2.1 The fixed point subalgebra

In this section, we introduce the fixed point subalgebra of $\mathfrak {gl}_{2n+1}(\mathbb {C})$ with respect to an involution on its Dynkin diagram. We will further give a presentation of this fixed point subalgebra. Let $\Gamma $ be the Dynkin diagram associated to the Lie algebra $\mathfrak {g}=\mathfrak {gl}_{2n+1}(\mathbb {C})$ . It has $2n$ vertices $1,2,\dots ,2n$ and a graph automorphism $\tau \in \operatorname {Aut}(\Gamma )$ such that $\tau (i)=2n+1-i$ . Let $E_1,\dots ,E_{2n},F_{1},\dots ,F_{2n},H_1,\dots ,H_{2n+1}$ be the Chevalley generators of $\mathfrak {g}$ and let $\mathfrak {h}\subset \mathfrak {g}$ be the Cartan subalgebra spanned by $H_i (1\leq i\leq 2n+1)$ . In other words, $E_i=E_{i,i+1}$ , $F_i=F_{i+1,i}$ , $H_i=E_{i,i}$ , where $E_{i,j}$ is the matrix unit with a single nonzero entry $1$ in the $(i,j)$ -position. The automorphism $\tau $ induces an automorphism of the Lie algebra $\mathfrak {gl}_{2n+1}(\mathbb {C})$ , which can be extended to an automorphism $\theta $ of $\mathfrak {g}=\mathfrak {gl}_{2n+1}(\mathbb {C})$ via

$$ \begin{align*} \theta(E_{i})=F_{\tau(i)}, \hspace{.2 in}\theta(F_{i})=E_{\tau(i)},\hspace{.2 in}\theta(H_{i})= H_{\tau(i)+1}. \end{align*} $$

In fact, for any subset X of the vertices in $\Gamma $ , and any graph automorphism $\tau $ on $\Gamma $ which leaves X invariant, Kolb [Reference Kolb8, Theorem 2.5] defined an automorphism $\theta (X,\tau )$ of the associated Lie algebra. When $X=\emptyset $ and $\tau $ is taken as above, the automorphism $\theta (X,\tau )$ specializes to the automorphism mentioned above.

Let

$$ \begin{align*} \mathfrak{g}^{\theta}=\{x\in \mathfrak{g}\hspace{.1 in} | \hspace{.1 in}x=\theta(x)\}, \hspace{.2 in} \mathfrak{h}^{\theta}= \{h\in \mathfrak{h} \hspace{.1 in} | \hspace{.1 in} h=\theta(h)\}. \end{align*} $$

The result in [Reference Kolb8, Lemma 2.8] specializes to the following

Lemma 2.1.1. [Reference Kolb8, Lemma 2.8] The subalgebra $\mathfrak {g}^{\theta }$ is generated by the following elements in $\mathfrak {h}^{\theta }$ and

$$ \begin{align*} \{F_i+\theta(F_i) \hspace{.1 in} | \hspace{.1 in}1\leq i\leq 2n\}. \end{align*} $$

We now give a description of generators for $\mathfrak {h}^{\theta }$ .

Proof. Let $h=c_1H_1+\cdots +c_{2n+1}H_{2n+1}$ be fixed by $\theta $ , then by the definition of $\theta $

$$ \begin{align*} c_1H_1+\cdots +c_{2n+1}H_{2n+1}=c_{2n+1}H_1+\cdots+ c_{1}H_{2n+1} \end{align*} $$

and $c_i=c_{2n+1-i}$ , hence h is a linear combination of $H_i+H_{2n+2-i}$ .

Let $U^j=U(\mathfrak {g}^{\theta })$ be the universal enveloping algebra of $\mathfrak {g}^{\theta }$ . By the two lemmas above, $U^j$ is generated by elements

(1) $$ \begin{align} \mathbf{e}_i & = E_i+F_{2n+1-i} ,\hspace{.5 in}1\leq i\leq n, \end{align} $$

(2) $$ \begin{align} \mathbf{f}_i & = F_i+E_{2n+1-i}, \hspace{.5 in}1\leq i\leq n, \end{align} $$

(3) $$ \begin{align} \mathbf{d}_i & = H_i+H_{2n+2-i}, \hspace{.5 in}1\leq i\leq n, \end{align} $$

(4) $$ \begin{align} \mathbf{d}_{n+1} & = 2 H_{n+1}. \end{align} $$

We aim to give a presentation of $U^j$ . The following relations are known to be true for the quantum case in [Reference Letzter10, Lemma 2.2].

Proposition 2.1.4. The algebra $U^j$ is generated by $\mathbf {e}_1,\dots ,\mathbf {e}_n$ , $\mathbf {f}_1,\dots ,\mathbf {f}_n$ , $\mathbf {d}_1,\dots ,\mathbf {d}_{n+1}$ subject to the following relations

(5) $$ \begin{align} \mathbf{d}_i\mathbf{d}_j&=\mathbf{d}_j\mathbf{d}_i, \end{align} $$

(6) $$ \begin{align} [\mathbf{d}_a,\mathbf{e}_j]&=(-\delta_{a,j+1}-\delta_{2n+2-a,j+1}+\delta_{a,j})\mathbf{e}_j, \end{align} $$

(7) $$ \begin{align} [\mathbf{d}_a,\mathbf{f}_j]&=(-\delta_{a,j}+\delta_{a,j+1}+\delta_{2n+2-a,j+1})\mathbf{f}_j, \end{align} $$

(8) $$\begin{align}\mathbf{e}_i\mathbf{f}_j-\mathbf{f}_j\mathbf{e}_i&=\delta_{ij}(\mathbf{d}_i-\mathbf{d}_{i+1}) \hspace{.5 in}i<n, \end{align} $$

(9) $$ \begin{align}\mathbf{e}_i^2\mathbf{e}_j-2\mathbf{e}_i\mathbf{e}_j\mathbf{e}_i+\mathbf{e}_j\mathbf{e}_i^2&=0\hspace{.5 in} |i-j|=1, \end{align} $$

(10) $$ \begin{align}\mathbf{f}_i^2\mathbf{f}_j-2\mathbf{f}_i\mathbf{f}_j\mathbf{f}_i+\mathbf{f}_j\mathbf{f}_i^2&=0 \hspace{.5 in} |i-j|=1, \end{align} $$

(11) $$ \begin{align}\mathbf{e}_i\mathbf{e}_j&=\mathbf{e}_j\mathbf{e}_i\hspace{.5 in} |i-j|>1, \end{align} $$

(12) $$\begin{align} \mathbf{f}_i\mathbf{f}_j&=\mathbf{f}_j\mathbf{e}_i\hspace{.5 in} |i-j|>1, \end{align} $$

(13) $$ \begin{align} mathbf{e}_n^2\mathbf{f}_n-2\mathbf{e}_n\mathbf{f}_n\mathbf{e}_n+\mathbf{f}_n\mathbf{e}_n^2=-4\mathbf{e}_n, \end{align} $$

(14) $$ \begin{align}\mathbf{f}_n^2\mathbf{e}_n-2\mathbf{f}_n\mathbf{e}_n\mathbf{f}_n+\mathbf{e}_n\mathbf{f}_n^2&=-4\mathbf{f}_n. \end{align} $$

Proof. These relations hold by a straightforward calculation.

Let B be the algebra generated by ${\mathbf {e}_1},\dots ,\mathbf {e}_n$ , $\mathbf {f}_1,\dots ,\mathbf {f}_n$ , $\mathbf {d}_1,\dots ,\mathbf {d}_{n+1}$ subject to Relations (5), (6), and (7), then B has a basis $\{w\mathbf {d}^{\mathbf {s}}\},$ where w is a word in $\mathbf {e}_1,\dots ,\mathbf {e}_n$ , $\mathbf {f}_1,\dots ,\mathbf {f}_n$ , and $\mathbf {s}=(s_1,\dots ,s_{n+1})\in \mathbb {Z}_{\geq 0}^{n+1}$ with $\mathbf {d}^{\mathbf {s}}=\mathbf {d}_1^{s_1}\cdots \mathbf {D}_{n+1}^{s_{n+1}}$ . Define a filtration on B by declaring $\operatorname {deg} \mathbf {e}_i = \operatorname {deg} \mathbf {f}_i=1$ and $\operatorname {deg} \mathbf {D}_i=0$ , then any element in B can be written as a sum of homogeneous parts, and declare its degree to be the highest among its homogeneous parts. On the other hand, the algebra $U(\mathfrak {gl}_{2n+1}(\mathbb {C}))$ has a Poincaré–Birkhoff–Witt (PBW) -basis and admits a filtration. The degree of each monomials is defined via $\operatorname {deg}E_i=1$ , $\operatorname {deg}F_i=0$ , $\operatorname {deg}H_i=0$ . As a subalgebra of $U(\mathfrak {gl}_{2n+1}(\mathbb {C}))$ , $U^j$ inherits a filtration where degrees on monomials are defined similarly.

Let I be the ideal in B generated by Relations (8)–(14). The previous lemma implies that there is a well-defined surjection $\phi : B\to U^j$ with I lying in the kernel. Now we claim that the kernel is exactly I. Assume by contradiction that

$$ \begin{align*} x=\displaystyle\sum_{w,\textbf{s}}a_{w,\textbf{s}}w\mathbf{d}^{\textbf{s}} \end{align*} $$

is in the kernel of $\phi $ , $x\not \in I$ , $a_{w,\textbf {s}}\in \mathbb {C}$ and x is of smallest degree in B. For notation purposes, rewrite $\mathbf {f}_i=\mathbf {e}_{2n+1-i}$ for $1\leq i\leq n$ . By the definition of $\phi $ , if $w=\mathbf {e}_{i_1}\mathbf {e}_{i_2}\dots \mathbf {e}_{i_k}$ , then

$$ \begin{align*} \phi(a_{w,\textbf{s}}w\mathbf{d}^{{\textbf{s}}})=a_{\textbf{s}} E_{i_1}E_{i_2}\ldots E_{i_k}\phi(\mathbf{d}^{{\textbf{s}}}) + \text{ lower terms in }U^j. \end{align*} $$

In the above case, let $E^w=E_{i_1}E_{i_2}\ldots E_{i_k}$ and let $I_0$ be the index set of $(w,\mathbf {s})$ for the top degree terms in x, then by degree considerations,

$$ \begin{align*} \sum_{\textbf{s}\in I_0}a_{w,\textbf{s}} E^{w}\phi(\mathbf{d}^{{\textbf{s}}}) =0 \in U(\mathfrak{gl}_{2n+1}(\mathbb{C})). \end{align*} $$

Moreover, based on the PBW basis of $U(\mathfrak {gl}_{2n+1}(\mathbb {C}))$ and the fact that $\phi (\mathbf {d}^{\mathbf {s}_1})=\phi (\mathbf {d}^{\mathbf {s}_2})$ if and only if $\mathbf {s}_1=\mathbf {s}_2$ , for each fixed $\mathbf {s}$ , we have

$$ \begin{align*} \sum_{w:a_{w,\mathbf{s}}\neq 0}a_{w,\textbf{s}} E^{w} =0 \in U(\mathfrak{gl}_{2n+1}(\mathbb{C})) \end{align*} $$

and the above element is in the ideal generated by the Serre relations for the positive part of $U(\mathfrak {gl}_{2n+1})$ . Comparing these relations and Relations (9)–(14), one can use the corresponding relations to rewrite

$$ \begin{align*} \displaystyle\sum_{w:a_{w,\mathbf{s}}\neq 0}a_{w,\textbf{s}} w \end{align*} $$

as an element of strictly less degree, resulting an element $x'$ in the kernel of $\phi $ , $x'\not \in I$ , whose degree is less than x, contradicting the choice of x.

2.2 A homomorphism between $U^j$ and $U(\mathfrak {gl}_{n+1}\oplus \mathfrak {gl}_n)$

Lemma 2.2.1. There is a well-defined algebra homomorphism $\psi : U^j \to U$ given by

(15) $$ \begin{align} \psi(\mathbf{e}_i) &= f_i+f_{\overline{i}} \hspace{.3 in}\psi(\mathbf{f}_i)= e_i+e_{\overline{i}} \hspace{.3 in} (i<n), \end{align} $$

(16) $$ \begin{align} \psi(\mathbf{e}_n) &= 2f_n \hspace{.5 in}\psi(\mathbf{f}_n)=e_n, \end{align} $$

(17) $$ \begin{align} \psi(\mathbf{d}_i) &= -h_i - h_{\overline{i}} \hspace{.1 in} (1\leq i\leq n) \hspace{.3 in} \psi(\mathbf{d}_{n+1})= -2h_{n+1}. \end{align} $$

Proof. The Relations (5), (9), (10), (11), and (12) are trivial to check. For (6), it boils down to

$$ \begin{align*} [\mathbf{d}_a,\mathbf{e}_j]&=(-\delta_{a,j+1}+\delta_{a,j})\mathbf{e}_j \end{align*} $$

unless $a=n+1$ and $j=n$ , which will be checked separately ( $2n+2-a=j+1$ is equivalent to $a+j=2n+1$ , and since $j \leq n$ and $a\leq n+1$ , this only holds when $j=n$ and $a=n+1$ .) The calculation is as follows:

$$ \begin{align*} [\psi(\mathbf{d}_a),\psi(\mathbf{e}_j)]&=[-h_a-h_{\overline{a}}, f_j+f_{\overline{j}}]\\ &=(-\delta_{a,j+1}+\delta_{a,j})f_j+f_{\overline{j}} =(-\delta_{a,j+1}+\delta_{a,j}) (\psi(\mathbf{e}_j)). \end{align*} $$

For the case $a=n+1$ and $j=n,$

$$ \begin{align*} [\psi(\mathbf{d}_{n+1}),\psi(\mathbf{e}_n)]=[-2h_{n+1},2f_{n}]=-4f_{n} =-2\psi(\mathbf{e}_n), \end{align*} $$

which is the value of $(-\delta _{a,j+1}-\delta _{2n+2-a,j+1}+\delta _{a,j})\psi (\mathbf {e}_j)$ .

