11institutetext: National Heart & Lung Institute, Faculty of Medicine,
Imperial College London, United Kingdom 22institutetext: Turing Research and
Innovation Cluster: Digital Twins, The Alan Turing Institute, London, United
Kingdom 33institutetext: Department of Computer Science, University College
London, United Kingdom
44institutetext: School of Computer Science and Engineering, University of New
South Wales, Sydney, Australia
44email<EMAIL_ADDRESS>
44email<EMAIL_ADDRESS><EMAIL_ADDRESS>
# Assessment of Left Atrium Motion Deformation Through Full Cardiac Cycle

Abdul Qayyum 1122 Moona Mazher 33 Angela Lee 11 Jose A Solis-Lemus 11
Imran Razzak 33 Steven A Niederer 1122

###### Abstract

Unlike Right Atrium (RA), Left Atrium (LA) presents distinctive challenges,
including much thinner myocardial walls, complex and irregular morphology, as
well as diversity in individual’s structure, making off-the-shelf methods
designed for the Left Ventricle (LV) may not work in the context of the left
atrium. To overcome aforementioned challenges, we are the first to present
comprehensive technical workflow designed for 4D registration modeling to
automatically analyze LA motion using high-resolution 3D Cine MR images. We
integrate segmentation network and 4D registration process to precisely
delineate LA segmentation throughout the full cardiac cycle. Additionally, an
image 4D registration network is employed to extract LA displacement vector
fields (DVFs). Our findings show the potential of proposed end to end
framework in providing clinicians with novel regional biomarkers for left
atrium motion tracking and deformation, carrying significant clinical
implications.

###### Keywords:

Left atrium Displacement vector fields 4D registration Full cardiac cycle.

##  1 Introduction

The operational and structural integrity of the left atrium (LA) is adversely
affected in conditions such as heart failure (HF) and atrial fibrillation
(AF). Currently assessment methods mainly rely on one-dimensional or global
changes in wall length of atria, which is estimated from single-plane view of
cardiac and quantified as left atrium strain. In contrast to LV, that benefits
from various dedicated 3D regional metrics for deformation and motion tracking
with known clinical correlates, LA lacks such evaluation and quantification
methods [6, 8]. The potential of 3D full coverage strain measurements for the
LA lies in their ability to capture crucial information about tissue bio-
mechanics, revealing features such as the presence of stiffness, fibrotic
regions in LA and expanding applications in clinics. Similar to LV, these
measurements can identify atrial dys-synchrony, offering early detection
before functional biomarkers i.e. ejection fraction show abnormalities. Thus,
the generation of 3D higher resolution spatio-temporal map of LA deformation
may give specific LA pathology signatures, enhancing diagnostic and prognostic
performance. Currently, two main imaging techniques have been used for LA
motion tracking: Feature Tracking (FT) [19] and 2D Speckle Tracking
Echocardiography (STE)[1]. Both FT and STE track the position of regions of
interest across the cardiac cycle, utilizing optical flow techniques such as
block matching or image registration for alignment . Ultimately, they estimate
Displacement Vector Fields (DVFs)[5], which link material points in myocardium
across different phases of cardiac cycle [18, 11]. Mathematical manipulation
of the DVFs allows for the estimation of strain tensors. Recently, deep
learning (DL) based image registration with applications in medical imaging
showed promising results for LA motion tracking [3, 13, 4, 3, 12]. Hence,
various DL based image registration methods have been proposed for analyzing
LV deformation across the cardiac cycle from Cine MR images, using
architectures such as U-Net [10, 20, 17], variational autoencoders [2, 15],
pyramid networks [22] or Siamese autoencoders [23, 14]. Typically,
displacement fields are learned in an unsupervised manner[5], employing a
spatial resampling module. To enhance performance, segmentation maps has been
integrated into cardiac motion tracking networks, either by incorporating
segmentation outputs as part of the input or by constraining DVFs to align
with target segmentation. Multiple cardiac MRI views (3D and 2D) is also being
considered to improve motion estimation. Biomechanics-informed modeling [16]
has also been used which leverages underlying deformation properties of the
heart to regularize DVFs toward more realistic motion [9]. While research to
estimate 3D based DVFs for LV from MR images matured over the years,
addressing specific challenges posed by the LA remains a challenging task i.e.
LA presents distinctive challenges, including much thinner myocardial walls
[21], complex and irregular morphology, and diversity in individual’s
structure [7], making off-the-shelf techniques designed for the LV unlikely to
work in the context of the LA. In this work, we are the first time to present
4D Registration model aided with motion tracking module for full cardiac cycle
using three different networks to predict DVF. The key contributions of this
work are: (i) propose 3D segmentation model (hybrid CNN and transformer) for
full cardiac cycle, trained based on ground-truth ED and ES frames of full
cardiac cycle of left atrium dataset, (ii) present 4D registration model for
full cardiac cycle using three various networks to predict DVF, (iii) propose
motion tracking module and spatiotemporal Mask autoencoder network to utilize
the representational capacity of conditional variational autoencoder (CVAE) as
a foundation for mapping organ deformations across a low-dimensional space
that encapsulates concise representations of respiratory states.

##  2 Method

Figure 2 illustrate the proposed novel framework for estimating left atrial
scar or fibrosis throughout the entire cardiac cycle. We utilize 24 frames of
3D short-axis view to segment scar in each 3D short-axis frame using hybrid
segmentation (described in sec. 2.1), followed by generating the mesh
encompassing the full cardiac cycle based on the segmented frames. We then
present 4D registration model (described in section 2.2) and construct DVF for
each frame in full cardiac cycle. Finally, DFV and segmented scare of each
frame passed to spatial-temporal network to obtain the meshes. Figure 1
demonstrate mesh generation through the integration of registration and
segmentation. Subsequently, we delve into the model components, offering
comprehensive explanations and elucidating their operational intricacies.

Figure 1: overall pipeline of proposed method for LA segmentation and
registration  Figure 2: 4D Registration method to estimate DVF for LA motion
tracking

###  2.1 Fibrosis Segmentation

We present 3D hybrid fibrosis segmentation framework which consist of
transformer aided with attention window-based mechanism in encoder, whereas
normal 3D-CNN based module has been used in decoder. We have used rectangular-
paralleled-piped windows based swin transformer which is based on computation
of attention within the partition of local windows (which are non-overlapping)
in low resolution features and original image. It uses paralleled-piped
rectangular windows to deal with non-square images through parallel-cross
attention. In order to obtain distinctive feature set from each convolutional
block in encoder, each block consists of 3D convolutional layers with ReLU
activation function and batch-normalization. The three-level deep-supervision
is applied to get the aggregated loss between prediction and ground truth.
Notice that we have trained model on two frame and used to predict
segmentation of full cardiac cycle.

###  2.2 4D Registration Model

Figure 2 illustrate schematic illustration of our framework, which comprises
(1) motion estimation using DVF inference network, (2) CVAE designed to learn
the DVF distribution concerning the surrogate and multi-time predictive
module. We formalized modeling task, emphasizing its utility for real-time
image-guided motion tracking and estimation for left atrium.

We consider a collection of $P$ time-resolved 3D acquisitions spanning $k$
respiratory cycles, generating $T$ temporal volumes. Within each subject’s
dataset $s\in SV_{t=m}$, we select $V_{t=m}$ at a specific respiratory state
$m$ as the reference volume which is utilized to estimate relative-
displacements ($\mathbf{dvf_{r}}$) to other temporal volumes
($V_{t^{\prime}}$). The motion noticed between fixed $V_{t}\in R^{H\times
W\times D}$ and moving image $V_{m}\in R^{H\times W\times D}$, is
characterized by a dense displacement field $dvf_{t}\in R^{H\times W\times
D}$, where $t\in[1,T]\neq m$ and $H$, $W$ and $D$ denotes the height, width,
and depth respectively. The objective of motion modelling task is three fold:
(i) it involves mapping each deformation to low-dimensional space
$dvf_{s}=dvf_{1}......dvf_{T-1}$, where ${}_{t}\rightarrow z_{t}\in R^{d}$ and
$d\leq\leq H\times W\times D$ thereby summarizing the observed inputs into
compact representation, (ii) it fulfills temporal requirements such that
deformations must be forecasted ahead-of-time, (iii) establish a relationship
between partial observations and the corresponding extrapolated-in-time dense
deformations.

Motion Tracking Module: In our scenario, the temporal volumes obtained during
free breathing exhibit variations due to tissue deformation caused by complex
respiratory motion. The deformations due to complex respiratory motion occur
in higher dimensional space, and can be determined by the number of motion and
voxels components, whereas spatio-temporal changes are primarily driven by a
small set of degrees of freedom, implying that the underlying structure can be
represented by a lower-dimensional subspace. Consequently, we can view these
variations as points on manifold with significantly fewer dimensions. Besides,
Autoencoders offer a solution by leveraging their ability to learn mapping
(nonlinear parametric) from volume deformation to latent representations,
thereby uncovering these embeddings.

We have a set of deformations $dvf=\\{dvf_{1}....dvf_{T-1}\\}$ , and a
sequence of 2D slices $I_{seq}=\\{I_{t},.....I_{t-m}\\}$ , and $V_{ref}$ as
the reference volume. The objective is to maximize the conditional probability
distribution $p_{true}(dvf(dvf|I_{seq})$ to obtain the deformations sequence
$dvf\in R^{H\times W\times D\times 3\times m}$ though, we have partially
available information and anatomy of subjects, where $H$, $W$, $D$, and m
represent height, width, depth and the number of predicted time steps,
respectively. Accordingly, our goal is to learn parameterized model (with
parameters $\theta$) in order to sample new phase-specific deformations,
resembling samples from unknown distribution $p$.

An encoder network is employed to approximate this distribution.

$q\phi\big{(}z_{i}|I_{seq},V_{ref}\big{)}=\mu\big{(}dvf_{i},I_{seq},V_{ref}\big{)},\sigma(dvf_{i},I_{seq},V_{ref}\big{)}$
(1)

The network structured with stacked 3D convolutional layers, learns both the
mean $\mu\in R^{d}$ and the diagonal covariance $\sigma\in R^{d}$ from the
data, as described in Figure 2. During the training, sampling of $z_{i}$ is
made differentiable with respect to $\mu$ and $\sigma$ by employing the
reparameterization trick. This is achieved by defining
$z_{t}=\mu+\epsilon*\sigma$ where $\epsilon$ is drawn from a standard normal
distribution $\epsilon\sim N(0,1)$.

KL-divergence is employed in the CVAE, to minimize the distance between the
distributions $p_{\theta}$ and $p_{\phi}$. This KL-divergence loss term is
incorporated into the total loss function, which also includes minimizing a
reconstruction loss. As part of the spatial warping block, the reference
volume undergoes warping (represented by the symbol $o$) using the
transformation provided by the decoder. This enables the model to compute a
reconstruction term $l_{rec}$ between the warped reference volume
($V_{ref}\quad o\quad dfv_{i}$) and expected volume $V_{t}$. We have used the
stochastic gradient descent to find the optimal parameters by minimizing the
following loss function.

