&&&\boxed{B_{1}}&0&\cdots&0\\\ \vdots&&&0&\boxed{B_{2}}&\cdots&0&&&\vdots\\\
\vdots&&&\vdots&\vdots&\ddots&\vdots&&&\vdots\\\
&&&0&0&\cdots&\boxed{B_{l}}\\\ &&\boxed{A_{k}}&&&&&0\\\
\vdots&\iddots&&&&&&&\ddots&\vdots\\\
\boxed{A_{1}}&0&\cdots&&\cdots&&\cdots&&\cdots&0\end{array}\right).$ (210)

Here, $\tilde{D}$ is the number of equivalent sets at level $N$. One can see
that this is actually equal to the number of BMS states at level $N$, denoted
by $\widetilde{\rm{dim}_{N}}$. Note that this is different from the number of
BMS$\oplus\mathfrak{u}(1)$ descendants at level $N$, ${dim_{N}}$.

The anti-diagonal blocks $A_{k}$ will be square, as they are the inner product
between conjugate states. Due to the nature of the commutation relations
between the generators, one has (for odd levels)

$\boxed{A_{(i)}}=\boxed{A_{(D-i+1)}^{T}}.$ (211)

One can verify this in the case of the level 2 Gram matrix explicitly. Here,
the equivalent sets of states are ordered as (208) and the corresponding Gram
matrix is

$\left(\setcounter{MaxMatrixCols}{14}\begin{smallmatrix}8\Delta^{2}+4\Delta&&\cdots&&\cdots&&\cdots&&\cdots&&\cdots&4\xi
p&2p^{2}&8\xi^{2}\\\
&\ddots&&&\cdots&&\cdots&&\cdots&&\cdots&2\kappa_{P}p&2\kappa_{P}^{2}&2p^{2}\\\
&&\ddots&&&&\cdots&&\cdots&&\cdots&2\kappa_{P}\xi+p^{2}&2\kappa_{P}p&4\xi p\\\
\vdots&\vdots&&\ddots&&&\cdots&&\cdots&2p&4\xi&0&0&0\\\
&&&&\ddots&&\cdots&&\cdots&2\kappa_{P}&2p&0&0&0\\\
\vdots&\vdots&\vdots&\vdots&&4\xi^{2}&2\xi p&2\xi p&p^{2}&0&0&0&0&0\\\
&&&&&4\xi p&p^{2}&2\xi\kappa_{P}&\kappa_{P}p&0&0&0&0&0\\\ &&&&&2\xi
p&2\xi\kappa_{P}&p^{2}&\kappa_{P}p&0&0&0&0&0\\\
&&&&&p^{2}&\kappa_{P}p&\kappa_{P}p&p^{2}&0&0&0&0&0\\\
\vdots&\vdots&\vdots&2Q_{K}&2\mathcal{K}_{1}&0&0&0&0&0&0&0&0&0\\\
&&&4\xi&2Q_{K}&0&0&0&0&0&0&0&0&0\\\ 4\xi
Q_{K}&2\mathcal{K}_{1}Q_{K}&2\mathcal{K}_{1}\xi+Q_{K}^{2}&0&0&0&0&0&0&0&0&0&0&0\\\
2Q_{K}^{2}&2\mathcal{K}_{1}^{2}&2\mathcal{K}_{1}Q_{K}&0&0&0&0&0&0&0&0&0&0&0\\\
8\xi^{2}&2Q_{K}^{2}&4\xi Q_{K}&0&0&0&0&0&0&0&0&0&0&0\end{smallmatrix}\right),$
(212)

which has 4 anti-diagonal blocks $A_{i}$ and one diagonal block $B_{1}$ in the
middle (because of self conjugate set $\\{L_{-1}M_{-1}\\}$).

### Inverse Gram Matrix

Because of the block-wise upper-anti-triangular structure of the odd level
Gram matrix, the inverse Gram matrix will be lower-anti-triangular, with the
elements on the anti diagonal being the inverse of the anti diagonal element
of the Gram matrix. So the inverse Gram matrix will look like

$\bigl{(}K^{(N=\text{odd})}_{\text{BMS$\oplus\mathfrak{u}(1)$}}\bigr{)}^{-1}\equiv
K_{(N=\text{odd})}=\left(\begin{array}[]{ccccc}$0$&\cdots&$0$&\cdots&\framebox{$A_{1}^{-1}$}\\\
$0$&&&\framebox{$A_{2}^{-1}$}&\vdots\\\ \vdots&&\iddots&&\vdots\\\
&\framebox{$A_{D-1}^{-1}$}&&&\\\
\framebox{$A_{D}^{-1}$}&\iddots&\cdots&\cdots&\framebox{$K^{D,D}$}\end{array}\right).$
(213)

Similarly for even levels, the inverse Gram matrix looks like

$K_{(N=\text{even})}=\left(\begin{array}[]{cccccccccc}0&\cdots&\cdots&&\cdots&&0&&\cdots&\framebox{$A_{1}^{-1}$}\\\
\vdots&\ddots&&&&&&&\iddots&\\\
&&\ddots&&&&&\framebox{$A_{k}^{-1}$}&&\vdots\\\
&&&\framebox{$B_{1}^{-1}$}&0&\cdots&0\\\
\vdots&&&0&\framebox{$B_{2}^{-1}$}&\cdots&0&&&\vdots\\\
0&&&\vdots&\vdots&\ddots&\vdots&&&\vdots\\\
&&&0&0&\cdots&\framebox{$B_{l}^{-1}$}\\\
&&\framebox{$A_{k+l}^{-1}$}&&&&&\ddots\\\
\vdots&\iddots&&&&&&&\ddots&\vdots\\\
\framebox{$A_{D}^{-1}$}&&\cdots&&\cdots&&\cdots&&\cdots&\framebox{$K^{D,D}$}\end{array}\right).$
(214)

### Character Computation

Due to the non-orthogonal nature of the Hilbert space, we define the character
of a highest weight state
$\ket{A}=\ket{\Delta_{A},\xi_{A},\mathcal{Q}_{J,A},\mathcal{Q}_{K,A}}$ as

