Proof. The only thing that is not trivial is that the function is well defined. This follows from the above presentations. ◻

Proof. Call $S(X)$ the $\mathbf{k}[\mathbf{GL}(X)]$-module with the above presentation. By Theorems 2.4, sending a generator to the generator with the same name gives a well-defined surjective map $f:S(X)\rightarrow \mathbf{St}(X)$. Pick an isomorphism $X\cong K^{n}$. To see that $f$ is injective, note that we can use relations (a), (b), and (c) to show that $S(K^{n})$ is generated as a $\mathbf{k}$-module by symbols $[v_{1},v_{2},\ldots ,v_{n}]$ where the vectors assemble to form a unit upper-triangular matrix with respect to the standard basis of $K^{n}$. The Solomon–Tits theorem (see [Reference BrownBro89, Theorem IV.5.2]) implies that $\mathbf{St}_{n}(K)$ has a basis given by unit upper-triangular matrices. Thus, $f$ is also injective.◻

Proof. This is the Solomon–Tits theorem (see [Reference BrownBro89, Theorem IV.5.2]). ◻

Proof. Because $\overline{{\mathcal{B}}}_{\ast }^{n}(\mathbf{St})$ is supported in homological degrees ${\leqslant}n$, it is enough to prove that $\text{H}_{s}(\overline{{\mathcal{B}}}_{\ast }^{n}(\mathbf{St}))=0$ for $s<n$.

We define a $\mathbf{k}$-linear map $\unicode[STIX]{x1D6F7}:\overline{{\mathcal{B}}}_{s}^{n}(\mathbf{St})\rightarrow \overline{{\mathcal{B}}}_{s+1}^{n}(\mathbf{St})$ on every $x=[a_{W_{1}}|\cdots |a_{W_{s}}]$ with $a_{W_{j}}\in \text{PBW}_{W_{j}}$ for each $j$ (such elements form a $\mathbf{k}$-basis of $\overline{{\mathcal{B}}}_{s}^{n}(\mathbf{St})$) as follows.

1. (a) Set $k$ to be the smallest index such that $\operatorname{rank}W_{k}>1$. (If such $k$ does not exist, $\unicode[STIX]{x1D6F7}(x)=0$ automatically because $x$ is in top degree.) We call this the widening index of $x$.

2. (b) If $S_{W_{j}}\prec S_{W_{j+1}}$, we call $j$ an order-preserving index of $x$. If there is an order-preserving index $j$ strictly smaller than $k$, set $\unicode[STIX]{x1D6F7}(x)=0$. For example, suppose $n=4$ and let $x=[a_{W_{1}}|a_{W_{2}}|a_{W_{3}}]\in \overline{{\mathcal{B}}}_{3}^{n}(\mathbf{St})$ be given by

   $$\begin{eqnarray}x=\left[\left[\begin{array}{@{}c@{}}1\\ 0\\ 0\\ 0\end{array}\right]\left|\left[\begin{array}{@{}c@{}}0\\ 1\\ 0\\ 0\end{array}\right]\right|\left[\begin{array}{@{}cc@{}}0 & 0\\ 0 & 0\\ 1 & 0\\ 0 & 1\end{array}\right]\right].\end{eqnarray}$$
   Then $k=3$ and $S_{W_{1}}=\{1\}\prec \{2\}=S_{W_{2}}$. So $j=1$ is an order-preserving index strictly smaller than $k$.

3. (c) Otherwise, let $m=\min S_{W_{k}}$ and $S$ be the complement of $m$ in $S_{W_{k}}$.

4. (d) By property (P3), the partition $\{m\}\sqcup S$ gives rise to a map

   $$\begin{eqnarray}\text{PBW}_{W_{k}}\rightarrow \coprod _{W\oplus W^{\prime }=W_{k}}\text{PBW}_{W}\times \text{PBW}_{W^{\prime }}.\end{eqnarray}$$
   Let $(b,b^{\prime })$ be the image of $a_{W_{k}}$ under this map.

5. (e) Set $\unicode[STIX]{x1D6F7}(x)=(-1)^{k-1}[a_{W_{1}}|\cdots |a_{W_{k-1}}|b|b^{\prime }|a_{W_{k+1}}|\cdots |a_{W_{s}}]$. For example, suppose $n=4$ and let $x=[a_{W_{1}}|a_{W_{2}}|a_{W_{3}}]\in \overline{{\mathcal{B}}}_{3}^{n}(\mathbf{St})$ be given by

   $$\begin{eqnarray}x=\left[\left[\begin{array}{@{}c@{}}0\\ 0\\ 0\\ 1\end{array}\right]\left|\left[\begin{array}{@{}c@{}}0\\ 0\\ 1\\ 0\end{array}\right]\right.\left|\left[\begin{array}{@{}cc@{}}1 & 0\\ 6 & 1\\ 3 & 0\\ 5 & 0\end{array}\right]\right.\!\right].\end{eqnarray}$$
   Then $k=3,S_{W_{1}}=\{4\},S_{W_{2}}=\{3\}$, and $S_{W_{3}}=\{1,2\}$. So $m=1$ and $S=\{2\}$. In this case $\unicode[STIX]{x1D6F7}(x)$ is given by
   $$\begin{eqnarray}\unicode[STIX]{x1D6F7}(x)=(-1)^{2}\left[\left[\begin{array}{@{}c@{}}0\\ 0\\ 0\\ 1\end{array}\right]\left|\left[\begin{array}{@{}c@{}}0\\ 0\\ 1\\ 0\end{array}\right]\right.\left|\left[\begin{array}{@{}c@{}}1\\ 6\\ 3\\ 5\end{array}\right]\right.\left|\left[\begin{array}{@{}c@{}}0\\ 1\\ 0\\ 0\end{array}\right]\right.\!\right].\end{eqnarray}$$

Let us furthermore define a filtration on $\overline{{\mathcal{B}}}_{s}^{n}(\mathbf{St})$.

1. (a) To every basis element $x=[a_{W_{1}}|\cdots |a_{W_{s}}]$ with $a_{W_{j}}\in \text{PBW}_{W_{j}}$ we associate a word $w_{x}$ in $[n]^{n}$ by concatenating $S_{W_{1}},\ldots ,S_{W_{s}}$, where the elements of $S_{W_{j}}$ are ordered as natural numbers.

2. (b) Let $w=(i_{1},\ldots ,i_{n})$ be a sequence in $[n]^{n}$. We say that a pair $(\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD})$ such that $\unicode[STIX]{x1D6FC}<\unicode[STIX]{x1D6FD}$ is an inversion for $w$ if $i_{\unicode[STIX]{x1D6FC}}\geqslant i_{\unicode[STIX]{x1D6FD}}$. Define a quasi-order ${<}$ on $[n]^{n}$ by $w<w^{\prime }$ if $w$ has more inversions than $w^{\prime }$.