$argmin_{\theta}\Bigg{[}\frac{1}{n}\sum_{i=1}^{n}l_{rec}\big{(}V_{ref}dfv_{t},V_{t}\big{)}+KL\big{(}q_{\phi}(Z_{i}|I_{seq},V_{ref})||P_{\theta(z_{i})}\big{)}\Bigg{]}$
(2)

$l_{rec}=l_{sim}\Big{(}(V_{t},V^{\prime}_{i})+\sigma
l_{smooth}(dvf_{t})\Big{)}$ (3)

Where $V^{\prime}{{}_{t}}$ results from warping $V_{ref}$ with estimated
motion $dvf$, $\sigma$ is a regularization parameter and
$l_{smooth}(dvf_{l})\sum_{p\in R^{3}}||\delta dvf(p)||^{2}$ computes the
differences between neighbouring 3D position $p$. We applied Fourier band-
limited deformation network (Eq.2) to calculate the displacement vector field
(DVF) between two frames. Subsequently, we fine-tuned this network to estimate
DVF for all frames, effectively serving as an inference network. Notice that,
proposed motion tracking framework is designed to be independent of the
specific method employed for deformable registration.

Spatio-temporal $2D+t$ Mask autoencoder Network: We employed a spatio-temporal
network that utilizes a 2D+time input sequence which leverages a mask
autoencoder pretrained model with an 8-time step configuration to extract
features similar to 3D predictive model.

##  3 Experiment

###  3.1 Dataset

We have collected forty-one (41) subjects, each aged thirty-two (32) on
average, with an average ejection fraction of 60. The dataset was acquired
using a Siemens Aera machine, with voxel sizes ranging from $1.6mm\times
1.6mm$ to $2.1mm\times 2.1mm$ and slice thickness varying between 4mm and 8mm.
The images were of dimensions ranging from $210\times 280$ to $400\times 300$,
with the $z$ dimension varying from 50 to 80 slices. For training of
segmentation model, manual segmentation of the left atrium was conducted on
end-diastolic (ED) and end-systolic (ES) frames. Additionally, manual
annotations were available for 5 cases covering the full cardiac cycle, which
were utilized to evaluate the segmentation and registration performance.

###  3.2 Results and Discussion

Table 1. shows average Dice coefficients and Hausdorff Distance (mm) on test-
set using full cardiac frames. We have compared different variations of
proposed framework. We can notice that Motion+MAE encoder achieved higher Dice
and lower HD.

Figure.3 shows the visualization of constructed meshes for first frame, mid
frame-12 and last frame-24 for subject-1, using proposed Motion Encoder with
single encoder (after dvf), proposed MAE encoder and Motion+MAE encoder (after
dvf) whereas the last column shows meshes produced after registration using
VoxelMorph. To assess the precision of the 4D registration algorithm, we
conduct a comparative analysis between the segmented LA following registration
and the ground truth LA segmentation for corresponding cardiac phases. We can
notice that our model with different variations shows smooth meshes as
compared the voxelmorph which showed less surface area.

Our primary focus is to utilize DVFs to characterize LA motion throughout the
entire cardiac cycle. Figure 3, 4 and Figure 6 shows the comparative analysis
of dice and HD score of subject 1 across the entire cardiac cycle,
encompassing 24 frames. The visualizations include the ground-truth
segmentation post-registration, alongside the moved or predicted segmentation
mask (generated after applying Deformation Vector Fields, DVFs). Additionally,
a 3D segmentation map of both the ground truth and predicted segmentation for
the full cardiac cycle is provided. The proposed 4D registration method
consistently outperforms in segmenting each frame, as demonstrated in Figure
3.

Results showed that by combining the Motion and mask autoencoder (MAE)
encoders consistently yielded higher Dice coefficients compared to using
individual encoders such as MAE only or Motion only. Conversely, VoxelMorph
did not achieve superior performance, as depicted in Figure 4 and Figure 6.
Furthermore, the combined Motion and MAE encoder resulted in lower Hausdorff
Distance (HD) compared to the individual encoders, as illustrated in Figure 4.
Integrating the segmentation and image registration tasks, given their
interdependence, enhanced the overall performance, as demonstrated for the LV.

Experiment on manual segmentated LA dataset shows that our proposed framework
has successfully achieved accurate estimation of high-resolution left atrial
(LA) displacement and meshes across the entire cardiac cycle for all 41
subjects. The segmentation model showcases outstanding precision in
delineating the LA across diverse phases of the cardiac cycle, displaying
superior performance in phases closely aligned with the manually annotated
segmentation used for training. Moreover, our 4D registration approach
exhibits remarkable accuracy in tracking the LA throughout the cardiac cycle,
particularly during phases characterized by substantial displacements.

Table 1: Average Dice and HD for validation cases using 4D registration Proposed and VoxelMorph methods. Methods | Dice Score | Hausdorff Distance (mm)
---|---|---
Motion Encoder | 0.841± 0.055 | 7.68±4.81
MAE encoder | 0.829± 0.046 | 8.42± 5.10
Motion+MAE encoder | 0.880 ± 0.027 | 6.59± 4.27
VoxelMorph | 0.751±0.100 | 9.57±6.10
Figure 3: Validation results for subject 1 over the entire cardiac cycle,
comparing the performance of the proposed and existing deep learning models.

Due to our use of a novel high-resolution imaging protocol, our dataset is
relatively small, making traditional AI analysis techniques unsuitable. Our
model selection addresses the challenges posed by limited data and effectively
manages the substantial structural and functional variability of the LA.
Training the networks on a subject-by-subject basis mitigates biases from the
training data, ensuring continued application in future studies analyzing
images from subjects with diverse clinical presentations. Moreover, our
proposed method is well-suited for future analyses of the motion tracking and
deformation of the right atrium or right ventricle, which is oftenly
approximated as thin surfaces.

Figure 4: HD for full cardiac cycle using validation subject1 based on
proposed and existing registration methods. Figure 5: Dice Figure 6:  Dice for
full cardiac cycle using validation subject1 based on proposed and existing
registration methods. Figure 7: Visualization of results on subject-1 for full
cardiac cycle

Regional analyses of LA motion and deformation hold promise for identifying
dense fibrosis or scar tissue in the LA, as these regions typically exhibit
greater stiffness than healthy myocardium. Current MRI-based techniques for LA
fibrosis detection, relying on late-gadolinium enhancement images, face
subjectivity and low reproducibility in analysis. Future investigations will
explore whether 3D Cine MRI could serve as an alternative or complementary
technique for identifying LA fibrosis, potentially influencing treatment
strategies for atrial fibrillation (AF) and the personalization of catheter
ablations in AF. Furthermore, leveraging biomechanics-informed regularization,
potentially within a physics-informed neural network setting, may offer
additional performance improvements. Collectively, these enhancements may
contribute to more reliable DVF and strain estimations.

##  4 Conclusion

In this work, we presented a segmentation and 4D registration tool aimed at
delivering robust 3D displacement vector fields (DVFs) and strain maps across
the entire left atrium (LA), to capture information throughout the full
cardiac cycle. An end to end framework demonstrated efficient performance in
accurately estimating meshes representing the LA, enabling the derivation of
strain and fibrosis metrics from meshes generated after DVFs are applied.
Rigorous testing and evaluation using our proprietary full cardiac dataset
specific to the LA confirm the effectiveness of our approach. Anticipating
that our methodology will facilitate the identification of novel clinical
prognostic and diagnostic biomarkers related to LA function, in future, we aim
to enhance our understanding of the atria’s role in cardiovascular disease
processes.

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List all orbits of length $t$ containing $r$ involutions | Fix $t|n$ and $r\leq t$ |  There are $|\frac{X_{n/t,r}}{r}|$ permutations with stabilizer $F_{x}=\langle a^{t}\rangle$ | Proposition 5.3.2
Count number of $t$-dimensional irreps of $J_{n}$ with indicator $+1$ | Fix $t|n$ with $t$ odd |  There are $I_{n/t}^{(+1)}$ irreps of $J_{n}$ of dimension $t$ with indicator $+1$ | Theorem 6.1.2
Count number of $t$-dimensional irreps of $J_{n}$ with indicator $0$ | Fix $t|n$ with $t$ odd |  There are $I_{n/t}^{(0)}=\frac{n}{t}|M_{n/t}|-I_{n/t}^{(+1)}$ irreps of $J_{n}$ of dimension $t$ with indicator $0$ | Theorem 6.1.2
Count number of $t$-dimensional irreps of $J_{n}$ with indicator $+1$ | Fix $t|n$ with $t$ even |  $I_{n/t}^{(+1)}$ is an underestimate for total irreps of $J_{n}$ of dimension $t$ with indicator $+1$ | Theorem 6.1.2
Describe indicator of $t$-dimensional irreps of $J_{n}$ | If $t=2$ |  Irreps of dimension $t$ with indicators $+1,-1,0$ given by $I_{n/2}^{(+1)},I_{n/2}^{(-1)},I_{n/2}^{(0)}$ respectively |  Theorem 6.2.2 Proposition 6.2.1

Note: A table summarizing some results which are now obtainable through a
counting formula.

We end this section with a corollary mentioned earlier.

######  Corollary 6.2.3.

If $4$ divides $n$ and $n\geq 12$, there exists an irrep of $J_{n}$ of
dimension $2$ of with indicator $-1.$

Proof. Assume $n\geq 12$. Let $t=2$. Then let $j=1\in E_{n/2}$ for any $n\geq
12$ (in fact, $1\in E_{n/t}$ for any $\frac{n}{t}>1$). Let $u=2.$ Then
$(u+1)(j-1)=(3)(0)=0\mod\frac{n}{2}$ for any $n\geq 12$. Thus, the permutation

$x=(1\,\,\,5\,\,\,9\,\,\,\cdots\,\,\,n-3)(3\,\,\,7\,\,\,11\,\,\,\cdots\,\,\,n-1)$

satisfies that $\mathcal{O}_{x}=\\{x,x^{-1}\\}$ and for $i=\frac{n}{4}$, the
(group) irrep

$\displaystyle\rho_{x,\frac{n}{4}}:\langle a^{2}\rangle$
$\displaystyle\to\mathbb{C}^{\times}$ $\displaystyle a^{2}$
$\displaystyle\mapsto(\zeta_{\frac{n}{2}})^{\frac{n}{4}}$

satisfies that $i(j+1)=2\frac{n}{4}=0\mod\frac{n}{2}$ but
$i(u+1)=3\frac{n}{4}\not=0\mod\frac{n}{2}$. Therefore, the induced $J_{n}$
module $\hat{V}_{x,\frac{n}{4}}$ has indicator $-1.$ ∎

##  Limiting Behavior

Theorem 6.1.2 suggests that the conditions that a permutation $x\in S_{n-1}$
must meet in order to induce an irrep of $J_{n}$ with indicator $+1$ are very
restrictive. This intuition turns out to be correct and can be formalized in
the following theorem. However, to prove it, we must give a few bounding
relationships.

