Proof. It is clear that the map is a filtered homomorphism with respect to both filtrations. As the map $\mathcal {Y}_N(\sigma ) \to \mathcal {Y}_N$ is filtered with respect to the loop filtration it induces a map $\operatorname {gr}_L \mathcal {Y}_N(\sigma ) \to \operatorname {gr}_L \mathcal {Y}_N$ . We will write $\bar {h}_{i,2r+1}, \ {}^{\sigma }\bar {b}_{i,j}^{(r)}$ for the images of $h_{i,2r+1}, \ {}^{\sigma }b_{i,j}^{(r)}$ in $\operatorname {gr}_L \mathcal {Y}_N(\sigma )$ . It then follows from [Reference Lu, Wang and ZhangLWZ23, (5.3)] that the map on the associated graded sends ${}^{\sigma }\bar {b}_{i,j}^{(r)} \in \operatorname {gr}_L \mathcal {Y}_N(\sigma )$ to $\bar {b}_{i,j}^{(r)} \in \operatorname {gr}_L \mathcal {Y}_N$ as defined in [Reference Lu, Wang and ZhangLWZ23, §5.4] but from the proof of [Reference Lu, Wang and ZhangLWZ23, Theorem 5.1] we see that the ordered monomials in the elements $\bar {h}_{i,2r+1}, \ \bar {b}_{i,j}^{(r)}$ are linearly independent in $\operatorname {gr}_L \mathcal {Y}_n$ . In particular the ordered monomials in the elements $h_{i,2r+1}, {}^{\sigma }b_{i,j}^{(r)} \in \mathcal {Y}_N(\sigma ) $ are linearly independent too.

Following the proof of [Reference Lu, Wang and ZhangLWZ23, Theorem 5.1] verbatim it is easy to see that the ordered monomials in the elements $h_{i,2r+1}, {}^{\sigma }b_{j,k}^{(s)}$ for $0 \leq i < n, \ 1 \leq k<j < n, \ r \in \mathbb {N}, \ s \geq s_{j,k}$ span $\mathcal {Y}_n(\sigma )$ . Hence the map is injective.

Proof. We let ${}^{\sigma }\bar {b}_{i,j}^{(r)}$ denote ${}^{\sigma }{b}_{i,j}^{(r)}+F^{L}_{r-1}\mathcal {Y}_n(\sigma )$ . Following the proof of [Reference Lu, Wang and ZhangLWZ23, (5.11)] verbatim we obtain the relation

(3.13) $$ \begin{align} [{}^{\sigma}\bar{b}_{i,j}^{(r)},{}^{\sigma}\bar{b}_{k,l}^{(s)}]=\delta_{j,k}{}^{\sigma}\bar{b}_{i,l}^{(r+s)}-\delta_{l,i} {}^{\sigma}\bar{b}_{k,j}^{(r+s)}-(-1)^r\delta_{k,i}{}^{\sigma}\bar{b}_{j,l}^{(r+s)}+\delta_{l,j}(-1)^{(r)}{}^{\sigma}\bar{b}_{k,i}^{(r+s)} \end{align} $$

and hence the map in (1) is a homomorphism of algebras. It is easy to see that $\mathfrak {c}_n(\sigma )^{\tau }$ has a basis consisting of $e_{i,j}z^r+(-1)^{r-1}e_{j,i}z^r, \ i>j, \ r \geq s_{i,j}$ together with $e_{i,i}z^{2r+1}$ and hence the ordered monomials in these elements form a basis for $U(\mathfrak {c}_n(\sigma )^{\tau })$ . By Theorem 3.2 the ordered monomials in the elements ${}^{\sigma }b_{i,j}^{(r)}, \ i>j, \ r \geq s_{i,j}$ together with $b_{i,i}^{(2r+1)}$ form a basis for $\mathcal {Y}_n(\sigma )$ and hence our map is an isomorphism. For part (2) using the definition of the twisted Yangian as in [Reference Lu, Wang and ZhangLWZ23, Definition 2.1] we define a filtration by putting the generators $s_{i,j}^{(r)}$ in degree r. By [Reference MolevMo07, Theorem 2.4.3] we see that the associated graded algebra is commutative with bracket in degree $-1$ . Using [Reference Lu, Wang and ZhangLWZ23, (3.1), (3.12)-(3.13)] it follows that this filtration lines up with the canonical filtration on $\mathcal {Y}_n$ and hence $\operatorname {gr}_C \mathcal {Y}_n$ is commutative. The map defined in Theorem 3.2 is a filtered injection and hence also $\operatorname {gr}_C \mathcal {Y}_n(\sigma )$ is commutative. The current theorem then follows by part (1).

Proof. Let $\widehat {y}_n(\sigma )$ be the algebra with generators and relations as in the statement of the Proposition. From relations (3.2)–(3.6) it follows that there is a canonical surjection $\widehat {y}_n(\sigma )$ . We inductively define elements in $\widehat {y}_n(\sigma )$

(3.19) $$ \begin{align} \begin{array}{ll} {}^\sigma b_{i+1,i}^{(r)} &:= b_{i,r} \in \widehat{y}_n(\sigma) \ \ \text{ for } \ \ 1< i < n, \ r \geq s_{i+1,i}\\ {}^\sigma b_{i+1, j}^{(r)}&= \{{}^\sigma b_{i+1, i}^{(s_{i,i+1})}, {}^\sigma b_{i,j}^{(r-s_{i+1,i})}\} \in \widehat{y}_n(\sigma) \ \ \text{ for } \ \ 1\le j < i < n, \ r \geq s_{i+1,j} \\ {}^{\sigma}b_{i,i}^{(r)}&:=h_{0,r}+\dots+h_{i-1,r} \in \widehat{y}_n(\sigma) \ \ \text{ for } \ \ 1\le i \le n, \ r \geq 0 \end{array} \end{align} $$

and let ${}^{\sigma }b_{i,j}^{(r)}:=(-1)^{(r-1)}{}^{\sigma }b_{j,i}^{(r)}$ for $i < j$ . We also define a loop filtration on $\widehat {y}_n(\sigma )$ by putting the generators in degree r and denote the counterparts of the $^\sigma b_{i,j}^{(r)}$ in $\operatorname {gr}_L\widehat {y}_n(\sigma )$ by $^\sigma \bar {b}_{i,j}^{(r)}$ . Once more following the proof of [Reference Lu, Wang and ZhangLWZ23, (5.11)] verbatim we get

(3.20) $$ \begin{align} \{{}^{\sigma}\bar{b}_{i,j}^{(r)},{}^{\sigma}\bar{b}_{k,l}^{(s)}\}=\delta_{j,k}{}^{\sigma}\bar{b}_{i,l}^{(r+s)}-\delta_{l,i} {}^{\sigma}\bar{b}_{k,j}^{(r+s)}-(-1)^r\delta_{k,i}{}^{\sigma}\bar{b}_{j,l}^{(r+s)}+\delta_{l,j}(-1)^{(r)}{}^{\sigma}\bar{b}_{k,i}^{(r+s)} \end{align} $$

so that $\widehat {y}_n(\sigma )$ is spanned by monomials in the elements $h_{i,2r+1},{}^{\sigma }b_{i,j}^{(r)} \in \widehat {y}_n(\sigma )$ . By Lemma 3.3 $\operatorname {gr}_C \mathcal {Y}_{n}(\sigma )$ is a polynomial ring in the images of these maps under the canonical surjection so that the canonical surjection is an isomorphism.

Proof. (3.24)-(3.29) are as in [Reference TopleyTo23, Theorem 3.7]. We claim that (3.30) is equivalent to [Reference TopleyTo23, (3.74)] which says

$$ \begin{align*} & \Big\{\theta_i^{(r)}, \big\{\theta_i^{(s)}, \theta_j^{(t)}\big\}\Big\} + \Big\{\theta_i^{(s)}, \big\{\theta_i^{(r)}, \theta_j^{(t)}\big\}\Big\} \\& \quad \nonumber = 2(-1)^{s+s_{i,i+1}-1} \delta_{i,j+1} \sum_{m_1=0}^{m-1} \sum_{m_2=0}^{m_1} \eta_{i+1}^{(2(m-m_1-1))} \widetilde{\eta}_i^{(2m_2)} \theta_j^{(2(m_1-m_2) + t)} \\& \qquad \nonumber + 2(-1)^{s+s_{i,i+1}-1} \delta_{i+1,j} \sum_{m_1=0}^{m-1} \sum_{m_2=0}^{m-m_1-1} \widetilde{\eta}_i^{(2m_1)}\eta_{i+1}^{(2m_2)} \theta_j^{(2(m-m_1-m_2 - 1) + t)} \\& \qquad \nonumber \text{for } |i - j| =1 \text{ and } r + s = 2m \text{ even} \end{align*} $$

Making the substitution $q:=m-m_1-1$ in the first sum this now reads as

$$ \begin{align*} & \Big\{\theta_i^{(r)}, \big\{\theta_i^{(s)}, \theta_j^{(t)}\big\}\Big\} + \Big\{\theta_i^{(s)}, \big\{\theta_i^{(r)}, \theta_j^{(t)}\big\}\Big\} \\& \quad \nonumber = 2(-1)^{s+s_{i,i+1}-1} \delta_{i,j+1} \sum_{q=0}^{m-1} \sum_{m_2=0}^{m-q-1} \eta_{i+1}^{(2(q))} \widetilde{\eta}_i^{(2m_2)} \theta_j^{(2(m-q-1-m_2) + t)} \\& \qquad \nonumber + 2(-1)^{s+s_{i,i+1}-1} \delta_{i+1,j} \sum_{m_1=0}^{m-1} \sum_{m_2=0}^{m-m_1-1} \widetilde{\eta}_i^{(2m_1)}\eta_{i+1}^{(2m_2)} \theta_j^{(2(m-m_1-m_2 - 1) + t)} \\& \qquad \nonumber \text{for } |i - j| =1 \text{ and } r + s = 2m \text{ even} \end{align*} $$

now changing the order of summation in the first sum this reads as

$$ \begin{align*} & \Big\{\theta_i^{(r)}, \big\{\theta_i^{(s)}, \theta_j^{(t)}\big\}\Big\} + \Big\{\theta_i^{(s)}, \big\{\theta_i^{(r)}, \theta_j^{(t)}\big\}\Big\} \\& \quad \nonumber = 2(-1)^{s+s_{i,i+1}-1} \delta_{i,j+1} \sum_{m_2=0}^{m-1} \sum_{q=0}^{m-m_2-1} \eta_{i+1}^{(2(q))} \widetilde{\eta}_i^{(2m_2)} \theta_j^{(2(m-q-1-m_2) + t)} \\& \qquad \nonumber + 2(-1)^{s+s_{i,i+1}-1} \delta_{i+1,j} \sum_{m_1=0}^{m-1} \sum_{m_2=0}^{m-m_1-1} \widetilde{\eta}_i^{(2m_1)}\eta_{i+1}^{(2m_2)} \theta_j^{(2(m-m_1-m_2 - 1) + t)} \\& \qquad \nonumber \text{for } |i - j| =1 \text{ and } r + s = 2m \text{ even} \end{align*} $$

which is as claimed.

