$N\,\text {=}\,2$ Supersymmetric Structures on Classical W-algebras

Abstract

We describe a $N=2$ supersymmetric Poisson vertex algebra structure of $N=1$ (resp. $N=0$) classical W-algebra associated with ${{\mathfrak {s}}}{{\mathfrak {l}}}(n+1|n)$ and the odd (resp. even) principal nilpotent element. This $N=2$ supersymmetric structure is connected to the principal ${{\mathfrak {s}}}{{\mathfrak {l}}}(2|1)$-embedding in ${{\mathfrak {s}}}{{\mathfrak {l}}}(n+1|n)$ superalgebras, which are the only basic Lie superalgebras that admit such a principal embedding.

Appendix A: Proofs of Propositions and Theorems

We consider the decomposition ${\mathfrak {g}}= \bigoplus _{i=1}^{2n} R_i$ of ${\mathfrak {g}}$ into irreducible $\mathfrak {osp}(1|2)$-module $R_i$ of dimension $2i+1$ together with the dual bases ${\mathcal {B}}= \{v_i^{(m)}|(i,m)\in {{\mathcal {J}}} \}$ and ${\mathcal {B}}^\perp = \{{\tilde{v}}^i_{(m)}|(i,m)\in {{\mathcal {J}}}\}$ of ${\mathfrak {g}}$ introduced in section 4.2, where ${{\mathcal {J}}}= \{(i,m)|i=1,2,\ldots , 2n, \, m=0,1,\ldots , 2i\}$. Suppose $E=v_2^{(0)}\in {\mathcal {B}}$ and $E,F,H,f, e, {\tilde{f}}, {\tilde{e}}, U \in R_1 \oplus R_2 \subset {\mathfrak {g}}$ satisfy (sl-1)–(sl-4) in section 2.2. We can assume

$$\begin{aligned}{}[{\tilde{f}}, v_{2i}^{(0)}]=-v_{2i-1}^{(0)}\end{aligned}$$

for any $i=1,2,\ldots , n$. Then, from $[{\tilde{f}}, f]=0$ and $(\text {ad}{\tilde{f}})^2=-(\text {ad}f)^2$, we get

$$\begin{aligned}{}[{\tilde{f}}, v_{2i-1}^{(t)}]= (-1)^t\,v_{2i}^{(t+2)}, \quad [{\tilde{f}}, v_{2i}^{(s)}]=(-1)^{s+1}\,v_{2i-1}^{(s)}, \end{aligned}$$

(A.1)

for $t=0,1,\ldots , 2i-2$ and $s= 2,3,\ldots , 2i$. In addition, since ${\tilde{v}}^\alpha _{(2\alpha )}$ is of parity $\alpha$ (mod 2) and belongs to the one dimensional space ${\mathfrak {g}}^e \cap {\mathfrak {g}}(\alpha )= {\mathbb {C}}\cdot \left( \sum _{\beta =1}^{2n-\alpha +1}e_{\beta , \alpha +\beta }\right)$, we can show $[{\tilde{v}}^i_{(2i)}, {\tilde{v}}^j_{(2j)}]= 0$ if i or j is even. Finally, we remind that

$$\begin{aligned} v_i^{(m)}\in {\mathfrak {g}}\Big (\frac{i-m}{2} \Big )\ \text { and }\ {\tilde{v}}^i_{(m)}\in {\mathfrak {g}}\Big (\frac{m-i}{2} \Big ). \end{aligned}$$

(A.2)

1.1 A.1. Proof of Proposition 5.1

1.1.1 A.1.1. Proof of (1) in Proposition 5.1

For such a purpose, let us compute $[\omega (\bar{{\tilde{f}}}){}_\Lambda \omega (\bar{{\tilde{f}}})]$.

Lemma A1

For $a=b= {\tilde{f}}\in {\mathfrak {g}}^{f}_{-1/2}$, the only nontrivial terms in (4.12) are $-\omega (\overline{[a,b]})$ and

$$\begin{aligned} - \big (k(b|v_{1}^{(0)})(D+\chi ) \big )\big (k({\tilde{v}}^{1}_{(1)}|v_{1}^{(1)})(D+\chi ) \big )\big (k({\tilde{v}}^{1}_{(2)}|a)(D+\chi ) \big )=2k^3\lambda \chi . \end{aligned}$$

(A.3)

Proof

Let us show the only nonzero terms in $[\omega (\bar{a}){}_\Lambda \omega (\bar{b})]+\omega (\overline{[a,b]})$ are given by (A.3. By relation (4.12), we have $v_{i_0}^{(m_0)}\in {\mathfrak {g}}(j)$ for $j=-1, -\frac{1}{2}, 0, \frac{1}{2}.$

• (i) Let $v_{i_0}^{(m_0)}\in {\mathfrak {g}}(-1)$. Then $(b|v_{i_0}^{(m_0)})=0$. Since $[b,v_{i_0}^{(m_0)}]\in {\mathfrak {g}}(-\frac{3}{2})$, then $[b,v_{i_0}^{(m_0)}]^\sharp \ne 0$ implies $[{\tilde{f}},v_{i_0}^{(m_0)}] \in {\mathbb {C}}v_3^0$ (see Lemma 2.3). This in turn shows that ${\tilde{v}}^{i_t}_{(m_t+1)} \in \bigoplus _{j\ge \frac{3}{2}}{\mathfrak {g}}(j)$ for $t\in {\mathbb {Z}}_{+}$, so that $({\tilde{v}}^{i_t}_{(m_t+1)}| a)=0$. From ${\mathfrak {g}}^{f}\cap {\mathfrak {n}}=\emptyset$, we also have $[{\tilde{v}}^{i_t}_{(m_t+1)}, a]^\sharp =0$. Hence $v_{i_0}^{(m_0)}\in {\mathfrak {g}}(-1)$ cannot give rise to nonzero terms.

• (ii) Let $v_{i_0}^{(m_0)}\in {\mathfrak {g}}(-\frac{1}{2})$. Then $(b|v_{i_0}^{(m_0)})=0$ and $[b,v_{i_0}^{(m_0)}]^\sharp \ne 0$ implies $v_{i_0}^{(m_0)} \in {\mathbb {C}}{\tilde{f}}$. However, if $v_{i_0}^{(m_0)} \in {\mathbb {C}}{\tilde{f}}$ then ${\tilde{v}}^{i_0}_{(m_0+1)}=0$. Hence we cannot get any nonzero terms.

• (iii) Let $v_{i_0}^{(m_0)}\in {\mathfrak {g}}(0)$. Then $(b|v_{i_0}^{(m_0)})=0$ and $[b,v_{i_0}^{(m_0)}]^\sharp \ne 0$ implies $v_{i_0}^{(m_0)} \in {\mathbb {C}}H$ and $v^{i_0}_{(m_0+1)} \in {\mathbb {C}}e$. In this case, $[{\tilde{v}}^{i_0}_{(m_0+1)}, g]^\sharp =0$ for any $g\in {\mathfrak {g}}$, since $[e,{\mathfrak {g}}] \oplus {\mathfrak {g}}^f=0$.

