Proof of Lemma 3·7. The module $U\,:=\,U_{a_1,a_2,a_3,a_4}$ is free, and a basis is obtained from the following list by deleting zeroes and fixing signs among repetitions, as explained in Remark 3·4:

(3·9) \begin{equation} \mathsf{\bigwedge}\!\left({a_{\sigma(1)}},{a_{\sigma(2)}},{a_i}\,;\,{b_i},{a_{\sigma(3)}},{a_{\sigma(4)}}\right)\quad \hbox{with } \sigma \in \mathfrak{S}_4, i\in \{1,\dots,g\}.\end{equation}

We choose such a basis of U and we denote it by $\{e_{J,i}\}_{J,i}$ , which is indexed by 2-element subsets $J\subset \{1,2,3,4\}$ together with an element $i \in \{1,\dots, g\}\setminus J$ ; specifically, we have $e_{J,i}= \pm\mathsf{\bigwedge}\!\left({a_{j}},{a_{j'}},{a_i}\,;\,{b_i},{a_{k}},{a_{k'}}\right)$ writing J and $\overline{J}=\{1,2,3,4\}\setminus J$ as $\{\,j,j'\}$ and $\{k,k'\}$ , respectively. Let U ^′ be the submodule of U (freely) generated by the previous elements $e_{J,i}$ for $i\leq 4$ (hence $J\subset \{1,2,3,4\}$ is a 2-element subset and $i\in \overline{J}$ ). When viewed as a relation in U, any element of type (i) identifies a basis element $e_{J,i}$ with $\pm e_{J,k}$ for $i\geq 5$ , $J\subset \{1,2,3,4\}$ of cardinality 2 and some $k\in \overline{J}$ . Similarly, for any $J\subset \{1,2,3,4\}$ of cardinality 2, writing $\overline{J}=\{i,k\}$ , the basis element $e_{J,i}$ is identified with $\pm e_{J,k}$ if an appropriate element of type (ii) is viewed as a relation in U. Therefore, the inclusion of U ^′ in U induces an isomorphism

\begin{equation*}\frac{U'}{\langle \,\hbox{(ii)}, \hbox{(iii)},\hbox{(iv)} \, \rangle} \stackrel{\simeq}{\longrightarrow}\frac{U}{\langle \, \hbox{(i)},\hbox{(ii)}, \hbox{(iii)},\hbox{(iv)} \, \rangle}.\end{equation*}

So, we can assume for the rest of the proof that $g=4$ , and we need to show that the map $B^{(0)}$ induces an injection from $U/{\langle \, \hbox{(ii)}, \hbox{(iii)}, \hbox{(iv)} \, \rangle}$ to ${S^2(\Lambda^2 H^\mathbb{Q})}/{\Lambda^4 H^\mathbb{Q}}$ . By the above discussion, the module $U/{\langle \, \hbox{(ii)} \, \rangle}$ is free of rank 6, with basis $\{e_J\}_J$ indexed by 2-element subsets J of $\{1,2,3,4\}$ where $e_J = \pm\mathsf{\bigwedge}\!\left({a_{j}},{a_{j'}},{a_i}\,;\,{b_i},{a_{i}},{a_{k}}\right)$ writing $J=\{\,j,j'\}$ and $\overline{J}=\{i,k\}$ . When viewed as a relation in $U/{\langle \, \hbox{(ii)} \, \rangle}$ , any element of type (iii) identifies a basis element $e_J$ with $\pm e_{\overline{J}}$ for a 2-element subset J of $\{1,2,3,4\}$ . Consequently, the module $U/{\langle \, \hbox{(ii)}, \hbox{(iii)}\, \rangle}$ is free of rank 3, with basis $\{e_P\}_P$ indexed by partitions $P=J \sqcup \overline{J}$ of $\{1,2,3,4\}$ into two 2-element subsets.

On the other hand, the $\mathbb{Q}$ -vector space $W\,:=\,W_{a_1,a_2,a_3,a_4}$ is generated by

(3·10) \begin{equation} \big\{(a_1 \wedge a_2) \cdot (a_3 \wedge a_4)\,,\,(a_1 \wedge a_3) \cdot (a_4 \wedge a_2)\,,\,(a_1 \wedge a_4) \cdot (a_2 \wedge a_3)\big\}\end{equation}

with the single relation

(3·11) \begin{equation} (a_1 \wedge a_2) \cdot (a_3 \wedge a_4)+(a_1 \wedge a_3) \cdot (a_4 \wedge a_2) +(a_1 \wedge a_4) \cdot (a_2 \wedge a_3).\end{equation}

The map $B^{(0)}:U/{\langle \, \hbox{(ii)}, \hbox{(iii)} \, \rangle} \to W$ sends the basis element $e_{P}$ to $\pm (a_{j} \wedge a_{j'})\cdot (a_{k} \wedge a_{k'})$ writing $P=\{\,j,j'\}\sqcup \{k,k'\}$ . Finally, when viewed as a relation in $U/{\langle \, \hbox{(ii)}, \hbox{(iii)} \, \rangle}$ , any element of type (iv) goes to (3·11). The conclusion follows.

Proof of Lemma 3·8. The module $U\,:=\,U_{a_1,a_2,a_3,a_3}$ is free, and a basis is obtained from the following list by deleting zeroes and fixing signs among repetitions, as explained in Remark 3·4:

\begin{eqnarray*}\mathsf{\bigwedge}\!\left({a_{\sigma(1)}},{a_{\sigma(2)}},{a_i}\,;\,{{b_i}},{a_{\sigma(3)}},{a_{\sigma(4)}}\right)&& \rm{\small with }\{1,2,3,4\} \stackrel{\sigma}{\longrightarrow} \{1,2,3\}\ \rm{\small surjective, reaching 3 twice}, \\&& \rm{\small and}\ i\in \{1,\dots,g\}.\end{eqnarray*}

We choose such a basis of U and denote it by $\{e_{j,i}\}_{j,i}$ , which is indexed by $j\in \{1,2\}$ and $i\not \in \{\,j,3\}$ ; specifically, we have $e_{j,i} = \pm \mathsf{\bigwedge}\!\left({a_3},{a_j},{a_i}\,;\,{b_i},{a_{\overline{j}}},{a_3}\right)$ , where $\overline{j}$ is the element of $\{1,2\}$ not equal to j. When viewed as a relation in U, any element of type (i) identifies a basis element $e_{j,i}$ with $\pm e_{j,\overline{j}}$ for $i\geq 4$ and $j\in \{1,2\}$ , and any element of type (ii) identifies $e_{1,2}$ with $\pm e_{2,1}$ . It follows that the module $U/{\langle \, \hbox{(i)}, \hbox{(ii)} \, \rangle}$ is free of rank 1.

On the other hand, the $\mathbb{Q}$ -vector space $W\,:=\,W_{a_1,a_2,a_3,a_3}$ is 1-dimensional, generated by $(a_1\wedge a_3) \cdot (a_2 \wedge a_3)$ . Since $B^{(0)}$ maps $e_{1,2}$ to $\pm (a_1\wedge a_3) \cdot (a_2 \wedge a_3)$ , it induces an injection $U/{\langle \, \hbox{(i)}, \hbox{(ii)} \, \rangle} \to W$ .

Proof of Lemma 3·9. The module $U\,:=\,U_{a_1,a_1,a_2,a_2}$ is free, and a basis is obtained from the following list by deleting zeroes and fixing signs among repetitions, as explained in Remark 3·4:

\begin{eqnarray*}\mathsf{\bigwedge}\!\left({a_{\sigma(1)}},{a_{\sigma(2)}},{a_i}\,;\,{b_i},{a_{\sigma(3)}},{a_{\sigma(4)}}\right)&& \rm{\small with }\{1,2,3,4\} \stackrel{\sigma}{\longrightarrow} \{1,2\}\ \rm{\small reaching each of 1 and 2 twice}, \\&& \rm{\small and}\ i\in \{1,\dots,g\}.\end{eqnarray*}

We choose such a basis of U and denote it by $\{e_i\}_{i\geq 3}$ , having $e_i\,=\,\pm \mathsf{\bigwedge}\!\left({a_1},{a_2},{a_i}\,;\,{b_i},{a_2},{a_1}\right)$ . When viewed as a relation in U, any element of type (i) identifies a basis element $e_i$ with $\pm e_3$ for $i\geq 3$ . It follows that the module $U/{\langle \, \hbox{(i)} \, \rangle}$ is free of rank 1.

Besides, the $\mathbb{Q}$ -vector space $W\,:=\,W_{a_1,a_1,a_2,a_2}$ is 1-dimensional, generated by $(a_1\,\wedge\,a_2)\,\cdot\,(a_1\,\wedge\,a_2)$ . Since $B^{(0)}$ maps $e_{3}$ to $\pm (a_1\wedge a_2) \cdot (a_1 \wedge a_2)$ , it induces an injection $U/{\langle \, \hbox{(i)} \, \rangle}\,\to\,W$ .

