Recently, Gaiotto and Rapčák (GR) proposed a new family of the vertex operator
algebra (VOA) as the symmetry appearing at an intersection of five-branes to which they
refer as

Article funded by SCOAP3

0})^{\otimes 3}$ for a general box. We assign \begin{equation} \label{eq:assignedvalue} h_{\fbox{}}=h_1x_1+h_2x_2+h_3x_3 \end{equation} to the box located at $(x_1,x_2,x_3)$. We introduce an extra parameter ``$q$'' to represent the shift of the spectral parameter. It gives an automorphism of affine Yangian. While it does not change the structure of representation, it represents the charge of $\U(1)$ factor. In the following, we use the representation with $q=0$ when we do not mention it explicitly.\footnote{We may cancel the redundancy of $h_1,h_2,h_3$ and the center $\psi_0$ by scaling $u$.} As we already mentioned, Proch\'azka's parametrization has an advantage that the reduction of the representation becomes manifest. In the example we mentioned (one of $\lambda_i$, say $\lambda_3$, becomes positive integer), the basis $|\Lambda\rangle$ becomes null when it contains a box at $(1, 1,N+1)$. With such condition, the height of the plane partition $\Lambda$ for the nonvanishing states is not greater than $N$. One may decompose the plane partition layer by layer into $N$-tuple Young diagrams $Y_1,\cdots, Y_N$ with the condition $Y_1\succeq\cdots \succeq Y_N$. They give a representation space of $W_N$ algebra with an extra $\U(1)$ factor. In general, when $\lambda_i$ satisfies the extra condition \ba\label{DegLMN} \frac{L}{\lambda_1}+\frac{M}{\lambda_2}+\frac{N}{\lambda_3}=1, \ea the basis $|\Lambda\rangle$ which contains a box with a coordinate $(L+1, M+1, N+1)$ becomes null. Following~\cite{bershtein2018plane}, we refer to such a position as a ``pit'' with the implication that we cannot place a box there. Proch\'azka and Rap\v{c}\'{a}k~\cite{Prochazka-ml-2017qum} claimed that the affine Yangian whose parameter is constrained by this condition is equivalent to the vertex operator algebra $Y_{L,M,N}[\Psi]$ in~\cite{Gaiotto-ml-2017euk}. We can derive these null state conditions from~(\ref{eq:addbox}) and~(\ref{eq:removebox}). The equation \begin{equation} {\rm res}_{u\to h_{\fbox{}}}\psi_{\Lambda}(u)=0\qquad(\raisebox{2.5pt}{\fbox{}}\in\Lambda^+). \end{equation} implies that the application of the Drinfeld current $e(z)$ cannot generate the state $\ket{\Lambda+\raisebox{3.5pt}{\fbox{}}}$ since the coefficient attached to the basis vanishes. Because $\psi_{\Lambda}(u)$ contains a factor $\frac{u+\psi_0\sigma_3}{u-h_{\fbox{}}}$ for any $\raisebox{2.3pt}{\fbox{}}\in\Lambda^+$ in generic parameters, this happens if \begin{equation} \label{eq:truncation} \psi_0\sigma_3=-h_{\fbox{}}, \end{equation} or equivalently \begin{equation} \label{eq:truncation2} \sum_{i=1}^3\frac{x_i}{\lambda_i}=1\,. \end{equation} We may interpret it as the null state condition. We note that the condition~(\ref{DegLMN}) has a shift symmetry, \begin{eqnarray} L\rightarrow L+k,\quad M\rightarrow M+k,\quad N\rightarrow N+k. \end{eqnarray} for $k\in \mathbb{Z}$ due to~(\ref{eq:sumhlam}). It allows the redefinition of the location of the pit such that the smallest element is one and others are greater or equal to one. The character of the plane partitions with a pit was derived in~\cite{bershtein2018plane}. The plane partition may have non-trivial asymptotes written in the form of three Young diagrams $\mu_1, \mu_2, \mu_3$ for each axis $x_1, x_2, x_3$. The partition function for nontrivial plane partition is related to the topological vertex~\cite{Okounkov-ml-2003sp}, \ba \chi[q]\propto C_{\mu_1,\mu_2,\mu_3}(q) \prod_{n=1}^\infty (1-q^n)^{-n}\,. \ea When there is a pit at $(L+1, M+1, N+1)$, the asymptotic Young diagrams have a similar pit, for example, $\mu_1$ has a pit at $(M+1, N+1)$. ]]>

](-2,1)--(-2,1.5); \draw[->](-2,1)--(-1.5,1); \node (x2) at (-1.2,1) {$x_2$}; \node (x2) at (-2,1.8) {$x_3$}; \end{tikzpicture} ]]>

M$. As a consequence, $\widehat{\rm U}(N|L)_{\Psi}$ affine-Kac-Moody algebra is reduced to $\mathcal{DS}_{N-M}[\widehat{\rm U}(N|L)_{\Psi}]$. Here, $\mathcal{DS}_{N-M}$ means Drinfeld-Sokolov reduction with the principal $su(2)$ embedding in $(N-M)\times(N-M)$ part in $\U(N|L)$. Finally, we obtain the algebra by $u(M|L)$ BRST reduction. As explained in~\cite{Gaiotto-ml-2017euk}, it is considered to be equivalent to take coset as follows:
\begin{equation}
\frac{\mathcal{DS}_{N-M}[\widehat{\rm U}(N|L)_{\Psi}]}{\widehat{\rm U}(M|L)_{\Psi-1}}.
\end{equation}
Note that the level of the subalgebra $\widehat{\rm U}(M|L)$ in $\widehat{\rm U}(N|L)$ is changed due to the contribution from the triangular constituent necessary to make it BRST-closed.
One can consider the $N

