# Bootstrapping Multilingual Intent Models
via Machine Translation for Dialog Automation

Nicholas Ruiz, Srinivas Bangalore, John Chen
Interactions, LLC
Murray Hill, NJ, USA
<EMAIL_ADDRESS>
###### Abstract

With the resurgence of chat-based dialog systems in consumer and enterprise
applications, there has been much success in developing data-driven and rule-
based natural language models to understand human intent. Since these models
require large amounts of data and in-domain knowledge, expanding an equivalent
service into new markets is disrupted by language barriers that inhibit dialog
automation.

This paper presents a user study to evaluate the utility of out-of-the-box
machine translation technology to (1) rapidly bootstrap multilingual spoken
dialog systems and (2) enable existing human analysts to understand foreign
language utterances. We additionally evaluate the utility of machine
translation in human assisted environments, where a portion of the traffic is
processed by analysts. In English$\rightarrow$Spanish experiments, we observe
a high potential for dialog automation, as well as the potential for human
analysts to process foreign language utterances with high accuracy.

##  1 Introduction

With the present advances in natural language understanding and speech
recognition technologies, unprecedented opportunities have been created for
realizing natural and sophisticated human-machine conversations to accomplish
routine tasks. There has been a resurgence of speech/text-based conversation
systems spanning multiple platforms, such as interactive voice recordings,
chat, and SMS, owing to the availability of communication platforms that make
it convenient to configure human-machine conversations. The potential
opportunities of speech/text driven human-machine systems, or virtual agents,
can only be realized if the user’s requests are understood by the virtual
agent and acted upon appropriately.

For practical applications, such conversational agents and speech/text
analytics systems, the meaning of a sentence may be approximated as one or
more actionable labels, or intents, associated with the input utterance. In
such cases, the natural language understanding (NLU) task is modeled as an
intent classification problem. Although ambiguity is present in natural
language, data-driven NLU systems have been successful in modeling user
intents in many application domains.

Many commercial and enterprise applications service customers from different
geographic locations and varying language proficiencies, requiring
multilingual NLU for human-machine interaction. In order to deploy an intent
classification system for a new language a new set of labeled training data is
conventionally required. This data is often unavailable before a solution is
deployed, instead requiring a human-driven dialog system depending on intent
analysts or live agents. In time, a sufficient amount of production data may
be collected to build a data-driven intent model; however this approach is
expensive to operate and ignores valuable knowledge present in other languages
that could be used to build an initial model.

In this paper, we evaluate the use of machine translation (MT) as a tool to
bridge the knowledge present in one or more intent models for the creation of
an intent classifier in a target language. Given MT’s capability to translate
the content of utterances in a source language to a target language, our goal
is to minimize the number of language proficient intent analysts needed to
support a production-scale multilingual dialog system in the absence of target
language data.

The remainder of the paper is organized as follows: in Section 2 we discuss
the details of the intent classification system and present the NLU model,
including our two MT architectures in a multilingual spoken dialog system. In
Section 3 we outline our experiment and describe our ASR, MT, and NLU models.
In Sections 4-7, we evaluate the ASR, MT, and NLU model performance. In
Section 8 we evaluate human agents’ ability to label the intents of translated
user utterances and summarize our findings in Section 10.

##  2 NLU for Customer Care

A single utterance is tagged with three types of labels: intents, entities,
and conversational handlers. Intents are domain-specific labels such as sales,
tech assistance, and billing. Entity labels represent the names of products or
services mentioned by the user. These include specific models of smart phones
or subscription services. Conversational handlers are labels which are similar
to speech acts to guide the conversation. For example, live agent, confused,
or foreign language. In our experiments, intents and entities are labeled as
“session variables” (SV), while conversation handles are partitioned into
“task names” (TN) and “event names” (EN). Examples of each in our experiments
are shown in Table 1.

A joint SVM classifier is trained by concatenating the TN, EN, and SV labels
into a unique label. The model comprises a set of binary SVM classifiers, with
each classifier predicting if the input is assigned or not assigned to a
particular label type. For a given input utterance $x$, the joint label is
computed as:

$y^{*}=\operatorname*{arg\,max}_{y\in\langle tn,sv,en\rangle}F_{y}(x,y).$ (1)

The feature set F may comprise application context, conversation context, and
the utterance. We use $n$-gram word-level features in this experiment.

Task Names (TN) | Session Variables (SV) | Event Names (EN)
---|---|---
COMPLAINT | ACCOUNT action | ANGRY
ENGLISH | ACTIVATE product | DON’T KNOW
FOREIGN | ADD service | GARBLED
NONE | APPOINTMENT type | LIVE AGENT
QUESTION | BILLING AUTO PAY | NOISE
… | BILLING DETAILS | NO MATCH
| PAY BILL service | NONE
| CHANGE ADDRESS | THANK YOU
| DISCONNECT service | …
| … |
10 | 707 | 18

Table 1: Examples and counts of Spanish intent labels by category and language
for the “How may I help you?” dialog state. Concatenating the labels yields
833 distinct intent annotations.

###  2.1 Confidence Measures

We boost the accuracy of our intent models by using human analysts’
predictions on unconfident decisions made by the classifier. We obtain
prediction probabilities from the classifier by computing the sigmoid on the
scores output for each label by the SVM classifier, computed as:

$P(y^{*})=\frac{1}{1+\exp(F_{y}(x,y^{*}))}.$ (2)

The confidence measure is obtained by computing the ratio of the probabilities
of the first and second best labels assigned by the classifier:

$cf(y^{*})=\frac{P(y^{*})}{P(y^{*-1})}.$ (3)

The rejection threshold is empirically determined to maximize the accuracy of
the human-assisted solution while minimizing human labeling costs.

Spanish audio Spanish ASR ASR OK? Spanish-English MT MT OK? Spanish IA English
NL NL OK? English IA Spanish Intent yesyesnononoyes

Figure 1: Online Spanish$\rightarrow$English bootstrap architecture. Spanish
audio is translated into English and processed by an English intent
classifier. If ASR or MT scores are not confident, a Spanish intent analyst
(IA) labels the segment. If the English intent model is not confident, an
English IA labels it.

###  2.2 Multilingual Bridging via Machine Translation

In the context of dialog systems, MT can be used in one of two ways: (1)
translating real-time target language data into the source language and
predicting the intent with a source language intent classification model (cf.
Fig. 1); or (2) translating source language data offline into the target
language and training a target language intent classification model. In the
first scenario, utterances with high translation quality may be processed by
an analyst that only speaks the source language, if the NLU confidence score
is too low.

##  3 Experimental setup

We evaluate the efficacy of bootstrapping a Spanish intent classifier using
the data and underlying models from an English spoken language dialog system.
The data set consists of customer voice responses to the message “How may I
help you?” in the customer’s native language at the beginning of a phone call
to an Interactive Voice Response (IVR) system. We assume that no Spanish
intent data is available during training time and evaluate the performance of
our bootstrapped Spanish models against an intent model trained with a
standard training set. Table 2 lists the data used in our experiment, The
training data consists of 5-best ASR hypotheses on audio segments for English
and Spanish. We assume that the intent labels covered by the target Spanish
model are the same as the English model. Although most of the intent labels
overlap one another there is a subset of $\langle tn,sv,en\rangle$ intent
triples that do not overlap. We discard the non-training examples with non-
overlapping triples from each data set, reducing the number of unique labels
in both data sets to 623 (cf. Table 1). As a result, 2.94% of the English
training examples and 1.95% of the Spanish training examples are discarded,
respectively.

Data set | # utts | # words | # unique labels
---|---|---|---
English training | 6.5M | 40.6M | 623
Spanish training | 0.8M | 4.6M | 623
Spanish test (ASR) | 1007 | 4696 | 178
Spanish test (human) | 1007 | 5153 | 178

Table 2: Utterance, word, and distinct label counts for the training and test
data used in our experiments.

Model performance is evaluated on a test set of 1007 Spanish audio segments
randomly extracted from production logs and labeled by Spanish-speaking intent
analysts (IAs) and verified by a supervisor.

###  3.1 Automatic Speech Recognition

Our Spanish ASR system consists of an n-gram language model and hybrid DNN
acoustic model trained with the cross-entropy criterion followed by the state-
level Minimum Bayes Risk (sMBR) objective. We use sequence-training with
smoothing and speed-perturb the training data. The acoustic model has general-
phone and head-body-tail based digit-specific triphones. The training data
consists of 500K training utterances (around 1000 hours of audio) without
speech perturbations and about 40K unique vocabulary words.

###  3.2 Machine Translation

We use a conventional neural machine translation (NMT) sequence-to-sequence
encoder-decoder with attention architecture [Bahdanau et al., 2015, Luong and
Manning, 2015, Sennrich et al., 2016] commonly used by MT practitioners. The
NMT models were trained with parallel English-Spanish data from Europarl v7,
CommonCrawl, and WMT News Commentary v8 from the WMT 2013 evaluation campaign
[Bojar et al., 2013], as well as the TED talks from IWSLT 2014 [Cettolo et
al., 2014]. The training data has a shared vocabulary size of 89,500 words
after byte-pair encoding [Sennrich et al., 2016]. The model is trained for 20
epochs with two bidirectional LSTM encoding and decoding layers with 512
units. In this experiment we assume to have no in-domain parallel data.

For the offline English$\rightarrow$Spanish model, we translate the English
intent model’s training data (6.5 million utterances) into Spanish using our
baseline NMT system. The number of words in the translated data set remains
roughly the same. The translated outputs are used to train a bootstrapped
Spanish intent classifier, using the same training parameters as the native
English model. The ASR outputs from the test set are processed by the
bootstrapped Spanish intent classifier. For the real-time
Spanish$\rightarrow$English model, insert punctuation and apply truecasing to
the ASR outputs from the test set and translate the outputs with our
Spanish$\rightarrow$English baseline NMT system. We strip the punctuation and
lowercase the machine translated output and pass it through the native English
intent classifier.

###  3.3 Intent classification

Our intent classifiers are trained using an implementation of SVMs in scikit-
learn111http://scikit-learn.org, using the approach described in Section 2. We
evaluate the performance of an intent classification model by plotting an
error-rejection curve, which measures the error rate of the intent classifier
as the number of utterances that are processed by the intent analysts
increases. For example, a 10% rejection rate corresponds means that only 90%
of the test set is evaluated by the model.

We compare the results of each bootstrapped NLU approach with a native Spanish
intent model trained on the held-out Spanish training data. We additionally
repeat the experiment with the human transcripts to measure the difference in
intent classification error that may be explained by ASR. Error-rejection
curves for each intent model are shown in Fig. 2 and the scores at 0%, 10%,
and 20% rejection are shown in Table 4.

##  4 ASR performance

We use sclite from the NIST Speech Recognition Scoring
Toolkit222https://www.nist.gov/itl/iad/mig/tools to compute the word error
rate (WER) and utterance error rate. After further clean-up and adjudication,
we observe a 33.2% WER, with 60.0% of the utterances containing errors (32%
substitutions, 22% deletions, 28% insertions). A majority of the substitution
errors were confusions between singular and plural (e.g.
problemas$\rightarrow$problema) and articles (e.g. del$\rightarrow$de). Other
errors included phonetic confusions (e.g. cuenta$\rightarrow$fuenta), named
entity misrecognitions (e.g. HBO$\rightarrow$yo), and a high frequency of
dropped articles (e.g. de, la, a) caused by speaker under-articulation. Of
these types of errors, the most detrimental are substitutions of named
entities, verbs, and nouns that are not lemmatization errors. Another issue
driving up the WER score caused by the IVR system truncating audio longer than
four seconds to reduce latency.

