# An Algebraic Roadmap of Particle Theories
Part III: Intersections

N. Furey
Iris Adlershof, Humboldt-Universität zu Berlin,
Zum Grossen Windkanal 2, Berlin, 12489
<EMAIL_ADDRESS>
HU-EP-23/66

###### Abstract

In this article, we bypass the detailed symmetry breaking pathways established
in [1]. Instead, a direct route from the Spin(10) model to the Standard Model
is enabled via a single algebraic constraint.

This single constraint, however, may be reconfigured as a requirement that
three $\mathfrak{so}(10)$ actions coincide on a fixed space of multi-vector
fermions. This $\mathfrak{so}(10)\hskip 1.42262pt\mapsto\hskip
1.42262pt\mathfrak{su}(3)_{\textup{C}}\oplus\mathfrak{su}(2)_{\textup{L}}\oplus\mathfrak{u}(1)_{\textup{Y}}$
breaking (from a three-way intersection) mirrors, in certain ways, the
$\mathfrak{so}(8)\hskip 1.42262pt\mapsto\hskip 1.42262pt\mathfrak{g}_{2}$
breaking (from a three-way intersection) in the context of octonionic
triality. By extending this result to include quaternions and complex numbers,
we find that a five-way intersection breaks $\mathfrak{so}(10)\hskip
1.42262pt\mapsto\hskip
1.42262pt\mathfrak{su}(3)_{\textup{C}}\oplus\mathfrak{u}(1)_{\textup{Q}}$.
These are the Standard Model’s unbroken gauge symmetries, post-Higgs-
mechanism.

###### pacs:

112.10.Dm, 2.60.Rc, 12.38.-t, 02.10.Hh, 12.90.+b

This article relies on results and notation established in [1]. It follows
companion papers [1] and [2].

##  I Spin(10) $\mapsto$ Standard Model, directly

At this time of writing, the Georgi-Glashow model is known to be in a state of
tension with experiment, given its unrealized prediction of proton decay, [3],
[4]. What if the Pati-Salam and Left-Right Symmetric models are destined to a
similar fate?

In considering such circumstances, we point out the existence of a direct
route from the Spin(10) model to the Standard Model.

Recall from [1] that $\mathfrak{spin}(10)$ may be realized on a left-handed
Weyl spinor as

$\begin{array}[]{lc}\ell_{10}\Psi_{\textup{L}}=&(r_{ab}L_{e_{a}}L_{e_{b}}+r^{\prime}_{mn}L_{\epsilon_{m}}L_{\epsilon_{n}}+r_{ma}^{\prime\prime}iL_{\epsilon_{m}}L_{e_{a}})\Psi_{\textup{L}},\vspace{2mm}\\\
&\mathfrak{spin}(10)\end{array}$ (1)

for $a,b\in\\{1,\dots 7\\}$ with $a\neq b,$ and $m,n\in\\{1,2,3\\}$ with
$m\neq n,$ while $r_{ab},r^{\prime}_{mn},r_{ma}^{\prime\prime}\in\mathbb{R}.$

Furthermore, standard model symmetries,
$\mathfrak{g}_{\textup{SM}}:=\mathfrak{su}(3)_{\textup{C}}\oplus\mathfrak{su}(2)_{\textup{L}}\oplus\mathfrak{u}(1)_{\textup{Y}}$,
may be realized as

$\begin{array}[]{rccll}\ell_{\textup{SM}}\Psi_{\textup{L}}=\big{(}r_{k}i\Lambda_{k}&+&r^{\prime}_{m}L_{\epsilon_{m}}s&+&irY\big{)}\Psi_{\textup{L}},\vspace{2mm}\\\
\mathfrak{su}(3)_{\textup{C}}&\oplus&\mathfrak{su}(2)_{\textup{L}}&\oplus&\mathfrak{u}(1)_{\textup{Y}}\end{array}$
(2)

for $k\in\\{1,\dots 8\\}$, $m\in\\{1,\dots 3\\}$, and
$r_{k},r^{\prime}_{m},r\in\mathbb{R}$. The operator $s$ is defined as
$s:=\frac{1}{2}(1+iL_{e_{7}})$. These $\mathfrak{su}(3)_{C}$ generators are
defined as

$\begin{array}[]{lll}i\Lambda_{1}:=\frac{1}{2}\left(L_{34}-L_{15}\right)&&i\Lambda_{2}:=\frac{1}{2}\left(L_{14}+L_{35}\right)\vspace{2mm}\\\
i\Lambda_{3}:=\frac{1}{2}\left(L_{13}-L_{45}\right)&&i\Lambda_{4}:=-\frac{1}{2}\left(L_{25}+L_{46}\right)\vspace{2mm}\\\
i\Lambda_{5}:=\frac{1}{2}\left(L_{24}-L_{56}\right)&&i\Lambda_{6}:=-\frac{1}{2}\left(L_{16}+L_{23}\right)\vspace{2mm}\\\
i\Lambda_{7}:=-\frac{1}{2}\left(L_{12}+L_{36}\right)&&i\Lambda_{8}:=\frac{-1}{2\sqrt{3}}\left(L_{13}+L_{45}-2L_{26}\right),\end{array}$
(3)

where $L_{ij}$ is shorthand for the octonionic $L_{e_{i}}L_{e_{j}}$. Recall
that the weak hypercharge generator $iY$ acts as

$iY\Psi_{\textup{L}}:=\left(\frac{1}{6}\left(L_{13}+L_{26}+L_{45}\right)-\frac{1}{2}L_{\epsilon_{3}}s^{*}\right)\Psi_{\textup{L}}.$
(4)

Now, it is not difficult to confirm that $\mathfrak{spin}(10)$ of equation (1)
is sent directly to
$\mathfrak{su}(3)_{\textup{C}}\oplus\mathfrak{su}(2)_{\textup{L}}\oplus\mathfrak{u}(1)_{\textup{Y}}$
of equation (2) via a single constraint:

$\left[\ell_{10},\Psi_{\textup{L}}\right]=\frac{1}{2}\left(\ell_{10}+\ell_{10}^{*_{\overrightarrow{\mathbb{O}}}}\right)\Psi_{\textup{L}}.$
(5)

Please see Figure 1.

Figure 1:  Spin(10) breaks directly to the Standard Model’s symmetries in a
single algebraic constraint.

We point out in passing that reducing the Lie algebra to its
$*_{{\mathbb{D}}}$-invariant subalgebra also affects the spinor space on which
it acts. (Recall that $\mathbb{D}$ represents
${{\overrightarrow{\mathbb{O}}}}$, ${{\overleftarrow{\mathbb{O}}}}$,
${{\overrightarrow{\mathbb{H}}}}$, or ${{\overrightarrow{\mathbb{C}}}}.$)
Namely, $\Psi_{\textup{L}}$ becomes further reducible as

$\Psi_{\textup{L}}=\Psi_{\textup{L}}^{+}\oplus\Psi_{\textup{L}}^{-}$ (6)

where

$\Psi_{L}^{\pm}:=\frac{1}{2}\left(\Psi_{\textup{L}}\pm\Psi_{\textup{L}}^{*_{{\mathbb{D}}}}\right).$
(7)

##  II An overarching theme

It should be apparent for readers that there are two types of algebraic
constraints at play in Figure 6 of reference [1]. Namely, in one direction we
find the effects of the multivector condition, and in the other direction, we
find the effects of division algebraic reflections. Might there be an
overarching theme to it all?

There is. In the case of the multivector condition, we found that special
unitary symmetries may be isolated from within spin symmetries by setting
spinor and multivector group actions equal on $\Psi_{\textup{L}}$. At the Lie
algebra level, we have that

$\begin{array}[]{c}\delta\Psi_{\textup{L}}=[r_{k_{1}k_{2}}\Gamma_{k_{1}}\Gamma_{k_{2}},\hskip
1.42262pt\Psi_{\textup{L}}]\hskip
2.84526pt=r_{k_{1}k_{2}}\Gamma_{k_{1}}\Gamma_{k_{2}}\Psi_{\textup{L}},\end{array}$
(8)

as per equation (18) of [1].

Similarly, invariance of the action under division algebraic reflections means
that we are again setting two different group actions equal on the same
fermionic representation space. Schematically,

$\begin{array}[]{c}\delta\Psi_{\textup{L}}=\ell\hskip
1.42262pt\Psi_{\textup{L}}\hskip 2.84526pt=\ell^{*_{\mathbb{D}}}\hskip
1.42262pt\Psi_{\textup{L}},\end{array}$ (9)

where $\mathbb{D}$ represents ${{\overrightarrow{\mathbb{O}}}}$,
${{\overleftarrow{\mathbb{O}}}}$, ${{\overrightarrow{\mathbb{H}}}}$, or
${{\overrightarrow{\mathbb{C}}}}.$

It is interesting to compare this to the usual spontaneous symmetry breaking
scheme employed in the Standard Model. With the Standard Model, a subgroup is
chosen by fixing a (vacuum) element of a particular representation under the
original group. In contrast, here we break symmetries by allowing the same
fermionic space to play different roles, and then insisting on self-
consistency. These two different perspectives need not necessarily be at odds
with one another. The reader is directed to Hughes’ Higgs in [5], and to a
recent paper of Hun Jang, [6].

So, an overarching theme that has emerged from viewing the Standard Model in
the context of its neighbouring theories is one of simultaneous group actions.

As alluded to earlier in [1], the idea of simultaneous group actions is not
new to group theory. In fact, it gives us a way to extract the automorphisms
of division algebras, e.g. $\mathfrak{aut}(\mathbb{O}),$ from their triality
symmetries, e.g. $\mathfrak{tri}(\mathbb{O})$.

In the remaining sections, we will give an explicit description of this
phenomenon in the case of octonionic $\mathfrak{so}(8)$ triality. That is, we
will show how the intersection of three distinct actions of $\mathfrak{so}(8)$
isolates the octonionic derivations,
$\mathfrak{der}(\mathbb{O})=\mathfrak{aut}(\mathbb{O})=\mathfrak{g}_{2}$.

Then, we will draw analogous structures in the context of $\mathfrak{so}(10)$,
and will demonstrate that they exhibit a hint of the same theme. We will show
how the intersection of three distinct actions of $\mathfrak{so}(10)$ isolates
$\mathfrak{su}(3)_{\textup{C}}\oplus\mathfrak{su}(2)_{\textup{L}}\oplus\mathfrak{u}(1)_{\textup{Y}}$.
Extending further, we will show how the intersection of five distinct actions
of $\mathfrak{so}(10)$ isolates
$\mathfrak{su}(3)_{\textup{C}}\oplus\mathfrak{u}(1)_{\textup{Q}}$. These are
the standard model’s internal gauge symmetries: before and after the Higgs
mechanism.

##  III Trivializing triality

It is known that Spin(8) happens to have vector, $V$, spinor, $\psi$, and
conjugate spinor, $\widetilde{\psi}^{\prime}$, irreps that are each $8\hskip
1.42262pt\mathbb{R}$ dimensional. Furthermore each of these three 8D
representations may be identified with a copy of the octonions. Here, the
tilde on $\psi^{\prime}$ refers to octonionic conjugation, which sends
$e_{j}\mapsto-e_{j}$ for $j=\\{1,\dots 7\\}$, and which reverses the order of
multiplication, $\widetilde{xy}=\widetilde{y}\hskip 2.27621pt\widetilde{x}$
$\forall x,y,\in\mathbb{O}.$ For further details on the phenomenon of
triality, please see [7], [8], [9].

At the infinitesimal level, [9], the vector, spinor, and conjugate spinor
actions may be described by

$\begin{array}[]{lcl}\delta_{V}f&=&\widehat{d}\hskip 1.99168ptf+L_{a}\hskip
1.42262ptf+R_{b}\hskip 1.42262ptf,\vspace{2mm}\\\
\delta_{\psi}f&=&\widehat{d}\hskip 1.99168ptf+L_{a}\hskip
1.42262ptf+R_{a-b}\hskip 1.42262ptf,\vspace{2mm}\\\
\delta_{\widetilde{\psi}}\hskip 1.42262ptf&=&\widehat{d}\hskip
1.99168ptf+L_{b-a}\hskip 1.42262ptf+R_{b}\hskip 1.42262ptf,\end{array}$ (10)

respectively. Here, $f\in\mathbb{O},$ and $a,b\in\mathfrak{Im}(\mathbb{O})$.
That is, $a=a_{j}e_{j}$ and $b=b_{j}e_{j}$ for $j\in\\{1,\dots 7\\}$, and
$a_{j},b_{j}\in\mathbb{R}.$ The symbol $\widehat{d}$ is shorthand, and
represents the octonionic derivations
$\mathfrak{aut}(\mathbb{O})=\mathfrak{g}_{2}$. Explicitly,

$\begin{array}[]{lll}\widehat{d}&=&r_{k}\hskip
1.42262pti\Lambda_{k}\vspace{2mm}\\\
&+&r_{9}\frac{1}{2\sqrt{3}}\left(L_{15}+L_{34}+2L_{27}\right)+r_{10}\frac{1}{2\sqrt{3}}\left(-L_{14}+L_{35}-2L_{67}\right)\vspace{2mm}\\\
&+&r_{11}\frac{1}{2\sqrt{3}}\left(L_{46}-L_{25}+2L_{17}\right)+r_{12}\frac{1}{2\sqrt{3}}\left(L_{24}+L_{56}-2L_{37}\right)\vspace{2mm}\\\
&+&r_{13}\frac{1}{2\sqrt{3}}\left(-L_{16}+L_{23}+2L_{47}\right)+r_{14}\frac{1}{2\sqrt{3}}\left(L_{12}-L_{36}-2L_{57}\right),\end{array}$
(11)

where $k\in\\{1,\dots 8\\}$, and $r_{i}\in\mathbb{R}$ for $i\in\\{1,\dots
14\\}.$

Now, given that the vector, spinor, and conjugate spinor irreps can each be
identified with a copy of the octonions, one might wonder what happens when we
ask that their three Lie algebra actions coincide. It is straightforward to
confirm that these three $\mathfrak{so}(8)$ actions coincide on $\widehat{d}$,
the octonionic derivations
$\mathfrak{der}(\mathbb{O})=\mathfrak{aut}(\mathbb{O})=\mathfrak{g}_{2}.$
Please see Figure 2.

Figure 2:  The spinor action, conjugate spinor action, and vector action of
$\mathfrak{so}(8)$ coincide on the octonionic derivations,
$\mathfrak{aut}(\mathbb{O})=\mathfrak{g}_{2}$.

