We fix a splitting $\rep$ of $\mult$ as a map of $k$-vector spaces:

\[\begin{tikzcd}[ampersand replacement = \&, column sep = 6em]
\ar[r, twoheadrightarrow, "{ \mult}"]
\&
\ar[l, bend left = 50, dashed, "{ \rep}"']
\end{tikzcd}\]

Let $M'$ be the sub-$A$-bimodule of $M$ generated by the elements $\rep(a_i n_l) - a_i \rep(n_l)$ and by the elements $\rep(n_l a_i) - \rep(n_l) a_i$ for $i \in \mathbf{I}$ and $l \in \mathbf{L}$. The maps $\mult$ and $\rep$ define maps $\overline{\mult}$ and $\overline{\rep}$:

\[\begin{tikzcd}[ampersand replacement = \&, column sep = 9em]
\quoz{M}{M'}
\ar[r, twoheadrightarrow, "{ \overline{\mult}}", "{\overline{m} \mapsto \mult(m)}"']
\&
\ar[l, bend left = 50, dashed, "{ \overline{\rep}}"', "\overline{\rep(n)} \mapsfrom n"]
\end{tikzcd}\]

with $\overline{\rep}$ being a splitting of $\overline{\mult}$, this time not only as a homomorphism of $k$-vector spaces, but as a homomorphism of $A$-bimodules. It follows that the $A$-bimodule $\ker \overline{\mult}$ (i.e., $\quoz{(\ker \mult)}{M'}$) is generated by the elements $\overline{m_j - (\rep \circ \mult)(m_j)}$ for $j \in \mathbf{J}$, and we thus conclude that $\ker \mult$ is the sub-$A$-bimodule of $M$ generated by the following elements:

* $\rep(a_i n_l) - a_i \rep(n_l)$, for $i \in \mathbf{I}$ and $l \in \mathbf{L}$;

* $\rep(n_l a_i) - \rep(n_l) a_i$, for $i \in \mathbf{I}$ and $l \in \mathbf{L}$;

* $m_j - (\rep \circ \mult)(m_j)$, for $j \in \mathbf{J}$.

§.§ The tensor algebra

As explained, we work with the tensor algebra $T_{E^0}^\ast E^1$ and the multiplication map
\funcInline
{T_{E^0}^\ast E^1}
We denote its graded pieces by

\[
\funcInline
{T_{E^0}^i E^1}
\]

Let us define maps $\Gamma_\varpi$ and $\invol$ on $T_{E^0}^\ast E^1$ that are compatible with the namesake maps on $E^\ast$. We start by defining a map

\[
\functor
{T_{E^0}^\ast E^1}
{T_{E^0}^\ast E^1}
{E^0 \ni x}
{\beta_1 \otimes \dots \otimes \beta_i}
{\Gamma_\varpi(\beta_1) \otimes \dots \otimes \Gamma_\varpi(\beta_i).}
\]

extending the map $\Gamma_\varpi$ defined on $E^0$ and $E^1$. We see that it is an involutive automorphism the graded $k$-algebra $T_{E^0}^\ast E^1$. Similarly, we define the map

\[
\functor
{T_{E^0}^\ast E^1}
{T_{E^0}^\ast E^1}
{E^0 \ni x}
{\beta_1 \otimes \dots \otimes \beta_i}
{(-1)^{\frac{(i-1)i}{2}} \invol(\beta_i) \otimes \dots \otimes \invol(\beta_1).}
\]

Note that the sign $(-1)^{\frac{(i-1)i}{2}}$ is the sign of the permutation $\begin{psmallmatrix} 1 & \dots & i \\ i & \dots & 1 \end{psmallmatrix}$. The map $\invol$ that we have just defined is an involutive anti-automorphism on $T_{E^0}^\ast E^1$.

The above definitions are given in such a way that the following diagrams commute:

\begin{align}\label{eqMultiplicationAndCCCNAndInvol}
&\begin{tikzcd}[ampersand replacement = \&]
T_{E^0}^\ast E^1
\ar[r, "{ \mult}"]
\ar[d, "{ \Gamma_\varpi}"']
\&
\ar[d, "{ \Gamma_\varpi}"]
\\
T_{E^0}^\ast E^1
\ar[r, "{ \mult}"']
\&
\end{tikzcd}
&&\begin{tikzcd}[ampersand replacement = \&]
T_{E^0}^\ast E^1
\ar[r, "{ \mult}"]
\ar[d, "{ \invol}"']
\&
\ar[d, "{ \invol}"]
\\
T_{E^0}^\ast E^1
\ar[r, "{ \mult}"']
\&
\end{tikzcd}
\end{align}

As explained in the outline of the strategy in <ref>, we want to compute a (quite simple) set of generators of $T_{E_0}^2 E^1$ as an $E^0$-bimodule.

We recall that the following are generators of $E^1$ as an $E^0$-bimodule (see Lemma <ref>):

\begin{align*}
\end{align*}

It follows that $E^1 \otimes_{E^0} E^1$ is generated by the following elements as a left $E^0$-module (in particular also as an $E^0$-bimodule):

\begin{align}\label{eqGeneratorsE1tensorE1OnTheLeft}
&\bm{1} \otimes \beta^{\sigma}_{w},
&&\qquad\bp{1} \otimes \beta^{\sigma}_{w},
&&\qquad\bz{s_0} \otimes \beta^{\sigma}_{w},
&&\qquad\bz{s_1} \otimes \beta^{\sigma}_{w}
\end{aligned}
\\&\qquad
\text{for $w \in \widetilde{W}$ and $\sigma \in \{-,0,+\}$ (with $\ell(w) \geqslant 1$ in the case $\sigma = 0$).}
\notag
\end{align}

In the next lemma, we compute a simpler set of $E^0$-bimodule generators.

The following elements generate $E^1 \otimes_{E^0} E^1$ as an $E^0$-bimodule:

\begin{align*}
&\bm{1} \otimes \bm{1}, &&\qquad\bp{1} \otimes \bm{1}, &&\qquad\bz{s_0} \otimes \bm{1}, &&\qquad\bz{s_1} \otimes \bm{1},
\\
&\bm{1} \otimes \bp{1}, &&\qquad\bp{1} \otimes \bp{1}, &&\qquad\bz{s_0} \otimes \bp{1}, &&\qquad\bz{s_1} \otimes \bp{1},
\\
&\bm{1} \otimes \bz{s_0}, &&\qquad\bp{1} \otimes \bz{s_0}, &&\qquad\bz{s_0} \otimes \bz{s_0}, &&\qquad\bz{s_1} \otimes \bz{s_0},
\\
&\bm{1} \otimes \bz{s_1}, &&\qquad\bp{1} \otimes \bz{s_1}, &&\qquad\bz{s_0} \otimes \bz{s_1}, &&\qquad\bz{s_1} \otimes \bz{s_1},
\end{aligned}
\\
&\bp{1} \otimes \bm{(s_1s_0)^i} = \bp{(s_1s_0)^i} \otimes \bm{1} &&\mkern-170mu\text{for $i \in \zpiu$,}
\\
&\bp{1} \otimes \bm{s_1(s_0s_1)^i} = - \bp{s_1(s_0s_1)^i} \otimes \bp{1} &&\mkern-170mu\text{for $i \in \nn$,}
\\
&\bm{1} \otimes \bp{(s_0s_1)^i} = \bm{(s_0s_1)^i} \otimes \bp{1} &&\mkern-170mu\text{for $i \in \zpiu$,}
\\
&\bm{1} \otimes \bp{s_0(s_1s_0)^i} = - \bm{s_0(s_1s_0)^i} \otimes \bm{1} &&\mkern-170mu\text{for $i \in \nn$.}
\end{align*}

Note that the equalities in the last four lines follow from Lemma <ref>.

We start from the generators in (<ref>), and we recall from Lemma <ref> that

\begin{align*}
\bm{w} &= \bm{1} \cdot \tau_{w}
&&\text{for $w \in \widetilde{W}$ such that $\ell(s_1w)=\ell(w)+1$,}
\\
\bp{w} &= \bp{1} \cdot \tau_{w}
&&\text{for $w \in \widetilde{W}$ such that $\ell(s_0w)=\ell(w)+1$,}
\\
\bz{s_1w} &= \bz{s_1} \cdot \tau_{w}
&&\text{for $w \in \widetilde{W}$ such that $\ell(s_1w)=\ell(w)+1$,}
\\
\bz{s_0w} &= \bz{s_0} \cdot \tau_{w}
&&\text{for $w \in \widetilde{W}$ such that $\ell(s_0w)=\ell(w)+1$.}
\end{align*}

From this (and from the behaviour of multiplication by $\tau_\omega$ for $\omega \in \quoz{T^0}{T^1}$, see (<ref>)), we obtain that the following elements generate $E^1 \otimes_{E^0} E^1$ as an $E^0$-bimodule:

\begin{align*}
&\gamma \otimes \bm{1}
&&\text{for $\gamma \in \left\{ \bm{1}, \bp{1}, \bz{s_0}, \bz{s_1} \right\}$,}
\\
&\gamma \otimes \bp{1}
&&\text{for $\gamma$ as above,}
\\
&\gamma \otimes \bz{s_0}
&&\text{for $\gamma$ as above,}
\\
&\gamma \otimes \bz{s_1}
&&\text{for $\gamma$ as above.}
\\
&\gamma \otimes \bm{(s_1s_0)^i}
&&\text{for $\gamma$ as above and for $i \in \zpiu$,}
\\
&\gamma \otimes \bm{s_1(s_0s_1)^i}
&&\text{for $\gamma$ as above and for $i \in \zpiu$,}
\\
&\gamma \otimes \bp{(s_0s_1)^i}
&&\text{for $\gamma$ as above and for $i \in \zpiu$,}
\\
&\gamma \otimes \bp{s_0(s_1s_0)^i}
&&\text{for $\gamma$ as above and for $i \in \zpiu$.}
\end{align*}

The first four of these lines consist exactly of the first four lines of the claimed generators in the statement of the lemma. Now, let us look at the remaining four lines: we certainly get the families of generators in the remaining four lines of the statement of the lemma, and we have to argue that the remaining generators are superfluous.
Up to changing signs if necessary, using the formulas of Lemma <ref>, we can rewrite the last four lines as:

\begin{align*}
&(\gamma \cdot \tau_{(s_1s_0)^i}) \otimes \bm{1}
&&\text{for $\gamma$ as above and for $i \in \zpiu$,}
\\
&(\gamma \cdot \tau_{s_1(s_0s_1)^i}) \otimes \bp{1}
&&\text{for $\gamma$ as above and for $i \in \zpiu$,}
\\
&(\gamma \cdot \tau_{(s_0s_1)^i}) \otimes \bp{1}
&&\text{for $\gamma$ as above and for $i \in \zpiu$,}
\\
&(\gamma \cdot \tau_{s_0(s_1s_0)^i}) \otimes \bm{1}
&&\text{for $\gamma$ as above and for $i \in \zpiu$.}
\end{align*}

From Lemma <ref>, one has

\begin{align*}
\bm{1} \cdot \tau_{(s_1s_0)^i} = 0,
\bm{1} \cdot \tau_{s_1(s_0s_1)^i} = 0,
\bp{1} \cdot \tau_{(s_0s_1)^i} = 0,
\bp{1} \cdot \tau_{s_0(s_1s_0)^i} = 0.
\end{align*}

This shows that some of the remaining generators are superfluous, and now it remains to study the cases where $\gamma \in \{\bz{s_0}, \bz{s_1}\}$.

Let us start with the following case:

\begin{align*}
(\bz{s_0} \cdot \tau_{(s_1s_0)^i}) \otimes \bm{1}
\bz{s_0(s_1s_0)^i} \otimes \bm{1}
\\&=
\tau_{(s_0s_1)^i} \bz{s_0} \otimes \bm{1},
\end{align*}

where we used the formulas (<ref>) and (<ref>). We see that the element $\bz{s_0} \otimes \bm{1}$ is a generator already appearing in the list in the statement of the lemma, and so the generator $(\bz{s_0} \cdot \tau_{(s_1s_0)^i}) \otimes \bm{1}$ is superfluous. The other three cases where lengths add up are similar.

Now it remains to prove that the generators

\begin{align*}
&(\bz{s_1} \cdot \tau_{(s_1s_0)^i}) \otimes \bm{1},
&&(\bz{s_1} \cdot \tau_{s_1(s_0s_1)^i}) \otimes \bp{1},
\\&(\bz{s_0} \cdot \tau_{(s_0s_1)^i}) \otimes \bp{1},
&&(\bz{s_0} \cdot \tau_{s_0(s_1s_0)^i}) \otimes \bm{1}
\end{align*}

are superfluous. To this end, we can use the formulas (<ref>) and (<ref>): we spell the details for the first generator, the other three being similar.

\begin{align*}
(\bz{s_1} \cdot \tau_{(s_1s_0)^i}) \otimes \bm{1}
\left( -e_1 \cdot \bz{(s_1s_0)^i} + e_{\idd} \cdot \bp{(s_1s_0)^i} \right) \otimes \bm{1}
\\&=
e_1 \cdot \tau_{(s_1s_0)^{i-1}s_1} \cdot \bz{s_0} \otimes \bm{1} + e_{\idd} \cdot \bp{(s_1s_0)^i} \otimes \bm{1},
\end{align*}

where, besides (<ref>), we have also used (<ref>).

§.§ The 2nd graded piece of the kernel

The multiplication map $\mult_2$ is surjective: indeed, Proposition <ref> shows that $H^2(I,\xx(w))$ is contained in its image for all $w \in \widetilde{W}$ with $\ell(w) \geqslant 1$, and for all $\omega \in T^0/T^1$ (i.e., $\omega \in \widetilde{W}$ with $\ell(\omega) = 0$) one concludes with the following consequence of formulas (<ref>) and (<ref>):

\begin{equation}\label{eqToQuoteDeg2E}\begin{aligned}
\tau_{s_0} \cdot \ap{s_0^{-1}\omega} + \am{\omega}
\bigoplus_{\substack{v \in \widetilde{W}\\\text{s.t. $\ell(v)=1$}}}
\\
\tau_{s_1} \cdot \am{s_1^{-1}\omega} + \ap{\omega}
\bigoplus_{\substack{v \in \widetilde{W}\\\text{s.t. $\ell(v)=1$}}}
\end{aligned}\end{equation}

The purpose of this subsection is to compute the kernel of $\mult_2$, and following the outline in <ref> we start by constructing a section of $\mult_2$ as a homomorphism of $k$-vector space.

The multiplication map $\funcInline{\mult_2}{T^2_{E^0} E^1}{E^2}$ is surjective, and the following is a section of it as a homomorphism of $k$-vector spaces.

\[\begin{tikzcd}[row sep = -0.6em]
\rep_2 \colon &[-3em] E^2 \ar[r] &[-1em] T_{E^0}^2E^1 = E^1 \otimes_{E^0} E^1 &[-4.5em]
\\
& \am{s_1v} \ar[r, mapsto] & - \bp{1} \otimes \bz{s_1v} &&\scriptstyle\text{if $\ell(s_1v) = \ell(v) + 1$,}
\\
& \az{s_1v} \ar[r, mapsto] & \bp{1} \otimes \bm{s_1v} &&\scriptstyle\text{if $\ell(s_1v) = \ell(v) + 1$,}
\\
& \ap{s_1v} \ar[r, mapsto] & \bz{s_1} \otimes \bp{v} &&\scriptstyle\text{if $\ell(s_1v) = \ell(v) + 1$,}
\\
& \am{s_0w} \ar[r, mapsto] & - \bz{s_0} \otimes \bm{w} &&\scriptstyle\text{if $\ell(s_0w) = \ell(w) + 1$,}
\\
& \az{s_0w} \ar[r, mapsto] & - \bm{1} \otimes \bp{s_0w} &&\scriptstyle\text{if $\ell(s_0w) = \ell(w) + 1$,}
\\
& \ap{s_0w} \ar[r, mapsto] & \bm{1} \otimes \bz{s_0w} &&\scriptstyle\text{if $\ell(s_0w) = \ell(w) + 1$,}
\\
& \am{\omega} \ar[r, mapsto] & \rep_2 \big( \am{\omega} + \tau_{s_0} \cdot \ap{s_0^{-1}\omega} \big) - \tau_{s_0} \cdot \rep_2 \big( \ap{s_0^{-1}\omega} \big) &&\scriptstyle\text{for $\omega \in \quoz{T^0}{T^1}$,}
\\
& \ap{\omega} \ar[r, mapsto] & \rep_2 \big( \ap{\omega} + \tau_{s_1} \cdot \am{s_1^{-1}\omega} \big) - \tau_{s_1} \cdot \rep_2 \big( \am{s_1^{-1}\omega} \big) &&\scriptstyle\text{for $\omega \in \quoz{T^0}{T^1}$.}
\end{tikzcd}\]

Note that the last two lines make sense: by (<ref>), on the right hand side of such lines only the values of $\rep_2$ on \( \bigoplus_{\substack{v \in \widetilde{W}\\\text{s.t. $\ell(v)=1$}}} H^2(I,\xx(v)) \) are used, and on such subspace $\rep_2$ is defined in the preceding lines.

Combining formula (<ref>) in the proof of Proposition <ref> with (<ref>), we obtain

\begin{align*}
\bz{(s_0s_1)^i} \cdot \bp{1}
\bz{(s_0s_1)^i} \cupprod \bp{(s_0s_1)^i}
\am{(s_0s_1)^i},
\\
\bm{1} \cdot \tau_{(s_0s_1)^i} \cdot \bp{1}
\bm{(s_0s_1)^i} \cupprod \bp{(s_0s_1)^i}
- \az{(s_0s_1)^i}.
\\
\bm{1} \cdot \bz{(s_0s_1)^i}
\bm{(s_0s_1)^i} \cupprod \bz{(s_0s_1)^i}
\ap{(s_0s_1)^i},
\end{align*}

We rewrite this as follows:

\begin{align*}
\am{(s_0s_1)^i}
\bz{(s_0s_1)^i} \cdot \bp{1}
\bz{s_0} \cdot \tau_{s_1(s_0s_1)^{i-1}} \cdot \bp{1}
- \bz{s_0} \cdot \bm{s_1(s_0s_1)^{i-1}},
\\
\az{(s_0s_1)^i}
- \bm{1} \cdot \tau_{(s_0s_1)^i} \cdot \bp{1}
- \bm{1} \cdot \bp{(s_0s_1)^i},
\\
\ap{(s_0s_1)^i}
\bm{1} \cdot \bz{(s_0s_1)^i},
\end{align*}

where we have used the formulas of Lemma <ref>.

This proves that the map $\mult_2 \circ \rep_2$ is the identity at $\am{s_0w}$, $\az{s_0w}$, $\ap{s_0w}$ for all $w$ of the form $s_1 (s_0s_1)^{i-1}$ for some $i \in \zpiu$. From the formulas (<ref>) it follows that the same is true for $w$ of the form $s_1 (s_0s_1)^{i-1} \omega$ for some $i \in \zpiu$ and some $\omega \in \quoz{T^0}{T^1}$. The proof that the same claim is true for all $w$ of the form $(s_1s_0)^{i-1}\omega$ for some $i \in \zpiu$ and some $\omega \in \quoz{T^0}{T^1}$ is similar, and this shows that the map $\mult_2 \circ \rep_2$ is the identity at $\am{s_0w}$, $\az{s_0w}$, $\ap{s_0w}$ for all $w \in \widetilde{W}$ such that $\ell(s_0w) = \ell(w) + 1$.

Similarly one shows that $\mult_2 \circ \rep_2$ is the identity at $\am{s_1v}$, $\az{s_1v}$, $\ap{s_1v}$ for all $v \in \widetilde{W}$ such that $\ell(s_1v) = \ell(v) + 1$. Finally, the fact that $\mult_2 \circ \rep_2$ is the identity at $\am{\omega}$ and $\ap{\omega}$ for all $\omega \in \Omega$ follows from the fact that $\mult_2 \circ \rep_2$ is the identity on \( \bigoplus_{\substack{v \in \widetilde{W}\\\text{s.t. $\ell(v)=1$}}} H^2(I,\xx(v)) \).

For the next lemma, we use the identification of $k$-algebras

\[
\spann_k \set{\tau_\omega}{\omega \in T^0/T^1} \cong k[T^0/T^1].
\]

The map $\rep_2$ is a homomorphism of $k[\quoz{T^0}{T^1}]$-bimodules.

From the definition of $\rep_2$ and from the formulas for the right action of $k[\quoz{T^0}{T^1}]$ on $E^1$ and $E^2$ (see (<ref>) (<ref>)), it is clear that $\rep_2$ is a homomorphism of right $k[\quoz{T^0}{T^1}]$-modules.

Regarding the left action, from the formulas (<ref>) and (<ref>) describing the action of $k[\quoz{T^0}{T^1}]$ on $E^2$, we see that for all $\omega \in \quoz{T^0}{T^1}$, for all $\sigma \in \{-,0,+\}$ and for all $w \in \widetilde{W}$ (of length greater or equal than $1$ if $\sigma = 0$) we have

\[
\tau_\omega \cdot \alpha_w^\sigma = c_{\omega, \sigma} \alpha_w^\sigma \cdot \tau_{\omega^{(-1)^{\ell(w)}}},
\]

for some $c_{\omega, \sigma} \in k^\times$. Assume now that $\ell(w) \geqslant 1$. Looking at the definition of $\rep_2$ in Lemma <ref>, and using the formulas (<ref>) and (<ref>) describing the action of $k[\quoz{T^0}{T^1}]$ on $E^1$, we see that we also have the completely analogous result

\[
\tau_\omega \cdot \rep_2(\alpha_w^\sigma) = c_{\omega, \sigma}' \rep_2(\alpha_w^\sigma) \cdot \tau_{\omega^{(-1)^{\ell(w)}}},
\]

for some $c_{\omega, \sigma}' \in k^\times$. Using that $\rep_2$ is a homomorphism of right $k[\quoz{T^0}{T^1}]$-modules, we thus see that

\[
\tau_\omega \cdot \rep_2(\alpha_w^\sigma) = c_{\omega, \sigma}' c_{\omega, \sigma}^{-1} \rep_2(\tau_\omega \cdot \alpha_w^\sigma),
\]

and by applying $\mult_2$ to both sides, we see that the coefficient $c_{\omega, \sigma}' c_{\omega, \sigma}^{-1}$ must be $1$.

