Proof We proceed by induction on $\dim V$ . If $\dim V = 0$ , then $k = 0$ , and $\text {Gr}_k(V)^{\text {iso}}$ is a singleton. The statement is obvious in this case.

Assume now that $\dim V> 0$ . If $\langle {-},{-}\rangle $ is degenerate, choose a one-dimensional space $L \subset \operatorname {\mathrm {rad}} V$ . Otherwise, let L be any one-dimensional subspace of $M_1$ . In both cases, L is automatically isotropic. Choose a complement W to L in $L^\perp $ , so that $L^\perp = L \oplus W$ . Let $p_1: L^\perp \to L$ and $p_2: L^\perp \to W$ be the projection maps. If $H \subset L^\perp $ , then

$$ \begin{align*} \dim p_2(H) \cap p_2(M_i) &= \dim p_2^{-1}(p_2(H) \cap p_2(M_i)) -1 = \dim (H+L) \cap (M_i + L) -1 \\ &= \dim (H+L) + \dim (M_i+L) - \dim (H+M_i+L) - 1. \end{align*} $$

Comparing this to $\dim H \cap M_i = \dim H + \dim M_i - \dim (H+M_i)$ , we find that

(2.1) $$ \begin{align} \dim H \cap M_i = \begin{cases} \dim p_2(H) \cap p_2(M_i) + 1& \text{ if } L \subset H \text{ and } L \subset M_i, \\ \dim p_2(H) \cap p_2(M_i) & \text{ if } L \not\subset H \text{ and } L \subset M_i, \\ \dim p_2(H) \cap p_2(M_i) & \text{ if } L \subset H \text{ and } L \not\subset M_i, \\ \dim p_2(H) \cap p_2(M_i) -1 & \text{ if } L \not\subset H, L \not \subset M_i, \text{ and } L \subset H + M_i, \\ \dim p_2(H) \cap p_2(M_i) & \text{ if } L \not\subset H + M_i. \end{cases} \end{align} $$

Now, let

$$ \begin{align*} X_1 &= \{H \in \text{Gr}_k(V)^{\text{iso}} \mid L \subset H \}, \\ X_2 &= \{H \in \text{Gr}_k(V)^{\text{iso}} \mid L \not\subset H,\ H \subset L^\perp\}, \\ X_3 &= \{H \in \text{Gr}_k(V)^{\text{iso}} \mid H \not\subset L^\perp\}. \end{align*} $$

It is enough to show that each of these locally closed subsets separately admits a suitable affine paving.

For $X_1$ , observe that an isotropic subspace H that contains L is necessarily contained in $L^\perp $ . The map $p_2$ induces an isomorphism

$$\begin{align*}X_1 \cong \text{Gr}_{k-1}(W)^{\text{iso}}. \end{align*}$$

By induction, $\text {Gr}_{k-1}(W)^{\text {iso}}$ admits an affine paving such that the function $F \mapsto \dim F \cap p_2(M_i)$ is constant on each stratum. We transfer this affine paving to $X_1$ . Since $\dim p_2(H)\cap p_2(M_i)$ is constant on each stratum and L and L is fixed, so the condition in (2.1) is also constant. Therefore, the function $H \mapsto \dim H \cap M_i$ is constant on each stratum in $X_1$ .

Next, we turn to $X_2$ . Consider the variety

$$\begin{align*}E = \{ (F, f) \mid F \in \text{Gr}_k(W)^{\text{iso}},\ f \in \operatorname{\mathrm{Hom}}(F,L) \}. \end{align*}$$

This is a vector bundle over $\text {Gr}_k(W)^{\text {iso}}$ . For $H \in X_2$ , the map $p_2$ induces an isomorphism $H \cong p_2(H)$ . Let $f_H: p_2(H) \to L$ be the composition

$$\begin{align*}f_H: p_2(H) \xrightarrow[\sim]{\text{ induced by }p_2} H \xrightarrow{p_1} L. \end{align*}$$

Then we have a map

$$\begin{align*}\theta: X_2 \to E \qquad\text{ given by }\qquad H \mapsto (p_2(H), f_H). \end{align*}$$

In fact, this is an isomorphism: the inverse map sends $(F,f)$ to $H = \{(f(v),v) \in L \oplus W \mid v \in F \}$ . Thus, $X_2$ is a vector bundle over $\text {Gr}_k(W)^{\text {iso}}$ . By induction, $\text {Gr}_k(W)^{\text {iso}}$ admits an affine paving such that $F \mapsto \dim F \cap p_2(M_i)$ is constant along strata. Taking preimages along $X_2 \to \text {Gr}_k(W)^{\text {iso}}$ , we obtain an affine paving of $X_2$ . By (2.1), the function $H \mapsto \dim H \cap M_i$ is constant on each stratum in $X_2$ .

Finally, we consider $X_3$ . If $X_3$ is nonempty, then $L^\perp \ne V$ , that is, L is not contained in the radical of $\langle {-},{-}\rangle $ . Our set-up implies that this can happen only when $\langle {-},{-}\rangle $ is nondegenerate. Assume this from now. We also have $L \subset M_1$ , and since all the $M_i$ are isotropic, we have

(2.2) $$ \begin{align} L \subset M_1 \subset \cdots \subset M_c \subset L^\perp. \end{align} $$

In addition, the restriction of $\langle {-},{-}\rangle $ to W is nondegenerate, and $W^\perp $ is two-dimensional and contains L. Choose an isotropic complement R to L in $W^\perp $ , and then choose some nonzero vector $r_0 \in R$ .

Let

$$\begin{align*}X_2' = \{ F \in \text{Gr}_{k-1}(V)^{\text{iso}} \mid L \not\subset F,\ F \subset L^\perp \}. \end{align*}$$

This variety is analogous to $X_2$ , and the same reasoning shows that $X_2'$ has an affine paving such that $F \mapsto \dim F \cap M_i$ is constant on strata. Let $u \in L^\perp /F$ , we define a linear map $\langle u, {-} \rangle : F \to \mathbb {C}$ . This map is well defined as $\langle u,v\rangle =0$ for all $u, v\in F$ , which is because $F\subset F^\perp $ . The assignment $u \mapsto \langle u,-\rangle $ is a linear map $L^\perp /F \to F^*$ . Consider the element $\langle -r_0, {-}\rangle \in F^*$ . Its preimage in $L^\perp /F$ , given concretely by

$$\begin{align*}\{ u \in L^\perp/F \mid \langle u,v \rangle = \langle -r_0,v \rangle \text{ for all } v \in F \}, \end{align*}$$

is an affine subspace of $L^\perp /F$ . We deduce that the variety

$$\begin{align*}E' = \{ (F,u) \mid F \in X_2', u \in L^\perp/F, \text{ and } \langle u,v \rangle = \langle -r_0,v \rangle \text{ for all } v \in F \} \end{align*}$$

is an affine bundle over $X_2'$ . The affine paving of $X_2'$ induces an affine paving of $E'$ .