$\chi=\sum_{N,i,j}({K^{(N)}}^{-1})_{ij}\bra{\psi^{(N)}_{i}}\hat{O}\ket{\psi^{(N)}_{j}}=\sum_{N,i,j}({K_{(N)}})_{ij}\hat{O}_{ij},$
(215)

where ${K^{(N)}}^{-1}\equiv K_{(N)}$ is the Gram matrix of the level N
descendants of $\ket{A}$, and the operator
$\hat{O}=e^{2\pi i\sigma(L_{0}-\frac{c_{L}}{2})}e^{2\pi
i\rho(M_{0}-\frac{c_{M}}{2})}e^{2\pi i\gamma J_{0}}e^{2\pi i\delta P_{0}}$.
Take the general descendant state as defined in (203). One can simplify the
character expression by noting the following properties of the generators

$L_{0}\ket{\psi^{(N)}}=(\Delta_{A}+N)\ket{\psi^{(N)}}\ ,\
J_{0}\ket{\psi^{(N)}}=\mathcal{Q}_{J,A}\ket{\psi^{(N)}}\ ,\
P_{0}\ket{\psi^{(N)}}=\mathcal{Q}_{K,A}\ket{\psi^{(N)}}.$ (216)

Using these expressions one obtains

$\hat{O}\ket{\psi^{(N)}}=\bigl{(}q_{L}^{(\Delta_{A}+N-\frac{c_{L}}{2})}q_{M}^{(-\frac{c_{M}}{2})}e^{2\pi
i\gamma\mathcal{Q}_{J,A}}e^{2\pi
i\delta\mathcal{Q}_{K,A}}\bigr{)}q_{M}^{M_{0}}\ket{\psi^{(N)}},$ (217)

where $q_{M}=e^{2\pi i\rho},q_{L}=e^{2\pi i\sigma}$. Thus, one can write the
character of the highest weight state $\ket{A}$ as

$\chi_{A}=\bigl{(}q_{L}^{(\Delta_{A}-\frac{c_{L}}{2})}q_{M}^{(-\frac{c_{M}}{2})}e^{2\pi
i\gamma Q_{J,A}}e^{2\pi i\delta
Q_{K,A}}\bigr{)}\sum_{N,i,j}q_{L}^{N}{K^{(N)}}^{-1}_{ij}\bigl{(}q_{M}^{M_{0}}\bigr{)}_{ij}.$
(218)

This means that the remaining computation boils down to computing the matrix
elements of the operator $q_{M}^{M_{0}}$. In particular, we will see that the
matrix of $q_{M}^{M_{0}}$ will be upper-anti-triangular, just like the Gram
matrix. As a result, only the anti diagonal matrix elements of the matrices
${K^{(N)}}^{-1}$ and $q_{M}^{M_{0}}$ will contribute to the sum in (218).

### Action of $q_{M}^{M_{0}}$ on Equivalent State Blocks

For the following calculation, we will assume $N$ to be odd for simplicity.
The fact that $N$ = even gives the same answer for the character contribution
can be shown in a similar fashion.

We first note that

$[M_{0},L_{-i}]\sim M_{-i}\ ;\ [M_{0},J_{-i}]\sim P_{-i},$ (219)

that is, $M_{0}$ converts $L_{-i}\rightarrow M_{-i},J_{-i}\rightarrow P_{-i}$
and destroys the operators $M_{-i},P_{-i}$. So its effect on a generic state
can be seen as

$\begin{split}&M_{0}\ket{A,N;\overrightarrow{l};\overrightarrow{m};\overrightarrow{j};\overrightarrow{k}}=M_{0}\ket{\psi}\\\
&=\xi_{A}\ket{\psi}+\sum_{i}\bigl{(}C_{i}\ket{A,N;\\{..l_{i}-1,..\\};\\{..,m_{i}+1..\\};\overrightarrow{j};\overrightarrow{k}}\\\
&+D_{i}\ket{A,N;\overrightarrow{l};\overrightarrow{m};\\{..,j_{i}-1,..\\};\\{..,P_{i}+1,..\\}},\end{split}$
(220)

where, in the sum the $M_{0}$ have converted one $L_{-i}\to M_{-i}$ or one
$J_{-i}\to P_{-i}$. Since the states inside the sum on the right-hand-side
have more $M$($P$) operator content, they are not equivalent states of
$\ket{\psi}$. Rather they belong to an equivalent set that is placed to the
‘right ’ of the set $\\{\ket{\psi}\\}$ according to the ordering rule. So we
can write this in terms of equivalent sets as

$q_{M}^{M_{0}}\\{\ket{\psi}\\}=q_{M}^{\xi}\\{\ket{\psi}\\}+\\{\ket{{\psi}_{R}}\\}.$
(221)

That is, $q^{M_{0}}$ acting on some state will give that state, plus a linear
combination of other states which belong to the equivalence sets placed to the
right of $\ket{\psi}$.

For example, let us take the state $J_{-1}P_{-1}\ket{a}$. Then one has

$q_{M}^{M_{0}}J_{-1}P_{-1}\ket{A}=q_{M}^{\xi_{A}}J_{-1}P_{-1}\ket{A}+\log(q_{M})q_{M}^{\xi_{A}}P_{-1}^{2}\ket{A}.$
(222)

Here, the state $P_{-1}^{2}\ket{A}$ is placed to the right of
$J_{-1}P_{-1}\ket{A}$ as per our ordering rule because it has more $M$($P$)
operator content.