0$) by applying such bilinear operators. The basis takes the form \ba \beta_{-n_1}\cdots \beta_{-n_g} \gamma_{-m_1} \cdots \gamma_{-m_g}|0\rangle \ea where $n_1\geq \cdots \geq n_g\geq 1$ and $m_1\geq m_2\geq \cdots \geq m_g\geq 0$. We note that the ghost number should vanish. One may construct two Young diagrams $(n_1,\cdots, n_g)$ and $(m_1,\cdots, m_g)$ and attach the second piece to the first one perpendicularly as~\ref{fig2}. \begin{figure} \centering \begin{tikzpicture}[scale=.95] \node (white) at (-7.4,0.5) {${\displaystyle\beta_{-n_1}\cdots\beta_{-n_g}\gamma_{-m_1}\cdots\gamma_{-m_g}\ket{0}\quad=}$}; \scriptsize \draw (0,0)--(0,2.4); \draw (0,0)--(3,-1.5)--(3,-0.7)--(0,0.8); \draw (2.4,-0.4)--(2.4,-0)--(0,1.2); \draw (1.2,0.6)--(1.2,1)--(0,1.6); \draw (0.8,1.2)--(0.8,2)--(0,2.4); \draw (0,2)--(0.8,1.6)--(1.2,1.8); \draw (3,-1.5)--(3.4,-1.3)--(3.4,-0.5)--(3,-0.7); \draw (2.4,-0.4)--(2.8,-0.2); \draw (2.4,0)--(2.8,0.2); \draw (1.2,0.6)--(1.6,0.8); \draw (1.2,1)--(1.6,1.2); \draw (0.8,1.2)--(1.2,1.4); \draw (0.8,2)--(1.2,2.2); \draw (0,2.4)--(0.4,2.6); \draw (3.4,-0.5)--(2.8,-0.2)--(2.8,0.2)--(1.6,0.8)--(1.6,1.2)--(1.2,1.4)--(1.2,2.2)--(0.4,2.6); \draw (0,0)--(-3,-1.5)--(-3,-0.7)--(0,0.8); \draw (-0.8,1.6)--(-0.8,2)--(0,2.4); \draw (-3,-1.5)--(-3.4,-1.3)--(-3.4,-0.5)--(-3,-0.7); \draw (-2,-0.2)--(-2.4,0); \draw (-0.8,1.6)--(-1.2,1.8); \draw (-0.8,2)--(-1.2,2.2); \draw (-3.4,-0.5)--(-2.4,0)--(-2.4,0.8)--(-2,1)--(-2,1.4)--(-1.2,1.8)--(-1.2,2.2)--(0,2.8)--(0.4,2.6); \draw (-2,1.4)--(-1.6,1.2)--(-1.6,0.8)--(-2,1); \draw (-1.6,1.2)--(0,2); \draw (-2,0.6)--(-2,-0.2); \draw (-3.4,-0.9)--(-3,-1.1)--(0,0.36); \draw (-2.4,0.4)--(-2,0.2)--(0,1.2); \draw (-2.4,0.8)--(-2,0.6)--(0,1.6); \draw (0,0.36)--(3,-1.14)--(3.4,-0.94); \draw[->] (0,2.8)--(0,3.7); \draw[->] (-3.4,-1.3)--(-4.2,-1.7); \draw[->] (3.4,-1.3)--(4.2,-1.7); \node (1) at (-4.4,-1.8) {1}; \node (2) at (4.4,-1.8) {2}; \node (3) at (0,4) {3}; \draw[dotted] (0.4,2.6)--(0.4,0.2); \draw[dotted] (0,0)--(0.4,0.21)--(3.4,-1.29); \draw[dotted] (0,0.8)--(0.4,1); \draw[dotted] (0,1.2)--(0.4,1.4); \draw[dotted] (0,1.6)--(0.4,1.8); \draw[dotted] (0,2)--(0.4,2.2); \draw[dotted] (0,0.4)--(0.4,0.6); \node (n1) at (-3.2,-1.25) {\rotatebox{-10}{$n_1$}}; \node (n2) at (-3.2,-0.85) {\rotatebox{-10}{$n_2$}}; \node (n3) at (-2.2,0.05) {\rotatebox{-10}{$n_3$}}; \node (n4) at (-2.2,0.45) {\rotatebox{-10}{$n_4$}}; \node (n5) at (-1.8,1.05) {\rotatebox{-10}{$n_5$}}; \node (n6) at (-1,1.85) {\rotatebox{-10}{$n_6$}}; \node (m1) at (3.23,-1.25) {\rotatebox{20}{$m_1$}}; \node (m2) at (3.23,-0.85) {\rotatebox{20}{$m_2$}}; \node (m3) at (2.63,-0.15) {\rotatebox{20}{$m_3$}}; \node (m4) at (1.43,0.85) {\rotatebox{20}{$m_4$}}; \node (m5) at (1.03,1.45) {\rotatebox{20}{$m_5$}}; \node (m6) at (1.03,1.85) {\rotatebox{20}{$m_6$}}; \draw[dotted] (0.2,2.9)--(0.2,3.4); \draw[dotted] (-0.2,2.9)--(-0.2,3.4); \end{tikzpicture} ]]>

\cdots > n_g\geq 1$ and $m_1> m_2> \cdots > m_g\geq 0$. Again the state is constructed out of bilinear of fermionic ghosts, and the ghost number is zero. In this case, one can associated a Young diagram by combining hooks $(n_l, 1^{m_l})$ in the order $l=1,2,\cdots, g$ as figure~\ref{fig3}. \begin{figure} \centering \begin{tikzpicture}[scale=.95] \node (white) at (-3.3,-1.5) {${\displaystyle b_{-n_1}\cdots b_{-n_g}c_{-m_1}\cdots c_{-m_g}\ket{0}\quad=}$}; \scriptsize \draw (0,0) rectangle (4.5,-0.4); \draw (0,-0.4) rectangle (0.4,-3.5); \draw (0.4,-0.4) rectangle (4.0,-0.8); \draw (0.4,-0.8) rectangle (0.8,-3.0); \draw[dotted] (0.9,-0.9)--(1.1,-1.1); \draw (1.2,-1.2) rectangle (3.5,-1.6); \draw (1.2,-1.6) rectangle (1.65,-2.4); \node (white) at (2.5,-0.25) {$n_1$}; \node (white) at (2.5,-0.65) {$n_2$}; \node (white) at (2.5,-1.45) {$n_g$}; \node (white) at (0.22,-2.5) {$m_1$}; \node (white) at (0.62,-2.3) {$m_2$}; \node (white) at (1.45,-2.0) {$m_g$}; \end{tikzpicture} ]]>

\cdots> n_g\geq 1}\sum_{m_1>\cdots > m_g\geq 0}q^{\sum_i (n_i+m_i)} =\sum_{n=0}^\infty \frac{q^{n^2/2}}{((q;q)_n)^2}=\prod_{n=1}^\infty (1-q^n)^{-1}\,. \end{eqnarray} We note that the formula in the third term resembles that of the bosonic ghost. \newpage In this case, the plane partition is truncated to a single Young diagram. The location of the pit is $(1,1,2)$. The condition to have a pit there implies $\lambda_3=1$. One may solve~(\ref{eq:sumhlam}) by $\lambda_2=-\frac{\lambda_1}{\lambda_1+1}$. The central charge~(\ref{eq:Winftyparameter}) gives $c=1$ for any $\lambda_1$. This is the central charge of free fermion. The self-duality condition is met if $\lambda_1=-1$ and $\lambda_2=\infty$. If there are $N$ fermions, the pit moves to $(1,1,N+1)$. There is the explicit form of the character for such a diagram~\cite{Awata-ml-1994tf} for the self-dual case. ]]>

