ScalingStacks

0P5Y

Proof. The non-zero coefficients of π′′\pi^{\prime\prime} are

π11′′\displaystyle\pi^{\prime\prime}_{11} =σ​E2∘E2​σ∘E22​π∘E2​τ2∘τ2​E2\displaystyle=\sigma E_{2}\circ E_{2}\sigma\circ E_{2}^{2}\pi\circ E_{2}\tau_{2}\circ\tau_{2}E_{2}
π22′′\displaystyle\pi^{\prime\prime}_{22} =σ​E1∘E2​τ1∘E2​E1​π∘E2​σ∘τ2​E1\displaystyle=\sigma E_{1}\circ E_{2}\tau_{1}\circ E_{2}E_{1}\pi\circ E_{2}\sigma\circ\tau_{2}E_{1}
π33′′\displaystyle\pi^{\prime\prime}_{33} =τ1​E2∘E1​σ∘E1​E2​π∘E1​τ2∘σ​E2\displaystyle=\tau_{1}E_{2}\circ E_{1}\sigma\circ E_{1}E_{2}\pi\circ E_{1}\tau_{2}\circ\sigma E_{2}
π44′′\displaystyle\pi^{\prime\prime}_{44} =τ1​E1∘E1​τ1∘E12​π∘E1​σ∘σ​E1\displaystyle=\tau_{1}E_{1}\circ E_{1}\tau_{1}\circ E_{1}^{2}\pi\circ E_{1}\sigma\circ\sigma E_{1}
π12′′=σ​E2∘E2​σ∘τ2​E1,π13′′=σ​E2,π24′′=σ​E1,π34′′=τ1​E2∘E1​σ∘σ​E1.\pi^{\prime\prime}_{12}=\sigma E_{2}\circ E_{2}\sigma\circ\tau_{2}E_{1},\ \pi^{\prime\prime}_{13}=\sigma E_{2},\ \pi^{\prime\prime}_{24}=\sigma E_{1},\ \pi^{\prime\prime}_{34}=\tau_{1}E_{2}\circ E_{1}\sigma\circ\sigma E_{1}.

Let a=E1​τ∘π′′a=E_{1}\tau\circ\pi^{\prime\prime} and b=π′′∘E2​τb=\pi^{\prime\prime}\circ E_{2}\tau. We have

