Proof. Let $T^{S_{j}}\subset \operatorname{SL}_{n}(\mathbb{C})$ denote the diagonal torus

$$\begin{eqnarray}T^{S_{j}}=\{\operatorname{diag}(a_{1},\ldots ,a_{n}):\unicode[STIX]{x1D6F1}_{i\in S_{j}}a_{i}=\unicode[STIX]{x1D6F1}_{i\notin S_{j}}a_{i}=1\}.\end{eqnarray}$$

The stabiliser of $p_{S}$ is the intersection of the stabilisers of its direct summands. The stabiliser in $\operatorname{SL}_{n}(\mathbb{C})$ of the direct summand $\wedge _{i\in S_{j}}e_{i}$ is the semidirect product

$$\begin{eqnarray}U_{S}^{j}=\langle U_{\unicode[STIX]{x1D6FC}_{a}-\unicode[STIX]{x1D6FC}_{b}}:(a\neq b,a,b\in S_{j})~\text{or}~(a\neq b,b\notin S_{j})\rangle \rtimes T^{S_{j}}.\end{eqnarray}$$

Now $j\in S_{j}$ for $1\leqslant j\leqslant n$ and by (3) the intersection is

$$\begin{eqnarray}\mathop{\bigcap }_{j=1}^{n}U_{S}^{j}=\langle U_{\unicode[STIX]{x1D6FC}_{a}-\unicode[STIX]{x1D6FC}_{b}}:a<b,a\in S_{b}\rangle \rtimes \biggl(\mathop{\bigcap }_{j=1}^{n}T^{S_{j}}\biggr).\end{eqnarray}$$

Since $j\in S_{j}$ and $S_{j}\subset \{1,\ldots ,j\}$ for all $1\leqslant j\leqslant n$ we have by induction on $n$ that

$$\begin{eqnarray}\mathop{\bigcap }_{j=1}^{n}T^{S_{j}}=\{\operatorname{diag}(a_{1},\ldots ,a_{n}):a_{1}=1,\unicode[STIX]{x1D6F1}_{i\in S_{2}}a_{i}=1,\ldots ,\unicode[STIX]{x1D6F1}_{i\in S_{n}}a_{i}=1\}=1,\end{eqnarray}$$

and therefore

$$\begin{eqnarray}\mathop{\bigcap }_{j=1}^{n}U_{S}^{j}=\langle U_{\unicode[STIX]{x1D6FC}_{a}-\unicode[STIX]{x1D6FC}_{b}}:b<a,b\in S_{a}\rangle =U_{S}\end{eqnarray}$$

by definition. ◻

First proof.

We use the following fact about solvable groups.

Let $H\subset \operatorname{GL}(V)$ be a solvable algebraic group and let $v\in V$ . Then there is an $f\in k[V]$ and a character $\unicode[STIX]{x1D712}:H\rightarrow k^{\ast }$ such that $f(h\cdot w)=\unicode[STIX]{x1D712}(h)f(w)$ holds for all $b\in B$ , $w\in V$ and $\overline{H\cdot v}\setminus H\cdot v=\{w\in \overline{H\cdot v}:f(w)=0\}$ .

We apply this result with the Borel $H=B_{n}\subset \operatorname{SL}(n)$ . Let $(b^{(m)})$ be a sequence such that

$$\begin{eqnarray}p^{\infty }=\lim _{m\rightarrow \infty }b^{(m)}p_{\tilde{S}}\in \overline{B_{n}\cdot p_{\tilde{S}}}\setminus B_{n}\cdot p_{\tilde{S}}.\end{eqnarray}$$

Let $f$ be as above, vanishing on $\overline{B_{n}\cdot p_{\tilde{S}}}\setminus B_{n}\cdot p_{\tilde{S}}$ . Then

$$\begin{eqnarray}0=f(p^{\infty })=\lim _{m\rightarrow \infty }f(b^{(m)}\cdot p_{\tilde{S}})=\lim _{m\rightarrow \infty }\unicode[STIX]{x1D712}(b^{(m)})f(p_{\tilde{S}}).\end{eqnarray}$$

Since $f(p_{\tilde{S}})\neq 0$ , we must have $\lim _{m\rightarrow \infty }\unicode[STIX]{x1D712}(b^{(m)})=0$ so $\lim _{m\rightarrow \infty }\unicode[STIX]{x1D712}_{ii}(b^{(m)})=0$ for some $1\leqslant i\leqslant n$ .◻

Second proof.

Let $p^{\infty }=\lim _{m\rightarrow \infty }b^{(m)}p_{\tilde{S}}\in {\mathcal{B}}_{\emptyset }$ be defined by the normalised sequence $(b^{(m)})$ such that $\operatorname{VSpec}(b^{(m)})=\emptyset$ . By definition for all $1\leqslant i\leqslant n$ either $b_{ii}^{\infty }=\lim _{m\rightarrow \infty }b_{ii}^{(m)}\in \mathbb{C}$ exists or $\lim _{m\rightarrow \infty }|b_{ii}^{(m)}|=\infty$ . Since

$$\begin{eqnarray}b_{11}^{(m)}\cdot \ldots \cdot b_{nn}^{(m)}=1~\text{holds for all}~m,\end{eqnarray}$$

if $b_{ii}^{(m)}\rightarrow \infty$ for some $1\leqslant i\leqslant n$ , then for some $j\neq i~\lim _{m\rightarrow \infty }b_{jj}^{(m)}=0$ so $j\in \operatorname{VSpec}(b^{(m)})$ , a contradiction. This proves that

$$\begin{eqnarray}b_{ii}^{\infty }=\lim _{m\rightarrow \infty }b_{ii}^{(m)}\in \mathbb{C}\setminus \{0\}\quad \text{for}~1\leqslant i\leqslant n.\end{eqnarray}$$

Assume that some off-diagonal entries of the normalised sequence $(b^{(m)})$ are not convergent and let

$$\begin{eqnarray}v=\min \{j:\exists i\notin S_{j}~\text{such that}~i<j~\text{and}~b_{ij}^{\infty }=\infty \}\end{eqnarray}$$

be the leftmost column containing such entries and choose a $u<v$ , $u\notin S_{v}$ such that $b_{uv}^{\infty }=\infty$ . Then due to (9) the coefficient of $e_{u}\wedge (\wedge _{i\in S_{v}\setminus \{v\}}e_{i})$ in $p_{S_{v}}^{\infty }=\lim _{m\rightarrow \infty }\wedge _{i\in S_{v}}b^{(m)}e_{i}$ would be

$$\begin{eqnarray}\lim _{m\rightarrow \infty }b_{uv}^{(m)}\cdot \mathop{\prod }_{i\in S_{v}\setminus \{v\}}b_{ii}^{\infty }=\infty ,\end{eqnarray}$$

a contradiction. Hence $b_{ij}^{\infty }:=\lim _{m\rightarrow \infty }b_{ij}^{(m)}\in \mathbb{C}$ exists for all $1\leqslant i\leqslant j\leqslant n$ and $b_{ii}^{\infty }\neq 0$ so $b^{\infty }=\lim _{m\rightarrow \infty }b^{(m)}\in B_{n}$ exists and

$$\begin{eqnarray}p^{\infty }=\lim _{m\rightarrow \infty }b^{(m)}\cdot p_{\tilde{S}}=b^{\infty }\cdot p_{\tilde{S}}\in B_{n}\cdot p_{\tilde{S}}.\Box\end{eqnarray}$$

