2. Non-degenerate unimodular sequences

In this section we review notation and a few results from [Reference Sarwar and SchlichtingSS25].

Throughout this paper, $n\geqslant 0$ will be an integer, R will be a commutative local ring with infinite residue field, $R^*$ its group of units, $\text{GL}_n(R)$ the group of invertible $n\times n$ matrices with entries in R,

$$\psi_{2n} = \psi_2 \perp \cdots \perp \psi_2 = \Bigg(\begin{smallmatrix}\psi_2 &&&\\ & \psi_2 && \\ && \ddots & \\ &&& \psi_2\end{smallmatrix}\Bigg) = \bigoplus_1^n\psi_2,\quad \psi_2 = \big(\begin{smallmatrix}0 & 1 \\ -1 & 0 \end{smallmatrix}\big),$$

the standard hyperbolic symplectic form of rank 2n, and

$$\text{Sp}_{2n}(R) =\{ A \in \text{GL}_{2n}(R)|\ {^t\!A}\, \psi_{2n}\, A = \psi_{2n}\}$$

the symplectic group or rank 2n, considered as a subgroup of $\text{Sp}_{2n+2}(R)$ by means of the embedding

(2.1) \begin{equation}\text{Sp}_{2n}(R) \subset \text{Sp}_{2n+2}(R): A \mapsto \Big(\begin{smallmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & A\end{smallmatrix}\Big).\end{equation}

For the purpose of this paper, the symplectic group of rank $2n+1$ is the subgroup

$$\text{Sp}_{2n+1}(R) = \{A \in \text{Sp}_{2n+2}(R)|\ Ae_1 = e_1\}$$

of $\text{Sp}_{2n+2}(R)$ fixing the first standard basis vector $e_1$ . This is the group of matrices

(2.2) \begin{equation}\Big(\begin{smallmatrix}1& c & {^t\!u}\psi M \\ 0 & 1 & 0 \\ 0 & u & M \end{smallmatrix}\Big)\end{equation}

where $\psi = \psi_{2n}$ , $M\in \text{Sp}_{2n}(R)$ , $u\in R^{2n}$ , $c\in R$ . The inclusions (2.1) refine to the sequence of inclusions of groups

(2.3) \begin{equation}1=\text{Sp}_0(R) \subset \text{Sp}_1(R) \subset \text{Sp}_2(R) \subset \dots \subset \text{Sp}_n(R) \subset \text{Sp}_{n+1}(R)\subset \cdots,\end{equation}

where

$$\text{Sp}_{2n}(R) \subset \text{Sp}_{2n+1}(R): M \mapsto \Big(\begin{smallmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0& 0 & M \end{smallmatrix}\Big), \quad\text{Sp}_{2n-1} (R)\subset \text{Sp}_{2n}(R): M \mapsto M.$$

In Theorem 7.1 below we study homology stability for the sequence of groups (2.3). We shall denote the inclusions $\text{Sp}_{r}(R) \subset \text{Sp}_s(R)$ by $\varepsilon^s_r$ , or simply by $\varepsilon$ if source and target group are understood, $r\leqslant s$ . Small-rank symplectic groups are as follows:

$$\text{Sp}_0(R) = \{1\},\quad\text{Sp}_1(R) =\big\{\big(\begin{smallmatrix}1 & c \\ 0 & 1\end{smallmatrix}\big)|\ c\in R\big\},\quad\text{Sp}_2(R) = \text{SL}_2(R).$$

Let $0\leqslant q$ be an integer. We denote by $\text{Skew}_q(R)$ the set of $q\times q$ skew-symmetric matrices with entries in R, that is, those matrices $A=(a_{ij})$ such that $a_{ij}=-a_{ji}$ , $a_{ii}=0$ , $a_{ij}\in R$ , $1 \leqslant i,j \leqslant q$ . We denote by

$$\text{Skew}_q^+(R) \subset \text{Skew}_q(R)$$

the subset of non-degenerate skew-symmetric matrices, that is, those matrices $A\in \text{Skew}_q(R)$ such that for all subsets $I\subset \{1,\ldots ,q\}$ of even cardinality the matrix $A_I$ , obtained from A by deleting all rows and columns not in I, is invertible.

The R-module $R^{2n}$ will always be equipped with the standard symplectic bilinear form $\langle x, y \rangle = \sum_{i=0}^{n-1} (x_{2i+1}y_{2i+2} - x_{2i+2}y_{2i+1})$ where $x={^t}(x_1,x_2,\ldots ,x_{2n})$ , $y={^t}(y_1,y_2,\ldots ,y_{2n}) \in R^{2n}$ . The Gram matrix $\Gamma(v)$ of a sequence $v=(v_1,\ldots ,v_q)$ of q vectors $v_1,\ldots ,v_q \in R^{2n}$ is the skew-symmetric $q\times q$ matrix

$$\Gamma(v)=(\langle v_i,v_j\rangle )^q_{i,j=1} = {^tv}\ \psi_{2n}\ v$$

with (i,j)th entry $\langle v_i,v_j\rangle$ . A sequence $v=(v_1,\ldots ,v_q)$ of q vectors in $R^{2n}$ is called unimodular if each subsequence of length $r \leqslant \min(q,2n)$ is a basis of a direct summand of $R^{2n}$ . A unimodular sequence $v=(v_1,\ldots ,v_q)$ of vectors in $R^{2n}$ is called non-degenerate if for all subsets $I\subset \{1,\ldots q\}$ of even cardinality $|I| \leqslant \min(q,2n)$ , the Gram matrix $\Gamma(v_I)$ is invertible, where $v_I$ is the sequence of vectors obtained from v by deleting all columns not in I. We denote by

$$U_q(R^{2n}) = \{v =(v_1,\ldots ,v_q)\ |\ v\ \text{non-degenerate unimodular in } R^{2n}\}$$

the set of non-degenerate unimodular sequences of length q in $R^{2n}$ . The set $U_0(R^{2n})$ is the singleton set consisting of the empty sequence, and the set $U_q(R^0)$ is the singleton set with unique element the sequence $(0,0,\ldots ,0)$ of length q. The symplectic group $\text{Sp}_{2n}(R)$ acts from the left on $U_q(R^{2n})$ by matrix multiplication $Av=(Av_1,\ldots ,Av_q)$ for $A\in \text{Sp}_{2n}(R)$ , $v=(v_1,\ldots ,v_q)\in U_{q}(R^{2n})$ . Note that the Gram matrix of v and Av are the same for all $A\in \text{Sp}_{2n}(R)$ . The following result was proved in [Reference Sarwar and SchlichtingSS25, Lemma 2.8].

Lemma 2.1. Let R be a local ring. Then for all integers $0 \leqslant q \leqslant 2n+1$ the Gram matrix defines a bijection

$$\Gamma:\text{Sp}_{2n}(R)\backslash U_q(R^{2n}) \stackrel{\cong}{\longrightarrow} \text{Skew}_q^+(R).$$

In this paper, we will identify $R^q$ with the subspace of $R^{2n}$ sending the standard basis vector $e_i$ of $R^q$ to the standard basis vector $e_i$ of $R^{2n}$ , $i=1,\ldots ,q$ . Note that if $q\leqslant 2n$ and $u\in U_q(R^{2n})$ is in normal form then u spans $R^q$ .

In the situation of Lemma 2.3, we will call u a normal form of A.

For a set S, we denote by $\mathbb{Z}[S]$ the free abelian group with basis S. We make the graded abelian group

(2.4) \begin{equation}\mathbb{Z}[U_*(R^{2n})] = \{\mathbb{Z}[U_q(R^{2n})],q\geqslant 0\}\end{equation}

into a chain complex with differential $d:\mathbb{Z}[U_q(R^{2n})] \to \mathbb{Z}[U_{q-1}(R^{2n})]$ defined on basis elements $v=(v_1,\ldots ,v_q)$ by

$$dv = \sum_{i=1}^q(-1)^{i+1}d_iv$$

where $d_iv=v^{\wedge}_{{i}}=(v_1,\ldots ,\hat{v}_i,\ldots ,v_q)$ is obtained from v by deleting the ith vector $v_i$ .

Proof. This is a special case of [Reference Sarwar and SchlichtingSS25, Lemma 2.5] in view of [Reference Sarwar and SchlichtingSS25, Remark 2.4].

The following result was proved in [Reference Sarwar and SchlichtingSS25, Corollary 2.6].

Lemma 2.5. Let R be a local ring with infinite residue field and $n\geqslant 0$ an integer. Then the chain complex $(\mathbb{Z}[U_*(R^{2n})],d_*)$ is acyclic. That is, for all $p\in \mathbb{Z}$ we have

$$H_p(\mathbb{Z}[U_*(R^{2n})])=0.$$

Similarly, we make the graded abelian group $\mathbb{Z}[\text{Skew}^+_*(R)]$ into a chain complex with differential $d:\mathbb{Z}[\text{Skew}^+_q(R)] \to \mathbb{Z}[\text{Skew}^+_{q-1}(R)]$ defined on basis elements $A\in \text{Skew}^+_q(R)$ by

$$dA = \sum_{i=1}^q(-1)^{i+1}d_iA$$

where $d_iA=A^{\wedge}_{{i}}$ is obtained from A by deleting the ith row and column. The following result was proved in [Reference Sarwar and SchlichtingSS25, Lemma 2.9].

Lemma 2.6. Let R be a local ring with infinite residue field. Then the chain complex of non-degenerate skew-symmetric matrices $(\mathbb{Z}[\text{Skew}_*^+(R)],d_*)$ is acyclic. That is, for all $p\in \mathbb{Z}$ we have

$$H_p(\mathbb{Z}[\text{Skew}_*^+(R)])=0.$$

As an acyclic complex of free $\mathbb{Z}$ -modules, the complex $(\mathbb{Z}[\text{Skew}_*^+(R)],d_*)$ is actually contractible, and the exact sequences

$$0 \to \text{Im}(d_{n+1}) \to \mathbb{Z}[\text{Skew}_n^+(R)] \to \text{Im}(d_{n}) \to 0$$

are split exact. Such properties are preserved under tensor product with arbitrary abelian groups.

3. The spectral sequence and its $E^1$ page

In this section we introduce the spectral sequence (3.1) which leads to our homology stability range in Theorem 1.1 and we identify its $E^1$ term.

For a complex $M_*$ of abelian groups with differential of degree $-1$ and an integer $r\in \mathbb{Z}$ , we denote by $M_{\leqslant r} \subset M_*$ the subcomplex which is $(M_{\leqslant r})_i=M_i$ for $i\leqslant r$ and $(M_{\leqslant r})_i=0$ for $i>r$ . We call the resulting filtration $\cdots \subset M_{\leqslant r-1} \subset M_{\leqslant r} \subset M_{\leqslant r+1} \subset \cdots$ of $M_*$ , the filtration by degree. The filtration by degree

$$C_{\leqslant 0}(R^{2n}) \subset C_{\leqslant 1}(R^{2n}) \subset \dots \subset C_{\leqslant 2n-1}(R^{2n}) \subset C_{\leqslant 2n}(R^{2n})=C_*(R^{2n})$$

of the complex

$$C_*(R^{2n}) = \mathbb{Z}[U_{\leqslant 2n}(R^{2n})]$$

of $\text{Sp}_{2n}(R)$ -modules yields the exact sequence of complexes

$$0 \to C_{\leqslant q-1}(R^{2n}) \to C_{\leqslant q}(R^{2n}) \to C_{\leqslant q}(R^{2n})/C_{\leqslant q-1}(R^{2n}) \to 0.$$

Upon applying the functor $H_*(\text{Sp}_{2n},\phantom{fd}) = \text{Tor}_*^{\text{Sp}_{2n}}(\mathbb{Z},\phantom{fd})$ , the exact sequences yield the exact couple $D^1_{p+1,q-1} \to D^1_{p,q} \to E^1_{p,q} \to D^1_{p,q-1}$ where

$$D^1_{p,q}=H_{p+q}(\text{Sp}_{2n}(R),C_{\leqslant q}(R^{2n})),\quad E^1_{p,q}=H_{p+q}(\text{Sp}_{2n}(R),C_{\leqslant q}(R^{2n})/C_{\leqslant q-1}(R^{2n})),$$

and hence the spectral sequence

(3.1) \begin{equation}E^1_{p,q}=H_p(\text{Sp}_{2n}(R),C_q(R^{2n})) \Rightarrow H_{p+q}(\text{Sp}_{2n}(R),C_*(R^{2n}))\end{equation}

with differential $d^r_{p,q}$ of bidegree $(r-1,-r)$ .

The following lemma shows that the abutment of the spectral sequence (3.1) vanishes in degrees $p+q<2n$ .

Lemma 3.1. Let R be a local ring with infinite residue field. Then

$$H_{i}(\text{Sp}_{2n}(R),C_*(R^{2n})) = 0,\quad i<2n.$$

Proof. Let M be the kernel of $d:C_{2n}(R^{2n}) \to C_{2n-1}(R^{2n})$ . By Lemma 2.5, the inclusion of complexes $M[2n] \to C_*(R^{2n})$ is a quasi-isomorphism. In particular,

$$H_{i}(\text{Sp}_{2n}(R),C_*(R^{2n})) = H_{i}(\text{Sp}_{2n}(R),M[2n])= H_{i-2n}(\text{Sp}_{2n}(R),M).$$

The result follows since for all G-modules M, we have $H_j(G,M)=0$ for $j<0$ .

