v^{0}_{1}$, $v^{0}_{1}\leq w^{\prime}_{1}$, $v^{0}_{2}\leq w_{2}$,
$rv^{0}_{1}\geq v^{0}_{2}$), $v^{0}_{1}=v_{1}$, $f(v,w)\geq 0$, then either
$D_{1}\geq 0$ or $D_{4}>0$.

Proof of Claim. Let $\bar{f}:[0,w^{\prime}_{1}]\to\mathbb{R}_{\geq 0}$ be the
function with graph $C_{f}$. For each fixed $v_{1}$, if we increase $v_{2}$
then both $D_{1}$ and $D_{2}$ will decrease. So we can assume
$v_{2}=\bar{f}(v_{1})$.

Denote $x=v_{1},y=v^{0}_{2}$. Then

$f(v,w)=-x^{2}+rx\bar{f}(x)-\bar{f}(x)^{2}+w^{\prime}_{1}x-w_{2}\bar{f}(x)=0$
$\displaystyle D_{1}$
$\displaystyle=(x^{2}-rxy+y^{2})+(w^{\prime}_{1}-2x)x+(2rx-2\bar{f}(x)-w_{2})y$
$\displaystyle=-x^{2}+rxy+y^{2}+w^{\prime}_{1}x-2\bar{f}(x)y-w_{2}y$
$\displaystyle=f(v,w)+(w_{2}-rx)(\bar{f}(x)-y)+(\bar{f}(x)-y)^{2}=(w_{2}-rx+\bar{f}(x)-y)(\bar{f}(x)-y)$

So $D_{1}\geq 0$ if $y\leq w_{2}-rx+\bar{f}(x)$. On the other hand, if
$y>w_{2}-rx+\bar{f}(x)$, then

$D_{4}/2=x^{2}+y^{2}-x^{2}+(rx-\bar{f}(x)-w_{2})y=y(y+rx-\bar{f}(x)-w_{2})\geq
0,$

and the equality holds if and only if $y=0$, that is, $v_{2}^{0}=0$. But then
$v_{1}^{0}\leq rv^{0}_{2}=0$ forces $v^{0}_{1}=0$. This contradicts the
assumption that $v^{0}=(v^{0}_{1},v^{0}_{2})\neq{\bf 0}$. So either $D_{1}\geq
0$ or $D_{4}>0$. The claim is proved.

(II) Assume $r=2$. Then $(w^{\prime}_{1},w_{2})\in\Phi^{im}_{+}$ implies
$a_{1}=a_{2}=w^{\prime}_{1}=w_{2}>0$; $f(v,w)\geq 0$ together with other
conditions on $v$ imply that $v$ lies in the triangle $OBC$ where $O=(0,0)$,
$B=(a_{1},0)$, $C=(a_{1},a_{1})$. In particular, $v_{1}\geq v_{2}$.

$v_{1}$$v_{2}$$B=(a_{1},0)$$C=(a_{1},a_{1})$$rw^{\prime}_{1}-w_{2}=a_{1}$$f(v,w)\geq
0$$O$

Like (I), it suffices to prove one of the following: $D_{1}\geq 0$, $D_{2}>0$,
$D_{3}>0$, or $D_{4}>0$. Define $v^{0}=(at,bt)$ (with $a=1$),
$Q_{0},A_{2},A_{3},A_{4}$ as in (I).

If $b=a(=1)$, then $Q_{0}=(a-b)^{2}=0$,
$A_{2}=-v_{1}+(2v_{1}-v_{2})=v_{1}-v_{2}$,
$A_{3}=(a_{1}-v_{1})+(2v_{1}-v_{2}-a_{1})=v_{1}-v_{2}$,
$A_{1}=A_{2}+A_{3}=2(v_{1}-v_{2})\geq 0$. So $D_{1}=A_{1}t\geq 0$, we are
done.

If $b\neq a(=1)$, then $Q_{0}=(a-b)^{2}>0$, it suffices to prove that
$A_{2}\geq 0$ or $A_{3}\geq 0$. The proof of Case 1 of (I) applies: we
consider two subcases.

Case 1a: $b<1$. We claim that $A_{3}\geq 0$. It suffices to show that the
slope of $A_{3}=0$ is less than the slope of the diagonal, that is, $2-1/b<1$.
This is obviously true.

Case 1b: $b>1$. We claim that $A_{2}\geq 0$. It suffices to show that the
slope of $A_{2}=0$ is greater than the slope of the diagonal, that is,
$2-1/b>1$. This is again true. ∎

Now we can complete the proof of the main theorem.

###### Proof of Theorem 1.2.

Let $w=(0,a_{1},ra_{1}-a_{2},0)$ (so $w={}^{\phi}w$ and satisfies
$w^{\prime}_{1}-w_{1}=a_{1}$, $w^{\prime}_{2}-w_{2}=a_{2}-r[a_{1}]_{+}$). Let
$v=(a_{1}-q,\,p)$.

Apply Lemma 8.1 to the morphism
$\pi:\tilde{\mathcal{F}}_{v,w}\to\mathbf{E}_{v,w}$. Note that

$2\pi^{-1}(0)-\dim\tilde{\mathcal{F}}_{v,w}=2d(v,w)-\tilde{d}(v,w)=f(v,w)$

Note that $a_{v,0;w}(t)=\sum s_{0,b}t^{b}$ using the notation in (8.1). By
Lemma 8.1, if $f(v,w)<0$, then $a_{v,0;w}(t)=0$, and there is nothing to
prove.

For the rest of the proof, we assume $f(v,w)\geq 0$. Then the degrees of the
Laurent monomials in $a_{v,0;w}(t)$ must be in the interval
$[-f(v,w),f(v,w)]$. Therefore, the degrees of the Laurent monomials in
corresponding $e(p,q)=a_{v,0;w}(\mathbf{v}^{r})$ must lie in the set

$\\{-rf(v,w),-rf(v,w)+r,-rf(v,w)+2r,...,rf(v,w)\\}.$

Compare the max-degrees of both sides of (8.4), we see that the degree of
right side is
$-v_{1}w_{1}-v_{2}w_{2}+v_{2}(w^{\prime}_{2}+rv_{1}-v_{2})+v_{1}(w^{\prime}_{1}-v_{1})=-v_{1}^{2}+rv_{1}v_{2}-v_{2}^{2}+v_{1}(w^{\prime}_{1}-w_{1})+v_{2}(w^{\prime}_{2}-w_{2})=f(v,w)$.
On the other hand, by Lemma 8.8, the summand
$a_{v,v^{\prime};w}P_{-}(v^{\prime},w)$ on the left side of (8.4) has degree
less than $f(v,w)$ for all $v^{\prime}\in\mathrm{Dom}(w)\setminus\\{{\bf
0}\\}$. This forces $\deg(a_{v,{\bf 0};w}P_{-}({\bf 0},w))=f(v,w)$. Since
$P_{-}({\bf 0},w)=1$, we conclude that $\deg a_{v,{\bf 0};w}=f(v,w)$. All
conclusions then follow from Lemma 8.1. ∎

###### Proof of Corollary 1.4.

If $(a_{1},a_{2})\notin\Phi_{+}^{im}$, the statement follows from Theorem 3.1.
If $(a_{1},a_{2})\in\Phi_{+}^{im}$, the statement follows from Theorem 1.2 and
the definition of $R(a_{1},a_{2})$. ∎

###### Proof of Corollary 1.5.

By (7.2), $a_{v,v^{\prime};w}=a_{v^{\perp},0;w^{\perp}}$ where
$v^{\perp}=(v_{1}-v^{\prime}_{1},v_{2}-v^{\prime}_{2})$,
$w^{\perp}=(-v^{\prime}_{1}+rv^{\prime}_{2},w_{1}^{\prime}-v^{\prime}_{1},w_{2}-v^{\prime}_{2},-v^{\prime}_{2}+rv^{\prime}_{1})$.
Correspondingly,
$a_{1}=(w^{\perp})^{\prime}_{1}-w^{\perp}_{1}=w^{\prime}_{1}-rv^{\prime}_{2}$,
$a_{2}=r[a_{1}]_{+}+(w^{\perp})^{\prime}_{2}-w^{\perp}_{2}=r[a_{1}]_{+}+rv^{\prime}_{1}-w_{2}$,
$p=v_{2}^{\perp}=v_{2}-v^{\prime}_{2}$,
$q=[a_{1}]_{+}-v_{1}^{\perp}=[a_{1}]_{+}-v_{1}+v^{\prime}_{1}$. Then
$\mathbf{E}_{v^{\prime},w}$ appears in the decomposition if the condition in
Corollary 1.4 is satisfied. ∎

##  9\. Analogue of Kazhdan-Lusztig polynomials

We define $P(v,w)$ similar to the definition of Kazhdan-Lusztig polynomial:

$P(v,w)=t^{\tilde{d}(v,w)}P_{-}(v,w)=\sum_{k}t^{\tilde{d}(v,w)+k}\dim
H^{k}(i_{0}^{*}IC_{w}(v))\in\mathbb{Z}[t^{\pm}]$

######  Proposition 9.1.

Assume $(v,w)$ is $l$-dominant. Then $P(v,w)$ is a polynomial in
$\mathbb{Z}[t]$ such that:

(i) all coefficients are nonnegative;

(ii) the constant term is $1$;

(iii) $\deg P(v,w)\leq\max(0,\tilde{d}(v,w)-1)$. More precisely, either
$P(v,w)=1$, or $\deg
P(v,w)\leq\tilde{d}(v,w)+\min\big{(}-1,f(v,w),f^{swap}(v,w),g(v,w)\big{)}$.

###### Proof.

(i) follows from the definition. (ii) and (iii) follow immediately from Lemma
8.4. ∎

Use (8.5) we can determine $P_{-}(v,w)$, thus $P(v,w)$, as follows: since

$P_{-}(v,w)=t^{d_{v,w}-\tilde{d}_{v,w}}{w^{\prime}_{2}+rv_{1}\brack
v_{2}}_{t}{w^{\prime}_{1}\brack
v_{1}}_{t}-\sum_{v^{0}\in\mathrm{Dom}(w)\setminus\\{v\\}}a_{v,v^{0};w}P_{-}(v^{0},w)$

we can compute $P_{-}(v,w)$ recursively on $v$, if we can compute
$a_{v,v^{0};w}$. On the other hand,
$a_{v,v^{0};w}=a_{v^{\perp},0;w^{\perp}}=a_{v^{\perp},0;{}^{\phi}(w^{\perp})}$
(the first equality follows from the slice theorem, the second equality
follows from Lemma 7.3), which are coefficients of certain triangular basis
elements according to Lemma 8.5, and the triangular basis can be recursively
computed as described in [4].

