Proof. (1) It can be easily verified that \(u\overline{u}^T \in \mathfrak{sp}_N \;\forall \;u \in \mathbb{C}^N\). On the other hand, we have \(X_{i,j} = (e_i + e_{n+j}) \overline{(e_i + e_{n+j})}^T - e_i\overline{e_i}^T - e_{n+j}\overline{e_{n+j}}^T, \;Y_{i,j} = (e_i + e_j) \overline{(e_i + e_j)}^T - e_i\overline{e_i}^T - e_{j}\overline{e_{j}}^T, \;Z_{i,j} = (e_{n+i} + e_{n+j}) \overline{(e_{n+i} + e_{n+j})}^T - e_{n+i}\overline{e_{n+i}}^T - e_{n+j}\overline{e_{n+j}}^T \;\forall \;1 \leqslant i \neq j \leqslant n\) and \(U_i = -e_i\overline{e_i}^T, \;V_i = e_{n+i}\overline{e_{n+i}}^T, \;H_i = (e_i + e_{n+i}) \overline{(e_i + e_{n+i})}^T - e_i\overline{e_i}^T - e_{n+i}\overline{e_{n+i}}^T \;\forall \;1 \leqslant i \leqslant n\), which proves (1).
(2) \(u\overline{v}^T + v\overline{u}^T = (u+v)(\overline{u+v})^T - u\overline{u}^T - v\overline{v}^T \;\forall \;u, v \in \mathbb{C}^N\) and so we are done using (1).
(3) Claim. \(\big(v\overline{v}^T(u\overline{v}^T + v\overline{u}^T) + (u\overline{v}^T + v\overline{u}^T)v\overline{v}^T\big)\) acts trivially on \(\mathcal{J}_{\mathcal{H}}(V)\) for each \(u, v \in \mathbb{C}^N\).
Pick any \(w \in \mathcal{J}_{\mathcal{H}}(V)\). By applying (1), we then have \[\begin{align} 0 = \big((u+v)\overline{(u+v)}^T)^2w = (u\overline{u}^T)(u\overline{v}^T + v\overline{u}^T)w + (u\overline{v}^T + v\overline{u}^T)(u\overline{u}^T)w + (v\overline{v}^T)(u\overline{v}^T + v\overline{u}^T)w \\ + (u\overline{v}^T + v\overline{u}^T)(v\overline{v}^T)w \;\forall \;u, v \in \mathbb{C}^N. \end{align}\] Now replace \(u\) by \(pu\) and \(v\) by \(qv\) for \(p, q \in \mathbb{Z}\) in the above equation. Henceforth, comparing the coefficient of \(pq^3\) and using Lemma 11, the claim follows.
Again, for any \(r, s \in \mathbb{Z}^N\), consider \[\begin{align} (s\overline{s}^T)^2(r\overline{r}^T)w = (s\overline{s}^T)[s\overline{s}^T, r\overline{r}^T] w + [s\overline{s}^T, r\overline{r}^T](s\overline{s}^T)w = (\overline{s}|r) \big(s\overline{s}^T(s\overline{r}^T + r\overline{s}^T) + (s\overline{r}^T + r\overline{s}^T)s\overline{s}^T\big)w \end{align}\] The desired result is now an immediate consequence of (1) and the above claim.
(4) \((u\overline{u}^T)^2\) acts trivially on \({\Lambda}^k\mathbb{C}^N \;(0 \leqslant k \leqslant N)\) for all \(u \in \mathbb{C}^N\) , which proves the “if part".
Conversely, assume that \(\mathcal{J}_{\mathcal{H}}(V) = V\). We first show that \(V\) must be finite-dimensional.
By hypothesis, \(U_i^2w = V_i^2w = 0 \;\forall \;w \in V\). Consequently, for each \(1 \leqslant i \leqslant n\), there exists \(0 \neq w_i, w_i^{\prime} \in V\) such that \((V_i)w_i = (U_i)w_i^{\prime} = 0\).
Claim 1. \(X_{i,j}\) does not act injectively on \(V\) for any \(1 \leqslant i \neq j \leqslant n\).
Taking \(r = e_i + e_{n+j} \in \mathbb{Z}^{N}\), we get \((X_{i,j})^2w_j = (r\overline{r}^T)^2w_j = 0\), as \([U_i, V_j] = 0\). Hence the claim.
Claim 2. \(Y_{i,j}\) does not act injectively on \(V\) for any \(1 \leqslant i \neq j \leqslant n\).
Taking \(r = e_i + e_j \in \mathbb{Z}^{N}\), we obtain \((Y_{i,j})^2w_i^{\prime} = (r\overline{r}^T)^2w_i^{\prime} = 0\), since \([U_i, U_j] = 0\). Hence the claim.
Claim 3. \(Z_{i,j}\) does not act injectively on \(V\) for any \(1 \leqslant i \neq j \leqslant n\).
Taking \(r = e_{n+i} + e_{n+j} \in \mathbb{Z}^{N}\), we get \((Z_{i,j})^2w_i = (r\overline{r}^T)^2w_i = 0\), as \([V_i, V_j] = 0\). Hence the claim.
We can now directly invoke Lemma 10 to conclude that \(V\) is finite-dimensional.
Assume that \(V \ncong V(\omega_k)\) for any \(0 \leqslant k \leqslant n\). Then \(V \cong V(\lambda)\) for some \(\lambda \in P_{\mathfrak{sp}_{N}}^{+} \setminus \{\omega_k\}_{k=0}^{n}\). So there exist \(a_1, \ldots, a_k \in \mathbb{Z}_{\geqslant 0}\) such that \(\lambda = \sum_{k=1}^{n} a_k \omega_k\) with \(a_i + a_j \geqslant 2\) for some \(1 \leqslant i \leqslant j \leqslant n\). Note that \(\omega_i(H_i) = 1 = \omega_j(H_i)\), which implies that \(\lambda(H_i) \geqslant 2\). Now pick \(0 \neq v_{\lambda} \in V(\lambda)_{\lambda}\) and consider the \(S_i\)-module \(\text{span} \{(E_{n+i,i})^rv_{\lambda} \;| \;r \in \mathbb{Z}_{\geqslant 0} \}\), where \(S_i := \text{span} \{U_i, V_i, H_i \} \cong \mathfrak{sl}_2\). Consequently, by \(\mathfrak{sl}_2\)-theory, it immediately follows that \((E_{n+i,i})^2v_{\lambda} \neq 0\) and thus \(v_{\lambda} \notin \mathcal{J}_{\mathcal{H}}(V(\lambda))\). ◻

Proof. Assume that \(V \ncong V(\omega_k)\) for any \(0 \leqslant k \leqslant n\). Let \(W\) be a non-zero \(\mathcal{H}_N\)-submodule of \(L_{\mathcal{H}}^{\beta}(V)\). Taking \(W_m = \{v \in V \;| \;v \otimes t^m \in W \}\), we obtain \(W = \bigoplus_{m \in \mathbb{Z}^N} W_m \otimes \mathbb{C}t^m\), since \(W\) is \(\mathbb{Z}^N\)-graded. Set \(\widetilde{W} = \bigcap_{m \in \mathbb{Z}^N} W_m\). We now claim that \(\widetilde{W} \neq (0)\).
By Lemma 12, \(\mathcal{J}_{\mathcal{H}}(V) = (0)\) and thus for any \(0 \neq v \in V\), there exists \(r \in \mathbb{Z}^N\) such that \((r\overline{r}^T)^2v \neq 0\). Fix any non-zero \(w \in W_m\) for some \(m \in \mathbb{Z}^N\). For any \(s \in \mathbb{Z}^N\), consider \[\begin{align} h_{-r}h_{r+s}(w \otimes t^m) = -(\overline{r}|m+r+s+\beta)(\overline{r+s}|m+\beta)w\otimes t^{m+s} + (\overline{r+s}|m+\beta)(r\overline{r}^T)w \otimes t^{m+s} \\ - (\overline{r}|m+s+\beta)(r+s)(\overline{r+s})^Tw \otimes t^{m+s} + (r\overline{r})^T(r+s)(\overline{r+s})^Tw \otimes t^{m+s} \in W_{m+s} \otimes \mathbb{C}t^{m+s}. \end{align}\] Replacing \(r\) by \(pr\) for \(p \in \mathbb{Z}\) in the above equation and comparing the coefficients of \(p^4\), we can then apply Lemma 11 to conclude that \((r\overline{r}^T)^2w \in W_{m+s} \;\forall \;s \in \mathbb{Z}^N\). Hence the claim follows.
Again, for any \(m, s \in \mathbb{Z}^N\) and \(v \in \widetilde{W}\), we have \[\begin{align} h_m(v \otimes t^{s-m}) = (\overline{m}|s-m+\beta)v \otimes t^s + (m\overline{m}^T)v \otimes t^s, \end{align}\] which gives \((m\overline{m}^T)v \in \widetilde{W}\). This proves \(\widetilde{W} = V\) using Lemma 12, and thus the result follows. ◻

Proof. Let \((k+\beta)(\overline{k+\beta})^Tv \otimes t^k \in W_{min}^{k,\beta}(\Lambda^p\mathbb{C}^N)\). For any \(r \in \mathbb{Z}^N\), we have \[\begin{align} h_r\big((k+\beta)(\overline{k+\beta})^Tv \otimes t^k\big) = (\overline{r}|k+\beta)(k+\beta)(\overline{k+\beta})^Tv \otimes t^{k+r} + r\overline{r}^T(k+\beta)(\overline{k+\beta})^Tv \otimes t^{k+r}. \end{align}\] So it suffices to show that \((\overline{r}|k+\beta)(k+\beta)(\overline{k+\beta})^Tv + r\overline{r}^T(k+\beta)(\overline{k+\beta})^Tv \in W_{min}^{(k+r),\beta}(\Lambda^p\mathbb{C}^N)\). To this end, let us consider \[\begin{align} (\overline{r}|k+\beta)((k+\beta+r)(\overline{k+\beta+r})^Tv \big) + (k+\beta+r)(\overline{k+\beta+r})^Tr\overline{r}^Tv \in W_{min}^{(k+r),\beta}(\Lambda^p\mathbb{C}^N) \\ =(\overline{r}|k+\beta)\big((k+\beta)(\overline{k+\beta})^T + r\overline{r}^T + (k+\beta)\overline{r}^T + r(\overline{k+\beta})^T \big)v + (k +\beta)(\overline{k+\beta})^Tr\overline{r}^Tv \\ + (r\overline{r}^T)^2v + \big((k+\beta)\overline{r}^T +r(\overline{k+\beta})^T\big)r\overline{r}^Tv \ldots\ldots (*) \end{align}\] Now by Claim of Lemma 12, we obtain \[\begin{align} 0 = r\overline{r}^T\big((k+\beta)\overline{r}^T + r(\overline{k+\beta})^T\big)v + \big((k+\beta)\overline{r}^T + r\overline{(}{k+\beta})^T\big)r\overline{r}^Tv \\ = [r\overline{r}^T,(k+\beta)\overline{r}^T + r(\overline{k+\beta})^T]v + 2\big((k+\beta)\overline{r}^T + r(\overline{k+\beta})^T\big)r\overline{r}^Tv \\ = 2\big((\overline{r}|k+\beta)r\overline{r}^Tv + \big(r(\overline{k+\beta})^T + (k+\beta)\overline{r}^T \big)r\overline{r}^Tv \big), \end{align}\] which thereby reduces \((*)\) to \[\begin{align} (\overline{r}|k+\beta)(k+\beta)(\overline{k+\beta})^Tv + (\overline{r}|k+\beta)\big((k+\beta)\overline{r}^T + r(\overline{k+\beta})^T \big)v + (k +\beta)(\overline{k+\beta})^Tr\overline{r}^Tv \\ = (\overline{r}|k+\beta)(k+\beta)(\overline{k+\beta})^Tv + (\overline{r}|k+\beta)\big((k+\beta)\overline{r}^T + r(\overline{k+\beta})^T\big)r\overline{r}^Tv \\ + [(k+\beta)(\overline{k+\beta})^T, r\overline{r}^T]v + r\overline{r}^T(k+\beta)(\overline{k+\beta})^Tv \\ = (\overline{r}|k+\beta)(k+\beta)(\overline{k+\beta})^Tv + r\overline{r}^T(k+\beta)(\overline{k+\beta})^Tv \in W_{min}^{(k+r),\beta}(\Lambda^p\mathbb{C}^N) \end{align}\] and gives us the desired result. ◻

