Proof. We only prove (1). The assertion (2) can be proved similarly. Let $\mathfrak U$ be the subalgebra of ${\mathcal U}^+_{\mathbb Q}(\widehat {\mathfrak {gl}}_{m|n})$ generated by the elements $\mathsf E^\vartriangle _{h,h+1}$ , $\mathsf E^\vartriangle _{h,h+l(m+n)}$ for $1\leqslant h\leqslant m+n$ and $l>0$ . It is enough to prove $\mathsf E^\vartriangle _{i,j}\in \mathfrak U$ for $i<j$ with $i\not \equiv j\! \mod \!(m+n)$ . Write $j-i=s(m+n)+k$ where $s\in \mathbb Z$ and $1\leqslant k<m+n$ . We apply induction on k. If $k=1$ , then we have

$$ \begin{align*} \mathsf E^\vartriangle_{i,s(m+n)+i+ 1}=\mathsf E^\vartriangle_{i,i+ 1}\mathsf E^\vartriangle_{i+ 1,s(m+n)+i+ 1} -\mathsf E^\vartriangle_{i+ 1,s(m+n)+i+ 1}\mathsf E^\vartriangle_{i,i+ 1}\in \mathfrak U. \end{align*} $$

Now suppose, $1<k<m+n$ , then by the inductive hypothesis,

$$ \begin{align*} \mathsf E^\vartriangle_{i,s(m+n)+i+ k}=\mathsf E^\vartriangle_{i,i+1}\mathsf E^\vartriangle_{i+1,s(m+n)+i+ k} -(-1)^{(\overline{i+ 1}+\overline{i+ k})(\overline i+\overline{i+ 1})}\mathsf E^\vartriangle_{i+ 1,s(m+n)+i+ k}\mathsf E^\vartriangle_{i,i+ 1}\in \mathfrak U. \end{align*} $$

The assertion follows.

Proof. Let $\mathscr D(m|n,r)=\{({\lambda }, d,\mu )\mid d\in {\mathscr D}^{\!\vartriangle \!}_{{\lambda },\mu },{\lambda },\mu \in \Lambda _{\!\vartriangle \!}(m+n,r)\}$ . By [Reference Varagnolo and Vasserot25, 7.4], there is a bijective map

(3.1.1) $$ \begin{align} {\jmath_{\!\vartriangle\!}}:\mathscr D(m|n,r)\longrightarrow\{A\in M_{{\!\vartriangle\!},m+n}(\mathbb N)\mid\sigma(A)=r\} \end{align} $$

sending $({\lambda }, d,\mu )$ to $A=(a_{k,l})$ , where $a_{k,l}=|R_k^{\lambda }\cap dR_l^\mu |$ for all $k,l\in \mathbb Z$ .

We claim that $\jmath _{\!\vartriangle \!}({\lambda },d,\mu )\in \Theta _{\!\vartriangle \!}(m|n,r)$ if and only if $({\lambda },d,\mu )\in \mathscr D^\circ (m|n,r)$ (or equivalently, $d^{-1}{\mathfrak S}_{{\lambda }^{(0)}} d\cap {\mathfrak S}_{\mu ^{(1)}}=d^{-1}{\mathfrak S}_{{\lambda }^{(1)}} d\cap {\mathfrak S}_{\mu ^{(0)}}=\{1\}$ ). Indeed, if $({\lambda },d,\mu )\in \mathscr D^\circ (m|n,r)$ , then for $\alpha =({\lambda }^{(0)},1^{r-\sigma ({\lambda }^{(1)})})$ and $\beta =(1^{r-\sigma (\mu ^{(0)})},\mu ^{(1)})$ , regarded as elements in $\Lambda _{\!\vartriangle \!}(N',r)$ for some $N'$ , ${\mathfrak S}_\alpha ^d\cap {\mathfrak S}_\beta ={\mathfrak S}_{{\lambda }^{(0)}}^d\cap {\mathfrak S}_{\mu ^{(1)}}=\{1\}$ and so, $|R_i^\alpha \cap dR_j^\beta |=0$ or 1 for all $i,j\in \mathbb Z$ . Since $R_i^{\lambda }=R_i^\alpha $ for all $i\in [1,m]$ and $R_{m+j}^\mu =R_{\sigma (\mu ^{(0)})+j}^\beta $ for all $j\in [1,n]$ , it follows that $R^\mu_{m+j+lN}=R_{m+j}^\beta +lr=R_{\sigma (\mu ^{(0)})+j}^\beta +lr=R^\beta_{\sigma (\mu ^{(0)})+j+lN}$ for all $l\in \mathbb Z$ (and $N=m+n$ ). Hence, for all $i\in [1,m]$ , $j\in [1,n]$ ,

$$ \begin{align*}a_{i,m+j+lN}=|R_i^{\lambda}\cap dR^\mu_{m+j+lN}|=|R^\alpha_i\cap dR^\beta_{\sigma(\mu^{(0)})+j+lN}|= 0\text{ or }1,\end{align*} $$

as desired. The converse can be proved by reversing the argument above and the claim is proven.

By the claim, $\jmath _{\!\vartriangle \!}$ restricts to a bijection. The last assertion follows from [Reference Deng, Du and Fu6, 3.2.3].

Proof. This is clear since, for $\nu ={\lambda } d\cap \mu $ ,

Proof. By Lemmas 3.2 and 3.3, we conclude that for $({\lambda },d,\mu )\in {\mathscr D}^\circ (m|n,r)$ . Let , where . By Lemma 3.2, we have $a_{t_1w}=a_w=a_{wt_2}$ for $t_1\in {\mathfrak S}_{{\lambda }^{(0)}}\cap S$ , $t_2\in {\mathfrak S}_{\mu ^{(0)}}\cap S$ and $a_{t_3w}=-a_w=a_{wt_4}$ for $t_3\in {\mathfrak S}_{{\lambda }^{(1)}}\cap S$ , $t_4\in {\mathfrak S}_{\mu ^{(1)}}\cap S$ . It follows that

$$ \begin{align*}h=\sum_{d\in{\mathscr D}^{\!\vartriangle\!}_{{\lambda},\mu}}a_d\sum_{w_1\in{\mathfrak S}_{\lambda},\,w_2\in{\mathfrak S}_\mu\cap {\mathscr D}^{\!\vartriangle\!}_{{\lambda} d\cap\mu}}(-1)^{\ell(w_1^{(1)})+\ell(w_2^{(1)})}w_1dw_2.\end{align*} $$

If $d^{-1}{\mathfrak S}_{{\lambda }^{(0)}} d\cap {\mathfrak S}_{\mu ^{(1)}}\not =\{1\}$ , then there exists some $s\in d^{-1}{\mathfrak S}_{{\lambda }^{(0)}} d\cap {\mathfrak S}_{\mu ^{(1)}}\cap S$ . Let $x=dsd^{-1}$ . Then we have $x\in {\mathfrak S}_{{\lambda }^{(0)}}\cap S$ . Hence, we have $-a_d=a_{ds}=a_{xd}=a_d$ . Since is an integral domain and we conclude that $a_d=0$ . Similarly, if $d^{-1}{\mathfrak S}_{{\lambda }^{(1)}} d\cap {\mathfrak S}_{\mu ^{(0)}}\not =\{1\}$ , then $a_d=0$ . Therefore . The assertion follows.

Proof. Clearly, we have . Now the assertion follows from Lemma 3.5.

Proof. By (3.0.8), it is clear in all four cases that $(u_{h,j}^{\lambda })^{-1}\in {\mathscr D}^{\!\vartriangle \!}_{\lambda }$ . By explicitly working out $(u_{h,j}^{\lambda })^{-1}$ in all four cases and noting that

(4.3.1) $$ \begin{align} \mu:={\lambda}+\boldsymbol e^\vartriangle_h-\boldsymbol e^\vartriangle_j= \begin{cases}(\cdots\cdots,{\lambda}_h+1,\ldots,{\lambda}_{a}-1,\cdots\cdots), &\text{ if } h<a;\\ {\lambda},&\text{ if }h=a\;\, (\text{so }l\neq0);\\ (\cdots\cdots,{\lambda}_{a}-1,\ldots,{\lambda}_h+1,\cdots\cdots), &\text{ if } a<h,\\ \end{cases} \end{align} $$

we see $u_{h,j}^{\lambda }\in {\mathscr D}^{\!\vartriangle \!}_{\mu }$ . Also, for the matrix $\jmath _{\!\vartriangle \!}(\mu ,u_{h,j}^{\lambda },{\lambda })$ , we simply compute $|R_k^{\mu }\cap u_{h,j}^{\lambda }(R_l^{\lambda })|$ case by case. For example, if u is the first element given in (4.2.1), then $R_h^{\mu }\cap u(R_j^{\lambda })=\{\widetilde {\lambda }_h+1\}$ and so $|R_h^{\mu }\cap u(R_j^{\lambda })|=1$ . This proves (1). (2) follows from (1) and [Reference Deng, Du and Fu6, 3.2.3] with all 0s removed. The last assertion follows from (2).

Proof. Throughout the proof, we write $N=m+n$ . Write $j=j'+j''N\in \mathbb Z$ as in (1.0.2). Then $h\neq j'\implies j''\geqslant 0$ and $h= j'\implies j''>0$ .

Case 1. $j'\neq h$ . With respect to the i and k, we set $B=A+E^\vartriangle _{h,k}-E^\vartriangle _{j,k}=(b_{s,t})$ . Then ${\lambda }^B\!\!:=\text {ro}(B)={\lambda }+\boldsymbol {e}_h^{\!\vartriangle \!}-\boldsymbol {e}_j^{\!\vartriangle \!}$ and so ${\lambda }^B_h={\lambda }_h+1$ , ${\lambda }^B_{j'}={\lambda }_{j'}-1$ . Define the cycle permutation:

$$ \begin{align*}u_{h,k}^B=\left(\begin{matrix} {\widetilde{\lambda}}^B_{h-1}+\sum_{t\leqslant k}a_{h,t}+1&{\widetilde {\lambda}}^B_{h-1}+\sum_{t\leqslant k}a_{h,t}+2 &\cdots & {\widetilde{\lambda}}^B_{h}\\ {\widetilde{\lambda}}^B_{h}&{\widetilde{\lambda}}^B_{h-1}+\sum_{t\leqslant k}a_{h,t}+1&\cdots& {\widetilde{\lambda}}^B_{h}-1\\ \end{matrix}\right) \in\mathfrak S_{{\lambda}+\boldsymbol e^\vartriangle_h-\boldsymbol e^\vartriangle_j}\end{align*} $$

We first assume $j'<h$ (so $j''>0$ ) and take

$$ \begin{align*}u=u_{h,k}:=u_{h,k-1}^B=\left(\begin{matrix} \widetilde{\lambda}_{h-1}+\sum_{t\leqslant k-1}a_{h,t}&\widetilde {\lambda}_{h-1}+\sum_{t\leqslant k-1}a_{h,t}+1 &\cdots & \widetilde{\lambda}_{h}\\ \widetilde{\lambda}_{h}&\widetilde{\lambda}_{h-1}+\sum_{t\leqslant k-1}a_{h,t}&\cdots& \widetilde{\lambda}_{h}-1\\ \end{matrix}\right).\end{align*} $$

We claim in this case that as sequences

(4.4.1) $$ \begin{align}u(\boldsymbol{r}_{s,t}(B))=u_{h,j}^{\lambda} w_{h,i,j}(\boldsymbol{r}_{s,t}(A)),\;\forall (s,t)\in[1,N]\times\mathbb Z\backslash\{(j',k-j''N),(h,k)\}. \end{align} $$

Indeed, in this case, $u_{h,j}^{\lambda }$ is the second element in (4.2.1) and

$$ \begin{align*}u^*=u_{h,j}^{\lambda} w_{h,i,j}=\begin{pmatrix}i&i+1&\cdots&\widetilde{\lambda}_h\\ \widetilde{\lambda}_h-j''r&i&\cdots&\widetilde{\lambda}_h-1\\\end{pmatrix}. \end{align*} $$

If $s\in [1,j'-1]$ , or $s=j',t<k-j''N$ , or $h<s<N$ , then $u^*(\boldsymbol {r}_{s,t}(A))=\boldsymbol {r}_{s,t}(A)=\boldsymbol {r}_{s,t}(B)=u\boldsymbol {r}_{s,t}(B)$ . If $s=j',t>k-j''N$ , or $j'<s<h$ , or $s=h, t\neq k$ , then $u^*(\boldsymbol {r}_{s,t}(A))=\rho ^{-1}\boldsymbol {r}_{s,t}(A)=u\boldsymbol {r}_{s,t}(B)$ , proving the claim.

