Signed Quiver Flag Convolution And Klr Ext Realization Examples

The data. For this quiver $E={⨁}_{h:i\to j}\mathrm{Hom}(V_i,V_j)=\{0\}$, so $X_{\nu }=F_{\nu }=\mathrm{Flag}({\mathbb{C}}^2)={\mathbb{P}}^1$ with $d_{\nu }={\dim }_{\mathbb{C}}{X}_{\nu }=1$ and $Z_{\nu ,\nu }=X_{\nu }{\times }_E{X}_{\nu }={\mathbb{P}}^1\times {\mathbb{P}}^1$. The convolution algebra is therefore the single corner $H^{\mathrm{B}\mathrm{M},G}({\mathbb{P}}^1\times {\mathbb{P}}^1;k)$, a free module over the polynomial ring $k[x_1,x_2]$ generated by the diagonal class and the elementary class, by [F3] with $S_2=\{1,s_1\}$.

The normalized elementary action. The relative-position-$s_1$ stratum in ${\mathbb{P}}^1\times {\mathbb{P}}^1$ is the complement of the diagonal, and it is dense, so its closure is all of ${\mathbb{P}}^1\times {\mathbb{P}}^1$; at the two torus-fixed points of a ${\mathbb{P}}^1$ fiber the tangent Euler classes are $x_1-x_2$ and $x_2-x_1$. Hence the standard fundamental class acts by $(s_{1}f-f)\mathrm{/}(x_1-x_2)$, and the normalized class ${\tau }_1=-{\left[C\right]}_{\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{a}\mathrm{r}\mathrm{d}}$ acts by the displayed divided difference $(f-s_{1}f)\mathrm{/}(x_1-x_2)$, which is polynomial because $f-s_{1}f$ is antisymmetric in $x_1,x_2$; this is the finite cellwise localization computation over a characteristic-zero field, using the cellular model of ${\mathbb{P}}^1$.

Direct evaluation gives ${\tau }_1^2=0$ because $D_{1}f$ is symmetric and hence killed by another divided difference; it also gives ${\tau }_1{x}_1-x_2{\tau }_1=1$ and ${\tau }_1{x}_2-x_1{\tau }_1=-1$. The degree is $-2$. In the abstract nil-Hecke presentation, these mixed relations move dots left and the square relation removes every repeated crossing, so it is spanned by $f+g{\tau }_1$ for $f,g\in k[x_1,x_2]$. The two polynomial summands map independently and onto the geometric corner by [F3] and step 1.1. Thus this presentation is isomorphic to the corner, which proves the asserted nil-Hecke identification.

Dimensions and normalizations. With $h_{ij}=2$, $h_{ji}=1$ the defining polynomial is $Q_{ij}(u,v)=(-1{)}^2(u-v{)}^3=(u-v{)}^3$, and the dimension formula gives $d_{(i,j,i)}=1+(h_{ji}+h_{ij})=4$, $d_{(j,i,i)}=1+2h_{ij}=5$ and $d_{(i,i,j)}=1+2h_{ji}=3$; hence $d_{s_1\nu }-d_{\nu }=1$ and $d_{s_2\nu }-d_{\nu }=-1$. The unequal crossing normalization exponent $d_{s_r\nu }-{\dim }_{\mathbb{C}}{C}_{s_r\nu ,\nu }=h_{{\nu }_r{\nu }_{r+1}}$ is instead $2$ for $r=1$ and $1$ for $r=2$; it must not be identified with the dimension difference $d_{s_r\nu }-d_{\nu }$.

The crossing operators. The action formulas of [F2] give, on the relevant words, ${\tau }_1{e}_{(i,j,i)}=(x_1-x_2{)}^2{s}_1$, ${\tau }_2{e}_{(i,j,i)}=(x_2-x_3)s_2$, ${\tau }_2{e}_{(i,i,j)}=(x_2-x_3{)}^2{s}_2$, ${\tau }_1{e}_{(j,i,i)}=(x_1-x_2)s_1$, together with the equal-color divided differences $D_1$ on $(i,i,j)$ and $D_2$ on $(j,i,i)$; this exhibits the two triple products as the displayed composite operators on $k[x_1,x_2,x_3]$.

Put $x=x_1$, $y=x_2$, $z=x_3$ and $w_0=s_2{s}_1{s}_2=s_1{s}_2{s}_1$. In the skew group algebra the two composite operators of step 2.1 expand to $A=[(y-z{)}^2(z-y)-(y-z{)}^2(x-y)w_0]\mathrm{/}(x-z)$ and $B=[(x-y)(y-x{)}^2-(x-y)(y-z{)}^2{w}_0]\mathrm{/}(x-z)$. Their $w_0$ coefficients coincide. Thus $A-B=((z-y{)}^3-(x-y{)}^3)\mathrm{/}(x-z)=-[(x-y{)}^2+(x-y)(z-y)+(z-y{)}^2]$, exactly the native correction. It is nonzero of total polynomial degree $2$, since its $x^2$ coefficient is $-1$. This explicit expansion retains both the square factor from $h_{ij}=2$ and the linear factor from $h_{ji}=1$; all divisions represent polynomial operators by [F2].