Relation (7) can be checked similarly. For Relation (8), the relation holds when $i\neq j$ . When $i=j<n$ ,

$$ \begin{align*} [\psi(\mathbf{e}_i),\psi(\mathbf{f}_i)]&=[f_i+f_{\overline{i}},e_i+e_{\overline{i}}]=[f_i,e_i]+[f_{\overline{i}},e_{\overline{i}}]\\ &=-h_i+h_{i+1}-h_{\overline{i}}+h_{\overline{i+1}}=\psi(\mathbf{d}_i-\mathbf{d}_{i+1}). \end{align*} $$

Relations (13) is checked as follows ((14) can be checked similarly.)

$$ \begin{align*} \psi(\mathbf{e}_n^2\mathbf{f}_n-2\mathbf{e}_n\mathbf{f}_n\mathbf{e}_n+\mathbf{f}_n\mathbf{e}_n^2)&=\psi(\mathbf{e}_n)([\psi(\mathbf{e}_n),\psi(\mathbf{f}_n)])+[\psi(\mathbf{f}_n),\psi(\mathbf{e}_n)]\psi(\mathbf{e}_n)\\ &=4(f_n[f_n,e_n]+[e_n,f_n]f_n)=4(f_n(-h_n+h_{n+1})+(h_n-h_{n+1})f_n)\\ &=4([h_n,f_n]-[h_{n+1},f_n])=4(-f_n-f_n)=-8f_n=-4\psi(\mathbf{e}_n). \end{align*} $$

Lemma is proved.

Let $\mathbf {i}$ be the involution on $U(\mathfrak {gl}_{n+1}(\mathbb {C}))\otimes U(\mathfrak {gl}_{n}(\mathbb {C}))$ induced by the following involution on each factor: $U(\mathfrak {gl}_{n+1}(\mathbb {C}))\simeq U(\mathfrak {gl}_{n+1}(\mathbb {C}))$ via $e_i \mapsto f_i$ ( $1\leq i\leq n$ ), $f_i \mapsto e_i$ ( $1\leq i\leq n$ ), $h_i \mapsto -h_i$ ( $1\leq i\leq n+1$ ), and $U(\mathfrak {gl}_{n}(\mathbb {C}))\simeq U(\mathfrak {gl}_{n}(\mathbb {C}))$ similarly. Let $\phi =\mathbf {i}\circ \psi $ , then $\phi $ is given by

(18) $$ \begin{align} \phi(\mathbf{e}_i) &= e_i+e_{\overline{i}} \hspace{.3 in}\phi(\mathbf{f}_i)= f_i+f_{\overline{i}} \hspace{.3 in} (i<n) ,\nonumber\\ \phi(\mathbf{e}_n) &= 2e_n \hspace{.5 in} \phi(\mathbf{f}_n)= f_n ,\nonumber\\ \phi(\mathbf{d}_i) &= h_i + h_{\overline{i}} \hspace{.1 in} (1\leq i \leq n) \hspace{.3 in} \phi(\mathbf{d}_{n+1}) = 2h_{n+1}. \end{align} $$

2.3 Root vectors in $U^j$

Next we aim to give a weight space decomposition of $\mathfrak {g}^{\theta }$ . First let $\mathbf {d}_{n+1}'=\frac {1}{2}\mathbf {d}_{n+1}$ . Let $\epsilon _i=\mathbf {d}_i^*$ be the element in $(\mathfrak {h}^{\theta })^*$ dual to $\mathbf {d}_i$ for $1\leq i\leq n$ , and $\epsilon _{n+1}=(\mathbf {d}_{n+1}')^*$ .

We shall define root vectors in $\mathfrak {g}^{\theta }$ similar to the well-known root vectors in type A. Specifically, recall $h_1,\dots ,h_{n+1}$ is a basis of the Cartan subalgebra of $\mathfrak {gl}_{n+1}(\mathbb {C})$ , and $h_{\overline {1}},\dots ,h_{\overline {n}}$ is a basis of the Cartan subalgebra of $\mathfrak {gl}_{n}(\mathbb {C})$ . When context is clear and index makes sense, we identify $\epsilon _i$ with $h_i^* $ or $h_{\overline {i}}^*$ . Let

(19) $$ \begin{align} \Phi_{n+1}&=\{\epsilon_i-\epsilon_j \hspace{.1 in}|\hspace{.1 in} 1\leq i,j\leq n+1, i\neq j\} \nonumber\\ \Pi_{n+1}&=\{\epsilon_i-\epsilon_j \hspace{.1 in}|\hspace{.1 in} 1\leq i<j\leq n+1, j-i=1\}, \end{align} $$

be the set of roots and simple roots in $\mathfrak {gl}_{n+1}(\mathbb {C})$ . For any $\alpha \in \Phi _{n+1}$ , the root vectors $Y_{\alpha }\in \mathfrak {gl}_{n+1}(\mathbb {C})$ are defined recursively as follows. Recall that $e_1,\dots ,e_n$ are the standard upper-triangular Chevalley generators in $\mathfrak {gl}_{n+1}(\mathbb {C})$ . On a simple root $\alpha _i=\epsilon _i-\epsilon _{i+1}$ , set $Y_{\alpha _i}=e_i$ for $1\leq i\leq n$ . In general, let $\alpha =\epsilon _i-\epsilon _{j}$ where $i<j$ , define $Y_{\alpha }$ recursively via

(20) $$ \begin{align} Y_{\alpha}=Y_{(\epsilon_i-\epsilon_{i+1} )+(\epsilon_{i+1}-\epsilon_j)}=[Y_{\epsilon_i-\epsilon_{i+1}},Y_{\epsilon_{i+1}-\epsilon_j}]=[e_i,Y_{\epsilon_{i+1}-\epsilon_j}]. \end{align} $$

Similarly, for $\alpha _i=\epsilon _i-\epsilon _{i+1}$ , set $Y_{-\alpha _i}=f_i$ , where $f_1,\dots ,f_n$ are the standard lower-triangular Chevalley generators in $\mathfrak {gl}_{n+1}(\mathbb {C})$ , and define $Y_{-\alpha }$ for negative roots recursively as follows

(21) $$ \begin{align} Y_{-\alpha}=Y_{-(\epsilon_i-\epsilon_{i+1})-(\epsilon_{i+1}-\epsilon_j)}=[Y_{-(\epsilon_i-\epsilon_{i+1})},Y_{-(\epsilon_{i+1}-\epsilon_j)}]=[f_i,Y_{-\epsilon_{i+1}+\epsilon_j}] . \end{align} $$

Similarly, let

(22) $$ \begin{align} \Phi_{n}&=\{\epsilon_i-\epsilon_j \hspace{.1 in}|\hspace{.1 in} 1\leq i,j\leq n, i\neq j\} \hspace{.2 in} \nonumber\\ \Pi_{n}&=\{\epsilon_i-\epsilon_j \hspace{.1 in}|\hspace{.1 in} 1\leq i<j\leq n, j-i=1\} \end{align} $$

be the set of roots and simple roots in $\mathfrak {gl}_{n}(\mathbb {C})$ . For any $\alpha \in \Phi _{n}$ , the root vectors $Z_{\alpha }\in \mathfrak {gl}_{n}(\mathbb {C})$ are defined similarly.

It is well known that $\{Y_{\alpha }\}_{\alpha \in \Phi _{n+1}}\cup \{h_i\}_{i=1}^{n+1}$ form a basis for $\mathfrak {gl}_{n+1}(\mathbb {C})$ , and $[h_i,Y_{\alpha }]=\alpha (h_i)Y_{\alpha }$ . Similar results also hold for $Z_{\alpha }$ .

In the same fashion, we construct root vectors in $\mathfrak {g}^{\theta }$ . Recall that $\{H_1,\dots ,H_{2n+1}\}$ is a basis of the Cartan subalgebra in $\mathfrak {gl}_{2n+1}(\mathbb {C})$ . Let $\mu _i=H_i^*$ be the element in $\mathfrak {h}^*$ dual to $H_i$ for $1\leq i \leq 2n+1$ , and let

(23) $$ \begin{align} \Phi_{2n+1}^+&=\{\mu_i-\mu_j \hspace{.1 in}| \hspace{.1 in} 1\leq i<j\leq 2n+1\}. \end{align} $$

We call $\Phi _{2n+1}^+$ the set of positive roots in $\mathfrak {gl}_{2n+1}$ . For every positive root $\alpha \in \Phi ^{+}_{2n+1}$ , we construct $X_{\alpha }\in \mathfrak {g}^{\theta }$ recursively as follows. Recall that $\mathbf {e}_i,\mathbf {f}_i,\mathbf {d}_i$ are elements in $\mathfrak {g}^{\theta }$ defined in (1)–(4). On a simple root, let

$$ \begin{align*} X_{\mu_i-\mu_{i+1}}&=\mathbf{e}_i \hspace{.2 in} (1\leq i\leq n)\\ X_{\mu_i-\mu_{i+1}}&=\mathbf{f}_{2n+1-i}\hspace{.2 in} (n+1\leq i\leq 2n). \end{align*} $$

In general, for $\alpha =\mu _i-\mu _j$ , define $X_{\alpha }$ similar to (20):

(24) $$ \begin{align} X_{\mu_i-\mu_j}=X_{(\mu_i-\mu_{i+1})-(\mu_{i+1}-\mu_j)}=[X_{\mu_i-\mu_{i+1}},X_{\mu_{i+1}-\mu_j}] . \end{align} $$

Even though the definition uses a fixed order of splitting a positive root into a sum of simple roots, we argue that the result is independent of the order of the sum and its associated order of taking Lie brackets. Specifically, it is helpful to give an alternative formulation of the root vectors:

Corollary 2.3.1. The following holds for any $1\leq i <j \leq 2n+1$ : if $\alpha =\mu _i-\mu _j$ , then

(25) $$ \begin{align} X_{\alpha}=X_{(\mu_{i}-\mu_{j-1})+(\mu_{j-1}-\mu_j)}=[X_{\mu_{i}-\mu_{j-1}},X_{\mu_{j-1}-\mu_j}]. \end{align} $$

Proof. First, for $x,y,z\in \mathfrak {g}$ , if $[x,z]=0$ , then

$$ \begin{align*} [x,[y,z]]=-[y,[z,x]]-[z,[x,y]]=[[x,y],z]. \end{align*} $$

Fix j and induct on i. The base cases $i=j-1$ and $i=j-2$ follow from definition. Suppose the statement is true for $k\leq i<j$ , then

$$ \begin{align*} [X_{\mu_{k-1}-\mu_j}]=[X_{\mu_{k-1}-\mu_k},X_{\mu_k-\mu_j}]=[X_{\mu_{k-1}-\mu_k},[X_{\mu_k-\mu_{j-1}},X_{\mu_{j-1}-\mu_j}]]\\ =[[X_{\mu_{k-1}-\mu_k},X_{\mu_k-\mu_{j-1}}],X_{\mu_{j-1}-\mu_j}]]=[X_{\mu_{k-1}-\mu_{j-1}},X_{\mu_{j-1}-\mu_j}]. \end{align*} $$

The third equality follows from the fact that for any two simple roots not adjacent to each other in the sequence $\mathbf {e}_1,\dots ,\mathbf {e}_n,\mathbf {f}_n,\dots ,\mathbf {f}_1$ , their associated root vectors commute with each other.

For the set of simple roots

$$ \begin{align*} \Pi_{2n+1} = \{ \alpha_i=\mu_i-\mu_{i+1} \hspace{.1 in}| \hspace{.1 in} 1\leq i\leq 2n \}. \end{align*} $$

We impose a total order based on the ordering on i. Moreover, two roots are said to be adjacent if they are adjacent on the associated Dynkin diagram of type $A_{2n}$ . Each simple root vector induces a Lie algebra endomorphism in $\mathfrak {g}^{\theta }$ by its adjoint action:

$$ \begin{align*} \operatorname{ad} X_{\alpha_i}: \mathfrak{g}^{\theta} \to \mathfrak{g}^{\theta} \hspace{.3 in} \operatorname{ad} X_{\alpha_i} (z)=[X_{\alpha_i},z]. \end{align*} $$

For short, we denote $\operatorname {ad} X_{\alpha _i}$ simply as $\underline {\alpha }_i$ . The following identity will be helpful in the future.

The sequence of simple roots $(\alpha _{i_1},\alpha _{i_2},\alpha _{i_3},\dots ,\alpha _{i_s})$ is called admissible if for any k such that $1\leq k\leq s-1$ , $\alpha _k$ is adjacent to one of $\alpha _{i_{k+1}}$ , $\alpha _{i_{k+2}}$ , …, $\alpha _{i_s}$ .

Lemma 2.3.2. Let $\alpha _{i_1}$ ,…, $\alpha _{i_s}$ be a sequence of simple roots. Then

$$ \begin{align*} \underline{\alpha}_{i_1}\circ \underline{\alpha}_{i_1}\circ \cdots \circ \underline{\alpha}_{i_s} (\mathbf{d}_k)\neq 0 \end{align*} $$

for some k, $1\leq k\leq n+1$ , only if the sequence $(\alpha _{i_1},\dots ,\alpha _{i_s})$ is admissible.

Proof. We prove by induction on s. Suppose the statement is true for $s\leq m$ . For $s=m+1$ , suppose there exists $\mathbf {d}_k$ such that

$$ \begin{align*} \underline{\alpha}_{i_1}\circ\underline{\alpha}_{i_2}\circ\cdots \circ \underline{\alpha}_{i_{m+1}}(\mathbf{d}_k)\neq 0 \end{align*} $$

then the sequence $(\alpha _{i_2},\dots ,\alpha _{i_{m+1}})$ is admissible by induction hypothesis. Suppose further that $\alpha _{i_1}$ is not adjacent to any of $\alpha _{i_2},\dots ,\alpha _{i_{m+1}}$ , then

$$ \begin{align*} \underline{\alpha}_{i_1}\circ\underline{\alpha}_{i_2}\circ\cdots \circ \underline{\alpha}_{i_{m+1}}(\mathbf{d}_k)=[[X_{\alpha_{i_1}},X_{\alpha_{i_2}}],\underline{\alpha}_{i_3}\circ\underline{\alpha}_{i_2}\circ\cdots \circ \underline{\alpha}_{i_{m+1}}(\mathbf{d}_k)]\\ +[\underline{\alpha}_{i_1}\circ\underline{\alpha}_{i_3}\circ\cdots \circ \underline{\alpha}_{i_{m+1}}(\mathbf{d}_k),X_{\alpha_{i_2}}]. \end{align*} $$

In the sum, the first term is zero because $\alpha _{i_1}$ is nonadjacent to $\alpha _{i_2}$ , and the second term is zero by induction hypothesis for $s=m$ . Now we have shown that the claim is true for $s=m+1$ .