Proof. Using Lemma 12 and the fact that \(E\) is an \(\mathfrak{sp}_N\)-module homomorphism, we have \[\begin{align} h_r \widetilde{E}(v \otimes t^k ) &=& (\overline{r}|k+\beta)\widetilde{E}(v \otimes t^{k}) + r\overline{r}^T \widetilde{E}(v \otimes t^{k}) \\ &=& (\overline{r}|k+\beta)E(v) \otimes t^{k+r} + r\overline{r}^T \big(E(v)\big) \otimes t^{k+r} \\ &=& (\overline{r}|k+\beta)E(v) \otimes t^{k+r} + E\big((r\overline{r}^T)v\big) \otimes t^{k+r} \\ &=&E\big((\overline{r}|k+\beta)v + (r\overline{r}^T)(v) \big) \otimes t^{k+r} \\ &=& \widetilde{E}\big(h_r(v \otimes t^k) \big) \;\forall \;v \in V \;\text{and} \;r, k \in \mathbb{Z}^N. \end{align}\] This shows that \(\widetilde{E}\) is an \(\mathcal{H}_N\)-module map. ◻

Proof. From Proposition 17, it follows that \(\widetilde{\theta}_p^{\beta}\) is an \(\mathcal{H}_N\)-module map, as \(\theta_p\) is an \(\mathfrak{sp}_N\)-module homomorphism. We now deal with the map \(T_p^{\beta}\). The well-definedness of \(T_p^{\beta}\) is immediate from the universal property of exterior products, as \(\sum_{i=1}^{p} (-1)^{i}(\overline{k+\beta}|v_i)v_1 \wedge \ldots \widehat{v_i} \ldots \wedge v_p = 0\), whenever \(v_i=v_j\) for some \(1 \leqslant i \neq j \leqslant p\). Consider
\(h_r T_p^{\beta}\big((v_1 \wedge \ldots \wedge v_p) \otimes t^k \big)\) \[\begin{align} &=& (\overline{r}|k+\beta)T_p^{\beta}\big((v_1 \wedge \ldots \wedge v_p) \otimes t^{k+r} \big) + r\overline{r}^TT_p^{\beta}\big((v_1 \wedge \ldots \wedge v_p) \otimes t^{k+r} \big) \\ &=& (\overline{r}|k+\beta) \big(\sum_{i=1}^{p} (-1)^{i}(\overline{k+\beta}|v_i) (v_1 \wedge \ldots \widehat{v_i} \ldots \wedge v_p) \otimes t^{k+r} \big)\\ &&+ r\overline{r}^T \big(\sum_{i=1}^{p} (-1)^{i}(\overline{k+\beta}|v_i) (v_1 \wedge \ldots \widehat{v_i} \ldots \wedge v_p ) \otimes t^{k+r} \big)\\ &=& (\overline{r}|k+\beta) \big(\sum_{i=1}^{p} (-1)^{i}(\overline{k+\beta+r}|v_i) (v_1 \wedge \ldots \widehat{v_i} \ldots \wedge v_p) \otimes t^{k+r} \big) \\ &&+ \sum_{i=1}^{p}(-1)^{i}(\overline{k+\beta}|v_i) \big(\sum_{j \neq i}(\overline{r}|v_j)(v_1 \wedge \ldots \wedge v_{j-1} \wedge r \wedge v_{j+1} \wedge \ldots \widehat{v_i} \ldots \wedge v_p) \otimes t^{k+r}\big) \\ && - (\overline{r}|k+\beta)\big(\sum_{i=1}^{p}(-1)^{i}(\overline{r}|v_i)(v_1 \wedge \ldots \widehat{v_i} \ldots \wedge v_p) \otimes t^{k+r} \big) \end{align}\] \[\begin{align} &=& (\overline{r}|k+\beta) \big(\sum_{i=1}^{p} (-1)^{i}(\overline{k+\beta +r}|v_i) (v_1 \wedge \ldots \widehat{v_i} \ldots \wedge v_p) \otimes t^{k+r} \big) \\ && + \sum_{i=1}^{p}(-1)^{i}\big(\sum_{j \neq i}(\overline{k+\beta +r}|v_i)(\overline{r}|v_j)(v_1 \wedge \ldots \wedge v_{j-1} \wedge r \wedge v_{j+1} \wedge \ldots \widehat{v_i} \ldots \wedge v_p) \otimes t^{k+r}\big) \\ &&- \sum_{i=1}^{p}(-1)^{i}\big(\sum_{j \neq i}(\overline{r}|v_i)(\overline{r}|v_j)(v_1 \wedge \ldots \wedge v_{j-1} \wedge r \wedge v_{j+1} \wedge \ldots \widehat{v_i} \ldots \wedge v_p) \otimes t^{k+r}\big) \\ &&- (\overline{r}|k+\beta) \big(\sum_{i=1}^{p} (-1)^{i}(\overline{r}|v_i) (v_1 \wedge \ldots \widehat{v_i} \ldots \wedge v_p) \otimes t^{k+r} \big) \\ &=& (\overline{r}|k+\beta) \big(\sum_{i=1}^{p} (-1)^{i}(\overline{k+\beta +r}|v_i) (v_1 \wedge \ldots \widehat{v_i} \ldots \wedge v_p) \otimes t^{k+r} \big) \\ && + \sum_{i=1}^{p}(-1)^{i}\big(\sum_{j \neq i}(\overline{k+\beta +r}|v_i)(\overline{r}|v_j)(v_1 \wedge \ldots \wedge v_{j-1} \wedge r \wedge v_{j+1} \wedge \ldots \widehat{v_i} \ldots \wedge v_p) \otimes t^{k+r}\big) \\ &&- (\overline{r}|k+\beta) \big(\sum_{i=1}^{p} (-1)^{i}(\overline{r}|v_i) (v_1 \wedge \ldots \widehat{v_i} \ldots \wedge v_p) \otimes t^{k+r} \big). \end{align}\] Using \((\overline{k+\beta +r}|r) = -(\overline{r}|k+\beta)\) and interchanging \(i\) and \(j\) in the first and last summands gives \[\begin{align} (\overline{r}|k+\beta) \big(\sum_{j=1}^{p} (-1)^{j}(\overline{k+\beta +r}|v_j) (v_1 \wedge \ldots \widehat{v_j} \ldots \wedge v_p) \otimes t^{k+r} \big) + \\ \sum_{j=1}^{p}(\overline{r}|v_j)\big(\sum_{i \neq j}(-1)^{i}(\overline{k+\beta +r}|v_i)(v_1 \wedge \ldots \wedge v_{j-1} \wedge r \wedge v_{j+1} \wedge \ldots \widehat{v_i} \ldots \wedge v_p) \otimes t^{k+r}\big) + \\ \big(\sum_{j=1}^{p}(-1)^{j}(\overline{r}|v_j)(\overline{k+\beta +r}|r) (v_1 \wedge \ldots \widehat{v_j} \ldots \wedge v_p) \otimes t^{k+r} \big) \\ = (\overline{r}|k+\beta) \big(\sum_{j=1}^{p} (-1)^{j}(\overline{k+\beta +r}|v_j) (v_1 \wedge \ldots \widehat{v_j} \ldots \wedge v_p) \otimes t^{k+r} \big) + \\ \sum_{j=1}^{p}(-1)^{j}(\overline{r}|v_j)\big(\sum_{i=1}^{j-1} (-1)^{i}(\overline{k+\beta+r}|v_i)(v_1 \wedge \ldots \widehat{v_i} \ldots \wedge v_{j-1} \wedge r \wedge v_{j+1} \wedge \ldots \wedge v_p) \otimes t^{k+r}\big) \\ + \sum_{j=1}^{p}(\overline{r}|v_j)(\overline{k+\beta+r}|r)(v_1 \wedge \ldots \wedge v_{j-1} \wedge v_{j+1} \wedge \ldots \wedge v_p) \otimes t^{k+r} + \\ \sum_{j=1}^{p}(-1)^{j}(\overline{r}|v_j)\big(\sum_{i=j+1}^{p} (-1)^{i}(\overline{k+\beta+r}|v_i)(v_1 \wedge \ldots \wedge v_{j-1} \wedge r \wedge v_{j+1} \wedge \ldots \widehat{v_i} \ldots \wedge v_p) \otimes t^{k+r}\big) \\ = T_p^{\beta}\big(h_r(v_1 \wedge \ldots \wedge v_p \otimes t^k) \big) \;\forall \;r, k \in \mathbb{Z}^N, \end{align}\] which proves that \(T_p^{\beta}\) is an \(\mathcal{H}_N\)-module homomorphism, as required. ◻

Proof. Let \(v \in \widehat{W}_{max}^{k,\beta}(\Lambda^p\mathbb{C}^N)\), which implies that \((k+\beta)(\overline{k+\beta})^Tv = 0\). We need to show that \((k + \beta + r)(\overline{k + \beta + r})^T\big(h_r(v \otimes t^k)\big) \in \widehat{W}_{max}^{(k+r),\beta}(\Lambda^p\mathbb{C}^N)\) for any \(r \in \mathbb{Z}^N\). To prove this, we consider \[\begin{align} (k + \beta + r)(\overline{k + \beta + r})^T\big((\overline{r}|k+\beta)v + r\overline{r}^Tv\big) \\ = (\overline{r}|k+\beta)\big(r\overline{r}^Tv + \{(k+\beta)\overline{r}^T + r(\overline{k+\beta})^T \}v \big) + \big((k+\beta)(\overline{k+\beta})^T+ (k+\beta)\overline{r}^T + r(\overline{k+\beta})^T\big)r\overline{r}^Tv \\ = (\overline{r}|k+\beta)\big(r\overline{r}^Tv + \{(k+\beta)\overline{r}^T + r(\overline{k+\beta})^T\}v \big) + r\overline{r}^T(k+\beta)(\overline{k+\beta})^Tv \\ + (\overline{k+\beta}|r)\big((k+\beta)\overline{r}^T + r(\overline{k+\beta})^T\big)v + \big((k+\beta)\overline{r}^T + r(\overline{k+\beta})^T\big)r\overline{r}^Tv \\ = (\overline{r}|k+\beta)r\overline{r}^Tv + \big((k+\beta)\overline{r}^T + r(\overline{k+\beta})^T\big)r\overline{r}^Tv. \end{align}\] Now by Claim of Lemma 12, we have \[\begin{align} 0 = r\overline{r}^T\big((k+\beta)\overline{r}^T + r(\overline{k+\beta})^T\big)v + \big((k+\beta)\overline{r}^T + r(\overline{k +\beta})^T\big)r\overline{r}^Tv \\ = [r\overline{r}^T,(k+\beta)\overline{r}^T + r(\overline{k+\beta})^T]v + 2\big((k+\beta)\overline{r}^T + r(\overline{k+\beta})^T\big)r\overline{r}^Tv \\ = 2\big((\overline{r}|k+\beta)r\overline{r}^Tv + \{(k+\beta)\overline{r}^T + r(\overline{k+\beta})^T \}r\overline{r}^Tv\big), \end{align}\] whence we finally get \((k + \beta + r)(\overline{k + \beta + r})^T\big((\overline{r}|k+\beta)w + r\overline{r}^Tv\big) = 0\) and the result follows. ◻

