Equivalences between three presentations of orthogonal and symplectic Yangians

Abstract

We prove the equivalence of two presentations of the Yangian $Y(\mathfrak {g})$ of a simple Lie algebra $\mathfrak {g}$, and we also show the equivalence with a third presentation when $\mathfrak {g}$ is either an orthogonal or a symplectic Lie algebra. As an application, we obtain an explicit correspondence between two versions of the classification theorem of finite-dimensional irreducible modules for orthogonal and symplectic Yangians.

A Proof of Theorem 2.6 when \(\mathfrak {g}=\mathfrak {s}\mathfrak {l}_2\)

In this appendix, we complete the proof of Theorem 2.6 in the case where $\mathfrak {g}=\mathfrak {s}\mathfrak {l}_2$. The extra difficulty when $\mathfrak {g}= \mathfrak {s}\mathfrak {l}_2$ is to check that $\varPhi _{\mathrm {cr},J}$ preserves the relation $[{\tilde{h}}_{i1},[x_{i1}^+,x_{i1}^-]]=0$ of $Y_\zeta ^\mathrm{cr}(\mathfrak {g})$.

We normalize our symmetric invariant non-degenerate bilinear form $(\cdot , \cdot )$ on $\mathfrak {s}\mathfrak {l}_2$ so that it is given by $(X,Y)=\mathrm {Tr}(XY)$ for all $X,Y\in \mathfrak {s}\mathfrak {l}_2$. With this choice, the single positive root $\alpha$ has length 2 and $\{x_{i0}^+,x_{i0}^-,h\}$ coincides with the standard basis $\{e,f,h\}$ of $\mathfrak {s}\mathfrak {l}_2$. We will primarily employ the latter notation throughout the remainder of this proof, but when dealing with equations involving e and f simultaneously, we will sometimes replace them with $x^+$ and $x^-$, respectively. An orthonormal basis with respect to $(\cdot ,\cdot )$ is $\mathcal {B}=\{E,F,H\}$, where

$$\begin{aligned} E=\tfrac{1}{\sqrt{2}}(e+f)\quad F=\tfrac{1}{\sqrt{-2}}(e-f),\quad H=\tfrac{1}{\sqrt{2}}h. \end{aligned}$$

In particular, we have

$$\begin{aligned}{}[H,E]= & {} e-f=\sqrt{-2}F,\quad [H,F]=-\tfrac{\sqrt{-1}}{2}[h,e-f]=\tfrac{1}{i}(e+f)=-\sqrt{-2}E,\\&\quad [E,F]=\sqrt{-1} h=\sqrt{-2}H. \end{aligned}$$

Step 1: Reducing the defining relations.

We begin by showing that the relation

$$\begin{aligned}{}[[J(e),J(f)],J(h)]=\zeta ^2\left( fJ(e)-J(f)\,e\right) h \end{aligned}$$

is satisfied in $Y_\zeta (\mathfrak {g})$. This relation is well known throughout the literature (see for instance (4.64) in [5] and Definition 2.22 in [44]), but to our knowledge the relevant computations have never appeared explicitly, so we have decided to include them.

Consider equation (2.4) with $X_1=E=X_3$ and $X_2=F=X_4$. Since $[J(E),J(F)]=\sqrt{-1}[J(e),J(f)]$ and $[E,J(F)]=\sqrt{-1}J(h)$, the left-hand side is equal to $-2\left[ [J(e),J(f)],J(h)\right]$. Thus, it suffices to show that the right-hand side of (2.4) with $X_1=E=X_3$ and $X_2=F=X_4$ is equal to $-2\zeta ^2\left( fJ(e)-J(f)\,e\right) h$. By definition, it is equal to

$$\begin{aligned} 2\zeta ^2 \sum _{X_\lambda ,X_\mu ,X_\nu \in \mathcal {B}}\left( [E,X_\lambda ],\big [[F,X_\mu ],[[E,F],X_\nu ] \big ]\right) \{X_\lambda ,X_\mu ,J(X_\nu )\}. \end{aligned}$$