Proof. Let $\hat{{\mathcal{W}}}_{\tilde{S}}={\mathcal{W}}_{\tilde{S}}\times \prod _{u\in \mathbf{u}}\mathbb{P}_{u}^{1}$ denote the product of the affine ambient space ${\mathcal{W}}_{\tilde{S}}$ with a copy of $\mathbb{P}^{1}$ for every element of  $\mathbf{u}$ . We endow $\mathbb{P}_{u}^{1}$ with a $B_{n}$ action as follows: for $b=b_{ij}\in B_{n}$ we let $b\cdot [x:y]=[x:b_{uu}y]$ for $[x:y]\in \mathbb{P}_{u}^{1}$ . Define the point

$$\begin{eqnarray}\hat{p}_{\tilde{S}}=p_{\tilde{S}}\times \mathop{\prod }_{u\in \mathbf{u}}[1:1]\in \hat{{\mathcal{W}}}_{\tilde{S}}\end{eqnarray}$$

and take the orbit closure $\overline{B_{n}\cdot \hat{p}_{\tilde{S}}}$ in $\hat{{\mathcal{W}}}_{\tilde{S}}$ . If $\unicode[STIX]{x1D70B}_{\mathbf{u}}:\hat{{\mathcal{W}}}_{\tilde{S}}\rightarrow \prod _{u\in \mathbf{u}}\mathbb{P}_{u}^{1}$ denotes the projection, then

$$\begin{eqnarray}\hat{{\mathcal{B}}}_{\mathbf{u}}=\unicode[STIX]{x1D70B}_{\mathbf{u}}^{-1}\biggl(\mathop{\prod }_{u\in \mathbf{u}}[1:0]\biggr)\end{eqnarray}$$

is a Zariski closed subset of $\hat{{\mathcal{W}}}_{\tilde{S}}$ and by definition its image under the projection $\unicode[STIX]{x1D70B}:\hat{{\mathcal{W}}}_{\tilde{S}}\rightarrow {\mathcal{W}}_{\tilde{S}}$  is equal to ${\mathcal{B}}_{\mathbf{u}}$ . This image is, however, constructible as the image of a closed set. Even more, since the projection is a closed morphism, it takes Zariski closed subsets onto Zariski closed sets. The $B_{n}$ -invariance of ${\mathcal{B}}_{\mathbf{u}}=\unicode[STIX]{x1D70B}(\unicode[STIX]{x1D70B}_{\mathbf{u}}^{-1}(\prod _{u\in \mathbf{u}}[1:0]))$ follows from $B_{n}$ -equivariance of the projections and $B_{n}$ -invariance of the point $\prod _{u\in \mathbf{u}}[1:0]$ (that is, this is a $B_{n}$ -fixed point).◻

Proof. Assume $\unicode[STIX]{x1D703}_{i}<u_{1}$ but $\lim _{m\rightarrow \infty }|b^{(m)}e_{i}|\rightarrow \infty$ and $i$ is the smallest index with this property. Then $e_{j}^{\infty }=\lim _{m\rightarrow \infty }b^{(m)}e_{j}\in \mathbb{C}^{n}$ for $1\leqslant j\leqslant \unicode[STIX]{x1D703}_{i}$ , because in this case $j\leqslant \unicode[STIX]{x1D703}_{i}<i$ so $\unicode[STIX]{x1D703}_{j}<j\leqslant \unicode[STIX]{x1D703}_{i}<u_{1}$ and the limit exists by the minimality of  $i$ . Then

$$\begin{eqnarray}p_{S_{i}}^{\infty }=\wedge _{j=1}^{\unicode[STIX]{x1D703}_{i}}e_{j}^{\infty }\wedge \lim _{m\rightarrow \infty }b^{(m)}e_{i}\end{eqnarray}$$

does not exist, a contradiction. ◻

Proof. Let $p^{\infty }=\lim _{m\rightarrow \infty }b^{(m)}p_{\tilde{S}}\in {\mathcal{B}}_{u}$ be a point defined by a normalised sequence $(b^{(m)})$ with $\operatorname{VSpec}(b^{(m)})=\{u_{1}<\cdots <u_{s}\}$ . Assume that $u_{s}>\max _{1\leqslant i\leqslant n}\unicode[STIX]{x1D703}_{i}$ . Then

$$\begin{eqnarray}V=\{1,\ldots ,n\}\setminus \{u_{s}\}=\mathop{\bigcup }_{i\in \{1,\ldots ,n\}\setminus \{u_{s}\}}S_{i}\in \tilde{S}\end{eqnarray}$$

and the coefficient of $\wedge _{i\in V}e_{i}$ in $p_{V}^{\infty }$ is

$$\begin{eqnarray}p_{V}^{\infty }[\wedge _{i\in V}e_{i}]=\lim _{m\rightarrow \infty }\mathop{\prod }_{i\in V}b_{ii}^{(m)}=\lim _{m\rightarrow \infty }\frac{1}{b_{u_{s}u_{s}}^{(m)}}=\infty ,\end{eqnarray}$$

a contradiction. Here we used that $b^{(m)}\in \operatorname{SL}_{n}$ and hence $\prod _{i=1}^{n}b_{ii}^{(m)}=1$ for all  $m$ .◻

Proof. By definition we can write $p^{\infty }=\lim _{m\rightarrow \infty }b^{(m)}p_{\tilde{S}}=\bigoplus _{U\in \tilde{S}}p_{U}^{\infty }\in {\mathcal{B}}_{\mathbf{u}}$ as a limit point where $(b^{(m)})$ is normalised and $\operatorname{VSpec}(b^{(m)})=\mathbf{u}$ . By definition $Z\in {\mathcal{V}}(p^{\infty })_{\text{min}}$ satisfies the following properties:

1. (i) $Z\in {\mathcal{V}}(p^{\infty })$ , that is, $p_{Z}^{\infty }\neq 0$ and $\unicode[STIX]{x1D703}_{Z}\geqslant u_{1}$ ;

2. (ii) if $p_{U}^{\infty }\neq 0$ and $\unicode[STIX]{x1D703}_{U}\geqslant u_{1}$ for some $U\in \tilde{S}$ , then $\unicode[STIX]{x1D703}_{U}\geqslant \unicode[STIX]{x1D703}_{Z}$ holds;

3. (iii) if $U\in \tilde{S}$ , $\unicode[STIX]{x1D703}_{U}\geqslant u_{1}$ and $U\subsetneq Z$ , then $p_{U}^{\infty }=0$ .