   Note that there is a unique maximal element in $[n]^{n}$ given by $(1,2,\ldots ,n)$ which corresponds, for example, to $x=[\mathbf{e}_{1}|\mathbf{e}_{2}|\cdots |\mathbf{e}_{n}]$ where $\mathbf{e}_{1},\ldots ,\mathbf{e}_{n}$ is the coordinate basis of $K^{n}$ (or to any basis element $x$ which is lower-triangular with the notation as in the example above). On the other hand, if $x=[\mathbf{e}_{n}|\mathbf{e}_{n}+\mathbf{e}_{1}|\mathbf{e}_{n}+\mathbf{e}_{2}|\cdots |\mathbf{e}_{n}+\mathbf{e}_{n-1}]$ then $w_{x}=(n,n,\ldots ,n)$, which is a minimal element in $[n]^{n}$. Intuitively, $w_{x}>w_{y}$ if $x$ is closer to being a lower-triangular matrix compared to $y$.

3. (c) Let $F_{{\leqslant}\unicode[STIX]{x1D6FC}}\overline{{\mathcal{B}}}_{s}^{n}(\mathbf{St})$ be generated by all basis elements $x$ such that the number of inversions in $x$ is at most $\unicode[STIX]{x1D6FC}$.

For an element $y\in \overline{{\mathcal{B}}}_{s}^{n}(\mathbf{St})$, we say that $w_{y}>w$ if $y$ can be written as a sum of basis elements $x$ (as in (a) above) satisfying $w_{x}>w$. A crucial property of this order is the following. Let $x=[a_{W_{1}}|\cdots |a_{W_{s}}]$ be a basis element and let $y=[a_{W_{1}}|\cdots |a_{W_{j-1}}|a_{W_{j}}a_{W_{j+1}}|a_{W_{j+2}}|\cdots |a_{W_{s}}]$, where $a_{W_{j}}a_{W_{j+1}}$ is the image of $a_{W_{j}}\otimes a_{W_{j+1}}$ under the map $\mathbf{St}(W_{j})\otimes \mathbf{St}(W_{j+1})\rightarrow \mathbf{St}(W_{1}\oplus W_{2})$. Then we have $w_{x}\leqslant w_{y}$ and the equality holds if and only if $S_{W_{j}}\prec S_{W_{j+1}}$ (if and only if $y$ is a basis element). In particular,

$$\begin{eqnarray}0=F_{{\leqslant}-1}\overline{{\mathcal{B}}}_{\ast }^{n}(\mathbf{St})\subset F_{{\leqslant}0}\overline{{\mathcal{B}}}_{\ast }^{n}(\mathbf{St})\subset F_{{\leqslant}1}\overline{{\mathcal{B}}}_{\ast }^{n}(\mathbf{St})\subset \cdots \subset F_{{\leqslant}\binom{n}{2}}\overline{{\mathcal{B}}}_{\ast }^{n}(\mathbf{St})=\overline{{\mathcal{B}}}_{\ast }^{n}(\mathbf{St})\end{eqnarray}$$

defines a finite increasing filtration on the chain complex $\overline{{\mathcal{B}}}_{\ast }^{n}(\mathbf{St})$. We also note here that $w_{\unicode[STIX]{x1D6F7}(x)}=w_{x}$ for every basis element $x$ for which $\unicode[STIX]{x1D6F7}(x)$ is non-zero.

In the remainder of the proof, we will verify that $\unicode[STIX]{x2202}\unicode[STIX]{x1D6F7}+\unicode[STIX]{x1D6F7}\unicode[STIX]{x2202}-\text{id}$ sends $F_{{\leqslant}\unicode[STIX]{x1D6FC}}\overline{{\mathcal{B}}}_{s}^{n}(\mathbf{St})$ to $F_{{<}\unicode[STIX]{x1D6FC}}\overline{{\mathcal{B}}}_{s}^{n}(\mathbf{St})$ for all $s<n$. This then shows that

$$\begin{eqnarray}\text{H}_{s}\big(F_{{\leqslant}\unicode[STIX]{x1D6FC}}\overline{{\mathcal{B}}}_{\ast }^{n}(\mathbf{St})/F_{{<}\unicode[STIX]{x1D6FC}}\overline{{\mathcal{B}}}_{\ast }^{n}(\mathbf{St})\big)=0\quad \text{for }s<n,\end{eqnarray}$$

because every cycle

$$\begin{eqnarray}c=(\unicode[STIX]{x2202}\unicode[STIX]{x1D6F7}+\unicode[STIX]{x1D6F7}\unicode[STIX]{x2202})(c)=\unicode[STIX]{x2202}(\unicode[STIX]{x1D6F7}(c))\end{eqnarray}$$

in degree less than $n$ is also a boundary. By induction on the filtration, we deduce that $\text{H}_{s}(\overline{{\mathcal{B}}}_{\ast }^{n}(\mathbf{St}))=0$ for $s<n$. We prove the claim by a complete case study. Let $x=[a_{W_{1}}|\cdots |a_{W_{s}}]$ with $a_{W_{j}}\in \text{PBW}_{W_{j}}$ and $s<n$. Set $k$ to be the smallest index such that $\operatorname{rank}W_{k}>1$.

Case 1. Assume there is a $j<k$ such that $S_{W_{j}}\prec S_{W_{j+1}}$. Also let $j_{0}$ be the smallest order-preserving index. Then $(\unicode[STIX]{x2202}\unicode[STIX]{x1D6F7}+\unicode[STIX]{x1D6F7}\unicode[STIX]{x2202})(x)=\unicode[STIX]{x1D6F7}\unicode[STIX]{x2202}(x)$. Let $y_{i}=[a_{W_{1}}|\cdots |a_{W_{i}}a_{W_{i+1}}|\cdots |a_{W_{s}}]$, so that

$$\begin{eqnarray}\unicode[STIX]{x2202}(x)=\mathop{\sum }_{i=1}^{s-1}(-1)^{i-1}y_{i}.\end{eqnarray}$$

First we see that $\unicode[STIX]{x1D6F7}(y_{j_{0}})=x$, because $j_{0}$ is the widening index of $y_{j_{0}}$ and there are no order-preserving indices smaller than $j_{0}$. For $i<j_{0}$, we get that $w_{y_{i}}>w_{x}$, because $i$ is not an order-preserving index of $x$. Hence $w_{\unicode[STIX]{x1D6F7}(y_{i})}=w_{y_{i}}>w_{x}$. If $i=j_{0}+1$, we have two cases. If $j_{0}$ is an order-preserving index of $y_{j_{0}+1}$, then $\unicode[STIX]{x1D6F7}(y_{j_{0}+1})=0$ because its widening index is $j_{0}+1$. It is possible that $j_{0}$ is not an order-preserving index if $S_{W_{j_{0}}}\not \prec S_{W_{j_{0}+2}}$, but then $j_{0}+1$ cannot be order-preserving and $w_{\unicode[STIX]{x1D6F7}(y_{j_{0}+1})}=w_{y_{j_{0}+1}}>w_{x}$. Finally, if $i\geqslant j_{0}+2$, then $j_{0}$ is order-preserving and smaller than the widening index of $y_{i}$. Therefore $\unicode[STIX]{x1D6F7}(y_{i})=0$.

