $i_k\neq i_{k+1}$, then it is more complicated: now $\delta_{k,\mathbf i}$ does not act as a divided difference, but instead acts as a (possibly scaled) permutation of variables (see the description before Proposition 2.3). In particular, as noted shortly before Corollary 2.6, the grading on $\mathcal{Pol}_{\nu}=\bigoplus_{|\mathbf i|=\nu} \mathcal{Pol}_{\mathbf i}$ is determined by fixing a particular $\mathbf i\in{\rm Seq}(\nu)$ and fixing $1_{\mathbf i}=1\in \mathcal{Pol}_{\mathbf i}$ to be in degree 0. The grading on $\mathcal{Pol}_{\nu}$ is the induced from this choice; in particular, the other units $1_{\mathbf j}$ may have non-zero degrees as elements of $\mathcal{Pol}_{\nu}$.
For example, consider $\S$ 2.2 (7): $\nu=i+j$ with $i\cdot j=-1$. Assume without loss of generality that the edge in $\Gamma$ is oriented $i\leftarrow j$. Then $\mathcal{Pol}_\nu=\mathcal{Pol}_{i,j}\oplus \mathcal{Pol}_{j,i}$. Let us set $\deg(1_{i,j})=0$. Then the unique crossing for $(i,j)$, $\delta_{1,(i,j)}=\delta_{i,j}$ acts by mapping e.g. $1_{i,j}\mapsto 1_{j,i}$, so in particular $\deg(1_{j,i})=1$ by definition. This is consistent because the reverse crossing $\delta_{j,i}$ maps $1_{j,i}\mapsto (x_1+x_2)1_{i,j}$ which has degree $2$ as expected.