##  IV A parallel in particle physics

Curiously enough, there happens to be a similar pattern in particle physics
that bears some resemblance to the three-way intersection of Figure 2. It
results from an alternative perspective on the constraint equation (5) that we
obtained in Section I:

$\left[\ell_{10},\Psi_{\textup{L}}\right]=\frac{1}{2}\left(\ell_{10}+\ell_{10}^{*_{\overrightarrow{\mathbb{O}}}}\right)\Psi_{\textup{L}}.$
(12)

Much like the three-way constraint that sends
$\mathfrak{so}(8)\mapsto\mathfrak{g}_{2}$, we find a three-way constraint that
sends $\mathfrak{so}(10)\mapsto\mathfrak{g}_{\textup{SM}}$. For all
$\Psi_{\textup{L}}$,

$\begin{array}[]{c}\delta\Psi_{\textup{L}}=\vspace{2mm}\\\
\left[\ell_{10},\hskip
1.42262pt\Psi_{\textup{L}}\right]=\ell_{10}\Psi_{\textup{L}}=\ell_{10}^{*_{{\overrightarrow{\mathbb{O}}}}}\Psi_{\textup{L}}\vspace{2mm}\\\
\Rightarrow\vspace{2mm}\\\ \ell_{10}\hskip 5.69054pt\mapsto\hskip
5.69054pt\ell_{\textup{SM}}\vspace{2mm}\\\ \mathfrak{so}(10)\hskip
5.69054pt\mapsto\hskip
5.69054pt\mathfrak{su}(3)_{\textup{C}}\oplus\mathfrak{su}(2)_{\textup{L}}\oplus\mathfrak{u}(1)_{\textup{Y}}.\end{array}$
(13)

Please see Figure 3. In the figure, $\mathfrak{so}(10)_{V}$ refers to the
action $\left[\ell_{10},\hskip 1.42262pt\Psi_{\textup{L}}\right],$ while
$\mathfrak{so}(10)_{\Psi}$ refers to the action $\ell_{10}\Psi_{\textup{L}},$
and $\mathfrak{so}(10)_{\Psi}^{\mathbb{O}}$ refers to the action
$\ell_{10}^{*_{{\overrightarrow{\mathbb{O}}}}}\Psi_{\textup{L}}.$

Figure 3:  The spinor action, octonionic-conjugate spinor action, and multi-
vector action of $\mathfrak{so}(10)$ coincide on the Standard Model’s internal
symmetries,
$\mathfrak{g}_{\textup{SM}}=\mathfrak{su}(3)_{\textup{C}}\oplus\mathfrak{su}(2)_{\textup{L}}\oplus\mathfrak{u}(1)_{\textup{Y}}$.

Those familiar with Figure 6 of reference [1] might now wonder: What occurs
when such intersections are made to include not only octonionic reflections,
but quaternionic and complex reflections as well.

In this case, we find a five-way intersection that maps
$\mathfrak{so}(10)\mapsto\mathfrak{g}_{\textup{UB}}:=\mathfrak{su}(3)_{\textup{C}}\oplus\mathfrak{u}(1)_{\textup{Q}}$,
the Standard Model’s internal gauge symmetries that remain after the Higgs
mechanism. For all $\Psi_{\textup{L}},$

$\begin{array}[]{c}\delta\Psi_{\textup{L}}=\vspace{2.5mm}\\\
\left[\ell_{10},\hskip
1.42262pt\Psi_{\textup{L}}\right]=\ell_{10}\Psi_{\textup{L}}=\ell_{10}^{*_{{\overrightarrow{\mathbb{O}}}}}\Psi_{\textup{L}}=\ell_{10}^{*_{{\overrightarrow{\mathbb{H}}}}}\Psi_{\textup{L}}=\ell_{10}^{*_{{\overrightarrow{\mathbb{C}}}}}\Psi_{\textup{L}}\vspace{2mm}\\\
\Rightarrow\vspace{2mm}\\\ \ell_{10}\hskip 5.69054pt\mapsto\hskip
5.69054pt\ell_{\textup{UB}}\vspace{2mm}\\\ \mathfrak{so}(10)\hskip
5.69054pt\mapsto\hskip
5.69054pt\mathfrak{su}(3)_{C}\oplus\mathfrak{u}(1)_{Q}.\end{array}$ (14)

Here, $\ell_{\textup{UB}}$ is defined as

$\begin{array}[]{lrcl}\ell_{\textup{UB}}\Psi_{\textup{L}}=&\big{(}r_{j}i\Lambda_{j}&+&r^{\prime}iQ\big{)}\Psi_{\textup{L}},\end{array}$
(15)

where $r^{\prime}\in\mathbb{R},$ and $iQ$ turns out to be the electric charge
generator,

$iQ:=\frac{1}{6}\left(L_{13}+L_{26}+L_{45}\right)-\frac{1}{2}L_{\epsilon_{3}}.$
(16)

Please see Figure 4. In the figure, $\mathfrak{so}(10)_{\Psi}^{\mathbb{H}}$
furthermore refers to the action
$\ell_{10}^{*_{{\overrightarrow{\mathbb{H}}}}}\Psi_{\textup{L}}$, and
$\mathfrak{so}(10)_{\Psi}^{\mathbb{C}}$ refers to the action
$\ell_{10}^{*_{{\overrightarrow{\mathbb{C}}}}}\Psi_{\textup{L}}$.

Figure 4:  The spinor action, the octonionic-conjugate spinor action,
quaternionic-conjugate spinor action, complex-conjugate spinor action, and
multivector action of $\mathfrak{so}(10)$ coincide on the Standard Model’s
unbroken symmetries,
$\mathfrak{g}_{\textup{UB}}=\mathfrak{su}(3)_{\textup{C}}\oplus\mathfrak{u}(1)_{\textup{Q}}$.

##  V Conclusion

In this article, we identified a single algebraic constraint that sends
$\mathfrak{so}(10)$ to the Standard Model’s internal gauge symmetries,
$\mathfrak{g}_{\textup{SM}}=\mathfrak{su}(3)_{\textup{C}}\oplus\mathfrak{su}(2)_{\textup{L}}\oplus\mathfrak{u}(1)_{\textup{Y}}$.

Reworking this constraint, however, allowed us to view it as a coincidence
between three inequivalent $\mathfrak{so}(10)$ actions. Such a coincidence
constraint is already familiar from octonionic triality, where three
$\mathfrak{so}(8)$ actions coincide on
$\mathfrak{aut}(\mathbb{O})=\mathfrak{g}_{2}$.

We extended the idea further to include not only octonionic, but also
quaternionic and complex conjugate type reflections. In this case, the five-
way intersection of $\mathfrak{so}(10)$ actions results in
$\mathfrak{g}_{\textup{UB}}:=\mathfrak{su}(3)_{C}\oplus\mathfrak{u}(1)_{Q}$ \-
those Standard Model gauge symmetries surviving the Higgs mechanism.

###### Acknowledgements.

The author is grateful for discussions, feedback, and encouragement from John
Baez, Sukṛti Bansal, John Barrett, Latham Boyle, Hilary Carteret, Olaf Hohm,
Mia Hughes, Kaushlendra Kumar, Agostino Patella, Beth Romano, Matthias
Staudacher, Shadi Tahvildar-Zadeh, Carlos Tamarit, Andreas Trautner, and Jorge
Zanelli. This work was graciously supported by the VW Stiftung Freigeist
Fellowship, and Humboldt-Universität zu Berlin.

## References

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  * [2] Furey, N., “An Algebraic Roadmap of Particle Theories, Part II: Theoretical Checkpoints”, arXiv:2312.12799 [hep-ph].
  * [3] Bajc, B., Hisano, J., Kuwahara, T., Omura, Y., “Threshold corrections to dimension-six proton decay operators in non-minimal SUSY SU(5) GUTs”, Nucl.Phys. B, vol 910 (2016) arXiv:1603.03568 [hep-ph]
  * [4] Ellis, J., Evans, J.L., Nagata, N., Nanopoulos, D.V., Olive, K.A., “No-scale SU(5) super-GUTs”, Eur.Phys.J. C, 77, no.4, 232 (2017) arXiv:1702.00379 [hep-ph]
  * [5] Furey, N., Hughes, M.J., “Division algebraic symmetry breaking,” Phys. Lett. B, 831 (2022). This article was widely circulated amongst colleagues on the 16th of February, 2021. Seminar recording available https://pirsa.org/21030013 arXiv:2210.10126 [hep-ph]
  * [6] Jang, H., “Gauge Theory on Fiber Bundle of Hypercomplex Algebras,”Nucl. Phys. B 993 (2023) 116281 arXiv:2303.08159
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increases (decreases) this cost, thus reducing (increasing) the incentive for
participation among participants. In other words, CAD increase of
$\mathscr{P}^{-1}$ has the opposite impact, compared to CAD increases of
$\mathscr{P}^{1}$, on the incentive of the participants ((IC:P-S)).

For nonparticipants ($x\notin P$), higher similarity in state $\bm{\theta}=1$
reduces the LHS in (IC:NP-S) while higher similarity in state $\bm{\theta}=-1$
increases it. Under the following assumption, the incentive constraint of the
nonparticipants is always satisfied regardless of $\mathscr{P}^{-1}$.

######  Assumption 2:

For any $\sigma$ with $\mathscr{V}(\sigma)>\mathscr{V}(\sigma^{*})$, and any
$x\notin P(\sigma)$,

$\mu(x)\mathscr{P}^{1}_{x}(P(\sigma))\Lambda_{-\text{o}}<c.$

We can again use CAD to characterize the effect of information similarity.

######  Proposition 6:

Suppose $\widehat{\mathscr{P}}^{1}$ satisfies assumption 2. Let
$\mathscr{P}:=(\widehat{\mathscr{P}}^{1},\mathscr{P}^{-1})$ be an information
structure such that
$\mathscr{P}^{-1}\succcurlyeq_{CAD}\widehat{\mathscr{P}}^{-1}$, then

  1. 1.

$\mathscr{V}^{*}(\mathscr{P})\geq\mathscr{V}^{*}(\widehat{\mathscr{P}})$ if
$\Lambda_{\text{o}}>\Lambda_{\text{b}}$. And,

  2. 2.

$\mathscr{V}^{*}(\mathscr{P})\leq\mathscr{V}^{*}(\widehat{\mathscr{P}})$ if
$\Lambda_{\text{b}}>\Lambda_{\text{o}}$.

###### Proof.

Let $\sigma^{*}$ be a maximal equilibrium under $\mathscr{P}$ with associated
participation and nonparticipation sets $P^{*}$ and $NP^{*}$ respectively.

Suppose that $\Lambda_{\text{o}}>\Lambda_{\text{b}}$. Then, for all $x\in
P^{*}$

$\displaystyle\mathscr{P}^{-1}_{x}(P^{*})\Lambda_{\text{b}}+(1-\mathscr{P}^{-1}_{x}(P^{*}))\Lambda_{\text{o}}\leq\widehat{\mathscr{P}}^{-1}_{x}(P^{*})\Lambda_{\text{b}}+(1-\widehat{\mathscr{P}}^{-1}_{x}(P^{*}))\Lambda_{\text{o}}$

And, for all $x\notin P^{*}$, $\mu(x)\mathscr{P}^{1}_{x}(P^{*})<c$ by
Assumption 2. Therefore, $\sigma^{*}\in\mathcal{E}(\mathscr{P})$, proving part
$(1)$.

For $(2)$, for all $x\in P^{*}$,

$\displaystyle\mathscr{P}^{-1}_{x}(P^{*})\Lambda_{\text{b}}+(1-\mathscr{P}^{-1}_{x}(P^{*}))\Lambda_{\text{o}}\geq\widehat{\mathscr{P}}^{-1}_{x}(P^{*})\Lambda_{\text{b}}+(1-\widehat{\mathscr{P}}^{-1}_{x}(P^{*}))\Lambda_{\text{o}}.$

Therefore, $\sigma^{*}$ may not be in $\mathcal{E}(\mathscr{P})$. Finally,
consider any strategy profile $\sigma$ such that
$\mathscr{V}(\sigma)\geq\mathscr{V}(\sigma^{*})$. By Assumption 2, (IC:NP-S)
holds for all $x\in P$ in both, $\widehat{\mathscr{P}}$ and $\mathscr{P}$
(Since $\mathscr{P}^{1}=\widehat{\mathscr{P}}^{1}$). Since $\sigma^{*}$ is a
maximal equilibrium under $\widehat{\mathscr{P}}$,
$\sigma\notin\mathcal{E}(\widehat{\mathscr{P}})$. Therefore, $\exists x\in P$
for whom (IC:P-S) fails. Finally, for all $x\in P$,

$\displaystyle\mathscr{P}^{-1}_{x}(P)\Lambda_{\text{b}}+(1-\mathscr{P}^{-1}_{x}(P))\Lambda_{\text{o}}\geq\widehat{\mathscr{P}}^{-1}_{x}(P)\Lambda_{\text{b}}+(1-\widehat{\mathscr{P}}^{-1}_{x}(P))\Lambda_{\text{o}}.$

Therefore, $\sigma\notin\mathcal{E}(\mathscr{P})$. The proposition follows.

∎

###  B.2 On more than 2 groups

To establish that the qualitative results extend to settings with more than
two groups, we consider a specialized environment in which the group sizes
$\mathbf{N}_{g}$ are drawn from Poisson distributions with means $N$ and the
resilience $\bm{\bar{n}}$ is a deterministic $\bar{n}$, and restrict ourselves
to symmetric equilibrium. Let $\psi(k,N)=\frac{\exp(-N)N^{k}}{k!}$ be the
probability that nature chooses a group to have $k$ agents. We use the
following strong notion of similarity for $n$-dimensional random variables
with $n>2$, given by Meyer (1990). Let $\mathcal{Y}$ be a finite subset of n,
and let $\mathscr{Y}$ be a set of $\mathcal{Y}$-valued random variables with
exchangeable distributions, i.e., the marginal distribution of $Y_{i}$ and
$Y_{j}$ are equal for any $i,j$.

######  Definition 7:

Let $Y,\widehat{Y}\in\mathscr{Y}$ be two random variables with distributions
$\mathscr{D},\widehat{\mathscr{D}}$ respectively. Then,
$Y\succcurlyeq_{sCAD}\widehat{Y}$ if,

$\mathscr{D}(\\{Y_{1}=y_{i_{1}},Y_{2}=y_{i_{2}},\ldots,Y_{n}=y_{i_{n}}\\})\leq\widehat{\mathscr{D}}(\\{\widehat{Y}_{1}=y_{i_{1}},\widehat{Y}_{2}=y_{i_{2}},\ldots,\widehat{Y}_{n}=y_{i_{n}}\\})$

for all $(i_{1},\ldots,i_{n})$ for which it is not the case that
$i_{1}=i_{2}=\ldots=i_{n}$.

With this order, we can obtain results analogous to Theorem 1 and 2.

######  Proposition 7:

Suppose that $\mathscr{P}^{1}\succcurlyeq_{sCAD}\widehat{\mathscr{P}}^{1}$.

  1. 1.

If $\psi(\bar{n},GN)\geq\psi(\bar{n},kN)$ for all $k<G-1$, then
$\mathscr{V}^{*}(\mathscr{P})\geq\mathscr{V}^{*}(\widehat{\mathscr{P}})$.

  2. 2.

If $\psi(\bar{n},GN)<\psi(\bar{n},kN)$ for all $k<G-1$, and
$\widehat{\mathscr{P}}$ satisfies condition M, then
$\mathscr{V}^{*}(\mathscr{P})\leq\mathscr{V}^{*}(\widehat{\mathscr{P}})$.