### Matrix Structure of ${q_{M}^{M_{0}}}$

We first look at the the action of $q^{M_{0}}$ on the anti-diagonal blocks
(for odd $N$). We take states $\ket{\psi},\ket{\phi}$ from the blocks denoted
by $\\{\ket{\psi}\\},\\{\ket{\phi}\\}$, such that $\bra{\phi}\ket{\psi}$ is
placed in one of the antidiagonal blocks in the Gram matrix. One then has

$\begin{split}\bra{\phi}q_{M}^{M_{0}}\ket{\psi}&=q_{M}^{\xi_{A}}\bra{\phi}\ket{\psi}+\bra{\phi}\ket{\psi_{R}}\\\
&=q_{M}^{\xi_{A}}K_{ij},\end{split}$ (223)

where $K_{ij}=\bra{\phi}\ket{\psi}$, which is an element of the Gram matrix.
Here, we used the fact that the $\bra{\phi}\ket{\psi}$ is an element of one
anti-diagonal block, and $\bra{\phi}\ket{\psi_{R}}$ consists of matrix
elements that are below the anti-diagonal, which are identically zero as one
can see from the Gram matrix (209).
So for the $d^{\textrm{th}}$ anti-diagonal block elements, denoted by
$\\{q_{M}^{M_{0}}(d)\\}_{ij}$

$\\{q_{M}^{M_{0}}(d)\\}_{ij}=q_{M}^{\xi_{A}}\\{K^{(N)}(d)\\}_{ij}=q_{M}^{\xi_{A}}\\{A_{d}\\}_{ij}.$
(224)

Here, ($K^{(N)}(d)$ is the $d^{\textrm{th}}$ anti-diagonal block/submatrix of
the Gram matrix, which is denoted as $A_{d}$ according to (209).)

Next let us look at the lower anti-triangular elements of $q_{M}^{M_{0}}$
(elements below the anti diagonal). We choose $\ket{\phi},\ket{\psi}$ such
that $\innerproduct{\phi}{\psi}$ is below the anti-diagonal. Just like before
one has

$\begin{split}\bra{\phi}q_{M}^{M_{0}}\ket{\psi}&=q_{M}^{\xi_{A}}\bra{\phi}\ket{\psi}+\bra{\phi}\ket{\psi_{R}}.\end{split}$
(225)

However, now, both of the terms fall below the anti-diagonal in the Gram
matrix, so both $\bra{\phi}\ket{\psi}=\bra{\phi}\ket{\psi_{R}}=0$ (see also
(209)).

So we see that the matrix of $q_{M}^{M_{0}}$ is block-wise upper-anti-
triangular, just like the Gram matrix, and its anti-diagonal blocks are
proportional to those of the Gram matrix. So in the expression
$(K_{(N)})_{ij}(q_{M}^{M_{0}})_{ij}$, only the anti-diagonal blocks will
contribute.

### Contribution to Character

From (218), we consider the nontrivial summation, at a given level $N$
(writing ${K^{(N)}}^{-1}$ as $K_{(N)}$ and $q_{M}^{M_{0}}$ as $\hat{O}$ as
shorthand, also taking $N$ to be odd for simplicity)

$\begin{split}\sum_{i,j}(K_{(N)})_{ij}(\hat{O})_{ij}=\sum_{d}\sum_{i^{\prime},j^{\prime}}(K_{(N)}(d))_{i^{\prime}j^{\prime}}(\hat{O}(d))_{i^{\prime}j^{\prime}},\end{split}$
(226)

where we used the fact that the sum had support only from the diagonal blocks.
Here $\sum_{d}$ runs over the diagonal blocks at level $N$, and $K_{(N)}(d)$
and $\hat{O}_{d}$ are the $d^{\textrm{th}}$ diagonal block/submatrix of the
corresponding matrices. Now we use (224) to replace $\hat{O}_{d}$

$\begin{split}\sum_{i,j}(K_{(N)})_{ij}(\hat{O})_{ij}=&\sum_{d}\sum_{i^{\prime},j^{\prime}}(K_{(N)}(d))_{i^{\prime}j^{\prime}}(\hat{O}(d))_{i^{\prime}j^{\prime}}\\\
=&\sum_{d}\sum_{i^{\prime},j^{\prime}}q_{M}^{\xi_{A}}(A_{(D-d+1)}^{-1})_{i^{\prime}j^{\prime}}(A_{d})_{i^{\prime}j^{\prime}}.\end{split}$
(227)

Here, we have used$(K_{(N)}(d))=A_{(D-d+1)}^{-1}=({A_{d}^{T}})^{-1}$ from
(214) and (211).

Now one can insert this into the sum, and obtain

$\begin{split}\sum_{d}\sum_{i^{\prime},j^{\prime}}q_{M}^{\xi_{A}}(K_{(N)}(d))_{i^{\prime}j^{\prime}}(A_{d})_{i^{\prime}j^{\prime}}=&\sum_{d}\sum_{i^{\prime},j^{\prime}}q_{M}^{\xi_{A}}({A_{d}}^{T})^{-1}_{i^{\prime}j^{\prime}}(A_{d})_{i^{\prime}j^{\prime}}\\\
=&\sum_{d}\sum_{i^{\prime},j^{\prime}}q_{M}^{\xi_{A}}({A_{d}})^{-1}_{j^{\prime}i^{\prime}}(A_{d})_{i^{\prime}j^{\prime}}\\\
=&\sum_{d}\sum_{j^{\prime}}q_{M}^{\xi_{A}}\delta_{j^{\prime}j^{\prime}}=q_{M}^{\xi_{A}}\sum_{d}dim(d)=q_{M}^{\xi_{A}}{dim_{N}}.\end{split}$
(228)

Using this and (218), the character of a highest weight state $\ket{A}$
becomes

$\begin{split}\chi_{A}=\bigl{(}q_{L}^{(\Delta_{A}-\frac{c_{L}}{2})}q_{M}^{(\xi_{A}-\frac{c_{M}}{2})}e^{2\pi
i\gamma\mathcal{Q}_{J,A}}e^{2\pi
i\delta\mathcal{Q}_{K,A}}\bigr{)}\sum_{N}q_{L}^{N}{dim_{N}},\end{split}$ (229)

as claimed in (29).

##  Appendix D Character for BMS Doublet States

In this appendix we are calculating the character for BMS doublet (multiplet
for rank 2). Then, we perform the same calculation for the
BMS$\oplus\mathfrak{u}(1)$ doublet.