Translation | BLEU $\uparrow$ | TER $\downarrow$ | Length
---|---|---|---
ASR outputs | 42.5 | 51.2 | 94.1
ASR outputs (+PE) | 48.0 | 44.6 | 96.3
Human transcripts | 51.8 | 41.3 | 111.4
Table 3: Machine translation quality measured in BLEU, TER, and utterance
length, evaluated against post-edited translations of human transcripts.

##  5 Machine Translation quality

In order to assess the translation quality, we post-edit the translations of
the human-transcribed utterances and report the case-insensitive BLEU and TER
scores in Table 3. ASR errors increase the required translation edits by 10%,
from 41.3% edits to 51.2%. The primary sources of errors are incomplete
sentences, lack of punctuation; lexical mistranslations of key words: (e.g.
equipo (equipment)$\rightarrow$team; dirección
(address)$\rightarrow$direction; reclamo (complaint)$\rightarrow$ claim);
ambiguous translations (e.g. factura$\rightarrow$bill/invoice); and duplicated
words during translation (e.g. payment arrangements$\rightarrow$payment
payment). Many of these issues are due to lack of in-domain MT training data
and low tolerance of ASR errors. Highly repetitive errors indicate that an
automatic post-editing system could substantially improve the translation
system’s quality. Table 3 also shows that post-edited translations of ASR
outputs are substantially worse than those of the human transcripts (41.3% TER
difference), showing that ASR errors are exacerbated through translation.

##  6 Native NLU performance

Our reference native Spanish intent classification model is trained on ASR
outputs since an insufficient amount of human-transcribed intent modeling data
is available. From Table 4, we see that at 0% rejection, the native model
yields a 19.2% classification error, while the human intent analysts (IAs)
yielded a 11.0% error while listening directly to the audio. At the same time,
if the IAs are presented with the ASR outputs, they produce an error rate of
24.4%. This demonstrates the intent model’s ability to tolerate a certain
degree of ASR errors by being trained on ASR errors.

Figure 2: Error-rejection curves for bootstrapped intent classification
performance on Spanish ASR outputs, versus a native Spanish model. Results are
reported for intent classifier predictions on ASR outputs.
Spa$\rightarrow$Eng**: performance on post-edited MT outputs.

Of the 1007 test utterances, 152 utterances have both ASR errors and were
classified incorrectly, although they were labeled properly by IAs, comprising
15.1% of the 19.2% intent classification errors. Of the utterances with ASR
errors that cause an intent classification error, six were cases where ASR
failed to produce a hypothesis, 26 were cases of audio truncation, and 39
utterances containing only ASR substitution errors. The latter cases often
caused underspecification during intent classification (e.g. lower my
bill$\rightarrow$billing problem; make a payment$\rightarrow$billing). In
isolation, insertion and deletion ASR errors did not have a significant impact
on NLU.

Evaluating NLU performance on human transcripts, we observe improvements that
rival human labeling performance at 10% rejection and above. At 20% rejection,
the model significantly outperforms human labeling.

| Reject
---|---
Configuration | 0% | 10% | 20%
Native Spanish (ASR) | 19.2 | 13.5 | 10.3
Native Spanish (human) | 19.1 | 12.3 | 7.4
Boot Eng$\rightarrow$Spa (ASR) | 35.3 | 30.7 | 25.8
Boot Eng$\rightarrow$Spa (human) | 31.6 | 26.3 | 21.4
Boot Spa$\rightarrow$Eng* (ASR) | 31.8 | 26.5 | 21.4
Boot Spa$\rightarrow$Eng* (human) | 33.4 | 26.7 | 20.2
Boot Spa$\rightarrow$Eng* (ASR+PE) | 28.3 | 23.6 | 18.3
Spanish IA (audio) | 11.0 | 10.6 | 9.5
Spanish IA (human) | 14.9 | 13.1 | 10.9
Spanish IA (ASR) | 24.4 | 21.6 | 17.6
English IA (ASR+MT) | 33.9 | 30.6 | 26.1
English IA (ASR+MT+PE) | 25.9 | 22.8 | 18.3
Table 4: Intent classification performance by machine learning models and
human intent analysts (IAs) at 0%, 10%, and 20% rejection.

##  7 Bootstrapped NLU performance

While the native Spanish model has a 13.5% error rate at 10% rejection when
processing ASR hypotheses, the bootstrapped models have double the error rate
due to the use of out-of-domain machine translation. On ASR hypotheses, the
English$\rightarrow$Spanish model yields an error rate of 30.7%, while the
Spanish$\rightarrow$English model yields 26.5% at 10% rejection.

To better understand how machine translation further corrupts the bootstrapped
Spanish$\rightarrow$English performance, we compare the errors it makes to the
Native Spanish model. In Table 5, we group the NLU errors by whether they are
present in the Native Spanish model, the bootstrapped model, or both. 14.2% of
the test set are examples where the Native Spanish model makes a correct
prediction, but the Spanish$\rightarrow$English model yields errors. By
comparing to post-edited ASR+MT data, only 3% of those errors are directly
attributed to ASR errors. The 11% of Spanish$\rightarrow$English model-
specific errors are mostly attributed to intent underspecification. For
example, 60% of the account errors are null misclassifications. For billing
issues, two-thirds of the errors are semantically similar misclassifications,
such as payment, lower my bill, and bill details.

Native | Spa$\rightarrow$Eng | ASR | ASR+PE |
---|---|---|---|---
+ | + | 66.6% | 69.6% |
+ | - | 14.2% | 11.2% |
- | + | 1.5% | 2.0% |
- | - | 17.7% | 17.2% |
Table 5: Comparison of Native Spanish intent model to bootstrapped
Spanish$\rightarrow$English models on ASR outputs. +/- indicate that whether
the corresponding model’s prediction was correct.

##  8 Analyst performance on translated text

Finally, we measure IA labeling performance on translated utterances. In our
conventional scenario, intent analysts listen to audio segments in their
native language and provide an intent label. Instead, we replace the original
audio with machine translated or translation post-edits of ASR hypotheses.
Fig. 3 provides error-rejection curves for IAs, with error rates at 0%, 10%
and 20% rejection reported in Table 4.

We first assess the labeling loss when humans annotate ASR outputs in the
absence of audio. Although their error rate increases from 11.0% to 24.4% when
annotating ASR transcripts, their performance on human transcripts is within
5% of listening directly to the audio at 0% rejection. As the rejection rate
increases, the difference becomes negligible. As we introduce
Spanish$\rightarrow$English machine translation, we observe that the labeling
error increases from 24.4% to 33.9% on ASR, which is incidentally worse than
the Spanish$\rightarrow$English intent classification model’s accuracy
(31.8%)! However, the IA labeling error rate drops to 25.9% on post-edited MT
outputs – only a 1.5% increase in NLU errors caused by translation. These
results suggest that with proper ASR and MT adaptation through in-domain data,
we could obtain similar English-speaking IA performance on machine translation
outputs as the Spanish-speaking IAs on their native language utterances.

Figure 3: Error-rejection curves for the intent analyst (IA) labeling accuracy
on Spanish audio, Spanish ASR and human transcripts, and their machine
translations into English (MT). Rejection is computed from the English model’s
confidence scores on MT outputs. “+PE”: performance on post-edited MT outputs.

##  9 Related Work

The use of MT to translate texts in other languages into English for sentiment
analysis was proposed in Denecke [Denecke, 2008]. Bautin et al. [Bautin et
al., 2008] show that sometimes MT performs inadequate translations on
essential parts of a text, affecting sentiment analysis performance. Our
results confirm this phenomena due to a lack of in-domain MT training data.
Schwenk and Douze [Schwenk and Douze, 2017] explore learning multilingual
sentence embeddings with neural MT, which can aid in multilingual search.
Prior to that, multilingual approaches leveraged lexical resources such as
MultiWordNet [Pianta et al., 2002] to bridge concepts from one language to
another.

##  10 Conclusions

We have executed an experiment to measure machine translation’s ability to
rapidly bootstrap intent classification models for new languages. In our
English$\rightarrow$Spanish experiments, we observe that although the initial
results appear to be substantially worse than a Native Spanish intent
classification model, we show that MT can provide a degree of automation that
supports human-assisted multilingual dialog systems that can be deployed to
production on day one, reducing the need for human agent support over a fully
manual solution. There is further promise that model improvements can be
obtained by improving the ASR and machine translation models to include in-
domain data. Finally, we observe it is better to use the online
Spanish$\rightarrow$English bootstrap in our production system rather than an
offline English$\rightarrow$Spanish intent model.

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Nahm sums are specific $q$-hypergeometric series associated with symmetric positive definite matrices.
In this paper we study Nahm sums associated with symmetrizable matrices.
We show that one direction of Nahm's conjecture, which was proven by Calegari, Garoufalidis, and Zagier for the symmetric case,
also holds for the symmetrizable case.
This asserts that the modularity of a Nahm sum implies that a certain element in a Bloch group associated with the Nahm sum is a torsion element.
During the proof, we investigate the radial asymptotics of Nahm sums.
Finally, we provide lists of candidates of modular Nahm sums for symmetrizable matrices based on numerical experiments.

§ INTRODUCTION

The Rogers-Ramanujan identities relate infinite sum and infinite product expressions of $q$-series:

\begin{equation}\label{eq:RR sum}
\sum_{n=0}^\infty \frac{q^{n^2}}{(q;q)_n} = \frac{1}{(q, q^4; q^5)_\infty},\quad
\sum_{n=0}^\infty \frac{q^{n^2+n}}{(q;q)_n} = \frac{1}{(q^2, q^3; q^5)_\infty},
\end{equation}

where $(x;q)_n \coloneqq (1-x) (1-q x) \cdots (1-q^{n-1} x)$,
$(x;q)_\infty \coloneqq \prod_{i=0}^\infty (1 - q^i x)$, and
$(x_1,\dots,x_n;q)_\infty \coloneqq (x_1;q)_\infty \cdots (x_n;q)_\infty$.
One of the remarkable consequences of the Rogers-Ramanujan identities is that
the infinite sums in (<ref>), which are $q$-hypergeometric series,
are also modular functions. Specifically, the vector-valued function on the upper-half plane

\begin{align*}
g(\tau) =
\begin{pmatrix}
q^{-1/60} \displaystyle{\sum_{n=0}^\infty \frac{q^{n^2}}{(q;q)_n}}, &
q^{11/60} \displaystyle{\sum_{n=0}^\infty \frac{q^{n^2 + n}}{(q;q)_n}}
\end{pmatrix}^\mathsf{T}
\end{align*}

with $q = e^{2 \pi i \tau}$ has the following transformation formulas:

\begin{align*}
g(\tau + 1) =
\begin{pmatrix}
\zeta_{60}^{-1} & 0 \\
0 & \zeta_{60}^{11}
\end{pmatrix}
g(\tau), \quad
g(-\frac{1}{\tau}) =
\frac{2}{\sqrt{5}}
\begin{pmatrix}
\sin \frac{2 \pi}{5} & \sin \frac{\pi}{5} \\
\sin \frac{\pi}{5} & -\sin \frac{2 \pi}{5}
\end{pmatrix}
\end{align*}

where $\zeta_N = e^{2 \pi i / N}$.

