Proof Set $G := \mathrm {GL}_d \times \mathrm {GL}_{r+d} \times \mathrm {GL}_2$ . We let $T_d, T_{r+d}$ and $\mathcal {W}$ denote the corresponding tautological bundles on $B := BG$ . Let $t_i = c_i(T_{d})$ , $u_i = c_i(T_{r+d})$ , and $w_i = c_i(\mathcal {W})$ be their Chern classes, which freely generate $A^*(B)$ . Next, define $M:=[M_{r,d}/G]$ , which is the total space of the vector bundle $\mathcal {H}om(T_d,T_{r+d}) \otimes \mathcal {W}^\vee $ over B. Theorem 3.2 says that $\mathcal {V}_{r,d}$ is an open substack of M. Let us write $X:=[\Omega _{r,d}^c/G]$ for its closed complement. Here, $\Omega _{r,d}^c \subset M_{r,d}$ is the space of matrices of linear forms that drop rank along some point on $\mathbb {P}^1$ . By excision, we have a right exact sequence

(4.1) $$ \begin{align} A^{*-\mathrm{codim}X}(X) \rightarrow A^*(M) \rightarrow A^*(\mathcal{V}_{r,d}) \rightarrow 0. \end{align} $$

We shall see soon that X is irreducible of codimension r, which immediately implies that there are no relations among generators of $A^*(M)$ restricted to $\mathcal {V}_{r,d}$ in degrees less than r. In what follows, we describe relations among the restrictions of these Chern classes in degrees r up to d.

First, we construct a space $\widetilde {X}$ whose total space maps properly to M with image X. Let $\mathbb {P}(T_d)$ be the projectivization of the tautological rank d bundle, and let $\sigma : \mathbb {P} \mathcal {W} \times _B \mathbb {P}(T_d) \rightarrow B$ be the map to the base. On $\mathbb {P} \mathcal {W} \times _B \mathbb {P}(T_d)$ , we have a surjection of vector bundles

(4.2) $$ \begin{align} \hspace{12pt}\sigma^*M = \sigma^*(\mathcal{H}om(T_d, T_{r+d}) \otimes \mathcal{W}^\vee) \rightarrow \mathscr{O}_{\mathbb{P}(T_d)}(1) \otimes \sigma^*T_{r+d} \otimes \mathscr{O}_{\mathbb{P} \mathcal{W}}(1), \end{align} $$

corresponding to the evaluation of the map along a one-dimensional subspace of the fiber of $T_d$ . Let $\widetilde {X} \subset \sigma ^*M$ denote the total space of the kernel vector bundle. Informally,

(4.3) $$ \begin{align} \widetilde{X} = \{(p, \Lambda, \psi) : p \in \mathbb{P} \mathcal{W}, \Lambda \subset (T_d)_{\pi(p)}, \psi \in M_{\pi(p)}, \Lambda \subset \ker \psi(p) \subset (T_d)_{\pi(p)}\}. \end{align} $$

We have a commutative diagram:

where $\rho , \rho ',$ and $\rho "$ are all vector bundle maps. Since X consists of maps that drop rank along some point on $\mathbb {P}^1$ , we have $\sigma '(\iota (\widetilde {X})) = X$ by construction. Considering (4.3), we see that the fiber of $\widetilde {X} \to X$ over a point corresponding to $\psi $ is $\{(p, \Lambda ) : p \in \mathbb {P}^1, \Lambda \subset \ker \psi (p)\}$ (in words, the projectivization of kernels over points on $\mathbb {P}^1$ where $\psi $ drops rank). Alternatively, we can think of $\psi $ as defining a degree $1$ map $\mathbb {P}^1 \to \mathbb {P} \mathrm {Mat}_{d \times d+r}$ (so long as $\psi (p)$ is not the zero matrix at some p, which is a high codimension condition). Let $\widetilde {Z} \to Z \subset \mathbb {P} \mathrm {Mat}_{d \times d+r}$ be the usual resolution of the locus of matrices that are not full rank by marking a one-dimensional subspace of the kernel. Then, the fiber of $\widetilde {X} \to X$ over $\psi $ is the fiber product

The loci of matrices in $\mathbb {P} \mathrm {Mat}_{d \times d+r}$ that are not full rank have codimension $r+1 \geq 2$ , so a general line $\mathbb {P}^1 \to \mathbb {P} \mathrm {Mat}_{d \times d+r}$ that meets Z meets it in a single general point of Z. In particular, $\widetilde {X} \to X$ is generically one-to-one and has projective fibers. This establishes that

$$\begin{align*}\dim X = \dim \widetilde{X} = \dim M + \dim (\mathbb{P} \mathcal{W} \times_B \mathbb{P}(T_d)) - \operatorname{\mathrm{rank}} T_{k+r} = \dim M - r.\end{align*}$$

Since $\sigma ' \circ \iota : \widetilde {X} \to X$ is surjective and projective, the push forward of rational Chow groups $A_{\mathbb {Q}}^*(\widetilde {X}) \to A_{\mathbb {Q}}^*(X)$ is surjective. (Take the preimage of any cycle. Since the map is surjective, the fibers are nonempty. If the fibers are generically finite, the push forward is some nonzero multiple of the original cycle. Otherwise, slice with enough copies of the hyperplane class to get a cycle mapping with generically finite, nonempty fibers to the same image.) It follows that the image of $A_{\mathbb {Q}}^{*-r}(X) \rightarrow A_{\mathbb {Q}}^*(M)$ via push forward of cycles is the same as the image of $(\sigma ' \circ \iota )_*: A_{\mathbb {Q}}^{*-r}(\widetilde {X}) \rightarrow A_{\mathbb {Q}}^*(M)$ .