Proof of Lemma 3·10. The module $U\,:=\,U_{a_1,a_2}$ is free of basis $\{u(x,y)\}_{x,y} \sqcup \{u'(q,p)\}_{q,p}$ where

(3·12) \begin{equation} u(x,y) \,:=\, \mathsf{\bigwedge}\!\left({a_1},{x},{y}\,;\,{a_2},{\widetilde{x}},{\widetilde{y}}\right)\hbox{}\end{equation}

is indexed by 2-element subsets $\{x,y\}$ of $S\setminus \{a_1,b_2\}$ , and

(3·13) \begin{equation} u'(q,p) \,:=\, \mathsf{\bigwedge}\!\left({a_1},{a_2},{q}\,;\,{\widetilde{q}},{p},{\widetilde{p}}\right)\end{equation}

is indexed by pairs (q, p) with $q\in S\setminus\{a_1,a_2\}$ and $p \in S\setminus\{a_1,a_2,b_1,b_2, q,\widetilde{q}\}$ . We claim that the quotient module $U/{\langle \, \hbox{(i), (ii)} \, \rangle}$ is generated by

(3·14) \begin{equation} u(a_3,b_3) \quad \hbox{and} \quad u(a_2,p) \hbox{ for $p\in S \setminus \{a_1,a_2,b_2\}$}.\end{equation}

We start by observing that any element u $'$ (q,p) of type (3·13) is, modulo (ii), equal to the element $\pm u(p,\widetilde{p})$ of type (3·12). Indeed, we have

\begin{equation*}\mathsf{IHX}_2\!\left({a_1},{a_2}\,;\,{p}\,;\,{q}\right)=-\varepsilon(p)\, u(p,\widetilde{p}) - \varepsilon(q)\, u'(q,p).\end{equation*}

Thus, to prove the above claim, it is enough to show that the elements (3·12) not among the elements (3·14) can be written in terms of the latter using the relations (i). For any $r\in S \setminus\{a_1,a_2,a_3,b_1,b_2,b_3\}$ , we have the following elements of type (i):

\begin{equation*}\mathsf{S}\!\left({a_{1}},{a_{2}}\,;\,\begin{array}{cc}{r} & {a_2}\!\\{a_3} & {b_3}\! \end{array}\right)=- u(r,a_2) -\varepsilon(r)\, u(a_2,b_3)+ \varepsilon(r)\, u(a_3,b_3) + u(r,a_3)\end{equation*}

\begin{equation*}\mathsf{S}\!\left({a_{1}},{a_{2}}\,;\,\begin{array}{cc}{r} & {a_2}\!\\{b_3} & {a_3}\! \end{array}\right)= - u(r,a_2) + \varepsilon(r)\, u(a_2,a_3)+ \varepsilon(r)\, u(b_3,a_3) - u(r,b_3)\end{equation*}

\begin{equation*}\mathsf{S}\!\left({a_{1}},{a_{2}}\,;\,\begin{array}{cc}{a_3} & {b_3}\!\\{r} & {\widetilde{r}}\! \end{array}\right)= - u(a_3,b_3) - \varepsilon(r)\, u(b_3,\widetilde{r})- u(r,\widetilde{r}) - \varepsilon(r)\, u(a_3,r).\end{equation*}

We deduce that, modulo (i), $u(r,\widetilde{r})$ is a linear combination of (3·14) for $r\in S \setminus\{a_1,a_2,b_1,b_2\}$ . Next, for any such r and $p\in S \setminus \{a_1,a_2,b_2, r,\widetilde{r}\} $ , considering

\begin{equation*}\mathsf{S}\!\left({a_{1}},{a_{2}}\,;\,\begin{array}{cc}{p} & {a_2}\!\\{r} & {\widetilde{r}}\! \end{array}\right)= - u(p,a_2) - \varepsilon(p)\varepsilon(r)\, u(a_2,\widetilde{r})+ \varepsilon(p)\, u(r,\widetilde{r}) +\varepsilon(r)\, u(p,r)\end{equation*}

shows that u(p, r) is also modulo (i) a linear combination of (3·14). This proves the above claim.

The $\mathbb{Q}$ -vector space $W\,:=\,W_{a_1,a_2}$ has a basis $\{w(x)\}_x$ defined by $w(x) \,:=\, (a_1\wedge x )\cdot (a_2\wedge \widetilde{x})$ for $x\in S \setminus \{a_1,b_2\}$ . The map $B^{(0)}\,:\,U/{\langle \, \hbox{(i)},\hbox{(ii)} \, \rangle} \to W$ sends u(x, y) to $\varepsilon(x)\, w(y)+ \varepsilon(y)\, w(x)$ . Hence, the images of the generators (3·14) are

\begin{equation*}-w(a_3) + w(b_3), \quad\varepsilon(p)\, w(a_2)+ w(p)\ \hbox{ for $p\in S \setminus \{a_1,a_2,b_2\}$},\end{equation*}

and they are easily checked to be $\mathbb{Q}$ -linearly independent. It follows that the map $B^{(0)}\,:\,U/{\langle \, \hbox{(i)}, \hbox{(ii)} \, \rangle} \to W$ is injective.

Proof of Lemma 3·11. The module $U\,:=\,U_{a_1,a_1}$ is free, and a basis is obtained from the following list by deleting zeroes and fixing signs among repetitions:

(3·15) \begin{equation} \mathsf{\bigwedge}\!\left({a_1},{x},{y}\,;\,{a_1},{\widetilde{x}},{\widetilde{y}}\right)\quad \hbox{with $x,y \in S\setminus \{a_1,b_1\}$}.\end{equation}

Thus, we get a basis $\{u(x,y)\}_{x,y}$ indexed by unordered pairs $\{x,y\}\subset S\setminus\{a_1,b_1\}$ such that $x\neq \widetilde{y}$ , up to the involution $\{x,y\} \mapsto \{\widetilde{x},\widetilde{y}\}$ , and where $u(x,y)= \pm \mathsf{\bigwedge}\!\left({a_1},{x},{y}\,;\,{a_1},{\widetilde{x}},{\widetilde{y}}\right)$ .

For any $q,r\in \{a_3,\dots,a_g,b_3,\dots,b_g\}$ with $q\neq \widetilde{r}$ , we have the following element of type $(*)$ :

\begin{equation*}\mathsf{T}\!\left({a_1}\,;\,{a_2} ,{q},{r}\right) =\pm u(a_2,q) \pm u(a_2,r) \pm u(q,\widetilde{r}).\end{equation*}

Hence, $U/\langle \,(*)\, \rangle$ is generated by the elements u(x,y) such that the pair $\{x,y\}$ contains $a_2$ or $b_2$ . Furthermore, for any $r\in \{b_4,\dots,b_g\}$ , the elements

\begin{equation*}\mathsf{T}\!\left({a_1}\,;\,{a_2} ,{b_3},{r}\right)= \pm u(a_2,b_3) \pm u(a_2,r) \pm u(b_3,\widetilde{r})\\[-18pt]\nonumber\end{equation*}

\begin{equation*}\mathsf{T}\!\left({a_1}\,;\,{\widetilde{r}} ,{a_2},{b_3}\right)= \pm u(\widetilde{r},a_2) \pm u(\widetilde{r},b_3) \pm u(a_2,a_3)\end{equation*}

show that, modulo $(*)$ , $ u(a_2,r)$ is a linear combination of

(3·16) \begin{equation} u(a_2,q) \quad \hbox{ for } q\in \{a_3,a_4,\dots,a_g,b_3\}.\end{equation}

Hence the quotient module $U/\langle \,(*)\, \rangle$ is generated by (3·16).

The $\mathbb{Q}$ -vector space $W\,:=\,W_{a_1,a_1}$ has the basis $\{w(x)\}_x$ defined by $w(x) \,:=\, (a_1\wedge x )\cdot (a_1\wedge \widetilde{x})$ and indexed by the elements $x\in S \setminus \{a_1,b_1\}$ , up to the involution $x\mapsto \widetilde{x}$ . The map $B^{(0)}\,:\,U/{\langle \, (*) \, \rangle} \to W$ sends u(x,y) to $\pm \big(\varepsilon(x)\, w(y)+ \varepsilon(y)\, w(x)\big)$ . Hence, the images of the generators (3·16) are

\begin{equation*}\pm\big (\varepsilon(q)\, w(a_2)+ w(q) \big)\ \hbox{ for } q\in \{a_3,a_4, \dots,a_g,b_3\}.\end{equation*}

Those vectors being linearly independent, the map $B^{(0)}\,:\,U/{\langle \, (*) \, \rangle} \to W$ is injective.