######  Example 9.2.

Assume $w=(0,w_{1}^{\prime},w_{2},0)\in\mathbb{Z}_{\geq 0}^{4}$. Using the
recursive method described in the previous paragraph, we compute a few
examples of $P_{-}(v,w)$ and $P(v,w)$:

For $r=2$:

$v$ | $w$ | $P_{-}(v,w)$ | $P(v,w)$
---|---|---|---
$(1,1)$ | any | $1$ | $1$
$(1,1)$ | $(0,1,1,0)$ | $t^{-2}$ | $1$
$(1,1)$ | $(0,1,i,0),i\geq 2$ | $t^{-i-1}+t^{-i+1}$ | $1+t^{2}$
$(1,1)$ | $(0,2,2,0)$ | $t^{-4}+2t^{-2}$ | $1+2t^{2}$
$(1,1)$ | $(0,2,i,0),i\geq 3$ | $t^{-i-2}+2t^{-i}+t^{-i+2}$ | $1+2t^{2}+t^{4}$
$(1,1)$ | $(0,3,3,0)$ | $t^{-6}+2t^{-4}+2t^{-2}$ | $1+2t^{2}+2t^{4}$
$(1,1)$ | $(0,3,i,0),i\geq 4$ | $t^{-i-3}+2t^{-i-1}+2t^{-i+1}+t^{-i+3}$ | $1+2t^{2}+2t^{4}+t^{6}$

For $r=3$:

$v$ | $w$ | $P_{-}(v,w)$ | $P(v,w)$
---|---|---|---
$(1,1)$ | $(0,2,3,0)$ | $t^{-6}+2t^{-4}+2t^{-2}$ | $1+2t^{2}+2t^{4}$
$(1,1)$ | $(0,3,3,0)$ | $t^{-7}+2t^{-5}+3t^{-3}+t^{-1}$ | $1+2t^{2}+3t^{4}+t^{6}$
$(1,1)$ | $(0,4,4,0)$ | $t^{-9}+2t^{-7}+3t^{-5}+3t^{-3}+t^{-1}$ | $1+2t^{2}+3t^{4}+3t^{6}+t^{8}$
$(1,2)$ | $(0,4,4,0)$ | $t^{-13}+2t^{-11}+3t^{-9}+3t^{-7}+2t^{-5}+t^{-3}$ | $1+2t^{2}+3t^{4}+3t^{6}+2t^{8}+t^{10}$
$(2,2)$ | $(0,4,4,0)$ |  $t^{-20}+2t^{-18}+5t^{-16}+6t^{-14}+8t^{-12}+6t^{-10}$ $+5t^{-8}+2t^{-6}+t^{-4}$ |  $1+2t^{2}+5t^{4}+6t^{6}+8t^{8}+6t^{10}$ $+5t^{12}+2t^{14}+t^{16}$

We expect that $P(v,w)$ to always be unimodal. Note that the last example
shows it is not necessarily symmetric or log-concave. It would be interesting
to look for a combinatorial model for $P(v,w)$.

##  10\. Examples

The examples in §10.1 are the triangular basis element $C[3,4]$ and $C[2,8]$
in $\mathcal{A}_{\mathbf{v}}(3,3)$. In §10.2 we use the geometry of the
Nakajima’s quiver varieties to compute $\chi(L(w))$ for $w=(0,1,1,0)$ and
$r\geq 2$ to show that it coincides with a triangular basis element. In §10.3
we use an example to illustrate the idea of the proof of the main theorem. In
§10.4, we give an example of Corollary 1.5.

###  10.1. Two examples of triangular basis elements

Consider the skew-symmetric cluster algebra $\mathcal{A}_{\mathbf{v}}(3,3)$.

(a) $C[3,4]$. Using computer, we get
$C[3,4]=X^{\left(-3,-4\right)}+\left(\mathbf{v}^{6}+1+\mathbf{v}^{-6}\right)X^{\left(-3,-1\right)}+\left(\mathbf{v}^{6}+1+\mathbf{v}^{-6}\right)X^{\left(-3,2\right)}+X^{\left(-3,5\right)}+\left(\mathbf{v}^{9}+\mathbf{v}^{3}+\mathbf{v}^{-3}+\mathbf{v}^{-9}\right)X^{\left(0,-4\right)}+\left(\mathbf{v}^{6}+1+\mathbf{v}^{-6}\right)X^{\left(0,-1\right)}+\left(\mathbf{v}^{12}+\mathbf{v}^{6}+2+\mathbf{v}^{-6}+\mathbf{v}^{-12}\right)X^{\left(3,-4\right)}+X^{\left(3,-1\right)}\\\
+\left(\mathbf{v}^{9}+\mathbf{v}^{3}+\mathbf{v}^{-3}+\mathbf{v}^{-9}\right)X^{\left(6,-4\right)}+X^{\left(9,-4\right)}$.

Below the table for the coefficients $e(p,q)$. We see that the max-degrees of
the coefficients match the corresponding values of $D(p,q)$ in Figure 9.

$q$$p$ | $0$ | $1$ | $2$ | $3$ | $4$
---|---|---|---|---|---
3 | $1$ |  |  |  |
2 | $\left(\mathbf{v}^{6}+1+\mathbf{v}^{-6}\right)$ |  |  |  |
1 | $\left(\mathbf{v}^{6}+1+\mathbf{v}^{-6}\right)$ | $\left(\mathbf{v}^{6}+1+\mathbf{v}^{-6}\right)$ | $1$ |  |
0 | $1$ | $\left(\mathbf{v}^{9}+\mathbf{v}^{3}+\mathbf{v}^{-3}+\mathbf{v}^{-9}\right)$ | $\left(\mathbf{v}^{12}+\mathbf{v}^{6}+2+\mathbf{v}^{-6}+\mathbf{v}^{-12}\right)$ | $\left(\mathbf{v}^{9}+\mathbf{v}^{3}+\mathbf{v}^{-3}+\mathbf{v}^{-9}\right)$ | $1$

091290$-15$660$-12$6$-3$$-18$0$-18$$-12$$q$$p$

Figure 9. Left: some values of $D(p,q)$; the shaded region is
$\\{(p,q)\in[0,4]\times[0,3]\ |\ D(p,q)\geq 0\\}$. Middle: the front view of
$D(p,q)$ for $(p,q)\in R(3,4)$. Right: the top view of the the graph of
$D(p,q)$ for $(p,q)\in R(3,4)$.

(b) $C[2,8]$. Using computer, we get
$C[2,8]=X^{\left(-2,-8\right)}+\left(\mathbf{v}^{3}+\mathbf{v}^{-3}\right)X^{\left(-2,-5\right)}+\cdots+X^{\left(22,-8\right)}$
.

Again, the max-degrees of the coefficients match the corresponding values of
$D(p,q)$; see Figure 10.

021364548453621031521211530$3$$0$$q$$p$

Figure 10. Left: some values of $D(p,q)$; the shaded region is
$\\{(p,q)\in[0,8]\times[0,2]\ |\ D(p,q)\geq 0\\}$. Middle: the front view of
$D(p,q)$ for $(p,q)\in R(2,8)$. Right: the top view of the the graph of
$D(p,q)$ for $(p,q)\in R(2,8)$.

###  10.2. An example to illustrate that $\\{\chi(L(w))\\}$ is identical with
the triangular basis

This example is to support Lemma 7.3. Let $r\geq 2$, $w=(0,1,1,0)$. For this
$w$, there are only two possibilities for $v^{0}$, namely $(0,0)$ and $(1,1)$,
such that $(v^{0},w)$ is $l$-dominant. The stratification of
$\mathbf{E}_{v,w}$ is:

$\mathbf{E}_{v,w}=\mathbf{E}^{\circ}_{(0,0),w}\cup\mathbf{E}^{\circ}_{(1,1),w}=\\{0\\}\cup(\mathbb{C}^{r}\setminus\\{0\\}).$

Since both strata are simply connected (if $r\geq 2$), the local systems
appeared in the BBDG decomposition must be trivial. We can easily check that
$a_{v,0;w}$ is nonempty for

$v=(0,0),(1,0),(1,1),(1,2),\dots,(1,r),$

and the corresponding $\tilde{\mathcal{F}}(v,w)$ are

a point, a point, $\widetilde{\mathbb{C}^{r}}$, the tautological bundle ${\rm
Taut}(Gr_{2}(\mathbb{C}^{r}))$,$\dots$, ${\rm Taut}(Gr_{r}(\mathbb{C}^{r}))$.

Consider $v=(1,i)$. Then $w-C_{q}v=(-1+ri,0,1-i,-i+r)$.

For $i\geq 1$, the BBDG decomposition is of the following form (note that
$\dim{\rm
Taut}(Gr_{i}(\mathbb{C}^{r}))=\dim(Gr_{i}(\mathbb{C}^{r}))+r=i(r-i)+i$):

$\pi_{*}(\mathbb{Q}_{{\rm
Taut}(Gr_{i}(\mathbb{C}^{r}))}[-i(r-i)-i])=(A)\mathbb{Q}_{\mathbb{C}^{r}}[-r]\oplus(B)\mathbb{Q}_{0}.$

Since $\pi^{-1}(x)\cong Gr_{i-1}(\mathbb{C}^{r-1})$ for $x\neq 0$, we get
$A={r-1\brack i-1}_{\mathbf{v}}$. Take cohomology over the point $0$, we get

${r\brack i}_{\mathbf{v}}\mathbf{v}^{i}={r-1\brack
i-1}_{\mathbf{v}}\mathbf{v}^{r}+B.$

Therefore

$a_{(1,i),0;w}=B={r\brack i}_{\mathbf{v}}\mathbf{v}^{i}-{r-1\brack
i-1}_{\mathbf{v}}\mathbf{v}^{r}={r-1\brack i}_{\mathbf{v}}.$

Then

$\chi(L(w))=X^{(-1,1)}+X^{(-1,-r+1)}+{r-1\brack
1}_{\mathbf{v}^{r}}X^{(r-1,r+1)}+\cdots+{r-1\brack
r-1}_{\mathbf{v}^{r}}X^{((r-1)r-1,-r+1)}$

As we expected, this coincides with the triangular basis element $C[1,r-1]$,
which also equals to the greedy element $X[1,r-1]$. It specializes to
$\frac{x_{2}^{r}+(1+x_{1}^{r})^{r-1}}{x_{1}x_{2}^{r-1}}$ when setting $q=1$.

###  10.3. An example to illustrate the proof of the main theorem

Let $r=3$, we illustrate that for $(a_{1},a_{2})=(4,8)$, $(p,q)=(2,2)$, our
strategy to show that the maximal degree of $e(2,2)$ is $\mathbf{v}^{D}$ where
$D=D(p,q)=r(a_{1}q+a_{2}p-p^{2}-rpq-q^{2})=3(4\times 2+8\times 2-2^{2}-3\times
2\times 2-2^{2})=12$.

By Lemma 8.5, we have $w^{\prime}_{1}-w_{1}=a_{1}=4$ and
$w^{\prime}_{2}-w_{2}=a_{2}-r[a_{1}]_{+}=-4$. Letting
$w_{1}=w^{\prime}_{2}=0$, we get $w=(0,4,4,0)$; also,
$v=([a_{1}]_{+}-q,p)=(2,2)$. Note that $(v,w)$ is $l$-dominant and
$w\in\Phi^{im}_{+}$.

The following $v^{\prime}$ satisfies that $(v^{\prime},w)$ is $l$-dominant and
$v^{\prime}\leq v$: $(0,0),(1,1),(1,2),(2,1),(2,2)$. We compute

$d_{v,w}=v_{2}(w_{2}^{\prime}+rv_{1}-v_{2})+v_{1}(w_{1}^{\prime}-v_{1})=2(0+3\times
2-2)+2(4-2)=12$,

$\tilde{d}_{v,w}=v_{2}(w_{2}+w_{2}^{\prime}+rv_{1}-v_{2})+v_{1}(w_{1}+w_{1}^{\prime}-v_{1})=2(4+0+3\times
2-2)+2(0+4-2)=20$.

So (8.4) becomes

(10.1) $\sum_{v^{\prime}}a_{v,v^{\prime};w}P_{-}(v^{\prime},w)=t^{-8}{6\brack
2}_{t}{4\brack 2}_{t}=t^{4}+\textrm{lower degree terms}$

Below is a table for the each term. Note that all except the first one have
max-degree $<4$. Then (10.1) forces the max-degree term of $a_{v,0;w}$ to be
$t^{4}$.

$v^{\prime}$ | $a_{v,v^{\prime};w}$ | $P_{-}(v,w)$ | max-deg of $a_{v,v^{\prime};w}P_{-}(v,w)$
---|---|---|---
$(0,0)$ | $t^{4}+2t^{2}+4+2t^{-2}+t^{-4}$ | $1$ | $4$
$(1,1)$ | $t+t^{-1}$ | $t^{-1}+3t^{-3}+\cdots+t^{-9}$ | $0$
$(1,2)$ | $0$ | $t^{-3}+2t^{-5}+\cdots+t^{-13}$ | $-\infty$
$(2,1)$ | $t+t^{-1}$ | $t^{-3}+2t^{-5}+\cdots+t^{-13}$ | $-2$
$(2,2)$ | $1$ | $t^{-4}+2t^{-6}+\cdots+t^{-20}$ | $-4$

###  10.4. An example to Corollary 1.5

We illustrate Corollary 1.5 by a special example where the morphism
$\pi:{\tilde{\mathcal{F}}_{v,w}}\to\mathbf{E}_{v,w}$ is not necessarily
birational, and we shall examine the generic fiber.