Proof. (1) and (3) directly follow from the definitions.
(2) Fix \(k \in \mathbb{Z}^N\) such that \(k+\beta \neq 0\). Now, for any \(v_1 \wedge \ldots \wedge v_p \in \Lambda^p\mathbb{C}^N\), we have \[\begin{align} (k+\beta)(\overline{k +\beta})^T\big(v_1 \wedge \ldots \wedge v_p \otimes t^k \big) \\ = \sum_{i=1}^{p}(-1)^{i-1}(\overline{k+\beta}|v_i)\big((k + \beta) \wedge \ldots \wedge v_{i-1} \wedge \widehat{v_i} \wedge v_{i+1} \wedge \ldots \wedge v_p \otimes t^k \big) \\ = T_{p+1}^{\beta}\big(k+ \beta \wedge v_1 \wedge v_2 \ldots \wedge v_p \otimes t^k \big), \end{align}\] which implies that \(W_{min}^{\beta}(\Lambda^p\mathbb{C}^N) \subseteq W_{int}^{\beta}(\Lambda^p\mathbb{C}^N)\).
Claim. \((k+\beta)(\overline{k +\beta})^T(\sum_{i=1}^{p} (-1)^{i}(\overline{k+\beta}|v_i) \big(v_1 \wedge \ldots \widehat{v_i} \ldots \wedge v_p) \big) = 0\).
We shall prove this claim by induction. For \(p= 1\) and \(p=2\), the claim follows trivially. Let us now assume that the claim holds good for \(p-1\) and consider \[\begin{align} (k+\beta)(\overline{k +\beta})^T(\sum_{i=1}^{p} (-1)^{i}(\overline{k+\beta}|v_i) \big(v_1 \wedge \ldots \widehat{v_i} \ldots \wedge v_p) \big) \\ = (k+\beta)(\overline{k +\beta})^T(\sum_{i=1}^{p-1} (-1)^{i}(\overline{k+\beta}|v_i) \big(v_1 \wedge \ldots \widehat{v_i} \ldots \wedge v_{p-1})\wedge v_p \big) \\ + \sum_{i=1}^{p-1} (-1)^{i}(\overline{k+\beta}|v_i) \big(v_1 \wedge \ldots \widehat{v_i} \ldots \wedge v_{p-1}\wedge (k+\beta)(\overline{k +\beta})^Tv_p \big)\\ + (k+\beta)(\overline{k +\beta})^T(-1)^p(\overline{k+\beta}|v_p)(v_1 \wedge \ldots \wedge v_{p-1}) \end{align}\] \[\begin{align} = \sum_{i=1}^{p-1} (-1)^{i}(\overline{k+\beta}|v_i)(\overline{k+\beta}|v_p)\big(v_1 \wedge \ldots \widehat{v_i} \ldots \wedge v_{p-1}\wedge k+\beta \big) \\ + (-1)^p (\overline{k+\beta}|v_p) \sum_{i=1}^{p-1}(\overline{k+\beta}|v_i)\big(v_1 \wedge \ldots \wedge v_{i-1} \wedge k+\beta \wedge v_{i+1} \wedge \ldots v_{p-1}\big) = 0, \end{align}\] by applying our induction hypothesis. Hence the claim follows, which consequently shows that \(W_{int}^{\beta}(\Lambda^p\mathbb{C}^N) \subseteq W_{max}^{\beta}(\Lambda^p\mathbb{C}^N)\) and so we get the desired inclusion. ◻

Proof. We clearly have \(W_{min}^{\beta}(\Lambda^p\mathbb{C}^N) \subseteq W^{\beta}(\Lambda^p\mathbb{C}^N)\), which gives \(W_{min}^{\beta}\big(V(\omega_p) \big) \subseteq W^{\beta} \big(V(\omega_p) \big)\). Let \[\begin{align} v = \sum_{i_1 < \ldots < i_{p-1}} a_{i_1, \ldots, i_{p-1}} \big(k + \beta \wedge w_{i_1}^{(k)} \wedge \ldots \wedge w_{i_{p-1}}^{(k)}\big) + a_{i_{1}^{\prime} \ldots , i_{p-1}^{\prime}}\big(k + \beta \wedge w_{i_1^{\prime}}^{(k)} \wedge \ldots \wedge w_{i_{p-1}^{\prime}}^{(k)}\big) \end{align}\] be a non-zero element of \(W^{k,\beta}\big(V(\omega_p)\big)\), where \(k \in \mathbb{Z}^N\) such that \(k+\beta \neq 0\) and the set containining all the summands of \(v\) are taken to be linearly independent.
Claim. None of the above summands of \(v\) contains \(w_1^{(k)}\).
If not, then there exists a summand of \(v\) of the form \(a_{1, j_1, \ldots, j_{p-2}}\big(k + \beta \wedge w_1^{(k)} \wedge w_{j_1}^{(k)} \wedge \ldots \wedge w_{j_{p-2}}^{(k)} \big)\) for some \(a_{1, j_1, \ldots, j_{p-2}} \neq 0\) and \(j_m \neq 1 \;\forall \;1 \leqslant m \leqslant p-2\). Without loss of generality, we can take \(a_{i_{1}^{\prime} \ldots , i_{p-1}^{\prime}} \neq 0\) and \(k + \beta \wedge w_{i_1^{\prime}}^{(k)} \wedge \ldots \wedge w_{i_{p-1}^{\prime}}^{(k)} = k + \beta \wedge w_1^{(k)} \wedge w_{j_1}^{(k)} \wedge \ldots \wedge w_{j_{p-2}}^{(k)}\). By hypothesis, \[\begin{align} 0 = \theta_p(v) = \theta_p\big(a_{i_{1}^{\prime} \ldots , i_{p-1}^{\prime}}(k + \beta \wedge w_1^{(k)} \wedge w_{j_1}^{(k)} \wedge \ldots \wedge w_{j_{p-2}}^{(k)}) + \ldots \ldots \ldots \big) \\ = a_{i_{1}^{\prime} \ldots , i_{p-1}^{\prime}}(w_{j_1}^{(k)} \wedge \ldots \wedge w_{j_{p-2}}^{(k)}) + \ldots \ldots \ldots, \end{align}\] which implies that \(w_{j_1}^{(k)} \wedge \ldots \wedge w_{j_{p-2}}^{(k)}\) occurs in \(\theta_p \big(\sum_{i_1 < \ldots < i_{p-1}} a_{i_1, \ldots, i_{p-1}} \big(k + \beta \wedge w_{i_1}^{(k)} \wedge \ldots \wedge w_{i_{p-1}}^{(k)})\big)\). But this is not possible, as \(k+\beta\) is present in every summand of \(v\). Hence the claim follows.
Finally, consider \[\begin{align} w = \sum_{i_1 < \ldots < i_{p-1}} a_{i_1, \ldots, i_{p-1}} \big(w_1^{(k)} \wedge w_{i_1}^{(k)} \wedge \ldots \wedge w_{i_{p-1}}^{(k)}\big) + a_{i_{1}^{\prime} \ldots , i_{p-1}^{\prime}}\big(w_1^{(k)} \wedge w_{i_1^{\prime}}^{(k)} \wedge \ldots \wedge w_{i_{p-1}^{\prime}}^{(k)}\big), \end{align}\] which again belongs to \(V(\omega_p)\), since \(v \in V(\omega_p)\). But then \((k+\beta)(\overline{k+\beta})^Tw = v\), whence we finally obtain \(W^{\beta} \big(V(\omega_p) \big) \subseteq W_{min}^{\beta}\big(V(\omega_p) \big)\) and thus the lemma follows. ◻

Proof. Define \[\begin{align} f_p^{\beta} : F^{\beta}(\Lambda^{p}\mathbb{C}^N) & \longrightarrow & W_{min}^{\beta}(\Lambda^{p}\mathbb{C}^N) \\ v_1 \wedge \ldots \wedge v_p \otimes t^k &\longmapsto & (k+\beta)(\overline{k+\beta})^T\big(v_1 \wedge v_2 \wedge \ldots \wedge v_{p}) \otimes t^k, \;k \in \mathbb{Z}^N. \end{align}\] It is easy to see that \(f_p^{\beta} = T_{p+1}^{\beta} \circ \pi_p^{\beta}\), whence by Proposition 19, it follows that \(f_p^{\beta}\) is an \(\mathcal{H}_N\)-module homomorphism. Also, \(\text{Ker}f_p^{\beta} = \widehat{W}_{max}^{\beta}(\Lambda^p\mathbb{C}^N)\) and \(\text{Im}f_p^{\beta} = W_{min}^{\beta}(\Lambda^p\mathbb{C}^N)\), which proves (1).
Let us now denote the restriction of \(f_p^{\beta}\) to \(L_{\mathcal{H}}^{\beta}\big(V(\omega_p)\big)\) by \(\widetilde{f}_p^{\beta}\). Then \(\text{Ker}\widetilde{f}_p^{\beta} = \widehat{W}_{max}^{\beta}\big(V(\omega_p)\big)\) and \(\text{Im}\widetilde{f}_p^{\beta} = W_{min}^{\beta}\big(V(\omega_p)\big)\) (since \((k+\beta)(\overline{k+\beta})^T\) commutes with \(\theta_p\)), from which (2) follows. ◻

Proof. Pick \(0 \neq v \in M_k \cap \mathrm{span} \{w_1^{(k)} \wedge w_{i_1}^{(k)} \wedge \ldots \wedge w_{i_{p-1}}^{(k)} \;| \;i_j \neq 1 \}\) and any \(r \in \mathbb{Z}^N \setminus \{0 \}\). Consider \[\begin{align} h_{-r}\big(h_{r+s}(v \otimes t^k)\big) = -(\overline{r}|k+r+s+\beta)(\overline{r} + \overline{s}|k + \beta)v \otimes t^{k+s} + (\overline{r} + \overline{s}|k + \beta)(r\overline{r}^T)v \otimes t^{k+s} \\ - (\overline{r}|k+r+s+\beta)(r+s)(\overline{r+s})^Tv \otimes t^{k+s} + (r\overline{r}^T) (r+s)(\overline{r+s})^Tv \otimes t^{k+s} \\ = -(\overline{r}|k+s+\beta)(\overline{r}|k + \beta)v \otimes t^{k+s} - (\overline{r}|k+s+\beta)(\overline{s}|k + \beta)v \otimes t^{k+s} + (\overline{r}|k + \beta)(r\overline{r}^T)v \otimes t^{k+s} \\ + (\overline{s}|k + \beta)(r\overline{r}^T)v \otimes t^{k+s} - (\overline{r}|k+s+\beta)(r\overline{r}^T)v \otimes t^{k+s} - (\overline{r}|k+s+\beta)(s\overline{s}^T)v \otimes t^{k+s} \\ - (\overline{r}|k+s+\beta)(r\overline{s}^T + s\overline{r}^T)v \otimes t^{k+s} + (r\overline{r}^T)^2v \otimes t^{k+s} - (r\overline{r}^T)(s\overline{s}^T)v \otimes t^{k+s} \\ + (r\overline{r}^T)(r\overline{s}^T + s\overline{r}^T)v \otimes t^{k+s} \in M_{k+s} \otimes \mathbb{C}t^{k+s} \;\forall \;s \in \mathbb{Z}^N \setminus \{0 \}. \end{align}\] Now replace \(r\) by \(qr\) for \(q \in \mathbb{Z}\) in the above equation. Henceforth, collecting the coefficients of \(q^2\) and using Lemma 11, we obtain \[\begin{align} (\overline{r}|k+s+\beta)(\overline{r}|k + \beta)v - (\overline{s}|k + \beta)(r\overline{r}^T)v - (r\overline{r}^T)(s\overline{s}^T)v \\ + (\overline{r}|k+s+\beta)(r\overline{s}^T + s\overline{r}^T)v \in M_{k+s} \;\forall \;s \in \mathbb{Z}^N \setminus \{0 \}. \end{align}\] Finally, putting \(r = w_1^{(k)}\) in the previous equation gives \[\begin{align} \big(\overline{w_1^{(k)}}|k+s+\beta \big)\big(\overline{w_1^{(k)}}|k + \beta \big)v - \big(\overline{s}|k + \beta \big) \big(w_1^{(k)}\overline{w_1^{(k)}}^T \big)v \\ - \big(w_1^{(k)}\overline{w_1^{(k)}}^T \big)\big(s\overline{s}^T \big)v + \big(\overline{w_1^{(k)}}|k+s+\beta \big) \big(w_1^{(k)}\overline{s}^T + s\overline{w_1^{(k)}}^T \big)v \\ = \big(\overline{w_1^{(k)}}|k+\beta \big)^2v - \big(\overline{s}|w_1^{(k)}\big)\big(\overline{w_1^{(k)}}|k+\beta \big)v + \big(\overline{s}|w_1^{(k)}\big)^2v + \big(\overline{w_1^{(k)}}|k+\beta \big)\big(\overline{s}|w_1^{(k)}\big)v + \big(\overline{w_1^{(k)}}|s \big)\big(\overline{s}|w_1^{(k)}\big)v \\ = \big(\overline{w_1^{(k)}}|k+\beta \big)^2v \in M_{k+s} \;\forall \;s \in \mathbb{Z}^N \setminus \{0 \} \ldots \ldots (**), \; \end{align}\] where we have used the following relations, which are straightforward to deduce. \[\begin{align} \big(w_1^{(k)}\overline{w_1^{(k)}}^T \big)v = 0, \,\,\;\big(w_1^{(k)}\overline{w_1^{(k)}}^T \big)\big(s\overline{s}^T \big)v = -\big(\overline{s}|w_1^{(k)}\big)^2v, \,\,\;\big(w_1^{(k)}\overline{s}^T + s\overline{w_1^{(k)}}^T \big)v = \big(\overline{s}|w_1^{(k)}\big)v. \end{align}\] Then (**) implies that \(v \in \cap_{s \in \mathbb{Z}^N}M_{k+s}\) and thus the proposition directly follows by providing a similar argument as in the proof of Theorem 14. ◻