Lemma 2.3.3. Let $(\beta _{1},\dots ,\beta _{s})$ be a sequence of consecutive simple roots in increasing order. If $(\alpha _{i_1},\dots ,\alpha _{i_s})$ is a permutation of the sequence $(\beta _{1},\dots ,\beta _{s})$ , then

$$ \begin{align*} \underline{\alpha}_{i_1}\circ\underline{\alpha}_{i_2} \circ \cdots\circ \underline{\alpha}_{i_s}(\mathbf{d}_k)\in \mathbb{C} \underline{\beta}_{1}\circ \underline{\beta}_{2}\circ\cdots\circ \underline{\beta}_{s}(\mathbf{d}_k) \end{align*} $$

for all $1\leq k\leq n+1$ .

Proof. First observe that if $\underline {\alpha }_{i_1}\circ \underline {\alpha }_{i_2}\circ \cdots \circ \underline {\alpha }_{i_{t+1}}(\mathbf {d}_k)=0$ , then the statement is automatically true. We now assume that this quantity is nonzero. It follows that the sequence $(\alpha _{i_2},\alpha _{i_{k+1}},\dots ,\alpha _{i_s})$ must be admissible, and therefore must be a permutation of consecutive roots by Lemma 2.3.2.

We proceed by induction on s: the base case when $s=2$ is a straightforward check. Now suppose the statement holds for $s\leq t$ . We claim it also holds for $s=t+1$ . Let $\beta _1,\dots ,\beta _t$ be the rearrangement of $\alpha _{i_2},\dots ,\alpha _{i_{k+1}}$ in increasing order, then

$$ \begin{align*} x=\underline{\alpha}_{i_1}\circ \underline{\alpha}_{i_2}\circ \cdots\circ \underline{\alpha}_{i_{t+1}}(\mathbf{d}_k)\in \mathbb{C} \underline{\alpha}_{i_1}\circ \underline{\beta}_{1}\circ \cdots\circ \underline{\beta}_{t}(\mathbf{d}_k). \end{align*} $$

Since $x\neq 0$ , $\alpha _{i_1}$ is adjacent to one of $\beta _1,\dots ,\beta _t$ . By assumption there are no repeated roots, therefore $\alpha _{i_1}$ must be immediately smaller than $\beta _1$ or immediately larger than $\beta _t$ . The former case yields the desired conclusion automatically. If the latter is true, then

$$ \begin{align*} &\underline{\alpha}_{i_1}\circ \underline{\beta}_{1}\circ \cdots\circ \underline{\beta}_{t}(\mathbf{d}_k)\\ &=[[X_{\alpha_1},X_{\beta_1}],\underline{\beta}_2\circ\cdots\circ \underline{\beta}_t(\mathbf{d}_k)+[\underline{\alpha}_{i_1}\circ\underline{\beta}_2 \circ \cdots\circ \underline{\beta}_t (\mathbf{d}_k),X_{\beta_1}]. \end{align*} $$

The first term is zero, and the second term is in $\mathbb {C}[X_{\beta _1},\underline {\beta }_2\circ \cdots \circ \underline {\beta }_t \circ \underline {\alpha }_{i_1}(\mathbf {d}_k)]$ by the induction hypothesis as desired.

Lemma 2.3.4. For any $a,b,c\in \mathfrak {g,}$

$$ \begin{align*} [a,[b,[a,c]]]=\frac{1}{2}[b,[a,[a,c]]]+\frac{1}{2}[c,[a,[a,b]]]+\frac{1}{2}[a,[a,[b,c]]]. \end{align*} $$

Proof. This is a straightforward check.

Proof. Let $(\mathfrak {g}^{\theta })'=[\mathfrak {g}^{\theta },\mathfrak {g}^{\theta }]$ . Recall $\mathfrak {h}^{\theta }$ is the subalgebra spanned by $\mathbf {d}_i$ , $1\leq i\leq n+1$ . Since $[\mathfrak {g}^{\theta },\mathfrak {h}^{\theta }]\subset (\mathfrak {g}^{\theta })'$ , it suffices to show root vectors span $(\mathfrak {g}^{\theta })'$ .

Given an arbitrary element

(26) $$ \begin{align} \underline{\alpha}_{i_1}\circ\underline{\alpha}_{i_2}\circ \cdots \circ \underline{\alpha}_{i_s}(\mathbf{d}_k) \in (\mathfrak{g}^{\theta})', \end{align} $$

where $\alpha _{i_1}$ , …, $\alpha _{i_s}$ are simple roots in $\mathfrak {gl}_{2n+1}(\mathbb {C})$ (with possible repetitions.) We claim it is in the span of root vectors whose expression only involves the roots $\alpha _{i_1}$ , …, $\alpha _{i_s}$ .

We prove this by induction on s. Suppose the statement is true for $s=t$ . Then for $s=t+1$ , by Lemma 2.3.2, if (26) is nonzero, then the sequence $(\alpha _{i_2},\dots ,\alpha _{i_{t+1}})$ is admissible. If there is no repetition of simple roots among them, then the admissible condition implies that they are a permutation of a sequence of consecutive simple roots, and by Lemma 2.3.3 is a root vector.

Now we discuss the case when there are repeated roots. List all roots which sit at the left endpoint of such a pair of repeated roots, and choose the rightmost one, $\alpha _{i_\ell }$ , among them. By design, this root has a twin counterpart to its right, but no pairs of repeated roots to its right. By the induction on length,

(27) $$ \begin{align} \underline{\alpha}_{i_{\ell+1}}\circ\cdots\circ \underline{\alpha}_{i_{t+1}}(\mathbf{d}_k) \end{align} $$

is in the span of root vectors concerning $\alpha _{i_{\ell +1}}$ , …, $\alpha _{i_{t+1}}$ for any k. Therefore, (26) is a linear combination of elements the form

(28) $$ \begin{align} \underline{\alpha}_{i_1}\circ\cdots\circ \underline{\alpha}_{i_\ell} \circ \beta_1 \circ \beta_2\circ \cdots\circ \beta_p(\mathbf{d}_k), \end{align} $$

where $(\beta _1,\beta _2,\ldots , \beta _p)$ is an increasing sequence of consecutive simple roots, and $\alpha =\alpha _{i_\ell }=\beta _j$ for some $1\leq j\leq p$ . Notice that if two simple roots $\gamma _1$ and $\gamma _2$ are nonadjacent, then for any $z\in \mathfrak {g}^{\theta }$ ,

$$ \begin{align*} [X_{\gamma_1},[X_{\gamma_2},z]]=[[X_{\gamma_1},X_{\gamma_2}],z]+[[X_{\gamma_1},z],X_{\gamma_2}]=-[X_{\gamma_2},[X_{\gamma_1},z]], \end{align*} $$

therefore up to a sign, one can move $\alpha _{i_{\ell }}$ past all nonadjacent roots to its right in (28), such that its rightmost end is of the form

1) $\underline {\alpha }\circ \underline {\alpha } \circ \underline {\gamma }(y)$ , where $\gamma $ is immediately larger than $\alpha $ and y is a root vector whose expression contains only roots nonadjacent to $\alpha $ ; or

2) $\underline {\alpha }\circ \underline {\beta } \circ \underline {\alpha } \circ \underline {\gamma }(y),$

where $(\beta ,\alpha ,\gamma )$ is an increasing sequence of consecutive simple roots, y is a root vector whose expression only contains roots nonadjacent to $\alpha $ or $\beta $ .

In Case 1),

$$ \begin{align*} [X_{\alpha},[X_{\alpha},[X_{\gamma},y]]]=[X_{\alpha},[[X_{\alpha},X_{\gamma}],y]]=[[X_{\alpha},[X_{\alpha},X_{\gamma}]],y]=c[X_{\alpha},y] . \end{align*} $$

Here,

$$ \begin{align*} c=\begin{cases} -4 \hspace{.2 in} \text{if } \{\alpha,\beta\}=\{\epsilon_n-\epsilon_{n+1},-\epsilon_n+\epsilon_{n+1}\}\\ 0 \hspace{.2 in}\text{otherwise}. \end{cases} \end{align*} $$

Therefore, (28) can be written as an element in the form of (26) of strictly lower length by the relations in $\mathfrak {g}^{\theta }$ .

In Case 2),

$$ \begin{align*} [X_{\alpha},[X_{\beta},[X_{\alpha},[X_{\gamma},y]]]]& =[X_{\alpha},[X_{\beta},[[X_{\alpha},X_{\gamma}],y]]]\\ &=[X_{\alpha},[[X_{\beta},[X_{\alpha},X_{\gamma}]],y]]=[[X_{\alpha},[X_{\beta},[X_{\alpha},X_{\gamma}]]],y]. \end{align*} $$

By Lemma 2.3.4,

(29) $$ \begin{align} [X_{\alpha},[X_{\beta},[X_{\alpha},X_{\gamma}]]]=\frac{1}{2}[X_{\beta},[X_{\alpha},[X_{\alpha},X_{\gamma}]]] ,\nonumber\\ +\frac{1}{2}[X_{\gamma},[X_{\alpha},[X_{\alpha},X_{\beta}]]]+\frac{1}{2}[X_{\alpha},[X_{\alpha},[X_{\beta},X_{\gamma}]]] . \end{align} $$

The third term is of the form in Case 1), therefore (29) is in

$$ \begin{align*} \mathbb{C}[X_{\alpha},X_{\beta}]\oplus \mathbb{C}[X_{\alpha},X_{\gamma}]. \end{align*} $$

After taking the bracket with y, one can use the Jacobi identity to rewrite the ending part in (28), so that it is of the form (26) of smaller length.

For $\alpha \in \Phi _{n+1}$ , an element $x\in \mathfrak {g}^{\theta }$ is a weight vector of weight $\alpha $ , under the adjoint action of $\mathbf {d}_1,\dots ,\mathbf {d}_n$ , $\mathbf {d}_{n+1}'$ , if and only if

$$ \begin{align*} [\mathbf{d}_i,x]=\alpha(\mathbf{d}_i)x \hspace{.2 in} (1\leq i\leq n) \hspace{.3 in} [\mathbf{d}^{\prime}_{n+1},x]=\alpha(\mathbf{d}^{\prime}_{n+1})x. \end{align*} $$

Define a map $s: \Phi _{2n+1}^+\to \Phi _{n+1}\cup \{0\}$ , such that on a simple root $\alpha \in \Pi _{2n+1}$ , $X_{\alpha }$ is a weight vector of weight $s(\alpha )$ . That is to say,

$$ \begin{align*} s(\mu_i-\mu_{i+1})&=\epsilon_i-\epsilon_{i+1} \hspace{.2 in} (1\leq i\leq n), \\ s(\mu_i-\mu_{i+1})&=-\epsilon_{2n+1-i}+\epsilon_{2n+2-i} \hspace{.2 in} (n+1\leq i\leq 2n). \end{align*} $$

Extend s linearly so that on a sum of simple roots $\alpha =\alpha _1+\cdots \alpha _i \in \Phi _{2n+1}^+$ ,

$$ \begin{align*} s(\alpha)=s(\alpha_1)+\cdots+s(\alpha_i). \end{align*} $$

The map s specifies the weight of each root vector.

Proof. The statement is clear on a simple root vector. On a sum of simple roots, it follows from the calculation

$$ \begin{align*} [\mathbf{d}_i,[X_{\alpha},X_{\beta}]]=[[\mathbf{d}_i,X_{\alpha}],X_{\beta}]+[[\mathbf{d}_i,X_{\beta}],X_{\alpha}]=s(\alpha+\beta)(\mathbf{d}_i)[X_{\alpha},X_{\beta}]. \end{align*} $$

The lemma is proved.

Recall $\phi $ is the homomorphism defined in Corollary 2.2.4. We can describe the stalks of s explicitly.

Proposition 2.3.8. For the map s defined above and $1\leq i<j\leq n$ ,

$$ \begin{align*} s^{-1}(0)&=\{\mu_i-\mu_{2n+2-i}\hspace{.1 in}|\hspace{.1 in} 1\leq i\leq n\}, \\ s^{-1}(\epsilon_i-\epsilon_{n+1})&=\{\mu_i-\mu_{n+1}\}, \hspace{.3 in} \\ s^{-1}(-\epsilon_i+\epsilon_{n+1})&=\{\mu_{n+1}-\mu_{2n+2-i}\}, \hspace{.3 in} \\ s^{-1}(\epsilon_i-\epsilon_j)&= \{\mu_i-\mu_j,\mu_i-\mu_{2n+2-j}\}, \\ s^{-1}(-\epsilon_i+\epsilon_j)&= \{\mu_{2n+2-j}-\mu_{2n+2-i},\mu_{j}-\mu_{2n+2-i}\}. \end{align*} $$

Proof. Notice that

$$ \begin{align*} 0&=(\epsilon_i-\epsilon_{i+1})+\cdots+(\epsilon_n-\epsilon_{n+1})+(-\epsilon_{n}+\epsilon_{n+1})+\cdots+(-\epsilon_i+\epsilon_{i+1}). \end{align*} $$

Therefore, the element simplifies to be

$$ \begin{align*} &s^{-1}(\epsilon_i-\epsilon_{i+1})+\cdots+s^{-1}(\epsilon_n-\epsilon_{n+1})+s^{-1}(-\epsilon_{n}+\epsilon_{n+1})+\cdots+s^{-1}(-\epsilon_i+\epsilon_{i+1})\\ &=(\mu_i-\mu_{i+1})+\cdots+(\mu_n-\mu_{n+1})+(\mu_{n+1}-\mu_{n+2})+\cdots+(\mu_{2n+1-i}-\mu_{2n+2-i})\\ &=\mu_i-\mu_{2n+2-i}, \end{align*} $$

which is in $s^{-1}(0)$ . Also, when $i<j<n+1$ ,

$$ \begin{align*} \epsilon_i-\epsilon_j=\,&(\epsilon_i-\epsilon_{i+1})+\cdots+(\epsilon_{j-1}-\epsilon_{j})\\ &+(\epsilon_j-\epsilon_{j+1})+\cdots+(\epsilon_n-\epsilon_{n+1})+(-\epsilon_n+\epsilon_{n+1})+\cdots+(-\epsilon_j+\epsilon_{j+1}). \end{align*} $$

Therefore, we have

$$ \begin{align*} &s^{-1}(\epsilon_i-\epsilon_{i+1})+\cdots+s^{-1}(\epsilon_n-\epsilon_{n+1})+s^{-1}(-\epsilon_n+\epsilon_{n+1})+\cdots+s^{-1}(-\epsilon_j+\epsilon_{j+1})\\ &=(\mu_i-\mu_{i+1})+\cdots+(\mu_n-\mu_{n+1})+(\mu_{n+1}-\mu_{n+2})+\cdots+(\mu_{2n+1-j}-\mu_{2n+2-j})\\ &=\mu_i-\mu_{2n+2-j}, \end{align*} $$

which is also in $s^{-1}(\epsilon _i-\epsilon _j)$ , in addition to $\mu _i-\mu _j$ . The other cases can be checked via a similar calculation.