Proof. We endow $U\,:=\,U_3$ and $W\,:=\,W_2$ with the following filtrations. First, observe that the map $N:S \to \{1,\dots,g\}$ defined by $N(a_i)=N(b_i)=i$ induces an increasing filtration

\begin{equation*}\varnothing = F_0 S \subset F_1 S\subset \cdots \subset F_{g-1} S \subset F_g S=S\end{equation*}

of the set S, where $F_k S$ consists of the $s\in S$ with $N(s)\leq k$ . Thus, we get increasing filtrations

\begin{eqnarray*}&& \{0\} = F_0U \subset F_1 U\subset \cdots \subset F_{g-1} U \subset F_g U=U\\[6pt]&& \{0\} = F_0W \subset F_1 W\subset \cdots \subset F_{g-1} W \subset F_g W=W\end{eqnarray*}

of the module U and the $\mathbb{Q}$ -vector space W, respectively. Specifically, for every $k\in\{1,\dots,g\}$ , the module $F_k U$ is generated by the elements

(3·17) \begin{equation} u(p,q,r) \,:=\, \mathsf{\bigwedge}\!\left({p},{q},{r}\,;\,{\widetilde{p}},{\widetilde{q}},{\widetilde{r}}\right)\quad \hbox{with $p,q,r\in F_k S$ pairwise different},\end{equation}

and

(3·18) \begin{equation} u'(r\,;\,p,q)\,:=\, \mathsf{\bigwedge}\!\left({p},{\widetilde{p}},{r}\,;\,{\widetilde{r}},{q},{\widetilde{q}}\right)\quad \hbox{with $p,q,r\in F_k S$ and $p,\widetilde{p},q,\widetilde{q},r,\widetilde{r}$ pairwise different}.\end{equation}

Similarly, for every $k\in\{1,\dots,g\}$ , the $\mathbb{Q}$ -vector space $F_kW$ is generated by the elements

\begin{equation*}w(p,q) \,:=\, (p \wedge q) \cdot (\widetilde{p} \wedge \widetilde{q})\quad \hbox{with $p,q\in F_k S$ and $p\neq q$}.\end{equation*}

For future use, we compute the map $B^{(0)}\,:\,U \to W$ in terms of those generators:

(3·19) \begin{align}B^{(0)}\big({u}(p,q,r)\big)&=\varepsilon(p)\, w(q,r) + \varepsilon(q)\, w(p,r)+ \varepsilon(r)\, w(p,q),\\[-10pt]\nonumber\end{align}

(3·20) \begin{align} B^{(0)}\big({u'}(r\,;\,p,q) \big)&=\varepsilon(r)\, w(p,q)- \varepsilon(r)\, w(p,\widetilde{q}).\end{align}

Since we have $B^{(0)}(F_kU)\subset F_k W $ , there is a map $\operatorname{Gr} B^{(0)}\,:\,\operatorname{Gr} U \to \operatorname{Gr} W$ on the associated graded, and we set $L\,:=\, \ker \big(\!\operatorname{Gr} B^{(0)}\big)$ . The filtration on U induces a filtration on $K\cap U$ , by setting $F_k(K \cap U) \,:=\, K \cap (F_k U)$ . By definition of K, the inclusion of $K\cap U$ in U induces an injective map

\begin{equation*}\kappa\,:\,\operatorname{Gr}(K \cap U) \longrightarrow L.\end{equation*}

Let I be the submodule of U generated by the elements (i),(ii),(iii),(iv) of the statement, and endow I with the filtration defined by $F_k I \,:=\, I \cap (F_k U)$ . The inclusion of I in $K \cap U$ induces an injective map

\begin{equation*}\iota\,:\,\operatorname{Gr} I \longrightarrow \operatorname{Gr}(K \cap U).\end{equation*}

We claim two things:

(3·21) \begin{align}&\hbox{the composition $\kappa_k \circ \iota_k:\operatorname{Gr}_k I \to L_k$ is surjective for $k\geq 4$;}\\[-8pt]\nonumber\end{align}

(3·22) \begin{align}& \hbox{we have $F_3 I =F_3( K\cap U)$.}\end{align}

Claim (3·21) will imply that $\iota_k$ is surjective for every $k\geq 4$ , from which we will deduce that $K\cap U=I + F_3(K\cap U)$ ; then, with claim (3·22), we will conclude that $K\cap U=I$ which will prove the lemma.

We first prove (3·21), fixing $k\in \{4,\dots,g\}$ . Since the maps $\iota$ and $\kappa$ are injective, the quotient modules $\operatorname{Gr}_k I=F_kI/F_{k-1} I$ and $\operatorname{Gr}_k (K \cap U) = F_k(K\cap U)/F_{k-1} (K \cap U)$ can be viewed as submodules of $\operatorname{Gr}_k U = F_k U/F_{k-1} U$ . Thus, claim (3·21) is equivalent to the injectivity of

(3·23) \begin{equation} \Phi\,:=\, \operatorname{Gr}_k B^{(0)}\,:\,\frac{\operatorname{Gr}_k U}{\operatorname{Gr}_k I} \longrightarrow \operatorname{Gr}_k W.\end{equation}

The quotient module $\operatorname{Gr}_k U$ is generated by the (classes of) elements $u(q,t,\widetilde{t})$ , u(q, r, t), $u'(t\,;\,q,r)$ , $u'(r\,;\,q,t)$ with $q,r,t \in S$ such that $N(t)=k$ , $N(q)\lt k$ , $N(r)\lt k$ . This induces a generating system

(3·24) \begin{align}\big\{\, \overline{u}(q,t,\widetilde{t})\ \big\vert \ q,t \in S,\ N(t)&=k,\ N(q)\lt k \, \big\}\\[-3pt]\nonumber\end{align}

(3·25) \begin{align}\cup \big\{\, \overline{u}(q,\widetilde{q},t)\ \big\vert \ q,t \in S,\ N(t)&=k,\ N(q)\lt k \, \big\}\\[-3pt]\nonumber\end{align}

(3·26) \begin{align}\cup \big\{\, \overline{u}(q,r,t)\ \big\vert \ q,r,t \in S,\ N(t)&=k,\ N(q)\lt k,\ N(r)\lt k,\ N(q)\neq N(r) \, \big\}\\[-3pt]\nonumber\end{align}

(3·27) \begin{align}\cup \big\{\, \overline{u'}(t\,;\,q,r)\ \big\vert \ q,r,t \in S,\ N(t)&=k,\ N(q)\lt k,\ N(r)\lt k,\ N(q)\neq N(r) \, \big\}\\[-3pt]\nonumber\end{align}

(3·28) \begin{align}\cup \big\{\, \overline{u'}(r\,;\,q,t)\ \big\vert \ q,r,t \in S,\ N(t)&=k,\ N(q)\lt k,\ N(r)\lt k,\ N(q)\neq N(r)\, \big\} \end{align}

of $\operatorname{Gr}_k U/\operatorname{Gr}_k I$ , which we reduce as follows:

1. (1) for any $t,q,r,s\in S$ such that $N(t)=k$ , $N(q)\lt k$ , $N(r)\lt k$ , $N(s)\lt k$ and N(q), N(r), N(s) are pairwise different, the element

   \begin{equation*}\mathsf{D}\!\left({q},{\widetilde{q}}\,;\,{t},{s}\,;\,{r},{\widetilde{r}}\right)=\varepsilon(t)\, u'(t\,;\,q,r) - \varepsilon(s)\, u'(s\,;\,q,r)\end{equation*}
   is of type (ii), which shows that $\overline{u'}(t\,;\,q,r)=0$ . Thus, all the generators of $\operatorname{Gr}_k U/\operatorname{Gr}_k I$ of type (3·27) are actually zero;

2. (2) for any $t,q,r\in S$ such that $N(t)=k$ , $N(q)\lt k$ , $N(r)\lt k$ and $N(q)\neq N(r)$ , the element

   \begin{equation*}\mathsf{IHX}_3\!\left({t},{q}\,;\,{r}\right) =\varepsilon(r)\, u'(q\,;\,t,r) + \varepsilon(q)\, u(r,\widetilde{r},t) - \varepsilon(r)\, u'(t\,;\,q,r) - \varepsilon(t)\, u(r,\widetilde{r},q)\end{equation*}
   is of type (iii), which (using the previous item) implies that
   (3·29) \begin{equation} \overline{u'}(q\,;\,t,r)= -\varepsilon(q)\varepsilon(r) \, \overline{u}(r,\widetilde{r},t).\end{equation}
   Thus, all the generators of type (3·28) are repeated in (3·25), so that they can be removed;