Proof. First note that \(L_{\mathcal{H}}^{\beta}\big(V(\omega_0)\big) \cong A_N\) for any \(\beta \in \mathbb{C}^N\). Now define \[\begin{align} \pi_0^{\beta} : L_{\mathcal{H}}^{\beta}\big(V(\omega_0)\big) & \longrightarrow & L_{\mathcal{H}}^{\beta}\big(V(\omega_1)\big) \\ t^k &\longmapsto & (k+\beta) \otimes t^k, \;k \in \mathbb{Z}^N. \end{align}\] Then \(\pi_0^{\beta}\) is an \(\mathcal{H}_N\)-module homomorphism and consequently (1) follows, thanks to Theorem 25. It is clear that \(V(\omega_0) \otimes \mathbb{C}t^{-\beta}\) is irreducible over \(\mathcal{H}_N\), for \(\beta \in \mathbb{Z}^N\). Moreover, by restricting \(\pi_0^{\beta}\) to \(\bigoplus_{k \in \mathbb{Z}^N, \;k \neq -\beta}\big( V(\omega_0) \otimes \mathbb{C}t^k\big)\), we also obtain (2). Finally, (3) is immediate by just observing that \(\mathrm{Ker}\pi_0^{\beta} = V(\omega_0) \otimes \mathbb{C}t^{-\beta}\) and \(\mathrm{Im}\pi_0^{\beta} = \bigoplus_{k \in \mathbb{Z}^N, \;k \neq -\beta}\big( V(\omega_0) \otimes \mathbb{C}t^k\big)\), for \(\beta \in \mathbb{Z}^N\). ◻

Proof. (1) Let us denote the restriction of \(T_p^{\beta}\) to \(W^{\beta}(\Lambda^p\mathbb{C}^N)\) by \(\widetilde{T}_p^{\beta}\). Then \[\begin{align} \widetilde{T}_p^{\beta}\big(k+\beta \wedge v_1 \wedge \ldots \wedge v_{p-1} \otimes t^k \big) = (k+\beta)(\overline{k+\beta})^T\big(v_1 \wedge v_2 \wedge \ldots \wedge v_{p-1}\big) \otimes t^k \;\forall \;k \in \mathbb{Z}^N. \end{align}\] Claim. \(\text{Ker}\widetilde{T}_p^{\beta} = W_{min}^{\beta}(\Lambda^p\mathbb{C}^N)\).
For any \(k \in \mathbb{Z}^N\), it is clear that \(\text{span} \{(k+\beta) \wedge w_{i_1}^{(k)} \wedge \ldots \wedge w_{i_{p-1}}^{(k)} \otimes t^k \;| \;i_m \neq 1 \} \subseteq \text{Ker}\widetilde{T}_p^{\beta}|_{W^{k,\beta}(\Lambda^p\mathbb{C}^N) \otimes \mathbb{C}t^k}\). Now for \((k+ \beta) \wedge w_1^{(k)} \wedge w_{j_1}^{(k)} \wedge \ldots \wedge w_{j_{p-1}}^{(k)} \in F^{\beta}(\Lambda^{p}\mathbb{C}^N)\), we have \[\begin{align} (k+\beta)(\overline{k+\beta})^T\big(w_1^{(k)} \wedge w_{j_1}^{(k)} \wedge \ldots \wedge w_{j_{p-1}}^{(k)} \big) \neq 0. \end{align}\] But again, \[\begin{align} W_{min}^{k,\beta}(\Lambda^p\mathbb{C}^N) = \text{span} \{(k+\beta) \wedge w_{i_1}^{(k)} \wedge \ldots \wedge w_{i_{p-1}}^{(k)} \;| \;i_m \neq 1 \}, \end{align}\] which proves our claim. It is also evident that \(\text{Im}\widetilde{T}_p^{\beta} = W_{min}^{\beta}(\Lambda^{p-1}\mathbb{C}^N)\), from which (1) and subsequently (2) follows.
(3) Let us consider \(\widehat{\pi}_p^{\beta} := \pi_p^{\beta}|_{\widehat{W}_{max}^{\beta}(\Lambda^p\mathbb{C}^N)}\). Note that \[\begin{align} \widehat{\pi}_p^{\beta}(w_1^{(k)} \wedge w_{i_1}^{(k)} \wedge \ldots \wedge w_{i_{p-1}}^{(k)}) \neq 0, \;\text{where} \;i_r \neq 1 \;\text{and} \\ \widehat{W}_{max}^{k,\beta}(\Lambda^{p}\mathbb{C}^N) = \text{span}\{(k+\beta) \wedge w_{j_1}^{(k)}\wedge \ldots \wedge w_{j_{p-1}}^{(k)}, w_{i_1}^{(k)} \wedge \ldots \wedge w_{i_{p}}^{(k)} \;| \;i_r \neq 1 \}. \end{align}\] for any \(k \in \mathbb{Z}^N\) satisfying \(k+\beta \neq 0\). Then it is easy to see that \(\mathrm{Im}\widehat{\pi}_p^{\beta} = W_{min}^{\beta}(\Lambda^{p+1}\mathbb{C}^N)\) and \(\mathrm{Ker}\widehat{\pi}_p^{\beta} = W^{\beta}(\Lambda^{p}\mathbb{C}^N)\), which proves (3) as well as (4). ◻

Proof. (1) We can check that \[\begin{align} \text{span}\{(k+\beta) \wedge w_{i_1}^{(k)} \wedge \ldots \wedge w_{i_{p-2}}^{(k)}, \;w_{j_1}^{(k)} \wedge \ldots \wedge w_{j_{p-1}}^{(k)} | \;i_r, j_s \neq 1 \} \subseteq W_{int}^{k,\beta}(\Lambda^{p-1}\mathbb{C}^N). \end{align}\] Also, note that neither \((k+\beta) \wedge w_1^{(k)} \wedge w_{i_1}^{(k)} \wedge \ldots \wedge w_{i_{p-3}}^{(k)} \;(i_r \neq 1)\) nor \(w_1^{(k)} \wedge w_{j_1}^{(k)} \wedge \ldots \wedge w_{j_{p-2}}^{(k)} \;(j_s \neq 1)\) belongs to \(W_{int}^{k,\beta}(\Lambda^{p-1}\mathbb{C}^N)\) and so we get \[\begin{align} W_{int}^{k,\beta}(\Lambda^{p-1}\mathbb{C}^N) = \text{span}\{(k+\beta) \wedge w_{i_1}^{(k)} \wedge \ldots \wedge w_{i_{p-2}}^{(k)}, \;w_{j_1}^{(k)} \wedge \ldots \wedge w_{j_{p-1}}^{(k)} | \;i_r, j_s \neq 1 \}. \end{align}\] Then it can be easily deduced that \(\text{Im}\big(\pi_p^{\beta}|_{W_{int}^{\beta}(\Lambda^{p-1}\mathbb{C}^N)}\big) = W_{min}^{\beta}(\Lambda^{p}\mathbb{C}^N)\) and \(\text{Ker}\big(\pi_p^{\beta}|_{W_{int}^{\beta}(\Lambda^{p-1}\mathbb{C}^N)}\big) = W_{min}^{\beta}(\Lambda^{p-1}\mathbb{C}^N)\), which establishes (1) and consequently (2).
To conclude the proof of this lemma, it suffices to prove (3). It is clear that \[\begin{align} W^{\beta}(\Lambda^{p-1}\mathbb{C}^N) + W^{\beta}_{int} (\Lambda^{p-1}\mathbb{C}^N) = \widehat{W}^{\beta}_{max}(\Lambda^{p-1}\mathbb{C}^N), \\ W^{\beta}(\Lambda^{p-1}\mathbb{C}^N) \cap W^{\beta}_{int} (\Lambda^{p-1}\mathbb{C}^N) = W^{\beta}_{min}(\Lambda^{p-1}\mathbb{C}^N). \end{align}\] Then by Proposition 31, \[\begin{align} \widehat{W}^{\beta}_{max}(\Lambda^{p-1}\mathbb{C}^N)/\widehat{W}^{\beta}_{int}(\Lambda^{p-1}\mathbb{C}^N) \cong W^{\beta}(\Lambda^{p-1}\mathbb{C}^N)/W^{\beta}_{min}(\Lambda^{p-1}\mathbb{C}^N) \cong W_{min}^{\beta}(\Lambda^{p-2}\mathbb{C}^N) \end{align}\] and therefore we are done. ◻