For $H \in X_3$ , we have $H \cap L^\perp \in X_2'$ . Moreover, the composition $H \to H/(H \cap L^\perp ) \cong (H+L^\perp )/L^\perp $ is surjective. As H is one-dimensional and not contained in $L^\perp $ , hence $H+L^\perp \cong V$ . This implies $H/(H \cap L^\perp ) \cong V/L^\perp $ . Now, we claim $V/L^\perp \cong R$ . The intersection $R\cap L^\perp $ must be $0$ . If not then $R\subset L^\perp $ . But $L^\perp \cap W^\perp $ is one-dimensional and hence $L=L^\perp \cap W^\perp $ . So R cannot be contained in both $W^\perp $ and $L^\perp $ and we are done with our claim. Therefore, $R\hookrightarrow V/L^\perp $ and by dimension equality $V/L^\perp \cong R$ . Finally, we get a surjective map $H\to R$ . so there is a vector $\tilde r_H \in H$ that maps to $r_0$ . We have $\tilde r_H - r_0 \in L^\perp $ . Since H is isotropic, for any $v \in H \cap L^\perp $ , we have $\langle \tilde r_H, v \rangle = 0$ , so

$$\begin{align*}\langle \tilde r_H - r_0, v \rangle = \langle -r_0, v\rangle. \end{align*}$$

We can thus consider the map

$$\begin{align*}\theta': X_3 \to E' \qquad\text{ given by }\qquad \theta'(H) = (H \cap L^\perp, \tilde r_H - r_0 + (H \cap L^\perp)). \end{align*}$$

The remainder of the argument is slightly different depending on whether $\langle {-},{-}\rangle $ is symmetric or skew-symmetric. In the skew-symmetric case, $\theta '$ is an isomorphism: the inverse map sends a pair $(F,u) \in E'$ to the space $H = F + \mathbb {C}(u + r_0)$ . Thus, $X_3$ has an affine paving induced by that $X_2'$ . In view of (2.2), we have

$$\begin{align*}\dim H \cap M_i = \dim (H \cap L^\perp) \cap M_i, \end{align*}$$

and thus the function $H \mapsto \dim H \cap M_i$ is constant on strata.

It remains to consider the symmetric case. We begin by describing $E'$ another way. For $F \in X_2'$ , the projection maps $p_1: L^\perp \to L$ and $p_2: L^\perp \to W$ induce an isomorphism

$$\begin{align*}L^\perp/F \cong L \oplus W/p_2(F). \end{align*}$$

For $u \in L^\perp /F$ , write $u = u_1 + u_2$ with $u_1 \in L$ and $u_2 \in W/p_2(F)$ . Then $\langle u, v\rangle = \langle u_2, v\rangle $ for all $v \in F$ , so we can identify

$$\begin{align*}E' = \left\{ (F,u_1,u_2) \,\Big|\, \begin{array}{c} F \in X_2', u_1 \in L, u_2 \in W/p_2(F), \\ \text{ and } \langle u_2,v \rangle = \langle -r_0,v \rangle \text{ for all } v \in F \end{array} \right\}. \end{align*}$$

In this description, there is no condition on $u_1$ . Consider the subset

$$\begin{align*}E" = \{ (F, u_1, u_2) \in E' \mid \langle u_1, r_0\rangle = \textstyle -\frac{1}{2}\langle u_2 + r_0, u_2 + r_0\rangle \}. \end{align*}$$

We need to check that the expression “ $\langle u_2+r_0, u_2+r_0\rangle $ ” makes sense, because $W/p_2(F) \oplus R$ does not have an induced bilinear form. Since we are starting with a point in $E'$ , for $v \in F$ , we have

$$\begin{align*}\langle u_2 + v +r_0, u_2 +v+r_0\rangle = \langle u_2 + r_0, u_2+r_0\rangle + 2\langle u_2+r_0,v\rangle + \langle v, v\rangle. \end{align*}$$

On the right-hand side, the middle term vanishes by the definition of $E'$ , and the last term vanishes because F is isotropic. We conclude that $E"$ is well-defined. Its definition says simple that the $u_1$ component is determined by $u_2$ . Thus, $E"$ is isomorphic to the variety

$$\begin{align*}\{ (F, u_2) \in E' \mid F \in X_2', u_2 \in W/p_2(F), \text{ and } \langle u_2,v \rangle = \langle -r_0,v \rangle \text{ for all } v \in F \}. \end{align*}$$

In particular, $E"$ is also an affine space bundle over $X_2'$ .

Note that

$$\begin{align*}\langle u_1 + u_2 + r_0, u_1 + u_2 + r_0\rangle = 2\langle u_1, r_0\rangle + \langle u_2 + r_0, u_r + r_0\rangle. \end{align*}$$

In view of this, the definition of $E"$ can be rewritten as

$$\begin{align*}E" = \{ (F,u) \in E' \mid \langle u + r_0, u+ r_0\rangle = 0 \}. \end{align*}$$

In the symmetric case, it is clear that the image of $\theta '$ is contained in $E"$ , since $\tilde r_H$ is an isotropic vector. In fact, $\theta '$ induces an isomorphism $X_3 \cong E"$ : the inverse map again sends $(F,u) \in E'$ to $H = F + \mathbb {C}(u + r_0)$ . We conclude that $X_3$ has an affine paving with the desired properties.

Proof Let i be an integer with $1 \le i \le c$ , and let

$$\begin{align*}F_i = \{(x, v) \in \mathbb{A}^d \times V \mid v \in f(x) + M_i \} \subset \mathbb{A}^d \times V. \end{align*}$$

Using the fact that $\dim (f(x) \cap M_i)$ is independent of x, one can check that $F_i$ is a vector bundle over $\mathbb {A}^d$ . Similarly,

$$\begin{align*}F_\perp = \{ (x, v) \in \mathbb{A}^d \times V \mid v \in f(x)^\perp \} \subset \mathbb{A}^d \times V \end{align*}$$

is also a vector bundle over $\mathbb {A}^d$ (because $\dim (f(x) \cap \operatorname {\mathrm {rad}} V)$ is independent of x). Note that $M_i \subset \operatorname {\mathrm {rad}} V \subset f(x)^\perp $ for all x, so we have containments of vector bundles

(2.3) $$ \begin{align} F_1 \subset F_2 \subset \cdots \subset F_c \subset F_\perp. \end{align} $$

By the Quillen–Suslin theorem, each of these vector bundles is a trivial bundle. Moreover, each quotient $F_i/F_{i-1}$ or $F_\perp /F_c$ is also trivial. It follows that we can choose compatible filtrations of each of the terms in (2.3). In concrete terms, this means that we can choose a point $x_0 \in \mathbb {A}^d$ and an isomorphism

$$\begin{align*}\theta: F_\perp \overset{\sim}{\to} f(x_0)^\perp \times \mathbb{A}^d \end{align*}$$

such that $\theta $ restricts to an isomorphism

$$\begin{align*}F_i \overset{\sim}{\to} f(x_0) + M_i. \end{align*}$$

Via these isomorphisms, we obtain an isomorphism

$$\begin{align*}Y \cong \text{Gr}_{k-r}(f(x_0)^\perp/ f(x_0))^{\text{iso}} \times \mathbb{A}^d. \end{align*}$$

Now consider the sequence of subspaces

$$\begin{align*}(f(x_0) + M_1)/f(x_0) \subset (f(x_0) + M_2)/f(x_0) \subset \cdots \subset (f(x_0) + M_c)/f(x_0). \end{align*}$$

By Lemma 2.1, $\text {Gr}_{k-r}(f(x_0)^\perp /f(x_0))^{\text {iso}}$ admits an affine paving with the property that the function $H \mapsto \dim H \cap \frac {f(x_0) + M_i}{f(x_0)}$ is constant on strata. The lemma follows.