### Character for BMS doublet

In case of rank two we need to start from two primary operators
$\mathcal{O}_{1}$ and $\mathcal{O}_{2}$. The corresponding primary states will
satisfy the following

$\displaystyle[L_{0},\mathcal{O}_{i}]$
$\displaystyle=\Delta\mathcal{O}_{i},\hskip
14.22636pt[M_{0},\mathcal{O}_{i}]=(\hat{\xi}\mathcal{O})_{i},$

where

$\hat{\xi}=\left(\begin{array}[]{cc}\xi&0\\\ 1&\xi\\\ \end{array}\right).$
(230)

In case of singlets, the lowest level of the Gram matrix was 1, since we had
only one primary to begin with. In this case, however, we are beginning from
two-component primaries resulting in two primary states and hence there will
be a level 0 $2\times 2$ Gramm matrix. As shown in Hao:2021urq , the inner
product of two components of a doublet is

$\displaystyle\bra{\mathcal{O}_{1}}\ket{\mathcal{O}_{1}}=\bra{\mathcal{O}_{2}}\ket{\mathcal{O}_{2}}=0\
,\ \bra{\mathcal{O}_{1}}\ket{\mathcal{O}_{2}}=1.$ (231)

Hence, at level 0, the Gram matrix will be

$\displaystyle{}\mathcal{K}_{0}=\left(\begin{array}[]{cc}0&1\\\ 1&0\\\
\end{array}\right).$ (234)

Now, in case of usual BMS singlets elements of the Gram matrices, for generic
level, are given by

$\displaystyle\bra{\psi}\Big{(}\prod_{i}L_{i}^{n_{i}}\prod_{j}M_{j}^{m_{j}}\Big{)}\Big{(}\prod_{i^{\prime}}M_{-i^{\prime}}^{m^{\prime}_{i^{\prime}}}\prod_{j^{\prime}}L_{-j^{\prime}}^{n^{\prime}_{j^{\prime}}}\Big{)}\ket{\psi}$
$\displaystyle=$
$\displaystyle\bra{\psi}\Big{[}\prod_{i}L_{i}^{n_{i}}\prod_{j}M_{j}^{m_{j}},\prod_{i^{\prime}}M_{-i^{\prime}}^{m^{\prime}_{i^{\prime}}}\prod_{j^{\prime}}L_{-j^{\prime}}^{n^{\prime}_{j^{\prime}}}\Big{]}\ket{\psi}$
$\displaystyle=$ $\displaystyle
F(\Delta,\xi)\bra{\psi}\ket{\psi}=F(\Delta,\xi),$ (235)

where $\ket{\psi}$ is the primary or highest weight state. In case of a
doublet, however, the Gram matrix elements for generic level change to

$\displaystyle\bra{\mathcal{O}_{a}}\Big{(}\prod_{i}L_{i}^{n_{i}}\prod_{j}M_{j}^{m_{j}}\Big{)}\Big{(}\prod_{i^{\prime}}M_{-i^{\prime}}^{m^{\prime}_{i^{\prime}}}\prod_{j^{\prime}}L_{-j^{\prime}}^{n^{\prime}_{j^{\prime}}}\Big{)}\ket{\mathcal{O}_{b}}$
$\displaystyle=$
$\displaystyle\bra{\mathcal{O}_{a}}\Big{[}\prod_{i}L_{i}^{n_{i}}\prod_{j}M_{j}^{m_{j}},\prod_{i^{\prime}}M_{-i^{\prime}}^{m^{\prime}_{i^{\prime}}}\prod_{j^{\prime}}L_{-j^{\prime}}^{n^{\prime}_{j^{\prime}}}\Big{]}\ket{\mathcal{O}_{b}}$
$\displaystyle=$ $\displaystyle
F(\Delta,\xi)\bra{\mathcal{O}_{a}}\ket{\mathcal{O}_{b}}=F(\Delta,\xi)(\mathcal{K}_{0})_{ab},$
(236)

where $\mathcal{K}_{0}$ is given by (234). Hence, each element of the BMS Gram
matrix for the singlet is replaced by a $2\times 2$ block.

Let us suppose that we have the following BMS Gram matrix for the singlet at
level $N$ (the upper-anti-triangular structure was discussed in Bagchi:2019unf
)

$\displaystyle\mathcal{K}^{s}_{N}=\left(\begin{array}[]{ccccccc}\cdots&\cdots&\cdots&C_{1,j}&\cdots&\cdots&C_{1,M}\\\
\cdots&\cdots&\cdots&\cdots&\cdots&C_{2,M-1}&0\\\ \vdots&&&&\iddots&0&0\\\
\vdots&&&\iddots&&\vdots&\vdots\\\ \vdots&&\iddots&&&\vdots&\vdots\\\
\vdots&C_{M-1,2}&0&&&\vdots&\vdots\\\ C_{M,1}&0&0&\cdots&\cdots&\cdots&0\\\
\end{array}\right),$ (244)

where $M=\widetilde{dim_{N}}$ is the number of independent BMS states at level
$N$.

Then, as per our discussion above, the Gram matrix of the doublet case will be

$\displaystyle(\mathcal{K}^{d}_{N})=\left(\begin{array}[]{ccccccc}\cdots&\cdots&\cdots&\framebox{$C_{1,j}$}&\cdots&\cdots&\framebox{$C_{1,M}$}\\\
\cdots&\cdots&\cdots&\cdots&\cdots&\framebox{$C_{2,M-1}$}&0\\\
\vdots&&&&\iddots&0&0\\\ \vdots&&&\iddots&&\vdots&\vdots\\\
\vdots&&\iddots&&&\vdots&\vdots\\\
\vdots&\framebox{$C_{M-1,2}$}&0&&&\vdots&\vdots\\\
\framebox{$C_{M,1}$}&0&0&\cdots&\cdots&\cdots&0\\\ \end{array}\right),$ (252)

where

$\displaystyle\framebox{$C_{i,j}$}=C_{i,j}\mathcal{K}_{0}=C_{i,j}\left(\begin{array}[]{cc}0&1\\\
1&0\\\ \end{array}\right).$ (255)