Here (ii) and (iii) together say that $Z\in {\mathcal{V}}(p^{\infty })_{\text{min}}$ . Assume there is a $V\in {\mathcal{V}}(p^{\infty })$ such that $[p_{Z}^{\infty }]\nsubseteq [p_{V}^{\infty }]$ . By definition

$$\begin{eqnarray}p_{Z}^{\infty }=\lim _{m\rightarrow \infty }\wedge _{z\in Z}b^{(m)}e_{z}\quad \text{and}\quad p_{V}^{\infty }=\lim _{m\rightarrow \infty }\wedge _{v\in V}b^{(m)}e_{v},\end{eqnarray}$$

therefore $Z\subset V$ would imply that $[p_{Z}^{\infty }]\subseteq [p_{V}^{\infty }]$ . So $Z\setminus V$ must be nonempty.

Fix a hermitian form $(\cdot ,\cdot )$ on $\mathbb{C}^{n}$ and let $\unicode[STIX]{x1D70B}_{V}:\mathbb{C}^{n}\rightarrow [p_{V}^{\infty }]$ denote the projection to the subspace $[p_{V}^{\infty }]$ . For $w\in \mathbb{C}^{n}$ let $w^{\bot }=w-\unicode[STIX]{x1D70B}_{V}(w)$ denote the orthogonal component.

In Corollary 4.17 below we show that if we drop any subset $\emptyset \neq \unicode[STIX]{x1D6E4}\subseteq Z\setminus V$ from $Z$ , then the smaller subset $Z\setminus \unicode[STIX]{x1D6E4}$ is still in  $\tilde{S}$ . We claim that

(12) $$\begin{eqnarray}p_{Z\setminus \unicode[STIX]{x1D6E4}}^{\infty }=0\quad \text{for all}~\emptyset \neq \unicode[STIX]{x1D6E4}\subseteq Z\setminus V.\end{eqnarray}$$

Indeed, if $\unicode[STIX]{x1D703}_{Z\setminus \unicode[STIX]{x1D6E4}}\geqslant u_{1}$ , then this is property (iii) above. If $\unicode[STIX]{x1D703}_{Z\setminus \unicode[STIX]{x1D6E4}}<u_{1}$ , then, by Lemma 4.10,

(13) $$\begin{eqnarray}e_{z}^{\infty }=\lim _{m\rightarrow \infty }b^{(m)}e_{z}\in \mathbb{C}^{n}\quad \text{exists for all}~z\in Z\setminus \unicode[STIX]{x1D6E4}.\end{eqnarray}$$

Moreover, $\unicode[STIX]{x1D703}_{Z}\geqslant u_{1}$ and hence $\{1,2,\ldots ,u_{1}\}\subset \{1,2,\ldots ,\unicode[STIX]{x1D703}_{Z}\}\subset Z$ . On the other hand property (ii) tells us that $\unicode[STIX]{x1D703}_{V}\geqslant \unicode[STIX]{x1D703}_{Z}$ and therefore

$$\begin{eqnarray}\{1,2,\ldots ,\unicode[STIX]{x1D703}_{Z}\}\subseteq \{1,2,\ldots ,\unicode[STIX]{x1D703}_{V}\}\subset V\end{eqnarray}$$

which implies that

$$\begin{eqnarray}\{1,2,\ldots ,u_{1}\}\subset Z\setminus \unicode[STIX]{x1D6E4}\end{eqnarray}$$

and therefore

$$\begin{eqnarray}p_{Z\setminus \unicode[STIX]{x1D6E4}}^{\infty }=e_{u_{1}}^{\infty }\wedge (\wedge _{z\in Z\setminus \{\unicode[STIX]{x1D6E4}\cup \{u_{1}\}\}}e_{z}^{\infty })=b_{u_{1}u_{1}}^{\infty }e_{u_{1}}\wedge b_{u_{1}-1u_{1}-1}^{\infty }e_{u_{1}-1}\wedge \cdots \wedge b_{11}^{\infty }e_{1}\wedge (\wedge _{z\in Z\setminus \{\unicode[STIX]{x1D6E4}\cup \{1,\ldots ,u_{1}\}\}}e_{z}^{\infty }).\end{eqnarray}$$

But $b_{u_{1}u_{1}}^{\infty }=0$ and by (13) all other terms are finite so this wedge product is 0 and (12) is proved.

Then (12) implies that

(14) $$\begin{eqnarray}0=\unicode[STIX]{x1D70B}_{V}(p_{Z\setminus \unicode[STIX]{x1D6E4}}^{\infty })=\lim _{m\rightarrow \infty }\wedge _{j\in Z\setminus \unicode[STIX]{x1D6E4}}\unicode[STIX]{x1D70B}_{V}(b^{(m)}\cdot e_{j})\quad \text{for all}~\emptyset \neq \unicode[STIX]{x1D6E4}\subseteq Z\setminus V.\end{eqnarray}$$

Hence

$$\begin{eqnarray}\displaystyle p_{Z}^{\infty } & = & \displaystyle {\lim _{m\rightarrow \infty }\,\bigoplus }_{\unicode[STIX]{x1D6E4}\subseteq Z\setminus V}\,\bigwedge _{j\in \unicode[STIX]{x1D6E4}}b^{(m)}e_{j}^{\bot }\wedge \bigwedge _{j\in Z\setminus \unicode[STIX]{x1D6E4}}\unicode[STIX]{x1D70B}_{V}(b^{(m)}\cdot e_{j})\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D70B}_{V}(p_{Z}^{\infty })\oplus {\lim _{m\rightarrow \infty }\,\bigoplus }_{\emptyset \neq \unicode[STIX]{x1D6E4}\subseteq Z\setminus V}\,\bigwedge _{j\in \unicode[STIX]{x1D6E4}}b^{(m)}e_{j}^{\bot }\wedge \bigwedge _{j\in Z\setminus \unicode[STIX]{x1D6E4}}\unicode[STIX]{x1D70B}_{V}(b^{(m)}\cdot e_{j}).\nonumber\end{eqnarray}$$

By (14) all terms corresponding to nonempty $\unicode[STIX]{x1D6E4}$ vanish and therefore $[p_{Z}^{\infty }]\subseteq [p_{V}^{\infty }]$ unless there is a $z\in Z\setminus V$ such that the limit norm $\lim _{m\rightarrow \infty }|b^{(m)}e_{z}^{\bot }|=\infty$ .