The pullback maps $(\rho ")^*$ , $(\rho ')^*$ , and $\rho ^*$ all induce isomorphisms on Chow rings. Recall that $\widetilde {X}$ is the kernel of (4.2), so it is defined inside the total space $\sigma ^*M$ by the vanishing of a section of a vector bundle $(\rho ')^* \mathscr {O}_{\mathbb {P}(T_d)}(1) \otimes \sigma ^*T_{r+d} \otimes \mathscr {O}_{\mathbb {P} \mathcal {W}}(1)$ . Therefore, the fundamental class of $\widetilde {X}$ in the Chow ring of $\sigma ^*M$ is equal to $(\rho ')^*\beta $ where

$$\begin{align*}\beta := c_{r+d}(\mathscr{O}_{\mathbb{P}(T_d)}(1) \otimes \sigma^*T_{r+d} \otimes \mathscr{O}_{\mathbb{P} \mathcal{W}}(1)).\end{align*}$$

We can write every class in $A^{*-r}(\widetilde {X})$ as $(\rho ")^* \alpha $ for a unique class $\alpha \in A^{*-r}(\mathbb {P} \mathcal {W} \times _B \mathbb {P}(T_d))$ . Then, the effect of $(\sigma ' \circ \iota )_*$ is

(4.4) $$ \begin{align} \sigma'_* \iota_* (\rho")^* \alpha = \sigma'_* \iota_* \iota^* (\rho')^* \alpha = \sigma'_*([\widetilde{X}] \cdot (\rho')^*\alpha) = \sigma'_*(\rho')^*(\beta \cdot \alpha) = \rho^*\sigma_*(\beta \cdot \alpha). \end{align} $$

Now, let $z = c_1(\mathscr {O}_{\mathbb {P} \mathcal {W}}(1))$ and $\zeta = c_1(\mathscr {O}_{\mathbb {P}(T_d)}(1))$ . By the projective bundle theorem,

(4.5) $$ \begin{align} A^*(\mathbb{P} \mathcal{W} \times_B \mathbb{P}(T_d)) = A^*(B)[z,\zeta]/(z^2 +w_1z+w_2, \zeta^{d} + \zeta^{d-1}t_1 + \cdots + \zeta t_{d-1} + t_{d}). \end{align} $$

Thus, each class $\alpha \in A^*(\mathbb {P} \mathcal {W} \times _B \mathbb {P}(T_d))$ is uniquely expressible as

(4.6) $$ \begin{align} \alpha = \sum_{i=0}^{d-1} (\sigma^*\gamma_i) \zeta^i + \sum_{i=1}^d (\sigma^*\gamma_i') z\zeta^{i-1}, \end{align} $$

where $\gamma _i \in A^*(B)$ with $\deg \gamma _i = \deg \gamma _i' = \deg \alpha - i$ .

By (4.4) and (4.6), the image of $(\sigma ' \circ \iota )_*: A_{\mathbb {Q}}^{*-r}(\widetilde {X}) \rightarrow A_{\mathbb {Q}}^*(X) \cong A_{\mathbb {Q}}^*(B)$ is the ideal generated by the classes

$$\begin{align*}f_{i,j} := \sigma_*(\beta \cdot z^i \zeta^j) \qquad \text{for }0 \leq i \leq 1, \ 0 \leq j \leq d - 1.\end{align*}$$

As $\rho ^*$ is an isomorphism on Chow, we omit it above and in what follows for ease of notation. By the excision sequence (4.1), we have

$$\begin{align*}A_{\mathbb{Q}}^*(\mathcal{V}_{r,d}) = \frac{A_{\mathbb{Q}}^*(B)}{\langle f_{i,j} : 0 \leq i \leq 1, 0 \leq j \leq d - 1\rangle} \cong \frac{\mathbb{Q}[w_1, w_2, t_1, \ldots, t_d, u_1, \ldots, u_{r+d}]}{\langle f_{i,j} : 0 \leq i \leq 1, 0 \leq j \leq d - 1\rangle}.\end{align*}$$

Since $\sigma $ has relative dimension d, the codimension of $f_{i,j}$ is $(r+d)+i+j - d = r+i+j$ . Recall that there are no relations among the generators of $A^*(M) \cong A^*(B)$ , so $f_{i,j}$ is a unique polynomial of codimension $i+j+r$ in the t’s, u’s, and w’s. Next, we shall determine the coefficients of $t_{i+j+r}$ and $u_{i+j+r}$ in this expression for $f_{i,j}$ . We shall then use this to obtain expressions for each of $t_{j+r}$ and $u_{j+r}$ in terms of $f_{0,j}, f_{1,j-1}$ , and the $t_n, u_m$ with $m, n < j + r$ . This will inductively allow us to write each $t_{j + r}$ and $u_{j+r}$ in terms of the $t_n, u_m$ with $n < r$ and $m \leq r$ .

Using the splitting principle as in [Reference Eisenbud and Harris13, Proposition 5.17], we have

$$ \begin{align*} \beta &= c_{r+d}(\mathscr{O}_{\mathbb{P}(T_d)}(1) \otimes \sigma^*T_{r+d} \otimes \mathscr{O}_{\mathbb{P} \mathcal{W}}(1)) = \sum_{i=0}^{r+d} (\zeta + z)^{r+d-i} \sigma^*u_i, \end{align*} $$

and applying (4.5), this becomes

$$ \begin{align*} &=(\zeta^{r+d} + (r+d)z\zeta^{r+d-1}) + (\zeta^{r+d-1}+(r+d-1)z\zeta^{r+d-2})\sigma^*u_1 \\ & \qquad + \cdots + (\zeta^d+dz\zeta^{d-1})\sigma^*u_r + \cdots + \sigma^*u_{r+d} + \langle \sigma^*w_1, \sigma^*w_2\rangle. \end{align*} $$

The push forward of any term involving $\sigma ^* w_1$ or $\sigma ^*w_2$ cannot contribute to the coefficient of $t_{i+j+r}$ or $u_{i+j+r}$ . Since $z^2 \in \langle \sigma ^*w_1, \sigma ^*w_2\rangle $ , after we multiply $z^i\zeta ^j$ with $\beta $ , we only care about the resulting terms where the power of z is $1$ (if the power of z is zero, then the push forward vanishes).