Using the fact that $\mathcal{K}_{0}$ is its own inverse, we find that the
inverse of Gramm matrix is

$\displaystyle(\mathcal{K}^{d}_{N})^{-1}=\left(\begin{array}[]{ccccccc}0&\cdots&\cdots&0&\cdots&\cdots&\framebox{$C^{-1}_{1,M}$}\\\
0&\cdots&\cdots&\cdots&0&\framebox{$C^{-1}_{2,M-1}$}&\vdots\\\
0&&&&\iddots&\vdots&\vdots\\\ \vdots&&&\iddots&&\vdots&\vdots\\\
\vdots&&\iddots&&&\vdots&\vdots\\\
0&\framebox{$C^{-1}_{M-1,2}$}&&&&\vdots&\vdots\\\
\framebox{$C^{-1}_{M,1}$}&\cdots&\cdots&\framebox{$C_{M,j}$}&\cdots&\cdots&\cdots\\\
\end{array}\right),$ (263)

where,

$\displaystyle(C_{i,j})^{-1}=\frac{1}{C_{i,j}}\left(\begin{array}[]{cc}0&1\\\
1&0\\\ \end{array}\right).$ (266)

Now, we want to determine the trace of the operator $\widehat{O}=e^{2\pi
i\sigma(L_{0}-c_{L}/12)}e^{2\pi i\rho(M_{0}-c_{M}/12)}$ over a primary doublet
and its descendants. In order to do so, one needs to know the action of the
operator $e^{2\pi i\rho M_{0}}\equiv q^{M_{0}}_{M}$ on the primary doublet
which is given by

$\displaystyle q^{M_{0}}_{M}\left(\begin{array}[]{c}\ket{\mathcal{O}_{1}}\\\
\ket{\mathcal{O}_{2}}\\\
\end{array}\right)=q_{M}^{\xi}\left(\begin{array}[]{cc}1&0\\\ \log{q_{M}}&1\\\
\end{array}\right)\left(\begin{array}[]{c}\ket{\mathcal{O}_{1}}\\\
\ket{\mathcal{O}_{2}}\\\ \end{array}\right).$ (273)

On the right-hand-side of the above expression, $\hat{\xi}$ has been
exponentiated as given in (230). For a BMS singlet, the operator
$q_{M}^{M_{0}}$ acts on a generic descendant as

$\displaystyle q_{M}^{M_{0}}\ket{\psi}=q_{M}^{\xi}\ket{\psi}+\ket{\psi_{R}}.$

Here too, one has

$\displaystyle{}q_{M}^{M_{0}}\ket{\psi_{a}}=q_{M}^{\xi}\ket{\psi_{a}}+\ket{\psi_{R,a}},$
(274)

where $\ket{\psi_{a}}$ is descendant of the doublet $\ket{\mathcal{O}_{a}}$

$\displaystyle\ket{\psi_{1}}$
$\displaystyle=\Big{(}\prod_{i}L^{n_{i}}_{-i}\prod_{j}M^{m_{j}}_{-j}\Big{)}\ket{\mathcal{O}_{1}},$
$\displaystyle\ket{\psi_{2}}$
$\displaystyle=\Big{(}\prod_{i^{\prime}}L^{n_{i^{\prime}}}_{-i^{\prime}}\prod_{j^{\prime}}M^{m_{j^{\prime}}}_{-j^{\prime}}\Big{)}\ket{\mathcal{O}_{2}}.$
(275)

Then, $q_{M}^{M_{0}}$ acts on such descendant states in the following manner

$\displaystyle{}q_{M}^{M_{0}}\ket{\psi_{1}}$
$\displaystyle=q_{M}^{\xi}\ket{\psi_{1}}+\ket{\psi_{R,1}},$ $\displaystyle
q_{M}^{M_{0}}\ket{\psi_{2}}$ $\displaystyle=\log{q_{M}}\hskip
1.42262ptq_{M}^{\xi}\ket{\psi_{1}}+q_{M}^{\xi}\ket{\psi_{2}}+\ket{\psi_{R,2}}.$
(276)

Similarly to the singlet case one has anti-diagonal blocks in the gram matrix
(the only difference here is that each anti-diagonal element of the singlet
Gram matrix will be replaced by $2\times 2$ blocks as in (255)). Let us define
two descendants states $\ket{\psi_{a}}$ and $\ket{\phi_{b}}$ in such a way
that $\bra{\phi_{b}}\ket{\psi_{a}}$ belongs to the anti-diagonal block of the
Gram matrix. As before, one has

$\displaystyle{}\bra{\phi_{b}}q_{M}^{M_{0}}\ket{\psi_{a}}=\bra{\phi_{b}}q_{M}^{\hat{\xi}}\ket{\psi_{a}}.$
(277)

In the above expression, we have used the fact that
$\bra{\phi_{b}}\ket{\psi_{R,a}}=0$, since they are below the anti-diagonal
blocks of the Gram matrices. Writing (277) component-wise, one obtains

$\displaystyle\bra{\phi_{1}}q_{M}^{M_{0}}\ket{\psi_{1}}$
$\displaystyle=q_{M}^{\xi}\bra{\phi_{1}}\ket{\psi_{1}}=q_{M}^{\xi}(C_{\phi,\psi})_{1,1}\,,$
$\displaystyle\bra{\phi_{1}}q_{M}^{M_{0}}\ket{\psi_{2}}$
$\displaystyle=\log{q_{M}}\hskip
2.84526ptq_{M}^{\xi}(C_{\phi,\psi})_{1,1}+q_{M}^{\xi}(C_{\phi,\psi})_{1,2}\,,$
(278) $\displaystyle\bra{\phi_{2}}q_{M}^{M_{0}}\ket{\psi_{1}}$
$\displaystyle=q_{M}^{\xi}(C_{\phi,\psi})_{2,1}\,,$
$\displaystyle\bra{\phi_{2}}q_{M}^{M_{0}}\ket{\psi_{2}}$
$\displaystyle=\log{q_{M}}\hskip
2.84526ptq_{M}^{\xi}(C_{\phi,\psi})_{2,1}+q_{M}^{\xi}(C_{\phi,\psi})_{2,2}\,.$