Consider $w=(0,w^{\prime}_{1},w_{2},0)$ with $rw_{2}\geq w^{\prime}_{1}$ and
$rw^{\prime}_{1}\geq w_{2}$ (that is, Case 1 in Figure 2). We assume
$v=(v_{1},v_{2})$ satisfying $0\leq v_{1}\leq w^{\prime}_{1}$ and $w_{2}\leq
v_{2}\leq rv_{1}$. Let $v^{\prime}=\bar{v}=(v_{1},w_{2})$ (Lemma 4.5). By
(1.4), $a_{1}=w^{\prime}_{1}-rw_{2}\leq 0$, $a_{2}=rv_{1}-w_{2}\geq 0$,
$p=v_{2}-w_{2}$, $q=0$. The triangular basis element $C[a_{1},a_{2}]$ is a
cluster monomial, so we have the following (for example, by [13, Corollary
3.4]):

$e(p,0)={a_{2}\brack p}_{\mathbf{v}^{r}}={rv_{1}-w_{2}\brack
v_{2}-w_{2}}_{\mathbf{v}^{r}}$

On the other hand, we compute $a^{d}_{v,v^{\prime};w}$ directly by considering
the geometry. The morphism
$\pi:{\tilde{\mathcal{F}}_{v,w}}\to\mathbf{E}_{v,w}=\mathbf{E}_{v^{\prime},w}$
is a $\mathcal{M}$-bundle when restricting over the stratum
$\mathbf{E}^{\circ}_{v^{\prime},w}$, where the generic fiber $\mathcal{M}$ is
a $Gr(v_{2}-v^{\prime}_{2},w_{2}^{\prime}+rv_{1}-v^{\prime}_{2})$-bundle over
$Gr(v_{1}-v^{\prime}_{1},w_{1}^{\prime}-v^{\prime}_{1})$ by Proposition 4.7
(b). These two Grassmannians are $Gr(v_{2}-w_{2},rv_{1}-w_{2})$ and
$Gr(0,w^{\prime}_{1}-v_{1})$ respectively. So $\mathcal{M}$ is isomorphic to
$Gr(v_{2}-w_{2},rv_{1}-w_{2})$. Since we know that the local systems appeared
in the BBDG decomposition must be trivial, $a^{d}_{v,v^{\prime};w}$ are
determined by the homology of the fiber $Gr(v_{2}-w_{2},rv_{1}-w_{2})$, more
precisely,

$\sum_{d}a^{d}_{v,v^{\prime};w}t^{d}={rv_{1}-w_{2}\brack v_{2}-w_{2}}_{t}.$

Thus $e(p,0)={rv_{1}-w_{2}\brack
v_{2}-w_{2}}_{\mathbf{v}^{r}}=\sum_{d}a^{d}_{v,v^{\prime};w}\mathbf{v}^{rd}$,
as predicted by Corollary 1.5.

##  11\. Some technical proofs

To ensure that the main theorem’s argument is easily readable, the proofs for
several statements in sections §4 and §5 have been deferred to this section.

###  11.1. Proof of Lemma 4.6

(a) $\mathbf{E}^{\circ}_{v^{\prime},w}\subseteq\mathbf{E}_{w}$ is locally
closed because it is the difference of closed subsets:

$\displaystyle\mathbf{E}^{\circ}_{v^{\prime},w}=$ $\displaystyle\\{({\bf
x},{\bf y})\in\mathbf{E}_{w}\,|\,\mathrm{rank}A\leq
v^{\prime}_{1},\mathrm{rank}B\leq v_{2}^{\prime}\\}-\\{({\bf x},{\bf
y})\in\mathbf{E}_{w}\,|\,\mathrm{rank}A\leq
v^{\prime}_{1}-1,\mathrm{rank}B\leq v_{2}^{\prime}\\}$ $\displaystyle-\\{({\bf
x},{\bf y})\in\mathbf{E}_{w}\,|\,\mathrm{rank}A\leq
v^{\prime}_{1},\mathrm{rank}B\leq v_{2}^{\prime}-1\\}$

To show $\mathbf{E}^{\circ}_{v^{\prime},w}$ is nonsingular, we consider the
following open covering

(11.1) $\mathbf{E}^{\circ}_{v^{\prime},w}=\bigcup U_{I,J}$

where $I\subseteq\\{1,\dots,w_{1}^{\prime}\\}$ and $|I|=v_{1}^{\prime}$,
$J\subseteq\\{1,\dots,w_{2}\\}$ and $|J|=v_{2}^{\prime}$,
$U_{I,J}\subseteq\mathbf{E}^{\circ}_{v^{\prime},w}$ consists of those $({\bf
x},{\bf y})$ such that the $v_{1}^{\prime}$ rows of $A$ indexed by $I$ are
linearly independent, and the $v_{2}^{\prime}$ columns of $B({\bf x},{\bf y})$
indexed by $J$ are linearly independent. We will show that $U_{I,J}$ is
isomorphic to an open subset $\tilde{\mathbb{A}}^{\circ}$ of an affine space
$\tilde{\mathbb{A}}$, hence is nonsingular, which implies that
$\mathbf{E}^{\circ}_{v^{\prime},w}$ is nonsingular.

Without loss of generality, we can assume $I=\\{1,\dots,v_{1}^{\prime}\\}$ and
$J=\\{1,\dots,v_{2}^{\prime}\\}$. Denote

$\tilde{\mathbb{A}}=\mathbb{A}^{(w^{\prime}_{1}-v_{1}^{\prime})\times
v_{1}^{\prime}}\times\mathbb{A}^{v_{2}^{\prime}\times(w_{2}-v_{2}^{\prime})}\times\mathbb{A}^{v_{1}^{\prime}\times
w_{1}}\times\mathbb{A}^{w_{2}^{\prime}\times
v_{2}^{\prime}}\times(\mathbb{A}^{v_{1}^{\prime}\times
v_{2}^{\prime}})^{\oplus r}$

and write elements in this affine space as
$(P,Q,Z_{1}^{\prime},Z_{2}^{\prime},C_{1},\dots,C_{r})$ where each component
is a matrix of indicated size. Then define a morphism

$\displaystyle\varphi:\mathbb{A}^{(w^{\prime}_{1}-v_{1}^{\prime})\times
v_{1}^{\prime}}\times\mathbb{A}^{v_{2}^{\prime}\times(w_{2}-v_{2}^{\prime})}\times\mathbb{A}^{v_{1}^{\prime}\times
w_{1}}\times\mathbb{A}^{w_{2}^{\prime}\times
v_{2}^{\prime}}\times(\mathbb{A}^{v_{1}^{\prime}\times
v_{2}^{\prime}})^{\oplus r}\to\mathbf{E}_{w}$
$\displaystyle(P,Q,Z_{1}^{\prime},Z_{2}^{\prime},C_{1},\dots,C_{r})\mapsto({\bf
x},{\bf y})=\Big{(}x_{1}=\begin{bmatrix}Z_{1}^{\prime}\\\
PZ_{1}^{\prime}\end{bmatrix},x_{2}=\begin{bmatrix}Z_{2}^{\prime}&Z_{2}^{\prime}Q\end{bmatrix},y_{i}=\begin{bmatrix}C_{i}&C_{i}Q\\\
PC_{i}&PC_{i}Q\end{bmatrix}\Big{)}.$

Note that ${\rm im}\varphi\subseteq\mathbf{E}_{v^{\prime},w}$. Indeed, define

$A^{\prime}=[Z_{1}^{\prime},C_{1},\dots,C_{r}]_{\text{hor}},\quad
B^{\prime}=[Z^{\prime}_{2},C_{1},\dots,C_{r}]_{\text{vert}}$

Then letting

$x_{1}=\begin{bmatrix}Z^{\prime}_{1}\\\ PZ^{\prime}_{1}\end{bmatrix},\quad
x_{2}=\begin{bmatrix}Z^{\prime}_{2}&Z^{\prime}_{2}Q\end{bmatrix},\quad
y_{i}=\begin{bmatrix}C_{i}&C_{i}Q\\\ PC_{i}&PC_{i}Q\end{bmatrix}\quad(1\leq
i\leq r),$

the corresponding matrices

(11.2) $A=[x_{1}\ y_{1}\ \cdots
y_{r}]=\begin{bmatrix}Z^{\prime}_{1}&C_{1}&C_{1}Q&\cdots&C_{r}&C_{r}Q\\\
PZ^{\prime}_{1}&PC_{1}&PC_{1}Q&\cdots&PC_{r}&PC_{r}Q\end{bmatrix},\quad
B=\begin{bmatrix}x_{2}\\\ y_{1}\\\ \vdots\\\
y_{r}\end{bmatrix}=\begin{bmatrix}Z^{\prime}_{2}&Z^{\prime}_{2}Q\\\
C_{1}&C_{1}Q\\\ PC_{1}&PC_{1}Q\\\ \vdots&\vdots\\\ C_{r}&C_{r}Q\\\
PC_{r}&PC_{r}Q\end{bmatrix}$

satisfy ${\rm rank}A={\rm rank}A^{\prime}\leq v^{\prime}_{1}$, ${\rm
rank}B={\rm rank}B^{\prime}\leq v^{\prime}_{2}$. Therefore ${\rm
im}\varphi\subseteq\mathbf{E}_{v^{\prime},w}$.

Define $\tilde{\mathbb{A}}^{\circ}$ to be the following open subset of
$\tilde{\mathbb{A}}$:

$\displaystyle\tilde{\mathbb{A}}^{\circ}=\Bigg{\\{}(P,Q,Z^{\prime}_{1},Z^{\prime}_{2},C_{1},\dots,C_{r})\in\tilde{\mathbb{A}}\;\Bigg{|}$
$A^{\prime}=[Z^{\prime}_{1}\ C_{1}\ \cdots\ C_{r}]_{\text{hor}}$ has full row
rank $v^{\prime}_{1}$, and
$\displaystyle\textrm{$B^{\prime}=[Z^{\prime}_{2},C_{1},\cdots,C_{r}]_{\text{vert}}$
has full column rank $v^{\prime}_{2}$.}\Bigg{\\}}$

Then $\varphi$ restricts to a morphism
$\varphi|_{\tilde{\mathbb{A}}^{\circ}}:\ \tilde{\mathbb{A}}^{\circ}\to
U_{I,J}$.

We claim that $\varphi|_{\tilde{\mathbb{A}}^{\circ}}$ is an isomorphism. To
prove this, it suffices to construct the inverse morphism
$\varphi|_{\tilde{\mathbb{A}}^{\circ}}^{-1}:U_{I,J}\to\tilde{\mathbb{A}}^{\circ}$
by sending $(x_{1},x_{2},y_{1},\dots,y_{r})$ to
$(P,Q,Z^{\prime}_{1},Z^{\prime}_{2},C_{1},\dots,C_{r})$ defined as follows.
For a matrix $X$, denote $X_{[1,\dots,a;1,\dots,b]}$ the submatrix of $X$ by
selecting the first $a$ rows and first $b$ columns. Define
$Z^{\prime}_{1}={x_{1}}_{[1,\dots,v^{\prime}_{1};1,\dots,w_{1}]}$,
$Z^{\prime}_{2}={x_{2}}_{[1,\dots,w^{\prime}_{2};1,\dots,v^{\prime}_{2}]}$,
$C_{i}={y_{i}}_{[1,\dots,v^{\prime}_{1};1,\dots,v^{\prime}_{2}]}$. By the
definition of $U_{I,J}$, the top $v^{\prime}_{1}$ rows of $A$ are linearly
independent, thus $A^{\prime}$ is of full rank $v^{\prime}_{1}$. Denote
$\begin{bmatrix}A^{\prime}\\\ A^{\prime\prime}\end{bmatrix}$ the
$w^{\prime}_{1}\times(w_{1}+rv^{\prime}_{2})$-submatrix of $A$ with
$A^{\prime}$ as the top block. Define $P$ to be the unique matrix satisfying
$PA^{\prime}=A^{\prime\prime}$. Since $A^{\prime}$ has full row rank, it has a
right inverse $A^{\prime T}(A^{\prime}A^{\prime T})^{-1}$. Thus,
$P=A^{\prime\prime}A^{\prime T}(A^{\prime}A^{\prime T})^{-1}$ whose entries
are rational functions that are regular on $U_{I,J}$. The matrix $Q$ is
defined similarly and has similar properties. Therefore
$\varphi|_{\tilde{\mathbb{A}}^{\circ}}^{-1}$ is a morphism. It is easy to see
that $\varphi|_{\tilde{\mathbb{A}}^{\circ}}^{-1}$ is indeed the inverse of
$\varphi|_{\tilde{\mathbb{A}}^{\circ}}$. This implies that
$\mathbf{E}^{\circ}_{v^{\prime},w}$ is nonsingular.