So far, we have shown that the maps $\tau_\omega \cdot \rep_2({}_-)$ and $\rep_2(\tau_\omega \cdot {}_-)$ coincide on the subspace \( \bigoplus_{\substack{v \in \widetilde{W}\\\text{s.t. $\ell(v) \geqslant 1$}}} H^2(I,\xx(v)) \), and looking at the definition of $\rep_2$ on the subspace \( \bigoplus_{\substack{v \in \widetilde{W}\\\text{s.t. $\ell(v)=0$}}} H^2(I,\xx(v)) \), we then deduce that the maps $\tau_\omega \cdot \rep_2({}_-)$ and $\rep_2(\tau_\omega \cdot {}_-)$ coincide on the whole $E^2$.

The maps $\rep_2$ and $\Gamma_\varpi$ commute. More precisely, the following diagram is commutative:

\[\begin{tikzcd}
\ar[r, "{ \rep_2}"]
\ar[d, "{ \Gamma_\varpi}"]
E^1 \otimes_{E^0} E^1
\ar[d, "{ \Gamma_\varpi}"]
\\
\ar[r, "{ \rep_2}"]
E^1 \otimes_{E^0} E^1.
\end{tikzcd}\]

Let $v \in \widetilde{W}$ with $\ell(s_1v) = \ell(v) + 1$. We compute

\begin{align*}
\Gamma_\varpi \big( \rep_2(\am{s_1v}) \big)
\Gamma_\varpi \big( - \bp{1} \otimes \bz{s_1v} \big)
\\&=
\bm{1} \otimes \bz{s_0 \varpi v \varpi^{-1}},
\\
\Gamma_\varpi \big( \rep_2(\az{s_1v}) \big)
\Gamma_\varpi \big( \bp{1} \otimes \bm{s_1v} \big)
\\&=
\bm{1} \otimes \bp{s_0 \varpi v \varpi^{-1}},
\\
\Gamma_\varpi \big( \rep_2(\ap{s_1v}) \big)
\Gamma_\varpi \big( \bz{s_1} \otimes \bp{v} \big)
\\&=
- \bz{s_0} \otimes \bm{\varpi v \varpi^{-1}}.
\end{align*}

Here we have used the definition of $\Gamma_\varpi$ on $T^\ast_{E^0} E^1$, the formulas for the action of $\Gamma_\varpi$ on $E^1$ (<ref>), and the fact that $\varpi s_1 \varpi^{-1} = s_0$.
On the other side, we compute

\begin{align*}
\rep_2 \big( \Gamma_\varpi(\am{s_1v}) \big)
\rep_2 (\ap{s_0 \varpi v \varpi^{-1}}),
\\
\rep_2 \big( \Gamma_\varpi(\az{s_1v}) \big)
- \rep_2 (\az{s_0 \varpi v \varpi^{-1}}),
\\
\rep_2 \big( \Gamma_\varpi(\ap{s_1v}) \big)
\rep_2 (\am{s_0 \varpi v \varpi^{-1}}),
\end{align*}

where we have used the formulas for the action of $\Gamma_\varpi$ on $E^2$ (<ref>). Comparing the two above computations with the definition of $\rep_2$, we see that the maps $\Gamma_\varpi \circ \rep_2$ and $\rep_2 \circ \Gamma_\varpi$ coincide at $\am{s_1v}$, $\az{s_1v}$ and $\ap{s_1v}$.

We also see that

\begin{align*}
\Gamma_\varpi(\rep_2(\ap{\omega}))
\Gamma_\varpi \left( \rep_2 \big( \ap{\omega} + \tau_{s_1\omega} \cdot \am{s_1^{-1}\omega} \big) - \tau_{s_1} \cdot \rep_2 \big( \am{s_1^{-1}\omega} \big) \right)
\\&=
\Gamma_\varpi \left( \rep_2 \big( \ap{\omega} + \tau_{s_1} \cdot \am{s_1^{-1}\omega} \big) \right) - \tau_{s_0} \cdot \Gamma_\varpi \left( \rep_2 \big( \am{s_1^{-1}\omega} \big) \right)
\\&=
\rep_2 \left( \Gamma_\varpi \big( \ap{\omega} + \tau_{s_1} \cdot \am{s_1^{-1}\omega} \big) \right) - \tau_{s_0} \cdot \rep_2 \left( \Gamma_\varpi \big( \am{s_1^{-1}\omega} \big) \right)
\\&\qquad\left( \substack{\text{since $\Gamma_\varpi \circ \rep_2$ and $\rep_2 \circ \Gamma_\varpi$ coincide on $\bigoplus_{\vartheta \in \quoz{T^0}{T^1}} H^2(I,\xx(s_1 \vartheta))$}} \right)
\\&=
\rep_2 \big( \am{\varpi \omega \varpi^{-1}} + \tau_{s_0} \cdot \ap{s_0^{-1}\varpi \omega \varpi^{-1}} \big) - \tau_{s_0} \cdot \rep_2 \big( \ap{s_0^{-1}\varpi \omega \varpi^{-1}} \big)
\\&=
\rep_2 \left( \am{\varpi \omega \varpi^{-1}} \right)
\\&=
\rep_2(\Gamma_\varpi(\ap{\omega})),
\end{align*}

where we have used the same formulas as before (and the fact that $\Gamma_\varpi(\tau_{s_1}) = \tau_{s_0}$, see (<ref>)).

So far, we have shown that the maps $\Gamma_\varpi \circ \rep_2$ and $\rep_2 \circ \Gamma_\varpi$ coincide on half of the elements of the $k$-basis of $E^2$ used in the definition of $\rep_2$.
To conclude the proof, we remark that, since $\Gamma_\varpi$ is an involution, if the maps $\Gamma_\varpi \circ \rep_2$ and $\rep_2 \circ \Gamma_\varpi$ coincide on an element $x \in E^2$, then they coincide also on $\Gamma_\varpi(x)$.

Let us consider the following $\invol$-invariant $k$-subspace of $E^2$:

\[
\defeq
\bigoplus_{\mathclap{\substack{w \in \widetilde{W} \\\text{s.t. } \ell(w) \geqslant 1}}} H^2(I,\xx(w)).
\]

One has that map $\restr{\rep_2}{F^1E^2}$ commutes with the anti-involution $\invol$. More precisely, the following diagram is commutative:

\[\begin{tikzcd}[row sep = 2.7em, column sep = 4em]
\ar[r, "{ \restr{\rep_2}{F^1E^2}}"]
\ar[d, "{ \invol}"']
E^1 \otimes_{E^0} E^1
\ar[d, "{ \invol}", "{\substack{ \beta \otimes \beta' \\ \downmapsto \\ - \invol(\beta') \otimes \invol(\beta) }}"']
\\
\ar[r, "{ \restr{\rep_2}{F^1E^2}}"']
E^1 \otimes_{E^0} E^1.
\end{tikzcd}\]

We first recall from (<ref>) that the involutions $\invol$ and $\Gamma_\varpi$ commute on $E^\ast$. Using this property, let us see that if the maps $\invol \circ \rep_2$ and $\rep_2 \circ \invol$ coincide at a certain $\alpha \in E^2$, then they also coincide at $\Gamma_\varpi(\alpha)$: indeed, one has

\begin{align*}
(\invol \circ \rep_2) (\Gamma_\varpi(\alpha))
(\invol \circ \Gamma_\varpi \circ \rep_2)(\alpha)
&&\text{as $\rep_2$ and $\Gamma_\varpi$ commute (Lemma \ref{lmmCommutesCCCN})}
\\&=
(\Gamma_\varpi \circ \invol \circ \rep_2)(\alpha)
&&\begin{matrix*} \text{as $\invol$ and $\Gamma_\varpi$ commute on $E^1$} \\ \text{and hence on $E^1 \otimes_{E^0} E^1$} \end{matrix*}
\\&=
(\Gamma_\varpi \circ \rep_2 \circ \invol)(\alpha)
&&\text{by assumption}
\\&=
(\rep_2 \circ \Gamma_\varpi \circ \invol)(\alpha)
&&\text{as $\rep_2$ and $\Gamma_\varpi$ commute (Lemma \ref{lmmCommutesCCCN})}
\\&=
(\rep_2 \circ \invol) (\Gamma_\varpi(\alpha))
&&\text{as $\invol$ and $\Gamma_\varpi$ commute on $E^2$.}
\end{align*}

Now, let us look at the definition of $\rep_2$ in Lemma <ref>. By what we have just remarked, we only need to show that $\invol \circ \rep_2$ and $\rep_2 \circ \invol$ coincide at $\am{s_1v}$, $\az{s_1v}$ and $\ap{s_1v}$ for all $v \in \widetilde{W}$ such that $\ell(s_1v) = \ell(v) + 1$. Since $\rep_2$ is a homomorphism of $k[\quoz{T^0}{T^1}]$-bimodules (Lemma <ref>) and since $\funcInline{\invol}{E^\ast}{E^\ast}$ and $\funcInline{\invol}{T^\ast_{E^0}E^1}{T^\ast_{E^0}E^1}$ are anti-multiplicative, it follows that $\invol \circ \rep_2 \circ \invol$ is a homomorphism of $k[\quoz{T^0}{T^1}]$-bimodules, and so (looking at formulas (<ref>)) we can assume that $v$ is of the form $(s_0s_1)^i$ or $s_0(s_1s_0)^i$ for some $i \in \nn$.
Using the formulas for the action of $\invol$ on $E^2$ (<ref>) as well as the definition of $\rep_2$ and recalling the definition of $c_{-1}$ from (<ref>), we compute

\begin{align*}
\rep_2 \big(\invol( \am{s_1(s_0s_1)^i} ) \big)
- \rep_2 \big( \ap{s_1(s_0s_1)^ic_{-1}} \big)
- \bz{s_1} \otimes \bp{(s_0s_1)^ic_{-1}}.
\end{align*}

Similarly, using the formulas for the action of $\invol$ on $E^1$ (<ref>) (together with the definition of $\invol$ on $E^1 \otimes_{E^0} E^1$ and the definition of $\rep_2$), we compute

\begin{align*}
\invol \big( \rep_2( \am{s_1(s_0s_1)^i} ) \big)
- \invol \big( \bp{1} \otimes \bz{s_1(s_0s_1)^i} \big)
- \bz{s_1(s_0s_1)^ic_{-1}} \otimes \bp{1}.
\end{align*}

And, using the formulas in Lemma <ref>, we conclude that the two simple tensors we have obtained are equal, and hence that $\rep_2 \circ \invol$ and $\invol \circ \rep_2$ coincide at $\am{s_1(s_0s_1)^i}$.

Similarly, one computes that $\rep_2 \circ \invol$ and $\invol \circ \rep_2$ coincide at $\az{s_1(s_0s_1)^i}$, $\ap{s_1(s_0s_1)^i}$, $\am{(s_1s_0)^{i+1}}$, $\az{(s_1s_0)^{i+1}}$ and $\ap{(s_1s_0)^{i+1}}$.

The above result leaves open the question of whether $\rep_2$ and $\invol$ could commute on the whole $E^2$. This is dealt with in the following lemma and remark.

One has the following equalities:

\begin{align*}
(\rep_2 \circ \invol)(\am{1}) - (\invol \circ \rep_2)(\am{1})
\bp{s_0} \otimes \bz{s_0^{-1}}
\bz{s_0} \otimes \bm{s_0^{-1}}
\\&=
- \tau_{s_0} \cdot \bm{1} \otimes \bz{s_0^{-1}}
\bz{s_0} \otimes \bm{1} \cdot \tau_{s_0^{-1}}
\\
(\rep_2 \circ \invol)(\ap{1}) - (\invol \circ \rep_2)(\ap{1})
- \bm{s_1} \otimes \bz{s_1^{-1}}
\bz{s_1} \otimes \bp{s_1^{-1}}
\\&=
\tau_{s_1} \cdot \bp{1} \otimes \bz{s_1^{-1}}
\bz{s_1} \otimes \bp{1} \cdot \tau_{s_1^{-1}}
\end{align*}

First note that the second formula can be obtained from the first one by applying $\Gamma_\varpi$: indeed $\Gamma_\varpi$ commutes with $\rep_2$ (Lemma <ref>) and with $\invol$ (see (<ref>)), and $\Gamma_\varpi(\am{1}) = \ap{1}$ (see (<ref>)); hence applying $\Gamma_\varpi$ on $(\rep_2 \circ \invol)(\am{1}) - (\invol \circ \rep_2)(\am{1})$ we obtain exactly $(\rep_2 \circ \invol)(\ap{1}) - (\invol \circ \rep_2)(\ap{1})$. Using the formulas for the action of $\Gamma_\varpi$ on $E^0$ (<ref>) and on $E^1$ (<ref>), we also see that the terms on the right of the equations correspond.

The equality
\bp{s_0} \otimes \bz{s_0^{-1}}
\bz{s_0} \otimes \bm{s_0^{-1}}
- \tau_{s_0} \cdot \bm{1} \otimes \bz{s_0^{-1}}
\bz{s_0} \otimes \bm{1} \cdot \tau_{s_0^{-1}}
is immediate from the formulas (<ref>) and (<ref>). Therefore, to prove the lemma it remains to check that

\[
(\rep_2 \circ \invol)(\am{1}) - (\invol \circ \rep_2)(\am{1})
- \tau_{s_0} \cdot \bm{1} \otimes \bz{s_0^{-1}}
\bz{s_0} \otimes \bm{1} \cdot \tau_{s_0^{-1}}
\]

Using (<ref>) and the definition of $\rep_2$, we see that

\begin{align*}
(\rep_2 \circ \invol)(\am{1})
\rep_2(\am{1})
\\&=
\rep_2 \big( \am{1} + \tau_{s_0} \cdot \ap{s_0^{-1}} \big) - \tau_{s_0} \cdot \rep_2 ( \ap{s_0^{-1}} ).
\end{align*}

On the other side, using that the maps $\invol \circ \rep_2$ and $\rep_2 \circ \invol$ coincide on $F^1E^2$, we deduce that

\begin{equation}\label{eqInvolRep2}\begin{aligned}
(\invol \circ \rep_2)(\am{1})
(\invol \circ \rep_2) \big( \am{1} + \tau_{s_0} \cdot \ap{s_0^{-1}} \big)
(\invol \circ \rep_2) ( \ap{s_0^{-1}} ) \cdot \tau_{s_0^{-1}}
\\&=
(\rep_2 \circ \invol) \big( \am{1} + \tau_{s_0} \cdot \ap{s_0^{-1}} \big)
(\rep_2 \circ \invol) ( \ap{s_0^{-1}} ) \cdot \tau_{s_0^{-1}}
\\&=
\rep_2 \big( \am{1} + \invol (\tau_{s_0} \cdot \ap{s_0^{-1}}) \big)
\rep_2 ( \am{s_0} ) \cdot \tau_{s_0^{-1}},
\end{aligned}\end{equation}

where we have also used that $\invol$ is anti-multiplicative, that $\invol(\tau_{s_0}) = \tau_{s_0^{-1}}$ by (<ref>) and, again, that $\invol(\am{1}) = \am{1}$. We thus deduce that

\begin{equation}\label{eqRepInvolInvolRep}\begin{aligned}
&(\rep_2 \circ \invol)(\am{1}) - (\invol \circ \rep_2)(\am{1})
\\&\qquad=
\rep_2 \big( \tau_{s_0} \cdot \ap{s_0^{-1}} - \invol (\tau_{s_0} \cdot \ap{s_0^{-1}}) \big)
\tau_{s_0} \cdot \rep_2 ( \ap{s_0^{-1}} )
\rep_2 ( \am{s_0} ) \cdot \tau_{s_0^{-1}}
\end{aligned}\end{equation}

We now focus on the first term in this sum. Using (<ref>), we compute

\[
\tau_{s_0} \cdot \ap{s_0^{-1}}
- e_1 \ap{s_0^{-1}} - e_{\idd} \az{s_0^{-1}} + e_{\idd^2} \am{s_0^{-1}} - \am{1}.
\]

Using formulas (<ref>) and (<ref>) and using the fact that $\invol(e_\lambda) = e_{\lambda^{-1}}$ for all $\lambda \in \widehat{\quoz{T^0}{T^1}}$ (which follows from (<ref>)), we compute

\begin{align*}
\invol \big( \tau_{s_0} \cdot \ap{s_0^{-1}} \big)
\invol \big( - e_1 \ap{s_0^{-1}} - e_{\idd} \az{s_0^{-1}} + e_{\idd^2} \am{s_0^{-1}} - \am{1} \big)
\\&=
\am{s_0} e_1 + \az{s_0} e_{\idd^{-1}} - \ap{s_0} e_{\idd^{-2}} - \am{1}
\\&=
e_{\idd^2} \am{s_0}  + e_{\idd} \az{s_0}  - e_1 \ap{s_0} - \am{1}
\\&=
e_{\idd^2} \am{s_0^{-1}}  - e_{\idd} \az{s_0^{-1}}  - e_1 \ap{s_0^{-1}} - \am{1}.
\end{align*}

In the last line we have used the formulas for the left action of $\tau_{c_{-1}}$ on $E^2$ (see (<ref>) and recall the definition of $c_{-1}$ from (<ref>)) and the fact $e_\lambda \tau_{c_{-1}} = \lambda(c_{-1}) e_\lambda$ for all $\lambda \in \widehat{\quoz{T^0}{T^1}}$ (see (<ref>)).

We have shown that $\tau_{s_0} \cdot \ap{s_0^{-1}} = \invol (\tau_{s_0} \cdot \ap{s_0^{-1}})$, and so going back to (<ref>), we obtain

\begin{align*}
(\rep_2 \circ \invol)(\am{1}) - (\invol \circ \rep_2)(\am{1})
\tau_{s_0} \cdot \rep_2 ( \ap{s_0^{-1}} )
\rep_2 ( \am{s_0} ) \cdot \tau_{s_0^{-1}}
\\&=
- \tau_{s_0} \cdot \big( \bm{1} \otimes \bz{s_0^{-1}} \big)
\big( - \bz{s_0} \otimes \bm{1} \big) \cdot \tau_{s_0^{-1}}
\end{align*}

This is the formula we had to check.

It can be proved that the elements of $E^1 \otimes_{E^0} E^1$ in the previous lemma are nonzero, thus showing that $\invol \circ \rep_2 \neq \rep_2 \circ \invol$. For the details see <cit.>. Furthermore, in loc.cit. it is shown that one can modify the definition of $\rep_2$ on \( \bigoplus_{\substack{w \in \widetilde{W}\\\text{s.t. $\ell(w)=0$}}} H^2(I, \xx(w)) \) in such a way that one gets a new section $\rep_2'$ of $\mult_2$ that is a homomorphism of $k[\quoz{T^0}{T^1}]$-bimodules, that commutes with $\Gamma_\varpi$ and that commutes with $\invol$. However, we do not change the definition of $\rep_2$ to reduce the amount of computations.

The sub-$E^0$-bimodule of $E^1 \otimes_{E^0} E^1$ generated by the elements of the form

\[
x - \rep_2(\mult_2(x)) \in \ker(\mult_2),
\]

where $x$ runs through the set of generators of $E^1 \otimes_{E^0} E^1$ as an $E^0$-bimodule computed in Lemma <ref>, is the sub-$E^0$-bimodule of $E^1 \otimes_{E^0} E^1$ generated by the following elements.

\begin{align*}
&\bm{1} \otimes \bm{1}, &&\qquad\qquad\qquad\qquad\bp{1} \otimes \bm{1}, &&\qquad\qquad\qquad\qquad\bz{s_1} \otimes \bm{1},
\\
&\bm{1} \otimes \bp{1}, &&\qquad\qquad\qquad\qquad\bp{1} \otimes \bp{1}, &&\qquad\qquad\qquad\qquad\bz{s_0} \otimes \bp{1},
\\
&\bp{1} \otimes \bz{s_0}, &&\qquad\qquad\qquad\qquad\bz{s_1} \otimes \bz{s_0},
\\
&\bm{1} \otimes \bz{s_1}, &&\qquad\qquad\qquad\qquad\bz{s_0} \otimes \bz{s_1},
\end{aligned}
\\
&\bz{s_0} \otimes \bz{s_0} + e_{\idd^{-1}} \cdot \bm{1} \otimes \bz{s_0} + e_{\idd} \cdot \bz{s_0} \otimes \bm{1} - e_1 \cdot \bm{1} \otimes \bp{s_0},
\\
&\bz{s_1} \otimes \bz{s_1} - e_{\idd} \cdot \bp{1} \otimes \bz{s_1} - e_{\idd^{-1}} \cdot \bz{s_1} \otimes \bp{1} - e_1 \cdot \bp{1} \otimes \bm{s_1}.
\end{align*}

In particular, all the above elements lie in $\ker(\mult_2)$.

Let us look at the generators of $E^1 \otimes_{E_0} E_1$ as an $E^0$-bimodule computed in Lemma <ref>.
From the definition of $\rep_2$ (Lemma <ref>), we immediately see that some of these lie in the image of $\rep_2$, and so we can discard them immediately, because if $x$ is one of these elements then $x - \rep_2(\mult_2(x)) = 0$. We are left with the following elements:

\begin{align*}
&\bm{1} \otimes \bm{1}, &&\bp{1} \otimes \bm{1}, &&//////// &&\bz{s_1} \otimes \bm{1},
\\
&\bm{1} \otimes \bp{1}, &&\bp{1} \otimes \bp{1}, &&\bz{s_0} \otimes \bp{1}, &&////////
\\
&//////// &&\bp{1} \otimes \bz{s_0}, &&\bz{s_0} \otimes \bz{s_0}, &&\bz{s_1} \otimes \bz{s_0},
\\
&\bm{1} \otimes \bz{s_1}, &&//////// &&\bz{s_0} \otimes \bz{s_1}, &&\bz{s_1} \otimes \bz{s_1}.
\end{align*}

We consider these remaining elements.

* The elements $\bm{1} \otimes \bm{1}$, $\bp{1} \otimes \bm{1}$, $\bm{1} \otimes \bp{1}$ and $\bp{1} \otimes \bp{1}$ are all in the kernel of $\mult_2$: indeed products are cup products by (<ref>), and then trivially $\bm{1} \cdot \bm{1}$ and $\bp{1} \cdot \bp{1}$ are both zero; moreover, the fact that $\bp{1} \cupprod \bm{1}$ and $\bm{1} \cupprod \bp{1}$ are both zero can be shown with a simple computation using Poincaré duality (see <cit.>).