2.4 An algebra isomorphism with $U(\mathfrak {gl}_{n+1}\oplus \mathfrak {gl}_n)$

Based on the previous lemma, we re-index the root vectors using weights of $\mathbf {d}_i$ . In particular, for $1\leq i<j\leq n$ , let

$$ \begin{align*} X^{\prime}_{\epsilon_i-\epsilon_j}&=X_{\mu_i-\mu_j} \hspace{.3 in} X_{\epsilon_i-\epsilon_j}=\frac{1}{2} X_{\mu_i-\mu_{2n+2-j}}\\ X^{\prime}_{-\epsilon_i+\epsilon_j}&=X_{\mu_{2n+2-j}-\mu_{2n+2-i}} \hspace{.3 in} X_{-\epsilon_i+\epsilon_j}=\frac{1}{2}X_{\mu_{j}-\mu_{2n+2-i}} . \end{align*} $$

In addition, for $1\leq i \leq n$ , let

$$ \begin{align*} X_{\epsilon_i-\epsilon_{n+1}}=\frac{1}{2}X_{\mu_i-\mu_{n+1}}, \hspace{.3 in} X_{-\epsilon_i+\epsilon_{n+1}}=X_{-\mu_i+\mu_{n+1}}. \end{align*} $$

Also, define $\mathbf {h}_i$ for $1\leq i\leq n+1$ recursively via $\mathbf {h}_{n+1}=\mathbf {d}_{n+1}'$ and

(30) $$ \begin{align} \mathbf{h}_i=\mathbf{h}_{i+1}+\frac{1}{2}X_{\mu_i-\mu_{2n+2-i}} \hspace{.1 in } (1\leq i\leq n). \end{align} $$

Recall $\Phi _{n+1}$ and $\Phi _n$ are the set defined in (19) and (22), and $\phi $ is the map defined after (18). The following lemma establishes a connection between the various root vectors.

Lemma 2.4.1. The following holds

(31) $$ \begin{align} \phi(X_{\alpha})&=Y_{\alpha} \hspace{.3 in}(\forall \alpha\in \Phi_{n+1} ), \end{align} $$

(32) $$ \begin{align} \phi(X^{\prime}_{\alpha})&=Y_{\alpha}+Z_{\alpha} \hspace{.3 in}(\forall \alpha\in \Phi_n), \end{align} $$

(33) $$ \begin{align} \phi(\mathbf{h}_i)&=h_i \hspace{.3 in}(1\leq i\leq n+1). \end{align} $$

Proof. We first prove (32) for a root $\alpha =\epsilon _i-\epsilon _j$ with $i<j$ by fixing j and induction on i. The base case $i=j-1$ is true by definition:

$$ \begin{align*} \phi(X^{\prime}_{\epsilon_{j-1}-\epsilon_j})=\phi(X_{\mu_{j-1}-\mu_j})=\phi(\mathbf{e}_{j-1})=e_{j-1}+e_{\overline{j-1}}=Y_{\epsilon_{j-1}-\epsilon_j}+Z_{\epsilon_{j-1}-\epsilon_j}. \end{align*} $$

Suppose the statement is true for $i=k$ , then

$$ \begin{align*} \phi(X_{\epsilon_{k-1}-\epsilon_j})&=\phi(X_{\mu_{k-1}-\mu_j})=\phi([X_{\mu_{k-1}-\mu_k},X_{\mu_k-\mu_j}]) \\ &=[e_{k-1}+e_{\overline{k-1}},Y_{\epsilon_k-\epsilon_j}+Z_{\epsilon_k-\epsilon_j} ]=Y_{\epsilon_{k-1}-\epsilon_j}+Z_{\epsilon_{k-1}-\epsilon_j}. \end{align*} $$

The arguments are similar for a negative root in (32).

To show (31) for a root $\alpha =\epsilon _i-\epsilon _{n+1}$ for $1\leq i\leq n$ , we induct on i. The base case when $i=n$ is as follows

$$ \begin{align*} \phi(X_{\epsilon_n-\epsilon_{n+1}})=\phi(\frac{1}{2}X_{\mu_n-\mu_{n+1}})=\frac{1}{2}\phi(\mathbf{e}_n)=e_n=Y_{\epsilon_n-\epsilon_{n+1}}. \end{align*} $$

Suppose the statement is true for $i=k$ ,

$$ \begin{align*} \phi(X_{\epsilon_{k-1}-\epsilon_{n+1}})&=\phi(\frac{1}{2}X_{\mu_{k-1}-\mu_{n+1}})=\frac{1}{2}\phi([X_{\mu_{k-1}-\mu_k},X_{\mu_k-\mu_{n+1}}])=\phi([\mathbf{e}_{k-1},X^{\prime}_{\epsilon_k-\epsilon_{n+1}}])\\ &=[e_{k-1}+e_{\overline{k-1}},Y_{\epsilon_k-\epsilon_{n+1}}]=Y_{\epsilon_{k-1}-\epsilon_{n+1}}. \end{align*} $$

The arguments are similar for a negative root in (32).

We now show that (33) is true by induction on i. The case when $i=n+1$ is given via definition. The next case when $i=n$ is verified as follows:

$$ \begin{align*} \phi({\mathbf{h}_n})&=\phi(\mathbf{h}_{n+1}+\frac{1}{2}X_{\mu_n-\mu_{n+2}})=\phi(\mathbf{h}_{n+1})+\frac{1}{2}[\phi(X_{\mu_n-\mu_{n+1}}),\phi(X_{\mu_{n+1}-\mu_{n+2}})]\\ &=\phi(\mathbf{h}_{n+1})+\frac{1}{2}[\phi(X_{\epsilon_n-\epsilon_{n+1}}),\phi(X_{-\epsilon_n+\epsilon_{n+1}})]=h_{n+1}+[e_n,f_n]=h_n. \end{align*} $$

Using $i=n$ as the base case, suppose the statement is true for i, then by Lemma 2.3.1,

$$ \begin{align*} \phi(\mathbf{h}_{i-1})&=\phi(\mathbf{h}_{i})+\phi(\frac{1}{2}X_{\mu_{i-1}-\mu_{2n+3-i}})\\ &=\phi(\mathbf{h}_i)+\frac{1}{2}\phi([X_{\mu_{i-1}-\mu_i},[X_{\mu_i-\mu_{2n+2-i}},X_{\mu_{2n+2-i}-\mu_{2n+3-i}}]])\\ &=h_i+\phi([\mathbf{e}_{i-1},[\mathbf{h}_i-\mathbf{h}_{i+1},X_{-\epsilon_{i-1}+\epsilon_i}]])=h_i+\phi([\mathbf{e}_{i-1},[\mathbf{h}_{i}-\mathbf{h}_{i+1},\mathbf{f}_{i-1}]])\\ &=h_i+[Y_{\epsilon_{i-1}-\epsilon_i}+Z_{\epsilon_{i-1}-\epsilon_i},[h_i-h_{i+1},Y_{\epsilon_{i-1}+\epsilon_i}+Z_{\epsilon_{i-1}+\epsilon_i}]]\\ &=h_i+[Y_{\epsilon_{i-1}-\epsilon_i}+Z_{\epsilon_{i-1}-\epsilon_i},Y_{-\epsilon_{i-1}+\epsilon_i}]=h_i+h_{i-1}-h_i=h_{i-1}, \end{align*} $$

and therefore the statement is also true for $i-1$ . Notice the root $\mu _i-\mu _{2n+2-i}=0$ if and only if $i=n+1$ , and the induction starts at $i=n$ .

We now show (31) via fixing j and induction on i. The base case for $(i,j)$ when $i=j-1$ (or $j=i+1$ ) is as follows

$$ \begin{align*} \phi(X^{\prime}_{\epsilon_i-\epsilon_{i+1}})&=\phi(\frac{1}{2}X_{\mu_i-\mu_{2n+2-(i+1)}})=\frac{1}{2}\phi([X_{\mu_i-\mu_{i+1}},X_{\mu_{i+1}-\mu_{2n+2-(i+1)}}])\\ &=\frac{1}{2}\phi([\mathbf{e}_i,2(\mathbf{h}_{i+1}-\mathbf{h}_{i+2})])=[Y_{\epsilon_i-\epsilon_{i+1}}+Z_{\epsilon_i-\epsilon_{i+1}},h_{i+1}-h_{i+2}]=Y_{\epsilon_i-\epsilon_{i+1}}. \end{align*} $$

Now suppose (31) holds for the index pair $(i,j)$ , then we claim it also holds for $(i-1,j)$ :

$$ \begin{align*} \phi(X_{\epsilon_{i-1}-\epsilon_j})&=\phi(\frac{1}{2}X_{\mu_{i-1}-\mu_{2n+2-j}})=\frac{1}{2}\phi([X_{\mu_{i-1}-\mu_i},X_{\mu_i-\mu_{2n+2-j}}])\\ &=\phi([\mathbf{e}_{i-1},X^{\prime}_{\epsilon_i-\epsilon_j}])=[Y_{\epsilon_{i-1}-\epsilon_i}+Z_{\epsilon_{i-1}-\epsilon_i},Y_{\epsilon_i-\epsilon_j}]=Y_{\epsilon_{i-1}-\epsilon_j}. \end{align*} $$

Therefore, we have shown (31) by induction. The negative roots follow a similar argument.

Recall the definition of $\mathbf {h}_i$ in (30).

Theorem 2.4.2. The morphism $\phi $ (resp. $\phi |_{\mathfrak {g}^{\theta }}$ ) in (18) is an isomorphism of (resp. Lie) algebras. Moreover, the following set is a basis of $\mathfrak {g}^{\theta }$ .

$$ \begin{align*} \{X_{\alpha}\}_{\alpha\in \Phi_{n+1}}\cup \{X^{\prime}_{\alpha}\}_{\alpha\in \Phi_n}\cup\{\mathbf{h}_1,\dots,\mathbf{h}_{n+1}\}\cup \{\mathbf{d}_1,\dots,\mathbf{d}_n\}, \end{align*} $$

and the elements $\mathbf {d}_1$ , …, $\mathbf {d}_{n}$ , $\mathbf {h}_1$ , …, $\mathbf {h}_{n+1}$ pairwise commute.

Proof. These elements are exactly those mentioned in Theorem 2.3.5, hence they span $\mathfrak {g}^{\theta }$ . By Lemma 2.4.1, they are also linearly independent, because their image forms a linearly independent set.

2.5 The type B Schur algebra

Recall the elements $\mathbf {h}_i$ , defined in terms of generators of $U^i$ in (30). For $1\leq i\leq n$ , let $ \mathbf {h}_{\overline {i}}=\mathbf {d}_{\overline {i}}-\mathbf {h}_{i} $ . By the definition of $\phi $ and Lemma 2.4.1, $\phi (\mathbf {h}_{\overline {i}})=h_{\overline {i}}$ . Now fix a positive integer d. Let S be a quotient of $U(\mathfrak {gl}_{n+1}(\mathbb {C})\oplus \mathfrak {gl}_n(\mathbb {C}))$ , by the following relations

(34) $$ \begin{align} h_1+\cdots+h_{n+1}+h_{\overline{1}}+\cdots+h_{\overline{n}}&=d, \end{align} $$

(35) $$ \begin{align} h_{i}(h_i-1)\cdots(h_{i}-d)&=0, \hspace{.5 in} 1\leq i\leq n+1, \end{align} $$

(36) $$ \begin{align} h_{\overline{i}}(h_{\overline{i}}-1)\cdots(h_{\overline{i}}-d)&=0, \hspace{.5 in} 1\leq i\leq n. \end{align} $$

These relations are analogous to the type A relations, that is, relations in a presentation for the type A Schur algebras given by Doty–Giaquinto [Reference Doty and Giaquinto5].

On the other hand, let $S^j$ be the quotient of $U^j$ under further relations

(37) $$ \begin{align} \mathbf{d}_1+\cdots+\mathbf{d}_n+\mathbf{h}_{n+1}&=d, \end{align} $$

(38) $$ \begin{align} \mathbf{h}_i(\mathbf{h}_i-1)\cdots(\mathbf{h}_i-d)&=0, \hspace{.5 in} 1\leq i\leq n+1, \end{align} $$

(39) $$ \begin{align} \mathbf{h}_{\overline{i}}(\mathbf{h}_{\overline{i}}-1)\cdots(\mathbf{h}_{\overline{i}}-d)&=0, \hspace{.5 in} 1\leq i\leq n. \end{align} $$

Note: once unraveling the notation of root vectors, these relations are purely in terms of generators $\mathbf {e}_i$ , $\mathbf {f}_i$ , and $\mathbf {d}_i$ of $U^j$ .

