Appendix A: Proof of Proposition 5.3
This appendix is devoted to a proof of Proposition 5.3, which concerns about the computation of \([(S'_i)^{\oplus m_1}\oplus (S'_{{{\tau }}i})^{\oplus n_1}]*[S'_j]* [(S'_i)^{\oplus m_2} \oplus (S'_{{{\tau }}i})^{\oplus n_2}]\).
1.1 The setup
By definition, we have
$$\begin{aligned}&[S_i^{\oplus m_1}\oplus S_{{{\tau }}i}^{\oplus n_1}]*[S_{j}]* [S_i^{\oplus m_2} \oplus S_{{{\tau }}i}^{\oplus n_2}] = [S_i^{\oplus m_1}\oplus S_{{{\tau }}i}^{\oplus n_1}]*[S_{j}\oplus S_i^{\oplus m_2} \oplus S_{{{\tau }}i}^{\oplus n_2}]\\&\quad ={\mathbf{v}}^{ c_{ij}m_1+c_{{{\tau }}i,j}n_1 -m_1m_2-n_1n_2 }\sum _{[L]\in {\text {Iso}}({\text {mod}}(\Lambda ^\imath ))}\big |{\text {Ext}}^1_{\Lambda ^\imath }( S_i^{\oplus m_1}\oplus S_{{{\tau }}i}^{\oplus n_1}, S_{j}\oplus S_i^{\oplus m_2} \oplus S_{{{\tau }}i}^{\oplus n_2})_L\big | \cdot [L]. \end{aligned}$$
For any \([L]\in {\text {Iso}}({\text {mod}}(\Lambda ^\imath ))\) such that
$$\begin{aligned} \big |{\text {Ext}}^1_{\Lambda ^\imath }( S_i^{\oplus m_1}\oplus S_{{{\tau }}i}^{\oplus n_1}, S_{j}\oplus S_i^{\oplus m_2} \oplus S_{{{\tau }}i}^{\oplus n_2})_L\big |\ne 0, \end{aligned}$$
(A.1)
there exist a unique \([M]\in {\text {Iso}}({\text {mod}}(kQ))\) and \(d,e\in {{\mathbb {N}}}\) such that \([L]=[M\oplus {{\mathbb {K}}}_i^{\oplus d} \oplus {{\mathbb {K}}}_{{{\tau }}i}^{\oplus e}]\) in \({{{\widetilde{{{\mathcal {H}}}}}({k}Q,{{\tau }})}}\). In this case, M admits the following exact sequence
$$\begin{aligned} 0\longrightarrow S_{j}\oplus S_i^{\oplus (m_2-e)}\oplus S_{{{\tau }}i}^{\oplus (n_2-d)} \longrightarrow M\longrightarrow S_i^{\oplus (m_1-d)} \oplus S_{{{\tau }}i}^{\oplus (n_1-e)} \longrightarrow 0. \end{aligned}$$
Fix \([M]\in {\text {Iso}}({\text {mod}}(kQ))\), \(0\le d\le \min (n_2,m_1)\) and \(0\le e\le \min (n_1,m_2)\). Then there exists a unique indecomposable kQ-module N such that \(M\cong N\oplus S_i^{\oplus t_1^M}\oplus S_{{{\tau }}i}^{\oplus t_3^M}\) for some unique \(t_1^M,t_3^M\). Denote by
$$\begin{aligned}&{{\mathcal {C}}}_M:=\{[\xi ]\in {\text {Ext}}^1_{\Lambda ^\imath }( S_i^{\oplus m_1}\oplus S_{{{\tau }}i}^{\oplus n_1}, S_{j}\oplus S_i^{\oplus m_2} \oplus S_{{{\tau }}i}^{\oplus n_2})_L{|} L\text { admits }\\&\quad \text {a short exact sequence } 0\rightarrow M\rightarrow L\rightarrow {{\mathbb {K}}}_i^{\oplus d}\oplus {{\mathbb {K}}}_{{{\tau }}i}^{\oplus e}\rightarrow 0\}. \end{aligned}$$
In this way, we have
$$\begin{aligned}&[S_i^{\oplus m_1}\oplus S_{{{\tau }}i}^{\oplus n_1}]*[S_{j}]* [S_i^{\oplus m_2} \oplus S_{{{\tau }}i}^{\oplus n_2}]\nonumber \\&\quad =\sum _{e=0}^{\min ( n_1,m_2)}\sum _{d=0}^{\min (n_2,m_1 )}\sum _{[M]\in {\mathcal {I}}'_{m_1+m_2-d-e,n_1+n_2-d-e}} {\mathbf{v}}^{ c_{ij}m_1+c_{{{\tau }}i,j}n_1 -m_1m_2-n_1n_2 } |{{\mathcal {C}}}_M| \cdot [M\oplus {{\mathbb {K}}}_i^{\oplus d}\oplus {{\mathbb {K}}}_{{{\tau }}i}^{\oplus e}]\nonumber \\&\quad =\sum _{e=0}^{\min ( n_1,m_2)}\sum _{d=0}^{\min (n_2,m_1 )}\sum _{[M]\in {\mathcal {I}}'_{m_1+m_2-d-e,n_1+n_2-d-e}} {\mathbf{v}}^{ c_{ij}m_1+c_{{{\tau }}i,j}n_1 -m_1m_2-n_1n_2 +(e-d)(m_1+m_2-n_1-n_2) } |{{\mathcal {C}}}_M| \nonumber \\&\qquad \times [M]*[{{\mathbb {K}}}_i]^{ d}* [{{\mathbb {K}}}_{{{\tau }}i}]^{e}. \end{aligned}$$
(A.2)
1.2 Computation of \( |{{\mathcal {C}}}_M|\)
We shall compute \( |{{\mathcal {C}}}_M|\). Let \(Q'\) and \(Q''\) be the full subquivers of Q formed by the vertices i, j and the vertices \({{\tau }}i,j\) respectively. Then we have two restriction functors \({\text {res}}_{ij}:{\text {mod}}(kQ)\rightarrow {\text {mod}}(kQ')\) and \({\text {res}}_{{{\tau }}i,j}: {\text {mod}}(kQ)\rightarrow {\text {mod}}(kQ'')\). Set
$$\begin{aligned} M_1:={\text {res}}_{ij}(M), \quad N_1:={\text {res}}_{ij}(N), \quad M_2:={\text {res}}_{{{\tau }}i,j}(M), \quad N_2:={\text {res}}_{{{\tau }}i,j}(N).\nonumber \\ \end{aligned}$$
(A.3)
Denote by
$$\begin{aligned} {{\mathcal {C}}}_1&:=\{[\xi ]\in {\text {Ext}}^1_{\Lambda ^\imath }( S_i^{\oplus m_1}, S_{j}\oplus S_i^{\oplus m_2} \oplus S_{{{\tau }}i}^{\oplus n_2})_{L_1}{|} L_1\text { admits a short }\\&\quad \text { exact sequence } 0\rightarrow M_1\rightarrow L_1\rightarrow {{\mathbb {K}}}_i^{\oplus d}\oplus S_i^{\oplus e} \oplus S_{{{\tau }}i}^{\oplus (n_2-d)} \rightarrow 0\},\\ {{\mathcal {C}}}_2&:=\{[\xi ]\in {\text {Ext}}^1_{\Lambda ^\imath }( S_{{{\tau }}i}^{\oplus n_1}, S_{j}\oplus S_i^{\oplus m_2} \oplus S_{{{\tau }}i}^{\oplus n_2})_{L_2}{|} L_2\text { admits a short }\\&\quad \text { exact sequence } 0\rightarrow M_2\rightarrow L_2\rightarrow {{\mathbb {K}}}_{{{\tau }}i}^{\oplus e}\oplus S_{{{\tau }}i}^{\oplus d} \oplus S_i^{\oplus (m_2-e)}\rightarrow 0\}. \end{aligned}$$
Lemma A.1
Retain the notations as above. Then \(|{{\mathcal {C}}}_M|=|{{\mathcal {C}}}_1|\cdot |{{\mathcal {C}}}_2|\).