In particular,  if $x$ is one of the above four elements, then, trivially,

\[ x - \rep_2(\mult_2(x)) = x. \]

* Now let us consider the four elements $\bz{s_1} \otimes \bm{1}$, $\bz{s_0} \otimes \bp{1}$, $\bp{1} \otimes \bz{s_0}$ and $\bm{1} \otimes \bz{s_1}$. To compute the product, one can use the formula (<ref>) relating cup product and (opposite of the) Yoneda product. But then we see that all such products are zero, because

\begin{align*}
\tau_{s_0} \cdot \bp{1} &= 0,
&\bp{1} \cdot \tau_{s_0} &= 0,
\\
\tau_{s_1} \cdot \bm{1} &= 0,
&\bm{1} \cdot \tau_{s_1} &= 0
\end{align*}

(see formulas (<ref>) and (<ref>)).

So, if $x$ is one of the above four elements, one has $x - \rep_2(\mult_2(x)) = x$.

* Now let us consider the two elements $\bz{s_1} \otimes \bz{s_0}$ and $\bz{s_1} \otimes \bz{s_0}$. We compute the first product using the formula (<ref>) relating cup product and (opposite of the) Yoneda product, and the formulas (<ref>) and (<ref>) for the action of $E^0$ on $E^1$ (the other product can be computed exactly in the same way, or alternatively one can use $\Gamma_\varpi$ or $\invol$):

\begin{align*}
\bz{s_1} \cdot \bz{s_0}
(\bz{s_1} \cdot \tau_{s_0}) \cupprod (\tau_{s_1} \cdot \bz{s_0})
\\&=
\bz{s_1s_0} \cupprod (-\bz{s_1s_0})
\\&=
\end{align*}

So, if $x$ is one of the above two elements, one has $x - \rep_2(\mult_2(x)) = x$.

* It remains to consider the two elements $\bz{s_0} \otimes \bz{s_0}$ and $\bz{s_1} \otimes \bz{s_1}$; we start with the first one and then we use $\Gamma_\varpi$ for the second one. In the end of the proof of <cit.> the following formula is computed:

\[
\mathbf{x}_0 \cdot \mathbf{x}_0 = e_1 \cdot \upalpha_{s_0}^\star,
\]

where $\mathbf{x}_0$ is defined as

\[
\mathbf{x}_0 \defeq - \bz{s_0} - e_{\idd^{-1}} \bm{1}
\]

(see <cit.>) and $\upalpha_{s_0}^\star$ is a certain element of $E^2$ (defined in <cit.>) with the property that

\[
\azstar{s_0} + \az{s_0} \in e_{\{\idd, \idd^{-1}\}} E^2
\]

(see <cit.>).
Therefore, we deduce that

\begin{align*}
- e_1 \cdot \az{s_0}
\big(-\bz{s_0} - e_{\idd^{-1}} \cdot \bm{1}\big) \cdot \big(-\bz{s_0} - e_{\idd^{-1}} \cdot \bm{1}\big)
\\&=
\bz{s_0} \cdot \bz{s_0} + e_{\idd^{-1}} \cdot \bm{1} \cdot \bz{s_0} + \bz{s_0} \cdot e_{\idd^{-1}} \cdot \bm{1}
\\&\qquad+ e_{\idd^{-1}} \cdot \bm{1} \cdot e_{\idd^{-1}} \cdot \bm{1}
\\&=
\bz{s_0} \cdot \bz{s_0} + e_{\idd^{-1}} \cdot \bm{1} \cdot \bz{s_0} + e_{\idd} \cdot \bz{s_0} \cdot \bm{1}
\\&\qquad+ e_{\idd^{-1}} \cdot e_{\idd^{-3}}  \cdot \bm{1} \cdot \bm{1}
\\&=
\bz{s_0} \cdot \bz{s_0} + e_{\idd^{-1}} \cdot \ap{s_0} - e_{\idd} \cdot \am{s_0}
\end{align*}

(here we have used the behaviour of the left and right action of the idempotents (<ref>); moreover, in the last step one can compute products explicitly, but actually we have already computed them – see the definition of $\rep_2$ in Lemma <ref>). Now we can compute $\rep_2(\bz{s_0} \cdot \bz{s_0})$:

\begin{align*}
\rep_2(\bz{s_0} \cdot \bz{s_0})
\rep_2 \left( -e_{\idd^{-1}} \cdot \ap{s_0} + e_{\idd} \cdot \am{s_0} - e_1 \cdot \az{s_0} \right)
\\&=
-e_{\idd^{-1}} \cdot \bm{1} \otimes \bz{s_0} - e_{\idd} \cdot \bz{s_0} \otimes \bm{1} + e_1 \cdot \bm{1} \otimes \bp{s_0}
\end{align*}

(here we have used that $\rep_2$ is a homomorphism of left $E^0$-modules, by Lemma <ref>). Hence, the value of $x - \rep_2(\mult_2(x))$ for $x = \bz{s_0} \otimes \bz{s_0}$ is

\[
\bz{s_0} \otimes \bz{s_0}
+ e_{\idd^{-1}} \cdot \bm{1} \otimes \bz{s_0} + e_{\idd} \cdot \bz{s_0} \otimes \bm{1} - e_1 \cdot \bm{1} \otimes \bp{s_0}.
\]

Now it remains to compute the value of $x - \rep_2(\mult_2(x))$ for $x = \bz{s_1} \otimes \bz{s_1}$, but since $\Gamma_\varpi(\bz{s_0}) = -\bz{s_1}$ and since $\rep_2$ is $\Gamma_\varpi$-invariant (Lemma <ref>), we see that such value can be obtained by applying $\Gamma_\varpi$ to the last displayed equality. Hence we get

\[
\bz{s_1} \otimes \bz{s_1}
- e_{\idd} \cdot \bp{1} \otimes \bz{s_1} - e_{\idd^{-1}} \cdot \bz{s_1} \otimes \bp{1} - e_1 \cdot \bp{1} \otimes \bm{s_1}
\]

(using the formulas for the action of $\Gamma_\varpi$ on $E^0$ (<ref>) and on $E^1$ (<ref>)).

Let us define $K_2$ as the sub-$E^0$-bimodule of $E^1 \otimes_{E^0} E^1$ generated by the following elements:

\begin{align*}
&\bm{1} \otimes \bm{1}, &&\qquad\qquad\qquad\qquad\bp{1} \otimes \bm{1}, &&\qquad\qquad\qquad\qquad\bz{s_1} \otimes \bm{1},
\\
&\bm{1} \otimes \bp{1}, &&\qquad\qquad\qquad\qquad\bp{1} \otimes \bp{1}, &&\qquad\qquad\qquad\qquad\bz{s_0} \otimes \bp{1},
\\
&\bp{1} \otimes \bz{s_0}, &&\qquad\qquad\qquad\qquad\bz{s_1} \otimes \bz{s_0},
\\
&\bm{1} \otimes \bz{s_1}, &&\qquad\qquad\qquad\qquad\bz{s_0} \otimes \bz{s_1},
\end{aligned}
\\
&\bz{s_0} \otimes \bz{s_0} + e_{\idd^{-1}} \cdot \bm{1} \otimes \bz{s_0} + e_{\idd} \cdot \bz{s_0} \otimes \bm{1} - e_1 \cdot \bm{1} \otimes \bp{s_0},
\\
&\bz{s_1} \otimes \bz{s_1} - e_{\idd} \cdot \bp{1} \otimes \bz{s_1} - e_{\idd^{-1}} \cdot \bz{s_1} \otimes \bp{1} - e_1 \cdot \bp{1} \otimes \bm{s_1},
\\
&\bp{s_0} \otimes \bz{s_0} + \bz{s_0} \otimes \bm{s_0} = - \tau_{s_0} \cdot \bm{1} \otimes \bz{s_0} + \bz{s_0} \otimes \bm{1} \cdot \tau_{s_0},
\\
&\bm{s_1} \otimes \bz{s_1} + \bz{s_1} \otimes \bp{s_1} = - \tau_{s_1} \cdot \bp{1} \otimes \bz{s_1} + \bz{s_1} \otimes \bp{1} \cdot \tau_{s_1}.
\end{align*}

The first twelve elements were obtained in Lemma <ref> (in particular, they lie in $\ker(\mult_2)$). The last two elements are slight modifications of those obtained in Lemma <ref>: more precisely, recalling the definition of $c_{-1}$ from (<ref>), they are respectively equal to

\begin{align*}
(\rep_2 \circ \invol) (\am{1}) \cdot \tau_{c_{-1}} &- (\invol \circ \rep_2) (\am{1}) \cdot \tau_{c_{-1}},
\\
- (\rep_2 \circ \invol) (\ap{1}) \cdot \tau_{c_{-1}} &+ (\invol \circ \rep_2) (\ap{1}) \cdot \tau_{c_{-1}}
\end{align*}

(in particular, they lie in $\ker(\mult_2)$, too, since $\invol$ and $\mult_2$ commute by (<ref>)).

We want to prove that $K_2$ is actually the full $\ker(\rep_2)$. Let us start with some lemmas.

One has that $K_2$ is $\Gamma_\varpi$-invariant and $\invol$-invariant.

It suffices to prove that the generators listed in the definition of $K_2$ (Remark <ref>) are $\Gamma_\varpi$-invariant and $\invol$-invariant. From the formulas for the action of $\Gamma_\varpi$ (<ref>) and of $\invol$ (<ref>) on $E^1$, it is immediate to see that applying $\Gamma_\varpi$ or $\invol$ to each of the first ten generators we get, up to a sign, again one of such generators. The same is true for the last two generators. It remains to study the behaviour of $\Gamma_\varpi$ and $\invol$ on the following two elements:

\begin{align*}
x_0 &\defeq \bz{s_0} \otimes \bz{s_0} + e_{\idd^{-1}} \cdot \bm{1} \otimes \bz{s_0} + e_{\idd} \cdot \bz{s_0} \otimes \bm{1} - e_1 \cdot \bm{1} \otimes \bp{s_0},
\\
x_1 &\defeq \bz{s_1} \otimes \bz{s_1} - e_{\idd} \cdot \bp{1} \otimes \bz{s_1} - e_{\idd^{-1}} \cdot \bz{s_1} \otimes \bp{1} - e_1 \cdot \bp{1} \otimes \bm{s_1}.
\end{align*}

In the proof of Lemma <ref>, the element $x_1$ was obtained by applying $\Gamma_\varpi$ to $x_0$: in other words $\Gamma_\varpi(x_0) = x_1$ and hence $\Gamma_\varpi(x_1) = x_0$. Now let us compute $\invol(x_0)$, using the formula (<ref>) for the action of $\invol$ on $E^1$ and the formula (<ref>) for the action of the idempotents on $E^1$:

\begin{align*}
\invol(x_0)
\invol \left(\bz{s_0} \otimes \bz{s_0} + e_{\idd^{-1}} \cdot \bm{1} \otimes \bz{s_0} + e_{\idd} \cdot \bz{s_0} \otimes \bm{1} - e_1 \cdot \bm{1} \otimes \bp{s_0}\right)
\\&=
- \bz{s_0^{-1}} \otimes \bz{s_0^{-1}} + \bz{s_0^{-1}} \otimes \bm{1} \cdot e_{\idd} + \bm{1} \otimes \bz{s_0^{-1}} \cdot e_{\idd^{-1}} - \bm{s_0^{-1}} \otimes \bm{1} \cdot e_1
\\&=
- \bz{s_0} \otimes \bz{s_0} + \bz{s_0^{-1}} \otimes e_{\idd^{-1}} \bm{1} + \bm{1} \otimes e_{\idd} \bz{s_0^{-1}} - \bm{s_0^{-1}} \otimes e_{\idd^{-2}} \bm{1}
\\&=
- \bz{s_0} \otimes \bz{s_0} + e_{\idd} \cdot \bz{s_0^{-1}} \otimes \bm{1} + e_{\idd^{-1}} \cdot \bm{1} \otimes \bz{s_0^{-1}} - e_{1} \cdot \bm{s_0^{-1}} \otimes \bm{1}
\\&=
- \bz{s_0} \otimes \bz{s_0} - e_{\idd} \cdot \bz{s_0} \otimes \bm{1} - e_{\idd^{-1}} \cdot \bm{1} \otimes \bz{s_0} + e_{1} \cdot \bm{1} \otimes \bp{s_0}
\\&=
\end{align*}

Now it remains to treat $x_1$: we have already recalled that $x_1 = \Gamma_\varpi(x_0)$. We can use the fact that $\Gamma_\varpi$ and $\invol$ commute (on $E^\ast$, as recalled in (<ref>), and hence also on $T_{E^0}^\ast E^1$), getting that

\[
\invol(x_1)
\invol(\Gamma_\varpi( x_0))
\Gamma_\varpi(\invol(x_0))
\Gamma_\varpi(-x_0)
\qedhere
\]

The two composite maps

\[\begin{tikzcd}[row sep = 0em]
E^2 \ar[r, "{\rep_2}"] & E^1 \otimes_{E^0} E^1 \ar[r, "{\invol}"] & E^1 \otimes_{E^0} E^1 \ar[r, "{\text{quot.}}"] & \quoz{(E^1 \otimes_{E^0} E^1)}{K_2},
\\
E^2 \ar[r, "{\invol}"] & E^2 \ar[r, "{\rep_2}"] & E^1 \otimes_{E^0} E^1 \ar[r, "{\text{quot.}}"] & \quoz{(E^1 \otimes_{E^0} E^1)}{K_2}
\end{tikzcd}\]

By Lemma <ref> the two maps coincide on $F^1E^2$, and by Remark <ref> they also coincide at $\am{1}$ and at $\ap{1}$. For $\omega \in \quoz{T^0}{T^1}$, both maps transform left multiplication by $\tau_\omega$ into right multiplication by $\tau_{\omega^{-1}}$ by Lemma <ref>, and so we conclude that the two maps coincide on the whole $E^2$.

One has that the composite map

\[
\begin{tikzcd}[column sep = 4em]
\ar[r, "{\rep_2}"]
E^1 \otimes_{E^0} E^1
\ar[r, "\text{quot. map}"]
\quoz{(E^1 \otimes_{E^0} E^1)}{K_2}
\end{tikzcd}
\]

is a homomorphism of $E^0$-bimodules.

We will show that the following equalities are true:

$\rep_2(\tau_{s_0} \cdot \alpha_{s_1v}^\sigma) \equiv \tau_{s_0} \cdot \rep_2(\alpha_{s_1v}^\sigma)$ modulo $K_2$ for $v \in \widetilde{W}$ with $\ell(s_1v) = \ell(v) + 1$ and for $\sigma \in \{-,0,+\}$ (we will see that in this case we actually have a true equality in $E^1 \otimes_{E^0} E^1$, with no need to consider the quotient modulo $K_2$);

$\rep_2(\tau_{s_1} \cdot \alpha_{s_1v}^\sigma) \equiv \tau_{s_1} \cdot \rep_2(\alpha_{s_1v}^\sigma)$ modulo $K_2$ for $v \in \widetilde{W}$ with $\ell(s_1v) = \ell(v) + 1$ and for $\sigma \in \{-,0,+\}$;

$\rep_2(\tau_{s_i} \cdot \ap{1}) \equiv \tau_{s_i} \cdot \rep_2(\ap{1})$ modulo $K_2$ for $i \in \{0,1\}$.

Before checking these three properties, let us show that, if they hold, then the lemma is proved. We saw in Lemma <ref> that $\Gamma_\varpi$ commutes with $\rep_2$ and in Lemma <ref> that $K_2$ is $\Gamma_\varpi$-invariant, and so by applying $\Gamma_\varpi$ to the equalities in (i), (ii) and (iii) (and using (<ref>) and (<ref>)), we get:

$\rep_2(\tau_{s_1} \cdot \alpha_{s_0w}^\sigma) \equiv \tau_{s_1} \cdot \rep_2(\alpha_{s_0w}^\sigma) \mod K_2$ for $w \in \widetilde{W}$ with $\ell(s_0w) = \ell(w) + 1$ and for $\sigma \in \{-,0,+\}$ (as before, we already have equality in $E^1 \otimes_{E^0} E^1$);

$\rep_2(\tau_{s_0} \cdot \alpha_{s_0w}^\sigma) \equiv \tau_{s_0} \cdot \rep_2(\alpha_{s_0w}^\sigma) \mod K_2$ for $w \in \widetilde{W}$ with $\ell(s_0w) = \ell(w) + 1$ and for $\sigma \in \{-,0,+\}$;

$\rep_2(\tau_{s_i} \cdot \am{1}) \equiv \tau_{s_i} \cdot \rep_2(\am{1}) \mod K_2$ for $i \in \{0,1\}$.

Note that, for $\omega \in \quoz{T^0}{T^1}$, in (iii) and (iii') we can consider the analogous congruences with $\ap{\omega}$ (respectively, of $\am{\omega}$) in place of $\ap{1}$ (respectively, of $\am{1}$): the new congruences are still true because $\rep_2$ is a homomorphism of left $k[\quoz{T^0}{T^1}]$-modules (by Lemma <ref>, or just by the definition of $\rep_2$ in Lemma <ref>).

All in all, this shows that the congruence

\begin{equation}\label{eqCongruence}
\rep_2(x \cdot \alpha) \equiv x \cdot \rep_2(\alpha) \mod K_2
\end{equation}

is true for $\alpha$ running through a $k$-basis of $E^2$ and for $x \in \{\tau_{s_0},\tau_{s_1}\}$. But since $\rep_2$ is a homomorphism of left $k[\quoz{T^0}{T^1}]$-modules, the same congruence is also true for $x = \tau_{\omega}$ for $\omega \in \quoz{T^0}{T^1}$. In other words (<ref>) is true for $\alpha$ running through a $k$-basis of $E^2$ and for $x$ running through a set of generators of $E^0$ as a $k$-algebra, and hence it follows that it must be true for all $\alpha \in E^2$ and all $x \in E^0$, completing the proof that $\rep_2$ becomes a homomorphism of left $E^0$-modules after modding out $K_2$.

To prove the same claim for the right $E^0$-action, since $K_2$ is $\invol$-invariant by Lemma <ref>, applying $\invol$ to (<ref>) we obtain that

\[
(\invol \circ \rep_2)(x \cdot \alpha) \equiv (\invol \circ \rep_2)(\alpha) \cdot \invol(x) \mod K_2
\]

for all $x \in E^0$ and all $\alpha \in E^2$. Since the maps $\invol \circ \rep_2$ and $\rep_2 \circ \invol$ coincide after modding out by $K_2$ (Lemma <ref>), we deduce that

\[
\rep_2(\invol(\alpha) \cdot \invol(x)) \equiv \rep_2(\invol(\alpha)) \cdot \invol(x) \mod K_2,
\]

completing the proof that $\rep_2$ becomes a homomorphism of right $E^0$-modules after modding out $K_2$.

Now, it remains to prove (i), (ii) and (iii).

One has

\begin{align*}
\tau_{s_0} \cdot \rep_2(\am{s_1v})
- \tau_{s_0} \cdot \bp{1} \otimes \bz{s_1v}
\\&=
&&\text{by \eqref{eqFormulasTauQp}}
\\&=
\rep_2(\tau_{s_0} \cdot \am{s_1v})
&&\text{by \eqref{eqExplicitDeg2AddUp}.}
\end{align*}

\begin{align*}
\tau_{s_0} \cdot \rep_2(\az{s_1v})
\tau_{s_0} \cdot \bp{1} \otimes \bm{s_1v}
\\&=
&&\text{by \eqref{eqFormulasTauQp}}
\\&=
\rep_2(\tau_{s_0} \cdot \az{s_1v})
&&\text{by \eqref{eqExplicitDeg2AddUp}.}
\end{align*}

And finally,

\begin{align*}
\tau_{s_0} \cdot \rep_2(\ap{s_1v})
\tau_{s_0} \cdot \bz{s_1} \otimes \bp{v}
\\&=
- \bz{s_0s_1} \otimes \bp{v}
&&\text{by \eqref{eqFormulasTauQp}}
\\&=
\bz{s_0} \otimes \bm{s_1v}
&&\text{by \eqref{eqFormulaRighteasyQp} and \eqref{eqFormulasTauQp}}
\\&=
- \rep_2(\am{s_0s_1v})
\\&=
\rep_2(\tau_{s_0} \cdot \ap{s_1v})
&&\text{by \eqref{eqExplicitDeg2AddUp}.}
\end{align*}

Let us prove that $\rep_2 \big( \tau_{s_1} \cdot \am{s_1v} \big) \equiv \tau_{s_1} \cdot \rep_2 ( \am{s_1v} )$ modulo $K_2$:
we first assume that $\ell(v) \geqslant 1$, and compute

\begin{align*}
\rep_2(\tau_{s_1} \cdot \am{s_1v})
- e_1 \rep_2(\am{s_1 v}) - \rep_2(\ap{c_{-1} v})
\\&\qquad\qquad\qquad\qquad\text{by \eqref{eqTauS1OnE2LeftNotAddUp} and Lemma \ref{lmmBimodulesGpAlgebraFiniteTorus}}
\\&=
e_1 \bp{1} \otimes \bz{s_1 v} - \bm{1} \otimes \bz{c_{-1} v},
\end{align*}

\begin{align*}
\tau_{s_1} \cdot \rep_2(\am{s_1v})
- \tau_{s_1} \cdot \bp{1} \otimes \bz{s_1v}
\\&=
\tau_{s_1} \cdot \bp{s_1} \otimes \bz{v}
\\&\qquad\qquad\qquad\qquad\text{by \eqref{eqFormulasTauQp} and \eqref{eqFormulaRighteasyQp}}
\\&=
-e_1 \bp{s_1}  \otimes \bz{v}
+ 2e_{\idd^{-1}} \bz{s_1}  \otimes \bz{v}
\\&\qquad+ e_{\idd^{-2}} \bm{s_1} \otimes \bz{v}
- \bm{c_{-1}} \otimes \bz{v}
\\&\qquad\qquad\qquad\qquad\text{by \eqref{eqTauS1OnE1LeftNotAddUp}}
\\&=
e_1 \bp{1}  \otimes \bz{s_1v}
+ 2e_{\idd^{-1}} \bz{s_1}  \otimes \bz{s_0} \cdot \tau_{s_0^{-1} v}
\\&\qquad- e_{\idd^{-2}} \tau_{s_1} \cdot \bp{1} \otimes \bz{s_0} \cdot \tau_{s_0^{-1} v}
- \bm{1} \otimes \bz{c_{-1} v}
\\&\qquad\qquad\qquad\qquad\text{by \eqref{eqFormulasTauQp}, \eqref{eqFormulaRighteasyQp}, \eqref{eqOmegaOnE1Left} and \eqref{eqOmegaOnE1Right}.}
\end{align*}

Since both $\bz{s_1}  \otimes \bz{s_0}$ and $\bp{1} \otimes \bz{s_0}$ lie in the list of generators defining $K_2$, we conclude that $\rep_2 \big( \tau_{s_1} \cdot \am{s_1v} \big) \equiv \tau_{s_1} \cdot \rep_2 ( \am{s_1v} )$ modulo $K_2$.