Proof. The first one of these relations follows immediately by relation (3.24) as $\bar {\eta }^{(2r)}_i$ is polynomial in $\eta _i^{2r},\eta _{i-1}^{2s}, \ 1 \leq r,s$ . The second relation is proved just as [Reference TopleyTo23, (3.56)]. For the third relation we have

$$ \begin{align*} \{ \bar{\eta}^{(2r)}_i,\theta_j^{(s)} \}&=\sum_{t=0}^{r} \{ \tilde{\eta}_{i-1}^{(2t)}{\eta}^{(2r-2t)}_i,\theta_j^{(s)} \}\\&=(\delta_{i-1,j+1}-\delta_{i-1,j})\sum_{t=1}^{r}\eta_{i}^{(2r-2t)}\sum_{t'=0}^{t-1}\tilde{\eta}_{i-1}^{(2t')}\theta_{j}^{(2t+s-1-2t')}\\ &\quad+(\delta_{i,j}-\delta_{i,j+1})\sum_{t=0}^{r-1}\tilde{\eta}_{i-1}^{(2t)}\sum_{t'=0}^{r-t-1}{\eta}_{i}^{(2t')}\theta_{j}^{(2r-2t+s-1-2t')} \end{align*} $$

where the second equality follows by (3.25) and (3.39). Thus the coefficient of $\theta _j^{(s+1+2k)}, \ 0 \leq k \leq r-1$ is given by $((\delta _{i-1,j+1}-\delta _{i-1,j})+(\delta _{i,j}-\delta _{i,j+1}))\sum _{w=0}^{r-j-1}\eta _{i}^{(w)}\tilde {\eta }_{i-1}^{(2r-2w-2j-2)}$ which proves the lemma.

Proof. Consider the map $ \operatorname {gr}_C \mathcal {Y}_n(\sigma ) \to R(y_n(\sigma ),\tau ) $ given by

(3.41) $$ \begin{align} \begin{aligned} h_{0,2r-1}+F_{r-1}\mathcal{Y}_n(\sigma) & \mapsto (2)^{2r+2}\bar{\eta}_{1}^{(2r)} \\ h_{i-1,2r-1}+F_{r-1}\mathcal{Y}_n(\sigma)& \mapsto (2)^{2r+2}\bar{\eta}_{i}^{(2r)} \text{ for } i>1 \\ h_{i,2r} &\mapsto 0 \\ b_{i,r-1}+F_{r-1}\mathcal{Y}_n(\sigma)& \mapsto (-1)^{\frac{s_{i,i+1} + 1}{2}}(-2)^{r}\theta_{i}^{(r)} \end{aligned} \end{align} $$

Using relations (3.25)-(3.30) together with (3.38) and (3.40) it is readily verified that the map is a morphism of Poisson algebras. Indeed to check (3.14) use (3.38), to check (3.15) use (3.40), to check (3.16) use (3.26) and (3.27), to check (3.17) use (3.28) and to check (3.18) use (3.29),(3.30) and (3.40). By the proof of [Reference TopleyTo23, Theorem 3.7] we know that $R(y_n(\sigma ))$ has a basis consisting of monomials in $\eta _{i}^{(2r)},\theta _{i,j}^{(s)}, 1 \leq i < j \leq n, r \geq 0, s> s_{i,j}$ where the $\theta _{i,j}^{(s)}$ are defined inductively via the recursion

(3.42) $$ \begin{align} \begin{array}{rcl} \theta_{i,i+1}^{(r)} := \theta_i^{(r)} & \text{ for } & 1\le i < n, \ 0 < r \\ \theta_{i,j}^{(r)} := \{\theta_{i,j-1}^{(r-s_{j-1, j})}, \theta_{j-1}^{(s_{j-1, j} + 1)}\} & \text{ for } & 1\le i < j\le n, \ s_{i,j} < r.\end{array} \end{align} $$

By definition of the map we see that ${}^{\sigma }b_{j,i}^{(r-1)}$ as defined in (3.9) maps to a nonzero scalar multiple of $\theta _{i,j}^{(r)}$ .

By Remark 3.7 it follows that the set of monomials in $\bar {\eta }_{i}^{(2r)},\theta _{i,j}^{(s)}, 1 \leq i < j \leq n, r \geq 0, s> s_{i,j}$ is a basis for $R(y_n(\sigma ),\tau )$ and thus our map sends a spanning set to a basis. Hence it is an isomorphism.

Proof. Part (i) is proven using the fact that $J^{-1} = \epsilon J$ , and multiplying matrices, bearing in mind the fact that the $\lambda _i$ all have the same parity. Now (ii) and (iii) follow by applying (i) to (4.3), (4.4) and (4.6).

Proof. The index bounds (4.14) ensure that $\mathcal {X}_{a,i,b,k, \bar {\nu }_t}^{(r,s)}$ is finite. Since $\upsilon ^2$ is the identity map it will suffice to show that $\upsilon $ maps the left hand side of (4.15) to the right. We fix $(a_u,i_u, j_u,b_u, k_u, l_u)_{u = 1,...,s}\in \mathcal {X}_{a,i,b,k; \bar {\nu }_t}^{(r,s)}$ and check that conditions (a)–(f) hold for $\upsilon (a_u,i_u, j_u, b_u, k_u, l_u)_{u=1,...,s}$ . For example,

$$ \begin{align*} \sum_{u=1}^s \big(2(\lambda(\nu)_{a_{s+1-u}} + 1 - j_{s+1-u}) - 2(\lambda(\nu)_{b_{s+1-u}} & + 1 - l_{s+1-u}) + (\lambda(\nu)_{b_{s+1-u}} - \lambda(\nu)_{a_{s+1-u}})\big)\\ & = \sum_{u=1}^s (2l_u - 2j_u + \lambda(\nu)_{a_u} - \lambda(\nu)_{b_u}) = 2(r - s) \end{align*} $$

This shows that when (a) is satisfied for $(a_u,i_u, j_u, b_u, k_u, l_u)_{u=1,...,s}$ , it follows that the formulation of (a) which defines $\mathcal {X}_{b, k' , a, i'; \bar {\nu }_t}^{(r,s)}$ holds for $\upsilon (a_u,i_u, j_u, b_u, k_u, l_u)_{u=1,...,s}$ . Conditions (b)–(f) can be checked similarly.

Proof. The assertion that the image of $\theta $ lies in $S(\mathfrak {g},e)$ is an immediate consequence of [Reference Brundan and KleshchevBK06, Theorem 10.1], upon taking the top graded term with respect to the Kazhdan filtration of the finite W-algebra.

Thanks to Lemma 4.2(iii) the $\tau $ -equivariance will follow from (4.17). Fix indices $1\le a, b \le m, 1 \le i \le \lambda (\nu )_a, 1 \le k \le \lambda (\nu )_b, \ 0\le x < n$ and $0 < r$ . Choose $s\in \{1,...,r\}$ . Let $(a_m,i_m, j_m, b_m, k_m,l_m)_{m=1}^s \in \mathcal {X}_{a,i,b,k; \bar {\nu }_t}^{(r,s)}$ . Thanks to Lemma 4.2(i) we have

(4.17) $$ \begin{align} \tau(\prod_{m=1}^s e_{a_m,i_m, j_m;b_m k_m, l_m}) = (\beta)^{\sum_{m=1}^s k_m+i_m}(-1)^{\sum_{m=1}^s (j_m - l_m + 1)} \prod_{t=1}^s e_{\upsilon(a_m,i_m, j_m;b_m k_m, s_m)} \end{align} $$

Conditions (a), (e), (f) imply that $\sum _m (j_m - l_m + 1) = \frac {\lambda _{i_1} - \lambda _{k_s}}{2} - r.$ Conditions (e), (f) imply $(-1)^{\sum _{m=1}^s k_m+i_m}=(-1)^{i+k}$ . Hence the sign on the right hand side of (4.17) is equal to $(\beta )^{i+k}(-1)^{r-\frac {\lambda _k - \lambda _i}{2}}$ . Furthermore condition (f) together with our condition on x ensures that $\# \{ m =1,...,s-1 \mid \nu _1+\dots + \nu _{b_m-1}+k_m \le \bar {\nu }_t\} = \# \{m =2,...,s \mid \nu _1+\dots + \nu _{a_m-1}+\lambda (\nu )_m + 1 - i_m \le \bar {\nu }_t\}.$ Combining these observations together with Lemma 4.4 we have

$$ \begin{align*} \tau(t_{a,i,b,k;\bar{\nu}_t}^{(r)}) &= \sum_{s=1}^r (-1)^{r-s} \sum_{(a_m,i_m, j_m, b_m, k_m, l_m)_m \in \mathcal{X}_{a,i,b,k; \bar{\nu}_t}^{(r,s)}} (-1)^{\#_{k,\bar{\nu}_t}}\tau (\prod_m e_{a_m,i_m, j_m, b_m, k_m, l_m}) \\ & = (\beta)^{i+k}(-1)^{r-\frac{\lambda_k - \lambda_i}{2}} \sum_{s=1}^r (-1)^{r-s} \sum_{(a_m,i_m, j_m,b_m, k_m, l_m)_m \in \mathcal{X}_{b,k',a,i';\bar{\nu}_t}^{(r,s)}} (-1)^{\#_{k,\bar{\nu}_t}}\prod_m e_{a_m,i_m, j_m, b_m, k_m, l_m}\\ & = (\beta)^{i+k}(-1)^{r-\frac{\lambda_k - \lambda_i}{2}} t_{b, k',a,i';\bar{\nu}_t}^{(r)} \end{align*} $$

This completes the proof.