Proof
Applying \({\text {Hom}}_{\Lambda ^\imath }(-,S_{j}\oplus S_i^{\oplus m_2}\oplus S_{{{\tau }}i}^{\oplus n_2})\) to the split short exact sequence
$$\begin{aligned} 0\longrightarrow S_i^{\oplus m_1} \longrightarrow S_i^{\oplus m_1}\oplus S_{{{\tau }}i}^{\oplus n_1}\longrightarrow S_{{{\tau }}i}^{\oplus n_1}\longrightarrow 0, \end{aligned}$$
(A.4)
we have the following short exact sequence
$$\begin{aligned}&0\longrightarrow {\text {Ext}}^1(S_{{{\tau }}i}^{\oplus n_1},S_{j}\oplus S_i^{\oplus m_2}\oplus S_{{{\tau }}i}^{\oplus n_2}){\mathop {\longrightarrow }\limits ^{\beta }} {\text {Ext}}^1( S_i^{\oplus m_1}\oplus S_{{{\tau }}i}^{\oplus n_1},S_{j}\oplus S_i^{\oplus m_2}\oplus S_{{{\tau }}i}^{\oplus n_2})\nonumber \\&\quad {\mathop {\longrightarrow }\limits ^{\alpha }}{\text {Ext}}^1( S_i^{\oplus m_1},S_{j}\oplus S_i^{\oplus m_2}\oplus S_{{{\tau }}i}^{\oplus n_2})\longrightarrow 0. \end{aligned}$$
(A.5)
Then \(\alpha ({{\mathcal {C}}}_M)={{\mathcal {C}}}_1\), and thus, \({{\mathcal {C}}}_M= \bigsqcup _{[\xi ]\in {{\mathcal {C}}}_1} (\alpha |_{{{\mathcal {C}}}_M})^{-1}([\xi ])\).
Since (A.4) is split, we have the following exact sequence
$$\begin{aligned}&0\longrightarrow {\text {Ext}}^1(S_i^{\oplus m_1},S_{j}\oplus S_i^{\oplus m_2}\oplus S_{{{\tau }}i}^{\oplus n_2}){\mathop {\longrightarrow }\limits ^{\delta }} {\text {Ext}}^1( S_i^{\oplus m_1}\oplus S_{{{\tau }}i}^{\oplus n_1},S_{j}\oplus S_i^{\oplus m_2}\oplus S_{{{\tau }}i}^{\oplus n_2})\nonumber \\&\quad {\mathop {\longrightarrow }\limits ^{\gamma }} {\text {Ext}}^1( S_{{{\tau }}i}^{\oplus n_1},S_{j}\oplus S_i^{\oplus m_2}\oplus S_{{{\tau }}i}^{\oplus n_2})\longrightarrow 0 \end{aligned}$$
(A.6)
such that \(\gamma \circ \beta ={\text {Id}}\) and \(\alpha \circ \delta ={\text {Id}}\). For any \([\eta _1],[\eta _2]\in {{\mathcal {C}}}_M\) such that \(\alpha ([\eta _1])=\alpha ([\eta _2])=[\xi ]\), if \(\gamma ([\eta _1])=\gamma ([\eta _2])\), then there exists a unique \([\xi ']\in {\text {Ext}}^1(S_i^{\oplus m_1},S_{j}\oplus S_i^{\oplus m_2}\oplus S_{{{\tau }}i}^{\oplus n_2})\) such that \(\delta ([\xi '])=[\eta _1] -[\eta _2]\). It follows that \([\xi ']=\alpha \circ \delta ([\xi '])= \alpha ([\eta _1])-\alpha ([\eta _2])=[0]\). So \(\delta ([\xi '])=0\) and \([\eta _1]=[\eta _2]\). Therefore, we obtain that \(\gamma |_{(\alpha |_{{{\mathcal {C}}}_M})^{-1} ([\xi ])}\) is injective for any \([\xi ]\in {{\mathcal {C}}}_1\).
By the above calculations,
$$\begin{aligned} {{\mathcal {C}}}_M=\bigsqcup _{[\xi ]\in {{\mathcal {C}}}_1} {{\mathcal {C}}}_\xi , \text { where }{{\mathcal {C}}}_\xi =\gamma (\alpha |_{{{\mathcal {C}}}_M})^{-1} ([\xi ]). \end{aligned}$$
(A.7)
One can show that \({{\mathcal {C}}}_\xi ={{\mathcal {C}}}_2\), which is independent of \([\xi ]\in {{\mathcal {C}}}_1\). Therefore, \(|{{\mathcal {C}}}_M|=|{{\mathcal {C}}}_1|\cdot |{{\mathcal {C}}}_2|\).\(\square \)
1.3 Computation of \( |{{\mathcal {C}}}_1|\) and \( |{{\mathcal {C}}}_2|\)
It remains to compute \( |{{\mathcal {C}}}_1|\) and \( |{{\mathcal {C}}}_2|\).