Proof If H is in the image of $\varpi $ , then it satisfies the condition $P\subset H\subset P^\perp $ with $\dim P= k-r$ . This implies $P\subset H^\perp \cap H$ . Hence, $k-r\leq \dim H\cap H^\perp $ and $H\in \overline {\mathcal {C}_{k,r}(V)}$ . The map $\varpi |_{\varpi ^{-1}(\mathcal {C}_{k,r}(V))}: \varpi ^{-1}(\mathcal {C}_{k,r}(V))\to \mathcal {C}_{k,r}(V)$ is clearly an isomorphism. Now consider the map, $\widetilde {\mathcal {C}}_{k,r}\to \text {Gr}_{k-r}(V)^{\text {iso}}$ by sending $(P,H)$ to P. The fibers of this map are isomorphic to $\{H \mid P\subset H \subset P^\perp \}\cong \text {Gr}_r(P^\perp /P)$ , which is smooth. Also $k-r< k-r/2 \leq n/2$ , hence by Lemma 2.4, $\text {Gr}_{k-r}(V)^{\text {iso}}$ is smooth. Therefore, $\widetilde {\mathcal {C}}_{k,r}(V)$ is smooth. We know the projection map $\text {Gr}_{k-r}(V)^{\text {iso}} \times \text {Gr}_r(V)\to \text {Gr}_k(V)$ is proper and $\widetilde {\mathcal {C}}_{k,r}(V)$ is a closed subset of $\text {Gr}_{k-r}(V)^{\text {iso}} \times \text {Gr}_r(V) $ . Hence, the map $\varpi $ is proper. Combining these facts, we deduce that $\varpi $ is a resolution of singularities.

Proof (1) There is a map

(4.1) $$ \begin{align} p: \mathcal{B}_{k,r}(V) \to \text{Gr}_{k-r}(V)^{\text{iso}} \end{align} $$

given by $(P,Q) \mapsto P$ . Since $2k -n < r$ , we have $k-r < k - r/2 < n/2$ , and so $\text {Gr}_{k-r}(V)^{\text {iso}}$ is smooth, connected, and simply connected by Lemma 2.4.

Suppose for now that r is even. Let $\mathcal {F}l(k-r,k-r/2)$ be the variety of partial flags $P \subset Q \subset V$ with $\dim P = k-r$ and $\dim Q = k-r/2$ . The projection $\mathcal {F}l(k-r,k-r/2) \to \text {Gr}_{k-r}(V)$ is a locally trivial fibration, and the map p in (4.1) is the restriction of this map to the preimage of $\text {Gr}_{k-r}(V)^{\text {iso}} \subset \text {Gr}_{k-r}(V)$ . Hence, p is a locally trivial fibration whose fiber over a point $P \in \text {Gr}_{k-r}(V)^{\text {iso}}$ is the variety

$$\begin{align*}\{ Q \mid {\dim Q = k - r/2 \text{ and } P \subset Q \subset Q^\perp \subset P^\perp} \} \cong \text{Gr}_{r/2}(P^\perp / P)^{\text{iso}}. \end{align*}$$

Since $2k -n < r$ , we have $\dim P^\perp /P = n - 2(k - r) = n - 2k + 2r> r$ , so $\text {Gr}_{r/2}(P^\perp /P)^{\text {iso}}$ is again smooth, connected, and simply connected. We conclude that $\mathcal {B}_{k,r}(V)$ is smooth, connected, and simply connected.

The case where r is odd is similar, but the fiber of p over $P \in \text {Gr}_{k-r}(V)^{\text {iso}}$ is given by

$$\begin{align*}\{ Q \mid {\textstyle\dim Q = k - \frac{r-1}{2} \text{ and } P \subset Q \subset Q^\perp_+ \subset P^\perp \oplus \mathbb{C}} \} \cong \text{Gr}_{\frac{r+1}{2}}(P^\perp / P \oplus \mathbb{C})^{\text{iso}}. \end{align*}$$

Since $\dim (P^\perp /P \oplus \mathbb {C} )> r+1$ , the variety $\text {Gr}_{\frac {r+1}{2}}(P^\perp / P \oplus \mathbb {C})^{\text {iso}}$ is smooth, connected, and simply connected, and we conclude as before.

(2) We have seen in part (1) that $\mathcal {B}_{k,r}$ is an isotropic Grassmannian bundle over an isotropic Grassmannian. The claim that its cohomology is even and free then follows from the Serre spectral sequence and Corollary 2.2.

(3) Suppose first r is even. From the definitions, the map $\widehat {\mathcal {C}}_{k,r}(V) \to \mathcal {B}_{k,r}(V)$ is a locally trivial fibration whose fiber over a point $(P,Q) \in \mathcal {B}_{k,r}(V)$ is identified with

$$\begin{align*}\text{Gr}_{r/2}(P^\perp/Q), \end{align*}$$

which is smooth, connected, and simply connected. In view of part (1), the claim for $\widehat {\mathcal {C}}_{k,r}(V)$ follows.

Now suppose that r is odd. For a point $(P,Q) \in \mathcal {B}_{k,r}(V)$ , let $\bar Q$ be the image of Q under the projection map $V^+ \to V$ . The kernel of this projection map contains no nonzero isotropic vectors. Since Q is isotropic, it follows that $\dim \bar Q = \dim Q = k - \frac {r-1}{2}$ . Note also that Q is contained in $P^\perp _+ = P^\perp \oplus \mathbb {C}$ , and hence that $\bar Q \subset P^\perp $ . Finally, in the definition of $\widehat {\mathcal {C}}_{k,r}(V)$ , the condition $Q \subset H^+ \subset P^\perp _+ \subset V^+$ is equivalent to $\bar Q \subset H \subset P^\perp \subset V$ . In view of these observations, we deduce that $\widehat {\mathcal {C}}_{k,r}(V) \to \mathcal {B}_{k,r}(V)$ is a locally trivial fibration whose fiber over a point $(P,Q) \in \mathcal {B}_{k,r}(V)$ is identified with

$$\begin{align*}\text{Gr}_{\frac{r-1}{2}}(P^\perp/\bar Q), \end{align*}$$

and then we conclude as before.