• (vi) Let $v_{i_0}^{(m_0)}\in {\mathfrak {g}}(\frac{1}{2})$. The only nonzero term arises only when $v_{i_0}^{(m_0)}\in {\mathbb {C}}{\tilde{e}}$ and $v_{i_1}^{(m_1)}\in {\mathbb {C}}U$. In this case we get relation (A.3.

$\square$

By Lemma A1, we have $\{\omega (\bar{{\tilde{f}}}{})_\Lambda \omega (\bar{{\tilde{f}}})\}=-2\omega (\bar{F})+2k^3 \lambda \chi$. Hence we proved Proposition 5.1 (1).

1.1.2 A.1.2 Additionnal lemmas

To prove Proposition 5.1 (2), let us introduce some more lemmas. We remind that ${{\mathcal {I}}}= \{ 1,2,\ldots , 2n+1\}$ and ${{\mathcal {J}}}= \{ (i,m)| i\in {{\mathcal {I}}}, \ m=0,1,\ldots , d_i\}$.

Lemma A2

Let $i, j \in {{\mathcal {I}}}$ and $(j,t)\in {{\mathcal {J}}}$. Then

• (1) $[{\tilde{v}}^i_{(2i)}, v_j^{(t)}]^\sharp =0$ for $t \le 2j-1$,

• (2) $[{\tilde{v}}^i_{(2i-1)}, v_j^{(t)}]^\sharp =0$ for $t \le 2j-2$.

Proof

For (1), observe that $v_j^{(t)}$ for $t \le 2j-1$ is in $[e,{\mathfrak {g}}]$. Let $u_j^{(t+1)} \in {\mathfrak {g}}$ satisfy $[e,u_j^{(t+1)}]= v_j^{(t)}$. Since ${\tilde{v}}_{(2i)}^i \in {\mathfrak {g}}^e$, we have $[e,[u_j^{(t+1)}, {\tilde{v}}_{(2i)}^i]]= [v_j^{(t)},{\tilde{v}}_{(2i)}^i ] \in [e,{\mathfrak {g}}].$ Hence, from $[e,{\mathfrak {g}}]\cap {\mathfrak {g}}^f=\emptyset$, we proved (1).

For (2), let $t \le 2j-2$. Observe there are ${u'}_j^{(t+2)} \in {\mathfrak {g}}$ and ${\tilde{u}}_{(2i)}^i \in {\mathfrak {g}}^e$ such that $v_j^{(t)}=[E, {u'}_j^{(t+2)}]$ and ${\tilde{v}}^i_{(2i-1)}=[f, {\tilde{u}}_{(2i)}^i]$. Now we have $[{\tilde{v}}^i_{(2i-1)}, v_j^{(t)}]^\sharp = [E,[[f,{\tilde{u}}_{(2i)}^i],$${u'}_j^{(t+2)}]]^\sharp + [[[f,{\tilde{u}}_{(2i)}^i], E],{u'}_j^{(t+2)}]^\sharp .$ Since $[[f,{\tilde{u}}_{(2i)}^i],E]=0$ and $[E,[[f,{\tilde{u}}_{(2i)}^i],{u'}_j^{(t+2)}]]^\sharp =0$, we conclude $[{\tilde{v}}^i_{(2i-1)},$ $v_j^{(t)}]^\sharp =0$. $\square$

Now, we can list nonzero terms in relation (4.12) when $b={\tilde{f}}$ or F. To lighten the notations, let us denote

$$\begin{aligned} {\widetilde{\omega }}(\overline{[x,y]}):= \omega (\overline{[x,y]}^\sharp )-k(x|y)(\chi +D). \end{aligned}$$

(A.4)

Lemma A3

Let $b=F$. For $a= {\tilde{f}}$, the bracket $[\omega (b){}_{\Lambda }\omega (a)]$ can be computed similarly to what has been done for (A.17). If $a=F$, the only nonzero terms in (4.12) can be listed as follows:

$$\begin{aligned} \begin{aligned}&{\widetilde{\omega }}(\overline{[b,v_1^{(0)}]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^1_{(1)},v^{(1)}_1]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^1_{(2)},a]})= \frac{1}{2}k \, \omega (\bar{{\tilde{f}}}) (\chi +D)\omega (\bar{{\tilde{f}}}),\\&{\widetilde{\omega }}(\overline{[b,v_1^{(0)}]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^1_{(1)},v^{(3)}_2]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^2_{(4)},a]})= -\frac{1}{4} k \, \omega (\bar{{\tilde{f}}})^2 \chi ,\\&{\widetilde{\omega }}(\overline{[b,v_2^{(0)}]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^2_{(1)},v^{(1)}_1]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^1_{(2)},a]})=-\frac{1}{4}k(\chi +D)\omega (\bar{{\tilde{f}}})^2,\\&{\widetilde{\omega }}(\overline{[b,v_2^{(0)}]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^2_{(1)},v^{(1)}_2]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^2_{(2)},v^{(2)}_2]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^2_{(3)},v^{(3)}_2]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^2_{(4)},a]})= \frac{1}{2}k^5\, (\lambda +\partial )^2(\chi +D),\\&{\widetilde{\omega }}(\overline{[b,v_2^{(0)}]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^2_{(1)},v^{(1)}_2]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^2_{(2)},a]})= k^2(\lambda +\partial )\omega (\bar{F}),\\&{\widetilde{\omega }}(\overline{[b,v^{(2)}_2]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^2_{(3)},v^{(3)}_2]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^2_{(4)},a]})= k^2 \omega (\bar{F})\lambda ,\\&{\widetilde{\omega }}(\overline{[b,v_2^{(0)}]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^2_{(1)},v^{(3)}_2]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^2_{(4)},a]})=\frac{1}{2}k^2 \, (\chi +D)\omega (\bar{F})\chi .\\ \end{aligned}\nonumber \\ \end{aligned}$$

(A.5)

Proof

Use Lemma A2 and track all the nonzero terms as in the proof of Lemma A1. Essentially, the computations done for relation (4.16) work. $\square$

By Eq. (A.17), and adding the contributions in relation (A.5), we get

$$\begin{aligned} \begin{aligned}&\{ G{}_\Lambda \omega ({\tilde{f}}) \}= (2\partial + \chi D + 2\lambda ) \omega ({\tilde{f}})\\&\{ G{}_\Lambda G\} = (2\partial + \chi D + 3\lambda ) G -2k \lambda ^2\chi \end{aligned} \end{aligned}$$

(A.6)

for $G:= -\frac{2}{k^2}\omega (\bar{F})$. Hence $\omega ({\tilde{f}})$ is G-primary of conformal weight 1.