###### Proof.

For any set $T\subset\mathcal{X}$, we define $I(T)$ as the random variable
denoting the number of other groups that receive a signal $x\in T$. Formally,
for any group $g=1$ (say), $I(T):=\sum_{g\neq 1}\mathbbm{1}_{\mathbf{X}_{g}\in
T}$. Abusing notation, we define
$\mathscr{P}^{1}_{x}(T,k)=\mathscr{P}(I(T)=k|\mathbf{X}_{1}=x,\bm{\theta}=1)$
as the probability that $k=0,1,\ldots G-1$ other groups see a signal in $T$
when group $g=1$ (say) sees the signal $x$ and the state is $\bm{\theta}=1$.
Notice that $\mathscr{P}^{1}\succcurlyeq_{sCAD}\widehat{\mathscr{P}}^{1}$
implies that

  1. 1.

For $x\in T$, $\mathscr{P}^{1}_{x}(T,k)\leq\widehat{\mathscr{P}}^{1}_{x}(T,k)$
for all $k=0,1,2\ldots G-2$ and
$\mathscr{P}^{1}_{x}(T,G-1)\geq\widehat{\mathscr{P}}^{1}_{x}(T,G-1)$.

  2. 2.

For $x\notin T$,
$\mathscr{P}^{1}_{x}(T,k)\leq\widehat{\mathscr{P}}^{1}_{x}(T,k)$ for all
$k=0,1,2\ldots G-1$

Let $\sigma^{*}$ be the maximal participation equilibrium in
$\widehat{\mathscr{P}}$, and let $P^{*}$ and $NP^{*}$ be the associated
participation and not-participation sets respectively. For more than 2 groups,
the IC for protesting and not-protesting can be modified as follows

$\displaystyle\sum_{k=0}^{G-1}\widehat{\mathscr{P}}^{1}_{x}(T,k)\psi(\bar{n},(k+1)N)$
$\displaystyle\geq\frac{c}{\mu(x)}\quad\text{ if }x\in P^{*}$ (IC:P-G)
$\displaystyle\sum_{k=0}^{G-1}\widehat{\mathscr{P}}^{1}_{x}(T,k)\psi(\bar{n},kN)$
$\displaystyle\leq\frac{c}{\mu(x)}\quad\text{ if }x\in NP^{*}$ (IC:NP-G)

It is easy to see that when the similarity increases
($\mathscr{P}^{1}\succcurlyeq_{sCAD}\widehat{\mathscr{P}}^{1}$), if
$\psi(\bar{n},GN)\geq\psi(\bar{n},kN)$ for all $k<G-1$, then the LHS increases
in (IC:P-G) and decreases in (IC:NP-G). Therefore,
$\sigma^{*}\in\mathcal{E}(\mathscr{P})$ and accordingly
$\mathscr{V}^{*}(\mathscr{P})\geq\mathscr{V}^{*}(\widehat{\mathscr{P}})$. On
the other hand, if $\psi(\bar{n},GN)\leq\psi(\bar{n},kN)$ for all $k<G-1$,
then under $\mathscr{P}^{1}$ the LHS decreases in (IC:P-G), which can violate
the incentive of the participant, and accordingly, $\sigma^{*}$ may no longer
be an equilibrium. Moreover, given that $\widehat{\mathscr{P}}$ satisfies
Condition M, given any $\sigma$ such that $P(\sigma)\geq P^{*}$, $\exists x\in
P(\sigma)$ such that,

$\displaystyle\sum_{k=0}^{G-1}\widehat{\mathscr{P}}^{1}_{x}(T,k)\psi(\bar{n},(k+1)N)$
$\displaystyle<\frac{c}{\mu(x)}$
$\displaystyle\implies\sum_{k=0}^{G-1}\mathscr{P}^{1}_{x}(T,k)\psi(\bar{n},(k+1)N)$
$\displaystyle<\frac{c}{\mu(x)}.$

Therefore, such $\sigma$ cannot constitute an equilibrium under
$\mathscr{P}^{1}$, and accordingly,
$\mathscr{V}^{*}(\mathscr{P})\leq\mathscr{V}^{*}(\widehat{\mathscr{P}})$ with
the inequality being strict whenever
$\sigma^{*}\notin\mathcal{E}(\mathscr{P})$. ∎

###  B.3 Informativeness of turnout

Consider our baseline model with two groups, two states, and a finite set of
signals. For simplicity, we assume that $\mathbf{N}_{1},\mathbf{N}_{2}$ are
Poisson distributed with mean $N$; that is, let
$\psi(k,N)=\frac{\exp(-N)N^{k}}{k!}$ be the probability that nature chooses a
group to have $k$ agents. However, now assume that a strategic policymaker
observes the realized turnout and then decides whether to change the regime.
We restrict attention to symmetric strategies, and as before, we denote the
associated participation and nonparticipation sets of any strategy by $P$ and
$NP$, respectively.

Given an information structure $\mathscr{P}$, agents’ strategy $\sigma$, and
aggregate turnout $\mathbf{A}$, the policymaker’s belief about the state of
the world is given by the likelihood function

$\beta(\cdot;\mathscr{P},P):=\frac{Prob(\bm{\theta}=1|\mathbf{A}=\cdot,P)}{1-Prob(\bm{\theta}=1|\mathbf{A}=\cdot,P)}.$

The policymaker changes the status quo only if she is sufficiently confident
that the state is $1$; that is, there is a cutoff belief, $\underline{\beta}$,
such that the policymaker changes the status quo if
$\beta(k)\geq\underline{\beta}>0$. Therefore, the policymaker’s preferences
are (ordinally) aligned with the citizens’.

We define informativeness of turnout, given a strategy $\sigma$, and its
associated participation set $P$, as in Ekmekci and Lauermann (2022):

$\displaystyle I(P):=\mathbb{P}^{1}(P)-\mathbb{P}^{0}(P)$

Define
$\bar{P}:=\\{x\in\mathcal{X}:\mathbb{P}^{1}(\\{x\\})>\mathbb{P}^{0}(\\{x\\})$.
It is easy to see that the informativeness of any strategy is bounded
$I(\bar{P})$. Therefore, for a given information structure $\mathscr{P}$ with
fixed marginals, we say that _information aggregates_ if
$\mathbbm{1}_{\bar{P}}$ is an equilibrium under $\mathscr{P}$.

We fix $\mathscr{P}^{0}$ to investigate when, if at all, increasing similarity
of information facilitates information aggregation. To this end, define

$\displaystyle\underline{l}$
$\displaystyle:=\frac{\mu}{1-\mu}\frac{\mathbb{P}^{1}(\bar{P})^{2}}{\mathscr{P}^{0}(\bar{P},\bar{P})}$
$\displaystyle\bar{l}$
$\displaystyle:=\frac{\mu}{1-\mu}\frac{\mathbb{P}^{1}(\bar{P})}{\mathscr{P}^{0}(\bar{P},\bar{P})}.$

######  Proposition 8:

Suppose that
$\mathbb{P}^{1}(\bar{P})\mathbb{P}^{0}(\bar{P})>\mathscr{P}^{0}(\bar{P},\bar{P})$.292929It
is easy to generate examples in which this inequality is satisfied. For
example, this inequality holds if the signals are independent conditional on
the state.

  1. 1.

If $\underline{l}\leq\underline{\beta}<\bar{l}$, then information does not
aggregate if $\mathscr{P}^{1}$ has conditionally independent signals (denoted
by $\mathscr{P}^{1,CI}$) as long as $c>0$; and $\exists c>0$ and a signal
$\mathscr{P}^{1}\succcurlyeq_{CAD}\mathscr{P}^{1,CI}$, such that information
aggregates under $\mathscr{P}^{1}$.

  2. 2.

If $\underline{\beta}>\bar{l}$, then information does not aggregate for any
$\mathscr{P}^{1}\succcurlyeq_{CAD}\mathscr{P}^{1,CI}$ and any $c>0$.

###### Proof of Proposition 8.

Substituting the expression for Poisson pdf, we get

$\displaystyle\beta(k,\mathscr{P},\bar{P})=$
$\displaystyle\frac{\mu}{1-\mu}\frac{\mathscr{P}^{1}(\bar{P},\bar{P})\psi(k,2N)+2\mathscr{P}^{1}(\bar{P},\bar{NP})\psi(k,N)}{\mathscr{P}^{0}(\bar{P},\bar{P})\psi(k,2N)+2\mathscr{P}^{0}(\bar{P},\bar{NP})\psi(k,N)}$
$\displaystyle=$
$\displaystyle\frac{\mu}{1-\mu}\frac{\mathbb{P}^{1}(\bar{P})+\mathscr{P}^{1}(\bar{P},\bar{P})(e^{-N}2^{k-1}-1)}{\mathbb{P}^{0}(\bar{P})+\mathscr{P}^{0}(\bar{P},\bar{P})(e^{-N}2^{k-1}-1)}$

Since $2^{k-1}$ is increasing in $k$,

$\displaystyle\operatorname{sign}\left(\frac{\partial\beta(k,\mathscr{P},\bar{P})}{\partial
k}\right)=\operatorname{sign}(\mathscr{P}^{1}(\bar{P},\bar{P})\mathbb{P}^{0}(\bar{P})-\mathbb{P}^{1}(\bar{P})\mathscr{P}^{0}(\bar{P},\bar{P})).$

When signals are conditionally independent in state $1$,
$\mathscr{P}^{1}(\bar{P},\bar{P})=\mathbb{P}^{1}(\bar{P})^{2}$. Moreover, for
any $\mathscr{P}^{1}\succcurlyeq_{CAD}\mathscr{P}^{1,CI}$,
$\mathscr{P}^{1}(\bar{P},\bar{P})\geq\mathbb{P}^{1}(\bar{P})^{2}$. This
implies

$\displaystyle\operatorname{sign}\left(\frac{\partial\beta(k,\mathscr{P},\bar{P})}{\partial
k}\right)=\operatorname{sign}(\mathbb{P}^{1}(\bar{P})\mathbb{P}^{0}(\bar{P})-\mathscr{P}^{0}(\bar{P},\bar{P}))>0.$

The last inequality is true by hypothesis. Therefore, for any $\mathscr{P}$
such that $\mathscr{P}^{1}\succcurlyeq_{CAD}\mathscr{P}^{1,CI}$,
$\beta(k,\mathscr{P},\bar{P})$ is increasing in $k$. Moreover,

$\displaystyle\lim_{k\to\infty}\beta(k,\mathscr{P},\bar{P})=\frac{\mu}{1-\mu}\frac{\mathscr{P}^{1}(\bar{P},\bar{P})}{\mathscr{P}^{0}(\bar{P},\bar{P})}$

When $\mathscr{P}^{1}=\mathscr{P}^{1,CI}$,

$\displaystyle\lim_{k\to\infty}\beta(k,\mathscr{P},\bar{P})=\underline{l}<\underline{\beta}$

Therefore, $\beta(k,\mathscr{P},\bar{P})<\underline{\beta}$ for all $k$
whenever $\mathscr{P}^{1}=\mathscr{P}^{1,CI}$. Therefore,
$\mathbbm{1}_{\bar{P}}\notin\mathcal{E}(\mathscr{P})$ if
$\mathscr{P}^{1}=\mathscr{P}^{1,CI}$ as long as $c>0$. That is, information
aggregation fails when signals are conditionally independent.

In contrast, when $\mathscr{P}^{1}=\mathscr{P}^{1,corr}$, where
$\mathscr{P}^{1,corr}$ means signals being fully correlated in state $1$,

$\lim_{k\to\infty}\beta(k,\mathscr{P},\bar{P})=\bar{l}.$

If $\bar{l}>\underline{\beta}$, $\exists k^{*}$ such that
$\beta(k,\mathscr{P},\bar{P})>\underline{\beta}$ for all $k\geq k^{*}$. For
any $\mathscr{P}^{1}\succcurlyeq_{CAD}\mathscr{P}^{1,CI}$, by Lemma 3,
$\mathscr{P}^{1}(\bar{P},\bar{P})=\mathscr{P}^{1,CI}(\bar{P},\bar{P})+\alpha$,
for some $\alpha\geq 0$. We know that, when $\alpha=0$,
$\beta(k,\mathscr{P},\bar{P})<\underline{\beta}$ for all $k$, and, when
$\alpha=\mathbb{P}^{1}(\bar{P})-\mathbb{P}^{1}(\bar{P})^{2}$, $\exists
k^{*}\in\mathbb{N}$ such that $\beta(k,\mathscr{P},\bar{P})>\underline{\beta}$
whenever $k\geq k^{*}$. Therefore, we can choose an $\alpha>0$ small enough to
construct $\mathscr{P}^{1}$ so that
$\mathscr{P}^{1}(\bar{P},\bar{P})=\mathscr{P}^{1,CI}(\bar{P},\bar{P})+\alpha$
and $\beta(k,\mathscr{P},\bar{P})\geq\underline{\beta}$ if and only if
$k>k^{*}>2N$. Therefore, the policymaker would use a threshold of
$\bar{n}=k^{*}>2N$ when $\mathscr{P}^{1}$ constructed using $\alpha$ described
above. It is easy to check that $\psi(\bar{n},2N)>\psi(\bar{n},N)$ in this
case. Claim 2 then establishes that,

$\displaystyle\min_{x\in\bar{P}}\mu(x)\left[\mathscr{P}^{1}_{x}(\bar{P})\psi(\bar{n},2N)+\mathscr{P}^{1}_{x}(\mathcal{X}\backslash\bar{P})\psi(\bar{n},N)\right]>\max_{x\in\mathcal{X}\backslash\bar{P}}\mu(x)\mathscr{P}^{1}_{x}(\bar{P})\psi(\bar{n},N)$

Therefore, by letting $c$ to be strictly between the LHS and the RHS of the
above, we get that $\mathbbm{1}_{\bar{P}}$ is an equilibrium, for it satisfies
(IC:P) for all $x\in\bar{P}$ and (IC:NP) for all
$x\in\mathcal{X}\backslash\bar{P}$. Therefore, information aggregates under
$\mathscr{P}$ wherein $\mathscr{P}^{1}\succcurlyeq_{CAD}\mathscr{P}^{1,CI}$.

Finally, if $\underline{\beta}>\bar{l}$, then the policymaker would not change
the status quo regardless of the turnout for any
$\mathscr{P}^{1}\succcurlyeq_{CAD}\mathscr{P}^{1,CI}$ establishing the last
part of the Proposition.

######  Claim 2:

If $\psi(\bar{n},2N)>\psi(\bar{n},N)$, then

$\displaystyle\min_{x\in\bar{P}}\mu(x)\left[\mathscr{P}^{1}_{x}(\bar{P})\psi(\bar{n},2N)+\mathscr{P}^{1}_{x}(\mathcal{X}\backslash\bar{P})\psi(\bar{n},N)\right]>\max_{x\in\mathcal{X}\backslash\bar{P}}\mu(x)\mathscr{P}^{1}_{x}(\bar{P})\psi(\bar{n},N)$

for any $\mathscr{P}^{1}\succcurlyeq_{CAD}\mathscr{P}^{1,CI}$.