Proof. Fix any \(k \in \mathbb{Z}^N\) satisfying \(k+\beta \neq 0\). Then \[\begin{align} (1) \;W_{int}^{k,\beta}\big(V(\omega_p)\big) = \mathrm{span}\{(k+\beta) \wedge w_{i_1}^{(k)} \wedge \ldots \wedge w_{i_{p-1}}^{(k)}, \;w_{j_1}^{(k)} \wedge \ldots \wedge w_{j_{p}}^{(k)} | \;i_r, j_s \neq 1 \} \cap L^{k,\beta}\big(V(\omega_p)\big), \end{align}\] which implies that \[\begin{align} W_{min}^{\beta}\big(V(\omega_{p})\big) \subseteq \mathrm{Ker}\big(\widetilde{\theta}_{p+1}^{\beta}\circ \pi_{p}^{\beta}|_{W_{max}^{\beta}\big(V(\omega_{p})\big)}\big), \\ \mathrm{Im}\big(\pi_p^{\beta}|_{W_{max}^{k,\beta}\big(V(\omega_{p})\big)}\big) \subseteq \mathrm{span}\{ (k+\beta) \wedge w_{j_1}^{(k)} \wedge \ldots \wedge w_{j_{p}}^{(k)} | \; j_s \neq 1 \}. \end{align}\] Now since \(\theta_{p+1}^{\beta}\big((k+\beta) \wedge w_{j_1}^{(k)} \wedge \ldots \wedge w_{j_{p}}^{(k)}\big) = (k+\beta) \wedge \big(\theta_{p}^{\beta}(w_{j_1}^{(k)} \wedge \ldots \wedge w_{j_{p}}^{(k)})\big) \;(j_s \neq 1)\), we obtain \[\begin{align} W_{int}^{\beta}\big(V(\omega_{p})\big) \subseteq \mathrm{Ker}\big(\widetilde{\theta}_{p+1}^{\beta}\circ \pi_{p}^{\beta}|_{W_{max}^{\beta}\big(V(\omega_{p})\big)}\big). \end{align}\] Finally, we can easily check that if \(v\) is a non-zero element of \(L_{\mathcal{H}}^{k,\beta}\big(V(\omega_p)\big)\) of the form \(y_i^{(k)}\) or \(z_i^{(k)}\) (see Lemma 26) or their linear combination, then \(\big(\widetilde{\theta}_{p+1}^{\beta}\circ \pi_{p}^{\beta}\big)(v \otimes t^k) \neq 0\) and hence we are done. (2) Put \(\widehat{T}_p^{\beta}:= T_p^{\beta}|_{\widehat{W}_{max}^{\beta}\big(V(\omega_{p})\big)}\). Now by Lemma 26, if \(0 \neq v \in \widehat{W}_{max}^{k,\beta}\big(V(\omega_{p})\big)\) contains \(w_1^{(k)}\) in one of its summands, then \(k+\beta\) must also occur in that same summand and in that case, \(\widehat{T}_p^{\beta}(v \otimes t^k) \neq 0\). Also, it is evident from our above description of \(\widehat{W}_{int}^{k,\beta}\big(V(\omega_p)\big)\) that \(\widehat{T}_p^{\beta}|_{\widehat{W}_{int}^{k,\beta}\big(V(\omega_p)\big) \otimes \mathbb{C}t^k} = 0\), which yields that \(\mathrm{Ker}\widehat{T}_p^{\beta} = \widehat{W}_{int}^{\beta}\big(V(\omega_p)\big)\). Next, we establish that \(\mathrm{Im}\widehat{T}_p^{\beta} = W_{min}^{\beta}\big(V(\omega_{p-1})\big)\). Pick \[\begin{align} w= (k+\beta \wedge w_1^{(k)}) \wedge (\sum_{i_1 < \ldots < i_{p-2}} w_{i_1}^{(k)} \wedge \ldots \wedge w_{i_{p-2}}^{(k)}) + \sum_{j_1 < \ldots < j_{p}} w_{j_1}^{(k)}\wedge \ldots \wedge w_{j_p}^{(k)} \\ + \sum_{l_1 < \ldots < l_{p-1}} (k + \beta) \wedge w_{l_1}^{(k)}\wedge \ldots \wedge w_{l_{p-1}}^{(k)} \in W_{max}^{k,\beta}\big(V(\omega_{p})\big), \end{align}\] where \(i_q, j_r, l_s \neq 1\). Then \(\widehat{T}_p^{\beta}(w \otimes t^k) = -(k + \beta) \wedge (\sum_{i_1 < \ldots < i_{p-2}} w_{i_1}^{(k)} \wedge \ldots \wedge w_{i_{p-2}}^{(k)}) \otimes t^k\), with \[\begin{align} \theta_{p-1}\big((k + \beta) \wedge (\sum_{i_1 < \ldots < i_{p-2}} w_{i_1}^{(k)} \wedge \ldots \wedge w_{i_{p-2}}^{(k)})\big) = (k+\beta) \wedge \theta_{p-2} \big(\sum_{i_1 < \ldots < i_{p-2}} w_{i_1}^{(k)} \wedge \ldots \wedge w_{i_{p-2}}^{(k)}\big) = 0, \end{align}\] since we have \(\theta_p(w) = 0\). This implies that \((0) \neq \mathrm{Im}\widehat{T}_p^{\beta} \subseteq L_{\mathcal{H}}^{\beta}\big(V(\omega_{p-1})\big) \cap W_{min}^{\beta}(\Lambda^{p-1}\mathbb{C}^N)\). The desired result now follows from Lemma 23 and Theorem 25, whereas a similar argument gives (4). ◻

Proof. Setting \(\widetilde{\pi}_p^{\beta} := \pi_p^{\beta}|_{W_{int}^{\beta}\big(V(\omega_{p})\big)}\) and using (1) of Lemma 33, we obtain \[\begin{align} \theta_{p+1}(k+\beta \wedge v) = (k+\beta)\wedge \theta_p(v) = 0 \;\forall \;v \in W_{int}^{\beta}\big(V(\omega_{p})\big) \end{align}\] and as a result, we have \((0) \neq \mathrm{Im}\widetilde{\pi}_p^{\beta} \subseteq L_{\mathcal{H}}^{\beta}\big(V(\omega_{p+1})\big)\). But again, \(\mathrm{Im}\widetilde{\pi}_p^{\beta} \subseteq W_{min}^{\beta}(\Lambda^{p+1}\mathbb{C}^N)\), which implies that \(\mathrm{Im}\widetilde{\pi}_p^{\beta} = W_{min}^{\beta}\big(V(\omega_{p+1})\big)\), by Lemma 23 and Theorem 25. Finally an analogous argument also gives \(\mathrm{Ker}\widetilde{\pi}_p^{\beta} = W_{min}^{\beta}\big(V(\omega_{p})\big)\), which proves (1) and consequently (2). ◻

Proof. Fix any \(k \in \mathbb{Z}^N\) with \(k+\beta \neq 0\).
(1) \(T_2^{\beta}(k+\beta \wedge w_1^{(k)}) = k + \beta\) and \(T_2^{\beta}(w_i^{(k)} \wedge w_1^{(k)}) = w_i^{(k)} \;\forall \;2 \leqslant i \leqslant N-1\), which implies that \[\begin{align} \widehat{W}_{max}^{k,\beta}\big(V(\omega_{1})\big) = \{v \in \mathbb{C}^N \;| \;(\overline{k+ \beta}|v) = 0 \} = \text{span}\{k+\beta, w_i^{(k)} \;| \;2 \leqslant i < N \} \\ \subseteq \widehat{W}_{int}^{k,\beta}\big(V(\omega_{1})\big) \subseteq \widehat{W}_{max}^{k,\beta}\big(V(\omega_{1})\big) \;(\text{by Lemma} \;\ref{Inclusion}). \end{align}\] (2) Using Lemma 33 and Remark 8, we obtain \[\begin{align} W_{int}^{k,\beta}\big(V(\omega_{n})\big) \otimes \mathbb{C}t^k = \mathrm{Ker}\big(\widetilde{\theta}_{n+1}^{\beta}|_{F^{\beta}(\Lambda^{n+1}\mathbb{C}^N) \otimes \mathbb{C}t^k} \circ \pi_{n}^{\beta}|_{L^{k,\beta}(V(\omega_{n})) \otimes \mathbb{C}t^k }\big) \\ = \mathrm{Ker}\big(\pi_{n}^{\beta}|_{L^{k,\beta}(V(\omega_{n})) \otimes \mathbb{C}t^k}\big) = W_{min}^{k,\beta}\big(V(\omega_{n})\big) \otimes \mathbb{C}t^k. \end{align}\] ◻

Proof. (1) For any \(v \in M_k\), consider \[\begin{align} h_{-s}h_{-r}\big(h_{r+s}(v \otimes t^k)\big) \\ = (\overline{r+s}|k+\beta)(\overline{r}|k+s+ \beta)\big[(\overline{s}|k+\beta)(v \otimes t^k) - (s\overline{s}^T)v \otimes t^k\big] - (\overline{r+s}|k+\beta)\big[(\overline{s}|k+ \beta)(r\overline{r}^T)v \otimes t^k \\ -(s\overline{s}^T)(r\overline{r}^T)v \otimes t^k \big]+ (\overline{r}|k+ s + \beta)\big[(\overline{s}|k + \beta)(r+s) (\overline{r+s})^Tv \otimes t^k - (s\overline{s}^T)(r+s)(\overline{r+s})^Tv \otimes t^k\big] \\ -(\overline{s}|k+\beta)\big[(r\overline{r}^T)(r+s)(\overline{r+s})^Tv \otimes t^k - (s\overline{s}^T)(r\overline{r}^T)(r+s)(\overline{r+s})^Tv \otimes t^k\big] \\ = - \Omega_{k,r,s}^{\mathcal{H},\beta}.(v \otimes t^k) + \ldots \ldots \ldots \in M_k \otimes \mathbb{C}t^k. \end{align}\] Replacing \(r\) by \(pr\) and \(s\) by \(qs\) for \(p, q \in \mathbb{Z}\) in the above equation, the result now follows by simply comparing the coefficients of \(p^2q^2\) and then applying Lemma 11.
(2) For all \(r, s \in \mathbb{Z}^N\) and \(v \in M_k\), we have \[\begin{align} \phi\bigg((h_{-s}h_{-r}\big(h_{r+s}(v \otimes t^k)\big)\bigg) = \bigg(h_{-s}h_{-r}\big(h_{r+s}(\phi(v \otimes t^k)\big)\bigg). \end{align}\] Now expand both sides using the action of \(\mathcal{H}_N\) and then replace \(r\) by \(pr\), \(s\) by \(qs\) for \(p, q \in \mathbb{Z}\). The result subsequently follows by just comparing the coefficients of \(p^2q^2\) and invoking Lemma 11. ◻

Proof. (1) If \((\overline{k+\beta})_i \neq 0\) for some \(1 \leqslant i \leqslant N\), then it can be shown using elementary arguments that \(\{\mathcal{T}_{k,e_i,e_j}^{\mathcal{H},\beta} \;| \;1 \leqslant j \neq i \leqslant N \}\) is a linearly independent subset of \(W_{max}^{k,\beta}\big(V(\omega_1)\big)\), which proves the assertion, as we already know that \(W_{max}^{k,\beta}\big(V(\omega_1)\big)\) is \((N-1)\)-dimensional.
(2) Without loss of generality, let \((\overline{k+\beta})_1 \neq 0\). Clearly \(W_{max}^{k,\beta}\big(V(\omega_1)\big) = \{ v \in \mathbb{C}^N \;| \;(\overline{k+\beta}|v) = 0 \}\). By (1), there exist \(a_2, \ldots, a_N \in \mathbb{C}\) such that \(w = \sum_{i=2}^{N}a_i\mathcal{T}_{k,e_1,e_i}^{\mathcal{H},\beta}\). Then for any \(v \in M_k\), we have \[\begin{align} (w\overline{w}^T).(v \otimes t^k) = \sum_{i=2}^{N}a_i^2\Omega_{k,e_1,e_i}^{\mathcal{H},\beta}.(v \otimes t^k) + 2\sum_{1 \leqslant i <j \leqslant N} a_ia_j \big(\mathcal{T}_{k,e_1,e_i}^{\mathcal{H},\beta}\overline{\mathcal{T}_{k,e_1,e_j}^{\mathcal{H},\beta}}^T + \mathcal{T}_{k,e_1,e_j}^{\mathcal{H},\beta}\overline{\mathcal{T}_{k,e_1,e_i}^{\mathcal{H},\beta}}^T \big).(v \otimes t^k) \\ = \sum_{i=2}^{N}a_i^2\Omega_{k,e_1,e_i}^{\mathcal{H},\beta}.(v \otimes t^k) + 2 \sum_{1 \leqslant i <j \leqslant N} a_ia_j \big(\Omega_{k,e_i,e_i+e_j}^{\mathcal{H},\beta} - \Omega_{k,e_1,e_i}^{\mathcal{H},\beta} - \Omega_{k,e_1,e_j}^{\mathcal{H},\beta}\big).(v \otimes t^k) \end{align}\] and thus we are done using Lemma 38.
(3) is a direct consequence of (2), as for every \(v \in M_k\), we have \[\begin{align} (u_1\overline{u_2}^T + u_2\overline{u_1}^T).(v \otimes t^k) = \big((u_1 + u_2)(\overline{u_1+u_2})^T - (u_1\overline{u_1}^T) - (u_2\overline{u_2}^T) \big).(v \otimes t^k) \in M_k \otimes \mathbb{C}t^k. \end{align}\] ◻