Lemma A4

Let $i\ge 3$ be an integer. Then

• (1) $[ {\tilde{e}}, v_{2j}^{(4j)}]=2j \, v^{(4j-2)}_{2j-1},$

• (2) $[ {\tilde{e}}, v_{2j-1}^{(4j-2)}]=-(2j-1) v^{(4j-2)}_{2j}.$

Proof

We know $[ {\tilde{e}}, v_{2j}^{(4j)}]= c\, v_{2j-1}^{(4j-2)}$ for a constant $c\in {\mathbb {C}}$. Since $[{\tilde{f}},[{\tilde{e}}, v_{2j}^{(4j)}]]=-[2H, v_{2j}^{(4j)}]= 2j \, v_{2j}^{(4j)}$ and $[{\tilde{f}},[{\tilde{e}}, v_{2j}^{(4j)}]]= c\,[{\tilde{f}}, v_{2j-1}^{(4j-2)}]= c\, v_{2j}^{(4j)}$, Hence $c=2j$ and (1) follows.

The proof for (2) is analogous to the one for (1) once one uses equation (A.1). $\square$

Lemma A5

Suppose $(j,m)\in {\mathcal {J}}$. Then the following properties hold.

• (1) If ${\widetilde{w}}(\overline{[F,v_j^{(m)}]})\ne 0$, then $(j,m)= (2,0)$ or $(j, 2j-2)$.

• (2) If ${\widetilde{w}}(\overline{[{\tilde{v}}^2_{(1)},v_j^{(m)}]})\ne 0$, then $(j,m)=(j,2j-1)$ or $(j,m)=(2,1)$.

• (3) If ${\widetilde{w}}(\overline{[{\tilde{v}}^i_{(2i-1)},v_j^{(m)}]})\ne 0$ for $i\in {\mathcal {I}}$, then

  $$\begin{aligned}(j,m)\in \{ (i',2i'-1), (i'',2i'') \in {\mathcal {J}} | i'\ge i, i''>i\}. \end{aligned}$$

• (4) If ${\widetilde{w}}(\overline{[{\tilde{v}}^i_{(2i)},v_j^{(m)}]})\ne 0$ for $i\in {\mathcal {I}}$, Then

  $$\begin{aligned} (j,m)\in \{ (i',2i') \in {\mathcal {J}} | i'\ge i\}. \end{aligned}$$

• (5) Let $(i_0, m_0), (i_1, m_2), \ldots (i_p,m_p) \in {\mathcal {J}}$ and $a,b\in {\mathfrak {g}}^f$. If

  $$\begin{aligned} {\widetilde{w}}(\overline{[b,v_{i_0}^{(m_0)}]}) {\widetilde{w}}(\overline{[{\tilde{v}}^{i_0}_{(m_0+1)},v_{i_1}^{(m_1)}]}) \ldots {\widetilde{w}}(\overline{[{\tilde{v}}^{i_p}_{(m_p+1)},a]})\ne 0, \end{aligned}$$

  then $m_t <2 i_t$ for $t=0,1,\ldots , p$.

Proof

(1), (2) directly follow from the $\mathfrak {osp}(1|2)$ representation theory and the fact that for $(i,m), (i',m')\in {\mathcal {J}}$ we have $({\tilde{v}}^i_{(m)}|v_{i'}^{(m')})\ne 0$ iff $i=i'$ and $m=m'$. (5) holds since if $m_t\ge 2i_t$ then ${\tilde{v}}^{i_t}_{(m_t+1)}=0$. For (3), Lemma A2 shows ${\widetilde{w}}(\overline{[{\tilde{v}}^i_{(2i-1)},v_j^{(m)}]})\ne 0$ only when $m=2j-1$ or 2j. Now, since ${\mathfrak {g}}^f \subset \bigoplus _{t<0} {\mathfrak {g}}(t)$, we need $j\ge i$. In addition, we have ${\widetilde{w}}(\overline{[{\tilde{v}}^i_{(2i-1)},v_i^{(2i)}]})=0$ since $[{\tilde{v}}^i_{(2i-1)},v_i^{(2i)}]^\sharp = (v_1^{(0)}|[{\tilde{v}}^i_{(2i-1)},v_i^{(2i)}]){\tilde{v}}^1_{(0)}$ and $(v_1^{(0)}|[{\tilde{v}}^i_{(2i-1)},v_i^{(2i)}])=(-1)^{i+1}({\tilde{v}}^i_{(2i-1)}|[{\tilde{e}},v_i^{(2i)}])=0$ by Lemma A4. (4) is proved similarly. $\square$

Lemma A6

Suppose $n,i\ge 2$ are integers and $i\le n$. Take $a=v^{(4i-2)}_{2i-1}$ and $b=F=-\frac{1}{2}\,v_2^{(4)}$ in ${\mathfrak {g}}={{\mathfrak {s}}}{{\mathfrak {l}}}(n+1|n)$. We restrict ourselves to the terms in (4.12) such that

$$\begin{aligned} i_0, i_1, \ldots , i_p\in \{1,2,2i-1, 2i\}. \end{aligned}$$

(A.7)

Then, the only possibly nonzero terms in (4.12) satisfying (A.7) are listed below:

$$\begin{aligned} \begin{aligned}&{\widetilde{\omega }}(\overline{[b,v^{(0)}_2]})\,{\widetilde{\omega }}(\overline{[{\tilde{v}}^2_{(1)},v^{(1)}_2]})\,{\widetilde{\omega }}(\overline{[{\tilde{v}}^2_{(2)},a]})= \frac{2i-1}{2} k^2 \,(\lambda +\partial )\, \omega (\bar{v}_{2i-1}^{(4i-2)}),\\&{\widetilde{\omega }}(\overline{[b,v^{(0)}_2]})\,{\widetilde{\omega }}(\overline{[{\tilde{v}}^2_{(1)},v^{(4i-3)}_{2i-1}]})\,{\widetilde{\omega }}(\overline{[{\tilde{v}}^{2i-1}_{(4i-2)},a]})=\frac{k^2}{2}(-\lambda +\chi D)\, \omega (\bar{v}_{2i-1}^{(4i-2)}),\\&{\widetilde{\omega }}(\overline{[b,v^{(4i-4)}_{2i-1}]})\,{\widetilde{\omega }}(\overline{[{\tilde{v}}^{2i-1}_{(4i-3)},v^{(4i-3)}_{2i-1}]})\,{\widetilde{\omega }}(\overline{[{\tilde{v}}^{2i-1}_{(4i-2)},a]})=k^2\, \omega (\bar{v}_{2i-1}^{(4i-2)}) \lambda . \end{aligned} \end{aligned}$$

(A.8)

Proof

Using Lemma A5, one shows that the only non-zero terms in (4.12) are given by (A.8). For instance, one can show that there is no nonzero term starting with ${\widetilde{\omega }}(\overline{[b,v^{(2)}_2]})$ in the following way. Since $[{\tilde{v}}^2_{(3)}, {\mathfrak {g}}]\oplus {\mathfrak {g}}^f={\mathfrak {g}}$, we have ${\widetilde{\omega }}(\overline{[{\tilde{v}}^2_{(3)},v_i^{(m)}]}) \ne 0$ iff $(i,m)=(2,3).$ Now, take $(i,m)=(2,3)$ and consider $(i',m')$ such that ${\widetilde{\omega }}(\overline{[{\tilde{v}}^i_{(m+1)},v_{i'}^{(m')}]})\ne 0$. Then $(i',m')$ should be (2, 4). Finally, Lemma A5 (5) tells that there is no nonzero term starting with ${\widetilde{\omega }}(\overline{[b,v^{(2)}_2]})$. Furthermore, the equalities in (A.8) are obtained by direct computations. $\square$

