2.1 Parabolic and stabilizer subgroups

Our proof of Theorem 1.1 depends on a careful study of various subgroups of $\mathbf{G}_{n}(\text{k})$ . In this section, we will introduce notation for these subgroups: a certain parabolic subgroup $\mathbf{PG}_{n}^{\ell }(\text{k})$ , its unipotent radical $\mathbf{UG}_{n}^{\ell }(\text{k})$ , a Levi component $\mathbf{LG}_{n}^{\ell }(\text{k})$ of $\mathbf{PG}_{n}^{\ell }(\text{k})$ , and another subgroup $\mathbf{FG}_{n}^{\ell }(\text{k})$ that lies in $\mathbf{PG}_{n}^{\ell }(\text{k})$ and fixes certain vectors.

General and special linear groups

Assume first that $\mathbf{G}_{n}$ is either $\operatorname{GL}_{n}$ or $\operatorname{SL}_{n}$ . The group $\mathbf{G}_{n}(\text{k})$ thus acts on the vector space $\text{k}^{n}$ , and the $\text{k}$ -parabolic subgroups of $\mathbf{G}_{n}(\text{k})$ are the stabilizers of flags of subspaces of $\text{k}^{n}$ . Let $(\vec{a}_{1},\ldots ,\vec{a}_{n})$ be the standard basis for $\text{k}^{n}$ . For $1\leqslant \ell \leqslant n$ , the group $\mathbf{PG}_{n}^{\ell }(\text{k})$ is defined to be the $\mathbf{G}_{n}(\text{k})$ -stabilizer of the flag

$$\begin{eqnarray}0\subsetneq \langle \vec{a}_{1},\ldots ,\vec{a}_{\ell }\rangle .\end{eqnarray}$$

The group $\mathbf{UG}_{n}^{\ell }(\text{k})$ is the subgroup of $\mathbf{PG}_{n}^{\ell }(\text{k})$ consisting of all $M\in \mathbf{PG}_{n}^{\ell }(\text{k})$ that act as the identity on both

$$\begin{eqnarray}\langle \vec{a}_{1},\ldots ,\vec{a}_{\ell }\rangle \quad \text{and}\quad \text{k}^{n}/\langle \vec{a}_{1},\ldots ,\vec{a}_{\ell }\rangle .\end{eqnarray}$$

The group $\mathbf{LG}_{n}^{\ell }(\text{k})$ is defined to be the $\mathbf{PG}_{n}^{\ell }(\text{k})$ -stabilizer of the flag

$$\begin{eqnarray}0\subsetneq \langle \vec{a}_{\ell +1},\ldots ,\vec{a}_{n}\rangle .\end{eqnarray}$$

If $\mathbf{G}_{n}=\operatorname{GL}_{n}$ then $\mathbf{LG}_{n}^{\ell }(\text{k})$ is the subgroup $\operatorname{GL}_{\ell }(\text{k})\times \operatorname{GL}_{n-\ell }(\text{k})$ of $\mathbf{G}_{n}$ , while if $\mathbf{G}_{n}=\operatorname{SL}_{n}$ then $\mathbf{LG}_{n}^{\ell }(\text{k})$ is the subgroup of $\operatorname{GL}_{\ell }(\text{k})\times \operatorname{GL}_{n-\ell }(\text{k})$ consisting of matrices of determinant $1$ . Finally, define

$$\begin{eqnarray}\mathbf{FG}_{n}^{\ell }(\text{k})=\{M\in \mathbf{G}_{n}(\text{k})|M(\vec{a}_{j})=\vec{a}_{j}\text{ for }1\leqslant j\leqslant \ell \}.\end{eqnarray}$$

We thus have $\mathbf{FG}_{n}^{\ell }(\text{k})\subset \mathbf{PG}_{n}^{\ell }(\text{k})$ .

Symplectic groups

Now assume that $\mathbf{G}_{n}=\operatorname{Sp}_{2n}$ . Letting $\unicode[STIX]{x1D714}(\cdot ,\cdot )$ be the standard symplectic form on $\text{k}^{2n}$ , the group $\mathbf{G}_{n}(\text{k})$ is the subgroup of $\operatorname{GL}_{n}(\text{k})$ consisting of elements that preserve $\unicode[STIX]{x1D714}(\cdot ,\cdot )$ . The $\text{k}$ -parabolic subgroups of $\mathbf{G}_{n}(\text{k})$ are the $\mathbf{G}_{n}(\text{k})$ -stabilizers of flags of isotropic subspaces of $\text{k}^{2n}$ , that is, subspaces on which $\unicode[STIX]{x1D714}(\cdot ,\cdot )$ vanishes identically. Let $(\vec{a}_{1},\ldots ,\vec{a}_{n},\vec{b}_{1},\ldots ,\vec{b}_{n})$ be the standard symplectic basis for $\text{k}^{2n}$ , so

$$\begin{eqnarray}\unicode[STIX]{x1D714}(\vec{a}_{j},\vec{a}_{j^{\prime }})=\unicode[STIX]{x1D714}(\vec{b}_{j},\vec{b}_{j^{\prime }})=0\quad \text{and}\quad \unicode[STIX]{x1D714}(\vec{a}_{j},\vec{b}_{j^{\prime }})=\unicode[STIX]{x1D6FF}_{jj^{\prime }}\end{eqnarray}$$

for $1\leqslant j,j^{\prime }\leqslant n$ , where $\unicode[STIX]{x1D6FF}_{jj^{\prime }}$ is the Kronecker delta function. For $1\leqslant \ell \leqslant n$ , the group $\mathbf{PG}_{n}^{\ell }(\text{k})$ is defined to be the $\mathbf{G}_{n}(\text{k})$ -stabilizer of the isotropic flag

$$\begin{eqnarray}0\subsetneq \langle \vec{a}_{1},\ldots ,\vec{a}_{\ell }\rangle .\end{eqnarray}$$

The group $\mathbf{UG}_{n}^{\ell }(\text{k})$ is the subgroup of $\mathbf{PG}_{n}^{\ell }(\text{k})$ consisting of all $M\in \mathbf{PG}_{n}^{\ell }(\text{k})$ that act as the identity on both

$$\begin{eqnarray}\displaystyle \langle \vec{a}_{1},\ldots ,\vec{a}_{\ell }\rangle \quad \text{and}\quad \langle \vec{a}_{1},\ldots ,\vec{a}_{\ell }\rangle ^{\bot }/\langle \vec{a}_{1},\ldots ,\vec{a}_{\ell }\rangle =\langle \vec{a}_{1},\ldots ,\vec{a}_{\ell },\vec{a}_{\ell +1},\vec{b}_{\ell +1},\ldots ,\vec{a}_{n},\vec{b}_{n}\rangle /\langle \vec{a}_{1},\ldots ,\vec{a}_{\ell }\rangle . & & \displaystyle \nonumber\end{eqnarray}$$

The group $\mathbf{LG}_{n}^{\ell }(\text{k})$ is defined to be the $\mathbf{PG}_{n}^{\ell }(\text{k})$ -stabilizer of the isotropic flag

$$\begin{eqnarray}0\subsetneq \langle \vec{b}_{1},\ldots ,\vec{b}_{\ell }\rangle .\end{eqnarray}$$

The group $\mathbf{LG}_{n}^{\ell }(\text{k})$ is thus isomorphic to $\operatorname{GL}_{\ell }(\text{k})\times \mathbf{G}_{n-\ell }(\text{k})$ . Finally, define

$$\begin{eqnarray}\mathbf{FG}_{n}^{\ell }(\text{k})=\{M\in \mathbf{G}_{n}(\text{k})\mid M(\vec{a}_{j})=\vec{a}_{j}\text{ for }1\leqslant j\leqslant \ell \}.\end{eqnarray}$$

We thus have $\mathbf{FG}_{n}^{\ell }(\text{k})\subset \mathbf{PG}_{n}^{\ell }(\text{k})$ .

Orthogonal groups

Finally, assume that $\mathbf{G}_{n}$ is either $\operatorname{SO}_{n,n}$ or $\operatorname{SO}_{n,n+1}$ . For an appropriate $m$ , the group $\mathbf{G}_{n}(\text{k})$ is then the subgroup of $\operatorname{SL}_{m}(\text{k})$ consisting of elements that preserve a quadratic form $q(\cdot )$ on $\text{k}^{m}$ .

