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The following relations in the definition of the quiver Hecke algebra in Brundan--Kleshchev's paper are $$ y_r \psi_r e(\underline{i}) =(\psi_r y_{r+1} - \delta_{i_r,i_{r+1}})e(\underline{i});\\ $$ $$ y_{r+1} \psi_r e(\underline{i}) =(\psi_r y_r + \delta_{i_r,i_{r+1}})e(\underline{i});\\ $$ However in Khovanov--Lauda's paper these are given as $$ y_r \psi_r e(\underline{i}) =(\psi_r y_{r+1} + \delta_{i_r,i_{r+1}})e(\underline{i});\\ $$ $$ y_{r+1} \psi_r e(\underline{i}) =(\psi_r y_r - \delta_{i_r,i_{r+1}})e(\underline{i});\\ $$ Does anyone know how to resolve these different definitions? I would like to do this without changing any of the other relations. Also, if they are isomorphic why did everyone `change their mind' as to what sign should be used in the first place?

In particular, I would like to use the following as my definition of the KLR algebra (which is just obtained from Brundan-Kleshchev by changing the two relations mentioned above). Is this okay?

Fix $e\in \{2,3,\ldots\} \cup\{\infty\}$ and let $R$ denote an arbitrary ring. The quiver Hecke algebra, $H_n(\kappa)$, is defined to be the unital, associative $R$-algebra with generators \begin{align}\label{gnrs}\tag{$\dagger$} \{e(\underline{i}) \ | \ \underline{i}=(i_1,\dots,i_n)\in (\mathbb{Z}/e\mathbb{Z})^n\}\cup\{y_1,\dots,y_n\}\cup\{\psi_1,\dots,\psi_{n-1}\}, \end{align} subject to the relations $$e(\underline{i})e(\underline{j}) =\delta_{\underline{i},\underline{j}} e(\underline{i}); $$ $$\sum_{\underline{i} \in (\mathbb{Z}/e\mathbb{Z})^n } e(\underline{i}) =1; y_r e(\underline{i}) =e(\underline{i})y_r; $$ $$ \psi_r e(\underline{i}) = e(s_r\underline{i}) \psi_r; $$ $$ y_ry_s =y_sy_r$$ $$\psi_ry _s = {y_s\psi_r} \;\; \text{if }\;\; s\neq r,r+1$$ $$ \psi_r\psi_s = \psi_s\psi_r \;\; \text{if }\;\; |r-s|>1; $$ $$ \color{red}{ y_r \psi_r e(\underline{i}) =(\psi_r y_{r+1} + \delta_{i_r,i_{r+1}})e(\underline{i})}$$ $$\color{red}{y_{r+1} \psi_r e(\underline{i}) =(\psi_r y_r - \delta_{i_r,i_{r+1}})e(\underline{i})};$$ $$\psi_r^2 e(\underline{i}) =\begin{cases} 0 & \text{if }i_r=i_{r+1},\\ e(\underline{i}) & \text{if }i_{r+1}\neq i_r, i_r\pm1,\\ (y_{r+1} - y_r) e(\underline{i}) & \text{if }i_{r+1}=i_r-1 \ \& \ e\neq2,\\ (y_r - y_{r+1}) e(\underline{i}) & \text{if }i_{r+1}=i_r+1 \ \& \ e\neq2,\\ (y_{r+1} - y_r)(y_r - y_{r+1}) e(\underline{i}) & \text{if } i_{r+1}\neq i_r \ \& \ e=2; \end{cases}$$ $$\psi_r \psi_{r+1} \psi_r =\begin{cases} (\psi_{r+1}\psi_r\psi_{r+1} + 1)e(\underline{i})& \text{if }i_r=i_{r+2}=i_{r+1}+1 \ \& \ e\neq2,\\ (\psi_{r+1}\psi_r\psi_{r+1} - 1)e(\underline{i})& \text{if }i_r=i_{r+2}=i_{r+1}-1 \ \& \ e\neq2,\\ (\psi_{r+1}\psi_r\psi_{r+1}+y_r-2y_{r+1}+y_{r+2})e(\underline{i})& \text{if }i_r=i_{r+2}\neq i_{r+1} \ \& \ e=2,\\ (\psi_{r+1}\psi_r\psi_{r+1})e(\underline{i})& \text{otherwise;} \end{cases} $$ for all admissible $r,s,i,j$.

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  • $\begingroup$ I can't solve your sign question, but just a minor correction: should $\kappa$ be playing a role in the definition, in particular controlling the indexing set for the $\underline i$? $\endgroup$ Feb 18, 2017 at 21:58
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    $\begingroup$ Yes, I suppose it should! Thanks Matt. Liron Speyer has answered my question for me: the isomorphism between the above definition and Brundan and Kleshchev's is given by sending $\psi_r \to -\psi_r$ and `reversing the orientation on the quiver' i.e. swapping the signs $i_{r+1}\pm1$ to become $i_{r+1}\mp1$. $\endgroup$ Feb 20, 2017 at 15:02
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    $\begingroup$ Rouquier has shown that, up to some mild assumptions, these algebras are independent of the choice of signs/orientation on the quiver. See Proposition 3.12 of 2-Kac-Moody algebras. $\endgroup$
    – Andrew
    Feb 23, 2017 at 21:56

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