######  Lemma 7.1.

Let $E_{n/t}$, where $\frac{n}{t}\geq 2$, be as from (1) in 1 ( (Big Table of
Commonly Cited Values and Sets).), $\varphi(m)$ be the Euler totient function
which is the number of integers smaller than $m$ which are coprime to $m$, and
finally let $\omega(m)$ is the number of distinct prime divisors of $m$. Then
for all $m\geq 3$

  1. (1)

$|E_{n/t}|\leq 2^{\omega(n/t)+2}$

  2. (2)

$\omega(m)\leq 1.3841\frac{\log m}{\log(\log m)}$

  3. (3)

$\varphi(m)\geq\frac{1}{\sqrt{2}}\sqrt{m}$

  4. (4)

$\displaystyle\lim_{m\to\infty}\frac{2^{1.3841\frac{\log(m)}{\log(\log(m))}}}{\sqrt{m}}=0$

Proof.

  1. (1)

This is a well known result which comes from the Chinese Remainder Theorem.
Note the special case if $\frac{n}{t}=2$ then $E_{n/t}=\\{1\\}$ and
$\omega(n/t)=\omega(2)=1$ so the bound provided holds.

  2. (2)

Robin proves this in [16].

  3. (3)

Let $m=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\cdots p_{u}^{\alpha_{u}}$. Then
since $\alpha_{i}\geq 1$

$\frac{\varphi(m)^{2}}{m}=\prod_{i=1}^{u}\frac{(p_{i}^{\alpha_{i}-1}(p_{i}-1))^{2}}{p_{i}^{\alpha_{i}}}=\prod_{i=1}^{u}p_{i}^{\alpha_{i}-2}(p_{i}-1)^{2}\geq\prod_{i=1}^{u}p_{i}^{-1}(p_{i}-1)^{2}=\prod_{i=1}^{u}\frac{(p_{i}-1)^{2}}{p_{i}}\geq\frac{1}{2}.$

Note the last step is because if $p_{i}>2$, then
$(p_{i}-1)^{2}=p_{i}^{2}-2p_{i}+1=p_{i}(p_{i}-2)+1\geq p_{i}+1\geq p_{i}$ and
so $\frac{(p_{i}-1)^{2}}{p_{i}}\geq 1$. Of course, if $p_{i}=2$ then
$\frac{(p_{i}-1)^{2}}{p_{i}}=\frac{1}{2}$.

Finally, we get that

$\frac{\varphi(m)^{2}}{m}\geq\frac{1}{2}\implies\varphi(m)\geq\frac{1}{\sqrt{2}}\sqrt{m}.$

  4. (4)

This is elementary calculus. Let
$L=\lim_{m\to\infty}\frac{2^{1.3841\frac{\log(m)}{\log(\log(m))}}}{\sqrt{m}}=0$.
Then a standard logarithm trick gives

$\displaystyle\log L$
$\displaystyle=\lim_{m\to\infty}\left(1.3841\frac{\log(m)}{\log(\log(m))}\log
2-\log\left(\sqrt{m}\right)\right)$
$\displaystyle=\lim_{m\to\infty}-\frac{1}{2}\log(m)=-\infty$

and so $L=0.$∎

Now, we can make precise what we mean by saying it is rare for a permutation
to satisfy the conditions we have discussed so far.

But first, we show that there are many more involutions than permutations with
stabilizer $\langle a^{t}\rangle.$

######  Lemma 7.2.

Fix $t$ and let $n=mt$ for $m\geq 2.$ Let $i_{n-1}$ be the number of
involutions in $S_{n-1}$. Then $\frac{|M_{n/t}|}{i_{n-1}}\to 0$ as
$m\to\infty.$

Proof. Fix $t$ and let $n=mt$ for $m\geq 2$. Note that
$|M_{n/t}|\leq\left(\frac{n}{t}\right)^{t}t!$. Now, in the proof of
Proposition 5.2.1, we show that

$i_{n-1}=\sum_{l=0}^{\lfloor\frac{n-1}{2}\rfloor}\frac{(n-1)!}{(n-1-2l)!2^{l}l!}.$

In fact, [2] show that $i_{n-1}\geq\sqrt{n-1}i_{i-2}$ and so inductively we
obtain that $i_{n-1}\geq\sqrt{(n-1)!}\geq 2^{\frac{n-1}{2}}$. Note that
$(n-1)!\geq 2^{n-1}$ for $n\geq 5$ which is easily provable by induction.

Thus,

$\lim_{m\to\infty}\frac{|M_{n/t}|}{i_{n-1}}\leq\lim_{m\to\infty}\frac{\left(m\right)^{t}t!}{2^{\frac{mt-1}{2}}}=0$

The last limit is zero using L’Hopital’s rule. ∎

Additionally, we can show that involutions with a given stabilizer are rare
when compared to general permutations with a given stabilizer.

######  Lemma 7.3.

Fix $t$ and let $n=mt$ for $m\geq 2.$ Then $\frac{|T_{n/t}|}{|M_{n/t}|}\to 0$
as $m\to\infty.$

###### Proof.

  1. (1)

To start, let $t=1$, then

$\lim_{m\to\infty}\frac{|T_{n/1}|}{|M_{n/1}|}=\lim_{m\to\infty}\frac{|E_{n/1}|}{\varphi(m)}\leq\lim_{m\to\infty}\frac{2^{1.3841\frac{\log(m)}{\log(\log(m))}}}{\sqrt{m}}=0$

by Lemma 7.1.

  2. (2)

Now, assume $t>1$. We will find an upper bound for the numerator and a lower
bound for the denominator to bound the entire ratio.

First, we set an upper bound for $|T_{n/t}|$. Let $n=mt$ for
$m=\frac{n}{t}\geq 2$.

$\displaystyle|T_{n/t}|$ $\displaystyle=\sum_{j\in
E_{n/t}}\sum_{l=0}^{\left\lfloor\frac{t-1}{2}\right\rfloor}(\alpha_{j,n/t})^{t-1-2l}\left(\frac{n}{t}\right)^{l}\frac{(t-1)!}{(t-1-2l)!2^{l}l!}-\sum_{\begin{subarray}{c}s|t\\\
s\not=t\end{subarray}}|T_{n/s}|$ $\displaystyle\leq\sum_{j\in
E_{n/t}}\sum_{l=0}^{\left\lfloor\frac{t-1}{2}\right\rfloor}(\alpha_{j,n/t})^{t-1-2l}\left(\frac{n}{t}\right)^{l}\frac{(t-1)!}{(t-1-2l)!2^{l}l!}\qquad\text{
forget }-\sum_{\begin{subarray}{c}s|t\\\ s\not=t\end{subarray}}|T_{n/s}|$
$\displaystyle\leq|E_{n/t}|\sum_{l=0}^{\left\lfloor\frac{t-1}{2}\right\rfloor}\left(\frac{n}{t}\right)^{t-1-l}\frac{(t-1)!}{(t-1-2l)!2^{l}l!}\qquad\alpha_{j,n/t}\leq\frac{n}{t}$
$\displaystyle\leq|E_{n/t}|\left(\frac{n}{t}\right)^{t-1}t!$
$\displaystyle\leq\left(\frac{n}{t}\right)^{t-1}t!2^{1.3841\frac{\log
m}{\log(\log m)}+2}\qquad\text{\hyperref@@ii[lem:some bounds
(2)]{\lx@cref{creftype}{lem:some bounds}\leavevmode\nobreak\
\lx@cref{refnum}{*}{lem:some bounds}\leavevmode\nobreak\ \ref*{lem:some bounds
(2)}} }$

Second, we set a lower bound for $|M_{n/t}|$ for $t>1$. Note that
$|M_{n/s}|\leq\left(\frac{n}{s}\right)^{s}(s-1)!$ and so

$\displaystyle\frac{n}{t}|M_{n/t}|$
$\displaystyle=\varphi\left(\frac{n}{t}\right)\left(\frac{n}{t}\right)^{t-1}(t-1)!-\sum_{\begin{subarray}{c}s|t\\\
s\not=t\end{subarray}}|M_{n/s}|$
$\displaystyle\geq\frac{1}{\sqrt{2}}\sqrt{\frac{n}{t}}\left(\frac{n}{t}\right)^{t-1}(t-1)!-\sum_{\begin{subarray}{c}s|t\\\
s\not=t\end{subarray}}|M_{n/s}|\qquad\text{ \hyperref@@ii[lem:some bounds
(3)]{\lx@cref{creftype}{lem:some bounds}\leavevmode\nobreak\
\lx@cref{refnum}{*}{lem:some bounds}\leavevmode\nobreak\ \ref*{lem:some bounds
(3)}} }$
$\displaystyle\geq\frac{1}{\sqrt{2}}\sqrt{\frac{n}{t}}\left(\frac{n}{t}\right)^{t-1}(t-1)!-\sum_{\begin{subarray}{c}s|t\\\
s\not=t\end{subarray}}\left(\frac{n}{s}\right)^{s}(s-1)!$

Finally, we get

$\displaystyle\lim_{m\to\infty}\frac{|T_{n/t}|}{|M_{n/t}|}$
$\displaystyle\leq\lim_{m\to\infty}\frac{\left(\frac{n}{t}\right)^{t-1}t!2^{1.3841\frac{\log
m}{\log(\log
m)}+2}}{\frac{1}{\sqrt{2}}\sqrt{\frac{n}{t}}\left(\frac{n}{t}\right)^{t-1}(t-1)!-\sum_{\begin{subarray}{c}s|t\\\
s\not=t\end{subarray}}\left(\frac{n}{s}\right)^{s}(s-1)!}$
$\displaystyle=\lim_{m\to\infty}\frac{\left(m\right)^{t-1}t!2^{1.3841\frac{\log
m}{\log(\log
m)}+2}}{\sqrt{m}\left(m\right)^{t-1}\left(\frac{1}{\sqrt{2}}(t-1)!-\left(\frac{1}{m}\right)^{t-1}\sum_{\begin{subarray}{c}s|t\\\
s\not=t\end{subarray}}\left(m\frac{t}{s}\right)^{s}(s-1)!\right)}$
$\displaystyle=\lim_{m\to\infty}4\sqrt{2}t\frac{2^{1.3841\frac{\log
m}{\log(\log m)}}}{\sqrt{m}}$
$\displaystyle=0\qquad\text{\hyperref@@ii[lem:some bounds
(4)]{\lx@cref{creftype}{lem:some bounds}\leavevmode\nobreak\
\lx@cref{refnum}{*}{lem:some bounds}\leavevmode\nobreak\ \ref*{lem:some bounds
(4)}} }\qed$

We now give one of our most powerful results from this section.