1. – If $\mathbf{G}_{n}=\operatorname{SO}_{n,n}$ , then let $m=2n$ and let $(\vec{a}_{1},\ldots ,\vec{a}_{n},\vec{b}_{1},\ldots ,\vec{b}_{n})$ be the standard basis for $\text{k}^{m}$ . The group $\mathbf{G}_{n}(\text{k})$ is the $\operatorname{SL}_{m}(\text{k})$ -stabilizer of the quadratic form $q(\cdot )$ on $\text{k}^{m}$ defined via the formula

   $$\begin{eqnarray}q\biggl(\mathop{\sum }_{j=1}^{n}(c_{j}\vec{a}_{j}+d_{j}\vec{b}_{j})\biggr)=\mathop{\sum }_{j=1}^{n}c_{j}d_{j}\quad (c_{j},d_{j}\in \text{k}).\end{eqnarray}$$

2. – If $\mathbf{G}_{n}=\operatorname{SO}_{n,n+1}$ , then let $m=2n+1$ and let $(\vec{a}_{1},\ldots ,\vec{a}_{n},\vec{b}_{1},\ldots ,\vec{b}_{n},\vec{e})$ be the standard basis for $\text{k}^{m}$ . The group $\mathbf{G}_{n}(\text{k})$ is the $\operatorname{SL}_{m}(\text{k})$ -stabilizer of the quadratic form $q(\cdot )$ on $\text{k}^{m}$ defined via the formula

   $$\begin{eqnarray}q\biggl(\unicode[STIX]{x1D706}\vec{e}+\mathop{\sum }_{j=1}^{n}(c_{j}\vec{a}_{j}+d_{j}\vec{b}_{j})\biggr)=\unicode[STIX]{x1D706}^{2}+\mathop{\sum }_{j=1}^{n}c_{j}d_{j}\quad (c_{j},d_{j},\unicode[STIX]{x1D706}\in \text{k}).\end{eqnarray}$$

In both cases, the $\text{k}$ -parabolic subgroups of $\mathbf{G}_{n}(\text{k})$ are the $\mathbf{G}_{n}(\text{k})$ -stabilizers of flags of isotropic subspaces of $\text{k}^{m}$ , that is, subspaces on which $q(\cdot )$ vanishes identically. For $1\leqslant \ell \leqslant n$ the group $\mathbf{PG}_{n}^{\ell }(\text{k})$ is defined to be the $\mathbf{G}_{n}(\text{k})$ -stabilizer of the isotropic flag

$$\begin{eqnarray}0\subsetneq \langle \vec{a}_{1},\ldots ,\vec{a}_{\ell }\rangle .\end{eqnarray}$$

For $\mathbf{G}_{n}=\operatorname{SO}_{n,n}$ , the group $\mathbf{UG}_{n}^{\ell }(\text{k})$ is the subgroup of $\mathbf{PG}_{n}^{\ell }(\text{k})$ consisting of all $M\in \mathbf{PG}_{n}^{\ell }(\text{k})$ that act as the identity on both

$$\begin{eqnarray}\displaystyle \langle \vec{a}_{1},\ldots ,\vec{a}_{\ell }\rangle \quad \text{and}\quad \langle \vec{a}_{1},\ldots ,\vec{a}_{\ell }\rangle ^{\bot }/\langle \vec{a}_{1},\ldots ,\vec{a}_{\ell }\rangle =\langle \vec{a}_{1},\ldots ,\vec{a}_{\ell },\vec{a}_{\ell +1},\vec{b}_{\ell +1},\ldots ,\vec{a}_{n},\vec{b}_{n}\rangle /\langle \vec{a}_{1},\ldots ,\vec{a}_{\ell }\rangle , & & \displaystyle \nonumber\end{eqnarray}$$

while if $\mathbf{G}_{n}=\operatorname{SO}_{n,n+1}$ , then the group $\mathbf{UG}_{n}^{\ell }(\text{k})$ is the subgroup of $\mathbf{PG}_{n}^{\ell }(\text{k})$ consisting of all $M\in \mathbf{PG}_{n}^{\ell }(\text{k})$ that act as the identity on both

$$\begin{eqnarray}\displaystyle \langle \vec{a}_{1},\ldots ,\vec{a}_{\ell }\rangle \quad \text{and}\quad & & \displaystyle \langle \vec{a}_{1},\ldots ,\vec{a}_{\ell }\rangle ^{\bot }/\langle \vec{a}_{1},\ldots ,\vec{a}_{\ell }\rangle \nonumber\\ \displaystyle & & \displaystyle \quad =\langle \vec{a}_{1},\ldots ,\vec{a}_{\ell },\vec{a}_{\ell +1},\vec{b}_{\ell +1},\ldots ,\vec{a}_{n},\vec{b}_{n},\vec{e}\rangle /\langle \vec{a}_{1},\ldots ,\vec{a}_{\ell }\rangle .\nonumber\end{eqnarray}$$

The group $\mathbf{LG}_{n}^{\ell }(\text{k})$ is defined to be the $\mathbf{PG}_{n}^{\ell }(\text{k})$ -stabilizer of the isotropic flag

$$\begin{eqnarray}0\subsetneq \langle \vec{b}_{1},\ldots ,\vec{b}_{\ell }\rangle .\end{eqnarray}$$

The group $\mathbf{LG}_{n}^{\ell }(\text{k})$ is thus isomorphic to $\operatorname{GL}_{\ell }(\text{k})\times \mathbf{G}_{n-\ell }(\text{k})$ . Finally, define

$$\begin{eqnarray}\mathbf{FG}_{n}^{\ell }(\text{k})=\{M\in \mathbf{G}_{n}(\text{k})\mid M(\vec{a}_{j})=\vec{a}_{j}\text{ for }1\leqslant j\leqslant \ell \}.\end{eqnarray}$$

We thus have $\mathbf{FG}_{n}^{\ell }(\text{k})\subset \mathbf{PG}_{n}^{\ell }(\text{k})$ .

2.2 Facts about the Steinberg representation

Let $R$ be a commutative ring. The following theorem of Reeder [Reference ReederRee91] will play an important role in our proof of Theorem 1.1.

Theorem 2.1 [Reference ReederRee91, Proposition 1.1].

Let $\mathbf{G}$ be a connected reductive group defined over a field $\text{k}$ , let $\mathbf{PG}(\text{k})$ be a $\text{k}$ -parabolic subgroup of $\mathbf{G}(\text{k})$ , and let $\mathbf{LG}(\text{k})$ be a Levi component of $\mathbf{PG}(\text{k})$ . Then there exists an $\mathbf{LG}(\text{k})$ -equivariant map

$$\begin{eqnarray}\operatorname{St}_{\mathbf{LG}}(\text{k};R)\longrightarrow \operatorname{Res}_{\mathbf{LG}(\text{k})}^{\mathbf{G}(\text{k})}\operatorname{St}_{\boldsymbol{ G}}(\text{k};R)\end{eqnarray}$$

such that the induced map

$$\begin{eqnarray}\operatorname{Ind}_{\mathbf{LG}(\text{k})}^{\mathbf{PG}(\text{k})}\operatorname{St}_{\boldsymbol{ LG}}(\text{k};R)\rightarrow \operatorname{Res}_{\mathbf{PG}(\text{k})}^{\mathbf{G}(\text{k})}\operatorname{St}_{\boldsymbol{ G}}(\text{k};R)\end{eqnarray}$$

is an isomorphism.

We wish to apply this to the distinguished parabolic subgroups $\mathbf{PG}_{n}^{\ell }(\text{k})$ that we introduced in § 2.1. To do this, we need to identify $\operatorname{St}_{\mathbf{LG}_{n}^{\ell }}(\text{k};R)$ .

Lemma 2.3. Let $\mathbf{G}_{n}$ be either $\operatorname{GL}_{n}$ , $\operatorname{SL}_{n}$ , $\operatorname{Sp}_{2n}$ , $\operatorname{SO}_{n,n}$ , or $\operatorname{SO}_{n,n+1}$ . Then for all fields $\text{k}$ and all commutative rings $R$ , we have

$$\begin{eqnarray}\operatorname{St}_{\mathbf{LG}_{n}^{\ell }}(\text{k};R)=\operatorname{St}_{\operatorname{GL}_{\ell }}(\text{k};R)\otimes \operatorname{St}_{\mathbf{G}_{n-\ell }}(\text{k};R)\end{eqnarray}$$

for $1\leqslant \ell \leqslant n$ .

These two results allow us to make the following definition.

Definition 2.4. Let $\mathbf{G}_{n}$ be either $\operatorname{GL}_{n}$ , $\operatorname{SL}_{n}$ , $\operatorname{Sp}_{2n}$ , $\operatorname{SO}_{n,n}$ , or $\operatorname{SO}_{n,n+1}$ . Also, let $\text{k}$ be a field and $R$ be a commutative ring. For $1\leqslant \ell \leqslant n$ , the Reeder product map is the map

$$\begin{eqnarray}\operatorname{St}_{\operatorname{GL}_{\ell }}(\text{k};R)\otimes \operatorname{St}_{\mathbf{G}_{n-\ell }}(\text{k};R)\longrightarrow \operatorname{St}_{\mathbf{G}_{n}}(\text{k};R)\end{eqnarray}$$

obtained by combining Lemma 2.3 and Theorem 2.1.

Remark 2.5. Identifying $\operatorname{St}_{\operatorname{GL}_{\ell }}(\text{k};R)\otimes \operatorname{St}_{\mathbf{G}_{n-\ell }}(\text{k};R)$ with its image in $\operatorname{St}_{\mathbf{G}_{n}}(\text{k};R)$ under the Reeder product map, one way of viewing Theorem 2.1 is that it asserts that

$$\begin{eqnarray}\operatorname{St}_{\mathbf{G}_{n}}(\text{k};R)=\bigoplus _{u\in \mathbf{UG}_{n}^{\ell }(\text{k})}u\cdot (\operatorname{St}_{\operatorname{GL}_{\ell }}(\text{k};R)\otimes \operatorname{St}_{\mathbf{G}_{n-\ell }}(\text{k};R)).\end{eqnarray}$$

We will also need the following lemma, which is precisely the case $i=0$ of Theorem 1.1. It generalizes [Reference Lee and SzczarbaLS76, Theorem 4.1]. Recall that if $G$ is a group and $M$ is a $G$ -module, then the coinvariants $M_{G}$ are the largest quotient of $M$ on which $G$ acts trivially. The coinvariants $M_{G}$ are isomorphic to $\operatorname{H}_{0}(G;M)$ .