(A.1)

As $[E,F]=\sqrt{-1}h$ and $(\cdot ,\cdot )$ is invariant and symmetric, we have

$$\begin{aligned} \left( [E,X_\lambda ],\big [[F,X_\mu ],[[E,F],X_\nu ] \big ]\right) =\sqrt{-1}\left( [h,X_\nu ],\big [[E,X_\lambda ],[F,X_\mu ] \big ] \right) . \end{aligned}$$

Thus, we can rewrite (A.1) as

$$\begin{aligned}&\sum _{X_\lambda ,X_\mu \in \mathcal {B}}\left( [h,E],\big [[E,X_\lambda ],[F,X_\mu ] \big ] \right) \{X_\lambda ,X_\mu ,J(E)\}\nonumber \\&\quad +\,\sum _{X_\lambda ,X_\mu \in \mathcal {B}}\left( [h,F],\big [[E,X_\lambda ],[F,X_\mu ] \big ] \right) \{X_\lambda ,X_\mu ,J(F)\} \end{aligned}$$

(A.2)

multiplied by $2\sqrt{-1}\zeta ^2$. Consider the different nonzero possibilities for $\big [[E,X_\lambda ],[F,X_\mu ] \big ]$. They are

$$\begin{aligned} \big [[E,H],[F,H]\big ],\quad \big [[E,H],[F,E]\big ]\quad \text {and}\quad \big [[E,F],[F,H]\big ]. \end{aligned}$$

Both $([H,E],\cdot )$ and $([H,F],\cdot )$ applied to the first of these is zero, while $([H,E],\cdot )$ applied to [[E, H], [F, E]] is zero, and $([H,F],\cdot )$ applied to [[E, F], [F, H]] is zero. Therefore, (A.2) reduces to

$$\begin{aligned} \left( [h,E],\left[ [E,F],[F,H] \right] \right) \{F,H,J(E)\}+\left( [h,F],\left[ [E,H],[F,E] \right] \right) \{H,E,J(F)\}. \end{aligned}$$

By invariance, $\left( [h,F],\left[ [E,H],[F,E] \right] \right) =-\left( [h,E],\left[ [E,F],[F,H] \right] \right)$, and moreover

$$\begin{aligned} \left( [h,E],\left[ [E,F],[F,H] \right] \right)= & {} -\sqrt{2}\left( e-f,[h,e+f] \right) \\= & {} -2\sqrt{2}\left( e-f,e-f\right) = 4\sqrt{2}. \end{aligned}$$

Hence, (A.2) multiplied by $2\sqrt{-1}\zeta ^2$, which is the right-hand side of (2.4) with $X_1=E=X_3$ and $X_2=F=X_4$, is equal to $8i\sqrt{2}\zeta ^2\left( \{F,H,J(E)\}-\{H,E,J(F)\}\right)$. Let us rewrite this in terms of e, f and h:

$$\begin{aligned}&8i\sqrt{2}\zeta ^2\left( \{F,H,J(E)\}-\{H,E,J(F)\}\right) \\&\quad = 4\zeta ^2\left( \{e-f,h,J(e+f)\}-\{h,e+f,J(e-f)\}\right) \\&\quad = 8\zeta ^2\left( \{e,h,J(f)\}-\{h,f,J(e)\}\right) . \end{aligned}$$

Expanding $\{h,f,J(e)\}-\{e,h,J(f)\}$ and using the defining relations of $\mathfrak {s}\mathfrak {l}_2$ to rewrite it only in terms of $fJ(e)\,h$ and $J(f)\,eh$, we find that it is equal to $\frac{1}{4}\big (J(f)\,eh-fJ(e)\,h\big )$, which gives the desired result.

Step 2: Checking the relation $\big [\tilde{h}_1,[x_1^-,x_1^+]\big ]=0$ is preserved.