3. (3) for any $t,p,q\in S$ such that $N(t)=k$ , $N(p)\lt k$ , $N(q)\lt k$ and $N(p)\neq N(q)$ , the element

   \begin{equation*}\mathsf{IHX}_3\!\left({p},{q}\,;\,{t}\right)=\varepsilon(t)\,u'(q\,;\,p,t) + \varepsilon(q)\, u(t,\widetilde{t},p)-\varepsilon(t)\, u'(p\,;\,q,t)- \varepsilon(p)\, u(t,\widetilde{t},q)\end{equation*}
   is of type (iii), which (using (3·29)), implies that
   \begin{equation*} \overline{u}(t,\widetilde{t},p)=\varepsilon(t)\varepsilon(p)\, \overline{u}(p,\widetilde{p},t)-\varepsilon(t)\varepsilon(p)\, \overline{u}(q,\widetilde{q},t)+ \varepsilon(p)\varepsilon(q)\, \overline{u}(t,\widetilde{t},q).\end{equation*}
   Thus, setting $q\,:=\,a_1$ and $t\,:=\,a_k$ in the above equation, we see that the submodule of $\operatorname{Gr}_k U/\operatorname{Gr}_k I$ generated by (3·24) and (3·25) is actually generated by the single element $u(a_1,a_k,b_k)$ of (3·24) and (3·25);

4. (4) for any $t,q,r,s\in S$ such that $N(t)=k$ , $N(q)\lt k$ , $N(r)\lt k$ , $N(s)\lt k$ and N(q), N(r), N(s) are pairwise different, the element

   \begin{eqnarray*}\mathsf{IHX}'_3\!\left({t},{q}\,;\,{r},{s}\right)&= &\varepsilon(q)\, u(t,r,s)- \varepsilon(t)\, u(q,r,s) -\varepsilon(s)\, u(\widetilde{r},\widetilde{t},{q})\\&& - \varepsilon(r)\, u(s,\widetilde{t},q) - \varepsilon(r)\, u'(q\,;\,s,t) + \varepsilon(s)\, u'(t\,;\,q,r)\end{eqnarray*}
   is of type (iv), which (using (3·29)) implies that
   (3·30) \begin{equation} \varepsilon(q)\ \overline{u}(t,r,s) +\varepsilon(s)\, \overline{u}({r},{t},\widetilde{q})+ \varepsilon(r)\, \overline{u}(\widetilde{s},t,\widetilde{q})+\varepsilon(r)\varepsilon(s)\varepsilon(q)\, \overline{u}(s,\widetilde{s},t) =0.\end{equation}
   Thus, setting $(t,r)\,:=\,(a_k,a_1)$ in the above equation, we get
   \begin{equation*}\overline{u}(\widetilde{s},\widetilde{q},a_k) = -\varepsilon(s)\, \overline{u}(\widetilde{q},a_1,a_k)- \varepsilon(q)\, \overline{u}(s,a_1,a_k)- \varepsilon(s)\varepsilon(q)\, \overline{u}(s,\widetilde{s},a_k),\end{equation*}
   and, setting $(t,s)\,:=\,(a_k,a_1)$ in the same equation, we get
   \begin{equation*} \overline{u}(\widetilde{q},b_1,a_k)= -\varepsilon(q)\, \overline{u}(b_1,a_1,a_k)- \varepsilon(r)\, \overline{u}({r},\widetilde{q},a_k)-\varepsilon(q)\varepsilon(r)\, \overline{u}(r,a_1,a_k).\end{equation*}
   The above two equations show that, in the presence of (3·24) and (3·25), the generators of type (3·26) can be reduced to those of the form $\overline{u}(q,a_1,a_k)$ with $q\in S$ such that $1\lt N(q)\lt k$ . Furthermore, by setting $(q,r,s,t)\,:=\,(a_j,a_1,a_2,a_k)$ in (3·30) for every $j\in \{3,\dots,k-1\}$ , we obtain
   \begin{equation*}\overline{u}(b_j,a_1,a_k)=-\overline{u}(a_2,b_2,a_k) - \overline{u}(b_2,a_k,b_j)- \overline{u}(a_2,a_1,a_k)\end{equation*}
   and, by setting $(q,r,s,t)\,:=\,(a_2,a_1,a_j,a_k)$ in (3·30), we also get
   \begin{equation*}\overline{u}(b_2,a_k,b_j) = -\overline{u}(a_j,a_1,a_k) - \overline{u}(b_2,a_1,a_k) - \, \overline{u}(a_j,b_j,a_k).\end{equation*}
   We deduce that, in the presence of (3·24) and (3·25), the generators of type (3·26) can be reduced to those of the form $\overline{u}(b_2,a_1,a_k)$ and $\overline{u}(a_j,a_1,a_k)$ with $j\in \{2,\dots,k-1\}$ .

It follows from the above discussion that $\operatorname{Gr}_k U/\operatorname{Gr}_k I$ is generated by the following elements:

\begin{equation*}\overline{u}(a_1,a_k,b_k), \quad\overline{u}(a_i,b_i,a_k) \hbox{ with $i\lt k$}, \quad\overline{u}(b_2,a_1,a_k), \quad\overline{u}(a_j,a_1,a_k) \hbox{ with $1\lt j\lt k$}.\end{equation*}

For $p,q\in S$ with $N(p)\leq k, N(q)\leq k$ , let $\overline{w}(p,q)$ be the class of w(p,q) in $\operatorname{Gr}_k W$ . According to (3·19), the values of the map (3·23) on this generating set are as follows:

\begin{eqnarray*}\Phi\big( \overline{u}(a_1,a_k,b_k)\big)&=& \overline{w}(a_k,b_k) + \overline{w}(a_1,b_k)- \overline{w}(a_1,a_k),\\ \Phi\big( \overline{u}(a_i,b_i,a_k) \big)&=& \overline{w}(b_i,a_k) -\overline{w}(a_i,a_k), \\ \Phi\big( \overline{u}(b_2,a_1,a_k) \big) &=&-\overline{w}(a_1,a_k) + \overline{w}(b_2,a_k),\\ \Phi\big( \overline{u}(a_j,a_1,a_k) \big) &=&\overline{w}(a_1,a_k) + \overline{w}(a_j,a_k).\end{eqnarray*}

Using the fact that $\big\{\overline{w}(a_i,a_k)\,\vert\, i\lt k\big\}\cup \big\{\overline{w}(b_i,a_k)\,\vert\, i\lt k\big\}\cup \big\{\overline{w}(a_k,b_k)\big\}$ is a basis of $\operatorname{Gr}_k W$ , we easily check that the above values of $\Phi$ are $\mathbb{Q}$ -linearly independent. Claim (3·21) follows.

We now prove (3·22). The quotient $Q\,:=\,F_3 U /F_3 I$ is generated by the classes

(3·31) \begin{equation} \overline{u}(p,q,r) \quad \hbox{with $p,q,r \in F_3 S$ pairwise different}\end{equation}

and the classes

(3·32) \begin{equation} \overline{u'}(r\,;\,p,q)\quad \hbox{with $p,q,r \in F_3 S$ such that}\,{{N}(p),\,}\,{{N}(q),\,}\,{{N}({r})\,{\rm are\,pairwise\,different}}\end{equation}

of elements in $F_3U$ of type (3·17) and (3·18), respectively. For any $p,q,r \in F_3S$ such that N(p), N(q), N(r) are pairwise different, the element

\begin{equation*}\mathsf{D}\!\left({p},{\widetilde{p}}\,;\,{q},{\widetilde{q}}\,;\,{r},{\widetilde{r}}\right)= \varepsilon(q)\, u'(q\,;\,p,r) + \varepsilon(q)\, u'(\widetilde{q}\,;\,p,r)\end{equation*}

is of type (i), so that we have $\overline{u'}(q;\,p,r) = - \overline{u'}(\widetilde{q};\,p,r) \in Q$ . Thus, among the generators of Q of type (3·32), we can reduce ourselves to $\overline{u'}(a_1;\,a_2,a_3)$ , $\overline{u'}(a_2;\,a_1,a_3)$ , $\overline{u'}(a_3;\,a_1,a_2)$ . Furthermore, for any $p,q,r \in F_3S$ such that N(p), N(q), N(r) are pairwise different, the element

\begin{equation*}\mathsf{IHX}_3\!\left({p},{q}\,;\,{r}\right)=\varepsilon(r)\,u'(q;\,p,r) + \varepsilon(q)\, u(r,\widetilde{r},p)-\varepsilon(r)\, u'(p;\,q,r)- \varepsilon(p)\, u(r,\widetilde{r},q)\end{equation*}

is of type (iii), so that $\varepsilon(q)\, \overline{u}(r,\widetilde{r},p)= -\varepsilon(r)\,\overline{u'}(q;\,p,r)+\varepsilon(r)\, \overline{u'}(p;\,q,r)+ \varepsilon(p)\, \overline{u}(r,\widetilde{r},q)\in Q$ ; hence, among the generators of Q of type (3·31), we can remove $\overline{u}(a_1,b_1,a_2)$ , $\overline{u}(a_2,b_2,a_3)$ and $\overline{u}(a_3,b_3,a_1)$ . Consequently, the module Q is generated by the following 10 elements:

\begin{eqnarray*}&\overline{u}(a_1,a_2,a_3), \overline{u}(b_1,a_2,a_3),\ \overline{u}(a_1,b_2,a_3),\ \overline{u}(a_1,a_2,b_3), \\ &\overline{u}(a_1,b_1,a_3),\ \overline{u}(a_2,b_2,a_1),\ \overline{u}(a_3,b_3,a_2), \overline{u'}(a_1;\,a_2,a_3),\ \overline{u'}(a_2;\,a_1,a_3),\ \overline{u'}(a_3\,;\,a_1,a_2).\end{eqnarray*}