However, $V\cup \{z\}=V\cup S_{z}$ because $z\in Z$ and $S_{z}=\{1,\ldots ,\unicode[STIX]{x1D703}_{z},z\}\subseteq \{1,\ldots ,\unicode[STIX]{x1D703}_{Z},z\}\subseteq \{1,\ldots ,\unicode[STIX]{x1D703}_{V},z\}$ since $\unicode[STIX]{x1D703}_{Z}\leqslant \unicode[STIX]{x1D703}_{V}$ by property (ii). Therefore $V\cup \{z\}\in \tilde{S}$ and then

$$\begin{eqnarray}p_{V\cup \{z\}}^{\infty }=\lim _{m\rightarrow \infty }b^{(m)}e_{z}^{\bot }\wedge p_{V}^{\infty }\end{eqnarray}$$

does not exist (the limit is not finite), which is a contradiction. So $[p_{Z}^{\infty }]\subseteq [p_{V}^{\infty }]$ holds, and Proposition 4.14 is proved.◻

Proof. As $Z_{1}$ and $Z_{2}$ are both in ${\mathcal{V}}(p^{\infty })$ , by Proposition 4.14 we have $[p_{Z_{1}}^{\infty }]\subseteq [p_{Z_{2}}^{\infty }]$ and $[p_{Z_{2}}^{\infty }]\subseteq [p_{Z_{1}}^{\infty }]$ .◻

Proof. It suffices to show that $Z\setminus \{z\}=\bigcup _{i\in Z\setminus \{z\}}S_{i}$ . The direction $\subseteq$ is clear as $i\in S_{i}$ for all  $i$ . For $\supseteq$ note that if $i\in Z\setminus \{z\}$ and $z>\unicode[STIX]{x1D703}_{Z}$ , then

$$\begin{eqnarray}S_{i}=\{1,2,\ldots ,\unicode[STIX]{x1D703}_{i}\}\cup \{i\}\subset \{1,2,\ldots ,\unicode[STIX]{x1D703}_{Z}\}\cup \{i\}\subseteq Z\setminus \{z\}.\Box\end{eqnarray}$$

Proof. Since $\unicode[STIX]{x1D703}_{V}\geqslant \unicode[STIX]{x1D703}_{Z}$ , any $z\in Z\setminus V$ must satisfy $z>\unicode[STIX]{x1D703}_{Z}$ and the statement follows from Lemma 4.16.◻

Proof. To prove (a) assume that $r\notin \mathbf{u}$ and $\unicode[STIX]{x1D703}_{r}<u_{1}$ . Let

$$\begin{eqnarray}p^{\infty }=\lim _{m\rightarrow \infty }b^{(m)}p_{\tilde{S}}=\bigoplus _{U\in \tilde{S}}p_{U}^{\infty }\in {\mathcal{B}}_{\boldsymbol{ u}}^{r}\end{eqnarray}$$

be a limit point such that $\operatorname{VSpec}(b^{(m)})=\mathbf{u}$ and $Z\in {\mathcal{V}}(p^{\infty })_{\text{min}}$ . By Remark 4.21 we can assume that $e_{r}\in [p_{Z}^{\infty }]$ .

According to Lemma 4.10  $\lim _{m\rightarrow \infty }b_{ii}^{(m)}\in \mathbb{C}$ exists whenever $\unicode[STIX]{x1D703}_{i}<u_{1}$ and, since $r\notin \mathbf{u}$ , this limit is nonzero for $i=r$ :

$$\begin{eqnarray}b_{rr}^{\infty }:=\lim _{m\rightarrow \infty }b_{rr}^{(m)}\in \mathbb{C}\setminus \{0\}.\end{eqnarray}$$

Define the modified sequence

(15) $$\begin{eqnarray}\tilde{b}_{ij}^{(m)}=\left\{\begin{array}{@{}ll@{}}b_{rr}^{\infty },\quad & (i,j)=(u_{1},u_{1}),\\ {\displaystyle \frac{1}{b_{rr}^{\infty }}}\cdot b_{rr}^{(m)}\cdot b_{u_{1}u_{1}}^{(m)},\quad & (i,j)=(r,r),\\ {\displaystyle \frac{1}{b_{rr}^{\infty }}}\cdot b_{rj}^{(m)}\cdot b_{u_{1},u_{1}}^{(m)}\quad & \text{for}~i=r~\text{and}~j>r~\text{with}~\unicode[STIX]{x1D703}_{j}\geqslant u_{1},\\ b_{ij}^{(m)}\quad & \text{otherwise.}\end{array}\right.\end{eqnarray}$$

In short, we fix the diagonal entry $b_{u_{1}u_{1}}^{(m)}$ to be the nonzero constant $b_{rr}^{\infty }$ and multiply the entries in the $r$ th row of those columns which cover $u_{1}$ by $(1/b_{rr}^{\infty })b_{u_{1}u_{1}}^{(m)}$ . Then the new sequence still sits in  $\operatorname{SL}_{n}(\mathbb{C})$ and part (a) of Proposition 4.22 follows from the following three statements:

1. (i) $\tilde{p}^{\infty }=\lim _{m\rightarrow \infty }\tilde{b}^{(m)}p_{\tilde{S}}$ exists and therefore the map $\tilde{\unicode[STIX]{x1D70C}}:p^{\infty }\mapsto \tilde{p}^{\infty }$ is well defined;

2. (ii) $\tilde{p}^{\infty }\in {\mathcal{B}}_{\{u_{2},\ldots ,u_{s}\}\cup \{r\}}$ ;

3. (iii) $\tilde{\unicode[STIX]{x1D70C}}:p^{\infty }\mapsto \tilde{p}^{\infty }$ is injective.

To prove (i) and (ii) we first show that

(16) $$\begin{eqnarray}\text{if}~\unicode[STIX]{x1D703}_{V}\geqslant u_{1}\quad \text{then}~\tilde{p}_{V}^{\infty }=p_{V}^{\infty }.\end{eqnarray}$$

Note that in this case $u_{1}\in \{1,2,\ldots ,\unicode[STIX]{x1D703}_{V}\}\subseteq V$ and therefore

$$\begin{eqnarray}p_{V}^{\infty }=\lim _{m\rightarrow \infty }\wedge _{v\in V}b^{(m)}e_{v}=\lim _{m\rightarrow \infty }\biggl(\unicode[STIX]{x1D6F1}_{i=1}^{\unicode[STIX]{x1D703}_{V}}b_{ii}^{(m)}\bigwedge _{i=1}^{\unicode[STIX]{x1D703}_{V}}e_{i}\wedge \bigwedge _{\unicode[STIX]{x1D703}_{V}<v\in V}b^{(m)}e_{v}\biggr).\end{eqnarray}$$

First we study the case when $r\leqslant \unicode[STIX]{x1D703}_{V}$ . If $v>\unicode[STIX]{x1D703}_{V}\geqslant r$ , then $b^{(m)}e_{v}=\tilde{b}^{(m)}e_{v}$ , and therefore the second product remains the same by changing $b^{(m)}$ to $\tilde{b}^{(m)}$ . The product $\unicode[STIX]{x1D6F1}_{i=1}^{\unicode[STIX]{x1D703}_{V}}b_{ii}^{(m)}$ of the first $\unicode[STIX]{x1D703}_{V}$ diagonal entries in $b^{(m)}$ and $\tilde{b}^{(m)}$ are again equal, so the first product does not change either, giving us $\tilde{p}_{V}^{\infty }=p_{V}^{\infty }$ .