The nonemptyness statement is proved in [21, Proposition 4.23]. Here we give
an alternative proof. First, assume $\mathbf{E}^{\circ}_{v^{\prime},w}$ is
nonempty. Then

  * •

$w_{1}^{\prime}-v^{\prime}_{1}\geq 0$ follows from $v^{\prime}_{1}={\rm
rank}A\leq\textrm{(number of rows of $A$)}=w_{1}^{\prime}$.

  * •

$w_{1}-v_{1}^{\prime}+rv_{2}^{\prime}\geq 0$ follows from $v_{1}^{\prime}={\rm
rank}A={\rm rank}[x_{1}\ y_{1}\ \cdots\
y_{r}]\leq\textrm{rank}x_{1}+\sum_{i=1}^{r}\textrm{rank}y_{i}\leq
w_{1}+\sum_{i=1}^{r}{\rm rank}B=w_{1}+rv_{2}^{\prime}$.

  * •

$w_{2}-v_{2}^{\prime}\geq 0$ follows from $v_{2}^{\prime}={\rm
rank}B\leq\textrm{(number of columns of $B$)}=w_{2}$.

  * •

$w_{2}^{\prime}-v_{2}^{\prime}+rv_{1}^{\prime}\geq 0$ follows from
$v_{2}^{\prime}={\rm
rank}B\leq\textrm{rank}x_{2}+\sum_{i=1}^{r}\textrm{rank}y_{i}\leq
w^{\prime}_{2}+\sum_{i=1}^{r}{\rm rank}A=w^{\prime}_{2}+rv_{1}^{\prime}$.

Therefore $v^{\prime}\in\mathrm{Dom}(w)$. Conversely, assume
$v^{\prime}\in\mathrm{Dom}(w)$. Let $I=\\{1,\dots,v_{1}^{\prime}\\}$ and
$J=\\{1,\dots,v_{2}^{\prime}\\}$, we shall show that $U_{I,J}\neq\emptyset$. A
generic choice of $Z_{1}^{\prime},Z_{2}^{\prime},C_{1},\dots,C_{r}$ will make
the matrices $A^{\prime}$ and $B^{\prime}$ full-rank (here we used the fact
that the base field $\mathbb{C}$ is infinite), so
$\mathrm{rank}A^{\prime}=v_{1}^{\prime}$ and
$\mathrm{rank}B^{\prime}=v_{2}^{\prime}$ thanks to the $l$-dominant
assumption. After choosing arbitrary $P,Q$, we obtain a point in
$\tilde{\mathbb{A}}^{\circ}$. Since $\tilde{\mathbb{A}}^{\circ}\cong
U_{I,J}\subseteq\mathbf{E}^{\circ}_{v^{\prime},w}$, we conclude that
$\mathbf{E}^{\circ}_{v^{\prime},w}$ is nonempty.

Now assume that $\mathbf{E}^{\circ}_{v^{\prime},w}$ is nonempty, and we shall
prove that this nonsingular variety is irreducible. For this, it suffices to
find a common point $({\bf x},{\bf y})$ in all $U_{I,J}$. Indeed, a generic
choice of $(P,Q,Z_{1}^{\prime},Z_{2}^{\prime},C_{1},\dots,C_{r})$ will give
such a point. To see this, let $A_{[I;-]}$ be the submatrix of $A$ by
selecting the rows with indices in $I$. Let $\tilde{P}=[I,P]_{\text{vert}}$.
Then

$A_{[I;-]}=\tilde{P}_{[I;-]}\;[Z^{\prime}_{1},C_{1},C_{1}Q,\dots,C_{r},C_{r}Q]_{\text{hor}}$

where the right side is the product of two matrices. Since $P$ is generic, we
can assume $\tilde{P}_{[I;-]}$ is of full rank $v^{\prime}_{1}$ (which equals
its number of columns). Also
$[Z^{\prime}_{1},C_{1},C_{1}Q,\dots,C_{r},C_{r}Q]_{\text{hor}}$ is of full
rank $v^{\prime}_{1}$ (which equals its number of rows). It follows that
$A_{[I;-]}$ has rank $v^{\prime}_{1}$. Similarly, $B_{[-;J]}$, the submatrix
of $B$ by selecting the columns with indices in $J$, has rank
$v^{\prime}_{2}$. Therefore $({\bf x},{\bf y})$ is in $U_{I,J}$.

Since $\mathbf{E}^{\circ}_{v^{\prime},w}$ is covered by open subsets $U_{I,J}$
of affine spaces and is irreducible, it is rational.

(b) To prove that
$\pi^{-1}(\mathbf{E}^{\circ}_{v^{\prime},w})\to\mathbf{E}^{\circ}_{v^{\prime},w}$
is a locally trivial bundle, we show that it is a pullback of another locally
trivial bundle $\mathcal{X}\to\mathcal{F}_{v^{\prime},w}$ that is easier to
deal with. Here is the diagram:

$\textstyle{\pi^{-1}(\mathbf{E}^{\circ}_{v^{\prime},w})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\textstyle{\mathcal{X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{q}$$\textstyle{\mathbf{E}^{\circ}_{v^{\prime},w}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{\mathcal{F}_{v^{\prime},w}}$

where

$\displaystyle\mathcal{X}=\\{(Y_{1},Y_{2},X_{1},X_{2})|(Y_{1},Y_{2})\in\mathcal{F}_{v^{\prime},w},(X_{1},X_{2})\in\mathcal{F}_{v,w},Y_{1}\subseteq
X_{1},Y_{2}\subseteq X_{2}\\}$ $\displaystyle f:({\bf x},{\bf
y})\mapsto(Y_{1},Y_{2})=(\text{im}A({\bf x},{\bf y}),\text{im}B({\bf x},{\bf
y}))$ $\displaystyle q:(Y_{1},Y_{2},X_{1},X_{2})\mapsto(Y_{1},Y_{2})$

We claim that the natural projection
$q:\mathcal{X}\to\mathcal{F}_{v^{\prime},w}$ gives a locally trivial fiber
bundle with fibers isomorphic to $\mathcal{M}$. To prove the claim, note that
$\mathcal{F}_{v^{\prime},w}$ can be covered by open subsets over which the
direct sum of universal subbundles
$S_{Gr(v^{\prime}_{1},W^{\prime}_{1})}\oplus
S_{Gr(v^{\prime}_{2},W^{\prime}_{2}\oplus{W^{\prime}_{1}}^{\oplus r})}$ is
trivial, so we only need to discuss fiberwisely. For fixed $Y_{1},Y_{2}$, we
show that the set $(X_{1},X_{2})$ satisfying $Y_{1}\subseteq X_{1}\subseteq
W^{\prime}_{1}$ and $Y_{2}\subseteq X_{2}\subseteq W^{\prime}_{2}\oplus
X_{1}^{\oplus r}$ is the variety $\mathcal{M}$. Indeed, choose a complement
$Y_{1}^{c}\subseteq W^{\prime}_{1}$ to $Y_{1}$, then $X_{1}\mapsto
X^{\prime}_{1}=X_{1}\cap Y_{1}^{c}$ gives an isomorphism between
$\\{X_{1}|Y_{1}\subseteq X_{1}\subseteq W^{\prime}_{1}\\}$ and
$Gr(v_{1}-v^{\prime}_{1},Y_{1}^{c})\cong
Gr(v_{1}-v^{\prime}_{1},w^{\prime}_{1}-v^{\prime}_{1})$. Choose a complement
$Y_{2}^{c}\subseteq W^{\prime}_{2}\oplus Y_{1}^{\oplus r}$ to $Y_{2}$, then
$\widetilde{Y}_{2}^{c}=Y_{2}^{c}+(0\oplus(X^{\prime}_{1})^{\oplus r})$ is a
complement to $Y_{2}$ as subspaces of $W^{\prime}_{2}\oplus X_{1}^{\oplus r}$,
and $X_{2}\mapsto X^{\prime}_{2}=X_{2}\cap\widetilde{Y}_{2}^{c}$ gives an
isomorphism between $\\{X_{2}|Y_{2}\subseteq X_{2}\subseteq
W^{\prime}_{2}\oplus X_{1}^{\oplus r}\\}$ and

$Gr(v_{2}-v^{\prime}_{2},\widetilde{Y}_{2}^{c})\cong
Gr(v_{2}-v^{\prime}_{2},Y_{2}^{c}+(0\oplus(X^{\prime}_{1})^{\oplus r}))\cong
Gr(v_{2}-v^{\prime}_{2},\mathbb{C}^{w^{\prime}_{2}+rv^{\prime}_{1}-v^{\prime}_{2}}\oplus(X^{\prime}_{1})^{\oplus
r})$

Now observe that $\pi^{-1}(\mathbf{E}^{\circ}_{v^{\prime},w})$ is isomorphic
to the pullback $f^{*}(\mathcal{X})$. Indeed, the fiber of
$f^{*}(\mathcal{X})$ over a point $({\bf x},{\bf
y})\in\mathbf{E}^{\circ}_{v^{\prime},w}$ is isomorphic to the fiber of
$q:\mathcal{X}\to\mathcal{F}_{v^{\prime},w}$ over $f({\bf x},{\bf
y})=(\text{im}A({\bf x},{\bf y}),\text{im}B({\bf x},{\bf y}))$, which by
definition of $\mathcal{X}$ is the set of $(X_{1},X_{2})$ satisfying
$(X_{1},X_{2})\in\mathcal{F}_{v,w}$, $\text{im}A({\bf x},{\bf y})\subseteq
X_{1},\text{im}B({\bf x},{\bf y})\subseteq X_{2}$. This coincides with the
conditions in the definition of $\tilde{\mathcal{F}}_{v,w}$.

Since the pullback of a locally trivial fiber bundle is again a locally
trivial fiber bundle with the same fiber, (b) is proved.

This completes the proof of Lemma 4.6.