Case 2. Assume there is no $j<k$ such that $S_{W_{j}}\prec S_{W_{j+1}}$. Let $y_{i}=[a_{W_{1}}|\cdots |a_{W_{i}}a_{W_{i+1}}|\cdots |a_{W_{s}}]$. Let us write $\unicode[STIX]{x1D6F7}(x)=(-1)^{k-1}[a_{W_{1}}|\cdots |a_{W_{k-1}}|a_{W_{k}^{\prime }}|a_{W_{k}^{\prime \prime }}|a_{W_{k+1}}|\cdots |a_{W_{s}}]$ as in its definition, that is, $(a_{W_{k}^{\prime }},a_{W_{k}^{\prime \prime }})$ is the image of $a_{W_{k}}$ under the map

$$\begin{eqnarray}\text{PBW}_{W_{k}}\rightarrow \coprod _{W_{k}^{\prime }\oplus W_{k}^{\prime \prime }=W_{k}}\text{PBW}_{W_{k}^{\prime }}\times \text{PBW}_{W_{k}^{\prime \prime }}.\end{eqnarray}$$

We now define

$$\begin{eqnarray}\,z_{i}=\left\{\begin{array}{@{}ll@{}}[a_{W_{1}}|\cdots |a_{W_{i}}a_{W_{i+1}}|\cdots |a_{W_{k-1}}|a_{W_{k}^{\prime }}|a_{W_{k}^{\prime \prime }}|a_{W_{k+1}}|\cdots |a_{W_{s}}]\quad & \text{if }i<k-1,\\ \,[a_{W_{1}}|\cdots |a_{W_{i}}a_{W_{i+1}}|\cdots |a_{W_{k-1}}a_{W_{k}^{\prime }}|a_{W_{k}^{\prime \prime }}|a_{W_{k+1}}|\cdots |a_{W_{s}}]\quad & \text{if }i=k-1,\\ \,[a_{W_{1}}|\cdots |a_{W_{k-1}}|a_{W_{k}^{\prime }}a_{W_{k}^{\prime \prime }}|a_{W_{k+1}}|\cdots |a_{W_{s}}]\quad & \text{if }i=k,\\ \,[a_{W_{1}}|\cdots |a_{W_{k-1}}|a_{W_{k}^{\prime }}|a_{W_{k}^{\prime \prime }}a_{W_{k+1}}|\cdots |a_{W_{s}}]\quad & \text{if }i=k+1,\\ \,[a_{W_{1}}|\cdots |a_{W_{k-1}}|a_{W_{k}^{\prime }}|a_{W_{k}^{\prime \prime }}|a_{W_{k+1}}|\cdots |a_{W_{i-1}}a_{W_{i}}|\cdots |a_{W_{s}}]\quad & \text{if }i>k+1.\end{array}\right.\end{eqnarray}$$

Then

$$\begin{eqnarray}\unicode[STIX]{x2202}(x)=\mathop{\sum }_{i=1}^{s-1}(-1)^{i-1}y_{i}\quad \text{and}\quad \unicode[STIX]{x2202}\unicode[STIX]{x1D6F7}(x)=\mathop{\sum }_{i=1}^{s}(-1)^{i-1}(-1)^{k-1}z_{i}.\end{eqnarray}$$

Then we have the following results.

• ∙ For $i<k$, $w_{\unicode[STIX]{x1D6F7}(y_{i})}=w_{y_{i}}$ is larger than $w_{x}$ as $S_{W_{i}}\not \prec S_{W_{i+1}}$ ($i$ is not an order-preserving index of $x$).

• ∙ For $i<k-1$, $w_{z_{i}}$ is larger than $w_{x}$ again because $S_{W_{i}}\not \prec S_{W_{i+1}}$ ($i$ is not an order-preserving index of $x$).

• ∙ $w_{z_{k-1}}$ is larger than $w_{x}$ as $S_{W_{k-1}}\not \prec S_{W_{k}^{\prime }}$ (recall that $S_{W_{k}^{\prime }}=\min S_{W_{k}}$ and $k-1$ is not an order-preserving index for $x$).

• ∙ $z_{k}=x$ as $S_{W_{k}^{\prime }}\prec S_{W_{k}^{\prime \prime }}$ by construction (in other words, $a_{W_{k}^{\prime }}a_{W_{k}^{\prime \prime }}=a_{W_{k}}$ is already a basis element in $\mathbf{St}(W_{k})$).

• ∙ If $z_{k+1}$ and $\unicode[STIX]{x1D6F7}(y_{k})$ do not cancel, $k$ cannot be an order-preserving index of $x$, which implies that both $w_{\unicode[STIX]{x1D6F7}(y_{k})}=w_{y_{k}}$ and $w_{z_{k+1}}$ are larger than $w_{x}$.

• ∙ $\unicode[STIX]{x1D6F7}(y_{i})$ cancels with $z_{i+1}$ in $(\unicode[STIX]{x2202}\unicode[STIX]{x1D6F7}+\unicode[STIX]{x1D6F7}\unicode[STIX]{x2202})(x)$ if $i>k$. To see this, just note that the widening index for each term in $y_{i}$ expressed in the basis (a) has widening index $k$ and order-preserving index ${\geqslant}k$, and so we have

  $$\begin{eqnarray}\unicode[STIX]{x1D6F7}(y_{i})=(-1)^{k-1}[a_{W_{1}}|\cdots |a_{W_{k-1}}|a_{W_{k}^{\prime }}|a_{W_{k}^{\prime \prime }}|a_{W_{k+1}}|\cdots |a_{W_{i}}a_{W_{i+1}}|\cdots |a_{W_{s}}]\end{eqnarray}$$
  even if $a_{W_{i}}a_{W_{i+1}}$ is not a basis element.

This finishes the case study and the proof. ◻

Question 3.7. Can one prove a version of Theorem 3.2 when $R$ is not a field, for example, when $R=\mathbb{Z}$? A version of Theorem 3.2 or even Theorem 4.2 for $R=\mathbb{Z}$ would establish the Church–Farb–Putman conjecture [Reference Church, Farb and PutmanCFP14, Conjecture 2] on vanishing of the high-dimensional rational cohomology of $\mathbf{SL}_{n}(\mathbb{Z})$ (except with a slightly worse range).