Let $0 \leqslant q \leqslant 2n$ be integers. Let $v\in U_q(R^{2n})$ be a non-degenerate unimodular sequence which spans the submodule $R^q$ of $R^{2n}$ . Note that for every $A\in \text{Skew}_q(R)$ there is such a v with $\Gamma(v)=A$ ; for instance, a normal form of A will do (see Lemma 2.3). As an ordered basis of $R^q$ , v defines an element of $\text{GL}_q(R)$ and as such has a determinant $\det(v)\in R^*$ . Using the standard functoriality of group homology as in [Reference BrownBro94, III.8], we define a map

$$f_v: H_p(\text{Sp}_{2n-q}(R);\mathbb{Z}) \longrightarrow H_p(\text{Sp}_{2n}(R);\mathbb{Z}[U_q(R^{2n})])$$

by

$$f_v = \Bigg\{\renewcommand\arraystretch{1.5}\begin{array}{@{}lll}(\varepsilon,v)_*, & \ 0 \leqslant q \leqslant 2n,& q \text{\ even},\\(\varepsilon \circ c_{\det v}, v)_*, & \ 0 \leqslant q \leqslant 2n,& q \text{ odd},\end{array}\Bigg.$$

where $\varepsilon:\text{Sp}_{2n-q} (R) \to \text{Sp}_{2n}(R)$ is the standard embedding, v denotes the homomorphism of abelian groups $\mathbb{Z} \to \mathbb{Z}[U_q(R^{2n})]$ sending 1 to v, and for $a\in R^*$ , $c_a:\text{Sp}_{2n-q}(R) \to \text{Sp}_{2n-q}(R)$ is conjugation $A \mapsto D A D^{-1}$ with the diagonal matrix

$$D= \bigg(\begin{smallmatrix}a & 0 & 0 \\ 0 & a^{-1} & 0 \\ 0 & 0 & 1_{2n-q-1}\end{smallmatrix}\bigg) \in \text{Sp}_{2n-q+1}(R)$$

for $0< q <2n$ odd.

Assume now $q=2r+1$ odd, $0 \leqslant r < n$ . We consider u,v as elements in $\text{GL}_q(R)$ . There are unique vectors $x,y\in R^{q+1}$ such that

$$\bigg(\begin{array}{c|c}{^t}u& 0 \\ \hline 0 & 1\end{array}\bigg)\psi_{q+1}\ x=\bigg(\begin{array}{c|c}{^t}v& 0 \\ \hline 0 & 1\end{array}\bigg)\psi_{q+1}\ y = e_q\in R^{q+1}$$

since the $(q+1)\times (q+1)$ matrices involved are invertible. Then $\Gamma(u,x)=\Gamma(v,y)$ , by definition of x and y. We show that (v,y) is a basis of $R^{q+1}$ . Indeed, let $V \subset R^{q+1}$ be the R-span of $v_1,\ldots ,v_{q-1}$ . Then V equipped with the symplectic form $\langle\phantom{X},\phantom{Y}\rangle$ is non-degenerate since the Gram matrix of $v_1,\ldots ,v_{q-1}$ is invertible. Therefore, there is a unique $w\in V$ such that $\langle w,v_i\rangle = \langle v_q,v_i\rangle$ for all $i=1,\ldots ,q-1$ . In the orthogonal decomposition $V \perp V^{\perp}=R^{q+1}$ of $R^{q+1}$ , the vectors $v_q-w,y\in V^{\perp}$ are a hyperbolic basis of $V^{\perp}$ since $\Gamma(v_q-w,y)=\psi_2$ . It follows that $(v_1,\ldots ,v_{q-1},v_q-w,y)$ is a basis of $R^{q+1}$ , hence $(v_1,\ldots ,v_{q-1},v_q,y)$ is a basis of $R^{q+1}$ . Similarly, (u,x) is also a basis of $R^{q+1}$ . The R-linear endomorphism $B = (v,y)\circ (u,x)^{-1}:R^{q+1} \to R^{q+1}$ sending (u,x) to (v,y) is an isometry and thus has determinant 1 as $\text{Sp}_{q+1}(R)\subset \text{SL}_{q+1}(R)$ . Since $u,v\in \text{GL}_q(R)$ , the matrices (u,x) and (v,y) have the form

$$(u,x) = \bigg(\begin{array}{c|c}u& \ast \\ \hline 0 & x_0\end{array}\bigg)\quad\text{and}\quad (v,y)=\bigg(\begin{array}{c|c}v& \ast \\ \hline 0 & y_0\end{array}\bigg).$$

Thus, $x_0\det u = \det(u,x) = \det(v,y) = y_0 \det v$ , and the matrix

$$ \bigg(\begin{smallmatrix}1_{2r} & 0 & 0 \\ 0 & \det^{-1} v& 0 \\ 0 & 0 & \det v \end{smallmatrix}\bigg) B \bigg(\begin{smallmatrix}1_{2r} & 0 & 0\\ 0 & \det u & 0\\ 0 & 0 & \det^{-1} u\end{smallmatrix}\bigg) \in \text{Sp}_{q+1}(R)$$

has last row equal to $^te_{q+1}=(0,0,\ldots ,0,1)$ . In particular, that matrix has the form

$$ \bigg(\begin{smallmatrix}P& 0 & g \\ ^th & 1& g_0 \\ 0 & 0 & 1\end{smallmatrix}\bigg)$$

for some $g,h\in R^{q-1}$ , $g_0\in R$ , and $P\in \text{Sp}_{q-1}(R)$ . We now extend the isometry B of $R^{q+1}$ to all of $R^{2n}$ by requiring $Be_i=e_i$ for $i = q+2,\ldots ,2n$ . Then $Bu=v$ , $B\in \text{Sp}_{2n}(R)$ , and for all $M \in \text{Sp}_{2n-q}(R)$ we have $c_{\det v}(M) = B \circ c_{\det u}(M)\circ B^{-1}$ since

$$ \bigg(\begin{smallmatrix}1_{2r} & 0 & 0 & 0 \\ 0 & \det^{-1} v& 0 & 0\\ 0 & 0 & \det v & 0 \\ 0 & 0 & 0 & 1_{2n-q-1}\end{smallmatrix} \bigg) B \bigg(\begin{smallmatrix}1_{2r} & 0 & 0 &0\\ 0 & \det u & 0 &0\\ 0 & 0 & \det^{-1} u& 0 \\ 0 & 0 & 0 & 1_{2n-q-1}\end{smallmatrix}\bigg) = \bigg(\begin{smallmatrix}P& 0 & g & 0 \\ ^th & 1& g_0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1_{2n-q-1}\end{smallmatrix}\bigg),$$

and any such matrix commutes with every matrix in $\text{Sp}_{2n-q}(R)$ because

$$\bigg(\begin{smallmatrix} P & 0 & g & 0 \\ ^th & 1 & g_0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{smallmatrix}\bigg)\bigg(\begin{smallmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & b_0 & ^tb \\ 0 & 0 & 1 & 0 \\ 0 & 0 & a & N\end{smallmatrix}\bigg)=\bigg(\begin{smallmatrix} P & 0 & g & 0 \\ ^th & 1 & g_0+b_0 & ^tb \\ 0 & 0 & 1 & 0\\ 0 & 0 & a & N\end{smallmatrix}\bigg)=\bigg(\begin{smallmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & b_0 & ^tb \\ 0 & 0 & 1 & 0 \\ 0 & 0 & a & N\end{smallmatrix}\bigg)\bigg(\begin{smallmatrix} P & 0 & g & 0 \\ ^th & 1 & g_0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{smallmatrix}\bigg)$$

for all $a,b\in R^{2n-q-1}$ , $b_0\in R$ , $N\in M_{2n-q-1}(R)$ . This finishes the proof.

Corollary 3.4. For $0 \leqslant q \leqslant 2n$ , the map

(3.2) \begin{equation}H_p(\text{Sp}_{2n-q}(R)) \otimes_{\mathbb{Z}} \mathbb{Z}[\text{Skew}^+_q(R)] \stackrel{\cong}{\longrightarrow} H_p(\text{Sp}_{2n}(R);\mathbb{Z}[U_q(R^{2n})]),\end{equation}

sending $\alpha \otimes A$ to $f_v(\alpha)$ , does not depend on the choice of v and is an isomorphism provided $v\in U_q(R^{2n})$ with $\Gamma(v)=A$ and v generates the submodule $R^q$ of $R^{2n}$ .

The following lemma identifies the $E^1$ page of the spectral sequence (3.1) and its $d^1$ differential.

If q is even then $q+1$ is odd and going first right then down gives the map

\begin{equation*}(\varepsilon^{2n}_{2n-q-1}\circ c_a, d_iu)_* = (\varepsilon^{2n}_{2n-q}, d_iu)_* \circ (c_a \circ \varepsilon_{2n-q-1}^{2n-q})_* = (\varepsilon,d_iu)_*\end{equation*}

where $c_a$ is conjugation with the diagonal matrix D in $\text{Sp}_{2n-q}(R)$ whose diagonal entries are $(a,a^{-1},1_{2n-q-2})$ and $a=\det u$ . Conjugation with any matrix $D\in \text{Sp}_{2n-q}(R)$ is the identity on $H_*(\text{Sp}_{2n-q}(R);\mathbb{Z})$ . Thus, this map equals the map obtained by going down then right.

If q is odd, then $q+1$ is even and going right then down is $(\varepsilon, d_iu)$ , whereas going down then right is $(\varepsilon^{2n}_{2n-q}\circ c_a \circ \varepsilon^{2n-q}_{2n-q-1}, d_iu) = (\varepsilon^{2n}_{2n-q-1}, d_iu) $ since $\ c_a \varepsilon^{2n-q}_{2n-q-1}= \varepsilon^{2n-q}_{2n-q-1}$ , where $c_a$ is conjugation with the diagonal matrix $(a,a^{-1},1_{2n-q-1})$ of $\text{Sp}_{2n-q+1}(R)$ and $a=\det d_iu$ .

4. The limit theorem

The goal of this section is to prove the limit theorem (Theorem 4.9) which is fundamental in our proof in ${\rm{\S}}$ 6 of the degeneration of the spectral sequence (3.1).

Let R be a commutative ring which, for now, need not be local. An R-module M carries a left action $R \times M \to M: (a,x) \mapsto ax$ of the multiplicative monoid $(R,\cdot,1)$ of R which is linear in M. In particular, it is a module over the associated integral monoid ring $\mathbb{Z}[R]=\mathbb{Z}[R,\cdot,1]$ . We denote by $\langle a \rangle$ the element of $\mathbb{Z}[R]$ corresponding to $a\in R$ and note that the abelian subgroup $\mathbb{Z}\langle 0 \rangle \subset \mathbb{Z}[R]$ generated by $\langle 0 \rangle$ is an ideal. Since $0\cdot M = 0$ , the R-module M is naturally a module over the quotient ring

$$\mathbb{Z}_0[R]=\mathbb{Z}[R]/\mathbb{Z}\langle 0 \rangle = \mathbb{Z}[R,\cdot,1]/\mathbb{Z}\langle 0 \rangle.$$

By functoriality, the action of the monoid $(R,\cdot,1)$ on M induces an action of that monoid on $H_q(M)$ , $M^{\otimes_{\mathbb{Z}} q}$ , $\Lambda_{\mathbb{Z}}^qM$ , and M(q) by means of abelian group homomorphisms where M(q) is M with action through the qth power of its natural action. For $q\geqslant 1$ , all those abelian groups are therefore $\mathbb{Z}_0[R]$ -modules. For instance, for $q\geqslant 1$ , the $\mathbb{Z}_0[R]$ -module structure on M(q) is

(4.1) \begin{equation}\bigg(\sum_{i=1}^rn_i\langle a_i\rangle \bigg) \cdot x = \sum_{i=1}^rn_i a^q_i \, x.\end{equation}

A $\mathbb{Z}_0[R]$ -module M is an R-module if and only if the multiplicative left action of R on M is also linear in R, that is, if for all $a,b\in R$ , the element $\langle a\rangle + \langle b \rangle - \langle a+b \rangle$ acts as zero on M. We may call such $\mathbb{Z}_0[R]$ -modules linear. The criterion for linearity is the $m=2$ case of the following generalization. For a sequence $x=(x_1,\ldots .,x_m)$ of m elements in R and a subset $J \subset \{1,\ldots ,m\}$ , we denote by $x_J$ the partial sum

$$x_J = \sum_{j\in J} x_j \in R$$

with the convention $x_{\emptyset}=0$ . Then a $\mathbb{Z}_0[R]$ -module M is linear if and only if the element

$$-\sum_{\emptyset \neq J \subset \{1,\ldots ,m\}} (-1)^{|J|}\ \langle x_J\rangle \in \mathbb{Z}_0[R]$$

acts as zero on M for all $m\geqslant 2$ and all sequences $x=(x_1,\ldots .,x_m)$ of m elements in R. More generally, we have the following lemma. Our convention is that $x^0=1$ for $x\in R$ even if $x=0$ .

Lemma 4.1. Let R be a ring, M an abelian group and let $t\geqslant 1$ be an integer. Let

$$[ \phantom{x,\ldots ,y}]:R^{\times t} \to M: (a_1,\ldots ,a_t) \mapsto [a_1,\ldots ,a_t]$$

be a $\mathbb{Z}$ -multilinear map, linear in each of the t variables. Let $x=(x_1,\ldots ,x_m)$ be a sequence of $m\geqslant 1$ elements in R. Let $p_1(X),\ldots ,p_t(X) \in R[X]$ be polynomials of degree $\gamma_1,\ldots ,\gamma_t$ with $\gamma_1+\cdots + \gamma_t <m$ . Then

(4.2) \begin{equation}-\sum_{\emptyset \neq J \subset \{1,\ldots ,m\}} (-1)^{|J|}\ [p_1(x_J), \ldots, p_t(x_J)] = [p_1(0), \ldots , p_t(0)]. \end{equation}

Proof. If one of the polynomials $p_i$ is the zero polynomial, then both sides of (4.2) are zero. So we can assume $\gamma_1,\ldots ,\gamma_t\geqslant 0$ . We first prove the lemma for $p_i(X)=a_iX^{\gamma_i}$ , $a_i\in R$ . If $\gamma_1=\cdots =\gamma_t=0$ then the left term in (4.2) is $[a_1,\ldots ,a_t]=[p_1(0), \dots , p_t(0)]$ because

$$1+\displaystyle{\sum_{\emptyset \neq J \subset \{1,\ldots ,m\}} (-1)^{|J|}} = \displaystyle{\sum_{J \subset \{1,\ldots ,m\}} (-1)^{|J|}} = (1-1)^m =0$$

for $m\geqslant 1$ .