Lemma A.2
Retain the notations as above. Then
$$\begin{aligned} |{{\mathcal {C}}}_1|=&q^{-c_{ij}d} \big |{\text {Ext}}^1_{kQ}(S_i^{\oplus (m_1-d)}, S_{j})_{N_1\oplus S_i^{\oplus (t_1^M+e-m_2)}}\big |\cdot \big |{\text {Ext}}_{\Lambda ^\imath }^1 (S_i^{\oplus m_1},S_{{{\tau }}i}^{\oplus n_2})_{{{\mathbb {K}}}_i^{\oplus d} \oplus S_i^{\oplus (m_1-d)} \oplus S_{{{\tau }}i}^{\oplus (n_2-d)} }\big |,\\ |{{\mathcal {C}}}_2|=&q^{-c_{{{\tau }}i,j}e} \big |{\text {Ext}}^1_{kQ}(S_{{{\tau }}i}^{\oplus (n_1-e)}, S_{j})_{N_2\oplus S_{{{\tau }}i}^{\oplus ( t_3^M+d-n_2 )}}\big |\cdot \big |{\text {Ext}}_{\Lambda ^\imath }^1 (S_{{{\tau }}i}^{\oplus n_1},S_i^{\oplus m_2})_{{{\mathbb {K}}}_{{{\tau }}i}^{\oplus e} \oplus S_i^{\oplus (m_2-e)} \oplus S_{{{\tau }}i}^{\oplus (n_1-e)} }\big |. \end{aligned}$$
Proof
The 2 formulas are equivalent, and we only prove the first one. Let \(M'_1:=N_1\oplus S_i^{\oplus (t_1^M-m_2)}\), and
$$\begin{aligned} {{\mathcal {C}}}'_1&:=\{[\xi ]\in {\text {Ext}}^1_{\Lambda ^\imath }( S_i^{\oplus m_1}, S_{j} \oplus S_{{{\tau }}i}^{\oplus n_2})_{L'_1}{|} L'_1\text { admits a short }\\&\quad \text { exact sequence } 0\rightarrow M'_1\rightarrow L'_1\rightarrow {{\mathbb {K}}}_i^{\oplus d}\oplus S_i^{\oplus e} \oplus S_{{{\tau }}i}^{\oplus (n_2-d)} \rightarrow 0\}. \end{aligned}$$
Consider the \(\imath \)quiver algebra with its quiver as in the right figure of (3.8) (the number of arrows from i to j is \( a=-c_{ij}\)), which is denoted by \({}^s\Lambda ^\imath \) to avoid confusions. Then any \({\varvec{s}}_i\Lambda ^\imath \)-module \(L=(L_k,L_\alpha ,L_{\varepsilon _k})\) supported at \(i,{{\tau }}i\) and j with \(L_{\alpha }=0\) for any \(\alpha :{{\tau }}i\rightarrow j\) can be viewed as a \({}^s\Lambda ^\imath \)-module G(L), that is, \(G(L)_i:=L_i\oplus L_{{{\tau }}i}\), \(G(L)_j:=L_j\). Let
$$\begin{aligned} {{\mathcal {C}}}''_1&:=\{[\xi ]\in {\text {Ext}}^1_{^s\Lambda ^\imath }( S_i^{\oplus m_1}, S_{j} \oplus S_{i}^{\oplus n_2})_{L''_1}{|} L''_1\text { admits a short }\\&\quad \text { exact sequence } 0\rightarrow G(M'_1)\rightarrow L''_1\rightarrow {{\mathbb {K}}}_i^{\oplus d}\oplus S_i^{\oplus (n_2-d+e)} \rightarrow 0\}. \end{aligned}$$
Then \(|{{\mathcal {C}}}'_1|=|{{\mathcal {C}}}''_1|\). By applying [24, Proposition 3.10] to compute \([S_i^{\oplus m_1}]\diamond [ S_{j} \oplus S_{i}^{\oplus n_2}]\) in \({{\mathcal {S}}}{{\mathcal {D}}}{{\mathcal {H}}}(^s\Lambda ^\imath )\), one obtains that
$$\begin{aligned} |{{\mathcal {C}}}''_1|=&q^{-c_{ij}d} \big |{\text {Ext}}^1_{kQ^s}(S_i^{\oplus (m_1-d)}, S_{j})_{G(N_1)\oplus S_i^{\oplus (t_1^M+e-m_2)}}\big |\cdot \big |{\text {Ext}}_{^s\Lambda ^\imath }^1 (S_i^{\oplus m_1},S_{ i}^{\oplus n_2})_{{{\mathbb {K}}}_i^{\oplus d} \oplus S_i^{\oplus (m_1+n_1-2d)} }\big |. \end{aligned}$$
Here \(Q^s\) is the quiver 
. Clearly, we have
$$\begin{aligned} \big |{\text {Ext}}^1_{kQ^s}(S_i^{\oplus (m_1-d)}, S_{j})_{G(N_1)\oplus S_i^{\oplus (t_1^M+e-m_2)}}\big |&= \big |{\text {Ext}}^1_{kQ}(S_i^{\oplus (m_1-d)}, S_{j})_{N_1\oplus S_i^{\oplus (t_1^M+e-m_2)}}\big |,\\ \big |{\text {Ext}}_{^s\Lambda ^\imath }^1 (S_i^{\oplus m_1},S_{ i}^{\oplus n_2})_{{{\mathbb {K}}}_i^{\oplus d} \oplus S_i^{\oplus (m_1+n_1-2d)} }\big |&=\big |{\text {Ext}}_{\Lambda ^\imath }^1 (S_i^{\oplus m_1},S_{{{\tau }}i}^{\oplus n_2})_{{{\mathbb {K}}}_i^{\oplus d} \oplus S_i^{\oplus (m_1-d)} \oplus S_{{{\tau }}i}^{\oplus (n_2-d)} }\big |, \end{aligned}$$
and then the desired formula follows.\(\square \)
Recall the notation \(N_1, N_2\) from (A.3).