###### Proof.

First, by definition of $\bar{P}$,
$\min_{x\in\bar{P}}\mu(x)>\max_{x\in\mathcal{X}\backslash\bar{P}}\mu(x)$.
Also, $\mathscr{P}^{1,CI}_{x}(\bar{P})=\mathbb{P}^{1}(\bar{P})$ is independent
of $x$, and, $\mathscr{P}^{1}_{x}(\bar{P})\geq\mathbb{P}^{1}(\bar{P})$ for all
$x\in\bar{P}$ and $\mathscr{P}^{1}_{x}(\bar{P})\leq\mathbb{P}^{1}(\bar{P})$
for all $x\notin\bar{P}$. Therefore,
$\min_{x\in\bar{P}}\mathscr{P}^{1}_{x}(\bar{P})\geq\max_{x\in\mathcal{X}\backslash\bar{P}}\mathscr{P}^{1}_{x}(\bar{P})$.
The claim, then, follows due to $\psi(\bar{n},2N)>\psi(\bar{n},N)$.

∎

∎

Given an information structure $\mathscr{P}$, we say an equilibrium
$\sigma^{*}$ has “ _maximally informative turnout”_ if $I(\sigma^{*})\geq
I(\sigma)$ for all $\sigma\in\mathcal{E}(\mathscr{P})$. We denote
$I(\sigma^{*})$ by $\mathcal{I}(\mathscr{P})$. Proposition 9 below, shows that
increasing similarity of information can reduce the informativeness of
turnout.

######  Proposition 9:

Suppose that $\mathscr{P}^{1}\succcurlyeq_{CAD}\widehat{\mathscr{P}}^{1}$ and
$\mathscr{P}^{0}=\widehat{\mathscr{P}}^{0}$. Let $\bar{n}^{*}$ be the optimal
participation threshold for the equilibrium with maximally informative turnout
under $\widehat{\mathscr{P}}$.

  1. 1.

If $\psi(\bar{n}^{*},2N)>2\psi(\bar{n}^{*},N)$, then
$\mathcal{I}(\mathscr{P})\geq\mathcal{I}(\widehat{\mathscr{P}})$ if
$\max_{T\subset\mathcal{X}}\mathscr{P}^{1}(T,T)-\widehat{\mathscr{P}}^{1}(T,T)$
is sufficiently small.

  2. 2.

If $\psi(\bar{n}^{*},2N)<\psi(\bar{n}^{*},N)$, then it is possible that
$\mathcal{I}(\mathscr{P})<\mathcal{I}(\widehat{\mathscr{P}})$.

###### Proof of Proposition 9.

Let $\sigma^{*}$ be the maximally informative equilibrium under
$\widehat{\mathscr{P}}$. If the policymaker continues to use $\bar{n}^{*}$ as
the cutoff, then $\sigma^{*}$ continues to remain an equilibrium under
$\mathscr{P}$ due to Theorem 1. While this takes care of the incentives of the
participants, unlike the earlier arguments, we also need to ensure that a
cutoff of $\bar{n}^{*}$ is indeed a best response for the policymaker. Since
$\bar{n}^{*}$ is the cutoff for the maximally informative equilibrium
$\beta(\bar{n}^{*};\widehat{\mathscr{P}})\geq\underline{\beta}$ and
$\beta(k;\widehat{\mathscr{P}})<\underline{\beta}$ for all $k<\bar{n}^{*}$. By
Bayes’ rule, we have,

$\displaystyle\frac{\beta(k;\widehat{\mathscr{P}})}{1-\beta(k;\widehat{\mathscr{P}})}=\frac{\mu}{1-\mu}\frac{\widehat{\mathscr{P}}^{1}(P,P)\psi(k,2N)+2\widehat{\mathscr{P}}^{1}(P,NP)\psi(k,N)}{\widehat{\mathscr{P}}^{0}(P,P)\psi(k,2N)+2\widehat{\mathscr{P}}^{0}(P,NP)\psi(k,N)}$

Since $\psi(\bar{n}^{*},2N)>2\psi(\bar{n}^{*},N)$,
$\mathscr{P}^{1}(P,P)=\widehat{\mathscr{P}}^{1}(P,P)+\alpha$ and
$\mathscr{P}^{1}(P,NP)=\widehat{\mathscr{P}}^{1}(P,NP)-\alpha$ for some
$\alpha>0$ by Lemma 3, and $\mathscr{P}^{0}=\widehat{\mathscr{P}}^{0}$,
$\beta(\bar{n}^{*};\mathscr{P})>\beta(\bar{n}^{*};\widehat{\mathscr{P}})\geq\underline{\beta}$.
However, it is now also possible that
$\beta(k;\mathscr{P})\geq\underline{\beta}$ for some $k<\bar{n}^{8}$. Simply
lowering the threshold in this case is not an option either as it affects the
incentives of the agents, possibly destroying $\sigma^{*}$ as an equilibrium.
However, when
$\max_{T\subseteq\mathcal{X}}\mathscr{P}^{(}T,T)-\widehat{\mathscr{P}}^{1}(T,T)$
is sufficiently small, $\beta(k;\mathscr{P})<\underline{\beta}$. Finally,
since $\mathcal{I}(\cdot)$ only depends on the marginal distributions, we
obtain the desired inequality.

For the second part, suppose that $N=20$, $\mathcal{X}=\\{0,1\\}$, $c=0.0368,$
and $\underline{\beta}=0.7281$. Signals are conditionally independent in state
$0$ with the marginal distribution $\mathbb{P}^{0}(1)=0.3$. In state $1$,
$\widehat{\mathscr{P}}^{1}(1,1)=0.66,\widehat{\mathscr{P}}^{1}(1,0)=0.15$.
$\mathscr{P}^{1}$ is constructed from $\widehat{\mathscr{P}}^{1}$ by using
$\alpha=0.05$. It is easy to see that $\bar{n}^{*}(\widehat{\mathscr{P}})=28$,
while the same no longer constitutes an equilibrium under $\mathscr{P}$. In
this case, if an informative equilibrium exists, it must involve mixing. It is
easy to check that mixing can only happen on signal $1$, and agents continue
to not participate when they receive a signal $0$. Therefore, informativeness
under $\mathscr{P}$ is strictly lower than under $\widehat{\mathscr{P}}$.

∎

###  B.4 On optimal information similarity

Consider two extreme cases: conditionally independent signals and perfectly
correlated signals. Suppose $Y=(Y_{1},Y_{2})$ where $Y_{i}$ is distributed
according to $\mathbb{P}^{1}$, and $Y_{1},Y_{2}$ are independent. Denote this
joint distribution by $\mathscr{P}^{1,CI}$. Analogously, let $Y=(Y_{1},Y_{2})$
be a random variable such that $Y_{1}=Y_{2}$ a.s., and $Y_{i}$ is distributed
according to $\mathbb{P}^{1}$. We denote this joint distribution by
$\mathscr{P}^{1,corr}$. Given a conditionally independent signal distribution
$\mathscr{P}^{1,CI}$, define

$CI^{\uparrow}:=\\{\mathscr{D}\in\Delta(\mathcal{X}\times\mathcal{X}):\mathscr{D}\succcurlyeq_{CAD}\mathscr{P}^{1,CI}\\}$

as all the signal distributions that are more similar (in the CAD sense) than
$\mathscr{P}^{1,CI}$. Recall that by definition of CAD, all such distributions
have the same marginal, and in this case, $\mathscr{P}^{0}$ does not affect
$\mathscr{V}(\cdot)$. Therefore, the designer solves the following problem:

$\displaystyle\sup_{\mathscr{P}^{1}\in
CI^{\uparrow}}\mathscr{V}^{*}(\mathscr{P}^{1},\mathscr{P}^{0})$

######  Proposition 10 (Optimal information similarity):

An optimal information structure exists. In encouragement environments, fully
correlated signals are optimal. In discouragement environments, if
conditionally independent signals satisfy Condition M, then they are optimal.
In other cases, intermediate levels of similarity can be optimal.

###### Proof.

We prove this using three steps. Steps 1 and 2 establish the existence of an
optimal information structure, while Step 3 describes it.

Step 1: We show that $CI^{\uparrow}$ is weak-∗ compact. Consider a sequence
$\\{\mathscr{D}_{m}\\}$ from $CI^{\uparrow}$ that converges to $\mathscr{D}$
in the sense that for all $f\in C(\mathcal{X}\times\mathcal{X})$, $\int
f\text{d}\mathscr{D}_{m}\to\int f\text{d}\mathscr{D}$. Consider a symmetric
$\alpha\in_{+}^{\mathcal{X}\times\mathcal{X}}$, i.e.,
$\alpha(i,j)=\alpha(j,i)$ for all $i,j$, and
$\widehat{\mathscr{D}}\in\Delta(\mathcal{X}\times\mathcal{X})$. Define,

$\displaystyle\mathscr{D}(i,j)=\widehat{\mathscr{D}}(i,j)-\alpha(i,j)\mathbbm{1}_{i\neq
j}+\sum_{k\neq i}\alpha(i,k)\mathbbm{1}_{i=j}$

If $\mathscr{D}\in\Delta(\mathcal{X}\times\mathcal{X})$, then we say that
“$\mathscr{D}$ is obtained from $\widehat{\mathscr{D}}$ by an ETI given by
$\alpha$”, denoted by $\mathscr{D}=\widehat{\mathscr{D}}\biguplus\alpha$.
Recall from the proof of Lemma 3 that an alternative characterization of the
CAD order (from Proposition 1 in Meyer (1990)) is

$\displaystyle\mathscr{D}\succcurlyeq_{CAD}\widehat{\mathscr{D}}\Longleftrightarrow\exists\alpha\in_{+}^{\mathcal{X}\times\mathcal{X}}\text{
such that }\mathscr{D}=\widehat{\mathscr{D}}\biguplus\alpha.$

Since $\mathscr{D}_{m}\succcurlyeq_{CAD}\mathscr{P}^{1,CI}$, we have a
sequence $(\alpha_{m})\in_{+}^{\mathcal{X}\times\mathcal{X}}$ such that
$\mathscr{D}_{m}=\mathscr{P}^{1,CI}\biguplus\alpha_{m}$. Due to finiteness of
$|\mathcal{X}\times\mathcal{X}|$, and boundedness of $\alpha_{m}$,
$\\{\alpha_{m}\\}$ has a convergent subsequence, $\\{\alpha_{m_{k}}\\}$. Let
$\alpha$ be a limit of one such convergent subsequence. Let
$\tilde{\mathscr{D}}:=\mathscr{P}^{1,CI}\biguplus\alpha$. Suppose, for
contradiction, that $\tilde{\mathscr{D}}\neq\mathscr{D}$. Then, there exists
some $(x,y)\in\mathcal{X}\times\mathcal{X}$ such that
$\tilde{\mathscr{D}}(x,y)\neq\mathscr{D}(x,y)$. Consider a continuous function
$f\in C(\mathcal{X}\times\mathcal{X})$ such that $f(x^{\prime},y^{\prime})=1$
if $x^{\prime}=x$ and $y^{\prime}=y$, and $f(x^{\prime},y^{\prime})=0$,
otherwise.303030Such a function obviously exists because of the finiteness of
$\mathcal{X}$. By construction,
$\int_{\mathcal{X}\times\mathcal{X}}f\text{d}\mathscr{D}_{m_{k}}\to\int_{\mathcal{X}\times\mathcal{X}}f\text{d}\tilde{\mathscr{D}}\neq\int_{\mathcal{X}\times\mathcal{X}}f\text{d}\mathscr{D}$.
Hence, a contradiction. Therefore,
$\tilde{\mathscr{D}}=\mathscr{D}=\mathscr{P}^{1,CI}\biguplus\alpha$, i.e.,
$\mathscr{D}\succcurlyeq_{CAD}\mathscr{P}^{1,CI}$, and therefore,
$\mathscr{D}\in CI^{\uparrow}$. Therefore, $CI^{\uparrow}$ is closed. Finally,
since $\mathcal{X}\times\mathcal{X}$ is compact,
$\Delta(\mathcal{X}\times\mathcal{X})$ is weak$-^{*}$ compact. This makes
$CI^{\uparrow}$ weak$-^{*}$ compact.

Step 2: Given compactness, consider a sequence $\\{\mathscr{D}_{m}\\}\in
CI^{\uparrow}$ such that
$\mathscr{V}(\mathscr{D}_{m})\to\bar{\mathscr{V}}=\sup_{\mathscr{D}\in
CI^{\uparrow}}\mathscr{V}^{*}(\mathscr{D})$. By compactness, we can (wlog)
assume $\mathscr{D}_{m}$ converges to some $\mathscr{D}\in CI^{\uparrow}$. For
each $\mathscr{D}_{m}$, let $\sigma^{*}_{m}$ be a maximal turnout equilibrium,
with participation and not-participation sets $(P^{*}_{m},NP^{*}_{m})$. Due to
the finiteness of $\mathcal{X}$, $|2^{\mathcal{X}}|$ is finite, and therefore,
wlog, $P^{*}_{m}=P^{*}$ for a sufficiently large $m$.313131To be precise, it
may be necessary to pass onto a subsequence for this to be true. Since the
marginals are unchanged, we have
$\mathscr{V}(\mathscr{D}_{m})=2N\mathbb{P}^{1}(P^{*})$ for a sufficiently
large $m$. Therefore, $\mathscr{V}(\mathscr{D}_{m})=\bar{\mathscr{V}}$ for a
sufficiently large $m$.

Step 3: Since more similarity increases maximal equilibrium turnout in
encouragement environments (when $\Lambda_{\text{b}}>\Lambda_{\text{o}}$), we
have
$\mathscr{V}^{*}(\mathscr{P}^{1,corr},\mathscr{P}^{0})\geq\mathscr{V}^{*}(\mathscr{P}^{1},\mathscr{P}^{0})$
for any $\mathscr{P}^{1}\in CI^{\uparrow}$. In discouragement environments
(when $\Lambda_{\text{o}}>\Lambda_{\text{b}}$), since $\mathscr{P}^{1,CI}$
satisfies Condition M, any
$\mathscr{P}^{1}\succcurlyeq_{CAD}\mathscr{P}^{1,CI}$ has
$\mathscr{V}^{*}(\mathscr{P}^{1},\mathscr{P}^{0})\leq\mathscr{V}^{*}(\mathscr{P}^{1,CI},\mathscr{P}^{0})$.
To see that intermediate levels of similarity can be optimal otherwise,
consider the example below.

######  Example 1:

$\mathbf{N}_{1},\mathbf{N}_{2}$ follow a Poisson distribution with mean $N=15$
and deterministic threshold $\bar{n}=20$. The cost of participation is
$c=0.009$ and the signal structure is as follows: $\mathcal{X}=\\{1,2,3\\}$,
$\mu_{0}=\frac{1}{2}$,
$\mathbb{P}^{1}=[0.25,0.3,0.45],\mathbb{P}^{0}=[0.6,0.35,0.05]$.