We now consider the case $\ell(v) = 0$: by definition of $\rep_2$ we have

\[
\rep_2(\ap{c_{-1}v})
\rep_2 \big( \ap{c_{-1}v} + \tau_{s_1} \cdot \am{s_1 v} \big) - \tau_{s_1} \cdot \rep_2 \big( \am{s_1 v} \big),
\]

and then the claimed congruence is actually an equality in $E^1 \otimes_{E^0} E^1$ by $k$-linearity of $\rep_2$.

Let us prove that $\rep_2 \big( \tau_{s_1} \cdot \az{s_1v} \big) \equiv \tau_{s_1} \cdot \rep_2 ( \az{s_1v} )$ modulo $K_2$: we compute

\begin{align*}
\rep_2(\tau_{s_1} \cdot \az{s_1v})
\begin{cases}
\begin{matrix*}[l]
- e_1 \rep_2(\az{s_1v}) -2 e_{\idd^{-1}} \rep_2(\ap{s_1v})
\\\qquad
- \rep_2(\az{c_{-1}v})
\end{matrix*}
&\text{if $\ell(v) \geqslant 1$}
\\
- e_1 \rep_2(\az{s_1v}) -2 e_{\idd^{-1}} \rep_2(\ap{s_1v})
&\text{if $\ell(v) = 0$}
\end{cases}
\\&\qquad\qquad\qquad\qquad\text{by \eqref{eqTauS1OnE2LeftNotAddUp} and Lemma \ref{lmmBimodulesGpAlgebraFiniteTorus}}
\\&=
\begin{cases}
\begin{matrix*}[l]
- e_1 \bp{1} \otimes \bm{s_1v} -2 e_{\idd^{-1}} \bz{s_1} \otimes \bp{v}
\\\qquad
+ \bm{1} \otimes \bp{c_{-1}v}
\end{matrix*}
&\text{if $\ell(v) \geqslant 1$}
\\
- e_1 \bp{1} \otimes \bm{s_1v} -2 e_{\idd^{-1}} \bz{s_1} \otimes \bp{v}
&\text{if $\ell(v) = 0$,}
\end{cases}
\end{align*}

\begin{align*}
\tau_{s_1} \cdot \rep_2(\az{s_1v})
\tau_{s_1} \cdot \bp{1} \otimes \bm{s_1v}
\\&=
- \tau_{s_1} \cdot \bp{s_1} \otimes \bp{v}
\\&\qquad\qquad\qquad\qquad\qquad\text{by \eqref{eqFormulasTauQp} and \eqref{eqFormulaRighteasyQp}}
\\&=
e_1 \bp{s_1} \otimes \bp{v}
- 2e_{\idd^{-1}} \bz{s_1} \otimes \bp{v}
\\&\qquad - e_{\idd^{-2}} \bm{s_1} \otimes \bp{v}
+ \bm{c_{-1}} \otimes \bp{v}
\\&\qquad\qquad\qquad\qquad\qquad\text{by \eqref{eqTauS1OnE1LeftNotAddUp}}
\\&=
- e_1 \bp{1} \otimes \bm{s_1v}
- 2e_{\idd^{-1}} \bz{s_1} \otimes \bp{v}
\\&\qquad - e_{\idd^{-2}} \bm{s_1} \otimes \bp{v}
+ \bm{1} \otimes \bp{c_{-1}v}
\\&\qquad\qquad\qquad\qquad\qquad\text{by \eqref{eqFormulasTauQp}, \eqref{eqFormulaRighteasyQp}, \eqref{eqOmegaOnE1Left} and \eqref{eqOmegaOnE1Right}.}
\end{align*}

Now, if $\ell(v) \geqslant 1$, then using (<ref>) and (<ref>) we see that

\[
\bm{s_1} \otimes \bp{v} = - (\bm{s_1} \tau_{s_0}) \otimes \bm{s_0^{-1} v} = 0,
\]

and hence we conclude that $\rep_2(\tau_{s_1} \cdot \az{s_1v}) = \tau_{s_1} \cdot \rep_2(\az{s_1v})$.

If instead $\ell(v) = 0$, then using (<ref>) and (<ref>) we see that

\begin{align*}
\bm{s_1} \otimes \bp{v}
- \tau_{s_1} \cdot \bp{1} \otimes \bp{1} \cdot \tau_v
\in
\\
\bm{1} \otimes \bp{c_{-1}v}
\bm{1} \otimes \bp{1} \cdot \tau_{c_{-1}v}
\in
\end{align*}

and hence we conclude that $\rep_2(\tau_{s_1} \cdot \az{s_1v}) \equiv \tau_{s_1} \cdot \rep_2(\az{s_1v})$ modulo $K_2$.

Let us prove that $\rep_2 \big( \tau_{s_1} \cdot \ap{s_1v} \big) \equiv \tau_{s_1} \cdot \rep_2 ( \ap{s_1v} )$ modulo $K_2$:
we compute

\begin{align*}
\rep_2 \big( \tau_{s_1} \cdot \ap{s_1v} \big)
- \rep_2 \big( e_1 \ap{s_1v} \big)
&&\text{by \eqref{eqTauS1OnE2LeftNotAddUp}}
\\&=
- e_1 \rep_2 \big( \ap{s_1v} \big)
&&\text{by Lemma \ref{lmmBimodulesGpAlgebraFiniteTorus}}
\\&=
- e_1 \bz{s_1} \otimes \bp{v},
\end{align*}

\begin{align*}
\tau_{s_1} \cdot \rep_2 ( \ap{s_1v} )
\tau_{s_1} \cdot \bz{s_1} \otimes \bp{v}
\\&=
-e_1 \bz{s_1} \otimes \bp{v} - e_{\idd^{-1}}\bm{s_1} \otimes \bp{v}
&&\text{by \eqref{eqTauS1OnE1LeftNotAddUp}.}
\end{align*}

If $\ell(v) \geqslant 1$, we can write $v$ as $s_0v'$ for some $v' \in \widetilde{W}$ such that lengths add up. Using formulas (<ref>) and (<ref>), we obtain

\begin{align*}
\bm{s_1} \otimes \bp{v}
- (\bm{s_1} \cdot \tau_{s_0}) \otimes \bm{v}
\end{align*}

If instead $\ell(v) = 0$, using formulas (<ref>) and (<ref>), we obtain

\begin{align*}
- e_{\idd^{-1}}\bm{s_1} \otimes \bp{v}
e_{\idd^{-1}} \tau_{s_1} \cdot \bp{1} \otimes \bp{1} \cdot \tau_v
\in%\\&\in
\end{align*}

Therefore, both in the case $\ell(v) \geqslant 1$ and in the case $\ell(v) = 0$, we obtain that $\rep_2 \big( \tau_{s_1} \cdot \ap{s_1v} \big) \equiv \tau_{s_1} \cdot \rep_2 ( \ap{s_1v} )$ modulo $K_2$.

Let us prove that $\rep_2 \big( \tau_{s_1} \cdot \ap{1} \big) \equiv \tau_{s_1} \cdot \rep_2 ( \ap{1} )$ modulo $K_2$:
we compute

\begin{align*}
\tau_{s_1} \cdot \rep_2(\ap{1})
\tau_{s_1} \cdot \left( \rep_2 \big( \ap{1} + \tau_{s_1} \cdot \am{s_1^{-1}} \big) - \tau_{s_1} \cdot \rep_2 \big( \am{s_1^{-1}} \big) \right)
\\&\equiv
\rep_2 \big( \tau_{s_1} \cdot (\ap{1} + \tau_{s_1} \cdot \am{s_1^{-1}}) \big) - \tau_{s_1}^2 \cdot \rep_2 \big( \am{s_1^{-1}} \big)
\mod K_2
\\&\qquad\qquad
\substack{\text{by (ii), since $\ap{1} + \tau_{s_1} \cdot \am{s_1^{-1}} \in \bigoplus_{\omega \in \quoz{T^0}{T^1}} H^2(I,\xx(s_1\omega))$}}
\\&\equiv
\rep_2 \big( \tau_{s_1} \cdot (\ap{1} + \tau_{s_1} \cdot \am{s_1^{-1}}) \big) - \rep_2 \big( \tau_{s_1}^2 \cdot \am{s_1^{-1}} \big)
\mod K_2
\\&\qquad\qquad
\substack{\text{again by (ii)}}
\\&=
\rep_2(\tau_{s_1} \cdot \ap{1}).
\end{align*}

Let us prove that $\rep_2 \big( \tau_{s_0} \cdot \ap{1} \big) \equiv \tau_{s_0} \cdot \rep_2 ( \ap{1} )$ modulo $K_2$. Since $K_2$ is $\invol$-invariant, this is equivalent to showing that

\[
(\invol \circ \rep_2) \big( \tau_{s_0} \cdot \ap{1} \big) \equiv (\invol \circ \rep_2) ( \ap{1} ) \cdot \tau_{s_0^{-1}}
\mod K_2.
\]

Recall from Lemma <ref> that the maps $\invol \circ \rep_2$ and $\rep_2 \circ \invol$ coincide after post-composing with the quotient map onto $\quoz{(E^1 \otimes_{E^0} E^1)}{K_2}$. Therefore our claim is equivalent to showing that

\[
(\rep_2 \circ \invol) \big( \tau_{s_0} \cdot \ap{1} \big) \equiv (\rep_2 \circ \invol) ( \ap{1} ) \cdot \tau_{s_0^{-1}}
\mod K_2.
\]

I.e., using the formula (<ref>) for the action of $\invol$ on $E^2$, this is equivalent to showing that

\begin{equation}\label{eqAp1Claim}
\rep_2 \big( \ap{1} \cdot \tau_{s_0^{-1}} \big) \equiv \rep_2 ( \ap{1} ) \cdot \tau_{s_0^{-1}}
\mod K_2.
\end{equation}

By definition of $\rep_2$, the right hand side of the above equation is

\begin{equation}\label{eqAp1Reversed}
\rep_2 ( \ap{1} ) \cdot \tau_{s_0^{-1}}
\rep_2 \big( \ap{1} + \tau_{s_1} \cdot \am{s_1^{-1}} \big)
\cdot \tau_{s_0^{-1}}
- \tau_{s_1} \cdot \rep_2 \big( \am{s_1^{-1}} \big)
\cdot \tau_{s_0^{-1}}.
\end{equation}

From (i) we know that

\[
\tau_{s_0} \cdot \rep_2(\alpha) = \rep_2(\tau_{s_0} \cdot \alpha)
\]

for all $\alpha \in \oplus_{\omega \in \quoz{T^0}{T^1}} H^2(I,\xx(\omega s_1))$, and therefore for the same $\alpha$'s we have

\[
(\invol \circ \rep_2)(\alpha) \cdot \tau_{s_0^{-1}} = (\invol \circ \rep_2)(\tau_{s_0} \cdot \alpha).
\]

As explained above, we can replace $\invol \circ \rep_2$ with $\rep_2 \circ \invol$ if we content ourselves with a congruence modulo $K_2$. As the subspace $\oplus_{\omega \in \quoz{T^0}{T^1}} H^2(I,\xx(\omega s_1))$ is $\invol$-invariant, we then obtain

\[
\rep_2(\alpha) \cdot \tau_{s_0^{-1}} \equiv \rep_2(\alpha \cdot \tau_{s_0^{-1}}) \mod K_2
\]

for all $\alpha \in \oplus_{\omega \in \quoz{T^0}{T^1}} H^2(I,\xx(\omega s_1))$. Since both $\ap{1} + \tau_{s_1} \cdot \am{s_1^{-1}}$ and $\am{s_1^{-1}}$ lie in such subspace, applying this latter fact to (<ref>) we obtain

\begin{equation}\label{eqAp1Reversed2}\begin{aligned}
\rep_2 ( \ap{1} ) \cdot \tau_{s_0^{-1}}
\rep_2 \big( \ap{1} + \tau_{s_1} \cdot \am{s_1^{-1}} \big)
\cdot \tau_{s_0^{-1}}
\\&\qquad - \tau_{s_1} \cdot \rep_2 \big( \am{s_1^{-1}} \big)
\cdot \tau_{s_0^{-1}}
\\&\equiv
\rep_2 \big( \ap{1} \cdot \tau_{s_0^{-1}} + \tau_{s_1} \cdot \am{s_1^{-1}} \cdot \tau_{s_0^{-1}} \big)
\\&\qquad - \tau_{s_1} \cdot \rep_2 \big( \am{s_1^{-1}} \cdot \tau_{s_0^{-1}} \big)
\mod K_2.
\end{aligned}\end{equation}

Now, since $\am{s_1^{-1}} \cdot \tau_{s_0^{-1}}$ lies in $H^2(I,\xx(s_1s_0))$, by (ii) we have that

\[
\tau_{s_1} \cdot \rep_2 \big( \am{s_1^{-1}} \cdot \tau_{s_0^{-1}} \big)
\equiv
\rep_2 \big( \tau_{s_1} \cdot \am{s_1^{-1}} \cdot \tau_{s_0^{-1}} \big)
\mod K_2.
\]

Therefore, continuing the computation in (<ref>), we obtain

\[
\rep_2 ( \ap{1} ) \cdot \tau_{s_0^{-1}}
\equiv
\rep_2 \big( \ap{1} \cdot \tau_{s_0^{-1}} \big)
\mod K_2,
\]

as we wanted to show in (<ref>).

The kernel of the degree $2$ multiplication map

\[
\funcInline{\mult_2}{T_{E_0}^2 E^1 = E^1 \otimes_{E^0} E^1}{E^2}
\]

is $K_2$ (defined in Remark <ref>).

We consider the maps

\[\begin{tikzcd}[column sep = huge]
\quoz{(E^1 \otimes_{E^0} E^1)}{K_2}
\ar[r, shift left, "{\overline{\mult_2}}"]
\ar[l, shift left, "{\overline{\rep_2}}"]
\end{tikzcd}\]

induced by $\mult_2$ and $\rep_2$ (note that $\overline{\mult_2}$ is well defined because in Remark <ref> we have observed that $K_2 \subseteq \ker(\mult_2)$).

To prove the proposition it suffices to show that $\overline{\rep_2} \circ \overline{\mult_2} = \id_{\quoz{(E^1 \otimes_{E^0} E^1)}{K_2}}$: this can be checked on a set of generators of $E^1 \otimes_{E^0} E^1$ as an $E^0$-bimodule, since Lemma <ref> states that $\overline{\rep_2}$ is a homomorphism of $E^0$-bimodules: then the claim is clear from Lemma <ref> and the definition of $K_2$.

§.§ The 3rd graded piece of the kernel

To study the kernel of the multiplication map $\funcInline{\mult_3}{T^3_{E^0} E^1}{E^3}$, we follow the same strategy as for the kernel of $\mult_2$. Since $E^3$ admits a somewhat simpler basis than $E^2$, some things will be easier.

The multiplication map $\funcInline{\mult_3}{T^3_{E^0} E^1}{E^3}$ is surjective, and the following is a section of it as a homomorphism of $k$-vector spaces.

\[\begin{tikzcd}[row sep = -0.6em]
\rep_3 \colon &[-3.5em] E^3 \ar[r] &[-1em] T_{E^0}^3E^1 &[-1em]
\\
& \phi_{(s_1s_0)^i\omega} \ar[r, mapsto] & - \bp{1} \otimes \bz{(s_1s_0)^i} \otimes \bm{1} \cdot \tau_\omega & \scriptstyle\text{for $i \in \zpiu$, $\omega \in \quoz{T^0}{T^1}$,}
\\
& \phi_{(s_0s_1)^i\omega} \ar[r, mapsto] & \bm{1} \otimes \bz{(s_0s_1)^i} \otimes \bp{1} \cdot \tau_\omega  & \scriptstyle\text{for $i \in \zpiu$, $\omega \in \quoz{T^0}{T^1}$,}
\\
& \phi_{s_0(s_1s_0)^i\omega} \ar[r, mapsto] & - \bm{1} \otimes \bz{s_0(s_1s_0)^i} \otimes \bm{1} \cdot \tau_\omega  & \scriptstyle\text{for $i \in \nn$, $\omega \in \quoz{T^0}{T^1}$,}
\\
& \phi_{s_1(s_0s_1)^i\omega} \ar[r, mapsto] & \bp{1} \otimes \bz{s_1(s_0s_1)^i} \otimes \bp{1} \cdot \tau_\omega  & \scriptstyle\text{for $i \in \nn$, $\omega \in \quoz{T^0}{T^1}$,}
\\
& \phi_{\omega} \ar[r, mapsto] & (\tau_{s_0} + e_1) \cdot \rep_3(\phi_{s_0^{-1}}) \cdot \tau_\omega  & \scriptstyle\text{for $\omega \in \quoz{T^0}{T^1}$.}
\end{tikzcd}\]

It can also be rewritten as

\[\begin{tikzcd}[row sep = -0.6em]
\rep_3 \colon &[-3.5em] E^3 \ar[r] &[-1em] T_{E^0}^3E^1 &[-1em]
\\
& \phi_{s_1v} \ar[r, mapsto] & \bp{1} \otimes \bz{s_1} \otimes \bp{v} &\scriptstyle\text{if $\ell(s_1v) = \ell(v) + 1$,}
\\
& \phi_{s_0w} \ar[r, mapsto] & - \bm{1} \otimes \bz{s_0} \otimes \bm{w} &\scriptstyle\text{if $\ell(s_0w) = \ell(w) + 1$,}
\\
& \phi_{\omega} \ar[r, mapsto] & (\tau_{s_0} + e_1) \cdot \rep_3(\phi_{s_0^{-1}\omega}) & \scriptstyle\text{for $\omega \in \quoz{T^0}{T^1}$.}
\end{tikzcd}\]

Moreover, $\rep_3$ is a homomorphism of $k[\quoz{T^0}{T^1}]$-bimodules. Finally, defining

\[
\defeq
\bigoplus_{\mathclap{\substack{w \in \widetilde{W} \\\text{s.t. } \ell(w) \geqslant 1}}} H^3(I,\xx(w)),
\]

one has that the following diagrams are commutative:

\begin{align*}
&\begin{tikzcd}[row sep = 2.7em, column sep = 4em, ampersand replacement = \&]
\ar[r, "{ \restr{\rep_3}{F^1E^3}}"]
\ar[d, "{ \Gamma_\varpi}"']
\&
T^3_{E^0} E^1
\ar[d, "{ \Gamma_\varpi}"]
\\
\ar[r, "{ \restr{\rep_3}{F^1E^3}}"']
\&
T^3_{E^0} E^1,
\end{tikzcd}
&&\begin{tikzcd}[row sep = 2.7em, column sep = 4em, ampersand replacement = \&]
\ar[r, "{ \restr{\rep_3}{F^1E^3}}"]
\ar[d, "{ \invol}"']
\&
T^3_{E^0} E^1
\ar[d, "{ \invol}", "{\substack{ \beta \otimes \beta' \\ \downmapsto \\ - \invol(\beta') \otimes \invol(\beta) }}"']
\\
\ar[r, "{ \restr{\rep_3}{F^1E^3}}"']
\&
T^3_{E^0} E^1.
\end{tikzcd}
\end{align*}

Note the last line in each of the definitions of $\rep_3$ makes sense because $\rep_3(\phi_{s_0^{-1}})$ is already defined lines before.

Let us first check that $\rep_3$, defined in the first way, is a section of $\mult_3$ (and hence, in particular, $\mult_3$ is surjective). Using the computations (<ref>), (<ref>), (<ref>) and (<ref>) in the proof of Proposition <ref>, one obtains

\begin{align*}
\bm{1} \cdot \bz{s_0(s_1s_0)^i} \cdot \bm{1}
- \bm{s_0(s_1s_0)^i} \cupprod \bz{s_0(s_1s_0)^i} \cupprod \bp{s_0(s_1s_0)^i},
\\
\bp{1} \cdot \bz{s_1(s_0s_1)^i} \cdot \bp{1}
- \bp{s_1(s_0s_1)^i} \cupprod \bz{s_1(s_0s_1)^i} \cupprod \bm{s_1(s_0s_1)^i},
\\
\bm{1} \cdot \bz{(s_0s_1)^i} \cdot \bp{1}
\bm{(s_0s_1)^i} \cupprod \bz{(s_0s_1)^i} \cupprod \bp{(s_0s_1)^i},
\\
\bp{1} \cdot \bz{(s_1s_0)^i} \cdot \bm{1}
\bp{(s_1s_0)^i} \cupprod \bz{(s_1s_0)^i} \cupprod \bm{(s_1s_0)^i}.
\end{align*}

Recalling from (<ref>) that $\phi_w = \bm{w} \cupprod \bz{w} \cupprod \bp{w}$ for all $w \in \widetilde{W}$ such that $\ell(w) \geqslant 1$, we see that $\mult_3 \circ \rep_3$ is the identity map on $F^1 E^3$. For the last line in the definition of $\rep_3$ one uses instead the formula $(\tau_{s_0} + e_1) \cdot \phi_{s_0^{-1}} = \phi_1$ (immediate from (<ref>)), and so we conclude that $\rep_3$ is indeed a section of $\mult_3$.

The claim about the second description of $\rep_3$ follows from the formulas of Lemma <ref>: for example, for all $i \in \zpiu$ and all $\omega \in \quoz{T^0}{T^1}$, we have

\begin{align*}
- \bp{1} \otimes \bz{(s_1s_0)^i} \otimes \bm{1} \cdot \tau_\omega
- \bp{1} \otimes \bz{s_1} \otimes (\tau_{s_0(s_1s_0)^{i-1}} \cdot \bm{1}) \cdot \tau_\omega
\\&=
\bp{1} \otimes \bz{s_1} \otimes \bp{s_0(s_1s_0)^{i-1}} \cdot \tau_\omega
\\&=
\bp{1} \otimes \bz{s_1} \otimes \bp{s_0(s_1s_0)^{i-1} \omega}.
\end{align*}

The fact that $\rep_3$ is a homomorphism of right $K[\quoz{T^0}{T^1}]$-modules is basically true by construction (using the formula $\phi_{w \omega} = \phi_w \cdot \tau_{\omega}$ for all $w \in \widetilde{W}$ and all $\omega \in \quoz{T^0}{T^1}$, see (<ref>)).