a11\displaystyle a_{11} =σ​E2∘E2​σ∘E1​τ2∘E22​π∘E2​τ2∘τ2​E2=σ​E2∘E2​σ∘E22​π∘τ2​E2∘E2​τ2∘τ2​E2\displaystyle=\sigma E_{2}\circ E_{2}\sigma\circ E_{1}\tau_{2}\circ E_{2}^{2}\pi\circ E_{2}\tau_{2}\circ\tau_{2}E_{2}=\sigma E_{2}\circ E_{2}\sigma\circ E_{2}^{2}\pi\circ\tau_{2}E_{2}\circ E_{2}\tau_{2}\circ\tau_{2}E_{2}
=σ​E2∘E2​σ∘E22​π∘E2​τ2∘τ2​E2∘E2​τ2=b11\displaystyle=\sigma E_{2}\circ E_{2}\sigma\circ E_{2}^{2}\pi\circ E_{2}\tau_{2}\circ\tau_{2}E_{2}\circ E_{2}\tau_{2}=b_{11}
a12=E1​τ2∘σ​E2∘E2​σ∘τ2​E1=0=b12a_{12}=E_{1}\tau_{2}\circ\sigma E_{2}\circ E_{2}\sigma\circ\tau_{2}E_{1}=0=b_{12}
a13=E1​τ2∘σ​E2=b13a_{13}=E_{1}\tau_{2}\circ\sigma E_{2}=b_{13}
a23\displaystyle a_{23} =E1​σ−1∘τ1​E2∘E1​σ∘E1​E2​π∘E1​τ2∘σ​E2\displaystyle=E_{1}\sigma^{-1}\circ\tau_{1}E_{2}\circ E_{1}\sigma\circ E_{1}E_{2}\pi\circ E_{1}\tau_{2}\circ\sigma E_{2}
=E1​σ−1∘τ1​E2∘E1​σ∘E1​E2​π∘σ​E2∘E2​σ∘τ2​E1∘E2​σ−1\displaystyle=E_{1}\sigma^{-1}\circ\tau_{1}E_{2}\circ E_{1}\sigma\circ E_{1}E_{2}\pi\circ\sigma E_{2}\circ E_{2}\sigma\circ\tau_{2}E_{1}\circ E_{2}\sigma^{-1}
=E1σ−1∘τ1E2∘E1σ∘σE2∘E2E1π∘∘E2σ∘τ2E1∘E2σ−1\displaystyle=E_{1}\sigma^{-1}\circ\tau_{1}E_{2}\circ E_{1}\sigma\circ\sigma E_{2}\circ E_{2}E_{1}\pi\circ\circ E_{2}\sigma\circ\tau_{2}E_{1}\circ E_{2}\sigma^{-1}
=σ​E1∘E2​τ1∘E2​E1​π∘E2​σ∘τ2​E1∘E2​σ−1=b23\displaystyle=\sigma E_{1}\circ E_{2}\tau_{1}\circ E_{2}E_{1}\pi\circ E_{2}\sigma\circ\tau_{2}E_{1}\circ E_{2}\sigma^{-1}=b_{23}
a24=E1​σ−1∘τ1​E2∘E1​σ∘σ​E1=σ​E1∘E2​τ1=b24a_{24}=E_{1}\sigma^{-1}\circ\tau_{1}E_{2}\circ E_{1}\sigma\circ\sigma E_{1}=\sigma E_{1}\circ E_{2}\tau_{1}=b_{24}
a44\displaystyle a_{44} =E1​τ1∘τ1​E1∘E1​τ1∘E12​π∘E1​σ∘σ​E1=τ1​E1∘E1​τ1∘τ1​E1∘E12​π∘E1​σ∘σ​E1\displaystyle=E_{1}\tau_{1}\circ\tau_{1}E_{1}\circ E_{1}\tau_{1}\circ E_{1}^{2}\pi\circ E_{1}\sigma\circ\sigma E_{1}=\tau_{1}E_{1}\circ E_{1}\tau_{1}\circ\tau_{1}E_{1}\circ E_{1}^{2}\pi\circ E_{1}\sigma\circ\sigma E_{1}
=τ1​E1∘E1​τ1∘E12​π∘E1​σ∘σ​E1∘E2​τ1=b44\displaystyle=\tau_{1}E_{1}\circ E_{1}\tau_{1}\circ E_{1}^{2}\pi\circ E_{1}\sigma\circ\sigma E_{1}\circ E_{2}\tau_{1}=b_{44}

All the other coefficients of aa and bb vanish. We deduce that a=ba=b, hence τ\tau is an endomorphism of E2​(m,π)E^{2}(m,\pi). It follows easily that τ\tau defines an endomorphism of E2E^{2}.

We have τ2=0\tau^{2}=0 and

d⁡(τ)\displaystyle d(\tau) =(d⁡(τ2)00000000000000d⁡(τ1))+τ∘∂+∂∘τ\displaystyle=\left(\begin{matrix}d(\tau_{2})&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&d(\tau_{1})\end{matrix}\right)+\tau\circ\partial+\partial\circ\tau
=(id2​E2​π∘τ2idσ∘E2​π∘τ22idτ12∘E1​π2​τ1∘E1​π∘σid)\displaystyle=\left(\begin{matrix}\operatorname{id}\nolimits&&&\\ 2E_{2}\pi\circ\tau_{2}&\operatorname{id}\nolimits&&\\ \sigma\circ E_{2}\pi\circ\tau_{2}^{2}&&\operatorname{id}\nolimits\\ &\tau_{1}^{2}\circ E_{1}\pi&2\tau_{1}\circ E_{1}\pi\circ\sigma&\operatorname{id}\nolimits\end{matrix}\right)
=id.\displaystyle=\operatorname{id}\nolimits.