With $\mathscr{P}^{1,CI}$, the unique equilibrium is
$\sigma=\mathbbm{1}_{s=3}$. However, if we perform ETI on the square
$\\{(1,1),(1,2),(2,1),(2,2)\\}$ with $\alpha=0.005$, we can support an
equilibrium $\sigma=\mathbbm{1}_{s\in\\{2,3\\}}$. Finally, it is easy to see
that $\mathscr{P}^{1,corr}$ also cannot support $\mathbbm{1}_{s\in\\{2,3\\}}$
as an equilibrium. We can verify that $\mu=[0.2941,0.4615,0.9000]$. When
$P=\\{2,3\\}$, $\mathbb{P}^{1}(P)=0.75=\mathscr{P}^{1,CI}_{x}(P)$ for all $x$
since signals are conditionally independent.
$\psi(\bar{n},N)=0.0418,\psi(\bar{n},2N)=0.0134$. ∎

###  B.5 Changing information similarity along with changing marginals

Throughout, we have assumed that the agents’ marginal distribution of signals
in any given state remains unchanged and it is only the joint distribution
that changes. However, one may wish to understand the effect of changing
information similarity without holding the marginal distribution the same. To
this end, it is useful to think of signals as posteriors, as is standard
following Kamenica and Gentzkow (2011). That is, we assume that $\mathcal{X}$
is the set of posteriors with the natural order, and
$\mathscr{P}^{1},\mathscr{P}^{0}$ are two feasible joint distributions over
posteriors. As before, we assume that $\mathcal{X}$ is finite. Similar to
Definition 2, we say that $Y\succcurlyeq_{CAD}\widehat{Y}$ if only part (2) of
that definition holds. That is, we require that, given own signal, an agent
assigns a higher probability to the other agent receiving the signal, but we
dispense away the requirement of the two signals having the same marginal
distributions. This enables us to capture situations such as ones where the
two groups exchange each other’s information with some probability. We make
the following additional assumption that maintain in this section.

######  Assumption 3:

Define $g(y,x):=\mathscr{P}^{1}(\mathbf{X}_{2}\geq y|\mathbf{X}_{1}=x)$. For
every $y$, $g(y,x_{1})\geq g(y,x_{2})$ if $y\leq x_{2}\leq x_{1}$.

######  Lemma 4:

Suppose that $\Lambda_{\text{b}}>\Lambda_{\text{o}}$ and $\mathscr{P}^{1}$
satisfies Assumption 3. Then, the maximal equilibrium $\sigma^{*}$ under
$\mathscr{P}$ has a cutoff structure, i.e., $P(\sigma^{*})=x^{\uparrow}$ for
some some $x\in\mathcal{X}$.323232$x^{\uparrow}$ is the upper contour set,
i.e., the set of signal realizations greater than or equal to $x$.

###### Proof.

The proof follows similar steps as the proof of Lemma 2. Let $\sigma$ be an
equilibrium. Define $\underline{x}$ to be the minimum element of $P(\sigma)$,
where $\mathcal{X}$ is endowed with the natural order of posteriors. Then, for
every $x\geq\underline{x}$,

$\displaystyle\mu(x)\left[\mathscr{P}^{1}_{x}(\underline{x}^{\uparrow})\Lambda_{\text{b}}+(1-\mathscr{P}^{1}_{x}(\underline{x}^{\uparrow}))\Lambda_{\text{o}}\right]$
$\displaystyle\geq\mu(x)\left[\mathscr{P}^{1}_{\underline{x}}(\underline{x}^{\uparrow})\Lambda_{\text{b}}+(1-\mathscr{P}^{1}_{\underline{x}}(\underline{x}^{\uparrow}))\Lambda_{\text{o}}\right]$
$\displaystyle\geq\mu(\underline{x})\left[\mathscr{P}^{1}_{\underline{x}}(\underline{x}^{\uparrow})\Lambda_{\text{b}}+(1-\mathscr{P}^{1}_{\underline{x}}(\underline{x}^{\uparrow}))\Lambda_{\text{o}}\right]\geq
c$

where the first inequality follows from Assumption 3 and the second inequality
is due to $\mu(x)\geq\mu(\underline{x})$ whenever $x\geq\underline{x}$.

Therefore, (IC:P) is satisfied for all $x\geq\underline{x}$. Hence, following
identical steps as in Lemma 2 with $P_{1}=P_{2}=\underline{x}^{\uparrow}$, we
obtain that the maximal equilibrium is in symmetric, cut-off strategies. ∎

For the proposition below, we let $\mathscr{P},\widehat{\mathscr{P}}$ be two
joint distributions and let
$\mathbb{P}^{\theta},\widehat{\mathbb{P}}^{\theta}$ be the joint distribution
of signals (and hence posteriors) they induce in state $\theta$. Following
Kuvalekar et al. (2023), we now define what it means for a random variable to
be “more spread out” than another. Similar notions have appeared in Johnson
and Myatt (2006).

######  Definition 8:

We say that a random variable $Y$ is “more spread out around $y$” than
$\widehat{Y}$ if, $G_{Y}(z)\geq G_{\widehat{Y}}(z)$ for all $z\leq y$ and
$G_{Y}(z)\leq G_{\widehat{Y}}(z)$ for all $z\geq y$.

######  Proposition 11:

Suppose that $\Lambda_{\text{b}}>\Lambda_{\text{o}}$,
$\mathscr{P}^{1}\succcurlyeq_{CAD}\widehat{\mathscr{P}}^{1}$, and
$\mathbb{P}^{1}$ is more spread out around $\tilde{\mu}$ than
$\widehat{\mathbb{P}}^{1}$ for some $\tilde{\mu}\leq c$. Then,
$\mathscr{V}^{*}(\mathscr{P})\geq\mathscr{V}^{*}(\widehat{\mathscr{P}})$.

###### Proof.

By Theorem 1, if $\sigma\in\mathcal{E}(\widehat{\mathscr{P}})$, then
$\sigma\in\mathcal{E}(\mathscr{P})$ whenever
$\Lambda_{\text{b}}>\Lambda_{\text{o}}$. Notice that the proof of this part in
Theorem 1 did not use the fact that the marginal distributions remained
unchanged therein. By Lemma 4, the maximal equilibrium is in cutoff strategies
for $\mathscr{P}$ and $\widehat{\mathscr{P}}$. Let
$\sigma(\widehat{\mathscr{P}})$ denote the maximal equilibrium under
$\widehat{\mathscr{P}}$. Since it has a cutoff structure, let $\widehat{x}$ be
the associated cutoff. An agent would never participate on a signal such that
$\mu(x)<c$. Therefore, $\mu(\widehat{x})\geq c$. Hence,

$\displaystyle\mathscr{V}^{*}(\widehat{\mathscr{P}})=\widehat{\mathbb{P}}^{1}(\widehat{x}^{\uparrow})\leq\mathbb{P}^{1}(\widehat{x}^{\uparrow})$

where the inequality follows due to the spread-out ranking of $\mathbb{P}^{1}$
and $\widehat{\mathbb{P}}^{1}$. Thus,
$\mathscr{V}^{*}(\mathscr{P})\geq\mathscr{V}^{*}(\widehat{\mathscr{P}})$. ∎

Finally, unlike in Theorem 1, more information can strictly lower
participation even in the encouragement environment without an additional
condition such as the spread-out order.333333An example is available on
request.

###  B.6 Example from the introduction

In this section, we analyze the example from the introduction in more detail.
Recall the example. The payoff matrix is as follows.

|  | Bob
---|---|---
|  | work | shirk
Abe | work | $\theta-c,\theta-c$ | $q\theta-c,q\theta$
shirk | $q\theta,q\theta-c$ | $0,0$

Abe and Bob each receives a binary signal $\mathbf{X}_{i}\in\\{0,1\\}$, such
that $\mathbf{X}_{i}=0$ if $\bm{\theta}=0$, and if $\bm{\theta}=1$, the
signals are drawn from some exchangeable joint distribution $\mathscr{P}^{1}$.
Table 2 below describes $\mathscr{P}^{1}$.

$\bm{\theta}=1$ | $\mathbf{X}_{-i}=0$ | $\mathbf{X}_{-i}=1$ | marg
---|---|---|---
$\mathbf{X}_{i}=0$ | $(1-p)^{2}+\alpha$ | $p(1-p)-\alpha$ | $1-p$
$\mathbf{X}_{i}=1$ | $p(1-p)-\alpha$ | $p^{2}+\alpha$ | $p$
marg | $1-p$ | $p$ | $1$
Table 2: Probability distribution of signals when $\bm{\theta}=1$

Notice that the marginal in state $\bm{\theta}=1$ is given by
$P(\mathbf{X}_{i}=1|\bm{\theta}=1)=p$, and $\mu(0)\in(\frac{1}{2},1)$. Higher
$\alpha$ makes Abe and Bob’s signals more similar conditional on the state
being $\theta=1$: Signals are independent when $\alpha=0$ and perfectly
correlated when $\alpha=p(1-p)$. Define

$\tilde{p}:=\mathscr{P}^{1}_{1}(\mathbf{X}_{-i}=1)=p+\frac{\alpha}{p}.$

Obviously, $\tilde{p}\geq p$ and the equality holds under conditional
independence. Given the primitive $p$, we can say that the signals are more
similar when $\tilde{p}$ is higher.

For simplicity, we assumed in the example that a player never works after
seeing $\mathbf{X}_{i}=0$, or formally,

$\frac{c}{\mu(0)}>1.$

This assumption holds when $c>\frac{1-p}{1+1-p}$, or $p>\frac{1-2c}{1-c}$. If
$c>\frac{1}{2}$, then this assumption always holds true, and when
$c<\frac{1}{2}$, this assumption requires $p$ to be sufficiently high. For
simplicity, we assume

$c>\frac{1}{2}.$

Let us start with a numerical example. Suppose that $c=0.6$ and
$Prob(\mathbf{X}_{i}=1|\theta=1)=\frac{1}{3}$.

Consider signals that are independent conditional on $\bm{\theta}=1$. Then,
$\mathscr{P}^{1}_{1}(\mathbf{X}_{-i}=1)=\frac{1}{3}$. If
$\mathscr{P}^{1}_{1}(\mathbf{X}_{-i}=1)<\frac{1}{2}$, a player’s incentive to
work is more influenced by whether she alone can make a difference, making the
LHS of the incentive constraint increasing in $q$. Therefore, if $\sigma^{1}$
is an equilibrium for some $q$, it is an equilibrium for any $q^{\prime}>q$.
This gives us the first result with $q^{*}=0.8$.

Next, we increase the similarity of the signal as in Figure 1, where
$\alpha=\frac{2}{9}$. This makes $\mathscr{P}^{1}_{1}(\mathbf{X}_{-i}=1)=1$.
If $\mathscr{P}^{1}_{1}(\mathbf{X}_{-i}=1)>\frac{1}{2}$, a player’s incentive
to work is more influenced by whether the other player works, making the LHS
on the incentive constraint decreasing in $q$. Therefore, if $\sigma^{1}$ is
an equilibrium for some $q$, it is an equilibrium for all $q^{\prime}<q$. This
gives us the second result with $q^{**}=0.4$.

Below, we show that the qualitative insight demonstrated in the introduction
is valid even if we allowed asymmetric strategies or mixing. For any
$\beta\in[0,1]$, define $I^{\beta}(\tilde{p},q)$ to be the probability of
being pivotal when working, given that the other player works with probability
$\beta$ when he receives a signal $\mathbf{X}_{-i}=1$.

$I^{1}(\tilde{p},q):=\tilde{p}(1-q)+(1-\tilde{p})q.$ $I^{0}(\tilde{p},q):=q.$

For any $\beta\in(0,1)$,

$I^{\beta}(\tilde{p},q):=\tilde{p}(\beta(1-q)+(1-\beta)q)+(1-\tilde{p})q.$

Note that $I^{\beta}(\tilde{p},q)$ is decreasing in $\beta$ if $q>\frac{1}{2}$
and increasing in $\beta$ if $q<\frac{1}{2}$. If
$I^{0}(\tilde{p},q),I^{1}(\tilde{p},q)<c$, then $I^{\beta}(\tilde{p},q)<c$ for
all $\beta$.

In the introduction, we only considered the symmetric pure strategy profile
$\sigma^{1}$ in which both Abe and Bob work with probability $1$ after seeing
$\mathbf{X}_{i}=1$. Now let $\sigma^{\beta}$ denote symmetric mixed strategy
profiles in which both players work with probability $\beta$ after seeing
$\mathbf{X}_{i}=1$. Let $\sigma^{a}$ denote an asymmetric strategy profile in
which exactly one of Abe and Bob works after seeing $\mathbf{X}_{i}=1$, and
let $\sigma^{0}$ denote the strategy profile with no effort even after
$\mathbf{X}_{i}=1$. We say an equilibrium is maximal if it maximizes the total
expected effort by the two players.

######  Proposition 12:

Assume $c>\frac{1}{2}$.

  1. 1.

If $q<\frac{1}{2}$ then

    1. (a)

for $\tilde{p}<c$, $\sigma^{0}$ is the maximal equilibrium

    2. (b)

for $\tilde{p}>c$, there exists a threshold $q^{**}$ such that, $\sigma^{1}$
is the maximal equilibrium whenever $q\in(0,q^{**})$ and $\sigma^{0}$ is the
maximal equilibrium in $q\in(q^{**},\frac{1}{2})$

  2. The threshold $q^{**}$ is increasing in $\tilde{p}$. Therefore, the region where $\sigma^{1}$ is the maximal equilibrium is increasing in the set order in $\tilde{p}$. Therefore, similarity helps participation.

  3. 2.

If $q>\frac{1}{2}$, then there exist two thresholds, $\widehat{q},q^{*}$ such
that the following holds:

    1. (a)

for $\tilde{p}<1-c$, $\sigma^{0}$ is the maximal equilibrium in
$q\in(\frac{1}{2},c)$, $\sigma^{a}$ is the maximal equilibrium in
$q\in(c,\widehat{q})$, $\sigma^{\beta}$ is the maximal equilibrium in
$q\in(\widehat{q},q^{*})$, $\sigma^{1}$ is the maximal equilibrium in
$q\in(q^{*},1)$

    2. (b)

for $\tilde{p}>1-c$, $\sigma^{0}$ is the maximal equilibrium in
$q\in(\frac{1}{2},c)$, $\sigma^{a}$ is the maximal equilibrium in
$q\in(c,\widehat{q})$, $\sigma^{\beta}$ is the maximal equilibrium in
$q\in(\widehat{q},1)$.

  4. The thresholds $\widehat{q},q^{*}$ are increasing in $\tilde{p}$, and the mixing probability $\beta$ is decreasing in $\tilde{p}$. Therefore, similarity hurts participation.