###  11.2. Proof of Lemma 5.3

(1) We will define $\psi$. To fix notations, for a point $q=({\bf x},{\bf y})$
in the domain $U$ of $\psi$, denote $(q^{0},q^{\perp})=\psi(q)$, and denote
$q^{0}=({\bf x}^{0},{\bf
y}^{0})=(x_{1}^{0},x_{2}^{0},y_{1}^{0},\dots,y_{r}^{0})\in U^{0}$,
$q^{\perp}=({\bf x}^{\perp},{\bf
y}^{\perp})=(x_{1}^{\perp},x_{2}^{\perp},y_{1}^{\perp},\dots,y_{r}^{\perp})\in
U^{\perp}$. Recall that the point $p$ determines two matrices as in (4.2),
denoted $A_{p}$ and $B_{p}$. Without loss of generality, assume the first
$v_{1}^{0}$ row vectors of $A_{p}$ (resp. the first $v_{2}^{0}$ column vectors
of $B_{p}$) are linearly independent. Then $A_{p}$ and $B_{p}$ must be in the
form of (11.2), where the the sizes of the blocks are

block | $Z_{1}^{\prime}$ | $Z_{2}^{\prime}$ | $C_{i}$ | $P$ | $Q$
---|---|---|---|---|---
size | $v_{1}^{0}\times w_{1}$ | $w^{\prime}_{2}\times v_{2}^{0}$ | $v_{1}^{0}\times v_{2}^{0}$ | $(w_{1}^{\prime}-v_{1}^{0})\times v_{1}^{0}$ | $v_{2}^{0}\times(w_{2}-v_{2}^{0})$

First, we define the open set $U$ and $U^{0}$. Define the set

$M=\\{1,\dots,w_{1}\\}\cup\bigcup_{i=1}^{r}\\{w_{1}+(i-1)w_{2}+1,\dots,w_{1}+(i-1)w_{2}+v_{2}^{0}\\}.$

The submatrix
$A^{\prime}_{p}=[Z^{\prime}_{1},C_{1},\cdots,C_{r}]_{\text{hor}}=(A_{p})_{[1,\dots,v_{1}^{0};M]}$
satisfies $\mathrm{rank}A^{\prime}_{p}=\mathrm{rank}A_{p}=v_{1}^{0}$, thus
there exists a subset $J\subseteq M$ of cardinality $v_{1}^{0}$ and
$(A_{p})_{[1,\dots,v_{1}^{0};J]}$ is an invertible $v_{1}^{0}\times
v_{1}^{0}$-matrix. Similarly, define the set

$N=\\{1,\dots,w_{2}^{\prime}\\}\cup\bigcup_{i=1}^{r}\\{w_{2}^{\prime}+(i-1)w_{1}^{\prime}+1,\dots,w_{2}^{\prime}+(i-1)w_{1}^{\prime}+v_{1}^{0}\\}.$

There is a subset $I\subseteq N$ of cardinality $v_{2}^{0}$ such that
$(B_{p})_{[I;1,\dots,v_{2}^{0}]}$ is an invertible $v_{2}^{0}\times v_{2}^{0}$
-matrix. Define an open subset $U$ of $\mathbf{E}_{v,w}$ as

$U=\\{q\in\mathbf{E}_{v,w}\ |\ \textrm{ $(A_{q})_{[1,\dots,v_{1}^{0};J]}$ and
$(B_{q})_{[I;1,\dots,v_{2}^{0}]}$ are invertible square matrices}\\}.$

Define $U^{0}=U\cap\mathbf{E}_{v^{0},w}$ which is an open subset of
$\mathbf{E}^{\circ}_{v^{0},w}$ because $q\in U$ implies
$\mathrm{rank}A_{q}\geq v_{1}^{0}$ and $\mathrm{rank}B_{q}\geq v_{2}^{0}$, and
furthermore the equalities must hold by the condition
$q\in\mathbf{E}_{v^{0},w}$.

Second, we construct $q^{0}$ as follows. Denote

$A_{q}=[x_{1}\ y_{1}\ \cdots
y_{r}]=\begin{bmatrix}Z^{\prime}_{1}&C_{1}&C_{1}^{\prime}&\cdots&C_{r}&C_{r}^{\prime}\\\
Z_{1}^{\prime\prime}&C_{1}^{\prime\prime}&C_{1}^{\prime\prime\prime}&\cdots&C_{r}^{\prime\prime}&C_{r}^{\prime\prime\prime}\end{bmatrix},\quad
B_{q}=\begin{bmatrix}x_{2}\\\ y_{1}\\\ \vdots\\\
y_{r}\end{bmatrix}=\begin{bmatrix}Z^{\prime}_{2}&Z^{\prime\prime}_{2}\\\
C_{1}&C_{1}^{\prime}\\\ C_{1}^{\prime\prime}&C_{1}^{\prime\prime\prime}\\\
\vdots&\vdots\\\ C_{r}&C_{r}^{\prime}\\\
C_{r}^{\prime\prime}&C_{r}^{\prime\prime\prime}\end{bmatrix}$

where the sizes of the blocks are:

block | $Z_{1}^{\prime}$ | $Z_{1}^{\prime\prime}$ | $Z_{2}^{\prime}$ | $Z_{2}^{\prime\prime}$
---|---|---|---|---
size | $v_{1}^{0}\times w_{1}$ | $(w_{1}^{\prime}-v_{1}^{0})\times w_{1}$ | $w^{\prime}_{2}\times v_{2}^{0}$ | $w^{\prime}_{2}\times(w_{2}-v_{2}^{0})$
block | $C_{i}$ | $C_{i}^{\prime}$ | $C_{i}^{\prime\prime}$ | $C_{i}^{\prime\prime\prime}$
---|---|---|---|---
size | $v_{1}^{0}\times v_{2}^{0}$ | $v_{1}^{0}\times(w_{2}-v_{2}^{0})$ | $(w_{1}^{\prime}-v_{1}^{0})\times v_{2}^{0}$ | $(w_{1}^{\prime}-v_{1}^{0})\times(w_{2}-v_{2}^{0})$

Since $(A_{q})_{[1,\dots,v_{1}^{0};J]}$ and $(B_{q})_{[I;1,\dots,v_{2}^{0}]}$
are invertible, there are unique $(w_{1}^{\prime}-v_{1}^{0})\times v_{1}^{0}$
-matrix $P$ and $v_{2}^{0}\times(w_{2}-v_{2}^{0})$ -matrix $Q$ such that

(11.3) $(A_{q})_{[-;J]}=\begin{bmatrix}(A_{q})_{[1,\dots,v_{1}^{0};J]}\\\
P(A_{q})_{[1,\dots,v_{1}^{0};J]}\end{bmatrix},\quad(B_{q})_{[I;-]}=\bigg{[}(B_{q})_{[I;1,\dots,v_{2}^{0}]}\;\;(B_{q})_{[I;1,\dots,v_{2}^{0}]}Q\bigg{]}.$

Define $q^{0}$ such that

$A_{q^{0}}=\begin{bmatrix}Z^{\prime}_{1}&C_{1}&C_{1}Q&\cdots&C_{r}&C_{r}Q\\\
PZ_{1}^{\prime}&PC_{1}&PC_{1}Q&\cdots&PC_{r}&PC_{r}Q\end{bmatrix},\quad
B_{q^{0}}=\begin{bmatrix}Z_{2}^{\prime}&Z_{2}^{\prime}Q\\\ C_{1}&C_{1}Q\\\
PC_{1}&PC_{1}Q\\\ \vdots&\vdots\\\ C_{r}&C_{r}Q\\\ PC_{r}&PC_{r}Q\\\
\end{bmatrix}.$

That is, $x_{1}^{0}=\begin{bmatrix}Z^{\prime}_{1}\\\
PZ^{\prime}_{1}\end{bmatrix}$,
$x_{2}^{0}=\begin{bmatrix}Z^{\prime}_{2}&Z^{\prime}_{2}Q\end{bmatrix}$,
$y_{i}^{0}=\begin{bmatrix}C_{i}&C_{i}Q\\\ PC_{i}&PC_{i}Q\end{bmatrix}$ ($1\leq
i\leq r$). It is easy to see that $q^{0}$ is in $U^{0}$. (To check that
$q_{0}\in\mathbf{E}_{v^{0},w}$, note that

$\mathrm{rank}A_{q^{0}}=\mathrm{rank}(\begin{bmatrix}{\bf I}\\\
P\end{bmatrix}\begin{bmatrix}Z_{1}^{\prime}&C_{1}&\cdots&C_{r}\end{bmatrix}\begin{bmatrix}{\bf
I}&0&0&\dots&0&0\\\ 0&{\bf I}&Q&\dots&{\bf
I}&Q\end{bmatrix})=\mathrm{rank}[Z_{1}^{\prime},C_{1},\cdots,C_{r}]_{\rm
hor}\leq v_{1}^{0},$

thus the equality holds since the first $v_{1}^{0}$ rows of $A_{q}$ are
linearly independent; similarly $\mathrm{rank}B_{q^{0}}=v_{2}^{0}$.)

Third, we construct $q^{\perp}$ as follows. Let

$\widetilde{A_{q}}=[\widetilde{P}x_{1},\widetilde{P}y_{1}\widetilde{Q},\cdots,\widetilde{P}y_{r}\widetilde{Q}]_{\rm
hor},\text{ where }\widetilde{P}=\begin{bmatrix}{\bf I}_{v_{1}^{0}}&0\\\
-P&{\bf
I}_{w_{1}^{\prime}-v_{1}^{0}}\end{bmatrix},\quad\widetilde{Q}=\begin{bmatrix}{\bf
I}_{v_{2}^{0}}&-Q\\\ 0&{\bf I}_{w_{2}-v_{2}^{0}}\end{bmatrix}.$

Then $\mathrm{rank}\widetilde{A_{q}}=\mathrm{rank}A_{q}=u_{1}$, and the
submatrix
$(\widetilde{A_{q}})_{[1,\dots,v_{1}^{0};J]}=(A_{q})_{[1,\dots,v_{1}^{0};J]}$,
while the submatrix
$(\widetilde{A_{q}})_{[v_{1}^{0}+1,\dots,w^{\prime}_{1};J]}=0$. (To see this:
for all $j\in J$, let ${\bf v}$ be the $j$-th column of the matrix. If
$j\in\\{1,\dots,w_{1}\\}$, then ${\bf v}$ is the $j$-column of $Px_{1}$, and
by (11.3), the $j$-column of $x_{1}$ is of the form $\begin{bmatrix}{\bf u}\\\
P{\bf u}\end{bmatrix}$, so ${\bf v}=\widetilde{P}\begin{bmatrix}{\bf u}\\\
P{\bf u}\end{bmatrix}=\begin{bmatrix}{\bf I}&0\\\ -P&{\bf
I}\end{bmatrix}\begin{bmatrix}{\bf u}\\\ P{\bf
u}\end{bmatrix}=\begin{bmatrix}{\bf u}\\\ 0\end{bmatrix}$. In the other case,
assume $j\in\\{w_{1}+(i-1)w_{2}+1,\dots,w_{1}+(i-1)w_{2}+v_{2}^{0}\\}$ for
some $1\leq i\leq r$, note that
$\widetilde{P}y_{i}\widetilde{Q}=\begin{bmatrix}{\bf I}&0\\\ -P&{\bf
I}\end{bmatrix}\begin{bmatrix}C_{i}&C_{i}^{\prime}\\\
C_{i}^{\prime\prime}&C_{i}^{\prime\prime\prime}\end{bmatrix}\begin{bmatrix}{\bf
I}&-Q\\\ 0&{\bf I}\end{bmatrix}=\begin{bmatrix}C_{i}&*\\\
-PC_{i}+C_{i}^{\prime\prime}&*\end{bmatrix}$. Since the $j$-th column of
$A_{q}$ is of the form $\begin{bmatrix}{\bf u}\\\ P{\bf u}\end{bmatrix}$, the
vector ${\bf v}$ is of the form $\begin{bmatrix}{\bf u}\\\ -P{\bf u}+P{\bf
u}\end{bmatrix}=\begin{bmatrix}{\bf u}\\\ 0\end{bmatrix}$.)