The fact that $\rep_3$ is also a homomorphism of left $K[\quoz{T^0}{T^1}]$-modules can be proved exactly as in Lemma <ref>, replacing the references to the formulas involving $E^2$ with those involving $E^3$ (namely, (<ref>)).

The fact that the restriction of $\rep_3$ to $F^1E^3$ commutes with $\invol$ and $\Gamma_\varpi$ can be proved in the same way as in Lemmas <ref> and <ref>, of course replacing the references to the formulas for $E^2$ with the formulas for $E^3$: namely, (<ref>), (<ref>) and (<ref>). Actually, the first description of $\rep_3$ makes the computations for $\invol$ easier than the analogous proof for $\rep_2$.

The map

\[
\func
{T_{E^0}^3 E^1 = E^1 \otimes_{E^0} E^1 \otimes_{E^0} E^1}
\rep_3(\phi_w)
&\text{if $\ell(w) \geqslant 1$,}
\\
\begin{matrix*}[l]
\scriptstyle \frac{1}{4} \big( \rep_3(\phi_w) + (\Gamma_\varpi \circ \rep_3 \circ \Gamma_\varpi)(\phi_w)
\\\scriptstyle \quad + (\invol \circ \rep_3 \circ \invol)(\phi_w) + (\Gamma_\varpi \circ \invol \circ \rep_3 \circ \invol \circ \Gamma_\varpi)(\phi_w) \big)
\\\scriptstyle \quad\quad=\frac{1}{4} \big( \rep_3(\phi_1) + \Gamma_\varpi(\rep_3(\phi_1))
\\\scriptstyle \quad\quad\quad + \invol(\rep_3(\phi_1)) + \Gamma_\varpi (\invol(\rep_3(\phi_1))) \big) \cdot \tau_w
\end{matrix*}
&\text{if $\ell(w) = 0$.}
\end{cases}}
\]

is well defined, is a section of the multiplication map $\mult_3$ as a homomorphism of $k \left[ \quoz{T^0}{T^1} \right]$-bimodules and commutes with $\Gamma_\varpi$ and with $\invol$.

Let us consider

\begin{align*}
\bigoplus_{\mathclap{\substack{w \in \widetilde{W} \\\text{s.t. } \ell(w) \geqslant 1}}} H^3(I,\xx(w)),
\\
\bigoplus_{\mathclap{\omega \in \quoz{T^0}{T^1}}} H^3(I,\xx(\omega)).
\end{align*}

The fact that $\restr{\rep_3'}{F^1E^3}$ is a homomorphism of $k[\quoz{T^0}{T^1}]$-bimodules, the fact that $\mult_3 \circ \restr{\rep_3'}{F^1E^3}$ is the identity and the fact that $\restr{\rep_3'}{F^1E^3}$ commutes with $\Gamma_\varpi$ and $\invol$ are all clear from the previous lemma, and we now have to study what happens on $F_0E^3$.

We will check in the end that the two definitions of $\rep_3'$ on $F_0E^3$ coincide, but for the moment we stick to the first definition: using that $\mult_3 \circ \restr{\rep_3}{F_0E^3}$ is the identity and that $\mult_3$ commutes with $\Gamma_\varpi$ and $\invol$, we see that $\mult_3 \circ \restr{\rep_3'}{F_0E^3}$ is the identity. Let $\omega \in \quoz{T^0}{T^1}$. Since $\restr{\rep_3}{F_0E^3}$ is a homomorphism of $k[\quoz{T^0}{T^1}]$-bimodules, since $\Gamma_\varpi$ transforms the left (respectively, right) action of $\tau_\omega$ into the left (respectively, right) action by $\tau_{\omega^{-1}}$ and since $\invol$ transforms the left (respectively, right) action of $\tau_\omega$ into the right (respectively, left) action by $\tau_{\omega^{-1}}$, we conclude that $\restr{\rep_3'}{F_0E^3}$ is a homomorphism of $k[\quoz{T^0}{T^1}]$-bimodules (also using (<ref>)).

The fact that $\restr{\rep_3'}{F_0E^3}$ commutes with $\Gamma_\varpi$ and $\invol$ is clear from the construction and the fact that $\Gamma_\varpi$ and $\invol$ commute by (<ref>) (on $E^\ast$ and hence on $T^\ast_{E^0} E^1$).

For $\omega \in \quoz{T^0}{T^1}$, the alternative description

\[
\rep_3'(\phi_\omega)
\frac{1}{4} \Big( \rep_3(\phi_1) + \Gamma_\varpi(\rep_3(\phi_1))
+ \invol(\rep_3(\phi_1)) + \Gamma_\varpi (\invol(\rep_3(\phi_1))) \Big) \cdot \tau_\omega
\]

is clear from the fact that $\restr{\rep_3'}{F_0E^3}$ is a homomorphism of right $k[\quoz{T^0}{T^1}]$-modules, together with the fact that both $\Gamma_\varpi$ and $\invol$ fix $\phi_1$ (see (<ref>) and (<ref>)).

Using the formulas for the action of $\Gamma_\varpi$ and $\invol$ on $E^1$ (respectively, (<ref>) and (<ref>)) one readily computes explicitly the four summands in the definition of $\rep_3'(\phi_1)$:

\begin{align*}
\rep_3(\phi_1)
%	(\tau_{s_0} + e_1) \cdot \rep_3(\phi_{s_0^{-1}})
%	\\&=
%	(\tau_{s_0} + e_1) \cdot (- \bm{1} \otimes \bz{s_0^{-1}} \otimes \bm{1})
%	\\&=
- (\tau_{s_0} + e_1) \cdot \bm{1} \otimes \bz{s_0^{-1}} \otimes \bm{1},
\\
\Gamma_\varpi(\rep_3(\phi_1))
%	\Gamma_\varpi \big( - (\tau_{s_0} + e_1) \cdot \bm{1} \otimes \bz{s_0^{-1}} \otimes \bm{1} \big)
%	\\&=
(\tau_{s_1} + e_1) \cdot \bp{1} \otimes \bz{s_1^{-1}} \otimes \bp{1},
\\
\invol(\rep_3(\phi_1))
%	\invol \big( - (\tau_{s_0} + e_1) \cdot \bm{1} \otimes \bz{s_0^{-1}} \otimes \bm{1} \big)
%	\\&=
- \bm{1} \otimes \bz{s_0} \otimes \bm{1} \cdot (\tau_{s_0^{-1}} + e_1),
\\
\Gamma_\varpi (\invol(\rep_3(\phi_1)))
%	\Gamma_\varpi \big( - \bm{1} \otimes \bz{s_0} \otimes \bm{1} \cdot (\tau_{s_0^{-1}} + e_1) \big)
%	\\&=
\bp{1} \otimes \bz{s_1} \otimes \bp{1} \cdot (\tau_{s_1^{-1}} + e_1).
\end{align*}

Let $K_{2,3}$ be the sub-$E^0$-bimodule of $T_{E^0}^3 E^1$ generated by the kernel of the degree $2$ multiplication map, i.e.,

\[
K_{2,3} \defeq \overline{\ker(\mult_2) \otimes_{E^0} E^1} + \overline{E^1 \otimes_{E^0} \ker(\mult_2)} \subseteq T_{E^0}^3 E^1,
\]

where $\overline{(?)}$ denotes the image of $(?)$ in $T_{E^0}^3 E^1$. One has the following congruences:

\begin{align*}
\Gamma_\varpi (\invol(\rep_3(\phi_1))) &\equiv \rep_3(\phi_1) \mod K_{2,3},
\\
\invol(\rep_3(\phi_1)) &\equiv \Gamma_\varpi(\rep_3(\phi_1)) \mod K_{2,3}.
\end{align*}

First of all we note that the second congruence follows from the first one, since $K_{2,3}$ is $\Gamma_\varpi$-invariant (which is true because the multiplication map commutes with $\Gamma_\varpi$, see (<ref>)). To show the first congruence, we first compute a couple of useful equalities and congruences: we have

\begin{align*}
(\tau_{s_0} + e_1) \cdot \bm{1} \cdot \tau_{s_0}
(\tau_{s_0} + e_1) \cdot \bm{s_0}
\\&=
- 2 e_{\idd} \bz{s_0} + e_{\idd^2} \bp{s_0} - \bp{c_{-1}},
\end{align*}

by (<ref>) and (<ref>).
Since both $\bz{s_0} \otimes \bz{s_1}$ and $\bp{s_0} \otimes \bz{s_1} = - \tau_{s_0} \cdot \bm{1} \otimes \bz{s_1}$ lie in $\ker(\mult_2)$ (see, e.g., Lemma <ref> and see (<ref>) for the claimed equality), tensoring both sides of the last equality by $\bz{s_1}$ on the right, we get the congruence

\begin{equation}\label{eqCongOne}
((\tau_{s_0} + e_1) \cdot \bm{1} \cdot \tau_{s_0}) \otimes \bz{s_1}
\equiv
- \bp{1} \otimes \bz{s_1^{-1}}
\mod \ker(\mult_2)
\end{equation}

(also using the formulas for the left and right action of $\tau_{c_{-1}}$ on $E^1$: see (<ref>) and (<ref>)).
Now we apply $\Gamma_\varpi$ to the last congruence, obtaining again a congruence since $\ker(\mult_2)$ is $\Gamma_\varpi$-invariant:

\[
- ((\tau_{s_1} + e_1) \cdot \bp{1} \cdot \tau_{s_1}) \otimes \bz{s_0}
\equiv
\bm{1} \otimes \bz{s_0^{-1}}
\mod \ker(\mult_2),
\]

where we have used the formulas for the action of $\Gamma_\varpi$ on $E^0$ and $E^1$ (see (<ref>) and (<ref>)).
Now we apply $\invol$ (or rather $-\invol$) to the last congruence, obtaining again a congruence since $\ker(\mult_2)$ is $\invol$-invariant (by (<ref>)):

\[
- \invol(\bz{s_0}) \otimes \invol\big((\tau_{s_1} + e_1) \cdot \bp{1} \cdot \tau_{s_1}\big)
\equiv
\invol(\bz{s_0^{-1}}) \otimes \invol(\bm{1})
\mod \ker(\mult_2),
\]

\[
\bz{s_0^{-1}} \otimes (\tau_{s_1^{-1}} \cdot \bp{1} \cdot (\tau_{s_1^{-1}} + e_1))
\equiv
- \bz{s_0} \otimes \bm{1}
\mod \ker(\mult_2),
\]

using the formulas for the action of $\invol$ on $E^0$ and $E^1$ (see (<ref>) and (<ref>)). Equivalently, using the formulas for the action of $\tau_{c_{-1}}$ ((<ref>) and (<ref>)),

\begin{equation}\label{eqCongTwo}
\bz{s_0} \otimes (\tau_{s_1} \cdot \bp{1} \cdot (\tau_{s_1} + e_1))
\equiv
- \bz{s_0^{-1}} \otimes \bm{1}
\mod \ker(\mult_2),
\end{equation}

We now get the desired congruence $\Gamma_\varpi (\invol(\rep_3(\phi_1))) \equiv \rep_3(\phi_1)$ modulo $K_{2,3}$, by putting together (<ref>) and (<ref>): indeed denoting by $\equiv$ congruences modulo $K_{2,3}$, we have

\begin{align*}
&\Gamma_\varpi (\invol(\rep_3(\phi_1)))
\\&\quad=
\bp{1} \otimes \bz{s_1} \otimes \bp{1} \cdot (\tau_{s_1^{-1}} + e_1)
&&\text{by Remark \ref{remRep3Phi1Explicit}}
\\&\quad=
\bp{1} \otimes \bz{s_1} \otimes \bp{1} \cdot \tau_{c_{-1}} \cdot (\tau_{s_1} + e_1)
\\&\quad=
\bp{1} \otimes \bz{s_1^{-1}} \otimes \bp{1} \cdot (\tau_{s_1} + e_1)
&&\text{by \eqref{eqOmegaOnE1Left} and \eqref{eqOmegaOnE1Right}}
\\&\quad\equiv
- ((\tau_{s_0} + e_1) \cdot \bm{1} \cdot \tau_{s_0}) \otimes \bz{s_1} \otimes \bp{1} \cdot (\tau_{s_1} + e_1)
&&\text{by \eqref{eqCongOne}}
\\&\quad=
(\tau_{s_0} + e_1) \cdot \bm{1} \otimes \bz{s_0} \otimes (\tau_{s_1} \cdot \bp{1} \cdot (\tau_{s_1} + e_1))
&&\text{by \eqref{eqFormulasTauQp} and \eqref{eqFormulaRighteasyQp}}
\\&\quad\equiv
- (\tau_{s_0} + e_1) \cdot \bm{1} \otimes \bz{s_0^{-1}} \otimes \bm{1}
&&\text{by \eqref{eqCongTwo}}
\\&\quad=
\rep_3(\phi_1).
\end{align*}

As in the last lemma, let $K_{2,3}$ be the sub-$E^0$-bimodule of $T_{E^0}^3 E^1$ generated by the kernel of the degree $2$ multiplication map, i.e.,

\[
K_{2,3} \defeq \overline{\ker(\mult_2) \otimes_{E^0} E^1} + \overline{E^1 \otimes_{E^0} \ker(\mult_2)} \subseteq T_{E^0}^3 E^1,
\]

where $\overline{(?)}$ denotes the image of $(?)$ in $T_{E^0}^3 E^1$. Furthermore, let $K_{\operatorname{extra},3}$ be the sub-left-$k \left[ \quoz{T^0}{T^1} \right]$-module generated by the following element:

\begin{align*}
&\Gamma_\varpi(\rep_3(\phi_1)) - \rep_3(\phi_1)
\\&\qquad=
(\tau_{s_1} + e_1) \cdot \bp{1} \otimes \bz{s_1^{-1}} \otimes \bp{1}
(\tau_{s_0} + e_1) \cdot \bm{1} \otimes \bz{s_0^{-1}} \otimes \bm{1}.
%	&\Gamma_\varpi (\invol(\rep_3(\phi_1))) - \invol (\rep_3(\phi_1))
%	&&\text{i.e.,}
%	&&\bp{1} \otimes \bz{s_1} \otimes \bp{1} \cdot (\tau_{s_1^{-1}} + e_1)
%	+
%	\bm{1} \otimes \bz{s_0} \otimes \bm{1} \cdot (\tau_{s_0^{-1}} + e_1).
\end{align*}

One has that the kernel $\ker(\mult_3)$ of the multiplication map in degree $3$ coincides with $K_{2,3} + K_{\operatorname{extra},3}$. Furthermore, $K_{\operatorname{extra},3}$ is also the sub-right-$k \left[ \quoz{T^0}{T^1} \right]$-module generated by the above element.

The last claim is easy to see using that $\rep_3$ is a homomorphism of $k \left[ \quoz{T^0}{T^1} \right]$-bimodules (Lemma <ref>), that $\Gamma_\varpi$ transforms multiplication by $\tau_\omega$ into multiplication by $\tau_{\omega^{-1}}$ and that $k \left[ \quoz{T^0}{T^1} \right]$ centralizes $\phi_1$.

We now turn to the proof of the main claim. Let us define

\[
V \defeq K_{2,3} + K_{\operatorname{extra},3} + \image(\rep_3') \subseteq T_{E^0}^3 E^1,
\]

where the map $\rep_3'$ was defined in Lemma <ref>.
To prove the proposition, it suffices to show that $V = T_{E^0}^3 E^1$ (i.e., the inclusion from right to left): indeed assuming this equality, we obtain

\[
K_{2,3} + K_{\operatorname{extra},3} + \image(\rep_3') = \ker(\mult_3) \oplus  \image(\rep_3'),
\]

and this, together with the easy inclusion $K_{2,3} + K_{\operatorname{extra},3} \subseteq \ker(\mult_3)$, shows that we have $K_{2,3} + K_{\operatorname{extra},3} = \ker(\mult_3)$, as we wanted (for the inclusion $K_{\operatorname{extra},3} \subseteq \ker(\mult_3)$, recall from (<ref>) that $\Gamma_\varpi$ commutes with $\mult$ and from (<ref>) that $\Gamma_\varpi$ fixes $\phi_1$).

To reduce the amount of computations, we first make some preliminary observations.

We remark that one has the congruences

\[
\Gamma_\varpi(\invol(\phi_1)) \equiv \rep_3(\phi_1) \equiv \Gamma_\varpi(\rep_3(\phi_1)) \equiv \invol(\rep_3(\phi_1))
\mod K_{2,3} + K_{\operatorname{extra},3}.
\]

Indeed the first and last congruence follow from Lemma <ref>, while the second congruence holds by definition of $K_{\operatorname{extra},3}$. Looking at the definition of $\rep_3'$, this shows that

\begin{equation}\label{eqRep3Rep3Prime}\begin{aligned}
\rep_3'(\phi_1)
\rep_3(\phi_1)
\equiv
\Gamma_\varpi(\rep_3(\phi_1))
\\&\equiv
\invol(\rep_3(\phi_1))
\equiv
\Gamma_\varpi(\invol(\rep_3(\phi_1)))
\mod K_{2,3} + K_{\operatorname{extra},3}.
\end{aligned}\end{equation}

We remark that $V$ is invariant for $\Gamma_\varpi$ and $\invol$.

* The term $K_{2,3}$ is invariant for $\Gamma_\varpi$ and $\invol$ because both involutions commute with $\mult_2$ (see (<ref>)).

* The term $\image(\rep_3')$ is invariant for $\Gamma_\varpi$ and $\invol$ because these involutions commute with $\rep_3'$ (Lemma <ref>).

* The term $K_{\operatorname{extra},3}$ is visibly invariant for $\Gamma_\varpi$. Moreover, applying $\invol$ to the difference $\Gamma_\varpi(\rep_3(\phi_1)) - \rep_3(\phi_1)$ and using that $\Gamma_\varpi$ and $\invol$ commute on $E^\ast$ by (<ref>) and hence also on $T_{E^0}^3 E^1$, we obtain the element $\Gamma_\varpi(\invol(\rep_3(\phi_1))) - \invol(\rep_3(\phi_1))$, which lies in $K_{2,3} + K_{\operatorname{extra},3}$ by part <ref>. Also considering the behaviour of $\invol$ with respect to multiplication by $\tau_\omega$ for $\omega \in \quoz{T^0}{T^1}$, we obtain that $\invol(K_{\operatorname{extra},3}) \subseteq K_{2,3} + K_{\operatorname{extra},3}$.

We further remark that $V$ is a sub-$k \left[ \quoz{T^0}{T^1} \right]$-bimodule of $T_{E^0}^3 E^1$: indeed the term $K_{2,3}$ is clearly even a sub-$E^0$-bimodule, the term $K_{\operatorname{extra},3}$ is a sub-$k \left[ \quoz{T^0}{T^1} \right]$-bimodule as already discussed and the term $\image(\rep_3')$ is a sub-$k \left[ \quoz{T^0}{T^1} \right]$-bimodule because $\rep_3'$ is a homomorphism of $k \left[ \quoz{T^0}{T^1} \right]$-bimodules (Lemma <ref>).

Now we can start the actual proof of the proposition. Let us consider $x \in T_{E^0}^3 E^1$ and let us prove that $x \in V$. Without loss of generality, we can of course assume that $x$ is of the form $y \otimes z$ for $y \in E^1$ and $z \in E^1 \otimes_{E^0} E^1$. Considering the equality

\[
x = y \otimes z = y \otimes (z - \rep_2(\mult_2(z))) + y \otimes \rep_2(\mult_2(z)),
\]

we see that the first summand on the right hand side lies in $\overline{E^1 \otimes_{E^0} \ker(\mult_2)}$ and hence in $V$. Therefore, without loss of generality, we can assume that $x$ is of the form $y \otimes z'$ for some $y \in E^1$ and some $z' \in \image(\rep_2)$. From Lemma <ref> we obtain that $\rep_2(F^1E^2) = \invol(\rep_2(F^1E^2))$. Using the formulas for the action of $\invol$ on $E^1$ (<ref>) and looking at the explicit definition of $\rep_2$ in Lemma <ref> we then see that every $z' \in \image(\rep_2)$ can be written as a sum of simple tensors of the following forms:

\begin{align*}
&u \otimes \bm{1},
&u \otimes \bp{1},
&u \otimes \bz{s_0},
&u \otimes \bz{s_1},
\end{align*}

for $u \in E^1$.
Hence, without loss of generality, we can assume that $z'$ is of one of those forms. To simplify a little further, we note that since $V$ is invariant for $\Gamma_\varpi$, it follows that it suffices to prove that the elements of the following forms lie in $V$:

$y \otimes u \otimes \bm{1}$ for some $u \in E^1$,

$y \otimes u \otimes \bz{s_0}$ for some $u \in E^1$,

because by applying $\Gamma_\varpi$ to such elements we immediately obtain what is left (see (<ref>)). We treat separately elements of the forms <ref> and <ref>.

Let us start the proof that every element of the form

\begin{align*}
&y \otimes u \otimes \bm{1} &&\text{(for some $y,u \in E^1$)}
\end{align*}

lies in $V$. Similarly to what we did before, we can consider the equality

\[
y \otimes u \otimes \bm{1} = (y \otimes u - \rep_2(\mult_2(y \otimes u)))  \otimes \bm{1} + \rep_2(\mult_2(y \otimes u)) \otimes \bm{1},
\]

from which we see that we may consider instead elements of the form $r \otimes \bm{1}$ for $r \in \image (\rep_2)$, and, looking at the explicit definition of $\rep_2$ (Lemma <ref>), we can further assume that $r \otimes \bm{1}$ is of one of the following forms:

\begin{align*}
&\bp{1} \otimes \bz{s_1v} \otimes \bm{1} &&\text{for $v \in \widetilde{W}$ such that $\ell(s_1v) = \ell(v) + 1$,}
\\
&\bp{1} \otimes \bm{s_1v} \otimes \bm{1} &&\text{for $v \in \widetilde{W}$ such that $\ell(s_1v) = \ell(v) + 1$,}
\\
&\bz{s_1} \otimes \bp{v} \otimes \bm{1} &&\text{for $v \in \widetilde{W}$ such that $\ell(s_1v) = \ell(v) + 1$,}
\\
&\bz{s_0} \otimes \bm{w} \otimes \bm{1} &&\text{for $w \in \widetilde{W}$ such that $\ell(s_0w) = \ell(w) + 1$,}
\\
&\bm{1} \otimes \bp{s_0w} \otimes \bm{1} &&\text{for $w \in \widetilde{W}$ such that $\ell(s_0w) = \ell(w) + 1$,}
\\
&\bm{1} \otimes \bz{s_0w} \otimes \bm{1} &&\text{for $w \in \widetilde{W}$ such that $\ell(s_0w) = \ell(w) + 1$,}
\\
&\tau_{s_0} \cdot \rep_2 \big( \ap{s_0^{-1}\omega} \big) \otimes \bm{1} &&\text{for $\omega \in \quoz{T^0}{T^1}$,}
\\
&\tau_{s_1} \cdot \rep_2 \big( \am{s_1^{-1}\omega} \big) \otimes \bm{1} &&\text{for $\omega \in \quoz{T^0}{T^1}$.}
\end{align*}

If $v$ is as above, from Lemma <ref> and from (<ref>) one sees that either $\bm{s_1v} \in k^\times \tau_{s_1v} \bm{1}$ or $\bm{s_1v} \in k^\times \tau_{s_1v} \bp{1}$. This justifies the first line below:

\begin{align*}
\bm{s_1v} \otimes \bm{1} &\in E^0 \cdot \{\bm{1} \otimes \bm{1}, \bp{1} \otimes \bm{1}\},
\\
\bp{v} \otimes \bm{1} &\in E^0 \cdot \{\bm{1} \otimes \bm{1}, \bp{1} \otimes \bm{1}\},
\\
\bm{w} \otimes \bm{1} &\in E^0 \cdot \{\bm{1} \otimes \bm{1}, \bp{1} \otimes \bm{1}\},
\\
\bp{s_0w} \otimes \bm{1} &\in E^0 \cdot \{\bm{1} \otimes \bm{1}, \bp{1} \otimes \bm{1}\},
\end{align*}

and the other lines can be justified with a completely analogous argument.