Assume now that $r>\unicode[STIX]{x1D703}_{V}$ and recall that $e_{r}\in [p_{Z}^{\infty }]\subset [p_{V}^{\infty }]$ . Hence the term $e_{r}$ must be selected in each term of the expansion of the second product, that is, if $\unicode[STIX]{x1D70B}^{r}:\mathbb{C}^{n}\rightarrow \mathbb{C}e_{r}$ denotes the projection of a vector to the line spanned by $e_{r}$ , then $e_{r}\subset [p_{V}^{\infty }]$ implies that

(17) $$\begin{eqnarray}\displaystyle p_{V}^{\infty } & = & \displaystyle \lim _{m\rightarrow \infty }\mathop{\sum }_{v\in V}\biggl(\unicode[STIX]{x1D70B}^{r}(b^{(m)}e_{v})\wedge \bigwedge _{i\in V\setminus v}b^{(m)}e_{i}\biggr)\nonumber\\ \displaystyle & = & \displaystyle \lim _{m\rightarrow \infty }\unicode[STIX]{x1D6F1}_{i=1}^{\unicode[STIX]{x1D703}_{V}}b_{ii}^{(m)}\bigwedge _{i=1}^{\unicode[STIX]{x1D703}_{V}}e_{i}\wedge \mathop{\sum }_{r\leqslant v\in V}\biggl(b_{rv}^{(m)}e_{r}\wedge \bigwedge _{\unicode[STIX]{x1D703}_{V}<i\in V\setminus \{v\}}b^{(m)}e_{i}\biggr).\end{eqnarray}$$

It is easy to see that if $r\leqslant v$ and $\unicode[STIX]{x1D703}_{v}<u_{1}$ , then the corresponding term in the direct sum on the right-hand side has zero contribution. Indeed, by Lemma 4.10

(18) $$\begin{eqnarray}b_{rv}^{\infty }=\lim _{m\rightarrow \infty }b_{rv}^{(m)}\in \mathbb{C}\quad \text{for}~r\leqslant v,\unicode[STIX]{x1D703}_{v}<u_{1}.\end{eqnarray}$$

On the other hand $v\geqslant r>\unicode[STIX]{x1D703}_{V}$ holds and therefore by Lemma 4.16  $V\setminus \{v\}\in \tilde{S}$ . Moreover, since $\unicode[STIX]{x1D703}_{V}\geqslant u_{1}$ but $\unicode[STIX]{x1D703}_{v}<u_{1}$ , we must have $\unicode[STIX]{x1D703}_{V\setminus \{v\}}\geqslant u_{1}$ . Now, we have the following.

1. – If $p_{V\setminus \{v\}}^{\infty }=0$ , then by (18) we have $b_{rv}^{\infty }e_{r}\wedge p_{V\setminus \{v\}}^{\infty }=0$ .

2. – If $p_{V\setminus \{v\}}^{\infty }\neq 0$ , then by Proposition 4.14  $e_{r}\in [p_{Z}^{\infty }]\subset [p_{V\setminus \{v\}}^{\infty }]$ and therefore by (18) we have $b_{rv}^{\infty }e_{r}\wedge p_{V\setminus \{v\}}^{\infty }=0$ again.

In both cases we get

$$\begin{eqnarray}0=b_{rv}^{\infty }e_{r}\wedge p_{V\setminus \{v\}}^{\infty }=\lim _{m\rightarrow \infty }(\unicode[STIX]{x1D6F1}_{i=1}^{\unicode[STIX]{x1D703}_{V}}b_{ii}^{(m)})\bigwedge _{i=1}^{\unicode[STIX]{x1D703}_{V}}e_{i}\wedge \biggl(b_{rv}^{\infty }e_{r}\wedge \bigwedge _{\unicode[STIX]{x1D703}_{V}<i\in V\setminus \{v\}}b^{(m)}e_{i}\biggr).\end{eqnarray}$$

So in (17) only those terms have nonzero contributions where $\unicode[STIX]{x1D703}_{v}\geqslant u_{1}$ and

$$\begin{eqnarray}p_{V}^{\infty }=\lim _{m\rightarrow \infty }\biggl(\unicode[STIX]{x1D6F1}_{i=1}^{\unicode[STIX]{x1D703}_{V}}b_{ii}^{(m)}\bigwedge _{i=1}^{\unicode[STIX]{x1D703}_{V}}e_{i}\wedge \mathop{\sum }_{\substack{ \unicode[STIX]{x1D703}_{v}\geqslant u_{1} \\ r\leqslant v\in V}}\biggl(b_{rv}^{(m)}e_{r}\wedge \bigwedge _{\unicode[STIX]{x1D703}_{V}<i\in V\setminus \{v\}}b^{(m)}e_{i}\biggr)\biggr).\end{eqnarray}$$

Replacing $b^{(m)}$ with $\tilde{b}^{(m)}$ clearly does not change the right-hand side, so $\tilde{p}_{V}^{\infty }=p_{V}^{\infty }$ is proved for this case too. We have completed the proof of (16).

Next, if $V\in \tilde{S}$ but $\unicode[STIX]{x1D703}_{V}<u_{1}$ , then $e_{v}^{\infty }=\lim _{m\rightarrow \infty }b^{(m)}e_{v}\in \mathbb{C}^{n}$ exists for all $v\in V$ by Lemma 4.10 and therefore

$$\begin{eqnarray}p_{V}^{\infty }=\lim _{m\rightarrow \infty }\wedge _{v\in V}b^{(m)}e_{v}=\wedge _{v\in V}e_{v}^{\infty }.\end{eqnarray}$$

Similarly

$$\begin{eqnarray}\tilde{p}_{V}^{\infty }=\lim _{m\rightarrow \infty }\wedge _{v\in V}\tilde{b}^{(m)}e_{v}=\wedge _{v\in V}\tilde{e}_{v}^{\infty }.\end{eqnarray}$$

where

(19) $$\begin{eqnarray}\tilde{e}_{v}^{\infty }=\lim _{m\rightarrow \infty }\tilde{b}^{(m)}e_{v}=\left\{\begin{array}{@{}ll@{}}e_{v}^{\infty },\quad & v\neq u_{1},r,\\ e_{v}^{\infty }+b_{rr}^{\infty }e_{u_{1}},\quad & v=u_{1},\\ \unicode[STIX]{x1D70B}^{r-1}(e_{r}^{\infty }),\quad & v=r,\end{array}\right.\end{eqnarray}$$

where $\unicode[STIX]{x1D70B}^{r-1}:\mathbb{C}^{n}\rightarrow \operatorname{Span}(e_{1},\ldots ,e_{r-1})$ is the projection. This is because $\lim _{m\rightarrow \infty }b_{u_{1},u_{1}}^{(m)}=0$ by definition and the last coordinate of $b^{(m)}e_{r}$ tends to 0:

$$\begin{eqnarray}\lim _{m\rightarrow \infty }\frac{1}{b_{rr}^{\infty }}\cdot b_{rr}^{(m)}\cdot b_{u_{1}u_{1}}^{(m)}=0.\end{eqnarray}$$

This means that $\tilde{p}_{V}^{\infty }=\lim _{m\rightarrow \infty }\tilde{b}^{(m)}p_{V}$ exists for all $V\in \tilde{S}$ and $\tilde{p}^{\infty }\in {\mathcal{B}}_{\{u_{2},\ldots ,u_{s}\}\cup \{r\}}$ .