Note that the space spanned by the top $v_{1}^{0}$ rows of $\widetilde{A_{q}}$
has zero intersection with the space spanned by the rest rows by considering
the columns indexed by $J$. Since
$\mathrm{rank}\widetilde{A_{q}}=\mathrm{rank}A_{q}\leq v_{1}$, and that
$\mathrm{rank}(\widetilde{A_{q}})_{[1,\dots,v_{1}^{0};-]}=v_{1}^{0}$ (because
its $J$ columns is equal to the invertible matrix
$(A_{q})_{[1,\dots,v_{1}^{0};J]}$), we have

$\mathrm{rank}(\widetilde{A_{q}})_{[v_{1}^{0}+1,\dots,w^{\prime}_{1};-]}=\mathrm{rank}\widetilde{A_{q}}-\mathrm{rank}(\widetilde{A_{q}})_{[1,\dots,v_{1}^{0};-]}=u_{1}^{\perp}\leq
v_{1}^{\perp}.$

We define $A_{q}^{\perp}$ to be
$(\widetilde{A_{q}})_{[v_{1}^{0}+1,\dots,w_{1}^{\prime};\\{1,\dots,w_{1}+rw_{2}\\}\setminus
J]}$ with an appropriate rearrangement of the columns; to be more precise:

$A_{q}^{\perp}=\Big{[}(\widetilde{A_{q}})_{[v_{1}^{0}+1,\dots,w_{1}^{\prime};M\setminus
J]}\ \vline\
PC_{1}Q-PC_{1}^{\prime}-C_{1}^{\prime\prime}Q+C_{1}^{\prime\prime\prime}\
\vline\ \cdots\Big{]}$

That is, if we denote $J^{\prime}=\iota_{M}^{-1}(J)$ where
$\iota_{M}:\\{1,\dots,w_{1}+rv_{2}^{0}\\}\to M$ is the order preserving
bijection, then

$\displaystyle x_{1}^{\perp}=\text{obtained from
$[-PZ_{1}^{\prime}+Z_{1}^{\prime\prime},-PC_{1}+C_{1}^{\prime\prime},\cdots,-PC_{r}+C_{r}^{\prime\prime}]_{\rm
hor}$ }$ $\displaystyle\hskip 30.0pt\text{by deleting columns $J^{\prime}$
(which are zero)},$ $\displaystyle
y_{i}^{\perp}=PC_{i}Q-PC_{i}^{\prime}-C_{i}^{\prime\prime}Q+C_{i}^{\prime\prime\prime}\quad(1\leq
i\leq r).$

Note that the column spaces of $A_{q}^{\perp}$ and
$(\widetilde{A_{q}})_{[v_{1}^{0}+1,\dots,w^{\prime}_{1};-]}$ are the same,
which implies

(11.4) $\mathrm{rank}A_{q}^{\perp}=u_{1}^{\perp}$

Similarly, we define

$\displaystyle x_{2}^{\perp}=\text{obtained from
$[-Z_{2}^{\prime}Q+Z_{2}^{\prime\prime},-C_{1}Q+C_{1}^{\prime},\cdots,-C_{r}Q+C_{r}^{\prime}]_{\rm
vert}$}$ $\displaystyle\hskip 30.0pt\text{by deleting rows $I^{\prime}$ (which
are zero)},$

where $I^{\prime}=\iota_{N}^{-1}(I)$ where
$\iota_{N}:\\{1,\dots,w_{2}^{\prime}+rv_{1}^{0}\\}\to N$ is the order
preserving bijection. Like (11.4), we have

(11.5) $\mathrm{rank}B_{q}^{\perp}=u_{2}^{\perp}$

This complete the construction of
$q^{\perp}=(x_{1}^{\perp},x_{2}^{\perp},y_{1}^{\perp},\dots,y_{r}^{\perp})$.

Note that $\psi$ is a morphism because it can be expressed in terms of matrix
additions, multiplications, and inverses, so can be expressed as rational
functions.

(2) We explain that $\psi$ is an isomorphism by constructing its inverse
$\psi^{-1}$. Assume $q^{0},q^{\perp}$ are given. From $q^{0}$ we can recover
$Z_{1}^{\prime},Z_{2}^{\prime},C_{i},P,Q$ by rational functions. From
$q^{\perp}$ we can recover
$Z_{1}^{\prime\prime},Z_{2}^{\prime\prime},C_{i}^{\prime},C_{i}^{\prime\prime},C_{i}^{\prime\prime\prime}$
by rational functions. Thus $\psi^{-1}$ is a morphism and it is easy to see
that it is indeed the two-sided inverse of $\psi$. Thus $\psi$ is an
isomorphism.

(3) We define $\varphi$. Given $(q=({\bf x},{\bf
y}),X_{1},X_{2})\in\pi^{-1}U$, define
$\varphi(q,X_{1},X_{2})=(q^{0},(q^{\perp},X_{1}^{\perp},X_{2}^{\perp}))$,
where $q^{0}$ and $q^{\perp}$ are defined in (1), and
$X_{1}^{\perp},X_{2}^{\perp}$ are defined as follows:

For ${\bf u}\in X_{1}$, define ${\bf u}^{0}=\begin{bmatrix}{\bf I}&0\\\
P&0\end{bmatrix}{\bf u}$ (where $P$ is determined by $q$), ${\bf
u}^{\perp}={\bf u}-{\bf u}^{0}=\begin{bmatrix}0&0\\\ -P&{\bf
I}\end{bmatrix}{\bf u}$ has all its top $v_{1}^{0}$ entries being 0. Define

$X_{1}^{0}=\\{{\bf u}^{0}\ |\ {\bf u}\in X_{1}\\},\quad X_{1}^{\perp}=\\{{\bf
u}^{\perp}\ |\ {\bf u}\in X_{1}\\}=\Big{\\{}\begin{bmatrix}0\\\
-P\alpha+\beta\end{bmatrix}\;\Big{|}\;\begin{bmatrix}{\alpha}\\\
{\beta}\end{bmatrix}\in X_{1}\Big{\\}}$

Then $\dim X_{1}^{0}=v_{1}^{0}$ (since the top $v_{1}^{0}$ rows of $A_{q}$ are
linearly independent which implies $\dim X_{1}^{0}\geq v_{1}^{0}$; meanwhile,
$\dim X_{1}^{0}\leq v_{1}^{0}$ because $\mathrm{rank}\begin{bmatrix}{\bf
I}&0\\\ P&0\end{bmatrix}=v_{1}^{0}$) , $\dim X_{1}^{\perp}=\dim X_{1}-\dim
X_{1}^{0}=v_{1}-v_{1}^{0}=v_{1}^{\perp}$. Note that
$X_{1}^{\perp}=X_{1}\cap(0\oplus\mathbb{C}^{w_{1}-v_{1}^{0}})$ is the set of
vectors in $X_{1}$ with the first $v_{1}^{0}$ entries being 0.

We define $X_{2}^{\perp}$ as follows.

$X_{2}^{\perp}=\Bigg{\\{}\gamma=\begin{bmatrix}\gamma_{1}\\\ \gamma_{2}\\\
\vdots\\\
\gamma_{w_{2}^{\prime}+rw_{1}^{\prime}}\end{bmatrix}\in\widetilde{X}_{2}=P^{*}X_{2},\
{\textrm{such that}}\ \gamma_{i}=0\;{\rm for}\;i\in
I\Bigg{\\}},\;P^{*}=\begin{bmatrix}{\bf I}&0&\cdots&0\\\
0&\widetilde{P}&\cdots&0\\\ \vdots&\vdots&\ddots&\vdots\\\
0&0&\cdots&\widetilde{P}\end{bmatrix}$

We now show that $\dim X_{2}^{\perp}=v_{2}^{\perp}$. Note that
$\dim\widetilde{X}_{2}=\dim X_{2}=v_{2}$. The $I$-rows of $\widetilde{X}_{2}$
is the same as the $I$-rows of $X_{2}$, so they have the full row rank
$v_{2}^{0}$. So there is a subspace $V$ of $\widetilde{X}_{2}$ of dimension
$v_{2}^{0}$ whose $I$-rows are full rank. Then $V\cap X_{2}^{\perp}=0$ and
$V+X_{2}^{\perp}=\widetilde{X}_{2}$, so $\dim
X_{2}^{\perp}=\dim\widetilde{X}_{2}-\dim V=v_{2}-v_{2}^{0}=v_{2}^{\perp}$.

Next we shall show that $X_{2}^{\perp}\subseteq
W_{2}^{\prime\perp}\oplus(X_{1}^{\perp})^{\oplus r}$, where
$W_{2}^{\prime\perp}\cong\mathbb{C}^{w_{2}^{\prime}+rv_{1}^{0}-v_{2}^{0}}$
consists of vectors whose $i$-th coordinates are 0 except for $i\in N\setminus
I$. Assume $P^{*}u\in X_{2}^{\perp}$ for $u=\begin{bmatrix}u_{0}\\\ u_{1}\\\
\vdots\\\ u_{r}\end{bmatrix}\in X_{2}$ where $u_{0}\in W^{\prime}_{1}$,
$u_{i}\in X_{1}$ for $i=1,\dots,r$, we have

$P^{*}u=\begin{bmatrix}u_{0}\\\ \widetilde{P}u_{1}\\\ \vdots\\\
\widetilde{P}u_{r}\end{bmatrix}=\begin{bmatrix}u_{0}\\\ \begin{bmatrix}{\bf
I}&0\\\ 0&0\end{bmatrix}u_{1}\\\ \vdots\\\ \begin{bmatrix}{\bf I}&0\\\
0&0\end{bmatrix}u_{r}\end{bmatrix}+\begin{bmatrix}u_{0}\\\
\begin{bmatrix}0&0\\\ -P&{\bf I}\end{bmatrix}u_{1}\\\ \vdots\\\
\begin{bmatrix}0&0\\\ -P&{\bf I}\end{bmatrix}u_{r}\end{bmatrix}$

where the second vector is in $(X_{1}^{\perp})^{\oplus r}$; the first vector
has zero entries at coordinates except $N\setminus I$, so is in
$W_{2}^{\prime\perp}$. This proves $X_{2}^{\perp}\subseteq
W_{2}^{\prime\perp}\oplus(X_{1}^{\perp})^{\oplus r}$.

(4) To show $\varphi$ is an isomorphism, we observe that its inverse
$\varphi^{-1}$ can be constructed as follows: given
$(q^{0},(q^{\perp},X_{1}^{\perp},X_{2}^{\perp}))$, to construct
$(q,X_{1},X_{2})$, we get $q$ by $\psi^{-1}$, and define

$\displaystyle X_{1}=(\text{ column space of }A_{q^{0}})+X_{1}^{\perp}=(\text{
column space of }\begin{bmatrix}{\bf I}\\\ P\end{bmatrix})+X_{1}^{\perp}$
$\displaystyle X_{2}=(\text{ column space of
}B_{q^{0}})+(P^{*})^{-1}X_{2}^{\perp}$

In the construction of $X_{1}$ (resp. $X_{2}$), the sum is an internal direct
sum since the first $v_{1}^{0}$ entries of vectors in $X_{1}^{\perp}$ are 0
(resp. the $I$-entries of vectors in $(P^{*})^{-1}X_{2}^{\perp}$ are 0). From
the construction of $X_{1}^{\perp}$ and $X_{2}^{\perp}$ it is easy to see that
$\phi$ and $\phi^{-1}$ are indeed inverse to each other.

(5) To check that the diagram commutes:

$\psi\circ\pi(q,X_{1},X_{2})=\psi(q)=(q^{0},q^{\perp})=(1\times\pi)(q^{0},(q^{\perp},X_{1}^{\perp},X_{2}^{\perp}))=(1\times\pi)\circ\varphi(q,X_{1},X_{2})$

(6) To check $\psi(p)=(p,0)$: let $q=p$, and use the same assumption as above,
we need to show $q^{0}=q$ and $q^{\perp}=0$. Since $A_{q}$ has rank
$v_{1}^{0}$, all other rows are linear combinations of its first $v_{1}^{0}$
rows, that is, $A_{q}=\begin{bmatrix}(A_{q})_{[1,\dots,v_{1}^{0};-]}\\\
P^{\prime}(A_{q})_{[1,\dots,v_{1}^{0};-]}\\\ \end{bmatrix}$ for some matrix
$P^{\prime}$. Comparing with (11.3) we see that $P^{\prime}=P$. Thus
$Z_{1}^{\prime\prime}=PZ_{1}^{\prime}$, $C_{i}^{\prime\prime}=PC_{i}$,
$C_{i}^{\prime\prime\prime}=PC_{i}^{\prime}$. Argue similarly for $B_{q}$. We
then conclude that $q=q_{0}$. Next, note that
$\widetilde{A_{q}}=\begin{bmatrix}Z^{\prime}_{1}&C_{1}&0&\cdots&C_{r}&0\\\
0&0&0&\cdots&0&0\end{bmatrix}$, so $A_{q}^{\perp}=0$, and similarly
$B_{q}^{\perp}=0$. Thus $q^{\perp}=0$.

The fact that $\psi({\bf E}^{\circ}_{u,w}\cap U)=U_{0}\times({\bf
E}^{\circ}_{u^{\perp},w^{\perp}}\cap U^{\perp})$, follows from (11.4) and
(11.5).

This completes the proof of Lemma 5.3.

###  11.3. The second proof of trivial local systems

In this subsection we give another proof of Theorem 5.1. We first study simply
connectedness of some varieties.