Proof. Theorem 2.1 implies that

$$\begin{eqnarray}\operatorname{Ind}_{\mathbf{LG}_{n}^{2}(\text{k})}^{\mathbf{PG}_{n}^{2}(\text{k})}\operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R)\otimes \operatorname{St}_{\mathbf{G}_{n-2}}(\text{k};R)\cong \operatorname{Res}_{\mathbf{PG}_{n}^{2}(\text{k})}^{\mathbf{G}_{n}(\text{k})}\operatorname{St}_{\boldsymbol{ G}_{n}}(\text{k};R).\end{eqnarray}$$

It is thus enough to prove that

$$\begin{eqnarray}(\operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R)\otimes \operatorname{St}_{\mathbf{G}_{n-2}}(\text{k};R))_{\mathbf{LG}_{n}^{2}(\text{k})}=0.\end{eqnarray}$$

Whatever $\mathbf{G}_{n}$ is, the group $\mathbf{LG}_{n}^{2}(\text{k})$ contains the subgroup $\operatorname{SL}_{2}(\text{k})\times 1$ . It is thus enough to prove that

$$\begin{eqnarray}(\operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R))_{\operatorname{SL}_{2}(\text{k})}=0.\end{eqnarray}$$

This is an easy exercise using the fact that

$$\begin{eqnarray}\operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R)=\widetilde{\operatorname{H}}_{0}({\mathcal{T}}_{2}(\text{k});R)=\widetilde{\operatorname{H}}_{0}(\mathbb{P}^{1}(\text{k});\mathbb{Q}),\end{eqnarray}$$

where $\mathbb{P}^{1}(\text{k})$ is the projective line over $\text{k}$ , regarded as a discrete set of points. For details, see [Reference Lee and SzczarbaLS76, Theorem 4.1].◻

5.1 Equivariant homology

In our proof of Theorem 1.1, we will need some basic facts about equivariant homology. A basic reference is [Reference BrownBro94, ch. VII.7].

Let $G$ be a group, let $X$ be a semisimplicial set on which $G$ acts, let $R$ be a ring, and let $M$ be an $R[G]$ -module. Let $EG$ be a contractible semisimplicial set on which $G$ acts freely and let $BG=EG/G$ , so $BG$ is a classifying space for $G$ . Denote by $EG\times _{G}X$ the quotient of $EG\times X$ by the diagonal action of $G$ . This is known as the Borel construction. The homotopy type of $EG\times _{G}X$ does not depend on the choice of $EG$ . The projection $EG\times _{G}X\rightarrow EG/G=BG$ induces a homomorphism $\unicode[STIX]{x1D70B}_{1}(EG\times _{G}X)\rightarrow \unicode[STIX]{x1D70B}_{1}(BG)=G$ . Via this homomorphism, we can regard $M$ as a local coefficient system on $EG\times _{G}X$ . The $G$ -equivariant homology groups of $X$ with coefficients in $M$ , denoted $\operatorname{H}_{\ast }^{G}(X;M)$ , are the homology groups of $EG\times _{G}X$ with respect to the local coefficient system $M$ .

Our main tool for understanding $\operatorname{H}_{\ast }^{G}(X;M)$ is the following spectral sequence, which is constructed in [Reference BrownBro94, Equation VII.7.7].

Lemma 5.2. For all $p\geqslant 0$ , let $\unicode[STIX]{x1D6F4}_{p}$ be a set containing exactly one representative for each orbit of the action of $G$ on the $p$ -simplices of $X$ . For $\unicode[STIX]{x1D70E}\in \unicode[STIX]{x1D6F4}_{p}$ , let $G_{\unicode[STIX]{x1D70E}}$ be the stabilizer of $\unicode[STIX]{x1D70E}$ . Then there is a first quadrant spectral sequence

$$\begin{eqnarray}E_{p,q}^{1}=\bigoplus _{\unicode[STIX]{x1D70E}\in \unicode[STIX]{x1D6F4}_{p}}\operatorname{H}_{q}(G_{\unicode[STIX]{x1D70E}};\operatorname{Res}_{G_{\unicode[STIX]{x1D70E}}}^{G}M)\Longrightarrow \operatorname{H}_{p+q}^{G}(X;M).\end{eqnarray}$$

5.2 Complexes of partial bases

Let $\text{k}$ be a field and let $\mathbf{G}_{n}$ be either $\operatorname{GL}_{n}$ , $\operatorname{SL}_{n}$ , $\operatorname{Sp}_{2n}$ , $\operatorname{SO}_{n,n}$ , or $\operatorname{SO}_{n,n+1}$ . To prove Theorem 1.1, we will need to construct a highly connected space $\mathbf{CG}_{n}(\text{k})$ on which $\mathbf{G}_{n}(\text{k})$ acts. The definition of this complex is as follows.

1. – If $\mathbf{G}_{n}=\operatorname{GL}_{n}$ or $\mathbf{G}_{n}=\operatorname{SL}_{n}$ , then define $\mathbf{CG}_{n}(\text{k})$ to be the complex of partial bases for $\text{k}^{n}$ , i.e. the semisimplicial complex whose $\ell$ -simplices are ordered sequences $[\vec{v}_{0},\ldots ,\vec{v}_{\ell }]$ of linearly independent elements of $\text{k}^{n}$ .

2. – If $\mathbf{G}_{n}=\operatorname{Sp}_{2n}$ or $\mathbf{G}_{n}=\operatorname{SO}_{n,n}$ or $\mathbf{G}_{n}=\operatorname{SO}_{n,n+1}$ and $\text{k}^{m}$ is the vector space upon which $\mathbf{G}_{n}(\text{k})$ acts (so $m$ is either $2n$ or $2n+1$ ), then define $\mathbf{CG}_{n}(\text{k})$ to be the complex of partial isotropic bases for $\text{k}^{m}$ , i.e. the semisimplicial complex whose $\ell$ -simplices are ordered sequences $[\vec{v}_{0},\ldots ,\vec{v}_{\ell }]$ of linearly independent elements of $\text{k}^{m}$ that span an isotropic subspace.

The following theorem summarizes the properties of $\mathbf{CG}_{n}(\text{k})$ .

5.3 The proof of Theorem 1.1

Let us first recall the statement of the theorem. Let $\mathbf{G}_{n}$ be either $\operatorname{GL}_{n}$ , $\operatorname{SL}_{n}$ , $\operatorname{Sp}_{2n}$ , $\operatorname{SO}_{n,n}$ , or $\operatorname{SO}_{n,n+1}$ . Also, let $\text{k}$ be a field and $R$ be a commutative ring. Our goal is to prove that $\operatorname{H}_{i}(\mathbf{G}_{n}(\text{k});\operatorname{St}_{\mathbf{G}_{n}}(\text{k};R))=0$ for $n\geqslant 2i+2$ and that there exists a surjection

(5.1) $$\begin{eqnarray}\displaystyle \operatorname{H}_{i}(\mathbf{G}_{2i}(\text{k});\operatorname{St}_{\mathbf{G}_{2i}}(\text{k};R))\rightarrow \operatorname{H}_{i}(\mathbf{G}_{2i+1}(\text{k});\operatorname{St}_{\mathbf{G}_{2i+1}}(\text{k};R)). & & \displaystyle\end{eqnarray}$$

Of course, this surjection will be induced by the stabilization map defined in § 3.

The proof is by induction on $i$ . We begin with the base case $i=0$ . Lemma 2.6 says that $\operatorname{H}_{0}(\mathbf{G}_{n}(\text{k});\operatorname{St}_{\mathbf{G}_{n}}(\text{k};R))=0$ for $n\geqslant 2$ , so we only need to show that the map (5.1) is a surjection for $i=0$ . For the domain, $\mathbf{G}_{0}(\text{k})$ is the trivial group. By our convention regarding the empty set discussed at the end of the introduction, we thus have $\operatorname{St}_{\mathbf{G}_{0}}(\text{k};R)=R$ , and hence $\operatorname{H}_{i}(\mathbf{G}_{0}(\text{k});\operatorname{St}_{\mathbf{G}_{0}}(\text{k};R))=R$ . To simplify the codomain, we have several cases.

1. – $\mathbf{G}_{1}=\operatorname{GL}_{1}$ or $\mathbf{G}_{1}=\operatorname{SO}_{1,1}$ . In fact, these groups are isomorphic and are commutative, so $\operatorname{St}_{\mathbf{G}_{1}}(\text{k};R)=R$ in these cases and (5.1) is an isomorphism.

2. – $\mathbf{G}_{1}=\operatorname{SL}_{1}$ . The group $\operatorname{SL}_{1}$ is the trivial group and thus $\operatorname{St}_{\mathbf{G}_{1}}(\text{k};R)=R$ and (5.1) is an isomorphism.