Proof. By the previous paragraph, we have

$$\begin{eqnarray}\dim _{\mathbf{k}}\operatorname{Tor}_{2}^{\mathbf{A}}(\mathbf{k},\mathbf{St})(\mathbb{F}_{q}^{4})\leqslant \dim _{\boldsymbol{ k}}(\operatorname{Sym}^{2}(\text{triv}_{1})\otimes \mathbf{St})(\mathbb{F}_{q}^{4}).\end{eqnarray}$$

Now note that

$$\begin{eqnarray}\displaystyle \dim _{\mathbf{k}}\operatorname{Sym}^{2}(\text{triv}_{1}) & = & \displaystyle \frac{|\mathbf{GL}_{2}(\mathbb{F}_{q})|}{|\mathbf{GL}_{1}(\mathbb{F}_{q})|^{2}|S_{2}|},\nonumber\\ \displaystyle \dim _{\mathbf{k}}\mathbf{St}(\mathbb{F}_{q}^{2}) & = & \displaystyle q,\nonumber\end{eqnarray}$$

and so

$$\begin{eqnarray}\dim _{\mathbf{k}}(\operatorname{Sym}^{2}(\text{triv}_{1})\otimes \mathbf{St})(\mathbb{F}_{q}^{4})=\frac{|\mathbf{GL}_{4}(\mathbb{F}_{q})|}{|\mathbf{GL}_{2}(\mathbb{F}_{q})|^{2}}\frac{|\mathbf{GL}_{2}(\mathbb{F}_{q})|}{|\mathbf{GL}_{1}(\mathbb{F}_{q})|^{2}|S_{2}|}q=\frac{(q^{4}-1)(q^{3}-1)q^{6}}{2(q-1)^{2}}.\end{eqnarray}$$

This completes the proof. ◻

Proof. Let $T_{n}$ denote $\dim _{\mathbf{k}}\operatorname{Tor}_{n}^{\mathbf{St}}(\mathbf{k},\mathbf{k})$, and let $ni_{q}$ denote the $q$-binomial coefficient. It is clear that $T_{0}=1$, and we have

$$\begin{eqnarray}\frac{|\mathbf{GL}_{n}(\mathbb{F}_{q})|}{|\mathbf{GL}_{i}(\mathbb{F}_{q})||\mathbf{GL}_{n-i}(\mathbb{F}_{q})|}=\left[\binom{n}{i}\right]_{q}q^{i(n-i)}.\end{eqnarray}$$

Since $\mathbf{St}$ is Koszul (Theorem 3.2), we have a Koszul resolution given by $\mathbf{St}\otimes \operatorname{Tor}_{\ast }^{\mathbf{St}}(\mathbf{k},\mathbf{k})\rightarrow \mathbf{k}\rightarrow 0$. So $T_{n}$ satisfies the recursion

$$\begin{eqnarray}T_{n}=\mathop{\sum }_{i=1}^{n}(-1)^{i}\left[\binom{n}{i}\right]_{q}q^{i(n-i)}q^{\binom{i}{2}}T_{n-i}.\end{eqnarray}$$

By induction, it suffices to verify that

$$\begin{eqnarray}q^{n^{2}-n}=\mathop{\sum }_{i=1}^{n}(-1)^{i}\left[\binom{n}{i}\right]_{q}q^{i(n-i)}q^{\binom{i}{2}}q^{(n-i)^{2}-(n-i)},\end{eqnarray}$$

which is equivalent, after canceling $q^{\binom{n}{2}}$ from both sides, to

$$\begin{eqnarray}q^{\binom{n}{2}}=\mathop{\sum }_{i=1}^{n}(-1)^{i}\left[\binom{n}{i}\right]_{q}q^{\binom{n-i}{2}}.\end{eqnarray}$$

This follows from [Reference NagpalNag19, Lemma 5.9], completing the proof. ◻

Question 3.11. Can one give a conceptual description of the Koszul dual of the Steinberg monoid, or explicitly describe the cycles in $\overline{{\mathcal{B}}}_{n}^{n}(\mathbf{St})$?

Proof. The Bykovskii presentation [Reference BykovskiĭByk03] (here Theorem 2.5) in the case of the integers and Theorem 2.8 in the case of fields imply that we have an exact sequence

$$\begin{eqnarray}\mathbf{A}\otimes M\rightarrow \mathbf{A}\rightarrow \mathbf{St}\rightarrow 0\end{eqnarray}$$

of $\mathbf{A}$-modules where $M$ is a $\text{VB}$-module supported only in degree two and is given by

$$\begin{eqnarray}M(X)=\langle [v_{1},v_{2}]-[v_{0},v_{2}]+[v_{0},v_{1}]\mid v_{0}=v_{1}+v_{2}\rangle \subset \mathbf{A}(X)\end{eqnarray}$$

in which $X$ is a free $R$-module of rank two. The assertion is immediate from this partial resolution.◻

Proof. Let $A\otimes V_{\ast }\rightarrow \mathbf{k}\rightarrow 0$ be the Koszul resolution. In particular, $V_{i}$ is concentrated in degree $i$. Now note that $\operatorname{Tor}_{i}^{A}(M,\mathbf{k})=\text{H}_{i}(M\otimes V_{\ast })$. Since $V_{i}$ is concentrated in degree $i$ and $M$ is supported in degrees ${\leqslant}d$, we see that $M\otimes V_{\ast }$ is supported in degrees ${\leqslant}(i+d)$. This finishes the proof.◻

Proof. Let $P_{\ast }\rightarrow \mathbf{k}$ and $Q_{\ast }\rightarrow B$ be free $A$-resolutions. Then $\operatorname{Tor}_{i}^{A}(\mathbf{k},B)=\text{H}_{i}(\mathbf{k}\otimes _{A}Q_{\ast })$. Since $A_{+}$ acts trivially on $\mathbf{k}\otimes _{A}Q_{i}=Q_{i}/(A_{+}Q_{i})$, we conclude that $A_{+}$ acts trivially on $\operatorname{Tor}_{i}^{A}(\mathbf{k},B)$. However, $A_{+}$ acts via its image $B_{+}$ on $\operatorname{Tor}_{i}^{A}(\mathbf{k},B)=\text{H}_{i}(P_{\ast }\otimes _{A}B)$. Thus $B_{+}$ acts trivially on $\operatorname{Tor}_{i}^{A}(\mathbf{k},B)$. The assertion is immediate from this.◻

Proof. Denote $\deg \operatorname{Tor}_{i}^{A}(\mathbf{k},B)$ by $s_{i}$. We prove by induction on $i$ that $s_{i}\leqslant 2i$. The base case $i=0$ is trivial. Let $i>0$. Consider the base change spectral sequence

$$\begin{eqnarray}E_{ab}^{2}=\operatorname{Tor}_{a}^{B}(\operatorname{Tor}_{b}^{A}(\mathbf{k},B),\mathbf{k})\;\Longrightarrow \;\operatorname{Tor}_{a+b}^{A}(\mathbf{k},\mathbf{k}).\end{eqnarray}$$

Let $t_{i}^{A}$ denote $\deg \operatorname{Tor}_{i}^{A}(\mathbf{k},\mathbf{k})$. By Koszulness of $A$, we know that $t_{i}^{A}\leqslant i$. Now suppose $b<i$. By induction, $s_{b}\leqslant 2b$. By Lemma 4.4 and Koszulness of $B$, we have

$$\begin{eqnarray}\deg E_{ab}^{2}\leqslant a+s_{b}\leqslant a+2b.\end{eqnarray}$$

The spectral sequence now implies that

$$\begin{eqnarray}\deg E_{0,i}^{2}\leqslant \max \Bigl(t_{i}^{A},\max _{a+b=i+1,b<i}\deg E_{ab}^{2}\Bigr)\leqslant 2i.\end{eqnarray}$$

Now note that $E_{0,i}^{2}=\operatorname{Tor}_{0}^{B}(\operatorname{Tor}_{i}^{A}(\mathbf{k},B),\mathbf{k})=\operatorname{Tor}_{i}^{A}(\mathbf{k},B)$ by Lemma 4.5. Thus $s_{i}\leqslant 2i$, completing the proof.◻

Proof of Theorem 4.2.