######  Theorem 7.4.

Fix $t$ an odd integer. Let $n=mt$ for some integer $m\geq 2$. Then the number
ratio of irreps of $J_{n}$ of dimension $t$ with indicator $+1$ to the total
number of irreps of dimension $t$ converges to $0$ as $m\to\infty$.

###### Proof.

The number of permutations $S_{n-1}$ which have stabilizer $F_{x}=\langle
a^{t}\rangle$ is given by $|M_{n/t}|$ by Proposition 5.1.1. There are
$\frac{n}{t}=|F_{x}|$ irreducible (group) representations of $F_{x}$ and by
Theorem 3.2 and Lemma 3.3, each of these group irreps of $F_{x}$ induce a
unique irrep of $J_{n}$ of dimension $t=[F:F_{x}]$.

Thus, there are $\frac{n}{t}|M_{n/t}|$ total irreps of $J_{n}$ of dimension
$t.$

From Theorem 6.1.2, $I_{n/t}^{(+1)}$ is the number of irreps of $J_{n}$ of
dimension $t$ with indicator $+1$ and so the ratio is given by

$\frac{I_{n/t}^{(+1)}}{\frac{n}{t}|M_{n/t}|}.$

To show convegence to $0$, we will place an upper bound on $I_{n/t}^{(+1)}$
and a lower bound on $\frac{n}{t}|M_{n/t}|$ and then show that this ratio
converges to zero. We do this in two cases, first for $t>1$ then for $t=1$.

  1. (1)

When $t=1$, $|M_{n/1}|=\varphi(n)\geq\frac{1}{\sqrt{2}}\sqrt{n}$ and

$I_{n/1}^{(+1)}=\sum_{j\in E_{n/1}}\alpha_{j,n/1}\leq
n2^{1.3841\frac{\log(m)}{\log(\log(m))}+2}$

so

$\lim_{n\to\infty}\frac{I_{n/1}^{(+1)}}{n|M_{n/1}|}\leq\lim_{n\to\infty}\frac{4n2^{1.3841\frac{\log(n)}{\log(\log(n))}}}{\frac{1}{\sqrt{2}}n\sqrt{n}}=\lim_{n\to\infty}4\sqrt{2}\frac{2^{1.3841\frac{\log(n)}{\log(\log(n))}}}{\sqrt{n}}=0$

  2. (2)

First, an upper bound on $I_{n/t}^{(+1)}$. Let $n=mt$ for $m\geq 2$ and $t>1$.

$\displaystyle I_{n/t}^{(+1)}$ $\displaystyle=t\sum_{j\in
E_{n/t}}\sum_{r=1}^{t}\alpha_{j,n/t}O_{n/t,r,j}$
$\displaystyle\leq\frac{t}{r}\sum_{j\in
E_{n/t}}\sum_{r=1}^{t}\alpha_{j,n/t}(\alpha_{j,n/t})^{r-1}\left(\frac{n}{t}\right)^{\frac{t-r}{2}}\frac{(t-1)!}{(r-1)!2^{\frac{t-r}{2}}\left(\frac{t-r}{2}\right)!}\qquad\text{
ignore }-\tilde{C}\text{ terms}.$ $\displaystyle\leq\sum_{j\in
E_{n/t}}\sum_{r=1}^{t}\left(\frac{n}{t}\right)^{r}\left(\frac{n}{t}\right)^{\frac{t-r}{2}}\frac{t!}{r!2^{\frac{t-r}{2}}\left(\frac{t-r}{2}\right)!}\qquad\alpha_{j,n/t}\leq\frac{n}{t}$
$\displaystyle\leq\sum_{r=1}^{t}2^{\omega(n/t)+2}\left(\frac{n}{t}\right)^{\frac{t+r}{2}}\frac{t!}{r!2^{\frac{t-r}{2}}\left(\frac{t-r}{2}\right)!}\qquad\text{
\hyperref@@ii[lem:some bounds (1)]{\lx@cref{creftype}{lem:some
bounds}\leavevmode\nobreak\ \lx@cref{refnum}{*}{lem:some
bounds}\leavevmode\nobreak\ \ref*{lem:some bounds (1)}} }$
$\displaystyle\leq\sum_{r=1}^{t}2^{1.3841\frac{\log(n/t)}{\log(\log(n/t))}+2}\left(\frac{n}{t}\right)^{\frac{t+r}{2}}\frac{t!}{r!2^{\frac{t-r}{2}}\left(\frac{t-r}{2}\right)!}\qquad\text{
\hyperref@@ii[lem:some bounds (2)]{\lx@cref{creftype}{lem:some
bounds}\leavevmode\nobreak\ \lx@cref{refnum}{*}{lem:some
bounds}\leavevmode\nobreak\ \ref*{lem:some bounds (2)}} }$

Now, using the same lower bound as from Lemma 7.3, we get

$\frac{n}{t}|M_{n/t}|\geq\frac{n}{t}\left(\frac{1}{\sqrt{2}}\sqrt{\frac{n}{t}}\left(\frac{n}{t}\right)^{t-1}(t-1)!-\sum_{\begin{subarray}{c}s|t\\\
s\not=t\end{subarray}}\left(\frac{n}{s}\right)^{s}(s-1)!\right).$

Finally, noting that $\frac{n}{t}=m$ and $\frac{n}{s}=m\frac{t}{s}$, the ratio
converges

$\displaystyle\lim_{m\to\infty}\frac{I_{n/t}^{(+1)}}{\frac{n}{t}|M_{n/t}|}$
$\displaystyle\leq\lim_{m\to\infty}\frac{\sum_{r=1}^{t}2^{1.3841\frac{\log(m)}{\log(\log(m))}+2}\left(m\right)^{\frac{t+r}{2}}\frac{t!}{r!2^{\frac{t-r}{2}}\left(\frac{t-r}{2}\right)!}}{m\left(\frac{1}{\sqrt{2}}\sqrt{m}\left(m\right)^{t-1}(t-1)!-\sum_{\begin{subarray}{c}s|t\\\
s\not=t\end{subarray}}\left(m\frac{t}{s}\right)^{s}(s-1)!\right)}$
$\displaystyle=\lim_{m\to\infty}\frac{2^{1.3841\frac{\log(m)}{\log(\log(m))}}\left(\left(m\right)^{t}+\sum_{r=1}^{t-1}4\left(m\right)^{\frac{t+r}{2}}\frac{t!}{r!2^{\frac{t-r}{2}}\left(\frac{t-r}{2}\right)!}\right)}{\sqrt{m}\left(m\right)^{t}\left(\frac{1}{\sqrt{2}}(t-1)!-\left(\frac{t}{n}\right)^{t-1}\sum_{\begin{subarray}{c}s|t\\\
s\not=t\end{subarray}}\left(m\frac{t}{s}\right)^{s}(s-1)!\right)}$
$\displaystyle=\lim_{m\to\infty}\frac{2^{1.3841\frac{\log(m)}{\log(\log(m))}}\left(m\right)^{t}\left(1+\sum_{r=1}^{t-1}4\left(m\right)^{\frac{r-t}{2}}\frac{t!}{r!2^{\frac{t-r}{2}}\left(\frac{t-r}{2}\right)!}\right)}{\sqrt{m}\left(m\right)^{t}\left(\frac{1}{\sqrt{2}}(t-1)!-\left(\frac{1}{m}\right)^{t-1}\sum_{\begin{subarray}{c}s|t\\\
s\not=t\end{subarray}}\left(m\frac{t}{s}\right)^{s}(s-1)!\right)}$
$\displaystyle=\lim_{m\to\infty}\frac{2^{1.3841\frac{\log(m)}{\log(\log(m))}}}{\sqrt{m}\frac{1}{\sqrt{2}}(t-1)!}=0\qquad\text{
\hyperref@@ii[lem:some bounds (4)]{\lx@cref{creftype}{lem:some
bounds}\leavevmode\nobreak\ \lx@cref{refnum}{*}{lem:some
bounds}\leavevmode\nobreak\ \ref*{lem:some bounds (4)}} }\qed$

Theorem 7.4 tells us that odd $t$-dimensional irreps of $J_{n}$ with nonzero
indicator become rare as $n$ grows. In fact, we conjecture that for any $t$,
not only odd $t$, the number of irreps of $J_{n}$ of dimension $t$ with
nonzero indicator approaches $0.$

We summarize the results of these limit results as follows.

######  Corollary 7.5.

Fix $t$ and let $n=mt$. Then for large $m$ (and so large $n$)

  1. (1)

$i_{n-1}$ is much less than $(n-1)!$

  2. (2)

$|M_{n/t}|$ is much less than $i_{n-1}$

  3. (3)

$|T_{n/t}|$ is much less than $|M_{n/t}|$

  4. (4)

if $t$ is odd, $I_{n/t}^{(+1)}$ is much less than $|M_{n/t}|$

Proof. The only claim we have not proved is the first part. However, this
follows from another inequality of [2] which states that for $n\geq 3$,
$i_{n-1}\leq(\sqrt{n-1}+1)i_{n-2}\leq\sqrt{n}i_{n-2}\leq\sqrt{n!}$.

Thus,

$\lim_{n\to\infty}\frac{i_{n-1}}{(n-1)!}\leq\lim_{n\to\infty}\frac{\sqrt{n!}}{(n-1)!}=\lim_{n\to\infty}\sqrt{\frac{n}{(n-1)!}}=0$
∎

Now, using the formulas from Theorem 6.2.2, we can again show that the
proportion of $2$-dimensional irreps of $J_{n}$ with nonzero indicator
approaches $0.$

######  Theorem 7.6.

Let $n=2m$ for some integer $m\geq 2$. Then the number ratio of irreps of
$J_{n}$ of dimension $2$ with either indicator $+1$ or $-1$ (so nonzero
indicator) to the total number of irreps of dimension $2$ converges to $0$ as
$m\to\infty$.

Proof. We follow the same outline as Theorem 7.4.

From Theorem 6.2.2, we can use the following bounds.

$\displaystyle I_{n/2}^{(+1)}$ $\displaystyle\leq\sum_{j\in E_{n/2}}\sum_{u\in
K_{j,n/t}}2(2\alpha_{j,n/t})\qquad|K_{j,n/t}|=\alpha_{j,n/t}\text{ and set
both }\delta\text{ functions to }1$ $\displaystyle\leq\sum_{j\in
E_{n/t}}4(\alpha_{j,n/t})^{2}\qquad|K_{j,n/t}|=\alpha_{j,n/t}$
$\displaystyle\leq 2^{\omega(n/2)+4}\left(\frac{n}{2}\right)^{2}\qquad\text{
Lemma }\hyperref@@ii[lem:some bounds (1)]{\lx@cref{creftype}{lem:some
bounds}\leavevmode\nobreak\
\lx@cref{refnum}{*}{lem:somebounds}\leavevmode\nobreak\ \ref*{lem:some bounds
(1)}}$

Then, letting $n=2m$ for some $m\geq 2$ and using the same lower bound for
$\frac{n}{2}|M_{n/2}|$ we obtain

$\lim_{m\to\infty}\frac{I_{n/2}^{(+1)}}{\frac{n}{2}|M_{n/2}|}\leq\lim_{m\to\infty}\frac{2^{\omega(m)+4}m^{2}}{\sqrt{m}m^{2}\frac{1}{\sqrt{2}}}\leq\lim_{m\to\infty}16\sqrt{2}\frac{2^{1.3841\frac{\log(m)}{\log(\log(m))}}}{\sqrt{m}}=0$

using Lemma LABEL:*lem:some bounds (4) .