According to (3·19) and (3·20), the values of the map $\overline{B}:Q \to F_3 W \oplus \mathbb{Q}$ (which is induced by $B:F_3 U \to F_3 W \oplus \mathbb{Q}$ ) on this generating set are as follows:

\begin{eqnarray*}\overline{B}\big(\overline{u}(a_1,a_2,a_3) \big) &=&\big(w(a_2,a_3)+ w(a_1,a_3)+w(a_1,a_2) \,,\, -1/4 \big) \\ \overline{B}\big(\overline{u}(b_1,a_2,a_3)\big) &=&\big(-w(a_2,a_3)+w(a_1,b_3)+w(a_1,b_2) \,,\, +1/4 \big) \\ \overline{B}\big(\overline{u}(a_1,b_2,a_3)\big) &=&\big(w(a_2,b_3)-w(a_1,a_3)+w(a_1,b_2) \,,\, +1/4 \big) \\ \overline{B}\big(\overline{u}(a_1,a_2,b_3)\big) &=&\big(w(a_2,b_3)+w(a_1,b_3)-w(a_1,a_2) \,,\, +1/4\big) \\ \overline{B}\big(\overline{u}(a_1,b_1,a_3)\big) &=&\big(w(a_1,b_3) -w(a_1,a_3)+w(a_1,b_1) \,,\, +1/4 \big) \\ \overline{B}\big(\overline{u}(a_2,b_2,a_1)\big) &=&\big(w(a_1,b_2)-w(a_1,a_2)+ w(a_2,b_2)\,,\, +1/4 \big) \\ \overline{B}\big(\overline{u}(a_3,b_3,a_2)\big) &=&\big(w(a_2,b_3) -w(a_2,a_3)+ w(a_3,b_3) \,,\, +1/4 \big) \\ \overline{B}\big(\overline{u'}(a_1\,;\,a_2,a_3)\big) &=&\big(w(a_2,a_3)-w(a_2,b_3) \,,\, 0\big) \\ \overline{B}\big(\overline{u'}(a_2\,;\,a_1,a_3)\big) &=&\big(w(a_1,a_3)-w(a_1,b_3) \,,\, 0\big) \\ \overline{B}\big(\overline{u'}(a_3\,;\,a_1,a_2)\big) &=&\big(w(a_1,a_2)-w(a_1,b_2) \,,\, 0\big).\end{eqnarray*}

Using that $\big\{w(a_i,a_j)\,\vert\, 1 \leq i \lt j \leq 3 \big\}\cup \big\{w(a_i,b_i)\,\vert\, 1 \leq i \leq 3 \big\}\cup \big\{w(a_i,b_j)\,\vert\, 1 \leq i \lt j \leq 3 \big\}$ is a basis of the $\mathbb{Q}$ -vector space $F_3W$ , it is easily checked that the above vectors of $F_3W\oplus \mathbb{Q}$ are $\mathbb{Q}$ -linearly independent. Therefore, the map $\overline{B}$ is injective.

Proof. Let $\eta \,:=\, \varepsilon(c_1)\, \varepsilon(c_2)$ . We claim that

\begin{equation*}C\,:=\, \mathsf{\bigwedge}\!\left({x},{y},{c_1+c_2}\,;\,{-\eta\, \widetilde{c_1} + \widetilde{c_2}},{\widetilde{u}},{\widetilde{v}}\right)\in \ker(J).\end{equation*}

Since $\mathsf{\bigwedge}\!\left({x},{y},{c_1}\,;\,{\widetilde{c_2}},{\widetilde{u}},{\widetilde{v}}\right)$ and $\mathsf{\bigwedge}\!\left({x},{y},{c_2}\,;\,{\widetilde{c_1}},{\widetilde{u}},{\widetilde{v}}\right)$ belong to $\ker(J)$ by Lemma 3·14, we deduce that $-\eta\,\mathsf{\bigwedge}\!\left({x},{y},{c_1}\,;\,{\widetilde{c_1}},{\widetilde{u}},{\widetilde{v}}\right)+\mathsf{\bigwedge}\!\left({x},{y},{c_2}\,;\,{\widetilde{c_2}},{\widetilde{u}},{\widetilde{v}}\right)= -\varepsilon(c_2)\, \mathsf{D}\!\left({x},{y}\,;\,{c_1},{c_2}\,;\,{\widetilde{u}},{\widetilde{v}}\right)$ does too.

We now prove the above claim. If ${c_2} \notin \{\widetilde{x},\widetilde{y}\}$ and $c_1\not\in \{\widetilde{u},\widetilde{v}\}$ , then we consider the symplectic transformation T that maps $(c_1,\widetilde{c_1},c_2,\widetilde{c_2})$ to $(c_1+c_2,\widetilde{c_1},c_2,-\eta\, \widetilde{c_1}+\widetilde{c_2})$ and fixes all other elements of S; then T maps

\begin{equation*}\mathsf{\bigwedge}\!\left({x},{y},{c_1}\,;\,{\widetilde{c_2}},{\widetilde{u}},{\widetilde{v}}\right) \in \ker(J)\end{equation*}

to C, so that $C\in \ker(J)$ . If we have ${c_1} \notin \{\widetilde{x},\widetilde{y}\}$ and $c_2\not\in \{\widetilde{u},\widetilde{v}\}$ , we can proceed similarly by exchanging the roles of $c_1$ and $c_2$ .

We now assume that none of the above two conditions occur, i.e $.$ we have $\{c_1,c_2\}=\{\widetilde{x},\widetilde{y}\}$ or (exclusively) $\{c_1,c_2\}=\{\widetilde{u},\widetilde{v}\}$ . For instance, we assume that $(c_1,c_2)=(\widetilde{x},\widetilde{y})$ , the other cases being treated similarly. Then

\begin{equation*}C=\mathsf{\bigwedge}\!\left({\widetilde{c_1}},{-\eta \widetilde{c_1}+\widetilde{c_2}},{c_1+c_2}\,;\,{-\eta \widetilde{c_1}+\widetilde{c_2}},{\widetilde{u}},{\widetilde{v}}\right)= T\cdot \mathsf{\bigwedge}\!\left({\widetilde{c_1}},{\widetilde{c_2}},{c_1}\,;\,{\widetilde{c_2}},{\widetilde{u}},{\widetilde{v}}\right),\end{equation*}

and we deduce from Lemma 3·15 that C belongs to $\ker(J)$ in the present case too.

Proof. Let T be the symplectic transformation that maps c to $c+\widetilde{c}$ and fixes all other elements of S. We have

\begin{align*} &T \cdot \mathsf{\bigwedge}\!\left({x},{y},{c}\,;\,{c},{\widetilde{u}},{\widetilde{v}}\right) -\mathsf{\bigwedge}\!\left({x},{y},{c}\,;\,{c},{\widetilde{u}},{\widetilde{v}}\right)-\mathsf{\bigwedge}\!\left({x},{y},{\widetilde{c}}\,;\,{\widetilde{c}},{\widetilde{u}},{\widetilde{v}}\right) \\&\quad = \varepsilon(c)\, \mathsf{D}\!\left({x},{y}\,;\,{c},{\widetilde{c}}\,;\,{\widetilde{u}},{\widetilde{v}}\right)\end{align*}

and we conclude with Lemma 3·14.

Proof. This has been proved by Gervais and Habegger [ Reference Gervais and Habegger7 ] but, for the sake of completeness, we repeat their arguments (with some variations). We focus on a subsurface of $\Sigma$ of genus 4 which contains the based loops $\alpha_1,\beta_1,\dots,\alpha_4,\beta_4$ of Figure 1. This subsurface (union a rectangle) is shown in Figure 3 as the cylinder $D\times [0,1]$ over a 4-holed disk D: the curves X, Y, Z (shown in Figure 3) lie on $D \times \{1\}$ and we denote by $X_0,Y_0,Z_0$ the corresponding curves on $D \times \{0\}$ (not shown in Figure 3).

Figure 3.

The curves X,Y,Z and U,V,W on $\partial(D\times [0,1])$ .