Since the two elements $\bm{1} \otimes \bm{1}$ and $\bp{1} \otimes \bm{1}$ lie in $\ker(\mult_2)$ (see, e.g., Lemma <ref>), we see that the elements $\bp{1} \otimes \bm{s_1v} \otimes \bm{1}$, $\bz{s_1} \otimes \bp{v} \otimes \bm{1}$, $\bz{s_0} \otimes \bm{w} \otimes \bm{1}$ and $\bm{1} \otimes \bp{s_0w} \otimes \bm{1}$ in the above list all lie in $V$. Also recalling that $V$ is a left-$k \left[ \quoz{T^0}{T^1} \right]$-submodule, we see that we are reduced to showing that the following elements lie in $V$:

\begin{equation}\label{eqNonSo}\begin{aligned}
&\bp{1} \otimes \bz{(s_1s_0)^i} \otimes \bm{1} && \text{for $i \in \zpiu$,}
\\
&\bp{1} \otimes \bz{s_1(s_0s_1)^i} \otimes \bm{1} && \text{for $i \in \nn$,}
\\
&\bm{1} \otimes \bz{(s_0s_1)^i} \otimes \bm{1} && \text{for $i \in \zpiu$,}
\\
&\bm{1} \otimes \bz{s_0(s_1s_0)^i} \otimes \bm{1} && \text{for $i \in \nn$,}
\\
&\tau_{s_0} \cdot \bm{1} \otimes \bz{s_0^{-1}} \otimes \bm{1}, &&
\\
&\tau_{s_1} \cdot \bp{1} \otimes \bz{s_1^{-1}} \otimes \bm{1}. &&
\end{aligned}\end{equation}

Up to a sign, the first line (respectively, the fourth line) is $\rep_3'(\phi_{(s_1s_0)^i})$ (respectively, $\rep_3'(\phi_{s_0(s_1s_0)^i})$), and hence they both lie in $V$. Moreover, we know that $\bz{s_1} \otimes \bm{1} \in \ker(\mult_2)$ (see for example Lemma <ref>), from which it follows that

\begin{align*}
\bz{(s_0s_1)^i} \otimes \bm{1} &= - \tau_{(s_0s_1)^{i-1}s_0} \cdot \bz{s_1} \otimes \bm{1} \in \ker(\mult_2) &&\text{for all $i \in \zpiu$,}
\\
\bz{s_1(s_0s_1)^i} \otimes \bm{1} &= \tau_{(s_1s_0)^i} \cdot \bz{s_1} \otimes \bm{1} \in \ker(\mult_2) &&\text{for all $i \in \nn$}
\end{align*}

(where we used (<ref>)).
Therefore, it follows that also the second, the third and the sixth line in (<ref>) lie in $V$. It remains to consider the element $\tau_{s_0} \cdot \bm{1} \otimes \bz{s_0^{-1}} \otimes \bm{1}$. First of all, we see that there is no harm in considering instead the element $(\tau_{s_0} + e_1) \cdot \bm{1} \otimes \bz{s_0^{-1}} \otimes \bm{1} = - \rep_3(\phi_1)$, since $e_1 \cdot \bm{1} \otimes \bz{s_0^{-1}} \otimes \bm{1} \in \image(\rep_3') \subseteq V$. But we have seen in formula (<ref>) that $\rep_3(\phi_1)$ is congruent to $\rep_3'(\phi_1)$ modulo $K_{2,3} + K_{\operatorname{extra},3}$, and hence $\rep_3(\phi_1) \in V$, thus completing the proof that all the lines in (<ref>) lie in $V$.

We have to consider the elements of the form

\begin{align*}
&y \otimes u \otimes \bz{s_0} = y \otimes u \otimes \bz{s_0} &&\text{(for some $y,u \in E^1$)}
\end{align*}

and prove that they lie in $V$. As before, we see that we may only treat the elements of the form $r \otimes \bz{s_0}$ for $r \in \image (\rep_2)$, and, looking at the explicit definition of $\rep_2$ (Lemma <ref>), we can further assume that $r \otimes \bz{s_0}$ is of one of the following forms:

\begin{align*}
&\bp{1} \otimes \bz{s_1v} \otimes \bz{s_0} &&\text{for $v \in \widetilde{W}$ such that $\ell(s_1v) = \ell(v) + 1$,}
\\
&\bp{1} \otimes \bm{s_1v} \otimes \bz{s_0} &&\text{for $v \in \widetilde{W}$ such that $\ell(s_1v) = \ell(v) + 1$,}
\\
&\bz{s_1} \otimes \bp{v} \otimes \bz{s_0} &&\text{for $v \in \widetilde{W}$ such that $\ell(s_1v) = \ell(v) + 1$,}
\\
&\bz{s_0} \otimes \bm{w} \otimes \bz{s_0} &&\text{for $w \in \widetilde{W}$ such that $\ell(s_0w) = \ell(w) + 1$,}
\\
&\bm{1} \otimes \bp{s_0w} \otimes \bz{s_0} &&\text{for $w \in \widetilde{W}$ such that $\ell(s_0w) = \ell(w) + 1$,}
\\
&\bm{1} \otimes \bz{s_0w} \otimes \bz{s_0} &&\text{for $w \in \widetilde{W}$ such that $\ell(s_0w) = \ell(w) + 1$,}
\\
&\tau_{s_0} \cdot \rep_2 \big( \ap{s_0^{-1}\omega} \big) \otimes \bz{s_0} &&\text{for $\omega \in \quoz{T^0}{T^1}$,}
\\
&\tau_{s_1} \cdot \rep_2 \big( \am{s_1^{-1}\omega} \big) \otimes \bz{s_0} &&\text{for $\omega \in \quoz{T^0}{T^1}$.}
\end{align*}

In part <ref> we have proved that all the elements of the form $x \otimes y \otimes \bm{1}$ (for some $x,y \in E^1$) lie in $V$. As we have already said that $V$ is $\invol$-invariant and $\Gamma_\varpi$-invariant, we deduce that the elements of the following forms lie in $V$:

\begin{align*}
&\bm{1} \otimes x \otimes y &&\text{for some $x,y \in E^1$,}
\\
&\bp{1} \otimes x \otimes y &&\text{for some $x,y \in E^1$.}
\end{align*}

This shows that most of the elements in the above list are in $V$. For the remaining ones, we first use Lemma <ref> and (<ref>) as in part <ref> to reduce to checking that the following elements lie in $V$:

\begin{align*}
& \bz{(s_1s_0)^i} \otimes \bm{1} \otimes \bz{s_0} && \text{for $i \in \zpiu$,}
\\
& \bz{s_1(s_0s_1)^i} \otimes \bp{1} \otimes \bz{s_0} && \text{for $i \in \nn$,}
\\
& \bz{(s_0s_1)^i} \otimes \bp{1} \otimes \bz{s_0} && \text{for $i \in \zpiu$,}
\\
& \bz{s_0(s_1s_0)^i} \otimes \bm{1} \otimes \bz{s_0} && \text{for $i \in \nn$,}
\\
&\tau_{s_0} \cdot \bm{1} \otimes \bz{s_0^{-1}} \otimes \bz{s_0}, &&
\\
&\tau_{s_1} \cdot \bp{1} \otimes \bz{s_1^{-1}} \otimes \bz{s_0}. &&
\end{align*}

We recall from Lemma <ref> that $\bp{1} \otimes \bz{s_0} \in \ker(\mult_2)$ and that the same is true for $\bz{s_1} \otimes \bz{s_0}$, and so it only remains to consider the following elements:

\begin{align}
& \bz{(s_1s_0)^i} \otimes \bm{1} \otimes \bz{s_0} && \text{for $i \in \zpiu$,}
\label{eqBetaS0One}
\\
& \bz{s_0(s_1s_0)^i} \otimes \bm{1} \otimes \bz{s_0} && \text{for $i \in \nn$,}
\label{eqBetaS0Two}
\\
&\tau_{s_0} \cdot \bm{1} \otimes \bz{s_0^{-1}} \otimes \bz{s_0}. &&
\label{eqBetaS0Three}
\end{align}

Let us treat the last line, where, multiplying by $\tau_{c_{-1}}$, we can replace $s_0^{-1}$ by $s_0$. We consider the following element of $\ker(\mult_2)$ (Lemma <ref>):

\[
\bz{s_0} \otimes \bz{s_0} + e_{\idd^{-1}} \cdot \bm{1} \otimes \bz{s_0} + e_{\idd} \cdot \bz{s_0} \otimes \bm{1} - e_1 \cdot \bm{1} \otimes \bp{s_0}.
\]

Tensoring by $\tau_{s_0} \cdot \bm{1}$ on the left we get

\begin{align*}
&\tau_{s_0} \cdot \bm{1} \otimes \bz{s_0} \otimes \bz{s_0}
+ \tau_{s_0} \cdot \bm{1} \otimes e_{\idd^{-1}} \bm{1} \otimes \bz{s_0}
\\&\qquad+ \tau_{s_0} \cdot \bm{1} \otimes e_{\idd} \bz{s_0} \otimes \bm{1}
- \tau_{s_0} \cdot \bm{1} \otimes e_1 \bm{1} \otimes \bp{s_0}
\\&\qquad\qquad\in K_{2,3}.
\end{align*}

Since $\bm{1} \otimes \bm{1}$ lies in $\ker(\rep_2)$, we can delete the two terms where this element appears, getting that

\[
\tau_{s_0} \cdot \bm{1} \otimes \bz{s_0} \otimes \bz{s_0}
+ \tau_{s_0} \cdot \bm{1} \otimes e_{\idd} \bz{s_0} \otimes \bm{1}
\in K_{2,3}.
\]

In part <ref> we have seen that the element $\tau_{s_0} \cdot \bm{1} \otimes \bz{s_0^{-1}} \otimes \bm{1}$ lies in $V$, and so this proves that (<ref>) lies in $V$.

It remains to prove that the elements in (<ref>) and in (<ref>) lie in $V$. We claim that $\bz{s_0} \otimes \bm{1} \otimes \bz{s_0}$ lies in $K_{2,3}$. If we show this, then we are done because $K_{2,3}$ is a sub-$E^0$-bimodule, and using (<ref>) we get that the elements in the lines (<ref>) and (<ref>) lie in $K_{2,3}$ as well and hence also to $V$. So, let us show that $\bz{s_0} \otimes \bm{1} \otimes \bz{s_0}$ lies in $K_{2,3}$.

\begin{equation}\label{eqbzbmbz}\begin{aligned}
\bz{s_0} \otimes \bm{1} \otimes \bz{s_0}
\rep_2(- \am{s_0}) \otimes \bz{s_0}
\\&=
\rep_2(\tau_{s_0} \cdot \ap{1}) \otimes \bz{s_0}
\\&\qquad\qquad\qquad\qquad\text{(by \eqref{eqExplicitDeg2AddUp})}
\\&\equiv
\big( \tau_{s_0} \cdot \rep_2(\ap{1}) \big) \otimes \bz{s_0}
\mod K_{2,3}
\end{aligned}\end{equation}

We now compute $\rep_2(\ap{1})$ explicitly

\begin{align*}
\rep_2(\ap{1})
\rep_2 \big( \ap{1} + \tau_{s_1} \cdot \am{s_1^{-1}} \big) - \tau_{s_1} \cdot \rep_2 \big( \am{s_1^{-1}} \big)
\\&\qquad\qquad\qquad\text{by Lemma \ref{lmmDefRep2}}
\\&=
\rep_2 \big( - e_1 \am{s_1^{-1}} + e_{\idd^{-1}} \az{s_1^{-1}} + e_{\idd^{-2}} \ap{s_1^{-1}} \big) - \tau_{s_1} \cdot \rep_2 \big( \am{s_1^{-1}} \big)
\\&\qquad\qquad\qquad\text{by \eqref{eqTauS1OnE2LeftNotAddUp}}
\\&=
e_1 \cdot \bp{1} \otimes \bz{s_1^{-1}}
+ e_{\idd^{-1}} \cdot \bp{1} \otimes \bm{s_1^{-1}}
\\&\qquad+ e_{\idd^{-2}} \cdot \bz{s_1} \otimes \bp{c_{-1}}
+ \tau_{s_1} \cdot \bp{1} \otimes \bz{s_1^{-1}}
\\&\qquad\qquad\qquad\text{by Lemma \ref{lmmDefRep2} and Lemma \ref{lmmBimodulesGpAlgebraFiniteTorus}}
\\&=
e_1 \cdot \bp{c_{-1}} \otimes \bz{s_1}
- e_{\idd^{-1}} \cdot (\bp{1} \cdot \tau_{s_1^{-1}}) \otimes \bp{1}
\\&\qquad+ e_{\idd^{-2}} \cdot \bz{s_1^{-1}} \otimes \bp{1}
+ \tau_{s_1} \cdot \bp{c_{-1}} \otimes \bz{s_1}
\\&\qquad\qquad\qquad\text{by \eqref{eqOmegaOnE1Left}, \eqref{eqOmegaOnE1Right}, \eqref{eqF}}
\end{align*}

Going back to (<ref>), and recalling from Lemma <ref> that both $\bp{1} \otimes \bz{s_0}$ and $\bz{s_1} \otimes \bz{s_0}$ lie in $\ker(\mult_2)$, we conclude that $\bz{s_0} \otimes \bm{1} \otimes \bz{s_0}$ lies in $K_{2,3}$, as claimed. This concludes the proof that all the elements of the forms (<ref>) and (<ref>) lie in $V$, and with it the proof that all the elements of the form <ref> lie in $V$.

§.§ The 4th graded piece of the kernel

Since $E^4 = 0$, the kernel of the multiplication map $\funcInline{\mult_4}{T_{E^0}^4 E^1}{E^4}$ is of course the whole $T_{E^0}^4 E^1$. As we computed generators for $\ker(\mult_2)$ as an $E^0$-bimodule (Proposition <ref>), and we computed $\ker(\mult_3)$ in terms of $\ker(\mult_2)$ and an additional generator (Proposition <ref>), we now would like to compute $\ker(\mult_4) = T_{E^0}^4 E^1$ in terms of $\ker(\mult_3)$ (and, a priori, some other generators). The following result achieves this, showing that no further generators are needed and hence that $\ker(\mult)$ is generated as a two-sided ideal by its $2$nd and $3$rd graded pieces.

Let $\funcInline{\mult}{T_{E^0}^\ast E^1}{E^\ast}$ be the multiplication map, and let

\[
K_{3,4} \defeq \overline{\ker(\mult_3) \otimes_{E^0} E^1} + \overline{E^1 \otimes_{E^0} \ker(\mult_3)} \subseteq T_{E^0}^4 E^1,
\]

where $\overline{(?)}$ denotes the image of $(?)$ in $T_{E^0}^4 E^1$. One has that $K_{3,4} = T_{E^0}^4 E^1$.

We have to prove that every element of $ T_{E^0}^4 E^1$ lies in $K_{3,4}$, and clearly it suffices to prove this for simple tensors of the form $x \otimes y$ for $x \in E^1$ and $y \in T_{E^0}^3 E^1$. Considering the equality

\[
x \otimes y = x \otimes \left( y - \rep_3(\mult_3(y)) \right) + x \otimes \rep_3(\mult_3(y)),
\]

we see that it suffices to prove our claim for elements of the form $x \otimes y'$ for $x \in E^1$ and $y' \in \image(\mult_3)$. Now looking at the explicit definition of $\rep_3$ in Lemma <ref>, we see that every $y' \in \image(\mult_3)$ can be written as a $k$-linear combination of tensors of the form $z \otimes t \otimes \bm{1}$ for some $z,t \in E^1$ and of the form $z \otimes t \otimes \bp{1}$ for some $z,t \in E^1$ (the presence of $\tau_\omega$ in the definition of $\rep_3$ is irrelevant, since it commutes, up to a scalar, with both $\bm{1}$ and $\bp{1}$). We are thus reduced to showing that the elements of the form $u \otimes \bm{1}$ and $u \otimes \bp{1}$, for $u \in T_{E^0}^3E^1$, lie in $K_{3,4}$. With the same trick as before, we see that we can assume that $u \in \image(\mult_3)$. Using the same property of $\image(\mult_3)$ as above, we are thus reduced to showing that the elements of the form

\begin{align*}
&z \otimes t \otimes \bm{1} \otimes \bm{1},
&z \otimes t \otimes \bp{1} \otimes \bm{1},
\\
&z \otimes t \otimes \bm{1} \otimes \bp{1},
&z \otimes t \otimes \bp{1} \otimes \bp{1},
\end{align*}

(for $z,t \in E^1$) all lie in $K_{3,4}$. This is clear, since all of the elements $\bm{1} \otimes \bm{1}$, $\bp{1} \otimes \bm{1}$, $\bm{1} \otimes \bp{1}$, and $\bp{1} \otimes \bp{1}$ lie in $\ker(\mult_2)$ (see, e.g., Lemma <ref>).

§.§ First main result: finite presentation over the tensor algebra of Eˆ1

We are now going to state the first main result of this section, consisting in a presentation of the $\ext$-algebra $E^\ast$ in terms of the tensor algebra $T_{E^0}^\ast E^1$.

Recall the assumptions stated at the beginning of <ref>.
The multiplication map

\[
\funcInline{\mult}{T_{E^0}^\ast E^1}{E^\ast}
\]

is surjectve and its kernel is generated by its $2$nd and $3$rd graded pieces. Furthermore, the kernel is finitely generated as a two-sided ideal and the following is an explicit list of generators:

\begin{align*}
&\bm{1} \otimes \bm{1},
&&\quad\bp{1} \otimes \bm{1},
&&\quad\bz{s_1} \otimes \bm{1},
&&\quad\bp{1} \otimes \bz{s_0},
&&\quad\bz{s_1} \otimes \bz{s_0},
\\
&\bm{1} \otimes \bp{1},
&&\quad\bp{1} \otimes \bp{1},
&&\quad\bz{s_0} \otimes \bp{1},
&&\quad\bm{1} \otimes \bz{s_1},
&&\quad\bz{s_0} \otimes \bz{s_1},
\end{aligned}
\\
&\bz{s_0} \otimes \bz{s_0} + e_{\idd^{-1}} \cdot \bm{1} \otimes \bz{s_0} + e_{\idd} \cdot \bz{s_0} \otimes \bm{1} - e_1 \cdot \bm{1} \otimes \bp{s_0},
\\
&\bz{s_1} \otimes \bz{s_1} - e_{\idd} \cdot \bp{1} \otimes \bz{s_1} - e_{\idd^{-1}} \cdot \bz{s_1} \otimes \bp{1} - e_1 \cdot \bp{1} \otimes \bm{s_1},
\\
&\bp{s_0} \otimes \bz{s_0} + \bz{s_0} \otimes \bm{s_0} = - \tau_{s_0} \cdot \bm{1} \otimes \bz{s_0} + \bz{s_0} \otimes \bm{1} \cdot \tau_{s_0},
\\
&\bm{s_1} \otimes \bz{s_1} + \bz{s_1} \otimes \bp{s_1} = - \tau_{s_1} \cdot \bp{1} \otimes \bz{s_1} + \bz{s_1} \otimes \bp{1} \cdot \tau_{s_1},
\\
&(\tau_{s_1} + e_1) \cdot \bp{1} \otimes \bz{s_1^{-1}} \otimes \bp{1}
(\tau_{s_0} + e_1) \cdot \bm{1} \otimes \bz{s_0^{-1}} \otimes \bm{1}.
\end{align*}

Surjectivity has been proved in Lemmas <ref> and <ref>, while the description of the kernel follows from Propositions <ref>, <ref> and <ref>.

The statement of the theorem does not exclude the possibility that $\ker(\mult)$ is actually generated by its $2$nd graded piece. It is actually possible to show that this is not the case (see <cit.>). However, the proof is quite long and we do not present it here.

§.§ Second main result: finite presentation as a k-algebra

The purpose of this section is to show that the $\ext$-algebra $E^\ast$ is finitely presented a $k$-algebra, and to compute an explicit presentation (Theorem <ref>). For this, besides Theorem <ref> we will need a presentation of $E^0$ as a $k$-algebra and a presentation of $E^1$ as an $E^0$-bimodule (where by $E^0$-bimodule we mean $E^0 \otimes_k (E^0)^{\opp}$-left-module and not $E^0 \otimes_{\zz} (E^0)^{\opp}$-left-module). We begin by studying the latter problem.