To compute the push forward of such terms, first note that $\sigma _*(z\zeta ^n) = 0$ for $n < d-1$ and $\sigma _*(z\zeta ^{d-1}) = 1$ (see, e.g., [Reference Eisenbud and Harris13, Lemma 9.7]). Therefore, $\sigma _*(z\zeta ^{d-1+i})$ is equal to the coefficient of $\zeta ^{d-1}$ in the expression for $\zeta ^{d-1+i}$ in terms of $\zeta ^{n}$ with $n \leq d-1$ . We find such an expression by using (4.5) repeatedly. Each contribution to the coefficient of $\zeta ^{d-1}$ in $\zeta ^{d - 1 + i}$ comes from an ordered partition $i = \lambda _1 + \lambda _2 + \cdots + \lambda _s$ with $\lambda _n \leq d$ ; the term for such a partition contributes the degree i class $\prod _{n = 1}^s (-t_{\lambda _n})$ to the coefficient of $\zeta ^{d-1}$ . We therefore conclude

$$\begin{align*}\sigma_*(z\zeta^{d-1+i}) = \begin{cases} 0,& \text{if}\ i < 0, \\ 1, &\text{if}\ i = 0, \\ \sum_{m_1 \cdot 1 + \cdots + m_d \cdot d = i} (-1)^{m_1 + \cdots + m_d} \cdot \frac{(m_1 + \cdots + m_d)!}{m_1! \cdots m_d!} \cdot t_1^{m_1} \cdots t_d^{m_d}, &\text{if}\ i \geq 1. \end{cases}\end{align*}$$

Above, the coefficient in front of a monomial for $(m_1, \ldots , m_d)$ is the number of ordered partitions of i so that j appears with multiplicity $m_j$ . In particular, we compute

(4.7) $$ \begin{align} f_{1, j-1} &= \sigma_*(\beta z \zeta^{j-1}) = - t_{j + r} + u_{j + r} + (\text{terms with subscript}\ < j+r), \end{align} $$

(4.8) $$ \begin{align}\hspace{12pt} f_{0, j} &= \sigma_*(\beta \zeta^j) = -(r+d)t_{j+r} + (d - j)u_{j+r} + (\text{terms with subscript}\ < j+r). \end{align} $$

When $j = 0$ , we can write

(4.9) $$ \begin{align} t_r = \frac{d}{r+d}u_r + \text{(terms with subscript}\ < r). \end{align} $$

When $0 < j \leq d-r$ , taking appropriate linear combinations of (4.7) and (4.8), we have

(4.10) $$ \begin{align} u_{j+r} &= \frac{1}{j+r} ((r+d) f_{1,j-1} - f_{0,j}) + (\text{terms with subscript}\ < j+r), \end{align} $$

(4.11) $$ \begin{align} t_{j+r} &= \frac{1}{r+j} ((d-j) f_{1,j-1} - f_{0,j}) + (\text{terms with subscript}\ < j+r). \end{align} $$

In $A_{\mathbb {Q}}^*(\mathcal {V}_{r,d})$ , the classes $f_{i, j}$ are zero. Thus, using (4.9)–(4.11) inductively, the classes $t_{j+r}$ for $0 \leq j \leq d-r$ and $u_{j + r}$ for $0 < j \leq d-r$ are expressible as polynomials in $w_1, w_2, t_1, \ldots , t_{r-1}, u_1, \ldots , u_r$ . The remaining $f_{0,j}$ and $f_{1,j-1}$ with $j> d - r$ give rise to relations in degrees greater than d. Hence, the map

(4.12) $$ \begin{align} \mathbb{Q}[w_1, w_2, t_1,t_2,\ldots, t_{r-1}, u_1, \ldots, u_{r}] \to A_{\mathbb{Q}}^*(\mathcal{V}_{r,d}) \end{align} $$

is an isomorphism in degrees $* \leq d$ . For dimension reasons, there can be no relations among the generators $w_1, w_2, a_1, \ldots , a_r, a_2', \ldots , a_r'$ in degrees less than d because they are a list of the same number of generators in the same degrees as (4.12).▪

Proof Recall that the codimension of the complement of $\mathcal {U}_{m, r, \ell }$ is $\ell + mr + 1$ . Thus, choosing m such that $\ell + mr + 1> i$ , we see

$$ \begin{align*} \dim A_{\mathbb{Q}}^i(\mathcal{B}_{r,\ell}) &= \dim A_{\mathbb{Q}}^i(\mathcal{U}_{m, r, \ell}) = \dim A_{\mathbb{Q}}^i(\mathcal{V}_{r,mr+\ell}) \\ &= \dim \mathbb{Q}[w_1, w_2, a_1, \ldots, a_r, a_2', \ldots, a_r']_i. \end{align*} $$

We already know $ \mathbb {Q}[w_1, w_2, a_1, \ldots , a_r, a_2', \ldots , a_r'] \twoheadrightarrow A^*(\mathcal {B}_{r,\ell })$ , so we conclude that it has no kernel for dimension reasons.▪