Proof. First note that \(\Omega_{k,r,s}^{\mathcal{H},\beta}\) acts trivially on the vector \(k+\beta\). The lemma then follows by applying Proposition 39, since \((\overline{k+\beta}|w_i^{(k)}) = 0 \;\forall \;2 \leqslant i \leqslant N-1\) and \(\mathfrak{sp}_{N-2}^{(k)} = \text{span} \{r\overline{r}^T \;| \;r \in \mathbb{Z}^{N-2} \}\), where we identify \(\mathbb{Z}^{N-2}\) with \(\bigoplus_{i=2}^{N-1}\mathbb{Z}w_i^{(k)}\). ◻

Proof. Let \(M\) be an irreducible \(\mathcal{H}_N\)-submodule of \(L_{\mathcal{H}}^{\beta}\big(V(\omega_p)\big)\). In view of Theorem 25, it suffices to prove that \(W_{min}^{\beta}\big(V(\omega_p)\big) \cap M \neq (0)\). Now since \(V(\omega_p)\) is not the trivial module, we can find \(k \in \mathbb{Z}^N\) with \(k+\beta \neq 0\) such that \(M_k \neq (0)\). Let \(0 \neq v \in M_k\) be arbitrary. By Lemma 26, there exist \(p_i, q_i, r_i, s_i \in \mathbb{C}\) such that \(v = \sum_{i=1}^{a} p_iu_i^{(k)} + \sum_{i=1}^{b} q_ix_i^{(k)} + \sum_{i=1}^{c} r_iy_i^{(k)} + \sum_{i=1}^{d} s_iz_i^{(k)}\), where \(\{u_1^{(k)}, \ldots, u_a^{(k)}, x_1^{(k)}, \ldots, x_b^{(k)}, y_1^{(k)}, \ldots, y_c^{(k)}, z_1^{(k)}, \ldots, z_d^{(k)} \}\) is a linearly independent subset of \(V(\omega_p)\).
Case 1. \(s_i \neq 0\) for some \(1 \leqslant i \leqslant d\).
By Proposition 39, we have \(0 \neq (k+\beta)(\overline{k+\beta})^Tv = \sum_{i=1}^{d} s_i(k+\beta)(\overline{k+\beta})^Tz_i^{(k)}\in M_k \cap W_{min}^{k,\beta}\big(V(\omega_p)\big)\).

Case 2. \(s_i = 0 \;\forall \;1 \leqslant i \leqslant d\) and \(p_i \neq 0\) for some \(1 \leqslant i \leqslant a\).
For \(p=1, \;v = b_0(k + \beta) + \sum_{j}a_jw_{i_j}^{(k)}\), where \(i_j \neq 1, \;b_0 \in \mathbb{C}^{\times}, \;a_j \in \mathbb{C}\) and the \(\mathfrak{h}_{\mathfrak{sp}_{N-2}}^{(k)}\)-weights of \(w_{i_j}^{(k)}\) are all different, which implies that \(\sum_{j}a_jw_{i_j}^{(k)} \in M_k\) and hence \(b_0(k + \beta) \in M_k \cap W_{min}^{k,\beta}\big(V(\omega_p)\big)\). So let us assume that \(2 \leqslant p \leqslant n\). In this case, it is easy to see that exactly \((p-1)\) elements of \(\mathbb{C}^N\) occurring in the exterior product of any summand of \(u_i^{(k)}\) have non-zero \(\mathfrak{h}_{\mathfrak{sp}_{N-2}}^{(k)}\)-weights, whereas each element of \(\mathbb{C}^N\) in the exterior product of every summand of \(x_i^{(k)}\) has non-zero \(\mathfrak{h}_{\mathfrak{sp}_{N-2}}^{(k)}\)-weight. Again, any summand of \(y_i^{(k)}\) either has exactly \((p-2)\) elements of \(\mathbb{C}^N\) with non-zero \(\mathfrak{h}_{\mathfrak{sp}_{N-2}}^{(k)}\)-weights in its exterior product or all the elements of \(\mathbb{C}^N\) occuring in the exterior product of a summand of \(y_i^{(k)}\) have non-zero \(\mathfrak{h}_{\mathfrak{sp}_{N-2}}^{(k)}\)-weights (see the discussion before Lemma 40). It can be now easily deduced using elementary arguments that the \(\mathfrak{h}_{\mathfrak{sp}_{N-2}}^{(k)}\)-weight of \(u_i^{(k)}\) differs from the \(\mathfrak{h}_{\mathfrak{sp}_{N-2}}^{(k)}\)-weights of both \(x_i^{(k)}\) and \(y_i^{(k)}\), from which we finally get \(0 \neq \sum_{i=1}^{a} p_iu_i^{(k)} \in M_k \cap W_{min}^{k,\beta}\big(V(\omega_p)\big)\).

Case 3. \(p_i = 0 \;\forall \;1 \leqslant i \leqslant a\) and \(s_i = 0 \;\forall \;1 \leqslant i \leqslant d\).
For \(p=1\), we have \(v = \sum_{j}a_jw_{i_j}^{(k)}\), where \(i_j \neq 1\) and \(a_j \in \mathbb{C}\) for all \(j\). Without loss of generality, let \(i_1 = \mathrm{min} \{i_j\}_j\) and consider \(s = i_1-1\) or \(s = i_1+1\), depending on whether \((\overline{w_{i_1-1}^{(k)}}|w_{i_1}^{(k)}) \neq 0\) or \((\overline{w_{i_1}^{(k)}}|w_{i_1+1}^{(k)}) \neq 0\). This unravels that \(0 \neq \big((k + \beta)\overline{w_s^{(k)}}^T + w_s^{(k)}(\overline{k +\beta})^T\big)v \in M_k\cap W_{min}^{k,\beta}\big(V(\omega_p)\big)\) by Lemma 23 and Proposition 39, as required. So let us suppose that \(2 \leqslant p \leqslant n\). In this case, \(v = \sum_{i=1}^{b} q_ix_i^{(k)} + \sum_{i=1}^{c} r_iy_i^{(k)}\). Consider the minimum of all the indices of the elements \(w_{i_j}^{(k)}\) of \(\mathbb{C}^N\), occuring in the exterior product of every summand of \(v\) that does not contain \(k+\beta\). For example, if \(v = w_2^{(k)}\wedge w_4^{(k)} +(k+\beta \wedge w_1^{(k)} - w_2^{(k)} \wedge w_3^{(k)})\), then the indices present in the exterior products are \(2,3,4\) and so their minimum will be \(2\). Denoting this minimum by \(i_l\), let us take \(m= i_l+1\)or \(m= i_l-1\), depending on whether\((\overline{w_{i_1-1}^{(k)}}|w_{i_1}^{(k)}) \neq 0\) or \((\overline{w_{i_1}^{(k)}}|w_{i_1+1}^{(k)}) \neq 0\). If \(r_i=0 \;\forall \;i\) or \(w_m\) does not occur in the exterior product of any summand containing \(k+\beta\), then it is easy to see that \(\big((k + \beta)\overline{w_m^{(k)}}^T + w_m^{(k)}(\overline{k +\beta})^T\big)v\) is non-zero. On the other hand, if \(r_i \neq 0\) for some \(i\) and \(w_m\) occurs in the exterior product of some summand containing \(k+\beta\), then it is not difficult to deduce that \(\big((k + \beta)\overline{w_m^{(k)}}^T + w_m^{(k)}(\overline{k +\beta})^T\big)v \neq 0\), using the fact that \(y_i^{(k)}\) belongs to \(V(\omega_p)\). The result now follows verbatim as in the \(p=1\) setting, since \(\big((k + \beta)\overline{w_m^{(k)}}^T + w_m^{(k)}(\overline{k +\beta})^T\big)v \in M_k\cap W_{min}^{k,\beta}\big(V(\omega_p)\big)\).

Let \(M^{\prime}\) be a proper non-zero \(\mathcal{H}_N\)-submodule of \(L_{\mathcal{H}}^{\beta}\big(V(\omega_p)\big)\). To prove that \(L_{\mathcal{H}}^{\beta}\big(V(\omega_p)\big)\) has a unique irreducible quotient, we shall show that \(M^{\prime}\) is contained in \(W_{max}\big(V(\omega_p)\big)\) if \(\beta \notin \mathbb{Z}^N\) and in \(\widehat{W}_{max}\big(V(\omega_p)\big)\) if \(\beta \in \mathbb{Z}^N\). Pick \(0 \neq v \in N_k\), where \(k+\beta \neq 0\). By Lemma 26, there exist \(p_i, q_i, r_i, s_i \in \mathbb{C}\) such that \(v = \sum_{i=1}^{a} p_iu_i^{(k)} + \sum_{i=1}^{b} q_ix_i^{(k)} + \sum_{i=1}^{c} r_iy_i^{(k)} + \sum_{i=1}^{d} s_iz_i^{(k)}\). Moreover, as \(W_{min}^{\beta}\big(V(\omega_p)\big) \subseteq M^{\prime}\), we are reduced to \(v = \sum_{i=1}^{b} q_ix_i^{(k)} + \sum_{i=1}^{c} r_iy_i^{(k)} + \sum_{i=1}^{d} s_iz_i^{(k)}\).
Claim. \(s_i = 0 \;\forall \;1 \leqslant i \leqslant d\).
If not, then there exists \(1 \leqslant i \leqslant d\) with \(s_i \neq 0\). For \(p =1\), it is evident that \(v = \sum_{j}a_jw_{i_j}^{(k)} + c_0w_1^{(k)}\), where \(i_j \neq 1\) and \(c_0 \in \mathbb{C}^{\times}, a_j \in \mathbb{C}\) for all \(j\). Now since the \(\mathfrak{h}_{\mathfrak{sp}_{N-2}}^{(k)}\)-weights of \(w_{i_j}^{(k)}\) differ from each other, we have \(\sum_{j}a_jw_{i_j}^{(k)} \in M^{\prime}_k\) and thus \(0 \neq c_0w_1^{(k)} \in M^{\prime}_k\), whence it follows that \(M^{\prime} = L_{\mathcal{H}}^{\beta}\big(V(\omega_1)\big)\) by Proposition 28. This contradiction establishes our claim for \(p=1\). For \(2 \leqslant p \leqslant n\), consider \(v = \sum_{i=1}^{b} q_ix_i^{(k)} + \sum_{i=1}^{c} r_iy_i^{(k)} + \sum_{i=1}^{d} s_iz_i^{(k)} \in M^{\prime}_k\). It is also clear that exactly \((p-1)\) elements of \(\mathbb{C}^N\) occuring in the exterior product of any summand of \(z_i^{(k)}\) have non-zero \(\mathfrak{h}_{\mathfrak{sp}_{N-2}}^{(k)}\)-weights. Then it is easy to deduce that \(0 \neq \sum_{i=1}^{d} s_iz_i^{(k)} \in M^{\prime}_k\), by means of \(\mathfrak{h}_{\mathfrak{sp}_{N-2}}^{(k)}\)-weight arguments, as in Case 2. This finally yields that \(M^{\prime} = L_{\mathcal{H}}^{\beta}\big(V(\omega_p)\big)\) by Proposition 28, which is a contradiction. Hence the claim. Consequently \(L_{\mathcal{H}}^{\beta}\big(V(\omega_p)\big)\) admits a unique irreducible \(\mathcal{H}_N\)-quotient, given by \(W_{min}\big(V(\omega_p)\big)\). ◻