In the study of rational conformal field theories,
Nahm <cit.> considered a higher rank generalization of the infinite sums appearing in the Rogers-Ramanujan identities (<ref>),
which we call Nahm sum.
Let $N$ be a natural number, and suppose that $A \in \mathbb{Q}^{N \times N}$ is a symmetric positive definite matrix,
$B \in \mathbb{Q}^N$ is a vector,
and $C \in \mathbb{Q}$ is a scalar.
The Nahm sum is defined by

\begin{equation}\label{eq:intro nahm}
f_{A, b, c} (q) \coloneqq \sum_{n \in \mathbb{N}^N}
\frac{q^{ \frac{1}{2} n^\mathsf{T} A n + n^\mathsf{T} b + c}}{(q;q)_{n_1} \cdots (q;q)_{n_N}}.
\end{equation}

In general, Nahm sums rarely have modularity.
However, as discussed in <cit.>, a character in rational conformal field theories is often expressed as a Nahm sum
and also possess modularity.
Motivated by such situations, Nahm conjectured that the modularity of
Nahm sums are related to torsion elements of the Bloch group.
Let us explain the conjecture in a little more detail. The matrix $A$ determines the Nahm's equation

\begin{align}\label{eq:intro nahm equation}
1 - z_i = \prod_{j=1}^N z_j^{A_{ij}},
\end{align}

and each solution of this equation determines an element in the Bloch group $B(F)$:

\begin{align}\label{eq:intro xi}
\xi_A \coloneqq \sum_{i=1}^N [z_i] ,
\end{align}

where $F$ is the number field generated by $z_1, \dots, z_N$.
The Nahm's conjecture states that, for any $A$, the Nahm sum $f_{A, B, C}(q)$ is modular
for some $B, C$ if and only if $\xi_A$ is a torsion in $B(F)$ for any solution $z_i$ <cit.>.
Although this conjecture itself is known to be false <cit.>,
Calegari, Garoufalidis, and Zagier <cit.> proved one direction of Nahm's conjecture:
If the Nahm sum $f_{A,B,C}(q)$ is modular, then the element (<ref>) associated with the solution of
Nahm's equation (<ref>)
in the real interval $(0, 1)^N$ is a torsion.
They construct a map

\begin{align*}
R_\zeta : B(F) / m B(F) \to F_m^\times / F_m^{\times m}
\end{align*}

where $m$ is an integer, $\zeta$ is a primitive $m$th root of unity, and $F_m = F(\zeta)$,
and they show that this map is injective for sufficiently many $m$.
The one direction of Nahm's conjecture follows from this injectivity result
combined with the formula of the asymptotic expansion of Nahm sums given in <cit.>.

In this paper, we study Nahm sums associated with symmetrizable matrices, which has the following form:

\begin{equation}\label{eq:intro nahm symmetrizable}
f_{A, b, c, d} (q) \coloneqq \sum_{n \in \mathbb{N}^N}
\frac{q^{ \frac{1}{2} n^\mathsf{T} A D n + n^\mathsf{T} b + c}}{(q^{d_1}; q^{d_1})_{n_1} \cdots (q^{d_N};q^{d_N})_{n_N}}.
\end{equation}

We have a new input $d = (d_1,\dots,d_N) \in \mathbb{Z}_{>0}^{N}$,
and the symmetric matrix $A$ in the original Nahm sum (<ref>) is replaced by a symmetrizable
matrix $A$ with the symmetrizer $D \coloneqq \diag(d_1,\dots,d_N)$.
Specifically, in (<ref>), we require that $A D$ is symmetric positive definite rather than $A$ itself.

Prototypical examples of this type of Nahm sums appear in the Kanade-Russell conjecture <cit.> (in the form by Kurşungöz <cit.>),
which gives the mod $9$ version of the Rogers-Ramanujan type identities.
The conjecture predicts the following $q$-series identities:

f_1 (q)    ∑_n_1,n_2 ≥0 q^n_1^2 + 3 n_1 n_2 + 3 n_2^2 + 0 n_2 + 0 n_2/(q;q)_n_1 (q^3;q^3)_n_2       ?= 1/(q,q^3,q^6,q^8;q^9)_∞,

f_2 (q)    ∑_n_1,n_2 ≥0 q^n_1^2 + 3 n_1 n_2 + 3 n_2^2 + n_1 + 3 n_2/(q;q)_n_1 (q^3;q^3)_n_2       ?= 1/(q^2,q^3,q^6,q^7;q^9)_∞,

f_3 (q)    ∑_n_1,n_2 ≥0 q^n_1^2 + 3 n_1 n_2 + 3 n_2^2 + 2 n_1 + 3 n_2/(q;q)_n_1 (q^3;q^3)_n_2       ?= 1/(q^3,q^4,q^5,q^6;q^9)_∞.

The Nahm sums in the left-hand sides are given by the matrix
$A = \begin{bsmallmatrix}
2 & 1 \\ 3 & 2
\end{bsmallmatrix}$
with the symmetrizer $D = \diag(1,3)$.
The three linear terms are given by $b = \begin{bsmallmatrix} 0 \\ 0 \end{bsmallmatrix}$, $\begin{bsmallmatrix} 1 \\ 3 \end{bsmallmatrix}$,
and $\begin{bsmallmatrix} 2 \\ 3 \end{bsmallmatrix}$, respectively.
Assuming these identities (<ref>)–(<ref>), we see that the vector-valued function

\begin{align*}
g(\tau) =
\begin{pmatrix}
q^{-1/18} f_1 (q) \\
q^{5/18} f_2 (q)\\
q^{11/18} f_3 (q)\\
\end{pmatrix}
\end{align*}

satisfies the following modular transformation formula:

\begin{align}\label{eq:intro KR modular}
g(\tau + 1) =
\begin{pmatrix}
\zeta_{18}^{-1} & 0 & 0 \\
0 & \zeta_{18}^5 & 0 \\
0 & 0 & \zeta_{18}^{11}
\end{pmatrix}
g(-\frac{1}{\tau}) =
\begin{pmatrix}
\alpha_1 & \alpha_2 & \alpha_4 \\
\alpha_2 & -\alpha_4 & -\alpha_1 \\
\alpha_4 & -\alpha_1 & \alpha_2
\end{pmatrix}
\end{align}

where $\alpha_k = \dfrac{1}{2 \sqrt{3} \sin \frac{k \pi}{9}}$.
The formula (<ref>) implies that $g(\tau)$ is a vector-valued modular function on the
congruence subgroup $\Gamma_0(3) \coloneqq \left\{ \begin{bsmallmatrix} a & b \\  c & d \end{bsmallmatrix} \mid c \equiv 0 \bmod 3 \right\} \subseteq \SL(2, \mathbb{Z})$.

Nahm sums for symmetrizable matrices also appear in various areas of mathematics, such as
partition identities <cit.>,
character formulas of principal subspaces for affine Lie algebras of twisted type <cit.>,
and $Y$-systems on cluster algebras associated with skew-symmetrizable matrices <cit.>.

The purpose of this paper is to demonstrate that the theory of Nahm sums for symmetric matrices
can be similarly applied to Nahm sums for symmetrizable matrices.
In section <ref>,
we see that one direction of Nahm's conjecture, proved by Calegari, Garoufalidis, and Zagier in <cit.> for the symmetric case,
also holds for symmetrizable case (Theorem <ref>).
As with the symmetric case, the proof uses the injectivity of the map $R_\zeta: B(F) / m B(F) \to F_m^\times / F_m^{\times m}$.
Assuming the injectivity result, the main step of the proof is to study the asymptotic behavior of the Nahm sum at roots of unity,
which is given in Section <ref>.
We provide the explicit formula for the asymptotic expansion of Nahm sums, which is a generalization of the symmetric case given in <cit.>.
We also need some number-theoretic properties of the constant term in the asymptotic expansion, which are provided in Proposition <ref>.
The main ingredient in the constant term is the cyclic quantum dilogarithm function.
In Appendix <ref>, we provide some properties of the cyclic quantum dilogarithm function that we use in the proof of Proposition <ref>.
Finally, in Section <ref>, we present some experiments on modular Nahm sums for symmetrizable matrices.
We use a numerical method from <cit.> to search for modular Nahm sum for rank $2$ and rank $3$ cases.
The resulting lists of modular candidates are given in Table <ref> for rank $2$ case,
and Table <ref> and <ref> for rank $3$ case.
For rank $2$ we might expect that Table <ref>, together with symmetric case in <cit.>, exhausts all modular Nahm sums.
In addition, many candidates in Table <ref> are already known in the literature, and some of them are known to be modular.
For rank $3$, however, the number of modular candidates explodes. We should emphasize that Table <ref> and <ref> are just the tip of the iceberg.
Providing a complete picture and a systematic understanding are left as a future task.
The only observation on the rank 3 lists in this paper is that there seem to be “Langlands dual” pairs of modular Nahm sums.
As an example, we verify numerically that two Nahm sums associated with
$A = \begin{bsmallmatrix}
1 & 0 & 1/2\\
0 & 2 & 1\\
1 & 2 & 2
\end{bsmallmatrix}$ and
$A^{\vee} = \begin{bsmallmatrix}
1 & 0 & 1/2\\
0 & 2 & 1\\
1 & 2 & 2
\end{bsmallmatrix}$, which are related by taking the transpose, are related by the modular $S$-transformation.

The author would like to thank Shunsuke Tsuchioka for fruitful discussions and for pointing out the identity (<ref>).
This work is supported by JSPS KAKENHI Grant Number JP21J00050.

§ ASYMPTOTICS OF NAHM SUMS

Let $N$ be a natural number.
Suppose that $Q = (A, b, c, d)$ is a quadruple where
$A \in \mathbb{Q}^{N \times N}$ is a matrix,
$b \in \mathbb{Q}^N$ is a vector,
$c \in \mathbb{Q}$ is a scalar,
and $d \in \mathbb{Z}^N$ is a vector,
such that $A D$ is symmetric positive definite where $D = \diag (d_1, \dots, d_N)$.
The Nahm sum associated with $Q$ is a $q$-series defined by

\begin{align}
f_Q (q) = \sum_{n \in \mathbb{N}^N} \frac{q^{Q(n)}}{(q^{d_1}; q^{d_1})_{n_1} \cdots (q^{d_N};q^{d_N})_{n_N}}
\end{align}

where we denote by, abuse of notation,

\begin{align}
Q(n) = \frac{1}{2} n^{\mathsf{T}} A D n + n^\mathsf{T} b + c
\end{align}

the quadratic form associated with $Q = (A, b, c, d)$.
We say that $Q$ is symmetric if $d_i = 1$ for any $i$.
We also use the term symmetrizable to mean not necessarily symmetric.

§.§ Asymptotics at roots of unity

In this section, we study the asymptotics of Nahm sums $f_Q (q)$ at roots of unity.
We think $f_Q (q)$ as a function on the upper half-plane by setting $q^\lambda = e^{2 \pi i \tau \lambda}$
for any $\lambda \in \mathbb{Q}$ and $\tau \in \mathbb{C}$ with $\Im \tau >0$.
We will consider the limit where $\tau$ tends to a rational number from above in the complex plane.

Let $(z_1, \dots, z_N) \in \mathbb{R}^N$ be the unique solution of the Nahm's equation

\begin{align}\label{eq:nahm equation}
1 - z_i = \prod_{j=1}^N z_j^{A_{ij}}
\end{align}

such that $0 < z_i < 1$ for any $i$.
In fact, the existence and the uniqueness are proved in the same way as in the symmetric case
(see e.g. <cit.>)
by using the fact that $A D$ is positive definite.
We define a real number

\begin{align}\label{eq:volume}
\Lambda = - \sum_{i=1}^N d_i^{-1} \mathrm{L}(z_i)
\end{align}

\begin{align}
\mathrm{L}(z) = \Li_2 (z) + \frac{1}{2} \log(z) \log(1-z) - \frac{\pi^2}{6}
\end{align}

is the Rogers dilogarithm function shifted by the constant ${\pi^2}/{6}$ so that $\mathrm{L}(1) = 0$.