(4) Let $H \in \mathcal {C}_{k,r}(V)$ , and let $(P,Q,H) \in \widehat {\varpi }^{-1}(H)$ . The definition of $\widehat {\mathcal {C}}_{k,r}(V)$ implies that $P \subset \operatorname {\mathrm {rad}} H$ , but the fact that $H \in \mathcal {C}_{k,r}$ means that $\dim \operatorname {\mathrm {rad}} H = k - r = \dim P$ . These observations let us describe $\widehat {\mathcal {C}}_{k,r}^\circ (V)$ as follows: if r is even, then

$$ \begin{align*} \widehat{\mathcal{C}}_{k,r}^\circ(V) &= \left\{ \operatorname{\mathrm{rad}} H \subset Q \subset H \,\Big|\, \begin{array}{@{}c@{}} {H \in \mathcal{C}_{k,r}(V), \dim Q = k - \frac{r}{2},}\\ \text{ and } Q \subset Q^\perp \end{array} \right\} &&\text{ if } r \text{ is even, } \\ \widehat{\mathcal{C}}_{k,r}^\circ(V) &= \left\{ \operatorname{\mathrm{rad}} H \subset Q \subset H^+ \,\Big|\, \begin{array}{@{}c@{}} {H \in \mathcal{C}_{k,r}(V), \dim Q = k - \frac{r-1}{2},}\\ \text{ and } Q \subset Q^\perp_+ \end{array} \right\} &&\text{ if } r \text{ is odd. } \end{align*} $$

In both cases, it follows that $\widehat {\mathcal {C}}_{k,r}^\circ (V) \to \mathcal {C}_{k,r}$ is a locally trivially fibration. The fiber over $H \in \mathcal {C}_{k,r}$ is given by

$$\begin{align*}\text{Gr}_{r/2}(H/\operatorname{\mathrm{rad}} H)^{\text{iso}}\quad\text{ if } r \text{ is even, } \qquad\qquad \text{Gr}_{\frac{r+1}{2}}(H^+/\operatorname{\mathrm{rad}} H)^{\text{iso}}\quad\text{ if } r \text{ is odd. } \end{align*}$$

(5) Since $\widehat {\mathcal {C}}_{k,r}(V)$ is smooth and connected, its Zariski-open subset $\widehat {\mathcal {C}}_{k,r}^\circ (V)$ is as well. Let $\Gamma = \pi _1(\mathcal {C}_{k,r}(V))$ , and let $\Gamma ' = \pi _1(\widehat {\mathcal {C}}^\circ _{k,r}(V))$ . (The group $\Gamma $ is described in Remark 3.11, but we will not need this information.)

Next, let F denote a fiber of $\widehat {\varpi }: \widehat {\mathcal {C}}^\circ _{k,r}(V) \to \mathcal {C}_{k,r}(V)$ . We have identified F in the proof of part (4). By Proposition 3.10(2), F has two connected components, and by Lemma 2.4, each connected component is simply connected. In other words, $\pi _0(F) \cong \mathbb {Z}/2\mathbb {Z}$ and $\pi _1(F) = 1$ . Then the long exact sequence of homotopy groups

$$\begin{align*}\cdots \to \pi_1(F) \to \underbrace{\pi_1(\widehat{\mathcal{C}}^\circ_{k,r}(V))}_{\Gamma'} \to \underbrace{\pi_1(\mathcal{C}_{k,r}(V))}_{\Gamma} \to \pi_0(F) \to \pi_0(\widehat{\mathcal{C}}^\circ_{k,r}(V)) \to \cdots \end{align*}$$

shows that $\Gamma ' \to \Gamma $ is injective, and that its image is a subgroup of index $2$ .

The category of local systems on $\mathcal {C}_{k,r}(V)$ , respectively, on $\widehat {\mathcal {C}}^\circ _{k,r}(V)$ , is equivalent to the category of $\Bbbk[\Gamma ]$ -modules, respectively, of $\Bbbk[\Gamma ']$ -modules. Under these equivalences, the pullback functor $\widehat {\varpi }^*$ corresponds to the forgetful functor

(4.2) $$ \begin{align} \Bbbk[\Gamma]\text{ -mod } \to \Bbbk[\Gamma']\text{ -mod }. \end{align} $$

Since $\widehat {\varpi }^\circ $ is a locally trivial fibration, the (non-derived) push-forward functor $\mathcal {H}^0 \circ \widehat {\varpi }_*$ sends local systems to local systems. It is right adjoint to $\widehat {\varpi }^*$ , so we can identify it with the right adjoint to (4.2), which is

$$\begin{align*}\operatorname{\mathrm{Hom}}_{\Bbbk[\Gamma']}(\Bbbk[\Gamma]^{\text{op}}, {-}): \Bbbk[\Gamma']\text{ -mod } \to \Bbbk[\Gamma]\text{ -mod }. \end{align*}$$

Therefore, the local system $\mathcal {H}^0(\widehat {\varpi }_*\Bbbk)$ corresponds to $\operatorname {\mathrm {Hom}}_{\Bbbk[\Gamma ']}(\Bbbk[\Gamma ]^{\text {op}},\Bbbk)$ , which can be identified with the regular representation of $\Bbbk[\Gamma /\Gamma '] = \Bbbk[\mathbb {Z}/2\mathbb {Z}]$ . This local system decomposes as in (3.6) because $2$ is invertible in $\Bbbk$ .

Proof We will first prove (2) by induction on m. If $m = 1$ , then $\mathcal {O}_{\underline {k},\underline {r}} = \mathcal {C}_{\underline {k},\underline {r}}$ , and there is nothing to prove. Otherwise, let $B' = B_{\le m-1}$ , and let $n' = n_1 + \cdots + n_{m-1}$ . Given $(\underline {k},\underline {r}) = ((k_1,\ldots , k_m),(r_1,\ldots , r_m))$ , let $k' = k_1 + \cdots + k_{m-1}$ , and let

$$\begin{align*}(\underline{k}',\underline{r}') = ((k_1,\ldots,k_{m-1}),(r_1,\ldots,r_{m-1})). \end{align*}$$

This pair labels a subset $\mathcal {O}_{\underline {k}',\underline {r}'} \subset \text {Gr}_{k'}(B')$ . The map $\mathcal {O}_{\underline {k},\underline {r}} \to \mathcal {C}_{\underline {k},\underline {r}}$ in the statement of the proposition factors as

(5.1) $$ \begin{align} & \mathcal{O}_{\underline{k},\underline{r}} \xrightarrow{H \mapsto (H \cap B', \text{pr}_m H)} \mathcal{O}_{\underline{k}',\underline{r}'} \times \mathcal{C}_{k_m,r_m}(B_m) \nonumber\\ &\qquad \xrightarrow{(H',K) \mapsto (\text{pr}_1 H', \ldots, \text{pr}_{m-1} H',K)} \mathcal{C}_{k_1,r_1}(B_1) \times \cdots \times \mathcal{C}_{k_m,r_m}(B_m) = \mathcal{C}_{\underline{k},\underline{r}}. \end{align} $$

The second map in (5.1) is an affine space bundle by induction, so it is enough to prove that the first map is an affine space bundle.