If $\gamma_1+ \cdots + \gamma_t\geqslant 1$ we write [n] for the set $\{1,\ldots ,n\}$ . Then the negative of the left term in (4.2) is

$$\renewcommand\arraystretch{2.5}\begin{array}{lll}&&\displaystyle{\sum_{\emptyset \neq J \subset \{1,\ldots ,m\}} (-1)^{|J|}\ [a_1(x_J)^{\gamma_1}, \ldots, a_t(x_J)^{\gamma_t}]}\\&&\quad =\displaystyle{\sum_{\emptyset \neq J \subset [m], \sigma_i:[\gamma_i] \to J, 1 \leqslant i \leqslant t} (-1)^{|J|}}a_1 x_{\sigma_1(1)}\cdots x_{\sigma_1(\gamma_1)}, \dots, a_t x_{\sigma_t(1)}\cdots x_{\sigma_t(\gamma_t)]}\\&&\quad =\displaystyle{\sum_{\sigma_i:[\gamma_i] \to [m], 1 \leqslant i \leqslant t} [a_1 x_{\sigma_1(1)}\cdots x_{\sigma_1(\gamma_1)}, \dots ,a_t x_{\sigma_t(1)}\cdots x_{\sigma_t(\gamma_t)}]\sum_{\bigcup_{i=1}^{t} \text{Im}(\gamma_i)\subset J \subset [m]} (-1)^{|J|}}\\&&\quad = 0\end{array}$$

since $\emptyset \neq \bigcup_{i=1}^{t} \text{Im}(\gamma_i) \subsetneq [m]$ as $1 \leqslant \gamma_1+ \cdots + \gamma_t<m$ , and for $S \subsetneq [m]$ we have

$$\sum_{S\subset J \subset [m]} (-1)^{|J|} =(-1)^{|S|} \sum_{J \subset [m]-S}(-1)^{|J|} = (-1)^{|S|}(1-1)^{m-|S|} =0.$$

We now assume that $p_1(X),\ldots ,p_t(X) \in R[X]$ are arbitrary polynomials of degree $\gamma_1,\ldots ,\gamma_t\geqslant 0$ with $\gamma_1+\cdots + \gamma_t <m$ . Each polynomial p(X) is the sum of p(0) and a $\mathbb{Z}$ -linear combination of polynomials $a_{\gamma}X^{\gamma}$ with $\gamma \geqslant 1$ and $a_{\gamma}\in R$ . It follows that the left term of (4.2) is the sum of

(4.3) \begin{equation}-\displaystyle{\sum_{\emptyset \neq J \subset \{1,\ldots ,m\}} (-1)^{|J|} \ [p_1(0),\dots, p_t(0)]}\end{equation}

and a $\mathbb{Z}$ -linear combination of terms

(4.4) \begin{equation}-\sum_{\emptyset \neq J \subset \{1,\ldots ,m\}} (-1)^{|J|}\ [a_1(x_J)^{\delta_1}, \ldots, a_t(x_J)^{\delta_t}]\end{equation}

for some $a_i\in R$ and where $0 \leqslant \delta_i $ and $1 \leqslant \delta_1+\cdots +\delta_t<m$ . By the first part of the proof, the terms (4.4) are zero and therefore, the left term of (4.2) equals (4.3) which is $[p_1(0), \dots p_t(0)]$ , again by the first part of the proof.

For a sequence $a=(a_1,\ldots ,a_m)$ of m elements in R and a polynomial $p(X) \in R[X]$ with coefficients in R, we write $s_p(a) \in \mathbb{Z}[R]$ for the element

$$s_p(a) = -\sum_{\emptyset \neq J \subset \{1,\ldots ,m\}} (-1)^{|J|}\ \langle p({a_J})\rangle \in \mathbb{Z}[R].$$

Remark 4.2. For $p(X)=X$ , the element $s_p(a)$ was first considered in [Reference SchlichtingSch17a] to prove optimal homology stability for special linear groups. Note that for $m\geqslant 1$ ,

$$s_1 = -\sum_{\emptyset \neq J \subset \{1,\ldots ,m\}} (-1)^{|J|} = 1.$$

Note that the category of quasi-linear $\mathbb{Z}_0[R]$ -modules is a Serre abelian subcategory of the abelian category of all $\mathbb{Z}_0[R]$ -modules, that is, subobjects, quotients, and extensions of quasi-linear $\mathbb{Z}_0[R]$ -modules in the category of $\mathbb{Z}_0[R]$ -modules are quasi-linear.

In order to state our limit theorem we introduce the following terminology.

Definition 4.7. Let R be a local ring, and $\mathcal{D} \subset R$ a region of R. A function $f:\mathcal{D} \to \mathbb{Z}_0[R]$ is called admissible if there are polynomials ${P(X_1,\ldots ,X_n)} \in \mathbb{Z}[X_1,\ldots ,X_n]$ , ${P_i(X),Q_i(X)} \in R[X]$ , $i=1,\ldots ,n$ , such that $Q_i(t)\in R^*$ for all $t\in \mathcal{D}$ and

(4.5) \begin{equation}f(t) = P\bigg(\bigg\langle \frac{P_1(t)}{Q_1(t)}\bigg\rangle, \dots, \bigg\langle \frac{P_n(t)}{Q_n(t)}\bigg\rangle\bigg) \in \mathbb{Z}_0[R]\end{equation}

for all $t\in \mathcal{D}$ where the evaluation of the polynomial P takes place in $\mathbb{Z}_0[R]$ . The polynomials $P,P_i,Q_i$ are called a presentation of f.

For $a \in R$ , we say that f is defined at a (relative to a presentation $(P,P_i,Q_i)$ ) if the elements $Q_i(a)\in R$ are units in R. Clearly, f is defined at all elements of $\mathcal{D}$ . Note that if f is defined at $a\in R$ then f(a) is a well-defined element in $\mathbb{Z}_0[R]$ , given by (4.5), though the value f(a) may depend on the presentation of f.

Definition 4.8. Let $\mathcal{D} \subset R$ be a dense region of a local ring R with infinite residue field. Let $f:\mathcal{D} \to \mathbb{Z}_0[R]$ be an admissible function represented by $(P,P_i,Q_i)$ as in (4.5). For $a \in R\cup \{\infty\}$ we say that the limit $\lim_{t\to a}f(t)$ of f when t tends to a exists and write

$$\lim_{t\to a}f(t)=L\in\mathbb{Z}_0[R]$$

if either of the following assertions holds.

1. (i) If $a\in R$ , then we require f to be defined at a and set $L = f(a)$ .

2. (ii) dfn:Lim:item2 If $a=\infty$ , then we require $\deg P_i \leqslant \deg Q_i$ and the coefficients of the highest-degree monomials of $Q_i(X)$ to be units in R, $i=1,\ldots ,n$ . Then

   $$\bar{Q}_i(X) = X^{\deg Q_i}Q_i(1/X),\quad\bar{P}_i(X) = X^{\deg Q_i}P_i(1/X)$$
   are polynomials with $\bar{Q}_i(0) \in R^*$ , $i=1,\ldots ,n$ . We note that
   $$f(1/t) = \bar{f}(t) = P\bigg(\biggl\langle \frac{\bar{P}_1(t)}{\bar{Q}_1(t)}\biggr\rangle, \dots, \biggl\langle \frac{\bar{P}_n(t)}{\bar{Q}_n(t)}\biggr\rangle\bigg)\in \mathbb{Z}_0[R]$$
   for $t\in R^*$ with $1/t\in \mathcal{D}$ , and that $\bar{f}$ is defined at 0 relative to the presentation $(P,\bar{P}_i,\bar{Q}_i)$ . We set
   $$L=\lim_{t\to \infty}f(t) = \lim_{t\to 0}f(1/t) = \lim_{t\to 0}\bar{f}(t) = \bar{f}(0).$$

We do not know if $\lim_t f(t)$ does or does not depend on the presentation of f. For the purpose of this paper, the limit will always be calculated relative to a given presentation of f. Recall that for a commutative ring S such as $\mathbb{Z}_0[R]$ and an S-module M, the radical of the annihilator of an element $x\in M$ is

$$\sqrt{\text{Ann}(x)} = \sqrt{\text{Ann}_M(x)} = \{s\in S|\ s^nx=0\ \text{for some}\ n\geqslant 1\}.$$

Theorem 4.9 (Limit theorem). Let R be a local ring with infinite residue field. Let M be a quasi-linear $\mathbb{Z}_0[R]$ -module, and let $x\in M$ . Let $\mathcal{D} \subset R$ be a dense region of R, and let $f:\mathcal{D} \to \mathbb{Z}_0[R]$ be an admissible function with given presentation. Assume that $f(t) \in \sqrt{\text{Ann}(x)} \subset \mathbb{Z}_0[R]$ for all $t\in \mathcal{D}$ . Then for all $a\in R \cup \{\infty\}$ , if $\lim_{t \to a}f(t)$ exists in $\mathbb{Z}_0[R]$ in the given presentation then that limit satisfies

$$\lim_{t \to a}f(t) \in \sqrt{\text{Ann}(x)}.$$

Proof of Theorem 4.9. Let $(P,P_i,Q_i)$ be the given presentation of f as in (4.5). We first consider the case $a=0$ . Since $\lim_{t\to 0}f(t)$ exists, we have $Q_i(0)\in R^*$ for all $i=1,\ldots ,n$ . Let $d_i$ be the highest power of $X_i$ occurring in $P(X_1,\ldots ,X_n)$ . Then $g(t)= \langle Q_1(t)^{d_1} \cdots Q_n(t)^{d_n}\rangle \ f(t)$ is an integer linear combination of expressions $\langle p_j(t) \rangle$ with $p_j(X)\in R[X]$ polynomials, $j=1,\ldots ,\ell$ , for some $\ell \in \mathbb{N}$ :

$$g(t) =\langle {Q_1(t)^{d_1} \cdots Q_n(t)^{d_n}}\rangle f(t)= \sum_{j=1}^{\ell}n_j\langle p_j(t) \rangle.$$

Since M is a quasi-linear $\mathbb{Z}_0[R]$ -module, we can choose $m_0\in \mathbb{N}$ such that $\sigma_j(a) = s_{p_j}(a) -\langle p_j(0)\rangle$ satisfies $\sigma_j(a)^{-1}M=0$ for all $j=1,\ldots ,\ell$ and all sequences $a=(a_1,\ldots ,a_m)$ of m elements in R with $m\geqslant m_0$ . In particular, $\sigma_j(a)\in \sqrt{\text{Ann}(x)}$ and hence

(4.6) \begin{equation}s_g(a)-g(0) = \sum_{j=1}^{\ell}n_j\sigma_j (a)\in \sqrt{\text{Ann}(x)}\end{equation}

where (abusing notation slightly)

$$s_g(a) := -\sum_{\emptyset \neq J \subset \{1,\ldots ,m\}} (-1)^{|J|}\ g(a_J) = \sum_{j=1}^{\ell}n_js_{p_j}(a).$$

Fix $m\geqslant m_0$ and choose a sequence $a=(a_1,\ldots ,a_m)$ of m elements in R such that $a_J\in \mathcal{D}$ for all $\emptyset \neq J \subset \{1,\ldots ,m\}$ . This is possible for if we denote by $\pi:R \to k$ the quotient map to the residue field of R, and if we have chosen $(a_1,\ldots ,a_t)$ such that $a_J\in \mathcal{D}$ for all $\emptyset \neq J \subset \{1,\ldots ,t\}$ , then $a_{t+1}\in R$ can be any element such that $\pi({a}_{t+1})$ is not the solution $x\in k$ to one of the finitely many non-trivial linear equations $x+\pi({a}_J) = y$ , $y\in k-\pi(\mathcal{D})$ , $J\subset \{1,\ldots ,t\}$ . Such $x\in k$ exists since k is infinite. Since $a_J\in \mathcal{D}$ , we have $f(a_J)\in \sqrt{\text{Ann}(x)}$ for all $\emptyset \neq J \subset \{1,\ldots ,m\}$ , by assumption. Then $g(a_J)\in \sqrt{\text{Ann}(x)}$ for all $\emptyset \neq J \subset \{1,\ldots ,m\}$ . As a $\mathbb{Z}$ -linear combination of the $g(a_J)$ we then have $s_g(a)\in \sqrt{\text{Ann}(x)}$ . By (4.6), we have $g(0)\in \sqrt{\text{Ann}(x)}$ and thus

$$\lim_{t\to 0}f(t) = f(0) = \langle Q_1(0)^{-d_1} \cdots Q_n(0)^{-d_n}\rangle\ g(0) \in \sqrt{\text{Ann}(x)}$$

since $Q_i(0)\in R^*$ , $i=1,\ldots ,n$ .

Now assume $a\in R$ arbitrary. Define $\bar{P}_i(X) = P_i(X+a)$ , $\bar{Q}_i(X) = Q_i(X+a)$ , $\bar{f}(t) = f(t+a)$ , $\bar{P} = P$ , $\bar{\mathcal{D}}=\mathcal{D}-a$ . Then $\bar{f}(t)\in \sqrt{\text{Ann}(x)}$ for all $t\in \bar{\mathcal{D}}$ , and the case of $t\to 0$ treated above shows that $\lim_{t\to a}f(t) = \lim_{t\to 0}\bar{f}(t)\in \sqrt{\text{Ann}(x)}$ .