Lemma A.3
Retain the notations as above. Then
$$\begin{aligned}&\big |{\text {Ext}}^1_{kQ}(S_i^{\oplus (m_1-d)}\oplus S_{{{\tau }}i}^{\oplus (n_1-e)}, S_{j})_{N\oplus S_i^{\oplus (t_1^M+e-m_2)} \oplus S_{{{\tau }}i}^{\oplus (t_3^M+d-n_2)}} \big |\\&\quad =\big |{\text {Ext}}^1_{kQ}(S_i^{\oplus (m_1-d)}, S_{j})_{N_1\oplus S_i^{\oplus (t_1^M+e-m_2)}}\big |\cdot \big |{\text {Ext}}^1_{kQ}(S_{{{\tau }}i}^{\oplus (n_1-e)}, S_{j})_{N_2\oplus S_{{{\tau }}i}^{\oplus ( t_3^M+d-n_2 )}}\big |. \end{aligned}$$
Proof
Note that N, \(N_1\) and \(N_2\) are indecomposable. Then the orders of their automorphism groups are \(q-1\). A direct computation shows that
$$\begin{aligned} F_{S_i^{\oplus (m_1-d)}\oplus S_{{{\tau }}i}^{\oplus (m_1-e)},S_j}^{N\oplus S_i^{\oplus (t_1^M+e-m_2)} \oplus S_{{{\tau }}i}^{\oplus (t_3^M+d-n_2)}}=1=F_{S_i^{\oplus (m_1-d)}, S_{j}}^{N_1\oplus S_i^{\oplus (t_1^M+e-m_2)}}=F_{S_{{{\tau }}i}^{\oplus (m_1-e)}, S_{j}}^{N_2\oplus S_{{{\tau }}i}^{\oplus (t_3^M+d-n_2) }}, \end{aligned}$$
if they are nonzero.
We have
$$\begin{aligned}&(q-1)|{\text {Aut}}(N\oplus S_i^{\oplus (t_1^M+e-m_2)} \oplus S_{{{\tau }}i}^{\oplus (t_3^M+d-n_2)})|\\&\quad =|{\text {Aut}}(N_1\oplus S_i^{\oplus (t_1^M+e-m_2)})|\cdot |{\text {Aut}}(N_2\oplus S_{{{\tau }}i}^{\oplus (t_3^M+d-n_2) })|. \end{aligned}$$
Using the Riedtman-Peng formula, we have
$$\begin{aligned}&\big |{\text {Ext}}^1_{kQ} (S_i^{\oplus (m_1-d)}\oplus S_{{{\tau }}i}^{\oplus (m_1-e)}, S_{j})_{N\oplus S_i^{\oplus (t_1^M+e-m_2)} \oplus S_{{{\tau }}i}^{\oplus (t_3^M+d-n_2)}} \big |\\&\quad = \frac{\big |{\text {Aut}}(S_i^{\oplus (m_1-d)}\oplus S_{{{\tau }}i}^{\oplus (m_1-e)})\big |\cdot |{\text {Aut}}(S_j)|}{\big |{\text {Aut}}(N\oplus S_i^{\oplus (t_1^M+e-m_2)} \oplus S_{{{\tau }}i}^{\oplus (t_3^M+d-n_2)})\big |}\\&\quad = \frac{\big |{\text {Aut}}(S_i^{\oplus (m_1-d)})\big |\cdot \big |{\text {Aut}}(S_{{{\tau }}i}^{\oplus (m_1-e)})\big |\cdot |{\text {Aut}}(S_j)|^2}{\big |{\text {Aut}}(N_1\oplus S_i^{\oplus (t_1^M+e-m_2)})\big |\cdot \big |{\text {Aut}}(N_2\oplus S_{{{\tau }}i}^{\oplus (t_3^M+d-n_2)})\big |}\\&\quad = \big |{\text {Ext}}^1_{kQ}(S_i^{\oplus (m_1-d)}, S_{j})_{N_1\oplus S_i^{\oplus (t_1^M+e-m_2)}}\big |\cdot \big |{\text {Ext}}^1_{kQ}(S_{{{\tau }}i}^{\oplus (m_1-e)}, S_{j})_{N_2\oplus S_{{{\tau }}i}^{\oplus (t_3^M+d-n_2) }}\big |. \end{aligned}$$
The lemma is proved.\(\square \)
1.4 The proof
Now we can complete the proof of Proposition 5.3. By Lemma A.3 we have
$$\begin{aligned}&\big |{\text {Ext}}^1_{kQ}(S_i^{\oplus (m_1-d)}\oplus S_{{{\tau }}i}^{\oplus (n_1-e)}, S_{j}\oplus S_i^{\oplus (m_2-e)}\oplus S_{{{\tau }}i}^{\oplus (n_2 -d) })_M \big |\\&\quad =\big |{\text {Ext}}^1_{kQ}(S_i^{\oplus (m_1-d)}\oplus S_{{{\tau }}i}^{\oplus (n_1-e)}, S_{j})_{N\oplus S_i^{\oplus (t_1^M+e-m_2)} \oplus S_{{{\tau }}i}^{\oplus (t_3^M+d-n_2)}} \big |\\&\quad =\big |{\text {Ext}}^1_{kQ}(S_i^{\oplus (m_1-d)}, S_{j})_{N_1\oplus S_i^{\oplus (t_1^M+e-m_2)}}\big |\cdot \big |{\text {Ext}}^1_{kQ}(S_{{{\tau }}i}^{\oplus (n_1-e)}, S_{j})_{N_2\oplus S_{{{\tau }}i}^{\oplus ( t_3^M+d-n_2 )}}\big |. \end{aligned}$$
Together with Lemmas A.1 and A.2, we have
$$\begin{aligned} |{{\mathcal {C}}}_M|&= q^{-c_{ij}d-c_{{{\tau }}i,j}e}\big |{\text {Ext}}^1_{kQ}(S_i^{\oplus (m_1-d)}\oplus S_{{{\tau }}i}^{\oplus (n_1-e)}, S_{j}\oplus S_i^{\oplus (m_2-e)}\oplus S_{{{\tau }}i}^{\oplus (n_2-d) })_M \big |\nonumber \\&\quad \cdot \big |{\text {Ext}}_{\Lambda ^\imath }^1 (S_i^{\oplus m_1},S_{{{\tau }}i}^{\oplus n_2})_{{{\mathbb {K}}}_i^{\oplus d} \oplus S_i^{\oplus (m_1-d)} \oplus S_{{{\tau }}i}^{\oplus (n_2-d)} }\big |\cdot \big |{\text {Ext}}_{\Lambda ^\imath }^1 (S_{{{\tau }}i}^{\oplus n_1},S_i^{\oplus m_2})_{{{\mathbb {K}}}_{{{\tau }}i}^{\oplus e} \oplus S_i^{\oplus (m_2-e)} \oplus S_{{{\tau }}i}^{\oplus (n_1-e)} }\big |. \end{aligned}$$