Example 2.5.4. When $n=1$ , $S^j$ is the algebra with generators $\mathbf {e}_1$ , $\mathbf {f}_1$ , $\mathbf {d}_1$ , $\mathbf {d}_2$ subject to the relations in Proposition 2.1.4 and the following additional relations.

$$ \begin{align*} \mathbf{d}_1+\frac{1}{2}\mathbf{d}_2&=d,\\ \mathbf{d}_2(\mathbf{d}_2-2)\cdots (\mathbf{d}_2-2d)&=0,\\ (\mathbf{d}_2+[\mathbf{e}_1,\mathbf{f}_1])(\mathbf{d}_2+[\mathbf{e}_1,\mathbf{f}_1]-2)\cdots(\mathbf{d}_2+[\mathbf{e}_1,\mathbf{f}_1]-2d)&=0,\\ (\mathbf{d}_2+\frac{1}{2}[\mathbf{e}_1,\mathbf{f}_1]-d)(\mathbf{d}_2+\frac{1}{2}[\mathbf{e}_1,\mathbf{f}_1] -d-1)\cdots (\mathbf{d}_2+\frac{1}{2}[\mathbf{e}_1,\mathbf{f}_1] -2d)&=0. \end{align*} $$

3.1 The fixed point subalgebra

We start by introducing the fixed point subalgebra and its presentation using generators similar to those in Section 2.1. We now discuss the case when the underlying Dynkin diagram has an odd number of vertices. We will abuse some of the previous notation despite this new assumption on the parity. In particular, let $\mathfrak {g}=\mathfrak {gl}_{2n}(\mathbb {C})$ and let $\Gamma $ be the Dynkin diagram of $\mathfrak {gl}_{2n}(\mathbb {C})$ . Then $\Gamma $ has vertices $1,2,\dots ,2n-1$ . Let $\tau \in \operatorname {Aut}(\Gamma )$ be the graph automorphism such that $\tau (i)=2n-i$ , then $\tau $ induces a Lie algebra automorphism on $\mathfrak {sl}_{2n}(\mathbb {C})$ , which extends to a Lie algebra automorphism $\theta $ on $\mathfrak {gl}_{2n}(\mathbb {C})$ via

$$ \begin{align*} \theta(E_i)=F_{\tau(i)},\hspace{.2 in} \theta(F_i)=E_{\tau(i)}, \hspace{.2 in} \theta(H_i)=H_{\tau(i)+1}. \end{align*} $$

Let $\mathfrak {g}^{\theta }=\mathfrak {gl}_{2n}(\mathbb {C})^{\theta }$ be the subalgebra of $\mathfrak {g}$ fixed by $\theta $ , and $E_i, F_i (1\leq i\leq 2n-1)$ , $H_i (1\leq i\leq 2n)$ be the Chevalley generators of $U(\mathfrak {gl}_{2n})$ . Define the following elements in $U(\mathfrak {gl}_{2n})$ :

$$ \begin{align*} \mathbf{e}_i&=E_i+F_{2n-i} \hspace{.3 in} (1\leq i\leq n-1),\\ \mathbf{f}_i&=F_i+E_{2n-i} \hspace{.3 in} (1\leq i\leq n-1),\\ \mathbf{t}&=E_n+F_n,\\ \mathbf{d}_i&=H_i+H_{2n+1-i} \hspace{.3 in} (1\leq i\leq n). \end{align*} $$

Let $\mathfrak {h}\subset \mathfrak {g}$ be the Cartan subalgebra of $\mathfrak {gl}_{2n}(\mathbb {C})$ , and $\mathfrak {h}^{\theta }=\{h\in \mathfrak {h}\hspace {.1 in}|\hspace {.1 in} \theta (h)=h\}$ the subalgebra of $\mathfrak {h}$ fixed by $\theta $ . Similar to Lemma 2.1.2, we have the following

By Lemma 2.1.1, the elements $\mathbf {e}_i$ , $\mathbf {f}_i$ , $\mathbf {t}$ , $\mathbf {d}_i$ generate $\mathfrak {g}^{\theta }$ . Let $U(\mathfrak {g}^{\theta })$ be the universal enveloping algebra of $\mathfrak {g}^{\theta }$ .

Proposition 3.1.2. The algebra $U(\mathfrak {g}^{\theta })$ is generated by $\mathbf {e}_1,\dots ,\mathbf {e}_{n-1}$ , $\mathbf {f}_1,\dots ,\mathbf {f}_{n-1}$ , $\mathbf {t}$ , $\mathbf {d}_1,\dots ,\mathbf {d}_n$ , subject to the following relations

(R1) $$ \begin{align} \mathbf{t}\mathbf{d}_i&=\mathbf{d}_i \mathbf{t} \hspace{.5 in}(1\leq i\leq n),\nonumber\\ \mathbf{d}_i\mathbf{e}_j-\mathbf{e}_j\mathbf{d}_i&=(\delta_{i,j}-\delta_{i,j+1})\mathbf{e}_j,\nonumber\\ \mathbf{d}_i\mathbf{f}_j-\mathbf{f}_j\mathbf{d}_i&=(-\delta_{i,j}+\delta_{i,j+1})\mathbf{f}_j, \end{align} $$

(R2) $$ \begin{align} \mathbf{t}\mathbf{e}_i & =\mathbf{e}_i \mathbf{t},\nonumber\\ \mathbf{t} \mathbf{f}_i & =\mathbf{f}_i \mathbf{t}\hspace{.5 in} (1\leq i \leq n-2) ,\nonumber\\ \mathbf{e}_i\mathbf{e}_j&=\mathbf{e}_j\mathbf{e}_i \hspace{.5 in} |i-j|>1,\nonumber\\ \mathbf{f}_i\mathbf{f}_j&=\mathbf{f}_j\mathbf{f}_i \hspace{.5 in} |i-j|>1,\nonumber\\ \mathbf{e}_i\mathbf{f}_j-\mathbf{f}_j\mathbf{e}_i&=\delta_{i,j}\mathbf{d}_i-\mathbf{d}_{i+1}, \end{align} $$

(R3) $$ \begin{align} \mathbf{e}_{n-1}^2\mathbf{t}-2\mathbf{e}_{n-1}\mathbf{t}\mathbf{e}_{n-1}+\mathbf{t}\mathbf{e}_{n-1}^2 &=0,\nonumber\\ \mathbf{f}_{n-1}^2\mathbf{t}-2\mathbf{f}_{n-1}\mathbf{t}\mathbf{f}_{n-1}+\mathbf{t}\mathbf{f}_{n-1}^2 &=0, \end{align} $$

(R4) $$ \begin{align} \mathbf{e}_{n-1}\mathbf{t}^2-2\mathbf{t}\mathbf{e}_{n-1}\mathbf{t}+\mathbf{t}^2\mathbf{e}_{n-1}&=\mathbf{e}_{n-1}, \nonumber \\ \mathbf{f}_{n-1}\mathbf{t}^2-2\mathbf{t}\mathbf{f}_{n-1}\mathbf{t}+\mathbf{t}^2\mathbf{f}_{n-1}&=\mathbf{f}_{n-1}. \end{align} $$

Proof. R1): $[\mathbf {t},\mathbf {d}_i]=[E_n+F_n, H_i+H_{2n+1-i}]$ is only nonzero when $i=n$ , in which case

$$ \begin{align*} [\mathbf{t},\mathbf{d}_{n-1}]&=[E_n+F_n,H_{n}+H_{n+1}]\\ &=[E_n,H_{n}]+[E_n,H_{n+1}]+[F_n,H_{n}]+[F_n,H_{n+1}]\\ &=E_n-E_n-F_n+F_n=0. \end{align*} $$

Also, for $1\leq i \leq n$ and $1\leq j\leq n-1$ , $2\leq i+j\leq 2n-1$ and

$$ \begin{align*} \mathbf{d}_i\mathbf{e}_j-\mathbf{e}_j\mathbf{d}_i&=[H_i+H_{2n+1-i},E_j+F_{2n-j}]\\ &=(\delta_{i,j}-\delta_{i,j+1}+\delta_{2n+1-i,j}-\delta_{2n+1-i,j+1})E_j\\ &\quad+(-\delta_{i,2n-j}+\delta_{i,2n-j+1}-\delta_{2n+1-i,2n-j}+\delta_{2n+1-i,2n-j+1})F_{2n-j}\\ &=(\delta_{i,j}-\delta_{i,j+1})(E_j+F_{2n-j})=(\delta_{i,j}-\delta_{i,j+1})\mathbf{e}_j. \end{align*} $$

Similarly,

$$ \begin{align*} \mathbf{d}_i\mathbf{f}_j-\mathbf{f}_j\mathbf{d}_i&=[H_i+H_{2n+1-i},F_j+E_{2n-j}]\\ &=(-\delta_{i,j}+\delta_{i,j+1}-\delta_{2n+1-i,j}+\delta_{2n+1-i,j+1})F_j\\ &\quad+(\delta_{i,2n-j}-\delta_{i,2n-j+1}+\delta_{2n+1-i,2n-j}-\delta_{2n+1-i,2n-j+1})E_{2n-j}\\ &=(-\delta_{i,j}+\delta_{i,j+1})(F_j+E_{2n-j})=(-\delta_{i,j}+\delta_{i,j+1})\mathbf{f}_j. \end{align*} $$

R2): The only nontrivial relation to check is $\mathbf {e}_i\mathbf {f}_j-\mathbf {f}_j\mathbf {e}_i=\delta _{i,j}\mathbf {d}_i-\mathbf {d}_{i+1}$ , and the calculation is as follows:

$$ \begin{align*} \mathbf{e}_i\mathbf{f}_j-\mathbf{f}_j\mathbf{e}_i&=[E_i+F_{2n-i},F_i+E_{2n-i}]\\ &=[E_i,F_i]-[E_{2n-i},F_{2n-i}]\\ &=H_{i}-H_{i+1}-H_{2n-i}+H_{2n-i+1}=\mathbf{d}_i-\mathbf{d}_{i+1}. \end{align*} $$

R3):

$$ \begin{align*} &\mathbf{e}_{n-1}^2\mathbf{t}-2\mathbf{e}_{n-1}\mathbf{t}\mathbf{e}_{n-1}+\mathbf{t}\mathbf{e}_{n-1}^2 \\ &=(E_{n-1}+F_{n+1})^2(E_n+F_n)-2(E_{n-1}+F_{n+1})(E_n+F_n)(E_{n-1}+F_{n+1})\\ &\quad-(E_n+F_n)(E_{n-1}+F_{n+1})^2, \end{align*} $$

where

$$ \begin{align*} &(E_{n-1}+F_{n+1})^2E_n-2(E_{n-1}+F_{n+1})E_n(E_{n-1}+F_{n+1})+E_n(E_{n-1}+F_{n+1})^2\\ &=(E_{n-1}^2+2E_{n-1}F_{n+1}+F_{n+1}^2)E_n\\ &\quad-2(E_{n-1}E_n E_{n-1}+F_{n+1}E_nE_{n-1}+E_{n-1}E_nF_{n+1}+F_{n+1}E_nF_{n+1})\\ &\quad+E_n(E_{n-1}^2+2E_{n-1}F_{n+1}+F_{n+1}^2)\\ &=(E_{n-1}^2E_n-2E_{n-1}E_n E_{n-1}+E_nE_{n-1}^2)\\ &\quad+(2E_{n-1}F_{n+1}E_n-2F_{n+1}E_nE_{n-1}-2E_{n-1}E_nF_{n+1}+2E_nE_{n-1}F_{n+1})\\ &\quad+(F_{n+1}^2E_n-2F_{n+1}E_nF_{n+1}+E_nF_{n+1}^2)=0. \end{align*} $$

Similarly,

$$ \begin{align*} (E_{n-1}+F_{n+1})^2F_n-2(E_{n-1}+F_{n+1})F_n(E_{n-1}+F_{n+1})+F_n(E_{n-1}+F_{n+1})^2=0. \end{align*} $$

and therefore the relation holds. The other relation can be checked in a similar way.

R4):

$$ \begin{align*} &\mathbf{e}_{n-1}\mathbf{t}^2-2\mathbf{t}\mathbf{e}_{n-1}\mathbf{t}+\mathbf{t}^2\mathbf{e}_{n-1}\\ &=(E_{n-1}+F_{n+1})(E_n+F_n)^2-2(E_n+F_n)(E_{n-1}+F_{n+1})(E_n+F_n)\\ &\quad+(E_n+F_n)^2(E_{n-1}+F_{n+1}), \end{align*} $$

where

$$ \begin{align*} &E_{n-1}(E_n+F_n)^2-2(E_n+F_n)E_{n-1}(E_n+F_n)+(E_n+F_n)^2E_{n-1}\\ &=E_{n-1}(E_n^2+E_nF_n+F_nE_n+F_n^2)\\ &\quad-2(E_nE_{n-1}E_n+E_nE_{n-1}F_n+F_nE_{n-1}E_n+F_nE_{n-1}F_n)\\ &\quad+(E_n^2+E_nF_n+F_nE_n+F_n^2)E_{n-1}\\ &=(E_{n-1}E_n^2-2E_nE_{n-1}E_n+E_n^2E_{n-1})\\ &\quad+(E_{n-1}F_n^2-2F_nE_{n-1}F_n+F_n^2E_{n-1})\\ &\quad+E_{n-1}(E_nF_n+F_nE_n)-2(E_nE_{n-1}F_n+F_nE_{n-1}F_n)+(E_nF_n+F_nE_n)E_{n-1}\\ &=E_{n-1}(2F_nE_n+H_n-H_{n+1})-2(E_nF_nE_{n-1}+E_{n-1}F_nF_n)\\ &\quad+(2E_nF_n-(H_n-H_{n+1}))E_{n-1}\\ &=[E_{n-1},H_n]=E_{n-1}. \end{align*} $$

On the other hand,

$$ \begin{align*} &F_{n+1}(E_n+F_n)^2-2(E_n+F_n)F_{n+1}(E_n+F_n)+(E_n+F_n)^2F_{n+1}\\ &=F_{n+1}(E_n^2+E_nF_n+F_nE_n+F_n^2)\\ &\quad-2(E_nF_{n+1}E_n+E_nF_{n+1}F_n+F_nF_{n+1}E_n+F_nF_{n+1}F_n)\\ &\quad+(E_n^2+E_nF_n+F_nE_n+F_n^2)F_{n+1}\\ &=(F_{n+1}F_n^2-2F_nF_{n+1}F_n+F_n^2F_{n+1})\\ &\quad+(F_{n+1}E_n^2-2E_nF_{n+1}E_n+E_n^2F_{n+1})\\ &\quad+F_{n+1}(E_nF_n+F_nE_n)-2(E_nF_{n+1}F_n+F_nF_{n+1}E_n)+(E_nF_n+F_nE_n)F_{n+1}\\ &=F_{n+1}(2E_nF_n-(H_n-H_{n+1}))-2(F_{n+1}E_nF_n+F_nE_nF_{n+1})\\ &\quad+(2F_nE_n+H_n-H_{n+1})F_{n+1}\\ &=[F_{n+1},H_{n+1}]=F_{n+1}. \end{align*} $$

Therefore,

$$ \begin{align*} \mathbf{e}_{n-1}\mathbf{t}^2-2\mathbf{t}\mathbf{e}_{n-1}\mathbf{t}+\mathbf{t}^2\mathbf{e}_{n-1}=E_{n-1}+F_{n+1}=\mathbf{e}_{n-1}. \end{align*} $$

The other relation can be checked similarly.