The theorem follows immediately from the previous proposition and Koszulness of the apartment and the Steinberg monoids (Theorems 3.1 and 3.2). ◻

Proof. We note that our tensor product is the same as Djament’s; see Remark 2.1. Part (a) is proven in [Reference DjamentDja15, Proposition 1.3, 1.4]. Now we prove part (b). In [Reference DjamentDja15, §1.5], it is shown that $\text{H}_{0}({\mathcal{G}};-)$ is lax monoidal. We claim that $\text{H}_{0}({\mathcal{G}};-)$ is in fact strong monoidal, which we verify as follows:

$$\begin{eqnarray}\displaystyle \text{H}_{0}({\mathcal{G}};M\otimes N)(X) & = & \displaystyle \text{H}_{0}({\mathcal{G}}(X);(M\otimes N)(X))\nonumber\\ \displaystyle & = & \displaystyle \text{H}_{0}\biggl({\mathcal{G}}(X);\bigoplus _{X_{1}\oplus X_{2}=X\text{ in }\text{VB}_{\mathbb{Z}}}M(X_{1})\otimes _{\mathbf{k}}N(X_{2})\biggr)\nonumber\\ \displaystyle & = & \displaystyle \text{H}_{0}\biggl({\mathcal{G}}(X);\bigoplus _{X_{1}\oplus X_{2}=X\text{ in }\text{VB}_{\mathbb{Z}}/{\mathcal{G}}}\text{Ind}_{{\mathcal{G}}(X)\cap (\mathbf{GL}(X_{1})\times \mathbf{GL}(X_{2}))}^{{\mathcal{G}}(X)}M(X_{1})\otimes _{\mathbf{k}}N(X_{2})\biggr)\nonumber\\ \displaystyle & = & \displaystyle \text{}\bigoplus _{X_{1}\oplus X_{2}=X\text{ in }\text{VB}_{\mathbb{Z}}/{\mathcal{G}}}\text{H}_{0}({\mathcal{G}}(X);\text{Ind}_{{\mathcal{G}}(X_{1})\times {\mathcal{G}}(X_{2})}^{{\mathcal{G}}(X)}M(X_{1})\otimes _{\mathbf{k}}N(X_{2}))\nonumber\\ \displaystyle & = & \displaystyle \text{}\bigoplus _{X_{1}\oplus X_{2}=X\text{ in }\text{VB}_{\mathbb{Z}}/{\mathcal{G}}}\text{H}_{0}({\mathcal{G}}(X_{1})\times {\mathcal{G}}(X_{2});M(X_{1})\otimes _{\mathbf{k}}N(X_{2}))\nonumber\\ \displaystyle & = & \displaystyle (\text{H}_{0}({\mathcal{G}};M)\otimes \text{H}_{0}({\mathcal{G}};N))(X),\nonumber\end{eqnarray}$$

where the third equality follows from the Mackey decomposition theorem, the fourth equality follows from the fact that ${\mathcal{G}}$ is strong, and the last equality follows from the Künneth formula for group homology. This completes the proof.◻

Proof. Part (a) follows immediately from the fact that $\text{H}_{0}({\mathcal{G}};-)$ is strong monoidal. Moreover, since the action of the symmetric group on tensors commutes with the action of the congruence subgroup, we see that parts (b) and (c) hold as well.◻

Proof. Let $I_{A}$ and $I_{B}$ be the right adjoints of $\mathbf{k}\otimes _{A}-$ and $\mathbf{k}\otimes _{B}-$, respectively. By our assumption on $A$, $I_{A}(M)$ is the same as $M$ regarded as an $A$-module via the augmentation map $A\rightarrow \mathbf{k}$. A similar statement is true for $I_{B}$. This shows that $\text{H}_{0}^{A}({\mathcal{G}};-)\circ I_{A}=I_{B}\circ \text{H}_{0}({\mathcal{G}};-)$.

Let $\unicode[STIX]{x1D716}:\text{id}\rightarrow I_{A}\circ (\mathbf{k}\otimes _{A}-)$ be the unit of the adjunction. By the previous paragraph, we have a natural transformation $\text{H}_{0}^{A}({\mathcal{G}};-)\rightarrow I_{B}\circ \text{H}_{0}({\mathcal{G}};-)\circ (\mathbf{k}\otimes _{A}-)$ obtained by composing $\text{H}_{0}^{A}({\mathcal{G}};-)$ with the unit $\unicode[STIX]{x1D716}$. By adjunction, there exists a natural transformation

$$\begin{eqnarray}(\mathbf{k}\otimes _{B}-)\circ \text{H}_{0}^{A}({\mathcal{G}};-)\rightarrow \text{H}_{0}({\mathcal{G}};-)\circ (\mathbf{k}\otimes _{A}-)\end{eqnarray}$$

which we claim is an isomorphism. Since all the functors involved are right exact, it suffices to prove the claim for objects of the form $A\otimes V$ (projective objects are of this form). We now check this as follows:

$$\begin{eqnarray}\displaystyle (\mathbf{k}\otimes _{B}-)\circ \text{H}_{0}^{A}({\mathcal{G}};-)(A\otimes V) & = & \displaystyle \mathbf{k}\otimes _{B}\text{H}_{0}^{A}({\mathcal{G}};A\otimes V)\nonumber\\ \displaystyle & = & \displaystyle \mathbf{k}\otimes _{B}(\text{H}_{0}({\mathcal{G}};A)\otimes \text{H}_{0}({\mathcal{G}};V))\quad \text{since }\text{H}_{0}({\mathcal{G}};-)\text{ is monoidal}\nonumber\\ \displaystyle & = & \displaystyle \mathbf{k}\otimes _{B}(B\otimes \text{H}_{0}({\mathcal{G}};V))\nonumber\\ \displaystyle & = & \displaystyle \text{H}_{0}({\mathcal{G}};V)=\text{H}_{0}({\mathcal{G}};\mathbf{k}\otimes _{A}(A\otimes V))\nonumber\\ \displaystyle & = & \displaystyle \text{H}_{0}({\mathcal{G}};-)\circ (\mathbf{k}\otimes _{A}-)(A\otimes V).\nonumber\end{eqnarray}$$

This completes the proof. ◻

Proof. We first verify that $\text{H}_{0}^{A}({\mathcal{G}};-)$ preserves projectives. It is an easy fact that every projective $A$-module is of the form $A\otimes V$ where $V$ is a projective $\text{VB}$-module, and the same holds for $B$-modules; see [Reference NagpalNag19, Proposition 3.2] for a proof in a similar setting. Since $\text{H}_{0}({\mathcal{G}};-)$ is monoidal, we see that $\text{H}_{0}^{A}({\mathcal{G}};A\otimes V)=B\otimes \text{H}_{0}({\mathcal{G}};V)$. By definition, $\text{H}_{0}({\mathcal{G}};-)$ is a left adjoint of an exact functor $\unicode[STIX]{x1D6F1}_{{\mathcal{G}}}^{\ast }$ and so it preserves projectives. Since $\text{H}_{0}^{A}({\mathcal{G}};A\otimes V)=B\otimes \text{H}_{0}({\mathcal{G}};V)$, we conclude that $\text{H}_{0}^{A}({\mathcal{G}};-)$ preserves projectives. This verifies our claim.