(A.8)
By a standard computation (see e.g. [33]), we have
$$\begin{aligned}&F_{S_i^{\oplus (m_1-d)}\oplus S_{{{\tau }}i}^{\oplus (n_1-e)}, S_{j}\oplus S_i^{\oplus (m_2-e)}\oplus S_{{{\tau }}i}^{\oplus (n_2-d) }}^{ M}\\&\quad ={\mathbf{v}}^{(t_1^M-(m_2-e))(m_2-e)} \begin{bmatrix} t_1^M\\ m_2-e \end{bmatrix} _{\mathbf{v}}\cdot {\mathbf{v}}^{(t_3^M-(n_2-d))(n_2-d)} \begin{bmatrix} t_3^M\\ n_2-d \end{bmatrix} _{\mathbf{v}}. \end{aligned}$$
Using (2.25), one can obtain that
$$\begin{aligned}&\big |{\text {Ext}}^1_{kQ}(S_i^{\oplus (m_1-d)}\oplus S_{{{\tau }}i}^{\oplus (n_1-e)}, S_{j}\oplus S_i^{\oplus (m_2-e)}\oplus S_{{{\tau }}i}^{\oplus (n_2-d) })_M \big |\\&\quad = \prod _{i=0}^{m_1-d-1}(q^{m_1-d}-q^i)\prod _{i=0}^{n_1-e-1}(q^{n_1-e}-q^i) \prod _{i=0}^{m_2-e-1}(q^{m_2-e}-q^i)\prod _{i=0}^{n_2-d-1}(q^{n_2-d}-q^i)\\ {}&\quad \times {\mathbf{v}}^{ 2(m_1-d)(m_2-e)+ 2(n_1-e)(n_2-d)}{\mathbf{v}}^{(t_1^M-(m_2-e))(m_2-e)+(t_3^M-(n_2-d))(n_2-d)}\\&\qquad \times \begin{bmatrix} t_1^M\\ m_2-e \end{bmatrix} _{\mathbf{v}}\begin{bmatrix} t_3^M\\ n_2-d \end{bmatrix} _{\mathbf{v}}\frac{(q-1)}{|{\text {Aut}}(M)|}\\&\quad = {\mathbf{v}}^{(m_1-d)^2+{ m_1-d \atopwithdelims ()2} + (n_1-e)^2 + { n_1-e \atopwithdelims ()2} + (m_2-e)^2 +{ m_2-e \atopwithdelims ()2} +(n_2-d)^2 + {n_2-d \atopwithdelims ()2} } \\&\qquad \times {\mathbf{v}}^{ 2(m_1-d)(m_2-e)+ 2(n_1-e)(n_2-d)}{\mathbf{v}}^{(t_1^M-(m_2-e))(m_2-e)+(t_3^M-(n_2-d))(n_2-d)} ({\mathbf{v}}-{\mathbf{v}}^{-1})^{m_1+m_2+n_1+n_2-2d-2e} \\&\qquad \times [m_1-d]_{\mathbf{v}}^! [n_1-e]_{\mathbf{v}}^! [m_2-e]_{\mathbf{v}}^! [n_2-d]_{\mathbf{v}}^!\begin{bmatrix} t_1^M\\ m_2-e \end{bmatrix} _{\mathbf{v}}\begin{bmatrix} t_3^M\\ n_2-d \end{bmatrix} _{\mathbf{v}}\frac{(q-1)}{|{\text {Aut}}(M)|}. \end{aligned}$$
Furthermore, we have
$$\begin{aligned}&\big |{\text {Ext}}_{\Lambda ^\imath }^1 (S_i^{\oplus m_1},S_{{{\tau }}i}^{\oplus n_2})_{{{\mathbb {K}}}_i^{\oplus d} \oplus S_i^{\oplus (m_1-d)} \oplus S_{{{\tau }}i}^{\oplus (n_2-d)} }\big | \nonumber \\&\quad = \frac{\prod _{i=0}^{d-1} (q^{m_1}-q^{i}) \prod _{i=0}^{d-1} (q^{n_2}-q^{i})}{\prod _{i=0}^{d-1}(q^d-q^i)}\nonumber \\&\quad = {\mathbf{v}}^{d(m_1+n_2-d)+ {d \atopwithdelims ()2} } ({\mathbf{v}}-{\mathbf{v}}^{-1})^d \begin{bmatrix} m_1\\ d \end{bmatrix} _{\mathbf{v}}\begin{bmatrix} n_2\\ d \end{bmatrix} _{\mathbf{v}}[d]_{\mathbf{v}}^!, \end{aligned}$$
(A.9)
and
$$\begin{aligned}&\big |{\text {Ext}}_{\Lambda ^\imath }^1 (S_{{{\tau }}i}^{\oplus n_1},S_i^{\oplus m_2})_{{{\mathbb {K}}}_{{{\tau }}i}^{\oplus e} \oplus S_i^{\oplus (m_2-e)} \oplus S_{{{\tau }}i}^{\oplus (n_1-e)} }\big |={\mathbf{v}}^{e(n_1+m_2-e)+ {e \atopwithdelims ()2} } ({\mathbf{v}}-{\mathbf{v}}^{-1})^e\begin{bmatrix} m_2\\ e \end{bmatrix} _{\mathbf{v}}\begin{bmatrix} n_1\\ e \end{bmatrix} _{\mathbf{v}}[e]_{\mathbf{v}}^!. \end{aligned}$$
Plugging into (A.8), we have
$$\begin{aligned} |{{\mathcal {C}}}_M|&={\mathbf{v}}^{-2c_{ij}d-2c_{{{\tau }}i,j}e} {\mathbf{v}}^{(m_1-d)^2+{ m_1-d \atopwithdelims ()2} + (n_1-e)^2 + { n_1-e \atopwithdelims ()2} + (m_2-e)^2 +{ m_2-e \atopwithdelims ()2} +(n_2-d)^2 + {n_2-d \atopwithdelims ()2} } \\&\quad \times {\mathbf{v}}^{d(m_1+n_2-d)+ {d \atopwithdelims ()2} +e(n_1+m_2-e)+ {e \atopwithdelims ()2}} {\mathbf{v}}^{ 2(m_1-d)(m_2-e)+ 2(n_1-e)(n_2-d)}\\&\quad \times {\mathbf{v}}^{(t_1^M-(m_2-e))(m_2-e)+(t_3^M-(n_2-d))(n_2-d)} ({\mathbf{v}}-{\mathbf{v}}^{-1})^{m_1+m_2+n_1+n_2-d-e}\\&\quad \times \frac{[m_1]_{\mathbf{v}}^![m_2]_{\mathbf{v}}^![n_1]_{\mathbf{v}}^![n_2]_{\mathbf{v}}^!}{[d]_{\mathbf{v}}^![e]_{\mathbf{v}}^!