Thus, by (4.1.4) and Lemma 4.2,

$$ \begin{align*}uy_B=\bigg(\begin{matrix}\underline{1,\,\cdots}&\boldsymbol{c}_{h-k''N,k'}(B)&\boldsymbol{c}_{h-k''N+1,k'}(B)&\cdots&\boldsymbol{c}_{j-k''N,k'}(B)&\underline{\cdots,r}\\ \underline{y_1,\cdots}&u\boldsymbol{r}_{h-k''N,k'}(B)&\underline{u\boldsymbol{r}_{h-k''N+1,k'}(B)}&\underline{\cdots}&u\boldsymbol{r}_{j-k''N,k'}(B)&\underline{\cdots,y_r}\\\end{matrix}\bigg)\end{align*} $$

and

$$ \begin{align*}u^*y_A=\bigg(\begin{matrix}\underline{1,\,\cdots}&\boldsymbol{c}_{h-k''N,k'}(A)&\boldsymbol{c}_{h-k''N+1,k'}(A)&\cdots&\boldsymbol{c}_{j-k''N,k'}(A)&\underline{\cdots,r}\\ \underline{y_1,\cdots}& u^*\boldsymbol{r}_{h-k''N,k'}(A)&\underline{u^*\boldsymbol{r}_{h-k''N+1,k'}(A)}&\underline{\cdots}&u^*\boldsymbol{r}_{j-k''N,k'}(A)&\underline{\cdots,y_r}\\\end{matrix}\bigg),\end{align*} $$

where the underline parts for both elements are identical. Now, a direct calculation shows that (noting $j'<h$ ), for $\boxed {\blacksquare :=\widetilde {\lambda }_h-k''r}$ , $p=i-\widetilde {\lambda }_{j'-1}-\sum _{t<k}a_{j,t}$ ,

$$ \begin{align*}\begin{aligned} u\big(\boldsymbol{r}_{h-k''N,k'}(B)\big)&=\big(\blacksquare,\rho^{-1}\boldsymbol{r}_{h,k}(A)-k''r\big);\\ u^*\big(\boldsymbol{r}_{h-k''N,k'}(A)\big)&=\;\rho^{-1}\boldsymbol{r}_{h,k}(A)-k''r;\\ u\big(\boldsymbol{r}_{j-k''N,k'}(B)\big)&=\boldsymbol{r}_{j',k-j''N}(B)+(j''-k'')r=\big(\widetilde{\lambda}_{j'-1,j''-k''}+\sum_{t<k}a_{j,t}+x\big)_{1\leqslant x\leqslant b_{j,k}};\\ u^*\!\big(\!\boldsymbol{r}_{\!j-k''N\!,k'}(\!A)\!\big) &=\!\Big(\!\!\big(\widetilde{\lambda}_{j'-1,j''-k''}+\!\sum_{t<k}\!a_{j,t}+x\big)_{\!1\leqslant x<p},\blacksquare, \big(\widetilde{\lambda}_{j'-1,j''-k''}+\!\!\sum_{t<k}\!a_{j,t}+x\big)_{\!p\leqslant x\leqslant b_{j,k}}\!\Big)\!. \end{aligned} \end{align*} $$

Let $a_k=(uy_{B})^{-1}(\widetilde {\lambda }_{h}-k''r)$ and $b_{k,i}=(u^*y_A)^{-1}(\widetilde {\lambda }_{h}-k''r)$ . Then

(4.4.2) $$ \begin{align} a_k=\widetilde\mu_{k'-1}+\sum_{s\leqslant h-1}a_{s,k}+1,\qquad b_{k,i}= \widetilde\mu_{k'-1}+\sum_{s<j}a_{s,k}+i-\widetilde{\lambda}_{j'-1}-\sum_{t< k}a_{j,t}. \end{align} $$

Then, by the choice of i, we have $a_k\leqslant b_{k,i}$ , and $uy_B$ and $u^*y_A$ have the form, respectively,

(4.4.3) $$ \begin{align} \bigg(\begin{matrix}\cdots&a_k&a_k+1&\cdots&b_{k,i}&\cdots\\ \cdots&\widetilde{\lambda}_h-k''r&j_1&\cdots&j_s&\cdots\\\end{matrix}\bigg),\quad \bigg(\begin{matrix}\cdots&a_k&\cdots&b_{k,i}-1&b_{k,i}&\cdots\\ \cdots&j_1&\cdots&j_s&\widetilde{\lambda}_h-k''r&\cdots\\\end{matrix}\bigg) \end{align} $$

Hence, $u^*y_A=uy_Bw$ if $w=w_{k,i}=\bigg (\begin {matrix}a_k&a_k+1&\cdots &b_{k,i}-1&b_{k,i}\\ a_k+1&a_k+2&\cdots &b_{k,i}&a_k\\\end {matrix}\bigg )\in {\mathfrak S}_\mu $ , proving the $j'<h$ case.

We now assume $h<j'$ . Then $u_{h,j}^{\lambda }$ is the first element in (4.2.1). Putting $u^*=u_{h,j}^{\lambda } w_{h,i,j}$ and

$$ \begin{align*}u=u_{h,k}:=u_{h,k}^B=\left(\begin{matrix} \widetilde{\lambda}_{h-1}+\sum_{t\leqslant k}a_{h,t}+1&\widetilde {\lambda}_{h-1}+\sum_{t\leqslant k}a_{h,t}+2 &\cdots & \widetilde{\lambda}_{h}+1\\ \widetilde{\lambda}_{h}+1&\widetilde{\lambda}_{h-1}+\sum_{t\leqslant k}a_{h,t}+1&\cdots& \widetilde{\lambda}_{h}\\ \end{matrix}\right),\end{align*} $$

one checks easily that (4.4.1) continues to hold. But the remaining cases become

$$ \begin{align*}\begin{aligned} u\big(\boldsymbol{r}_{h-k''N,k'}(B)\big)&=\big(\boldsymbol{r}_{h,k}(A)-k''r,\blacktriangle\big), \text{ where }\boxed{\blacktriangle:=\widetilde{\lambda}_h+1-k''r};\\ u^*\big(\boldsymbol{r}_{h-k''N,k'}(A)\big)&=\;\boldsymbol{r}_{h,k}(A)-k''r;\\ u\big(\boldsymbol{r}_{j-k''N,k'}(B)\big)&=\boldsymbol{r}_{j',k-j''N}(B)+(j''-k'')r=\big(\widetilde{\lambda}_{j'-1,j''-k''}+\sum_{t<k}a_{j,t}+x\big)_{2\leqslant x\leqslant a_{j,k}}\\ u^*\!\big(\boldsymbol{r}_{\!j-k''N\!,k'}(\!A)\!\big) &=\!\Big(\!\!\big(\widetilde{\lambda}_{j'-1,j''-k''}+\!\sum_{t<k}\!a_{j,t}+x\big)_{\!2\leqslant x\leqslant p},\blacktriangle, \big(\widetilde{\lambda}_{j'-1,j''-k''}+\!\!\sum_{t<k}\!a_{j,t}+x\!\big)_{\!p< x\leqslant a_{j,k}}\!\Big)\!, \end{aligned} \end{align*} $$

where $p=i-\widetilde {\lambda }_{j'-1}-\sum _{t<k}a_{j,t}$ . Thus, if we set $a_k=(uy_{B})^{-1}(\widetilde {\lambda }_{h}+1-k''r)$ and $b_{k,i}=(u^*y_A)^{-1}(\widetilde {\lambda }_{h}+1-k''r)$ , then (cf. (4.4.2))

(4.4.4) $$ \begin{align} a_k=\widetilde\mu_{k'-1}+\sum_{s\leqslant h}a_{s,k}+1,\quad b_{k,i}= \widetilde\mu_{k'-1}+\sum_{s<j}a_{s,k}+i-\widetilde{\lambda}_{j'-1}-\sum_{t< k}a_{j,t}. \end{align} $$

Thus, $uy_B$ and $u^*y_A$ has the form as in (4.4.3), respectively, with $\blacksquare $ replaced by $\blacktriangle $ . Defining $w=w_{k,i}$ as above, we obtain $u^*y_A=uy_Bw$ in this case. This proves Case 1.

Case 2. $j'=h$ . (So $j''>0$ and $u_{h,j}^{\lambda }$ is the second element in (4.2.1).) With $B=A+E^\vartriangle _{h,k}-E^\vartriangle _{j,k}$ , we have ${\lambda }^B=\text {ro}(B)={\lambda }$ . By taking $u^*=u_{h,j}^{\lambda } w_{h,i,j}$ and

$$ \begin{align*}u=u_{h,k}=\left(\begin{matrix} \widetilde{\lambda}_{h-1}+\sum_{s\leqslant k}a_{h,s}& \widetilde{\lambda}_{h-1}+\sum_{s\leqslant k}a_{h,s}+1 &\cdots & \widetilde{\lambda}_{h}\\ \widetilde{\lambda}_{h}&\widetilde{\lambda}_{h-1}+\sum_{s\leqslant k}a_{h,s}&\cdots& \widetilde{\lambda}_{h}-1\\ \end{matrix}\right)\in \mathfrak S_{{\lambda}},\end{align*} $$

we see that (4.4.1) continues to hold with $j'=h$ and $\ell (u)=\sum _{s> k}a_{h,s}$ . The remaining cases become

$$ \begin{align*}\begin{aligned} u\big(\boldsymbol{r}_{h-k''N,k'}(B)\!\big)&=\big(\rho^{-1}\boldsymbol{r}_{h,k}(A)-k''r,\blacksquare,\big);\\ u^*\!\big(\!\boldsymbol{r}_{h-k''N,k'}(\!A)\!\big)&=\;\rho^{-1}\boldsymbol{r}_{h,k}(A)-k''r;\\ u\big(\boldsymbol{r}_{j-k''N,k'}(B)\!\big)&=\boldsymbol{r}_{j',k-j''N}(B)+(j''-k'')r=\big(\widetilde{\lambda}_{j'-1,j''-k''}+\sum_{t<k}a_{j,t}+x\big)_{1\leqslant x\leqslant a_{j,k}-1}\\ u^*\!\big(\!\boldsymbol{r}_{\!j-k''N\!,k'}(\!A)\!\big) &=\!\Big(\!\!\big(\widetilde{\lambda}_{j'-1,j''-k''}+\!\!\sum_{t<k}\!a_{j,t}+x\!\big)_{\!\!1\leqslant x<p},\blacksquare,\! \big(\widetilde{\lambda}_{j'-1,j''-k''}\!+\!\!\sum_{t<k}\!a_{j,t}+x\!\big)_{\!\!p\leqslant x\leqslant a_{j,k}-1}\!\!\Big)\!, \end{aligned} \end{align*} $$

where $p=i-\widetilde {\lambda }_{j'-1}-\sum _{t<k}a_{j,t}$ . The rest of the argument with $a_k, b_{k,i}$ as in (4.4.4) (with $j'=h$ ) is similar to the $j'<h$ case.

Finally, the length formulas are clear for cycle permutations and can be easily derived from the discussion above.