Proof. (1) and (2) follow by proceeding similarly as in Theorem 41. ◻

Proof. (1) For any \(v \in M_k\), it is easy to see that \[\theta_{p+1}^{\beta}\big((k+\beta)\wedge \Omega_{k,r,s}^{\mathcal{H},\beta}.v\big) = \begin{cases} (k+\beta)(\overline{k+\beta})^T(\Omega_{k,r,s}^{\mathcal{H},\beta}.v \big) + \theta_{p}^{\beta}\big(\Omega_{k,r,s}^{\mathcal{H},\beta}.v \big)(k+\beta), & p=2 \\ (k+\beta)(\overline{k+\beta})^T(\Omega_{k,r,s}^{\mathcal{H},\beta}.v \big) + (k+\beta) \wedge \theta_{p}^{\beta}\big(\Omega_{k,r,s}^{\mathcal{H},\beta}.v\big), & 2 < p < n. \end{cases}\] Then (1) follows using Lemma 33 and Lemma 38, as \(v \in W_{max}^{k,\beta}\big(V(\omega_p)\big)\).
To complete the proof of the lemma, it is enough to establish (2), as (3) can be handled in a more or less similar manner. Let us fix any \(k \in \mathbb{Z}^N\) with \(M_k \neq (0)\). We proceed by considering 2 cases.
Case 1. \(p = 2\).
If \(M + W_{int}^{\beta}\big(V(\omega_p)\big) = M\) or \(M + W_{int}^{\beta}\big(V(\omega_p)\big) = W_{int}^{\beta}\big(V(\omega_p)\big)\), then the result follows from Theorem 36. So it suffices to consider \(M + W_{int}^{\beta}\big(V(\omega_p)\big) = W_{max}^{\beta}\big(V(\omega_p)\big)\), which implies that \(\{k+\beta \wedge w_1^{(k)} - w_{2i-2}^{(k)} \wedge w_{2i-1}^{(k)} \;| \;2 \leqslant i \leqslant N \} \subseteq M_k + W_{int}^{k,\beta}\big(V(\omega_p)\big)\). Therefore, we can find \(v = \sum_{i=1}^{a}p_iu_i^{(k)} + \sum_{i=1}^{b}q_ix_i^{(k)} + \sum_{i=2}^{n}r_{2i-2,2i-1}(k+\beta \wedge w_1^{(k)} - w_{2i-2}^{(k)} \wedge w_{2i-1}^{(k)}) \in M_k\), where \(r_{2i-2,2i-1} = 0\) for some \(2 \leqslant i \leqslant N\), with either \(\sum_{i=2}^{n}r_{2i-2,2i-1}(k+\beta \wedge w_1^{(k)} - w_{2i-2}^{(k)} \wedge w_{2i-1}^{(k)}) \neq 0\) or \(\sum_{i=1}^{b}q_ix_i^{(k)} \neq 0\) (see Lemma 26). If \(\sum_{i=2}^{n}r_{2i-2,2i-1}(k+\beta \wedge w_1^{(k)} - w_{2i-2}^{(k)} \wedge w_{2i-1}^{(k)}) = 0\), then \(\sum_{i=1}^{b}q_ix_i^{(k)} \in M_k\), thanks to Theorem 41, which gives us the desired result. Now suppose that \(\sum_{i=2}^{n}r_{2i-2,2i-1}(k+\beta \wedge w_1^{(k)} - w_{2i-2}^{(k)} \wedge w_{2i-1}^{(k)}) \neq 0\) and \(r_{2j-2,2j-1} = 0\) for some \(2 \leqslant j \leqslant N\). Without loss of generality, let us take \(r_{2,3} \neq 0\). Consequently, if \(w_{2j-2}^{(k)} \wedge w_{2j-1}^{(k)} \neq x_i^{(k)}\) for any \(1 \leqslant i \leqslant b\), then \((w_{2j-1}^{(k)}\overline{w_2^{(k)}}^T + w_2^{(k)}\overline{w_{2j-1}^{(k)}}^T)v = -r_{2,3}(w_2^{(k)} \wedge w_{2j-1}^{(k)}) + \ldots\) is a non-zero element of \(M_k\), which proves the lemma. On the other hand, if \(w_{2j-2}^{(k)} \wedge w_{2j-1}^{(k)}= x_i^{(k)}\) for some \(1 \leqslant i \leqslant b\) , then \(w_{2m-2}^{(k)} \wedge w_{2m-1}^{(k)}\) must also appear in one of the summands of \(\sum_{i=1}^{b}q_ix_i^{(k)}\) for some \(2 \leqslant m \leqslant N\) satisfying \(r_{2m-2,2m-1} = 0\), which yields that \(0 \neq (w_{2m-2}^{(k)}\overline{w_{2j-2}^{(k)}}^T + w_{2j-2}^{(k)}\overline{w_{2m-2}^{(k)}}^T)v \notin W_{min}^{k,\beta}\big(V(\omega_p)\big)\) and thus we are done using Proposition 39, Theorem 41 and (1).
Case 2. \(2 < p <n\).
Let, if possible, \(M \cap W_{int}^{\beta}\big(V(\omega_p)\big) = W_{min}^{\beta}\big(V(\omega_p)\big)\). Consider the \(\mathcal{H}_N\)-module map \(T_p^{\beta}|_M\). From Lemma 33, it is evident that \(\mathrm{Ker}(T_p^{\beta}|_M) = M \cap W_{int}^{\beta}\big(V(\omega_p)\big)\), which implies that \(T_p^{\beta}|_M\) is a non-zero map, due to our hypothesis. But again, since \(W_{min}^{\beta}\big(V(\omega_{p-1})\big)\) is irreducible, we obtain \(\mathrm{Im}(T_p^{\beta}|_M) = W_{min}^{\beta}\big(V(\omega_{p-1})\big)\), as \(\mathrm{Im}(T_p^{\beta}|_{W_{max}^{\beta}\big(V(\omega_p)\big)}) = W_{min}^{\beta}\big(V(\omega_{p-1})\big)\) by Lemma 33.
Claim. There exists \(v \in M_k\) such that \(\Omega_{k,r,s}^{\mathcal{H},\beta}.v \notin W_{min}^{k,\beta}\big(V(\omega_p)\big)\) for some \(r,s \in \mathbb{Z}^N\).
If not, then \(\Omega_{k,r,s}^{\mathcal{H},\beta}M_k \subseteq W_{min}^{k,\beta}\big(V(\omega_p)\big) \;\forall \;r,s \in \mathbb{Z}^N\). Now since \(0 \neq v^{\prime} = (k+\beta) \wedge w_2^{(k)} \wedge w_4^{(k)} \wedge \ldots \wedge w_{2(p-3)}^{(k)} \wedge w_{2(p-2)}^{(k)} \in W_{min}^{\beta}\big(V(\omega_{p-1})\big)\), there exists \(x \in M_k\) such that \(T_p(x \otimes t^k) = v^{\prime} \otimes t^k\). This implies that \(T_p\bigg(\big(w_{2p-3}^{(k)}\overline{w_{2p-3}^{(k)}}^T\big)(x \otimes t^k)\bigg) = 0\). But on the other hand, we have \(\big(w_{2p-3}^{(k)}\overline{w_{2p-3}^{(k)}}^T\big)T_p(x \otimes t^k) \neq 0\), which contradicts Lemma 38. Hence the claim.
The assertion now follows from Lemma 38, Theorem 41 and (1). ◻

Proof. (1) is an immediate consequence of Corollary 29.
(2), (3), (4), (5) follow by simply combining Theorem 36, Theorem 41 and Lemma 43. ◻

Proof. Follows from Corollary 29, Theorem 36 and Theorem 44. ◻

Proof. For (1) and (2), it suffices to show that \(\mathrm{Ker}T_p^{k, \beta} = W_{int}^{k,\beta}(\Lambda^p\mathbb{C}^N)\) for any \(k \in \mathbb{Z}^N\) satisfying \(k+\beta \neq 0\). To prove this, first observe that \[\begin{align} T_p^{k,\beta}\big((w_{i_1}^{(k)} \wedge \ldots \wedge w_{i_p}^{(k)}) \otimes t^k \big) =T_p^{k,\beta}\big((k +\beta \wedge w_{j_1}^{(k)} \wedge \ldots \wedge w_{j_{p-1}}^{(k)}) \otimes t^k \big) = 0, \end{align}\] where \(i_r, j_s \neq 1\). But again, we have \(T_p^{k,\beta}\big((w_1^{(k)} \wedge w_{i_1}^{(k)} \wedge \ldots \wedge w_{i_{p-1}}^{(k)}) \otimes t^k \big) \neq 0 \;(i_r \neq 1)\), along with \(W_{int}^{k,\beta}(\Lambda^p\mathbb{C}^N) = \mathrm{span} \{w_{i_1}^{(k)} \wedge \ldots \wedge w_{i_p}^{(k)}, \;k +\beta \wedge w_{j_1}^{(k)} \wedge \ldots \wedge w_{j_{p-1}}^{(k)} \;| \;i_r, j_s \neq 1 \}\) and so we are done.
(3) follows by a similar computation. ◻

Proof. (1) follows by simply observing that \(E_{i,i} - E_{j,j} = (e_i + e_j)(e_i - e_j)^T + e_ie_j^T - e_je_i^T\) and \(E_{i,j} = e_ie_j^T \;\forall \;1 \leqslant i \neq j \leqslant N\) and \(E_{i,i} = e_ie_i^T \;\forall \;1 \leqslant i \leqslant N\).
(2) can be deduced using Lemma 11, by providing a similar argument as in Lemma 12.
(3) If \(V \cong V(\delta_k,k)\) for some \(0 \leqslant k \leqslant N\), then a direct checking readily yields \(\mathcal{J}_{\mathcal{W}}(V) = V\).
Conversely, assume that \(\mathcal{J}_{\mathcal{W}}(V) = V\). The irreducibility of \(V\) thereby implies that \(V\) is countable-dimensional and so by Dixmier’s lemma, the central elements (which are scalar multiples of \(\mathrm{Id}\)) act by fixed scalars on \(V\). Consequently, \(V\) remains irreducible over \(\mathfrak{sl}_N\). Now putting \(r=e_i\) and \(u=e_j\) for any \(1 \leqslant i \neq j \leqslant N\), we get \((E_{ij})^2v = 0 \;\forall \;v \in V\). So we can directly invoke Lemma 10 to conclude that \(V\) is finite-dimensional. Then by the representation theory of \(\mathfrak{gl}_N\), we have:
(i) \(V \cong V(\overline{\lambda}) \cong V(\lambda, \overline{\lambda}(\mathrm{Id}))\) as \(\mathfrak{gl}_N\)-modules for some \(\overline{\lambda} \in P_{\mathfrak{gl}_N}^{+}\), where \(\lambda = \overline{\lambda}|_{\mathfrak{h}_{\mathfrak{sl}_N}^*} \in P_{\mathfrak{sl}_N}^{+}\).
(ii) There exists \(0 \neq v_{\overline{\lambda}} \in V(\overline{\lambda})\) such that \(E_{i,i}v_{\overline{\lambda}} = \overline{\lambda}(E_{i,i})v_{\overline{\lambda}} \;\forall \;1 \leqslant i \leqslant N\).
Now taking \(r=u=e_i\) for any \(1 \leqslant i \leqslant N\), we obtain \((E_{i,i})^2v = E_{i,i}v \;\forall \;v \in V(\overline{\lambda})\). This implies that \((\overline{\lambda}(E_{i,i})^2 - \overline{\lambda}(E_{i,i}))v_{\overline{\lambda}} = 0\), which gives \(\overline{\lambda}(E_{i,i}) \in \{0, 1 \} \;\forall \;1 \leqslant i \leqslant N\). But again since \(\overline{\lambda}(E_{i,i}) - \overline{\lambda}(E_{i+1,i+1}) \in \mathbb{Z}_{\geqslant 0} \;\forall \;1 \leqslant i \leqslant N-1\), it is clear that \(\overline{\lambda} = \overline{\delta_k}\) for some \(0 \leqslant k \leqslant N\). The result now immediately follows by observing that \(\overline{\delta_k}(\mathrm{Id}) = k \;\forall \;0 \leqslant k \leqslant N\). ◻