For an invertible matrix $A$ and an analytic function $f(x) = f(x_1, \dots, x_N)$, we consider the formal Gaussian integral:

\begin{align}
\mathbf{I}_A [f] = \frac{\int_{\mathbb{R}^N} e^{-\frac{1}{2} x^\mathsf{T} A x} f(x)\, dx}
{\int_{\mathbb{R}^N} e^{-\frac{1}{2} x^\mathsf{T} A x}\, dx}.
\end{align}

For a positive integer $m$, an $m$th root of unity $\zeta$, and a complex number $w$ with $\lvert w \rvert < 1$,
we define a formal power series

\begin{align}
\psi_{w, \zeta} (\nu, \varepsilon) = - \sum_{r\geq2} \sum_{t=1}^{m} \Bigl( B_r \Bigl(1 - \frac{t + \nu}{m} \Bigr) - \delta_{2,r} \frac{\nu^2}{m^2} \Bigr)
\Li_{2-r} (\zeta^t w) \frac{\varepsilon^{r-1}}{r!},
\end{align}

where $B_r(x)$ is the Bernoulli polynomial of degree $r$ defined by $t e^{tx}/(e^t - 1) = \sum_{r=0}^\infty B_r(x) t^r / r !$,
and $\Li_r (w) = \sum_{k=1}^\infty w^k / k^r$ is the polylogarithm function.
We then consider the formal Gaussian integral

\begin{align}\label{eq:def of I}
I_{Q, \zeta}(k, \varepsilon) =
\mathbf{I}_{\frac{1}{m} \widetilde{A} D} \left[ e^{- \frac{1}{m} x^\mathsf{T} b \varepsilon^{1/2} - \frac{1}{m} (c + \frac{\tr D}{24}) \varepsilon}
\prod_{i=1}^{N} \exp \Bigl(\psi_{\zeta_i^{k_i} z_i^{\frac{1}{m}}, \zeta_i} (x_i \varepsilon^{-1/2}, d_i \varepsilon) \Bigr) \right],
\end{align}

for any $k \in (\mathbb{Z}/m \mathbb{Z})^N$ where

\begin{align}
\widetilde{A} = A + \diag (z / (1-z))
\end{align}

and $\zeta_i = \zeta^{d_i}$.
The expression (<ref>) has a well-defined meaning as a formal power series in $\mathbb{C} \llbracket \varepsilon \rrbracket$.

For any coprime integers $p$ and $q$, we define the dedekind sum by

\begin{align}
s (p, q) \coloneqq \sum_{r \bmod q} \dedekind{\frac{r}{q}} \dedekind{\frac{pr}{q}}
\end{align}

where we use the saw wave function defined by

\begin{align}
((x)) =
\begin{cases}
x - \floor{x} - \frac{1}{2} &\text{if $x \notin \mathbb{Z}$} \\
0 &\text{if $x \in \mathbb{Z}$}.
\end{cases}
\end{align}

For example, we have $s(1, m) = \frac{(m-1)(m-2)}{12m}$.

We say that a positive integer $\delta$ is a strong denominator of $Q$ if
the value of $Q(k)$ modulo $1$ for $k \in \mathbb{Z}^N$ depends only on the residue class of $k$ modulo $\delta$.
We define the quadratic Gauss sum

\begin{equation}
G(Q, \alpha) = \frac{1}{\delta^N} \sum_{k \in (\mathbb{Z}/\delta \mathbb{Z})^N} \mathbf{e} (\bar{\alpha} Q(k))
\end{equation}

where $\mathbf{e}(x) = e^{2 \pi i x}$,
$\bar{\alpha}$ is the reduction of $\alpha$ modulo $\delta$,
and $\delta$ is a chosen strong denominator of $Q$.
The $G(Q, \alpha)$ does not depend on the choice of $\delta$.

Let $\tilde{f}_Q(\tau) = f_Q(e^{2 \pi i \tau})$.
We now have the following asymptotic formula for the Nahm sum, whose proof is obtained as a minor modification of the proof for the symmetric case in <cit.>.

Let $Q = (A, b, c, d)$ be a quadruple as above,
and $\delta$ be a strong denominator of $Q$.
Suppose that $\alpha$ is a rational number whose denominator $m$ is prime to $\delta$ and $d_1, \dots, d_N$.
Let $\zeta = \mathbf{e}(\alpha)$ be the corresponding $m$th root of unity.
Let $\theta_i$ be the $m$th root of $z_i$ such that $\theta_i \in (0, 1)$, where
$(z_1, \dots, z_N)$ is the unique solution of the Nahm's equation (<ref>)
such that $z_i \in (0,1)$. Then we have the asymptotic formula

\begin{align}\label{eq:nahm asymptotic}
e^{-\frac{\Lambda}{m\varepsilon}} \tilde{f}_Q \Bigl(\alpha + \frac{i\varepsilon}{2 \pi m} \Bigr) \sim
m^{-\frac{N}{2}} \chi(d, \alpha) c(Q) G(Q, \alpha) S_{Q,\zeta} (\varepsilon)
\end{align}

as $\varepsilon$ tends to $0$ from the right-half in the complex plane.
$\chi(d, \alpha) = \prod_{i=1}^N \chi_i$ is the product of
the $12$th root of $\zeta$ defined by $\chi_i = \mathbf{e}(s(p_i,q_i)/2)$ where $p_i/q_i = \alpha d_i$,

\begin{align}
c(Q) = (\det \widetilde{A})^{- \frac{1}{2}} \prod_{i=1}^{N} \theta_i^{b_i/d_i} (1 - z_i)^{\frac{1}{2} - \frac{1}{m}},
\end{align}

\begin{align}
S_{Q, \zeta}(\varepsilon) =
\biggl(\prod_{i=1}^{N} D_{\zeta_i} (\zeta_i \theta_i)^{- \frac{1}{m}} \biggr)
\sum_{k \in (\mathbb{Z}/m\mathbb{Z})^N} \zeta^{\overline{Q(k)}}
\biggl(\prod_{i=1}^{N} \frac{\theta_i^{(k^\mathsf{T}A)_i}}{(\zeta_i \theta_i ; \zeta_i)_{k_i}}\biggr)  I_{Q, \zeta}(k, \varepsilon),
\end{align}

where $D_{\zeta} (x)$ is the cyclic quantum dilogarithm function defined by (<ref>),
and $\overline{Q(k)}$ is the reduction of $Q(k)$ modulo $m$.

We have corrected the following errors in <cit.>:

* the sign of $- \frac{1}{m} x^\mathsf{T} b \varepsilon^{1/2}$ in (<ref>),

* the missing term $\frac{\tr D}{24}$ (which reduces $\frac{N}{24}$ for symmetric case) in (<ref>),

* the absence of the Dedekind sum in $\chi(d, \alpha)$.

When $\alpha = 0$, the asymptotic formula for symmetrizable Nahm sums was previously obtained in <cit.>.

§ NAHM SUMS AND BLOCH GROUPS

§.§ Bloch groups

Let $F$ be a number field.
We denote by $Z(F)$ the (additive) free abelian group on $\mathbb{P}^1(F) = F \sqcup \{ \infty \}$.
We denote by $[X] \in Z(F)$ the element corresponding to $X \in \mathbb{P}^1(F)$.
The Bloch group of $F$ is the quotient

\begin{align}
B(F) = A(F) / C(F)
\end{align}

where $A(F)$ is the kernel of the map

\begin{align}
d : Z(F) \to \wedge^2 F^\times,\quad
[X] \to X \wedge (1 - X)
\end{align}

(and $[0], [1], [\infty] \mapsto 0$), and $C(F) \subseteq A(F)$ is the subgroup generated by
the five-term relation

\begin{align}\label{eq:five term}
[X] - [Y] + \left[ \frac{Y}{X} \right] - \left[ \frac{1-X^{-1}}{1-Y^{-1}} \right] + \left[ \frac{1-X}{1-Y} \right]
\end{align}

for any $X, Y \in \mathbb{P}^1 (F)$ except where $\frac{0}{0}$ or $\frac{\infty}{\infty}$ appears in (<ref>).
There are several conventions for the definition of the Bloch group, which agree up to $6$-torsion.
We use the definition in <cit.>.

Let $F_m$ be a field obtained by adjoining to $F$ a primitive $m$th root of unity $\zeta = \zeta_m$.
The extension $F_m / F$ is Galois, and we have an injective morphism

\begin{align*}
\chi : \Gal(F_m/F) \to (\mathbb{Z}/m \mathbb{Z})^\times
\end{align*}

determined by $\sigma(\zeta) = \zeta^{\chi (\sigma)}$ for any $\sigma$.
The group $F_m^\times / F_m^{\times m}$ is equipped with the (multiplicative) $\mathbb{Z}/m \mathbb{Z}$-module
structure given by $k \mapsto (x \mapsto x^k)$.
We define the $\chi^{-1}$-eigenspace by

\begin{align*}
(F_m^\times / F_m^{\times m})^{\chi^{-1}} \coloneqq
\{ x \in F_m^\times / F_m^{\times m} \mid \forall \sigma \in \Gal(F_m/F),\ \sigma(x) = x^{\chi(\sigma)^{-1}} \}.
\end{align*}

For a number field $F$ that does not contain any non-trivial $m$th root of unity,
Calegari-Garoufalidis-Zagier <cit.> defined a map

\begin{align}\label{eq:CGZ map}
R_\zeta : B(F) / n B(F) \to F_m^\times / F_m^{\times m}
\end{align}

such that its image is contained in the $\chi^{-1}$-eigenspace.
This map is defined by using cyclic dilogarithm function (<ref>).
They prove the following injectivity result:
Let $F$ be a number field.
There exists a integer $M$ (depending only on $F$) such that,
for any integer $m$ prime to $M$
and any $m$th root of unity $\zeta$,
the map $R_\zeta$ is injective.

The proof in <cit.> uses the compatibility between $R_\zeta$ and
a Chern class map in algebraic $K$-theory, and the injectivity of this Chern class map.

§.§ Nahm's conjecture for symmetrizable case

We first observe number-theoretic features of the constant term in the asymptotic expansion (<ref>).
Let $Q = (A, b, c, d)$ be a quadruple as in Section <ref>.
We fix a strong denominator $\delta$ of $Q$ such that $\delta d_i^{-1} b_i$ and $\delta d_i^{-1} A_{ij}$ are integers for any $i,j$.
Let $F$ be a number field obtained by adjoining $z_1^{1/\delta}, \dots, z_N^{1/\delta}$ to $\mathbb{Q}$,
where $z_i$ is the solution of the Nahm's equation (<ref>) such that $0 < z_i < 1$.
Let $m$ be an integer prime to $\delta$ and $d_1, \dots, d_N$.
Recall that $F_m$ is a field obtained by adjoining to $F$ a primitive $m$th root of unity $\zeta$.
We define a number

\begin{align}\label{eq:def of u}
u =  \biggl( \prod_{i = 1}^{N} \theta_i^{b_i / d_i} (1 - z_i)^{-\frac{1}{m}}
D_{\zeta_i} (\zeta_i \theta_i)^{-\frac{1}{m}} \biggr)
\sum_{k \in (\mathbb{Z}/m\mathbb{Z})^N} \zeta^{\overline{Q(k)}}
\prod_{i=1}^{N} \frac{\theta_i^{(k^\mathsf{T}A)_i}}{(\zeta_i \theta_i ; \zeta_i)_{k_i}}
\end{align}

which is a part of the constant term in the asymptotic formula (<ref>).
We see that the $m$th power of $u$ satisfies the following properties,
which is stated (without a proof) in <cit.> for the symmetric case.
We will provide the proof for the sake of completeness.

We have $u^m \in F_m$.
Moreover, if $u^m \neq 0$, then its image in $F_m^\times / F_m^{\times m}$
belongs to the $\chi^{-1}$-eigenspace.