Finally, assume $a=\infty$ . Set $\bar{P}_i(X) $ , $\bar{Q}_i(X)$ , $\bar{f}(t)=f(1/t)$ as in Definition 4.8(ii). Note that $\bar{\mathcal{D}} = \{t \in R^*|\ t^{-1}\in \mathcal{D}\}$ is a dense region of R since $\mathcal{D}$ is. Then $\bar{f}(t)=f(1/t) \in \sqrt{\text{Ann}(x)}$ for $t\in \bar{\mathcal{D}}$ . By the case $a=0$ treated above, we have

$$\lim_{t \to \infty}f(t) = \lim_{t \to 0} \bar{f}(t) \in\sqrt{\text{Ann}(x)}.$$

5. Quasi-linear modules and group homology

The goal of this section is to prove in Proposition 5.3 that the relative homology groups $H_s(G,K)$ are quasi-linear for certain $(R,\cdot,1)$ -equivariant inclusions of groups $K \subset G$ . This will be applied to show that the relative homology groups $H_s(\text{Sp}_{2r+1}(R),\text{Sp}_{2r}(R))$ are quasi-linear $\mathbb{Z}_0[R]$ -modules. At the end of the section we will give a few first applications of Theorem 4.9.

For an integer $t\geqslant 1$ , we consider the ring homomorphism

(5.1) \begin{equation}\varphi_t:\mathbb{Z}[R,\cdot,1] \to R^{\otimes t}: \langle a\rangle \mapsto a^{\otimes t} = a \otimes \cdots \otimes a\end{equation}

where $a\in R$ . Assume that the multiplicative monoid $(R,\cdot,1)$ of R acts on a group G from the left through group homomorphisms. By functoriality, $(R,\cdot,1)$ acts on the homology group $H_q(G)$ from the left through abelian group homomorphisms, that is, $H_q(G)$ is a left $\mathbb{Z}[R]$ -module. Recall from (4.1) the $\mathbb{Z}_0[R]$ -module M(q) associated with an R-module M and an integer $q\geqslant 1$ where the $(R,\cdot,1)$ -action is through the qth power of its natural action on the module M.

Lemma 5.1. Let R be a commutative ring whose underlying abelian group $(R,+,0)$ is torsion-free. Let A, B be R-modules, and let $r,\alpha,\beta \geqslant 1$ be integers. Let

(5.2) \begin{equation}1 \to B(\beta)\to N \to A(\alpha) \to 1\end{equation}

be an $(R,\cdot,1)$ -equivariant central extension of groups. Let $\sigma\in \mathbb{Z}[R]$ be such that $\varphi_t(\sigma)=0$ for $1 \leqslant t \leqslant r$ . Then $\sigma^{-1}H_s(N)=0$ whenever $1 \leqslant s\cdot \max(\alpha,\beta) \leqslant r$ .

Note that the group N in the lemma need not be abelian.

Recall that a $\mathbb{Z}[R]$ -module A is the same as an abelian group A together with an action of the monoid $(R,\cdot,1)$ on A through abelian group homomorphisms. For two $\mathbb{Z}[R]$ -modules A,B, we endow their tensor product $A\otimes B=A\otimes_{\mathbb{Z}}B$ with the diagonal $(R,\cdot,1)$ action. This defines the $\mathbb{Z}[R]$ -module $A\otimes B$ . If A and B are $\mathbb{Z}_0[R]$ -modules then so is $A\otimes B$ .

Proof of Lemma 5.1. We will first prove the lemma when A is torsion-free as an abelian group. To do so we show that in this case

(5.3) \begin{equation}\sigma^{-1}(H_p(A{(\alpha)})\otimes H_q(B{(\beta)})) =0 \quad\text{for }1 \leqslant \alpha p + \beta q \leqslant r,\end{equation}

and then apply the Hochschild–Serre spectral sequence to (5.2). To prove (5.3) we first also assume that B is torsion-free as an abelian group. Then $H_p(A)\otimes H_q(B) \cong \Lambda_{\mathbb{Z}}^p(A) \otimes \Lambda_{\mathbb{Z}}^q(B)$ , functorial in A and B. In particular, the result of the action of $a\in R \subset \mathbb{Z}[R]$ on $(x_1\wedge \cdots \wedge x_p) \otimes (y_1\wedge \cdots \wedge y_q) \in H_p(A({\alpha}))\otimes H_q(B({\beta}))$ is $(a^{\alpha}x_1\wedge \cdots \wedge a^{\alpha}x_p) \otimes (a^{\beta}y_1\wedge \cdots \wedge a^{\beta}y_q)$ . This is the image of $\varphi_{\alpha p + \beta q}(a)$ under the $\mathbb{Z}$ -linear map

(5.4) \begin{equation}R^{\otimes \alpha p} \otimes R^{\otimes \beta q} \longrightarrow \Lambda_{\mathbb{Z}}^p(A) \otimes \Lambda_{\mathbb{Z}}^q(B)\end{equation}

which uniquely extends the $\mathbb{Z}$ -multilinear map

$$R^{\alpha p} \times R^{\beta q} \longrightarrow \Lambda_{\mathbb{Z}}^p(A) \otimes \Lambda_{\mathbb{Z}}^q(B)$$

sending $(M, N) \in R^{\alpha p} \times R^{\beta q} = M_{\alpha,p}(R)\times M_{\beta,q}(R)$ to

$$\bigg(\bigg(\prod_{i=1}^{\alpha}M_{i,1}\bigg) x_1\wedge \cdots \wedge \bigg(\prod_{i=1}^{\alpha}M_{i,p}\bigg) x_p\bigg) \otimes \bigg(\bigg(\prod_{i=1}^{\beta} N_{i,1}\bigg) y_1 \wedge \cdots \wedge \bigg(\prod_{i=1}^{\beta} N_{i,q}\bigg)y_q\bigg).$$

In particular, the result of the action of $\sigma \in \mathbb{Z}[R]$ on $(x_1\wedge \cdots \wedge x_p) \otimes (y_1\wedge \cdots \wedge y_q)$ is the image of $\varphi_{\alpha p + \beta q}(\sigma)$ under the $\mathbb{Z}$ -linear map (5.4). But $\varphi_{t}(\sigma)=0$ for $1 \leqslant t \leqslant r$ . Hence, $\sigma( \Lambda_{\mathbb{Z}}^p(A) \otimes \Lambda_{\mathbb{Z}}^q(B)) =0$ for $1 \leqslant \alpha p + \beta q \leqslant r$ . In particular, (5.3) holds when A and B are torsion-free.

We now prove (5.3) when A is torsion-free as an abelian group and B is an arbitrary R-module. Choose a surjective weak equivalence of simplicial R-modules $B_* \to B$ with $B_i$ a projective R-module for all $i\in \mathbb{N}$ . For instance, the simplicial R-module corresponding to an R-projective resolution of B under the Dold–Kan correspondence will do. Each $B_i$ is a torsion-free abelian group since R is. The classifying space functor $\mathcal{B}$ induces an $(R,\cdot,1)$ -equivariant weak equivalence of simplicial sets $\mathcal{B}(B_*(\beta)) \to \mathcal{B} B(\beta)$ . Tensoring the spectral sequence of the simplicial space $n \mapsto \mathcal{B} B_n$ ,

$$E^1_{s,t} = H_t(\mathcal{B} B_s) \Rightarrow H_{s+t}(\mathcal{B} B_*) = H_{s+t}(\mathcal{B} B)= H_{s+t}(B),$$

with the flat $\mathbb{Z}$ -module $H_p(A)\cong\Lambda_{\mathbb{Z}}^pA$ yields the spectral sequence of $\mathbb{Z}[R]$ -modules

$$H_p(A(\alpha))\otimes E^1_{s,t} = H_p(A(\alpha))\otimes H_t(B_s(\beta)) \Rightarrow H_p(A(\alpha)) \otimes H_{s+t}(B(\beta)).$$

Localizing at $\sigma$ , this yields a spectral sequence with trivial $E^1_{s,t}$ -term for $1\leqslant \alpha p +\beta t \leqslant r$ . Since $t \leqslant s+t$ for $0 \leqslant s,t$ , the $E^1_{s,t}$ -term of the localized spectral sequence is trivial for $1\leqslant \alpha p +\beta (s+t) \leqslant r$ (and $p,s,t\geqslant 0$ ). This proves (5.3) when A is torsion-free as an abelian group.

We now prove the lemma when A is torsion-free as an abelian group. In this case, the integral homology groups $H_*(A)\cong\Lambda_{\mathbb{Z}}^*(A)$ are torsion-free and the natural map $H_p(A) \otimes F \to H_p(A;F)$ is an isomorphism for any abelian group F, by the universal coefficient theorem. Since the extension (5.2) is central, the group A acts trivially on $H_*(B)$ and the Hochschild–Serre spectral sequence of the group extension has the form

$$E^2_{p,q}= H_p(A;H_q(B)) \cong H_p(A)\otimes H_q(B) \Rightarrow H_{p+q}(N).$$

The spectral sequence is functorial in the exact sequence (5.2). In particular, it is equivariant for the $(R,\cdot,1)$ -action and thus a spectral sequence of $\mathbb{Z}[R]$ -modules. Localizing the spectral sequence at $\sigma$ yields a spectral sequence with $E^2$ term $\sigma^{-1}E^2_{p,q}=0$ for $1 \leqslant \alpha p+\beta q \leqslant r$ , by (5.3). This implies the lemma in case A is torsion-free.

Finally, we prove the lemma for arbitrary R-modules A and B. As above, we choose a surjective weak equivalence $A_* \to A$ of simplicial R-modules with $A_n$ a projective R-module for all n. Then each $A_n$ is flat as an abelian group since R is. Let $N_n = N\times_{A(\alpha)}A_n(\alpha)$ . The action of $(R,\cdot,1)$ on N, $A(\alpha)$ , and $A_n(\alpha)$ defines an action of $(R,\cdot,1)$ on $N_n$ . We obtain a simplicial $(R,\cdot,1)$ -equivariant central extension

$$1 \to B(\beta) \to N_{\ast} \to A_{\ast}(\alpha) \to 1$$

with degreewise torsion-free base $A_n$ . The surjection $N_* \to N$ of simplicial groups has contractible kernel as it equals the kernel of the surjective weak equivalence $A_{\ast} \to A$ . In particular, the map on classifying spaces $\mathcal{B} |s\mapsto N_s| = |s \mapsto \mathcal{B} N_s| \to \mathcal{B} N$ is an $(R,\cdot ,1)$ -equivariant weak equivalence. By the torsion-free case treated above, we have $\sigma^{-1}H_q(\mathcal{B} N_s) = 0$ for $1\leqq\cdot \max(\alpha,\beta) \leqslant r$ and for all $s\geqslant 0$ . Therefore, the spectral sequence of the simplicial space $s \mapsto \mathcal{B} N_s$ ,

$$E^2_{p,q}=\pi_p|s \mapsto H_q(\mathcal{B} N_s)| \Rightarrow H_{p+q}(\mathcal{B} N_*) = H_{p+q}(\mathcal{B} N),$$

localized at $\sigma$ has trivial $E^2_{p,q}$ -term for $1\leqslant q\cdot \max(\alpha,\beta) \leqslant r$ and for all p. In particular, $\sigma^{-1}E^2_{p,q}=0$ whenever $1\leqslant (p+q)\cdot \max(\alpha,\beta) \leqslant r$ (and $0\leqslant p,q$ ). This proves the lemma.

Lemma 5.2. Let $a=(a_1,\ldots ,a_m)$ be a sequence of m elements in R, and let $p(X) \in R[X]$ be a polynomial of degree d with coefficients in R. Then for $1 \leqslant td < m$ , the element $s_p(a) - \langle p(0)\rangle \in \mathbb{Z}[R]$ is sent to zero under the homomorphism $\varphi_t$ defined in (5.1):

$$\varphi_t ( s_p(a) - \langle p(0)\rangle) = 0 \in R^{\otimes t}.$$

Proof. The image of $s_p(a)$ in $R^{\otimes t}$ is

$$-\sum_{\emptyset \neq J \subset \{1,\ldots ,m\}} (-1)^{|J|}\ P(x_J)^{\otimes t}.$$

We apply Lemma 4.1 to the canonical $\mathbb{Z}$ -multilinear map $R^{\times t} \to R^{\otimes t}:(x_1,\ldots ,x_t) \mapsto [x_1,\ldots ,x_t] = x_1 \otimes \cdots \otimes x_t$ and find that

\begin{align*}\varphi_t(s_p(a)) &= -\sum_{\emptyset \neq J \subset \{1,\ldots ,m\}} (-1)^{|J|}\ [p(a_J), \ldots, p(a_J)]\\& = [p(0), \dots , p(0)] =p(0)^{\otimes t} = \varphi_t(\langle p(0)\rangle ). \end{align*}

Proposition 5.3. Let R be a commutative ring, let A, B be R-modules, and let $\alpha,\beta \geqslant 1$ be integers. Let G, K, N be groups with left $(R,\cdot,1)$ -actions where the action on K is trivial. Given $(R,\cdot,1)$ -equivariant exact sequences of groups

$$1 \to B(\beta)\to N \to A(\alpha) \to 1, \quad 1 \to N \to G \stackrel{\rho}{\to} K \to 1,$$

in which the first sequence is a central extension, the second sequence is equipped with an $(R,\cdot,1)$ -equivariant group homomorphism $i:K \to G$ such that $\rho i=1_K$ and $i \rho:G \to G$ is the action of $0\in R$ on G. Then for all $s\in \mathbb{Z}$ the relative homology groups $H_s(G,K)$ are quasi-linear $\mathbb{Z}_0[R]$ -modules where K is considered a subgroup of G by means of the inclusion i.

Proof. The result $i\rho$ of the action of $\langle 0 \rangle\in \mathbb{Z}[R]$ on $H_s(G,K)$ factors through $H_s(K,K)=0$ . Hence, the $\mathbb{Z}[R]$ -module $H_s(G,K)$ is a $\mathbb{Z}_0[R]$ -module. We will prove that for every sequence $a=(a_1,\ldots ,a_m)$ of m elements in R and every polynomial $p(X) \in R[X]$ of degree d with coefficients in R, the element $\sigma = s_p(a)-\langle p(0)\rangle \in \mathbb{Z}[R]$ satisfies

(5.5) \begin{equation}\sigma^{-1}H_s(G,K)=0\quad \text{provided }sd\max(\alpha,\beta)<m.\end{equation}

This establishes that $H_s(G,K)$ is quasi-linear with $m_0=sd\max(\alpha,\beta)$ in Definition 4.3.