We have E3​(m,π)=([m′′′,δ′],π′′′)E^{3}(m,\pi)=([m^{\prime\prime\prime},\delta^{\prime}],\pi^{\prime\prime\prime}), where

m′′′=E23​(m)⊕E22​E1​(m)⊕E2​E1​E2​(m)⊕E2​E12​(m)⊕E1​E22​(m)⊕E1​E2​E1​(m)⊕E12​E2​(m)⊕E13​(m).m^{\prime\prime\prime}=E_{2}^{3}(m)\oplus E_{2}^{2}E_{1}(m)\oplus E_{2}E_{1}E_{2}(m)\oplus E_{2}E_{1}^{2}(m)\oplus E_{1}E_{2}^{2}(m)\oplus E_{1}E_{2}E_{1}(m)\oplus E_{1}^{2}E_{2}(m)\oplus E_{1}^{3}(m).

We have

τ​E=(τ2​E200000000τ2​E10000000000σ−1​E200000000σ−1​E1000000000000000000000000τ1​E200000000τ1​E1)\tau E=\left(\begin{matrix}\tau_{2}E_{2}&0&0&0&0&0&0&0\\ 0&\tau_{2}E_{1}&0&0&0&0&0&0\\ 0&0&0&0&\sigma^{-1}E_{2}&0&0&0\\ 0&0&0&0&0&\sigma^{-1}E_{1}&0&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&\tau_{1}E_{2}&0\\ 0&0&0&0&0&0&0&\tau_{1}E_{1}\end{matrix}\right)

and

E​τ=(E2​τ2000000000E2​σ−10000000000000000E2​τ100000000E1​τ2000000000E1​σ−10000000000000000E1​τ1)E\tau=\left(\begin{matrix}E_{2}\tau_{2}&0&0&0&0&0&0&0\\ 0&0&E_{2}\sigma^{-1}&0&0&0&0&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&E_{2}\tau_{1}&0&0&0&0\\ 0&0&0&0&E_{1}\tau_{2}&0&0&0\\ 0&0&0&0&0&0&E_{1}\sigma^{-1}&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&E_{1}\tau_{1}\end{matrix}\right)

Let a=(E​τ)∘(τ​E)∘(E​τ)a=(E\tau)\circ(\tau E)\circ(E\tau) and b=(τ​E)∘(E​τ)∘(τ​E)b=(\tau E)\circ(E\tau)\circ(\tau E). We have

a11=E2​τ2∘τ2​E2∘E2​τ2=τ2​E2∘E2​τ2∘τ2​E2=b11a_{11}=E_{2}\tau_{2}\circ\tau_{2}E_{2}\circ E_{2}\tau_{2}=\tau_{2}E_{2}\circ E_{2}\tau_{2}\circ\tau_{2}E_{2}=b_{11}
a44=E1​τ1∘τ1​E1∘E1​τ1=τ1​E1∘E1​τ1∘τ1​E1=b44a_{44}=E_{1}\tau_{1}\circ\tau_{1}E_{1}\circ E_{1}\tau_{1}=\tau_{1}E_{1}\circ E_{1}\tau_{1}\circ\tau_{1}E_{1}=b_{44}
a25=E2​σ−1∘σ−1​E2∘E1​τ2=τ2​E1∘E2​σ−1∘σ−1​E2=b25a_{25}=E_{2}\sigma^{-1}\circ\sigma^{-1}E_{2}\circ E_{1}\tau_{2}=\tau_{2}E_{1}\circ E_{2}\sigma^{-1}\circ\sigma^{-1}E_{2}=b_{25}
a47=E2​τ1∘σ−1​E1∘E1​σ−1=σ−1​E1∘E1​σ−1∘τ1​E2=b47a_{47}=E_{2}\tau_{1}\circ\sigma^{-1}E_{1}\circ E_{1}\sigma^{-1}=\sigma^{-1}E_{1}\circ E_{1}\sigma^{-1}\circ\tau_{1}E_{2}=b_{47}

and all the other coefficients of aa and bb vanish. It follows that a=ba=b. This completes the proof of the theorem. ∎

Original mathematics by the credited authors. Source collection and HTML conversion remain in progress.

Andrew Manion, Raphael Rouquier

Original source: arXiv:2009.09627v2