Following (255) one has

$\displaystyle(C_{\phi,\psi})_{1,1}$ $\displaystyle=(C_{\phi,\psi})_{2,2}=0\
,\ (C_{\phi,\psi})_{1,2}=(C_{\phi,\psi})_{2,1}=1.$ (279)

Hence, the entries of $q_{M}^{M_{0}}$ of the singlet picture, for any level,
are being replaced by $2\times 2$ blocks as given below

$\displaystyle(q_{M}^{M_{0}})_{\phi,\psi}=q_{M}^{\xi}\bra{\phi}\ket{\psi}\left(\begin{array}[]{cc}0&1\\\
1&\log{q_{M}}\end{array}\right).$ (282)

The matrix form of $q_{M}^{M_{0}}$ looks like

$\displaystyle
q_{M}^{M_{0}}=q_{M}^{\xi}\left(\begin{array}[]{ccccccc}\cdots&\cdots&\cdots&\framebox{$\widehat{C}_{1\hskip
1.42262ptj}$}&\cdots&\cdots&\framebox{$\widehat{C}_{1\hskip 1.42262ptM}$}\\\
\cdots&\cdots&\cdots&\cdots&\cdots&\framebox{$\widehat{C}_{2\hskip
1.42262pt(M-1)}$}&0\\\ \vdots&&&&\iddots&0&0\\\
\vdots&&&\iddots&&\vdots&\vdots\\\ \vdots&&\iddots&&&\vdots&\vdots\\\
\vdots&\framebox{$\widehat{C}_{(M-1)\hskip 1.42262pt2}$}&0&&&\vdots&\vdots\\\
\framebox{$\widehat{C}_{M\hskip 1.42262pt1}$}&0&0&\cdots&\cdots&\cdots&0\\\
\end{array}\right),$ (290)

where

$\displaystyle\widehat{C}_{ij}=C_{ij}\left(\begin{array}[]{cc}0&1\\\
1&\log{q_{M}}\end{array}\right).$ (293)

The character of the highest weight doublet, is

$\displaystyle\chi=\sum_{N,i,j}(\mathcal{K}^{d}_{N})^{-1}_{ij}\bra{\phi^{N}_{i}}\widehat{O}\ket{\phi^{N}_{j}}.$
(294)

As argued previously, one can see that only the anti-diagonal blocks of both
$(\mathcal{K}_{N})^{-1}$ and $q_{M}^{M_{0}}$ contribute. Using (263), (266),
(282),(290) and (293), the character of a primary doublet
$\\{\mathcal{O}_{1},\mathcal{O}_{2}\\}$ with weights ($\Delta_{A},\xi_{A}$)
turns out the be

$\displaystyle\chi^{\\{\mathcal{O}_{1},\mathcal{O}_{2}\\}}_{A}$
$\displaystyle=\sum_{N}q_{M}^{\Delta_{A}-\frac{c_{L}}{2}}q_{M}^{-\frac{c_{M}}{2}}\sum_{a,i,j}q_{M}^{\xi_{A}}(C_{a,M-a})\left(\begin{array}[]{cc}0&1\\\
1&\log{q_{M}}\end{array}\right)_{ij}\frac{1}{(C_{a,M-a})}\left(\begin{array}[]{cc}0&1\\\
1&0\end{array}\right)_{ij}$ (299)
$\displaystyle=q_{M}^{\Delta_{A}-\frac{c_{L}}{12}}q_{M}^{\xi_{A}-\frac{c_{M}}{12}}\sum_{N}q^{N}_{L}\hskip
1.42262pt(2\widetilde{dim_{N}}),$ (300)

where $\widetilde{dim_{N}}$ represents the number of independent BMS
descendant states at level $N$. Hence, the character contribution at any level
of the doublet primary states, is twice the contribution of the singlet
states. The factor of 2 represents the fact that we have taken a doublet
primary state.

### Character for BMS$\oplus\mathfrak{u}(1)$ Doublet

In the case of pure BMSFT it was found that $L_{0}$ and $M_{0}$ cannot be
simultaneously diagonalized. Hence, one had to start from a basis where
$L_{0}$ was diagonalized but $M_{0}$ had a Jordan cell structure. As
elucidated previously, for BMS$\oplus$ $\mathfrak{u}(1)$, we work in a basis
where $L_{0}$, $P_{0}$ and $J_{0}$ are simultaneously diagonalized but $M_{0}$
has a Jordan structure (see (64), (65), (71)). Hence, the algorithm for
computing the character of a BMS$\oplus\mathfrak{u}(1)$ doublet primary is
very similar to that of BMS doublet primary. The only difference here will be
due to the presence of the charges $P_{0}$ and $J_{0}$. As a result, (D) will
be replaced by

$\displaystyle\ket{\psi_{1}}$
$\displaystyle=\Big{(}\prod_{p}L^{n_{p}}_{-p}\prod_{q}M^{m_{q}}_{-q}\prod_{r}P^{k_{r}}_{-r}\prod_{s}J^{l_{s}}_{-s}\Big{)}\ket{\mathcal{O}_{1}},$
$\displaystyle\ket{\psi_{2}}$
$\displaystyle=\Big{(}\prod_{p^{\prime}}L^{n^{\prime}_{p^{\prime}}}_{-p^{\prime}}\prod_{q^{\prime}}M^{m^{\prime}_{q^{\prime}}}_{-q^{\prime}}\prod_{r^{\prime}}P^{k^{\prime}_{r^{\prime}}}_{-r^{\prime}}\prod_{s^{\prime}}J^{l^{\prime}_{s^{\prime}}}_{-s^{\prime}}\Big{)}\ket{\mathcal{O}_{2}}.$
(301)

Hence, each element of the BMS$\oplus\mathfrak{u}(1)$ Gram matrices is given
by the following expression