Proof We induct on s. The $s=1$ case is immediate. Let $G' = \operatorname {\mathrm {Aut}}(\oplus _{i=1}^{s-1} \mathscr {O}(d_i)^{\oplus r_i})$ and $H = G' \times \mathrm {GL}_{r_s} \subset \operatorname {\mathrm {Aut}}(\mathscr {O}(\vec {e}))$ . There is a quotient map $\operatorname {\mathrm {Aut}}(\mathscr {O}(\vec {e})) \rightarrow H$ defined by forgetting blocks not in H, which expresses $\operatorname {\mathrm {Aut}}(\mathscr {O}(\vec {e}))$ as a semidirect product $N \rtimes H$ where $N \cong H^0(\mathscr {O}_{\mathbb {P}^1}(d_s - d_1))^{\oplus (r_1r_s)} \oplus \cdots \oplus H^0(\mathscr {O}_{\mathbb {P}^1}(d_s - d_{s-1}))^{\oplus (r_{s-1}r_s)}$ is affine. Moreover,

$$\begin{align*}H_{\vec{e}} = (N \rtimes H) \rtimes \mathrm{GL}_2 = N \rtimes ((G' \rtimes \mathrm{GL}_2) \times \mathrm{GL}_{r_s}).\end{align*}$$

Now, let $H_{\vec {e}}$ act on N where elements of N act by left multiplication and elements of $(G' \rtimes \mathrm {GL}_2) \times \mathrm {GL}_{r_s}$ act by conjugation. This action is affine linear, so the quotient, which is $B((G' \rtimes \mathrm {GL}_2) \times \mathrm {GL}_{r_s})$ , is an affine bundle over $BH_{\vec {e}}$ . By the homotopy property, the Chow ring and higher Chow groups of $BH_{\vec {e}}$ agree with that of $B(G' \rtimes \mathrm {GL}_2) \times \mathrm {BGL}_{r_s}$ , which, by induction, are isomorphic to those of $(\mathrm {BGL}_{r_1} \times \cdots \times \mathrm {BGL}_{r_{s-1}} \times \mathrm {BGL}_2) \times \mathrm {BGL}_{r_s}$ . The vanishing of the first higher Chow group over $\mathbb {C}$ is equation (2.2).▪

Proof Suppose that U is a finite union of strata and $\Sigma _{\vec {e}}$ is a disjoint stratum which is closed in $X = U \cup \Sigma _{\vec {e}}$ . By Lemma 5.1, we know $A^{i}(\Sigma _{\vec {e}})$ is a finitely generated free $\mathbb {Z}$ -module for all i. By induction, we may also assume that $A^i(U)$ is a finitely generated free $\mathbb {Z}$ -module. It suffices to show that $A^i(X)$ is also a finitely generated free $\mathbb {Z}$ -module.

Let us first deduce the result over $\mathbb {C}$ . We have a localization long exact sequence

$$ \begin{align*} \cdots &\rightarrow \mathrm{CH}^{*-u(\vec{e})}(\Sigma_{\vec{e}}, 1, \mathbb{Z}/p) \rightarrow \mathrm{CH}^*(X,1,\mathbb{Z}/p) \rightarrow \mathrm{CH}^*(U,1,\mathbb{Z}/p) \\ &\rightarrow A^{*-u(\vec{e})}(\Sigma_{\vec{e}}) \otimes \mathbb{Z}/p\mathbb{Z} \rightarrow A^*(X) \otimes \mathbb{Z}/p\mathbb{Z} \rightarrow A^*(U) \otimes \mathbb{Z}/p\mathbb{Z} \rightarrow 0. \end{align*} $$

By Lemma 5.1, we have $\mathrm {CH}^{*-u(\vec {e})}(\Sigma _{\vec {e}}, 1, \mathbb {Z}/p)=0$ , so if $\mathrm {CH}^{*}(U, 1, \mathbb {Z}/p\mathbb {Z}) = 0$ , then we have $\mathrm {CH}^*(X,1,\mathbb {Z}/p) = 0$ too. It follows by induction that $\mathrm {CH}^*(U,1,\mathbb {Z}/p) = 0$ for any finite union of strata. Hence, we have an exact sequence

(5.2) $$ \begin{align} 0 \rightarrow A^{*-u(\vec{e})}(\Sigma_{\vec{e}}) \otimes \mathbb{Z}/p\mathbb{Z} \rightarrow A^*(X) \otimes \mathbb{Z}/p\mathbb{Z} \rightarrow A^*(U) \otimes \mathbb{Z}/p\mathbb{Z} \rightarrow 0 \end{align} $$

for each prime p. Because the $\mathbb {Z}$ -modules involved are finitely generated, the exactness of (5.2) for all p implies that

(5.3) $$ \begin{align} 0 \rightarrow A^{i-u(\vec{e})}(\Sigma_{\vec{e}}) \rightarrow A^i(X) \rightarrow A^i(U) \rightarrow 0 \end{align} $$

is exact. Since $A^{i-u(\vec {e})}(\Sigma _{\vec {e}})$ and $A^i(U)$ are finitely generated free $\mathbb {Z}$ -modules, so is $A^i(X)$ . This concludes the proof over $\mathbb {C}$ .

The exactness of (5.3) over $\mathbb {C}$ also tells us that

(5.4) $$ \begin{align} \operatorname{\mathrm{rank}} A^i(\mathcal{B}_{r,\ell}) = \sum_{\vec{e}} \operatorname{\mathrm{rank}} A^{i-u(\vec{e})}(\Sigma_{\vec{e}}). \end{align} $$

We claim that (5.4) in fact holds over any ground field. Indeed, Corollary 4.2 holds over any field, so the left-hand side is independent of the ground field. Similarly, using Lemma 5.1, we have $A^*(\Sigma _{\vec {e}}) = A^*(\prod \mathrm {BGL}_{r_i} \times \mathrm {BGL}_2)$ and the latter is independent of the ground field, so the right-hand side of (5.4) is independent of the ground field.

