On the Feigin–Tipunin conjecture

Abstract

We prove the Feigin–Tipunin conjecture (Feigin and Tipunin in Logarithmic CFTs connected with simple Lie algebras. arXiv:1002.5047) on the geometric construction of the logarithmic W-algebras $W(p)_Q$ associated with a simply-laced simple Lie algebra ${\mathfrak {g}}$ and $p\in {\mathbb {Z}}_{\ge 2}$.

Appendix A: A proof of Lemma 3.7

For $1\le i_1,\ldots ,i_n\le l$, denote $\sigma _{i_n\cdots i_1}$ by $\sigma _{i_n}\cdots \sigma _{i_1}$. We use the letter $\sigma _{i_0}$ for the identity map ${\text {id}}_W$ of W throughout this subsection.

Lemma A.1

The condition Lemma 3.7 (1) is equivalent to the following condition: for $1\le n\le l(w_0)$, we have

$$\begin{aligned} \epsilon _\lambda (\sigma _{i_{n}\cdots i_0})=\sum _{j=0}^{n-1}\epsilon _{\sigma _{i_{j}\cdots i_0}*\lambda }(\sigma _{i_{j+1}}). \end{aligned}$$

(155)

Proof

Let us write $\sigma$ for $\sigma _{i_{n}\cdots i_1}$ and i for $i_{n+1}$, respectively. Then we have

$$\begin{aligned} \epsilon _{\lambda }(\sigma )&=\sigma _i\epsilon _{\lambda }(\sigma )+(\epsilon _{\lambda }(\sigma ),\alpha _i)\alpha _i \end{aligned}$$

(156)

$$\begin{aligned}&=\epsilon _\lambda (\sigma _i\sigma )+\epsilon _{\sigma _i\sigma *\lambda }(\sigma _i)+\alpha _i+\delta _{(\sqrt{p}\overline{\sigma *\lambda },\alpha _i),p-1}\alpha _i+(\epsilon _{\lambda }(\sigma ),\alpha _i)\alpha _i \end{aligned}$$

(157)

where the second equality follows from (61). If $(\sqrt{p}\overline{\sigma *\lambda },\alpha _i)=p-1$, then by (60), (157) is equal to

$$\begin{aligned} \epsilon _\lambda (\sigma _i\sigma )+\alpha _i+(\epsilon _{\lambda }(\sigma ),\alpha _i)\alpha _i=\epsilon _\lambda (\sigma _i\sigma )-\epsilon _{\sigma *\lambda }(\sigma _i)+(\epsilon _{\lambda }(\sigma ),\alpha _i)\alpha _i. \end{aligned}$$

(158)

Thus, we obtain that

$$\begin{aligned} \epsilon _\lambda (\sigma _i\sigma )=\epsilon _\lambda (\sigma )+\epsilon _{\sigma *\lambda }(\sigma _i)-(\epsilon _{\lambda }(\sigma ),\alpha _i)\alpha _i. \end{aligned}$$

(159)

On the other hand, if $(\sqrt{p}\overline{\sigma *\lambda },\alpha _i)\le p-2$, then by (60), (157) is equal to

$$\begin{aligned} \epsilon _\lambda (\sigma _i\sigma )-\epsilon _{\sigma *\lambda }(\sigma _i)+(\epsilon _{\lambda }(\sigma ),\alpha _i)\alpha _i \end{aligned}$$

(160)

Thus, we obtain that

$$\begin{aligned} \epsilon _\lambda (\sigma _i\sigma )=\epsilon _\lambda (\sigma )+\epsilon _{\sigma *\lambda }(\sigma _i)-(\epsilon _{\lambda }(\sigma ),\alpha _i)\alpha _i. \end{aligned}$$

(161)

If $\lambda \in \Lambda$ satisfies the condition in Lemma 3.7 (1), then by (159) and (161), for any $0\le n\le l(w_0)-1$, we have

$$\begin{aligned} \epsilon _{\lambda }(\sigma _{i_{n+1}\cdots i_0})=\epsilon _\lambda (\sigma _{i_{n}\cdots i_0})+\epsilon _{\sigma _{i_{n}\cdots i_0}*\lambda }(\sigma _{i_{n+1}}). \end{aligned}$$

(162)

By solving the recurrence relation (162) for n, we obtain (155) for $1\le n\le l(w_0)$. The converse is clear from (159) and (161). $\square$

Remark A.2

By combining Lemma 3.7 (1) with Lemma A.1, the condition Lemma 3.7 (1) and Lemma A.1 are equivalent to the following: For $1\le n\le l(w_0)-1$, we have

$$\begin{aligned} (\sum _{j=0}^{n-1}\epsilon _{\sigma _{i_{j}\cdots i_0}*\lambda }(\sigma _{i_{j+1}}),\alpha _{i_{n+1}})=0. \end{aligned}$$

(163)

Lemma A.3

The conditions in Lemma 3.7 and Lemma A.1 are independent of the choice of a minimal expression of the longest element $w_0\in W$.