Proof. The linear part of (4.12) corresponds to $s=1$ . Conditions (b)-(d) as well as condition (f) on the indices in (4.12) become redundant when $s=1$ whereas combining conditions (a) and (e) gives the sum

(4.20) $$ \begin{align} (-1)^{r-1}\sum_{ 2r-1 = 2(l-j) + \lambda(\nu)_{a}- \lambda(\nu)_{b}} e_{a,i,j;b,k,l} \end{align} $$

which is equal to $(-1)^{r-1}c_{a,i,b,k}^{(r-1)}$ by definition 4.6.

Proof. Write $\underline {0}$ for the $n\times n$ zero matrix. It follows from (2.1) that the linear map $^{\nu }\mathfrak {u}_n(\sigma ) \to ^{\nu }\mathfrak {u}_n(\underline {0})$ defined by $e_{a,i,b,j}t^r \mapsto e_{a,i,b,j}t^{r-s_{a,b}}$ is a Lie algebra isomorphism and so it suffices to prove the current lemma when $\sigma = \underline {0}$ .

Let $^{\nu }\widehat {\mathfrak {u}}_n$ be the Lie algebra with generators (5.3) and relations (5.4)–(5.6), with $\sigma = \underline {0}$ . We inductively define elements $e_{a,i,b,j;r} \in {}^{\nu }\widehat {\mathfrak {u}}_n$ by setting $e_{a,i,a+1,j;r} := e_{a,i,j;r}$ and $e_{a,i,b,j;r} := [e_{a,i,k,b-1; r}, e_{b-1,k,j; 0}]$ for $1\le a < b \le m$ and all admissible $i,j$ . There is a homomorphism $^{\nu }\widehat {\mathfrak {u}}_n \twoheadrightarrow {}^{\nu }\mathfrak {u}_n(\underline 0)$ given by $e_{a,i,j;r} \mapsto e_{a,i,a+1,j}t^r$ and, in order to show that it is an isomorphism, we show that $^{\nu }\widehat {\mathfrak {u}}_n$ is spanned by the elements $\{e_{a,i,b,j;r} \mid 1\le a<b\le m, \ 1 \le i \le \nu _a, \ 1\le j \le \nu _b, \ 0 \le r\}$ .

Following (1)–(7) in the proof of [Reference Brundan and KleshchevBK05, Lemma 6.7] verbatim we have for all $a,b,c,d,j,k,l,r,s$

(5.7) $$ \begin{align} [e_{a,i,b,j;r} , e_{c,k,d,l;s}] = \delta_{b,c}\delta_{j,k} e_{a,i,d,l;r+s} - \delta_{a,d}\delta_{i,l} e_{c,k,b,j;r+s}. \end{align} $$

Define an ascending filtration on $\widehat {\mathfrak {u}}_n = \bigcup _{d> 0} \operatorname {F}_d \widehat {\mathfrak {u}}_n$ satisfying $\operatorname {F}_{d} \widehat {\mathfrak {u}}_n = \sum _{\substack {d_1 + d_2 = d}} [\operatorname {F}_{d_1} \widehat {\mathfrak {u}}_n, \operatorname {F}_{d_2} \widehat {\mathfrak {u}}_n]$ by placing $e_{a,i,j;r}$ in degree 1. We prove by induction that $\operatorname {F}_d\widehat {\mathfrak {u}}_n$ is spanned by elements $e_{a,i,b,j;r}$ with $j-i \le d$ . The base case $d = 1$ holds by definition. For $d_1 + d_2 = d> 1$ we know by the inductive hypothesis that $\operatorname {F}_{d_1} \widehat {\mathfrak {u}}_n$ and $\operatorname {F}_{d_2} \widehat {\mathfrak {u}}_n$ are spanned by elements $e_{a,i,b,j;r}$ . Using (5.7) we complete the induction, which finishes the proof.

Proof. By [Reference TopleyTo23, (3.1)] it suffices to show that $\mathfrak {c}_n(\sigma )$ is generated as a Lie algebra by (5.8) subject to relations (5.9)–(5.18). Let $(\widehat {\mathfrak {c}}_n(\sigma ), \{\cdot , \cdot \})$ denote the Lie algebra with these generators and relations. Define a map from the set (5.8) to $\mathfrak {c}_n(\sigma )$ by $d_{a,i,j;r} \mapsto e_{a,i,a,j}t^r, \ e_{a,i,j;r} \mapsto e_{a,i,a+1,j}t^r, \ f_{a,i,j;r} \mapsto e_{a+1,i, a,j} t^r$ . One can easily verify using (5.7) that this extends to a surjective Lie algebra homomorphism $\widehat {\mathfrak {c}}_n(\sigma ) \twoheadrightarrow \mathfrak {c}_n(\sigma )$ . To show that this is an isomorphism it suffices to show that the elements

(5.19) $$ \begin{align} \{e_{a,i,b,j;r} \mid 1 \le a,b\le m, \ 1 \le i \le \nu_a, \ 1 \le b \le \nu_{b} \ s_{a,b}\le r\} \subseteq \widehat{\mathfrak{c}}_n(\sigma) \end{align} $$

defined inductively by setting $e_{a,i,a+1,j; r} := e_{a,i,j;r}$ , $e_{a+1, i,a,j; r} := f_{a,i,j;r}$ and

(5.20) $$ \begin{align} e_{a,i,b,j;r} & := \{e_{a,i,b-1,k;r-s_{b-1,b}}, e_{b-1,k,j; s_{b,b-1}}\} \text{ for } a < b; \end{align} $$

(5.21) $$ \begin{align} e_{a,i,b,j;r} & := \{f_{a-1,i,k; s_{a,a-1}}, e_{a-1,k,b, j; r-s_{a,a-1}}\} \text{ for } a> b, \end{align} $$

form a spanning set. Using (5.9),(5.11), (5.12) and a simple inductive argument one can see that $\widehat {\mathfrak {c}}_n(\sigma )$ is a direct sum of three subalgebras: the diagonal subalgebra, spanned by the elements $d_{a,i,j;r}$ , and the upper and lower triangular subalgebras $^{\nu }\mathfrak {u}_n^+(\sigma )$ and $^{\nu }\mathfrak {u}_n^-(\sigma )$ generated by the elements $e_{a,i,j; r}$ , respectively by the elements $f_{a,i,j; r}$ .

In order to complete the proof it suffices to show that $\mathfrak {u}_n^+(\sigma )$ and $\mathfrak {u}_n^-(\sigma )$ are spanned by the elements defined in (5.20) and (5.21) respectively. Since the argument is identical for $\mathfrak {u}_n^+(\sigma )$ and $\mathfrak {u}_n^-(\sigma )$ we only need to consider the former, where the claim follows from Lemma 5.1.

Proof. Relation (5.7) implies that $\tau $ gives a Poisson automorphism. By [Reference TopleyTo23, Remark 2.3] we can identify $R(S(\mathfrak {c}_n(\sigma )), \tau )$ with $S(\mathfrak {c}_n(\sigma )^\tau )$ , so it suffices to check that $S(\mathfrak {c}_n(\sigma )^\tau )$ has the stated Poisson presentation. Use [Reference TopleyTo23, (3.1)] to reduce the claim to a statement about the presentation of $\mathfrak {c}_n(\sigma )^\tau $ .

Consider the Lie algebra $\widehat {\mathfrak {c}}_n(\sigma )^\tau $ which is generated by the set (5.23) subject to relations (5.25)–(5.32). Make the notation $\theta _{a,i,a+1,j}t^r := e_{a,i,a+1,j}t^r + \tau (e_{a,i,a+1,j}t^r), \ \eta _{a,i,a,j}t^r := e_{a,i,a,j}t^r + \tau (e_{a,i,a,j}t^r) \in \mathfrak {c}_n(\sigma )^\tau $ . We define a map from the set (5.34) to $\mathfrak {c}_n(\sigma )^\tau $ by sending $\eta _{a,i,j;r} \mapsto \eta _{a,i,a,j}t^{r} \in \mathfrak {c}_n(\sigma )^\tau $ , and sending $\theta _{a,i,j;r} \mapsto \theta _{a,i,a+1,j}t^r \in \mathfrak {c}_n(\sigma )^\tau $ . One can check that this determines a surjective Lie algebra homomorphism $\widehat {\mathfrak {c}}_n(\sigma )^\tau \twoheadrightarrow \mathfrak {c}_n(\sigma )^\tau $ , indeed, checking that relations (5.25)–(5.32) hold amongst the corresponding elements of $\mathfrak {c}_n(\sigma )^\tau $ is a routine calculation using (5.7) and relation 5.24 follows immediately by definition of $\tau $ .