Now, working over any ground field, we claim that the map $A^{i-u(\vec {e})}(\Sigma _{\vec {e}}) \to A^i(X)$ for attaching each stratum is injective. If not, then the image of $A^{i-u(\vec {e})}(\Sigma _{\vec {e}}) \to A^i(X)$ would have rank strictly less than $\operatorname {\mathrm {rank}} A^{i - u(\vec {e})}(\Sigma _{\vec {e}})$ . Then, $\operatorname {\mathrm {rank}} A^i(\mathcal {B}_{r,\ell })$ would be less than the sum $\sum _{\vec {e}} \operatorname {\mathrm {rank}} A^{i-u(\vec {e})}(\Sigma _{\vec {e}})$ , violating (5.4). Hence, we must have an exact sequence as in (5.3) for each $\vec {e}$ . Arguing as before, we know $A^{i-u(\vec {e})}(\Sigma _{\vec {e}})$ and $A^i(U)$ are finitely generated free $\mathbb {Z}$ -modules, so because (5.3) is exact, $A^i(X)$ is also a finitely generated free $\mathbb {Z}$ -module.▪

Proof First, consider the case $k = \mathbb {C}$ . As before, suppose that U is a finite union of strata and $\Sigma _{\vec {e}}$ is a disjoint stratum which is closed in $X = U \cup \Sigma _{\vec {e}}$ . Because $\Sigma _{\vec {e}}$ and X are smooth, the cohomology of the pair $H^*(X, U)$ is the reduced cohomology of the Thom space of the normal bundle of $\Sigma _{\vec {e}} \subset X$ . By the Thom isomorphism, the cohomology of this pair is then $H^*(X, U) \cong H^{* - 2u(\vec {e})}(\Sigma _{\vec {e}})$ . By Lemma 5.1, we have $H^*(\Sigma _{\vec {e}}) = H^*(\prod \mathrm {BGL}_{r_i} \times \mathrm {BGL}_2)$ , which vanishes in odd degrees. By induction, we may assume that the odd cohomology of U vanishes. Thus, the long exact sequence for the pair $(X, U)$ gives an exact sequence

(5.5) $$ \begin{align} 0 \rightarrow H^{2(i - u(\vec{e}))}(\Sigma_{\vec{e}}) \rightarrow H^{2i}(X) \rightarrow H^{2i}(U) \rightarrow 0 \end{align} $$

for each i. The cycle class map sends the short exact sequence (5.3) to (5.5). Because $A^*(\mathrm {BGL}_{r_i}) \to H^{2i}(\mathrm {BGL}_{r_i})$ is an isomorphism, using Lemma 5.1, we see that $A^{i - u(\vec {e})}(\Sigma _{\vec {e}}) \to H^{2(i - u(\vec {e}))}(\Sigma _{\vec {e}})$ is an isomorphism. By induction, we may assume that $A^i(U) \to H^{2i}(U)$ is an isomorphism. By the five lemma, we see that $A^i(X) \to H^{2i}(X)$ is also an isomorphism.

In étale cohomology, there is a Gysin long exact sequence for $\Sigma _{\vec {e}} \subset X$ and the open compliment $U = X \smallsetminus \Sigma _{\vec {e}}$ . Moreover, we have $H_{\acute{{\rm e}}\text{t}}^*(\Sigma _{\vec {e}}, \mathbb {Z}_\ell ) = H_{\acute{{\rm e}}\text{t}}^*(\prod \mathrm {BGL}_{r_i} \times \mathrm {BGL}_2, \mathbb {Z}_\ell )$ , which vanishes in odd degrees. Thus, upon tensoring Chow with $\mathbb {Z}_\ell $ the same proof establishes that $A^*(\mathcal {B}_{r,d}) \otimes \mathbb {Z}_{\ell } \to H^{2*}_{\acute{{\rm e}}\text{t}}(\mathcal {B}_{r,d}, \mathbb {Z}_\ell )$ is an isomorphism.▪

Proof Corollary 4.2 determines $A_{\mathbb {Q}}^*(\mathcal {B}_{r,\ell })$ . Lemma 5.2 shows that $A^*(\mathcal {B}_{r,\ell })$ is a subring of $A_{\mathbb {Q}}^*(\mathcal {B}_{r,\ell })$ , and Lemma 3.5 identifies it as the subring generated by the Chern classes of the sheaves $\pi _*\mathcal {E}(m)$ . Lemma 5.3 shows that the cycle class map is an isomorphism.▪

Proof We will use Grothendieck–Riemann–Roch to compute the Chern characters and make use of formal manipulations that turn power sums (Chern characters) into elementary symmetric functions (Chern classes). It is convenient to package this information in generating functions.

Given some $\alpha _1, \ldots , \alpha _r$ , let $\sigma _j := \sigma _j(\alpha _1, \ldots , a_r) = \sum _{i_1 < i_2 < \cdots < i_j} \alpha _{i_1}\alpha _{i_2} \cdots a_{i_j}$ denote the jth elementary symmetric function in the $\alpha _i$ . For each $j \geq 0$ , there is a polynomial $p_j(x_1, \ldots , x_r)$ such that $p_j(\sigma _1, \ldots , \sigma _r) = \alpha _1^j + \cdots + \alpha _r^j$ . These polynomials satisfy

(5.8) $$ \begin{align} \log (1+x_1t +\cdots + x_rt^r) = \sum_{j=1}^\infty \frac{(-1)^{j+1}}{j} p_j(x_1,\ldots,x_r) t^j. \end{align} $$

For any vector bundle E, the jth Chern character is related to the Chern classes by

$$\begin{align*}\mathrm{ch}_j(E) = \frac{1}{j!} p_j(c_1(E), \ldots, c_r(E)).\end{align*}$$

In what follows, we work modulo the ideal $\langle w_1, w_2 \rangle $ , so that $z^2 = 0$ . Recall that, for any polynomial $f(x_1, \ldots , x_r)$ , we have

$$\begin{align*}f(x_1, \ldots, x_i + \epsilon, \ldots, x_r) = f(x_1, \ldots, x_r) + \epsilon \frac{\partial f}{\partial x_i}(x_1, \ldots, x_r) + \langle \epsilon^2 \rangle.\end{align*}$$