The fact that these are the only relations follows from the proof of Proposition 2.1.4.

3.2 An algebra homomorphism between $U^i$ and $U(\mathfrak {gl}_n\oplus \mathfrak {gl}_n)$

Denote $U^i=U(\mathfrak {g}^{\theta })$ based on the parity assumption in this section. On the other hand, let $e_1,\dots ,e_{n-1}$ , $f_1,\dots ,f_{n-1}$ , $h_1,\dots ,h_n$ be the Chevalley generators of the first copy of $\mathfrak {gl}_n(\mathbb {C})$ in $\mathfrak {gl}_n(\mathbb {C})\oplus \mathfrak {gl}_n(\mathbb {C})$ , and $e_{\overline {1}},\dots ,e_{\overline {n-1}}$ , $f_{\overline {1}},\dots ,f_{\overline {n-1}}$ , $h_{\overline {1}},\dots ,h_{\overline {n}}$ the Chevalley generators of the second copy.

Lemma 3.2.1. There is an algebra homomorphism: $\rho : U^i \to U(\mathfrak {gl}_n \oplus \mathfrak {gl}_n)$ via the following assignment

$$ \begin{align*} \mathbf{e}_i &\mapsto e_i+e_{\overline{i}} \hspace{.2 in}(1\leq i\leq n-1),\\ \mathbf{f}_i &\mapsto f_i+f_{\overline{i}}\hspace{.2 in} (1\leq i\leq n-1),\\ \mathbf{d}_i &\mapsto h_i+h_{\overline{i}} \hspace{.2 in} (1\leq i\leq n),\\ \mathbf{t} & \mapsto h_n-h_{\overline{n}}. \end{align*} $$

Proof. We check that all the relations are satisfied in $U(\mathfrak {gl}_n \oplus \mathfrak {gl}_n)$ after applying $\rho $ .

R1) The first relation becomes $[h_i+h_{\overline {i}},h_n+h_{\overline {n}}]=0$ which is true. The other two relations are straightforward to check.

R2) becomes $[e_i+e_{\overline {i}}, h_n-h_{\overline {n}}]=0$ for $1\leq i\leq n-2$ which is true. The other relation in R2) are straightforward to check.

For R3), we have

$$ \begin{align*} &\rho(\mathbf{e}_{n-1}^2\mathbf{t}-2\mathbf{e}_{n-1}\mathbf{t}\mathbf{e}_{n-1}+\mathbf{t}\mathbf{e}_{n-1}^2)\\ &=(e_{n-1}+e_{\overline{n-1}})^2(h_{n}-h_{\overline{n}})-2(e_{n-1}+e_{\overline{n-1}})(h_{n}-h_{\overline{n}})(e_{n-1}+e_{\overline{n-1}})+(h_{n}-h_{\overline{n}})(e_{n-1}+e_{\overline{n-1}})^2. \end{align*} $$

Since generators of the first block commute with those of the second block, it yields

$$ \begin{align*} &(e_{n-1}+e_{\overline{n-1}})^2h_{n}-2(e_{n-1}+e_{\overline{n-1}})h_{n}(e_{n-1}+e_{\overline{n-1}})+h_{n}(e_{n-1}+e_{\overline{n-1}})^2\\ &=(e_{n-1}^2+2e_{n-1}e_{\overline{n-1}}+e_{\overline{n-1}}^2)h_{n}\\ &\quad-2(e_{n-1}h_ne_{n-1}+e_{\overline{n-1}}h_{n}e_{n-1}+e_{n-1}h_ne_{\overline{n-1}}+e_{\overline{n-1}}h_ne_{\overline{n-1}})\\ &\quad+h_n(e_{n-1}^2+2e_{n-1}e_{\overline{n-1}}+e_{\overline{n-1}}^2)\\ &=e_{n-1}^2h_n-2e_{n-1}h_ne_{n-1}+h_ne_{n-1}^2\\ &=e_{n-1}(h_ne_{n-1}-[h_n,e_{n-1}])-2e_{n-1}h_ne_{n-1}+(e_{n-1}h_n+[h_n,e_{n-1}])e_{n-1}=0. \end{align*} $$

By symmetry,

$$ \begin{align*} (e_{n-1}+e_{\overline{n-1}})^2h_{\overline{n}}-2(e_{n-1}+e_{\overline{n-1}})h_{\overline{n}}(e_{n-1}+e_{\overline{n-1}})+h_{\overline{n}}(e_{n-1}+e_{\overline{n-1}})^2=0 \end{align*} $$

and so R3) is satisfied. The other relation in R3) can be checked similarly.

For R4), we have

$$ \begin{align*} &\rho(\mathbf{e}_{n-1}\mathbf{t}^2-2\mathbf{t}\mathbf{e}_{n-1}\mathbf{t}+\mathbf{t}^2\mathbf{e}_{n-1})\\ &=(e_{n-1}+e_{\overline{n-1}})(h_{n}-h_{\overline{n}})^2-2(h_{n}-h_{\overline{n}})(e_{n-1}+e_{\overline{n-1}})(h_{n}-h_{\overline{n}})+(h_{n}-h_{\overline{n}})^2(e_{n-1}+e_{\overline{n-1}}). \end{align*} $$

Since generators of different blocks commute, it becomes

$$ \begin{align*} &e_{n-1}(h_{n}-h_{\overline{n}})^2-2(h_{n}-h_{\overline{n}})e_{n-1}(h_{n}-h_{\overline{n}})+(h_{n}-h_{\overline{n}})^2e_{n-1}\\ &=e_{n-1}(h_{n}^2-2h_{n}h_{\overline{n}}+h_{\overline{n}}^2)\\ &\quad-2(h_{n}e_{n-1}h_{n}-h_{\overline{n}}e_{n-1}h_n-h_ne_{n-1}h_{\overline{n}}+h_{\overline{n}}e_{n-1}h_{\overline{n}})\\ &\quad+(h_{n}^2-2h_{n}h_{\overline{n}}+h_{\overline{n}}^2)e_{n-1}\\ &=e_{n-1}h_{n}^2-2h_{n}e_{n-1}h_{n}+h_{n}^2e_{n-1}\\ &=(h_ne_{n-1}+e_{n-1})h_n-2h_{n}e_{n-1}h_{n}+h_n(e_{n-1}h_n-e_{n-1})\\ &=[e_{n-1},h_n]=e_{n-1}. \end{align*} $$

On the other hand, there is

$$ \begin{align*} &e_{\overline{n-1}}(h_{n}-h_{\overline{n}})^2-2(h_{n}-h_{\overline{n}})e_{\overline{n-1}}(h_{n}-h_{\overline{n}})+(h_{n}-h_{\overline{n}})^2e_{\overline{n}-1}\\ &=e_{\overline{n-1}}(h_{n}^2-2h_{n}h_{\overline{n}}+h_{\overline{n}}^2)\\ &\quad-2(h_{n}e_{\overline{n-1}}h_{n}-h_{\overline{n}}e_{\overline{n-1}}h_n-h_ne_{\overline{n-1}}h_{\overline{n}}+h_{\overline{n}}e_{\overline{n-1}}h_{\overline{n}})\\ &\quad+(h_{n}^2-2h_{n}h_{\overline{n}}+h_{\overline{n}}^2)e_{\overline{n-1}}\\ &=e_{\overline{n-1}}h_{\overline{n}}^2-2h_{\overline{n}}e_{\overline{n-1}}h_{\overline{n}}+h_{\overline{n}}^2e_{\overline{n-1}}\\ &=(h_{\overline{n}}e_{\overline{n-1}}+e_{\overline{n-1}})h_{\overline{n}}-2h_{\overline{n}}e_{\overline{n-1}}h_{\overline{n}}+h_{\overline{n}}(e_{\overline{n-1}}h_{\overline{n}}-e_{\overline{n-1}})\\ &=[e_{\overline{n-1}},h_{\overline{n}}]=e_{\overline{n-1}}. \end{align*} $$

Therefore, we must have

$$ \begin{align*} &\rho(\mathbf{e}_{n-1}\mathbf{t}^2-2\mathbf{t}\mathbf{e}_{n-1}\mathbf{t}+\mathbf{t}^2\mathbf{e}_{n-1})=e_{n -1}+e_{\overline{n-1}}=\rho(\mathbf{e}_{n-1}). \end{align*} $$

The other relation in R4) can be checked similarly. Lemma is proved.

3.3 Compatibility of two actions

In this section, we will see that the homomorphism $\rho $ defined in Section 3.2 is compatible with the actions of $U^i$ and $U(\mathfrak {gl}_n\oplus \mathfrak {gl}_n)$ on $V^{\otimes }$ . In particular, there is a vector space automorphism on $V^{\otimes d}$ which intertwines these two actions. We will also show that this linear map intertwines two (ostensibly different) Weyl group actions, which are known to be Schur–Weyl dual of $U^i$ and $U(\mathfrak {gl}_n\oplus \mathfrak {gl}_n)$ , respectively. Let V be the vector space over $\mathbb {C}$ with ordered basis $\mathcal {B}_1=\{v_1,\dots ,v_n,v_{\overline {n}},\dots ,v_{\overline {1}}\}$ . Since V is $2n$ -dimensional, it is isomorphic to the $\mathbb {C}$ -vector space with ordered basis $\mathcal {B}_2=\{w_1,\dots ,w_n,w_{\overline {1}},\dots ,w_{\overline {n}}\}$ . Let $L:V\to V$ be the linear map such that $L(v_i)=w_i+w_{\overline {i}}$ , $L(v_{\overline {i}})=w_i-w_{\overline {i}}$ for $1\leq i\leq n$ . Notice L is an isomorphism if and only if the characteristic of the underlying field is odd. Let $L^{\otimes d}:V^{\otimes d}\to V^{\otimes d}$ be the linear map such that

$$ \begin{align*} L^{\otimes d}(u_1\otimes \cdots\otimes u_d)=L(u_1)\otimes \cdots \otimes L(u_d) \hspace{.3 in} u_i\in V, \forall i. \end{align*} $$

We first show that $L^{\otimes d}$ intertwines two Weyl group actions. The type B Weyl group $B_d$ is the group with generators $s_1,\dots , s_{d}$ , subject to relations

$$ \begin{align*} &s_i^2=1 \hspace{.2 in} (1\leq i,j\leq d) \hspace{.5 in} s_is_j=s_js_i \hspace{.2 in} (|i-j|>1),\\ &s_is_js_i=s_js_is_j \hspace{.2 in}(|i-j|=1, 1\leq i,j\leq d-1),\\ &s_{d-1}s_ds_{d-1}s_d=s_ds_{d-1}s_ds_{d-1}. \end{align*} $$

The first action of $B_d$ on $V^{\otimes d}$ , denoted by $f_1:\mathbb {C}B_d\to \operatorname {End}(V^{\otimes d})$ , was first given by Green [Reference Green6] and defined as follows:

$$ \begin{align*} f_1(s_i) ( v_{k_1} \otimes \cdots \otimes v_{k_i} \otimes v_{k_{i+1}} \otimes \cdots \otimes v_{k_d}) &= v_{k_1} \otimes \cdots \otimes v_{k_{i+1}} \otimes v_{k_{i}} \otimes \cdots \otimes v_{k_d} \\ &\quad(1\leq i\leq d-1),\\ f_1(s_d) ( v_{k_1} \otimes \cdots \otimes v_{k_d}) &= v_{k_1} \otimes \cdots \otimes v_{\overline{k_d}}. \end{align*} $$

Here, $k_1,\dots , k_d \in \{1,\dots ,n,\overline {n},\dots ,\overline {1}\}$ , and $\overline {\overline {i}}=i$ .

On the other hand, another action $f_2$ , was defined in Hu–Stoll [Reference Hu and Stoll7] and Mazorchuk–Stroppel [Reference Mazorchuk and Stroppel13], with its quantum version in Shoji–Sakamoto [Reference Sakamoto and Shoji14]. Using the basis element $w_i$ , this is the action below:

$$ \begin{align*} f_2(s_i) ( w_{k_1} \otimes \cdots \otimes w_{k_i} \otimes w_{k_{i+1}} \otimes \cdots \otimes w_{k_d}) &= w_{k_1} \otimes \cdots \otimes w_{k_{i+1}} \otimes w_{k_{i}} \otimes \cdots \otimes w_{k_d} \\ &\quad(1\leq i\leq d-1),\\ f_2(s_d) ( w_{k_1} \otimes \cdots \otimes w_{k_d}) &= (-1)^{\epsilon(k_d)} w_{k_1} \otimes \cdots \otimes w_{{k_d}}, \end{align*} $$

where $k_1,\dots , k_d \in \{1,\dots ,n,\overline {1},\dots ,\overline {n}\}$ , and $\epsilon (k_d)$ is $1$ if $k_d$ is unbarred, and $-1$ if $k_d$ is barred.