To prove (5.5), assume first that the underlying abelian group $(R,+,0)$ of R is torsion-free. By functoriality, the Hochschild–Serre spectral sequence

(5.6) \begin{equation}E^2_{p,q}=H_p(K;H_q(N)) \Rightarrow H_{p+q}(G)\end{equation}

of the extension $1 \to N \to G \to K\to 1$ carries an action of the monoid $(R,\cdot,1)$ induced from the action of that monoid on the extension. Section $i:K\to G$ and projection $\rho:G \to K$ make the extension $1 \to 1 \to K \to K \to 1$ of groups with (trivial) $(R,\cdot,1)$ -action a direct factor of $1 \to N \to G \to K\to 1$ , hence its Hochschild–Serre spectral sequence of (trivial) $\mathbb{Z}[R]$ -modules (which degenerates at $E^2$ ) is a direct factor of that of (5.6). Its complement yields the strongly convergent spectral sequence

$$\tilde{E}^2_{p,q}=H_p(K;\tilde{H}_q(N)) \Rightarrow H_{p+q}(G,K)$$

where $\tilde{H}_q(N)=H_q(N)$ for $q\geqslant 1$ and 0 otherwise. The action of $g\in K$ on $H_q(N)$ is induced by conjugation with i(g) on N. Since $(R,\cdot,1)$ acts trivially on K and i is equivariant, the action of $(R,\cdot,1)$ on N and the action of K on N commute. It follows that $\sigma^{-1}\tilde{E}^2_{p,q}=\sigma^{-1}H_p(K;\tilde{H}_q(N)) = H_p(K;\sigma^{-1}\tilde{H}_q(N)) = 0$ for $0 \leqslant q \max(\alpha,\beta) \leqslant r$ and any p, by Lemmas 5.1 and 5.2. Hence, $\sigma^{-1}H_s(G,K)=0$ for $0 \leqslant s\cdot \max(\alpha,\beta) \leqslant r$ .

We now prove (5.5) when $(R,+,0)$ is not assumed torsion-free. Choose a surjection of commutative rings $\pi: \bar{R} \twoheadrightarrow R$ such that the abelian group $(\bar{R},+,0)$ of $\bar{R}$ is torsion-free, for instance, $\mathbb{Z}[R] \twoheadrightarrow R:\langle a\rangle \mapsto a$ . Choose a sequence $\bar{a}=(\bar{a}_1,\ldots ,\bar{a}_m)$ in $\bar{R}$ and a polynomial $\bar{p}(X)\in \bar{R}[X]$ such that $\pi(\bar{a}) = a$ and $\pi(\bar{p}(X)) = p(X)$ . The ring homomorphism $\pi$ makes A and B into $\bar{R}$ -modules, and the action of $(\bar{R},\cdot,1)$ on $H_s(G,K)$ is induced from the $({R},\cdot,1)$ -action via the map $\pi$ . Therefore, multiplication by the element $\bar{\sigma} = s_{\bar{p}}(\bar{a}) - \bar{p}(0)$ on $H_s(G,K)$ equals multiplication by the element $\sigma=s_{{p}}({a}) - {p}(0)$ . In particular, $\bar{\sigma}^{-1}H_s(G,K)={\sigma}^{-1}H_s(G,K)$ . By the torsion-free case above, we have $\bar{\sigma}^{-1}H_s(G,K) = 0$ for $sd\max(\alpha,\beta)<m$ . This finishes the proof of (5.5) and hence that of the proposition.

Example 5.4. Let G be a group that acts from the left on an R-module M through R-module homomorphisms. Then for all $q\geqslant 0$ , the semi-direct product $M(q)\rtimes G$ carries an action of $(R,\cdot,1)$ defined by $a(x,g)=(a^qx,g)$ such that the exact sequence

$$1 \to M(q) \to M(q)\rtimes G \to G \to 1$$

is $(R,\cdot,1)$ -equivariant with trivial action on the base G and equivariant section $G \to M(q)\rtimes G: g\mapsto (0,g)$ . By Proposition 5.3 with $B=0$ , $\alpha =q$ , $A=M$ , $N=M(q)$ , the relative homology groups $H_s(M(q)\rtimes G,G)$ are quasi-linear $\mathbb{Z}_0[R]$ -modules whenever $q\geqslant 1$ .

Now comes the most relevant example for this paper.

Example 5.7. Let R be a commutative ring and let $n\geqslant 0$ be an integer. For $a\in R^*$ , the conjugation action $c_a$ of the $(2n+2)\times (2n+2)$ diagonal matrix $D_a\in \text{Sp}_{2n+2}(R)$ with diagonal entries $(a,a^{-1},1,1,\ldots ,1)$ on the group $\text{Sp}_{2n+2}(R)$ induces an action

$$\bigg(\begin{smallmatrix}1& c & {^tu}\psi M \\ 0 & 1 & 0 \\ 0 & u & M \end{smallmatrix}\bigg)\stackrel{c_a}{\mapsto}\bigg(\begin{smallmatrix}a& 0 & 0 \\ 0 & a^{-1} & 0 \\ 0 & 0 & 1 \end{smallmatrix}\bigg)\bigg(\begin{smallmatrix}1& c & {^tu}\psi M \\ 0 & 1 & 0 \\ 0 & u & M \end{smallmatrix}\bigg)\bigg(\begin{smallmatrix}a^{-1}& 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & 1 \end{smallmatrix}\bigg)=\bigg(\begin{smallmatrix}1& a^2c & {^t(au)}\psi M \\ 0 & 1 & 0 \\ 0 & au & M \end{smallmatrix}\bigg)$$

on the subgroup $\text{Sp}_{2n+1}(R)$ which extends to an action

$$\bigg(\begin{smallmatrix}1& c & {^tu}\psi M \\ 0 & 1 & 0 \\ 0 & u & M \end{smallmatrix}\bigg)\stackrel{\langle a \rangle}{\mapsto}\bigg(\begin{smallmatrix}1& a^2c & {^t(au)}\psi M \\ 0 & 1 & 0 \\ 0 & au & M \end{smallmatrix}\bigg)$$

of the monoid $(R,\cdot,1)$ on the group $\text{Sp}_{2n+1}(R)$ , $a\in R$ . Denote by $N\subset \text{Sp}_{2n+1}(R)$ the subgroup of matrices with $M=1$ , by $A\subset \text{Sp}_{2n+1}(R)$ the subgroup of matrices with $M=1$ and $c=0$ , and by $B\subset \text{Sp}_{2n+1}(R)$ the subgroup with $M=1$ , $u=0$ , then $(A,\cdot,1)=(R^{2n},+,0)$ , $(B,\cdot,1)=(R,+,0)$ , and we have $(R,\cdot,1)$ -equivariant exact sequences

$$1 \to B(2) \to N \to A(1) \to 1, \quad 1 \to N \to \text{Sp}_{2n+1}(R) \to \text{Sp}_{2n}(R)\to 1,$$

with left sequence central and $\langle 0 \rangle :\text{Sp}_{2n+1}(R) \to \text{Sp}_{2n+1}(R)$ the projection $\rho:\text{Sp}_{2n+1}(R) \to \text{Sp}_{2n}(R)$ followed by the inclusion $\varepsilon:\text{Sp}_{2n}(R) \to \text{Sp}_{2n+1}(R)$ . By Proposition 5.3, the image of the projector $1-(\varepsilon\rho)_*$ of $H_p(\text{Sp}_{2n+1}(R))$ , which is the relative homology group

$$\tilde{H}_p(\text{Sp}_{2n+1}(R)):=H_p(\text{Sp}_{2n+1}(R),\text{Sp}_{2n}(R))=\text{Im}(1-(\varepsilon\rho)_*), $$

is a quasi-linear $\mathbb{Z}_0[R]$ -module for all $p\in \mathbb{Z}$ . We have a canonical decomposition

$$H_p(\text{Sp}_{2n+1}(R)) = \text{Im}((\varepsilon\rho)_*) \oplus \text{Im}(1-(\varepsilon\rho)_*) = H_p(\text{Sp}_{2n}(R)) \oplus \tilde{H}_p(\text{Sp}_{2n+1}(R)).$$

Lemma 5.8. Let R be a local ring with infinite residue field. Then for all integers $n,p\geqslant 0$ , the $\mathbb{Z}_0[R]$ -module $\tilde{H}_p(\text{Sp}_{2n+1}(R))=H_p(\text{Sp}_{2n+1}(R),\text{Sp}_{2n}(R))$ is quasi-linear, and the composition

$$\tilde{H}_p(\text{Sp}_{2n+1}(R)) \subset H_p(\text{Sp}_{2n+1}(R)) \to H_p(\text{Sp}_{2n+2}(R))$$

is zero. Moreover, the map $H_p(\text{Sp}_{2n+1}(R)) \to H_p(\text{Sp}_{2n+2}(R))$ is surjective if and only if the map $H_p(\text{Sp}_{2n}(R)) \to H_p(\text{Sp}_{2n+2}(R))$ is surjective.

The composition $\tilde{H}_p(\text{Sp}_{2n+1}(R)) \to H_p(\text{Sp}_{2n+2}(R))$ is $R^*$ -equivariant where $R^*$ acts through conjugation with $D_a\in \text{Sp}_{2n+2}$ and thus acts trivially on the target. Since the source is quasi-linear, there is an integer $m_0\geqslant 1$ such that for all sequences $a=(a_1,\ldots ,a_m)$ of $m\geqslant m_0$ elements in R we have $s_{X}^{-1}(a)\tilde{H}_p(\text{Sp}_{2n+1}R) =0$ . If R is local with infinite residue field, we can find a sequence $a=(a_1,\ldots ,a_m)$ such that $a_J\in R^*$ for all $\emptyset \neq J \subset \{1,\ldots ,m\}$ . Since $R^*$ acts on $H_p(\text{Sp}_{2n+2}(R))$ trivially, for such an a, $s_{X}(a)$ acts as the identity on $H_p(\text{Sp}_{2n+2}(R))$ and thus $s_{X}^{-1}(a)H_p(\text{Sp}_{2n+2}R) = H_p(\text{Sp}_{2n+2}(R))$ . In particular, the $R^*$ -equivariant map $\tilde{H}_p(\text{Sp}_{2n+1}(R)) \to H_p(\text{Sp}_{2n+2}(R))$ factors through $s_{X}^{-1}(a)\tilde{H}_p(\text{Sp}_{2n+1}R) =0$ , hence that map is zero. For the last statement we note that $H_p(\text{Sp}_{2n}(R)) \to H_p(\text{Sp}_{2n+2}(R))$ is the localization of $H_p(\text{Sp}_{2n+1}(R)) \to H_p(\text{Sp}_{2n+2}(R))$ at $s_{X}(a)$ . In particular, surjectivity of the second map implies surjectivity of the first. The converse is obvious.

and for $a\in R$ , the action of $\langle a \rangle \in \mathbb{Z}[R]$ on $H_p(\text{Sp}_{2n+1}R)$ defined in Example 5.7 becomes

$$\big(\begin{smallmatrix} 1 & 0 \\ 0 & \langle a \rangle \end{smallmatrix}\!\big): H_p(\text{Sp}_{2n}) \oplus \tilde{H}_p(\text{Sp}_{2n+1}) \to H_p(\text{Sp}_{2n}) \oplus \tilde{H}_p(\text{Sp}_{2n+1}).$$

Proof. This follows from Lemma 5.8 and the fact that $\varepsilon$ and $\rho$ are $(R,\cdot,1)$ -equivariant.

6. Degeneration at $E^2$

In this section we will prove that the spectral sequence (3.1) degenerates at $E^2$ . Our strategy for degeneration is to construct a map of spectral sequences $\tilde{E} \to E$ from a spectral sequence $\tilde{E}$ to (3.1). The spectral sequence $\tilde{E}$ will trivially degenerate at ${\tilde{E}}^2$ , and the main point will be to show that $\tilde{E}^2 \to E^2$ is surjective in all bidegrees. This will ensure that (3.1) degenerates at $E^2$ as well. The spectral sequence $\tilde{E}$ will be a direct sum of spectral sequences E(r), $r=0,\ldots ,n$ , which we will now introduce.

For $0\leqslant r < n$ and $i=1,\ldots ,2r+2$ , we consider the $\text{Sp}_{2n-2r}(R)$ -set

$$U_{2r+2}^{(i)}(R^{2n}) =\{ (\begin{smallmatrix}u \\ w \end{smallmatrix})\in M_{2n,2r+2}(R)|\ u \in U_{2r+2}(R^{2r}), w_i \in U_1(R^{2n-2r}), d_iw=0\}$$

where $N\in \text{Sp}_{2n-2r}(R)$ acts by $N\cdot (\begin{smallmatrix}u \\ w \end{smallmatrix}) = (\begin{smallmatrix}u \\ Nw \end{smallmatrix})$ . The equation $d_iw=(w_1,\ldots ,\hat{w}_i,\ldots ,w_{2r+2})=0$ means that $w={(0,\ldots ,0,w_i,0,\ldots ,0)}$ only has a potentially non-zero entry in the ith column. We have the bijection

$$\text{Sp}_{2n-2r}(R)\backslash U_{2r+2}^{(i)}(R^{2n}) \stackrel{\cong}{\longrightarrow} U_{2r+2}(R^{2r}): (\begin{smallmatrix}u \\ w \end{smallmatrix}) \mapsto u.$$

The stabilizer of the action on $U^{(i)}_{2r+2}(R)$ at $(\begin{smallmatrix}u \\ (e_{1})_i \end{smallmatrix})$ is $\text{Sp}_{2n-2r-1}(R)$ , where $(e_1)_i={(0,\ldots ,0,e_1,0,\ldots ,0)}$ with $e_1\in R^{2n-2r}$ in the ith column. Note that if $v = (\begin{smallmatrix}u \\ w \end{smallmatrix})\in U_{2r+2}^{(i)}$ then $d_jv\in U_{2r+1}(R^{2n})$ for all $1\leqslant j\leqslant 2r+2$ with $j\neq i$ , and $d_iv \in U_{2r+1}(R^{2r})$ . We define the complex ${C}_*(R^{2n};r)$ as

with $\mathbb{Z} [U_{2r+1}(R^{2r})]$ placed in degree 2r, and the ith component of the differential is $(-1)^id_i$ . This is a complex of $\text{Sp}_{2n-2r}(R)$ -modules where $\text{Sp}_{2n-2r}(R)$ acts trivially on the degree 2r piece $\mathbb{Z} [U_{2r+1}(R^{2r})] $ . Since $d\circ d=0$ , the diagram

commutes where the second from left vertical map $d_i^ \wedge$ : $\mathbb{Z}[U^{(i)}_{2r+2}(R^{2n})] \to \mathbb{Z}[U_{2r+2}(R^{2n})]$ is defined on basis elements $v\in U^{(i)}_{2r+2}(R^{2n})$ by

$$d_i^ \wedge (v) = \sum\limits_{j = 1,j \ne i}^{2r + 2} {{{( - 1)}^{j + 1}}} {d_j}v$$

and can informally be thought of as $d_i^ \wedge$ $= d + (-1)^id_i$ . This defines the map of complexes $\varphi: C_*(R^{2n};r) \to C_{*}(R^{2n})$ of $\text{Sp}_{2n-2r}(R)$ -modules (where we have suppressed some of the entries $R^{2n}$ ).