Applying this repeatedly, one sees more generally that

$$\begin{align*}f(x_1 + h_1 \epsilon, x_2 + h_2 \epsilon, \ldots, x_r + h_r \epsilon) = f(x_1, \ldots, x_r) + \epsilon \cdot \sum_{i=1}^r h_i \frac{\partial f}{\partial x_i} (x_1, \ldots, x_r) + \langle \epsilon^2 \rangle.\end{align*}$$

In particular,

$$ \begin{align*} \mathrm{ch}_j(\mathcal{E}) &= \frac{1}{j!}p_j(\pi^*a_1 + \pi^*a_1' z, \ldots, \pi^*a_r + \pi^* a_r' z) \\ &= \frac{1}{j!} p_j(\pi^*a_1, \ldots, \pi^*a_r) + z \cdot \frac{1}{j!} \left(\sum_{i=1}^r \pi^*a_i' \frac{\partial p_j}{\partial x_i}(\pi^*a_1, \ldots, \pi^*a_r)\right). \end{align*} $$

Now, let

(5.9) $$ \begin{align} c_j = \frac{1}{j!} p_j(a_1, \ldots, a_r) \qquad \text{and} \qquad c_j' = \frac{1}{j!} \sum_{i=1}^r a_i' \frac{\partial p_j}{\partial x_i}(a_1, \ldots, a_r), \end{align} $$

so that $\mathrm {ch}_j(\mathcal {E}) = \pi ^*(c_j) + \pi ^*(c_j')z$ , with $c_j \in A_{\mathbb {Q}}^{j}(\mathcal {B}_{r,\ell })$ and $c_j' \in A_{\mathbb {Q}}^{j-1}(\mathcal {B}_{r,\ell })$ . In addition, let us write $c = c_0 + c_1 + c_2 + \cdots $ and $c' = c_1' + c_2' + \cdots $ , so that $\mathrm {ch}(\mathcal {E}) = \pi ^*(c) + \pi ^*(c')z$ .

The relative tangent bundle of $\pi : \mathbb {P} \mathcal {W} \rightarrow \mathcal {B}_{r,\ell }$ has Todd class $\mathrm {td}(T_\pi ) = 1 + \frac {1}{2} c_1(T_\pi ) = 1+z.$ Moreover, $\mathrm {ch}(\mathcal {E}(m)) = \mathrm {ch}(\mathcal {E})\mathrm {ch}(\mathscr {O}_{\mathbb {P} \mathcal {W}}(m)) = \mathrm {ch}(\mathcal {E})(1+mz)$ . On $\mathcal {U}_m$ , we have $R^1\pi _* \mathcal {E}(m) =0$ , so Grothendieck–Riemann–Roch tells us

$$ \begin{align*} \mathrm{ch}(\pi_*\mathcal{E}(m)) &= \pi_*(\mathrm{ch}(\mathcal{E}(m))\mathrm{td}(T_\pi)) \\ &= \pi_*((\pi^*(c)+\pi^*(c')z)(1+(m+1)z)) = c' + (m+1)c. \end{align*} $$

To recover the Chern classes, by (5.8), we have to evaluate

(5.10) $$ \begin{align} \hspace{15pt}&\exp\left(\sum_{j=1}^\infty (j-1)! (-1)^{j+1}\mathrm{ch}_j(\pi_*\mathcal{E}(m)) t^j\right) \notag\\&\quad= \exp\left(\sum_{j=1}^\infty (j-1)! (-1)^{j+1}(c_{j+1}' + (m+1)c_j )t^j\right) \notag \\ &\quad =\exp\left(\sum_{j=1}^\infty\left((j-1)!(-1)^{j+1} \cdot \frac{1}{j!}\sum_{i=1}^r a_i' \frac{\partial p_j}{\partial x_i}(a_1, \ldots, a_r)\right) t^j\right) \notag\\&\qquad{\times}\,\exp\left(\sum_{j=1}^\infty \frac{(-1)^{j+1}}{j!}p_j(a_1, \ldots, a_r)t^j\right)^{m+1} \notag \\ &\quad =\exp\left(\sum_{i=1}^r a_i'\sum_{j=1}^\infty \frac{(-1)^{j+1}}{j(j+1)} \frac{\partial p_{j+1}}{\partial x_i}(a_1,\ldots,a_r)t^j\right) (1+a_1t+\cdots+a_rt^r)^{m+1}.\notag\\ \end{align} $$

To evaluate the infinite sums inside (5.10), we consider

(5.11) $$ \begin{align} \frac{d}{d t} \left(\sum_{j=1}^\infty \frac{(-1)^{j+1}}{j(j+1)} \frac{\partial p_{j+1}}{\partial x_i}(a_1, \ldots, a_r) t^j\right) &= \sum_{j=1}^\infty \frac{(-1)^{j+1}}{j+1} \frac{\partial p_{j+1}}{\partial x_i}(a_1, \ldots, a_r) t^{j-1}\end{align} $$

(5.12) $$ \begin{align} & = \sum_{j=2}^\infty \frac{(-1)^j}{j} \frac{\partial p_j}{\partial x_i}(a_1, \ldots, a_r)t^{j-2}. \end{align} $$

Note that $\partial p_j/\partial x_i = 0$ if $j < i$ . Therefore taking the partial derivative of (5.8) with respect to $x_i$ , we see that (5.12) is equal to

(5.13) $$ \begin{align} &\qquad\qquad\qquad\qquad = \frac{-1}{t^2} \left(\frac{\partial}{\partial x_i} \log(1 + a_1 t + \cdots + a_r t^r) - \delta_{i1} t\right) \notag \\ &\qquad\qquad\qquad \qquad = \begin{cases} (a_1 + a_2 t + \cdots + a_rt^{r-1})/(1 + a_1 t + \cdots + a_r t^r), & \text{if}\ i = 1, \\ -t^{i-2}/(1 + a_1t + \cdots + a_rt^r), & \text{if}\ i \geq 2. \end{cases} \end{align} $$