0$, one may fill the unbalanced rows by an infinite leg with hight $h$ on the right. On the other hand, with $g_4-g_3=h$, one may similarly fill the other wing of the Young diagram with an infinite leg similarly. See figures~\ref{fig4} and~\ref{fig5}. \begin{figure} \centering \begin{tikzpicture}[scale=0.83] \scriptsize \draw (0,0)--(0,2.4); \draw (1.2,0.6)--(1.2,1)--(0,1.6); \draw (0.8,1.2)--(0.8,2)--(0,2.4); \draw (0,2)--(0.8,1.6)--(1.2,1.8); \draw (1.2,0.6)--(1.6,0.8); \draw (1.2,1)--(1.6,1.2); \draw (0.8,1.2)--(1.2,1.4); \draw (0.8,2)--(1.2,2.2); \draw (0,2.4)--(0.4,2.6); \draw (1.6,0.8)--(1.6,1.2)--(1.2,1.4)--(1.2,2.2)--(0.4,2.6); \draw (0,0)--(-3,-1.5)--(-3,-0.7)--(0,0.8); \draw (-0.8,1.6)--(-0.8,2)--(0,2.4); \draw (-3,-1.5)--(-3.4,-1.3)--(-3.4,-0.5)--(-3,-0.7); \draw (-2,-0.2)--(-2.4,0); \draw (-0.8,1.6)--(-1.2,1.8); \draw (-0.8,2)--(-1.2,2.2); \draw (-3.4,-0.5)--(-2.4,0)--(-2.4,0.8)--(-2,1)--(-2,1.4)--(-1.2,1.8)--(-1.2,2.2)--(0,2.8)--(0.4,2.6); \draw (-2,1.4)--(-1.6,1.2)--(-1.6,0.8)--(-2,1); \draw (-1.6,1.2)--(0,2); \draw (-2,0.6)--(-2,-0.2); \draw (-3.4,-0.9)--(-3,-1.1)--(0,0.36); \draw (-2.4,0.4)--(-2,0.2)--(0,1.2); \draw (-2.4,0.8)--(-2,0.6)--(0,1.6); \draw[->] (0,2.8)--(0,3.7); \draw[->] (-3.4,-1.3)--(-4.2,-1.7); \draw[->] (3.4,-1.3)--(4.2,-1.7); \node (1) at (-4.4,-1.8) {1}; \node (2) at (4.4,-1.8) {2}; \node (3) at (0,4) {3}; \draw[dotted] (0.4,2.6)--(0.4,0.2); \draw[dotted] (0,0)--(0.4,0.21)--(3.4,-1.29); \draw[dotted] (0,0.8)--(0.4,1); \draw[dotted] (0,1.2)--(0.4,1.4); \draw[dotted] (0,1.6)--(0.4,1.8); \draw[dotted] (0,2)--(0.4,2.2); \draw[dotted] (0,0.4)--(0.4,0.6); \node (n1) at (-3.2,-1.25) {\rotatebox{-10}{$n_1$}}; \node (n2) at (-3.2,-0.85) {\rotatebox{-10}{$n_2$}}; \node (n3) at (-2.2,0.05) {\rotatebox{-10}{$n_3$}}; \node (n4) at (-2.2,0.45) {\rotatebox{-10}{$n_4$}}; \node (n5) at (-1.8,1.05) {\rotatebox{-10}{$n_5$}}; \node (n6) at (-1,1.85) {\rotatebox{-10}{$n_6$}}; \node (m4) at (1.43,0.85) {\rotatebox{20}{$m_1$}}; \node (m5) at (1.03,1.45) {\rotatebox{20}{$m_2$}}; \node (m6) at (1.03,1.85) {\rotatebox{20}{$m_3$}}; \draw[dotted] (0.2,2.9)--(0.2,3.4); \draw[dotted] (-0.2,2.9)--(-0.2,3.4); \draw (0,1.2)--(3.4,-0.5); \draw (0,0)--(3.2,-1.6); \draw (1.6,0.8)--(3.6,-0.2); \fill[pattern=north east lines] (0,1.2)--(3.2,-0.4)--(3,-1.5)--(0,0)--cycle; \fill[pattern=north west lines] (1.2,0.6)--(3.2,-0.4)--(3.6,-0.2)--(1.6,0.8)--cycle; \large \node (times) at (4.8,1) {$\otimes$}; \scriptsize \begin{scope}[yshift=3.3cm] \fill[pattern=north east lines] (6,0) rectangle (6.4,-0.4); \draw(6,-0.4) rectangle (6.4,-4.7); \fill[pattern=north east lines] (6.4,-0) rectangle (11.4,-0.8); \draw (6.4,-0.8) rectangle (6.8,-4.3); \draw (7.2,-1.2) rectangle (9.9,-1.6); \draw (6.8,-1.2) rectangle (7.2,-4); \draw (7.2,-1.6) rectangle (7.6,-3.6); \draw[dotted] (7.7,-1.7)--(7.9,-1.9); \draw (8,-2) rectangle (9.4,-2.4); \draw (8,-2.4) rectangle (8.45,-3); \node (white) at (6.2,-3.25) {$\bar{n}_1$}; \node (white) at (6.6,-3.1) {$\bar{n}_2$}; \node (white) at (8.5,-1.45) {$\bar{m}_1$}; \node (white) at (7.0,-2.95) {$\bar{n}_3$}; \node (white) at (7.4,-2.85) {$\bar{n}_4$}; \node (white) at (8.25,-2.7) {$\bar{n}_{g_4}$}; \node (white) at (8.9,-2.25) {$\bar{m}_{g_3}$}; \draw(6,-0)--(6,-0.4); \draw[->](6,0)--(12.1,0); \draw(9.9,-1.2)--(11.6,-1.2); \fill[pattern=north east lines] (6.8,-0.8) rectangle (11.4,-1.2); \draw[->] (6,-4.7)--(6,-5.1); \node (1) at (12.4,0) {2}; \node (2) at (6,-5.4) {1}; \end{scope} \end{tikzpicture} ]]>

0$. The rows with infinite length and height $g_1-g_2$ are inserted. The above case corresponds to $g_1-g_2=3$.]]>

] (0,2.8)--(0,3.7); \draw[->] (-3.2,-1.2)--(-4.2,-1.7); \draw[->] (3.4,-1.3)--(4.2,-1.7); \node (1) at (-4.4,-1.8) {1}; \node (2) at (4.4,-1.8) {2}; \node (3) at (0,4) {3}; \draw[dotted] (0.4,2.6)--(0.4,0.2); \draw[dotted] (0,0)--(0.4,0.21)--(3.4,-1.29); \draw[dotted] (0,0.8)--(0.4,1); \draw[dotted] (0,1.2)--(0.4,1.4); \draw[dotted] (0,1.6)--(0.4,1.8); \draw[dotted] (0,2)--(0.4,2.2); \draw[dotted] (0,0.4)--(0.4,0.6); \node (n4) at (-2.2,0.45) {\rotatebox{-10}{$n_1$}}; \node (n5) at (-1.8,1.05) {\rotatebox{-10}{$n_2$}}; \node (n6) at (-1,1.85) {\rotatebox{-10}{$n_3$}}; \node (m1) at (3.23,-1.25) {\rotatebox{20}{$m_1$}}; \node (m2) at (3.23,-0.85) {\rotatebox{20}{$m_2$}}; \node (m3) at (2.63,-0.15) {\rotatebox{20}{$m_3$}}; \node (m4) at (1.43,0.85) {\rotatebox{20}{$m_4$}}; \node (m5) at (1.03,1.45) {\rotatebox{20}{$m_5$}}; \node (m6) at (1.03,1.85) {\rotatebox{20}{$m_6$}}; \draw[dotted] (0.2,2.9)--(0.2,3.4); \draw[dotted] (-0.2,2.9)--(-0.2,3.4); \draw (-2,0.2)--(-3.4,-0.5); \draw (-2.4,0.4)--(-3.6,-0.2); \draw (0,0)--(-3.2,-1.6); \fill[pattern=north west lines] (-3.2,-0.4)--(0,1.2)--(0,0)--(-3,-1.5)--cycle; \fill[pattern=north east lines] (-2,0.2)--(-2.4,0.4)--(-3.5,-0.15)--(-3.2,-0.4)--cycle; \large \node (times) at (4.5,1) {$\otimes$}; \scriptsize \begin{scope}[xshift=6cm,yshift=3.3cm] \scriptsize \draw (0,0) rectangle (4.5,-0.4); \fill[pattern=north east lines] (0,-0.8) rectangle (0.8,-4.6); \draw (0.4,-0.4) rectangle (4.0,-0.8); \fill[pattern=north east lines] (0,-0.4) rectangle (0.4,-0.8); \draw (0.8,-0.8) rectangle (3.5,-1.2); \draw (1.2,-1.2) rectangle (1.6,-3.2); \fill[pattern=north east lines] (0.8,-1.2) rectangle (1.2,-4.6); \draw (1.2,-3.2)--(1.2,-4.8); \draw[dotted] (1.7,-1.3)--(1.9,-1.5); \draw (2,-1.6) rectangle (3.4,-2.0); \draw (2,-2.0) rectangle (2.45,-2.6); \node (white) at (2.5,-0.25) {$\bar{m}_1$}; \node (white) at (2.5,-0.65) {$\bar{m}_2$}; \node (white) at (2.5,-1.05) {$\bar{m}_3$}; \node (white) at (1.4,-2.0) {$\bar{n}_1$}; \node (white) at (2.25,-2.3) {$\bar{n}_{g_4}$}; \node (white) at (2.9,-1.85) {$\bar{m}_{g_3}$}; \draw[->](0,-0.4)--(0,-5.3); \draw[->](4.5,0)--(5.3,0); \node (1) at (5.5,0) {2}; \node (2) at (0,-5.5) {1}; \end{scope} \end{tikzpicture} ]]>