Similarly,

$I_{n/2}^{(-1)}\leq 2^{\omega(n/2)+3}\left(\frac{n}{2}\right)^{2}$

and so

$\lim_{m\to\infty}\frac{I_{n/2}^{(-1)}}{\frac{n}{2}|M_{n/2}|}\leq\lim_{m\to\infty}8\sqrt{2}\frac{2^{1.3841\frac{\log(m)}{\log(\log(m))}}}{\sqrt{m}}=0.$

Thus,

$\lim_{m\to\infty}\frac{I_{n/2}^{(+1)}+I_{n/2}^{(-1)}}{\frac{n}{2}|M_{n/2}|}=\lim_{m\to\infty}\frac{I_{n/2}^{(+1)}}{\frac{n}{2}|M_{n/2}|}+\lim_{m\to\infty}\frac{I_{n/2}^{(-1)}}{\frac{n}{2}|M_{n/2}|}=0$

and so the ratio of irreps of dimension $2$ of $J_{2m}$ which have nonzero
indicator to the total number of irreps of dimension $2$ approaches $0$ as
$m\to\infty.$ ∎

## Appendices

##  Appendix A Discussing the Python Code

Here we walk through examples that illustrate how use the theorems and
propositions of the previous section. We provide graphs of the various
formulas and show some direct computations to aid with visualizing the
indicators.

Throughout, we will let $t$ be a divisor of $n$.

The process of writing the recursive counting formulas of Section 5 went hand
in hand with directly computing examples using Python. In fact, all counting
formulas proved in Section 5 were cross checked against what was reasonable to
compute directly using Python.

At first, efforts were made to do direct computation by actually multiplying
permutations in Python using a series of combinatorics packages similar to
those used in GAP. Using this method, we ran out of processing power around
$n=12.$ Using the reformulation of the action described in Section 4, we were
able to drastically increase the computations possible. For example, when
$n<50$ and $t<9$, we can easily describe all indicators of irreps of $J_{n}$
of dimension $t.$

We are limited by the factorial growth of this problem and so when doing
computations it is important to choose either that $n$ is small ($n<20$
computes quickly) or that $t$ is small. For example, if we fix $t=2$, then it
took less than 30 minutes to describe all irreps of $J_{n}$ of dimension $2$
for $n\leq 150$ (see Appendix F).

Let us understand why the value of $t$ is so closely tied with computational
speed. In Python, each permutation is treated as a tuple. For example, if
$n=12$, the permutation

$x=(1\,\,\,3\,\,\,5)(2\,\,\,4)(7\,\,\,9\,\,\,11)(8\,\,\,10)$

is written in Python as

$x=[x(1),x(2),...,x(n)]=[3,4,5,2,1,6,9,10,11,8,7,12].$

Now, using Proposition 4.1.3, if we want to compute the orbit of $x$ we simply
take shifts of all the values of $x$. For example, the permutation
$y_{1}=a^{-x(1)}xa\in\mathcal{O}_{x}$ is characterized by the fact that
$y_{1}(i)=x(1+i)-x(1)$ and so using the previous example, we obtain easily
that

$y_{1}=[y_{1}(1),y_{1}(2),...,y_{1}(n)]=[x(2)-x(1),x(3)-x(1),...,x(1)-x(1)]=[1,2,11,10,3,6,7,8,5,4,9,12]$

we can construct all permutation in the orbit of $x$ by taking shifts of the
values of $x$. Namely, given a permutation $x\in S_{n-1}$ it is
computationally easy to construct the orbit of $x$.

######  Fig. A.1.

Excerpt from python code which calculates the orbit of a given permutation for
fixed $n.$

Now, the smaller $t$ gets, the more restrictions there are on $x$ in order for
it to have stabilizer $\langle a^{t}\rangle$. For example, when $t=1$, $x$ is
completely determined by where it sends $1$ and so there are $\varphi(n)$ (the
Euler totient function of $n$) permutation which have stabilizer $\langle
a\rangle$ (by Proposition 5.1.1). If $t=2$ then $x$ is determined by where it
sends $1$ and $2$, etc. Thus, as $t$ grows closer to $n$, the number of $x$
with stabilizer $\langle a^{t}\rangle$ approaches $(n-1)!$.

Lets say now that we want to compute all orbits of size $t$ for a given $n$
which contain $r$ involutions. First, we generate all the permutation in
$S_{n-1}$, then we compute all their orbits, then we pick out those of length
$t,$ then of those we pick out the ones which contain $r$ involutions. To
count the number of involutions in an orbit, we need to manually check if each
permutation in the set is its own inverse. As $t$ grows, not only there more
permutations within the orbit to compare with their inverses, but also the
total number of orbits of length $t$ is approaching $(n-1)!$ which cannot be
done efficiently for large $n.$

For this reason, we demanded that $t$ stayed small (usually $t<8$) throughout
our computations, especially as $n$ grew large. Note the $y$-axis scaling on
many of the graphs is nonlinear. This is because the values differ by so much
that they cannot be visualized on the same graph without logarithmically
scaling the $y$-axis.

##  Appendix B Using Proposition 5.1.1

Recall from Proposition 5.1.1, that $M_{n/t}$ is the set of permutations in
$S_{n-1}$ with stabilizer $F_{x}=\langle a^{t}\rangle$.

Using the proof of Proposition 5.1.1, we can explicitly write down which
permutations have stabilizer $\langle a^{t}\rangle$.

choose $x(t)=jt$ for $j\in E_{n/t}$ choose $x(1)$ in $\\{u\,|\,u\not=0\mod
t\\}$ choose $x(2)$ in $\\{u\,|\,u\not=0,x(1)\mod t\\}$ $\cdots$ choose
$x(n-1)$ in $\\{u\,|\,u\not=0,x(1),x(2),...,x(n-2)\mod t\\}$

######  Fig. B.1.

A flow chart illustrating the process of Proposition 5.1.1. This will
construct all permutations which are stabilized by $a^{t}$. To get the exact
set of permutations which have stabilizer $\langle a^{t}\rangle$, this process
must be done recursively so that all $x$ which are stabilized by $a^{s}$ for
some $s|t$ can be removed from the set.

######  Example B.2.

For example, lets say $n=12$ we want to write down all permutations $x\in
S_{11}$ with stabilizer $\langle a^{2}\rangle$. We break this process into
steps as follows: we determine all permutations with full stabilizer, we
determine all permutations stabilized by $a^{2}$, then we remove those we have
already found.

Since $n=12$ and $t=2$, we will use (5) from Proposition 5.1.1.

  1. (1)

First need to determine all permutations $x\in S_{11}$ with stabilizer
$\langle a\rangle=F$.

All $x$ with full stabilizer are determined by where they send $1$. Since the
number of values coprime to $12$ are $1,5,7,11$ the number of such $x$ is $4$.
Furthermore, they are given by

$\displaystyle j=1:x(i)=i\implies x$ $\displaystyle=(1)$ $\displaystyle
j=5:x(i)=5i\implies x$
$\displaystyle=(1\,\,\,5)(2\,\,\,10)(4\,\,\,8)(7\,\,\,11)$ $\displaystyle
j=7:x(i)=7i\implies x$ $\displaystyle=(1\,\,\,7)(3\,\,\,9)(5\,\,\,11)$
$\displaystyle j=11:x(i)=11i\implies x$
$\displaystyle=(1\,\,\,11)(2\,\,\,10)(3\,\,\,9)(4\,\,\,8)(5\,\,\,7)$

  2. (2)

Now, we compute all permutations with are stabilized by $2$. Since the only
$j$ which are coprime to $\frac{12}{2}=6$ are $1$ and $5$, we have that
$x(2)=2$ or $x(2)=10$. Then, $x(1)$ can be any odd value and since there are
six choices $\\{1,3,5,7,9,11\\}$, we have $12$ total permutations.

$\displaystyle j=1;x(1)=1:x=(1)$ $\displaystyle j=1;x(1)=3:x$
$\displaystyle=(1\,\,\,3\,\,\,5\,\,\,7\,\,\,9\,\,\,11)$ $\displaystyle
j=1;x(1)=5:x$ $\displaystyle=(1\,\,\,5\,\,\,9)(3\,\,\,7\,\,\,11)$
$\displaystyle j=1;x(1)=7:x=(1\,\,\,7)(3\,\,\,9)(5\,\,\,11)$ $\displaystyle
j=1;x(1)=9:x$ $\displaystyle=(1\,\,\,9\,\,\,5)(3\,\,\,11\,\,\,7)$
$\displaystyle j=1;x(1)=11:x$
$\displaystyle=(1\,\,\,11\,\,\,9\,\,\,7\,\,\,5\,\,\,3)$ $\displaystyle
j=5;x(1)=1:x$ $\displaystyle=(2\,\,\,10)(3\,\,\,11)(4\,\,\,8)(5\,\,\,9)$
$\displaystyle j=5;x(1)=3:x$
$\displaystyle=(1\,\,\,3)(2\,\,\,10)(4\,\,\,8)(5\,\,\,11)(7\,\,\,9)$
$\displaystyle j=5;x(1)=5:x=(1\,\,\,5)(2\,\,\,10)(4\,\,\,8)(7\,\,\,11)$
$\displaystyle j=5;x(1)=7:x$
$\displaystyle=(1\,\,\,7)(2\,\,\,10)(3\,\,\,5)(4\,\,\,8)(9\,\,\,11)$
$\displaystyle j=5;x(1)=9:x$
$\displaystyle=(1\,\,\,9)(2\,\,\,10)(3\,\,\,7)(4\,\,\,8)$ $\displaystyle
j=5;x(1)=11:x=(1\,\,\,11)(2\,\,\,10)(3\,\,\,9)(4\,\,\,8)(5\,\,\,7)$

  3. (3)

Of course, the highlighted permutations were those we found previously to have
full stabilizer and so those should be removed from our list. This leaves us
with our final total of $M_{12/2}=8$ permutations as counted by Remark 5.1.2.

######  Fig. B.3.

Graph of $M_{n/t}$: number of permutations $x\in S_{n-1}$ with stabilizer
$\langle a^{t}\rangle$.