3. – $\mathbf{G}_{1}=\operatorname{Sp}_{2}\cong \operatorname{SL}_{2}$ or $\mathbf{G}_{1}=\operatorname{SO}_{2,1}\cong \operatorname{PSL}_{2}$ . These groups have isomorphic Steinberg representations and the action of $\operatorname{SL}_{2}(\text{k})$ on $\operatorname{St}_{\operatorname{SL}_{2}}(\text{k};R)$ factors through $\operatorname{PSL}_{2}(\text{k})$ . This case thus follows from Lemma 2.6, which says that $\operatorname{H}_{0}(\operatorname{SL}_{2}(\text{k});\operatorname{St}_{\operatorname{SL}_{2}}(\text{k};R))=0$ .

This completes the base case.

Assume now that $i>0$ and that the desired result is true for all smaller values of $i$ . We will prove that the stabilization map

(5.2) $$\begin{eqnarray}\operatorname{H}_{i}(\mathbf{G}_{n-1}(\text{k});\operatorname{St}_{\mathbf{G}_{n-1}}(\text{k};R))\rightarrow \operatorname{H}_{i}(\mathbf{G}_{n}(\text{k});\operatorname{St}_{\mathbf{G}_{n}}(\text{k};R))\end{eqnarray}$$

is surjective for $n\geqslant 2i+1$ . Lemma 3.1 will then imply that $\operatorname{H}_{i}(\mathbf{G}_{n}(\text{k});\operatorname{St}_{\mathbf{G}_{n}}(\text{k};R))=0$ for $n\geqslant 2i+2$ , and the theorem will follow.

Fix some $n\geqslant 2i+1$ and let $\text{k}^{m}$ be the standard vector space representation of $\mathbf{G}_{n}(\text{k})$ (so $m$ is either $n$ , $2n$ , or $2n+1$ ). Let $\{\vec{a}_{1},\ldots ,\vec{a}_{n}\}$ be the vectors in $\text{k}^{m}$ such that

$$\begin{eqnarray}\mathbf{FG}_{n}^{\ell }(\text{k})=\{M\in \mathbf{G}_{n}(\text{k})\mid M(\vec{a}_{j})=\vec{a}_{j}\text{ for }1\leqslant j\leqslant \ell \}\end{eqnarray}$$

for $1\leqslant \ell \leqslant n$ . Combining the second conclusion of Theorem 5.4 with Lemma 5.1, we have a surjection

(5.3) $$\begin{eqnarray}\operatorname{H}_{i}^{\mathbf{G}_{n}(\text{k})}(\mathbf{CG}_{n}(\text{k});\operatorname{St}_{\mathbf{G}_{n}}(\text{k};R))\rightarrow \operatorname{H}_{i}(\mathbf{G}_{n}(\text{k});\operatorname{St}_{\mathbf{G}_{n}}(\text{k};R)).\end{eqnarray}$$

We will analyze $\operatorname{H}_{i}^{\mathbf{G}_{n}(\text{k})}(\mathbf{CG}_{n}(\text{k});\operatorname{St}_{\mathbf{G}_{n}}(\text{k};R))$ using the spectral sequence from Lemma 5.2. To calculate its $E^{1}$ -page, observe that the first conclusion of Theorem 5.4 says that $\mathbf{G}_{n}(\text{k})$ acts transitively on the $p$ -simplices of $\mathbf{CG}_{n}(\text{k})$ for $0\leqslant p<n-1$ . The stabilizer of the $(p-1)$ -simplex $[\vec{a}_{1},\ldots ,\vec{a}_{p}]$ is $\mathbf{FG}_{n}^{p}(\text{k})$ , so the spectral sequence in Lemma 5.2 has

(5.4) $$\begin{eqnarray}E_{p,q}^{1}=\operatorname{H}_{q}(\mathbf{FG}_{n}^{p+1}(\text{k});\operatorname{Res}_{\mathbf{FG}_{n}^{p+1}(\text{k})}^{\mathbf{G}_{n}(\text{k})}\operatorname{St}_{\boldsymbol{ G}_{n}}(\text{k};R)),\end{eqnarray}$$

for $0\leqslant p<n-1$ .

We will prove that all of the terms on the $p+q=i$ line of the $E^{\infty }$ -page of our spectral sequence vanish except for possibly the term $E_{0,i}^{\infty }$ . To do this, consider $p,q\geqslant 0$ with $p+q=i$ and $p\geqslant 1$ . The case $p=1$ and $n=2i+1$ is exceptional and must be treated separately. To avoid getting bogged down here, we postpone this calculation until § 6 below, where it appears as Lemma 6.1.Footnote ¹

We thus can assume that either $p\geqslant 2$ or that $n\geqslant 2i+2$ . Since $n\geqslant 2i+1$ , we certainly have $p<n-1$ , so $E_{p,q}^{1}$ is in the regime where the above description of the $E^{1}$ -page holds. Applying Corollary 4.2 to (5.4), we see that

$$\begin{eqnarray}E_{p,q}^{1}=\operatorname{St}_{\operatorname{ GL}_{p+1}}(\text{k};R)\otimes \operatorname{H}_{q}(\mathbf{G}_{n-p-1}(\text{k});\operatorname{St}_{\mathbf{G}_{n-p-1}}(\text{k};R)).\end{eqnarray}$$

To see that this vanishes, it is enough to show that $\operatorname{H}_{q}(\mathbf{G}_{n-p-1}(\text{k});\operatorname{St}_{\mathbf{G}_{n-p-1}}(\text{k};R))=0$ . This is a consequence of our inductive hypothesis; to see that it applies, observe that if $p\geqslant 2$ then

$$\begin{eqnarray}n-p-1\geqslant (2i+1)-p-1=2(p+q)-p=2q+p\geqslant 2q+2,\end{eqnarray}$$

while if $p=1$ and $n\geqslant 2i+2$ then

$$\begin{eqnarray}n-p-1\geqslant (2i+2)-1-1=2(p+q)=2q+2.\end{eqnarray}$$

This implies that $E_{p,q}^{1}=0$ , and thus that $E_{p,q}^{\infty }=0$ .

The $p+q=i$ line of the $E^{\infty }$ -page of our spectral sequence thus only has a single potentially nonzero entry, namely $E_{0,i}^{\infty }$ , and this is a quotient of

$$\begin{eqnarray}E_{0,i}^{1}=\operatorname{H}_{i}(\mathbf{FG}_{n}^{1}(\text{k});\operatorname{Res}_{\mathbf{FG}_{n}^{1}(\text{k})}^{\mathbf{G}_{n}(\text{k})}\operatorname{St}_{\boldsymbol{ G}_{n}}(\text{k};R)).\end{eqnarray}$$

This entry thus surjects onto $\operatorname{H}_{i}^{\mathbf{G}_{n}(\text{k})}(\mathbf{CG}_{n}(\text{k});\operatorname{St}_{\mathbf{G}_{n}}(\text{k};R))$ . Combining this with the surjection (5.3), we obtain a surjection

(5.5) $$\begin{eqnarray}\operatorname{H}_{i}(\mathbf{FG}_{n}^{1}(\text{k});\operatorname{Res}_{\mathbf{FG}_{n}^{1}(\text{k})}^{\mathbf{G}_{n}(\text{k})}\operatorname{St}_{\boldsymbol{ G}_{n}}(\text{k};R))\longrightarrow \operatorname{H}_{i}(\mathbf{G}_{n}(\text{k});\operatorname{St}_{\mathbf{G}_{n}}(\text{k};R)).\end{eqnarray}$$

Examining the construction of our spectral sequence in [Reference BrownBro94, ch. VII.7], it is easy to see that this comes from the map induced by the inclusion $\mathbf{FG}_{n}^{1}(\text{k}){\hookrightarrow}\mathbf{G}_{n}(\text{k})$ . Combining (5.5) with the isomorphism

$$\begin{eqnarray}\operatorname{H}_{i}(\mathbf{G}_{n-1}(\text{k});\operatorname{St}_{\mathbf{G}_{n-1}}(\text{k};R))\xrightarrow[{}]{\cong }\operatorname{H}_{i}(\mathbf{FG}_{n}^{1}(\text{k});\operatorname{Res}_{\mathbf{FG}_{n}^{1}(\text{k})}^{\mathbf{G}_{n}(\text{k})}\operatorname{St}_{\boldsymbol{ G}_{n}}(\text{k};R))\end{eqnarray}$$

given by the $\ell =1$ case of Corollary 4.2, we conclude that (5.2) is a surjection, as desired.