Lemma A7

Suppose $n,i\ge 2$ are integers and $i\le n$. Take $a=v^{(4i)}_{2i}$ and $b=F=-\frac{1}{2}\,v_2^{(4)}$ in ${\mathfrak {g}}={{\mathfrak {s}}}{{\mathfrak {l}}}(n+1|n)$. The only possibly nonzero terms in (4.12) satisfying (A.7) are listed below:

$$\begin{aligned} \begin{aligned}&{\widetilde{\omega }}(\overline{[b,v_1^{(0)}]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^1_{(1)},v^{(1)}_1]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^1_{(2)},a]})=-ik\, \omega (\bar{v}_1^{(2)})(\chi +D) \omega (\bar{v}_{2i-1}^{(4i-2)}),\\&{\widetilde{\omega }}(\overline{[b,v_1^{(0)}]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^1_{(1)},v^{(4i-1)}_{2i}]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^{2i}_{(4i)},a]})= \frac{1}{2}k\,\omega (\bar{v}_1^{(2)})\omega (\bar{v}_{2i-1}^{(4i-2)}) \chi ,\\&{\widetilde{\omega }}(\overline{[b,v_2^{(0)}]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^2_{(1)},v^{(1)}_1]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^1_{(2)},a]})=\frac{i}{2}k\, (\chi +D)\omega (\bar{v}_1^{(2)})\omega (\bar{v}_{2i-1}^{(4i-2)}),\\&{\widetilde{\omega }}(\overline{[b,v_2^{(0)}]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^2_{(1)},v^{(4i-3)}_{2i-1}]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^{2i-1}_{(4i-2)},a]})= \frac{i}{2} k\, (\chi +D)\omega (\bar{v}_{2i-1}^{(4i-2)}) \omega (\bar{v}_1^{(2)}),\\&{\widetilde{\omega }}(\overline{[b,v^{(4i{-}4)}_{2i{-}1}]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^{2i-1}_{(4i-3)},v^{(4i-3)}_{2i-1}]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^{2i-1}_{(4i-2)},a]}){=}{-}ik\, \omega (\bar{v}_{2i-1}^{(4i-2)}) (\chi {+}D) \omega (\bar{v}_1^{(2)}),\\&{\widetilde{\omega }}(\overline{[b,v^{(4i-4)}_{2i-1}]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^{2i-1}_{(4i-3)},v^{(4i-1)}_{2i}]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^{2i}_{(4i)},a]})=\frac{2i-1}{2}k \, \omega (\bar{v}_{2i-1}^{(4i-2)}) \omega (\bar{v}_1^{(2)})\chi ,\\&{\widetilde{\omega }}(\overline{[b,v_2^{(0)}]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^2_{(1)},v^{(1)}_2]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^2_{(2)},a]})=i k^2 \, (\lambda +\partial ) \omega (\bar{v}^{(4i)}_{2i}),\\&{\widetilde{\omega }}(\overline{[b,v_2^{(0)}]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^2_{(1)},v^{(4i-1)}_{2i}]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^{2i}_{(4i)},a]})= -\frac{k^2}{2}(\lambda -\chi D) \omega (\bar{v}_{2i}^{(4i)}),\\&{\widetilde{\omega }}(\overline{[b,v^{(4i-2)}_{2i}]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^{2i}_{(4i-1)},v^{(4i-1)}_{2i}]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^{2i}_{(4i)},a]})=k^2 \, \omega (\bar{v}^{(4i)}_{2i})\lambda .\\ \end{aligned}\nonumber \\ \end{aligned}$$

(A.9)

Proof

The proof of the lemma is similar to the one of Lemma A6. We use Lemma A5 and check the equalities by computations. $\square$

Corollary A8

We have

$$\begin{aligned} \begin{aligned}&[\omega (\bar{v}^{(6)}_{3}){}_\Lambda \omega (\bar{F})]= k^2 \bigg ( 2\lambda + \frac{1}{2}\chi D + \frac{3}{2}\partial \bigg )\omega (\bar{v}^{(6)}_{3}),\\&[\omega (\bar{v}^{(8)}_{4}){}_\Lambda \omega (\bar{F})]= -k^2 \bigg ( \frac{5}{2}\lambda + \frac{1}{2}\chi D + 2\partial \bigg )\omega (\bar{v}^{(8)}_{4}). \end{aligned} \end{aligned}$$

(A.10)

Hence for $n=2$, $G= -\frac{2}{k^2}\omega (\bar{F})$ is a superconformal vector of ${\mathcal {W}}^k({\overline{{{\mathfrak {s}}}{{\mathfrak {l}}}}}(3|2), f)$. Furthermore, when $n\ge 2$, the two elements $\omega (\bar{v}^{(6)}_{3})$ and $\omega (\bar{v}^{(8)}_{4})$ are G-primary of conformal weight 2 and $\frac{5}{2}$ respectively.

Proof

For the case ${\mathfrak {g}}= {{\mathfrak {s}}}{{\mathfrak {l}}}(3|2)$, $i=2$ is the only possibility and (A.7) does not bring any further constrain. Hence it suffices to consider Lemmas A6 and A7. $\square$

In order to compute $[\omega (\bar{v}^{(2i)}_{i}){}_\Lambda \omega (\bar{F})]$ for $i\ge 5,$ we have to find all nontrivial terms in relation (4.12) when $b=F$ and $a= \bar{v}^{(2i)}_{i}$ (see Lemmas A12 and A13). We first use the following lemma and corollaries (see Lemma A9, Corollaries A10 and A11).

Lemma A9

Let $j\in {\mathbb {Z}}_{+}$ be an integer such that $j\le n$. We have the following properties:

• (1) $v_j^{(2j-1)} = \frac{1}{j}[e,v_j^{(2j)}]$,

• (2) ${\tilde{v}}_{(2j-1)}^j = (-1)^j[f,{\tilde{v}}^j_{(2j)}]$,

• (3) $[{\tilde{v}}_{(2j-1)}^{j}, v_{k}^{(2k-1)}]^\sharp = \frac{-j}{k}[{\tilde{v}}^{j}_{(2j)},v_{k}^{(2k)}]^\sharp$.