##  Appendix C Using Proposition 5.2.1

Recall from Proposition 5.2.1, that $T_{n/t}$ is the set of involutions in
$S_{n-1}$ with stabilizer $F_{x}=\langle a^{t}\rangle$.

choose $x(t)=jt$ for $j\in E_{n/t}$ choose an involution $\sigma_{x}\in
S_{t-1}$ if $\sigma_{x}(i)=i$, choose $u_{i}$ such that
$u_{i}(j+1)=0\mod\frac{n}{t}$ if $\sigma_{x}(i)\not=i$, choose any $u_{i}$
which determines $u_{\sigma_{x}(i)}=-ju_{i}\mod\frac{n}{t}$ construct $x$

######  Fig. C.1.

A flow chart illustrating the process of Proposition 5.2.1. This will
construct all involutions which are stabilized by $a^{t}$. To get the exact
set of involutions which have stabilizer $\langle a^{t}\rangle$, this process
must be done recursively so that all $x$ which are stabilized by $a^{s}$ for
some $s|t$ can be removed from the set.

######  Example C.2.

As an example, say we want to find $T_{6/3}$. As always, the first step is to
compute $T_{6/1}$ since these permutations must be removed. Of course, any
permutation which is stabilized by $a^{3}$ and is not stabilized by $a$ must
have stabilizer $\langle a^{3}\rangle.$ Again we break this process into
steps.

Since $n=6$ and $t=1$, we will use (6) from Proposition 5.2.1

  1. (1)

First, we find all $x$ with full stabilizer. $1$ is the only divisor of $3$
not equal to $3$. Each $j\in E_{6/1}=\\{1,5\\}$ gives a single involution with
full stabilizer:

$\displaystyle j=1:x$ $\displaystyle=(1)$ $\displaystyle j=5:x$
$\displaystyle=(1\,\,\,5)(2\,\,\,4)$

  2. (2)

Second, we find all permutations stabilized by $a^{3}.$ There are two
involutions in $S_{2}$ which can act as remainder permutations. They are
$\sigma_{1}=(1)$, $\sigma_{2}=(1\,\,\,2)$. Here $E_{6/3}=\\{1\\}$ so $j=1$ is
fixed. Now, we group by remainder permutations.

    1. (i)

Say $\sigma_{x}=(1)$ so every $i$ is a fixed point of $\sigma_{x}$. Then,
since $u_{i}(j+1)=2u_{i}=0\mod 2=\frac{6}{3}$, $u_{i}$ can take any value
$\\{0,1\\}$ for $i=1,2$. This gives the possible permutations

$\displaystyle\sigma_{x}=(1);u_{1}=0,u_{2}=0;x(1)=1,x(2)=2:x=(1)$
$\displaystyle\sigma_{x}=(1);u_{1}=0,u_{2}=1;x(1)=1,x(2)=5:x$
$\displaystyle=(2\,\,\,5)$
$\displaystyle\sigma_{x}=(1);u_{1}=1,u_{2}=0;x(1)=4,x(2)=2:x$
$\displaystyle=(1\,\,\,4)$
$\displaystyle\sigma_{x}=(1);u_{1}=1,u_{2}=1;x(1)=4,x(2)=5:x$
$\displaystyle=(1\,\,\,4)(2\,\,\,5)$

    2. (ii)

On the other hand, if $\sigma_{x}=(1\,\,\,2)$ then $\sigma_{x}$ has no fixed
points and so $u_{1}=-u_{2}=u_{2}\mod 2$. Again, we use these values to write
$x(i)=tu_{i}+\sigma_{x}(i)$. This gives the possible permutations

$\displaystyle\sigma_{x}=(1\,\,\,2);u_{1}=0,u_{2}=0;x(1)=2,x(2)=1:x$
$\displaystyle=(1\,\,\,2)(4\,\,\,5)$
$\displaystyle\sigma_{x}=(1\,\,\,2);u_{1}=1,u_{2}=1;x(1)=5,x(2)=4:x=(1\,\,\,5)(2\,\,\,4)$

  3. (3)

Again, the highlighted involutions are those we already found to have full
stabilizer. Remove those leaves us with $4$ involutions with stabilizer
$\langle a^{3}\rangle.$

Note how in (2).(i) all permutations $x$ are composed of the two $(2\,\,\,5)$
and $(1\,\,\,4)$ and their product. These two permutations are in fact all of
the involutions of $S_{5}$ which consist of transpositions $(i\,\,\,i+3)$. Of
course $(3\,\,\,6)$ is not a valid involution in $S_{5}$.

Throughout our study of the sorts of permutations which satisfy various
requirements (such as number of fixed points, size of stabilizer, etc.), we
have stumbled upon many such “patterns.” The percieved lack of randomness
invites endless questions. Are these patterns indicative of some structure,
perhaps induced from $S_{t-1}$?

Of course, the pattern of $(i\,\,\,i+3)$ does not hold for (2).(ii), nor does
$(i\,\,\,i+1)$ so are these patterns coincidental?

It is our belief that there is a much easier way to compute the sets defined
in Section 5 using some deeper understanding of these patterns. However, we
have not explored these questions in this work.

######  Fig. C.3.

Graph of $T_{n/t}$: number of involutions $x\in S_{n-1}$ with stabilizer
$\langle a^{t}\rangle$.

##  Appendix D Using Proposition 5.2.3

Recall from Proposition 5.2.3, that $R_{n/t,r}$ is the set of involutions in
$S_{n-1}$ with stabilizer $F_{x}=\langle a^{t}\rangle$ and exactly $r$ fixed
points (including $x(n)=n$ as a fixed point).

######  Example D.1.

The more restriction we place on $x$, the more complicated our formulas must
be for constructing them. However, we will again showcase how to use
Proposition 5.2.3 to construct the set $R_{8/4,2}$ which is the number of
involutions in $S_{7}$ with stabilizer $\langle a^{4}\rangle$ and $2$ fixed
points (including $x(8)=8$ as a fixed point).

Since $n=8$, $t=4$, and $r=2$, we use (10) from Proposition 5.2.3

First, we need $R_{8/1,2}$ and $R_{8/2,2}$ so the process becomes more
complicated here.

  1. (1)

First, we find all involutions with full stabilizer. For $t=1$,
$E_{8/1}=\\{1,3,5,7\\}$. Since $\beta_{1,8/1}=8$, $\beta_{3,8/1}=2$,
$\beta_{5,8/1}=4$, $\beta_{7,8/1}=2$ there are only two permutations with $2$
fixed points and full stabilizer coming from $j=3$ and $j=7$. This is because
$\delta_{j,n/t,r}=0$ unless $\beta_{j,n/t}$ divides $r$.

$\displaystyle j=3:x$ $\displaystyle=(1\,\,\,3)(2\,\,\,6)(5\,\,\,7)$
$\displaystyle j=7:x$ $\displaystyle=(1\,\,\,7)(2\,\,\,6)(3\,\,\,5)$

  2. (2)

For $t=2$, $E_{8/2}=\\{1,3\\}$. $\beta_{1,8/2}=4$, $\beta_{3,8/2}=2$ so only
$j=3$ is viable here. Since the only $\sigma\in S_{1}$ is the identity,
$u_{1}\in P_{3,8/2}^{(c)}=\\{1,3\\}$ and so again we get two permutations

$\displaystyle u_{1}=1;x(1)=3:x=(1\,\,\,3)(2\,\,\,6)(5\,\,\,7)$ $\displaystyle
u_{1}=3;x(1)=7:x=(1\,\,\,7)(2\,\,\,6)(3\,\,\,5)$

Of course, both of these we found to have full stabilizer for $R_{8/2,2}=0.$

  3. (3)

Now we compute $R_{8/4,2}$. First, $E_{8/4}=\\{1\\}$ so $j=1$ and $x(4)=4$ is
fixed. Now, $\beta_{1,8/4}=\text{gcd}(1-1,\frac{8}{4})=\text{gcd}(0,2)=2$.
Thus, $m_{2,1,8/4}=1$ and $\frac{t-m_{2,1,8/4}}{2}=\frac{3}{2}$ so $l=0$ or
$l=1$ in the second sum of $R_{n/t,r}$. This tells us that $\sigma_{x}\in
S_{3}$ can be any involution comprised of either $0$ or $1$ transpositions. In
fact, this is all such permutations so we have possible choices for
$\sigma_{x}$ are $(1),(1\,\,\,2),(1\,\,\,3),$ and $(2\,\,\,3)$.

The sets that we will be pulling from are

$\displaystyle K_{1,8/4}$ $\displaystyle=\\{0,1\\}$ $\displaystyle P_{1,8/4}$
$\displaystyle=\\{0\\}$ $\displaystyle P_{1,8/4}^{(c)}$
$\displaystyle=\\{1\\}$

    1. (i)

Say $\sigma_{x}=(1)$, then this permutation has $t-1-2l=3$ fixed points and of
these we choose $m_{2,1,8/4}-1=0$ of the $u_{i}$ to be in $P_{1,8/4}$ so all
$u_{i}$ are in $P_{1,8/4}^{(c)}$.

    2. (ii)

Similarly, if $\sigma_{x}$ is one of the other transpositions. There is one
just one fixed point $i$ of $\sigma_{x}$ which forces $u_{i}\in
P_{1,8/4}^{(c)}$ for that $i$. The two values swapped by the transposition of
$\sigma_{x}$ equal (via the relation
$u_{\sigma_{x}(i)}=-ju_{i}\mod\frac{n}{t}$).

    3. (iii)

Finally, we can write out all possible involutions.

$\displaystyle\sigma_{x}=(1);u_{1}=1,u_{2}=1,u_{3}=1;x(1)=5,x(2)=6,x(3)=7:x$
$\displaystyle=(1\,\,\,5)(2\,\,\,6)(3\,\,\,7)$
$\displaystyle\sigma_{x}=(1\,\,\,2);u_{1}=0,u_{2}=0,u_{3}=1;x(1)=2,x(2)=1,x(3)=7:x$
$\displaystyle=(1\,\,\,2)(3\,\,\,7)(5\,\,\,6)$
$\displaystyle\sigma_{x}=(1\,\,\,2);u_{1}=1,u_{2}=1,u_{3}=1;x(1)=6,x(2)=5,x(3)=7:x$
$\displaystyle=(1\,\,\,6)(2\,\,\,5)(3\,\,\,7)$
$\displaystyle\sigma_{x}=(1\,\,\,3);u_{1}=0,u_{2}=1,u_{3}=0;x(1)=3,x(2)=6,x(3)=1:x=(1\,\,\,3)(2\,\,\,6)(5\,\,\,7)$
$\displaystyle\sigma_{x}=(1\,\,\,3);u_{1}=1,u_{2}=1,u_{3}=1;x(1)=7,x(2)=6,x(3)=5:x=(1\,\,\,7)(2\,\,\,6)(3\,\,\,5)$
$\displaystyle\sigma_{x}=(2\,\,\,3);u_{1}=1,u_{2}=0,u_{3}=0;x(1)=5,x(2)=3,x(3)=2:x$
$\displaystyle=(1\,\,\,5)(2\,\,\,3)(6\,\,\,7)$
$\displaystyle\sigma_{x}=(2\,\,\,3);u_{1}=1,u_{2}=1,u_{3}=1;x(1)=5,x(2)=7,x(3)=6:x$
$\displaystyle=(1\,\,\,5)(2\,\,\,7)(3\,\,\,6)$

  4. (4)

So there are five involutions in $S_{7}$ with stabilzier $\langle
a^{4}\rangle$ which have exactly two fixed points.