For later purposes, we fix a generator $\omega_0$ of the cyclic group $T^0/T^1 \cong \ff_p^\times$:

\begin{equation}\label{eqOmegaZero}
T^0/T^1 = \{ 1, \omega_0, \omega_0^2, \dots, \omega_0^{p-2} \}.
\end{equation}

Let $M$ be the free $E^0 \otimes_k (E^0)^{\opp}$-left-module of rank $4$ generated by the four symbols $\widehat{\bm{1}}$, $\widehat{\bp{1}}$, $\widehat{\bz{s_0}}$ and $\widehat{\bz{s_1}}$. Let us consider the submodule $N$ generated by the following elements:

\begin{align}
&\tau_{s_1} \cdot \widehat{\bm{1}},
&\tau_{s_0} \cdot \widehat{\bp{1}},
\label{eqBigListLine1}
\\
&\widehat{\bp{1}} \cdot \tau_{s_0},
&\widehat{\bm{1}} \cdot \tau_{s_1},
\label{eqBigListLine2}
\\
(\tau_{s_0} + e_1) \cdot \widehat{\bm{1}} \cdot (\tau_{s_0} + e_1)
\\\qquad
{} + 2 e_{\idd} \widehat{\bz{s_0}} + \tau_{\omega_0}^{\frac{p-1}{2}} \cdot \widehat{\bp{1}},
\end{matrix*}
(\tau_{s_1} + e_1) \cdot \widehat{\bp{1}} \cdot (\tau_{s_1} + e_1)
\\\qquad
{} - 2 e_{\idd^{-1}} \widehat{\bz{s_1}} + \tau_{\omega_0}^{\frac{p-1}{2}} \cdot \widehat{\bm{1}},
\end{matrix*}
\label{eqBigListLine3}
\\
&\tau_{s_0} \cdot \widehat{\bz{s_1}} + \widehat{\bz{s_0}} \cdot \tau_{s_1},
&\tau_{s_1} \cdot \widehat{\bz{s_0}} + \widehat{\bz{s_1}} \cdot \tau_{s_0},
\label{eqBigListLine4}
\\
&(\tau_{s_0} + e_1) \cdot \widehat{\bz{s_0}} + e_{\idd} \tau_{s_0} \cdot \widehat{\bm{1}},
&(\tau_{s_1} + e_1) \cdot \widehat{\bz{s_1}} - e_{\idd^{-1}} \tau_{s_1} \cdot \widehat{\bp{1}},
\label{eqBigListLine5}
\\
&\widehat{\bz{s_0}} \cdot (\tau_{s_0} + e_1) + e_{\idd^{-1}} \widehat{\bm{1}} \cdot \tau_{s_0},
&\widehat{\bz{s_1}} \cdot (\tau_{s_1} + e_1) - e_{\idd} \widehat{\bp{1}} \cdot \tau_{s_1},
\label{eqBigListLine6}
\\
&\tau_{\omega_0} \cdot \widehat{\bm{1}} - u_{\omega_0}^{-2} \widehat{\bm{1}} \cdot \tau_{\omega_0},
&\tau_{\omega_0} \cdot \widehat{\bp{1}} - u_{\omega_0}^{2} \widehat{\bp{1}} \cdot \tau_{\omega_0},
\label{eqBigListLine7}
\\
&\tau_{\omega_0} \cdot \widehat{\bz{s_0}} - \widehat{\bz{s_0}} \cdot \tau_{\omega_0^{-1}},
&\tau_{\omega_0} \cdot \widehat{\bz{s_1}} - \widehat{\bz{s_1}} \cdot \tau_{\omega_0^{-1}}.
\label{eqBigListLine8}
\end{align}

For $\sigma \in \{ \bm{1}, \bp{1}, \bz{s_0}, \bz{s_1} \}$, we denote by $\widetilde{\sigma}$ the image of $\widehat{\sigma}$ in $\quoz{M}{N}$. One has an isomorphism of $E^0 \otimes_k (E^0)^{\opp}$-left-modules

\[
\funcNN
\begin{array}{c}
\widetilde{\sigma}
\\%[-0.5em]
\scriptstyle\text{(for $\sigma \in \{ \bm{1}, \bp{1}, \bz{s_0}, \bz{s_1} \}$)}
\end{array}
\]

To show that the claimed isomorphism is well-defined, we need to show that the elements of $E^1$ that we obtain from the elements in (<ref>)-(<ref>) by replacing $\widehat{\sigma}$ with $\sigma$ (for $\sigma \in \{ \bm{1}, \bp{1}, \bz{s_0}, \bz{s_1} \}$) are all zero. For line (<ref>) the claim follows from (<ref>), for line (<ref>) the claim follows from (<ref>). Regarding line (<ref>), using again such results and also (<ref>), (<ref>) and (<ref>), we compute

\begin{align*}
&(\tau_{s_0} + e_1) \cdot \bm{1} \cdot (\tau_{s_0} + e_1)
\\&\qquad\qquad=
\tau_{s_0} \cdot \bm{1} \cdot \tau_{s_0}
e_1 \cdot \bm{1} \cdot \tau_{s_0}
\tau_{s_0} \cdot \bm{1} \cdot e_1
e_1 \cdot \bm{1} \cdot e_1
\\&\qquad\qquad=
\tau_{s_0} \cdot \bm{s_0}
e_1 \bm{s_0}
e_{\idd^2} \bp{s_0}
\\&\qquad\qquad=
\left(
e_1 \bm{s_0}
- 2
e_{\idd} \bz{s_0}
e_{\idd^2} \bp{s_0}
\bp{c_{-1}}
\right)
e_1 \bm{s_0}
e_{\idd^2} \bp{s_0}
\\&\qquad\qquad=
- 2
e_{\idd} \bz{s_0}
\tau_{\omega_0}^{\frac{p-1}{2}} \cdot \bp{1}.
\end{align*}

This shows the claim for the first part of line (<ref>), and for the second part one can just apply $\Gamma_\varpi$ and use the pertaining formulas.

To show that the elements in line (<ref>) are zero it suffices to use again the formulas (<ref>) and (<ref>). Similarly, for line (<ref>) we use (<ref>) and (<ref>) and for line (<ref>) we use instead (<ref>) and (<ref>). Finally, for lines (<ref>) and (<ref>) we use (<ref>) and (<ref>).

Now, it remains to prove injectivity and surjectivity of our homomorphism. Surjectivity is clear from the fact that the elements $\bm{1}$, $\bp{1}$, $\bz{s_0}$ and $\bz{s_1}$ generate $E^1$ as an $E^0$-bimodule (see Lemma <ref>). To prove injectivity, we fix a $k$-basis $\mathcal{B}$ of $E^1$, and, using surjectivity, for all $b \in \mathcal{B}$ we consider a preimage $m_b \in \quoz{M}{N}$, to be chosen later. If we prove that the elements $m_b$ generate $\quoz{M}{N}$ as a $k$-vector space then injectivity of $P$ follows.

To pursue the above strategy, let us consider the following table: the elements of $\quoz{M}{N}$ in the first column get mapped to the elements of the second column by the formulas in Lemma <ref> and (<ref>). Moreover, the elements in the second column form a basis of $E^1$.

\[\begin{tikzcd}[row sep = -0.6em]
\widetilde{\bz{s_1}} \cdot \tau_{w}
\ar[r, mapsto]&
\bz{s_1w}
&\scriptstyle \text{for $w \in \widetilde{W}$ such that $\ell(s_1w) = \ell(w) + 1$,}
\\
\widetilde{\bz{s_0}} \cdot \tau_{w}
\ar[r, mapsto]&
\bz{s_0w}
&\scriptstyle \text{for $w \in \widetilde{W}$ such that $\ell(s_0w) = \ell(w) + 1$,}
\\
\widetilde{\bm{1}} \cdot \tau_{w}
\ar[r, mapsto]&
\bm{w}
&\scriptstyle \text{for $w \in \widetilde{W}$ such that $\ell(s_1w) = \ell(w) + 1$,}
\\
\widetilde{\bp{1}} \cdot \tau_{w}
\ar[r, mapsto]&
\bp{w}
&\scriptstyle \text{for $w \in \widetilde{W}$ such that $\ell(s_0w) = \ell(w) + 1$,}
\\
\tau_{(s_1s_0)^j} \cdot \widetilde{\bm{1}} \cdot \tau_{\omega}
\ar[r, mapsto]&
\bm{(s_1s_0)^j\omega}
&\scriptstyle \text{for $\omega \in \quoz{T^0}{T^1}$ and $j \in \zpiu$,}
\\
\tau_{s_0(s_1s_0)^j} \cdot \widetilde{\bm{1}} \cdot \tau_{\omega}
\ar[r, mapsto]&
- \bp{s_0(s_1s_0)^j\omega}
&\scriptstyle \text{for $\omega \in \quoz{T^0}{T^1}$ and $j \in \nn$,}
\\
\tau_{(s_0s_1)^j} \cdot \widetilde{\bp{1}} \cdot \tau_{\omega}
\ar[r, mapsto]&
\bp{(s_0s_1)^j\omega}
&\scriptstyle \text{for $\omega \in \quoz{T^0}{T^1}$ and $j \in \zpiu$,}
\\
\tau_{s_1(s_0s_1)^j} \cdot \widetilde{\bp{1}} \cdot \tau_{\omega}
\ar[r, mapsto]&
- \bm{s_1(s_0s_1)^j\omega}
&\scriptstyle \text{for $\omega \in \quoz{T^0}{T^1}$ and $j \in \nn$.}
\end{tikzcd}\]

Therefore, it remains to show that the elements in the first column generate $\quoz{M}{N}$ as a $k$-vector space. For this, it suffices to prove that, for all $v, w \in \widetilde{W}$, the following elements lie in the sub-$k$-vector space $V$ generated by the first column:

\begin{align*}
&\tau_v \cdot \widetilde{\bm{1}} \cdot \tau_w,
&\tau_v \cdot \widetilde{\bp{1}} \cdot \tau_w,
&\tau_v \cdot \widetilde{\bz{s_0}} \cdot \tau_w,
&\tau_v \cdot \widetilde{\bz{s_1}} \cdot \tau_w.
\end{align*}

We are going to prove this claim by induction on $\ell(v) + \ell(w)$.

Let us consider the case $\ell(v) + \ell(w) = 0$. This condition means that $v$ and $w$ lie in $\quoz{T^0}{T^1}$, and so that they are powers of $\omega_0$. Using (<ref>), we see that the elements $\tau_v \cdot \widetilde{\bm{1}} \cdot \tau_w$ (respectively, $\tau_v \cdot \widetilde{\bp{1}} \cdot \tau_w$) can be rewritten, up to a coefficient, in the form $\widetilde{\bm{1}} \cdot \tau_\omega$ (respectively, $\widetilde{\bp{1}} \cdot \tau_\omega$) for a suitable $\omega \in \quoz{T^0}{T^1}$, and so it lies in $V$, as we wanted to show. In the same way one deals with the elements $\tau_v \cdot \widetilde{\bz{s_0}} \cdot \tau_w$ and $\tau_v \cdot \widetilde{\bz{s_1}} \cdot \tau_w$, this time using (<ref>).

Now it remains to consider the induction step. We distinguish some cases.

* Let us consider first $\tau_v \cdot \widetilde{\bm{1}} \cdot \tau_w$ and $\tau_v \cdot \widetilde{\bp{1}} \cdot \tau_w$.

Since both $\widehat{\bp{1}} \cdot \tau_{s_0}$ and $\widehat{\bm{1}} \cdot \tau_{s_1}$ lie in $N$ by (<ref>), we see that it suffices to consider the following cases:

\begin{align*}
&\tau_v \cdot \widetilde{\bm{1}} \cdot \tau_\omega &&\text{for some $\omega \in \quoz{T^0}{T^1}$,}
\\
&\tau_v \cdot \widetilde{\bp{1}} \cdot \tau_\omega &&\text{for some $\omega \in \quoz{T^0}{T^1}$,}
\\
&\tau_v \cdot \widetilde{\bm{1}} \cdot \tau_{s_0 w'} &&\text{for some $w' \in \widetilde{W}$ such that $\ell(s_0 w') = \ell(w') + 1$,}
\\
&\tau_v \cdot \widetilde{\bp{1}} \cdot \tau_{s_1 w'} &&\text{for some $w' \in \widetilde{W}$ such that $\ell(s_1 w') = \ell(w') + 1$.}
\end{align*}

Let us look at the first two elements. Since both $\tau_{s_1} \cdot \widehat{\bm{1}}$ and $\tau_{s_0} \cdot \widehat{\bp{1}}$ lie in $N$ by (<ref>), it suffices to consider the following cases:

\begin{align*}
&\tau_{(s_1s_0)^j \omega'} \cdot \widetilde{\bm{1}} \cdot \tau_\omega &&\text{for some $j \in \nn$ and some $\omega,\omega' \in \quoz{T^0}{T^1}$,}
\\
&\tau_{s_0(s_1s_0)^j \omega'} \cdot \widetilde{\bm{1}} \cdot \tau_\omega &&\text{for some $j \in \nn$ and some $\omega,\omega' \in \quoz{T^0}{T^1}$,}
\\
&\tau_{(s_0s_1)^j \omega'} \cdot \widetilde{\bp{1}} \cdot \tau_\omega &&\text{for some $j \in \nn$ and some $\omega,\omega' \in \quoz{T^0}{T^1}$,}
\\
&\tau_{s_1(s_0s_1)^j \omega'} \cdot \widetilde{\bp{1}} \cdot \tau_\omega &&\text{for some $j \in \nn$ and some $\omega,\omega' \in \quoz{T^0}{T^1}$.}
\end{align*}

If $\omega'=1$ then these elements are in $V$ because they are in the list of generators of $V$, and in the general case we can reduce to this situation exactly as we did in the case $\ell(v) + \ell(w) = 0$.

Now we have to consider the elements

\begin{align*}
&\tau_v \cdot \widetilde{\bm{1}} \cdot \tau_{s_0} \cdot \tau_{w'} &&\text{for some $w' \in \widetilde{W}$ such that $\ell(s_0 w') = \ell(w') + 1$,}
\\
&\tau_v \cdot \widetilde{\bp{1}} \cdot \tau_{s_1} \cdot \tau_{w'} &&\text{for some $w' \in \widetilde{W}$ such that $\ell(s_1 w') = \ell(w') + 1$.}
\end{align*}

If $\ell(v) = 0$, we reduce as usual to the case $v=1$, in which case we see that the above elements are in the list of generators of $V$. So we might assume that $\ell(v) \geqslant 1$. In this case, using again that both $\tau_{s_1} \cdot \widehat{\bm{1}}$ and $\tau_{s_0} \cdot \widehat{\bp{1}}$ lie in $N$, we are reduced to considering elements of the following forms:

\begin{align*}
&\tau_{v'} \cdot \tau_{s_0} \cdot \widetilde{\bm{1}} \cdot \tau_{s_0} \cdot \tau_{w'},
\\
&\tau_{v'} \cdot \tau_{s_1} \cdot \widetilde{\bp{1}} \cdot \tau_{s_1} \cdot \tau_{w'},
\end{align*}

where $v'$ and $w'$ are elements of $\widetilde{W}$ such that lengths add up in the corresponding multiplications (i.e., $\tau_{v'} \cdot \tau_{s_0}$ and $\tau_{s_0} \cdot \tau_{w'}$ in the case of the first line).
Now, looking at (<ref>), we see that we can write these two elements as linear combinations of elements of the form

\begin{align*}
&\tau_{v''} \cdot \widetilde{\bm{1}} \cdot \tau_{w''},
&\tau_{v''} \cdot \widetilde{\bp{1}} \cdot \tau_{w''},
&\tau_{v''} \cdot \widetilde{\bz{s_0}} \cdot \tau_{w''},
&\tau_{v''} \cdot \widetilde{\bz{s_1}} \cdot \tau_{w''}
\end{align*}

for $v'',w'' \in \widetilde{W}$ such that $\ell(v'') + \ell(w'') < \ell(v) + \ell(w)$. The induction hypothesis then allow us to conclude that the elements $\tau_{v' s_0} \cdot \widetilde{\bm{1}} \cdot \tau_{s_0 w'}$ and $\tau_{v' s_1} \cdot \widetilde{\bp{1}} \cdot \tau_{s_1 w'}$ lie in $V$.

* Now we consider the elements $\tau_v \cdot \widetilde{\bz{s_i}} \cdot \tau_w$ (for $i \in \{0,1\}$) under the additional assumption that $\ell(vs_i) = \ell(v) + 1$ and $\ell(s_iw) = \ell(w) + 1$.

Making $v$ explicit, we see that we are dealing with the following elements:

\begin{align*}
\tau_{(s_0s_1)^i\omega} \cdot \widetilde{\bz{s_0}} \cdot \tau_w
&&\scriptstyle{\text{for $i \in \nn$, $\omega \in \quoz{T^0}{T^1}$, $w \in \widetilde{W}$ with $\ell(s_0w) = \ell(w) + 1$,}}
\\
\tau_{s_1(s_0s_1)^i\omega} \cdot \widetilde{\bz{s_0}} \cdot \tau_w
&&\scriptstyle{\text{for $i \in \nn$, $\omega \in \quoz{T^0}{T^1}$, $w \in \widetilde{W}$ with $\ell(s_0w) = \ell(w) + 1$,}}
\\
\tau_{(s_1s_0)^i\omega} \cdot \widetilde{\bz{s_1}} \cdot \tau_w
&&\scriptstyle{\text{for $i \in \nn$, $\omega \in \quoz{T^0}{T^1}$, $w \in \widetilde{W}$ with $\ell(s_1w) = \ell(w) + 1$,}}
\\
\tau_{s_0(s_1s_0)^i\omega} \cdot \widetilde{\bz{s_1}} \cdot \tau_w
&&\scriptstyle{\text{for $i \in \nn$, $\omega \in \quoz{T^0}{T^1}$, $w \in \widetilde{W}$ with $\ell(s_1w) = \ell(w) + 1$.}}
\end{align*}

As before, we see that we can assume without loss of generality that $\omega = 1$, and then, using (<ref>) repeatedly, we see that, up to a sign, the above elements can be rewritten in the form $\widetilde{\bz{s_i}} \cdot \tau_v$ for some $i \in \{0,1\}$ and some $v \in \widetilde{W}$ such that $\ell(s_i v) = \ell(v) + 1$.

* Now we consider the elements $\tau_v \cdot \widetilde{\bz{s_i}} \cdot \tau_w$ (for $i \in \{0,1\}$) under the additional assumption that $\ell(s_iw) = \ell(w) - 1$.

We only deal with the case $i=0$, the other case being completely analogous. Using (<ref>), we see that

\begin{align*}
\tau_v \cdot \widetilde{\bz{s_0}} \cdot \tau_w
\tau_v \cdot \widetilde{\bz{s_0}} \cdot \tau_{s_0} \cdot \tau_{s_0^{-1}w}
\\&=
\tau_v \cdot \left( - \widetilde{\bz{s_0}} \cdot e_1 - e_{\idd^{-1}} \widetilde{\bm{1}} \cdot \tau_{s_0} \right) \cdot \tau_{s_0^{-1}w}
\\&=
- \tau_v \cdot \widetilde{\bz{s_0}} \cdot \tau_{s_0^{-1}w}e_1
- \tau_v e_{\idd^{-1}}  \cdot \widetilde{\bm{1}} \cdot \tau_{w}.
\end{align*}

The first term in the last line is a linear combination of terms of the form $\tau_{v'} \cdot \widetilde{\bz{s_0}} \cdot \tau_{w'}$ for $v', w' \in \widetilde{W}$ such that $\ell(v') + \ell(w') = \ell(v) + \ell(w) - 1$, and so by inductive hypothesis it lies in $V$. On the other hand, the second term is a linear combination of terms of the form $\tau_{v'} \cdot \widetilde{\bm{1}} \cdot \tau_{w'}$ for $v', w' \in \widetilde{W}$ such that $\ell(v') + \ell(w') = \ell(v) + \ell(w)$, and these also lie in $V$, as already proved.

* Now it remains to consider the elements $\tau_v \cdot \widetilde{\bz{s_i}} \cdot \tau_w$ (for $i \in \{0,1\}$) under the additional assumption that $\ell(vs_i) = \ell(v) - 1$. This case is completely analogous to the previous one, using (<ref>) in place of (<ref>).

In the next lemma we compute a presentation of the Hecke algebra $E^0=H$. This is straightforward (and probably well-known) under our assumptions. Also compare the finite presentation of $H$ in the case $G = \GL_n(\field)$ determined in <cit.>. The following lemma does not use anything about the $\ext$-algebra, and it is actually valid in the case $G = \SL_2(\field)$, without assumptions on $\field$ or $p$, but replacing the occurrences of $p$ with $q$.

The pro-$p$ Iwahori–Hecke algebra $H=E^0$ can be expressed by generators and relations as follows. Let

\[
k \left\langle \widehat{\tau_{\omega_0}}, \widehat{\tau_{s_0}}, \widehat{\tau_{s_1}} \right\rangle
\]

be the ring of non-commutative polynomials in three indeterminates called $\widehat{\tau_{\omega_0}}$, $\widehat{\tau_{s_0}}$ and $\widehat{\tau_{s_1}}$. Furthermore, let $I$ be the two-sided ideal of $k \left\langle \widehat{\tau_{\omega_0}}, \widehat{\tau_{s_0}}, \widehat{\tau_{s_1}} \right\rangle$ generated by the following elements

\begin{align}
&\widehat{\tau_{\omega_0}}^{p-1} - 1,
\label{eqSmallList1}
\\
&\widehat{\tau_{\omega_0}} \cdot \widehat{\tau_{s_0}} - \widehat{\tau_{s_0}} \cdot \widehat{\tau_{\omega_0}}^{p-2},
&\widehat{\tau_{\omega_0}} \cdot \widehat{\tau_{s_1}} - \widehat{\tau_{s_1}} \cdot \widehat{\tau_{\omega_0}}^{p-2},
\label{eqSmallList2}
\\
&\widehat{\tau_{s_0}}^2 - \sum_{i=0}^{p-2} \widehat{\tau_{\omega_0}}^{i} \cdot \widehat{\tau_{s_0}},
&\widehat{\tau_{s_1}}^2 - \sum_{i=0}^{p-2} \widehat{\tau_{\omega_0}}^{i} \cdot \widehat{\tau_{s_1}}.
\label{eqSmallList3}
\end{align}

Let $\widetilde{\tau_{\omega_0}}$, $\widetilde{\tau_{s_0}}$ and $\widetilde{\tau_{s_1}}$ be respectively the images of $\widehat{\tau_{\omega_0}}$, $\widehat{\tau_{s_0}}$ and $\widehat{\tau_{s_1}}$ in the quotient algebra $\quoz{k \left\langle \widehat{\tau_{\omega_0}}, \widehat{\tau_{s_0}}, \widehat{\tau_{s_1}} \right\rangle}{I}$. One has an isomorphism of $k$-algebras

\[
\funcAboveTriple
{\quoz{k \left\langle \widehat{\tau_{\omega_0}}, \widehat{\tau_{s_0}}, \widehat{\tau_{s_1}} \right\rangle}{I}}
\]

The braid relations, the quadratic relations and the properties of the group $\widetilde{W}$ imply that the homomorphism in the statement is indeed well-defined and surjective.