###### Proof.

Define

$\displaystyle q^{*}(\tilde{p}):=$
$\displaystyle\max\left\\{\min\left\\{\frac{c-\tilde{p}}{1-2\tilde{p}},1\right\\},0\right\\}$
$\displaystyle q^{**}(\tilde{p}):=$
$\displaystyle\max\\{\min\left\\{\frac{\tilde{p}-c}{2\tilde{p}-1},1\right\\},0\\}$
$\displaystyle\widehat{q}(\tilde{p}):=$
$\displaystyle\max\left\\{\min\left\\{\frac{1}{2}\left(\frac{2c-1}{1-\tilde{p}}+1\right),1\right\\},0\right\\}.$

All three functions are non-decreasing in $\tilde{p}$. Since
$q^{*}(0)=\widehat{q}(0)=c$, for any $\tilde{p}$,
$q^{*}(\tilde{p}),\widehat{q}(\tilde{p})\geq c$. Recall that $c>\frac{1}{2}$.
On the other hand, for any $\tilde{p}$, $q^{**}(\tilde{p})\leq
q^{**}(1)=1-c<\frac{1}{2}$. Also, note that $q^{*}(1-c)=1$ and $q^{**}(c)=0$.

######  Claim 3:

For any $\tilde{p}\in[p,1]$, $\sigma^{1}$ is the maximal equilibrium if $q\geq
q^{*}(\tilde{p})$ or if $q\leq q^{**}(\tilde{p})$. Moreover, if
$\tilde{p}\in(1-c,c)$, then $q^{*}(\tilde{p})=1$ and $q^{**}(\tilde{p})=0$,
and $\sigma^{1}$ is not an equilibrium for any $q\in[0,1]$.

###### Proof.

$\sigma^{1}$ is an equilibrium if $I^{1}(\tilde{p},q)\geq c$. If $\sigma^{1}$
is an equilibrium, then it is the maximal equilibrium. Suppose that
$\tilde{p}\leq\frac{1}{2}$. Then, $I^{1}(\tilde{p},q)$ is increasing in $q$.
Therefore, there exists $q^{*}(\tilde{p})$ (defined above) such that for
$q\geq q^{*}(\tilde{p})$, $I^{1}(\tilde{p},q)\geq c$, i.e, $\sigma^{1}$ is the
maximal equilibrium. Next, suppose that $\tilde{p}\geq\frac{1}{2}$. Then,
$I^{1}(\tilde{p},q)$ is decreasing in $q$. Therefore, there exists
$q^{**}(\tilde{p})$ (defined above) such that for $q\leq q^{**}(\tilde{p})$,
$I^{1}(\tilde{p},q)\geq c$, i.e, $\sigma^{1}$ is the maximal equilibrium.

If $\tilde{p}<1-c$, then $q^{**}(\tilde{p})=0$ and $q^{*}(\tilde{p})\in(c,1)$.
Therefore, $\sigma^{1}$ is not an equilibrium for $q<\frac{1}{2}$, whereas it
is an equilibrium only if $q$ is sufficiently high. If $\tilde{p}=1-c$, then
$q^{*}(\tilde{p})=\frac{c-\tilde{p}}{1-2\tilde{p}}=1$, and $\sigma^{1}$ is an
equilibrium only if $q=1$. For $\tilde{p}\in(1-c,c)$, $\sigma^{1}$ is not an
equilibrium for any $q$. If $\tilde{p}=c$, then
$q^{**}(\tilde{p})=\frac{\tilde{p}-c}{2\tilde{p}-1}=0$, and $\sigma^{1}$ is an
equilibrium only if $q=0$. If $\tilde{p}>c$, then
$q^{**}(\tilde{p})\in(0,1-c)$ and $q^{*}(\tilde{p})=1$. Therefore,
$\sigma^{1}$ is not an equilibrium for $q>\frac{1}{2}$, whereas it is an
equilibrium only if $q$ is sufficiently low.

∎

######  Claim 4:

For any $\tilde{p}\in[p,1]$, $\sigma^{0}$ is the maximal equilibrium if
$q\in(q^{**}(\tilde{p}),c)$.

###### Proof.

If $q<c\leq q^{*}(\tilde{p})$ and $q>q^{**}(\tilde{p})$, then $\sigma^{1}$ is
not an equilibrium, i.e., $I^{1}(\tilde{p},q)<c$. Moreover
$I^{0}(\tilde{p},q)=q<c$, which means $\sigma^{a}$ cannot be an equilibrium
either. Recall that if $I^{1}(\tilde{p},q)<c$ and $I^{0}(\tilde{p},q)<c$, then
$I^{\beta}(\tilde{p},q)<c$ for any $\beta$, which means $\sigma^{\beta}$
cannot be an equilibrium either. Note that $\sigma^{0}$ is always an
equilibrium for $q<c$. Therefore, it is the only equilibrium, and hence, the
maximal equilibrium. ∎

Since $c>\frac{1}{2}$, it follows from the above two claims that for
$q<\frac{1}{2}$, the maximal equilibrium is either $\sigma^{1}$ or
$\sigma^{0}$. Recall that $q^{**}(\tilde{p})=0$ for all $\tilde{p}<c$.
Therefore, for $\tilde{p}<c$ the maximal equilibrium is $\sigma^{0}$ for all
$q\in[0,\frac{1}{2}]$ and for $\tilde{p}>c$ the maximal equilibrium is
$\sigma^{1}$ for all $q\in[0,q^{**}(\tilde{p})]$ and $\sigma^{0}$ for all
$q\in[q^{**}(\tilde{p}),\frac{1}{2}]$. This proves the first part of the
proposition.

######  Claim 5:

For any $\tilde{p}\in[p,1]$, $\sigma^{a}$ is the maximal equilibrium if
$q\in[c,\widehat{q}(\tilde{p})]$ and $\sigma^{\beta}$ is the maximal
equilibrium if $q\in[\widehat{q}(\tilde{p}),q^{*}(\tilde{p})]$.

###### Proof.

For any $\tilde{p}$, $\widehat{q}(\tilde{p})\in[c,q^{*}(\tilde{p})]$. It
follows from Claim 3 that $\sigma^{1}$ is not an equilibrium in the interval
$[c,q^{*}(\tilde{p})]$. In this interval, $\sigma^{a}$ is an equilibrium since
$I^{0}(\tilde{p},q)=q\geq c$. In this interval, there is also a mixed strategy
equilibrium $\sigma^{\beta}$ where the players are indifferent between
exerting effort and not exerting effort after seeing signal $1$, i.e.,
$I^{\beta}(\tilde{p},q)=c$. This gives us the equilibrium probability of
effort

$\beta=\frac{q-c}{(2q-1)\tilde{p}}=\frac{1}{\tilde{p}}\left(\frac{1}{2}-\frac{c-\frac{1}{2}}{2q-1}\right).$

The expected effort under $\sigma^{a}$ is $p$ whereas that under
$\sigma^{\beta}$ is $2\beta p$. Therefore, in this interval, $\sigma^{\beta}$
is the maximal equilibrium if $\beta\geq\frac{1}{2}$ and $\sigma^{a}$ is the
maximal equilibrium, otherwise. $\beta\geq\frac{1}{2}$ is equivalent to
$q\geq\widehat{q}(\tilde{p})$. Since $\beta$ is decreasing in $\tilde{p}$ and
increasing in $q$, $\widehat{q}(\tilde{p})$ increasing in $\tilde{p}$.

∎

Recall that for any $\tilde{p}$, $\widehat{q}(\tilde{p}),q^{*}(\tilde{p})\geq
c>\frac{1}{2}$ and $q^{*}(\tilde{p})=1$ for $\tilde{p}>1-c$. Therefore, under
$\tilde{p}<1-c$, for $q>\frac{1}{2}$, $\sigma^{0}$ is the maximal equilibrium
in $q\in(\frac{1}{2},c)$, $\sigma^{a}$ is the maximal equilibrium in
$q\in(c,\widehat{q}(\tilde{p}))$, $\sigma^{\beta}$ is the maximal equilibrium
in $q\in(\widehat{q},(\tilde{p})q^{*}(\tilde{p}))$, $\sigma^{1}$ is the
maximal equilibrium in $q\in(q^{*}(\tilde{p}),1)$. Finally, under
$\tilde{p}>1-c$, $q^{*}(\tilde{p})=1$ and $\sigma^{1}$ is not an equilibrium
at all. This proves the second part of the proposition.

∎

Figure 4 provides a visual description of how the maximal equilibrium changes
with information similarity. In this simple example, under conditionally
independent signals $\tilde{p}=p<1-c<\frac{1}{2}$, $I^{1}(\tilde{p},q)$
increases in $q$. That is, a player has a higher incentive to work if he is
more likely to complete the project on his own. Therefore, there is a
$q^{*}>c$ such that $\sigma^{1}$ is an equilibrium for $q>q^{*}$. For
$q<q^{*}$, $\sigma^{1}$ cannot be sustained as an equilibrium. However, if
$q>c$, there is an asymmetric and mixed equilibrium. A higher $\tilde{p}$
means $I^{1}(\tilde{p},q)$ is lower (since $q>\frac{1}{2}$). Thus, similar
information reduces a player’s incentive to work for high $q$, and it becomes
more difficult to sustain $\sigma^{1}$ as an equilibrium ($q^{*}$ increases).
In fact, if $\tilde{p}>1-c$, then $q^{*}=1$, i.e., $\sigma^{1}$ cannot be an
equilibrium for any $q>\frac{1}{2}$. Even for the mixed strategy equilibrium,
the probability of working must decrease to sustain an equilibrium. As
$\tilde{p}$ increases further ($\tilde{p}>c$), then $I^{1}(\tilde{p},q)>c$ for
sufficiently small values of $q$ ($q<q^{**}$). That is, players who are
unlikely to complete the project on their own, are now willing to work because
it is very likely ($\tilde{p}>c$), that when they see $1$ the other also sees
$1$. As $\tilde{p}$ increases further, $I^{1}(\tilde{p},q)$ increases, that
is, it becomes easier to sustain $\sigma^{1}$ as an equilibrium for low values
of $q$, and accordingly $q^{**}$ increases.

$0$$1$$1/2$$c$$\widehat{q}$$q^{*}$$\sigma^{0}$$\sigma^{0}$$\sigma^{a}$$\sigma^{\beta}$$\sigma^{1}$$\tilde{p}<1-c$$0$$1$$1/2$$c$$\widehat{q}$$\sigma^{0}$$\sigma^{0}$$\sigma^{a}$$\sigma^{\beta}$$\tilde{p}\in(1-c,c)$$0$$1$$1/2$$c$$\widehat{q}$$q^{**}$$\sigma^{1}$$\sigma^{0}$$\sigma^{0}$$\sigma^{0}$$\sigma^{a}$$\sigma^{\beta}$$\tilde{p}>c$
Figure 4: Information Similarity and Maximal Equilibrium


# Dynamics of a Towed Cable with Sensor-Array for Underwater Target Motion
Analysis

Rohit Kumar Singh  Department of Electrical Engineering
Indian Institute of Technology Patna
Bihar, India
<EMAIL_ADDRESS>Subrata Kumar  Department of Mechanical Engineering
Indian Institute of Technology Patna
Bihar, India
<EMAIL_ADDRESS>Shovan Bhaumik  Department of Electrical Engineering
Indian Institute of Technology Patna
Bihar, India
<EMAIL_ADDRESS>
###### Abstract

During a war situation, many times an underwater target motion analysis (TMA)
is performed using bearing-only measurements, obtained from a sensor array,
which is towed by an own-ship with the help of a connected cable. It is well
known that the own-ship is required to perform a manoeuvre in order to make
the system observable and localise the target successfully. During the
maneuver, it is important to know the location of the sensor array with
respect to the own-ship. This paper develops a dynamic model of a cable-sensor
array system to localise the sensor array, which is towed behind a sea-surface
vessel. We adopt a lumped-mass approach to represent the towed cable. The
discretized cable elements are modelled as an interconnected rigid body,
kinematically related to one another. The governing equations are derived by
balancing the moments acting on each node. The derived dynamics are solved
simultaneously for all the nodes to determine the orientation of the cable and
sensor array. The position of the sensor array obtained from this proposed
model will further be used by TMA algorithms to enhance the accuracy of the
tracking system.

##  I Introduction

Target motion analysis (TMA) is used to track and predict the motion of a
target, such as enemy submarines or surface ships [1]. Often, a passive TMA is
preferred due to some tactical advantages and it uses passive SONAR to collect
acoustic signals emitted by the target. A passive SONAR operates by listening
without transmitting, and it can either be mounted on a hull or towed behind a
ship [2]. If the sensor array is not hull mounted, it is attached to a long
cable that is towed by the ship [3] in order to minimize the interference of
ship’s sound during sensing [3]. This configuration ensures that the sensor
array remains distant from the ship’s inherent noise sources [4] as shown in
Fig. 1.

The Kalman filtering techniques are then applied to predict and update the
target’s state based on the collected measurements [5]. In the passive mode,
SONAR can only measure bearing angle [6]. Tracking a target with only bearing
measurement is known as bearing-only tracking (BOT). The BOT is indeed a
challenging task, and the own-ship moving with constant velocity needs to
execute a manoeuvre to ensure the observability of the tracking system [7] so
that the tracking filter tracks the target and the estimation converges [8].

To track a target, filters require an accurate location of the sensor array
_w.r.t._ ship [9]. It is easily obtained if the own-ship maintains a steady,
level course during measurement sampling intervals. However, any manoeuvres or
changes in the course of the ship can disrupt the path of the sensor-array,
and its position becomes untraceable [10]. So, a dynamic model of towed array
system is warranted which will provide sensor location accurately for all
sorts of maneuvers of the ship. In this paper, we tried to develop a dynamic
model using Newtonian mechanics which when solved will provide the position of
the sensor array at any point in time.

Research on the configuration of towed cable-array systems (in different
contexts) traces its origins back to the 1950s and has witnessed substantial
development over the last few decades [2]. This evolution can be broadly
categorized into two aspects: a simulation-based approach [11], and
experimental observation [12, 13, 14]. In the simulation-based approach, the
towed cable is conceptualized as a flexible body, while the sensor array is
treated as rigid [15]. The flexible modelling of the cable involves
discretizing it into numerous small finite elements, where the mass of each
element is concentrated into a node [16]. This approach is also known as the
lumped mass method. The lumped mass method [17, 18, 19, 9] uses Newton’s
second law of motion to model the dynamic behaviour of towed cable. It solves
the resulting non-linear ordinary differential equation using finite
difference approximations, transforming the continuous problem into a discrete
one [20]. All these existing works modelled the marine towing cable for
different applications such as marine seismic exploration for oil and gas
detection [21], towed submarine for mineral exploration [4], sea-terrain
mapping [22], establishing the communication relay with the onshore station
[23], and offshore energy exploration [24]. To best of our knowledge, none of
the existing work has performed the dynamics modelling of the towed cable
sensor-array system (TCSAS) which could be useful for the purpose of
underwater TMA.

