formulate a geometric interpretation of our double bracket from a quiver so
that a special choice of constants $A_{\bullet}$ and $A_{\circ}$ realizes this
Goldman bracket. First we define $\hat{\pi}$ to be the set of conjugacy
classes in $\pi$ (or free homotopy classes of loops). Goldman observes that
the bracket above induces a Lie bracket on the vector space spanned by
$\hat{\pi}$. Formally, we simply remove the “$\mathrm{tr}$” in the formula
above. So for $\alpha,\beta\in\hat{\pi}$, we have

$[\alpha,\beta]=\sum_{p\in\alpha\cap\beta}\varepsilon_{p}(\alpha,\beta)\,\alpha_{p}\beta_{p}$

With this setup, we have the following

######  Theorem 10.

[Gol86] $\mathbb{Z}\hat{\pi}$ is a Lie algebra with the bracket shown above,
and the map $\mathrm{tr}\colon\mathbb{Z}\hat{\pi}\to
C^{\infty}(\mathrm{Rep}_{n}(\pi))$ given by $\alpha\mapsto\mathrm{tr}(\alpha)$
is a Lie algebra homomorphism.

In fact, our proof of Theorem 8 is essentially the same as Goldman’s original
proof ([Gol86], Theorem 5.3).

In the literature, fat graphs are also commonly called “ribbon graphs”. From a
fat graph $\Gamma$, one can construct an oriented surface with boundary,
$S_{\Gamma}$, by replacing the edges with rectangular strips (ribbons) and the
vertices with discs, where the ribbons are glued to the discs according to the
cyclic ordering prescribed by the fat graph structure. We call this surface
the _ribbon surface_ associated to the fat graph. It is clear that the
original graph $\Gamma$ is a deformation retract of $S_{\Gamma}$. We say that
$\Gamma$ is a _spine_ of the surface $S=S_{\Gamma}$.

Given a quiver $Q$ (oriented fat graph) with underlying unoriented graph
$\Gamma$, we consider, as before, the subspace
$\mathscr{L}\subset\mathcal{F}_{Q}$ of loops. Then in a natural way we
identify the cyclic space $\mathscr{L}^{\natural}$ with $\mathbb{Q}\hat{\pi}$,
the space generated by free homotopy classes of loops on $S_{\Gamma}$. In the
commutative case, we considered the case $A_{\bullet}=A_{\circ}=-\frac{1}{2}$,
which gives $\varepsilon_{p}(f,g)=1$ when $f$ and $g$ touch and
$\varepsilon_{p}(f,g)=0$ when $f$ and $g$ cross. We will again be primarily
concerned with this specific choice of coefficients in the non-commutative
case. This is opposite from what $\varepsilon_{p}(\alpha,\beta)$ means for
Goldman’s bracket, however. This is because if paths $f$ and $g$ touch in the
quiver, then on $S_{\Gamma}$, they are homologous to paths which do not touch
at all, and so their Goldman bracket should be zero. The way around this is to
consider the “dual” surface.

######  Definition 16.

We define the _dual surface_ to $S_{\Gamma}$, which we denote
$\widetilde{S}_{\Gamma}$, to be glued out of ribbons like $S_{\Gamma}$, except
that whenever an edge joins vertices of different colors, the corresponding
ribbon is given a half-twist.

######  Theorem 11.

For the choice of coefficients $A_{\bullet}=A_{\circ}=-\frac{1}{2}$, the
induced bracket $\left<-,-\right>$ on $\mathscr{L}^{\natural}$ coincides with
Goldman’s bracket under the identification of $\mathscr{L}^{\natural}$ with
$\mathbb{Q}\hat{\pi}$, where $\pi=\pi_{1}(\widetilde{S}_{\Gamma})$.