To show injectivity, we proceed as in Lemma <ref>. Let us consider the following table, where the elements in the first column are mapped to the elements on the right column by our $k$-algebra homomorphism
\funcInlineNN
{\quoz{k \left\langle \widehat{\tau_{\omega_0}}, \widehat{\tau_{s_0}}, \widehat{\tau_{s_1}} \right\rangle}{I}}
$, and where the elements in the right column form a $k$-basis of $E^0$:

\[\begin{tikzcd}[row sep = -0.6em]
\widetilde{\tau_{\omega_0}}^i
\ar[r, mapsto]
&\substack{\text{for $i \in \{0, \dots, p-2\}$,}}
\\
\widetilde{\tau_{\omega_0}}^i \cdot \widetilde{\tau_{s_1}} \cdot (\widetilde{\tau_{s_0}} \cdot \widetilde{\tau_{s_1}})^j
\ar[r, mapsto]
&\tau_{\omega_0^i s_1(s_0s_1)^j}
&\substack{\text{for $i \in \{0, \dots, p-2\}$ and $j \in \nn$,}}
\\
\widetilde{\tau_{\omega_0}}^i \cdot \widetilde{\tau_{s_0}} \cdot (\widetilde{\tau_{s_1}} \cdot \widetilde{\tau_{s_0}})^j
\ar[r, mapsto]
&\tau_{\omega_0^i s_0(s_1s_0)^j}
&\substack{\text{for $i \in \{0, \dots, p-2\}$ and $j \in \nn$,}}
\\
\widetilde{\tau_{\omega_0}}^i \cdot (\widetilde{\tau_{s_0}} \cdot \widetilde{\tau_{s_1}})^j
\ar[r, mapsto]
&\tau_{\omega_0^i (s_0s_1)^j}
&\substack{\text{for $i \in \{0, \dots, p-2\}$ and $j \in \zpiu$,}}
\\
\widetilde{\tau_{\omega_0}}^i \cdot (\widetilde{\tau_{s_1}} \cdot \widetilde{\tau_{s_0}})^j
\ar[r, mapsto]
&\tau_{\omega_0^i (s_1s_0)^j}
&\substack{\text{for $i \in \{0, \dots, p-2\}$ and $j \in \zpiu$.}}
\end{tikzcd}\]

As in Lemma <ref>, to show injectivity of our homomorphism, it suffices to show that the $k$-vector space $V$ generated by the elements in the left column is the full $\quoz{k \left\langle \widehat{\tau_{\omega_0}}, \widehat{\tau_{s_0}}, \widehat{\tau_{s_1}} \right\rangle}{I}$. So, it suffices to prove that every element of the form

\[
\widetilde{\tau_{w_1}} \cdots \widetilde{\tau_{w_n}},
\]

for $n \in \nn$ and $w_1, \dots, w_n \in \{ \omega_0, s_0, s_1 \}$, lies in $V$. Using the relations (<ref>) and (<ref>), we see that we may further reduce to elements of the form

\[
\widetilde{\tau_{\omega_0}}^i \cdot \widetilde{\tau_{s_{l_1}}} \cdots \widetilde{\tau_{s_{l_m}}}
\]

for $i \in \{0, \dots, p-2\}$, for $m \in \nn$ and for $l_1, \dots, l_m \in \{0,1\}$. We now prove that an element of this form lies in $V$ by induction on $m$. If $m = 0$, then the result is clear. Furthermore, for general $m$ the result is clear in the case that there are no consecutive indices $l_{j}$ and $l_{j+1}$ both equal to $0$ or both equal to $1$. If instead there is at least one such pair of indices, then we use (<ref>) (together with (<ref>) and (<ref>) again) to reduce to the case of smaller $m$.

Let $A^\ast$ be a positively graded algebra over a field $F$ (associative, with $1$, not necessarily commutative) such that the natural map
\funcInlineNN{T^\ast_{A^0} A^1}{A^\ast}
is surjective. Let us consider generators the kernel of the above map (as a two-sided ideal), obtaining a presentation

\[
\funcInlineNN{\frac{T^\ast_{A^0} A^1}{(a_i)_{i \in \mathbf{I}}}}{A^\ast}.
\]

Let us consider a presentation of $A^1$ as an $A^0 \otimes_k (A^0)^{\opp}$-left module

\[
\funcInlineNN{\frac{\bigoplus_{j \in \mathbf{J}} \big( A^0 \otimes_k (A^0)^{\opp} \big) X_j}{(b_k)_{k \in \mathbf{K}}}}{A^1},
\]

and let us consider a presentation of $A^0$ as a $k$-algebra

\[
\funcInlineNN{\frac{F \langle Y_l \rangle_{l \in \mathbf{L}}}{(c_m)_{m \in \mathbf{M}}}}{A^0}.
\]

For all $i \in \mathbf{I}$ (respectively, for all $k \in \mathbf{K}$) let us choose a lift $\dot{a}_i$ (respectively, $\dot{b}_k$) of $a_i$ (respectively, of $b_k$) in $F \langle X_j, Y_l \rangle_{j \in \mathbf{J}, l \in \mathbf{L}}$.

One then has an induced presentation of $A^\ast$ as a $k$-algebra

\[
\funcInlineNN
{F \langle X_j, Y_l \rangle_{j \in \mathbf{J}, l \in \mathbf{L}}}
{\ideal{ \dot{a}_i, \dot{b}_k, c_m }{ i \in \mathbf{I}, k \in \mathbf{K}, m \in \mathbf{M} }}
\]

For the moment let us not assume that the map
\funcInlineNN{T^\ast_{A^0} A^1}{A^\ast}
is surjective.

Let $\funcInline{\iota_0}{A^0}{T_{A^0}^\ast A^1}$ and $\funcInline{\iota_1}{A^1}{T_{A^0}A^1}$ be the canonical inclusions. It is easy to check that the triple $(T^\ast_{A^0}A^1, \iota_0, \iota_1)$ enjoys the following universal property: for all triples $(R,\xi_0,\xi_1)$ consisting of a $F$-algebra $R$ (associative, not necessarily commutative), a homomorphism of $F$-algebras $\funcInline{\xi_0}{A^0}{R}$ and a homomorphism of left $A^0 \otimes_F (A^0)^{\opp}$-modules $\funcInline{\xi_1}{A^1}{R}$ (where $R$ is a left $A^0 \otimes_F (A^0)^{\opp}$-module via $\xi_0$), there exists a unique homomorphism of $F$-algebras $\funcInline{\eta}{T_{A^0}^\ast A^1}{R}$ making the following diagrams commute:

\begin{align*}
&\begin{tikzcd}[row sep = 0.2em, ampersand replacement = \&]
\&
T_{A^0}^\ast A^1
\ar[dd, "{\eta}"]
\\
\ar[ur, "{\iota_0}"]
\ar[dr, "{\xi_0}"']
\&
\\
\&
\end{tikzcd}
&\begin{tikzcd}[row sep = 0.2em, ampersand replacement = \&]
\&
T_{A^0}^\ast A^1
\ar[dd, "{\eta}"]
\\
\ar[ur, "{\iota_1}"]
\ar[dr, "{\xi_1}"']
\&
\\
\&
\end{tikzcd}
\end{align*}

This universal property provides an inverse to the obvious map

\[
\funcInlineNN
{F \langle X_j, Y_l \rangle_{j \in \mathbf{J}, l \in \mathbf{L}}}
{\ideal{ \dot{b}_k, c_m }{ k \in \mathbf{K}, m \in \mathbf{M} }}
\]

and the claim of the lemma follows (assuming that the map
\funcInlineNN{T^\ast_{A^0} A^1}{A^\ast}
is surjective).

The preceding lemma directly allows us to compute a presentation of $E^\ast$, since we have already computed the three presentations required in the statement of the lemma, in Theorem <ref>, Lemma <ref> and Lemma <ref> respectively.

Let us work with the ring of noncommutative polynomials in seven indeterminates
k \left\langle \widehat{\tau_{\omega_0}}, \widehat{\tau_{s_0}}, \widehat{\tau_{s_1}}, \widehat{\bm{1}}, \widehat{\bp{1}}, \widehat{\bz{s_0}}, \widehat{\bz{s_1}} \right\rangle
(where the indeterminates have been labelled in accordance with the notation of Lemma <ref> and Lemma <ref>).

For all $\lambda \in \widehat{\quoz{T^0}{T^1}}$, let us give the following definition, mimicking the definition of $e_\lambda$:

\[
\varepsilon_\lambda \defeq - \sum_{i=0}^{p-1} \lambda(\omega_0^{-i}) \widehat{\tau_{\omega_0}}^i \in k[\widehat{\tau_{\omega_0}}].
\]

Let us consider the following elements of
k \left\langle \widehat{\tau_{\omega_0}}, \widehat{\tau_{s_0}}, \widehat{\tau_{s_1}}, \widehat{\bm{1}}, \widehat{\bp{1}}, \widehat{\bz{s_0}}, \widehat{\bz{s_1}} \right\rangle

\begin{equation}\label{eqList3}\begin{aligned}
&\widehat{\bm{1}} \cdot \widehat{\bm{1}}, &&\qquad\qquad\qquad\widehat{\bp{1}} \cdot \widehat{\bm{1}}, &&\qquad\qquad\qquad\widehat{\bz{s_1}} \cdot \widehat{\bm{1}},
\\
&\widehat{\bm{1}} \cdot \widehat{\bp{1}}, &&\qquad\qquad\qquad\widehat{\bp{1}} \cdot \widehat{\bp{1}}, &&\qquad\qquad\qquad\widehat{\bz{s_0}} \cdot \widehat{\bp{1}},
\\
&\widehat{\bp{1}} \cdot \widehat{\bz{s_0}}, &&\qquad\qquad\qquad\widehat{\bz{s_1}} \cdot \widehat{\bz{s_0}},
\\
&\widehat{\bm{1}} \cdot \widehat{\bz{s_1}}, &&\qquad\qquad\qquad\widehat{\bz{s_0}} \cdot \widehat{\bz{s_1}},
\end{aligned}
\\
&\widehat{\bz{s_0}} \cdot \widehat{\bz{s_0}} + \varepsilon_{\idd^{-1}} \cdot \widehat{\bm{1}} \cdot \widehat{\bz{s_0}} + \varepsilon_{\idd} \cdot \widehat{\bz{s_0}} \cdot \widehat{\bm{1}} + \varepsilon_1 \cdot \widehat{\bm{1}} \cdot \widehat{\tau_{s_0}} \cdot \widehat{\bm{1}},
\\
&\widehat{\bz{s_1}} \cdot \widehat{\bz{s_1}} - \varepsilon_{\idd} \cdot \widehat{\bp{1}} \cdot \widehat{\bz{s_1}} - \varepsilon_{\idd^{-1}} \cdot \widehat{\bz{s_1}} \cdot \widehat{\bp{1}} + \varepsilon_1 \cdot \widehat{\bp{1}} \cdot \widehat{\tau_{s_1}} \cdot \widehat{\bp{1}},
\\
&\widehat{\bz{s_0}} \cdot \widehat{\bm{1}} \cdot \widehat{\tau_{s_0}} - \widehat{\tau_{s_0}} \cdot \widehat{\bm{1}} \cdot \widehat{\bz{s_0}},
\\
&\widehat{\bz{s_1}} \cdot \widehat{\bp{1}} \cdot \widehat{\tau_{s_1}} - \widehat{\tau_{s_1}} \cdot \widehat{\bp{1}} \cdot \widehat{\bz{s_1}},
\\
&(\widehat{\tau_{s_1}} + \varepsilon_1) \cdot \widehat{\bp{1}} \cdot \widehat{\bz{s_1}} \cdot \widehat{\bp{1}}
(\widehat{\tau_{s_0}} + \varepsilon_1) \cdot \widehat{\bm{1}} \cdot \widehat{\bz{s_0}} \cdot \widehat{\bm{1}}.
\end{aligned}\end{equation}

We claim that this is a set of lifts to
k \left\langle \widehat{\tau_{\omega_0}}, \widehat{\tau_{s_0}}, \widehat{\tau_{s_1}}, \widehat{\bm{1}}, \widehat{\bp{1}}, \widehat{\bz{s_0}}, \widehat{\bz{s_1}} \right\rangle
of a set of generators of the kernel of the multiplication map $\funcInline{\mult}{T^\ast_{E^0} E^1}{E^\ast}$ (as a two-sided ideal). Indeed, this list has been obtained in the following way from the list of generators of the $\ker(\mult)$ given in Theorem <ref>: the occurrences of $\bm{1}$ have been replaced by $\widehat{\bm{1}}$, the occurrences of $\bp{1}$ have been replaced by $\widehat{\bp{1}}$, the occurrences of $\bz{s_i}$ have been replaced by $\widehat{\bz{s_i}}$ for $i \in \{0,1\}$, the occurrences of $\tau_{s_i}$ have been replaced by $\widehat{\tau_{s_i}}$ for $i \in \{0,1\}$ and the occurrences of $e_\lambda$ have been replaced by $\varepsilon_\lambda$ for $\lambda \in \{ 1, \idd, \idd^{-1} \}$. Furthermore, the occurrences of $\beta^\pm_{s_i}$ (for $i \in \{0,1\}$) have first been replaced according to the following rules (see the formulas in Lemma <ref>):

\begin{align*}
\bp{s_0} &= - \tau_{s_0} \cdot \bm{1},
\bm{s_1} &= - \tau_{s_1} \cdot \bp{1},
\\
\bm{s_0} &= \bm{1} \cdot \tau_{s_0},
\bp{s_1} &= \bp{1} \cdot \tau_{s_1},
\end{align*}

and then, of course, according to the rules above. Finally, as regards the generator
(\tau_{s_1} + e_1) \cdot \bp{1} \otimes \bz{s_1^{-1}} \otimes \bp{1}
(\tau_{s_0} + e_1) \cdot \bm{1} \otimes \bz{s_0^{-1}} \otimes \bm{1}
for simplicity we have also deleted the exponent $-1$ (which can be achieved up to multiplying by $\tau_{c_{-1}}$, thus not affecting the fact that we are dealing with a set of generators).
This shows that (<ref>) is a set of lifts to
k \left\langle \widehat{\tau_{\omega_0}}, \widehat{\tau_{s_0}}, \widehat{\tau_{s_1}}, \widehat{\bm{1}}, \widehat{\bp{1}}, \widehat{\bz{s_0}}, \widehat{\bz{s_1}} \right\rangle
of a set of generators of $\ker(\mult)$, as claimed. In the notation of the previous lemma, the list (<ref>) corresponds to the family $(\dot{a}_i)_{i \in \mathbf{I}}$.

Furthermore, let consider the following list of elements, obtaining from the list in Lemma <ref> by replacing $\tau_{\omega_0}$ with $\widehat{\tau_{\omega_0}}$, by replacing $\tau_{\omega_0^{-1}}=\tau_{\omega_0}^{p-2}$ with $\widehat{\tau_{\omega_0}}^{p-2}$ by replacing $\tau_{s_i}$ with $\widehat{\tau_{s_i}}$ for $i \in \{0,1\}$ and by replacing $e_\lambda$ by $\varepsilon_\lambda$ for $\lambda \in \{ 1, \idd, \idd^{-1} \}$:

\begin{equation}\label{eqListE1}\begin{aligned}
&\widehat{\tau_{s_1}} \cdot \widehat{\bm{1}},
&\qquad\qquad\widehat{\tau_{s_0}} \cdot \widehat{\bp{1}},
\\
&\widehat{\bp{1}} \cdot \widehat{\tau_{s_0}},
&\qquad\qquad\widehat{\bm{1}} \cdot \widehat{\tau_{s_1}},
\\
\begin{matrix*}[l]
(\widehat{\tau_{s_0}} + \varepsilon_1) \cdot \widehat{\bm{1}} \cdot (\widehat{\tau_{s_0}} + \varepsilon_1)
\\\qquad
+ 2 \varepsilon_{\idd} \widehat{\bz{s_0}} + \widehat{\tau_{\omega_0}}^{\frac{p-1}{2}} \cdot \widehat{\bp{1}},
\end{matrix*}
\begin{matrix*}[l]
(\widehat{\tau_{s_1}} + \varepsilon_1) \cdot \widehat{\bp{1}} \cdot (\widehat{\tau_{s_1}} + \varepsilon_1)
\\\qquad
- 2 \varepsilon_{\idd^{-1}} \widehat{\bz{s_1}} + \widehat{\tau_{\omega_0}}^{\frac{p-1}{2}} \cdot \widehat{\bm{1}},
\end{matrix*}
\\
&\widehat{\tau_{s_0}} \cdot \widehat{\bz{s_1}} + \widehat{\bz{s_0}} \cdot \widehat{\tau_{s_1}},
&\qquad\qquad\widehat{\tau_{s_1}} \cdot \widehat{\bz{s_0}} + \widehat{\bz{s_1}} \cdot \widehat{\tau_{s_0}},
\\
&(\widehat{\tau_{s_0}} + \varepsilon_1) \cdot \widehat{\bz{s_0}} + \varepsilon_{\idd} \widehat{\tau_{s_0}} \cdot \widehat{\bm{1}},
&\qquad\qquad(\widehat{\tau_{s_1}} + \varepsilon_1) \cdot \widehat{\bz{s_1}} - \varepsilon_{\idd^{-1}} \widehat{\tau_{s_1}} \cdot \widehat{\bp{1}},
\\
&\widehat{\bz{s_0}} \cdot (\widehat{\tau_{s_0}} + \varepsilon_1) + \varepsilon_{\idd^{-1}} \widehat{\bm{1}} \cdot \widehat{\tau_{s_0}},
&\qquad\qquad\widehat{\bz{s_1}} \cdot (\widehat{\tau_{s_1}} + \varepsilon_1) - \varepsilon_{\idd} \widehat{\bp{1}} \cdot \widehat{\tau_{s_1}},
\\
&\widehat{\tau_{\omega_0}} \cdot \widehat{\bm{1}} - u_{\omega_0}^{-2} \widehat{\bm{1}} \cdot \widehat{\tau_{\omega_0}},
&\qquad\qquad\widehat{\tau_{\omega_0}} \cdot \widehat{\bp{1}} - u_{\omega_0}^{2} \widehat{\bp{1}} \cdot \widehat{\tau_{\omega_0}},
\\
&\widehat{\tau_{\omega_0}} \cdot \widehat{\bz{s_0}} - \widehat{\bz{s_0}} \cdot \widehat{\tau_{\omega_0}}^{p-2},
&\qquad\qquad\widehat{\tau_{\omega_0}} \cdot \widehat{\bz{s_1}} - \widehat{\bz{s_1}} \cdot \widehat{\tau_{\omega_0}}^{p-2}.
\end{aligned}\end{equation}

In the notation of the previous lemma, this list corresponds to the family $(\dot{b}_k)_{k \in \mathbf{K}}$.

Furthermore, let us recall the list of elements Lemma <ref> (which we simply rewrite using the notation $\varepsilon_1$):

\begin{equation}\label{eqListE0}\begin{aligned}
&\widehat{\tau_{\omega_0}}^{p-1} - 1,
\\
&\widehat{\tau_{\omega_0}} \cdot \widehat{\tau_{s_0}} - \widehat{\tau_{s_0}} \cdot \widehat{\tau_{\omega_0}}^{p-2},
&\widehat{\tau_{\omega_0}} \cdot \widehat{\tau_{s_1}} - \widehat{\tau_{s_1}} \cdot \widehat{\tau_{\omega_0}}^{p-2},
\\
&\widehat{\tau_{s_0}}^2 + \varepsilon_1 \widehat{\tau_{s_0}},
&\widehat{\tau_{s_1}}^2 + \varepsilon_1 \widehat{\tau_{s_1}}.
\end{aligned}\end{equation}

In the notation of the previous lemma, this list corresponds to the family $(c_m)_{m \in \mathbf{M}}$.

Recall the assumptions stated at the beginning of the current <ref>.
The $\ext$-algebra $E^\ast$ is finitely presented as a $k$-algebra. More precisely, it can be expressed as the quotient of the ring of non-commutative polynomials in seven indeterminates
k \left\langle \widehat{\tau_{\omega_0}}, \widehat{\tau_{s_0}}, \widehat{\tau_{s_1}}, \widehat{\bm{1}}, \widehat{\bp{1}}, \widehat{\bz{s_0}}, \widehat{\bz{s_1}} \right\rangle
by the two-sided ideal generated by the elements in the lists (<ref>), (<ref>) and (<ref>).

This follows from Lemma <ref> and the subsequent discussion (which relied on the presentations obtained in Theorem <ref>, Lemma <ref> and Lemma <ref>).

The relation

\[(\tau_{s_1} + e_1) \cdot \bp{1} \cdot (\tau_{s_1} + e_1) - 2 e_{\idd^{-1}} \bz{s_1} + \tau_{\omega_0}^{\frac{p-1}{2}} \cdot \bm{1} = 0\]

(which we used in to produce the corresponding element in the list (<ref>)) allows us to express $\bm{1}$ in terms of $\bp{1}$, $\bz{s_1}$, $\tau_{s_1}$ and $\tau_{\omega_0}$.
This shows that $E^\ast$ is actually generated by $\tau_{\omega_0}$, $\tau_{s_0}$, $\tau_{s_1}$, $\bp{1}$, $\bz{s_0}$ and $\bz{s_1}$ as a $k$-algebra, without the need to add $\bm{1}$. Using the presentation we obtained in the last theorem, it is then immediate (but cumbersome) to obtain a presentation of $E^\ast$ as a quotient of $k \left\langle \widehat{\tau_{\omega_0}}, \widehat{\tau_{s_0}}, \widehat{\tau_{s_1}}, \widehat{\bp{1}}, \widehat{\bz{s_0}}, \widehat{\bz{s_1}} \right\rangle$ by replacing $\widehat{\bm{1}}$ with

\[ - \widehat{\tau_{\omega_0}}^{\frac{p-1}{2}} \cdot (\widehat{\tau_{s_1}} + \varepsilon_1) \cdot \widehat{\bp{1}} \cdot (\widehat{\tau_{s_1}} + \varepsilon_1) + 2 \widehat{\tau_{\omega_0}}^{\frac{p-1}{2}} \cdot \varepsilon_{\idd^{-1}} \widehat{\bz{s_1}}\]

wherever it appears in the lists (<ref>), (<ref>) and (<ref>).

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