We first see that $u$ can be also written as

\begin{align*}
u = \biggl( \prod_{i = 1}^{N} \theta_i^{b_i / d_i}
D_{\zeta_i} (\theta_i)^{-\frac{1}{m}} \biggr)
a_\zeta (\theta)
\end{align*}

where we define

\begin{equation}\label{eq:in Fm sum part}
a_\zeta (\theta) = \sum_{k \in (\mathbb{Z}/m\mathbb{Z})^N}  \zeta^{\overline{Q(k)}}
\prod_{i=1}^{N} \frac{\theta_i^{(k^\mathsf{T}A)_i}}{(\theta_i ; \zeta_i)_{k_i + 1}}.
\end{equation}

This follows from the formula

\begin{align}\label{eq:D zeta x / D x}
\frac{D_\zeta (\zeta x)}{D_\zeta (x)} = \frac{(1-x)^m}{1-x^m}.
\end{align}

Also note that the summand in (<ref>) is well-defined over the set $(\mathbb{Z}/m \mathbb{Z})^N$,
which follows from the fact that $\theta_i^m$ is a solution of the Nahm's equation (<ref>) (see Lemma 3.1 in <cit.>).

Let $H$ be the Kummer extension of $F_m$ obtained by adjoining the $m$th roots
of $z_1^{1/\delta}, \dots, z_N^{1/\delta}$ to $F_m$.
We see that $u^m \in H$ by definition. We now see that $u^m$ belongs to the smaller field $F_m$ by showing that it is fixed by any automorphism $\tau \in \Gal(H / F_m)$.
Let $e_i$ be integers such that $\zeta_i^{e_i} = \tau(\theta_i)/\theta_i$. Then we see that

\begin{align}\label{eq:u^m in Fm 1}
\tau \biggl( \prod_{i = 1}^{N}
D_{\zeta_i} (\theta_i)^{-1} \biggr) =
\prod_{i = 1}^{N}
D_{\zeta_i} (\theta_i)^{-1} \frac{\theta_i^{m(e^\mathsf{T} A)_i}}{(\theta_i; \zeta_i)_{e_i}^m}
\end{align}

by repeatedly using (<ref>) and also using the fact that $\theta_i^m$ is a solution of the Nahm's equation.
We also see that

\begin{align}\label{eq:u^m in Fm 2}
\biggl(\prod_{i=1}^{N} \frac{\theta_i^{m(e^\mathsf{T} A)_i}}{(\theta_i; \zeta_i)_{e_i}^m} \biggr) \cdot \tau(a_\zeta(\theta)^m) =
\biggl(\sum_{k \in (\mathbb{Z}/m\mathbb{Z})^N}  \zeta^{\overline{Q(k)}}
\prod_{i=1}^{N} \frac{\theta_i^{((k+e)^\mathsf{T} A)_i}}{(\theta_i ; \zeta_i)_{k_i + e_i + 1}} \biggr)^m
= a_\zeta(\theta)^m.
\end{align}

The equations (<ref>) and (<ref>) imply that $\tau(u^m) = u^m$.

Now we assume that $u^m \neq 0$.
Suppose that $\sigma \in \Gal(F_m / F)$.
Let $q = \chi(\sigma) \in (\mathbb{Z}/m \mathbb{Z})^\times$, and let $p \in \mathbb{Z}_{>0}$ be a lift of $q^{-1}$.
Let $\tilde{\sigma} \in \Gal(H / F)$ be a lift of $\sigma$,
and define $\tilde{\theta}_i \coloneqq \tilde{\sigma}(\theta_i)$.
Let $s_i$ be integers such that $\zeta_i^{s_i} = \tilde{\theta}_i / \theta_i$.
We define an element

\begin{equation}\label{eq:in Fm def of v}
v =
\biggl(\prod_{i = 1}^{N} h_i \biggr) \cdot
\frac{a_{\zeta^q} (\tilde{\theta})}{a_\zeta(\theta)^p}  \quad \in H^\times
\end{equation}

\begin{equation}\label{eq:in Fm def of h}
h_i = \frac{\tilde{\theta}_i^{b_i / d_i}} {\theta_i^{b_i p / d_i}} \cdot
\frac{\theta_i^{p (s^\mathsf{T} A)_i}}{(\theta_i;\zeta_i^q)_{ps_i}} \cdot
\prod_{t=1}^{p-1} \frac{1}{(\theta_i; \zeta_i)_{qt}} \quad \in H^\times.
\end{equation}

Then we have $v^m \equiv \sigma(u^m) / u^{mp} \bmod F_m^{\times m}$ by Proposition <ref> in Appendix <ref>.
Thus, to prove the proposition, it suffices to see that $v \in F_m$ by showing that $v$ is fixed by any automorphism $\tau \in \Gal(H/F_m)$.
Again, let $e_i$ be integers such that $\zeta_i^{e_i} = \tau(\theta_i)/\theta_i$.
We see that

\begin{equation}\label{eq:in Fm def tau h}
\begin{split}
\prod_{i=1}^{N} \frac{\tau(h_i)}{h_i} &=
\zeta^{ p s^\mathsf{T} AD e + b e^\mathsf{T} - p b e^\mathsf{T}}
\prod_{i=1}^N
\frac{(\theta_i; \zeta_i)_{e_i}^p (\theta_i; \zeta_i^q)_{ps_i}}{(\theta_i; \zeta_i^q)_{pe_i} (\zeta^{e_i} \theta_i; \zeta_i^q)_{ps_i}} \\
&= \zeta^{p s^\mathsf{T} AD e + b e^\mathsf{T} - p b e^\mathsf{T}}
\prod_{i=1}^N
\frac{(\theta_i; \zeta_i)_{e_i}^p}
{(\tilde{\theta}_i; \zeta_i^q)_{pe_i}} \\
&= \zeta^{b e^\mathsf{T} - p b e^\mathsf{T}}
\prod_{i=1}^N
\frac{\tilde{\theta}_i^{p(e^\mathsf{T} A)_i} (\theta_i; \zeta_i)_{e_i}^p}
{\theta_i^{p(e^\mathsf{T} A)_i} (\tilde{\theta}_i; \zeta_i^q)_{pe_i}},
\end{split}
\end{equation}

where we use (<ref>) in the first equality.
We now compute

\begin{equation}\label{eq:in Fm def tau tilde a}
\begin{split}
\zeta^{b e^\mathsf{T}}
\biggl(\prod_{i=1}^N \frac{\tilde{\theta}_i^{p (e^\mathsf{T} A)_i}}{(\tilde{\theta}_i; \zeta_i^q)_{pe_i}} \biggr)
\tau(a_{\zeta^q}(\tilde{\theta}))
\sum_{k \in (\mathbb{Z}/m\mathbb{Z})^N} \zeta^{q \overline{Q(pk)} + p k^\mathsf{T} A D e + b e^\mathsf{T}}
\prod_{i=1}^{N} \frac{\tilde{\theta}_i^{p((k+e)^\mathsf{T} A)_i}}{(\tilde{\theta}_i ; \zeta_i^q)_{p(k_i+e_i) + 1}} \\
\sum_{k \in (\mathbb{Z}/m\mathbb{Z})^N} \zeta^{q \overline{Q(p(k-e))} + p (k-e)^\mathsf{T} A D e + b e^\mathsf{T}}
\prod_{i=1}^{N} \frac{\tilde{\theta}_i^{p(k^\mathsf{T}A)_i}}{(\tilde{\theta}_i ; \zeta_i^q)_{pk_i + 1}} \\
&= \zeta^{- p e^\mathsf{T} AD e} a_{\zeta^q}(\tilde{\theta}).
\end{split}
\end{equation}

Similarly, we also compute

\begin{equation}\label{eq:in Fm def tau a}
\begin{split}
\zeta^{p b e^\mathsf{T}}
\biggl(\prod_{i=1}^N \frac{\theta_i^{p (e^\mathsf{T} A)_i}}{(\theta_i; \zeta_i)_{e_i}^p} \biggr)
\tau(a_{\zeta}(\theta))^p
\biggl(\sum_{k \in (\mathbb{Z}/m\mathbb{Z})^N} \zeta^{\overline{Q(k)} + k^\mathsf{T} A D e + b e^\mathsf{T}}
\prod_{i=1}^{N} \frac{\theta_i^{((k+e)^\mathsf{T} A)_i}}{(\theta_i ; \zeta_i)_{k_i+e_i + 1}} \biggr)^p \\
\biggl(\sum_{k \in (\mathbb{Z}/m\mathbb{Z})^N} \zeta^{\overline{Q(k-e)} + (k-e)^\mathsf{T} A D e + b e^\mathsf{T}}
\prod_{i=1}^{N} \frac{\theta_i^{(k^\mathsf{T} A)_i}}{(\theta_i ; \zeta_i)_{k_i + 1}} \biggr)^p \\
&= \zeta^{- p e^\mathsf{T} AD e} a_\zeta(\theta)^p.
\end{split}
\end{equation}

Combining (<ref>)–(<ref>),
we see that $\tau(v) = v$, which completes the proof.

We now prove the one direction of Nahm's conjecture for symmetrizable case.
We define an element in the Bloch group $B(F)$ by

\begin{align*}
\xi_{A,d} = \sum_{i=1}^N d_i^{-1} [z_i].
\end{align*}

To be precise, $\xi_{A,d}$ itself does not belong to $B(F)$,
rather a multiple of $\xi_{A,d}$ by at least $\lcm(d_1, \dots, d_N)$ belongs to $B(F)$.
But we shall not concern ourselves too much with that since we will only care whether $\xi_{A,d}$ is a torsion or not.
Finally, we say that the Nahm sum $f_{A,b,c,d}(q)$ is modular to mean that the function $\tilde{f}(\tau) = f_{A,b,c,d}(e^{2 \pi i \tau})$ is
invariant with respect to some finite index subgroup of $\SL(2, \mathbb{Z})$.

Suppose that the Nahm sum $f_Q(q)$ associated with $Q = (A, b, c, d)$ is modular.
Then $\xi_{A, d}$ is a torsion in the Bloch group $B(F)$.

Suppose that $\tilde{f}(\tau) = f_{A,b,c,d}(e^{2 \pi i \tau})$ is modular with respect to a finite index subgroup $\Gamma \subseteq \SL(2,\mathbb{Z})$.
We can assume that $\Gamma$ is contained in the principal congruence subgroup $\Gamma(M)$ for a fixed integer $M$
by replacing $\Gamma$ by its intersection with $\Gamma(M)$.

By the asymptotic formula (<ref>) for $\alpha = 0$, we have

\begin{align}\label{eq:nahm asymptotic alpha 0}
\tilde{f} \Bigl(\frac{i \varepsilon}{2 \pi}\Bigr) = e^{-\frac{\Lambda}{\varepsilon}} (K + O(\varepsilon))
\end{align}

as $\varepsilon$ tends to $0$ from the right, where $K$ is an algebraic number such that $K^2 \in F^\times$.
The number $\lambda = \Lambda / (2 \pi)^2$ must be rational since the modularity of $\tilde{f}(\tau)$ implies that
the function $\tilde{f} (-1/\tau)$ is invariant under some power of $\begin{bsmallmatrix}
1 & 1 \\ 0 & 1
\end{bsmallmatrix}$.
We denote by $\ell$ the denominator of $\lambda$.

For any positive real number $h$ and any $\begin{bsmallmatrix}
p & q \\ r & s
\end{bsmallmatrix} \in \Gamma$, taking $\varepsilon = \frac{s h}{1 - irh/2\pi}$, we find that

\begin{align}\label{eq:proof of torsion of modular}
\tilde{f}\Bigl( \frac{i \varepsilon}{2 \pi} \Bigr)
= \tilde{f}\Bigl( \frac{p i \varepsilon/2\pi + q}{ri\varepsilon/2\pi + s} \Bigr)
=\tilde{f}\Bigl(\alpha + \frac{i h}{2 \pi m} \Bigr)
\end{align}

where we set $\alpha = q/s$ and $m = s$.
Comparing the asymptotics of the left and the right expressions in (<ref>)
for $h$ tending to $0$ on the positive real line by using (<ref>) and (<ref>),
we have

\begin{align}\label{eq:proof of torsion of modular 2}
K \mathbf{e}(\lambda r / m) = \mu u_\zeta
\end{align}

where $u_\zeta = u$ is defined by (<ref>), and $\mu$ is an algebraic number such that $\mu^{12} \in F_m^{\times}$.
The equation (<ref>) implies that $u_\zeta^{12\ell} \in F_m^{\times}$.
Since $\Gamma$ is contained in $\Gamma(M)$, we see that $m$ is prime to $M$.