Note that, formally, the equation for $\tilde{e}_{v}^{\infty }$ depends on the sequence $b^{(m)}$ and in particular on $b_{rr}^{\infty }$ . However: (i) since $\unicode[STIX]{x1D703}_{r}\leqslant u_{1}$ we have $p_{1,\ldots ,\unicode[STIX]{x1D703}_{r}}^{\infty }\neq 0$ and; (ii)  $r\notin \mathbf{u}$ by assumption and therefore $p_{S_{r}}^{\infty }=p_{1,\ldots ,\unicode[STIX]{x1D703}_{r},r}^{\infty }\in \mathbb{C}\setminus \{0\}$ . Now

$$\begin{eqnarray}b_{rr}^{\infty }=\frac{\lim _{m\rightarrow \infty }b_{11}^{(m)}\ldots b_{\unicode[STIX]{x1D703}_{r}\unicode[STIX]{x1D703}_{r}}^{(m)}b_{rr}^{(m)}}{\lim _{m\rightarrow \infty }b_{11}^{(m)}\ldots b_{\unicode[STIX]{x1D703}_{r}\unicode[STIX]{x1D703}_{r}}^{(m)}}\end{eqnarray}$$

is a rational function of $p^{\infty }$ , where the numerator is a coordinate of $p_{S_{r}}^{\infty }$ and therefore finite by (ii) and the denominator is the coefficient of $p_{1,\ldots ,\unicode[STIX]{x1D703}_{r}}^{\infty }$ and therefore nonzero and finite by (i). This proves the map $p^{\infty }\mapsto \tilde{p}^{\infty }$ is well defined and continuous.

To prove (iii) (the injectivity of $\tilde{\unicode[STIX]{x1D70C}}:p^{\infty }\mapsto \tilde{p}^{\infty }$ ) note that by (16)

$$\begin{eqnarray}p_{V}^{\infty }=\tilde{p}_{V}^{\infty }\quad \text{whenever}~\unicode[STIX]{x1D703}_{V}\geqslant u_{1}\end{eqnarray}$$

so $\tilde{\unicode[STIX]{x1D70C}}$ is the identity (and therefore injective) on these coordinates. It remains to check injectivity on the other coordinates.

Take two points $p^{\infty }\neq (p^{\prime })^{\infty }$ in ${\mathcal{B}}_{\mathbf{u}}^{r}$ such that $p_{V}^{\infty }\neq (p_{V}^{\prime })^{\infty }$ for some $V$ with $\unicode[STIX]{x1D703}_{V}<u_{1}$ . This means that $e_{v}^{\infty }\neq (e_{v}^{\prime })^{\infty }$ for some $v\in V$ satisfying $\unicode[STIX]{x1D703}_{v}<u_{1}$ . Let $v$ be minimal with this property. Then

$$\begin{eqnarray}p_{S_{v}}^{\infty }-(p_{S_{v}}^{\prime })^{\infty }=(e_{v}^{\infty }-(e_{v}^{\prime })^{\infty })\wedge \bigwedge _{i=1}^{\unicode[STIX]{x1D703}_{v}}b_{ii}^{\infty }e_{i}\neq 0\end{eqnarray}$$

and using (19) we have the following cases.

1. – If $v\neq u_{1},r$ , then $\tilde{p}_{S_{v}}^{\infty }-(\tilde{p}_{S_{v}}^{\prime })^{\infty }=p_{S_{v}}^{\infty }-(p_{S_{v}}^{\prime })^{\infty }\neq 0$ .

2. – If $v=u_{1}$ , then $\tilde{p}_{S_{u_{1}}}^{\infty }-(\tilde{p}_{S_{u_{1}}}^{\prime }\!)^{\infty }=p_{S_{u_{1}}}^{\infty }-(p_{S_{u_{1}}}^{\prime }\!)^{\infty }+(b_{rr}^{\infty }-(b_{rr}^{\prime })^{\infty })e_{u_{1}}\wedge \bigwedge _{i=1}^{\unicode[STIX]{x1D703}_{u_{1}}}b_{ii}^{\infty }e_{i}\neq 0$ because $p_{S_{u_{1}}}^{\infty }-(p_{S_{u_{1}}}^{\prime }\!)^{\infty }$ does not contain $e_{u_{1}}$ due to the fact that $b_{u_{1},u_{1}}^{\infty }=0$ .

3. – If $v=r$ , then $\tilde{p}_{S_{r}}^{\infty }-(\tilde{p}_{S_{r}}^{\prime })^{\infty }=\unicode[STIX]{x1D70B}^{r-1}(e_{r}^{\infty }-(e_{r}^{\prime })^{\infty })\wedge \bigwedge _{i=1}^{\unicode[STIX]{x1D703}_{r}}b_{ii}^{\infty }$ , and if this is $0$ , then the $e_{r}$ coordinate of $e_{r}^{\infty }$ and $(e_{r}^{\prime })^{\infty }$ are not equal, that is, $b_{rr}^{\infty }-(b_{rr}^{\prime })^{\infty }\neq 0$ . But then again, as in the previous case we have

   $$\begin{eqnarray}\tilde{p}_{S_{u_{1}}}^{\infty }-(\tilde{p}_{S_{u_{1}}}^{\prime }\!)^{\infty }=p_{S_{u_{1}}}^{\infty }-(p_{S_{u_{1}}}^{\prime }\!)^{\infty }+(b_{rr}^{\infty }-(b_{rr}^{\prime })^{\infty })e_{u_{1}}\wedge \bigwedge _{i=1}^{\unicode[STIX]{x1D703}_{v}}b_{ii}^{\infty }e_{i}\neq 0.\end{eqnarray}$$

In any case, $\tilde{p}^{\infty }$ and $(\tilde{p}^{\prime })^{\infty }$ differ in at least one term, proving injectivity of  $\tilde{\unicode[STIX]{x1D70C}}$ .

Next we prove (b) and (c) of Proposition 4.22 simultaneously. Assume that $r\in \mathbf{u}$ or $\unicode[STIX]{x1D703}_{r}\geqslant u_{1}$ . The problem with this case is that $b_{rr}^{\infty }:=\lim _{m\rightarrow \infty }b_{rr}^{(m)}\in \mathbb{C}\setminus \{0\}$ does not necessarily hold any more and the limit can be $\infty$ or 0. In both cases $\tilde{b}^{(m)}$ is ill defined in (15).