######  Fig. D.2.

Graph of $R_{n/t,r}$: the number of involutions in $S_{n-1}$ with stabilier
$\langle a^{t}\rangle$ and $r$ fixed points.

##  Appendix E Using Proposition 5.3.1 and Counting Indicators Directly

Recall from Proposition 5.3.1, that $X_{n/t,r}$ is the set of involutions in
$x\in S_{n-1}$ with stabilizer $\langle a^{t}\rangle$ such that $x$ has $r$
fixed points smaller than $t$ (equivalently, such that $\mathcal{O}_{x}$
contains $r$ involutions).

######  Example E.1.

Let us again go through an example. Say we want to find $X_{12/3,1}$ the
number of involutions whose orbit contains one involution (and so contain $0$
involutions other than $x$ itself). These are all the involutions with $r-1=0$
fixed points smaller than $t=3$.

Since $n=12$, $t=3,$ and $r=1$, we will use (13) from Proposition 5.3.1

  1. (1)

First we compute $E_{n/t}$. Since $E_{12/3}=\\{1,3\\}$ we have two choices for
$j\in E_{12/3}$.

  2. (2)

Because there is only one $\sigma\in S_{2}$ with $r-1=0$ fixed points, we must
have $\sigma_{x}=(1\,\,\,2)$. Because $\sigma_{x}$ has no fixed points, $1\leq
u_{1}\leq\frac{12}{3}=4$ can take any value but
$u_{2}=-ju_{1}\mod\frac{12}{3}=4$. This gives us the following possibilities
for $x$

$\displaystyle j=1;u_{1}=0,u_{2}=0;x(1)=2,x(2)=1:x$
$\displaystyle=(1\,\,\,2)(4\,\,\,5)(7\,\,\,8)(10\,\,\,11)$ $\displaystyle
j=1;u_{1}=1,u_{2}=3;x(1)=5,x(2)=10:x$
$\displaystyle=(1\,\,\,5)(2\,\,\,10)(4\,\,\,8)(7\,\,\,11)$ $\displaystyle
j=1;u_{1}=2,u_{2}=2;x(1)=8,x(2)=7:x$
$\displaystyle=(1\,\,\,8)(2\,\,\,7)(4\,\,\,11)(5\,\,\,10)$ $\displaystyle
j=1;u_{1}=3,u_{2}=1;x(1)=11,x(2)=4:x$
$\displaystyle=(1\,\,\,11)(2\,\,\,4)(5\,\,\,7)(8\,\,\,10)$ $\displaystyle
j=3;u_{1}=0,u_{2}=0;x(1)=2,x(2)=1:x$
$\displaystyle=(1\,\,\,2)(3\,\,\,9)(4\,\,\,11)(5\,\,\,10)(7\,\,\,8)$
$\displaystyle j=3;u_{1}=1,u_{2}=1;x(1)=5,x(2)=4:x$
$\displaystyle=(1\,\,\,5)(2\,\,\,4)(3\,\,\,9)(7\,\,\,11)(8\,\,\,10)$
$\displaystyle j=3;u_{1}=2,u_{2}=2;x(1)=8,x(2)=7:x$
$\displaystyle=(1\,\,\,8)(2\,\,\,7)(3\,\,\,9)(4\,\,\,5)(10\,\,\,11)$
$\displaystyle j=3;u_{1}=3,u_{2}=3;x(1)=11,x(2)=10:x$
$\displaystyle=(1\,\,\,11)(2\,\,\,10)(3\,\,\,9)(4\,\,\,8)(5\,\,\,7)$

  3. (3)

Now, we must remove those we have over counted. However, for the first time,
our recursion does not require us to do the same computation but for a smaller
$t$. Because $t$ very much matters here, (we are looking for $x$ with no fixed
points below $t$), we use a different formula now to remove overcount.

We must subtract $C_{12,3,1,2}$. In this case, $E_{3/1}=\\{1,2\\}$ so we have
two choices for $j_{\sigma_{x}}$. When $j_{\sigma_{x}}=1$, $\beta_{1,3/1}=3$
which does not divide $r$ so we ignore this case. When $j_{\sigma_{x}}=2$,
$\beta_{2,3/1}=1$, $j_{\sigma_{x}}+\overline{m}\frac{3}{1}$ are $2,2+3,2+6,$
and $2+9$ so $\overline{E}_{2,12,3,1}=\\{5,11\\}$. Thus, there are two
permutations we must remove:

$\displaystyle j^{\prime}=5:x$
$\displaystyle=(1\,\,\,5)(2\,\,\,10)(4\,\,\,8)(7\,\,\,11)$ $\displaystyle
j^{\prime}=11:x$
$\displaystyle=(1\,\,\,11)(2\,\,\,10)(3\,\,\,9)(4\,\,\,8)(5\,\,\,7)$

  4. (4)

So there are $6$ involutions whose orbit contains one involution which have
stabilizer $\langle a^{3}\rangle$ in $S_{11}.$ Namely, since each of these
must be in a different orbit, there are $6$ orbits of length $3$ which contain
exactly one involution. Thus, we have also computed $O_{12/3,1}=6$ by
Proposition 5.3.2.

######  Fig. E.2.

Graph of $O_{n/t,r}$: the number of orbits of length $t$ which contain $r$
involutions.

Now, let us explore how using direct computation proved helpful in the case
where $t$ is even.

######  Fig. E.3.

Graph showing $I_{n/t}^{(+1)}$ the number of irreps of $J_{n}$ of dimension
$t$ with indicator $+1$. Notice that here, when $t$ is even, our formula is an
underestimate of the actual value.

######  Fig. E.4.

Graph showing a close up of E.3 which highlights the underestimate of our
formula whenever $t$ is even.

Using Python, it was easy to conjecture that while there exist irreps with
indicator $+1$ which are induced from permutations whose orbit does not
contain an involution, they are not common. While we have not been able to
formalize this in Section 7 because we could not come up for a formula to
count such irreps, all computational evidence points toward this being true.

######  Fig. E.5.

Graph showing the difference between direct computation and our formula for
irreps of $J_{n}$ of dimension $t$ (for even $t$) with indicator $+1$

Recall Theorem 6.1.2 tells us exactly how many irreps of $J_{n}$ of dimension
$t$ with indicator $+1$ are induced from permutations $x$ whose orbit contains
at least one involution. When the dimension $t$ is odd, the indicator can only
be $+1$ if the permutation which induces it shares its orbit with an
involution (see Proposition 4.3.5). However, when $t$ is even, there may be
orbits which are closed under inversion whose elements induce irreps with
indicator $+1$.

E.5 suggests that most of the irreps of $J_{n}$ of dimension $t$ with
indicator $+1$ are induced from permutations $x$ whose orbit contains at least
one involution.

For example, E.3 shows that there are $235,872$ irreps of $J_{36}$ of
dimension $6$ with indicator $+1$ and of these, $227,280$ are induced from
permutations whose orbit contains at least one involution. Namely, only
$3.6\%$ of the irreps of $J_{36}$ of dimension $6$ are induced from
permutations whose orbit contains no involutions.

Since Theorem 7.4 tells us that for fixed odd $t$, the percent of irreps of
$J_{mt}$ of dimension $t$ with indicator $+1$ approaches $0$ as $m\to\infty$,
it seems logical to purpose that the same holds true for even $t$.

######  Conjecture E.6.

Fix $t$ an even integer. Let $n=mt$ for some integer $m\geq 2$. Then the ratio
of irreps of $J_{n}$ of dimension $t$ with indicator $+1$ to the total number
of irreps of dimension $t$ converges to $0$ as $m\to\infty$.

Of course, Theorem 7.6 tells us this is true when $t=2$, but for general even
$t>2$, we have only a lower bound for the number of irreps with indicator $+1$
and this is not helpful for bounding the ratio from above.

######  Conjecture E.7.

Fix $t$ an even integer. Let $n=mt$ for some integer $m\geq 2$. Then the ratio
of irreps of $J_{n}$ of dimension $t$ with indicator $+1$ which are induced
from permutations which do not share their orbit with an involution to the
total number of irreps of dimension $t$ with indicator $+1$ convertes to $0$
as $m\to\infty$.

######  Fig. E.8.

Graph showing percent of $4$-dimensional irreps of $J_{n}$ with indicator $+1$
to total number of $4$-dimensional irreps of $J_{n}$

##  Appendix F Using Theorem 6.2.2

$x(2)=2j$; choose $j\in E_{n/2}$ $x=x^{-1}$; choose $u$ such that $u(j+1)=0$
remove case $\mathcal{O}_{x}=\\{x\\}$; ignore $u$ such that $j=2u+1$
$x^{-1}\not=x$; choose $u$ such that $(u+1)(j-1)=0$ remove case
$\mathcal{O}_{x}=\\{x\\}$; ignore $u$ such that $j=2u+1$ ensure $j\not=2u+1$,
$x(1)=2u+1$, $x(2)=2j$ $x^{-1}=x$ $x\not=x^{-1}$ $x^{-1}\in\mathcal{O}_{x}$
choose $i$ if $i(j+1)=0$, then $\nu(\hat{\chi}_{i})=+1$ if $i(j+1)\not=0$,
then $\nu(\hat{\chi}_{i})=0$ $i(j+1)=0$ $i(u+1)=0$, then
$\nu(\hat{\chi}_{i})=+1$ $i(u+1)\not=0$, then $\nu(\hat{\chi}_{i})=-1$
$i(j+1)\not=0$, then $\nu(\hat{\chi}_{i})=0$

######  Fig. F.1.

A flow chart illustrating the process of Theorem 6.2.2. The red entries
indicate the construction of involutions $x$ and the process of determining
which indicators are $+1$ and which are $0$. The green entries indicate the
construction of $x$ such that $\mathcal{O}_{x}=\\{x,x^{-1}\\}$ (so namely,
$x\not=x^{-1}$) and the process of determining which indicators are $+1,-1,0$.
The rectangles illustrate that $F_{x}=F$ if $j=2u+1\mod\frac{n}{2}$, the
diamonds illustrate intermediate computation, and the circles illustrate the
final conclusion of the indicator.

######  Example F.2.

For example, let $n=12$. We will to compute $I_{12/2}^{(+1)},I_{12/2}^{(-1)},$
and $I_{12/2}^{(0)}$.

  1. (1)

First we compute $E_{n/t}.$ Since $n=12$ and $t=2$, then $E_{12/2}=\\{1,5\\}$.
Now, we go case by case.