Note that $\text{Mod}_{B}$ has enough projectives. We conclude that there exists a Grothendieck spectral sequence

$$\begin{eqnarray}E_{ab}^{2}(-)=\operatorname{Tor}_{a}^{B}(\mathbf{k},\text{H}_{b}^{A}({\mathcal{G}};-))\;\Longrightarrow \;\text{L}_{a+b}((\mathbf{k}\otimes _{B}-)\circ \text{H}_{0}^{A}({\mathcal{G}};-)).\end{eqnarray}$$

By the previous proposition $(\mathbf{k}\otimes _{B}-)\circ \text{H}_{0}^{A}({\mathcal{G}};-)=\text{H}_{0}({\mathcal{G}};-)\circ (\mathbf{k}\otimes _{A}-)$, and so the first assertion follows.

It is clear from the last paragraph that $\mathbf{k}\otimes _{A}-$ preserves projectives. Also, $\text{Mod}_{\text{VB}}$ has enough projectives. This shows that we have another Grothendieck spectral sequence

$$\begin{eqnarray}\text{H}_{a}({\mathcal{G}};\operatorname{Tor}_{b}^{A}(\mathbf{k},-))\;\Longrightarrow \;\text{L}_{a+b}(\text{H}_{0}({\mathcal{G}};-)\circ (\mathbf{k}\otimes _{A}-)).\end{eqnarray}$$

If $\max _{i\leqslant n}(\deg \operatorname{Tor}_{i}^{A}(\mathbf{k},M))\leqslant d$, then we have $\deg \text{L}_{a+b}(\text{H}_{0}({\mathcal{G}};-)\circ (\mathbf{k}\otimes _{A}-))\leqslant d$ for all $a+b\leqslant n$ (see also Remark 5.6). This proves the second assertion.◻

Proof. Note that $\text{H}_{i}^{\mathbf{A}}({\mathcal{G}};M)$ is calculated by first taking an $\mathbf{A}$-module projective resolution $P_{\ast }\rightarrow M\rightarrow 0$ of $M$, applying ${\mathcal{G}}$-coinvariants to the resolution, and then taking the $i$th homology of the resulting complex $(P_{\ast })_{{\mathcal{G}}}$. Thus, to prove the isomorphism in the assertion, it suffices to show that a projective $\mathbf{A}$-module is $\text{H}_{0}({\mathcal{G}};-)$-acyclic when regarded as a $\text{VB}_{\mathbb{Z}}$-module (recall that $\text{Mod}_{\text{VB}_{\mathbb{ Z}}}$ is equivalent to the product category $\prod _{{\geqslant}0}\text{Mod}_{\mathbf{k}[\mathbf{GL}_{n}(\mathbb{Z})]}$, and so a module is projective if and only if it is pointwise projective). Let $P$ be a projective $\mathbf{A}$-module. Then $P$ is of the form $\mathbf{A}\otimes V$ for some projective $\text{VB}_{\mathbb{Z}}$-module $V$. We claim that $\text{H}_{i}({\mathcal{G}}(X);P(X))=0$ for all $X$ and all $i>0$. Note that we have

$$\begin{eqnarray}\displaystyle \text{H}_{i}({\mathcal{G}}(X);P(X)) & = & \displaystyle \text{H}_{i}({\mathcal{G}};\mathbf{A}\otimes V)(X)\nonumber\\ \displaystyle & = & \displaystyle (\text{H}_{i}({\mathcal{G}};\mathbf{A})\otimes \text{H}_{0}({\mathcal{G}};V))(X)\nonumber\end{eqnarray}$$

where the first equality follows from Remark 5.6. The second equality is obtained by taking a projective resolution $Q_{\ast }\rightarrow \mathbf{A}$ in $\text{Mod}_{\text{VB}_{\mathbb{ Z}}}$, and then calculating $\text{H}_{i}({\mathcal{G}};\mathbf{A}\,\otimes \,V)$ using the projective resolution $Q_{\ast }\otimes V\rightarrow \mathbf{A}\otimes V$ and monoidality of $\text{H}_{0}({\mathcal{G}};-)$. Thus, it suffices to show that $\text{H}_{i}({\mathcal{G}};\mathbf{A})=0$ for $i>0$. We have that $\mathbf{A}(\mathbb{Z}^{n})$ is isomorphic to $\text{H}_{0}(\mathbb{Z}/2\wr S_{n};\mathbf{k}[\mathbf{GL}_{n}(\mathbb{Z})])$ where $\mathbb{Z}/2\wr S_{n}$ acts on the right as in the definition of the apartment monoid. Since $\mathbb{Z}/2\wr S_{n}$ is a finite group, the stabilizer of ${\mathcal{G}}(\mathbb{Z}^{n})$ acting on $\mathbf{A}(\mathbb{Z}^{n})$ must have finite order. Since ${\mathcal{G}}(\mathbb{Z}^{n})$ is torsion free, this stabilizer is trivial. Thus, $\mathbf{A}(X)$ is a free $\mathbf{k}[{\mathcal{G}}(X)]$-module. This shows that $\text{H}_{i}({\mathcal{G}};\mathbf{A})(X)=0$ for $i>0$. This verifies our claim, and the proof is complete.◻

Proof. The natural map $\text{H}_{\ast }({\mathcal{G}};M)\otimes \text{H}_{\ast }({\mathcal{G}};N)\rightarrow \text{H}_{\ast }({\mathcal{G}};M\otimes N)$ given by composing the Künneth map with the Shapiro isomorphism provides the required lax monoidal structure. The details are very similar to those in Proposition 5.5 part (b). We get lax monoidality instead of strong monoidality because the Künneth map is not an isomorphism in general. ◻

Proof. We begin with the proof of part (a). It is clear from the $\text{H}_{\ast }({\mathcal{G}};-)$-acyclicity of projective $A$-modules that $\text{H}_{\ast }^{A}({\mathcal{G}};M)$ and $\text{H}_{\ast }({\mathcal{G}};M)$ are isomorphic as $\text{Mod}_{\text{VB}_{\mathbb{ Z}}/{\mathcal{G}}}^{\operatorname{gr}}$-modules. In particular, an $A$-projective resolution $P_{\ast }\rightarrow M$ of $M$ can be used to calculate both $\text{H}_{\ast }({\mathcal{G}};M)$ and $\text{H}_{\ast }^{A}({\mathcal{G}};M)$. The $\text{H}_{0}({\mathcal{G}};A)$ action on $\text{H}_{0}({\mathcal{G}};P_{i})$ is given by composing the Künneth map and the Shapiro isomorphism. Since these two maps are functorial, we see that the action of $\text{H}_{0}({\mathcal{G}};A)$ on

$$\begin{eqnarray}\text{H}_{i}(\text{H}_{0}({\mathcal{G}};P_{\ast }))=\text{H}_{i}^{A}({\mathcal{G}};M)\end{eqnarray}$$

is given by composing the Künneth map and the Shapiro isomorphism. This completes the proof of part (a).