Proof

By Lemma A.1, it is enough to show that the condition Lemma 3.7 (1) is independent of the choice of a minimal expression $w_0$. Let $w_0=\sigma _{i_{l(w_0)}\cdots {i_1}}$ be a minimal expression of $w_0$ that satisfies the condition Lemma 3.7 (1). For $\sigma =\sigma _{i_{n-1}\cdots {i_1}}$, we have

$$\begin{aligned} (\epsilon _\lambda (\sigma ),\alpha _{i_n})=0,~(\epsilon _\lambda (\sigma _{i_n}\sigma ),\alpha _{i_{n+1}})=(\epsilon _\lambda (\sigma )+\epsilon _{\sigma *\lambda }(\sigma _{i_{n}}),\alpha _{i_{n+1}})=0. \end{aligned}$$

(164)

When $(\alpha _{i_n},\alpha _{i_{n+1}})=0$, the second equation in (164) leads $(\epsilon _\lambda (\sigma ),\alpha _{i_{n+1}})=0$. Thus, by (60) and (61), we have

$$\begin{aligned} \epsilon _{\lambda }(\sigma _{i_{n+1}}\sigma )=\epsilon _\lambda (\sigma )+\epsilon _{\sigma *\lambda }(\sigma _{i_{n+1}}). \end{aligned}$$

(165)

By (165) and the first equation in (164), we also obtain that $(\epsilon _{\lambda }(\sigma _{i_{n+1}}\sigma ),\alpha _{i_n})=(\epsilon _\lambda (\sigma )+\epsilon _{\sigma *\lambda }(\sigma _{i_{n+1}}),\alpha _{i_n})=0$. Therefore, we have

$$\begin{aligned} (\epsilon _{\lambda }(\sigma ),\alpha _{i_{n+1}})=(\epsilon _{\lambda }(\sigma _{i_{n+1}}\sigma ),\alpha _{i_n})=0. \end{aligned}$$

(166)

When $i_{n\pm 1}=j$ and $(\alpha _{i_n},\alpha _j)=-1$, for $\sigma =\sigma _{i_{n-2}}\cdots \sigma _{i_1}$, we have

$$\begin{aligned} (\epsilon _\lambda (\sigma ),\alpha _{j})=(\epsilon _\lambda (\sigma _j\sigma ),\alpha _{i_n}) =(\epsilon _\lambda (\sigma _{i_n}\sigma _j\sigma ),\alpha _j)=0. \end{aligned}$$

(167)

It is straightforward to verify that (167) holds if and only if

$$\begin{aligned} \sqrt{p}((\overline{\sigma *\lambda },\alpha _{i_n}+\alpha _j))\le p-2. \end{aligned}$$

(168)

Since (168) is symmetrical with respect to $i_n$ and j, we also have

$$\begin{aligned} (\epsilon _\lambda (\sigma ),\alpha _{i_n})=(\epsilon _\lambda (\sigma _{i_n}\sigma ),\alpha _{j}) =(\epsilon _\lambda (\sigma _{j}\sigma _{i_n}\sigma ),\alpha _{i_n})=0. \end{aligned}$$

(169)

Therefore, by (166) and (169), the condition Lemma 3.7 (1) is preserved under relations in the Weyl group W. $\square$

Before the proof of Lemma 3.7, we set up some notations. For a sequence $\mathbf{m}=(m_1,\dots ,m_i)$ of elements in $\Pi =\{1,\ldots , l\}$, let $r_{\mathbf{m}}$ denote the corresponding element $\sigma _{m_1}\cdots \sigma _{m_i}$ in W. Also, for sequences $\mathbf{m}=(m_1,\dots ,m_i)$ and $\mathbf{n}=(n_1,\dots ,n_j)$ of elements in $\Pi$, we write $\mathbf{m}{} \mathbf{n}$ for the sequence $(m_1,\dots ,m_i,n_1,\dots ,n_j)$. Let us take the sequences $\mathbf{s}_k$ of elements in $\Pi$ as in Table 1, and set $(i_{l(w_0)},\dots ,i_1)=\mathbf{s}_1\cdots \mathbf{s}_l$. Then we obtain the minimal expression

$$\begin{aligned} w_0=r_{\mathbf{s}_1\cdots \mathbf{s}_l}=\sigma _{i_{l(w_0)}\cdots i_1} \end{aligned}$$