Let $p: B \to B'$ be the projection map with respect to the decomposition $B = B' \oplus B_m$ . For any $H \in \mathcal {O}_{\underline {k},\underline {r}}$ , the map $p|_H: H \to B'$ induces a map $\bar p_H: H/H \cap B' \to B'/H \cap B'$ . The assignment $H \mapsto (H \cap B', \text {pr}_m H, \bar p_H)$ defines an isomorphism of $\mathcal {O}_{\underline {k},\underline {r}}$ with the variety $\mathcal {V}$ given by

$$\begin{align*}\mathcal{V} = \left\{(H',K,\varphi) \,\Big|\, \begin{array}{@{}c@{}} {H' \in \mathcal{O}_{\underline{k}',\underline{r}'}, K \in \mathcal{C}_{k_m,r_m}(B_m),} \\ \text{ and } \varphi: K \to B'/H' \text{ a linear map} \end{array} \right\}. \end{align*}$$

Since $\mathcal {V}$ is a vector bundle over $\mathcal {O}_{\underline {k}',\underline {r}'} \times \mathcal {C}_{k_m,r_m}(B_m)$ , we are done.

Next, we prove (1). It is clear that the extra factors copies of $\mathbb {C}^\times $ that appear in $GQ_B$ preserve each $\mathcal {O}_{\underline {k},\underline {r}}$ , so it is enough to show that each $\mathcal {O}_{\underline {k},\underline {r}}$ is a single $Q_B$ -orbit. For this claim, we again proceed by induction on m. If $m = 1$ , the statement is trivial. Otherwise, let $B'$ be as above. Let

$$\begin{align*}U_{B'} = \{ g \in U_B \mid g(v) = v \text{ for } v \in B_m \}, \qquad X = \{ g \in U_B \mid g(v) = v \text{ for } v \in B' \}. \end{align*}$$

Then X is a normal subgroup of $U_B$ , and $U_B = U_{B'} \ltimes X$ . By induction, the variety $\mathcal {O}_{\underline {k}',\underline {r}} \times \mathcal {C}_{k_m,r_m}(B_m)$ is a single orbit under the group $I_B \ltimes U_{B'} = Q_{B'} \times \text {Iso}^\circ (B_m)$ .

On the other hand, given $H' \in \mathcal {O}_{\underline {k}',\underline {r}'}$ and $K \in \mathcal {C}_{k_m,r_m}(B_m)$ , the group X acts transitively on the set of linear maps $K \to B'/H'$ . That is, X acts transitively on the fibers of the first map in (5.1). The desired conclusion follows.

For part (3), the statement for $Q_B$ follows from Propositions 3.8(3) and 3.10(3). This statement implies the version for $GQ_B$ by Remark 3.4.

Similarly, part (4) follows from Propositions 3.8(4) and 3.10(4).

Proof Let j be an integer with $0 \le j \le m$ . We define varieties $Y_j$ and $Y^{\prime }_j$ as follows: a point is a collection of vector spaces as indicated below, subject to the same conditions as those in the definition of $X(\underline {k},\underline {r})$ .

There is an obvious sequence of maps

(6.1) $$ \begin{align} X(\underline{k},\underline{r}) = Y^{\prime}_m \to Y_m \to Y^{\prime}_{m-1} \to \cdots \to Y^{\prime}_1 \to Y_1, \end{align} $$

where each map forgets one subspace from the data.

Let us describe the fibers of the maps in (6.1). Let $1 \le j \le m$ , and fix a point in $Y_j$ . The fiber of $Y^{\prime }_j \to Y_j$ over the given point is the variety of subspaces $H_j$ such that

$$\begin{align*}P_j \subset H_j \subset B_{<j}\oplus \tilde P_j^\perp \qquad\text{ and }\qquad \dim H_j = k_1 + \cdots + k_j. \end{align*}$$

The space of choices for $H_j$ is isomorphic to the Grassmannian

$$\begin{align*}\textstyle\text{Gr}_{r_j}((B_{<j} \oplus \tilde P_j^\perp)/P_j), \end{align*}$$

and we conclude that $Y^{\prime }_j \to Y_j$ is a Grassmannian bundle, that is, a locally trivial fiber bundle whose fibers are Grassmannians. Similar arguments show that $Y_{j+1} \to Y^{\prime }_j$ is also a Grassmannian bundle.

For each variety Y appearing in (6.1), we will prove the following claims:

1. (1) Y is smooth, connected, and simply connected.

2. (2) We have $\mathsf {H}^{\bullet }(Y,\Bbbk)$ is even and free.

We proceed by induction, starting with $Y_1$ . Since $P_1 = \tilde P_1$ , and since each $\tilde P_i$ is isotropic, we may identify

(6.2) $$ \begin{align} Y_1 = Y_{11} \times \cdots \times Y_{1m} \quad\text{ where }\quad Y_{1i} = \begin{cases} \mathcal{C}_{k_i,r_i}(B_i) & \text{ if } r_i = 0' \text{ or } 0", \\ \text{Gr}_{k_i - r_i}(B_i)^{\text{iso}} & \text{ otherwise. } \end{cases} \end{align} $$

Recall that when $r_i = 0'$ or $0"$ , the variety $\mathcal {C}_{k_i,r_i}(B)$ is a connected component of $\text {Gr}_{k_i}(B_i)^{\text {iso}}$ . Using Lemma 2.4, we see that each $Y_{1i}$ is smooth, connected, and simply connected, and by Corollary 2.2, $\mathsf {H}^{\bullet }(Y_{1i},\Bbbk) = 0$ is even and free. We deduce that $Y_1$ has the desired properties as well.

For the inductive step, let $Y' \to Y$ be a map in (6.1), and assume that the claims are known for Y. We have seen that $Y' \to Y$ is a Grassmannian bundle. It follows immediately that $Y'$ is smooth, connected, and simply connected. Since Y is simply connected, the Serre spectral sequence shows that $\mathsf {H}^{\bullet }(Y') = 0$ is even and free.

Proof For $x \in X(\underline {k},\underline {r})$ , we have

$$ \begin{align*} \dim \text{pr}_i P_i(x) &\le k_i - r_i, \\ \dim \text{pr}_i H_i(x) &=\dim H_i/{B_{<i}\cap H_i}\le \dim H_i/P_i \le k_i. \end{align*} $$

Let U be the open subset of $X(\underline {k},\underline {r})$ where these inequalities are equalities for all i. The definitions imply that for $x \in U$ , we have

$$\begin{align*}\text{pr}_i P_i = \tilde P_i \qquad \text{ and }\qquad \tilde P_i \subset \text{pr}_i H_i \subset (\tilde P_i)^\perp. \end{align*}$$

In other words, $\tilde P_i$ is contained in the radical of $\text {pr}_i H_i$ . Therefore, the first assertion of the proposition is true.