] (1.1,0.2)--(0.2,1.1); \draw[->] (2,-0.1)--(0.2,1.7); \draw[->] (0,0.6)--(0,2.6); \draw[->] (-1.8,-1.8)--(-2.1,-2.1); \draw[->] (4.2,0)--(4.8,0); \node (h2) at (-2.4,-2.4) {1}; \node (h3) at (5,0) {2}; \node (h1) at (0,3) {3}; \begin{scope}[yshift=-7.5cm] \draw (-1.8,-1.8)--(-1.2,-1.8)--(-0.9,-1.5)--(-0.3,-1.5)--(0.9,-0.3)--(3.9,-0.3)--(4.2,0); \draw (-1.8,-1.2)--(-1.2,-1.2)--(-0.9,-0.9)--(-0.3,-0.9)--(0.9,0.3)--(3.9,0.3)--(4.2,0.6); \draw (-1.2,-0.6)--(-0.6,-0.6)--(-0.3,-0.3)--(0.3,-0.3); \draw (2.1,0.3)--(2.4,0.6); \draw (-1.2,0)--(-0.6,0)--(-0.3,0.3)--(0.3,0.3)--(0.9,0.9)--(2.1,0.9)--(2.4,1.2); \draw (-0.6,0.6)--(0,0.6)--(0.3,0.9)--(0.9,0.9)--(1.2,1.2); \draw (-0.6,1.2)--(0,1.2)--(0.3,1.5)--(0.9,1.5)--(1.2,1.8); \draw (0,1.8)--(-0.6,1.2)--(-0.6,0.6)--(-1.2,0)--(-1.2,-0.6)--(-1.8,-1.2)--(-1.8,-1.8); \draw (0,1.8)--(1.2,1.8)--(1.2,1.2)--(2.4,1.2)--(2.4,0.6)--(4.2,0.6)--(4.2,0); \draw (0.6,1.8)--(0.3,1.5)--(-0.3,1.5); \draw (0.3,1.5)--(0.3,0.9); \draw (0.9,1.5)--(0.9,0.9); \draw (0,1.2)--(0,0.6); \draw (-0.3,0.3)--(0,0.6)--(0.6,0.6); \draw (-0.9,0.3)--(-0.3,0.3)--(-0.3,-0.3); \draw (-0.6,0)--(-0.6,-0.6)--(-0.9,-0.9)--(-1.5,-0.9); \draw (-1.2,-1.8)--(-1.2,-1.2); \draw (-0.9,-1.5)--(-0.9,-0.9); \draw (-0.3,-1.5)--(-0.3,-0.9); \draw (0,-1.2)--(0,-0.6)--(-0.6,-0.6); \draw (0.3,-0.9)--(0.3,0.3); \draw (0.6,-0.6)--(0.6,0.6); \draw (0.9,-0.3)--(0.9,0.9); \draw (1.5,-0.3)--(1.5,0.9)--(1.8,1.2); \draw (2.1,-0.3)--(2.1,0.9); \draw (2.7,-0.3)--(2.7,0.3)--(3,0.6); \draw (3.3,-0.3)--(3.3,0.3)--(3.6,0.6); \draw (3.9,-0.3)--(3.9,0.3); \draw[->] (0,1.8)--(0,2.6); \draw[->] (-1.8,-1.8)--(-2.1,-2.1); \draw[->] (4.2,0)--(4.8,0); \node (h2) at (-2.4,-2.4) {1}; \node (h3) at (5,0) {2}; \node (h1) at (0,3) {3}; \node (a) at (0,4.5) {\LARGE$\Updownarrow$}; \end{scope} \end{tikzpicture} $$ \newpage \noindent The first figure is the case when we fill the pit at $(2,3,1)$. We cannot fill the other pit at $(1,1,2)$, and we obtain a plane partition whose height is one. The periodicity constraint implies that one can slice the diagram at $(k+1,2k+1,1)$ $k=1,2,\cdots$ in the shape of hook and pile the $(k+1)$-th piece on the $k$-th piece, which gives a diagram where we identify $(2,3,1)$ as the pit. In this simple example $n=1$, one may take the first diagram as the arbitrary partition. The second diagram, obtained in this way, belongs to a restricted set of the plane partition with the pit at $(2,3,1)$. For the general $n$, we need restrictions on both diagrams as we will see in the minimal model of $W_n$ algebra. ]]>

(p-N)k, y>(q-N)k\right\}$.
The dual plane partition after the translation~becomes,
$$
Y_1^{(0)}\succeq Y_{2}^{(0)}\succeq \cdots Y_N^{(0)}\succeq Y_1^{(1)}\succeq \cdots \succeq Y_N^{(1)}\succeq Y_1^{(2)}\succeq \cdots.
$$
Such a plane partition is sensible only if $Y_N^{(k)}\succeq Y_1^{(k+1)}$, ($k\geq 0$). They give extra constraints to the original $N$-tupple Young diagrams.
We note that the primary fields (or the irreducible representations) are parameterized by two set of positive integers $\vec n=(n_1,\cdots, n_{N-1})$ and $\vec n'=(n'_1,\cdots, n'_{N-1})$, $n_i, n'_i\geq 1$ with $\sum_{i=1}^{N-1} n_i] (-0.3,0.3)--(-0.3,-0.2);
\draw[->] (-0.3,0.3)--(0.2,0.3);
\node (1) at (-0.3,-0.4) {1};
\node (2) at (0.4,0.3) {2};
\node (y1) at (1,-1.1) {$Y_N$};
\node (y2) at (2.5,-2) {$Y_{N-1}$};
\node (y3) at (3.5,-3) {$Y_{N-2}$};
\node (y4) at (5.7,-4.5) {$Y_{2}$};
\node (y5) at (7,-5.3) {$Y_1$};
\footnotesize
\draw[<->] (0,-4.4)--(1.2,-4.4);
\node (n1) at (0.7,-4.8) {$\tilde{n}_{N-1}$};
\draw[<->] (1.2,-5.2)--(2.4,-5.2);
\node (n2) at (1.9,-5.6) {$\tilde{n}_{N-2}$};
\draw[<->] (4.8,-7.2)--(6,-7.2);
\node (n3) at (5.5,-7.6) {$\tilde{n}_1$};
\node (dot) at (3.5,-6.3) {$\ddots$};
\draw[<->] (5.2,0)--(5.2,-0.8);
\node (n1t) at (5.8,-0.4) {$\tilde{n}_{N-1}'$};
\draw[<->] (6.4,-0.8)--(6.4,-1.2);
\node (n2t) at (7,-1) {$\tilde{n}_{N-2}'$};
\draw[<->] (9.2,-2.8)--(9.2,-4);
\node (n3t) at (9.6,-3.4) {$\tilde{n}_1'$};
\node (dot) at (7.9,-1.8) {$\ddots$};
\end{tikzpicture}
]]>

] (0,0)--(1); \draw[->] (0,0)--(2); \draw[->] (0,0)--(3); \draw (-1.2,-1.2)--(-1.2,-0.6)--(0,0.6); \draw (0,0.6)--(3.6,0.6); \draw (-1.2,-0.6)--(2.4,-0.6); \draw (-1.2,-1.2)--(2.4,-1.2); \fill[pattern=north west lines] (0,0.6)--(-1.2,-0.6)--(2.2,-0.6)--(3.4,0.6)--cycle; \fill[pattern=north east lines] (-1.2,-0.6)--(2.2,-0.6)--(2.2,-1.2)--(-1.2,-1.2)--cycle; \node (mu) at (-1.4,-1.2) {$\mu$}; \node (comment1) at (0.3,-2.8) {(i)\ $\mu>0$}; \node (33) at (13,1) {2}; \node (11) at (10,3) {3}; \node (22) at (7,-2) {1}; \draw[dotted][->] (9,0)--(9,2); \draw[->] (9,0)--(22); \draw[dotted][->] (9,0)--(12,0); \draw (9,0)--(10,1); \draw[->] (10,1)--(33); \draw[->] (10,1)--(11); \node (comment1) at (9.3,-2.8) {(ii)\ $\mu<0$}; \node (pit1) at (0,0.75) {$\times$}; \node (pit2) at (10,1.75) {$\times$}; \end{tikzpicture} ]]>

0$ case where asymptotic Young diagram $(\mu)$ is imposed. The right one shows $\mu<0$ case where the origin is shifted to the $x_1$ direction by $\mu$. The crosses mean~pits. ]]>