Fix a nonzero $\unicode[STIX]{x1D6FF}\in \mathbb{C}$ and define the sequence

(20) $$\begin{eqnarray}\tilde{b}_{ij}^{(m),\unicode[STIX]{x1D6FF}}=\left\{\begin{array}{@{}ll@{}}\unicode[STIX]{x1D6FF},\quad & (i,j)=(u_{1},u_{1}),\\ {\displaystyle \frac{1}{\unicode[STIX]{x1D6FF}}}b_{rr}^{(m)}\cdot b_{u_{1}u_{1}}^{(m)},\quad & (i,j)=(r,r),\\ {\displaystyle \frac{1}{\unicode[STIX]{x1D6FF}}}b_{rj}^{(m)}\cdot b_{u_{1},u_{1}}^{(m)}\quad & \text{if}~i=r,j>r~\text{and}~\unicode[STIX]{x1D703}_{j}\geqslant u_{1},\\ b_{ij}^{(m)}\quad & \text{otherwise}.\end{array}\right.\end{eqnarray}$$

In short, we increase the diagonal entry $b_{u_{1}u_{1}}^{(m)}$ to be constant $\unicode[STIX]{x1D6FF}$ and multiply the entries in the $r$ th row above the diagonal by $(1/\unicode[STIX]{x1D6FF})b_{u_{1}u_{1}}^{(m)}$ whenever $\unicode[STIX]{x1D703}_{j}\geqslant u_{1}$ . Then the new sequence still sits in $\operatorname{SL}_{n}(\mathbb{C})$ . If $r\in \mathbf{u}$ , then $\operatorname{VSpec}(\tilde{b}^{(m),\unicode[STIX]{x1D6FF}})=\{u_{2},\ldots ,u_{s}\}$ . If $r\notin \mathbf{u}$ and $\unicode[STIX]{x1D703}_{r}\geqslant u_{1}$ , then

$$\begin{eqnarray}\operatorname{VSpec}(\tilde{b}^{(m),\unicode[STIX]{x1D6FF}})=\left\{\begin{array}{@{}ll@{}}\{u_{2},\ldots ,u_{s}\}\quad & \displaystyle \text{if}~\lim _{m\rightarrow \infty }b_{rr}^{(m)}b_{u_{1}u_{1}}^{(m)}\neq 0,\\ \{u_{2},\ldots ,u_{s}\}\cup \{r\}\quad & \displaystyle \text{if}~\lim _{m\rightarrow \infty }b_{rr}^{(m)}b_{u_{1}u_{1}}^{(m)}=0.\end{array}\right.\end{eqnarray}$$

In any case, if the limit exists, then $\tilde{p}^{\infty ,\unicode[STIX]{x1D6FF}}=\lim _{m\rightarrow \infty }\tilde{b}^{(m),\unicode[STIX]{x1D6FF}}p_{\tilde{S}}\in {\mathcal{B}}_{u_{2},\ldots ,u_{s}}$ or $\tilde{p}^{\infty ,\unicode[STIX]{x1D6FF}}\in {\mathcal{B}}_{\{u_{2},\ldots ,u_{s}\}\cup \{r\}}$ .

The same argument as for part (a) shows the following.

1. – If $V\in \tilde{S}$ with $\unicode[STIX]{x1D703}_{V}\geqslant u_{1}$ , then $\tilde{p}_{V}^{\infty ,\unicode[STIX]{x1D6FF}}=p_{V}^{\infty }$ .

2. – If $V\in \tilde{S}$ with $\unicode[STIX]{x1D703}_{V}<u_{1}$ , then $\lim _{m\rightarrow \infty }b^{(m)}e_{v}\in \mathbb{C}^{n}$ and $\lim _{m\rightarrow \infty }\tilde{b}^{(m),\unicode[STIX]{x1D6FF}}e_{v}\in \mathbb{C}^{n}$ exists for all $v\in V$ by Lemma 4.10 and therefore

   $$\begin{eqnarray}p_{V}^{\infty }=\lim _{m\rightarrow \infty }\wedge _{v\in V}b^{(m)}e_{v}=\wedge _{v\in V}\lim _{m\rightarrow \infty }b^{(m)}e_{v}\end{eqnarray}$$
   and the same holds with $b^{(m)}$ replaced by $\tilde{b}^{(m),\unicode[STIX]{x1D6FF}}$ . But for $v\in V$ we have
   $$\begin{eqnarray}\lim _{m\rightarrow \infty }\tilde{b}^{(m),\unicode[STIX]{x1D6FF}}e_{v}=\left\{\begin{array}{@{}ll@{}}\displaystyle \lim _{m\rightarrow \infty }b^{(m)}e_{v},\quad & v\neq u_{1},\\ \displaystyle \lim _{m\rightarrow \infty }b^{(m)}e_{u_{1}}+\unicode[STIX]{x1D6FF}e_{u_{1}},\quad & v=u_{1}.\end{array}\right.\end{eqnarray}$$
   Note that the first line holds for $v=r$ too, that is, $\lim _{m\rightarrow \infty }\tilde{b}^{(m),\unicode[STIX]{x1D6FF}}e_{r}=\lim _{m\rightarrow \infty }b^{(m)}e_{r}$ . We assumed that $v\in V$ , but if $\unicode[STIX]{x1D703}_{r}>u_{1}$ , then $r\notin V$ since $\unicode[STIX]{x1D703}_{V}<u_{1}$ by assumption. So the other condition must hold, that is, $r\in \mathbf{u}$ , and then $\tilde{b}_{rr}^{\infty }=b_{rr}^{\infty }=0$ and therefore $\lim _{m\rightarrow \infty }\tilde{b}^{(m),\unicode[STIX]{x1D6FF}}e_{r}=\lim _{m\rightarrow \infty }b^{(m)}e_{r}=\unicode[STIX]{x1D70B}^{r-1}(e_{r}^{\infty })$ .

   In particular, when $\unicode[STIX]{x1D6FF}\rightarrow 0$ the point $\tilde{p}_{V}^{\infty ,\unicode[STIX]{x1D6FF}}$ tends to $p_{V}^{\infty }$ .

In short,

(21) $$\begin{eqnarray}\lim _{\unicode[STIX]{x1D6FF}\rightarrow 0}\tilde{p}^{\infty ,\unicode[STIX]{x1D6FF}}=p^{\infty }\end{eqnarray}$$

and therefore

$$\begin{eqnarray}p^{\infty }\in \left\{\begin{array}{@{}ll@{}}\overline{{\mathcal{B}}_{\{u_{2},\ldots ,u_{s}\}}}\quad & \text{if}~r\in \mathbf{u},\\ \overline{{\mathcal{B}}_{\{u_{2},\ldots ,u_{s}\}}}~\text{or}~\overline{{\mathcal{B}}_{\{u_{2},\ldots ,u_{s}\}\cup \{r\}}}\quad & \text{if}~\unicode[STIX]{x1D703}_{r}\geqslant u_{1}.\end{array}\right.\end{eqnarray}$$

But $p^{\infty }\in {\mathcal{B}}_{\{u_{1},\ldots ,u_{s}\}}$ so $p^{\infty }\notin {\mathcal{B}}_{(u_{2},\ldots ,u_{s})}$ and $p^{\infty }\notin {\mathcal{B}}_{\{u_{2},\ldots ,u_{s}\}\cup \{r\}}$ and we are done.◻