In order to complete the proof of (2) it is sufficient to show that this map is an isomorphism. This proof is similar to [Reference TopleyTo23, Theorem 3.3] but in our case there is a linear relation between the $\eta _{a,i,j;r}$ . For all admissible $a,b,i,j,$ and $r \geq s_{a,b}$ we inductively define elements

(5.33) $$ \begin{align} \theta_{a,i,b,j;r} := \{\theta_{a,i,a+1,k;s_{i,i+1}}, \theta_{a+1,k,b, j; r-s_{i,i+1}}\} \in \widehat{\mathfrak{c}}_n(\sigma)^\tau. \end{align} $$

where $\theta _{a,i,a+1,j;r} := \theta _{a,i,j;r}$ . Let $\{v_1, v_2,\dots \}$ be a subset of $\{\eta _{a,i,j;r} \mid 1 \le a \le m, \ 1 \le i,j \le \nu _{a}, \ r \geq 0 \}$ such that it is a basis for the vector space spanned by $\{\eta _{a,i,j;r} \mid 1 \le a \le m, \ 1 \le i,j \le \nu _{a}, \ r \geq 0 \}$ together with the linear relation 5.24. It remains to check that $\widehat {\mathfrak {c}}_n(\sigma )^\tau $ is spanned by

(5.34) $$ \begin{align} \begin{array}{ll} &\{v_1,v_2, \dots \} \cup \{\theta_{a,i,b,j;r} \mid 1 \le a,b \le m, \ 1 \le i \le \nu_{a}, \ 1 \le j \le \nu_{b}, \ r \geq s_{a,b} \}. \end{array} \end{align} $$

We define a filtration $\widehat {\mathfrak {c}}_n(\sigma )^\tau = \bigcup _{i> 0} \operatorname {F}_i \widehat {\mathfrak {c}}_n(\sigma )^\tau $ by placing the generators (5.23) in degree 1 and satisfying $\operatorname {F}_{d} \widehat {\mathfrak {c}}_n(\sigma )^\tau = \sum _{d_1 + d_2 = d} \{\operatorname {F}_{d_1} \widehat {\mathfrak {c}}_n(\sigma )^\tau , \operatorname {F}_{d_2} \widehat {\mathfrak {c}}_n(\sigma )^\tau \}$ . By convention $\operatorname {F}_{0} \widehat {\mathfrak {c}}_n(\sigma )^\tau = 0$ . The associated graded Lie algebra $\operatorname {gr} \widehat {\mathfrak {c}}_n(\sigma )^\tau = \bigoplus _{i> 0} \operatorname {F}_i \widehat {\mathfrak {c}}_n(\sigma )^\tau / \operatorname {F}_{i-1} \widehat {\mathfrak {c}}_n(\sigma )^\tau $ is generated by elements

(5.35) $$ \begin{align} & \overline{\eta}_{a,i,j;r} := \eta_{a,i,j;r} + \operatorname{F}_{0} \widehat{\mathfrak{c}}_n(\sigma)^\tau \text{ for } 1\le a \le m, \ 1 \le i,j \le \nu_a, \ 0 \le r \end{align} $$

(5.36) $$ \begin{align} & \overline{\theta}_{a,i,j;r} := \theta_{a,i,j;r} + \operatorname{F}_0 \widehat{\mathfrak{c}}_n(\sigma)^\tau \text{ for } 1\le a < m-1, \ 1 \le i \le \nu_a, \ 1 \le j \le \nu_{a+1}, \ \ s_{a,a+1} \le r. \end{align} $$

These generators of $\operatorname {gr} \widehat {\mathfrak {c}}_n(\sigma )^\tau $ satisfy the top graded components of the relations (5.24)–(5.32).

Let $\mathfrak {a}$ be an abelian Lie algebra with spanning set $\{d_{a,i,j} t^r \mid 1\le a \le m, \ 1 \le i,j \le \nu _{a}, \ 0 \leq r\}$ and relations $d_{a,i,j} t^r=(\beta )^{i+j}(-1)^{r-1}d_{a,\nu _a+1-j,\nu _a+1-i}$ . Let $\mathfrak {u}_n(\sigma )$ be the Lie subalgebra of $\mathfrak {c}_n(\sigma )$ described in Lemma 5.1. Then $\mathfrak {a} \oplus \mathfrak {u}_n(\sigma )$ is a Lie algebra with $\mathfrak {a}$ an abelian ideal. Comparing the top components (5.24)–(5.32) to (5.4)–(5.6) we see that there is a surjective Lie algebra homomorphism $\mathfrak {a} \oplus \mathfrak {u}_n(\sigma ) \twoheadrightarrow \operatorname {gr} \widehat {\mathfrak {c}}_n(\sigma )^\tau $ defined by $d_{a,i,j} t^{r} \mapsto \overline {\eta }_{a,i,j; r}$ and $e_{a,i,a+1,j}t^r \mapsto \overline {\theta }_{a,i,j;r}$ . The algebra $\mathfrak {a} \oplus \mathfrak {u}_n(\sigma )$ has basis consisting of elements $e_{a,h,b,k}t^s$ where $ 1\le a < b \le m, 1 \le h \le \nu _{a}, \ 1 \le k \le \nu _{b}, \ s \ge s_{a,b}$ together with a basis $\{w_1,w_2,\dots \}$ for the vector space $\mathfrak {a}$ which can be picked such that $w_1 \mapsto v_1$ . It follows that $\operatorname {gr} \widehat {\mathfrak {c}}_n(\sigma )^\tau $ is spanned by elements $\overline {v_1},\overline {v_2},\dots , \overline {\theta }_{a,i,b,j;s}$ where the indices vary in the ranges specified in (5.34), and $\overline {\theta }_{a,i,b,j;s}$ defined inductively from (5.36), analogously to (5.33). We deduce that $\widehat {\mathfrak {c}}_n(\sigma )^\tau $ is spanned by the required elements, which completes the proof.

Proof. Part (1) is essentially the same as [Reference Goodwin and TopleyGT19c, Lemma 2.6].

Compare (5.22) with Lemma 4.2(iii) to see that the map is $\tau $ -equivariant, hence it is a well-defined map $\mathfrak {c}_n(\sigma )^\tau \twoheadrightarrow (\mathfrak {g}^e)^\tau $ .

We go on to describe the kernel in (2). First of all observe that the elements listed there are $\tau $ -fixed elements of the kernel of $\mathfrak {c}_n(\sigma )\twoheadrightarrow \mathfrak {g}^e$ . Now observe that $\mathfrak {g}_0^e$ has a spanning set consisting of elements of the form $c_{a,i,b,j}^{(r)} + \tau (c_{a,i,b,j}^{(r)})$ . By Lemma 4.1 we can show that the elements in (2) generate the kernel by checking that the ideal $\mathfrak {i}$ which they generate contains

(5.39) $$ \begin{align} \begin{array}{l} \{\eta_{a,i,j;r} \mid a=1,...,m, \ , 1\le i,j \le \nu_a \ , \ r \ge \lambda_a\} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \cup \,\{\theta_{a,i,j;r} \mid a=1,..,m-1, \ 1\le i\le \nu_a, \, 1\le j \le \nu_{a+1}, \ r \ge s_{a,a+1} + \lambda_a \}. \end{array} \end{align} $$

Consider the case $\epsilon = 1$ in (2)(i) and note that in this case for each i there is a j such that we have $\eta _{1,i,j;\lambda _1} \neq 0$ . This is obvious if $\nu _1 \neq 1$ whereas if $\nu _1=1$ we have that $\lambda _1$ is odd by $(\beta , \sigma )$ admissibility as $\beta =-\varepsilon (-1)^{\lambda _1}$ and hence $\eta _{1,1,1,\lambda _{1}}=d_{1,1,1,\lambda _{1}}+(-1)^{\lambda _1-1}d_{1,1,1,\lambda _{1}} \neq 0$ . Using relation (5.26) we see that $\mathfrak {i}$ contains $\{\eta _{1,i,j;\lambda _1}, \theta _{1,j,k;r}\} = \theta _{1,i,k; \lambda _1 + r}$ for $r \ge s_{1,2}$ and all admissible $i,k$ whilst (5.28) together with the fact that $\eta _{1,i,j,\lambda _2+r} \in \mathfrak {i}$ for $r \geq 0$ shows that $\eta _{2,i,j;\lambda _2+r} \in \mathfrak {i}$ for $r\ge 0$ and all admissible $i,j$ . By Theorem 5.3 the subalgebra of $\mathfrak {c}_n(\sigma )^\tau $ generated by $\eta _{a,i,j;r}, \theta _{a,i,j;r}$ with $2\le a$ is isomorphic to a current Lie algebra associated with a lower rank general linear Lie algebra and a shape of strictly smaller length. The description of the kernel follows by induction.

In the second case we see that if $\nu _1 \neq 1$ for each i there is some j such that $\eta _{1,i,j;\lambda _1}\neq 0$ . Hence the ideal $\mathfrak {i}$ contains $\{\eta _{1,i,j;\lambda _1}, \theta _{1,j,k; s}\} = \theta _{1,i,k;\lambda _1+s}$ for $s \ge s_{1,2}$ , proving (ii).

If $\nu _1=1$ we see that $\eta _{1,1,1;\lambda _1+1} \neq 0$ . Hence the ideal $\mathfrak {i}$ contains $\{\eta _{1,1,1;\lambda _1+1}, \theta _{1,1,k; s}\} = \theta _{1,1,k;\lambda _1 + 1 + s}$ for $s \ge s_{1,2}$ . Together with the additional generators $\theta _{1,i,j;\lambda _1 + s_{1,2}}$ this gives all elements of the form $\theta _{1,i,j;r}$ listed in (5.39).

Furthermore if $\nu _1=1$ we have by $(\beta , \sigma )$ admissibility that $\lambda _1$ is even. In particular by (5.32) we have that we have

$$ \begin{align*} \{\theta_{1,i,j;\mathfrak{s}_{1,2}+\lambda_1}, \{\theta_{1,i',k;\mathfrak{s}_{1,2}}, \theta_{2,k,g;\mathfrak{s}_{2,3}}\}\} &+\{\theta_{1,i,j;\mathfrak{s}_{1,2}}, \{\theta_{1,i',k;\mathfrak{s}_{1,2}+\lambda_1}, \theta_{2,k,g;\mathfrak{s}_{2,3}}\}\} \\ &= \frac{1}{2}(\beta)^{i+j}(-1)^{\lambda_1+2\mathfrak{s}_{1,2}}\theta_{2,j',g;\lambda_2+\mathfrak{s}_{2,3}}. \end{align*} $$

as we have $\lambda _2=\lambda _1+2s_{1,2}$ . So all additional generators of the form $\theta _{2,i,j;\lambda _2 + s_{2,3}}$ also lie in the ideal. Now the inductive proof of (iii) proceeds in the same way as in the first case.