###### Proof.

We think of the quiver as a planar projection picture of the ribbon surface
$\widetilde{S}_{\Gamma}$, where one of the colors is the “top” of the ribbon,
and the other color is the “bottom”. If we were to “untwist” the surface
$\widetilde{S}_{\Gamma}$, and try to view it without twists, we would see that
paths which touch in the quiver end up crossing in $\widetilde{S}_{\Gamma}$,
and paths which cross in the quiver end up not touching in
$\widetilde{S}_{\Gamma}$. Now that touching and crossing have been
interchanged, we see that the Goldman bracket on $\widetilde{S}_{\Gamma}$
coincides with the induced bracket $\left<-,-\right>$ on
$\mathscr{L}^{\natural}$ when $A_{\bullet}=A_{\circ}=-\frac{1}{2}$. ∎

We will sometimes want to choose a vertex in the quiver $Q$, and only consider
loops in the fundamental group of $S_{\Gamma}$ based at that point. After
choosing a vertex, we let $\mathscr{L}_{\bullet}$ denote the subspace of loops
which start and end at this point. Then $\mathscr{L}_{\bullet}$ is naturally
identified with the group algebra of $\pi=\pi_{1}(S_{\Gamma},\bullet)$. The
induced bracket is independent of this choice of basepoint, since conjugate
elements are equivalent in $\mathscr{L}_{\bullet}^{\natural}$. Choosing a
basepoint gives some nice properties. For example, $\mathscr{L}_{\bullet}$ is
a subalgebra, unlike $\mathscr{L}$, since the common basepoint allows us to
compose paths. Also, we have the following property for the double bracket.

######  Proposition 12.

The subalgebra $\mathscr{L}_{\bullet}$ is closed under the double bracket.
That is,
$\left\\{\\!\\!\left\\{\mathscr{L}_{\bullet},\mathscr{L}_{\bullet}\right\\}\\!\\!\right\\}\subseteq\mathscr{L}_{\bullet}\otimes\mathscr{L}_{\bullet}$.

###### Proof.

Let $f,g\in\mathscr{L}_{\bullet}$. Write $f$ and $g$ as monomials,
$f=f_{1}\cdots f_{k}$ and $g=g_{1}\cdots g_{\ell}$, where each $f_{i}$ and
$g_{j}$ are arrows in the quiver. Note that since they are both based at the
point $\bullet$, we have $s(f_{1})=s(g_{1})=t(f_{k})=t(g_{\ell})=\bullet$.
Now, using the Leibniz rule:

$\left\\{\\!\\!\left\\{f,g\right\\}\\!\\!\right\\}=\sum_{i,j}(g_{1}\cdots
g_{j-1}\otimes f_{1}\cdots
f_{i-1})\left\\{\\!\\!\left\\{f_{i},g_{j}\right\\}\\!\\!\right\\}(f_{i+1}\cdots
f_{k}\otimes g_{j+1}\cdots g_{\ell})$

Using the formulas from Lemma 2, and examining the three cases, we see that
each term in the sum above is of the form $\alpha\otimes\beta$, where $\alpha$
and $\beta$ are both in $\mathscr{L}_{\bullet}$. ∎

We now point out some similar and related work which was brought to the
author’s attention while working on the paper.

######  Remark.

In [MT12], Turaev and Massuyeau construct a _quasi-Poisson bracket_ on the
character variety $\mathrm{Rep}_{n}(\pi)$, which is induced by a _double
quasi-Poisson bracket_ on the group algebra of $\pi$. As was mentioned above,
we may identify the group algebra of $\pi$ with $\mathscr{L}_{\bullet}$, and
so they are both double brackets on $\mathscr{L}_{\bullet}$. However, the
double bracket considered in this paper is _not_ a double quasi-Poisson
bracket, and so the two constructions are not exactly the same. However, the
bracket of Turaev and Massuyeau also projects to the Goldman bracket in
$\mathscr{L}_{\bullet}^{\natural}$. Also, Semeon Artamonov, in his recent
thesis [Art18], constructed a much more abstract and categorical version of
the quasi-Poisson structure of Turaev and Massuyeau.