In this study, we have adopted the first physics principles, such as Newton’s
laws of motion and principles of conservation of momentum, to derive the
governing equations for the dynamics modelling of the towed cable sensor-array
system (TCSAS) without relying on empirical data. This approach offers a more
rigorous and comprehensive understanding of the system’s underlying physics.
This study uses a lumped-mass model to represent the towed cable as a flexible
body with discretized cable elements serving as interconnected rigid bodies,
kinematically related to one another [25]. The connecting flexible cable is
assumed to consist of two interconnected rigid bodies, and the sensor array is
hinged at the end of the second segment. The governing equations account for
gravity, buoyancy, lift, towed force, reaction force, material properties, and
hydrodynamics acting on the lumped mass point and they are derived by
balancing moments at each node, which are solved simultaneously for all the
nodes at every instant to determine the orientation of each discretized cable
element and sensor array. The proposed model is implemented to simulate the
dynamics of the towed sensor-array for an engagement scenario involving a
target and an own-ship manoeuvre [1] and the simulation results are presented
in the paper.

Figure 1:  Towed cable sensor-array system.

##  II Mathematical Model

Fig. 1 shows a surface vessel towing a sensor-array which is connected through
a flexible cable. This section aims to derive a mathematical model to provide
the location of towed sensor array when the ship’s motion is known. The
dynamics modelling of the towed cable sensor-array system is accomplished
using the fundamental laws of Newtonian mechanics, by considering the
following assumptions: (i) the sensor array is hinged at the end of cable,
(ii) the cable is a flexible body, (iii) ship motion is independent of the
towed cable senor-array because the ship’s mass is several order of magnitude
higher than the cable sensor-array system, (iv) water surface is still, (v)
the deformation of cable is negligible under the influence of flowing water,
(vi) and drag coefficient is constant irrespective of angle of attack between
the cable-array system and fluid surface.

The flexible rope is modelled by discretizing a single rigid body into
multiple interconnected rigid elements, with each element’s mass concentrated
into a node. Fig. 2 illustrates the free body diagram, depicting a rope
connected to the ship through a spherical joint at point A, and a rigid sensor
array segment is hinged to the end of the rope at point B. The tow point A
experiences a towing force from the surface ship. Additional forces, such as
gravity, buoyancy, lift, reaction force, and hydrodynamics, act on the towed
body. The dynamics of TCSAS are obtained using the moment balance equation at
each node, and they depend on various factors, including the forces acting on
the towed body, the number of segments used for modelling the rigid body,
types of joints, the medium of motion, towed body structure, and its material
composition.

###  II-A General/Governing equation of motions

Figure 2: Free body diagram representation of a rigid body hinged to a ship.

The angular momentum about point A, visualised with the aid of Fig. 2 is

$L_{A}=L_{G}+\vec{r}_{AG}\times m\vec{v},$ (1)

where $L_{G}$ is the angular momentum about the centre of gravity (CG) $G$,
$\vec{r}_{AG}$ is a vector from towing point $A$ to CG of the cable, $G$, and
$m\vec{v}$ is the linear momentum acting at G. The resultant of all the
external forces acting on the system is represented as resultant force
$\textstyle\textstyle\sum F$ through $G$ and its corresponding moment is
$\textstyle\sum M_{G}$. The sum of moments about point $A$ of Fig. 2, due to
all the forces external to the system, must be equal to the moment of their
resultants, which can be expressed as

$\textstyle\textstyle\textstyle\sum{M_{A}}=\dot{H}_{G}+\vec{r}_{AG}\times
m\vec{a}_{A},$ (2)

where $\dot{H}_{G}=I\ddot{\theta}$ is the moment due to the resultant force at
$G$, $I$ is the moment of inertia, and $\ddot{\theta}$ is the angular
acceleration. $\textstyle\textstyle\sum{M_{A}}$ is the sum of external moments
about point $A$. The motion of the moving coordinate system in Newtonian
mechanics is specified with respect to the inertial frame of reference, as
shown in Fig. 2.

The TCSAS achieves a steady-state equilibrium configuration when the surface
ship maintains a stable operating speed [21]. The depth of the towed body at
this configuration is determined using the tension balance equation at the tow
point. The ship is moving with constant velocity so the tension balance is
performed in an inertial reference frame as illustrated in Fig. 3. The pitch
angle orientation of the towed system under steady state condition is
$\psi_{1}$. The sensor-array is acted upon by the following forces: a vertical
force $m_{2}g$ due to gravity, a horizontal force $F_{d,2}$ due to drag, a
buoyant force $F_{buoy,2}$, and a tension $T_{2}$ exerted by cable. The
tension force on sensor-array is at an angle of $\psi_{2}$ away from the
vertical axis. The force balance equation is

$\begin{split}m_{2}g=&T_{2}\cos\psi_{2}-F_{buoy,2},\\\
F_{d,2}=&T_{2}\sin\psi_{2}.\end{split}$ (3)

Figure 3: Lumped mass model for TCSAS in steady state condition.

The following forces act on the cable: a vertical force of $m_{1}g$ due to
gravity, a horizontal force $F_{d,1}$ due to drag, a buoyant force
$F_{buoy,1}$, a tension force $T_{1}$ exerted by ship, and a tension force
$-T_{2}$ exerted by sensor-array. The tension force on the cable is at an
angle of $\psi_{1}$ away from the vertical axis. The force balance equation
for the rope is

$\begin{split}m_{1}g+T_{2}cos\psi_{2}=&T_{1}\cos\psi_{1}-F_{buoy,1}\\\
F_{d,1}+T_{2}\sin\psi_{2}=&T_{1}\sin\psi_{1}.\end{split}$ (4)

Using Eqn. (3) and (4), we get a steady state pitch angle orientation of
cable-array system _w.r.t._ to the z-axis, given as

$\tan\psi_{1}=\dfrac{F_{d,1}+F_{d,2}}{m_{1}g+m_{2}g+F_{buoy,1}+F_{buoy,2}}.$
(5)

The depth control of a conventional towed system can be achieved by adjusting
either the towing speed or the physical parameters of the towing cable or the
towed sensor array [23].

###  II-B Moment balance equation for a single rigid body

Figure 4: Free body diagram of a single body towed to ship.

Firstly, we derive the moment balance equation for a single rigid body while
considering the hydrodynamics associated with it in $xy$ inertial frame of
reference and then extend it for interconnected bodies. The illustration is
shown in Fig. 4, where a single rigid body AB is towed to the surface ship at
point $A$. The moment balance equation at point $A$ is given as

$\textstyle\textstyle\sum M_{e,A}=\dot{H}_{G}+\vec{r}_{AG}\times
m\vec{a}_{A},$ (6)

where $\textstyle\sum M_{e,A}$ is the moments generated due to the sum of
external forces, which in this case will be drag force.
$\vec{r}_{AG}=\frac{L}{2}\cos\theta\hat{i}+\frac{L}{2}\sin\theta\hat{j}$ and
$\vec{a}_{A}=\ddot{x}_{A}\hat{i}+\ddot{y}_{A}\hat{j}$. Substituting these in
Eqn. (6), the moment balance equation at point $A$ is

$\textstyle\sum
M_{d,A}=I\ddot{\theta}+\frac{ml}{2}[\ddot{y}_{A}\cos\theta-\ddot{x}_{A}\sin\theta],$
(7)

where $\textstyle\sum M_{d,A}$ is the sum of moment due to drag force acting
at point $A$. There are two components of drag force, _i.e._ pressure drag
($F_{D,n}$) and shear friction drag ($F_{D,t}$) acting on the body AB along
normal and tangential directions, respectively as shown in Fig. 4.

It is to be noted that the $\textstyle\sum{M_{d,A}}$ is the moments generated
due to the resultant of pressure drag force (normal drag) acting perpendicular
to the rigid body motion. The drag force due to shear friction (tangential
drag) won’t generate any moment as it acts along the tangential direction to
the motion of rigid body.

Now we proceed to find the components of Eqn. (7) using free body diagram Fig.
4. A fractional element $X$ of area $dA=Ddr$, with a velocity $\vec{V}_{X}$ at
an angle $\phi$ _w.r.t._ $x$-axis, is shown in Fig. 4. The drag force
$d\vec{F}_{d}$ (combined tangential and normal) acting on element $X$ is in
the opposite direction to velocity $\vec{V}_{X}$. The normal component of
$\vec{V}_{X}$ is $V_{X,n}=V_{X}sin(\phi-\theta)$. The moment generated at
point $A$, due to normal drag force $d\vec{F}_{d,n}$ acting on fractional
element $X$ is given as

$M_{d,A}=\vec{r}\times d\vec{F}_{d,n},$ (8)

where

$\begin{split}d\vec{F}_{d,n}=&-\dfrac{1}{2}\rho
C_{D,n}dA\vec{V}_{X,n}|\vec{V}_{X,n}|,\\\ =&-\dfrac{1}{2}\rho
C_{D,n}D\vec{V}_{X}\sin(\phi-\theta)|\vec{V}_{X}\sin(\phi-\theta)|dr.\end{split}$
(9)

In the above equation, $\vec{V}_{X}$ and $\phi$ are unknown and have to be
substituted with the known quantities. The velocity $\vec{V}_{X}$ is the
vector sum of its translational part $\vec{V}_{A}$ and the rotational part
$\vec{V}_{X/A}$. Thus,

$\vec{V}_{X}=\vec{V}_{A}+\vec{V}_{X/A},$ (10)

where,

$\vec{V}_{A}=\dot{x}_{A}\hat{i}+\dot{y}_{A}\hat{j},\
\vec{V}_{X/A}=\vec{\theta}\times\vec{r}_{XA},\
\vec{\dot{\theta}}=\dot{\theta}\hat{k}.$ (11)

On substituting, we get

$\begin{split}\vec{V}_{X}&=(\dot{x}_{A}\hat{i}+\dot{y}_{A}\hat{j})+\dot{\theta}\hat{k}\times(r\cos\theta\hat{i}+r\sin\theta\hat{j}),\\\
&=(\dot{x}_{A}-\dot{\theta}r\sin\theta)\hat{i}+(\dot{y}_{A}+\dot{\theta}r\cos\theta)\hat{j},\end{split}$
(12) $\
|\vec{V}_{X}|=\sqrt{\dot{x}_{A}^{2}+\dot{y}_{A}^{2}+\dot{\theta}^{2}r^{2}+2\dot{\theta}r(\dot{x}_{A}\sin\theta-\dot{y}_{A}\cos\theta)},$
(13)
$\phi=\tan^{-1}(\dfrac{\dot{y}_{A}+\dot{\theta}r\cos\theta}{\dot{x}_{A}-\dot{\theta}r\sin\theta}).$
(14)

Substituting Eqn. (12), (13) and (14) into Eqn. (9) gives the normal drag
force acting on fractional element $X$, given as

$\begin{split}dF_{d,n}=&-\frac{1}{2}\rho
C_{D,n}D[(\dot{x}_{A}-\dot{\theta}r\sin\theta)\hat{i}+(\dot{y}_{A}+\dot{\theta}r\cos\theta)\hat{j}]\\\
&[\dot{x}_{A}\sin\theta-\dot{y}_{A}\cos\theta-\dot{\theta}r]^{2}(|\vec{V}_{X}(r)|)^{-1}dr.\end{split}$
(15)

Similarly, the tangential drag force acting on fractional element $X$ is given
as

$\begin{split}dF_{d,t}=&-\frac{1}{2}\rho C_{D,t}\pi
D[(\dot{x}_{A}-\dot{\theta}r\sin\theta)\hat{i}+(\dot{y}_{A}+\dot{\theta}r\cos\theta)\hat{j}]\\\
&[\dot{x}_{A}\cos\theta-\dot{y}_{A}\sin\theta]^{2}(|\vec{V}_{X}(r)|)^{-1}dr.\end{split}$
(16)

The moment generated at point $A$ due to normal drag force acting over the
entire segment AB is given as

$M_{d,A}=\int_{0}^{L}\vec{r}\times d\vec{F}_{d,n}.$ (17)

Substituting $d\vec{F}_{d,n}$ from Eqn (15), we get

$\begin{split}M_{d,A}=&-\frac{1}{2}\rho
C_{D,n}D\int_{0}^{L}(r\cos\theta\hat{i}+r\sin\theta\hat{j})\times[(\dot{x}_{A}-\\\
&\dot{\theta}r\sin\theta)\hat{i}+(\dot{y}_{A}+\dot{\theta}r\cos\theta)\hat{j}][\dot{x}_{A}\sin\theta-\dot{y}_{A}\cos\theta\\\
&-\dot{\theta}r]^{2}(|\vec{V}_{X}(r)|)^{-1}dr\\\ =&-\frac{1}{2}\rho
C_{D,n}D\int_{0}^{L}r(\dot{y}_{A}\cos\theta-\dot{x}_{A}\sin\theta-\dot{\theta}r)^{3}\\\
&(|\vec{V}_{X}(r)|)^{-1}dr.\end{split}$ (18)

The moment balance equation at point $A$ using Eqns. (7) and (18) is

$\begin{split}&\int_{0}^{L}M_{d,A}dr=I\ddot{\theta}_{1}+\frac{ml}{2}(\ddot{y}_{A}\cos\theta-\ddot{x}_{A}\sin\theta),\\\
&-\frac{1}{2}\rho
C_{D,n}D\int_{0}^{L}r(\dot{y}_{A}\cos\theta-\dot{x}_{A}\sin\theta-\dot{\theta}r)^{3}(|\vec{V}_{X}(r)|)^{-1}dr\\\
&=I\ddot{\theta}_{1}+\frac{ml}{2}(\ddot{y}_{A}\cos\theta-\ddot{x}_{A}\sin\theta).\end{split}$
(19)

The yaw orientation $\theta$ of body AB is obtained by solving the $2^{nd}$
order ODE of $\ddot{\theta}$ from Eqn. (LABEL:thetadd). The kinematics
equations given below use the yaw orientation to obtain the position, velocity
and acceleration at the end of the segment AB _i.e._ at point B.

$\begin{split}x_{B}=&x_{A}+l\cos\theta_{1},\ y_{B}=y_{A}+l\sin\theta_{1},\\\
\dot{x}_{B}=&\dot{x}_{A}-l\dot{\theta}_{1}\sin\theta,\
\dot{y}_{B}=\dot{y}_{A}+l\dot{\theta}_{1}\cos\theta,\\\
\ddot{x}_{B}=&\ddot{x}_{A}-l\ddot{\theta_{1}}\sin\theta_{1}-l\dot{\theta_{1}}^{2}\cos\theta_{1},\\\
\ddot{y}_{B}=&\ddot{y}_{A}+l\ddot{\theta_{1}}\cos\theta_{1}-l\dot{\theta_{1}}^{2}\sin\theta_{1}.\\\
\end{split}$ (20)

Similarly, the moment balance equation can be derived for each interconnected
rigid segment, yielding a second-order ODE for yaw orientation at each node.

###  II-C Modelling of cable-array system

Figure 5: Free body diagram of a towed cable sensor-array system.