Proposition 3.3.1. The automorphism $L^{\otimes d}$ on $V^{\otimes d}$ induced by L, intertwines the two actions of the Weyl group $B_d$ , in the following sense

(40) $$ \begin{align} L^{\otimes d} (f_1(s)(u))=f_2(s)(L^{\otimes d}(u)) \hspace{.5 in} \forall s\in B_d, u\in V^{\otimes d}. \end{align} $$

In other words, the following diagram holds

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

Proof. It is enough to show equation (40) holds for any $s_i$ with $1\leq i\leq d$ and when u is any pure tensor. When $1 \leq i\leq d-1$ , this is a straightforward check. When $i=d$ :

1) When $u=v_{i_1}\otimes \cdots \otimes v_{i_d}$ , where $i_1,\dots ,i_{d-1}$ can be either barred or unbarred, and $i_d$ is unbarred,

$$ \begin{align*} L^{\otimes d} (f_1(s_d)(u))&=L^{\otimes d}(v_{i_1}\otimes \cdots\otimes v_{i_{d-1}}\otimes v_{\overline{i_d}})\\ &=L(e_{i_1})\otimes \cdots \otimes L(e_{i_d})\otimes (w_{i_d}-w_{\overline{i_d}})\\ f_2(s_d)(L^{\otimes d}(u))&= f_2(s_d)(L(v_{i_1})\otimes \cdots\otimes L(v_{i_{d-1}})\otimes (w_{i_d}+w_{\overline{i_d}}))\\ &=L(e_{i_1})\otimes \cdots \otimes L(e_{i_d})\otimes (w_{i_d}-w_{\overline{i_d}}). \end{align*} $$

2) On the other hand, when $u=v_{i_1}\otimes \cdots \otimes v_{\overline {i_d}}$ , where $i_1,\dots ,i_{d-1}$ can be either barred or unbarred, and $i_d$ is unbarred,

$$ \begin{align*} L^{\otimes d} (f_1(s_d)(u))&=L^{\otimes d}(v_{i_1}\otimes \cdots\otimes v_{i_{d-1}}\otimes v_{{i_d}})\\ &= L(e_1)\otimes \cdots L(e_{i_{d-1}})\otimes (w_{i_d}+w_{\overline{i_d}})\\ f_2(s_d)(L^{\otimes d}(u))&=f_2(s_d)(L(e_{i_1})\otimes \cdots \otimes L(e_{i_{d-1}})\otimes (w_{i_d}-w_{\overline{i_d}}))\\ &=L(e_{i_1})\otimes \cdots\otimes L(e_{i_{d-1}})\otimes (w_{i_d}+w_{\overline{i_d}}). \end{align*} $$

The space V can be regarded as the defining representation for $\mathfrak {gl}(V)=\mathfrak {gl}_{2n}(\mathbb {C})$ , using the ordered basis $\mathcal {B}_1$ . For future convenience, we will relabel the barred vectors as

$$ \begin{align*} \mathcal{B}_1=\{v_1,\dots,v_n,v_{n+1}=v_{\overline{n}},\dots,v_{2n}=v_{\overline{1}}\}. \end{align*} $$

As a subalgebra of $\mathfrak {gl}_{2n}(\mathbb {C})$ , $\mathfrak {gl}_{2n}^{\theta }$ also acts on V, and denote this action as $g_1: U(\mathfrak {gl}_{2n}^{\theta })\to \operatorname {End}(V)$ . It induces an action on $V^{\otimes d}$ , denoted as $g_1$ by an abuse of notation, via

$$ \begin{align*} g_1(x)(v_{i_1}\otimes \cdots \otimes v_{i_d}) = g_1(x)(v_{i_1})\otimes \cdots \otimes g_1(x)(v_{i_d}). \end{align*} $$

This action is known to commute with the action $f_1$ of $B_d$ on $V^{\otimes d}$ , when $\mathfrak {gl}_{2n}^{\theta }$ gets replaced by its group analogue in [Reference Green6], or in the quantum case in [Reference Bao, Kujawa, Li and Wang1].

Using the ordered basis $\mathcal {B}_2$ , one can define an action of $\mathfrak {gl}_n(\mathbb {C})\oplus \mathfrak {gl}_n(\mathbb {C})$ on V, by regarding the vector space $V_1$ spanned by $\{w_1,\dots ,w_n\}$ as the defining representation for the first copy $\mathfrak {gl}(V_1)=\mathfrak {gl}_n$ and the vector space $V_{-1}$ spanned by $\{w_{\overline {1}},\dots ,w_{\overline {n}}\}$ as the defining representation for the second copy $\mathfrak {gl}(V_{-1})=\mathfrak {gl}_n$ . We denote this action as $g_2 : U(\mathfrak {gl}_n(\mathbb {C})\oplus \mathfrak {gl}_n(\mathbb {C}))\to \operatorname {End}(V)$ , and by an abuse of notation also denote by $g_2$ the induced action on $V^{\otimes d}$ . This action is known to commute with the $f_2$ action of $B_d$ on $V^{\otimes d}$ (see [Reference Hu and Stoll7], [Reference Mazorchuk and Stroppel13].)

Proposition 3.3.2. The automorphism $L^{\otimes d}$ intertwines the action of $U^i$ and $U(\mathfrak {gl}_n\oplus \mathfrak {gl}_n)$ , with respect to the homomorphism $\rho $ , in the following sense

$$ \begin{align*} g_2(\rho(x))(L^{\otimes d}(u))=L^{\otimes d}(g_1(x)(u)) \hspace{.5 in} \forall x\in U^i,u\in V^{\otimes d}. \end{align*} $$

In other words, the following diagram commutes

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

Proof. It is enough to check the following holds for any x that is a generator among $\mathbf {e}_i, \mathbf {f}_i, \mathbf {t}, \mathbf {d}_i$ , and for any $v \in V$ .

(41) $$ \begin{align} g_2(\rho(x))(L(v))=L(g_1(x)(v)). \end{align} $$

The main claim then follows from extending the action of $U^i$ and $U(\mathfrak {gl}_n\oplus \mathfrak {gl}_n)$ from V to $V^{\otimes d}$ :

1) To check that (41) holds for $x=\mathbf {e}_j$ , $1\leq j\leq n-1$ , we first check it when $v=v_i$ for $1\leq i\leq n$ :

$$ \begin{align*} &g_2(\rho(\mathbf{e}_j))L(v_i)=g_2(\rho(\mathbf{e}_j))(w_i+w_{\overline{i}})=g_2(e_j+e_{\overline{j}})(w_i+w_{\overline{i}})=\delta_{j+1,i}(w_{i-1}+w_{\overline{i-1}}),\\ &L(g_1(\mathbf{e}_j)(v_i))=L(g_1(E_j+F_{2n-j})v_i)=L(\delta_{j+1,i}v_{i-1})=\delta_{j+1,i}(w_{i-1}+w_{\overline{i-1}}). \end{align*} $$

Here, the last equality holds because $n+1\leq 2n-j\leq 2n-1$ and $F_{2n-j}=E_{2n-j+1,2n-j}$ , where $E_{pq}$ is the matrix unit with $1$ in the $(p,q)$ -position and $0$ elsewhere. Since $2n-j\neq i$ , the action of $F_{2n-j}$ is zero on $v_i$ .

When $v=v_{\overline {i}}$ for $1\leq i\leq n$ :

$$ \begin{align*} g_2(\rho(\mathbf{e}_j))(L(v_{\overline{i}}))&=g_2(\rho(\mathbf{e}_j))(w_i-w_{\overline{i}})=g_2(e_j+e_{\overline{j}})(w_i-w_{\overline{i}})=\delta_{j+1,i}(w_{i-1}-w_{\overline{i-1}}),\\ L(g_1(\mathbf{e}_j)(v_{\overline{i}}))&=L(g_1(E_j+F_{2n-j})(v_{2n+1-i}))=L(\delta_{2n-j,2n+1-i}v_{2n+2-i}),\\ &=\delta_{i,j+1}L(v_{\overline{i-1}})=\delta_{i,j+1}(w_{i-1}-w_{\overline{i-1}}). \end{align*} $$

The second equality holds because $n+1\leq 2n+1-i\leq 2n$ , $2\leq j+1\leq n$ , $j+1\neq 2n+1-i$ . Hence the action of $E_j=E_{j,j+1}$ is zero on $v_{2n+1-i}$ .

2) We now check that (41) holds for $x=\mathbf {f}_j$ , $1\leq j\leq n-1$ and $v=v_i$ , $1\leq i\leq n$ :

$$ \begin{align*} &g_2(\rho(\mathbf{f}_j))(L(v_i))=g_2(\rho(\mathbf{f}_j))(w_i+w_{\overline{i}})=g_2(f_j+f_{\overline{j}})(w_i+w_{\overline{i}})=\delta_{j,i}(w_{i+1}+w_{\overline{i+1}}),\\ &g_1(\mathbf{f}_j)(v_i)=g_1(F_j+E_{2n-j})(v_i)=\delta_{j,i}v_{i+1},\\ &L(g_1(\mathbf{f}_j)(v_i))=\delta_{j,i}(w_{i+1}+w_{\overline{i+1}}). \end{align*} $$

Here, $E_{2n-j}$ acts on $v_i$ by zero. For a more careful discussion, please refer to case 1).

When $v=v_{\overline {i}}$ , $1\leq i\leq n$ ,

$$ \begin{align*} &g_2(\rho(\mathbf{f}_j))L(v_{\overline{i}})=g_2(f_j+f_{\overline{j}})(w_i-w_{\overline{i}})=\delta_{j,i}(w_{i+1}-w_{\overline{i+1}}),\\ &g_1(\mathbf{f}_j)(v_{\overline{i}})=g_1(F_j+E_{2n-j})(v_{2n+1-i})=\delta_{2n+1-j,2n+1-i}v_{2n-i}=\delta_{i,j}v_{\overline{i+1}},\\ &L(g_1(\mathbf{f}_j)(v_{\overline{i}}))=\delta_{i,j}(w_{i+1}-w_{\overline{i+1}}). \end{align*} $$

Here, $F_j$ acts on $v_{2n+1-i}$ by zero.

3) We now check the case when $x=\mathbf {t}$ , $v=v_i$ for $1\leq i\leq n$ :

$$ \begin{align*} &g_2(\rho(\mathbf{t}))L(v_i)=g_2(h_n-h_{\overline{n}})(w_i+w_{\overline{i}})=\delta_{n,i}(w_i-w_{\overline{i}}),\\ & g_1(\mathbf{t})=(E_n+F_n)v_i=\delta_{n+1,i}v_{i-1}+\delta_{n,i}v_{i+1}=\delta_{n+1,i}v_n+\delta_{n,i}v_{n+1}\\ &=\delta_{n+1,i}v_n+\delta_{n,i}v_{\overline{n}}=\delta_{n,i}v_{\overline{n}},\\ &L(g_1(\mathbf{t})(v_i))=\delta_{n,i}(w_n-w_{\overline{n}}). \end{align*} $$

When $v=v_{\overline {i}}$ for $1\leq i\leq n$ :

$$ \begin{align*} &g_2(\rho(\mathbf{t}))L(v_{\overline{i}})=g_2(h_n-h_{\overline{n}})(w_i-w_{\overline{i}})=\delta_{i,n}(w_n+w_{\overline{n}}),\\ &g_1(\mathbf{t})(v_{\overline{i}})=g_1(E_n+F_n)(v_{2n+1-i})=\delta_{n+1,2n+1-i}v_{2n-i}=\delta_{i,n}v_n,\\ & L(g_1(\mathbf{t})(v_{\overline{i}}))=\delta_{i,n}(w_n+w_{\overline{n}}). \end{align*} $$

4) We now prove the case when $x=\mathbf {d}_j$ and $v=v_i$ , for $1\leq i,j\leq n$ :

$$ \begin{align*} &g_2(\rho(\mathbf{d}_j))(L(v_i))=g_2(h_j+h_{\overline{j}})(w_i+w_{\overline{i}})=\delta_{i,j}(w_i+w_{\overline{i}}),\\ &g_1(\mathbf{d}_j)(v_i)=g_1(H_j+H_{2n+1-j})(v_i)=\delta_{i,j}v_i,\\ &L(g_1(\mathbf{d}_j)(v_i))=\delta_{i,j}(w_i+w_{\overline{i}}). \end{align*} $$

When $v=v_{\overline {i}}$ , $1\leq i\leq n$ :

$$ \begin{align*} &g_2(\rho(\mathbf{d}_j))(L(v_{\overline{i}}))=g_2(h_j+h_{\overline{j}})(w_i-w_{\overline{i}})=\delta_{i,j}(w_i-w_{\overline{i}}),\\ &g_1(\mathbf{d}_j)(v_{\overline{i}})=g_1(H_j+H_{2n+1-j})(v_{2n+1-i})=\delta_{i,j}v_{2n+1-i}=\delta_{i,j}v_{\overline{i}},\\ &L(g_1(\mathbf{d}_j)(v_{\overline{i}}))=\delta_{i,j}(w_i-w_{\overline{i}}). \end{align*} $$

Hence we have checked all cases.

3.4 Root vectors in $U^i$

We also introduce the notion of root vectors for $\mathfrak {gl}_{2n}(\mathbb {C})^{\theta }$ as we did prior to Lemma 2.3.1. Similar to the notation earlier, we distinguish between the first and second copy in $\mathfrak {gl}_n(\mathbb {C})\oplus \mathfrak {gl}_n(\mathbb {C})$ by letting $\mathfrak {g}_1=\mathfrak {gl}_n(\mathbb {C})$ , $\mathfrak {g}_2=\mathfrak {gl}_n(\mathbb {C})$ , and for $\alpha \in \Phi _n$ a root in $\mathfrak {gl}_n$ , let $Z_{\alpha }\in \mathfrak {g}_1,\overline {Z}_{\alpha }\in \mathfrak {g}_2$ be the associated root vectors. We adopt the previous notation that $\epsilon _i=\mathbf {d}_i^*$ for $1\leq i \leq n$ and $\mu _i=H_i^*$ for $1\leq i\leq 2n$ . Let

$$ \begin{align*} \Phi_{2n}^+&=\{\mu_i-\mu_j\hspace{.1 in}| \hspace{.1 in} 1\leq i<j\leq 2n \},\\ \Pi_{2n}&=\{\mu_i-\mu_{i+1}\hspace{.1 in}| \hspace{.1 in} 1\leq i\leq 2n-1 \}, \end{align*} $$

be the set of positive roots and simple roots in $\mathfrak {gl}_{2n}(\mathbb {C})$ , respectively. For any $\alpha \in \Phi _{2n}^+$ , define root vectors $X_{\alpha }$ recursively as follows. For the simple roots:

$$ \begin{align*} X_{\mu_i-\mu_{i+1}}&=\mathbf{e}_i \hspace{.2 in} (1\leq i\neq n-1),\\ X_{\mu_n-\mu_{n+1}}&=\mathbf{t} \hspace{.2 in} (1\leq i\neq n-1),\\ X_{\mu_i-\mu_{i+1}}&=\mathbf{f}_{2n-i} \hspace{.2 in} (n+1\leq i\neq 2n-1). \end{align*} $$

On $\alpha $ that is a sum of simple roots, $X_{\alpha }$ is defined recursively as (24). Versions of Lemmas 2.3.1 through 2.3.5 is also true for root vectors in $\mathfrak {g}^{\theta }$ . In particular, we have the following.