Taking the formal integral of the left-hand side of (5.11), which equals (5.13), we have

$$ \begin{align*} \sum_{i=1}^r a_i'\sum_{j=1}^\infty \frac{(-1)^{j+1}}{j(j+1)} \frac{\partial p_{j+1}}{\partial x_i}(a_1,\ldots,a_r)t^j &= \int a_1' \cdot \frac{a_1 + a_2t + \cdots + a_r t^{r-1}}{1 + a_1 t + \cdots + a_rt^r}\\ &\quad- \frac{a_2' + a_3't + \cdots + a_r't^{r-2}}{1 + a_1 t + \cdots + a_rt^r}. \end{align*} $$

Exponentiating this quantity and multiplying by $ (1+a_1t+\cdots +a_rt^r)^{m+1} $ shows that (5.10) is equal to the claimed formula (5.7).▪

Proof

By Vistoli’s theorem, we have $A^*(\mathcal {B}_{r,\ell }^\dagger ) = A^*(\mathcal {B}_{r,\ell })/\langle w_1, w_2 \rangle $ . By Lemma 5.4, after setting $w_1 = w_2 = 0$ , the Chern classes of $\pi _*\mathcal {E}(m)$ for all m are expressible in terms of the $e_i$ in (5.6) and $a_1, \ldots , a_r$ . To see that the cycle class map is an isomorphism, one may argue as in Lemmas 5.1 and 5.3, but using the splitting loci $\Sigma ^{\dagger }_{\vec {e}}$ on $\mathcal {B}_{r,\ell }^\dagger $ . One has $\Sigma _{\vec {e}}^{\dagger } \cong B\operatorname {\mathrm {Aut}}(\mathscr {O}(\vec {e}))$ , so the proof is essentially the same after dropping the $\mathrm {BGL}_2$ factor from Lemma 5.1.▪

Proof Since $A^*(\mathcal {B}_{r,\ell })$ surjects onto $A^*(\mathcal {B}_{r,\ell }^\dagger )$ , it suffices to prove the claim for $\mathcal {B}_{r,\ell }^\dagger $ . Suppose to the contrary that $A^*(\mathcal {B}_{r,\ell }^\dagger )$ were finitely generated as a $\mathbb {Z}$ -algebra. Then, there would exist a finite set of primes $\{p_1, \ldots , p_n\}$ such that for all $\eta \in A^*(\mathcal {B}_{r,\ell }^\dagger )$ , there exist $m_1, \ldots , m_n \in \mathbb {Z}$ such that $p_1^{m_1} \cdots p_n^{m_n} \cdot \eta \in \mathbb {Z}[a_1, \ldots , a_r, a_2', \ldots , a_r']$ . Let q be a prime not in $\{p_1, \ldots , p_n\}$ . Let us consider the coefficient $f_q$ in (5.6). Performing the formal integral, we see that $F(t) = \exp (-a_2' t + \cdots )$ , so $(a_2')^q$ appears in $f_q$ with coefficient $\frac {1}{q!}$ . Hence, $p_1^{m_1} \cdots p_n^{m_n} \cdot f_q \notin \mathbb {Z}[a_1, \ldots , a_r, a_2', \ldots , a_r']$ for any $m_1, \ldots , m_n$ .▪

Proof Let $\sum e_i t^i = F(t)$ as in (5.6). Then, $A^*(\mathcal {B}_{r,\ell }^\dagger ) \subset \mathbb {Q}[a_1, \ldots , a_r, a_2', \ldots , a_r']$ is the subring generated by the classes $e_i$ together with $a_1, \ldots , a_r, a_2', \ldots , a_r'$ . Let

$$\begin{align*}S = A^1(\mathcal{B}_{r,\ell}^\dagger) &\cdot A^{r-1}(\mathcal{B}_{r,\ell}^\dagger) + A^2(\mathcal{B}_{r,\ell}^\dagger) \cdot A^{r-1}(\mathcal{B}_{r,\ell}^\dagger) + \cdots + A^{\lfloor r/2 \rfloor}(\mathcal{B}_{r,\ell}^\dagger)\\ &\cdot A^{\lceil r/2 \rceil}(\mathcal{B}_{r,\ell}^\dagger) \subset A^r(\mathcal{B}_{r,\ell}^\dagger),\end{align*}$$

so S is the subspace of $A^r(\mathcal {B}_{r,\ell }^\dagger )$ generated by products of integral elements from lower codimension. Taking $\frac {d}{dt}$ of (5.6), we have

$$\begin{align*}(1 + a_1t + \cdots + a_rt^r) \sum_{i=1}^\infty i e_i t^{i-1} = (\ell(a_1 + \cdots + a_rt^{r-1}) - (a_2' + \cdots + a_r't^{r-2})) \sum_{j=0}^\infty e_jt^j.\end{align*}$$

Equating the coefficient of $t^{i-1}$ on both sides for $i \leq r$ , we find

(5.14) $$ \begin{align} ie_i + (i-1)e_{i-1}a_1 + \cdots + e_1a_{i-1} &= \ell(e_{i-1}a_1 + \cdots + e_1a_{i-1} + a_i)\nonumber\\&\quad - (e_{i-1}a_2' + \cdots + e_1a_i' + a_{i+1}'). \end{align} $$

(Above, $a_{i+1}' = 0$ if $i = r$ .) In particular, setting $i = r$ , we find

$$\begin{align*}re_r - \ell a_r \in S.\end{align*}$$

This establishes that $A^r(\mathcal {B}_{r,\ell }^\dagger )/S$ is a quotient of $\mathbb {Z} \oplus \mathbb {Z}/m\mathbb {Z}$ where $m = \gcd (r, \ell )$ .

