maximal flag $(x_0,\ldots x_n)$ in $M_n$ is a witness for this.

This follows immediately from the description of $\operatorname{acl}$ in Lemma \[r:acl\] and of forking in Corollary \[c:forking\].

\[T:notample\] The is not $n+1$-ample.

Suppose towards a contradiction that $a_0,\ldots a_{n+1}$ are witnesses for $T_n$ being $n+1$-ample over some set of parameters $A$. We have $${
 a_{n+1}\mathop{\mathpalette{
 \setbox0=\hbox{$x$}\kern\wd0\hbox to 0pt{\mathchardef
 \nn=12854\hss$\nn$\kern1.4\wd0\hss}\hbox to 0pt{\hss$\mid$\hss}\lower.9\ht0
 \hbox to 0pt{\hss$\smile$\hss}\kern\wd0
}{}}_{A}a_0
},$$ $${
 a_{n+1}\mathop{\mathpalette{
 \setbox0=\hbox{$x$}\kern\wd0\hbox to 0pt{\hss$
 \mid$\hss}\lower.9\ht0\hbox to 0pt{\hss$\smile$\hss}\kern\wd0
}{}}_{Aa_i}a_0\ldots a_i
},i=0,\ldots n.$$

By the first condition there are vertices in $\operatorname{acl}(a_0)$ and in $ \operatorname{acl}(a_{n+1})$ which are in the same connected component. Put $f_0=\operatorname{proj}(a_{n+1}/a_0A)\in\operatorname{acl}(a_0A)$ and $f_{n+1}=\operatorname{proj}(f_0/a_{n+1}A)\in\operatorname{acl}(a_{n+1}A)$.

Since $${
 a_{n+1}\mathop{\mathpalette{
 \setbox0=\hbox{$x$}\kern\wd0\hbox to 0pt{\hss$
 \mid$\hss}\lower.9\ht0\hbox to 0pt{\hss$\smile$\hss}\kern\wd0
}{}}_{Aa_i}a_0\ldots a_i
}, i=1,\ldots n$$ using Corollary \[c:forking\] we inductively find flags $$f_i=\operatorname{proj}(f_{n+1}/f_0f_1\ldots f_{i-1}a_iA)=\operatorname{proj}(f_{n+1}/a_iA)\in\operatorname{acl}(a_iA),i=1,\ldots n$$ such that $${
 f_{n+1}\mathop{\mathpalette{
 \setbox0=\hbox{$x$}\kern\wd0\hbox to 0pt{\hss$
 \mid$\hss}\lower.9\ht0\hbox to 0pt{\hss$\smile$\hss}\kern\wd0
}{}}_{f_i}f_0f_1\ldots f_i
}.$$

For $i=1,\ldots n$ we clearly have $$\operatorname{acl}(f_0,f_1.\ldots f_{i-1},f_i)\cap \operatorname{acl}(f_0,.\ldots f_{i-1},f_{i+1})\subseteq \operatorname{acl}(a_0,.\ldots a_{i-1}A).$$

By construction, there is a reduced path $\gamma=(f_0,x_1,\ldots x_k=f_{n+1})$ containing a vertex of each of the $f_i$ in ascending order. Since we cannot have a flag containing more than $n$ elements, there must be some vertex $x$ in $\gamma$ where $\gamma$ changes direction. For some $i$ we then 