Consequently, we see that there are infinitely many $m$ such that $u_\zeta^{12\ell}$ for a primitive $m$th root of unity $\zeta$ belongs to $F_m^{\times}$.
By Proposition <ref> and <cit.>,
this implies that $R_\zeta(\xi_{A,d})^{12\ell}$ is trivial in $F_m^\times / F_m^{\times m}$ for infinitely many $m$.
Thus, by the injectivity result Theorem <ref>, we see that $12 \ell \xi_{A,d}$ has a trivial image in $B(F)/m B(F)$ for infinitely many $m$.
If $\xi_{A,d}$ is not a torsion, this contradicts the fact that $B(F)$ is a finitely generated abelian group.

§ EXPERIMENTS ON MODULAR NAHM SUMS

In this section, we present some experiments on modular Nahm sums for symmetrizable matrices
by using the numerical method explained in <cit.>.
The method aims to detect, for a given $q$-series $f(q)$, whether $q^c f(q)$ is likely modular for some $c$.
The actual procedure is as follows:

* Compute $\phi(N) = N \log f(e^{-1/N})$ for four successive values (say $N = 20, 21, 22, 23$).

* Take the third difference of the values computed in (1). If this value is extremely small, then the Nahm sum $f(q)$ is likely modular.

This method is justified by the following reasoning:
If $f(q)$ is modular, $\phi(N)$ is approximated to high precision by a quadratic polynomial in $n$ (see <cit.>),
and thus its third difference should be extremely small.
For Nahm sums we actually only need three values in the step (1)
since we know the constant term of the quadratic polynomial by the formula (<ref>).
We use this method to give lists of candidates of modular Nahm sums in rank $2$ (Table <ref>), and
rank $3$ with specific symmetrizers (Table <ref> for $d = (1,1,2)$ and Table <ref> for $d = (2,2,1)$).

colspec = Q[l,m,mode=math]Q[c,m,mode=math]Q[r,m,mode=math]Q[r,m,mode=math],
row3,4,14,15,16,17,18 = rowsep+=2pt,

A     b     c     d
2     1

2     2

[ 0; 0 ]     -1/24

[ 0; 1 ]     1/24

[ -1; -1 ]     -1/24

[ -1/2;    0 ]     -1/48

[ 1; 2 ]     7/24

1     1/2

1     1

[ 0; 0 ]     -3/56

[ 0; 1 ]     1/56

[ 1; 1 ]     9/56

4     2

6     4

[ 0; 0 ]     -1/24

2     1

3     2

[ 0; 0 ]     -1/18

[ 1; 3 ]     5/18

[ 2; 3 ]     11/18

3     2

4     4

[ -1/2;    0 ]     -1/60

[ 1/2;   2 ]     11/60

3/2     1/2

1       1

[ -1;  1 ]     9/40

[ -1/2;    0 ]     -1/40

[ 0; 1 ]     1/40

3     1

4     2

[ -1/2;    0 ]     -1/16

[ 3/2;   4 ]     7/16

colspec = Q[l,m,mode=math]Q[c,m,mode=math]Q[r,m,mode=math]Q[r,m,mode=math],

A     b     c     d
1     -1/2

-1     1

[ 0; 0 ]     -1/12

[ -1/2;    1 ]     1/12

[ -1/2;    0 ]     1/12

[ -1/2;  1/2 ]     1/48

[ 0; 1 ]     1/12

2     -1

-2     2

[ 0; 0 ]     -1/14

[ -1;  2 ]     5/14

[ 1; 0 ]     3/14

1     -1/2

-3/2     1

[ 0; 0 ]     -1/8

2     -1

-3     2

[ 0; 0 ]     -1/9

[ -1;  3 ]     5/9

[ 1; 0 ]     2/9

1     -1/2

-1     3/4

[ -1/2;  1/2 ]     1/60

[ -1/2;    1 ]     1/15

1     -1/2

-1     3/2

[ -3/2;  5/2 ]     41/40

[ -1/2;  1/2 ]     1/40

[ -1/2;  3/2 ]     9/40

1     -1/2

-2     3/2

[ -1/2;    1 ]     -1/48

[ -1/2;    3 ]     23/48

Symmetrizable modular candidates for rank $2$.

We provide some comments on the lists.
The matrix $A$ in the right table in Table <ref> is the inverse of the matrix $A$
of the same row in the left table. More precisely, we expect that the following holds in general:

Suppose that the Nahm sum associated with $(A, b, c, d)$ is modular.
Then the Nahm sum associated with $(A^*, b^*, c^*, d^*)$ is also modular, where

\begin{align*}
A^* = A^{-1},\quad
b^* = A^{-1} b, \quad
c^* = \frac{1}{2} b^{\mathsf{T}} (AD)^{-1} b - \frac{\tr D}{24} - c, \quad
d^* = d.
\end{align*}

Next, we provide case-by-case comments on several Nahm sums in Table <ref>.

* $A = \begin{bsmallmatrix}
1 & 1/2 \\
1 & 1
\end{bsmallmatrix}$.
For $\sigma = 0,1$, we define

\begin{align}
\label{eq:B2 inv -3/56}
f_{-3/56, \sigma} (q) \coloneqq &\sum_{\substack{n_1,n_2 \geq 0 \\ n_1 \equiv \sigma \bmod 2}} \frac{q^{\frac{1}{2} n_1^2 + n_1 n_2 + n_2^2 }}{(q;q)_{n_1} (q^2;q^2)_{n_2}} \\
\label{eq:B2 inv 1/56}
f_{1/56, \sigma} (q) \coloneqq &\sum_{\substack{n_1,n_2 \geq 0 \\ n_1 \equiv \sigma \bmod 2}} \frac{q^{\frac{1}{2} n_1^2 + n_1 n_2 + n_2^2 + n_2}}{(q;q)_{n_1} (q^2;q^2)_{n_2}}  \\
\label{eq:B2 inv 9/56}
f_{9/56, \sigma} (q) \coloneqq &\sum_{\substack{n_1,n_2 \geq 0 \\ n_1 \equiv \sigma \bmod 2}} \frac{q^{\frac{1}{2} n_1^2 + n_1 n_2 + n_2^2 + n_1 + n_2}}{(q;q)_{n_1} (q^2;q^2)_{n_2}}
\end{align}

so that $f_{c, 0}(q) + f_{c, 1} (q)$ is the Nahm sum associated with $A$.
We conjecture that the following modular transformation formula holds:

\begin{align}
g(-\frac{1}{\tau}) =
\begin{pmatrix}
S & S \\
S & -S
\end{pmatrix}
\end{align}

where $g(\tau) = (f_{-3/56, 0} (q), f_{1/56, 1} (q), f_{9/56, 1} (q), f_{-3/56, 1} (q), f_{1/56, 0} (q), f_{9/56, 0} (q) )^\mathsf{T}$ and

\begin{align}
S =
\begin{pmatrix}
\alpha_3 & \alpha_2 & \alpha_1 \\
\alpha_2 & -\alpha_1 & -\alpha_3 \\
\alpha_1 & -\alpha_3 & \alpha_2
\end{pmatrix},
\quad
\alpha_k = \sqrt{\frac{2}{7}} \sin \frac{k \pi}{7}
\end{align}

* $A = \begin{bsmallmatrix} 2 & 1 \\ 3 & 2 \end{bsmallmatrix}$.
The modularity follows from the Kanade-Russell conjecture (<ref>)–(<ref>), as we mentioned in the introduction.

* $A = \begin{bsmallmatrix} 3 & 2 \\ 4 & 4 \end{bsmallmatrix}$.
We have the following identities:

∑_n_1, n_2 ≥0
q^3/2 n_1^2 + 4 n_1n_2 + 4 n_2^2 - 1/2 n_1 + 0n_2/(q;q)_n_1 (q^2;q^2)_n_2
= ∑_n ≥0 q^n^2/(q;q)_n
= 1/(q^2,q^3;q^5)_∞,

∑_n_1, n_2 ≥0
q^3/2 n_1^2 + 4 n_1n_2 + 4 n_2^2 + 1/2 n_1 + 2 n_2/(q;q)_n_1 (q^2;q^2)_n_2
= ∑_n ≥0 q^n^2 + n/(q;q)_n
= 1/(q^2,q^3;q^5)_∞.

The second equalities in (<ref>) and (<ref>) are the Rogers-Ramanujan identities.
For the first equalities, the author learned the following proof from Shunsuke Tsuchioka.
We can verify that the identities

\begin{align}\label{eq:double sum RR x}
\sum_{n_1, n_2 \geq 0}
\frac{q^{\frac{3}{2} n_1^2 + 4 n_1n_2 + 4 n_2^2 - \frac{1}{2} n_1}}{(q;q)_{n_1} (q^2;q^2)_{n_2}} x^{n_1 + 2 n_2}
= \sum_{n \geq 0} \frac{q^{n^2}}{(q;q)_n} x^n
\end{align}

holds by showing that both sides satisfy the same $q$-difference equation,
since the coefficients of $x^0$ (resp. $x^n$ for $n < 0$) of the both sides are $1$ (resp. $0$).
The first equalities in (<ref>) and (<ref>) are obtained by substituting $x$ by $1$ and $q$, respectively,
into (<ref>).
In fact, the desired $q$-difference equation can be found algorithmically by using the $q$-version of
Sister Celine's technique (e.g.,  package <cit.> can handle such computations).

* $A = \begin{bsmallmatrix} 3/2 & 1/2 \\ 1 & 1 \end{bsmallmatrix}$.
We have the following identities:

∑_n_1, n_2 ≥0
n_1 = 0 2
q^3/4 n_1^2 + n_1n_2 + n_2^2 - n_1 + n_2/(q;q)_n_1 (q^2;q^2)_n_2

=∑_n_1, n_2 ≥0
n_1 = 1 2
q^3/4 n_1^2 + n_1n_2 + n_2^2 - 1/2 n_1/(q;q)_n_1 (q^2;q^2)_n_2
= ∑_n ≥0 q^n^2+n/(q;q)_2n+1
= (q,q^9, q^10; q^10)_∞(q^8,q^12, q^20)/(q;q)_∞,

∑_n_1, n_2 ≥0
n_1 = 1 2
q^3/4 n_1^2 + n_1n_2 + n_2^2 - n_1 + n_2/(q;q)_n_1 (q^2;q^2)_n_2

=∑_n_1, n_2 ≥0
n_1 = 0 2
q^3/4 n_1^2 + n_1n_2 + n_2^2 - 1/2 n_1/(q;q)_n_1 (q^2;q^2)_n_2
= ∑_n ≥0 q^n^2/(q;q)_2n
= (q^2,q^8, q^10; q^10)_∞(q^6,q^14, q^20)/(q;q)_∞,

∑_n_1, n_2 ≥0
q^3/4 n_1^2 + n_1n_2 + n_2^2 + n_2/(q;q)_n_1 (q^2;q^2)_n_2
= ∑_n ≥0 q^n^2 + n/(q;q)_2n
= (q^3,q^7, q^10; q^10)_∞(q^4,q^16, q^20)/(q;q)_∞,

∑_n_1, n_2 ≥1
q^3/4 n_1^2 + n_1n_2 + n_2^2 + n_2/(q;q)_n_1 (q^2;q^2)_n_2
= ∑_n ≥0 q^n^2 + 2n/(q;q)_2n+1
= (q^4,q^6, q^10; q^10)_∞(q^2,q^18, q^20)/(q;q)_∞.

The rightmost equalities in
(<ref>), (<ref>), (<ref>), and (<ref>) are
(99), (98), (94), and (96), respectively, in the Slater's list <cit.>.
Other equalities can be easily verified in the same way as for the left equalities in (<ref>) and (<ref>) explained earlier.