}\begin{bmatrix} t_1^M\\ m_2-e \end{bmatrix} _{\mathbf{v}}\begin{bmatrix} t_3^M\\ n_2-d \end{bmatrix} _{\mathbf{v}}\frac{(q-1)}{|{\text {Aut}}(M)|}\\&= {\mathbf{v}}^{-2c_{ij}d-2c_{{{\tau }}i,j}e} {\mathbf{v}}^{ 2(m_1-d)(m_2-e)+ 2(n_1-e)(n_2-d)} ({\mathbf{v}}-{\mathbf{v}}^{-1})^{m_1+m_2+n_1+n_2-d-e}\\&\quad \times {\mathbf{v}}^{p'(t_3^M,d,m_1,n_2)} \frac{[m_1]_{\mathbf{v}}^![n_2]_{\mathbf{v}}^!}{[d]_{\mathbf{v}}^!} \begin{bmatrix} t_3^M\\ n_2-d \end{bmatrix} _{\mathbf{v}}{\mathbf{v}}^{p'(t_1^M,e,n_1,m_2)} \frac{[m_2]_{\mathbf{v}}^![n_1]_{\mathbf{v}}^!}{[e]_{\mathbf{v}}^!} \begin{bmatrix} t_1^M\\ m_2-e \end{bmatrix} _{\mathbf{v}}\frac{(q-1)}{|{\text {Aut}}(M)|}. \end{aligned}$$
The desired formula (5.7) follows from the above formula and (A.2). This completes the proof of Proposition 5.3.
Appendix B: Proof of the formula (3.7)
In this appendix, we provide the details for the proof of the formula (3.7).
1.1 Computation of \([S'_i]_{{\bar{1}}}^{(r)}*[S'_j] *[S'_i]_{\overline{1}+\overline{a}}^{(s)}\)
Let us first compute \([S'_i]_{{\bar{1}}}^{(r)}*[S'_j] *[S'_i]_{\overline{1}+\overline{a}}^{(s)}\), depending on the parity of r.
1.1.1
r is even
For any \(s\ge 0\) such that \(r+s+2t=a\) with \(t\ge 0\), we have by Lemma 3.1 and Lemma 3.3
$$\begin{aligned} {[}S'_i]_{{\bar{1}}}^{(r)}*[S'_j] *[S'_i]_{\overline{1}+\overline{a}}^{(s)}&= \sum _{k=0}^{\frac{r}{2}}\frac{{\mathbf{v}}^{k(k+1)-\left( {\begin{array}{c}r-2k\\ 2\end{array}}\right) } \cdot ({\mathbf{v}}-{\mathbf{v}}^{-1})^k }{[r-2k]_{\mathbf{v}}^{!}[2k]_{\mathbf{v}}^{!!}} [(r-2k)S'_i]*[{{\mathbb {K}}}'_i]^k*[S'_j] \\&\quad * \sum _{m=0}^{\lfloor \frac{s}{2}\rfloor } \frac{{\mathbf{v}}^{m(m+1)-\left( {\begin{array}{c}s-2m\\ 2\end{array}}\right) }\cdot ({\mathbf{v}}-{\mathbf{v}}^{-1})^{m}}{[s-2m]_{\mathbf{v}}^{!} [2m]_{\mathbf{v}}^{!!}} [(s-2m)S'_i]*[{{\mathbb {K}}}'_i]^m \\&= \sum _{k=0}^{\frac{r}{2}}\sum _{m=0}^{\lfloor \frac{s}{2}\rfloor } \frac{{\mathbf{v}}^{k(k+1)+m(m+1)-\left( {\begin{array}{c}r-2k\\ 2\end{array}}\right) -\left( {\begin{array}{c}s-2m\\ 2\end{array}}\right) }\cdot ({\mathbf{v}}-{\mathbf{v}}^{-1})^{k+m} }{[r-2k]_{\mathbf{v}}^![s-2m]_{\mathbf{v}}^![2k]_{\mathbf{v}}^{!!}[2m]_{\mathbf{v}}^{!!}}\\&\quad \times [(r-2k)S'_i]*[S'_j]*[(s-2m)S'_i]*[{{\mathbb {K}}}'_i]^{k+m}\\&= \sum _{k=0}^{\frac{r}{2}}\sum _{m=0}^{\lfloor \frac{s}{2}\rfloor } \sum _{n=0}^{\min \{r-2k,s-2m\}} \sum _{[M]\in {\mathcal {I}}_{r+s-2k-2m-2n}} \frac{{\mathbf{v}}^{k(k+1)+m(m+1)-\left( {\begin{array}{c}r-2k\\ 2\end{array}}\right) -\left( {\begin{array}{c}s-2m\\ 2\end{array}}\right) } }{[r-2k]_{\mathbf{v}}^![s-2m]_{\mathbf{v}}^![2k]_{\mathbf{v}}^{!!}[2m]_{\mathbf{v}}^{!!}}\\&\quad \times {\mathbf{v}}^{p(a,n,r-2k,s -2m)} ({\mathbf{v}}-{\mathbf{v}}^{-1})^{r+s -k -m -n+1 } \frac{[r-2k]_{\mathbf{v}}^{!} [s -2m]_{\mathbf{v}}^{!}}{[n]_{\mathbf{v}}^{!}} \\&\quad \times \begin{bmatrix} u_M\\ s -2m-n \end{bmatrix} _{\mathbf{v}}\frac{[M]}{|{\text {Aut}}(M)|}*[{{\mathbb {K}}}'_i]^{n+k+m}. \end{aligned}$$
This can be simplified to be
$$\begin{aligned}{}[S'_i]_{{\bar{1}}}^{(r)}*[S'_j] *[S'_i]_{{\bar{1}}}^{(s)}&= \sum _{k=0}^{\frac{r}{2}}\sum _{m=0}^{\lfloor \frac{s}{2}\rfloor } \sum _{n=0}^{r-2k} \sum _{[M]\in {\mathcal {I}}_{r+s-2k-2m-2n}} \frac{{\mathbf{v}}^{{{\mathcal {Z}}}(a,r,s,k,m,n)+2k+2m} ({\mathbf{v}}-{\mathbf{v}}^{-1})^{r+s-k-m-n+1} }{[n]_{\mathbf{v}}^{!} [2k]_{\mathbf{v}}^{!!}[2m]_{\mathbf{v}}^{!!}}\\&\quad \times \begin{bmatrix} u_M\\ s -2m-n \end{bmatrix} _{\mathbf{v}}\frac{[M] *[{{\mathbb {K}}}'_i]^{n+k+m}}{|{\text {Aut}}(M)|}. \end{aligned}$$
1.1.2
r is even
For any \(s\ge 0\) such that \(r+s+2t=a\) with \(t\ge 0\), we have by Lemma 3.1 and Lemma 3.3