By definition of $\varPhi _{\mathrm {cr},J}$, this amounts to checking that

$$\begin{aligned}{}[J(h)-\zeta v,[J(f)-\zeta w^-,J(e)-\zeta w^+]]=0, \end{aligned}$$

where

$$\begin{aligned} v=\tfrac{1}{2}(\{e,f\}-h^2),\quad w^+=-\tfrac{1}{4}\{e,h\},\quad w^-=-\tfrac{1}{4}\{f,h\}. \end{aligned}$$

Expanding the left-hand side, we obtain

$$\begin{aligned} \begin{aligned} \left[ J(h),[J(f),J(e)]\right] -\zeta \left[ J(h),[J(f),w^+]+[w^-,J(e)]\right] +\zeta ^2\left[ J(h),[w^-,w^+]\right] \\ -\zeta \left[ v,[J(f),J(e)]\right] +\zeta ^2\left[ v,[J(f),w^+]+[w^-,J(e)]\right] -\zeta ^3 \left[ v,[w^-,w^+]\right] \end{aligned}\nonumber \\ \end{aligned}$$

(A.3)

Step 2.1: $-\zeta ^3 \left[ v,[w^-,w^+]\right] =0$.

We have

$$\begin{aligned}{}[w^+,w^-]= & {} \tfrac{1}{16}([eh,fh]+[eh,hf]+[he,fh]+[he,hf])\nonumber \\= & {} \tfrac{1}{16}(4h^3-2\{e,f\}h-2e\{f,h\}-2\{f,h\}e-2h\{e,f\}). \end{aligned}$$

(A.4)

Now, since $[v,h]=0$ and $[\{e,f\},h]=0$, we have that $[v,\{e,f\}h+h\{e,f\}]=0$. Hence,

$$\begin{aligned}{}[v,[w^+,w^-]]=-\tfrac{1}{8}\,[v,2e\{f,h\}+2\{f,h\}e]=-\tfrac{1}{16}\,[\{e,f\}-h^2,e\{f,h\}+\{f,h\}e]. \end{aligned}$$

Using the commutator relations of $\mathfrak {s}\mathfrak {l}_2$, we can write $e\{f,h\}+\{f,h\}e=4efh-2h-2h^2$. Note that $[h,efh]=0$. Thus,

$$\begin{aligned}{}[v,[w^+,w^-]]=-\tfrac{1}{4}[\{e,f\},efh]=\tfrac{1}{4}([ef,efh]+[fe,efh])=\tfrac{1}{4}([fe,ef]h)=0. \end{aligned}$$

Step 2.2: The terms in (A.3) involving $\zeta ^2$.

After rewriting $\left[ J(h),[J(f),J(e)]\right] =\zeta ^2(fJ(e)-J(f)e)h$, the sum of the terms involving $\zeta ^2$ in (A.3) is

$$\begin{aligned} \zeta ^2\left( (fJ(e)-J(f)e)h+\left[ J(h),[w^-,w^+]\right] +\left[ v,[J(f),w^+]+[w^-,J(e)]\right] \right) \nonumber \\ \end{aligned}$$

(A.5)

Consider first $\left[ J(h),[w^-,w^+]\right]$. Recall that (A.4) and $\{e,f\}=2v+h^2$. Hence, since J(h) commutes with h and v, we have

$$\begin{aligned} \left[ J(h),[w^-,w^+]\right] = {}&\tfrac{1}{8}\left( [J(h),2vh+2hv]+[J(h),\{e,\{f,h\}\}]\right) \\ = {}&\tfrac{1}{8}\left( 4[J(h),v]h+[J(h),\{e,\{f,h\}\}]\right) . \end{aligned}$$

Observe that $e\{f,h\}+\{f,h\}e=4efh-2h-2h^2$ and $[J(h),\{e,\{f,h\}\}]=8(J(e)fh-eJ(f)h)$. Therefore, we have

$$\begin{aligned} \left[ J(h),[w^-,w^+]\right] =\tfrac{1}{2}\left( [J(h),v]h+2(J(e)fh-eJ(f)h)\right) . \end{aligned}$$

(A.6)