We now prove part (b). The commutativity of the top triangle is exactly part (a). The bottom triangle commutes because of the proposition above and functoriality of the Künneth and the Shapiro maps. This finishes the proof. ◻

Proof. When $p=3$, the spectral sequence (3) becomes

$$\begin{eqnarray}E_{ab}^{2}=\operatorname{Tor}_{a}^{\mathbf{A}_{\unicode[STIX]{x1D6E4}(3)}}(\mathbf{k},\text{H}_{b}(\unicode[STIX]{x1D6E4}(3);\mathbf{St}))\;\Longrightarrow \;\text{L}_{a+b}(\text{H}_{0}(\unicode[STIX]{x1D6E4}(3);-)\circ (\mathbf{k}\otimes _{\mathbf{A}}-))(\mathbf{St}).\end{eqnarray}$$

Note that $\deg \operatorname{Tor}_{i}^{\mathbf{A}}(\mathbf{k},\mathbf{St})\leqslant 2i$ for $i=0,1$ (by Proposition 4.1). Thus, by the second assertion of Theorem 5.9, we see that $E_{ab}^{\infty }$ is supported in degrees ${\leqslant}2$ for $a+b\leqslant 1$. Now note that $E_{0,1}^{\infty }$ is the cokernel of the map

$$\begin{eqnarray}\operatorname{Tor}_{2}^{\mathbf{A}_{\unicode[STIX]{x1D6E4}(3)}}(\mathbf{k},\text{H}_{0}(\unicode[STIX]{x1D6E4}(3);\mathbf{St}))=\operatorname{Tor}_{2}^{\mathbf{A}_{\unicode[STIX]{x1D6E4}(3)}}(\mathbf{k},\mathbf{St}_{\unicode[STIX]{x1D6E4}(3)})\rightarrow \operatorname{Tor}_{0}^{\mathbf{A}_{\unicode[STIX]{x1D6E4}(3)}}(\mathbf{k},\text{H}_{1}(\unicode[STIX]{x1D6E4}(3);\mathbf{St})).\end{eqnarray}$$

By the previous proposition, $\operatorname{Tor}_{2}^{\mathbf{A}_{\unicode[STIX]{x1D6E4}(3)}}(\mathbf{k},\mathbf{St}_{\unicode[STIX]{x1D6E4}(3)})$ is supported in degrees ${\leqslant}4$, and so we conclude that

$$\begin{eqnarray}\deg \operatorname{Tor}_{0}^{\mathbf{A}_{\unicode[STIX]{x1D6E4}(3)}}(\mathbf{k},\text{H}_{1}(\unicode[STIX]{x1D6E4}(3);\mathbf{St}))\leqslant 4,\end{eqnarray}$$

completing the proof of the proposition. ◻

Proof. This is immediate from the proposition above and Proposition 5.12 part (b). ◻

Proof of Theorem 5.1.

Proposition 5.14 implies that

$$\begin{eqnarray}\bigoplus _{\substack{ X_{1}\oplus X_{2}=X\text{ in }\text{VB}_{\mathbb{Z}}/\unicode[STIX]{x1D6E4}(3), \\ \operatorname{rank}X_{2}=4}}\mathbf{A}_{\unicode[STIX]{x1D6E4}(3)}(X_{1})\otimes _{\mathbf{k}}\text{H}_{1}(\unicode[STIX]{x1D6E4}(3);\mathbf{St})(X_{2})\rightarrow \text{H}_{1}(\unicode[STIX]{x1D6E4}(3);\mathbf{St})(X)\end{eqnarray}$$

is surjective for $\operatorname{rank}X\geqslant 4$. Setting $X=\mathbb{Z}^{n}$ for $n\geqslant 4$, we obtain the following surjection:

$$\begin{eqnarray}\text{Ind}_{\operatorname{Aut}_{\text{VB}_{\mathbb{Z}}/\unicode[STIX]{x1D6E4}(3)}(\mathbb{Z}^{n-4})\times \operatorname{Aut}_{\text{VB}_{\mathbb{ Z}}/\unicode[STIX]{x1D6E4}(3)}(\mathbb{Z}^{4})}^{\operatorname{Aut}_{\text{VB}_{\mathbb{ Z}}/\unicode[STIX]{x1D6E4}(3)}(\mathbb{Z}^{n})}\mathbf{A}_{\unicode[STIX]{x1D6E4}(3)}(\mathbb{Z}^{n-4})\otimes _{\boldsymbol{ k}}\text{H}_{1}(\unicode[STIX]{x1D6E4}(3);\mathbf{St})(\mathbb{Z}^{4})\rightarrow \text{H}_{1}(\unicode[STIX]{x1D6E4}(3);\mathbf{St})(\mathbb{Z}^{n}).\end{eqnarray}$$

As noted in Example 5.4, we have $\operatorname{Aut}_{\text{VB}_{\mathbb{ Z}}/\unicode[STIX]{x1D6E4}_{3}}(\mathbb{Z}^{n})=\mathbf{GL}_{n}(\mathbb{F}_{3})$. Using the monoidal equivalence $\unicode[STIX]{x1D704}^{\ast }$ as in that example, we see that the following map is a surjection:

(4)$$\begin{eqnarray}\text{Ind}_{\mathbf{GL}_{n-4}(\mathbb{F}_{3})\times \mathbf{GL}_{4}(\mathbb{F}_{3})}^{\mathbf{GL}_{n}(\mathbb{F}_{3})}\mathbf{A}_{\unicode[STIX]{x1D6E4}(3)}(\mathbb{Z}^{n-4})\otimes _{\boldsymbol{ k}}\text{H}_{1}(\unicode[STIX]{x1D6E4}(3);\mathbf{St})(\mathbb{Z}^{4})\rightarrow \text{H}_{1}(\unicode[STIX]{x1D6E4}(3);\mathbf{St})(\mathbb{Z}^{n}).\end{eqnarray}$$

We also have a natural surjection:

$$\begin{eqnarray}\mathbf{k}[\operatorname{Aut}_{\text{VB}_{\mathbb{ Z}}/\unicode[STIX]{x1D6E4}(3)}(\mathbb{Z}^{n-4})]=\mathbf{k}[\mathbf{GL}_{n-4}(\mathbb{F}_{3})]\rightarrow \mathbf{A}_{\unicode[STIX]{x1D6E4}(3)}(\mathbb{Z}^{n-4}).\end{eqnarray}$$

Combining it with (4), we obtain the desired surjection:

$$\begin{eqnarray}\text{Ind}_{\mathbf{GL}_{4}(\mathbb{F}_{3})}^{\mathbf{GL}_{n}(\mathbb{F}_{3})}\text{H}_{1}(\unicode[STIX]{x1D6E4}(3);\mathbf{St})(\mathbb{Z}^{4})\rightarrow \text{H}_{1}(\unicode[STIX]{x1D6E4}(3);\mathbf{St})(\mathbb{Z}^{n}).\end{eqnarray}$$

The first assertion now follows from the equality $\text{H}_{1}(\unicode[STIX]{x1D6E4}(3);\mathbf{St})(\mathbb{Z}^{n})=\text{H}_{1}(\unicode[STIX]{x1D6E4}_{n}(3);\mathbf{St}_{n}(\mathbb{Z}))$.