] (0,0)--(1); \draw[->] (0,0)--(2); \draw[->] (0,0)--(3); \draw (-0.3,-0.3)--(-0.3,2.2)--(0,2.5)--(4,2.5); \draw (-0.3,2.2)--(3.7,2.2); \draw (-0.3,-0.3)--(3.7,-0.3); \node (comment1) at (0.3,-4) {(i)\ $\mu>0$}; \fill[pattern=north west lines] (3.6,2.2)--(-0.3,2.2)--(0,2.5)--(3.9,2.5)--cycle; \fill[pattern=north east lines] (3.6,2.2)--(-0.3,2.2)--(-0.3,-0.3)--(3.6,-0.3)--cycle; \node (33) at (12,-1.5) {2}; \node (11) at (9,2) {3}; \node (22) at (6.3,-4.2) {1}; \draw[->] (9,0)--(11); \draw[dotted][->] (9,0)--(6,-3); \draw[dotted][->] (9,0)--(12,0); \draw (7.8,-2.4)--(10.2,0)--(10.2,-1.5)--(7.95,-3.75); \fill[pattern=north east lines] (8.1,-2.1)--(10.2,0)--(10.2,-1.5)--(8.1,-3.6)--cycle; \fill[pattern=north east lines] (8.1,-2.1)--(10.2,0)--(9,0)--(6.9,-2.1)--cycle; \fill[pattern=north west lines] (8.1,-2.1)--(6.9,-2.1)--(6.9,-3.6)--(8.1,-3.6)--cycle; \draw[->] (9,0)--(9,-1.5)--(33); \draw[->] (9,-1.5)--(22); \draw (6.6,-2.4)--(9,0)--(10.2,0); \node (comment1) at (9.3,-4) {(ii)\ $\mu<0$}; \node (pit1) at (1,-0.43) {$\times$}; \node (pit2) at (10,-1.95) {$\times$}; \end{tikzpicture} ]]>

0$ case where asymptotic Young diagram is $(\mu)$. The right one shows $\mu<0$ case. The origin is moved to the $x_3$ direction and asymptotic Young diagram is~$(|\mu|,|\mu|)$.]]>

0$ case and $\mu<0$ case. One may feel that the figures for $\mu<0$ look strange because it does not match the picture that two plane partitions share an infinite leg. In the free field case, the sign of the weight changes only the direction in which a shared leg extends and not the shape. That reflects the fact that the sign comes just from that of the supercurrents $G^{\pm}(z)$. From the point, it is better to consider the above figures for $\mu<0$ case as ``analytic continuation''. Although it does not manifestly keep such a picture, we will see that it gives a correct description. ]]>

0)\\ \prod_{i=1}^{-\mu}G_{-i-1/2}^-\ket{0}\quad({\rm for\ }\mu<0). \end{cases} \ea \sloppy{As pointed out in~\cite{Prochazka-ml-2017qum}, it corresponds to the primary field $(\partial^{|\mu|-1}G^{\pm}(\partial^{|\mu|-2}G^{\pm}(\cdots$ $(\partial G^{\pm}G^{\pm})\cdots)))(z)$. This gives further confirmation of our convention of $\U(1)$ current.} ]]>

0$ (resp. $\mu<0$) which describe the action of the supercurrents to the vacuum. These legs have the dual descriptions as in section~\ref{subsec:truncation}, which impose further constraints on the plane partition. For $\mu>0$ case, the leg $(\mu)$ is translated to the height one Young table of the following~shape, $$ \begin{tikzpicture} \footnotesize \draw[->] (0,0)--(0,-4); \draw[->] (0,0)--(5.3,0); \draw (0,-0.5)--(5,-0.5); \draw (1,-0.5)--(1,-1)--(5,-1); \draw (2,-1)--(2,-1.5)--(5,-1.5); \draw (3,-1.5)--(3,-2)--(5,-2); \draw (4,-2)--(4,-2.5)--(5,-2.5); \node (3) at (5.6,0) {$2$}; \node (2) at (0,-4.4) {$1$}; \node (dot) at (4.9,-2.8) {$\vdots$}; \fill[pattern=north east lines] (0,0)--(0,-0.5)--(5,-0.5)--(5,0)--cycle; \fill[pattern=north east lines] (1,-0.5)--(1,-1)--(5,-1)--(5,-0.5)--cycle; \fill[pattern=north east lines] (2,-1)--(2,-1.5)--(5,-1.5)--(5,-1)--cycle; \fill[pattern=north east lines] (3,-1.5)--(3,-2)--(5,-2)--(5,-1.5)--cycle; \fill[pattern=north east lines] (4,-2)--(4,-2.5)--(5,-2.5)--(5,-2)--cycle; \draw (5,-3.5) arc(270:340:0.5 and 1); \draw (5,0) arc(90:20:0.5 and 1); \node (m) at (5.5,-1.5) {$\mu$}; \draw (0,-0.5) arc(180:240:0.5 and 0.25); \draw (1,-0.5) arc(360:300:0.5 and 0.25); \node (m) at (0.5,-0.7) {\tiny $2$}; \end{tikzpicture} $$ Obviously, one can not regard it as a Young diagram with asymptotes unless we fill the vacant boxes located on the left side of the leg factors. It implies that the plane partition with the simple leg $(\mu)$ becomes null and we have to add $2+4+\cdots+2(\mu-1)$ boxes in order to realize the non-vanishing states. That can also be seen by explicitly computing the norm. See appendix~\ref{app:a} for detail. Similarly for $\mu<0$, the infinite leg with the shape $(|\mu|,|\mu|)$ is translated into the following height one diagram, $$ \begin{tikzpicture} \footnotesize \node (3) at (5.6,0) {$2$}; \node (2) at (0,-4.8) {$1$}; \draw[->] (0,0)--(2); \draw[->] (0,0)--(3); \draw (1,0)--(1,-4); \draw (1,-0.5)--(2,-0.5)--(2,-4); \draw (2,-1)--(3,-1)--(3,-4); \draw (3,-1.5)--(4,-1.5)--(4,-4); \node (dot) at (4.5,-3.5) {$\dots$}; \fill[pattern=north east lines] (0,0)--(0,-4)--(1,-4)--(1,0)--cycle; \fill[pattern=north east lines] (1,-0.5)--(1,-4)--(2,-4)--(2,-0.5)--cycle; \fill[pattern=north east lines] (2,-1)--(2,-4)--(3,-4)--(3,-1)--cycle; \fill[pattern=north east lines] (3,-1.5)--(3,-4)--(4,-4)--(4,-1.5)--cycle; \draw (0,-4) arc(180:240:2 and 0.5); \draw (5,-4) arc(360:300:2 and 0.5); \node (m) at (2.8,-4.5) {$|2\mu|$}; \draw (0,0) arc(180:120:0.5 and 0.25); \draw (1,0) arc(0:60:0.5 and 0.25); \node (m) at (0.5,0.2) {\tiny $2$}; \draw (1,-0.5) arc(270:330:0.15 and 0.3); \draw (1,0) arc(90:30:0.15 and 0.3); \node (m) at (1.2,-0.25) {\tiny $1$}; \end{tikzpicture} $$ Again the translated legs are not consistent as a Young diagram unless we fill the vacant boxes on the top of the shaded region. The necessity of adding extra boxes are essential to reproduce the CFT parameters and the characters of the irreducible representations. The distinction between the different representation is described by the modification of the leg factor in $x_1$ direction.\footnote{One cannot add the infinite leg in $x_3$ direction since there is a pit at $(1,1,2)$. $x_2$ direction is used to describe the intermediate Young diagrams. The freedom exists only in $x_1$ direction.} ]]>