Proof

(1) Let $x_j$ be the constant such that $v_j^{(2j-1)}=x_j\,[e,v_j^{(2j)}]$. Then $v_j^{(2j)}=[f,v_j^{(2j-1)}]= x_j\, [f,[e,v_j^{(2j)}]]=x_j\,[-2H, v_j^{(2j)}]= x_j\, j \, v_j^{(2j)}$. Hence $x_j= \frac{1}{j}.$

(2) Let $y_j$ be the constant such that ${\tilde{v}}_{(2j-1)}^j = y_j\, [f,{\tilde{v}}^j_{(2j)}]$. By (1), we have $(v_j^{(2j-1)}|{\tilde{v}}^j_{(2j-1)})$$= \frac{y_j}{j}\big ( [e,v_j^{(2j)}]| [f,{\tilde{v}}^j_{(2j)}] \big )=1.$ Since

$$\begin{aligned} \frac{y_j}{j}\big ( [e,v_j^{(2j)}]| [f,{\tilde{v}}^j_{(2j)}] \big )= (-1)^j\frac{y_j}{j}([e,f]|[v_j^{(2j)}, {\tilde{v}}^j_{(2j)}])=(-1)^j y_j (v_j^{(2j)}|{\tilde{v}}^j_{(2j)})=1, \end{aligned}$$

we have $y_j=(-1)^j$.

(3) Using the above relations (1), (2) and the Jacobi identity, we have

$$\begin{aligned} \begin{aligned}&[{\tilde{v}}^{j}_{(2j-1)}, v_k^{(2k-1)}]^\sharp \\&\quad = (-1)^j \frac{1}{k}[[f,{\tilde{v}}^j_{(2j)}],[e,v_k^{(2k)}]]^\sharp = (-1)^j \frac{1}{k}[[[f,{\tilde{v}}^j_{(2j)}], e], v_k^{(2k)}]^\sharp \\&\quad =(-1)^j \frac{1}{k} [[(-1)^{j+1}2H,{\tilde{v}}^j_{(2j)}], v_k^{(2k)}]^\sharp =-\frac{j}{k}[{\tilde{v}}^{j}_{(2j)}, v_k^{(2k)}]^\sharp . \end{aligned} \end{aligned}$$

(A.11)

Here, the second equality in (A.11) holds since $[e,{\mathfrak {g}}]$ and ${\mathfrak {g}}^f$ intersect trivially. $\square$

Corollary A10

Let $i,j\in {\mathcal {I}}$ and $j>i$. Then

$$\begin{aligned}^\sharp =-\frac{i}{j}[{\tilde{v}}_{(2i)}^{i}, v_{j}^{(2j)}]^\sharp =0\end{aligned}$$

when i and j have the same parity or j is odd.

Proof

We have $[{\tilde{v}}_{(2i)}^{i}, v_{j}^{(2j)}]^\sharp = (-1)^{j-i} ({\tilde{v}}^{j-i}_{(2j-2i)}|[{\tilde{v}}_{(2i)}^{i}, v_{j}^{(2j)}])v_{j-i}^{(2j-2i)}$. Since $[{\tilde{v}}^{j-i}_{(2j-2i)}, {\tilde{v}}_{(2i)}^{i}]=0$ if i or $j-i$ is even, the second equality follows. The first equality follows from Lemma A9. $\square$

Combining the result of Lemma A5 and Corollary A10, we can narrow down the number of terms in (4.12).

Corollary A11

Let $b= F$ and $a= v_{i}^{(2i)}\in {\mathfrak {g}}$ for $i\ge 5$. The longest nonzero terms in (4.12) have one of the following forms:

$$\begin{aligned}&{\widetilde{w}}(\overline{[b,v_j^{(2j-2)}]}) {\widetilde{w}}(\overline{[{\tilde{v}}^j_{(2j-1)},v_l^{(2l-1)}]}){\widetilde{w}}(\overline{[{\tilde{v}}^l_{(2l)},a]}) , \end{aligned}$$

(A.12)

$$\begin{aligned}&{\widetilde{w}}(\overline{[b,v_2^{(0)}]}) {\widetilde{w}}(\overline{[{\tilde{v}}_{(1)}^2,v_l^{(2l-1)}]}){\widetilde{w}}(\overline{[{\tilde{v}}^l_{(2l)},a]}). \end{aligned}$$

(A.13)

Moreover, (A.12$\ne 0$ only if $i,j,l \in {\mathcal {I}})$ obey one of the three following constraints:

• $i=l$ is even and j is odd such that $j\le l$,

• i is even and $j=l$ is odd such that $j=l<i$,

• $j=l=i$,

and $(A.13) = -\frac{1}{2} k(\chi +D)\omega (\bar{v}_l^{(2\,l)}) {\widetilde{\omega }}(\overline{[{\tilde{v}}_{(2\,l)}^l, v_i^{(2i)}]})\ne 0$ only if

• $i=l$,

• i is even and l is odd such that $l<i$.

Lemma A12

Suppose $n\ge 3$ and $2<i\le n$. For $b=F$ and $a=v^{(4i-2)}_{2i-1}$, the followings are all nontrivial terms in (4.12):

• (1) terms in relation (A.8);

• (2) for j such that $1<j< i$,

  $$\begin{aligned}&{\widetilde{\omega }}(\overline{[b,v^{(4j-2)}_{2j}]})\,{\widetilde{\omega }}(\overline{[{\tilde{v}}^{2j}_{(4j-1)},a]}) \nonumber \\ {}&\quad =-w(\bar{v}_{2j}^{(4j)}) \omega (\overline{[{\tilde{v}}_{(4j-1)}^{2j}, a]}^\sharp ) =\omega (\bar{v}_{2i-2j}^{(4i-4j)})\, \omega (\overline{[{\tilde{v}}_{(4i-4j-1)}^{2i-2j}, a]}^\sharp ); \end{aligned}$$

  (A.14)
  $$\begin{aligned}&{\widetilde{\omega }}(\overline{[b,v^{(4j-4)}_{2j-1}]})\,{\widetilde{\omega }}(\overline{[{\tilde{v}}^{2j-1}_{(4j-3)},a]}) \nonumber \\ {}&\quad =-\omega (\bar{v}_{2j-1}^{(4j-2)})\, \omega (\overline{[{\tilde{v}}^{2j-1}_{(4j-3)}, a]}^\sharp ) =\omega (\bar{v}_{2i-2j+1}^{(4i-4j+2)}) \,\omega (\overline{[{\tilde{v}}^{2i-2j+1}_{(4i-4j+1)}, a]}^\sharp ). \end{aligned}$$

  (A.15)

Proof

By Lemma A5 and Corollary A11, we can find all possible nonzero terms. For equalities (A.14), we use Lemma A9 (2). Since

$$\begin{aligned}{}[{\tilde{v}}_{(4j-1)}^{2j}, a]^\sharp = ([[f,{\tilde{v}}^{2j}_{(4j)}], a]|{\tilde{v}}^{2i-2j}_{(4i-4j)})\,v_{2i-2j}^{(4i-4j)} =({\tilde{v}}^{2j}_{(4j)}| [{\tilde{v}}^{2i-2j}_{(4i-4j-1)}, a])\,v_{2i-2j}^{(4i-4j)}, \end{aligned}$$

we have

$$\begin{aligned} \begin{aligned} \omega (\bar{v}_{2j}^{(4j)})\, \omega (\overline{[{\tilde{v}}_{(4j-1)}^{2j}, a]}^\sharp )&= ({\tilde{v}}^{2j}_{(4j)}| [{\tilde{v}}^{2i-2j}_{(4i-4j-1)}, a])\,\omega (\bar{v}_{2j}^{(4j)})\, \omega (\bar{v}_{2i-2j}^{(4i-4j)})\\&= - \omega (\bar{v}_{2i-2j}^{(4i-4j)})\, \omega (\overline{[{\tilde{v}}_{(4i-4j-1)}^{2i-2j}, a]}^\sharp ). \end{aligned} \end{aligned}$$