$\displaystyle\bra{\mathcal{O}_{a}}\Big{(}\prod_{s}J_{s}^{l_{s}}\prod_{r}P_{r}^{k_{r}}\prod_{q}M^{m_{q}}_{q}\prod_{p}L^{n_{p}}_{p}\Big{)}\Big{(}\prod_{p^{\prime}}L_{-p^{\prime}}^{n^{\prime}_{p^{\prime}}}\prod_{q^{\prime}}M_{-q^{\prime}}^{m^{\prime}_{q^{\prime}}}\prod_{r^{\prime}}P^{k^{\prime}_{r^{\prime}}}_{-r^{\prime}}\prod_{s^{\prime}}J^{l^{\prime}_{s^{\prime}}}_{-s^{\prime}}\Big{)}\ket{\mathcal{O}_{b}}$
$\displaystyle=$
$\displaystyle\bra{\mathcal{O}_{a}}\Big{[}\prod_{s}J_{s}^{l_{s}}\prod_{r}P_{r}^{k_{r}}\prod_{q}M^{m_{q}}_{q}\prod_{p}L^{n_{p}}_{p},\prod_{p^{\prime}}L_{-p^{\prime}}^{n^{\prime}_{p^{\prime}}}\prod_{q^{\prime}}M_{-q^{\prime}}^{m^{\prime}_{q^{\prime}}}\prod_{r^{\prime}}P^{k^{\prime}_{r^{\prime}}}_{-r^{\prime}}\prod_{s^{\prime}}J^{l^{\prime}_{s^{\prime}}}_{-s^{\prime}}\Big{]}\ket{\mathcal{O}_{b}}$
$\displaystyle=$ $\displaystyle
F(\Delta,\xi,p,j)\bra{\mathcal{O}_{a}}\ket{\mathcal{O}_{b}}=F(\Delta,\xi,p,j)(\mathcal{K}_{0})_{ab}.$
(302)

This means again that each element of the Gram matrix is getting replaced by 2
$\times$ 2 block. The action of $q_{M}^{M_{0}}$ on the primary and descendant
states, however, will be same as (273) and (D), respectively. Hence, the
matrix form of $q_{M}^{M_{0}}$ will have a structure similar to (290) and
(293). The operator whose trace we are going to calculate this time will be

$\displaystyle\widehat{O}=e^{2\pi i\sigma(L_{0}-c_{L}/12)}e^{2\pi
i\rho(M_{0}-c_{M}/12)}e^{2\pi i\alpha P_{0}}e^{2\pi i\gamma J_{0}}.$ (303)

Hence, following the same algorithm for BMS$\oplus\mathfrak{u}(1)$ one ends up
with the following expression for the character of a primary doublet labeled
by ($\Delta_{A},\xi_{A},p_{A},j_{A}$),

$\displaystyle\chi^{\\{\mathcal{O}_{1},\mathcal{O}_{2}\\}}_{A}=2q_{L}^{\Delta_{A}-\frac{c_{L}}{12}}q_{M}^{\xi_{A}-\frac{c_{M}}{12}}q_{K}^{p_{A}}q_{J}^{j_{A}}\sum_{N}q^{N}_{L}\hskip
1.42262pt{dim_{N}}.$ (304)

Here $dim_{N}$ is the number of independent BMS$\oplus\mathfrak{u}(1)$
descendants at level $N$ (which is different from $\widetilde{dim_{N}}$).
Hence, again, the character contribution of the doublet primary states for any
level turns out to be twice than the one of the singlet.

##  Appendix E Character for $r$-Multiplets

As already mentioned previously, in the space of multiplets the action of
$L_{0},J_{0},P_{0}$ are diagonal, but the action of $M_{0}$ is given by the
$r\times r$ dimensional Jordan matrix. The operator relations for the primary
fields $\\{\mathcal{O}_{i}\\}$ of the $r$-multiplet is given by

$\displaystyle[L_{0},\mathcal{O}_{i}]=\Delta\mathcal{O}_{i},\hskip
14.22636pt[M_{0},\mathcal{O}_{i}]=(\hat{\xi}_{r}\mathcal{O})_{i},\hskip
14.22636pt[P_{0},\mathcal{O}_{i}]=p,\mathcal{O}_{i}\hskip
14.22636pt[J_{0},\mathcal{O}_{i}]=j\mathcal{O}_{i},$ (305)

where highest weight conditions are

$\displaystyle[L_{n},\mathcal{O}_{i}]=0,\hskip
14.22636pt[M_{n},\mathcal{O}_{i}]=0,\hskip
14.22636pt[P_{n},\mathcal{O}_{i}]=0,\hskip
14.22636pt[J_{n},\mathcal{O}_{i}]=0,\hskip 14.22636pt\forall n>0.$ (306)

Here, $\hat{\xi}_{r}$ is a Jordan cell of rank $r$

$\displaystyle\hat{\xi}_{r}=\left(\begin{array}[]{ccccc}\xi&0&\cdots&&0\\\
1&\xi&0&\cdots\vdots\\\ 0&1&\xi&\ddots&\vdots\\\ \vdots&\vdots&&\ddots&0\\\
0&\cdots&&1&\xi\\\ \end{array}\right)_{r\times r}.$ (312)

The highest weight states of the multiplet follow the inner product relation
Hao:2021urq

$\mathcal{K}^{r}_{0}=\left(\begin{array}[]{cccc}0&\dots&\dots&1\\\
0&\dots&1&0\\\ \vdots&\iddots&\dots&0\\\ 1&\dots&\dots&0\\\
\end{array}\right).$ (313)

One can also obtain the action of $q_{M}^{M_{0}}$ within a multiplet as

$(q_{M}^{M_{0}})_{multiplet}=(q_{M})^{\hat{\xi}_{r}}=\left(\begin{array}[]{ccccc}(q_{M})^{\xi}&0&\cdots&\cdots&0\\\
\log(q_{M})(q_{M})^{\xi}&(q_{M})^{\xi}&0&\cdots&0\\\
\frac{1}{2!}(\log(q_{M}))^{2}(q_{M})^{\xi}&\log(q_{M})(q_{M})^{\xi}&(q_{M})^{\xi}&0&\vdots\\\
\vdots&\vdots&\vdots&\ddots&0\\\
\frac{1}{(r-1)!}(\log(q_{M}))^{r}(q_{M})^{\xi}&\frac{1}{(r-2)!}(\log(q_{M}))^{r-1}(q_{M})^{\xi}&\dots&\log(q_{M})(q_{M})^{\xi}&(q_{M})^{\xi}\\\
\end{array}\right).\\\ $ (314)