Proof. Comparing the top graded components of the relations (5.41)–(5.52) with respect to the loop filtration, with relations (5.9)–(5.18), it is straightforward to see that (5.58) gives a surjective Poisson homomorphism. To prove the theorem we demonstrate that the ordered monomials in the elements (5.59) are linearly independent.

Consider the set

(5.60) $$ \begin{align} \begin{array}{rcl} X &:= & \{E_{a,i,j}^{(r)} \mid 1 \le a < m, \ 1 \le i \le \nu_a, \ 1 \le j \le \nu_{a+1}, \ s_{a,a+1} < r\} \\ & & \hspace{40pt}\cup \{F_{a,i,j}^{(r)} \mid 1 \le a < m, \ 1 \le i \le \nu_{a+1}, \ 1 \le j \le \nu_{a}, \ s_{a+1,a} < r\} \\ & & \hspace{80pt}\cup \{D_{a,i,j}^{(r)} \mid 1 \le a < m, \ 1 \le i,j \le \nu_a, \ 0 < r\}. \end{array} \end{align} $$

In [Reference Brundan and KleshchevBK06, §3] the shifted Yangian $^{\nu }Y_n(\sigma )$ is defined as a quotient of the free algebra $\mathbb {C}\langle X \rangle $ by the ideal generated by the relations [Reference Brundan and KleshchevBK06, (3.3)–(3.14)]. Let $L := L_X$ be the free Lie algebra on X and define a grading $L = \bigoplus _{i \ge 0} L_i$ by placing $E_{a,i,j}^{(r)}, F_{a,i,j}^{(r)}, D_{a,i,j}^{(r)}$ in degree $r-1$ . Then we place a filtration on the enveloping algebra $U(L)$ so that $L_i$ lies in degree $i+1$ . By the PBW theorem for $U(L)$ we see that $\operatorname {gr} U(L) \cong S(L)$ .

The universal property of $U(L)$ ensures that there is a surjective algebra homomorphism $U(L) \twoheadrightarrow \mathbb {C}\langle X \rangle $ and by [Reference SerreSer06, I, Ch. IV, Theorem 4.2] this is an isomorphism. Identifying these algebras, the filtration on $U(L)$ descends to $^{\nu }Y_n(\sigma ) = \bigcup _{i\ge 0} \operatorname {F}^{\prime }_i \, {}^{\nu }Y_n(\sigma )$ , and this resulting filtration is commonly referred to as the canonical filtration [Reference Brundan and KleshchevBK06, §5]. The associated graded algebra $\operatorname {gr} {}^{\nu }Y_n(\sigma )$ is equipped with a Poisson structure in the usual manner. Comparing relations (5.41)–(5.52) with the top graded components of relations [Reference Brundan and KleshchevBK06, (3.3)–(3.14)] we see that the Poisson surjection $S(L) \twoheadrightarrow \operatorname {gr} {}^{\nu }Y_n(\sigma )$ factors through $S(L) \to {}^{\nu }y_n(\sigma )$ . As a result there is a surjective Poisson homomorphism $\pi : {}^{\nu }y_n(\sigma ) \twoheadrightarrow \operatorname {gr} {}^{\nu }Y_n(\sigma )$ given by $e_{a,i,j}^{(r)} \mapsto E_{a,i,j}^{(r)} + \operatorname {F}^{\prime }_{r-1} {}^{\nu }Y_n(\sigma ), f_{a,i,j}^{(r)} \mapsto F_{a,i,j}^{(r)} + \operatorname {F}^{\prime }_{r-1} {}^{\nu }Y_n(\sigma ), d_{a,i,j}^{(r)} \mapsto D_{a,i,j}^{(r)} + \operatorname {F}^{\prime }_{r-1} {}^{\nu }Y_n(\sigma )$ . Following [Reference Brundan and KleshchevBK06, (3.15), (3.16)] we introduce elements $E_{a,i,b,j}^{(r)}, F_{a,i,b,j}^{(r)}$ of $^{\nu }Y_n(\sigma )$ lying in filtered degree r. By the definition of the filtration and the elements (5.56), (5.57) we have

$$ \begin{align*} \begin{array}{rcl} \pi(e_{a,i,b,j}^{(r)}) & = & E_{a,i,b,j}^{(r)} + \operatorname{F}^{\prime}_{r-1} \ ^{\nu}Y_n(\sigma);\\ \pi(f_{a,i,b,j}^{(r)}) & = & F_{a,i,b,j}^{(r)} + \operatorname{F}^{\prime}_{r-1} \ ^{\nu}Y_n(\sigma).\\ \end{array} \end{align*} $$

By [Reference Brundan and KleshchevBK06, Theorem 3.2(iv)] the ordered monomials in $E_{a,i,b,j}^{(r)}, F_{a,i,b,j}^{(r)}, D_{a,i,j}^{(r)}$ are linearly independent in $^{\nu }Y_n(\sigma )$ and so we deduce that the images of these monomials in $\operatorname {gr} ^{\nu }Y_n(\sigma )$ are linearly independent. This completes the proof.

Proof. We only sketch (5.61), as the proof of (5.62) is almost identical. The argument is by induction based on (5.53). Note that $\widetilde {d}_{a,i,j}^{(1)} = -d_{a,i,j}^{(1)}, \widetilde {d}^{(0)}_{a,i,j}=d^{(0)}_{a,i,j}=\delta _{i,j}$ and so (5.61) is equivalent to (5.43) in this case. We have $\{\widetilde {d}_{a,i,j}^{(r)}, e_{b,h,k}^{(s)}\} = \{-\sum _{g=1}^{\nu _a}\sum _{t=1}^{r} d_{a,i,g}^{(t)} \widetilde {d}_{a,g,j}^{(r-t)}, e_{b,h,k}^{(s)}\}$ and relation (5.43) together with the inductive hypothesis imply that the coefficient of $e_{b,m,n}^{(s+t)}$ is

$$ \begin{align*} & \delta_{a,b+1}\delta_{m,h}(\sum_{w=t+1}^{r-1}\widetilde{d}_{a,n,j}^{(r-w)} d_{a,i,k}^{(w-t-1)}-\widetilde{d}_{a,n,j}^{(w-t-1)} d_{a,i,k}^{(r-w)}+d_{a,i,k}^{(r-t-1)}\widetilde{d}_{a,n,j}^{(0)}) \\ &\hspace{20pt}-(\delta_{a,b}\delta_{n,k}(\sum_{w=t+1}^{r-1}\widetilde{d}_{a,h,j}^{(r-w)} d_{a,i,m}^{(w-t-1)}-\widetilde{d}_{a,h,j}^{(w-t-1)} d_{a,i,m}^{(r-w)}+d_{a,i,m}^{(r-t-1)}\widetilde{d}_{a,h,j}^{(0)})). \end{align*} $$

Using $d_{a,i,j}^{(0)} = \widetilde {d}_{a,i,j}^{(0)} = \delta _{i,j}$ we see that all quadratic terms cancel, whilst the linear term equates to $\delta _{a,b+1}\delta _{m,h}\widetilde {d}_{a,n,j}^{(r-t-1)}\delta _{i,k} - \delta _{a,b}\delta _{n,k} \delta _{i,m}\widetilde {d}_{a,h,j}^{(r-t-1)}$ , which concludes the induction.

Proof. We will prove the result for $GL_{\mu _1} \times \dots \times GL_{\mu _m} \subseteq GL_{\nu _1} \times \dots \times GL_{\nu _{m'}}$ . This suffices as for $\beta =1$ the shape $(1,1,\dots ,1,1)$ is always $(1,\sigma )$ -admissible and if $\sigma $ admits a $\beta =-1$ admissible shape then the shape $(2,2,\dots ,2,2)$ is always $(-1,\sigma )$ -admissible. The general result then follows by composing isomorphisms. Let ${}^{\nu }Y_n(\sigma )$ denote the shifted Yangian of shape $\nu $ as defined in [Reference Brundan and KleshchevBK06, §3]. Also write $Y_n$ for the Yangian for $\mathfrak {gl}_n$ . Recall the definition of parabolic presentations of $Y_n$ from [Reference Brundan and KleshchevBK05, §6] where the elements $E_{a,b,i,j}^{(r)},F_{a,b,i,j}^{(r)},D_{a,i,j}^{(r)}$ are defined in terms of a Gauss decomposition of the T-matrix or more explicitly by quasi determinants [Reference Brundan and KleshchevBK05, (6.2)-(6.4)]. By [Reference Brundan and KleshchevBK06, Corollary 3.3] the canonical map $i_{\nu }:{}^{\nu }Y_n(\sigma ) \to Y_n$ given by sending the generators to the elements with the same name is injective and its image is independent of the choice of $\nu $ . Furthermore by [Reference Brundan and KleshchevBK05, §5] the map is filtered with respect to the canonical filtration. Consider now the map $\tau _{\nu }:Y_n \to Y_n: T(u) \mapsto J_{\nu }^{-1}T(-u)^tJ_{\nu }$ where

(5.64) $$ \begin{align} J_{\nu}:= \sum_{a=1}^{m} \sum_{i=1}^{\nu_a} (\beta)^{i}(-1)^{s_{1,a}(\nu)} e_{a,i;a,\nu_a + 1 - i} \in GL_{\nu_1} \times \dots \times GL_{\nu_m} \end{align} $$