######  Lemma 11.1.

Let $(v,w)$ be $l$-dominant. Then the quasi-affine variety ${\bf
E}^{\circ}_{v,w}$ is simply connected unless $v=(w^{\prime}_{1},w_{2})$,
$(w^{\prime}_{1},(w^{\prime}_{1}-w_{1})/r)$, or
$((w_{2}-w^{\prime}_{2})/r,w_{2})$.

###### Proof.

If $v=(0,0)$ then ${\bf E}^{\circ}_{v,w}$ is a point, which is simply
connected. In the rest, we assume $v\neq(0,0)$.

Consider the following diagram, where $\pi^{\prime}$ is an isomorphism by
Proposition 4.7 (e).

${\pi^{-1}(\mathbf{E}^{\circ}_{v,w})}$${\widetilde{\mathcal{F}}_{v,w}}$${\mathbf{E}^{\circ}_{v,w}}$${\mathbf{E}_{v,w}}$$\scriptstyle{j}$$\scriptstyle{\pi^{\prime}}$$\scriptstyle{\cong}$$\scriptstyle{\pi}$$\scriptstyle{i}$

Our idea to study the simply connectedness of ${\bf E}^{\circ}_{v,w}$ is:
first show that $\widetilde{\mathcal{F}}_{v,w}$ is simply connected; next,
show that $\pi^{-1}(\mathbf{E}^{\circ}_{v,w})$ is simply connected because the
complement has codimension greater than 1; last, show that
$\mathbf{E}^{\circ}_{v,w}$ is simply connected.

We first show that $\widetilde{\mathcal{F}}_{v,w}$ is simply connected. By
Proposition 4.7 (b), ${\mathcal{F}}_{v,w}=\pi^{-1}(0)$ is a fiber bundle over
a Grassmannian whose fibers are Grassmannian. It is well known that a complex
Grassmannian is simply connected. It follows from the long exact sequence of
homotopy groups that ${\mathcal{F}}_{v,w}$ is simply connected. By Lemma 4.3,
$\widetilde{\mathcal{F}}_{v,w}$ is a vector bundle over ${\mathcal{F}}_{v,w}$,
so $\widetilde{\mathcal{F}}_{v,w}$ is also simply connected.

Next, we compute the codimension of
$\widetilde{\mathcal{F}}_{v,w}\setminus\pi^{-1}(\mathbf{E}^{\circ}_{v,w})$ in
$\widetilde{\mathcal{F}}_{v,w}$. Since

$\widetilde{\mathcal{F}}_{v,w}\setminus\pi^{-1}(\mathbf{E}^{\circ}_{v,w})=\bigcup_{v^{\prime}<v,v^{\prime}\in\mathrm{Dom}(w)}\pi^{-1}({\bf
E}^{\circ}_{v^{\prime},w}).$

So the codimension of the complement of $\pi^{-1}(\mathbf{E}^{\circ}_{v,w})$
is

$\mathrm{Codim}=\min\\{\dim\pi^{-1}({\bf
E}^{\circ}_{v^{\prime},w})-\dim\pi^{-1}({\bf E}^{\circ}_{v^{\prime},w})\\}$

Note the following (where the equalities follow from Proposition 4.7):

$\displaystyle\dim\pi^{-1}({\bf E}^{\circ}_{v^{\prime},w})$
$\displaystyle=\dim({\bf E}^{\circ}_{v^{\prime},w})+\dim
Gr(v_{2}-v^{\prime}_{2},w^{\prime}_{2}+rv_{1}-v^{\prime}_{2})+\dim
Gr(v_{1}-v^{\prime}_{1},w^{\prime}_{1}-v^{\prime}_{1})$
$\displaystyle=\big{(}v^{\prime}_{2}(w_{2}+w_{2}^{\prime}+rv^{\prime}_{1}-v^{\prime}_{2})+v^{\prime}_{1}(w_{1}+w_{1}^{\prime}-v^{\prime}_{1})\big{)}$
$\displaystyle\quad+(v_{2}-v^{\prime}_{2})(w^{\prime}_{2}+rv_{1}-v_{2})+(v_{1}-v^{\prime}_{1})(w^{\prime}_{1}-v_{1})$

We define the following function, which satisfies
$f(v^{\prime})=\dim\pi^{-1}({\bf E}^{\circ}_{v^{\prime},w})$:

$\displaystyle f(x,y)$
$\displaystyle=\big{(}y(w_{2}+w_{2}^{\prime}+rx-y)+x(w_{1}+w_{1}^{\prime}-x)\big{)}$
$\displaystyle\quad+(v_{2}-y)(w^{\prime}_{2}+rv_{1}-v_{2})+(v_{1}-x)(w^{\prime}_{1}-v_{1})$
$\displaystyle=-x^{2}+rxy-y^{2}+(w_{1}+v_{1})x+(w_{2}-rv_{1}+v_{2})y+v_{1}w^{\prime}_{1}+v_{2}w^{\prime}_{2}-v_{1}^{2}+rv_{1}v_{2}-v_{2}^{2}$

with domain being the closed region

$S=\\{(x,y)\in\mathbb{R}^{2}\ |\ 0\leq x\leq v_{1},0\leq y\leq
v_{2},w_{1}-x+ry\geq 0\\}$

Thus

$\mathrm{Codim}=\min\\{f(v)-f(v^{\prime})\ |\
v^{\prime}<v,v^{\prime}\in\mathrm{Dom}(w)\\}$

Compute the partial derivatives:

$f_{x}=(w_{1}-x+ry)+(v_{1}-x)\geq 0,\quad
f_{y}=(w_{2}-y)+(v_{2}-y)-r(v_{1}-x),$

and note that $w_{2}\geq v_{2}\geq y$, and the directional derivative of $f$
along the direction $\langle r,1\rangle$ is

$\langle f_{x},f_{y}\rangle\cdot\langle
r,1\rangle=r(w_{1}-x+ry)+(w_{2}-y)+(v_{2}-y)\geq 0,$

it is then easy to check that $f$ is strictly increasing in $S$ along each
line of slope $0$ or $1/r$, an along the line $x=v_{1}$.

Next we give a lower bound for $f(v)-f(v^{\prime})$. Define a path $C$ from
$v^{\prime}$ to $v$ as follows: start $v^{\prime}$, walk along direction
$(r,1)$; if it hits the point $v$ we say the path is of type $1$; if hits the
horizontal line $y=v_{2}$ then walk east to $v$ and we say $C$ is of type 2;
if hits the vertical line $x=v_{1}$ then walk north to $v$ and we say $C$ is
of type 3. The fundamental theorem for line integrals asserts
$f(v)-f(v^{\prime})=\int_{C}\langle f_{x},f_{y}\rangle\cdot d{\bf r}$.

For a type 2 path $C$, it must contains the point $v^{\prime\prime}=v-(1,0)$.

$f(v)-f(v^{\prime})\geq
f(v)-f(v^{\prime\prime})=\int_{v_{1}-1}^{v_{1}}f_{x}dx=w_{1}-v_{1}+rv_{2}+1\geq
1$

A necessary condition for the equality $f(v)-f(v^{\prime})=1$ to hold is
$w_{1}-v_{1}+rv_{2}=0$ and $(v_{1}-1,v_{2})$ is $l$-dominant.

For a path $C$ of type 1 or 3, let $v^{\prime\prime}$ be the point where it
hits the vertical line $x=v_{1}$. Then

$\displaystyle f(v)-f(v^{\prime})$
$\displaystyle=(f(v)-f(v^{\prime\prime}))+(f(v^{\prime\prime})-f(v^{\prime}))$
$\displaystyle=\int_{v^{\prime\prime}_{2}}^{v_{2}}(w_{2}+v_{2}-2y)dy+\int_{v^{\prime}_{2}}^{v^{\prime\prime}_{2}}r(w_{1}-v^{\prime}_{1}+rv^{\prime}_{2})+(w_{2}+v_{2}-2y)dy$
$\displaystyle\geq\int_{v^{\prime}_{2}}^{v_{2}}(w_{2}+v_{2}-2y)dy\geq\int_{v_{2}-1}^{v_{2}}(w_{2}+v_{2}-2y)dy=w_{2}-v_{2}+1\geq
1$

For the equality $f(v)-f(v^{\prime})=1$ to hold, we need $v_{2}=w_{2}$; in
other cases, $f(v)-f(v^{\prime})\geq 2$.

So, $\pi^{-1}({\bf E}^{\circ}_{v,w})$ is simply connected unless either
“$w_{1}-v_{1}+rv_{2}=0$ and $(v_{1}-1,v_{2})$ is $l$-dominant” or
$v_{2}=w_{2}$.

Now we make an important observation: if we simultaneously swap the role of
$v_{1}$ with $v_{2}$, $w_{1}$ with $w^{\prime}_{2}$, $w_{2}$ with
$w^{\prime}_{1}$, the same argument above would imply that $\pi^{-1}({\bf
E}^{\circ}_{v,w})$ is simply connected unless either
“$w^{\prime}_{2}-v_{2}+rv_{1}=0$ and $(v_{1},v_{2}-1)$ is $l$-dominant” or
$v_{1}=w^{\prime}_{1}$.

Therefore, $\pi^{-1}({\bf E}^{\circ}_{v,w})$ is simply connected except the
following four cases:

Case 1: $w_{1}-v_{1}+rv_{2}=w^{\prime}_{2}-v_{2}+rv_{1}=0$, and
$(v_{1}-1,v_{2})$, $(v_{1},v_{2}-1)$ are $l$-dominant. Then
$w_{1}=w^{\prime}_{2}=0$, $r=1$ and $v_{1}=v_{2}$. In this case, the
$l$-dominant lattices lies in a line segment, a degeneration of the hexagon in
Figure 2 (Left). So neither $(v_{1}-1,v_{2})$ nor $(v_{1},v_{2}-1)$ is
$l$-dominant. Thus this case is impossible.

Case 2: $w_{1}-v_{1}+rv_{2}=0$ and $v_{1}=w^{\prime}_{1}$. In this case
$v=(w^{\prime}_{1},(w^{\prime}_{1}-w_{1})/r)$.

Case 3: $w^{\prime}_{2}-v_{2}+rv_{1}=0$ and $v_{2}=w_{2}$. In this case
$v=((w_{2}-w^{\prime}_{2})/r,w_{2})$.

Case 4: $v=(w^{\prime}_{1},w_{2})$.

Last, assume $\pi^{-1}({\bf E}^{\circ}_{v,w})$ is simply connected; since it
is a fiber bundle over ${\bf E}^{\circ}_{v,w}$, by the long exact sequence of
homotopy groups we conclude that ${\bf E}^{\circ}_{v,w}$ is also simply
connected. This proves the lemma. ∎

Next, we study the simply-connectedness of open determinantal varieties. Let
$\mathbb{C}^{m\times n}=\mathrm{Hom}(\mathbb{C}^{n},\mathbb{C}^{m})$ denote
the space of $m\times n$ matrices over $\mathbb{C}$. For nonnegative integers
$m,n,t$ satisfying $0\leq t\leq\min(m,n)$, the open determinantal variety
$Y^{\circ}_{t}$ (or denoted $Y^{\circ}_{t;m\times n}$ to specify $m,n$) is
defined as

$Y^{\circ}_{t}=Y^{\circ}_{t;m\times n}=\\{A\in\mathbb{C}^{m\times n}\,|\,{\rm
rank}A=t\\}.$

######  Lemma 11.2.

The open determinantal variety $Y^{\circ}_{t;m\times n}$ is simply connected
unless $t=m=n$.