Proof Proof of Theorem 4.1

We only prove the case for $h<j$ ; the case for $h>j$ is similar. By assumption, ${\lambda }=\text {ro}(A)\geqslant \boldsymbol e^\vartriangle _j$ . Let $\mu =\text {co}(A)$ , $y_A=d\in \mathscr D^{\!\vartriangle \!}_{{\lambda },\mu }$ , $\mathfrak S_\nu =d^{-1}\mathfrak S_{\lambda } d\cap \mathfrak S_\mu $ and $\mathfrak S_\beta =(u_{h,j}^{\lambda })^{-1}\mathfrak S_{{\lambda }+\boldsymbol e^\vartriangle _h-\boldsymbol e^\vartriangle _j} u_{h,j}^{{\lambda }}\cap {\mathfrak S}_{\lambda }$ . By Lemma 4.3(1),

. By (3.4.1) and Lemma 4.3(3),

Thus, putting $B^{(i)}=(b_{s,t}^{(i)})=A+E^\vartriangle _{h,k}-E^\vartriangle _{j,k}$ and

$$ \begin{align*}R^{\lambda}_{j'}[a_{j,k}]=\{\widetilde{\lambda}_{j'-1}+\sum_{t<k}a_{j,t}+x\mid x\!\in\![1, a_{j,k}]\},\end{align*} $$

Lemmas 4.3(3) and 4.4 imply

where

$$ \begin{align*}\Sigma_0=\sum_{ k\in\mathbb Z,\,\overline k=0 \atop a_{j,k}>0}\sum_{i\in R_{j'}^{\lambda}[a_{j,k}]}(-1)^{\ell(w_{h,i,j}^{(1)})+\ell(u_{h,k}^{(1)})+\ell(w_{k,i}^{(1)})} \frac{a_{h,k}+1}{a_{j,k}} \phi_{A+E^\vartriangle_{h,k}-E^\vartriangle_{j,k}}\end{align*} $$

and

$$ \begin{align*}\Sigma_1=\sum_{ k\in\mathbb Z,\,\overline k=1 \atop a_{j,k}>0}\sum_{i\in R_{j'}^{\lambda}[a_{j,k}]}(-1)^{\ell(w_{h,i,j}^{(1)})+\ell(u_{h,k}^{(1)})+\ell(w_{k,i}^{(1)})} \frac{a_{h,k}+1}{a_{j,k}} \phi_{A+E^\vartriangle_{h,k}-E^\vartriangle_{j,k}}.\end{align*} $$

It remains to prove that $\Sigma _{0}+\Sigma _{1}$ equals the right-hand side of (4.1.1).

Since $w_{h,i,j}^{(1)}=(w_{h,i,j})^{\delta _{\bar j,1}}$ , where $\widetilde {\lambda }_{j'-1}+1\leqslant i\leqslant \widetilde {\lambda }_{j'}$ , $u_{h,k}^{(1)}=(u_{h,k})^{\delta _{\overline h,1}}$ , $w_{k,i}^{(1)}=(w_{k,i})^{\delta _{\overline k,1}}$ , where $i\in R_{j'}^{\lambda }[a_{j,k}]$ , and $h<j$ , we have by Lemmas 4.3 and 4.4

(4.5.1) $$ \begin{align} \ell(w_{h,i,j}^{(1)})={\overline j(i-1-\widetilde{\lambda}_{j'-1})},\quad {\ell(u_{h,k}^{(1)})}= \begin{cases}{\overline{h}\sum_{s> k}a_{h,s}}, &\text{if}\ h\leqslant j';\\ {\overline{h}\sum_{s\geqslant k}a_{h,s}}, &\text{if}\ h>j', \end{cases} \end{align} $$

and

$$ \begin{align*} {\ell(w_{k,i}^{(1)})}= \begin{cases}{\overline k(\sum_{h<s<j}a_{s,k}+i-\widetilde{\lambda}_{j'-1}-\sum_{s< k}a_{j,s}-1)},&\text{if}\ j''\geqslant 0, h\leqslant j';\\ {\overline k(\sum_{h-1<s<j}a_{s,k}+i-\widetilde{\lambda}_{j'-1}-\sum_{s< k}a_{j,s}-1)},&\text{if}\ j''>0, h>j'. \end{cases} \end{align*} $$

If $\overline h+\overline k=1$ , then $A+E^\vartriangle _{h,k}-E^\vartriangle _{j,k}\in \Theta _{\!\vartriangle \!}(m|n,r)$ implies $a_{h,k}=0$ . It follows that the sum $\ell (u_{h,k}^{(1)})+\ell (w_{k,i}^{(1)})$ can be combined as

(4.5.2) $$ \begin{align} {\ell(u_{h,k}^{(1)})+\ell(w_{k,i}^{(1)})} ={\overline{h}\sum_{s> k}a_{h,s}+\overline k(\sum_{h<s<j}a_{s,k}+i-\widetilde{\lambda}_{j'-1}-\sum_{s< k}a_{j,s}-1)}. \end{align} $$

We now compute $\Sigma _0,\Sigma _1$ under the assumptions $\overline j=0$ and $\overline j=1$ . If $\overline j=0$ then, by (4.5.1) and (4.5.2)

$$ \begin{align*}\Sigma_0(\overline j=0)&=\sum_{k\in\mathbb Z,\,\overline k=0\atop a_{j,k}>0}(-1)^{\overline h\sum_{s>k}a_{h,s}} \bigg(\sum_{i\in R_{j'}^{\lambda}[a_{j,k}]}\frac{a_{h,k}+1}{a_{j,k}}\bigg) \phi_{A+E^\vartriangle_{h,k}-E^\vartriangle_{j,k}}\\ &=\sum_{k\in\mathbb Z,\,\overline k=0\atop a_{j,k}>0}(-1)^{\overline h\sum_{s>k}a_{h,s}+\overline k\sum_{h<s<j}a_{s,k}} (a_{h,k}+1) \phi_{A+E^\vartriangle_{h,k}-E^\vartriangle_{j,k}}. \end{align*} $$

Note that if $\overline j+\overline k=1$ , then $A+E^\vartriangle _{h,k}-E^\vartriangle _{j,k}\in \Theta _{\!\vartriangle \!}(m|n,r)$ implies $a_{j,k}=1$ . Hence, $i\in R_{j'}^{\lambda }[a_{j,k}]\iff i=\widetilde {\lambda }_{j'-1}+\sum _{t<k}a_{j,t}+1$ . Thus,

$$ \begin{align*}\Sigma_1(\overline j=0)= \sum_{k\in\mathbb Z,\,\overline k=1\atop a_{j,k}>0}(-1)^{\overline h\sum_{s>k}a_{h,s}+\overline k\sum_{h<s<j}a_{s,k}} (a_{h,k}+1) \phi_{A+E^\vartriangle_{h,k}-E^\vartriangle_{j,k}}.\end{align*} $$

Therefore, $\Sigma _0(\overline j=0)+\Sigma _1(\overline j=0) =\sum _{k\in \mathbb Z,a_{j,k}>0}(-1)^{g_{h,j,k}(A)}(a_{h,k}+1) \phi _{A+E^\vartriangle _{h,k}-E^\vartriangle _{j,k}}, $ as desired in this case.

Similarly, for $\overline k=0$ and $\overline j=1$ , we have $R^{\lambda }_{j'}[a_{j,k}]=\{\widetilde {\lambda }_{j'-1}+\sum _{t<k}a_{j,t}+1\}$ . Thus, by (4.5.1) and (4.5.2),

$$ \begin{align*}\Sigma_0(\overline j=1) =\sum_{k\in\mathbb Z,\,\overline k=0\atop a_{j,k}>0}(-1)^{\overline h\sum_{s>k}a_{h,s}+\overline j\sum_{s<k}a_{j,s}} (a_{h,k}+1) \phi_{A+E^\vartriangle_{h,k}-E^\vartriangle_{j,k}}.\end{align*} $$

For $\overline k=1$ and $\overline j=1$ , $(-1)^{\ell (w_{h,i,j}^{(1)})+\ell (u_{h,k}^{(1)})+\ell (w_{k,i}^{(1)})}=(-1)^{\overline {h}\sum _{s> k}a_{h,s}+ \sum _{h<s<j}a_{s,k}-\sum _{s< k}a_{j,s}}$ . Thus,

$$ \begin{align*} \Sigma_1(\overline j=1)&=\sum_{k\in\mathbb Z,\,\overline k=1\atop a_{j,k}>0}(-1)^{\overline{h}\sum_{s> k}a_{h,s}+ \sum_{h<s<j}a_{s,k}-\sum_{s< k}a_{j,s}} \sum_{i\in R_{j'}^{\lambda}[a_{j,k}]}\frac{a_{h,k}+1}{a_{j,k}} \phi_{A+E^\vartriangle_{h,k}-E^\vartriangle_{j,k}}\\ &=\sum_{k\in\mathbb Z,\,\overline k=1\atop a_{j,k}>0}(-1)^{\overline{h}\sum_{s> k}a_{h,s}+\overline k \sum_{h<s<j}a_{s,k}+\overline j\sum_{s< k}a_{j,s}} (a_{h,k}+1) \phi_{A+E^\vartriangle_{h,k}-E^\vartriangle_{j,k}}. \end{align*} $$

Hence, $\Sigma _0(\overline j=1)+\Sigma _1(\overline j=1)=\sum _{k\in \mathbb Z, a_{j,k}>0}(-1)^{g_{h,j,k}(A)}(a_{h,k}+1) \phi _{A+E^\vartriangle _{h,k}-E^\vartriangle _{j,k}},$ as desired. This completes the proof.

Proof. For $A\in \Theta _{\!\vartriangle \!}^\pm (m|n)$ , let $\Lambda _{m|n,r-\sigma (A)}=\{{\mathbf {j}}\in \mathbb N_{\!\vartriangle \!}^{m+n}\mid \,j_{i_0}=0,\,\sigma ({\mathbf {j}})\leqslant r-\sigma (A)\}$ . Then $\mathcal B_{i_0}=\bigcup _{A\in \Theta _{\!\vartriangle \!}^\pm (m|n)}{\mathcal B}_{i_0,A}$ , where ${\mathcal B}_{i_0,A}=\{A[{\mathbf {j}},r]\mid {\mathbf {j}}\in \Lambda _{m|n,r-\sigma (A)}\}$ . Then, by (5.0.2), $A[{\mathbf {j}},r]\in S(A):=\text {span}\{\widetilde \phi _{A+\operatorname {diag}({\lambda })}\mid {\lambda }\in \Lambda _{\!\vartriangle \!}(m|n,r-\sigma (A))\}$ for all ${\mathbf {j}}\in \Lambda _{m|n,r-\sigma (A)}$ . Since the coefficient matrix $({\lambda }^{{\mathbf {j}}})_{{\lambda }\in \Lambda _{\!\vartriangle \!}(m|n,r-\sigma (A)),{\mathbf {j}}\in \Lambda _{m|n,r-\sigma (A)}}$ from (5.0.2) is invertible (see [Reference Fu17, 4.2]), it follows that $\mathcal B_{i_0,A}$ forms a basis for $S(A)$ . Hence, $\mathcal B_{i_0}$ forms a basis for ${\widehat S}_{\mathbb Q}(m|n,r)$ .

Proof. We first prove Theorem 5.2(1). By definition, we have

$$ \begin{align*} 0[{\mathbf{j}},r] A[{\mathbf{j}}',r] &=\sum_{\beta\in\Lambda_{\!\vartriangle\!}(m|n,r-\sigma(A))} (\text{ro}(A)+\beta)^{\mathbf{j}}\beta^{{\mathbf{j}}'}\widetilde\phi_{A+\operatorname{diag}(\beta)}\\ &=\sum_{\beta\in\Lambda_{\!\vartriangle\!}(m|n,r-\sigma(A))}\bigg(\sum_{{\mathbf{j}}''\in\mathbb N_{\!\vartriangle\!}^{m+n}\atop {\mathbf{j}}''\leqslant{\mathbf{j}}} \left({{\mathbf{j}}\atop{\mathbf{j}}''}\right)\text{ro}(A)^{{\mathbf{j}}''}\beta^{{\mathbf{j}}-{\mathbf{j}}''}\bigg)\beta^{{\mathbf{j}}'}\widetilde\phi_{A+\operatorname{diag}(\beta)}\\ &=\sum_{{\mathbf{j}}''\in\mathbb N_{\!\vartriangle\!}^{m+n}\atop {\mathbf{j}}''\leqslant{\mathbf{j}}} \left({{\mathbf{j}}\atop{\mathbf{j}}''}\right)\text{ro}(A)^{{\mathbf{j}}''} A[{\mathbf{j}}+{\mathbf{j}}'-{\mathbf{j}}'',r]. \end{align*} $$

as desired. The proof for Theorem 5.2(2) is similar.