By [Reference MolevMo07, §1.3] this is an automorphism. Note that $J_{\nu }$ is a block diagonal matrix that can be described as follows. For each index $a = 1,...,m$ there is a block of size $\nu _a$ , each of these blocks has entries $(-1)^{s_{1,a}(\nu )}$ on the antidiagonal and zeros elsewhere when $\beta =1$ . When $\beta =-1$ it has alternating entries $(-1)^{s_{1,a}(\nu )}(\mp 1)$ on the antidiagonal and zeroes elsewhere. Thus when $\beta = 1$ the matrix J is a block diagonal matrix where each block is symmetric and when $\beta = -1$ each block is antisymmetric. It follows that

(5.65) $$ \begin{align} J_{\nu}^{-1} = \beta J_{\nu}. \end{align} $$

We claim that

(5.66) $$ \begin{align} \tau_{\nu}(E_{a,a+1,i,j}^{(r)}) &= (\beta)^{i+j}(-1)^{r+s_{a,a+1}(\nu)}F_{a+1,a,j',i'}^{(r)} \end{align} $$

(5.67) $$ \begin{align} \tau_{\nu}(F_{a+1,a,i,j}^{(r)}) &= (\beta)^{i+j}(-1)^{r+s_{a,a+1}(\nu)}E_{a+1,a,j',i'}^{(r)} \end{align} $$

(5.68) $$ \begin{align} \tau_{\nu}(D_{a,a,i,j}^{(r)}) &= (\beta)^{i+j}(-1)^{r}D_{a,a,j',i'}^{(r)}\end{align} $$

and in particular the image of $i_{\nu }$ is stabilised by $\tau _{\nu }$ . Indeed by [Reference Brundan and KleshchevBK05, (6.6)-(6.8)] we have that for the automorphism $\omega :Y_n \to Y_n: T(u) \mapsto T(-u)^\top $

(5.69) $$ \begin{align} \begin{array}{rcl} \omega(E_{a,a+1,i,j}^{(r)}) &=& (-1)^{r}F_{a+1,a,j,i}^{(r)} \\ \omega(F_{a+1,a,i,j}^{(r)})&=&(-1)^r E_{a,a+1,j,i}^{(r)} \\ \omega(D_{a,a,i,j}^{(r)})&=&(-1)^r D_{a,a,j,i}^{(r)} \end{array} \end{align} $$

Furthermore using the explicit description of $E_{a,a+1,i,j}^{(r)}$ , $F_{a+1,a,i,j}^{(r)}$ , $D_{a,a,i,j}^{(r)}$ in terms of quasi determinants and [Reference Gelfand, Gelfand, Retakh and WilsonGGRW04, §1.3(i),(ii)] it follows immediately that for any matrix $A=(A_{\nu _1}, \dots , A_{\nu _m}) \in GL_{\nu _1} \times \cdots \times GL_{\nu _m}$ and for the automorphism given by $c_A:T(u) \mapsto A^{-1}T(u)A$ we have

(5.70) $$ \begin{align} \begin{array}{rcl} c_A(E_{a,a+1}^{(r)}) &= & A_{\nu_a}^{-1}E_{a,a+1}^{(r)}A_{\nu_{a+1}} \\ c_A(F_{a+1,a}^{(r)})&=& A_{\nu_{a+1}}^{-1}F_{a+1,a,}^{(r)}A_{\nu_a} \\ c_A(D_{a,a}^{(r)})&=& A_{\nu_a}^{-1}D_{a,a}^{(r)}A_{\nu_a} \end{array} \end{align} $$

where $(E_{a,a+1}^{(r)})_{i,j}=E_{a,a+1,i,j}^{(r)}$ , $(F_{a+1,a}^{(r)})_{i,j}=F_{a+1,a,i,j}^{(r)}\, (D_{a,a}^{(r)})_{i,j}=D_{a,a,i,j}^{(r)}$ . The claim then follows by direct calculation.

From the description of the matrices $J_{\nu }$ and $J_{\mu }$ it follows that there exists $B \in GL_{\nu _1} \times \dots \times GL_{\nu _m}$ such that $J_{\mu }=B^{t}J_{\nu }B$ . It follows from (5.70) that the automorphism $c_{B}:Y_n \to Y_n : T(u) \mapsto B^{-1}T(u)B$ stabilises the image of $i_{\nu }$ and that it is filtered with respect to the canonical filtration as defined in [Reference Brundan and KleshchevBK06, §5]. It thus follows that $i_{\mu }^{-1} \circ c_{B} \circ i_{\nu }:{}^{\nu }Y_n(\sigma ) \to {}^{\mu }Y_n(\sigma ) $ is well defined and filtered.

Recall from the proof of Theorem 5.6 that the map $\pi _{\nu }: {}^{\nu }y_n(\sigma ) \twoheadrightarrow \operatorname {gr} \ ^{\nu }Y_n(\sigma )$ given by $e_{a,i,j}^{(r)} \mapsto E_{a,i,j}^{(r)} + \operatorname {F}^{\prime }_{r-1} \ ^{\nu }Y_n(\sigma ), f_{a,i,j}^{(r)} \mapsto F_{a,i,j}^{(r)} + \operatorname {F}^{\prime }_{r-1} \ ^{\nu }Y_n(\sigma ), d_{a,i,j}^{(r)} \mapsto D_{a,i,j}^{(r)} + \operatorname {F}^{\prime }_{r-1} \ ^{\nu }Y_n(\sigma )$ is an isomorphism of Poisson algebras, where $\operatorname {F}'$ denotes the canonical filtration. In particular $\pi _{\mu }^{-1} \circ \operatorname {gr}( i_{\mu }^{-1} \circ c_{B} \circ i_{\nu }) \circ \pi _{\nu }$ is the desired isomorphism.

Proof. In this proof we will frequently use the following two basic facts: $(\beta )^{i+j}=(\beta )^{i'+j'}$ , which follows from $(\beta , \sigma )$ -admissibility, and $xy+\tau (xy)=\frac {1}{2}(x+\tau (x))(y+\tau (y))$ for any $x,y \in R(y_n(\sigma ),\tau )$ .

First note that $\widetilde {\eta }_{a,i,j}^{(r)}=\frac {1}{2}(\tilde {d}_{a,i,j}^{(r)}+\tau (\tilde {d}_{a,i,j}^{(r)}))$ . The proof is by induction on r

(5.84) $$ \begin{align} \begin{aligned} \widetilde{\eta}_{a,i,j}^{(r)}&=-\sum_{g=1}^{\nu_a} \sum_{t=1}^{r} \eta_{a,i,g}^{(t)} \tilde{\eta}_{a,g,j}^{(r-t)}\\ &=-\frac{1}{4}\sum_{g=1}^{\nu_a} \sum_{t=1}^{r} (d_{a,i,g}^{(t)}+\tau(d_{a,i,g}^{(t)})) (\tilde{d}_{a,g,j}^{(r-t)}+\tau(\tilde{d}_{a,g,j}^{(r-t)}))\\ &=-\frac{1}{2} \sum_{g=1}^{\nu_a} \sum_{t=1}^{r} d_{a,i,g}^{(t)}\tilde{d}_{a,g,j}^{(r-t)}+\tau(d_{a,i,g}^{(t)})\tau(\tilde{d}_{a,g,j}^{(r-t)}) \end{aligned} \end{align} $$

Relation (5.76) follows immediately. For relation (5.77) we have

$$ \begin{align*} \{\eta_{a,i,j}^{(r)},\eta_{b,h,k}^{(s)}\}&=\frac{1}{4} \left(\{d_{a,i,j}^{(r)},d_{b,h,k}^{(s)}\}+ (\beta)^{i+j+h+k}(-1)^{r+s}\{d_{a,j',i'}^{(r)},d_{b,k',h'}^{(s)}\} \right)\\&\quad +\frac{1}{4}\left(((\beta)^{i+j}-1)^{r}\{d_{a,j',i'}^{(r)},d_{b,h,k}^{(s)}\}+(\beta)^{h+k}(-1)^{s}\{\eta_{a,i,j}^{(r)},\eta_{b,k',h'}^{(s)}\}\right) \\&= \frac{1}{4}\delta_{a,b} \sum_{t=0}^{\min(r,s)-1} \left( d_{a;i,k}^{(r+s-1-t)}d_{a;h,j}^{(t)} -d_{a;i,k}^{(t)}d_{a;h,j}^{(r+s-1-t)} \right)\\&\quad +\frac{1}{4} \delta_{a,b}(\beta)^{i+j+h+k}(-1)^{r+s}\sum_{t=0}^{\min(r,s)-1} \left( d_{a;j',h'}^{(r+s-1-t)}d_{a;k',i'}^{(t)} -d_{a;j',h'}^{(t)}d_{a;k',i'}^{(r+s-1-t)}\right) \\&\quad +\frac{1}{4}\delta_{a,b}(\beta)^{i+j}(-1)^{r}\sum_{t=0}^{\min(r,s)-1} \left( d_{a;j',k}^{(r+s-1-t)}d_{a;h,i'}^{(t)} -d_{a;j',k}^{(t)}d_{a;h,i'}^{(r+s-1-t)}\right) \\&\quad +\frac{1}{4} \delta_{a,b}(\beta)^{h+k}(-1)^{s}\sum_{t=0}^{\min(r,s)-1} \left( d_{a;i,h'}^{(r+s-1-t)}d_{a;k',j}^{(t)} -d_{a;i,h'}^{(t)}d_{a;k',j}^{(r+s-1-t)}\right)\\&= \frac{1}{2}\delta_{a,b} \sum_{t=0}^{\min(r,s)-1} \left( \eta_{a;i,k}^{(r+s-1-t)}\eta_{a;h,j}^{(t)} -\eta_{a;i,k}^{(t)}\eta_{a;h,j}^{(r+s-1-t)} \right) \\&\quad +\frac{1}{2}\delta_{a,b}(\beta)^{h+k}(-1)^{s}\sum_{t=0}^{\min(r,s)-1}\left(\eta_{a;i,h'}^{(r+s-1-t)}\eta_{a;k',j}^{(t)} -\eta_{a;i,h'}^{(t)}\eta_{a;k',j}^{(r+s-1-t)} \right) \end{align*} $$