* $A = \begin{bsmallmatrix} 4 & 2 \\ 6 & 4 \end{bsmallmatrix}$.
The modularity follows from the Capparelli identities <cit.> in the form presented in <cit.>:

\begin{align*}
\sum_{n_1,n_2 \geq 0} \frac{q^{2 n_1^2 + 6 n_1 n_2 + 6 n_2^2}}{(q;q)_{n_1} (q^3;q^3)_{n_2}} = (-q^2,-q^3,-q^4,-q^6; q^6)_\infty.
\end{align*}

* $A = \begin{bsmallmatrix} 3 & 1 \\ 4 & 2 \end{bsmallmatrix}$.
The modularity follows from the Göllnitz-Gordon identities in the form presented by Kurşungöz <cit.>:

\begin{align*}
\sum_{n_1,n_2 \geq 0}
\frac{q^{\frac{3}{2} n_1^2 + 4 n_1 n_2 + 4 n_2^2 - \frac{1}{2}n_1}}{(q;q)_{n_1} (q^4;q^4)_{n_2}} &= \frac{1}{(q,q^4,q^7;q^8)_\infty}, \\
\sum_{n_1,n_2 \geq 0}
\frac{q^{\frac{3}{2} n_1^2 + 4 n_1 n_2 + 4 n_2^2 + \frac{3}{2}n_1 + 4n_2}}{(q;q)_{n_1} (q^4;q^4)_{n_2}} &= \frac{1}{(q^3,q^4,q^5;q^8)_\infty}.
\end{align*}

* $A = \begin{bsmallmatrix} 1 & -1/2 \\ -1 & 1 \end{bsmallmatrix}$ or $\begin{bsmallmatrix} 1 & -3/2 \\ -1 & 1 \end{bsmallmatrix}$.
If $b = \begin{bsmallmatrix} 0 \\ 0 \end{bsmallmatrix}$, the Nahm sum coincides, up to a multiplication by a product of the dedekind eta function,
with a certain character of the integrable highest weight module $L(2 \Lambda_0)$ of the affine Lie algebra of type $D_3^{(2)}$ or $D_4^{(3)}$.
Such a character formula was conjectured in <cit.> and recently proved in <cit.> for any twisted type affine Lie algebra.
Then the modularity follows from the result by Kac and Peterson <cit.>.

colspec = Q[l,m,mode=math]Q[c,m,mode=math]Q[r,m,mode=math],
column2 = colsep+=-5pt,

A     b     c
1     1     0

1     2     1

0     2     4

[ 1/2;   1;   0;     ]   7/120

[ 1/2;   1;   4;     ]   103/120

1     1     0

1     3     1

0     2     2

[ 1/2;  -1;   0;     ]   3/56

[ 1/2;   0;   0;     ]   -1/56

[ 1/2;   1;   2;     ]   19/56

1     1     1

1     3     2

2     4     4

[    0; -1/2;    0;      ]   -1/32

[  1/2; -3/2;   -2;      ]   1/8

[  1/2; -1/2;    0;      ]   0

[ 1/2; 1/2;   2;     ]   1/8

[   1; 1/2;   2;     ]   7/32

1     1     0

1     4     1

0     2     1

[ 1/2;  -2;   1;     ]   7/24

[ 1/2;   0;   0;     ]   -1/48

[ 1/2;   0;   1;     ]   1/24

[ 1/2;   2;   1;     ]   7/24

[ 1/2;   2;   2;     ]   23/48

1     1     0

1     4     2

0     4     4

[ 1/2;   0;   0;     ]   -1/168

[ 1/2;   0;   2;     ]   47/168

[ 1/2;   2;   4;     ]   143/168

1     1     1

1     5     4

2     8     8

[  1/2; -1/2;    0;      ]   1/120

[ 1/2; 3/2;   4;     ]   49/120

1     1     0

1     8     1

0     2     1

[ 1/2;   0;   1;     ]   7/120

[ 1/2;   4;   1;     ]   103/120

1     1     0

1     8     2

0     4     2

[ 1/2;   0;   0;     ]   -1/168

[ 1/2;   2;   0;     ]   47/168

[ 1/2;   4;   2;     ]   143/168

colspec = Q[l,m,mode=math]Q[c,m,mode=math]Q[r,m,mode=math],
column2 = colsep+=-10pt,

A     b     c
2     1     1

1     2     1

2     2     2

[ -1;  0;  1;    ]   1/8

[ -1/2;    0;    0;      ]   -1/32

[  0; -1;  1;    ]   1/8

[    0; -1/2;    0;      ]   -1/32

[ 0; 0; 1;   ]   0

[ 0; 1; 1;   ]   1/8

[ 1; 0; 1;   ]   1/8

[ 1/2;   1;   2;     ]   7/32

[   1; 1/2;   2;     ]   7/32

2     1     1

1     4     1

2     2     2

[ 0; 0; 1;   ]   1/120

[ 0; 2; 1;   ]   49/120

2     2     1

2     4     2

2     4     3

[ -1; -1; -1;    ]   1/48

[ 0; 0; 0;   ]   -1/24

[ 0; 0; 1;   ]   1/48

[ 0; 1; 2;   ]   5/24

[ 1; 2; 3;   ]   25/48

1     1     1/2

1     2     1

1     2     3/2

[ 0; 0; 0;   ]   -1/18

[ 0; 0; 1;   ]   0

[ 0; 1; 1;   ]   1/9

[ 1; 1; 2;   ]   5/18

1     0     1/2

0     2     1

1     2     2

[ 0; 0; 0;   ]   -5/88

[ 0; 0; 1;   ]   -1/88

[ 0; 1; 2;   ]   19/88

[ 1; 1; 2;   ]   35/88

colspec = Q[l,m,mode=math]Q[c,m,mode=math]Q[r,m,mode=math],
column2 = colsep+=-10pt,

A     b     c
2     2     2

2     5     4

4     8     8

[    0; -1/2;    0;      ]   -1/42

[   0; 1/2;   2;     ]   5/42

[   1; 3/2;   4;     ]   17/42

1     1     0

1     3     1/2

0     1     3/2

[  1/2; -1/2;  1/2;      ]   1/40

[  1/2;  1/2; -1/2;      ]   1/40

[ 1/2; 1/2; 3/2;     ]   9/40

[ 1/2; 3/2; 1/2;     ]   9/40

1     1     0

1     4     1

0     2     3/2

[ 1/2;   0;  -1;     ]   3/56

[ 1/2;   0;   0;     ]   -1/56

[ 1/2;   2;   1;     ]   19/56

3     2     2

2     4     2

4     4     4

[ -1/2;    0;    0;      ]   -1/42

[ -1/2;    1;    0;      ]   5/42

[ 1/2;   2;   2;     ]   17/42

3/2     1     1

1     3     2

2     4     4

[ -1/2; -1/2;    0;      ]   1/168

[    0; -1/2;    0;      ]   -5/168

[ 1/2; 1/2;   2;     ]   25/168

2     1     1

1     5/2     3/2

2     3     3

[  0; -1; -1;    ]   1/168

[    0; -1/2;    0;      ]   -5/168

[ 1; 0; 1;   ]   25/168

tableSymmetrizable modular candidates with $d = (1,1,2)$.

colspec = Q[l,m,mode=math]Q[c,m,mode=math]Q[r,m,mode=math],
column2 = colsep+=-13pt,

A     b     c
1     0     1

0     2     2

1/2     1     2

[ 0; 0; 0;   ]   -7/88

[ 1; 0; 0;   ]   1/88

[ 1; 0; 1;   ]   9/88

[ 1; 2; 2;   ]   49/88

1     1     1

1     2     2

1/2     1     2

[    0;    0; -1/2;      ]   -3/56

[    1;    0; -1/2;      ]   1/56

[   1;   2; 1/2;     ]   9/56

1     1     0

1     4     4

0     2     3

[    1;    0; -1/2;      ]   1/48

[   1;   4; 3/2;     ]   25/48

2     1     0

1     2     2

0     1     2

[ 0; 1; 0;   ]   -1/36

[ 2; 1; 0;   ]   11/36

[ 2; 3; 1;   ]   23/36

2     2     2

2     4     4

1     2     3

[   -1;   -2; -3/2;      ]   1/24

[  -1;   0; 1/2;     ]   1/24

[  0; -1; -1;    ]   -1/48

[    0;    0; -1/2;      ]   -1/24

[ 0; 0; 0;   ]   -5/96

[   0;   2; 3/2;     ]   7/24

[    1;    0; -1/2;      ]   1/24

[   1;   0; 1/2;     ]   1/24

[   2;   2; 1/2;     ]   7/24

[   3;   4; 5/2;     ]   25/24

4     4     2

4     5     2

1     1     1

[   0;   0; 1/2;     ]   -1/40

[   4;   4; 1/2;     ]   31/40

colspec = Q[l,m,mode=math]Q[c,m,mode=math]Q[r,m,mode=math],
column2 = colsep+=-13pt,

A     b     c
1     1     1

1     2     2

1/2     1     3/2

[ 0; 0; 0;   ]   -5/72

[   1;   0; 1/2;     ]   1/72

[   1;   2; 3/2;     ]   25/72

1     0     1

0     2     2

1/2     1     5/2

[    0;    0; -1/2;      ]   -5/72

[  1;  0; -1;    ]   1/8

[ 1; 0; 0;   ]   1/72

[ 1; 2; 1;   ]   25/72

1     1     1

1     3     3

1/2     3/2     5/2

[  0; -1; -1;    ]   -1/40

[    1;   -2; -3/2;      ]   9/40

[    1;    0; -1/2;      ]   1/40

3/2     1/2     1

1/2     3/2     1

1/2     1/2     1

[ -3/2; -1/2;  1/2;      ]   1/8

[ -1/2; -3/2;  1/2;      ]   1/8

[ -1/2;  1/2;    0;      ]   -1/32

[ -1/2;  1/2;  1/2;      ]   0

[  1/2; -1/2;    0;      ]   -1/32

[  1/2; -1/2;  1/2;      ]   0

[ 1/2; 3/2; 1/2;     ]   1/8

[ 1/2; 3/2;   1;     ]   7/32

[ 3/2; 1/2; 1/2;     ]   1/8

[ 3/2; 1/2;   1;     ]   7/32

4     4     4

4     6     6

2     3     4

[  0; -1; -1;    ]   -1/60

[ 2; 1; 0;   ]   11/60

colspec = Q[l,m,mode=math]Q[c,m,mode=math]Q[r,m,mode=math],
column2 = colsep+=-13pt,

A     b     c
3/2     1/2     1

1/2     3/2     1

1/2     1/2     3/2

[ -1/2;  1/2; -1/2;      ]   1/168

[ -1/2;  1/2;    0;      ]   -5/168

[  1/2; -1/2; -1/2;      ]   1/168

[  1/2; -1/2;    0;      ]   -5/168

[ 1/2; 3/2; 1/2;     ]   25/168

[ 3/2; 1/2; 1/2;     ]   25/168

3/2     1/2     2

1/2     3/2     2

1     1     4

[ -1/2;  1/2;    0;      ]   -1/42

[ -1/2;  1/2;    1;      ]   5/42

[  1/2; -1/2;    0;      ]   -1/42

[  1/2; -1/2;    1;      ]   5/42

[ 1/2; 3/2;   2;     ]   17/42

[ 3/2; 1/2;   2;     ]   17/42

5/2     3/2     1

3/2     5/2     1

1/2     1/2     1

[ -1/2;  1/2;  1/2;      ]   1/120

[  1/2; -1/2;  1/2;      ]   1/120

[ 3/2; 5/2; 1/2;     ]   49/120

[ 5/2; 3/2; 1/2;     ]   49/120

5/2     3/2     2

3/2     5/2     2

1     1     2

[ -1/2;  1/2;    0;      ]   -1/42

[  1/2; -1/2;    0;      ]   -1/42

[ 1/2; 3/2;   0;     ]   5/42

[ 3/2; 1/2;   0;     ]   5/42

[ 3/2; 5/2;   1;     ]   17/42

[ 5/2; 3/2;   1;     ]   17/42

tableSymmetrizable modular candidates with $d = (2,2,1)$.