###  6.7 The $X,Y$ Variables

Just as in the commutative case, we will use gauge transformations to obtain
new weights on the graph. Start with the quiver $Q_{n}$ just as in Figure 6.
Give the edge weights the same names as in the commutative case, but now they
are formal noncommutative variables in $\mathcal{F}_{Q}$. Perform guage
transformations to obtain variables $a_{i},b_{i},c_{i},d_{i}$ just as before.
Note that the pictures in Section 3.2 are actually noncommutative gauge
transformations, as they take into account whether multiplication happens on
the left or right. In this case, the double bracket induced on the $a,b,c,d$
variables is given by:

$\displaystyle\left\\{\\!\\!\left\\{b_{i},a_{i}\right\\}\\!\\!\right\\}$
$\displaystyle=\frac{1}{2}b_{i}\otimes a_{i}$ $\displaystyle\hskip
28.45274pt\left\\{\\!\\!\left\\{b_{i},c_{i}\right\\}\\!\\!\right\\}$
$\displaystyle=\frac{1}{2}c_{i}\otimes b_{i}$
$\displaystyle\left\\{\\!\\!\left\\{a_{i},d_{i}\right\\}\\!\\!\right\\}$
$\displaystyle=\frac{1}{2}1\otimes a_{i}d_{i}$ $\displaystyle\hskip
28.45274pt\left\\{\\!\\!\left\\{c_{i},d_{i}\right\\}\\!\\!\right\\}$
$\displaystyle=\frac{1}{2}d_{i}c_{i}\otimes 1$

Define the following monomials in the $a,b,c,d$ variables:

$\displaystyle x_{i}$
$\displaystyle=a_{i}c_{i-1}^{-1}d_{i-1}^{-1}c_{i-2}^{-1}$ $\displaystyle
y_{i}$
$\displaystyle=b_{i}c_{i}^{-1}d_{i}^{-1}c_{i-1}^{-1}d_{i-1}^{-1}c_{i-2}^{-1}$
$\displaystyle z_{k}$ $\displaystyle=d_{1}c_{1}\cdots d_{k-1}c_{k-1}d_{k}$

These $x_{i}$ and $y_{i}$ are noncommutative versions of the same monomials
given in the commutative case, and $z_{k}$ are paths connecting the upper-left
corner of the first square face to the upper-right corner of the
$k^{\mathrm{th}}$ square face. We may again perform the exact same sequence of
gauge transformations as in the commutative case to arrive at the weights
depicted in Figure 7. However, these weights will now be noncommutative
versions of the same monomials. In terms of the monomials defined above, the
noncommutative weights we obtain in Figure 7 are given by

$\displaystyle X_{i}$ $\displaystyle=z_{i-2}\,x_{i}\,z_{i-2}^{-1}$
$\displaystyle Y_{i}$ $\displaystyle=z_{i-2}\,y_{i}\,z_{i-2}^{-1}$
$\displaystyle Z$ $\displaystyle=z_{n}c_{n}$

Just as in the commutative case, these can be interpreted as closed loops in
the quiver. The difference now in the non-commutative case is that all these
loops share a common basepoint. This common basepoint is the upper-left corner
of the first square face. As in the previous section, let
$\mathscr{L}_{\bullet}\subset\mathscr{L}$ denote the subspace of loops based
at this point. In fact, it is the group algebra of the fundamental group of
the ribbon surface of the graph, and it is generated by the $X_{i}$, $Y_{i}$,
and $Z$.

A simple calculation shows that $Z$ is a Casimir of the double bracket. The
bracket is then well-defined on the quotient
$\mathscr{L}_{\bullet}^{(1)}:=\mathscr{L}_{\bullet}/\left<Z-1\right>$, where
$Z$ is set to $1$. The induced brackets in $\mathscr{L}_{\bullet}^{(1)}$ are
given by:

$\displaystyle\left<X_{i+1},X_{i}\right>$ $\displaystyle=X_{i+1}X_{i}$
$\displaystyle\hskip 28.45274pt\left<Y_{i+1},Y_{i}\right>$
$\displaystyle=Y_{i+1}Y_{i}$ $\displaystyle\left<Y_{i+2},Y_{i}\right>$
$\displaystyle=Y_{i+2}Y_{i}$ $\displaystyle\hskip
28.45274pt\left<Y_{i},X_{i}\right>$ $\displaystyle=Y_{i}X_{i}$
$\displaystyle\left<Y_{i+1},X_{i}\right>$ $\displaystyle=Y_{i+1}X_{i}$
$\displaystyle\hskip 28.45274pt\left<X_{i+1},Y_{i}\right>$
$\displaystyle=X_{i+1}Y_{i}$ $\displaystyle\left<X_{i+2},Y_{i}\right>$
$\displaystyle=X_{i+2}Y_{i}$