Using relation (5.41). For relation (5.78) we have

$$ \begin{align*} \{\eta_{a,i,j}^{(r)}, \theta_{b,h,k}^{(s)}\}&=\frac{1}{4}\left(\{d_{a,i,j}^{(r)},e_{b,h,k}^{(s)}\}+(\beta)^{i+j+h+k}(-1)^{r+s+s_{b,b+1}} \{d_{a,j',i'}^{(r)}, f_{b,k',h'}^{(s)}\}\right) \\&\quad +\frac{1}{4}\left((\beta)^{i+j}(-1)^{r} \{d_{a,j',i'}^{(r)}, e_{b,h,k}^{(s)}\}+(\beta)^{h+k}(-1)^{s+s_{b,b+1}} \{d_{a,i,j}^{(r)}, f_{b,k',h'}^{(s)}\}\right)\\&= \frac{1}{4} \left( \delta_{a,b} \sum_{t=0}^{r-1} \sum_{g=1}^{\nu_a}d_{a;i,g}^{(t)} e_{a;g,k}^{(r+s-1-t)}\delta_{h,j} - \delta_{a,b+1} \sum_{t=0}^{r-1} d_{b+1;i,k}^{(t)} e_{b;h,j}^{(r+s-1-t)} \right) \\&\quad +\frac{1}{4}(\beta)^{i+j+h+k}(-1)^{r+s+s_{b,b+1}}\left( \delta_{a,b+1} \sum_{t=0}^{r-1} d_{b+1;k',i'}^{(t)} f_{b;j',h'}^{(r+s-1-t)}- \delta_{a,b} \sum_{t=0}^{r-1} \sum_{g=1}^{\nu_a}d_{a;g,i'}^{(t)} f_{a;k',g}^{(r+s-1-t)}\delta_{h',j'} \right)\\&\quad + \frac{1}{4}(\beta)^{i+j}(-1)^{r}\left( \delta_{a,b} \sum_{t=0}^{r-1} \sum_{g=1}^{\nu_a}d_{a;j',g}^{(t)} e_{a;g,k}^{(r+s-1-t)}\delta_{h,i'} - \delta_{a,b+1} \sum_{t=0}^{r-1} d_{b+1;j',k}^{(t)} e_{b;h,i'}^{(r+s-1-t)}\right)\\&\quad +\frac{1}{4}(\beta)^{k+h}(-1)^{s+s_{b,b+1}}\left( \delta_{a,b+1} \sum_{t=0}^{r-1} d_{b+1;k',j}^{(t)} f_{b;i,h'}^{(r+s-1-t)}- \delta_{a,b} \sum_{t=0}^{r-1} \sum_{g=1}^{\nu_a}d_{a;g,j}^{(t)} f_{a;k',g}^{(r+s-1-t)}\delta_{h',i} \right)\\&= \frac{1}{2} \left( \delta_{h,j} \delta_{a,b} \sum_{t=0}^{r-1} \sum_{g=1}^{\nu_{\alpha}} \eta_{a,i,g}^{(t)}\theta_{a,g,k}^{(r+s-1-t)}-\delta_{a,b+1} \sum_{t=0}^{r-1}\eta_{b+1,i,k}^{(t)}\theta_{b,h,j}^{(r+s-1-t)}\right)\\&\quad + \frac{1}{2}(\beta)^{i+j}(-1)^{r} \left( \delta_{h,i'} \delta_{a,b} \sum_{t=0}^{r-1} \sum_{g=1}^{\nu_{\alpha}} \eta_{a,j',g}^{(t)}\theta_{a,g,k}^{(r+s-1-t)}-\delta_{a,b+1} \sum_{t=0}^{r-1}\eta_{b+1,j',k}^{(t)}\theta_{b,h,i'}^{(r+s-1-t)}\right) \end{align*} $$

Using relations (5.43) and (5.44). For $r<s$ we have

$$ \begin{align*} \{\theta_{a,i,j}^{(r)}, \theta_{a,h,k}^{(s)}\}&= \frac{1}{4}\left( \{e_{a,i,j}^{(r)}, e_{a,h,k}^{(s)}\}- (\beta)^{i+j+h+k}(-1)^{r+s}\{f_{a,k',h'}^{(s)} f_{a,j',i'}^{(r)}\} \right)\\&\quad + \frac{1}{4}\left((\beta)^{h+k}(-1)^{s+s_{a,a+1}}\{e_{a,i,j}^{(r)}, f_{a,k',h'}^{(s)}\}- (\beta)^{i+j}(-1)^{r+s_{a,a+1}}\{e_{a,h,k}^{(s)} f_{a,j',i'}^{(r)}\}\right)\\&=\frac{1}{4}\sum_{t=r}^{s-1} e_{a;i,k}^{(t)} e_{a;h,j}^{(r+s-1-t)} \\&\quad -\frac{1}{4} (\beta)^{i+j+k+h}(-1)^{r+s}\sum_{t=r}^{s-1} f_{a;k',i'}^{(t)} f_{a;h',j'}^{(r+s-1-t)} \\&\quad +\frac{1}{4}(\beta)^{h+k}(-1)^{s+s_{a,a+1}} \sum_{t=0}^{r+s-1} \widetilde{d}_{a;i,h'}^{(r+s-1-t)}d_{a+1;k',j}^{(t)} \\&\quad -\frac{1}{4}(\beta)^{i+j}(-1)^{r+s_{a,a+1}}\sum_{t=0}^{r+s-1} \widetilde{d}_{a;h,i'}^{(r+s-1-t)}d_{a+1;j',k}^{(t)} \\&=\frac{1}{2} \left( \sum_{t=r}^{s-1} \theta_{a,i,k}^{(t)} {\theta}_{a,k,j}^{(r+s-1-t)}+(\beta)^{h+k}(-1)^{s_a+s} \sum_{t=0}^{r+s-1}\widetilde{\eta}_{a,i,h'}^{(r+s-1-t)}\eta_{a+1,k',j}^{(t)} \right)\ \end{align*} $$

Where we used relations (5.42), (5.45) and (5.46). The next relation follows from using relations (5.42), (5.47) and (5.48).

$$ \begin{align*} \{\theta_{a,i,j}^{(r)}, \theta_{a+1,h,k}^{(s+1)}\}-\{\theta_{a,i,j}^{(r+1)}, \theta_{a+1,h,k}^{(s)}\}&=\frac{1}{4}\left(\{e_{a,i,j}^{(r)}, e_{a+1,h,k}^{(s+1)}\}-\{e_{a,i,j}^{(r+1)}, e_{a+1,h,k}^{(s)}\}\right)\\&\quad +\frac{1}{4}(\beta)^{i+j+k+h}(-1)^{s_{a,a+2}+r+s} \{f_{a,j',i'}^{(r+1)}, f_{a+1,k',h'}^{(s)}\}\\&\quad -\frac{1}{4}(\beta)^{i+j+k+h}(-1)^{s_{a,a+2}r+s}\{f_{a,j',i'}^{(r)}, f_{a+1,k',h'}^{(s+1)}\} \\&=-\frac{1}{4}\delta_{h,j}\sum_{g=1}^{\nu_{a+1}}e_{a;i,g}^{(r)} e_{a+1;g,k}^{(s)}\\&\quad -\frac{1}{4}(\beta)^{i+j+h+k}(-1)^{s_{a,a+2}+r+s}\delta_{h',j'}\sum_{g=1}^{\nu_{a+1}}f_{a+1;k',g}^{(s)}f_{a;g,i'}^{(r)} \\&=-\frac{1}{2}\delta_{h,j}\left( \sum_{g=1}^{\nu_{a+1}}\theta^{(r)}_{a,i,g}\theta^{(s)}_{a+1,g,k}\right) \end{align*} $$

Finally the last relation follows from using relations (5.42), (5.51), (5.52), (5.43), (5.44) and Lemma 5.7.

$$ \begin{align*} &\{\theta_{a,i,j}^{(r)}, \{\theta_{a,h,k}^{(s)}, \theta_{b,f,g}^{(t)}\}\} +\{\theta_{a,i,j}^{(s)}, \{\theta_{a,h,k}^{(r)}, \theta_{b,f,g}^{(t)}\}\} \\&\quad = \frac{1}{8}(\beta)^{h+k+f+g}(-1)^{\mathfrak{s}_{a,a+1}+\mathfrak{s}_{b,b+1}+s+t}\{e_{a,i,j}^{(r)}, \{f_{a,k',h'}^{(s)}, f_{b,g',f'}^{(t)}\}\}\\&\qquad +\frac{1}{8}(\beta)^{h+k+f+g}(-1)^{\mathfrak{s}_{a,a+1}+\mathfrak{s}_{b,b+1}+r+t}\{e_{a,i,j}^{(s)}, \{f_{a,k',h'}^{(r)}, f_{b,g',f'}^{(t)}\}\} \\&\qquad +\frac{1}{8} (\beta)^{i+j}(-1)^{\mathfrak{s}_{a,a+1}+r}\{f_{a,j',i'}^{(r)}, \{e_{a,h,k}^{(s)}, e_{b,f,g}^{(t)}\}\}\\&\qquad +\frac{1}{8} (\beta)^{i+j}(-1)^{\mathfrak{s}_{a,a+1}+s}\{f_{a,j',i'}^{(s)}, \{e_{a,h,k}^{(r)}, e_{b,f,g}^{(t)}\}\} \\&\quad =\frac{1}{8}(\beta)^{h+k+f+g}(-1)^{s+t + \mathfrak{s}_{a,a+1}+ \mathfrak{s}_{b,b+1}}(1 + (-1)^{r+s}) \{\{e_{a,i,j}^{(r)}, f_{a,k',h'}^{(s)}\}, f_{b,g',f'}^{(t)}\} \\&\qquad -\frac{1}{8} (\beta)^{i+j}(-1)^{r+ \mathfrak{s}_{a,a+1}}(1 + (-1)^{r+s}) \{\{e_{a,h,k}^{(r)}, f_{a,j',i'}^{(s)}\}, e_{b,f,g}^{(t)}\} \\&\quad =\frac{1}{8}(\beta)^{h+k+f+g}(-1)^{s+t + \mathfrak{s}_{a,a+1}+ \mathfrak{s}_{b,b+1}}(1 + (-1)^{r+s})\left\{ \sum_{t'=0}^{r+s-1} \widetilde{d}_{a;i,h'}^{(r+s-1-t')}d_{a+1;k',j}^{(t')},f_{b,g',f'}^{(t)}\right\}\\&\qquad -\frac{1}{8}(\beta)^{i+j}(-1)^{r + \mathfrak{s}_{a,a+1}}(1 + (-1)^{r+s}) \left \{\sum_{t'=0}^{r+s-1} \widetilde{d}_{a;h,i'}^{(r+s-1-t')}d_{a+1;j',k}^{(t')},e_{b,f,g}^{(t)}\right \} \end{align*} $$