The same proof works for equality (A.15). Hence the lemma follows. $\square$

Lemma A13

Suppose $n\ge 4$ and $2<i\le n$. For $b=F$ and $a=v^{(4i)}_{2i}$, the followings are all nontrivial terms in (4.12):

• (1) terms in relation (A.9);

• (2) for j such that $1<j<i$, we have the five following possibilities:

  $$\begin{aligned}{} & {} {\widetilde{\omega }}(\overline{[b,v^{(4j-4)}_{2j-1}]})\,{\widetilde{\omega }}(\overline{[{\tilde{v}}^{2j-1}_{(4j-3)},v^{(4i-1)}_{2i}]})\,{\widetilde{\omega }}(\overline{[{\tilde{v}}^{2i}_{(4i)},a]})\nonumber \\{} & {} \qquad \qquad \qquad \qquad = - \frac{2j-1}{2i}k\, \omega (\bar{v}_{2j-1}^{(4j-2)})\,\omega (\overline{[{\tilde{v}}^{2j-1}_{(4j-2)}, v_{2i}^{(4i)}]}^{\sharp })\chi ,\nonumber \\{} & {} {\widetilde{\omega }}(\overline{[b,v^{(4j-4)}_{2j-1}]})\,{\widetilde{\omega }}(\overline{[{\tilde{v}}^{2j-1}_{(4j-3)},v^{(4j-3)}_{2j-1}]})\,{\widetilde{\omega }}(\overline{[{\tilde{v}}^{2j-1}_{(4j-2)},a]})\nonumber \\{} & {} \quad \qquad \qquad \qquad =k\, \omega (\bar{v}_{2j-1}^{(4j-2)})(\chi +D)\omega (\overline{[{\tilde{v}}^{2j-1}_{(4j-2)},v_{2i}^{(4i)}]}^{\sharp }),\nonumber \\{} & {} {\widetilde{\omega }}(\overline{[b,v^{(1)}_{2}]})\,{\widetilde{\omega }}(\overline{[{\tilde{v}}^{2}_{(1)},v^{(4j-3)}_{2j-1}]})\,{\widetilde{\omega }}(\overline{[{\tilde{v}}^{2j-1}_{(4j-2)},a]})\nonumber \\ {}{} & {} \qquad \qquad \qquad \quad =-\frac{1}{2}k\, (\chi +D)\omega (\bar{v}_{2j-1}^{(4j-2)})\omega (\overline{[{\tilde{v}}^{2j-1}_{(4j-2)},v_{2i}^{(4i)}]}^{\sharp }); \end{aligned}$$

  (A.16)
  $$\begin{aligned}{} & {} {\widetilde{\omega }}(\overline{[b,v^{(4j-4)}_{2j-1}]})\,{\widetilde{\omega }}(\overline{[{\tilde{v}}^{2j-1}_{(4j-3)},a]}) =-\omega (\bar{v}^{(4j-2)}_{2j-1})\,\omega (\overline{[{\tilde{v}}^{2j-1}_{(4j-3)},a]}^\sharp )\nonumber \\{} & {} \qquad \qquad \qquad =\omega (\bar{v}^{(4i-4j+4)}_{2i-2j+2})\,\omega (\overline{[{\tilde{v}}^{2i-2j+2}_{(4i-4j+3)},a]}^\sharp ), \end{aligned}$$

  (A.17)
  $$\begin{aligned}{} & {} {\widetilde{\omega }}(\overline{[b,v^{(4j-2)}_{2j}]})\,{\widetilde{\omega }}(\overline{[{\tilde{v}}^{2j}_{(4j-1)},a]}) =-\omega (\bar{v}^{(4j)}_{2j})\,\omega (\overline{[{\tilde{v}}^{2j}_{(4j-1)},a]}^\sharp )\nonumber \\{} & {} \qquad \qquad \qquad = \omega (\bar{v}^{(4i-4j+2)}_{2i-2j+1})\,\omega (\overline{[{\tilde{v}}^{2i-2j+1}_{(4i-4j+1)},a]}^\sharp ). \end{aligned}$$

  (A.18)

Proof

Again, by Lemma A5 and Corollary A11, we can find all possible nonzero terms. Equalities (A.17) and (A.18) can be proved similarly to the proof of (A.14). The equalities in (A.16) are obtained by direct computations using Lemma A9 (3). $\square$

1.1.3 A.1.3 Proof of (2) in Proposition 5.1

When $n=2$, the proof is done in Corollary A.10. Let us assume $n\ge 3$. For the cases, Proposition 5.1 (2) is proved by the following lemmas.

Lemma A14

For any $i\in {\mathcal {I}}$, we have

$$\begin{aligned}{}[\omega (\bar{v}^{(4i-2)}_{2i-1}){}_\Lambda \omega (\bar{F})]= k^2 \bigg ( i\lambda + \frac{1}{2}\chi D + \frac{2i-1}{2}\partial \bigg )\omega (\bar{v}^{(4i-2)}_{2i-1}). \end{aligned}$$

Proof

Recall Lemmas A3, A6 and A12. Denote $b=F$ and $a=v_{2i-1}^{(4i-2)}$ as in Lemma A12. It is enough to show that $\sum _{j=2}^{i-1}(A.14)-1(2) + \sum _{j=2}^{i-1}(A.15)=0$ for all $i\ge 5$. Indeed, we have

$$\begin{aligned}{} & {} \ 2 \ \sum _{j=2}^{i-1}{\widetilde{\omega }}(\overline{[b,v^{(4j-2)}_{2j}]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^{2j}_{(4j-1)}, a]}) \nonumber \\{} & {} \quad = \sum _{j=2}^{i-1} \bigg ( -w(\bar{v}_{2j}^{(4j)})\, \omega (\overline{[{\tilde{v}}_{(4j-1)}^{2j}, a]}^\sharp )+ \omega (\bar{v}_{2i-2j}^{(4i-4j)})\, \omega (\overline{[{\tilde{v}}_{(4i-4j-1)}^{2i-2j}, a]}^\sharp ) \bigg )\nonumber \\{} & {} \quad = - w(\bar{v}_{2i-2}^{(4i-4)})\, \omega (\overline{[{\tilde{v}}_{(4i-5)}^{2i-2}, a]}^\sharp ) + \omega (\bar{v}_{2}^{(4)})\, \omega (\overline{[{\tilde{v}}_{(3)}^{2}, a]}^\sharp ) \end{aligned}$$

(A.19)

and $[{\tilde{v}}_{(4i-5)}^{2i-2}, a]^\sharp =[{\tilde{v}}_{(3)}^{2}, a]^\sharp =0$. Hence $\sum _{j=2}^{i-1}(A.14)-1(2)=0$. Similarly, we also have $\sum _{j=2}^{i-1}(A.15)=0$. $\square$