In the rest of this proof, the Dehn twist $T_C$ along a simple closed curve C is simply denoted by C and, if C is oriented, the homology class $\{C\}\in H$ is denoted by the lower case roman letter c. By the Hall–Witt identity, we know that

\begin{equation*}\big[{}^ZX,[Y,Z]\big] \,\big[{}^YZ,[X,Y]\big] \,\big[{}^XY,[Z,X]\big] = 1 \ \in \mathcal{M}.\end{equation*}

and, since the Dehn twists along $X_0,Y_0,Z_0$ commute with those along X, Y, Z, we obtain that

\begin{equation*}\big[{}^Z(XX_0^{-1}),[Y,Z]\big] \,\big[{}^Y(ZZ_0^{-1}),[X,Y]\big] \,\big[{}^X(YY_0^{-1}),[Z,X]\big] = 1 \ \in \mathcal{M}.\end{equation*}

Note that $XX_0^{-1},YY_0^{-1},ZZ_0^{-1}$ are BP-maps (of genus 1) while [Y, Z], [X, Y], [Z, X] are PB-maps: hence the above identity gives a degree 2 relation in $\operatorname{Gr}^{\Gamma} \mathcal{I}$ . (See [ Reference Putman29 ] for a systematic use of Hall–Witt identities in presentations of the group $\mathcal{I}$ .) Consequently, the sum

\begin{align*}&\big(Z_* \cdot \tau_1\big(XX_0^{-1}\big)\big) \wedge \tau_1([Y,Z])\\ &\quad+\big(Y_* \cdot \tau_1\big(ZZ_0^{-1}\big)\big) \wedge \tau_1([X,Y])\\&\quad+\big(X_* \cdot \tau_1\big(YY_0^{-1}\big)\big) \wedge \tau_1([Z,X])\end{align*}

belongs to $\ker(J)$ . (Here $X_*\in \operatorname{Sp}\!(H)$ denotes the action of the Dehn twist along X.) Using (3·33) and (3·34), we obtain

\begin{eqnarray*} Z_* \cdot \tau_1\big(XX_0^{-1}\big)&=& \varepsilon\, b_1 \wedge Z_*(u) \wedge b_4 \\& = & \varepsilon\, b_1 \wedge (u+z) \wedge b_4 \\& = & \varepsilon\, b_1 \wedge ((a_1-a_4)+(b_3+b_4)) \wedge b_4\ = \ \varepsilon\, b_1 \wedge ( a_1-a_4 +b_3 ) \wedge b_4\end{eqnarray*}

and

\begin{equation*}\tau_1([Y,Z]) = \eta\, b_2 \wedge b_4 \wedge b_3\end{equation*}

respectively. (Here $\varepsilon$ and $\eta$ are fixed signs, which we do not need to determine.) Next, it follows from Lemma 3·14 that

\begin{equation*}\big(Z_* \cdot \tau_1\big(XX_0^{-1}\big)\big) \wedge \tau_1([Y,Z])\equiv - \varepsilon \eta \,\mathsf{\bigwedge}\!\left({b_1},{a_4},{b_4}\,;\,{b_2},{b_4},{b_3}\right)\mod \ker(J);\end{equation*}

similarly, we obtain

\begin{equation*}\big(Y_* \cdot \tau_1\big(ZZ_0^{-1}\big)\big) \wedge \tau_1([X,Y])\equiv - \varepsilon \eta \,\mathsf{\bigwedge}\!\left({b_3},{a_4},{b_4}\,;\,{b_1},{b_4},{b_2}\right)\mod \ker(J)\end{equation*}

and

\begin{equation*}\big(X_* \cdot \tau_1\big(YY_0^{-1}\big)\big) \wedge \tau_1([Z,X])\equiv - \varepsilon \eta \,\mathsf{\bigwedge}\!\left({b_2},{a_4},{b_4}\,;\,{b_3},{b_4},{b_1}\right)\mod \ker(J).\end{equation*}

Therefore the element

\begin{equation*}\underbrace{\mathsf{\bigwedge}\!\left({b_4},{b_1},{a_4}\,;\,{b_4},{b_3},{b_2}\right)+ \mathsf{\bigwedge}\!\left({b_4},{b_3},{a_4}\,;\,{b_4},{b_2},{b_1}\right)+ \mathsf{\bigwedge}\!\left({b_4},{b_2},{a_4}\,;\,{b_4},{b_1},{b_3}\right)}_{\mathsf{IHX}_1\!\left({b_4};{b_1},{b_3},{b_2}\,;\,{a_4}\right)}\end{equation*}

belongs to $\ker(J)$ , and we conclude using the G-action.

Proof. For all $h\in \{3,\dots, g\}$ , the element $\sum_{i=1}^{h-1} \mathsf{\bigwedge}\!\left({a_1},{a_2},{b_2}\,;\,{a_h},{a_i},{b_i}\right)$ of $\Lambda^2(\Lambda^3 H)$ belongs to $\ker(J)$ , since ${a_1}\wedge{a_2}\wedge{b_2}$ and $\sum_{i=1}^{h-1} {a_h}\wedge{a_i}\wedge{b_i}$ can be realised by BP maps (of genus 1 and $h-1$ , respectively) whose supports are disjoint one from the other. Setting $h\,:=\,3$ , we obtain that

(3·35) \begin{equation} W\,:=\, \mathsf{\bigwedge}\!\left({a_1},{a_2},{b_2}\,;\,{a_3},{a_1},{b_1}\right)+ \mathsf{\bigwedge}\!\left({a_1},{a_2},{b_2}\,;\,{a_3},{a_2},{b_2}\right)\end{equation}

belongs to $\ker(J)$ . The element of $G \simeq \mathbb{Z}_4^g \rtimes \mathfrak{S}_g$ corresponding to the transposition (23) maps W to $\mathsf{IHX}_2\!\left({a_1},{a_2}\,;\,{a_3}\,;\,{b_1}\right)$ which, therefore, is in $\ker(J)$ too.

Next, the element of G corresponding to the cycle (1423) transforms W to

\begin{equation*}E \,:=\, \mathsf{\bigwedge}\!\left({a_4},{a_3},{b_3}\,;\,{a_1},{a_3},{b_3}\right) + \mathsf{\bigwedge}\!\left({a_4},{a_3},{b_3}\,;\,{a_1},{a_4},{b_4}\right)\end{equation*}

which, therefore, belongs to $\ker(J)$ . Let also T be the symplectic transformation that maps $(a_4,b_2)$ to $( a_4 + a_2 , b_2 - b_4)$ and leaves all other elements of S fixed. We have

\begin{eqnarray*} {T} \cdot E - E & \equiv & \mathsf{\bigwedge}\!\left({a_2},{a_3},{b_3}\,;\,{a_1},{a_3},{b_3}\right) + \mathsf{\bigwedge}\!\left({a_4},{a_3},{b_3}\,;\,{a_1},{a_2},{b_4}\right) \mod K \cap U_0. \end{eqnarray*}

Hence, using Lemma 3·14 and the G-action, we obtain that

\begin{equation*}\mathsf{IHX}_2\!\left({a_1},{a_2}\,;\,{a_3}\,;\,{a_4}\right) =\mathsf{\bigwedge}\!\left({a_1},{a_3},{b_3}\,;\,{a_2},{a_3},{b_3}\right)- \mathsf{\bigwedge}\!\left({a_1},{a_2},{a_4}\,;\,{a_3},{b_3},{b_4}\right)\ \in \ker(J).\end{equation*}

Proof. Consider the element $W\in \ker(J)$ defined by (3·35), and let $T_0$ be the symplectic transformation that maps $(a_2,a_3)$ to $(a_2+b_3,a_3+b_2)$ . We have

\begin{equation*}T_0\cdot W-W = \mathsf{S}\!\left({a_1},{b_2}\,;\begin{array}{cc}{b_2} & {b_1}\!\\{b_3} & {a_3}\! \end{array}\right)+ \mathsf{D}\!\left({b_3},{b_2}\,;\,{a_1},{a_3}\,;\,{b_2},{a_1}\right)+ W'\end{equation*}

where $W'\,:=\,\mathsf{\bigwedge}\!\left({a_1},{a_2},{b_2}\,;\,{a_3},{b_3},{b_2}\right)-\mathsf{\bigwedge}\!\left({a_1},{b_3},{a_3}\,;\,{b_2},{a_3},{b_3}\right)$ . But, an appropriate element of G transforms W ^′ to W, and we deduce from Lemma 3·15 that $\mathsf{S}\!\left({a_1},{b_2}\,;\begin{array}{cc}{b_2} & {b_1}\!\\{b_3} & {a_3}\! \end{array}\right)$ belongs to $\ker(J)$ . Using again the G-action, we conclude that $\mathsf{S}\!\left({a_1},{a_2}\,;\begin{array}{cc}{a_2} & {b_1}\!\\{a_3} & {b_3}\! \end{array}\right)$ is in $\ker(J)$ too.