For $r=n$ , we let $C_*(R^{2n},n)$ be the complex $\mathbb{Z}[U_{2n+1}(R^{2n})][2n]$ concentrated in degree 2n and define the map of complexes $\varphi:C_*(R^{2n},n) \to C_*(R^{2n})$ in degree n as the map $d:\mathbb{Z}[U_{2n+1}(R^{2n})]\to \mathbb{Z}[U_{2n}(R^{2n})]$ . For $0 \leqslant r \leqslant n$ , the homomorphism of groups and coefficient complexes

$$(\varepsilon,\varphi):(\text{Sp}_{2n-2r}(R), C_*(R^{2n};r)) \longrightarrow (\text{Sp}_{2n}(R), C_{*}(R^{2n}))$$

defines a map of associated group homology spectral sequences

(6.1) \begin{equation}E_{p,q}^s(R^{2n};r) \longrightarrow E_{p,q}^s(R^{2n})\end{equation}

resulting from the filtrations by degree $C_{\leqslant q}(R^{2n};r)$ and $C_{\leqslant q}(R^{2n})$ of the coefficient complexes $C_*(R^{2n};r)$ $C_*(R^{2n};r)$ and $C_{*}(R^{2n})$ . By definition, we have

$$E^1_{p,q}(R^{2n};r)=H_p(\text{Sp}_{2n-2r}R, C_q(R^{2n};r)).$$

Since $C_q(R^{2n};r)=0$ for $q\neq 2r,2r+1$ , we have $E^s_{p,q}(R^{2n};r)=0$ for $q\neq 2r,2r+1$ and all $s\geqslant 1$ . In particular, $d^s_{p,q}=0$ for $s\geqslant 2$ , and the spectral sequences $E(R^{2n};r)$ degenerate at the $E^2$ page.

The following result shows that the spectral sequence (3.1) degenerates at $E^2$ .

Proposition 6.1. Let R be a local ring with infinite residue field. For all integers $0 \leqslant r \leqslant n$ , $s=2$ , $q=2r, 2r+1$ and all $p\in \mathbb{Z}$ , the map (6.1) is surjective:

$$E_{p,q}^2(R^{2n};r) \twoheadrightarrow E_{p,q}^2(R^{2n}),\quad q=2r, 2r+1.$$

In particular, the spectral sequence (3.1) degenerates at $E^2$ .

Proof of Proposition 6.1 for $q=2r$ . The map $E^1_{p,2r}(R^{2n};r) \to E^1_{p,2r}(R^{2n})$ is the first map in the following complex.

see Corollary 3.4. In view of Lemma 3.5, the second map in that complex is $d^1_{p,2r}: E^1_{p,2r}(R^{2n}) \to E^1_{p,2r-1}(R^{2n})$ . Since $\varepsilon_*:H_p(\text{Sp}_{2n-2r}) \to H_p(\text{Sp}_{2n-2r+1})$ is (split) injective, Lemmas 2.1 and 2.6 imply that this complex is exact. It follows that $E^1_{p,2r}(R^{2n};r)$ surjects onto the kernel of the right vertical map which surjects onto $E^2_{p,2r}(R^{2n})$ . In particular, its quotient $E^2_{p,2r}(R^{2n};r)$ surjects onto $E^2_{p,2r}(R^{2n})$ .

The case $q=2r+1$ of Proposition 6.1 is somewhat more involved except when $r=n$ , in which case the map $0=E^1_{p,2n+1}(R^{2n};n) \to E^1_{p,2n+1}(R^{2n})=0$ is clearly surjective. So assume $0\leqslant r<n$ . For $i=1,\ldots , 2r+2$ , consider the map

(6.2) \begin{equation}\gamma_i:H_p(\text{Sp}_{2n-2r-1}) \otimes \mathbb{Z}[U_{2r+2}(R^{2r})]\longrightarrow H_p(\text{Sp}_{2n-2r-1}) \otimes \mathbb{Z}[\text{Skew}^+_{2r+1}]\end{equation}

which for $u\in U_{2r+2}(R^{2r})$ and $\alpha \in H_p(\text{Sp}_{2n-2r-1})$ is defined by

$$\gamma_i(\alpha \otimes u) = \sum_{1\leqslant j \neq i \leqslant 2r+2}(-1)^{j+1} \big(c^{-1}_{\delta_{ij}\det (u^{\text{${\wedge}$}}_{ij})}\big)_*(\alpha)\otimes d_j\Gamma(u)$$

where $u_{ij}^ \wedge$ is obtained from u by omitting the ith and jth columns, $c_a$ is conjugation with the diagonal matrix $(a,a^{-1},1,\ldots ,1) \in \text{Sp}_{2n-2r}(R)$ for $a\in R^*$ , and $\delta_{ij}$ is defined by

$$\delta_{ij}= \Bigg\{\begin{array}{llc} (-1)^{i+1},& & i\lt j,\\ 0,& & i=j, \\ (-1)^{i},& & i\gt j,\end{array}\quad(\delta_{ij})=\Bigg(\begin{smallmatrix}0 & + & + & + & + & \cdots & +\\+ & 0 & - & - & - & \cdots & -\\- & - & 0 & + & + & \cdots & +\\+ & + & + & 0 & - & \cdots & -\\ & & & & & \ddots &\end{smallmatrix}\Bigg).$$

Lemma 6.2. The commutative diagram

is isomorphic to the following commutative diagram.

Proof. The right vertical and the lower horizontal map have already been identified in Corollary 3.4 and Lemma 3.5. For the other two maps, we note that

$$E_{p,2r+1}^1(R^{2n};r) = \bigoplus_{i=1}^{2r+2} H_p(\text{Sp}_{2n-2r},\mathbb{Z}[U^{(i)}_{2r+2}(R^{2n})]).$$

By Shapiro’s lemma, we obtain the isomorphism

$$ \sum_u \big(\varepsilon, \big(\begin{smallmatrix}u \\ (e_1)_i \end{smallmatrix}\big)\big)_*:\bigoplus_{u \in U_{2r+2}(R^{2r})} H_p(\text{Sp}_{2n-2r-1}) \stackrel{\cong}{\longrightarrow} H_p(\text{Sp}_{2n-2r},\mathbb{Z}[U^{(i)}_{2r+2}(R^{2n})]).$$

This yields the identification of the top horizontal map. Composing with the map $E^1_{p,2r+1}(R^{2n};r) \to E^1_{p,2r+1}(R^{2n})$ yields the map

(6.3) \begin{equation} \bigoplus_{u \in U_{2r+2}(R^{2r})} H_p(\text{Sp}_{2n-2r-1}) \longrightarrow H_p(\text{Sp}_{2n},\mathbb{Z}[{U}_{2r+1}(R^{2n})])\end{equation}

which is

$$\sum_{1\leqslant j \neq i \leqslant 2r+2}(-1)^{j+1} \big(\varepsilon, d_j\big(\begin{smallmatrix}u \\ (e_1)_i \end{smallmatrix}\big)\big)_*$$

on the component corresponding to $u\in U_{2r+2}(R^{2r})$ . We recall the isomorphism

(6.4) \begin{equation}\bigoplus_{A \in \text{Skew}_{2r+1}^+} H_p(\text{Sp}_{2n-2r-1}) \stackrel{\cong}{\longrightarrow} H_p(\text{Sp}_{2n},\mathbb{Z}[{U}_{2r+1}(R^{2n})]) \end{equation}

from Corollary 3.4 which is $(\varepsilon\circ c_{\det v},v)_*$ on the component corresponding to $A \in \text{Skew}_{2r+1}^+(R)$ where $v\in U_{2r+1}(R^{2n})$ satisfies $\Gamma(v)=A$ and generates $R^{2r+1}$ . For $u\in U_{2r+2}(R^{2r})$ and $j\neq i$ , the unimodular sequence $w=d_j (\begin{smallmatrix} u \\ (e_1)_i\end{smallmatrix})$ generates $R^{2r+1}$ . Since

$$\det w = \det \big( d_j \big(\begin{smallmatrix} u \\ (e_1)_i\end{smallmatrix}\big) \big) = \delta_{ij} \det u^{\text{${\wedge}$}}_{ij},$$

the diagram

commutes. Since

$$\Gamma(w) = \Gamma\big( d_j \big(\begin{smallmatrix} u \\ (e_1)_i\end{smallmatrix}\big) \big) = \Gamma( d_ju), $$

we apply Lemma 3.3 to identify the left vertical map in the lemma with $\gamma$ .

Proof of Proposition 6.1 for $q=2r+1$ . We need to show that the map of horizontal complexes

is surjective on homology at the middle term. By Corollary 5.9 and Lemma 6.2, this map of complexes is isomorphic to the direct sum of

(6.5)

and

(6.6)

where $A=H_p(\text{Sp}_{2n-2r-2})$ and $B=H_p(\text{Sp}_{2n-2r})$ . Proposition 6.1 now follows from Lemmas 6.3 and 6.4 below.

Lemma 6.3. The map of complexes (6.6) is surjective in homology.

Proof. Let F be the image of the map $\Gamma(d):\mathbb{Z}[U_{2r+1}(R^{2r})] \to \mathbb{Z}[\text{Skew}_{2r}^+]$ . This is a free $\mathbb{Z}$ -module, and it is also the image of $d:\mathbb{Z}[\text{Skew}^+_{2r+1}] \to \mathbb{Z}[\text{Skew}_{2r}^+]$ , by Lemma 2.1. In diagram (6.6), we can replace $\mathbb{Z}[\text{Skew}_{2r}^+]$ with F and the lower left horizontal arrow $1\otimes d$ with its cokernel $0 \to \text{coker} (1\otimes d)$ without changing homology since that cokernel is $A \otimes F$ , by Lemma 2.6 and the comment thereafter. Thus, we can replace diagram (6.6) with the diagram

(6.7)

without changing homology where the vertical maps are induced by those in diagram (6.6). The right-hand square is obtained by tensoring the diagram of free abelian groups

(6.8)

with the map $\varepsilon_*:A \to B$ . The top horizontal arrow in (6.8) is surjective because the maps $d_i:U_{2r+2}(R^{2r}) \to U_{2r+1}(R^{2r})$ are surjective, by Lemma 2.4. Since all abelian groups in diagram (6.8) are free, that diagram is isomorphic to

where $M= \mathbb{Z}[U_{2r+1}(R^{2r})]$ and N is the kernel of the top horizontal arrow in (6.8). It follows that diagram (6.7) is isomorphic to the following.

Since M and F are free abelian groups, the map on homology (kernels of right horizontal maps) is

$$ (1 \otimes f, 0): (\ker(\varepsilon_*) \otimes M) \oplus (A \otimes N) \longrightarrow \ker(\varepsilon_*) \otimes F,$$

which is surjective since f is.

For a $\mathbb{Z}_0[R]$ -module H, define the homomorphism of $\mathbb{Z}_0[R]$ -modules

(6.9) \begin{equation}\gamma = (\gamma_1,\gamma_2,\ldots ,\gamma_{2r+2}):\bigoplus_{i=1}^{2r+2}H \otimes_{\mathbb{Z}} \mathbb{Z}[U_{2r+2}(R^{2r})] \to H\otimes_{\mathbb{Z}} \mathbb{Z}[\text{Skew}^+_{2r+1}(R)]\end{equation}

by

(6.10) \begin{equation}\gamma_i(h\otimes u) = \sum_{1\leqslant j \neq i \leqslant 2r+2}(-1)^{j+1}\langle \delta_{ij}{\det}^{-1} u^{\text{${\wedge}$}}_{ij}\rangle \cdot h\otimes \Gamma(d_ju)\end{equation}

for $u\in U_{2r+2}(R^{2r})$ and $h\in H$ , generalizing (6.2). Recall from Lemma 5.8 that the relative homology groups $\widetilde{H}_p(\text{Sp}_{2n+1}(R))=H_p(\text{Sp}_{2n+1}(R),\text{Sp}_{2n}(R))$ are quasi-linear $\mathbb{Z}_0[R]$ -modules.