If $r+s$ is odd then this vanishes whereas if $r+s$ is even this simplifies too

$$ \begin{align*} &-\frac{1}{4}(\beta)^{h+k+f+g}(-1)^{s+t + \mathfrak{s}_{a,a+1} + \mathfrak{s}_{b,b+1}}\sum_{t'=0}^{r+s-1} \left( \widetilde{d}_{a;i,h'}^{(r+s-1-t')}\delta_{a+1,b} \delta_{k',f'}\sum_{t"=0}^{t'-1}\sum_{m=1}^{\nu_{a+1}} f_{a+1;g',m}^{(t'+t-1-t")}d_{a+1;m,j}^{(t")} \right) \\&-\frac{1}{4}(\beta)^{h+k+f+g}(-1)^{s+t + \mathfrak{s}_{a,a+1} + \mathfrak{s}_{b,b+1}}\sum_{t'=0}^{r+s-1} \left(d_{a+1;k',j}^{(t')} \delta_{a,b+1} \delta_{g,h}\sum_{t"=0}^{r+s-t'-2} \sum_{m=1}^{\nu_a}\widetilde{d}_{a;i,m}^{(t")} f_{b;m,f'}^{(r+s-t'+t-2-t")} \right) \\&-\frac{1}{4}(\beta)^{i+j}(-1)^{r + \mathfrak{s}_{a,a+1}}\sum_{t'=0}^{r+s-1} \left(\widetilde{d}_{a;h,i'}^{(r+s-1-t')}\delta_{a+1,b} \delta_{f,k} \sum_{t"=0}^{t'-1} \sum_{m=1}^{\nu_{a+1}}d_{a+1;j',m}^{(t")} e_{a+1;m,g}^{(t'+t-1-t")} \right)\\&-\frac{1}{4}(\beta)^{i+j}(-1)^{r+ \mathfrak{s}_{a,a+1}}\sum_{t'=0}^{r+s-1} \left(d^{(t')}_{a+1,j',k}\delta_{a,b+1}\delta_{h,g}\sum_{t"=0}^{r+s-t'-2}\sum_{m=1}^{\nu_a} e_{b;f,m}^{(r+s-t'+t-2-t")} \widetilde{d}_{a;m,i'}^{(t")} \right) \\&=\frac{1}{2}(\beta)^{i+j}(-1)^{r+ \mathfrak{s}_{a,a+1}+1}\sum_{t'=0}^{r+s-1} \left(\widetilde{\eta}_{a;h,i'}^{(r+s-1-t')}\delta_{a+1,b} \delta_{f,k} \sum_{t"=0}^{t'-1} \sum_{m=1}^{\nu_{a+1}}\eta_{a+1;j',m}^{(t")} \theta_{a+1;m,g}^{(t'+t-1-t")} \right)\\&+\frac{1}{2}(\beta)^{i+j}(-1)^{r+ \mathfrak{s}_{a,a+1}+1}\sum_{t'=0}^{r+s-1} \left(\eta^{(t')}_{a+1,j',k}\delta_{a,b+1}\delta_{h,g}\sum_{t"=0}^{r+s-t'-2}\sum_{m=1}^{\nu_a} \theta_{b;f,m}^{(r+s-t'+t-2-t")} \widetilde{\eta}_{a;m,i'}^{(t")} \right) \end{align*} $$

Let $\widehat {R}(^{\nu }y_n(\sigma ), \tau )$ denote the Poisson algebra with generators (5.75) and relations (5.76)-(5.83). We have shown that there is a surjection $\widehat {R}(^{\nu }y_n(\sigma ), \tau ) \twoheadrightarrow {R}(^{\nu }y_n(\sigma ), \tau ) $ . To conclude the proof we show that it sends a spanning set to a basis. We define a loop filtration on $\widehat {R}(^{\nu }y_n(\sigma ), \tau )=\bigcup _{i \geq 0}F_{i}\widehat {R}(^{\nu }y_n(\sigma ), \tau )$ by placing the generators $\theta _{a,i,j}^{(r)}, \eta _{a,i,j}^{(r)}$ in degree $r-1$ and the bracket in degree 0.

Examining the top filtered components of (5.77)-(5.83) we see from Theorem 5.3 that there is a surjective Poisson morphism $R(S(\mathfrak {c}_n(\sigma )), \tau ) \twoheadrightarrow \operatorname {gr} \widehat {R}(^{\nu }y_n(\sigma ), \tau ) $ . We thus see that $\operatorname {gr} \widehat {R}(^{\nu }y_n(\sigma ), \tau )$ is generated as a commutative algebra by the elements $\eta _{a,i,j}^{(r)}+F_{r-2}\widehat {R}(^{\nu }y_n(\sigma ), \tau ),\theta _{a,i,b,j}^{(r)}+F_{r-2}\widehat {R}(^{\nu }y_n(\sigma ), \tau )$ which are again defined via the recursion (5.74).

By a standard filtered algebra argument $ \widehat {R}(^{\nu }y_n(\sigma ), \tau )$ is generated as a commutative algebra by the elements with the same names in $\operatorname {gr} \widehat {R}(^{\nu }y_n(\sigma ), \tau )$ . We have thus shown that $\widehat {R}(^{\nu }y_n(\sigma ), \tau ) \twoheadrightarrow {R}(^{\nu }y_n(\sigma ), \tau ) $ maps a spanning set to a basis, which concludes the proof.

Proof. The existence of the homomorphism from the shifted Yangian to the quantum finite W-algebra is proven by Brundan and Kleshchev in [Reference Brundan and KleshchevBK06, Theorem 10.1]. The result stated here follows from theirs by taking the associated graded algebras with respect to the canonical filtration and the Kazhdan filtration. The $\tau $ -equivariance follows by comparing Proposition 4.5 with formula (5.63).

The fact that the elements (6.1) lie in the kernel of the map is an immediate consequence of [Reference Brundan and KleshchevBK06, Theorem 10.1]. The fact that these elements generate the kernel can be proven using an argument identical to [Reference TopleyTo23, Proposition 4.7], deploying our Lemma 5.5(1).

Proof. The existence of this surjective map follows by [Reference TopleyTo23, (2.6) &Theorem 2.6] and Theorem 6.1. For the kernel consider first the case (i). It follows from Theorem 6.1 that the elements $\{\eta _{1,i,j}^{(r)} \mid r> \lambda _1 \ , 1 \le i,j \le \nu _1\}$ lie in the kernel of $\varphi $ . Let I denote the Poisson ideal generated by 6.3. By Lemma 5.5(2)(i) we see that the associated graded with respect to the loop filtration $\operatorname {gr}(R(^\nu y_n(\sigma ), \tau )/I)$ is generated as a commutative algebra by the top filtered component of the elements

(6.5) $$ \begin{align} \begin{aligned} &\{\theta_{a,b,i,j}^{(r)} \mid 1 \leq a <b \leq m, \ 1 \leq i \leq \nu_a, \ 1 \leq j \leq \nu_b, \ s_{a,b}(\nu)<r \leq s_{a,b}(\nu)+\lambda_a(\nu) \} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \cup \{\eta_{a,i,j}^{(r)} \mid 1 \leq a \leq m, \ 1 \leq i,j \leq \nu_a, \ 0<r \leq \lambda_a(\nu)\} \end{aligned} \end{align} $$

modulo the linear relation (5.76). In particular $R(^\nu y_n(\sigma ), \tau )/I$ is generated by these elements as a commutative algebra.

By [Reference TopleyTo23, Theorem 2.4] we have that $S(\mathfrak {g}_0, e)$ is a polynomial algebra modulo the same linear relation generated by the images of these elements under $\varphi $ and, hence, $\varphi $ is an isomorphism.

Now for cases (ii) and (iii) it follows by Theorem 6.1 that the elements 6.4 lie in the kernel of $\varphi $ . Again let I denote the poisson ideal generated by the elements (6.3) and (6.4). As before it is enough to show that $R(^\nu y_n(\sigma ), \tau )/I$ is generated as a commutative algebra by the elements (6.5) modulo (5.76). A calculation in $^\nu y_n(\sigma )$ using (5.16) shows that the top filtered component of $\dot {\theta }_{1,i,j}^{( \lambda _1(\nu ) + s_{1,2}(\nu )+1)}$ with respect to the loop filtration is $\theta _{1,i,j; \lambda _1(\nu ) + s_{1,2}(\nu )}$ .

Using Lemma 5.5(2)(ii) and (iii) we see that the associated graded with respect to the loop filtration $\operatorname {gr}(R(^\nu y_n(\sigma ), \tau )/I)$ is generated as a commutative algebra by the top filtered component of the elements (6.5) modulo (5.76). The proof now proceeds as in case (i).