6.1 Identifying the differential

The notation is as in the beginning of § 6. In this section, we identify the differential $E_{2,i-1}^{1}\rightarrow E_{1,i-1}^{1}$ . Since $n=2i+1$ and $i>0$ , we have $1<n-1$ , so $E_{1,i-1}^{1}$ is as described in (5.4), i.e. 

$$\begin{eqnarray}E_{1,i-1}^{1}\cong \operatorname{H}_{i-1}(\mathbf{FG}_{n}^{2}(\text{k});\operatorname{Res}_{\mathbf{FG}_{n}^{2}(\text{k})}^{\mathbf{G}_{n}(\text{k})}\operatorname{St}_{\boldsymbol{ G}_{n}}(\text{k};R)).\end{eqnarray}$$

If $i=1$ , then we do not have $2<n-1$ , so in this case $E_{2,i-1}^{1}$ is not as described in (5.4). The issue is that $\mathbf{G}_{n}(\text{k})$ might not act transitively on the $2$ -simplices of $\mathbf{CG}_{n}(\text{k})$ (this is actually only a problem for $\mathbf{G}_{n}=\operatorname{SL}_{n}$ ). However, for all values of $i$ it is still the case that $E_{2,i-1}^{1}$ contains

$$\begin{eqnarray}\operatorname{H}_{i-1}(\mathbf{FG}_{n}^{3}(\text{k});\operatorname{Res}_{\mathbf{FG}_{n}^{3}(\text{k})}^{\mathbf{G}_{n}(\text{k})}\operatorname{St}_{\boldsymbol{ G}_{n}}(\text{k};R))\end{eqnarray}$$

as a summand. The restriction of the differential $E_{2,i-1}^{1}\rightarrow E_{1,i-1}^{1}$ to this summand is a map

(6.2) $$\begin{eqnarray}\unicode[STIX]{x2202}:\operatorname{H}_{i-1}(\mathbf{FG}_{n}^{3}(\text{k});\operatorname{Res}_{\mathbf{FG}_{n}^{3}(\text{k})}^{\mathbf{G}_{n}(\text{k})}\operatorname{St}_{\boldsymbol{ G}_{n}}(\text{k};R))\longrightarrow \operatorname{H}_{i-1}(\mathbf{FG}_{n}^{2}(\text{k});\operatorname{Res}_{\mathbf{FG}_{n}^{2}(\text{k})}^{\mathbf{G}_{n}(\text{k})}\operatorname{St}_{\boldsymbol{ G}_{n}}(\text{k};R)).\end{eqnarray}$$

To prove Lemma 6.1, it is enough to prove that $\unicode[STIX]{x2202}$ is surjective.

We can describe $\unicode[STIX]{x2202}$ using the recipe described in [Reference BrownBro94, ch. VII.8]. Recall that $\mathbf{FG}_{n}^{3}(\text{k})$ is the $\mathbf{G}_{n}(\text{k})$ -stabilizer of the ordered sequence of vectors $\unicode[STIX]{x1D70E}=[\vec{a}_{1},\vec{a}_{2},\vec{a}_{3}]$ . For $1\leqslant m\leqslant 3$ , let $\unicode[STIX]{x1D70E}_{m}$ be the ordered sequence obtained by deleting $\vec{a}_{m}$ from $\unicode[STIX]{x1D70E}$ and let $(\mathbf{G}_{n}(k))_{\unicode[STIX]{x1D70E}_{m}}$ denote the $\mathbf{G}_{n}(k)$ -stabilizer of $\unicode[STIX]{x1D70E}_{m}$ . We then have $\unicode[STIX]{x2202}=\unicode[STIX]{x2202}_{1}-\unicode[STIX]{x2202}_{2}+\unicode[STIX]{x2202}_{3}$ , where $\unicode[STIX]{x2202}_{m}$ is the composition

$$\begin{eqnarray}\displaystyle & & \displaystyle \operatorname{H}_{i-1}(\mathbf{FG}_{n}^{3}(\text{k});\operatorname{Res}_{\mathbf{FG}_{n}^{3}(\text{k})}^{\mathbf{G}_{n}(\text{k})}\operatorname{St}_{\boldsymbol{ G}_{n}}(\text{k};R))\nonumber\\ \displaystyle & & \displaystyle \quad \stackrel{\unicode[STIX]{x2202}_{m}^{\prime }}{\longrightarrow }\operatorname{H}_{i-1}((\mathbf{G}_{n}(\text{k}))_{\unicode[STIX]{x1D70E}_{m}};\operatorname{Res}_{(\mathbf{G}_{n}(\text{k}))_{\unicode[STIX]{x1D70E}_{m}}}^{\mathbf{G}_{n}(\text{k})}\operatorname{St}_{\boldsymbol{ G}_{n}}(\text{k};R))\nonumber\\ \displaystyle & & \displaystyle \quad \stackrel{\unicode[STIX]{x2202}_{m}^{\prime \prime }}{\longrightarrow }\operatorname{H}_{i-1}(\mathbf{FG}_{n}^{2}(\text{k});\operatorname{Res}_{\mathbf{FG}_{n}^{2}(\text{k})}^{\mathbf{G}_{n}(\text{k})}\operatorname{St}_{\boldsymbol{ G}_{n}}(\text{k};R))\nonumber\end{eqnarray}$$

of the following two maps.

1. – The map $\unicode[STIX]{x2202}_{m}^{\prime }$ is induced by the inclusion $\mathbf{FG}_{n}^{3}(\text{k}){\hookrightarrow}(\mathbf{G}_{n}(\text{k}))_{\unicode[STIX]{x1D70E}_{m}}$ .

2. – Define $\unicode[STIX]{x1D705}_{m}\in \mathbf{G}_{n}(\text{k})$ as follows. First, $\unicode[STIX]{x1D705}_{3}=\text{id}$ . For $m\in \{1,2\}$ , we do the following.

   1. * If $\mathbf{G}_{n}=\operatorname{GL}_{n}$ or $\mathbf{G}_{n}=\operatorname{SL}_{n}$ , then $\unicode[STIX]{x1D705}_{m}\in \operatorname{SL}_{n}(\text{k})$ is the map $\text{k}^{n}\rightarrow \text{k}^{n}$ that takes $\vec{a}_{m}$ to $\vec{a}_{3}$ , takes $\vec{a}_{3}$ to $-\vec{a}_{m}$ , and fixes all the other basis vectors.

   2. * If $\mathbf{G}_{n}=\operatorname{Sp}_{2n}$ or $\mathbf{G}_{n}=\operatorname{SO}_{n,n}$ or $\mathbf{G}_{n}=\operatorname{SO}_{n,n+1}$ , then $\unicode[STIX]{x1D705}_{m}\in \mathbf{G}_{n}(\text{k})$ is the map $\text{k}^{m}\rightarrow \text{k}^{m}$ defined as follows. Let $\vec{b}_{1},\ldots ,\vec{b}_{n}$ be the standard basis vectors for $\text{k}^{m}$ that pair with the $\vec{a}_{j}$ (there is one additional standard basis vector if $\mathbf{G}_{n}=\operatorname{SO}_{n,n+1}$ ). Then $\unicode[STIX]{x1D705}_{m}$ takes $\vec{a}_{m}$ to $\vec{a}_{3}$ , takes $\vec{a}_{3}$ to $-\vec{a}_{m}$ , takes $\vec{b}_{m}$ to $\vec{b}_{3}$ , takes $\vec{b}_{3}$ to $-\vec{b}_{m}$ , and fixes all the other basis vectors.

   Then $\unicode[STIX]{x2202}_{m}^{\prime \prime }$ is induced by the map $(\mathbf{G}_{n}(\text{k}))_{\unicode[STIX]{x1D70E}_{m}}\rightarrow \mathbf{FG}_{n}^{2}(\text{k})$ that takes $g\in (\mathbf{G}_{n}(\text{k}))_{\unicode[STIX]{x1D70E}_{m}}$ to $\unicode[STIX]{x1D705}_{m}g\unicode[STIX]{x1D705}_{m}^{-1}$ and the map $\operatorname{St}_{\mathbf{G}_{n}}(\text{k};R)\rightarrow \operatorname{St}_{\mathbf{G}_{n}}(\text{k};R)$ that takes $x\in \operatorname{St}_{\mathbf{G}_{n}}(\text{k};R)$ to $\unicode[STIX]{x1D705}_{m}(x)\in \operatorname{St}_{\mathbf{G}_{n}}(\text{k};R)$ . We remark that easier choices of $\unicode[STIX]{x1D705}_{m}$ (without the signs) could be used for $\mathbf{G}_{n}\neq \operatorname{SL}_{n}$ , but we chose the ones above to make our later formulas more uniform.

This is summarized in the following lemma.

Lemma 6.2. Let the notation be as above. Then the map $\unicode[STIX]{x2202}$ in (6.2) equals $\unicode[STIX]{x2202}_{1}-\unicode[STIX]{x2202}_{2}+\unicode[STIX]{x2202}_{3}$ , where $\unicode[STIX]{x2202}_{m}$ is induced by the map $\mathbf{FG}_{n}^{3}(\text{k})\rightarrow \mathbf{FG}_{n}^{2}(\text{k})$ defined via the formula

$$\begin{eqnarray}g\mapsto \unicode[STIX]{x1D705}_{m}g\unicode[STIX]{x1D705}_{m}^{-1}\quad (g\in \mathbf{FG}_{n}^{3}(\text{k}))\end{eqnarray}$$

and the map $\operatorname{St}_{\mathbf{G}_{n}}(\text{k};R)\rightarrow \operatorname{St}_{\mathbf{G}_{n}}(\text{k};R)$ defined via the formula

$$\begin{eqnarray}x\mapsto \unicode[STIX]{x1D705}_{m}(x)\quad (x\in \operatorname{St}_{\mathbf{G}_{n}}(\text{k};R)).\end{eqnarray}$$