###### Proof.

For simplicity we denote $Y^{m\times n}_{t}$ by $Y_{t}$. The closure $Y_{t}$
of $Y^{\circ}_{t}\subseteq\mathbb{C}^{m\times n}$ is usually called the
determinantal variety, and is equal to

$Y_{t}=\\{A\in\mathbb{C}^{m\times n}\,|\,{\rm rank}A\leq
t\\}=\bigcup_{i=0}^{t}Y^{\circ}_{i}.$

Consider the well-known desingularization $p:\tilde{Y}_{t}\to Y_{t}$ (see [23,
(6.1.1)])

$\tilde{Y}_{t}=\\{(A,X)\in Y_{t}\times Gr(t,m)\,|\,{\mathrm{im}}A\subseteq
X\\},$

where $p:(A,X)\mapsto A$. Since the projection $p^{\prime}:\tilde{Y}_{t}\to
Gr(t,m),\,(A,X)\mapsto X$ is a vector bundle of rank $nt$, we see that
$\tilde{Y}_{t}$ is nonsingular, simply connected since $Gr(t,m)$ is simply
connected. Moreover,

$\dim Y_{t}=\dim Y^{\circ}_{t}=\dim\tilde{Y}_{t}=\dim
Gr(t,m)+nt=t(m-t)+nt=t(m+n-t).$

The projection $p$ is stratified by $Y_{t}=\bigcup_{i=0}^{t}Y^{\circ}_{i}$.
The fibers over $Y^{\circ}_{i}$ are isomorphic to $\\{X\in
Gr(t,m)\,|\,\mathbb{C}^{r}\subseteq X\\}\cong Gr(t-i,m-i)$. So

$\dim p^{-1}(Y^{\circ}_{i})=\dim Y^{\circ}_{i}+\dim
Gr(t-i,m-i)=i(m+n-i)+(t-i)(m-t).$

Then the codimension of $p^{-1}(Y^{\circ}_{i})$ in $\tilde{Y}_{t}$ is

$\dim\tilde{Y}_{t}-\dim
p^{-1}(Y^{\circ}_{i})=t(m+n-t)-i(m+n-i)-(t-i)(m-t)=(t-i)(n-i),$

which is $\geq 2$ for all $i<t$, unless $t-i=n-i=1$. So we conclude that
$Y^{\circ}_{t}$ is simply connected unless $t=n$.

Now swap $m$ with $n$ (i.e., transpose $A$) and make a similar argument, we
conclude that $Y^{\circ}_{t}$ is simply connected unless $t=m$. This completes
the proof. ∎

For $s$ satisfying $t\leq s\leq m$, define

$\tilde{Y}_{t}^{s}=\\{(A,X)\in Y_{t}\times Gr(s,m)\,|\,{\mathrm{im}}A\subseteq
X\\}.$

######  Lemma 11.3.

For $s$ satisfying $t\leq s\leq m$, the local systems appeared in the
decomposition of the natural projection $\pi:\tilde{Y}_{t}^{s}\to Y_{t}$ are
all trivial.

###### Proof.

Note that $\pi$ is the same as the natural map
$\widetilde{\mathcal{F}}_{v,w}\to{\bf E}_{v,w}$ for $r=1$, $w=(0,m,n,0)$,
$v=(t,m)$. Using the analytic version of the Transversal Slice Theorem we see
that each summand appeared in the decomposition of
$\pi_{*}(IC_{\tilde{Y}_{t}^{s}})$ is of the form $IC_{Y_{t^{\prime}}}(L)[i]$
where $0\leq t^{\prime}\leq t$. If $t^{\prime}=m=n$ does not hold then
$Y^{\circ}_{t^{\prime}}$ is simply connected, thus $L$ is trivial. So we only
need to consider the case $t^{\prime}=m=n$, which forces $t^{\prime}=t=s=m=n$.
Since $\pi$ restricts to an isomorphism
$\pi^{-1}(Y^{\circ}_{m})\stackrel{{\scriptstyle\cong}}{{\to}}Y^{\circ}_{m}$,
the only summand that is supported on $Y_{m}$ must be $IC_{Y_{m}}$ with a
trivial local system. ∎

Now we are ready to give the second proof of trivial local systems:

###### The Second Proof of Theorem 5.1.

Here we only use the analytic version of the Transversal Slice Theorem. Unlike
the First Proof, now we can only conclude that each summand appeared in the
decomposition theorem is of the form

(11.6) $IC_{{\bf E}_{v^{0},w}}(L)[n]$ for some local system $L$ on ${\bf
E}^{\circ}_{v^{0},w}$

because the open neighborhood $U$ in the analytic Transversal Slice Theorem is
not necessarily open in Zariski topology.

To prove $L$ is trivial, by Lemma 11.1 we only need to consider the cases when
$v^{0}$ equals $w$, $(w^{\prime}_{1},(w^{\prime}_{1}-w_{1})/r)$, or
$((w_{2}-w^{\prime}_{2})/r,w_{2})$. We consider two cases:

Case 1: $v^{0}_{1}=w^{\prime}_{1}$. Since
$v^{0}\leq\bar{v}\leq(w^{\prime}_{1},w_{2})$, $\bar{v}\leq v$, and $v_{1}\leq
w^{\prime}_{1}$ (by (4.5)), we must have
$v^{0}_{1}=\bar{v}_{1}=v_{1}=w^{\prime}_{1}$, $v^{0}_{2}\leq\bar{v}_{2}\leq
w_{2}$. Let $U=\bigcup_{v^{\prime}}{\bf E}^{\circ}_{v^{\prime},w}$ where
$v^{\prime}$ satisfies $v^{0}\leq v^{\prime}\leq\bar{v}$. Consider the
following diagram:

${W=\\{({\bf x},{\bf y},X_{2})|\dim X_{2}=v_{2},{\rm im}B({\bf x},{\bf
y})\subseteq
X_{2}\\}}$${\pi^{-1}(U)}$${\widetilde{\mathcal{F}}_{v,w}}$${V=\\{({\bf x},{\bf
y})|\mathrm{rank}B({\bf x},{\bf
y})\leq\bar{v}_{2}\\}}$${U}$${\mathbf{E}_{\bar{v},w}}$$\scriptstyle{\pi^{\prime\prime}}$$\scriptstyle{j}$$\scriptstyle{j^{\prime}}$$\scriptstyle{\pi^{\prime}}$$\scriptstyle{\pi}$$\scriptstyle{i}$$\scriptstyle{i^{\prime}}$

Note that $i^{\prime}$ is a Zariski open embedding because
$V=\bigcup_{v^{\prime}}{\bf E}^{\circ}_{v^{\prime},w}$ where $v^{\prime}$
satisfies $v^{\prime}_{2}\leq\bar{v}_{2}$ (so $V\setminus
U=\bigcup_{v^{\prime}}{\bf E}^{\circ}_{v^{\prime},w}$ where $v^{\prime}$
satisfies $v^{\prime}_{1}\leq v_{1}-1$ or $v^{\prime}\leq v^{0}-(0,1)$; thus
$V\setminus U$ is Zariski closed in $V$). Also note that $j^{\prime}$ is a
Zariski open embedding because for any point $({\bf x},{\bf
y},X_{1},X_{2})\in\pi^{-1}(U)$, we must have $X_{1}=W^{\prime}_{1}$ by the
fact that $X_{1}\subset W^{\prime}_{1}$ and $\dim
X_{1}=v_{1}=w^{\prime}_{1}=\dim W^{\prime}_{1}$.

Now $V=\\{x_{1}\in\mathbb{C}^{w_{1}^{\prime}w_{1}}\\}\times
Y_{\bar{v}_{2};(w^{\prime}_{2}+rw^{\prime}_{1})\times w_{2}}$,
$W=\\{x_{1}\in\mathbb{C}^{w_{1}^{\prime}w_{1}}\\}\times\tilde{Y}^{v_{2}}_{\bar{v}_{2};(w^{\prime}_{2}+rw^{\prime}_{1})\times
w_{2}}$. By Lemma 11.3, the locally system appeared in the decomposition of
$\pi^{\prime\prime}_{*}(IC_{W})$ must be trivial, thus the same is true for
$\pi^{\prime}_{*}(IC_{\pi^{-1}(U)})$. This implies that $L$ is trivial.

Case 2: $v^{0}_{2}=w_{2}$. Since $v^{0}\leq\bar{v}\leq v$ and
$\bar{v}\leq(w^{\prime}_{1},w_{2})$, we must have
$v^{0}_{2}=\bar{v}_{2}=w_{2}$, $v_{1}=\bar{v}_{1}$, so
$\bar{v}=(v_{1},w_{2})$. The point $v$ is in the triangle which is the
intersection of three half-planes

(11.7) $\\{(x,y)\in\mathbb{R}^{2}\ |\ x\leq w^{\prime}_{1},y\geq
w_{2},w^{\prime}_{2}-y+rx\geq 0\\}$

As seen in (6.2), $\pi$ factors as $p_{1}p_{2}$ as shown in the following
diagram, where all the hooked arrows are Zariski open embedding:

${\pi^{-1}(U)}$${\widetilde{\mathcal{F}}_{v,w}}$${p_{1}^{-1}(U)}$${\tilde{\mathcal{H}}_{v,w}}$${U=\\{({\bf
x},{\bf y})\in\mathbf{E}_{\bar{v},w}\ |\ \mathrm{rank}B({\bf x},{\bf
y})=w_{2}\\}}$${\mathbf{E}_{\bar{v},w}}$$\scriptstyle{p_{2}|_{U}}$$\scriptstyle{p_{2}}$$\scriptstyle{p_{1}|_{U}}$$\scriptstyle{p_{1}}$

Note that

$\bullet$ ${\bf E}^{\circ}_{v^{0},w}\subseteq U$ because $v^{0}_{2}=w_{2}$;

$\bullet$ $p_{2}|_{U}$ is a fiber bundle with fiber

$\\{X_{2}|\dim X_{2}=v_{2},\mathrm{im}B({\bf x},{\bf y})\subseteq
X_{2}\subseteq W^{\prime}_{2}\oplus X_{1}^{\oplus r}\\}\cong
Gr(v_{2}-w_{2},w^{\prime}_{2}+rv_{1}-w_{2});$

$\bullet$ $p_{1}^{-1}(U)\to Gr(v_{1},w^{\prime}_{1})$ defined by $({\bf
x},{\bf y},X_{1})\mapsto X_{1}$ is a fiber bundle whose fiber is

$\displaystyle\\{({\bf x},{\bf y})\ |\ \mathrm{im}A({\bf x},{\bf y})\subseteq
X_{1},\mathrm{rank}B({\bf x},{\bf y})=w_{2}\\}$
$\displaystyle\cong\\{x_{1}\in\mathbb{C}^{v_{1}w_{1}}\\}\times\\{B\textrm{ is
a $(w^{\prime}_{2}+rv_{1})\times w_{2}$ matrix}\ |\ \mathrm{rank}B=w_{2}\\}$

so is simply connected unless the exceptional case
$w^{\prime}_{2}+rv_{1}=w_{2}$, by Lemma 11.2. Now consider the exceptional
case $w^{\prime}_{2}+rv_{1}=w_{2}$. In this case, $\bar{v}$ is on the line
$w^{\prime}_{2}-y+rx=0$. Thus $\bar{v}$ is the left-most point on the
horizontal line $y=w_{2}$ such that $(v^{0},w)$ is $l$-dominant, which implies
that $\bar{v}=v^{0}$. Since $v$ is in the triangle (11.7) and
$v_{1}=\bar{v}_{1}$, we have $v=\bar{v}=v^{0}$, and $p_{2}|_{U}$ is an
isomorphism. So in either situation, the direct summand appeared in the
decomposition of $(p_{2}|_{U})_{*}IC_{\pi^{-1}(U)}$ is (a shift) of
$IC_{p_{1}^{-1}(U)}$, with the trivial local system.

Next consider $p_{1}$. Up to an identify map on the affine space factor
$\\{{\bf x_{2}}\in\mathbb{C}^{w^{\prime}_{2}w_{2}}\\}$, $p_{1}$ is the same as
the map $\tilde{Y}_{t}\to Y_{t}$ with $(m,n)=(w^{\prime}_{1},w_{1}+rw_{2})$
and $t=v_{1}$. So all direct summand of $(p_{1})_{*}IC_{p_{1}^{-1}(U)}$ has
trivial local system by Lemma 11.3. After restricting to the open subset $U$,
we have a similar conclusion for $p_{1}|_{U}$. Now composing $p_{1}|_{U}$ with
$p_{2}|_{U}$, we see that all local system appeared in the decomposition of
$(\pi|_{U})_{*}IC_{\pi^{-1}(U)}$ are trivial. In particular, the local system
$L$ in (11.6) must be trivial since ${\bf E}^{\circ}_{v^{0},w}\subseteq U$.
This completes the proof. ∎

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