(170)

of $w_0$ given in [12]. For convenience, set $\sigma _{i_0}=r_{\mathbf{s}_{l+1}}={\text {id}}_W$. For $0\le m<n\le l(w_0)$, $\tau =\sigma _{i_m}\cdots \sigma _{i_0}$, $\tau '=\sigma _{i_n}\cdots \sigma _{i_0}$, let $(\epsilon _{\sigma *\lambda })_{\tau \le _L\sigma \le _L\tau '}$ be the sequence

$$\begin{aligned} (\epsilon _{\sigma *\lambda })_{\tau \le _L\sigma \le _L\tau '}=(\epsilon _{\tau *\lambda }(\sigma _{i_{m+1}}),\epsilon _{\sigma _{i_{m+1}}\tau *\lambda }(\sigma _{i_{m+2}}),\ldots ,\epsilon _{\sigma _{i_{n-1}\cdots i_{m+1}}\tau *\lambda }(\sigma _{i_{n}})) \end{aligned}$$

of elements in P. For $1\le i\le l$, we set

$$\begin{aligned} (\epsilon _{\sigma *\lambda })_{(i+1,i)}=(\epsilon _{\sigma *\lambda })_{r_{\mathbf{s}_{i+1}\cdots \mathbf{s}_{l+1}}\le _L\sigma \le _Lr_{\mathbf{s}_{i}\cdots \mathbf{s}_{l+1}}}. \end{aligned}$$

(171)

Proof of Lemma 3.7

We show that the condition Lemma 3.7 (2) and that in Lemma A.1 are equivalent. By Lemma A.3, it is enough to consider the minimal expression (170) of $w_0$.

It is straightforward to show the following: first, when $(\sqrt{p}{\bar{\lambda }}+\rho ,\theta )<p$, the list of $(\epsilon _{\sigma *\lambda })_{{\text {id}}\le _L\sigma \le _Lw_0}$ is given by Table 2. Second, when $(\sqrt{p}{\bar{\lambda }}+\rho ,\theta )=p$, the list of $(\epsilon _{\sigma *\lambda })_{{\text {id}}\le _L\sigma \le _Lw_0}$ is given by changing Table 2 as

$$\begin{aligned} {\left\{ \begin{array}{ll} \epsilon _{\sigma _1r_{\mathbf{s}_l}*\lambda }(\sigma _1)=-2\omega _1+\omega _2,~\epsilon _{\sigma _2r_{\mathbf{s}_{l-1}}r_{\mathbf{s}_l}*\lambda }(\sigma _2)=\omega _1-\omega _2+\omega _3,&{}(A_l)\\ \epsilon _{\sigma _1r_{\mathbf{s}_l}*\lambda }(\sigma _1)=-2\omega _1+\omega _2,~\epsilon _{r_{\mathbf{s}_l}*\lambda }(\sigma _2)=\omega _1-\omega _2,&{}(D_l)\\ \epsilon _{\sigma _6r_{\mathbf{s}_6}*\lambda }(\sigma _6)=-2\omega _6+\omega _5,~\epsilon _{\sigma _{321}r_{\mathbf{s}_6}*\lambda }(\sigma _5)=\omega _3-\omega _5+\omega _6,&{}(E_6)\\ \epsilon _{\sigma _7r_{\mathbf{s}_7}*\lambda }(\sigma _7)=-2\omega _7+\omega _6,~\epsilon _{\sigma _{5321}r_{\mathbf{s}_7}*\lambda }(\sigma _6)=-\omega _6+\omega _5+\omega _7,&{}(E_7)\\ \epsilon _{r_{\mathbf{s}_7}\sigma _8*\lambda }(\sigma _8)=-2\omega _8+\omega _7,~\epsilon _{\sigma _8r_{\mathbf{s}_7}\sigma _8*\lambda }(\sigma _7)=-\omega _7+\omega _8,&{}(E_8) \end{array}\right. }\nonumber \\ \end{aligned}$$

(172)

and others are the same as Table 2. Third, when $(\sqrt{p}{\bar{\lambda }}+\rho ,\theta )\le p$, by Table 2 and (172), the equation (163) holds for any $1\le n\le l(w_0)$. Finally, when $(\sqrt{p}{\bar{\lambda }}+\rho ,\theta )>p$, we have $(\sum _{j=0}^{n-1}\epsilon _{\sigma _{i_{j}\cdots i_0}*\lambda }(\sigma _{i_{j+1}}),\alpha _{i_{n+1}})>0$ for some $1\le n\le l(w_0)-1$. Thus, Lemma 3.7 is proved.

Corollary A.4

For $\lambda \in \Lambda$ such that $(\sqrt{p}{\bar{\lambda }}+\rho ,\theta )\le p$, we have $\epsilon _\lambda (w_0)=-\rho$.

Proof

By Lemma A.1, for the minimal expression (170) of $w_0$, we have $\epsilon _{\lambda }(w_0)=\sum _{j=1}^{l(w_0)}\epsilon _{i_{j-1}\cdots i_0*\lambda }(\sigma _{i_j})$. Then by Table 2 and (172), the claim is proved. $\square$

Table 1 The list of $\mathbf{s}_i$ for $1\le i\le l$ [12, Table 1]

Table 2 The list of $(\epsilon _{\sigma *\lambda })_{{\text {id}}\le _L\sigma \le _Lw_0}$ when $(\sqrt{p}{\bar{\lambda }}+\rho ,\theta )<p$