The cable is modelled by discretizing the rigid cable in two segments, and the
array is hinged to the end of cable $2^{nd}$ segment as shown in free body
diagram Fig. 5.
The moment balance equation at node $A$ for cable-segment 1 at time instant
$k$ is

$\textstyle\sum
M_{e,A}^{k}=I_{1}\ddot{\theta}_{1}^{k}+\frac{m_{1}l_{1}}{2}(\ddot{y}_{A}^{k}\cos\theta_{1}^{k}-\ddot{x}_{A}^{k}\sin\theta_{1}^{k}),$
(21)

where $\theta_{1}^{k}$ is the orientation of the cable-segment 1 at $k^{th}$
instant, $(\ddot{x}_{A}^{k},\ddot{y}_{A}^{k})$ is acceleration along $(x,y)$
at node A which is obtained from ship input, $\textstyle\sum M_{e,A}^{k}$ is
the sum of moment due to the drag force acting on segment AB and the moment
due to the reaction force from the segment CD acting on segment AB at $k^{th}$
instant, given as

$\begin{split}\textstyle\sum M_{d,A}^{k}-\textstyle\sum
M_{rtn,A}^{k}=&I_{1}\ddot{\theta}_{1}^{k}+\frac{m_{1}l_{1}}{2}(\ddot{y}_{A}^{k}\cos\theta_{1}^{k}-\ddot{x}_{A}^{k}\sin\theta_{1}^{k}),\end{split}$
(22)

where the moment generated due to reaction force at point $A$ is

$\textstyle\sum
M_{rtn,A}^{k}=R_{x,1}^{k}(l_{1}\sin{\theta}_{1}^{k})-R_{y,1}^{k}(l_{1}\cos{\theta}_{1}^{k}).$
(23)

$(R_{x,1}^{k},R_{y,1}^{k})$ is the reaction force acting at node $B$ of cable-
segment 1 which is generated due to cable-segment 2 and sensor-array along
$(x,y)$ direction, respectively, expressed as

$\begin{split}R_{x,1}^{k}=&-(m_{2}\ddot{x}_{G2}^{k}-D_{n,x,2}^{k}-D_{t,x,2}^{k}+F_{x,3}^{k}),\\\
R_{y,1}^{k}=&-(m_{2}\ddot{y}_{G2}^{k}-D_{n,y,2}^{k}-D_{t,y,2}^{k}+F_{y,3}^{k}),\end{split}$
(24)

where $(m_{2}\ddot{x}_{G2}^{k},m_{2}\ddot{y}_{G2}^{k})$ are force components
of cable-segment 2 along $(x,y)$ at $k^{th}$ instant,
$D_{n,2}^{k}=(D_{n,x,2}^{k},D_{n,y,2}^{k})$ and
$D_{t,2}^{k}=(D_{t,x,2}^{k},D_{t,y,2}^{k})$ are normal and tangential drag
force along $(x,y)$, respectively acting on cable-segment 2 at $k^{th}$
instant, given as

$\begin{split}D_{n,x,2}^{k}=&-\frac{1}{2}\rho
C_{D,n}D\int_{0}^{L}(\dot{x}_{C}^{k}-\dot{\theta}_{2}^{k}r\sin\theta_{2}^{k})(\dot{x}_{C}^{k}\sin\theta_{2}^{k}-\\\
&\dot{y}_{C}^{k}\cos\theta_{2}^{k}-\dot{\theta}_{2}^{k}r)^{2}(|\vec{V}_{2}^{k}(r)|)^{-1}dr,\\\
D_{n,y,2}^{k}=&-\frac{1}{2}\rho
C_{D,n}D\int_{0}^{L}(\dot{y}_{C}^{k}+\dot{\theta}_{2}^{k}r\cos\theta_{2}^{k})(\dot{x}_{C}^{k}\sin\theta_{2}^{k}-\\\
&\dot{y}_{C}^{k}\cos\theta_{2}^{k}-\dot{\theta}_{2}^{k}r)^{2}(|\vec{V}_{2}^{k}(r)|)^{-1}dr,\\\
D_{t,x,2}^{k}=&-\frac{1}{2}\rho C_{D,t}\pi
D\int_{0}^{L}(\dot{x}_{C}^{k}-\dot{\theta}_{2}^{k}r\sin\theta_{2}^{k})(\dot{x}_{C}^{k}\cos\theta_{2}^{k}-\\\
&\dot{y}_{C}^{k}\sin\theta_{2}^{k})^{2}(|\vec{V}_{2}^{k}(r)|)^{-1}dr,\\\
D_{t,y,2}^{k}=&-\frac{1}{2}\rho C_{D,t}\pi
D\int_{0}^{L}(\dot{y}_{C}^{k}+\dot{\theta}_{2}^{k}r\cos\theta_{2}^{k})(\dot{x}_{C}^{k}\cos\theta_{2}^{k}-\\\
&\dot{y}_{C}^{k}\sin\theta_{2}^{k})^{2}(|\vec{V}_{2}^{k}(r)|)^{-1}dr,\end{split}$
(25)

where
$(|\vec{V}_{2}(r)|)^{-1}=[\dot{x}_{C}^{2}+\dot{y}_{C}^{2}+\dot{\theta_{2}}^{2}r^{2}+2\dot{\theta_{2}}r(\dot{x}_{C}\sin\theta_{2}-\dot{y}_{C}\cos\theta_{2})]^{-\frac{1}{2}}$,
such that $\vec{V}_{2}$ is the velocity of cable-segment 2. The above drag
force components are obtained using Eqns. (15) and (16) and the integrations
are performed using the numerical integration method. The final expression for
the moment balance equation at point A is

$\begin{split}&\textstyle\sum
M_{d,A}^{k}-(m_{2}\ddot{x}_{G2}^{k}-D_{n,x,2}^{k}-D_{t,x,2}^{k}+F_{x,3}^{k})l_{1}\sin\theta_{1}^{k}-\\\
&(m_{2}\ddot{y}_{G2}^{k}-D_{n,y,2}^{k}-D_{t,y,2}^{k}+F_{x,3}^{k})l_{1}\cos\theta_{1}^{k}=I_{1}\ddot{\theta}_{1}^{k}+\frac{m_{1}l_{1}}{2}\\\
&(\ddot{y}_{A}^{k}\cos\theta_{1}^{k}-\ddot{x}_{A}^{k}\sin\theta_{1}^{k}).\end{split}$
(26)

$\textstyle\sum
M_{d,A}^{k}=f(\theta_{1}^{k},\dot{\theta}_{1}^{k},\theta_{2}^{k},\dot{x}_{A}^{k},\dot{y}_{A}^{k},C_{D,n},D,l_{1},m_{1})$
is obtained using Eqn. (18). The second-order ODE of $\ddot{\theta}_{1,k}$
from Eqn. (26) is solved using the numerical method to obtain the yaw
orientation of cable-segment 1 at instant $k$.

The moment balance equation at node C for the cable-segment 2 is

$\begin{split}\textstyle\sum M_{d,C}^{k}-\textstyle\sum
M_{rtn,C}^{k}=&I_{2}\ddot{\theta}_{2}^{k}+\frac{m_{2}l_{2}}{2}(\ddot{y}_{C}^{k}\cos\theta_{2}^{k}-\ddot{x}_{C}^{k}\\\
&\sin\theta_{2}^{k}),\end{split}$ (27)

where $\textstyle\sum
M_{d,C}^{k}=f(\theta_{2}^{k},\dot{\theta}_{2}^{k},\theta_{3}^{k},\dot{x}_{C}^{k},\dot{y}_{C}^{k},C_{D,n},D,l_{2},m_{2})$
is the moment generated at node $C$ due to drag force acting on segment CD
which is obtained using Eqn. (18). $\textstyle\sum M_{rtn,C}^{k}$ is the
moment generated at node $C$ due to reaction force from sensor-array EF, given
as

$\begin{split}\textstyle\sum
M_{rtn,C}^{k}=&(m_{2}\ddot{x}_{G3}^{k}-D_{n,x,3}^{k}-D_{t,x,3}^{k})l_{2}\sin\theta_{2}^{k}-(m_{2}\ddot{y}_{G3}^{k}\\\
&-D_{n,y,3}^{k}-D_{t,y,3}^{k})l_{2}\cos\theta_{1}^{k},\end{split}$ (28)

where $D_{n,3}^{k}=(D_{n,x,3}^{k},D_{n,y,3}^{k})$ and
$D_{t,3}^{k}=(D_{t,x,3}^{k},D_{t,y,3}^{k})$ are the $(x,y)$ components of
normal and tangential drag force, respectively, acting on the sensor-array EF.
They are the function of
$(D_{n,3}^{k},D_{t,3}^{k})=f(\theta_{3}^{k},\dot{\theta}_{3}^{k},\dot{x}_{E}^{k},\dot{y}_{E}^{k},C_{D,n,array},C_{D,t,array},D_{2},l_{3},m_{3})$
and can be obtained using Eqn. (25). The second-order ODE of
$\ddot{\theta}_{2,k}$ given in Eqn. (27) is solved to obtain the yaw
orientation of cable-segment 2 at any instant $k$.

The sensor-array end is free, so there is no reaction force at the end. The
moment balance equation at node E of sensor-array is given as

$\textstyle\sum
M_{d,E}^{k}=\dfrac{m_{3}l_{3}^{2}}{3}\ddot{\theta}_{3}^{k}+\frac{m_{3}l_{3}}{2}(\ddot{y}_{E}^{k}\cos\theta_{3}^{k}-\ddot{x}_{E}^{k}\sin\theta_{3}^{k}),$
(29)

where $m_{3}$ and $l_{3}$ are the mass and length of the sensor-array.
$\textstyle\sum
M_{d,E}^{k}=f(\theta_{3}^{k},\dot{\theta}_{3}^{k},\dot{x}_{E}^{k},\dot{y}_{E}^{k},C_{D,n,array},D_{3},l_{3},m_{3})$
is obtained using equation (15). The orientation of sensor-array
$\theta_{3}^{k}$ is obtained by solving the second-order ODE of
$\ddot{\theta}_{3}^{k}$ given in Eqn. (29).

The yaw orientation of the discretized cable segment and sensor-array _i.e._
$\theta_{1},\theta_{2},\theta_{3}$ are obtained by solving the $2^{nd}$ order
ODEs from Eqns. (26), (27), and (29), respectively, which are then utilized to
obtain the position, velocity and acceleration of each node using kinematic
relations from Eqn. (20). The boundary condition at the cable’s top end _i.e._
position, velocity and acceleration of node A at $k^{th}$ time instant, are
the same as those of the ship, which are known functions of time.

##  III Simulation Results

###  III-A Scenario

TABLE I: Physical properties of towing cable sensor-array system [21]. Parameters |  Towing cable |  Sensor-array
---|---|---
Length $(m)$ |  723 |  273.9
Diameter $(m)$ |  0.041 |  0.079
Mass/length $(Kg/m)$ |  2.33 |  5.07
$C_{D,n}$ |  2 |  1.8
$C_{D,t}$ |  0.015 |  0.009

The surface vessel is moving with constant velocity motion with a cable
sensor-array system connected to it through a spherical joint. The dynamics of
the towed cable sensor-array system is simulated for a scenario involving the
engagement of a target and an own-ship, which is popularly used in many
existing TMA problems [1],[26],[8]. The initial coordinate of the vessel is at
origin $(0,0,0)km$ moving with a constant velocity of $5\ knots$ for a period
of $30\ min$. The vessel starts at a course of $140^{o}$ with respect to
$y-axis$ and continues moving at that path for a period of first $12\ min$.
Then, the vessel performs a manoeuvre from $13^{th}$ to $17^{th}$ time steps
with a constant turn rate of $30^{o}\ per\ min$ and ends up at $20^{o}$ course
with respect to y-axis. The vessel continues to move at a course of $20^{o}$
with respect to y-axis from $18^{th}$ to $30^{th}$ minute. The course followed
by the target and ownship is shown in Fig. 6. The physical properties of the
towing cable-array system are given in Table I, which is taken from Ref. [21].
The density of fluid is taken to be $1000\ kg/m^{3}$.

Figure 6: Dynamics of towed cable- sensor array system. Figure 7: Enlarged
view showing the dynamics of towed cable-sensor array system during a
manoeuvre of the ship.

###  III-B Results and Discussion

The depth attained by towed cable sensor-array system is determined using Eqn.
(5). The steady-state orientation _w.r.t._ sea-surface _i.e._ $xy$ plane is
$5.64^{o}$, which is relatively small, such that the TCSAS can be considered
to be floating on the sea surface. The simulation is performed using MATLAB
R2018a software. The second order ODEs given by Eqns. (26), (27), and (29) are
solved simultaneously using the Runge-Kutta $4^{th}$ order method [27] to
obtain the orientation _i.e._ $\theta_{1}$, $\theta_{2}$, and $\theta_{3}$.
Subsequently, these orientations are inputted into the kinematic Eqn. (20) to
obtain the position, velocity and acceleration of the TCSAS at each instant
$k$. The solution for the second order ODE requires the initial condition. The
initial yaw for the two cable segments and one sensor-array are taken to be
same as that ship initial yaw orientation, and the initial yaw rate is taken
to be zero, such that the TCSAS is aligned with the surface ship initially.
The definite integral involved in the calculations of moment due to forces,
reaction force, and drag force from Eqns. (18), (24), and (25), respectively
are solved using the numerical integration method called Gauss-Legendre five
point rule [27]. For example, the integration of moment due to drag force is
approximated using the points $\zeta_{i}$ and their corresponding weights
$w_{i}$, given as
$\int_{0}^{L}M_{d,X}dr=\textstyle\sum_{i=1}^{n}w_{i}M_{d,X}(\zeta_{i})$. The
dynamics of 2 cable-segment and sensor-array for the simulated ship’s
trajectory is shown in Fig. 6. The red, black and cyan color plots show the
path traced by cable segments and sensor-array, respectively at each time
instant. Initially, it is evident that the path of the towed body aligns with
the ship’s trajectory during straight-line motion. However, upon initiation of
the ship’s manoeuvre, the towed body deviates from its original path.
Following the completion of the manoeuvre at the $17^{th}$ time step, the
towed body gradually stabilizes due to its inertia and reaches a steady-state
position after some delay. The enhanced view of the dynamics of the towed
array during the manoeuvring of ship is shown in Fig. 7, which shows the path
traced by the centre of gravity (CG) of the cable segment and sensor array.
The location of CG of sensor array is utilized to perform the TMA efficiently,
especially during the manoeuvring of the own-ship.

##  IV Conclusions

This paper develops a dynamic model of a towed cable sensor-array system using
the lumped mass approach. The developed model when solved at each node,
provides the location and orientation of the towed array sensor _w.r.t._ own-
ship motion. The coordinates of CG of the sensor-array are to be fed to state
estimation model to perform an accurate TMA of an underwater target,
especially during manoeuvres. As a future work, it is intended to develop a
generalized dynamics model which can consider the towed system as a
combination of any user defined number of rigid bodies. The position of the
towed array obtained out of the developed model shall be fed to a state
estimation algorithm, which will recommend using optimal control or potential
field algorithms to steer a tracking vessel into a dynamically determined
preferred tracking position. With the use of heuristics on ship movement and
detection parameters, a means to safely track very quiet submerged targets
could be derived.

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