Since $\mu $ is proper and $Q_B$ -equivariant, its image is closed, irreducible, and stable under $Q_B$ , and is therefore the closure of a single $Q_B$ -orbit.

Let $H\in \mathcal {O}_{\underline {k},\underline {r}}$ . We can define $H_m=H$ . Clearly, $\dim \text {pr}_i H\cap \text {pr}_i H^\perp = k_i-r_i$ . So we define $\tilde P_i= \text {pr}_i H\cap \text {pr}_i H^\perp $ which has dimension $k_i-r_i $ and an isotropic subspace. Define $H_i=H\cap B_{\leq i}$ . Then $\dim H_1=\dim H\cap B_{\leq 1}=\dim \text {pr}_1 H = k_1$ . Now, we prove $\dim H_i=k_1+\dots +k_i$ by induction. Let the statement is true for $H_{i-1}$ . Then $\dim H_i=\dim \text {pr}_i H + \dim H\cap B_{\leq i-1}=\dim \text {pr}_i H + \dim H_{i-1}=k_i+k_{i-1}+\dots +k_1 $ . Also $\text {pr}_i H_i=\text {pr}_i H \subset \text {pr}_i H +\text {pr}_i H ^\perp = \tilde P_i^\perp $ . We have the projection map $\pi : H_i\to \text {pr}_i H$ , where $\text {pr}_i H$ contains $\tilde P_i$ . We define $P_i$ to be the preimage of $\tilde P_i$ under the map $\pi $ . Then $\dim P_i=\dim \tilde P_i +\ker \pi =k_i-r_i+ \dim H\cap B_{<i}=k_i-r_i+k_{i-1}+\dots +k_1$ . Also $\ker \pi = H_{i-1}\subset P_i$ . The i-th projection, $\text {pr}_i P_i \subset \tilde P_i$ . So it is clear that the choice of $H\in \mathcal {O}_{\underline {k},\underline {r}}$ determines the element in $X_{\underline {k},\underline {r}}$ , hence $\mu |_{\mu ^{-1}(\mathcal {O}_{\underline {k},\underline {r}})}: \mu ^{-1}(\mathcal {O}_{\underline {k},\underline {r}} )\to \mathcal {O}_{\underline {k},\underline {r}}$ is a bijection. Since a bijective morphism of smooth complex varieties is an isomorphism, we are done.

Proof Let $1 \le j \le m$ . Define varieties $F_j$ and $F_j'$ as follows: a point is a collection of subspaces as indicated below, subject to the same conditions as in the definition of $F(\underline {k},\underline {r})$ , along with the additional conditions indicated below.

We have a sequence of maps

$$\begin{align*}\mu_F^{-1}(M) = F^{\prime}_m \to F_m \to F^{\prime}_{m-1} \to \cdots \to F_2 \to F^{\prime}_1 \to F_1. \end{align*}$$

We will show that every variety in this sequence admits an affine paving with the property that the functions mentioned in the statement of the lemma are all constant on each piece of the paving. The proof is by induction along the sequence of maps above.

Let us prove the claim for $F_j$ , assuming that $F^{\prime }_{j-1}$ has an appropriate affine paving. (This paragraph also covers the start of the induction: interpret $F^{\prime }_0$ to mean a singleton, and take $P_0 = H_0 = 0$ .) Consider the bilinear form $({-},{-})_{B_{\le j}}$ on $B_{\le j}$ defined by declaring $B_{<j}$ to be in its radical, and by declaring its restriction to $B_j$ to coincide with $\langle {-},{-}\rangle _{B_j}$ . Then the condition $\text {pr}_j P_j$ be isotropic with respect to $\langle {-},{-}\rangle _{B_j}$ is equivalent to requiring $P_j$ to be isotropic with respect to $({-},{-})_{B_{\le j}}$ . Fix a stratum S in $F^{\prime }_{j-1}$ . Each point $s \in S$ includes, as part of its data, a subspace $H_{j-1}(s) \subset B_{<j}$ . The quotient $B_{\le j}/H_{j-1}(s)$ has a bilinear form induced by $({-},{-})_{B_{\le j}}$ , and

$$\begin{align*}F_j \times_{F^{\prime}_{j-1}} S = \left\{ (s, P_j) \,\Big|\, \begin{array}{@{}c@{}} s \in S, H_{j-1}(s) \subset P_j \subset M \cap B_{\le j}, \text{ and } \\ {P_j \text{ is isotropic with respect to } ({-},{-})_{B_{\le j}}} \end{array} \right\}. \end{align*}$$

This variety is an instance of the variety called “Y” in the statement of Lemma 2.3, using the vector space $M \cap B_{\le j}$ and the subspaces

$$\begin{align*}B_{< j} \qquad\text{ and }\qquad \operatorname{\mathrm{rad}} \text{pr}_j M + B_{<j}. \end{align*}$$

(Note that the latter is the radical of $M \cap B_{\le j}$ with respect to $({-},{-})_{B_{\le j}}$ .) By Lemma 2.3, $F_j \times _{F^{\prime }_{-1}} S$ admits an affine paving such that the functions

$$\begin{align*}\dim P_j \cap B_{<j}, \qquad \dim P_j \cap (\operatorname{\mathrm{rad}} \text{pr}_j M + B_{<j}) \end{align*}$$

are constant on strata. It follows that $F_j$ itself admits an affine paving such that the desired functions are constant on strata.

Next, we prove the claim for $F^{\prime }_j$ , assuming that $F_j$ has an appropriate affine paving. Fix a stratum S in $F_j$ . Each point $s \in S$ determines a subspace $P_j(s) \subset B_{\le j}$ . We have

$$\begin{align*}F^{\prime}_j \times_{F_j} S = \{ (s, H_j) \mid {s \in S \text{ and } P_j(s) \subset H_j \subset M \cap B_{\le j}} \}. \end{align*}$$

Similar reasoning shows that $F^{\prime }_j \times _{F_j} S$ admits an affine paving such that $\dim H_j \cap B_{<i}$ is constant on strata, so $F^{\prime }_j$ admits the desired kind of affine paving as well.