We now prove Theorem 5.2(3). We abbreviate a diagonal matrix of the form $\operatorname {diag}({\lambda })$ as $({\lambda })$ . By the definitions (5.0.1) and (5.0.2), we have

$$ \begin{align*}E^\vartriangle_{h,h+1}[{\mathbf{0}},r]=\sum_{\alpha\in\Lambda_{\!\vartriangle\!}(m|n,r-1)}(-1)^{f_{h,\alpha}}\phi_{E^\vartriangle_{h,h+1}+ (\alpha)},\end{align*} $$

where $f_{h,\alpha }=f_{E^\vartriangle _{h,h+1}+ (\alpha )}=0$ if $h<m+n$ and $f_{{{\scriptstyle \mathrm {N}}},\alpha }= \sum _{s=1}^{n}\alpha _{m+s}=|\alpha ^{(1)}|$ . (Recall ${{\scriptstyle \mathrm {N}}}=m+n$ .)

For any ${\lambda }\in \Lambda _{\!\vartriangle \!}(m|n,r-\sigma (A))$ , note that $\phi _{E^\vartriangle _{h,h+1}+(\alpha )}\phi _{A+({\lambda })}\neq 0$ implies $\alpha ={\lambda }+\text {ro}(A)-\boldsymbol e^\vartriangle _{h+1}=:\alpha _{\lambda }$ , and, for $k\in \mathbb Z$ , let

(5.2.1) $$ \begin{align} f_{h,k,{\lambda},A}=f_{A+({\lambda})}+f_{h,\alpha_{\lambda}}-f_{A+E^\vartriangle_{h,k}-E^\vartriangle_{h+1,k}+({\lambda})} +g_{h,h+1,k}(A+({\lambda})). \end{align} $$

Then, by setting $A+({\lambda })=(a_{s,t}^{{\lambda }})_{s,t\in {\mathbb Z}}$ , and applying Theorem 4.1, we obtain

$$ \begin{align*} \begin{aligned} E^{\vartriangle}_{h,h+1}[{\mathbf{0}},r]A[{\mathbf{j}},r] &=\sum_{{\lambda}\in\Lambda_{\!\vartriangle\!}(m|n,r-\sigma(A))} (-1)^{f_{A+({\lambda})}+f_{h,\alpha_{\lambda}}}{\lambda}^{{\mathbf{j}}} \phi_{E^{\vartriangle}_{h,h+1}+(\alpha_{\lambda})}\phi_{A+({\lambda})}\\ &=\sum_{{\lambda}\in\Lambda_{\!\vartriangle\!}(m|n,r-\sigma(A))} (-1)^{f_{A+({\lambda})}+f_{h,\alpha_{\lambda}}}{\lambda}^{{\mathbf{j}}} \\ &\quad \times \sum_{k\in{\mathbb Z},a_{h+1,k}^{\lambda}>0}(-1)^{g_{h,h+1,k}(A+({\lambda}))} (a_{h,k}^{{\lambda}}+1) \phi_{A+E^{\vartriangle}_{h,k}-E^{\vartriangle}_{h+1,k}+({\lambda})}\\ &=\sum_{{\lambda}\in\Lambda_{\!\vartriangle\!}(m|n,r-\sigma(A))\atop k\in{\mathbb Z},a_{h+1,k}^{\lambda}>0} (-1)^{f_{h,k,{\lambda},A}} (a_{h,k}^{{\lambda}}+1) {\lambda}^{{\mathbf{j}}}\widetilde{\phi}_{A+E^{\vartriangle}_{h,k}-E^{\vartriangle}_{h+1,k}+({\lambda})}. \end{aligned}\end{align*} $$

However, we have the following parity relation with the numbers $\gamma $ defined in (5.1.1) and (5.1.2). This relation is proved in Appendix A.

(5.2.2) $$ \begin{align} {f_{h,k,{\lambda},A}}\equiv{\gamma_{h,h+1,k}(A)}\;\;(\text{{mod}}\;2)\;\;\text{ for all }{\lambda}\in\Lambda_{\!\vartriangle\!}(m|n,r-\sigma(A)). \end{align} $$

Thus, $(-1)^{f_{h,k,{\lambda },A}}=(-1)^{\gamma _{h,h+1,k}(A)}$ is independent of ${\lambda }$ . Hence,

(5.2.3) $$ \begin{align} E^\vartriangle_{h,h+1}[{\mathbf{0}},r]A[{\mathbf{j}},r] &=\sum_{k\in\mathbb Z,a_{h+1,k}^{\lambda}>0} (-1)^{\gamma_{h,h+1,k}(A)} (a_{h,k}^{{\lambda}}+1) \nonumber \\ &\quad \times \sum_{{\lambda}\in\Lambda_{\!\vartriangle\!}(m|n,r-\sigma(A))} {\lambda}^{{\mathbf{j}}}\widetilde\phi_{A+E^\vartriangle_{h,k}-E^\vartriangle_{h+1,k}+({\lambda})}. \end{align} $$

Let

$$ \begin{align*}\Sigma_k:=\sum_{{\lambda}\in\Lambda_{\!\vartriangle\!}(m|n,r-\sigma(A))} {\lambda}^{{\mathbf{j}}}\widetilde\phi_{A+E^\vartriangle_{h,k}-E^\vartriangle_{h+1,k}+({\lambda})}.\end{align*} $$

If $k\neq h,h+1$ (and $a_{h+1,k}>0$ ), we clearly have $\sigma (A+E^\vartriangle _{h,k}-E^\vartriangle _{h+1,k}+({\lambda }))=\sigma (A+({\lambda }))$ and so, by definition, $\Sigma _k=(A+E^\vartriangle _{h,k}-E^\vartriangle _{h+1,k})[{\mathbf {j}},r].$

If $k=h$ and $a_{h+1,h}>0$ , then $a_{h,h}^{\lambda }={\lambda }_h$ and

$$ \begin{align*} ({\lambda}_h+1)\Sigma_{h}&=\sum_{{\lambda}\in \Lambda_{\!\vartriangle\!}(m|n,r-\sigma(A))}({\lambda}_h+1)[({\lambda}_h+1)-1]^{j_h}\prod_{s\not= h}{\lambda}_s^{j_s}\cdot \widetilde\phi_{A-E^\vartriangle_{h+1,h}+({\lambda}+\boldsymbol e^\vartriangle_h)}\\ &=\sum_{{\lambda}\in \Lambda_{\!\vartriangle\!}(m|n,r-\sigma(A))\atop 0\leqslant i\leqslant j_h}(-1)^i\left({j_h\atop i}\right)({\lambda}+\boldsymbol e^\vartriangle_h)^{{\mathbf{j}}+(1-i)\boldsymbol e^\vartriangle_h} \cdot \widetilde\phi_{A-E^\vartriangle_{h+1,h}+({\lambda}+\boldsymbol e^\vartriangle_h)}\\ &=\sum_{0\leqslant i\leqslant j_h}(-1)^i\left({j_h\atop i}\right) (A-E^\vartriangle_{h+1,h})[{\mathbf{j}}+(1-i)\boldsymbol e^\vartriangle_h,r]. \end{align*} $$

Finally, if $k=h+1$ , then $a^{\lambda }_{h+1,h+1}={\lambda }_{h+1}>0$ and the map ${\lambda }\mapsto {\lambda }-\boldsymbol e^\vartriangle _{h+1}$ is a bijection from $\{{\lambda }\in \Lambda _{\!\vartriangle \!}(m|n,r-\sigma (A))\mid {\lambda }_{h+1}>0\}$ to $\Lambda _{\!\vartriangle \!}(m|n,r-\sigma (A+E^\vartriangle _{h,h+1}))$ . Hence,

$$ \begin{align*} \Sigma_{h+1}&= \sum_{{\lambda}\in \Lambda_{\!\vartriangle\!}(m|n,r-\sigma(A))\atop{\lambda}_{h+1}>0} [({\lambda}_{h+1}-1)+1]^{j_{h+1}}\prod_{s\not= h+1}{\lambda}_s^{j_s} \widetilde\phi_{A+E^\vartriangle_{h,h+1}+({\lambda}-\boldsymbol e^\vartriangle_{h+1})}\\ &=\sum_{0\leqslant i\leqslant j_{h+1}}\left({j_{h+1}\atop i}\right) (A+E^\vartriangle_{h,h+1})[{\mathbf{j}}-i\boldsymbol e^\vartriangle_{h+1},r]. \end{align*} $$

Substituting all three cases into (5.2.3) gives Theorem 5.2(3). Theorem 5.2(4) can be proved similarly.

Finally, we prove Theorem 5.2(5). Again we put ${{\scriptstyle \mathrm {N}}}=m+n$ . By (5.0.1), $f_{E^\vartriangle _{h,h+l{{\scriptstyle \mathrm {N}}}}+(\alpha )}=\overline h\,\alpha _h$ . Thus,

$$ \begin{align*}E^\vartriangle_{h,h+l{{\scriptscriptstyle \mathrm{N}}}}[{\mathbf{0}},r]=\sum_{\alpha\in\Lambda_{\!\vartriangle\!}(m|n,r-1)}(-1)^{\overline h\alpha_h} \phi_{E^\vartriangle_{h,h+l{{\scriptscriptstyle \mathrm{N}}}}+(\alpha)}\end{align*} $$

and, for ${\lambda }\in \Lambda _{\!\vartriangle \!}(m|n,r-\sigma (A))$ , $\phi _{E^\vartriangle _{h,h+l{{\scriptstyle \mathrm {N}}}}+(\alpha )}\phi _{A+({\lambda })}\neq 0$ implies $\alpha ={\lambda }+\text {ro}(A)-\boldsymbol e^\vartriangle _{h}$ and so $\alpha _h={\lambda }_{h}-1+\sum _{s\in \mathbb Z}a_{h,s}$ . Put $A+({\lambda })=(a_{s,t}^{{\lambda }})_{s,t\in \mathbb Z}$ and, for $k\in \mathbb Z$ , let

(5.2.4) $$ \begin{align} \begin{aligned} f_{h,l,k,{\lambda},A}&=f_{A+({\lambda})}-f_{A+E^\vartriangle_{h,k}-E^\vartriangle_{h+l{{\scriptscriptstyle \mathrm{N}}},k}+({\lambda})} +g_{h,h+l{{\scriptscriptstyle \mathrm{N}}},k}(A+({\lambda}))+\overline h({\lambda}_{h}-1+\sum_{s\in\mathbb Z}a_{h,s}). \end{aligned} \end{align} $$

Like the previous case, this number has the same parity as the number $\gamma $ defined in (5.1.3):

(5.2.5) $$ \begin{align} f_{h,l,k,{\lambda},A}\equiv \gamma_{h,h+l{{\scriptscriptstyle \mathrm{N}}},k}(A)\;\;(\text{mod}\;2)\;\;\text{ for all } {\lambda}\in\Lambda_{\!\vartriangle\!}(m|n,r-\sigma(A)). \end{align} $$

(See Appendix A for a proof.) Hence, by Theorem 4.1 and (5.2.4),

$$ \begin{align*} E^\vartriangle_{h,h+l{{\scriptscriptstyle \mathrm{N}}}}&[{\mathbf{0}},r]A[{\mathbf{j}},r]=\sum_{{\lambda}\in\Lambda_{\!\vartriangle\!}(m|n,r-\sigma(A))\atop k\in\mathbb Z,a_{h+l{{\scriptscriptstyle \mathrm{N}}},k}>0} (-1)^{f_{h,l,k,{\lambda},A}} (a_{h,k}^{{\lambda}}+1) {\lambda}^{{\mathbf{j}}}\widetilde\phi_{A+E^\vartriangle_{h,k}-E^\vartriangle_{h+l{{\scriptscriptstyle \mathrm{N}}},k}+({\lambda})}\\ &=\sum_{ k\in\mathbb Z\atop a_{h+l{{\scriptscriptstyle \mathrm{N}}},k}>0} (-1)^{\gamma_{h,h+l{{\scriptscriptstyle \mathrm{N}}},k}(A)} (a_{h,k}^{{\lambda}}+1)\sum_{{\lambda}\in\Lambda_{\!\vartriangle\!}(m|n,r-\sigma(A))} {\lambda}^{{\mathbf{j}}}\widetilde\phi_{A+E^\vartriangle_{h,k}-E^\vartriangle_{h+l{{\scriptscriptstyle \mathrm{N}}},k}+({\lambda})} \end{align*} $$

Now, the rest of the argument is parallel (and similar) to the previous argument.

Proof. Clearly, $\tau _r$ is a linear isomorphism. Thus, it is enough to prove that $\tau _r$ is a ${\widehat {{\mathfrak S}}_{r}}$ -module homomorphism. This follows easily from the fact that, for $u\in {\mathfrak S}_{\lambda }$ , $\omega _{{\mathbf {i}}_{\lambda }}u=(-1)^{\ell (u^{(1)})}\omega _{{\mathbf {i}}_{\lambda }}$ (see (6.0.2) and (3.0.7)).