0$, the infinite legs in the picture with a pit at $(1,1,2)$ appears as $$ \begin{tikzpicture} \footnotesize \draw[->] (0,0)--(0,-4); \draw[->] (0,0)--(5.3,0); \draw (0,-0.5)--(5,-0.5); \draw (1,-0.5)--(1,-1)--(5,-1); \draw (2,-1)--(2,-1.5)--(5,-1.5); \draw (3,-1.5)--(3,-2)--(5,-2); \draw (4,-2)--(4,-2.5)--(5,-2.5); \draw (0.5,0)--(0.5,-3.7); \node (3) at (5.6,0) {$2$}; \node (2) at (0,-4.4) {$1$}; \node (dot) at (4.9,-2.8) {$\vdots$}; \fill[pattern=north east lines] (0,0)--(0,-0.5)--(5,-0.5)--(5,0)--cycle; \fill[pattern=north east lines] (1,-0.5)--(1,-1)--(5,-1)--(5,-0.5)--cycle; \fill[pattern=north east lines] (2,-1)--(2,-1.5)--(5,-1.5)--(5,-1)--cycle; \fill[pattern=north east lines] (3,-1.5)--(3,-2)--(5,-2)--(5,-1.5)--cycle; \fill[pattern=north east lines] (4,-2)--(4,-2.5)--(5,-2.5)--(5,-2)--cycle; \fill[pattern=north east lines] (0.5,-0.5)--(0,-0.5)--(0,-3.6)--(0.5,-3.6)--cycle; \draw (5,-3.5) arc(270:340:0.5 and 1); \draw (5,0) arc(90:20:0.5 and 1); \node (m) at (5.5,-1.5) {$\mu$}; \end{tikzpicture} $$ As in $l=m=0$ case, we have to fill the open boxes on the right of $\mu$ infinite rows which amounts to $\sum_{i=1}^{\mu-1}(2i-1)=(\mu-1)^2$ boxes. When $\mu<0$, the state can be identified as follows: $$ \begin{tikzpicture} \footnotesize \node (3) at (5.6,0) {$2$}; \node (2) at (0,-4.8) {$1$}; \draw[->] (0,0)--(2); \draw[->] (0,0)--(3); \draw (1,0)--(1,-4); \draw (1,-0.5)--(2,-0.5)--(2,-4); \draw (3,-1.5)--(4,-1.5)--(4,-4); \draw (4,-2)--(4.5,-2)--(4.5,-4); \node (dot) at (2.5,-2.5) {$\dots$}; \fill[pattern=north east lines] (0,0)--(0,-4)--(1,-4)--(1,0)--cycle; \fill[pattern=north east lines] (1,-0.5)--(1,-4)--(2,-4)--(2,-0.5)--cycle; \fill[pattern=north east lines] (3,-1.5)--(3,-4)--(4,-4)--(4,-1.5)--cycle; \fill[pattern=north east lines] (4,-2)--(4,-4)--(4.5,-4)--(4.5,-2)--cycle; \draw (0,-4) arc(180:240:2 and 0.5); \draw (4,-4) arc(360:300:2 and 0.5); \node (m) at (2.1,-4.5) {\scriptsize $-2\mu$}; \draw (4,-4) arc(180:240:0.25 and 0.25); \draw (4.5,-4) arc(360:300:0.25 and 0.25); \node (m) at (4.25,-4.5) {\scriptsize $1$}; \draw (0,0) arc(180:120:0.5 and 0.25); \draw (1,0) arc(0:60:0.5 and 0.25); \node (m) at (0.5,0.2) {\tiny $2$}; \draw (1,-0.5) arc(270:330:0.15 and 0.3); \draw (1,0) arc(90:30:0.15 and 0.3); \node (m) at (1.2,-0.25) {\tiny $1$}; \end{tikzpicture} $$ The number of boxes which we have to fill is $\mu^2$. $\U(1)$ charge is computed from~(\ref{eq:formula}) and~(\ref{eq:u1normalization}) as \begin{equation} J=\mu-\frac{1}{3}. \end{equation} We claim that the conformal dimension after filling the boxes is \ba \label{eq:coformaldim1} h=\frac{3\mu^2}{2}-\mu+\frac{1}{3}. \ea We demonstrate the formula in $\mu>0$ case. First, the contribution from the infinite leg is $\frac{1}{3}$ by~(\ref{eq:formula}). Second, the contribution from the intermediate Young diagram is $\frac{\mu^2}{2}+\mu$ as in~(\ref{eq:totaldim}). Third, we need to subtract the number of overlapping boxes from the sum of each factor, which is the box at the origin in $Y_{1,2,0}$ side in this case. Finally, we have to fill $(\mu-1)^2$ boxes. The computation for $\mu<0$ case is similar. From~(\ref{eq:coformaldim1}), we see that the highest weight state is in $\mu=0$ whose conformal dimension and $\U(1)$ charge are $h=\frac{1}{3}$ and $J=-\frac{1}{3}$ respectively. To explain the discrepancy of the conformal weight in the minimal model, we need to decouple $\U(1)$ factor. It is given by~(\ref{eq:smallj}), and its Virasoro algebra is as usual given by \ba \label{eq:u1sugawara} L_0^{\U(1)}=\frac{(\psi_1^{(1)}-\psi_1^{(2)})^2}{2(\psi_0^{(1)}+\psi_0^{(2)})}. \ea In the case under consideration, \ba L_0^{\U(1)}=\frac{1}{4h_1^2\psi_0^{(1)}}=\frac{h_2h_3}{4\psi_0^{(1)}\sigma_3h_1}=-\frac{\lambda_1^{(1)}}{4\lambda_2^{(1)}\lambda_3^{(1)}}=\frac{1}{6}. \ea After removing the contribution from $\U(1)$ factor, the conformal dimension and $\U(1)$ charge of the highest weight state are equal to \ba h=\frac{1}{6},\quad J=-\frac{1}{3}, \ea which agree with the unitary minimal model. The character can be computed in the similar way with the vacuum case. \begin{equation} \begin{aligned} \chi(\tau,z)&=\frac{\sum_{\mu=-\infty}^{\infty}q^{\frac{3}{2}(\mu-\frac{1}{3})^2}y^{\mu-\frac{1}{3}}}{\eta(\tau)}\cdot\frac{q^{\frac{1}{6}}}{\eta(\tau)}\\ &=\frac{(\Theta_{-2,6}(\tau,\frac{z}{3})+\Theta_{4,6}(\tau,\frac{z}{3}))}{\eta(\tau)}\cdot\frac{q^{\frac{1}{6}}}{\eta(\tau)} \end{aligned} \end{equation} which agrees with literature. ]]>

0$ case and $\mu^2-\mu$ in $\mu<0$ case. After short computation, we have \begin{equation} J=\mu-\frac{2}{3},\qquad L_0^{\U(1)}=\frac{2}{3},\qquad h=\frac{3\mu^2}{2}-2\mu+\frac{4}{3}. \end{equation} We see that the highest weight state corresponds to $\mu=1$. The conformal dimension and $\U(1)$ charge after decoupling $\U(1)$ factor is $h=\frac{1}{6}$ and $J=\frac{1}{3}$, and these results indeed correspond to the primary field in NS sector. The character given below agrees with the~literature. \begin{equation} \begin{aligned} \chi(\tau,z)&=\frac{\sum_{\mu=-\infty}^{\infty}q^{\frac{3}{2}(\mu-\frac{2}{3})^2}y^{\mu-\frac{2}{3}}}{\eta(\tau)}\cdot\frac{q^{\frac{2}{3}}}{\eta(\tau)}\\ &=\frac{(\Theta_{2,6}(\tau,\frac{z}{3})+\Theta_{-4,6}(\tau,\frac{z}{3}))}{\eta(\tau)}\cdot\frac{q^{\frac{2}{3}}}{\eta(\tau)}. \end{aligned} \end{equation} ]]>