    1. (i)

For $j=1$:

      1. (a)

Again, $j=1$ so $u(j+1)=2u\mod\frac{12}{2}=6$ so $u=0\mod 3$ so $u=6,3\mod 6$.
Since $1=2(6)+1\mod 6$ and $1=2(3)+1\mod 6$, both values of $u$ are out, so
there are no involutions $x$ satisfying $x(2)=2$ with stabilizer $\langle
a^{2}\rangle$.

      2. (b)

However, we instead we look at $(u+1)(j-1)=0\mod 6$ we get that $u$ can take
any value. Since $j=1\not=2u+1$, we cannot use $u=6,u=3$ so there are four
$(u=1,2,4,5)$ possible permutations which are not involutions. Explicitly,
they are

$\displaystyle j=1;u=1;x(1)=3:x_{1}$
$\displaystyle=(1\,\,\,3\,\,\,5\,\,\,7\,\,\,9\,\,\,11)$ $\displaystyle
j=1;u=2;x(1)=5:x_{2}$ $\displaystyle=(1\,\,\,5\,\,\,9)(3\,\,\,7\,\,\,11)$
$\displaystyle j=1;u=4;x(1)=9:x_{3}$
$\displaystyle=(1\,\,\,9\,\,\,5)(3\,\,\,11\,\,\,7)$ $\displaystyle
j=1;u=5;x(1)=11:x_{4}$ $\displaystyle=(1\,\,\,11\,\,\,9\,\,\,7\,\,\,5\,\,\,3)$

    2. (ii)

Now, consider $j=5$.

      1. (a)

Then $u(j+1)=0\mod 6$ for any $1\leq u\leq 6$. Since we cannot have
$5=2u+1\mod 6$ we toss out $u=2,5$. This leaves four possible involutions.
Now, $K_{5,12/2}=\\{1,2,3,4,5,6\\}$ since $i(j+1)=0\mod 6$ for any $i$. Thus,
all indicators are $+1$ for all of these involutions. The involutions here are

$\displaystyle j=5;u=6;x(1)=1:x_{5}$
$\displaystyle=(2\,\,\,10\,\,\,)(3\,\,\,11)(4\,\,\,8)$ $\displaystyle
j=5;u=1;x(1)=3:x_{6}$
$\displaystyle=(1\,\,\,3)(2\,\,\,10)(4\,\,\,8)(5\,\,\,11)(7\,\,\,9)$
$\displaystyle j=5;u=3;x(1)=7:x_{7}$
$\displaystyle=(1\,\,\,7)(2\,\,\,10)(3\,\,\,5)(4\,\,\,8)(9\,\,\,11)$
$\displaystyle j=5;u=4;x(1)=9:x_{8}$
$\displaystyle=(1\,\,\,9)(2\,\,\,10)(3\,\,\,7)(4\,\,\,8)$

      2. (b)

For the case where $(u+1)(j-1)=4(u+1)=0\mod 6$, we require that $u=-1\mod 3$
so $u=2,5$. However, because for both of these $j=2u+1$ so there are no
permutations which are not involutions satisfying that $x(2)=10$ and
$F_{x}=\langle a^{2}\rangle$.

    3. (iii)

Finally, we count the indicators of the irreps induced from each $x$. $x_{1}$
and $x_{4}$ each contribute $2$ indicators of value $+1$ and $4$ indicators of
value $0$. $x_{2}$ and $x_{3}$ each contribute $1$ indicator of value $+1$,
$1$ indicator of value $-1$, and $4$ indicators of value $0$.
$x_{5},x_{6},x_{7}$ and $x_{8}$ each contribute $6$ indicators of value $+1$.
In total, we have

$\displaystyle I_{12/2}^{(+1)}$ $\displaystyle=2\cdot 2+2\cdot 1+4\cdot
6=4+2+24=30$ $\displaystyle I_{12/2}^{(-1)}$ $\displaystyle=2\cdot 0+2\cdot
1+4\cdot 0=2$ $\displaystyle I_{12/2}^{(0)}$ $\displaystyle=2\cdot 4+2\cdot
4+4\cdot 0=16$

Before concluding, let us go more in depth about the actual representations
induced from the permutations we’ve found.

Consider the four permutations from (1).(i).(b). They are grouped into two
orbits, where each orbit contains $x$ and $x^{-1}$. Now,
$K_{1,12/2}=\\{i\,|\,2i=0\mod 6\\}=\\{0,3\\}$. Since $6(u+1)=0$ for any
$u=1,2,4,5$ this gives four irreps of $J_{n}$ of dimension $2$ and indicator
$+1$. Similarly, $3(u+1)=0\mod 6$ for $u=1,5$ which adds two more irreps with
indicator $+1$. However, for $u=2,4$ we have two irrpes of $J_{n}$ with
dimension $2$ which have indicator $-1$ while the rest are of indicator $0$.

Let’s be even more specific. We use the information we found about $u$ and $j$
to construct the associated irreducible representations of $J_{n}$ which have
the indicators described. Consider $x_{2}=(1\,\,\,5\,\,\,9)(3\,\,\,7\,\,\,11)$
which has stabilizer $\langle a^{2}\rangle$. Let $\zeta_{6}$ be a primitive
$6^{\text{th}}$ root of unity. There are $6$ irreducible group representations
associated to $x_{2}$ defined by

$\displaystyle\rho_{x_{2},i}:\langle a^{2}\rangle$
$\displaystyle\to\mathbb{C}^{\times}$ $\displaystyle a^{2}$
$\displaystyle\mapsto\zeta_{6}^{i}$

for $1\leq i\leq 6$. Consider the equivalent $F_{x}$-module $V_{x_{2},i}$ to
the representation $\rho_{x_{2},i}$. Then the induced $J_{n}$ module is

$\hat{V}_{x_{2},i}=\mathbb{C}\langle a\rangle\otimes_{\mathbb{C}\langle
a^{2}\rangle}V_{x_{2},i}$

which is defined by the action

$(p_{y}\\#a^{l})\cdot(a^{q}\otimes v)=\begin{cases}a^{l+q}\otimes v&\text{ if
}x=a^{-y(l+q)}ya^{l+q}\\\ 0&\text{ otherwise}\end{cases}.$

The associated character $\hat{\chi}_{x_{2},i}$ gives rise the following
indicators:

$\displaystyle\nu(\hat{\chi}_{x_{2},1})$ $\displaystyle=0$
$\displaystyle\nu(\hat{\chi}_{x_{2},2})$ $\displaystyle=0$
$\displaystyle\nu(\hat{\chi}_{x_{2},3})$ $\displaystyle=-1$
$\displaystyle\nu(\hat{\chi}_{x_{2},4})$ $\displaystyle=0$
$\displaystyle\nu(\hat{\chi}_{x_{2},5})$ $\displaystyle=0$
$\displaystyle\nu(\hat{\chi}_{x_{2},6})$ $\displaystyle=+1$

$x_{3}$ induces the same irreps as $x_{2}$ and so they have the same indicator
table. For $x_{1}$ and $x_{4}$, $i=3$ gives indicators which are $+1$ instead
of $-1$.

######  Fig. F.3.

Graph of $I_{n/2}^{(+1)}$ showing the number of irreps of $J_{n}$ of dim $2$
with indicator $+1.$

######  Fig. F.4.

Graph of $I_{n/2}^{(-1)}$ showing the number of irreps of $J_{n}$ of dim $2$
with indicator $-1.$

######  Fig. F.5.

Graph of $I_{n/2}^{(0)}$ showing the number of irreps of $J_{n}$ of dim $2$
with indicator $0.$

To emphasize the difference in scale, we lay multiple graphs on a single plot.

######  Fig. F.6.

Plot showing all three of F.3, F.4, and F.5 on a single graph.

######  Fig. F.7.

Plot showing Figures F.3 and F.4 on a single graph.

##  Appendix G Counting Irreps with Indicator $-1$ Directly

While we do not have any explicit bounds for the number of irreps with
indicator $-1$ as we do for irreps with indicator $+1$, Python was still able
to count irreps with indicator $-1$ for some $n$ and $t$.

######  Table 3 (Table of Negative Indicators).

$n$ | $12$ | $16$ | $20$ | $24$ | $28$ | $32$ | $36$ | $40$ | $44$ | $48$
---|---|---|---|---|---|---|---|---|---|---
$t$ | $2$ | $6$ | $2$ | $4$ | $2$ | $2$ | $4$ | $6$ | $2$ | $2$ | $4$ | $2$ | $6$ | $2$ | $4$ | $2$ | $2$ | $4$
$\displaystyle I_{n/t}^{(-1)}$ | $2$ | $42$ | $2$ | $20$ | $4$ | $8$ | $64$ | $816$ | $6$ | $6$ | $124$ | $8$ | $2976$ | $16$ | $208$ | $10$ | $24$ | $648$

Note: Table showing total number of irreps of $J_{n}$ of dimension $t$ with
indicator $-1$. This value was computed directly in Python. Note how $t$ was
kept small to make computation possible.

######  Fig. G.1.

Graph showing number of irreps of $J_{n}$ of dimension $t$ with indicator
$-1$. This value was computed directly in Python.

In comparison with the number of irreps of dimension $t$ with indicator $+1$
in E.3, the values in G.1 appear phenominally smaller.

Recall from the discussion near the end of Appendix D, that there are
$235,872$ irreps of $J_{36}$ of dimension $6$ with indicator $+1$ and of
these, $227,280$ are induced from permutations whose orbit contains at least
one involution. G.1 shows that there are $2,976$ irreps of $J_{36}$ of
dimension $6$ with indicator $-1$, and there are
$\frac{36}{6}|M_{36/6}|=11,189,952$ total irreps of $J_{36}$ of dimension $6$.

Namely, when $n=36$ and $t=6$, about $2.1\%$ of the irreps of $J_{36}$ of
dimension $6$ have indicator $+1$ and $0.027\%$ have indicator $-1$. Based on
our direct analysis, this appears to be a trend, not a fluke and so we end
with one final conjecture.

######  Conjecture G.2.

Fix $t$. Let $n=mt$ for some integer $m\geq 2$. Then the number ratio of
irreps of $J_{n}$ of dimension $t$ with nonzero indicator to the total number
of irreps of dimension $t$ converges to $0$ as $m\to\infty$.

######  Fig. G.3.

Graph showing percent of $2$-dimensional irreps of $J_{n}$ with indicator $-1$
to total number of $2$-dimensional irreps of $J_{n}$

######  Fig. G.4.

Graph showing percent of $4$-dimensional irreps of $J_{n}$ with indicator $-1$
to total number of $4$-dimensional irreps of $J_{n}$

Supplementary Information No supplementary files are included.

Acknowledgements Many thanks to my advisor Susan Montgomery for being a
patient and thoughtful advisor. Also thank you to Patrick M. Reardon for
teaching me the basics of Python code. This work is part of a PhD thesis at
the University of Southern California.

Declarations There are no sources of funding or conflicts of interest in this
work. This work is part of a PhD thesis at University of Southern California.

Additionally, the Python notebook used to generate the graphs used in this
work is available on GitHub.

https://github.com/KaylaOrlinsk/USCProject

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