Proof. We first observe that combining (4.2.1) and Lemma 4.3(3) gives

$$ \begin{align*} u_{h,j}^{\lambda} w_{h,t,j}= \begin{cases} \left(\begin{matrix} \widetilde{\lambda}_{h}+1&\widetilde{\lambda}_{h}+2&\cdots &t-1 &t\\ \widetilde{\lambda}_{h}+2&\widetilde{\lambda}_{h}+3&\cdots &t &\widetilde{\lambda}_{h}-lr+1 \end{matrix}\right)\begin{matrix}=e_{\widetilde{\lambda}_{h}+1}^{-l}s_{\widetilde{\lambda}_{h}+1}s_{\widetilde{\lambda}_{h}+2}\cdots s_{{t-1}},\\ \hspace{2.4cm}\text{ if}\ h<a, l\geqslant 0;\end{matrix}&\\ \left(\begin{matrix} t&t+1&t+2 &\cdots &\widetilde{\lambda}_{h}\\ \widetilde{\lambda}_{h}-lr&t&t+1& \cdots &\widetilde{\lambda}_{h}-1 \end{matrix}\right) \begin{matrix}=e_{\widetilde{\lambda}_{h}}^{-l}s_{\widetilde{\lambda}_{h}-1}s_{\widetilde{\lambda}_{h}-2}\cdots s_{t}\qquad\qquad\\ \hspace{3.2cm}\text{ if}\ h\geqslant a, l>0;\end{matrix}&\\ \left(\begin{matrix} \widetilde{\lambda}_{h-1}+1&\cdots &t-1 &t\\ \widetilde{\lambda}_{h-1}+2&\cdots &t &\widetilde{\lambda}_{h-1}-lr+1 \end{matrix}\right) \begin{matrix}=e_{\widetilde{\lambda}_{h-1}+1}^{-l}s_{\widetilde{\lambda}_{h-1}+1}s_{\widetilde{\lambda}_{h-1}+2}\cdots s_{{t-1}},\\ \hspace{2.5cm}\text{ if}\ h\leqslant a, l< 0;\end{matrix} &\\ \left(\begin{matrix} t&t+1&t+2&\cdots &\widetilde{\lambda}_{h-1}\\ \widetilde{\lambda}_{h-1}-lr&t&t+1&\cdots &\widetilde{\lambda}_{h-1}-1 \end{matrix}\right)\begin{matrix}=e_{\widetilde{\lambda}_{h-1}}^{-l}s_{\widetilde{\lambda}_{h-1}-1}s_{\widetilde{\lambda}_{h-1}-2}\cdots s_{{t}},\\ \hspace{2.4cm}\text{ if}\ h>a, l\leqslant0.\end{matrix} &\\ \end{cases} \end{align*} $$

By Lemma 4.3 and definition (3.7.1), we have

(6.3.1) $$ \begin{align} \phi_{E^\vartriangle_{h,j}+\operatorname{diag}(\alpha)}(z_{\lambda}) =\sum_{\widetilde{\lambda}_{a-1}+1\leqslant t\leqslant\widetilde{\lambda}_{a}}(-1)^{\ell(w_{h,t,j}^{(1)})}z_\mu u_{h,j}^{\lambda} w_{h,t,j}, \end{align} $$

where $\mu ={\lambda }+\boldsymbol e^\vartriangle _h-\boldsymbol e^\vartriangle _j$ . Applying $\tau _r$ to (6.3.1) reduces to compute $\omega _{{\mathbf {i}}_\mu }u_{h,j}^{\lambda } w_{h,t,j}$ for every $\widetilde {\lambda }_{a-1}+1\leqslant t\leqslant \widetilde {\lambda }_{a}$ . Thus, a case-by-case argument shows that ${{\mathbf {i}}_\mu u_{h,j}^{\lambda } w_{h,t,j}}={\mathbf {i}}_{\lambda }-(i_t-h+l(m+n))\boldsymbol e^\vartriangle _t={\mathbf {i}}_{\lambda }+(h-j)\boldsymbol e^\vartriangle _t$ (noting (4.3.1)). Hence, by (6.0.2),

$$ \begin{align*}\omega_{{\mathbf{i}}_\mu}u_{h,j}^{\lambda} w_{h,t,j} =(-1)^{\chi_{h,j,t,{\lambda}}} \omega_{i_1}\cdots\omega_{i_{t-1}}\omega_{h-l(m+n)}\omega_{i_{t+1}} \cdots\omega_{i_r}, \end{align*} $$

where, by writing ${\mathbf {i}}_\mu $ as ${\mathbf {i}}_\mu =(j_k)_{k\in \mathbb Z}$ ,

$$ \begin{align*} \chi_{h,j,t,{\lambda}}= \begin{cases} \overline{j}_{\widetilde{\lambda}_h+1}\cdot\sum_{\widetilde{\lambda}_h+1<s\leqslant t}\overline{j_s}, &\text{if}\ h<a\ \text{and}\ l\geqslant 0\text{;}\\ \overline{j}_{\widetilde{\lambda}_h}\cdot\sum_{t\leqslant s<\widetilde{\lambda}_h}\overline{j_s},&\text{if}\ h\geqslant a\ \text{and}\ l>0\text{;}\\ \overline{j}_{\widetilde{\lambda}_{h-1}+1}\cdot\sum_{\widetilde{\lambda}_{h-1}+1<s\leqslant t}\overline{j_s},&\text{if}\ h\leqslant a\ \text{and}\ l<0\text{;}\\ \overline{j}_{\widetilde{\lambda}_{h-1}}\cdot\sum_{t\leqslant s<\widetilde{\lambda}_{h-1}}\overline{j_s},&\text{if}\ h>a\ \text{and}\ l\leqslant 0\text{.}\\ \end{cases} \end{align*} $$

Note that the first factor in all four cases is $\overline h$ . Since $\widetilde {\lambda }_{a-1}+1\leqslant t\leqslant \widetilde {\lambda }_{a}$ , we may rewrite

$$ \begin{align*} \chi_{h,j,t,{\lambda}}= \begin{cases} \overline{h}\cdot(\sum_{h<s<a}{\lambda}_s\overline{s}+(t-\widetilde{\lambda}_{a-1}-1)\overline{j}), &\text{if}\ h<a, l\geqslant 0\text{;}\\ \overline{h}\cdot(\sum_{a< s\leqslant h}{\lambda}_s\overline{s}+(\widetilde{\lambda}_{a}-t)\overline{j}),&\text{if}\ h\geqslant a, l>0\text{;}\\ \overline{h}\cdot(\sum_{h\leqslant s<a}{\lambda}_s\overline{s}+(t-\widetilde{\lambda}_{a-1}-1)\overline{j}),&\text{if}\ h\leqslant a, l<0\text{;}\\ \overline{h}\cdot(\sum_{a< s< h}{\lambda}_s\overline{s}+(\widetilde{\lambda}_{a}-t)\overline{j}),&\text{if}\ h>a, l\leqslant 0\text{.}\\ \end{cases} \end{align*} $$

On the other hand, by Lemma 4.3(3), we have

$$ \begin{align*} {\ell(w_{h,t,j}^{(1)})}={\overline{j}\cdot\ell(w_{h,t,j})}= \begin{cases} \overline{j}\cdot (t-\widetilde{\lambda}_{a-1}-1))\;&\text{if}\ l> 0, \text{or}\ l=0, h<a;\\ \overline{j}\cdot(\widetilde{\lambda}_a-t)&\text{if}\ l<0, \text{or}\ l=0, h>a, \end{cases} \end{align*} $$

Hence, by the relation $\alpha ={\lambda }-\boldsymbol e^\vartriangle _j$ , one checks easily that ${f_{h,j,t,{\lambda }}+\ell (w_{h,t,j}^{(1)})}\equiv {\chi _{h,j,t,{\lambda }}}(\text {mod}\,2).$ Consequently, $\omega _{{\mathbf {i}}_\mu }u_{h,j}^{\lambda } w_{h,t,j} =(-1)^{f_{h,j,t,{\lambda }}+\ell (w_{h,t,j}^{(1)})} \omega _{i_1}\cdots \omega _{i_{t-1}}\omega _{h-l(m+n)}\omega _{i_{t+1}} \cdots \omega _{i_r} $ , and the required formula follows.

Proof. The $h=j$ case is clear since, for ${\lambda }\in \Lambda _{\!\vartriangle \!}(m|n,r)$ , we have $\zeta _r(\mathsf E^\vartriangle _{h,h})(\omega _{{\mathbf {i}}_{\lambda }})={\lambda }_h\omega _{{\mathbf {i}}_{\lambda }}={\lambda }^{\boldsymbol e^\vartriangle _h}\phi _{\operatorname {diag}({\lambda })}(\omega _{{\mathbf {i}}_{\lambda }})$ . Thus, $\zeta _r(\mathsf E^\vartriangle _{h,h})=\sum _{{\lambda }\in \Lambda _{\!\vartriangle \!}(m|n,r)}{\lambda }^{\boldsymbol e^\vartriangle _h}\phi _{\operatorname {diag}({\lambda })}=0[\boldsymbol e^\vartriangle _h,r]$ .

Assume now that $h\neq j$ and write $j=a+l(m+n)\in \mathbb Z$ as in (1.0.2). Recall from (5.0.2)

$$ \begin{align*}E^\vartriangle_{h,j}[{\mathbf{0}},r]=\sum_{\alpha\in\Lambda_{\!\vartriangle\!}(m|n,r-1)} (-1)^{f_{E^\vartriangle_{h,j}+\operatorname{diag}(\alpha)}}\phi_{E^\vartriangle_{h,j}+\operatorname{diag}(\alpha)},\end{align*} $$

where, by (5.0.1),

$$ \begin{align*}f_{E^\vartriangle_{h,j}+\operatorname{diag}(\alpha)}= \begin{cases} \sum_{h<s<a,\, s>m} \alpha_s=\sum_{h<s<a}\overline s\alpha_s, &\text{if}\ h<a, l\geqslant 0\\ \sum_{a\leqslant s\leqslant h,\,s>m} \alpha_s=\sum_{a\leqslant s\leqslant h}\overline s\alpha_s, &\text{if}\ h\geqslant a, l>0\\ \sum_{h\leqslant s\leqslant a,\,s>m} \alpha_s=\sum_{h\leqslant s\leqslant a}\overline s\alpha_s, &\text{if}\ h\leqslant a, l<0.\\ \sum_{a<s<h,\,s>m} \alpha_s=\sum_{a<s<h}\overline s\alpha_s, &\text{if}\ h>a, l\leqslant 0\\ \end{cases} \end{align*} $$

If we put

$$ \begin{align*}g_{h,j,\alpha}= \begin{cases} \overline h\sum_{h<s<a}\overline s\alpha_s+(\overline h+\overline j)\sum_{1\leqslant s<a}\overline s\alpha_s &\text{if}\ h<a, l\geqslant 0\\ \overline h\sum_{a\leqslant s\leqslant h}\overline s\alpha_s+(\overline h+\overline j)\sum_{1\leqslant s<a}\overline s\alpha_s&\text{if}\ h\geqslant a, l>0\\ \overline h\sum_{h\leqslant s\leqslant a}\overline s\alpha_s+ (\overline h+\overline j)\sum_{1\leqslant s<a}\overline s\alpha_s +\alpha_a(\overline j+\overline h\overline j)&\text{if}\ h\leqslant a, l<0,\\ \overline h\sum_{a<s<h}\overline s\alpha_s+ (\overline h+\overline j)\sum_{1\leqslant s<a}\overline s\alpha_s+\alpha_a(\overline j+\overline h\overline j)&\text{if}\ h>a, l\leqslant 0\\ \end{cases} \end{align*} $$

then ${g_{h,j,\alpha }}\equiv f_{E^\vartriangle _{h,j}+\operatorname {diag}(\alpha )}\;(\text {mod}\, 2)$ .