For $\alpha \in \Phi _{n}$ , an element $x\in \mathfrak {g}^{\theta }$ is a weight vector of weight $\alpha $ , under the adjoint action of $\mathbf {d}_1,\dots ,\mathbf {d}_n$ , if and only if

$$ \begin{align*} [\mathbf{d}_i,x]=\alpha(\mathbf{d}_i)x \hspace{.2 in} (1\leq i\leq n). \end{align*} $$

Define a map $t: \Phi _{2n}^+\to \Phi _{n}\cup \{0\}$ , such that on the simple roots $\alpha \in \Pi _{2n}$ , $X_{\alpha }$ is a weight vector of weight $t(\alpha )$ . That is to say,

$$ \begin{align*} t(\mu_i-\mu_{i+1})&=\epsilon_i-\epsilon_{i+1} \hspace{.2 in} (1\leq i\leq n-1), \\ t(\mu_n-\mu_{n+1})&=0, \\ t(\mu_i-\mu_{i+1})&=-\epsilon_{2n-i}+\epsilon_{2n+1-i} \hspace{.2 in} (n+1\leq i\leq 2n-1). \end{align*} $$

Extend t linearly so that on a sum of simple roots $\alpha =\alpha _1+\cdots \alpha _i \in \Phi _{2n}^+$ , define

$$ \begin{align*} t(\alpha)=t(\alpha_1)+\cdots+t(\alpha_i). \end{align*} $$

Then a version of Lemma 2.3.6 is also true. Moreover, the stalks of t are described below.

Lemma 3.4.2. For the map t defined above,

$$ \begin{align*} t^{-1}(0)&=\{\mu_i-\mu_{2n+1-i} \hspace{.1 in} | \hspace{.1 in} 1\leq i \leq n \}, \\ t^{-1}(\epsilon_i-\epsilon_n)&=\{\mu_i-\mu_n\} \hspace{.5 in} (1\leq i \leq n-1),\\ t^{-1}(-\epsilon_i+\epsilon_n)&=\{\mu_n-\mu_{2n+1-i}\} \hspace{.5 in} (1\leq i \leq n-1),\\ t^{-1}(\epsilon_i-\epsilon_j)&=\{\mu_i-\mu_j,\mu_i-\mu_{2n+1-j}\} \hspace{.5 in} (1\leq i<j \leq n-1),\\ t^{-1}(-\epsilon_i+\epsilon_j)&=\{\mu_j-\mu_{2n+1-i}, \mu_{2n+1-j}-\mu_{2n+1-i}\} \hspace{.5 in} (1\leq i<j \leq n-1). \end{align*} $$

Proof. The proof is very similar to that of Proposition 2.3.8 and is left to the reader.

3.5 An algebra isomorphism with $U(\mathfrak {gl}_n\oplus \mathfrak {gl}_n)$

We now re-index the root vectors using their weights. Specifically, define $\mathbf {t}_1,\dots ,\mathbf {t}_{n-1},\mathbf {t}_n$ such that $\mathbf {t}_n=\mathbf {t}$ ,

(42) $$ \begin{align} \mathbf{t}_{i-1}=\mathbf{t}_i+X_{\mu_{i-1}-\mu_{2n+1-i}} \hspace{.2 in} (2\leq i\leq n). \end{align} $$

For positive root vectors,

$$ \begin{align*} X_{\epsilon_i-\epsilon_n}&=X_{\mu_i-\mu_n} \hspace{.2 in } (1\leq i\leq n-1),\\ X_{\epsilon_i-\epsilon_j}&=X_{\mu_i-\mu_j}, \hspace{.2 in} X^{\prime}_{\epsilon_i-\epsilon_j}=X_{\mu_i-\mu_{2n+1-j}} \hspace{.2 in } (1\leq i<j\leq n-1). \end{align*} $$

Recall that $e_i,f_i,h_i\in \mathfrak {g}_1$ and $e_{\overline {i}},f_{\overline {i}},h_{\overline {i}} \in \mathfrak {g}_2$ are Chevalley generators of each copy of $\mathfrak {gl}_n(\mathbb {C})$ .

Lemma 3.5.1. Under the map $\rho $ , for $\alpha \in \Phi _{n}$

(43) $$ \begin{align} \rho(\mathbf{t}_i)&=h_i-h_{\overline{i}} \hspace{.2 in}(1\leq i\leq n), \end{align} $$

(44) $$ \begin{align} \rho(X_{\alpha})&=Z_{\alpha}+\overline{Z}_{\alpha}, \end{align} $$

(45) $$ \begin{align} \rho(X_{\alpha}')&=Z_{\alpha}-\overline{Z}_{\alpha}. \end{align} $$

In particular, $\rho $ is an isomorphism between Lie algebras.

Proof. We first prove (44) for a positive root $\alpha =\epsilon _i-\epsilon _j$ , by inducting on i. The base case when $i=j-1$ is given by the definition of $\rho $ on $\mathbf {e}_{j-1}=X_{\epsilon _{j-1}-\epsilon _{j}}$ . Now we claim if (43) is true for $(i,j)$ , it is also true for $(i-1,j)$ :

$$ \begin{align*} \rho(X_{\epsilon_{i-1}-\epsilon_j})&=\rho([X_{\epsilon_{i-1}-\epsilon_i},X_{\epsilon_i-\epsilon_j}])=[\rho(\mathbf{e}_{i-1}),Z_{\epsilon_i-\epsilon_j}+\overline{Z}_{\epsilon_i-\epsilon_j}]\\ &=[Z_{\epsilon_{i-1}-\epsilon_i}+\overline{Z}_{\epsilon_{i-1}-\epsilon_{i}},Z_{\epsilon_i-\epsilon_j}+\overline{Z}_{\epsilon_i-\epsilon_j}]=Z_{\epsilon_{i-1}-\epsilon_j}+\overline{Z}_{\epsilon_{i-1}-\epsilon_{j}}. \end{align*} $$

Now let us prove (43). The base case when $i=n$ follows from the definition of $\rho $ on $\mathbf {t}$ . Suppose (43) is true for i, then we claim it is also true for $i-1$ :

$$ \begin{align*} \rho(\mathbf{t}_{i-1})&=\rho([X_{\mu_{i-1}-\mu_{2n+2-i}})+\rho(\mathbf{t}_i)\\ &=\rho([X_{\mu_i-1-\mu_i},[X_{\mu_i-\mu_{2n+1-i}},X_{\mu_{2n+1-i}-\mu_{2n+2-i}}]])+\rho(\mathbf{t}_i)\\ &=\rho([\mathbf{e}_{i-1},[\mathbf{t}_i,X_{-\epsilon_{i-1}+\epsilon_i}]])+\rho(\mathbf{t}_i)=[\mathbf{e}_{i-1},[\mathbf{t}_{i},\mathbf{f}_{i-1}]]+\rho(\mathbf{t}_i)\\ &=[Z_{\epsilon_{i-1}-\epsilon_i}+\overline{Z}_{\epsilon_{i-1}-\epsilon_i},[h_i-h_{\overline{i}},Z_{-\epsilon_{i-1}+\epsilon_i}+\overline{Z}_{-\epsilon_{i-1}+\epsilon_i}]]+h_i-h_{\overline{i}}\\ &=[Z_{\epsilon_{i-1}-\epsilon_i}+\overline{Z}_{\epsilon_{i-1}-\epsilon_i},Z_{-\epsilon_{i-1}+\epsilon_i}-\overline{Z}_{-\epsilon_{i-1}+\epsilon_i}]+h_i-h_{\overline{i}}\\ &=h_{i-1}-h_i-h_{\overline{i-1}}+h_{\overline{i}}+h_i-h_{\overline{i}}=h_{i-1}-h_{\overline{i-1}}. \end{align*} $$

Now we prove (45) by induction on i. The base case when $i=j-1$ (or $j=i+1$ ) is as follows

$$ \begin{align*} \rho(X_{\epsilon_i-\epsilon_{i+1}})&=\rho(X_{\mu_i-\mu_{2n-i}})=\rho([X_{\mu_i-\mu_{i+1}},X_{\mu_{i+1}-\mu_{2n-i}}])=\rho([\mathbf{e}_i,\mathbf{t}_{i+1}-\mathbf{t}_{i+2}])\\ &=[X_{\epsilon_i-\epsilon_{i+1}}+\overline{X}_{\epsilon_i-\epsilon_{i+1}},h_{i+1}-h_{\overline{i+1}}]=X_{\epsilon_i-\epsilon_{i+1}}-\overline{X}_{\epsilon_i-\epsilon_{i+1}}. \end{align*} $$

Now suppose (45) is true for $(i,j)$ with $i<j<2n+1-j$ , we claim it is also true for $(i-1,j)$ :

$$ \begin{align*} \rho(X_{\epsilon_{i-1}-\epsilon_j})&=\rho(X_{\mu_{i-1}-\mu_{2n+1-j}})=\rho([X_{\mu_{i-1}-\mu_i},X_{\mu_i-\mu_{2n+1-j}}])=\rho([\mathbf{e}_i,X^{\prime}_{\epsilon_i-\epsilon_j}])\\&=[X_{\epsilon_{i-1}-\epsilon_i}+\overline{X}_{\epsilon_{i-1}-\epsilon_i},X_{\epsilon_{i}-\epsilon_j}-\overline{X}_{\epsilon_{i}-\epsilon_j}]=X_{\epsilon_{i-1}-\epsilon_j}-\overline{X}_{\epsilon_{i-1}-\epsilon_j}. \end{align*} $$

Hence we have proved the induction step for positive roots. The arguments for negative roots are similar.

3.6 The type B Schur algebra

Similar to the previous case, fix a positive integer d, and let S be the quotient of $U(\mathfrak {gl}_n(\mathbb {C})\oplus \mathfrak {gl}_n(\mathbb {C}))$ under the following relations

$$ \begin{align*} h_1+\cdots+h_{n}+h_{\overline{1}}+\cdots+h_{\overline{n}}&=d,\\ h_{i}(h_i-1)\cdots(h_{i}-d)&=0, \hspace{.5 in} 1\leq i\leq n,\\ h_{\overline{i}}(h_{\overline{i}}-1)\cdots(h_{\overline{i}}-d)&=0, \hspace{.5 in} 1\leq i\leq n . \end{align*} $$

Further, recall that $\mathbf {t}_i$ is defined recursively using generators of $U^i$ in (42). Define the following element

$$ \begin{align*} \mathbf{h}_i=\frac{1}{2}(\mathbf{t}_i+\mathbf{d}_i), \hspace{.3 in}\mathbf{h}_{\overline{i}}=\frac{1}{2}(-\mathbf{t}_i+\mathbf{d}_i), \hspace{.5 in} (1\leq i\leq n). \end{align*} $$

Then Lemma 3.5.1 and the definition of $\rho $ implies that $\rho (\mathbf {h}_i)=h_i$ and $\rho (\mathbf {h}_{\overline {i}})=h_{\overline {i}}$ . Therefore, let $S^i$ be the quotient of $U(\mathfrak {g}^{\theta })$ under the following relations

(46) $$ \begin{align} \mathbf{d}_1+\cdots+\mathbf{d}_n=d, \end{align} $$

(47) $$ \begin{align} \mathbf{h}_i(\mathbf{h}_i-1)\cdots(\mathbf{h}_i-d)=0, \hspace{.5 in} 1\leq i\leq n, \end{align} $$

(48) $$ \begin{align} \mathbf{h}_{\overline{i}}(\mathbf{h}_{\overline{i}}-1)\cdots(\mathbf{h}_{\overline{i}}-d)=0, \hspace{.5 in} 1\leq i\leq n. \end{align} $$

Note: once unraveling the notation of root vectors, these relations are purely in terms of generators $\mathbf {e}_i$ , $\mathbf {f}_i$ , $\mathbf {t}$ and $\mathbf {d}_i$ of $U^i$ .

Example 3.6.2. When $n=1$ , $S^i$ is the quotient of $\mathbb {C}[\mathbf {t}]$ under the following relations, where d is a positive integer,

$$ \begin{align*} (\mathbf{t}+d)(\mathbf{t}+d-2)\cdots(\mathbf{t}-d+2)(\mathbf{t}-d)&=0. \end{align*} $$

We refer to [Reference Li11] for a presentation in the quantum case.

Example 3.6.3. When $n=2$ , $U(\mathfrak {gl}_{4}(\mathbb {C})^{\theta })$ is generated by $\mathbf {e},\mathbf {f},\mathbf {t},\mathbf {d}_1,\mathbf {d}_2$ . Relations (46)–(48) become explicitly as follows.

$$ \begin{align*} \mathbf{d}_1+\mathbf{d}_2&=d,\\ ([\mathbf{e},\mathbf{f}]+\mathbf{d}_1)([\mathbf{e},\mathbf{f}]+\mathbf{d}_1-2)\cdots([\mathbf{e},\mathbf{f}]+\mathbf{d}_1-2d)&=0,\\ (-[\mathbf{e},\mathbf{f}]+\mathbf{d}_1)(-[\mathbf{e},\mathbf{f}]+\mathbf{d}_1-2)\cdots(-[\mathbf{e},\mathbf{f}]+\mathbf{d}_1-2d)&=0,\\ (\mathbf{t}+\mathbf{d}_2)(\mathbf{t}+\mathbf{d}_2-2)\cdots(\mathbf{t}+\mathbf{d}_2-2d)&=0,\\ (-\mathbf{t}+\mathbf{d}_2)(-\mathbf{t}+\mathbf{d}_2-2)\cdots(-\mathbf{t}+\mathbf{d}_2-2d)&=0. \end{align*} $$