Proof. Consider the map

$$\begin{eqnarray}\unicode[STIX]{x1D711}:\operatorname{SL}_{n}(\mathbb{C})\times {\mathcal{B}}\rightarrow {\mathcal{W}}_{\tilde{S}},\quad \unicode[STIX]{x1D711}(g,w)\mapsto g\cdot w.\end{eqnarray}$$

Choose $w\in {\mathcal{B}}$ and let $T^{i,w}(\unicode[STIX]{x1D706})$ be the corresponding one-parameter subgroup as in Definition 4.23. Since ${\mathcal{B}}\subset {\mathcal{W}}_{\tilde{S}}$ is Borel-invariant, the fibre $\unicode[STIX]{x1D711}^{-1}(g\cdot w)$ contains $(g(bT^{i,w}(\unicode[STIX]{x1D706}))^{-1},(bT^{i,w}(\unicode[STIX]{x1D706}))\cdot w)$ for $b\in B_{n}$ , $\unicode[STIX]{x1D706}\in \mathbb{C}$ . Since $\{T^{i,w}(\unicode[STIX]{x1D706}):\unicode[STIX]{x1D706}\in \mathbb{C}\}\cap B_{n}=\{1\}$ ,

$$\begin{eqnarray}\dim (\{bT^{i,w}(\unicode[STIX]{x1D706}):b\in B_{n},\unicode[STIX]{x1D706}\in \mathbb{C}\})=\dim (B_{n})+1\end{eqnarray}$$

and we get

$$\begin{eqnarray}\displaystyle \dim (\operatorname{Im}(\unicode[STIX]{x1D711})) & = & \displaystyle \dim \operatorname{SL}_{n}(\mathbb{C})+\dim {\mathcal{B}}-\dim (\text{fibre})\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \dim \operatorname{SL}_{n}(\mathbb{C})+\dim \overline{B_{n}\cdot p_{\tilde{S}}}-1-(\dim (B_{n})+1)\nonumber\\ \displaystyle & = & \displaystyle \dim \operatorname{SL}_{n}(\mathbb{C})/U_{S}-2=\dim \overline{\operatorname{SL}_{n}(\mathbb{C})\cdot p_{\tilde{S}}}-2.\Box \nonumber\end{eqnarray}$$

Proof. Let $p^{\infty }\in {\mathcal{B}}_{u}^{r}$ and $Z\in {\mathcal{V}}(p^{\infty })_{\text{min}}$ . By Proposition 4.14 we have:

1. (i) $[p_{Z}^{\infty }]\subset \bigcap _{V\in {\mathcal{V}}(p^{\infty })}p_{V}^{\infty }$ where ${\mathcal{V}}(p^{\infty })=\{U\in \tilde{S}:p_{U}^{\infty }\neq 0,\unicode[STIX]{x1D703}_{U}\geqslant u\}$ ;

2. (ii) $\unicode[STIX]{x1D714}(p^{\infty })=r$ , and hence there is a vector $w=e_{r}+w_{r-1}e_{r-1}+\cdots +w_{1}e_{1}\in [p_{Z}^{\infty }]$ .

By Lemma 4.10

$$\begin{eqnarray}e_{j}^{\infty }=\lim _{m\rightarrow \infty }b^{(m)}e_{j}=\unicode[STIX]{x1D707}_{jj}e_{j}+\cdots +\unicode[STIX]{x1D707}_{1j}e_{1}\quad \text{exists when}~\unicode[STIX]{x1D703}_{j}<u\end{eqnarray}$$

and in particular, since only $b_{uu}^{(m)}$ tends to zero among the diagonal entries, we have

$$\begin{eqnarray}\unicode[STIX]{x1D707}_{jj}\neq 0\quad \text{when}~\unicode[STIX]{x1D703}_{j}<u.\end{eqnarray}$$

Therefore the linear base change

$$\begin{eqnarray}A:\tilde{e}_{j}:=\left\{\begin{array}{@{}ll@{}}e_{j}^{\infty },\quad & j\neq u,\unicode[STIX]{x1D703}_{j}<u,\\ e_{u},\quad & j=u,\\ w,\quad & j=r,\\ e_{j}\quad & \text{otherwise},\end{array}\right.\end{eqnarray}$$

sits in the Borel $B_{\text{SL}_{n}}$ . Since $\{1,\ldots ,u\}\subset Z$ , by (i) and (ii) above we have

(22) $$\begin{eqnarray}\operatorname{Span}(e_{1},\ldots ,e_{u},w)=\operatorname{Span}(\tilde{e}_{1},\ldots ,\tilde{e}_{u},\tilde{e}_{r})\subseteq [p_{Z}^{\infty }]\subseteq [p_{V}^{\infty }]\quad \text{for all}~V\in \tilde{S}~\text{with}~p_{V}^{\infty }\neq 0,\unicode[STIX]{x1D703}_{V}\geqslant u.\end{eqnarray}$$

If $\unicode[STIX]{x1D703}_{V}<u$ , then by Lemma 4.10 again $p_{V}^{\infty }=\wedge _{v\in V}e_{v}^{\infty }$ . But $\lim _{m\rightarrow \infty }b_{uu}^{(m)}=0$ and hence

$$\begin{eqnarray}\lim _{m\rightarrow \infty }b^{(m)}e_{u}\subseteq \operatorname{Span}(e_{1},\ldots ,e_{u-1})=\operatorname{Span}(\tilde{e}_{1},\ldots ,\tilde{e}_{u-1}).\end{eqnarray}$$

Note that since $\unicode[STIX]{x1D703}_{V}<u$ , we have $e_{v}^{\infty }=\tilde{e}_{v}$ for all $v\in V\setminus \{u\}$ . Therefore

(23) $$\begin{eqnarray}[p_{V}^{\infty }]\subset \operatorname{Span}(\tilde{e}_{1},\ldots ,\tilde{e}_{u-1},\tilde{e}_{u+1},\ldots ,\tilde{e}_{n})\quad \text{for all}~V\in \tilde{S}~\text{with}~p_{V}^{\infty }\neq 0,\unicode[STIX]{x1D703}_{V}<u.\end{eqnarray}$$

From (23) and (22) it follows that the one-parameter subgroup

$$\begin{eqnarray}\tilde{T}^{u,\tilde{e}_{r}}(\unicode[STIX]{x1D706}):\tilde{e}_{j}\mapsto \left\{\begin{array}{@{}ll@{}}\tilde{e}_{j},\quad & j\neq u,\\ \tilde{e}_{u}+\unicode[STIX]{x1D706}\tilde{e}_{r},\quad & j=u,\end{array}\right.\quad \text{for}~\unicode[STIX]{x1D706}\in \mathbb{C}\end{eqnarray}$$

stabilises $p^{\infty }$ so $T^{u,A^{-1}\tilde{e}_{r}}$ stabilises $p^{\infty }$ in the old basis, proving that it is $u$ -fixed.◻