Proof. We may write h[B] in place of $h\otimes B$ for $h\in H$ and $B\in \text{Skew}^+_{2r+1}(R)$ . Denote by $N=\text{coker}(\gamma)$ the cokernel of $\gamma$ . We have to show that $N=0$ . As a cokernel of a $\mathbb{Z}_0[R]$ -linear map of quasi-linear modules, N is also quasi-linear. In N, the expressions on the right-hand side of (6.10) are zero. For $h\in H$ and $u \in U_{2r+2}(R^{2r})$ , the system of equations, expressing the right-hand side of (6.10) as zero, can be written in matrix form as $M(u)\cdot X(u)=0$ where

$$M(u)=(\langle\delta_{ij} {\det}^{-1}u^{\wedge}_{i,j}\rangle )$$

is the $(2r+2)\times (2r+2)$ matrix with entries in $\mathbb{Z}_0[R]$ which has 0s on the diagonal and $\langle\delta_{ij}{\det}^{-1}u^{\wedge}_{i,j}\rangle$ at the (i,j) spot, and $X(u)=( (-1)^{j+1}h\ [\Gamma(u^{\wedge}_{j})])$ is the column vector which has $(-1)^{j+1}h\ [\Gamma(u^{\wedge}_{j})]$ at its jth entry. Multiplying the matrix equation with the adjoint of M(u) yields the equations $(\det M(u))\ h\ [\Gamma(u^{\wedge}_{j})] = 0 \in N$ for $j=1,\ldots ,2r+2$ . In particular, for $j=2r+2$ , we obtain the following result.

• For all $h\in H$ , $B\in \text{Skew}_{2r+1}^+(R)$ , $U\in U_{2r+1}(R^{2r})$ , $x\in R^{2r}$ such that $\Gamma(U)=B$ and $(U,x)\in U_{2r+2}(R^{2r})$ , the following equation holds in N:

(6.11) \begin{equation}(\det M(U,x))\cdot h\cdot [B] =0 \in N.\end{equation}

It turns out that equations (6.11) force N to be zero. More precisely, the following Lemma 6.5 shows that $h [B]= 0\in N$ for all $h\in H$ and $B\in \text{Skew}_{2r+1}^+(R)$ , that is, the map $\gamma$ in Lemma 6.4 is surjective.

Lemma 6.5. For all $B\in \text{Skew}_{2r+1}^+(R)$ and $h\in H$ , the radical of the annihilator ideal

$$\sqrt{\text{Ann}_N(h [B])} \subset \mathbb{Z}_0[R]$$

of $h[B] \in N=\text{coker}(\gamma)$ is the unit ideal.

For $B\in \text{Skew}_{2r+1}^+(R)$ , choose a normal form $U=(u_1,\ldots ,u_{2r+1}) \in U_{2r+1}(R^{2r})$ of B, that is, $\Gamma(U)=B$ , $(u_1,\ldots ,u_{2r})$ is upper triangular, $(u_{2i-1})_{2i-1}=1$ and $(u_{2i})_{2i-1}=0$ for $i=1,\ldots,r$ ; see Lemma 2.3. For $\ell = 1,\ldots ,r$ , the matrix $U(\ell)$ obtained from U by deleting the first $2r - 2\ell$ rows and columns is in $U_{2\ell+1}(R^{2\ell})$ . Indeed, the sequence $U(\ell)$ is unimodular in $R^{2\ell}$ because $(u_1,\ldots ,u_{2r})$ is upper triangular and $(u_1,\ldots ,u_{2r},u_{2r+1})$ is unimodular. It is non-degenerate as for $I\subset \{2r-2\ell +1,\ldots ,{2r+1}\}$ of even cardinality, the sequence $U(\ell)_{I-2r+2\ell}$ generates the orthogonal complement of $u_1,\ldots,u_{2\ell}$ in the non-degenerate space generated by $(u_1,\ldots,u_{2\ell},U_I)$ and is thus non-degenerate.

We will show by descending induction on $\ell = 1,\ldots ,r$ that

(6.12) \begin{equation}\det M(U(\ell),x) \in \sqrt{\text{Ann}(h [B])}\end{equation}

for all $x\in R^{2\ell}$ such that $(U(\ell),x) \in U_{2\ell+2}(R^{2\ell})$ .

The case $\ell = r$ is (6.11). Let $\ell \in \{1,\ldots ,r-1\}$ and assume (6.12) holds for $\ell +1$ in place of $\ell$ . We want to show that (6.12) holds for $\ell$ . Fix $x\in R^{2\ell}$ such that $(U(\ell),x) \in U_{2\ell+2}(R^{2\ell})$ . For $\xi = (s,t,x)\in R \times R \times R^{2\ell}$ , the matrix

$$(U(\ell+1),\xi) =\left(\renewcommand\arraystretch{1.5}\begin{array}{cc|ccc|c}1 & 0 & \cdots & \ast & \cdots & s\\0 & \alpha & \cdots & \ast & \cdots& t \\\hline0 & 0 & &&& x_1\\\vdots & \vdots & & U(\ell) && \vdots \\0 & 0 & &&& x_{2\ell}\end{array}\right)$$

is in $U_{2\ell+4}(R^{2\ell +2})$ if and only if for all $1 \leqslant i<j \leqslant 2\ell+3$ , the square matrix

$$(U(\ell+1)^{\wedge}_{ij},\xi)$$

is invertible, and for all $I \subset \{1,\ldots ,2\ell+3\}$ of odd cardinality less than $2\ell+2$ , the subspace spanned by $(U(\ell+1)_I,\xi)$ is non-degenerate. This happens if and only if $\bar{s},\bar{t}\in F$ is not a solution to any of the equations in F,

(6.13) \begin{equation}L_{ij}(s,t):=\det (U(\ell+1)^{\wedge}_{ij},\xi)=0\quad\text{and}\quad\text{Pf}(\Gamma(U(\ell+1)_I,\xi))=0,\end{equation}

where $1 \leqslant i<j \leqslant 2\ell+3$ and $I \subset \{1,\ldots ,2\ell+3\}$ is of odd cardinality less than $2\ell+2$ . Here, $\text{Pf}(A)$ denotes the Pfaffian of a skew-symmetric matrix A. The equations in (6.13) are linear and homogeneous in $\xi$ , hence, linear (possibly inhomogeneous) in $(s,t)\in R^2$ .

We check that every equation in (6.13) is non-trivial in (s,t), that is, that for each equation in (6.13), there is $(s,t)\in R^2$ for which the left-hand side of that equation does not vanish in F. We start by investigating the Pfaffian equations. Let $I \subset \{1,\ldots ,2\ell+3\}$ be a subset of odd cardinality less than $2\ell+2$ . By abuse of notation we will label the columns of $U(\ell)$ by $(U_3(\ell),\ldots ,U_{2r+3}(\ell))$ so that $U(\ell)_J$ is obtained from $U(\ell+1)_J$ by deleting the first two rows provided $J\subset \{3,4,\ldots ,2\ell+3\}$ . If $1,2 \in I$ , then the subspace spanned by $(U(\ell+1)_I,\xi)$ is non-degenerate as it equals the subspace generated by

which is non-degenerate since $(U(\ell),x)\in U_{2\ell+2}(R^{2\ell})$ . Hence, $\text{Pf}(U(\ell+1)_I,\xi)$ is a unit in R for all $s,t\in R$ . If $1\in I$ but $2 \notin I$ then the subspace spanned by $(U(\ell+1)_I,\xi)$ equals the subspace spanned by

which has Gram matrix

with Pfaffian $t\text{Pf}(\Gamma(U(\ell)_{I-\{1\}})) + c$ where c does not depend on t. Since $\text{Pf}(\Gamma(U(\ell)_{I-\{1\}}))\neq 0 \in F$ , for all $s\in R$ there is a $t\in R$ such that $\text{Pf}(U(\ell+1)_I,\xi)$ is a unit in R. If $2\in I$ but $1 \notin I$ then the subspace spanned by $(U(\ell+1)_I,\xi)$ equals the subspace spanned by

since $\alpha \in R^*$ . This has Gram matrix

with Pfaffian $-\alpha s\text{Pf}(\Gamma(U(\ell)_{I-\{2\}})) + c$ where c does not depend on s. Since $\alpha \text{Pf}(\Gamma(U(\ell)_{I-\{2\}}))\neq 0 \in F$ , for all t there is s such that $\text{Pf}(U(\ell+1)_I,\xi)$ is a unit in R. If $1,2\notin I$ , assume first that $|I| \neq 2\ell +1$ , hence $1 \leqslant |I|\leqslant 2\ell -1$ . Let $J \subset I$ be the subset obtained from I by deleting its maximal element. Then

(6.14) \begin{equation}(U(\ell+1)_I,\xi)=\left(\renewcommand\arraystretch{1.5}\begin{array}{ccc|c|c} & v & &a & s\\ & w & & b & t \\\hline & U(\ell)_J && y & x\end{array}\right).\end{equation}

We need to find $s,t\in R$ such that the Pfaffian of (6.14) is a unit in R, that is, such that the columns of (6.14) span a non-degenerate subspace of $R^{2\ell +2}$ . Since $U(\ell)_J$ spans a non-degenerate subspace of $R^{2\ell}$ , there are unique $a_j,b_j\in R$ , $ j \in J$ , such that

$$\langle U_i(\ell),x\rangle = \sum_{j\in J}a_j\langle U_i(\ell),U_j(\ell)\rangle, \quad\langle U_i(\ell),y\rangle = \sum_{j\in J}b_j\langle U_i(\ell),U_j(\ell)\rangle$$

for all $i\in J$ . Set

$$x_0=\sum_{j\in J}a_jU_j(\ell),\quad s_0 = \sum_{j\in J}a_jv_j,\quad t_0 = \sum_{j\in J}a_jw_j,$$

$$y_0=\sum_{j\in J}b_jU_j(\ell),\quad a_0 = \sum_{j\in J}b_jv_j,\quad b_0 = \sum_{j\in J}b_jw_j.$$

Then the columns of (6.14) and those of

(6.15) \begin{equation}\left(\renewcommand\arraystretch{1.5}\begin{array}{ccc|c|c} & v & &a-a_0 & s-s_0\\ & w & & b-b_0 & t-t_0 \\\hline & U(\ell)_J && y-y_0 & x-x_0\end{array}\right)\end{equation}

span the same subspace of $R^{2\ell+2}$ . Moreover, $y-y_0, x-x_0$ is a basis of the orthogonal complement of $U(\ell)_J$ inside the non-degenerate subspace of $R^{2\ell}$ generated by $(U(\ell)_J,y,x) = (U(\ell)_I,x)$ . In particular, $c:=\langle y-y_0,x-x_0\rangle\in R^*$ . For $(s,t)=(s_0,t_0)$ , the Gram-matrix of (6.15) is

$$\left(\renewcommand\arraystretch{1.5}\begin{array}{c|c|c} \Gamma (U(\ell+1)_J)& \ast & 0\\ \hline \ast & 0 & c \\ \hline 0 & -c & 0 \end{array}\right)$$

which has Pfaffian $c\text{Pf}(U(\ell +1)_J) \in R^*$ . In particular, the Pfaffian of (6.14) is a unit for $(s,t)=(s_0,t_0)$ . If $|I|=2\ell+1$ (and $1,2\notin I$ ) then the space generated by $(U(\ell+1)_I,\xi)$ is non-degenerate if and only if the determinant $L_{12}(s,t)$ of $(U(\ell+1)_I,\xi)$ is a unit. This is a special case of the linear equations $L_{ij}(s,t)$ , $1 \leqslant i<j \leqslant 2\ell+3$ , which we now investigate. We have

$$L_{12}(s,t) = \det (U(\ell+1),\xi)^{\wedge}_{12} = as+bt +c$$

for some $c\in R$ where $a=-\det(A)$ , $b=\det B$ , A is obtained from $U(\ell+1)^{\wedge}_{12}$ by deleting the first row, and B is obtained from $U(\ell+1)^{\wedge}_{12}$ by deleting the second row. The matrices A and B are invertible because $U(\ell +1) \in U_{2\ell+3}(R^{2\ell+2})$ , $U_1(\ell +1)=e_1$ and $U_2(\ell +1)=\alpha e_2$ , $\alpha\in R^*$ . In particular, a and b are units, and there is $(s,t)\in R^2$ such that $L_{12}(s,t)\in R^*$ . For $i=1,2$ and $3 \leqslant j \leqslant 2\ell+3$ , we have

$$L_{1j}(s,t)= a_{1j}s + c_{1j}\quad\text{and}\quad L_{2j}(s,t) = b_{2j}t +c_{2i},$$

where $a_{1j}=-\alpha\det U(\ell)^{\wedge}_{j}$ and $b_{2j}=\det U(\ell)^{\wedge}_{j}$ are units in R, and $c_{1j}, c_{2j}\in R$ . In particular, for $i=1,2$ and $3 \leqslant j \leqslant 2\ell+3$ , there is $(s,t)\in R^2$ such that $L_{ij}(s,t)\in R^*$ . For $3 \leqslant i<j\leqslant 2r+1$ ,

$$L_{ij}(s,t)=\alpha L_{i,j}(U(\ell),x)$$

does not depend on $s,t\in R$ and is a unit since $(U(\ell),x) \in U_{2\ell+2}(R^{2\ell})$ . Summarizing, for every equation in (6.13), there is $(s,t)\in R^2$ for which the left-hand side of that equation does not vanish in F.