If $\ell \neq 0$ mod r, then $m < r$ , so $A^r(\mathcal {B}_{r,\ell }^\dagger )/S$ has no element of order r. On the other hand, if $\ell = 0$ , then $m = r$ , and we claim that $e_r$ is an element of order r in $A^r(\mathcal {B}_{r,0}^\dagger )/S$ .

If $\ell = 0$ , we have $e_1 = -a_2'$ , and, more generally, no copy of $a_{r-1}$ appears in $e_{r-1}$ . In particular, using (5.14), we find that $e_r = -\frac {1}{r}a_2'a_{r-1} + \cdots $ . Meanwhile, since $a_{r-1}$ does not appear in $e_{r-1}$ , elements of $A^{r-1}(\mathcal {B}_{r,0}^\dagger )$ necessarily have integer coefficient of $a_{r-1}$ . Hence, elements of S always have integer coefficient of $a_2'a_{r-1}$ . It follows that $ne_r \notin S$ for any $n < r$ .▪

Proof The codimension of the complement of $\mathcal {U}_{3, 2, \ell } \subset \mathcal {B}_{2,\ell }$ is $6 + \ell $ . In particular, for $\ell = 0, 1$ , we have $A^4(\mathcal {B}_{2,\ell }) = A^4(\mathcal {U}_{3,2,\ell })$ . By Lemma 3.3, we have that $A^4(\mathcal {B}_{2,\ell })/(A^1(\mathcal {B}_{2,\ell }) \cdot A^3(\mathcal {B}_{2,\ell }) + A^2(\mathcal {B}_{2,\ell })^2)$ is generated by $t_4 := c_4(\pi _*\mathcal {E}(2))$ and $u_4 := c_4(\pi _*\mathcal {E}(3))$ . These classes are determined by the splitting principle and Grothendieck–Riemann–Roch, and one can calculate them quickly using the Schubert2 package in Macaulay2. When $\ell =0$ , we have

$$ \begin{align*} t_4 &= -\frac{1}{4}a_1^3a_2' + \frac{11}{24} a_1^2a_2^{\prime 2} - \frac{1}{4} a_1a_2^{\prime 3} + \frac{1}{24}a_2^{\prime 4} + 3a_1^3w_1 - 4a_1^2a_2'w_1 + a_1a_2^{\prime 2}w_1 + \frac{1}{12}a_1^{\prime 3}w_1 + 15a_1^2w_1^2 \\ &\quad - \frac{25}{4}a_1a_2'w_1^2 + \frac{13}{24}a_2^{\prime 2}w_1^2 + 18a_1w_1^3 - \frac{3}{2}a_2'w_1^3 + 4w_1^4 + 3a_1^2a_2 - 4a_1a_2a_2' + \frac{7}{6}a_2a_2^{\prime 2} + 18a_1a_2w_1 \\ &\quad- 5a_2a_2'w_1 + 18a_2w_1^2 + 12a_1^2w_2 + a_1a_2'w_2 - \frac{19}{6}a_2^{\prime 2}w_2 + 48a_1w_1w_2 + 2a_2'w_1w_2 + 40w_1^2w_2 + 3a_2^2 + 16w_2^2, \\ u_4 &= a_1^4 - \frac{25}{12}a_1^3a_2' + \frac{35}{24}a_1^2a_2^{\prime 2} - \frac{5}{12}a_1a_2^{\prime 3} + \frac{1}{24}a_2^{\prime 4} + 30a_1^3aw_1 - \frac{53}{2}a_1^2a_2'w_1 + \frac{20}{3}a_1a_2^{\prime 2}w_1 - \frac{5}{12}a_2^{\prime 3}w_1 + 185 a_2^2w_1^2 \\ &\quad -\frac{845}{12}a_2a_2'w_1^2 + \frac{133}{24}a_2^{\prime 2}w_1^2 + 360 a_1w_1^3 - 45a_2'w_1^3 + 193w_1^4 + 12a_1^2a_2 - \frac{55}{6}a_1a_2a_2' + \frac{5}{3}a_2a_2^{\prime 2} + 90 a_1a_2w_1 - \frac{43}{2}a_2a_2'w_1 \\ &\quad + 130a_2w_1^2 + 70a_1^2w_2 - \frac{65}{6}a_1a_2'w_2 - \frac{11}{3}a_2^{\prime 2}w_2 + 450a_1w_1w_2 - \frac{45}{2}a_2'w_1w_2 + 616w_1^2w_2 + 6a_2^2 + 20a_2w_2 + 118w_2^2. \end{align*} $$

From this, we observe that $t_4$ and $u_4$ are expressible as the sum of a class in

$$\begin{align*}\mathbb{Z}[a_1, a_2', a_2, w_1,w_2]_4 \subset A^1(\mathcal{B}_{2,\ell}) \cdot A^3(\mathcal{B}_{2,\ell}) + A^2(\mathcal{B}_{2,\ell})^2,\end{align*}$$

plus a class divisible by $a_2'$ . That is, when $\ell = 0$ , we can choose generators of

$$\begin{align*}A^4(\mathcal{B}_{2,\ell})/(A^1(\mathcal{B}_{2,\ell}) \cdot A^3(\mathcal{B}_{2,\ell}) + A^2(\mathcal{B}_{2,\ell})^2),\end{align*}$$

so that a multiple lies in $A^1(\mathcal {B}_{2,\ell }) \cdot A^3(\mathcal {B}_{2,\ell })$ . On the other hand, if $\ell = 1$ , then

$$ \begin{align*} t_4 = \frac{35}{8}a_2^2 + \cdots \qquad \text{and} \qquad u_4 = \frac{63}{8}a_2^2 + \cdots. \end{align*} $$

However, all codimension $2$ classes have denominators at most $2$ . Thus, there is no adjustment of $t_4$ or $u_4$ by integral classes from lower codimension so that the result is divisible by a codimension $1$ class.▪