Alternatively, one can write

$(q_{M}^{M_{0}})_{multiplet}=(q_{M})^{\hat{\xi}_{r}}=(q_{M})^{\xi}\mathcal{Q}^{r}_{0}.$
(315)

Here,

$\mathcal{Q}^{r}_{0}=\left(\begin{array}[]{ccccc}1&0&\cdots&\cdots&0\\\
\log(q_{M})&1&0&\cdots&0\\\
\frac{1}{2!}(\log(q_{M}))^{2}&\log(q_{M})&1&0&\vdots\\\
\vdots&\vdots&\vdots&\ddots&0\\\
\frac{1}{(r-1)!}(\log(q_{M}))^{r}&\frac{1}{(r-2)!}(\log(q_{M}))^{r-1}&\dots&\log(q_{M})&1\\\
\end{array}\right).$ (316)

Using the Gram matrix of descendants of a singlet BMS primary (denoted by
${K^{(1)}}_{(N)}$, where $1$ denotes singlet), and the inner product matrix of
the multiplets $\mathcal{K}^{r}_{0}$ (or the zero level Gram matrix), one can
write the $N$-level Gram matrix of the multiplet primaries as a simple tensor
product of the two matrices,

${K^{(r)}}_{(N)}={K^{(1)}}_{(N)}\otimes\mathcal{K}^{r}_{0}.$ (317)

This will be an ($rN\times rN$)-dimensional matrix, and each element of the
singlet Gram matrix will be replaced by the multiplet inner product matrix
$\mathcal{K}_{0}$,

$(K^{(1)}_{(N)})_{ij}\to\framebox{$\bigl{(}K^{(r)}_{(N)}\bigr{)}_{ij}$}=(K^{(1)}_{(N)})_{ij}\times\mathcal{K}^{r}_{0}.$
(318)

The action of $q_{M}^{M_{0}}$ can also be written as

$q_{M(r)}^{M_{0}}=q_{M(1)}^{M_{0}}\otimes\bigl{(}\mathcal{K}^{r}_{0}\mathcal{Q}^{r}_{0}\bigr{)}.$
(319)

Here, 1 and $r$ in the subscript mean singlet and r-multiplet respectively.

Every element of the previously calculated singlet matrix of $(q_{M})^{M_{0}}$
will be replaced by the nontrivial matrix to take into account the multiplet
structure of the Hilbert space. One can verify this result by looking at the
case of $r=2$.

Writing the new matrices in terms of tensor products makes computing the
character very straightforward, as we will show below. One has

$({K^{(r)}}_{(N)})^{-1}=({K^{(1)}}_{(N)})^{-1}\otimes\mathcal{[}{K}^{r}_{0}]^{-1}=({K^{(1)}}_{(N)})^{-1}\otimes\mathcal{K}^{r}_{0},$
(320)

where we have used $\mathcal{[}{K}^{r}_{0}]^{-1}=\mathcal{K}^{r}_{0}$.

Now, let us go back to the nontrivial part of the character contribution which
is given in terms of the following expression

$\begin{split}\sum_{i,j}\bigl{(}{K^{(r)}}_{(N)}\bigr{)}^{-1}_{ij}((q_{M})^{M_{0}}_{r})_{ij}&=\sum_{i,j}\Bigl{(}({K^{(1)}}_{(N)})^{-1}\otimes\mathcal{K}^{r}_{0}\Bigr{)}_{ij}\Bigl{(}q_{M(1)}^{M_{0}}\otimes\bigl{(}\mathcal{K}^{r}_{0}\mathcal{Q}^{r}_{0}\bigr{)}\Bigr{)}_{ij}\\\
&=\Bigl{[}\sum_{i^{\prime},j^{\prime}}\bigl{(}{K^{(1)}}_{(N)}\bigr{)}^{-1}_{i^{\prime}j^{\prime}}((q_{M})^{M_{0}}_{(1)})_{i^{\prime}j^{\prime}}\Bigr{]}\times\Bigl{[}\sum_{k,l}\bigl{(}(\mathcal{K}^{r}_{0})_{kl}(\mathcal{K}^{r}_{0}\mathcal{Q}^{r}_{0})_{kl}\bigr{)}\Bigr{]}\\\
&=q_{M}^{\xi}(dim_{N})\times\Bigl{(}Tr\bigl{(}(\mathcal{K}^{r}_{0})^{2}\mathcal{Q}^{r}_{0}\bigr{)}\Bigr{)}=q_{M}^{\xi}(dim_{N})\times\Bigl{(}Tr\bigl{(}\mathcal{Q}^{r}_{0}\bigr{)}\Bigr{)}\\\
&=q_{M}^{\xi}(dim_{N})\times r.\end{split}$ (321)

In the third line, the first expression is the usual expression for the
singlet case, which we have derived in (228). Thus, we have replaced it by
$q_{M}^{\xi}dim_{N}$. Also in the calculation, we used that
$\mathcal{K}^{r}_{0}$ is symmetric, and $(\mathcal{K}^{r}_{0})^{2}=I$.

Since the other operators are diagonal in the multiplet Hilbert space, the
character formula is given by

$\chi_{(r)}^{\\{\mathcal{O}_{i}\\}}=r\bigl{(}q_{L}^{(\Delta-\frac{c_{L}}{2})}q_{M}^{(\xi-\frac{c_{M}}{2})}e^{2\pi
i\gamma j}e^{2\pi i\alpha p}\bigr{)}\sum_{N}q_{L}^{N}\times({dim_{N}}).$ (322)

This expression is same as (229), except for a factor of $r$. For $r=2$ one
again obtains (304).

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