On the other hand, for ${\lambda }\in \Lambda _{\!\vartriangle \!}(m|n,r)$ with ${\lambda }\geqslant \boldsymbol e^\vartriangle _j$ , we have, by (2.1.2) and (2.1.1),

$$ \begin{align*} \mathsf E^\vartriangle_{h,j}.\omega_{{\mathbf{i}}_{\lambda}}=\sum_{\widetilde{\lambda}_{a-1}+1\leqslant t\leqslant\widetilde{\lambda}_{a}}(-1)^{f_{h,j,t,{\lambda}}'} \omega_{i_1} \cdots \omega_{i_{t-1}} \omega_{h-l(m+n)} \omega_{i_{t+1}} \cdots \omega_{i_r}, \end{align*} $$

where $f_{h,j,t,{\lambda }}'=(\overline h+\overline j) (\sum _{1\leqslant s\leqslant a-1}{\lambda }_s\overline s+(t-1-\widetilde {\lambda }_{a-1})\overline j)$ (cf. $f_{h,j,t,{\lambda }}$ in Lemma 6.3).

Finally, it is easy to check that ${f_{h,j,t,{\lambda }}'}\equiv {g_{h,j,\alpha }+f_{h,j,t,{\lambda }}}\;(\text {mod}\,2)$ for ${\lambda }\in \Lambda _{\!\vartriangle \!}(m|n,r)$ with ${\lambda }\geqslant \boldsymbol e^\vartriangle _j$ and $\alpha ={\lambda }-\boldsymbol e^\vartriangle _j$ . Thus, by Lemma 6.3, we have $\mathsf E^\vartriangle _{h,j}.\omega _{{\mathbf {i}}_{\lambda }}=(-1)^{f_{E^\vartriangle _{h,j}+\operatorname {diag}(\alpha )}} \tau _r(\phi _{E^\vartriangle _{h,j}+\operatorname {diag}(\alpha )} (z_{\lambda })).$ Consequently, the actions of $\zeta _r(\mathsf E^\vartriangle _{h,j})$ and $E^\vartriangle _{h,j}[0,r]$ on the tensor space are identical. Hence, they are equal.

Proof. By (4.1.1), we have

$$ \begin{align*} &E^\vartriangle_{h,k}[{\mathbf{0}},r]A[{\mathbf{0}},r] \\ & \quad = \sum_{\mu\in\Lambda_{\!\vartriangle\!}(m|n,r-\sigma(A)),\,{\lambda}\in\Lambda_{\!\vartriangle\!}(m|n,r-1) \atop{\lambda}+\boldsymbol e^\vartriangle_k=\mu+\text{ro}(A)}\mu^{\mathbf{j}}\widetilde\phi_{E^\vartriangle_{h,k}+\operatorname{diag}({\lambda})}\widetilde\phi_{A+\operatorname{diag}(\mu)}\\ & \quad =(a_{h,k}+1)\sum_{\mu\in\Lambda_{\!\vartriangle\!}(m|n,r-\sigma(A))\atop\mu_k\geqslant 1}(-1)^{s_{A,h,k,\mu}}(\mu-\boldsymbol e^\vartriangle_k+\boldsymbol e^\vartriangle_k)^{\mathbf{j}}\widetilde\phi_{A+E^\vartriangle_{h,k}+\operatorname{diag}(\mu-\boldsymbol e^\vartriangle_k)}+\Phi\\ & \quad =(a_{h,k}+1)\sum_{\mu\in\Lambda_{\!\vartriangle\!}(m|n,r-\sigma(A))\atop\mu_k\geqslant 1,0\leqslant t\leqslant j_k}(-1)^{s_{A,h,k,\mu}}{j_k\choose t}(\mu-\boldsymbol e^\vartriangle_k)^{{\mathbf{j}}-t\boldsymbol e^\vartriangle_k}\widetilde\phi_{A+E^\vartriangle_{h,k}+\operatorname{diag}(\mu-\boldsymbol e^\vartriangle_k)}+\Phi, \end{align*} $$

where $\Phi $ is a $\mathbb Z$ -linear combination of $\widetilde \phi _{B+\operatorname {diag}(\nu )}$ with $B\in \Theta _{\!\vartriangle \!}^\pm (m|n)$ and $\sigma (B)<\sigma (A+E^\vartriangle _{h,k})$ , and $s_{h,k,A,\mu }=g_{h,k,k}(A+\operatorname {diag}(\mu ))+f_{A+\operatorname {diag}(\mu )} +f_{E^\vartriangle _{h,k}+\operatorname {diag}(\mu +\text {ro}(A)-\boldsymbol e^\vartriangle _k)} -f_{A+E^\vartriangle _{h,k}+\operatorname {diag}(\mu -\boldsymbol e^\vartriangle _k)}.$

Clearly we have $g_{h,k,k}(A+\operatorname {diag}(\mu ))=g_{h,k,k}(A)$ . Furthermore, by definition, we have

$$ \begin{align*} f_{A+E^\vartriangle_{h,k}+\operatorname{diag}(\mu-\boldsymbol e^\vartriangle_k)} &= f_{A+\operatorname{diag}(\mu-\boldsymbol e^\vartriangle_k)}+f_{E^\vartriangle_{h,k}+\operatorname{diag}(\mu-\boldsymbol e^\vartriangle_k)}\\ &=f_{A+\operatorname{diag}(\mu)}+ f_{A-\operatorname{diag}(\boldsymbol e^\vartriangle_k)}+f_{E^\vartriangle_{h,k}+\operatorname{diag}(\mu-\boldsymbol e^\vartriangle_k)}\\ &=f_{A+\operatorname{diag}(\mu)}+ f_{A-\operatorname{diag}(\boldsymbol e^\vartriangle_k)}+ f_{E^\vartriangle_{h,k}+\operatorname{diag}(\mu+\text{ro}(A)-\boldsymbol e^\vartriangle_k)} -f_{E^\vartriangle_{h,k}+\operatorname{diag}(\text{ro}(A))}. \end{align*} $$

It follows that

(7.2.1) $$ \begin{align} s_{h,k,A}:=s_{h,k,A,\mu}=g_{h,k,k}(A)+f_{E^\vartriangle_{h,k}+ \operatorname{diag}(\text{ro}(A))}-f_{A-\operatorname{diag}(\boldsymbol e^\vartriangle_k)} \end{align} $$

is independent of $\mu $ (hence, of r), and the desired formula follows.

Proof. Since $E^\vartriangle _{i,j}[{\mathbf {0}}]$ is the homomorphic image of the root vector $\mathsf E^\vartriangle _{i,j}$ by Proposition 6.4, Lemma 2.1 implies that we may write $E^\vartriangle _{i,j}[{\mathbf {0}}]$ as a linear combination of monomials in $E^\vartriangle _{h,h\pm 1}[{\mathbf {0}}]$ and $E^\vartriangle _{h,h+l(m+n)}[{\mathbf {0}}]$ for $1\leqslant h\leqslant m+n$ and $l\not =0$ . Thus, by the multiplication formulas in Theorem 5.2, we see

$$ \begin{align*} E^\vartriangle_{h,k}[{\mathbf{0}}]A[{\mathbf{j}}]\in\operatorname{span}\{B[{\mathbf{j}}']\mid B\in\Theta_{\!\vartriangle\!}^\pm(m|n),\, {\mathbf{j}}'\in\mathbb N_{\!\vartriangle\!}^{m+n}\}. \end{align*} $$

Now the assertion follows from Lemma 7.2.

Proof. In a way similar to the proof of [Reference Deng, Du and Fu6, 6.3.4], we see that the set $\mathfrak B$ is linear independent.

Let $j_i=a_{i,i}$ and ${\mathbf {j}}=(j_i)$ . Then $\prod _{i=1}^{m+n}0(\boldsymbol e^\vartriangle _i)^{a_{i,i}}=0({\mathbf {j}})$ . Applying Corollary 7.4, Theorem 5.2(1) and repeatedly applying Proposition 7.3 yield

(7.5.1) $$ \begin{align} \begin{aligned} {\mathfrak m}^{(A)}&=\prod_{(i,j)\in{\mathcal L}^+} E^\vartriangle_{i,j}[{\mathbf{0}}]^{(a_{i,j})}0[{\mathbf{j}}]((-1)^{s(A^-)}A^-[{\mathbf{0}}]+\Psi)\\ &=\prod_{(i,j)\in{\mathcal L}^+} E^\vartriangle_{i,j}[{\mathbf{0}}]^{(a_{i,j})}((-1)^{s(A^-)}A^-[{\mathbf{j}}]+\sum_{{\mathbf{j}}'\in\mathbb N_{\!\vartriangle\!}^{n}\atop{\mathbf{j}}'<{\mathbf{j}}}c_{{\mathbf{j}}'}A^-[{\mathbf{j}}']+\Psi')\\ &=(-1)^{s(A)}A[{\mathbf{j}}]+\sum_{{\mathbf{j}}'\in\mathbb N_{\!\vartriangle\!}^{n}\atop{\mathbf{j}}'<{\mathbf{j}}}h_{A,{\mathbf{j}}}^{{\mathbf{j}}'} A[{\mathbf{j}}'] +\sum_{B\in\Theta_{\!\vartriangle\!}^\pm(n),\,{\mathbf{j}}'\in\mathbb N_{\!\vartriangle\!}^{n}\atop\sigma(B)<\sigma(A)} h_{A,{\mathbf{j}}}^{B,{\mathbf{j}}'}B[{\mathbf{j}}'], \end{aligned} \end{align} $$

for some $s(A)\in \mathbb Z$ , where $h_{A,{\mathbf {j}}}^{{\mathbf {j}}'},h_{A,{\mathbf {j}}}^{B,{\mathbf {j}}'}\in \mathbb Q$ . Here, $\Psi ,\Psi '$ is a linear combination of $B^-({\mathbf {j}}')$ with $\sigma (B^-)<\sigma (A^-)$ . Thus, by (1), ${\mathfrak M}$ forms a basis.

Finally, by Corollary 7.4, we have

$$ \begin{align*}{\mathfrak m}^{(A)}=((-1)^{s(A^+)}A^+[{\mathbf{0}}]+\Psi^+)0[{\mathbf{j}}]((-1)^{s(A^-)}A^-[{\mathbf{0}}]+\Psi^-),\end{align*} $$

where $\Psi ^+$ is a linear combination of $B[{\mathbf {0}}]$ for $B\in \Theta _{\!\vartriangle \!}^+(m|n)$ with $\sigma (B)<\sigma (A)$ and $\Psi ^-$ is a linear combination of $C[{\mathbf {0}}]$ for $C\in \Theta _{\!\vartriangle \!}^-(m|n)$ with $\sigma (C)<\sigma (A)$ . It follows that

$$ \begin{align*}{\mathfrak m}^{(A)}= (-1)^{s(A^+)+s(A^-)}A^+[{\mathbf{0}}]0[{\mathbf{j}}]A^-[{\mathbf{0}}]+\Psi,\end{align*} $$

where $\Psi $ is a linear combination of $B^+[{\mathbf {0}}]0[{\mathbf {j}}]B^-[{\mathbf {0}}]$ for $B\in \Theta _{\!\vartriangle \!}^\pm (m|n)$ with $\sigma (B)<\sigma (A^++A^-)$ . This implies that

$$ \begin{align*}A^+[{\mathbf{0}}]0[{\mathbf{j}}]A^-[{\mathbf{0}}]=(-1)^s{\mathfrak m}^{(A)}+\text{(lower terms)}.\end{align*} $$

Here the “lower terms” is a linear combination of ${\mathfrak m}^{(B+\operatorname {diag}({\mathbf {j}}))}$ for $B\in \Theta _{\!\vartriangle \!}^\pm (m|n)$ with $\sigma (B^++B^-)<\sigma (A^++A^-)$ . Hence, $\mathfrak L$ forms a basis.

Proof of Theorem 7.1

By [Reference Ross24, 2.1], the set $\mathsf U=\big \{{\mathsf u}^{(A)}\mid A\in \Theta _{\!\vartriangle \!}(m|n)\big \}$ forms a $\mathbb Q$ -basis for ${\mathcal U}_{\mathbb Q}(\widehat {\mathfrak {gl}}_{m|n})$ , a PBW basis. Since $\zeta ({\mathsf u}^{(A)})={\mathfrak m}^{(A)}$ , by Proposition 7.5(3), the set $\zeta (\mathsf U)$ forms a $\mathbb Q$ -basis for $\widehat {{\mathcal V}}_{\mathbb Q}(m|n)$ . Hence, $\zeta $ is injective and $\zeta ({\mathcal U}_{\mathbb Q}(\widehat {\mathfrak {gl}}_{m|n}))=\widehat {{\mathcal V}}_{\mathbb Q}(m|n)$ . The rest of the assertion is clear.