Proof Let S be a piece of the affine paving of $\mu _F^{-1}(M) \subset F(\underline {k},\underline {r})$ from Lemma 6.4. It is enough to show that

$$\begin{align*}X(\underline{k},\underline{r}) \times_{F(\underline{k},\underline{r})} S \end{align*}$$

admits an affine paving. For $s \in S$ , we have a collection of spaces $P_i(s), H_i(s) \subset B$ satisfying (6.3), and with $H_m(s) = M$ . For $1 \le i \le m$ , let $Y_{1i}$ be as in (6.2). We have an identification

$$\begin{align*}X(\underline{k},\underline{r}) \times_{F(\underline{k},\underline{r})} S = \left\{ (s, \tilde P_1, \ldots, \tilde P_m) \,\Big|\, \begin{array}{@{}c@{}} s \in S, \tilde P_i \in Y_{1i}, \text{ and } \\ {\text{pr}_i P_i(s) \subset \tilde P_i \subset B_i, 1\leq i\leq m-1}, \\ {\text{pr}_m P_m(s)\subset \tilde P_m\subset (\text{pr}_m M)^\perp}\end{array}\right\}. \end{align*}$$

The argument of Lemma 2.3 shows that $X(\underline {k},\underline {r}) \times _{F(\underline {k},\underline {r})} X \to S$ is a trivial fibration whose fibers are of the form

$$\begin{align*}\text{Gr}_{a_1}((\text{pr}_1M)^\perp/\text{pr}_1P_1(s))\times \dots \times \text{Gr}_{a_m}((\text{pr}_mM)^\perp/\text{pr}_mP_m(s)), \end{align*}$$

where $a_i$ is $\dim \tilde {P}_i-\dim \text {pr}_i P_i$ . (Note that the dimensions involved are independent of the choice of $s\in S$ , as $\dim \text {pr}_i P_i=\dim P_i-\dim P_i\cap B_{<i}$ and $\dim P_i\cap B_{<i}$ is constant on S by Lemma 6.4.) By Lemma 2.1, $X(\underline {k},\underline {r}) \times _{F(\underline {k},\underline {r})} S $ admits an affine paving such that the desired functions are constant on strata, and hence so does $X(\underline {k},\underline {r})$ .

Proof In the definition of $\widehat {X}(\underline {k},\underline {r})$ , the subspaces $Q_j$ and $\tilde Q_j$ are defined only for certain j. In this proof, it will be convenient to adopt the convention that

$$\begin{align*}Q_j = P_j,\ \tilde Q_j = \tilde P_j \qquad\text{ if } \begin{cases} \langle{-},{-}\rangle_{B_j} \text{ is skew-symmetric, or } \\ {\langle{-},{-}\rangle_{B_j} \text{ is symmetric and } r_j = \max \{0, 2k_j - n_j\}.} \end{cases} \end{align*}$$

The structure of the proof is very similar to that of Lemma 6.1. Let j be an integer with $0 \le j \le m$ . We define three varieties $Y_j$ , $Y^{\prime }_j$ , and $Y^{\prime \prime }_j$ as follows: a point is a collection of vector spaces as indicated below, subject to the same conditions as those in the definition of $\widehat {X}(\underline {k},\underline {r})$ .

In place of (6.1), we now have a sequence of maps

(7.1) $$ \begin{align} \widehat{X}(\underline{k},\underline{r}) = Y^{\prime\prime}_m \to Y^{\prime}_m \to Y_m \to Y^{\prime\prime}_{m-1} \to \cdots \to Y^{\prime}_1 \to Y_1. \end{align} $$

Let $1 \le j \le m$ . The same reasoning as in Lemma 6.1 shows that $Y^{\prime }_j \to Y_j$ and $Y_j \to Y^{\prime \prime }_{j-1}$ are locally trivial Grassmannian bundles. This reasoning also applies verbatim to $Y^{\prime \prime }_j \to Y^{\prime }_j$ except in the case where $\langle {-},{-}\rangle _{B_j}$ is symmetric, $r_j> \max \{0, 2k_j - n_j\}$ , and $r_j$ is odd. In this case, an additional comment is required because the definition involves $H_j \oplus \mathbb {C}$ rather than $H_j$ . Specifically, the fiber of this map over a given point of $Y^{\prime }_j$ is the set of subspaces $H_j$ such that

$$\begin{align*}Q_j \subset H_j \oplus \mathbb{C} \subset B_{<j}\oplus\tilde P_j^\perp \oplus \mathbb{C} \qquad\text{ and }\qquad \dim H_j = k_1 + \cdots + k_j. \end{align*}$$

Let $\bar Q_j$ be the image of $Q_j$ under the projection map $B \oplus \mathbb {C} \to B$ . The condition above is equivalent to

$$\begin{align*}\bar Q_j \subset H_j \subset B_{<j} \oplus \tilde P_j^\perp \qquad\text{ and }\qquad \dim H_j = k_1 + \cdots + k_j, \end{align*}$$

so the fiber over the given point of $Y^{\prime }_j$ is identified with

$$\begin{align*}\text{Gr}_{\frac{r_i-1}{2}}((B_{<j} \oplus \tilde P_j^\perp)/\bar Q_j). \end{align*}$$

The subtlety is that in order for these fibers to assemble into a locally trivial bundle, we must check that the dimension of $\bar Q_j$ is constant on $Y^{\prime }_j$ . This is true: indeed, the argument in the proof of Proposition 4.2 shows that $\dim \bar Q_j = \dim Q_j$ . We conclude that $Y^{\prime \prime }_j \to Y^{\prime }_j$ is also a locally trivial Grassmannian bundle.

Next, from the definition, we have

$$\begin{align*}Y_1 = Y_{11} \times \cdots \times Y_{1m}, \end{align*}$$

where

$$\begin{align*}Y_{1i} = \begin{cases} \mathcal{B}_{k_i,r_i}(B) & \text{ if } \langle{-},{-}\rangle_{B_i} \text{ is symmetric and } \max\{0,2k_i-n_i\} < r_i, \\ \mathcal{C}_{r_i}(B_i) & \text{ if } r_i = 0' \text{ or } 0", \\ \text{Gr}_{k_i-r_i}(B_i)^{\text{iso}} & \text{ in all other cases. } \end{cases} \end{align*}$$

Combining Corollary 2.2, Lemma 2.4, and Proposition 4.2, we conclude that $Y_1$ is smooth, connected, and simply connected, and that is cohomology is even and free. Since each map in (7.1) is a locally trivial Grassmannian bundle, the same conclusions hold for $\widehat {X}(\underline {k},\underline {r})$ .

Proof Let H be a point in $\mathcal {O}_{\underline {k},\underline {r}}$ , and let $F = \widehat {\mu }^{-1}(H)$ . A point in F consists of data $(\tilde P_i, P_i, H_i, \tilde Q_i, Q_i)$ satisfying varying conditions. By Proposition 6.2, the terms $\tilde P_i$ , $P_i$ , and $H_i$ are determined by H. To describe $\widehat {\mu }^{-1}(H)$ , we must describe the space of choices for $Q_i$ and $\tilde Q_i$ .