$$\begin{aligned} {[}S'_i]_{{\bar{1}}}^{(r)}*[S'_j] *[S'_i]_{\overline{1}+\overline{a}}^{(s)}&=\sum _{k=0}^{\lfloor \frac{r}{2}\rfloor }\frac{{\mathbf{v}}^{k(k-1)-\left( {\begin{array}{c}r-2k\\ 2\end{array}}\right) } \cdot ({\mathbf{v}}-{\mathbf{v}}^{-1})^k }{[r-2k]_{\mathbf{v}}^![2k]_{\mathbf{v}}^{!!}} [(r-2k)S'_i]*[{{\mathbb {K}}}'_i]^k*[S'_j]\\&\quad * \sum _{m=0}^{\lfloor \frac{s}{2}\rfloor } \frac{{\mathbf{v}}^{m(m-1)-\left( {\begin{array}{c}s-2m\\ 2\end{array}}\right) }\cdot ({\mathbf{v}}-{\mathbf{v}}^{-1})^{m}}{[s-2m]_{\mathbf{v}}^{!}[2m]_{\mathbf{v}}^{!!}} [(s-2m)S'_i]*[{{\mathbb {K}}}'_i]^m\\&=\sum _{k=0}^{\lfloor \frac{r}{2}\rfloor }\sum _{m=0}^{\lfloor \frac{s}{2}\rfloor } \sum _{n=0}^{\min \{r-2k,s-2m\}} \sum _{[M]\in {\mathcal {I}}_{r+s-2k-2m-2n}} \frac{{\mathbf{v}}^{k(k-1)+m(m-1)-\left( {\begin{array}{c}r-2k\\ 2\end{array}}\right) -\left( {\begin{array}{c}s-2m\\ 2\end{array}}\right) } }{[r-2k]_{\mathbf{v}}^![s-2m]_{\mathbf{v}}^![2k]_{\mathbf{v}}^{!!}[2m]_{\mathbf{v}}^{!!}}\\&\quad \times {\mathbf{v}}^{p(a,n,r-2k, s -2m)} ({\mathbf{v}}-{\mathbf{v}}^{-1})^{r+s -k -m -n+1 } \frac{[r-2k]_{\mathbf{v}}^{!} [s -2m]_{\mathbf{v}}^{!}}{[n]_{\mathbf{v}}^{!}}\\&\quad \times \begin{bmatrix} u_M\\ s -2m-n \end{bmatrix} _{\mathbf{v}}\frac{[M]}{|{\text {Aut}}(M)|}*[{{\mathbb {K}}}'_i]^{n+k+m}. \end{aligned}$$
This can be simplified to be
$$\begin{aligned}&[S'_i]_{{\bar{1}}}^{(r)}*[S'_j] *[S'_i]_{\overline{1}+\overline{a}}^{(s)} = \sum _{k=0}^{\lfloor \frac{r}{2}\rfloor }\sum _{m=0}^{\lfloor \frac{s}{2}\rfloor } \sum _{n=0}^{r-2k} \sum _{[M]\in {\mathcal {I}}_{r+s-2k-2m-2n}}\\&\frac{{\mathbf{v}}^{{{\mathcal {Z}}}(a,r,s,k,m,n)} ({\mathbf{v}}-{\mathbf{v}}^{-1})^{r+s -k -m -n+1} }{[n]_{\mathbf{v}}^{!} [2k]_{\mathbf{v}}^{!!}[2m]_{\mathbf{v}}^{!!}} \begin{bmatrix} u_M\\ s -2m-n \end{bmatrix} _{\mathbf{v}}\frac{[M] *[{{\mathbb {K}}}'_i]^{n+k+m}}{|{\text {Aut}}(M)|}. \end{aligned}$$
1.2 Reduction for (3.7)
Summing up the above two cases, we obtain
$$\begin{aligned} \text {RHS}(3.7)&= \sum _{r=0,2{|} r}^{a} \sum _{k=0}^{\frac{r}{2}}\sum _{m=0}^{\lfloor \frac{a-r}{2}\rfloor } \sum _{n=0}^{r-2k} \sum _{[M]\in {\mathcal {I}}_{a-2k-2m-2n}} (-1)^a ({\mathbf{v}}-{\mathbf{v}}^{-1})^{ -k -m -n+1}\nonumber \\&\quad \times \frac{{\mathbf{v}}^{r+{{\mathcal {Z}}}(a,r,a-r,k,m,n)+2k+2m-a} }{[n]_{\mathbf{v}}^{!} [2k]_{\mathbf{v}}^{!!}[2m]_{\mathbf{v}}^{!!}} \begin{bmatrix} u_M\\ a-r -2m-n \end{bmatrix} _{\mathbf{v}}\frac{[M] *[{{\mathbb {K}}}'_i]^{n+k+m}}{|{\text {Aut}}(M)|} \nonumber \\&\quad -\sum _{r=0,2\not \mid r}^{a}\sum _{k=0}^{\lfloor \frac{r}{2}\rfloor }\sum _{m=0}^{\lfloor \frac{a-r}{2}\rfloor } \sum _{n=0}^{r-2k} \sum _{[M]\in {\mathcal {I}}_{a-2k-2m-2n}} (-1)^a ({\mathbf{v}}-{\mathbf{v}}^{-1})^{ -k -m -n+1}\nonumber \\&\quad \times \frac{{\mathbf{v}}^{r+{{\mathcal {Z}}}(a,r,a-r,k,m,n)-a} }{[n]_{\mathbf{v}}^{!} [2k]_{\mathbf{v}}^{!!}[2m]_{\mathbf{v}}^{!!}} \begin{bmatrix} u_M\\ a-r -2m-n \end{bmatrix} _{\mathbf{v}}\frac{[M] *[{{\mathbb {K}}}'_i]^{n+k+m}}{|{\text {Aut}}(M)|} \nonumber \\&\quad -\sum _{t\ge 1}\sum _{r=0,2\not \mid r}^{a-2t}\sum _{k=0}^{\lfloor \frac{r}{2}\rfloor }\sum _{m=0}^{\lfloor \frac{a-r}{2}\rfloor -t} \sum _{n=0}^{r-2k} \sum _{[M]\in {\mathcal {I}}_{a-2t-2k-2m-2n}} (-1)^a ({\mathbf{v}}-{\mathbf{v}}^{-1})^{ -k -m -n+1} \nonumber \\&\quad \times \frac{{\mathbf{v}}^{r+ {{\mathcal {Z}}}(a,r,a-2t-r,k,m,n) -a+2t} }{[n]_{\mathbf{v}}^{!} [2k]_{\mathbf{v}}^{!!}[2m]_{\mathbf{v}}^{!!}} \begin{bmatrix} u_M\\ a-2t-r -2m-n \end{bmatrix} _{\mathbf{v}}\frac{[M] *[{{\mathbb {K}}}'_i]^{n+k+m+t}}{|{\text {Aut}}(M)|}. \end{aligned}$$
(B.1)
Fix
$$\begin{aligned} d =t+k+m+n. \end{aligned}$$
We have reduced the proof of (3.7) to proving the coefficient of \(\frac{[M] *[{{\mathbb {K}}}'_i]^{d}}{|{\text {Aut}}(M)|}\) of the RHS of (B.1) is 0 for any given \([M] \in {\mathcal {I}}_{a-2d}\) if not both d and \(u_M\) are zeros.