By Lemma 3·15, $D\,:=\,\mathsf{D}\!\left({a_1},{a_3}\,;\,{a_2},{a_4}\,;\,{a_2},{a_4}\right)$ belongs to $\ker(J)$ . For the symplectic transformation $T_1$ that maps $(a_3,a_4)$ to $(a_3+b_4,a_4+b_3)$ and fixes all other elements of S, we have

\begin{eqnarray*}T_1\cdot D -D & \equiv & D_1 \mod (K\cap U_0 + K\cap U_1),\end{eqnarray*}

where

\begin{eqnarray*}D_1 &\,:=\,&\mathsf{\bigwedge}\!\left({a_1},{a_3},{a_2}\,;\,{b_2},{a_2},{b_3}\right) + \mathsf{\bigwedge}\!\left({a_1},{b_4},{a_2}\,;\,{b_2},{a_2},{a_4}\right) \\&& \quad - \mathsf{\bigwedge}\!\left({a_1},{a_3},{a_4}\,;\,{b_4},{a_2},{b_3}\right) - \mathsf{\bigwedge}\!\left({a_1},{b_4},{a_4}\,;\,{b_4},{a_2},{a_4}\right)\ = \ -\mathsf{S}\!\left({a_1},{a_2}\,;\begin{array}{cc}{a_2} & {a_3}\!\\{b_4} & {a_4}\! \end{array}\right).\end{eqnarray*}

Hence, by Lemma 3·14 and Lemma 3·18, $D_1$ belongs to $\ker(J)$ . Using the G-action, we deduce that $\mathsf{S}\!\left({a_1},{a_2}\,;\begin{array}{cc}{a_2} & {a_4}\!\\{a_3} & {b_3}\! \end{array}\right)$ does too.

Proceeding with the same strategy, we consider the symplectic transformation $T_3$ mapping $(a_2,b_5)$ to $(a_2+a_5,b_5-b_2)$ and fixing all other elements of S. We have

\begin{eqnarray*}&& T_3\cdot \mathsf{S}\!\left({a_1},{a_2}\,;\begin{array}{cc}{a_2} & {a_4}\!\\{a_3} & {b_3}\! \end{array}\right)- \mathsf{S}\!\left({a_1},{a_2}\,;\begin{array}{cc}{a_2} & {a_4}\!\\{a_3} & {b_3}\! \end{array}\right) \\=&& D_3 + \mathsf{D}\!\left({a_1},{a_5}\,;\,{a_3},{a_4}\,;\,{a_2},{b_2}\right)+ \mathsf{D}\!\left({a_1},{a_5}\,;\,{a_3},{a_4}\,;\,{a_5},{b_2}\right),\end{eqnarray*}

where

\begin{eqnarray*}D_3 &\,:=\,& - \mathsf{\bigwedge}\!\left({a_1},{a_2},{a_4}\,;\,{a_5},{b_2},{b_4}\right)-\mathsf{\bigwedge}\!\left({a_1},{a_4},{b_3}\,;\,{a_5},{b_4},{a_3}\right)\\&& \quad + \mathsf{\bigwedge}\!\left({a_1},{a_3},{b_3}\,;\,{a_5},{b_3},{a_3}\right)+ \mathsf{\bigwedge}\!\left({a_1},{a_2},{a_3}\,;\,{a_5},{b_2},{b_3}\right)\ = \ \mathsf{S}\!\left({a_1},{a_5}\,;\begin{array}{cc}{a_2} & {a_4}\!\\{a_3} & {b_3}\! \end{array}\right).\end{eqnarray*}

Therefore, using the G-action, we deduce that $\mathsf{S}\!\left({a_1},{a_2}\,;\begin{array}{cc}{a_4} & {a_5}\!\\{a_3} & {b_3}\! \end{array}\right)$ is in $\ker(J)$ .

We continue with the symplectic transformation $T_4$ mapping $(a_2,b_4)$ to $(a_2+a_4,b_4-b_2)$ and fixing all other elements of S. We have

\begin{eqnarray*}&& T_4\cdot \mathsf{S}\!\left({a_1},{a_2}\,;\begin{array}{cc}{a_2} & {b_1}\!\\{a_3} & {b_3}\! \end{array}\right)- \mathsf{S}\!\left({a_1},{a_2}\,;\begin{array}{cc}{a_2} & {b_1}\!\\{a_3} & {b_3}\! \end{array}\right) \\=&& D_4- \mathsf{D}\!\left({a_1},{a_4}\,;\,{a_3},{b_1}\,;\,{a_2},{b_2}\right)- \mathsf{D}\!\left({a_1},{a_4}\,;\,{a_3},{b_1}\,;\,{a_4},{b_2}\right),\end{eqnarray*}

where

\begin{eqnarray*}D_4 &\,:=\,& - \mathsf{\bigwedge}\!\left({a_1},{a_2},{b_1}\,;\,{a_4},{b_2},{a_1}\right)-\mathsf{\bigwedge}\!\left({a_1},{b_1},{b_3}\,;\,{a_4},{a_1},{a_3}\right)\\&& \quad - \mathsf{\bigwedge}\!\left({a_1},{a_3},{b_3}\,;\,{a_4},{b_3},{a_3}\right)- \mathsf{\bigwedge}\!\left({a_1},{a_2},{a_3}\,;\,{a_4},{b_2},{b_3}\right)\ = \ \mathsf{S}\!\left({a_1},{a_4}\,;\begin{array}{cc}{a_2} & {b_1}\!\\{a_3} & {b_3}\! \end{array}\right).\end{eqnarray*}

Therefore, using the G-action, we deduce that $\mathsf{S}\!\left({a_1},{a_2}\,;\begin{array}{cc}{b_1} & {a_4}\!\\{a_3} & {b_3}\! \end{array}\right)$ is in $\ker(J)$ .

Finally, using again the above transformation $T_1$ , an elementary computation shows that the difference $ T_1\cdot \mathsf{\bigwedge}\!\left({a_1},{a_3},{a_4}\,;\,{a_2},{a_3},{a_4}\right)- \mathsf{\bigwedge}\!\left({a_1},{a_3},{a_4}\,;\,{a_2},{a_3},{a_4}\right)$ decomposes as

\begin{align*} &\mathsf{\bigwedge}\!\left({a_1},{b_4},{b_3}\,;\,{a_2},{a_3},{a_4}\right)+\mathsf{\bigwedge}\!\left({a_1},{b_4},{a_4}\,;\,{a_2},{b_4},{a_4}\right)\\&\quad+\mathsf{\bigwedge}\!\left({a_1},{a_3},{b_3}\,;\,{a_2},{a_3},{b_3}\right)+\mathsf{\bigwedge}\!\left({a_1},{a_3},{a_4}\,;\,{a_2},{b_4},{b_3}\right)\end{align*}

plus a sum of 11 terms which belongs to $K\cap U_0 + K\cap U_1$ . Using Lemma 3·14 and Lemma 3·18, we conclude that $\mathsf{S}\!\left({a_1},{a_2}\,;\begin{array}{cc}{a_4} & {b_4}\!\\{a_3} & {b_3}\! \end{array}\right)$ belongs to $\ker(J)$ .

Proof. Let T be the symplectic transformation that maps $(a_2,b_1)$ to $(a_2 + a_1, b_1 - b_2)$ and leaves all other elements of S fixed. We have

\begin{eqnarray*}&&T \cdot \mathsf{S}\!\left({a_1},{a_2}\,;\begin{array}{cc}{a_2} & {a_4}\!\\{a_3} & {b_3}\! \end{array}\right) - \mathsf{S}\!\left({a_1},{a_2}\,;\begin{array}{cc}{a_2} & {a_4}\!\\{a_3} & {b_3}\! \end{array}\right)\\=&&\mathsf{\bigwedge}\!\left({a_1},{a_2},{a_3}\,;\,{a_1},{b_2},{b_3}\right)- \mathsf{\bigwedge}\!\left({a_1},{a_2},{a_4}\,;\,{a_1},{b_2},{b_4}\right)- \mathsf{\bigwedge}\!\left({a_1},{a_4},{b_3}\,;\,{a_1},{b_4},{a_3}\right),\end{eqnarray*}

which is $-\mathsf{T}\!\left({a_1}\,;\,{a_2} ,{a_4},{a_3}\right)$ . Using the G-action, we obtain that $\mathsf{T}\!\left({a_1}\,;\,{a_2} ,{a_3},{a_4}\right) \in \ker(J)$ .

