Wall-crossing functors do not commute

The two class computations are $c[{\theta }_t{\theta }_s\mathrm{\Delta }(\lambda )]=(1+s)(1+t)=1+s+t+st$ and $c[{\theta }_s{\theta }_t\mathrm{\Delta }(\lambda )]=(1+t)(1+s)=1+t+s+ts$, that is $[{\theta }_t{\theta }_s\mathrm{\Delta }(\lambda )]=[\mathrm{\Delta }(\lambda )]+[\mathrm{\Delta }(s\cdot \lambda )]+[\mathrm{\Delta }(t\cdot \lambda )]+[\mathrm{\Delta }(st\cdot \lambda )]$ and $[{\theta }_s{\theta }_t\mathrm{\Delta }(\lambda )]=[\mathrm{\Delta }(\lambda )]+[\mathrm{\Delta }(t\cdot \lambda )]+[\mathrm{\Delta }(s\cdot \lambda )]+[\mathrm{\Delta }(ts\cdot \lambda )]$.

The two products of [F1], $(1+s)(1+t)$ and $(1+t)(1+s)$, differ: the coefficient of the length-two element $st$ is one in the first and zero in the second, and vice versa for $ts$, while $st\mathrm{\ne }ts$ in $S_3$ and the standard classes form a $\mathbb{Z}$-basis, so the induced class operators on $K_0({\mathcal{O}}_{\lambda })$ are different and already differ on the single class $[\mathrm{\Delta }(\lambda )]$.

By the determination theorem, two projective functors whose class operators on $K_0$ differ are not naturally equivalent; hence ${\theta }_s{\theta }_t\cong ̸{\theta }_t{\theta }_s$.

Feeding one factor at a time gives $c[{\theta }_s^{2}X]=c[X](1+s{)}^2=2c[X](1+s)=2c[{\theta }_{s}X]$ for every $X$, so the class operators of ${\theta }_s^2$ and of ${\theta }_s\oplus {\theta }_s$ agree and the determination theorem gives ${\theta }_s^2\cong {\theta }_s\oplus {\theta }_s\cong ̸\mathrm{id}$, since ${\theta }_s\mathrm{\Delta }(\lambda )$ has nonzero class $1+s$; hence $s\mapsto {\theta }_s$ is not a representation of $W$, the wall-crossing classes are the specialized self-dual Hecke generators $1+s$, whose square is $2(1+s)$; the group generator is recovered on classes by $s=[{\theta }_s]-1$, and 'generated by wall-crossing' cannot be read as the assignment $s\mapsto {\theta }_s$ being a representation of $W$.