Set \(u=u_M\), and set
$$\begin{aligned} A'(a,d,u)&= \sum _{t\ge 0}\sum _{r=0,2\not \mid r}^{a-2t}\sum _{k=0}^{\lfloor \frac{r}{2}\rfloor }\sum _{m=0}^{\lfloor \frac{a-r}{2}\rfloor -t} \delta \{0\le n \le r-2k\} \nonumber \\&\quad \times \frac{{\mathbf{v}}^{r+{{\mathcal {Z}}}(a,r,a-2t-r,k,m,n)-a+2t} ({\mathbf{v}}-{\mathbf{v}}^{-1})^{-k-m-n+1} }{[n]_{\mathbf{v}}^{!} [2k]_{\mathbf{v}}^{!!}[2m]_{\mathbf{v}}^{!!}} \begin{bmatrix} u\\ a-2t-r -2m-n \end{bmatrix} _{\mathbf{v}}\nonumber \\&\quad -\sum _{r=0,2{|} r}^{a} \sum _{k=0}^{\frac{r}{2}}\sum _{m=0}^{\lfloor \frac{a-r}{2}\rfloor } \delta \{0\le n \le r-2k\} \nonumber \\&\quad \times \frac{{\mathbf{v}}^{r+{{\mathcal {Z}}}(a,r,a-r,k,m,n)+2k+2m-a} ({\mathbf{v}}-{\mathbf{v}}^{-1})^{ -k -m -n+1} }{[n]_{\mathbf{v}}^{!} [2k]_{\mathbf{v}}^{!!}[2m]_{\mathbf{v}}^{!!}} \begin{bmatrix} u\\ a -r-2m-n \end{bmatrix} _{\mathbf{v}}. \end{aligned}$$
(B.2)
See (3.12) for \({{\mathcal {Z}}}(\cdot , \cdot , \cdot , \cdot , \cdot , \cdot )\) and also see (3.10) for \(p(\cdot , \cdot , \cdot , \cdot )\).
Then the coefficient of \(\frac{[M] *[{{\mathbb {K}}}'_i]^{d}}{|{\text {Aut}}(M)|}\) of the RHS of (B.1) is \((-1)^a A'(a,d,u_M)\). Summarizing, we have established the following (which is the counterpart of Proposition 3.4).
Proposition B.1
The formula (3.7) is equivalent to the following identity
$$\begin{aligned} A'(a,d,u) =0, \end{aligned}$$
(B.3)
for non-negative integers a, d, u subject to the constraints:
$$\begin{aligned} 0\le d\le \frac{a}{2}, \quad 0\le u \le a-2d, \quad d \text { and}\, u\,\text { not both zero}. \end{aligned}$$
(B.4)
1.3 Reduction for the identity (B.3)
We shall denote the 2 summands in \(A' =A'(a,d,u)\) in (B.2) as \(A'_0, A'_1\), and thus
$$\begin{aligned} A' =A'_0 -A'_1. \end{aligned}$$
Compare with (3.14). We also denote
$$\begin{aligned} w&=r-2k-n. \end{aligned}$$
(B.5)
Set
$$\begin{aligned} d= k+m+n +t \end{aligned}$$
(B.6)
in the \(A'_0\) side, and \(d =k+m+n\) in the \(A'_1\) side. Using the same argument as in §3.6, for \(d>0\), we see that the identity (B.3) for \(d>0\) is equivalent to the following identity
$$\begin{aligned} \sum _{{\mathop {t+k+m+n =d}\limits ^{w+n {\text { odd}}}}} \frac{{\mathbf{v}}^{t^2 -2dt +t +2nt +\left( {\begin{array}{c}n+1\\ 2\end{array}}\right) -2km -2m}}{[n]^!_{\mathbf{v}}[2k]^{!!}_{\mathbf{v}}[2m]^{!!}_{\mathbf{v}}} ({\mathbf{v}}-{\mathbf{v}}^{-1})^{t} - \sum _{{\mathop {k+m+n =d}\limits ^{w+n {\text { even}}}}} \frac{{\mathbf{v}}^{\left( {\begin{array}{c}n+1\\ 2\end{array}}\right) -2km +2k}}{[n]^!_{\mathbf{v}}[2k]^{!!}_{\mathbf{v}}[2m]^{!!}_{\mathbf{v}}} =0. \end{aligned}$$
(B.7)
The identity (B.7) is clearly equivalent to the identity (3.23) (by switching the parity of w), which was established in Sect. 4.
The identity (B.3) for \(d=0\) holds exactly in the same way as for (3.24) (up to an irrelevant overall sign). Therefore, the identity (B.3) is fully established, and then the formula (3.7) follows by Proposition B.1.
. Clearly, we have