Finally, we provide a single comment on Table <ref> and <ref>.
There is no apparent relationship between the candidates in these two lists,
which are associated with symmetrizers $d = (1,1,2)$ and $d = (2,2,1)$, except that there are several “Langlands dual” pairs:

\begin{gather*}
\begin{pmatrix}
2 & 2 & 1\\
2 & 4 & 2\\
2 & 4 & 3
\end{pmatrix}
\leftrightarrow
\begin{pmatrix}
2 & 2 & 2\\
2 & 4 & 4\\
1 & 2 & 3
\end{pmatrix},\quad
\begin{pmatrix}
1 & 1 & 1/2\\
1 & 2 & 1\\
1 & 2 & 3/2
\end{pmatrix}
\leftrightarrow
\begin{pmatrix}
1 & 1 & 1\\
2 & 2 & 2\\
1/1 & 1 & 3/2
\end{pmatrix},\\
\begin{pmatrix}
1 & 0 & 1/2\\
0 & 2 & 1\\
1 & 2 & 2
\end{pmatrix}
\leftrightarrow
\begin{pmatrix}
1 & 0 & 1\\
0 & 2 & 2\\
1/2 & 1 & 2
\end{pmatrix}.
\end{gather*}

These pairs consist of matrices that are transposed to each other.
The second (resp. first) pair consists of the inverses of (resp. the twice the inverse of) the Cartan matrices of type $B_3$ and $C_3$.
The third pair is more mysterious. We expect that two Nahm sums for the third pair are related by the modular $S$-transformation.
More precisely, according to numerical experiments, we conjecture that the following modular transformation formulas hold.
For matrices
$A = \begin{bsmallmatrix}
1 & 0 & 1/2\\
0 & 2 & 1\\
1 & 2 & 2
\end{bsmallmatrix}$ and
$A^{\vee} = \begin{bsmallmatrix}
1 & 0 & 1/2\\
0 & 2 & 1\\
1 & 2 & 2
\end{bsmallmatrix}$ with symmetrizers $d = (1,1,2)$ and $d^{\vee}=(2,2,1)$, respectively,
we define vector-valued functions

\begin{align*}
g (\tau) =
\begin{+pmatrix}[rows={l}]
f_{A, \begin{bsmallmatrix} 0\\0\\0 \end{bsmallmatrix}, -5/88; \, 0} (q) \\
f_{A, \begin{bsmallmatrix} 0\\0\\1 \end{bsmallmatrix}, -1/88; \, 1} (q) \\
f_{A, \begin{bsmallmatrix} 0\\1\\1 \end{bsmallmatrix}, 7/88; \, 1} (q) +
q f_{A, \begin{bsmallmatrix} 1\\2\\3 \end{bsmallmatrix}, 7/88; \, 1} (q) \\
f_{A, \begin{bsmallmatrix} 0\\1\\2 \end{bsmallmatrix}, 19/88; \, 0} (q) \\
f_{A, \begin{bsmallmatrix} 1\\1\\2 \end{bsmallmatrix}, 35/88; \, 0} (q) \\
\end{+pmatrix},\
g^{\vee} (\tau) =
\begin{+pmatrix}[rows={l}]
f_{A^{\vee}, \begin{bsmallmatrix} 0\\0\\0 \end{bsmallmatrix}, -7/88} (q) \\
f_{A^{\vee}, \begin{bsmallmatrix} 0\\0\\1 \end{bsmallmatrix}, 25/88} (q) +
f_{A^{\vee}, \begin{bsmallmatrix} 1\\2\\3 \end{bsmallmatrix}, 25/88} (q) \\
f_{A^{\vee}, \begin{bsmallmatrix} 1\\0\\0 \end{bsmallmatrix}, 1/88} (q) \\
f_{A^{\vee}, \begin{bsmallmatrix} 1\\0\\1 \end{bsmallmatrix}, 9/88} (q) \\
f_{A^{\vee}, \begin{bsmallmatrix} 1\\2\\2 \end{bsmallmatrix}, 49/88} (q) \\
\end{+pmatrix},
\end{align*}

where $f_{A, b, c, d;\, \sigma}(q)$ for $\sigma = 0,1$
is the partial sum of the Nahm sum where the sum is taken over $n \in \mathbb{N}^3$ such that $n_1 \equiv 0 \bmod \sigma$,
and we omit $d$ and $d^{\vee}$ from the notation.
Then we have

\begin{align*}
g(\tau + 1) = \diag(\zeta_{88}^{-5}, \zeta_{88}^{43},\zeta_{88}^{51},\zeta_{88}^{19},\zeta_{88}^{35}) g(\tau), \quad
g^{\vee} (\tau + 1) = \diag(\zeta_{88}^{-7}, \zeta_{88}^{25},\zeta_{88}^{1},\zeta_{88}^{9},\zeta_{88}^{49}) g^{\vee} (\tau),
\end{align*}

and we conjecture that

\begin{align}
g (-\frac{1}{\tau}) = S\ g^{\vee} (\frac{\tau}{2}), \quad
g^{\vee} (-\frac{1}{\tau}) =
2 S\ g (\frac{\tau}{2}),
\end{align}

\begin{align*}
\begin{pmatrix}
\alpha_{5} & \alpha_{4} & \alpha_{3} & \alpha_{2} & \alpha_{1} \\
\alpha_{4} & \alpha_{1} & -\alpha_{2} & -\alpha_{5} & -\alpha_{3} \\
\alpha_{3} & -\alpha_{2} & -\alpha_{4} & \alpha_{1} & \alpha_{5} \\
\alpha_{2} & -\alpha_{5} & \alpha_{1} & \alpha_{3} & -\alpha_{4} \\
\alpha_{1} & -\alpha_{3} & \alpha_{5} & -\alpha_{4} & \alpha_{2}
\end{pmatrix}, \quad
\alpha_k = \sqrt{\frac{2}{11}} \sin \frac{k\pi}{11}.
\end{align*}

§ CYCLIC QUANTUM DILOGARITHM FUNCTION

The cyclic quantum dilogarithm function is defined by

\begin{align}
\label{eq:cyclic dilog}
D_\zeta (x) = \prod_{t=1}^{m-1} (1 - \zeta^t x)^t
\end{align}

where $\zeta$ is a primitive $m$th root of unity.
Noting that the function $x \mapsto x - \floor{x}$ is periodic with period $1$,
we have the following expression:

\begin{align}
D_\zeta (x) = \prod_{t \bmod m} (1 - \zeta^t x)^{m (\frac{t}{m} - \floor*{\frac{t}{m}})}.
\end{align}

Let $p, q$ be integers coprime to $m$.
Suppose that $p > 0$ and $pq \equiv 1 \bmod m$. Then we have

\begin{align}\label{eq:cyclic dilog change of zeta 1}
\frac{D_\zeta(x)^p} {D_{\zeta^q}(x)} =
\left((1-x^m)^{p-1} \prod_{t=1}^{p-1} \frac{1}{(x; \zeta)_{\floor*{\frac{mt}{p}} + 1}} \right)^m.
\end{align}

Moreover, we also have

\begin{align}\label{eq:cyclic dilog change of zeta 2}
\frac{D_\zeta(x)^p} {D_{\zeta^q}(x)} \equiv
\left(\prod_{t=1}^{p-1} \frac{1}{(x; \zeta)_{tq}}\right)^m
\quad\mod (1 - x^m)^m.
\end{align}

We first see that

\begin{align*}
\frac{D_\zeta(x)^p} {D_{\zeta^q}(x)} =
\prod_{t=1}^{m-1} (1 - \zeta^t x)^{m \floor*{\frac{t p}{m}}}
\end{align*}

\begin{align*}
\prod_{t=1}^{m-1} (1 - \zeta^t x)^{m(\frac{pt}{m} - \floor*{\frac{pt}{m}})}
% &= \prod_{t \bmod m} (1 - \zeta^t x)^{m(\frac{pt}{m} - \floor*{\frac{pt}{m}})} \\
= \prod_{t=1}^{m-1} (1 - \zeta^{qt} x)^{m(\frac{pqt}{m} - \floor*{\frac{pqt}{m}})}
% &= \prod_{t=1}^{m-1} (1 - \zeta^{qt} x)^{m(\frac{t}{m} - \floor*{\frac{t}{m}})} \\
= \prod_{t=1}^{m-1} (1 - \zeta^{qt} x)^t.
\end{align*}

Let $N_r$ for $r = 1, \dots, p$ be the number of integer $t$ such that $0 < t < m-1$ and
$\floor*{\frac{tp}{m}} \leq r - 1$. The $N_r$ is exactly the number of integer $t$
such that $0 < t < \frac{mr}{p}$, and thus we have the following formulas:

\begin{align*}\label{eq:floor tp/m 3}
N_r = \floor*{\frac{mr}{p}} \quad\text{for $r = 1, \dots, p-1$}, \quad
N_p = m - 1.
\end{align*}

Now (<ref>) follows from the following computation:

\begin{align*}
\prod_{t=1}^{m-1} (1 - \zeta^t x)^{\floor*{\frac{t p}{m}}} =
\prod_{r=1}^{p} \frac{(x; \zeta)_{N_r+1}^{r-1}}{(x; \zeta)_{N_{r-1}+1}^{r-1}} &=
(x; \zeta)_{N_p+1}^{p-1} \prod_{r=1}^{p-1} \frac{1}{(x; \zeta)_{N_r+1}} \\ &=
(1-x^m)^{p-1} \prod_{t=1}^{p-1} \frac{1}{(x; \zeta)_{\floor*{\frac{mt}{p}} + 1}}.
\end{align*}

To prove (<ref>), it suffices to show that the sets $M(r)$ and $M'(r)$ defined by

\begin{align*}
M(r) &= \{ t \in [1, p-1] \mid \floor*{\frac{mt}{p}} +1 = r \} \\
M'(r) &= \{ t \in [1, p-1] \mid tq \equiv r \bmod m \}
\end{align*}

have the same cardinality for all $r = 1, \dots, m-1$, where $[1, p-1] \coloneqq \{1, 2, \dots, p-1\}$.
The equation $tq \equiv r \bmod m$ holds if and only if $t \equiv pr \bmod m$.
The number of such $t$ is the same as the number of $s \in [1,p-1]$ such that
$0 < pr - ms <p$, which is equivalent to the condition $\floor*{\frac{ms}{p}} = r-1$.

We also see that

\begin{equation}\label{eq:cyclic dilog change of zeta 3}
\prod_{t=0}^{p-1} \frac{(x;\zeta)_{tq}}{(\zeta^e x;\zeta)_{tq}}
= \prod_{t=0}^{p-1} \frac{(x;\zeta)_{e}}{(\zeta^{tq} x;\zeta)_{e}}
= \frac{(x; \zeta)_k^e}{(x; \zeta^q)_{pe}},
\end{equation}

for any $e \in \mathbb{Z}$,
where the left-hand side is the ratio of the $m$th root of the right-hand side of (<ref>)
for $\zeta^e x$ and $x$.
The second equality in (<ref>) follows from

\begin{align*}
\prod_{t=0}^{p-1} (\zeta^{tq} x;\zeta)_{e}
= \prod_{i=0}^{e-1} \prod_{t=0}^{p-1} (1-\zeta^{tq+i} x)
= \prod_{t=0}^{pe-1} (1-\zeta^{tq} x)
= (x; \zeta^q)_{pe}.
\end{align*}