ScalingStacks

Proof. Let $i\in\{1,\ldots,n\}$ and $j\in\{1,\ldots,n-1\}$. We have

$$\beta_{i}t_{j}=\begin{cases}t_{j+1}\beta_{i}&\text{ if }j<i-1\\ \beta_{i-1}&\text{ if }j=i-1\\ 0&\text{ if }j=i\\ t_{j}\beta_{i}&\text{ if }j>i.\end{cases}$$

Consider $I\subset\{1,\ldots,n\}$ non-empty with elements $1\leq i_{1}<\cdots<i_{r}\leq n$. We put $i_{0}=0$ and $i_{r+1}=n+1$.

Consider $j\in\{1,\ldots,n-1\}$. Fix $k\in\{0,\ldots,r\}$ such that $i_{k}\leq j<i_{k+1}$. Let us show that

(3.2.2)

$$\gamma_{I}t_{j}=\begin{cases}t_{j+r-k}\gamma_{I}&\text{ if }i_{k}<j<i_{k+1}-1\\ 0&\text{ if }i_{k}=j<i_{k+1}-1\\ \gamma_{\{i_{1}<\cdots<i_{k}<i_{k+1}-1<i_{k+2}<\cdots<i_{r}\}}&\text{ if }i_{k}<j=i_{k+1}-1\\ t_{k}\gamma_{I}&\text{ if }i_{k}=j=i_{k+1}-1.\end{cases}$$

We have

$$\gamma_{I}t_{j}=\beta_{i_{1}+r-1}\cdots\beta_{i_{k+1}+r-k-1}t_{j+r-k-1}\beta_{i_{k+2}+r-k-2}\cdots\beta_{i_{r}}.$$

If $j<i_{k+1}-1$, then $\beta_{i_{k+1}+r-k-1}t_{j+r-k-1}=t_{j+r-k}\beta_{i_{k+1}+r-k-1}$ and we deduce the first two equalities in (3.2.2). Assume now $j=i_{k+1}-1$. We have $\beta_{i_{k+1}+r-k-1}t_{j+r-k-1}=\beta_{i_{k+1}+r-k-2}$ and the third equality in (3.2.2) follows. The last equality from the fact that given $i\in\{1,\ldots,n-1\}$, we have

	$\displaystyle\beta_{i+1}\beta_{i}$	$\displaystyle=b^{2}t_{n-2}\cdots t_{i}t_{n-1}\cdots t_{i}=b^{2}t_{n-1}\cdots t_{i}t_{n-1}\cdots t_{i+1}=t_{1}b^{2}t_{n-2}\cdots t_{i}t_{n-1}\cdots t_{i+1}$
		$\displaystyle=t_{1}\beta_{i+1}^{2}.$

We deduce that $\gamma_{I}t_{j}=u\gamma_{I^{\prime}}$ for some $I^{\prime}\subset\{1,\ldots,n\}$ with $|I^{\prime}|=|I|$ and $\max(I^{\prime})\leq\max(I)$ and $u\in\{0,1,t_{1},\ldots,t_{n-1}\}$.

Fix $s\in\{1,\ldots,n\}$ with $s\geq\max(I)$. We have

$$\gamma_{I}\beta_{s}=\begin{cases}\beta_{r}\gamma_{\{i_{2}-1,\ldots,i_{r}-1,s\}}&\text{ if }1\in I\\ \gamma_{(I-1)\cup\{s\}}&\text{ otherwise.}\end{cases}$$

Consider $I_{1},\ldots,I_{m}$ as in the lemma. Let $k$ be minimal such that $1{\not\in}I_{k}$. We put $k=m+1$ if there is no such $k$. Define $u=\gamma_{\{|I_{m}|\}}$ if $k=m+1$ and $u=1$ otherwise. Put $I_{0}=\{1,\ldots,n\}$. Recall that $b=\beta_{n}$. We have

$$\gamma_{I_{m}}\cdots\gamma_{I_{1}}b=u\gamma_{I^{\prime}_{m}}\cdots\gamma_{I^{\prime}_{1}}$$

where $I^{\prime}_{r}=\{i-1|i\in I_{r}\setminus\{1\}\}\cup\{|I_{r-1}|\}$ for $1\leq r<k$, $I^{\prime}_{k}=\{i-1|i\in I_{k}\}\cup\{|I_{k-1}|\}$ and $I^{\prime}_{r}=I_{r}$ for $r>k$.

We deduce that the set $B=\{t_{w}\gamma_{I_{m}}\cdots\gamma_{I_{1}}\}$ of the lemma is stable under right multiplication by $t_{j}$ for $j\in\{1,\ldots,n-1\}$ and by $b$. Since $B$ contains $1$, it follows that $B$ is a generating family for $A_{n}$ as an ${\mathbf{F}}_{2}$-vector space. ∎