Let us now turn to the last term in (A.5). The identity

$$\begin{aligned}{}[J(x^\pm ),w^{\mp }]=-\tfrac{1}{4}[J(x^\pm ),x^\mp h +h x^\mp ]=\mp \tfrac{1}{2}(J(h)h-x^\mp J(x^\pm )-J(x^\pm )x^\mp ) \end{aligned}$$

implies that we can expand $[J(f),w^+]+[w^-,J(e)]$ in the form

$$\begin{aligned}{}[J(f),w^+]+[w^-,J(e)] = {}&\tfrac{1}{2}\left( J(h)h-eJ(f)-J(f)e)\right. \\&\left. +(J(h)h-fJ(e)-J(e)f)\right) \\ = {}&J(h)h-\tfrac{1}{2}(\{e,J(f)\}+\{f,J(e)\}). \end{aligned}$$

Hence,

$$\begin{aligned}{}[v,[J(f),w^+]+[w^-,J(e)]]=[v,J(h)]h-\tfrac{1}{2}([v,\{e,J(f)\}+\{f,J(e)\}]). \nonumber \\ \end{aligned}$$

(A.7)

We would like to simplify the last term. Since $[h,x^\pm J(x^\mp )+J(x^\mp )x^\pm ]=0$, we can replace v by $\tilde{v}$ ($= \nu + \frac{1}{2} h^2 = \frac{1}{2}\{e,f\}$). Next, the sequence of equalities

$$\begin{aligned}{}[{\tilde{v}},\{x^\pm ,J(x^\mp )\}]= {}&[{\tilde{v}}, x^\pm ]J(x^\mp )+x^\pm [{\tilde{v}},J(x^\mp )]+[{\tilde{v}},J(x^\mp )]x^\pm +J(x^\mp )[{\tilde{v}}, x^\pm ]\\ = {}&\tfrac{1}{2}\left( \mp x^\pm hJ(x^\mp )\mp hx^\pm J(x^\mp )\pm x^\pm x^\mp J(h)\pm x^\pm J(h)x^\mp \right. \\&\left. \pm x^\mp J(h)x^\pm \pm J(h)x^\mp x^\pm \mp J(x^\mp )x^\pm h\mp J(x^\mp )hx^\pm \right) \end{aligned}$$

allow us to express $2[{\tilde{v}},\{e,J(f)\}+\{f,J(e)\}]$ as

$$\begin{aligned}&-\,e hJ(f)- he J(f)+ efJ(h)+ e J(h)f + f J(h)e + J(h)f e - J(f)e h- J(f)he \\&\quad +\, f hJ(e)+ hf J(e)- f e J(h)- f J(h)e - e J(h)f - J(h)e f + J(e)f h+ J(e)hf. \end{aligned}$$

The terms involving J(h) all cancel out, and we are left with

$$\begin{aligned} 2[{\tilde{v}},\{e,J(f)\}+\{f,J(e)\}]= & {} -e hJ(f)- he J(f)- J(f)e h- J(f)he\\&+\, f hJ(e)+ hf J(e) + J(e)f h+ J(e)hf. \end{aligned}$$

We would like to express the right-hand side in terms of fJ(e)h and J(f)eh only. Using the defining relations of $\mathfrak {s}\mathfrak {l}_2$, we find that

$$\begin{aligned} 2\left[ \tilde{v},\{e,J(f)\}+\{f,J(e)\}\right] =4\,(fJ(e)\,h-J(f)\,eh). \end{aligned}$$

Multiplying by $\frac{1}{4}$ and substituting the result back into (A.7), we obtain

$$\begin{aligned}{}[v,[J(f),w^+]+[w^-,J(e)]]=[v,J(h)]\,h-(fJ(e)\,h-J(f)\,eh). \end{aligned}$$