Proof. Let \(V \ncong V(\delta_k,k)\) for any \(0 \leqslant k \leqslant N\) and \(W\) be a non-zero \(\mathcal{W}_N\)-submodule of \(F^{\beta}_{\mathcal{W}}(V)\). As in Theorem 14, put \(\widetilde{W} = \bigcap_{m \in \mathbb{Z}^N} W_m\) and apply Lemma 48 to obtain \(r \in \mathbb{Z}^N,\;u \in \mathbb{C}^N\) such that \((ru^T)^2v \neq (u|r)(ru^T)v\) for every non-zero \(v \in V\). Fix \(0 \neq w \in W_m\) for some \(m \in \mathbb{Z}^N\) and consider \[\begin{align} D(u,-r)D(u,r+s)(w \otimes t^m) = [(u|r)(ru^T) -(ru^T)^2]w \otimes t^{m+s}+ \ldots \ldots \in W_{m+s} \otimes \mathbb{C}t^{m+s}. \end{align}\] Replacing \(r\) by \(pr\) and \(u\) by \(qu\) for \(p,q \in \mathbb{Z}\) in the above equation and comparing the coefficients of \(p^2q^2\), we can then apply Lemma 11 to infer that \([(ru^T)^2 -(u|r)(ru^T)]w \in W_{m+s} \;\forall \;s \in \mathbb{Z}^N\). This shows that \(\widetilde{W} \neq (0)\). Again for any \(m, s \in \mathbb{Z}^N, \;x \in \mathbb{C}^N\) and \(v \in \widetilde{W}\), we have \[\begin{align} D(x,m)(v \otimes t^{s-m}) = (x|s-m+\beta)v \otimes t^s + (mx^T)v \otimes t^s, \end{align}\] which gives \((mx^T)v \in \widetilde{W}\). Applying Lemma 48, we then get \(\widetilde{W}= V\) and so we are done. ◻

Proof. (1) follows from Lemma 48, whereas (2) can be deduced using Lemma 11, by providing essentially an analogous argument as presented in Lemma 12.
(3) If \(V \cong V(\delta_k)\) for some \(0 \leqslant k \leqslant N-1\), then a direct checking readily yields \(\mathcal{J}_{\mathcal{S}}(V) = V\).
Conversely, assume that \(\mathcal{J}_{\mathcal{S}}(V) = V\). Now putting \(r=e_i\) and \(u=e_j\) for any \(1 \leqslant i \neq j \leqslant N\), we get \((E_{i,j})^2v = 0 \;\forall \;v \in V\). This shows that \(V\) is finite-dimensional by Lemma 10. Let \(V \cong V(\lambda)\) for some \(\lambda \in P_{\mathfrak{sl}_N}^{+} \setminus \{\delta_k \}_{k=0}^{N-1}\). Consequently, there exist \(a_1, \ldots, a_{N-1} \in \mathbb{Z}_{\geqslant 0}\) such that \(\lambda = \sum_{k=1}^{N-1} a_k \delta_k\) with \(a_i + a_j \geqslant 2\) for some \(1 \leqslant i \leqslant j < N\). Note that \(\delta_i(E_{i,i} - E_{j+1,j+1}) = 1 = \delta_j(E_{i,i} - E_{j+1,j+1})\), which implies that \(\lambda(E_{i,i} - E_{j+1,j+1}) \geqslant 2\). Now pick a non-zero \(v_{\lambda} \in V(\lambda)_{\lambda}\) and consider the \(S_{i,j+1}\)-module \(\text{span} \{(E_{j+1,i})^rv_{\lambda} \;| \;r \in \mathbb{Z}_{\geqslant 0} \}\), where \(S_{i,j+1} := \text{span} \{E_{i,j+1}, E_{j+1,i}, E_{i,i} - E_{j+1,j+1} \}\). Then by \(\mathfrak{sl}_2\)-theory, it immediately follows that \((E_{j+1,i})^2v_{\lambda} \neq 0\) and thus \(v_{\lambda} \notin \mathcal{J}_{\mathcal{S}}(V(\lambda))\). ◻

Proof. Let \(V \ncong V(\delta_k)\) for any \(0 \leqslant k < N\) and \(W\) be a non-zero \(\mathcal{S}_N\)-submodule of \(F^{\beta}_{\mathcal{S}}(V)\). As in Theorem 14, put \(\widetilde{W} = \bigcap_{m \in \mathbb{Z}^N} W_m\) and then invoke Lemma 51 to obtain \(r \in \mathbb{Z}^N,\;u \in \mathbb{C}^N\) with \((u|r)=0\) such that \((ru^T)^2v \neq 0\) for every non-zero \(v \in V\). Fix \(0 \neq w \in W_m\) for some \(m \in \mathbb{Z}^N\). Consequently, proceeding exactly as in Theorem 50 yields \((ru^T)^2w \in W_{m+s} \;\forall \;s \in \mathbb{Z}^N\) with \((u|s)=0\). Let us now pick any \(s \in \mathbb{Z}^N\) such that \((u|s) \neq 0\). Setting \(u_1 = u\), extend \(u_1\) to a basis \(\{u_i \}_{i=1}^{N}\) of \(\mathbb{C}^N\). Then \((u_i|r) \neq 0\) for some \(2 \leqslant i \leqslant N\). So without loss of generality, we can take \((u_2|r) \neq 0\) and define \(u_s^{\prime} := (u_1|s)u_2 - (u_2|r+s)u_1 \neq 0\). Note that \((u_s^{\prime}|r+s) =0\) and consider \[\begin{align} D(u,-r)D(u_s^{\prime},r+s)(w \otimes t^m) = -(u_2|r)[(u|r)(ru^T) + (ru^T)^2](w \otimes t^{m+s}) + \ldots \in W_{m+s} \otimes \mathbb{C}t^{m+s}. \end{align}\] Now replacing \(r\) by \(pr\) and \(u\) by \(qu\) for \(p,q \in \mathbb{Z}\) in the above equation and comparing the coefficients of \(p^3q^2\), we can apply Lemma 11 to obtain \((ru^T)^2w \in W_{m+s} \;\forall \;s \in \mathbb{Z}^N\) satisfying \((u|s) \neq 0\). This implies that \(\widetilde{W} \neq (0)\) and so by providing the same argument as in Theorem 50, we get \((mx^T)v \in \widetilde{W} \;\forall \;m \in \mathbb{Z}^N, \;x \in \mathbb{C}^N\) satisfying \((x|m)=0\). Applying Lemma 51 finally gives \(\widetilde{W}= V\) and thus we obtain the desired result. ◻

Proof. (1) Choose \(w \in \mathbb{C}^N\) such that \((r|w) \neq 0\). Now for any \(v^{\prime} \in M_k\), consider \[\begin{align} D(v,-(r+s))D(w,s)D(u,r).(v^{\prime} \otimes t^k) = - (r|w)[(k+\beta|u)rv^T - (k+\beta|v)ru^T].(v^{\prime} \otimes t^k) + \ldots \end{align}\] Replacing \(r\) by \(pr, s\) by \(qs\) and \(u\) by \(p^{\prime}u, v\) by \(q^{\prime}v\) for \(p, p^{\prime}, q, q^{\prime} \in \mathbb{Z}\) in the above equation and comparing the coeffecients of \(p^2p^{\prime}q^{\prime}\), we can then apply Lemma 11 to get \[\begin{align} [(k+\beta|u)rv^T - (k+\beta|v)ru^T]v^{\prime} \in W_k \;\forall \;r \in \mathbb{Z}^N \;\text{and} \;u, v \in \mathbb{C}^N. \\ \implies (k+\beta|r)[(k+\beta|u)sv^T - (k+\beta|v)su^T].v^{\prime} - (k+\beta|s)[(k+\beta|u)rv^T - (k+\beta|v)ru^T ].v^{\prime} \in W_k, \end{align}\] which shows that \(\Omega_{r,s,u,v}^{\mathcal{W},k,\beta}.v^{\prime} \in M_k\) and therefore (1) is proved.
(2) follows as \(\{\mathcal{T}_{k,e_i,e_j}^{\mathcal{W},\beta} \;| \;1 \leqslant j \neq i \leqslant N \}\) is a linearly independent subset of \(\{k+\beta \}^{\perp}\), since \(\{k+\beta \}^{\perp}\) is \((N-1)\)-dimensional. ◻

Proof. The lemma follows by just applying Proposition 57, since \(v_i^{(k)} \in \{k+\beta \}^{\perp} \;\forall \;1 \leqslant i < N\). ◻

Proof. (1) Let \(M = \oplus_{k \in \mathbb{Z}^N}M_k \otimes \mathbb{C}t^k\) be a non-zero proper \(\mathcal{W}_N\)-submodule of \(F^{\beta}_{\mathcal{W}}\big(V(\overline{\delta_p})\big)\). In view of Theorem 60, we must have \(M \subseteq \oplus_{k \in \mathbb{Z}^N, k \neq -\beta}V(\overline{\delta_p}) \otimes \mathbb{C}t^k \cong \oplus_{k \in \mathbb{Z}^N, k \neq -\beta} \mathbb{C}t^k\). So there exists \(k \in \mathbb{Z}^N\) such that \(k+\beta \neq 0\) and \(M_k \neq (0)\). But this gives \[\begin{align} D(k+\beta,-k-\beta).t^k = -(k+\beta|k+\beta)t^{-\beta} \neq 0, \end{align}\] which is a contradiction and hence the assertion is proved.
(2) Let \(W = \oplus_{k \in \mathbb{Z}^N}W_k \otimes \mathbb{C}t^k\) be an irreducible \(\mathcal{W}_N\)-submodule of \(F^{\beta}_{\mathcal{W}}\big(V(\overline{\delta_p})\big)\) and \(k \in \mathbb{Z}^N\) satisfying \(k+\beta \neq 0\) and \(W_k \neq (0)\). By Theorem 60, there exist \(a_{i_1, \ldots,i_{p-1}} \in \mathbb{C}\) and \(b_{j_1, \ldots,j_{p}} \in \mathbb{C}\) such that \[\begin{align} v = \sum_{(i_1, \ldots, i_{p-1})}a_{i_1, \ldots,i_{p-1}}(k+\beta) \wedge v_{i_1}^{(k)} \wedge \ldots \wedge v_{i_{p-1}}^{(k)} + \sum_{(j_1, \ldots, j_p)}b_{j_1, \ldots,j_{p}}v_{j_1}^{(k)} \wedge \ldots \wedge v_{j_{p}}^{(k)} \in W_k, \end{align}\] with \(\sum_{(i_1, \ldots, i_{p-1})}a_{i_1, \ldots,i_{p-1}}(k+\beta) \wedge v_{i_1}^{(k)} \wedge \ldots \wedge v_{i_{p-1}}^{(k)} \neq 0\), where \(1 \leqslant i_r, j_s < N\). By Lemma 59 and the discussion preceding this lemma, it now follows that the \(\mathfrak{h}_{\mathfrak{sl}_{N-1}}^{(k)}\)-weights of \((k+\beta) \wedge v_{i_1}^{(k)} \wedge \ldots \wedge v_{i_{p-1}}^{(k)}\) and \(v_{j_1}^{(k)} \wedge \ldots \wedge v_{j_{p}}^{(k)}\) are different. This implies that \(\sum_{(i_1, \ldots, i_{p-1})}a_{i_1, \ldots,i_{p-1}}(k+\beta) \wedge v_{i_1}^{(k)} \wedge \ldots \wedge v_{i_{p-1}}^{(k)} \in W_k\), which thereby proves the lemma, due to the irreducibility of \(W^{\beta}(\Lambda^p\mathbb{C}^N)\) over \(\mathcal{W}_N\). ◻

Proof. The proof follows from Remark 62, as the Loewy series of \(F^{\beta}_{\mathcal{W}}\big(V(\overline{\delta_p})\big)\) has length \(2\), by Theorem 60. ◻