Proof. Let $T_1$ be the symplectic transformation that maps $(a_1,a_2)$ to $( a_1 + b_2,a_2 +b_1)$ , and set

\begin{eqnarray*}E_1 &\,:=\,& \mathsf{\bigwedge}\!\left({b_2},{a_3},{b_3}\,;\,{b_1},{a_3},{b_3}\right)+ \mathsf{\bigwedge}\!\left({b_2},{a_2},{b_1}\,;\,{a_3},{b_3},{b_2}\right),\\F_1 &\,:=\,& \mathsf{\bigwedge}\!\left({a_1},{a_3},{b_3}\,;\,{b_1},{a_3},{b_3}\right)+ \mathsf{\bigwedge}\!\left({b_2},{a_3},{b_3}\,;\,{a_2},{a_3},{b_3}\right) \\&&+ \ \mathsf{\bigwedge}\!\left({a_1},{a_3},{b_3}\,;\,{b_1},{a_2},{b_2}\right) + \mathsf{\bigwedge}\!\left({b_2},{a_3},{b_3}\,;\,{a_1},{b_1},{a_2}\right).\end{eqnarray*}

We have

\begin{eqnarray*} T_1 \cdot \mathsf{IHX}_2\!\left({a_1},{a_2}\,;\,{a_3}\,;\,{b_1}\right) - \mathsf{IHX}_2\!\left({a_1},{a_2}\,;\,{a_3}\,;\,{b_1}\right)&= & E_1 + F_1 .\end{eqnarray*}

Since $E_1$ belongs to $K \cap(V_{1,1}+V_2)=K \cap U_2$ , it follows from Lemma 3·19 and Lemma 3·22 that $F_1$ belongs to $\ker\!(J)$ . Besides, we have

\begin{equation*}\mathsf{IHX}_3\!\left({a_1},{a_2}\,;\,{a_3}\right)= - F_1 +\underbrace{\mathsf{\bigwedge}\!\left({a_3},{b_3},{a_1}\,;\,{b_1},{a_2},{b_2}\right)+ \mathsf{\bigwedge}\!\left({a_3},{b_3},{b_1}\,;\,{a_1},{a_2},{b_2}\right)}_{\mathsf{D}\left({a_3},{b_3}\,;\,{a_1},{b_1}\,;\,{a_2},{b_2}\right)}.\end{equation*}

Thus, an application of Lemma 3·16 shows that J vanishes on $\mathsf{IHX}_3\!\left({a_1},{a_2}\,;\,{a_3}\right)$ .

Finally, consider the symplectic transformation $T_2$ that maps $(a_2,a_3)$ to $(a_2 + b_3,a_3 +b_2)$ and fixes all other elements of S. We have

\begin{equation*}T_2\cdot \mathsf{S}\!\left({a_3},{a_2}\,;\begin{array}{cc}{a_2} & {a_4}\!\\{a_1} & {b_1}\! \end{array}\right)- \mathsf{S}\!\left({a_3},{a_2}\,;\begin{array}{cc}{a_2} & {a_4}\!\\{a_1} & {b_1}\! \end{array}\right)\equiv E_2 \mod (K\cap V_1+K\cap V_2),\end{equation*}

where

\begin{eqnarray*}E_2& \,:=\, &-\mathsf{\bigwedge}\!\left({b_2},{a_2},{a_4}\,;\,{a_2},{b_2},{b_4}\right)-\mathsf{\bigwedge}\!\left({a_3},{b_3},{a_4}\,;\,{a_2},{b_2},{b_4}\right)-\mathsf{\bigwedge}\!\left({a_3},{a_2},{a_4}\,;\,{b_3},{b_2},{b_4}\right)\\&& + \mathsf{\bigwedge}\!\left({b_2},{a_2},{a_1}\,;\,{a_2},{b_2},{b_1}\right)+ \mathsf{\bigwedge}\!\left({a_3},{b_3},{a_1}\,;\,{a_2},{b_2},{b_1}\right)+ \mathsf{\bigwedge}\!\left({a_3},{a_2},{a_1}\,;\,{b_3},{b_2},{b_1}\right)\\&& -\mathsf{\bigwedge}\!\left({b_2},{a_4},{b_1}\,;\,{a_2},{b_4},{a_1}\right) - \mathsf{\bigwedge}\!\left({a_3},{a_4},{b_1}\,;\,{b_3},{b_4},{a_1}\right)\\&& +\mathsf{\bigwedge}\!\left({b_2},{a_1},{b_1}\,;\,{a_2},{b_1},{a_1}\right) + \mathsf{\bigwedge}\!\left({a_3},{a_1},{b_1}\,;\,{b_3},{b_1},{a_1}\right).\end{eqnarray*}

It follows from Lemma 3·18 and Lemma 3·22 that $E_2$ is in $\ker(J)$ . Next, we observe that:

\begin{eqnarray*}E_2&=& \mathsf{IHX}'_3\!\left({a_1},{a_4}\,;\,{a_2},{a_3}\right)+ \mathsf{\bigwedge}\!\left({a_3},{b_3},{a_4}\,;\,{b_4},{a_1},{b_1}\right) - \mathsf{\bigwedge}\!\left({a_4},{b_4},{a_1}\,;\,{b_1},{a_2},{b_2}\right)\\&& -\mathsf{\bigwedge}\!\left({b_2},{a_2},{a_4}\,;\,{a_2},{b_2},{b_4}\right)+ \mathsf{\bigwedge}\!\left({b_2},{a_2},{a_1}\,;\,{a_2},{b_2},{b_1}\right) \\&& +\mathsf{\bigwedge}\!\left({b_2},{a_1},{b_1}\,;\,{a_2},{b_1},{a_1}\right)+ \mathsf{\bigwedge}\!\left({a_3},{a_1},{b_1}\,;\,{b_3},{b_1},{a_1}\right) \\&& -\mathsf{\bigwedge}\!\left({a_3},{b_3},{a_4}\,;\,{a_2},{b_2},{b_4}\right)+ \mathsf{\bigwedge}\!\left({a_3},{b_3},{a_1}\,;\,{a_2},{b_2},{b_1}\right) \\&=& \mathsf{IHX}'_3\!\left({a_1},{a_4}\,;\,{a_2},{a_3}\right)+ \mathsf{\bigwedge}\!\left({a_3},{b_3},{a_4}\,;\,{b_4},{a_1},{b_1}\right) - \mathsf{\bigwedge}\!\left({a_4},{b_4},{a_1}\,;\,{b_1},{a_2},{b_2}\right)\\&& + \mathsf{IHX}_3\!\left({a_1},{a_4}\,;\,{a_2}\right) - \mathsf{\bigwedge}\!\left({a_1},{b_1},{a_4}\,;\,{b_4},{a_2},{b_2}\right)+ \mathsf{\bigwedge}\!\left({a_4},{b_4},{a_1}\,;\,{b_1},{a_2},{b_2}\right)\\&& -\mathsf{IHX}_3\!\left({a_2},{a_3}\,;\,{a_1}\right)+\mathsf{\bigwedge}\!\left({a_2},{b_2},{a_3}\,;\,{b_3},{a_1},{b_1}\right)-\mathsf{\bigwedge}\!\left({a_3},{b_3},{a_2}\,;\,{b_2},{a_1},{b_1}\right)\\&& -\mathsf{\bigwedge}\!\left({a_3},{b_3},{a_4}\,;\,{a_2},{b_2},{b_4}\right)+ \mathsf{\bigwedge}\!\left({a_3},{b_3},{a_1}\,;\,{a_2},{b_2},{b_1}\right) \\&=& \mathsf{IHX}'_3\!\left({a_1},{a_4}\,;\,{a_2},{a_3}\right)+ \mathsf{IHX}_3\!\left({a_1},{a_4}\,;\,{a_2}\right) -\mathsf{IHX}_3\!\left({a_2},{a_3}\,;\,{a_1}\right)\\&& + \mathsf{D}\!\left({a_3},{b_3}\,;\,{a_4},{a_2}\,;\,{a_1},{b_1}\right)+ \mathsf{D}\!\left({a_2},{b_2}\,;\,{a_3},{b_4}\,;\,{a_1},{b_1}\right)\\&&+\mathsf{D}\!\left({a_3},{b_3}\,;\,{a_1},{a_4}\,;\,{a_2},{b_2}\right).\end{eqnarray*}

From the previous paragraph, we know that $\mathsf{IHX}_3\!\left({a_1},{a_2}\,;\,{a_3}\right)$ belongs to $\ker(J)$ : so, by the G-action, $\mathsf{IHX}_3\!\left({a_1},{a_4}\,;\,{a_2}\right)$ and $\mathsf{IHX}_3\!\left({a_2},{a_3}\,;\,{a_1}\right)$ do too. Then, it follows from Lemma 3·15 and the above decomposition of $E_2$ that $\mathsf{IHX}'_3\!\left({a_1},{a_4}\,;\,{a_2},{a_3}\right)$ is in $\ker(J)$ : using the G-action, so is $\mathsf{IHX}'_3\!\left({a_1},{a_2}\,;\,{a_3},{a_4}\right)$ .