Substituting this and (A.6) back into (A.5) gives

$$\begin{aligned}&\zeta ^2 \left( (fJ(e)-J(f)e)h+\left[ J(h),[w^-,w^+]\right] \right. \\&\left. \qquad +\left[ v,[J(f),w^+]+[w^-,J(e)]\right] \right) \\&\quad =\zeta ^2\left( (fJ(e)-J(f)e)h+\tfrac{1}{2}\left( [J(h),v]h\right. \right. \\&\left. \left. \qquad +2(J(e)fh-eJ(f)h)\right) +[v,J(h)]h-(fJ(e)h-J(f)eh) \right) \\&\quad =\tfrac{\zeta ^2}{2}\left( [v,J(h)]h+2(J(e)fh-eJ(f)h)\right) . \end{aligned}$$

Since $[v,J(h)]\,h=\frac{1}{2}[ef+fe,J(h)]\,h=\left( eJ(f)-J(e)f+J(f)e-fJ(e)\right) h=2\left( eJ(f)-J(e)f\right) h$, the above expression is zero, hence (A.5) is also zero.

Step 2.2: The terms involving $\zeta$.

In order to see that (A.3) vanishes, it remains to see that

$$\begin{aligned} \left[ J(h),[J(f),w^+]+[w^-,J(e)]\right] +\left[ v,[J(f),J(e)]\right] =0. \end{aligned}$$

(A.8)

First observe that $[h^2,[J(f),J(e)]]=0$, so we may replace v by ${\tilde{v}}$ in (A.8). We then have

$$\begin{aligned} \left[ {\tilde{v}},[J(f),J(e)]\right] =-[J(f),[J(e),\tilde{v}]]-[J(e),[{\tilde{v}},J(f)]]. \end{aligned}$$

Since $2w^+=[{\tilde{v}}, e]$ and $-2w^-=[{\tilde{v}},f]$, we can rewrite the left-hand side of (A.8) as

$$\begin{aligned}&-[J(f),[J(e),{\tilde{v}}]]-[J(e),[\tilde{v},J(f)]]+\tfrac{1}{2}[J(h),[J(f),[{\tilde{v}}, e]]]\nonumber \\&\quad -\tfrac{1}{2}[J(h),[[{\tilde{v}},f],J(e)]]. \end{aligned}$$

(A.9)

Consider the last two terms. Since $[J(h),[J(x^\pm ),[\tilde{v},x^\mp ]]]=[J(h),[[J(x^\pm ),{\tilde{v}}],x^\mp ]]\pm [J(h),[\tilde{v},J(h)]]$, we have

$$\begin{aligned}&[J(h),[J(f),[{\tilde{v}}, e]]]-[J(h),[[{\tilde{v}},f],J(e)]]\\&\quad =[J(h),[[J(f),{\tilde{v}}],e]]-[J(h),[[{\tilde{v}},J(e)],f]]\\&\quad =[[J(h),[J(f),{\tilde{v}}]],e]+2[[J(f),\tilde{v}],J(e)]\\&\qquad -[[J(h),[{\tilde{v}},J(e)]],f]+2[[{\tilde{v}}, J(e)],J(f)]. \end{aligned}$$

Substituting these new expression back into (A.9), we obtain that the left-hand side of (A.8) is equal to

$$\begin{aligned} \tfrac{1}{2}\left( [[J(h),[J(f),{\tilde{v}}]],e]-[[J(h),[\tilde{v},J(e)]],f] \right) . \end{aligned}$$

(A.10)

Let us show that this vanishes. Since ${\tilde{v}}=\frac{1}{2}\{e,f\}$, we have

$$\begin{aligned}{}[J(x^\pm ),\tilde{v}]=\tfrac{1}{2}[J(x^\pm ),x^+f+fx^+]=\pm \tfrac{1}{2}(x^\pm J(h)+J(h)x^\pm ). \end{aligned}$$

Thus, $[J(h),[J(f),{\tilde{v}}]]=J(f)J(h)+J(h)J(f)$ and, similarly, $[J(h),[{\tilde{v}},J(e)]]=-J(e)J(h)-J(h)J(e)$. Substituting these back into (A.10), we see that it vanishes, and hence (A.8) holds. This completes the proof that $\varPhi _{\mathrm {cr},J}$ preserves the relation $[h_1,[x_1^-,x_1^+]]=0$. $\square$