Proof. For $r\in \mathbb N$ , let $\pi _r:{\widehat S}_{\mathbb Q}(m|n)\rightarrow {\widehat S}_{\mathbb Q}(m|n,r)$ be the algebra homomorphism defined by $\pi _r((a_r)_{r\geqslant 0})=a_r$ . The restriction of $\pi _r$ to $\widehat {{\mathcal V}}_{\mathbb Q}(m|n)$ gives a superalgebra homomorphism $\pi _r:\widehat {{\mathcal V}}_{\mathbb Q}(m|n)\rightarrow {\widehat S}_{\mathbb Q}(m|n,r)$ . Clearly, we have the following commutative diagram:

Note that $\pi _r(A[{\mathbf {j}}])=A[{\mathbf {j}},r]$ for all $A,{\mathbf {j}}$ . Hence, by Theorem 7.1 and Lemma 5.1,

$$ \begin{align*}\zeta_r({\mathcal U}_{\mathbb Q}(\widehat{\mathfrak{gl}}_{m|n}))=\pi_r(\widehat{{\mathcal V}}_{\mathbb Q}(m|n))=\operatorname{span}\{A[{\mathbf{j}},r]\mid A\in\Theta_{\!\vartriangle\!}^\pm(m|n),\,{\mathbf{j}}\in\mathbb N_{\!\vartriangle\!}^{m+n}\}={\widehat S}_{\mathbb Q}(m|n,r),\end{align*} $$

as desired.

Proof. As $m+n\geqslant r$ , let $\omega =(\underbrace {1,\ldots 1}_{r},\underbrace {0,\ldots ,0}_{m+n-r})$ . Then $\omega \in \Lambda _{\!\vartriangle \!}(m|n,r)$ . Consider the idempotent $e_\omega =\phi _{\omega ,\omega }^1$ ; see (3.7.1). It is easy to see that . Thus, by [Reference Buchweitz3, Corollary 1.10], it suffices to prove that . This is clear since $\phi _{{\lambda },\omega }^1\phi _{\omega ,{\lambda }}^1=|{\mathfrak S}_{\lambda }|\phi _{{\lambda },{\lambda }}^1$ for all ${\lambda }\in \Lambda _{\!\vartriangle \!}(m|n,r)$ implies , and hence, .

Proof. Clearly, there is a

-

-bimodule homomorphism

Since the set $\{w\boldsymbol {e}_1^{a_1}\cdots \boldsymbol {e}_r^{a_r}\mid w\in {\mathfrak S}_r,\,a_i\in \mathbb Z,\,1\leqslant i\leqslant r\}$ forms a

-basis for

, it follows that the set

$$ \begin{align*}{\mathcal X}:=\{\omega_{\mathbf{i}}\otimes \boldsymbol{e}_1^{a_1}\cdots \boldsymbol{e}_r^{a_r}\mid{\mathbf{i}}\in I(m|n,r),\,a_i\in\mathbb Z,\,1\leqslant i\leqslant r\}\end{align*} $$

forms a

-basis for

, where $I(m|n,r)=\{{\mathbf {i}}=(i_1,i_2,\ldots ,i_r)\in \mathbb Z^r\mid 1\leqslant i_k\leqslant m+n,\text { for } 1\leqslant k\leqslant r\}$ . By (6.0.1), we may regard $I(m|n,r)$ as a subset of $I_{\!\vartriangle \!}(m|n,r),$ and we have $I_{\!\vartriangle \!}(m|n,r)=I(m|n,r)+(m+n)\mathbb Z^r$ . In addition, we have

$$ \begin{align*}\varphi(\omega_{\mathbf{i}}\otimes \boldsymbol{e}_1^{a_1}\cdots \boldsymbol{e}_r^{a_r}) =\omega_{\mathbf{i}} \boldsymbol{e}_1^{a_1}\cdots \boldsymbol{e}_r^{a_r}=\omega_{{\mathbf{i}}+(m+n)(a_1,a_2,\ldots,a_r)},\end{align*} $$

for all $(a_1,a_2,\ldots ,a_r)\in \mathbb Z^r$ and ${\mathbf {i}}\in I(m|n,r)$ . Thus, the set $\varphi ({\mathcal X})$ forms a

-basis for

. The lemma is proved.

Proof. Since the group algebra

has a cofinite dimensional center, every irreducible

-module is finite dimensional. Assume first $m+n\geqslant r$ . Then, by Lemma 6.1, we have

and

. Thus, for an irreducible

-module M, Lemma 8.6 implies that $\mathsf F(M)$ is isomorphic to

Hence, $\mathsf F(M)$ is finite dimensional. Now, the general case follows from a well-known Schur functor argument, see, for example, [Reference Deng, Du and Fu6, Theorem 4.1.6]. Let V be an irreducible

-supermodule. It is easy to see that each

-submodule of V is a

-subsupermodule. Therefore, V is irreducible as an

-module.

Proof. For any $a,c\in \mathbb Z$ , $t\in \mathbb N$ , since $h_{i,r}-a=\sum _{\mu }(\mu _i-a)\widetilde \phi _{\operatorname {diag}(\mu )}$ , it follows that

(9.1.1) $$ \begin{align} \bigg({h_{i,r}+c\atop t}\bigg)=\sum_{\mu\in\Lambda_{\!\vartriangle\!}(m|n,r)} \bigg({\mu_i+c\atop t}\bigg)\widetilde\phi_{\operatorname{diag}(\mu)}, \end{align} $$

Thus, noting $\widetilde \phi _{\operatorname {diag}(\mu )}=\phi _{\operatorname {diag}(\mu )}$ ,

$$ \begin{align*}\begin{aligned} \zeta_r\bigg({H\atop{\lambda}}\bigg)&=\prod_{i=1}^{m+n}\bigg({h_{i,r}\atop {\lambda}_i}\bigg)=\sum_{\mu\in\Lambda_{\!\vartriangle\!}(m|n,r)} \bigg(\prod_{i=1}^{m+n}\bigg({\mu_i\atop {\lambda}_i}\bigg)\bigg)\phi_{\operatorname{diag}(\mu)}=\sum_{\mu\in\Lambda_{\!\vartriangle\!}(m|n,r)}{\mu\choose{\lambda}}\phi_{\operatorname{diag}(\mu)}. \end{aligned} \end{align*} $$

The assertion now follows.

Proof. Assertion (1) is well known. For (2), we only prove that the ${\mathcal U}_{\mathbb Z}^+(\widehat {\mathfrak {gl}}_{m|n})$ case. We first claim that $\mathfrak {B}^+=\{A[{\mathbf {0}}]\mid A\in \Theta _{\!\vartriangle \!}^+(m|n)\}$ forms a $\mathbb Q$ -basis for ${\mathcal U}^+_{\mathbb Q}(\widehat {\mathfrak {gl}}_{m|n})$ . By the last assertion from Corollary 7.4, we have, for $A\in \Theta _{\!\vartriangle \!}^+(m|n)$ , the following triangular relation in ${\widehat S}_{\mathbb Q}(m|n):$

$$ \begin{align*}\zeta(\mathsf E^{(A)})=\prod_{(i,j)\in{\mathcal L}^+} E^\vartriangle_{i,j}[{\mathbf{0}}]^{(a_{i,j})}=(-1)^s A[{\mathbf{0}}]+\sum_{B\in\Theta_{\!\vartriangle\!}^+(m|n) \atop\sigma(B)<\sigma(A)}f_{B,A}B[{\mathbf{0}}] \end{align*} $$

for some $s\in \mathbb Z$ , where $f_{B,A}\in \mathbb Q$ . Thus, every $A[{\mathbf {0}}]$ is a linear combination of monomials in $\mathsf E^\vartriangle _{i,j}=E^\vartriangle _{ij}[{\mathbf {0}}]\in \mathfrak {B}^+$ . Thus, by Theorem 5.2, ${\mathcal U}^+_{\mathbb Q}(\widehat {\mathfrak {gl}}_{m|n})\subseteq \text {span}_{\mathbb Q}(\mathfrak {B}^+)\subseteq {\mathcal U}^+_{\mathbb Q}(\widehat {\mathfrak {gl}}_{m|n})$ , proving the claim. It remains to prove that, for $A,B\in \Theta _{\!\vartriangle \!}^+(m|n)$ , $A[{\mathbf {0}}]B[{\mathbf {0}}]\in {\mathcal U}_{\mathbb Z}^+(\widehat {\mathfrak {gl}}_{m|n})$ . By the claim, we may write

$$ \begin{align*}A[{\mathbf{0}}]B[{\mathbf{0}}]=\sum_{C\in\Theta_{\!\vartriangle\!}^+(m|n)}g_{A,B,C}C[{\mathbf{0}}],\end{align*} $$

where $g_{A,B,C}\in \mathbb Q$ . Let ${\mathcal I}=\{C\in \Theta _{\!\vartriangle \!}^+(m|n)\mid g_{A,B,C}\not =0\}.$ Since ${\mathcal I}$ is a finite set, there exists $r\in \mathbb N$ such that $r>\sigma (C)$ for all $C\in {\mathcal I}$ . Then we have

$$ \begin{align*}A[{\mathbf{0}},r]B[{\mathbf{0}},r]=\sum_{C\in{\mathcal I}}g_{A,B,C}C[{\mathbf{0}},r]= \sum_{C\in{\mathcal I}\atop{\lambda}\in\Lambda_{\!\vartriangle\!}(m|n,r-\sigma(C))}g_{A,B,C} \widetilde\phi_{C+\operatorname{diag}({\lambda})}.\end{align*} $$

Since ${\widehat S}_{\mathbb Z}(m|n,r)$ is a $\mathbb Z$ -subalgebra of ${\widehat S}_{\mathbb Q}(m|n,r)$ we have $A[{\mathbf {0}},r]B[{\mathbf {0}},r]\in {\widehat S}_{\mathbb Z}(m|n,r)$ . Furthermore, by Proposition 3.8, we know that the set $\{\widetilde \phi _X\mid X\in \Theta _{\!\vartriangle \!}(m|n,r)\}$ forms a $\mathbb Z$ -basis for ${\widehat S}_{\mathbb Z}(m|n,r)$ . Therefore, we have $g_{A,B,C}\in \mathbb Z$ for all C. Hence, ${\mathcal U}_{\mathbb Z}^+(\widehat {\mathfrak {gl}}_{m|n})$ is a $\mathbb Z$ -subalgebra of ${\mathcal U}_{\mathbb Q}(\widehat {\mathfrak {gl}}_{m|n})$ . Similarly, one can show that ${\mathcal U}_{\mathbb Z}^-(\widehat {\mathfrak {gl}}_{m|n})$ is a $\mathbb Z$ -subalgebra of ${\mathcal U}_{\mathbb Q}(\widehat {\mathfrak {gl}}_{m|n})$ . Finally, assertion (3) follows from the claim and (1).

Proof. (1) By (9.1.1),

$$ \begin{align*}\bigg({h_{i,r}\atop t}\bigg)A[{\mathbf{0}},r]=\sum_{\mu\in\Lambda_{\!\vartriangle\!}(m|n,r-\sigma(A))} \bigg({\mu_i+x_i\atop t}\bigg)\widetilde\phi_{A+\operatorname{diag}(\mu)}= A[{\mathbf{0}},r]\bigg({h_{i,r}+c\atop t}\bigg),\end{align*} $$

where $x_i=\sum _{s\in \mathbb Z}a_{i,s}$ . Assertion (1) follows.

(2) For $r\in \mathbb N$ , we have

$$ \begin{align*} \bigg({h_{i,r}+c\atop t}\bigg)&=\sum_{{\lambda}\in\Lambda_{\!\vartriangle\!}(m|n,r)} \bigg({{\lambda}_i+c\atop t}\bigg)\widetilde\phi_{\operatorname{diag}({\lambda})}\\ &=\sum_{{\lambda}\in\Lambda_{\!\vartriangle\!}(m|n,r)\atop 0\leqslant s\leqslant t} \bigg({c\atop t-s}\bigg)\bigg({{\lambda}_i\atop s}\bigg)\widetilde\phi_{\operatorname{diag}({\lambda})}=\sum_{0\leqslant s\leqslant t}\bigg({c\atop t-s}\bigg)\left({h_{i,r}\atop s}\right). \end{align*} $$

Assertion (2) follows.