Lemma A15

For any $i\in {\mathcal {I}}$, we have

$$\begin{aligned}{}[\omega (\bar{v}^{(4i)}_{2i}){}_\Lambda \omega (\bar{F})]= -k^2 \bigg ( \frac{2i+1}{2}\lambda + \frac{1}{2}\chi D + i\partial \bigg )\omega (\bar{v}^{(4i)}_{2i}). \end{aligned}$$

Proof

We aim to show $\sum _{j=2}^{i-1}(A.16)=0$ and $\sum _{j=2}^{i-1}(A.17)+(A.18)=0$. The first assertion follows from the fact that

$$\begin{aligned} \omega (\bar{v}_{2j-1}^{(4j-2)})\,\omega (\overline{[{\tilde{v}}^{2j-1}_{(4j-2)},v_{2i}^{(4i)}]}^{\sharp })= \omega (\bar{v}_{2i-2j+1}^{(4i-4j+2)})\,\omega (\overline{[{\tilde{v}}^{2i-2j+1}_{(4i-4j+2)},v_{2i}^{(4i)}]}^{\sharp }), \end{aligned}$$

which can be deduced by a computation similar to the proof of (A.14). For the second assertion, we use equalities in (A.17) and (A.18). More precisely, observe that

$$\begin{aligned}{} & {} \sum _{j=2}^{i-1}{\widetilde{\omega }}(\overline{[b,v^{(4j-2)}_{2j}]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^{2j}_{(4j-1)},v_{2i}^{(4i)}]})+ \sum _{j=2}^{i-1}{\widetilde{\omega }}(\overline{[b,v^{(4j-4)}_{2j-1}]}){\widetilde{\omega }}(\overline{[{\tilde{v}}^{2j-1}_{(4j-3)},v_{2i}^{(4i)}]})\nonumber \\{} & {} \quad = -\sum _{l=3}^{2i-2}\omega (\bar{v}^{(2l)}_{l})\,\omega (\overline{[{\tilde{v}}^{l}_{(2l-1)},a]}^\sharp )= \sum _{l=3}^{2i-2}\omega (\bar{v}^{(2l)}_{l})\,\omega (\overline{[{\tilde{v}}^{l}_{(2l-1)},a]}^\sharp ). \end{aligned}$$

(A.20)

Here, the first equality in (A.20) corresponds to the first equality in the formulas (A.17) and (A.18). The second equality in (A.20) corresponds to the second equality of formulas (A.17) and (A.18). Hence $(A.20)=0$. $\square$

1.2 A.2 Proof of Theorem 5.2

We use Theorem 3.10 to prove Theorem 5.2.

Take ${\mathcal {C}}=\{v_i^{(2i)}, v_i^{(2i-1)}\,|\,i=1, \ldots 2n\}$ for $\{v_i^{(m)}\,|\,(i,m)\in {{\mathcal {J}}}\}$ as in section 6. Let us denote $J=\frac{\sqrt{-1}}{k}\omega (\bar{{\tilde{f}}})$ and recall that $G= -\frac{2}{k^2} \omega (\bar{F})$ is a $N=1$ superconformal vector. Then $\{ J {}_\Lambda J \}= -G+ 2k^2 \lambda \chi$ and hence $-J_{(0|0)}J=G$.

In order to prove that J is a $N=2$ superconformal vector and that all the elements of ${W}_{\mathfrak {osp}(1|2)}^{N=2}$ are J-primary, it is enough to show that $\{J {}_\Lambda \omega (\bar{v}^{(4i-2)}_{2i-1})\}= J_{(0|0)}\omega (\bar{v}^{(4i-2)}_{2i-1})$ for $i=2,3,\ldots , n$ and that ${\mathcal {W}}^k({\bar{{\mathfrak {g}}}},f)$ is freely generated by $\{\omega (\bar{v}^{(4i-2)}_{2i-1}), J_{(0|0)}\omega (\bar{v}^{(4i-2)}_{2i-1})|i=2,3,\ldots , n\}\cup \{J,G\}$ as a ${\mathbb {C}}[\nabla ]$-algebra.

Lemma A16

For $i=2, \ldots , n$, we have

$$\begin{aligned} \{ \omega (\bar{{\tilde{f}}}){}_\Lambda \omega (\bar{v}^{(4i-2)}_{2i-1})\}= -\omega (\bar{v}^{(4i)}_{2i})+ \sum _{j=2}^{i-1} c_{j,i} \omega (\bar{v}^{(4j)}_{2j}) \omega (\bar{v}^{(4i-4j)}_{2i-2j}), \end{aligned}$$

for some constants $c_{j,i} \in {\mathbb {C}}$.

Proof

Using relation (4.12) and Lemma A2, we have

$$\begin{aligned} \begin{aligned} \{ \omega (\bar{v}^{(4i-2)}_{2i-1}) {}_\Lambda \omega (\bar{{\tilde{f}}})\}&= -\omega (\bar{v}^{(4i)}_{2i})+\sum _{j=2}^{i-1} \omega (\overline{[{\tilde{f}}, v_{2j}^{(4j-2)}]})\omega (\overline{[{\tilde{v}}_{(4j-1)}^{2j}, v^{(4i-2)}_{2i-1}]^\sharp })\\&=-\omega (\bar{v}_{2i}^{(4i)})- \sum _{j=2}^{i-1}\omega (\bar{v}^{(4j)}_{2j}) \big ({\tilde{v}}^{2i-2j}_{(4i-4j)}| [{\tilde{v}}^{2j}_{(4i-1)}, v^{(4i-2)}_{2i-1}] \big )v^{(4i-4j)}_{2i-2j}. \end{aligned} \end{aligned}$$

(A.21)

We let $c_{j,i}=-\big ({\tilde{v}}^{2i-2j}_{(4i-4j)}| [{\tilde{v}}^{2j}_{(4j)}, v^{(4i-2)}_{2i-1}] \big )$ and get the lemma by skewsymmetry. $\square$

By Lemma A16, we conclude that J is a $N=2$ superconformal vector. Moreover, the set $W_{\mathfrak {osp}(1|2)}^{N=2}=\{\omega (\bar{v}^{(4i-2)}_{2i-1})|i=2,3,\ldots , n\}$ satisfies the second statement of Theorem 5.2.

1.3 A.3 Proof of Theorem 5.5

We use Theorems 3.9 and 3.10 to prove Theorem 5.5.

Lemma A17

The following relations hold:

$$\begin{aligned} \begin{aligned}&\{ \nu (f){}_\lambda \nu (f)\}= -2\nu (F)+\frac{1}{2}(\nu (U))^2-k^2 \lambda ^2, \\&\{ \nu ({\tilde{f}}){}_\lambda \nu ({\tilde{f}})\}= 2\nu (F)-\frac{1}{2}(\nu (U))^2+k^2 \lambda ^2. \end{aligned} \end{aligned}$$

(A.22)

Hence if we consider $G= \sqrt{-\frac{1}{k}}\nu (f)$ or $\sqrt{\frac{1}{k}}\nu ({\tilde{f}})$ then $G_{(0)}G= \frac{2}{k}\left( \nu (F)-\frac{1}{4}(\nu (U))^2\right)$, and G is a conformal vector of ${\mathcal {W}}^k({\mathfrak {g}}, F)$.