For the second assertion, note that by Corollary 5.15 we have the following analogue of (4):

(5)$$\begin{eqnarray}\text{Ind}_{\mathbf{GL}_{n-4}(\mathbb{F}_{3})\times \mathbf{GL}_{4}(\mathbb{F}_{3})}^{\mathbf{GL}_{n}(\mathbb{F}_{3})}\mathbf{St}_{\unicode[STIX]{x1D6E4}(3)}(\mathbb{Z}^{n-4})\otimes _{\boldsymbol{ k}}\text{H}_{1}(\unicode[STIX]{x1D6E4}(3);\mathbf{St})(\mathbb{Z}^{4})\rightarrow \text{H}_{1}(\unicode[STIX]{x1D6E4}(3);\mathbf{St})(\mathbb{Z}^{n}).\end{eqnarray}$$

The second assertion now follows from the fact that $\dim _{\mathbf{k}}\mathbf{St}_{\unicode[STIX]{x1D6E4}(3)}(\mathbb{Z}^{n-4})=3^{\binom{n-4}{2}}$.◻

Proof. By [Reference Lee and SzczarbaLS76b, Theorem 1.4], we have $\dim _{\mathbf{k}}\operatorname{Hom}_{\mathbb{Z}}(\text{H}_{2}(\unicode[STIX]{x1D6E4}_{3}(3)),\mathbf{k})\leqslant 27$. Moreover, by [Reference Lee and SzczarbaLS76b, Lemma 12.1], we have

$$\begin{eqnarray}\dim _{\mathbf{k}}\text{Ext}_{\mathbb{Z}}^{1}(\text{H}_{2}(\unicode[STIX]{x1D6E4}_{3}(3)),\mathbf{k})=\left\{\begin{array}{@{}ll@{}}8\quad & \text{if }\text{Char}(\mathbf{k})=3,\\ 0\quad & \text{otherwise}.\end{array}\right.\end{eqnarray}$$

Thus, by the universal coefficient theorem, we have $\dim _{\mathbf{k}}\text{H}^{2}(\unicode[STIX]{x1D6E4}_{3}(3);\mathbf{k})\leqslant 35$. By Borel–Serre duality, we conclude that $\dim _{\mathbf{k}}\text{H}_{1}(\unicode[STIX]{x1D6E4}_{3}(3);\text{St}_{3}(\mathbb{Z})\otimes _{\mathbb{Z}}\mathbf{k})\leqslant 35$.◻

Proof. Denote the $\mathbf{A}_{\unicode[STIX]{x1D6E4}(3)}$-module $\text{H}_{1}(\unicode[STIX]{x1D6E4}(3);\mathbf{St})$ by $M$, and let $M^{\prime }$ be the maximal submodule of $M$ generated in degrees ${\leqslant}3$. Let $V$ be the degree-four piece of the $\text{VB}_{\mathbb{Z}}/\unicode[STIX]{x1D6E4}(3)$-module $\operatorname{Tor}_{0}^{\mathbf{A}_{\unicode[STIX]{x1D6E4}(3)}}(\mathbf{k},\text{H}_{1}(\unicode[STIX]{x1D6E4}(3);\mathbf{St}))$. Then, by the definition of $\operatorname{Tor}_{0}^{\mathbf{A}_{\unicode[STIX]{x1D6E4}(3)}}(\mathbf{k},-)$, we have an isomorphism

$$\begin{eqnarray}V(\mathbb{Z}^{4})\cong (M/M^{\prime })(\mathbb{Z}^{4}).\end{eqnarray}$$

This shows that

$$\begin{eqnarray}\dim _{\mathbf{k}}M(\mathbb{Z}^{4})=\dim _{\boldsymbol{ k}}M^{\prime }(\mathbb{Z}^{4})+\dim _{\boldsymbol{ k}}V(\mathbb{Z}^{4}).\end{eqnarray}$$

The proof of Proposition 5.14 shows that the map $\operatorname{Tor}_{2}^{\mathbf{A}_{\unicode[STIX]{x1D6E4}(3)}}(\mathbf{k},\mathbf{St}_{\unicode[STIX]{x1D6E4}(3)})\rightarrow \operatorname{Tor}_{0}^{\mathbf{A}_{\unicode[STIX]{x1D6E4}(3)}}(\mathbf{k},\text{H}_{1}(\unicode[STIX]{x1D6E4}(3);\mathbf{St}))$ is surjective in degrees 3 and 4. In particular, we have

$$\begin{eqnarray}\dim _{\mathbf{k}}V(\mathbb{Z}^{4})\leqslant \dim _{\boldsymbol{ k}}\operatorname{Tor}_{2}^{\mathbf{A}_{\unicode[STIX]{x1D6E4}(3)}}(\mathbf{k},\mathbf{St}_{\unicode[STIX]{x1D6E4}(3)})(\mathbb{Z}^{4}).\end{eqnarray}$$

By Proposition 5.19 and Lemma 3.9 (for $q=3$), we have

$$\begin{eqnarray}\dim _{\mathbf{k}}\operatorname{Tor}_{2}^{\mathbf{A}_{\unicode[STIX]{x1D6E4}(3)}}(\mathbf{k},\mathbf{St}_{\unicode[STIX]{x1D6E4}(3)})(\mathbb{Z}^{4})\leqslant \frac{(3^{4}-1)(3^{3}-1)3^{6}}{8}=189\,540.\end{eqnarray}$$

By the previous lemma, $\dim _{\mathbf{k}}M^{\prime }(\mathbb{Z}^{3})\leqslant 35$. Since $M^{\prime }$ is generated in degrees ${\leqslant}3$, we see that

$$\begin{eqnarray}\dim _{\mathbf{k}}M^{\prime }(\mathbb{Z}^{4})\leqslant \frac{35|\mathbf{GL}_{4}(\mathbb{F}^{3})|}{|\mathbf{GL}_{3}(\mathbb{F}^{3})\times \mathbf{GL}_{1}(\mathbb{F}^{3})|}\leqslant 37\,800.\end{eqnarray}$$

Finally, we conclude that $\dim _{\mathbf{k}}M(\mathbb{Z}^{4})\leqslant 189\,540+37\,800=227\,340$. This completes the proof.◻

Proof. The first assertion follows immediately from Proposition 5.7 and the fact that the apartment monoid is an exterior algebra.

Let us denote the apartment and the Steinberg monoids in $\text{Mod}_{\text{VB}_{\mathbb{ F}_{3}}}$ by $\overline{\mathbf{A}}$ and $\overline{\mathbf{St}}$, respectively. By Theorems 2.5 and 2.8 (see also the proof of Proposition 4.1), we have an exact sequence of $\mathbf{A}$-modules