Proof. Let ${\mathcal I}=\{(B,\mu )\mid B\in \Theta _{\!\vartriangle \!}^\pm (m|n),\,\mu \in \mathbb N_{\!\vartriangle \!}^{m+n},\, f_{B,\mu }\not \in \mathbb Z\}$ . Then ${\mathcal I}$ is a finite set. It is enough to prove ${\mathcal I}=\emptyset $ . Suppose this is not the case. Let

$$ \begin{align*}y=\sum_{(B,\mu)\in{\mathcal I}} f_{B,\mu}B^+[{\mathbf{0}}]B^-[{\mathbf{0}}] \bigg({H\atop\mu}\bigg).\end{align*} $$

Since, by Lemma 9.1, all $\zeta _r(B^+[{\mathbf {0}}]B^-[{\mathbf {0}}] \big ({H\atop \mu }\big ))\in {\widehat S}_{\mathbb Z}(m|n,r)$ , we have $\zeta _r(y)\in {\widehat S}_{\mathbb Z}(m|n,r)$ for all $r\in \mathbb N$ . Let ${\mathcal J}=\{B\in \Theta _{\!\vartriangle \!}^\pm (m|n)\mid (B,\mu )\in {\mathcal I}, \text { for some } \mu \in \mathbb N_{\!\vartriangle \!}^{m+n}\}$ . Let $m_0=\max \{\sigma (B)\mid B\in {\mathcal J}\}$ and fix $A_0\in {\mathcal J}$ such that $m_0=\sigma (A_0)$ . By Corollary 7.6, we have

$$ \begin{align*}B^+[{\mathbf{0}}]B^-[{\mathbf{0}}]=(-1)^{s_B}B[{\mathbf{0}}]+\sum_{C\in\Theta_{\!\vartriangle\!}^\pm(m|n),\,{\mathbf{j}}\in\mathbb N_{\!\vartriangle\!}^{m+n}\atop \sigma(C)<\sigma(B)} h_{B,C,{\mathbf{j}}}C[{\mathbf{j}}],\end{align*} $$

where $s_B\in \mathbb Z$ , $h_{B,C,{\mathbf {j}}}\in \mathbb Q$ . Thus, by Lemma 9.1, we have, for all $r\in \mathbb N$ ,

$$ \begin{align*} \zeta_r(y) &= \sum_{(B,\mu)\in{\mathcal I},\,\mu\leqslant{\lambda}\atop{\lambda}\in\Lambda_{\!\vartriangle\!}(m|n,r)} f_{B,\mu} \bigg({{\lambda}\atop\mu}\bigg)B^+[{\mathbf{0}},r]B^-[{\mathbf{0}},r]\phi_{\operatorname{diag}({\lambda})}\\ &=\sum_{(B,\mu)\in{\mathcal I},\,\mu\leqslant{\lambda}\atop{\lambda}\in\Lambda_{\!\vartriangle\!}(m|n,r)} f_{B,\mu} \bigg({{\lambda}\atop\mu}\bigg)\bigg((-1)^{s_B}B[{\mathbf{0}},r]+\sum_{C\in\Theta_{\!\vartriangle\!}^\pm(m|n),\,{\mathbf{j}}\in\mathbb N_{\!\vartriangle\!}^{m+n}\atop \sigma(C)<\sigma(B)} h_{B,C,{\mathbf{j}}}C[{\mathbf{j}},r]\bigg) \phi_{\operatorname{diag}({\lambda})}. \end{align*} $$

Since $B[{\mathbf {0}},r]\phi _{\operatorname {diag}({\lambda })}=\widetilde \phi _{B+\operatorname {diag}({\lambda }-\text {co}(B))}$ , $C[{\mathbf {j}},r]\phi _{\operatorname {diag}({\lambda })}=({\lambda }-\text {co}(C))^{{\mathbf {j}}}\widetilde \phi _{C+\operatorname {diag}({\lambda }-\text {co}(C))}$ , we have

$$ \begin{align*}\zeta_r(y)=\sum_{(A_0,\mu)\in{\mathcal I},\,\mu\leqslant{\lambda}\atop{\lambda}\in\Lambda_{\!\vartriangle\!}(m|n,r)} f_{A_0,\mu}\bigg({{\lambda}\atop\mu}\bigg)X_{A_0,{\lambda},\mu}+\sum_{(B,\mu)\in{\mathcal I},B\neq A_0\atop{\lambda}\in\Lambda_{\!\vartriangle\!}(m|n,r),\,\mu\leqslant{\lambda}} f_{B,\mu}\bigg({{\lambda}\atop\mu}\bigg)X_{B,{\lambda},\mu}\end{align*} $$

where

$$\begin{align*}X_{B,{\lambda },\mu }=(-1)^{s_B}\widetilde \phi _{B+\operatorname {diag}({\lambda }-\text {co}(B))} +\sum _{C\in \Theta _{\!\vartriangle \!}^\pm (m|n),\,{\mathbf {j}}\in \mathbb N_{\!\vartriangle \!}^{m+n}\atop \sigma (C)<\sigma (B)} h_{B,C,{\mathbf {j}}}({\lambda }-\text {co}(C))^{{\mathbf {j}}} \widetilde \phi _{C+\operatorname {diag}({\lambda }-\text {co}(C))}.\end{align*}$$

Note that the first summation has the leading term

$$ \begin{align*}\sum_{{\lambda}\in\Lambda_{\!\vartriangle\!}(m|n,r)} \bigg(\sum_{(A_0,\mu)\in{\mathcal I}\atop\mu\leqslant{\lambda}} f_{A_0,\mu} \bigg({{\lambda}\atop\mu}\bigg)(-1)^{s_{A_0}}\bigg)\widetilde\phi_{A_0+\operatorname{diag}({\lambda}-\text{co}(A_0))}.\end{align*} $$

Now $\zeta _r(y)\in {\widehat S}_{\mathbb Z}(m|n,r)$ together with Proposition 3.8 implies $\sum _{(A_0,\mu )\in {\mathcal I}\atop \mu \leqslant {\lambda }} f_{A_0,\mu } \big ({{\lambda }\atop \mu }\big )(-1)^{s_{A_0}}\in \mathbb Z$ , for all $r\in \mathbb N$ and any ${\lambda }\in \Lambda _{\!\vartriangle \!}(m|n,r)$ .

Since the set $\{\mu \in \mathbb N_{\!\vartriangle \!}^{m+n}\mid (A_0,\mu )\in {\mathcal I}\}$ is a finite set, we may choose a minimal element ${\lambda }$ in the set $\{\mu \in \mathbb N_{\!\vartriangle \!}^{m+n}\mid (A_0,\mu )\in {\mathcal I}\}$ with respect to the order $\leqslant $ defined in (3.0.4) and choose $r=|{\lambda }|$ . Then $\sum _{(A_0,\mu )\in {\mathcal I}\atop \mu \leqslant {\lambda }} f_{A_0,\mu } \big ({{\lambda }\atop \mu }\big )(-1)^{s_{A_0}}=(-1)^{s_{A_0}}f_{A_0,{\lambda }}\in \mathbb Z$ , a contradiction.

Proof of Theorem 9.2

Let $A,B\in \Theta _{\!\vartriangle \!}^\pm (m|n)$ , ${\lambda },\mu \in \mathbb N_{\!\vartriangle \!}^{m+n}$ . We have to show that

$$ \begin{align*} A^+[{\mathbf{0}}]\bigg({H\atop{\lambda}}\bigg)A^-[{\mathbf{0}}]B^+[{\mathbf{0}}]\bigg({H\atop\mu}\bigg)B^-[{\mathbf{0}}]\in{\mathcal U}_{\mathbb Z}(\widehat{\mathfrak{gl}}_{m|n}),\end{align*} $$

and equivalently, by Lemma 9.3 and Corollary 9.5, to show that $A^-[{\mathbf {0}}]B^+[{\mathbf {0}}]\in {\mathcal U}_{\mathbb Z}(\widehat {\mathfrak {gl}}_{m|n})$ . By Lemma 9.3(3) and 9.5, we may write $A^-[{\mathbf {0}}]B^+[{\mathbf {0}}]$ as a $\mathbb Q$ -linear combination of the elements $C^+[{\mathbf {0}}]C^-[{\mathbf {0}}] \big ({H\atop \nu }\big )$ for $C\in \Theta _{\!\vartriangle \!}^\pm (m|n)$ and $\nu \in \mathbb N_{\!\vartriangle \!}^{m+n}$ . Since $\zeta _r(A^-[{\mathbf {0}}]A^+[{\mathbf {0}}])\in {\widehat S}_{\mathbb Z}(m|n,r)$ for all $r\in \mathbb N$ , Lemma 9.6 implies that $A^-[{\mathbf {0}}]B^+[{\mathbf {0}}]\in {\mathcal U}_{\mathbb Z}^+(\widehat {\mathfrak {gl}}_{m|n}){\mathcal U}_{\mathbb Z}^-(\widehat {\mathfrak {gl}}_{m|n}){\mathcal U}_{\mathbb Z}^0(\widehat {\mathfrak {gl}}_{m|n})= {\mathcal U}_{\mathbb Z}(\widehat {\mathfrak {gl}}_{m|n})$ .

Proof. Since $\zeta_r({\mathcal U}_{\mathbb Z}(\widehat {\mathfrak {gl}}_{m|n}))=\text{span}(\zeta _r(\mathfrak {B}_{\mathbb Z}))$ , Lemma 9.1 implies

$$ \begin{align*}\zeta_r({\mathcal U}_{\mathbb Z}(\widehat{\mathfrak{gl}}_{m|n}))&=\operatorname{span}_{\mathbb Z}\{A^+[{\mathbf{0}},r] \widetilde\phi_{\operatorname{diag}({\lambda})}A^-[{\mathbf{0}},r] \mid A\in\Theta_{\!\vartriangle\!}^\pm(m|n),\,{\lambda}\in\Lambda_{\!\vartriangle\!}(m|n,r)\} \\ &\quad \subseteq{\widehat S}_{\mathbb Z}(m|n,r).\end{align*} $$

By Corollary 7.6, for $A\in \Theta _{\!\vartriangle \!}^\pm (m|n)$ and ${\lambda }\in \Lambda _{\!\vartriangle \!}(m|n,r)$ with ${\lambda }\geqslant \text {co}(A^+)+\text {ro}(A^-)$ , we have

$$ \begin{align*} A^+[{\mathbf{0}},r]\widetilde\phi_{\operatorname{diag}({\lambda})}A^-[{\mathbf{0}},r]& = A^+[{\mathbf{0}},r]A^-[{\mathbf{0}},r]\widetilde\phi_{\operatorname{diag}({\lambda}-\text{ro}(A^-)+\text{co}(A^-))}\\ &=(-1)^sA[{\mathbf{0}},r]\widetilde\phi_{\operatorname{diag}({\lambda}-\text{ro}(A^-)+\text{co}(A^-))}+\Phi\\ &=(-1)^s\widetilde\phi_{A+\operatorname{diag}({\lambda}-\text{co}(A^+)-\text{ro}(A^-))}+\Phi, \end{align*} $$

where $s\in \mathbb Z$ , $\Phi $ is a $\mathbb Z$ -linear combination of $\widetilde \phi _{B+\operatorname {diag}(\mu )}$ for $B\in \Theta _{\!\vartriangle \!}^\pm (m|n)$ and $\mu \in \Lambda _{\!\vartriangle \!}(m|n,r-\sigma (B))$ with $\sigma (B)<\sigma (A)$ . Therefore, by induction on $\sigma (A)$ , we conclude that $\widetilde \phi _{A+\operatorname {diag}(\mu )}\in \zeta _r({\mathcal U}_{\mathbb Z}(\widehat {\mathfrak {gl}}_{m|n}))$ for $A\in \Theta _{\!\vartriangle \!}^\pm (m|n)$ and $\mu \in \Lambda _{\!\vartriangle \!}(m|n,r-\sigma (A))$ . Hence, we have $\zeta _r({\mathcal U}_{\mathbb Z}(\widehat {\mathfrak {gl}}_{m|n}))={\widehat S}_{\mathbb Z}(m|n,r)$ .