] (0,0)--(0,-3.7); \draw[->] (0,0)--(5.3,0); \draw (0,-0.5)--(5,-0.5); \draw (1,-0.5)--(1,-1)--(5,-1); \draw (3,-1.5)--(3,-2)--(5,-2); \draw (4,-2)--(4,-2.5)--(5,-2.5); \draw (0.5,-0.5)--(0.5,-3.5); \draw (1,-1)--(1,-3.5); \node (3) at (5.6,0) {$2$}; \node (2) at (0,-4) {$1$}; \node (dot) at (4,-1.2) {$\vdots$}; \fill[pattern=north east lines, opacity=0.5] (0,0)--(0,-0.5)--(5,-0.5)--(5,0)--cycle; \fill[pattern=north east lines, opacity=0.5] (1,-0.5)--(1,-1)--(5,-1)--(5,-0.5)--cycle; \fill[pattern=north east lines, opacity=0.5] (3,-1.5)--(3,-2)--(5,-2)--(5,-1.5)--cycle; \fill[pattern=north east lines, opacity=0.5] (4,-2)--(4,-2.5)--(5,-2.5)--(5,-2)--cycle; \fill[pattern=north east lines, opacity=0.5] (0,-0.5)--(0.5,-0.5)--(0.5,-3.5)--(0,-3.5)--cycle; \fill[pattern=north east lines, opacity=0.5] (0.5,-0.5)--(1,-0.5)--(1,-3.5)--(0.5,-3.5)--cycle; \draw (5,-2) arc(270:340:0.3 and 0.6); \draw (5,0) arc(90:20:0.3 and 0.6); \node (m) at (5.4,-1) {$k$}; \node (n1) at (2.5,-0.3) {$n$}; \node (n2) at (3.1,-0.8) {$n$}; \node (n3) at (4,-1.75) {$n$}; \node (n4) at (4.5,-2.3) {$r$}; \node (nu1) at (0.25,-2) {$\nu_1$}; \node (nu2) at (0.75,-2) {$\nu_2$}; \end{tikzpicture} ]]>

] (0,0)--(22); \draw[->] (0,0)--(33); \draw (1,0)--(1,-4); \draw (1,-0.5)--(2,-0.5)--(2,-4); \draw (3,-1.5)--(4,-1.5)--(4,-4); \draw (4,-2)--(5,-2)--(5,-4); \draw (4.5,-2)--(4.5,-4); \node (dot) at (2.5,-2.5) {$\dots$}; \fill[pattern=north east lines, opacity=0.5] (0,0)--(0,-4)--(1,-4)--(1,0)--cycle; \fill[pattern=north east lines, opacity=0.5] (1,-0.5)--(1,-4)--(2,-4)--(2,-0.5)--cycle; \fill[pattern=north east lines, opacity=0.5] (3,-1.5)--(3,-4)--(4,-4)--(4,-1.5)--cycle; \fill[pattern=north east lines, opacity=0.5] (4,-2)--(4,-4)--(5,-4)--(5,-2)--cycle; \draw (0,-4) arc(180:240:1.5 and 0.5); \draw (4,-4) arc(360:300:1.5 and 0.5); \node (m) at (2,-4.5) {\scriptsize$2(-k-1)$}; \draw (0,0) arc(180:120:0.5 and 0.25); \draw (1,0) arc(0:60:0.5 and 0.25); \node (m) at (0.5,0.2) {\tiny $2$}; \draw (1,-0.5) arc(270:330:0.15 and 0.3); \draw (1,0) arc(90:30:0.15 and 0.3); \node (m) at (1.2,-0.25) {\tiny $1$}; \node (n1) at (0.5,-1.8) {n}; \node (n2) at (1.5,-2.1) {n}; \node (n3) at (3.5,-2.7) {n}; \node (4) at (4.25,-3.05) {\scriptsize\rotatebox{-90}{$n+\nu_1-r$}}; \node (5) at (4.75,-3.05) {\scriptsize\rotatebox{-90}{$n+\nu_2-r$}}; \node (i) at (2.5,-6) {(i) $\nu_2\leq\nu_1\leq r$}; \begin{scope}[xshift=7.8cm] \node (33) at (6.3,0) {$2$}; \node (22) at (0,-4.8) {$1$}; \draw[->] (0,0)--(22); \draw[->] (0,0)--(33); \draw (1,0)--(1,-4); \draw (1,-0.5)--(2,-0.5)--(2,-4); \draw (3,-1.5)--(4,-1.5)--(4,-4); \draw (4,-2)--(5,-2)--(5,-4); \draw (4.5,-2)--(4.5,-4); \draw (5,-2.5)--(5.5,-2.5)--(5.5,-4); \node (dot) at (2.5,-2.5) {$\dots$}; \fill[pattern=north east lines, opacity=0.5] (0,0)--(0,-4)--(1,-4)--(1,0)--cycle; \fill[pattern=north east lines, opacity=0.5] (1,-0.5)--(1,-4)--(2,-4)--(2,-0.5)--cycle; \fill[pattern=north east lines, opacity=0.5] (3,-1.5)--(3,-4)--(4,-4)--(4,-1.5)--cycle; \fill[pattern=north east lines, opacity=0.5] (4,-2)--(4,-4)--(5,-4)--(5,-2)--cycle; \fill[pattern=north east lines, opacity=0.5] (5,-2.5)--(5.5,-2.5)--(5.5,-4)--(5,-4)--cycle; \draw (0,-4) arc(180:240:1.5 and 0.5); \draw (4,-4) arc(360:300:1.5 and 0.5); \node (m) at (2,-4.5) {\scriptsize$2(-k-1)$}; \draw (0,0) arc(180:120:0.5 and 0.25); \draw (1,0) arc(0:60:0.5 and 0.25); \node (m) at (0.5,0.2) {\tiny $2$}; \draw (1,-0.5) arc(270:330:0.15 and 0.3); \draw (1,0) arc(90:30:0.15 and 0.3); \node (m) at (1.2,-0.25) {\tiny $1$}; \node (n1) at (0.5,-1.8) {n}; \node (n2) at (1.5,-2.1) {n}; \node (n3) at (3.5,-2.7) {n}; \node (n4) at (4.25,-2.9) {n}; \node (4) at (5.25,-3.35) {\scriptsize\rotatebox{-90}{$\nu_1-r$}}; \node (5) at (4.75,-3.05) {\scriptsize\rotatebox{-90}{$n+\nu_2-r$}}; \node (i) at (2.5,-6) {(ii) $\nu_2\leq r<\nu_1$}; \end{scope} \begin{scope}[xshift=15.5cm] \node (33) at (6.6,0) {$2$}; \node (22) at (0,-4.8) {$1$}; \draw[->] (0,0)--(22); \draw[->] (0,0)--(33); \draw (1,0)--(1,-4); \draw (1,-0.5)--(2,-0.5)--(2,-4); \draw (3,-1.5)--(4,-1.5)--(4,-4); \draw (4,-2)--(5,-2)--(5,-4); \draw (5,-2.5)--(6,-2.5)--(6,-4); \draw (5.5,-2.5)--(5.5,-4); \node (dot) at (2.5,-2.5) {$\dots$}; \fill[pattern=north east lines, opacity=0.5] (0,0)--(0,-4)--(1,-4)--(1,0)--cycle; \fill[pattern=north east lines, opacity=0.5] (1,-0.5)--(1,-4)--(2,-4)--(2,-0.5)--cycle; \fill[pattern=north east lines, opacity=0.5] (3,-1.5)--(3,-4)--(4,-4)--(4,-1.5)--cycle; \fill[pattern=north east lines, opacity=0.5] (4,-2)--(4,-4)--(5,-4)--(5,-2)--cycle; \fill[pattern=north east lines, opacity=0.5] (5,-2.5)--(6,-2.5)--(6,-4)--(5,-4)--cycle; \draw (0,-4) arc(180:240:1.5 and 0.5); \draw (4,-4) arc(360:300:1.5 and 0.5); \node (m) at (2,-4.5) {\scriptsize$2(-k-1)$}; \draw (0,0) arc(180:120:0.5 and 0.25); \draw (1,0) arc(0:60:0.5 and 0.25); \node (m) at (0.5,0.2) {\tiny $2$}; \draw (1,-0.5) arc(270:330:0.15 and 0.3); \draw (1,0) arc(90:30:0.15 and 0.3); \node (m) at (1.2,-0.25) {\tiny $1$}; \node (n1) at (0.5,-1.8) {n}; \node (n2) at (1.5,-2.1) {n}; \node (n3) at (3.5,-2.7) {n}; \node (n4) at (4.5,-3) {n}; \node (4) at (5.25,-3.25) {\scriptsize\rotatebox{-90}{$\nu_1-r$}}; \node (5) at (5.75,-3.25) {\scriptsize\rotatebox{-90}{$\nu_2-r$}}; \node (i) at (2.5,-6) {(iii) $r<\nu_2\leq\nu_1$}; \end{scope} \end{tikzpicture} ]]>