6.2 Bringing in the stabilization map

The notation is as in the beginning of § 6. Fix some $1\leqslant m\leqslant 3$ , and let $\unicode[STIX]{x2202}_{m}$ and $\unicode[STIX]{x1D705}_{m}$ be as in Lemma 6.2. Applying the isomorphism in Corollary 4.2 to the domain and codomain of $\unicode[STIX]{x2202}_{m}$ , we obtain a homomorphism

$$\begin{eqnarray}\displaystyle \widehat{\unicode[STIX]{x2202}}_{m}: & & \displaystyle \operatorname{St}_{\operatorname{GL}_{3}}(\text{k};R)\otimes \operatorname{H}_{i-1}(\mathbf{G}_{n-3}(\text{k});\operatorname{St}_{\mathbf{G}_{n-3}}(\text{k};R))\nonumber\\ \displaystyle & & \displaystyle \quad \rightarrow \operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R)\otimes \operatorname{H}_{i-1}(\mathbf{G}_{n-2}(\text{k});\operatorname{St}_{\mathbf{G}_{n-2}}(\text{k};R)).\nonumber\end{eqnarray}$$

Our goal in this section is to prove that $\widehat{\unicode[STIX]{x2202}}_{m}$ is the tensor product of the stabilization map

(6.3) $$\begin{eqnarray}\operatorname{H}_{i-1}(\mathbf{G}_{n-3}(\text{k});\operatorname{St}_{\mathbf{G}_{n-3}}(\text{k};R))\rightarrow \operatorname{H}_{i-1}(\mathbf{G}_{n-2}(\text{k});\operatorname{St}_{\mathbf{G}_{n-2}}(\text{k};R))\end{eqnarray}$$

with the map $\unicode[STIX]{x1D701}_{m}:\operatorname{St}_{\operatorname{GL}_{3}}(\text{k};R)\rightarrow \operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R)$ defined as follows. Let $\{\vec{a}_{1},\vec{a}_{2},\vec{a}_{3}\}$ be the standard basis for $\text{k}^{3}$ . Let $\widehat{\unicode[STIX]{x1D705}}_{3}=\text{id}\in \operatorname{SL}_{3}(\text{k})$ , and for $m\in \{1,2\}$ let $\widehat{\unicode[STIX]{x1D705}}_{m}\in \operatorname{SL}_{3}(\text{k})$ be the element that takes $\vec{a}_{m}$ to $\vec{a}_{3}$ , takes $\vec{a}_{3}$ to $-\vec{a}_{m}$ , and fixes all the other basis vectors. Then $\unicode[STIX]{x1D701}_{m}$ is the composition

$$\begin{eqnarray}\operatorname{St}_{\operatorname{GL}_{3}}(\text{k};R)\stackrel{\widehat{\unicode[STIX]{x1D705}}_{m}}{\longrightarrow }\operatorname{St}_{\operatorname{GL}_{3}}(\text{k};R)\longrightarrow \operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R)\otimes \operatorname{St}_{\operatorname{GL}_{1}}(\text{k};R)\cong \operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R),\end{eqnarray}$$

where the second arrow is the Reeder projection map (see § 4) and the final isomorphism comes from the fact that $\operatorname{St}_{\operatorname{GL}_{1}}(\text{k};R)=R$ .

The main result of this section is then as follows.

Proof. By construction, $\widehat{\unicode[STIX]{x2202}}_{m}$ equals the composition

$$\begin{eqnarray}\displaystyle & & \displaystyle \operatorname{St}_{\operatorname{GL}_{3}}(\text{k};R)\otimes \operatorname{H}_{i-1}(\mathbf{G}_{n-3}(\text{k});\operatorname{St}_{\mathbf{G}_{n-3}}(\text{k};R))\nonumber\\ \displaystyle & & \displaystyle \quad \stackrel{\cong }{\longrightarrow }\operatorname{H}_{i-1}(1\times \mathbf{G}_{n-3}(\text{k});\operatorname{St}_{\operatorname{GL}_{3}}(\text{k};R)\otimes \operatorname{St}_{\mathbf{G}_{n-3}}(\text{k};R))\nonumber\\ \displaystyle & & \displaystyle \quad \stackrel{\cong }{\longrightarrow }\operatorname{H}_{i-1}(\mathbf{FG}_{n}^{3}(\text{k});\operatorname{St}_{\mathbf{G}_{n}}(\text{k};R))\nonumber\\ \displaystyle & & \displaystyle \quad \stackrel{\unicode[STIX]{x2202}_{m}}{\longrightarrow }\operatorname{H}_{i-1}(\mathbf{FG}_{n}^{2}(\text{k});\operatorname{St}_{\mathbf{G}_{n}}(\text{k};R))\nonumber\\ \displaystyle & & \displaystyle \quad \stackrel{\cong }{\longrightarrow }\operatorname{H}_{i-1}(1\times \mathbf{G}_{n-2}(\text{k});\operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R)\otimes \operatorname{St}_{\mathbf{G}_{n-2}}(\text{k};R))\nonumber\\ \displaystyle & & \displaystyle \quad \stackrel{\cong }{\longrightarrow }\operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R)\otimes \operatorname{H}_{i-1}(\mathbf{G}_{n-2}(\text{k});\operatorname{St}_{\mathbf{G}_{n-2}}(\text{k};R)),\nonumber\end{eqnarray}$$

where the various maps are as follows.

1. – The first and last arrows use the fact that $\operatorname{St}_{\operatorname{GL}_{3}}(\text{k};R)$ and $\operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R)$ are free $R$ -modules (cf. the proof of Corollary 4.2).

2. – The second arrow is the map described in Lemma 4.1, that is, the map induced by the inclusion $1\times \mathbf{G}_{n-3}(\text{k}){\hookrightarrow}\mathbf{FG}_{n}^{3}(\text{k})$ and the Reeder product map $\operatorname{St}_{\operatorname{GL}_{3}}(\text{k};R)\otimes \operatorname{St}_{\mathbf{G}_{n-3}}(\text{k};R)\rightarrow \operatorname{St}_{\mathbf{G}_{n}}(\text{k};R)$ .

3. – The third arrow is the map $\unicode[STIX]{x2202}_{m}$ described in Lemma 6.2, that is, the map induced by the map $\mathbf{FG}_{n}^{3}(\text{k})\rightarrow \mathbf{FG}_{n}^{2}(\text{k})$ given by conjugation by $\unicode[STIX]{x1D705}_{m}$ and the map $\operatorname{St}_{\mathbf{G}_{n}}(\text{k};R)\rightarrow \operatorname{St}_{\mathbf{G}_{n}}(\text{k};R)$ induced by $\unicode[STIX]{x1D705}_{m}$ .

4. – The fourth arrow is the map described in Lemma 4.3, that is, the map induced by the projection $\mathbf{FG}_{n}^{2}(\text{k})\rightarrow 1\times \mathbf{G}_{n-2}(\text{k})$ together with the Reeder projection map $\operatorname{St}_{\mathbf{G}_{n}}(\text{k};R)\rightarrow \operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R)\otimes \operatorname{St}_{\mathbf{G}_{n-2}}(\text{k};R)$ .

We must show that this composition equals the indicated tensor product of maps. This will take some work.

Define $\unicode[STIX]{x1D6F9}$ to be the composition

$$\begin{eqnarray}\displaystyle \operatorname{St}_{\operatorname{GL}_{3}}(\text{k};R)\otimes \operatorname{St}_{\mathbf{G}_{n-3}}(\text{k};R) & \longrightarrow & \displaystyle \operatorname{St}_{\mathbf{G}_{n}}(\text{k};R)\nonumber\\ \displaystyle & \stackrel{\unicode[STIX]{x1D705}_{m}}{\longrightarrow } & \displaystyle \operatorname{St}_{\mathbf{G}_{n}}(\text{k};R)\nonumber\\ \displaystyle & \longrightarrow & \displaystyle \operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R)\otimes \operatorname{St}_{\mathbf{G}_{n-2}}(\text{k};R),\nonumber\end{eqnarray}$$

where the first map is the Reeder product map and the last map is the Reeder projection map. Also, define $\unicode[STIX]{x1D6F7}$ to be the composition

$$\begin{eqnarray}\displaystyle \operatorname{St}_{\operatorname{GL}_{3}}(\text{k};R)\otimes \operatorname{St}_{\mathbf{G}_{n-3}}(\text{k};R) & \longrightarrow & \displaystyle \operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R)\otimes \operatorname{St}_{\operatorname{GL}_{1}}(\text{k};R)\otimes \operatorname{St}_{\mathbf{G}_{n-3}}(\text{k};R)\nonumber\\ \displaystyle & \longrightarrow & \displaystyle \operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R)\otimes \operatorname{St}_{\mathbf{G}_{n-2}}(\text{k};R),\nonumber\end{eqnarray}$$

where the maps are as follows.

1. – The first map is the tensor product of the Reeder projection map $\operatorname{St}_{\operatorname{GL}_{3}}(\text{k};R)\rightarrow \operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R)\otimes \operatorname{St}_{\operatorname{GL}_{1}}(\text{k};R)$ and the identity map $\operatorname{St}_{\mathbf{G}_{n-3}}(\text{k};R)\rightarrow \operatorname{St}_{\mathbf{G}_{n-3}}(\text{k};R)$ .

2. – The second map is the tensor product of the identity map $\operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R)\rightarrow \operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R)$ and the Reeder product map $\operatorname{St}_{\operatorname{GL}_{1}}(\text{k};R)\otimes \operatorname{St}_{\mathbf{G}_{n-3}}(\text{k};R)\rightarrow \operatorname{St}_{\mathbf{G}_{n-2}}(\text{k};R)$ .