Proof

Applying Theorem 4.7, we get the relations (A.22). By Theorem 4.8, the last assertion directly follows from equations (A.22). $\square$

Let $G= \sqrt{-\frac{1}{k}}\nu (f)$ and $L:=G_{(0)}G= \frac{2}{k}(\nu (F)-\frac{1}{4}(\nu (U))^2)$. Then, for $s=2i,2i-1$ and $i\ne 2$, the elements $\nu (v_i^{(s)})$ in $W_{{{\mathfrak {s}}}{{\mathfrak {l}}}_2}^{N=0}$ and G are L-primary by Theorem 4.8. We also have $\{G_\lambda \nu (v_{i}^{(2i-1)})\}= G_{(0)} \nu (v_{i}^{(2i-1)})$ for $i\ne 2$ since

$$\begin{aligned} \{\nu (f){}_\lambda \nu (v_{i}^{(2i-1)})\}= \nu (v^{(2i)}_{i})-\sum _{j=2}^{i-1}(-1)^{j}\nu ([f, v_{j}^{(2j-2)}])\nu ([{\tilde{v}}^{j}_{(2j)}, v_{i}^{(2i-1)}]^{\natural }),\nonumber \\ \end{aligned}$$

(A.23)

and thus $\{G, L\}\cup \{\nu (v_{i}^{(2i-1)}), G_{(0)}\nu (v_{i}^{(2i-1)})|i=1,3,4,\ldots , 2n \}$ freely generates ${\mathcal {W}}^k({\mathfrak {g}},F)$ as a ${\mathbb {C}}[\partial ]$-algebra. For $D:=G_{(0)}$, by Lemma A17 and Theorem 3.9, we conclude that (i) the element G is a $N=1$ superconformal vector, (ii) $\nu (v_{i}^{(2i-1)})$ for $i\ne 2$ is G-primary, (iii) the set $\{\nu (v_{i}^{(2i-1)})|i=1,2,\ldots , 2n\}$ freely generates ${\mathcal {W}}^k({\mathfrak {g}},F)$ as a ${\mathbb {C}}[\nabla ]$-algebra.

Similarly, we can show ${\tilde{G}}=\sqrt{\frac{1}{k}}\nu ({\tilde{f}})$ is another $N=1$ superconformal vector. Since

$$\begin{aligned} \{\nu ({\tilde{f}})_\lambda \nu (v^{(4i'-2)}_{2i'-1})\}= \nu (v^{(4i')}_{2i'}),\quad \{\nu ({\tilde{f}})_\lambda \nu (v_{2i-1}^{(4i-3)})\}= {}-\nu (v_{2i}^{(4i-1)}), \end{aligned}$$

(A.24)

for $i=1,2, \ldots , n$ and $i'=2,3, \ldots ,n$, the set

$$\begin{aligned} \{{\tilde{G}}, L\}\cup \{\nu (v^{(4i'-2)}_{2i'-1}), {\tilde{G}}_{(0)}\nu (v^{(4i'-3)}_{2i'-1})\}_{i'=2}^n \cup \{\nu (v_{2i-1}^{(4i-3)}), {\tilde{G}}_{(0)}\nu (v_{2i-1}^{(4i-3)})\}_{i=1}^n\end{aligned}$$

freely generates ${\mathcal {W}}^k({\mathfrak {g}},F)$ as a ${\mathbb {C}}[\partial ]$-algebra. If we denote ${\tilde{D}}:={\tilde{G}}_{(0)}$ then (i) ${\tilde{G}}$ is a $N=1$ superconformal vector (ii) $\nu (v^{(4i'-2)}_{2i'-1})$ and $\nu (v_{2i-1}^{(4i-3)})$ are ${\tilde{G}}$-primary (iii) $\{{\tilde{G}}\}\cup \{\nu (v^{(4i'-2)}_{2i'-1}), \nu (v_{2i-1}^{(4i-3)})|i'=2,3, \ldots , n \text { and } \, i=1,2,\ldots , n \}$ freely generates ${\mathcal {W}}^k({\mathfrak {g}}, F)$ as a ${\mathbb {C}}[{\tilde{\nabla }}]$-algebra.

In the rest of this section, we show that $J:=-\sqrt{-1}\,\nu (U)$ is a $N=2$ superconformal vector of ${\mathcal {W}}^k({\mathfrak {g}},F)$. Let us first show that the odd derivations D and ${\tilde{D}}$ give rise to $N=2$ SUSY structure on ${\mathcal {W}}^k({\mathfrak {g}},F)$. Observe that

$$\begin{aligned} \begin{aligned}&DJ= \sqrt{-\frac{1}{k}}\nu (f)_{(0)}\,J=\sqrt{\frac{1}{k}}\nu ({\tilde{f}})={\tilde{G}}, \\&{\tilde{D}}J= \sqrt{\frac{1}{k}}\nu ({\tilde{f}})_{(0)}\, J=-\sqrt{-\frac{1}{k}}\nu (f)=-G. \end{aligned} \end{aligned}$$

Now, using relations (A.23) and (A.24), we get

$$\begin{aligned} \begin{aligned}&\big \{DJ{}_\lambda \nu (v_{2i-1}^{(4i-3)})\big \}= \sqrt{\frac{1}{k}}\nu ({\tilde{f}})_{(0)}\nu (v_{2i-1}^{(4i-3)})={\tilde{D}}\,\nu (v_{2i-1}^{(4i-3)}),\\&\big \{{\tilde{D}}J{}_\lambda \nu (v_{2i-1}^{(4i-3)})\big \}=-\sqrt{-\frac{1}{k}}\nu (v_{2i-1}^{(4i-3)})=-D\, \nu (v_{2i-1}^{(4i-3)}).\\ \end{aligned} \end{aligned}$$

(A.25)

and

$$\begin{aligned} D{\tilde{D}}J= G_{(0)}(-G)=\sqrt{-\frac{1}{k}} \nu (f)_{(0)}\bigg (-\sqrt{\frac{-1}{k}}\nu (f)\bigg )=-L. \end{aligned}$$

(A.26)

We can also check $\{J{}_\lambda J\}=2k\lambda$ and $\big \{J{}_\lambda \nu (v_{2i-1}^{(4i-3)})\big \}=0$. Therefore, from Theorem 3.12, we deduce that J is a $N=2$ superconformal vector and $\nu (v_{2i-1}^{(4i-3)})$ for $i=2,\ldots , n$ are J-primary. Moreover, relations (A.23) and (A.24) show that $\{J\}\cup \{\nu (v_{2i-1}^{(4i-3)})|i=2,\ldots , n\}$ freely generates ${\mathcal {W}}^k({\mathfrak {g}},F)$ as a ${\mathbb {C}}[{\varvec{\nabla }}]$-algebra.