From the computation of $L_{ij}(s,t)$ above, the matrix $M(U(\ell+1),\xi)$ is

$$\left(\renewcommand\arraystretch{2}\begin{array}{cc|ccc}0 & \langle as+bt +c\rangle^{-1} & \cdots & \langle\delta_{1j}\rangle \langle a_{1j}s + c_{1j}\rangle^{-1} \cdots &\langle c_{1}\rangle^{-1}\\\langle as+bt +c\rangle^{-1} & 0 & \cdots & \langle\delta_{2j}\rangle\langle b_{2j}t +c_{2j} \rangle^{-1} \cdots &\langle -c_{2}\rangle^{-1}\\\hline\vdots&\vdots&&&\\\langle\delta_{i1}\rangle \langle a_{1i}s + c_{1i}\rangle^{-1}&\langle\delta_{i2}\rangle\langle b_{2i}t +c_{2i} \rangle^{-1} &&\\\vdots&\vdots&& \langle \alpha \rangle^{-1}\cdot M(U(\ell),x) &\\\langle c_1\rangle^{-1}&\langle c_2\rangle^{-1}&&&\end{array}\right).$$

By assumption, it has determinant $g(s,t)=\det M(U(\ell+1),\xi)$ in $\sqrt{\text{Ann}(h[B])}$ for all $(\bar{s},\bar{t})\in F^2-S$ where S is a finite union of affine subspaces of dimension no greater than 1 defined by the equations (6.13). For $\gamma \in R^*$ , consider the equation $\gamma = as+bt+c$ and note that for all but finitely many $\bar{\gamma}\in F^*$ the hyperplane $\bar{\gamma} = \bar{a}\bar{s}+\bar{b}\bar{t}+\bar{c}$ in $F^2$ is not entirely in S. Then $s=a^{-1}(\gamma-c -bt)$ and $M(U(\ell+1),\xi)$ becomes

$$\left(\renewcommand\arraystretch{2}\begin{array}{cc|ccc}0 & \langle \gamma\rangle^{-1} & \cdots & \langle\delta_{1j}\rangle\langle \tilde{a}_{1j}t + \tilde{c}_{1j}\rangle^{-1} \cdots &\langle c_{1}\rangle^{-1}\\\langle \gamma\rangle^{-1} & 0 & \cdots & \langle\delta_{2j}\rangle\langle b_{2j}t +c_{2j} \rangle^{-1} \cdots &\langle -c_{2}\rangle^{-1}\\\hline\vdots&\vdots&&&\\\langle\delta_{i1}\rangle\langle \tilde{a}_{1i}t + \tilde{c}_{1i}\rangle^{-1} &\langle\delta_{i2}\rangle\langle b_{2i}t +c_{2i}\rangle^{-1}&&&\\\vdots&\vdots&& \langle \alpha \rangle^{-1}\cdot M(U(\ell),x) &\\\langle c_1\rangle^{-1}&\langle c_2 \rangle^{-1}&&&\end{array}\right)$$

where $\tilde{a}_{1j} = -a_{1j}b/a$ and $\tilde{c}_{1j} = c_{1j} + a_{1j}(\gamma-c)/a$ . Its determinant $f(t,\gamma) = g(a^{-1}(\gamma-c -bt),t)$ is in $\sqrt{\text{Ann}(z[B])}$ for all $\bar{t}\in F-S'$ for a finite set $S'\subset F$ (for fixed $\gamma$ ). Since the coefficients $\tilde{a}_{1j}$ and $b_{2j}$ of t are units in R, we can apply Theorem 4.9 and find that $\lim_{t\to \infty}f(t,\gamma) \in \sqrt{\text{Ann}(h[B])}$ where

$$f(\gamma) = \lim_{t\to \infty}f(t,\gamma) = \det\left(\renewcommand\arraystretch{2}\begin{array}{cc|ccc}0 & \langle \gamma \rangle^{-1} &0 \cdots &0 \cdots 0&\langle c_{1}\rangle^{-1}\\\langle \gamma \rangle^{-1} & 0 &0 \cdots & 0 \cdots 0&\langle -c_{2}\rangle^{-1}\\\hline0&0&&&\\\vdots&\vdots&& \langle \alpha \rangle^{-1}\cdot M(U(\ell),x) &\\0&0&&&\\\langle c_{1}\rangle^{-1}&\langle c_{2}\rangle^{-1}&&&\end{array}\right)$$

for all but finitely many $\bar{\gamma} \in F$ . Then

$$\langle \gamma\rangle^2 f(\gamma) = \det\left(\renewcommand\arraystretch{2}\begin{array}{cc|ccc}0 & 1 &0 \cdots &0 \cdots 0&\langle c_{1}\rangle^{-1}\langle \gamma \rangle\\1 & 0 &0 \cdots & 0 \cdots 0&\langle -c_{2}\rangle^{-1}\langle \gamma \rangle\\\hline0&0&&&\\\vdots&\vdots&& \langle \alpha \rangle^{-1}\cdot M(U(\ell),x) &\\0&0&&&\\\langle c_{1}\rangle^{-1}&\langle c_{1}\rangle^{-1}&&&\end{array}\right)$$

is in $\sqrt{\text{Ann}(h[B])}$ for all but finitely many $\bar{\gamma} \in F$ . By Theorem 4.9, the element

$$\lim_{\gamma \to 0}\langle \gamma\rangle^2 f(\gamma) =\det \left(\renewcommand\arraystretch{2}\begin{array}{cc|ccc}0 & 1 &0 \cdots &0 \cdots 0&0\\1 & 0 &0 \cdots & 0 \cdots 0& 0\\\hline0&0&&&\\\vdots&\vdots&& \langle \alpha \rangle^{-1}\cdot M(U(\ell),x) &\\0&0&&&\\\langle c_{1}\rangle^{-1}&\langle c_{1}\rangle^{-1}&&&\end{array}\right)$$

is also in $\sqrt{\text{Ann}(h[B])}$ . Hence, we have $-\langle \alpha\rangle^{-2\ell}\det M(U(\ell),x) \in \sqrt{\text{Ann}(z[B])}$ which implies $\det M(U(\ell),x) \in \sqrt{\text{Ann}(h[B])}$ since $-\langle \alpha\rangle^{-2\ell}$ is a unit in $\mathbb{Z}_0[R]$ . This finishes the proof of (6.12) for $\ell = 1,\ldots ,r$ . In particular, it holds for $\ell =1$ .

Finally, we investigate what (6.12) means for $\ell = 1$ . The given matrix

$$U(1)=\begin{pmatrix} 1 & 0 & b \\ 0 & a & c \end{pmatrix}$$

has $a,b,c\in R^*$ since it is in $U_3(R^2)$ . For $x=(s,t)\in R^2$ , the matrix

$$(U(1),x) = \begin{pmatrix} 1 & 0 & b & s\\ 0 & a & c & t \end{pmatrix}$$

is in $U_4(R^2)$ if and only if $s,t,bt-cs\in R^*$ . Then $M(U(1),x) = (\langle\delta_{ij}{\det}^{-1}(U(1),x)^{\wedge}_{ij}\rangle)$ has determinant

$$f(s,t) = \det M(U(1),x)= \det \begin{pmatrix}0 & \langle bt-cs\rangle^{-1} & \langle -as\rangle^{-1} & \langle -ab\rangle^{-1}\\\langle bt-cs\rangle^{-1} & 0 & \langle -t\rangle^{-1} & \langle -c\rangle ^{-1}\\ \langle as\rangle^{-1} & \langle -t\rangle^{-1} & 0 & \langle a \rangle^{-1}\\ \langle -ab\rangle^{-1} & \langle c\rangle ^{-1} & \langle a \rangle^{-1} & 0\end{pmatrix}$$

in $\sqrt{\text{Ann}(h [B])}$ for all $s,t,\in R^*$ such that $bt-cs\in R^*$ . Setting $s=1$ , every $t\in R$ such that $\bar{t}\neq 0, \bar{c}/\bar{b}\in F$ has

$$f(1,t)= \det \begin{pmatrix}0 & \langle bt-c\rangle^{-1} & \langle -a\rangle^{-1} & \langle -ab\rangle^{-1}\\\langle bt-c\rangle^{-1} & 0 & \langle -t\rangle^{-1} & \langle -c\rangle ^{-1}\\ \langle a\rangle^{-1} & \langle -t\rangle^{-1} & 0 & \langle a \rangle^{-1}\\ \langle -ab\rangle^{-1} & \langle c\rangle ^{-1} & \langle a \rangle^{-1} & 0\end{pmatrix}$$

in $\sqrt{\text{Ann}(h [B])}$ . Since the coefficients b and $-1$ of t are units in R, we can apply Theorem 4.9 and find that the element

$$\lim_{t \to \infty}f(1,t) =\det \begin{pmatrix}0 &0 & \langle -a\rangle^{-1} & \langle -ab\rangle^{-1}\\0 & 0 & 0 & \langle -c\rangle ^{-1}\\ \langle a\rangle^{-1} & 0 & 0 & \langle a \rangle^{-1}\\ \langle -ab\rangle^{-1} & \langle c\rangle ^{-1} & \langle a \rangle^{-1} & 0\end{pmatrix}=\langle ac\rangle^{-2}$$

is in $\sqrt{\text{Ann}(h [B])}$ . Since $\langle ac\rangle^{-2}$ is a unit in $\mathbb{Z}_0[R]$ , the ideal $\sqrt{\text{Ann}(h [B])}$ is the unit ideal.

7. Homology stability

In this section we prove the results announced in the Introduction. The following result proves Theorem 1.1.

Theorem 7.1. Let R be a commutative local ring with infinite residue field and $n\geqslant 0$ an integer. Then in the sequence

$$H_{2n}(\text{Sp}_{2n}R) \stackrel{\cong}{\longrightarrow} H_{2n}(\text{Sp}_{2n+1}R) \stackrel{\cong}{\longrightarrow} H_{2n}(\text{Sp}_{2n+2}R) \stackrel{\cong}{\longrightarrow} \cdots$$

of integral homology groups, all maps are isomorphisms, and in the sequence

$$H_{2n+1}(\text{Sp}_{2n+1}R) \twoheadrightarrow H_{2n+1}(\text{Sp}_{2n+2}R) \stackrel{\cong}{\longrightarrow} H_{2n+1}(\text{Sp}_{2n+3}R) \stackrel{\cong}{\longrightarrow} \cdots.$$

of integral homology groups, the first map is a surjection and all other maps are isomorphisms. Moreover, inclusion of groups induces a surjection

$$H_{2n+1}(\text{Sp}_{2n}(R)) \twoheadrightarrow H_{2n+1}(\text{Sp}_{2n+2}(R)).$$

In particular, $H_i(\text{Sp}_{2n}(R),\text{Sp}_{2n-2}(R))=0$ for all $i<2n$ .

Proof. The case $n=0$ is clear, so assume $n\geqslant 1$ . The spectral sequence (3.1) degenerates at the $E^2$ page (Proposition 6.1). By Lemma 3.1, we have $E^2_{p,q}(R^{2n})=0$ for $p+q<2n$ . Moreover, $0=d:\mathbb{Z}[\text{Skew}^+_2(R)] \to \mathbb{Z}[\text{Skew}_1^+(R)]$ forcing $d^1_{p,2}=0$ for all $p\in \mathbb{Z}$ (Lemma 3.5). Therefore, $d^1_{p,1}:E^1_{p,1}(R^{2n}) \to E^1_{p,0}(R^{2n})$ is an isomorphism for $p\leqslant 2n-2$ and a surjection for $p=2n-1$ . Hence,

$$H_{p}(\text{Sp}_{2n-2}) \oplus \widetilde{H}_{p}(\text{Sp}_{2n-1}) = H_{p}(\text{Sp}_{2n-1}) \longrightarrow H_{p}(\text{Sp}_{2n})$$

is an isomorphism for $p\leqslant 2n-2$ and a surjection for $p=2n-1$ . By Lemma 5.8, the map is zero on the second summand. In particular,

$$\widetilde{H}_{p}(\text{Sp}_{2n-1}) = 0 \quad\text{for }p\leqslant 2n-2$$

and

$$H_{p}(\text{Sp}_{2n-2}) \stackrel{\cong}{\longrightarrow} H_{p}(\text{Sp}_{2n})\quad\text{for }p\leqslant 2n-2.$$

This proves the first string of isomorphisms and the second string of a surjection followed by isomorphisms in the theorem. Using Lemma 5.8, the surjectivity of $H_{2n-1}(\text{Sp}_{2n-1}) \longrightarrow H_{2n-1}(\text{Sp}_{2n})$ implies the surjectivity of $H_{2n-1}(\text{Sp}_{2n-2}) \longrightarrow H_{2n-1}(\text{Sp}_{2n})$ .

Let $K^{MW}_*(R)$ be the Milnor–Witt K-theory ring of R [Reference MorelMor12, Definition 3.1], [Reference SchlichtingSch17a, Definition 4.10]. The following result proves Theorem 1.2.

Theorem 7.2. Let R be a local ring with infinite residue field and $n\geqslant 1$ an integer. Then the inclusions of groups $\text{Sp}_{2r} \subset SL_{2r} \subset SL_{2r+1}$ induce a surjection

$$H_{2n}(\text{Sp}_{2n}(R),\text{Sp}_{2n-2}(R)) \twoheadrightarrow H_{2n}(SL_{2n}(R),SL_{2n-1}(R))=K^{MW}_{2n}(R).$$

Proof. Consider the string of maps

$$H_2(\text{Sp}_2(R))^{\otimes n} \to H_{2n}(\text{Sp}_{2n}(R)) \to H_{2n}(\text{Sp}_{2n}(R),\text{Sp}_{2n-2}(R)) \to H_{2n}(SL_{2n}(R),SL_{2n-1}(R))$$

in which the first map is induced by the block sum of matrices. By [Reference SchlichtingSch17a, Theorem 5.37 and proof], the composition is the surjective multiplication map

$$K_2^{MW}(R)^{\otimes n} \twoheadrightarrow K_{2n}^{MW}(R).$$

It follows that the last map in the composition is surjective.

Remark 7.3. Let k be an infinite perfect field of characteristic not equal to 2 which is finitely generated over its prime field. Then neither of the surjective maps

(7.1) \begin{equation}H_3(\text{Sp}_2(k)) \twoheadrightarrow H_3(\text{Sp}_4(k))\quad\text{and}\quad H_4(\text{Sp}_4(k),\text{Sp}_2(k))\twoheadrightarrow K^{MW}_4(k)\end{equation}

is injective. For the first map, this is [Reference Sarwar and SchlichtingSS25, Remark 8.4] and follows from [Reference Hutchinson and WendtHW15, Theorem 7.4]. If the second map were an isomorphism, then the map

$$H_4(\text{Sp}_4(k)) \to H_4(\text{Sp}_4(k),\text{Sp}_2(k))$$

would be surjective (see the proof of Theorem 7.2), and the long homology exact sequence for the pair $(\text{Sp}_4(k),\text{Sp}_2(k))$ would force the first map in (7.1) to be injective.