By the above, it is enough to prove that $\unicode[STIX]{x1D6F9}=\unicode[STIX]{x1D6F7}\circ (\widehat{\unicode[STIX]{x1D705}}_{m}\otimes \text{id})$ .

Define $\unicode[STIX]{x1D6F9}^{\prime }$ to be the composition

$$\begin{eqnarray}\displaystyle \operatorname{St}_{\operatorname{GL}_{3}}(\text{k};R)\otimes \operatorname{St}_{\mathbf{G}_{n-3}}(\text{k};R) & \longrightarrow & \displaystyle \operatorname{St}_{\mathbf{G}_{n}}(\text{k};R)\nonumber\\ \displaystyle & \longrightarrow & \displaystyle \operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R)\otimes \operatorname{St}_{\mathbf{G}_{n-2}}(\text{k};R),\nonumber\end{eqnarray}$$

where the first map is the Reeder product map and the second map is the Reeder projection map. From its definition, we see that $\unicode[STIX]{x1D6F9}=\unicode[STIX]{x1D6F9}^{\prime }\circ (\widehat{\unicode[STIX]{x1D705}}_{m}\otimes \text{id})$ . We thus see that it is enough to prove that $\unicode[STIX]{x1D6F9}^{\prime }=\unicode[STIX]{x1D6F7}$ .

Define $U=\mathbf{UGL}_{3}^{2}(\text{k})$ to be the unipotent radical of the parabolic subgroup $\mathbf{PGL}_{3}^{2}(\text{k})$ of $\operatorname{GL}_{3}(\text{k})$ (despite the bad notation, this is not the projective general linear group). Using Theorem 2.1 as in Remark 2.5, we see that

$$\begin{eqnarray}\operatorname{St}_{\operatorname{GL}_{3}}(\text{k};R)=\bigoplus _{u\in U}u\cdot (\operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R)\otimes \operatorname{St}_{\operatorname{GL}_{1}}(\text{k};R)).\end{eqnarray}$$

Consider $u\in U$ and $x\in \operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R)$ and $y\in \operatorname{St}_{\operatorname{GL}_{1}}(\text{k};R)$ and $z\in \operatorname{St}_{\mathbf{G}_{n-3}}(\text{k};R)$ . Examining the definition of $\unicode[STIX]{x1D6F9}^{\prime }$ , we see that

$$\begin{eqnarray}\unicode[STIX]{x1D6F9}^{\prime }((u\cdot (x\otimes y))\otimes z)=x\otimes (y\otimes z),\end{eqnarray}$$

where $y\otimes z\in \operatorname{St}_{\operatorname{GL}_{1}}(\text{k};R)\otimes \operatorname{St}_{\mathbf{G}_{n-3}}(\text{k};R)$ is identified with an element of $\operatorname{St}_{\mathbf{G}_{n-2}}(\text{k};R)$ using the Reeder product map. But this equals $\unicode[STIX]{x1D6F7}((u\cdot (x\otimes y))\otimes z)$ , as desired.◻

6.3 Summary of where we are

The notation is as in the beginning of § 6. Recall that Lemma 6.1 asserts that the differential $E_{2,i-1}^{1}\rightarrow E_{1,i-1}^{1}$ is surjective. Let $\unicode[STIX]{x2202}$ be as in § 6.1. Also, let $\unicode[STIX]{x1D701}_{m}$ and $\widehat{\unicode[STIX]{x1D705}}_{m}$ be as in § 6.2. Define

$$\begin{eqnarray}\unicode[STIX]{x1D701}:\operatorname{St}_{\operatorname{GL}_{3}}(\text{k};R)\rightarrow \operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R)\end{eqnarray}$$

via the formula $\unicode[STIX]{x1D701}=\unicode[STIX]{x1D701}_{1}-\unicode[STIX]{x1D701}_{2}+\unicode[STIX]{x1D701}_{3}$ . Combining Lemmas 6.2 and 6.3, we see that to prove Lemma 6.1, it is enough to show that the map

$$\begin{eqnarray}\operatorname{St}_{\operatorname{GL}_{3}}(\text{k};R)\otimes \operatorname{H}_{i-1}(\mathbf{G}_{n-3}(\text{k});\operatorname{St}_{\mathbf{G}_{n-3}}(\text{k};R))\rightarrow \operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R)\otimes \operatorname{H}_{i-1}(\mathbf{G}_{n-2}(\text{k});\operatorname{St}_{\mathbf{G}_{n-2}}(\text{k};R))\end{eqnarray}$$

obtained as the tensor product of $\unicode[STIX]{x1D701}$ and the stabilization map

$$\begin{eqnarray}\operatorname{H}_{i-1}(\mathbf{G}_{n-3}(\text{k});\operatorname{St}_{\mathbf{G}_{n-3}}(\text{k};R))\rightarrow \operatorname{H}_{i-1}(\mathbf{G}_{n-2}(\text{k});\operatorname{St}_{\mathbf{G}_{n-2}}(\text{k};R))\end{eqnarray}$$

is surjective. One of the assumptions in Lemma 6.1 is that this stabilization map is surjective. To prove that lemma, it is thus enough to prove the following.

6.4 Apartments

Before we prove Lemma 6.4, we need to discuss some background material on the Steinberg representation. Unlike the previous sections, in this section $n\geqslant 1$ is arbitrary. Recall that $\operatorname{St}_{\operatorname{GL}_{n}}(\text{k};R)=\widetilde{\operatorname{H}}_{n-2}({\mathcal{T}}_{\operatorname{GL}_{n}}(\text{k});R)$ , where ${\mathcal{T}}_{\operatorname{GL}_{n}}(\text{k})$ is the Tits building associated to $\operatorname{GL}_{n}(\text{k})$ . This building can be described as the simplicial complex whose $r$ -simplices are flags

$$\begin{eqnarray}0\subsetneq V_{0}\subsetneq \cdots \subsetneq V_{r}\subsetneq \text{k}^{n}\end{eqnarray}$$

of nonzero proper subspaces of $\text{k}^{n}$ .

The Solomon–Tits theorem [Reference BrownBro89, Theorem IV.5.2] says that the $R$ -module $\operatorname{St}_{\operatorname{GL}_{n}}(\text{k};R)$ is spanned by apartment classes, which are defined as follows. Consider an $n\times n$ matrix $B$ with entries in $\text{k}^{n}$ none of whose columns are identically $0$ . Let $(\vec{v}_{1},\ldots ,\vec{v}_{n})$ be the columns of $B$ . Let $S_{n}$ be the simplicial complex whose $r$ -simplices are chains

$$\begin{eqnarray}0\subsetneq I_{0}\subsetneq \cdots \subsetneq I_{r}\subsetneq \{1,\ldots ,n\}.\end{eqnarray}$$

The complex $S_{n}$ is isomorphic to the barycentric subdivision of the boundary of an $(n-1)$ -simplex; in particular, $S_{n}$ is homeomorphic to an $(n-2)$ -sphere. There is a simplicial map $f:S_{n}\rightarrow {\mathcal{T}}_{\operatorname{GL}_{n}}(\text{k})$ defined via the formula

$$\begin{eqnarray}f(I)=\langle \vec{v}_{i}\mid i\in I\rangle \quad (\emptyset \subsetneq I\subsetneq \{1,\ldots ,n\}).\end{eqnarray}$$

The apartment class corresponding to $B$ , denoted $\Vert \begin{array}{@{}c@{}}B\end{array}\Vert$ , is the image of the fundamental class $[S_{n}]\in \widetilde{\operatorname{H}}_{n-2}(S_{n};R)=R$ under the map $f_{\ast }:\widetilde{\operatorname{H}}_{n-2}(S_{n};R)\rightarrow \widetilde{\operatorname{H}}_{n-2}({\mathcal{T}}_{\operatorname{GL}_{n}}(\text{k});R)=\operatorname{St}_{\operatorname{GL}_{n}}(\text{k};R)$ .

Permuting the columns of $B$ changes $\Vert \begin{array}{@{}c@{}}B\end{array}\Vert$ by the sign of the permutation, and multiplying a column of $B$ by a nonzero scalar does not change $\Vert \begin{array}{@{}c@{}}B\end{array}\Vert$ . The apartment classes also satisfy the following more interesting relation.

The Solomon–Tits theorem [Reference BrownBro89, Theorem IV.5.2] gives the following basis for $\operatorname{St}_{\operatorname{GL}_{n}}(\text{k};R)$ .

Figure 1. As we illustrate here in the case $n=3$ , the apartment classes corresponding to the $B_{i}$ can be placed on the boundary of an $n$ -dimensional simplex such that their simplices cancel in pairs. In the picture, the vertices labeled with the vectors $\vec{v}_{i}$ are taken to the lines spanned by the $\vec{v}_{i}$ while the unlabeled vertices are taken to the two-dimensional subspaces spanned by the vectors on their two neighbors.