The formulas in Remark 6.3 imply that $P_i \cap B_{<i} = H_i \cap B_{<i}$ for all i, and that these spaces have dimension $k_1 + \cdots + k_{i-1}$ . From the definition of $\widehat {X}(\underline {k},\underline {r})$ , we see that $Q_i \cap B_{<i}$ must also equal $H_i \cap B_{<i}$ . This implies that $\dim \text {pr}_i Q_i = \dim \tilde Q_i$ , and hence that $\tilde Q_i = \text {pr}_i Q_i$ . Phrased another way, $Q_i$ must be the preimage of $\tilde Q_i$ under the map $H \cap B_i \to \text {pr}_i H$ (or if $r_i$ is odd, under $(H \cap B_i) \oplus C \to \text {pr}_i H \oplus \mathbb {C}$ ). Thus, it remains to determine the space of choices for $\tilde Q_i$ . The quotient $\tilde Q_i/\tilde P_i$ is an isotropic subspace of $\text {pr}_i H/\tilde P_i$ (or if $r_i$ is odd, of $(\text {pr}_i H \oplus \mathbb {C})/\tilde P_i$ ), which carries a nondegenerate bilinear form. We conclude that the fiber $F = \widehat {\mu }^{-1}(H)$ is described by

$$ \begin{align*} &F \cong \prod_{\substack{1 \le i \le m \\ \langle{-},{-}\rangle_{B_i} \text{ symmetric } \\ \max\{0,2k_i - n_i\} < r_i \\ r_i \text{ even }}} \text{Gr}_{r_i/2}(\text{pr}_i H/ \operatorname{\mathrm{rad}}(\text{pr}_i H))^{\text{iso}} \\& \qquad\qquad\qquad\qquad \times \prod_{\substack{1 \le i \le m \\ \langle{-},{-}\rangle_{B_i} \text{ symmetric } \\ \max\{0,2k_i - n_i\} < r_i \\ r_i \text{ odd }}} \text{Gr}_{(r_i+1)/2}((\text{pr}_i H \oplus \mathbb{C})/ \operatorname{\mathrm{rad}}(\text{pr}_i H))^{\text{iso}}. \end{align*} $$

More generally, $\widehat {\mu }^{-1}(\mathcal {O}_{\underline {k},\underline {r}}) \to \mathcal {O}_{\underline {k},\underline {r}}$ is a locally trivial fibration with fibers isomorphic to F.

Using Lemma 2.4, we see that $\pi _0(F) \cong (\mathbb {Z}/2\mathbb {Z})^d$ , where

$$\begin{align*}d = |\{ i \mid \langle{-},{-}\rangle_{B_i} \text{ is symmetric and } \max\{0,2k_i - n_i\} < r_i \}, \end{align*}$$

and $\pi _1(F) = 1$ . So we get the following short exact sequence of homotopy groups:

$$\begin{align*}1 \to \pi_1(\widehat{\mu}^{-1}(\mathcal{O}_{\underline{k},\underline{r}})) \to \pi_1(\mathcal{O}_{k,r}(V)) \to (\mathbb{Z}/2\mathbb{Z})^d\to 1. \end{align*}$$

Proposition 5.1 also identifies the middle term with $(\mathbb {Z}/2\mathbb {Z})^d$ , so $\widehat {\mu }^{-1}(\mathcal {O}_{\underline {k},\underline {r}})$ must be simply connected. Then the same reasoning as in Proposition 4.2(5) shows that $\mathcal {H}^0(\widehat {\mu }_*^\circ \Bbbk)$ is the local system corresponding to the regular representation $\Bbbk[(\mathbb {Z}/2\mathbb {Z})^d]$ . Since $2$ is invertible in $\Bbbk$ , the regular representation decomposes as a direct sum of representations that are each free of rank $1$ over $\Bbbk$ . Under the additional assumptions from (1.1), every indecomposable $\Bbbk[(\mathbb {Z}/2\mathbb {Z})^d]$ -module that is free over $\Bbbk$ occurs as a direct summand of the regular representation. This proves the second assertion in the proposition.

Proof Let S be a piece of the affine paving of $\mu ^{-1}(M) \subset X(\underline {k},\underline {r})$ from Lemma 6.5. It is enough to show that

$$\begin{align*}\widehat{F}(\underline{k},\underline{r}) \times_{X(\underline{k},\underline{r})} S \end{align*}$$

admits an affine paving. For $s \in S$ , we have subspaces $P_i(s), H_i(s), \tilde P_i(s) \subset B$ satisfying the condition in $X(\underline {k},\underline {r})$ , and with $H_m(s) = M$ . We have an identification

$$\begin{align*}\widehat{F}(\underline{k},\underline{r}) \times_{X(\underline{k},\underline{r})} S = \left\{ (s, (Q_i)_{\text{ certain }1\leq i\leq m}) \,\Big|\, \begin{array}{@{}c@{}} s \in S, \text{ and } \\ { P_i(s) \subset Q_i \subset H_i \text{ or } H_i\oplus \mathbb{C}}\end{array}\right\}. \end{align*}$$

In more detail, suppose $\langle {-},{-}\rangle _{B_i}$ is symmetric and $r_i> \max \{0,2k_i - n_i\}$ . If $r_i$ is even, then the space of choices for $ Q_i$ over $s \in S$ is identified with

$$\begin{align*}\text{Gr}( H_i(s)/( P_i(s))^{\text{iso}}. \end{align*}$$

If $r_i$ is odd, then we instead have

$$\begin{align*}\text{Gr}((H_i(s)\oplus \mathbb{C} )/P_i(s))^{\text{iso}}. \end{align*}$$

Combining these cases and using the argument of Lemma 2.3, we conclude that the map $\widehat {F}(\underline {k},\underline {r}) \times _{X(\underline {k},\underline {r})} S \to S$ is a locally trivial fibration whose fibers are products of isotropic Grassmannians. The lemma follows.

Proof Let S be a piece of the affine paving of $(\widehat {\mu }_F)^{-1}(M)$ from Lemma 7.3. We then have

$$ \begin{align*} \widehat{X}(\underline{k},\underline{r}) &\times_{\widehat{F}(\underline{k},\underline{r})} S = \\ &\left\{ (s, (\tilde Q_i)_{\text{ certain }} i \in \{1,\ldots,m\}) \,\Big|\, \begin{array}{@{}c@{}} s \in S \text{ and }\\ \tilde P_i(s) + \text{pr}_i Q_i(s) \subset \tilde Q_i \subset (\tilde Q_i)^\perp \text{ or } (\tilde Q_i)^\perp_+ \end{array} \right\}. \end{align*} $$

In more detail, suppose $\langle {-},{-}\rangle _{B_i}$ is symmetric and $r_i> \max \{0,2k_i - n_i\}$ . If $r_i$ is even, then the space of choices for $\tilde Q_i$ over $s \in S$ is identified with

$$\begin{align*}\text{Gr}(\tilde P_i(s)^\perp/(\tilde P_i(s) + \text{pr}_iQ_i(s)))^{\text{iso}}. \end{align*}$$

If $r_i$ is odd, then we instead have

$$\begin{align*}\text{Gr}((\tilde P_i(s)^\perp \oplus \mathbb{C})/(\tilde P_i(s) + \text{pr}_iQ_i(s)))^{\text{iso}}. \end{align*}$$

Combining these cases and using the argument of Lemma 2.3, we conclude that the map $\widehat {X}(\underline {k},\underline {r}) \times _{\widehat {F}(\underline {k},\underline {r})} S \to S$ is a locally trivial fibration whose fibers are products of isotropic Grassmannians. The lemma follows.