0$, the number of necessary boxes is
\ba
\sum_{i=1}^{k-1}\Bigl(2n(i-1)-(\nu_1+\nu_2)\Bigr)+(2k-2)r+(r-\nu_1)\theta(r>\nu_1)+(r-\nu_2)\theta(r>\nu_2),
\ea
where $\theta(P)$ is equal to 1 if $P$ is true and 0 if false. The last three terms correspond to boxes inserted in $(k+1)$th row. In the following, we set for simplicity
\ba
\theta=(r-\nu_1)\theta(r>\nu_1)+(r-\nu_2)\theta(r>\nu_2).
\ea
After adding boxes, the conformal dimension and $\U(1)$ charge (including the contribution from $Y_{1,0,0}$ side and decoupling $\U(1)$ factor) are computed from~(\ref{eq:totaldim}) and the above discussion as
\begin{align}
h&=\frac{-2\nu_1\nu_2+\nu_1+(2n+3)\nu_2}{2(n+2)}+\frac{n(n+2)}{2}k^2+\Bigl((n+2)r-(\nu_1+\nu_2)\Bigr)k
+\frac{r^2}{2}-r+\theta,\nonumber\\
J&=kn+r-\frac{\nu_1+\nu_2}{n+2}.
\end{align}
Note that we need to consider the overlapping boxes in the above computation. One can check that this expression is also true of $k<0$ case. In terms of the parameters $l,m$ introduced in $(\ref{eq:n2primary})$, it can be rewritten as follows:
\ba
\label{eq:619}
\begin{cases}
\displaystyle h=\frac{n(n+2)}{2}\biggl(k+\frac{m+(r-1)(n+2)}{n(n+2)}\biggr)^2+\frac{l(l+2)}{4(n+2)}-\frac{(m+2(r-1))^2}{4n}+\theta\\[3mm]
\displaystyle J=\frac{m}{n+2}+kn+r-1&\hspace{-3cm}\raisebox{0pt}{($\nu_2\geq1$, or $l\neq-m$),}\\
\\[5mm]
\displaystyle h=\frac{n(n+2)}{2}\biggl(k+\frac{m+r(n+2)}{n(n+2)}\biggr)^2+\frac{l(l+2)}{4(n+2)}-\frac{(-l+2r)^2}{4n}+(r-l)\theta(r>l)\\[3mm]
\displaystyle J=\frac{m}{n+2}+kn+r&\hspace{-3cm}\raisebox{0pt}{($\nu_2=0$, or $l=-m$).}
\end{cases}
\ea
We can obtain the character by multiplying the factor from the highest weight~(\ref{eq:619}) by the character of the plane partition obtained after the process of adding boxes.
To do that, we need to know the explicit expression for the plane partition's character.\footnote{We note that we need to consider the periodicity rule here.} As we have already seen, it corresponds to the character for some degenerate module of $W_n$ ($+ \U(1)$) minimal model up to the factor of the highest weight. In the language of section~\ref{sec:w-minimal}, it is parametrized by $p=n+1$ and $q=n+2$. One can also see it as the character of
para-fermion ($+ \U(1)$ factor).\footnote{The relation between $W$ algebra and para-fermion was studied in~\cite{Quijano1,Quijano2}.} Let's clarify which module corresponds to the given asymptotic Young diagram from the viewpoint of para-fermion. We first note that we can adjust the asymptotic Young diagram in $x_2$ direction to trivial one; one can change the order of Young diagrams in figure~\ref{fig:minimalW} cyclically so that $\tilde{n}_i'=0$ for $i=1,\cdots n-1$, so we only have to consider the asymptotic condition in $x_1$ direction.
When the asymptotic Young diagram is parametrized by $(l,m)$ in the way of~(\ref{eq:mapping}),
its conformal dimension (after decoupling $\U(1)$ factor) is,
\ba
\label{eq:paraweight}
h=\begin{cases}
h_{l,m-2}^{\rm PF}&(l\neq-m)\\
h_{l,m}^{\rm PF}&(l=-m)
\end{cases}
\ea
where
\ba
\label{eq:paraferdim}
h_{l,m}^{\rm PF}=\frac{l(l+2)}{4(n+2)}-\frac{m^2}{4n}.
\ea
The character for the product of this module and $\U(1)$ factor is given by string function $c_{l,m}^{(n)}(\tau)$ (see appendix~\ref{app:stringfunc} for the definition). To avoid confusion, we denote the parameters associated with SCA by $(l,m)_{\rm SCA}$ and the one associated with para-fermion by $(l,m)_{\rm PF}$ in the following.
The remaining thing we have to do is to clarify to which module the plane partition obtained after adding boxes to figure {~\ref{fig:identification1}},{~\ref{fig:identification2}} corresponds.
If $k<0$, the configuration is, up to shift of the origin, the plane partition with the asymptotic Young diagram in $x_1$ direction with two rows whose length are $n\theta(\nu_1

n$ become~null. ]]>

0),\quad J_0^+\ket{l}=0,\quad J_0^3\ket{l}=\frac{l}{2}\ket{l}. \ea The parameter $l$ can take the integer value satisfying $0\leq l\leq n$. In each value of $l$, there are several primary fields $\ket{l,m}\propto(J_0^-)^{\frac{l-m}{2}}\ket{l}$ for $-l\leq m\leq l$ and $l-m\equiv 0 \ ({\rm mod} 2)$. It is decomposed into the primary field of parafermion and that of $\widehat{\rm U}(1)$. The conformal dimension of the primary field for parafermion parametrized by $(l,m)$ can be computed by Sugawara construction and then obtain~(\ref{eq:paraferdim}). There are highest weight states for $\widehat{\rm U}(1)$ also in the descendant of $\widehat{\rm SU}(2)_n$. If the state has $\widehat{\rm U}(1)$ charge $\frac{j}{2}$ satisfying $j\equiv m$ (mod $n$), the parafermion parametrized by $(l,m)$ acts on it. This should be interpreted in the meaning of~(\ref{eq:paraidentification}) if $m$ takes the value out of the region $-l\leq m\leq l$. The character can be computed by decomposing $\widehat{\rm SU}(2)_n$ character $\chi^{\SU(2)}_{l,n}(\tau,z):={\rm Tr}\ q^{L_0-\frac{c}{24}}y^{J_0^3}$ ($q={\rm e}^{2\pi i\tau},y={\rm e}^{2\pi iz}$) into the parafermion character $\chi_{l,m}^{\rm PF}(\tau)$ and $\U(1)$ character as follows: \ba \chi_{l,n}^{\SU(2)}(\tau,z)=\sum_{m\in\mathbb{Z}_n}\chi_{l,m}^{\rm PF}(\tau)\frac{\Theta_{m,n}(\tau,z)}{\eta(\tau)}, \ea where \ba \label{eq:etatheta} \eta(\tau)=q^{\frac{1}{24}}\prod_{i=1}^{\infty}(1-q^i),\quad\Theta_{m,n}(\tau,z)=\sum_{k=-\infty}^{\infty}q^{n(k+\frac{m}{2n})^2}y^{n(k+\frac{m}{2n})}. \ea String function is defined by \ba \chi_{l,n}^{\SU(2)}(\tau,z)=\sum_{m\in\mathbb{Z}_n}c_{l,m}^{(n)}(\tau)\Theta_{m,n}(\tau,z). \ea It corresponds to the character for the product of parafermion and $\U(1)$ factor. As is expected from~(\ref{eq:paraidentification}), it satisfies \ba c_{l,m+2n}^{(n)}(\tau)=c_{n-l,m+n}^{(n)}(\tau)=c_{l,m}^{(n)}(\tau). \ea Ww note that string function is equal to $\frac{1}{\eta(\tau)}$ when $n=1$. ]]>