6.5 The proof of Lemma 6.4

We finally prove Lemma 6.4, which as discussed in § 6.3 suffices to prove Lemma 6.1. First, we recall its statement. For $1\leqslant m\leqslant 3$ , let $\unicode[STIX]{x1D701}_{m}$ and $\widehat{\unicode[STIX]{x1D705}}_{m}$ be as in § 6.2. Define

$$\begin{eqnarray}\unicode[STIX]{x1D701}:\operatorname{St}_{\operatorname{GL}_{3}}(\text{k};R)\rightarrow \operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R)\end{eqnarray}$$

via the formula $\unicode[STIX]{x1D701}=\unicode[STIX]{x1D701}_{1}-\unicode[STIX]{x1D701}_{2}+\unicode[STIX]{x1D701}_{3}$ . Our goal is to prove that $\unicode[STIX]{x1D701}$ is surjective.

Before we do that, we introduce some formulas. Let $\unicode[STIX]{x1D70B}:\operatorname{St}_{\operatorname{GL}_{3}}(\text{k};R)\rightarrow \operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R)$ be the composition

$$\begin{eqnarray}\operatorname{St}_{\operatorname{GL}_{3}}(\text{k};R)\longrightarrow \operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R)\otimes \operatorname{St}_{\operatorname{GL}_{1}}(\text{k};R)\stackrel{\cong }{\longrightarrow }\operatorname{St}_{\operatorname{GL}_{2}}(\text{k};R),\end{eqnarray}$$

where the first arrow is the Reeder projection map and the second arrow comes from the fact that $\operatorname{St}_{\operatorname{GL}_{1}}(\text{k};R)=R$ . From its definition, we see that

$$\begin{eqnarray}\unicode[STIX]{x1D70B}\left(\left\Vert \begin{array}{@{}ccc@{}}1 & x & y\\ 0 & 1 & z\\ 0 & 0 & 1\end{array}\right\Vert \right)=\left\Vert \begin{array}{@{}cc@{}}1 & x\\ 0 & 1\end{array}\right\Vert\end{eqnarray}$$

for all $x,y,z\in \text{k}$ . What is more, for all $3\times 3$ matrices $B$ none of whose columns are identically $0$ we have

$$\begin{eqnarray}\unicode[STIX]{x1D701}(\Vert \begin{array}{@{}c@{}}B\end{array}\Vert )=\unicode[STIX]{x1D70B}(\widehat{\unicode[STIX]{x1D705}}_{1}(B))-\unicode[STIX]{x1D70B}(\widehat{\unicode[STIX]{x1D705}}_{2}(B))+\unicode[STIX]{x1D70B}(\widehat{\unicode[STIX]{x1D705}}_{3}(B)).\end{eqnarray}$$

Here the $\widehat{\unicode[STIX]{x1D705}}_{m}$ act on $B$ via matrix multiplication.

We now turn to proving that $\unicode[STIX]{x1D701}$ is surjective. Consider $a\in \text{k}$ , and set

$$\begin{eqnarray}A_{a}=\left\Vert \begin{array}{@{}cc@{}}1 & a\\ 0 & 1\end{array}\right\Vert .\end{eqnarray}$$

By Theorem 6.7, it is enough to prove that $A_{a}\in \operatorname{Im}(\unicode[STIX]{x1D701})$ . We have

(6.4) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D701}\left(\left\Vert \begin{array}{@{}ccc@{}}1 & a & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}\right\Vert \right) & = & \displaystyle \unicode[STIX]{x1D70B}\left(\left\Vert \begin{array}{@{}ccc@{}}0 & 0 & -1\\ 0 & 1 & 0\\ 1 & a & 0\end{array}\right\Vert \right)-\unicode[STIX]{x1D70B}\left(\left\Vert \begin{array}{@{}ccc@{}}1 & a & 0\\ 0 & 0 & -1\\ 0 & 1 & 0\end{array}\right\Vert \right)+\unicode[STIX]{x1D70B}\left(\left\Vert \begin{array}{@{}ccc@{}}1 & a & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}\right\Vert \right)\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D70B}\left(-\left\Vert \begin{array}{@{}ccc@{}}-1 & 0 & 0\\ 0 & 1 & 0\\ 0 & a & 1\end{array}\right\Vert \right)-\unicode[STIX]{x1D70B}\left(-\left\Vert \begin{array}{@{}ccc@{}}1 & 0 & a\\ 0 & -1 & 0\\ 0 & 0 & 1\end{array}\right\Vert \right)+\unicode[STIX]{x1D70B}\left(\left\Vert \begin{array}{@{}ccc@{}}1 & a & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}\right\Vert \right)\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D70B}\left(-\left\Vert \begin{array}{@{}ccc@{}}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & a & 1\end{array}\right\Vert \right)-\unicode[STIX]{x1D70B}\left(-\left\Vert \begin{array}{@{}ccc@{}}1 & 0 & a\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}\right\Vert \right)+\unicode[STIX]{x1D70B}\left(\left\Vert \begin{array}{@{}ccc@{}}1 & a & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}\right\Vert \right)\nonumber\\ \displaystyle & = & \displaystyle -\unicode[STIX]{x1D70B}\left(\left\Vert \begin{array}{@{}ccc@{}}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & a & 1\end{array}\right\Vert \right)+\left\Vert \begin{array}{@{}cc@{}}1 & 0\\ 0 & 1\end{array}\right\Vert +\left\Vert \begin{array}{@{}cc@{}}1 & a\\ 0 & 1\end{array}\right\Vert ,\end{eqnarray}$$

where the second equality uses the fact that permuting the columns of a matrix changes the associated apartment by the sign of the permutation and the third equality uses the fact that multiplying a column by a nonzero scalar does not change the associated apartment.

If $a=0$ , then the right-hand side of (6.4) simplifies to

$$\begin{eqnarray}-\left\Vert \begin{array}{@{}cc@{}}1 & 0\\ 0 & 1\end{array}\right\Vert +\left\Vert \begin{array}{@{}cc@{}}1 & 0\\ 0 & 1\end{array}\right\Vert +\left\Vert \begin{array}{@{}cc@{}}1 & 0\\ 0 & 1\end{array}\right\Vert =\left\Vert \begin{array}{@{}cc@{}}1 & 0\\ 0 & 1\end{array}\right\Vert =A_{0},\end{eqnarray}$$

so $A_{0}\in \operatorname{Im}(\unicode[STIX]{x1D701})$ . Assume now that $a\neq 0$ . Plugging the matrix

$$\begin{eqnarray}\left(\begin{array}{@{}cccc@{}}1 & 0 & 0 & 0\\ 0 & 1 & 1 & 0\\ 0 & 0 & a & 1\end{array}\right)\end{eqnarray}$$

into Lemma 6.6, we get the relation

$$\begin{eqnarray}\displaystyle 0 & = & \displaystyle \left\Vert \begin{array}{@{}ccc@{}}0 & 0 & 0\\ 1 & 1 & 0\\ 0 & a & 1\end{array}\right\Vert -\left\Vert \begin{array}{@{}ccc@{}}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & a & 1\end{array}\right\Vert +\left\Vert \begin{array}{@{}ccc@{}}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}\right\Vert -\left\Vert \begin{array}{@{}ccc@{}}1 & 0 & 0\\ 0 & 1 & 1\\ 0 & 0 & a\end{array}\right\Vert \nonumber\\ \displaystyle & = & \displaystyle 0-\left\Vert \begin{array}{@{}ccc@{}}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & a & 1\end{array}\right\Vert +\left\Vert \begin{array}{@{}ccc@{}}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}\right\Vert -\left\Vert \begin{array}{@{}ccc@{}}1 & 0 & 0\\ 0 & 1 & a^{-1}\\ 0 & 0 & 1\end{array}\right\Vert ,\nonumber\end{eqnarray}$$

where the equality uses the fact that the columns of the first matrix are not linearly independent and the fact that multiplying a column of a matrix by a nonzero scalar does not change the associated apartment. Plugging this relation into (6.4), we see that the right-hand side of (6.4) equals

$$\begin{eqnarray}\displaystyle & & \displaystyle -\left(\unicode[STIX]{x1D70B}\left(\left\Vert \begin{array}{@{}ccc@{}}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}\right\Vert \right)-\unicode[STIX]{x1D70B}\left(\left\Vert \begin{array}{@{}ccc@{}}1 & 0 & 0\\ 0 & 1 & a^{-1}\\ 0 & 0 & 1\end{array}\right\Vert \right)\right)+\left\Vert \begin{array}{@{}cc@{}}1 & 0\\ 0 & 1\end{array}\right\Vert +\left\Vert \begin{array}{@{}cc@{}}1 & a\\ 0 & 1\end{array}\right\Vert \nonumber\\ \displaystyle & & \displaystyle \quad =-\left\Vert \begin{array}{@{}cc@{}}1 & 0\\ 0 & 1\end{array}\right\Vert +\left\Vert \begin{array}{@{}cc@{}}1 & 0\\ 0 & 1\end{array}\right\Vert +\left\Vert \begin{array}{@{}cc@{}}1 & 0\\ 0 & 1\end{array}\right\Vert +\left\Vert \begin{array}{@{}cc@{}}1 & a\\ 0 & 1\end{array}\right\Vert \nonumber\\ \displaystyle & & \displaystyle \quad =A_{0}+A_{a}.\nonumber\end{eqnarray}$$

Since we have already seen that $A_{0}\in \operatorname{Im}(\unicode[STIX]{x1D701})$ , we deduce that $A_{a}\in \operatorname{Im}(\unicode[STIX]{x1D701})$ , as desired.