If we don’t set $Z=1$, then the bracket relations in $\mathscr{L}_{\bullet}$
are mostly the same, except some exceptions when $i=1$ or $i=2$:

$\displaystyle\left<X_{3},X_{2}\right>$ $\displaystyle=X_{3}Z^{-1}X_{2}Z$
$\displaystyle\hskip 28.45274pt\left<Y_{3},Y_{2}\right>$
$\displaystyle=Y_{3}Z^{-1}Y_{2}Z$ $\displaystyle\left<Y_{3},X_{2}\right>$
$\displaystyle=Y_{3}Z^{-1}X_{2}Z$ $\displaystyle\hskip
28.45274pt\left<X_{3},Y_{2}\right>$ $\displaystyle=X_{3}Z^{-1}Y_{2}Z$
$\displaystyle\left<Y_{3},Y_{1}\right>$ $\displaystyle=Y_{3}Z^{-1}Y_{1}Z$
$\displaystyle\hskip 28.45274pt\left<Y_{4},Y_{2}\right>$
$\displaystyle=Y_{4}Z^{-1}Y_{2}Z$ $\displaystyle\left<X_{3},Y_{1}\right>$
$\displaystyle=X_{3}Z^{-1}Y_{1}Z$ $\displaystyle\hskip
28.45274pt\left<X_{4},Y_{2}\right>$ $\displaystyle=X_{4}Z^{-1}Y_{2}Z$

If we perform the same sequence of Postnikov moves as in the commutative case,
followed by gauge transformations, we get a graph isomorphism, with the same
edges having weight 1 as before. This gives the transformation

$\displaystyle X_{i}$
$\displaystyle\mapsto(X_{i}+Y_{i})^{-1}X_{i}(X_{i+2}+Y_{i+2})$ $\displaystyle
Y_{i}$ $\displaystyle\mapsto(X_{i+1}+Y_{i+1})^{-1}Y_{i+1}(X_{i+3}+Y_{i+3})$

Just as in the commutative case, this is almost the expression derived earlier
for the pentagram map. It differs only by a shift in the $Y$-indices.

######  Remark.

In the formula above, the indices are _not_ read cyclically. If one of the
indices is greater than $n$, we must conjugate by $Z$. For instance, if
$i=n-2$, and $i+3=n+1$, then by $X_{i+3}$ we really mean $ZX_{1}Z^{-1}$, and
similarly for $Y$.

###  6.8 The $P,Q$ Variables

We now define a non-commutative version of the $p,q$ variables from the
classical case. They are given by

$\displaystyle p_{i}$ $\displaystyle=b_{i}c_{i}^{-1}d_{i}^{-1}a_{i}^{-1}$
$\displaystyle\hskip 28.45274ptq_{i}$
$\displaystyle=c_{i-2}d_{i-1}a_{i+1}b_{i}^{-1}$

Conjugating by the same $z_{k}$ paths as for the $X,Y$ variables, we obtain
based versions:

$\displaystyle P_{i}$ $\displaystyle=z_{i-2}\,p_{i}\,z_{i-2}^{-1}$
$\displaystyle\hskip 28.45274ptQ_{i}$
$\displaystyle=z_{i-2}\,q_{i}\,z_{i-2}^{-1}$

Similar to the commutative case, we have the following relation with the $X,Y$
variables:

$\displaystyle P_{i}$ $\displaystyle=Y_{i}X_{i}^{-1}$ $\displaystyle\hskip
28.45274ptQ_{i}$ $\displaystyle=X_{i+1}Y_{i}^{-1}$

Their induced brackets are given by

$\displaystyle\left<Q_{i},P_{i}\right>$ $\displaystyle=Q_{i}P_{i}$
$\displaystyle\hskip 28.45274pt\left<P_{i},Q_{i-1}\right>$
$\displaystyle=P_{i}Q_{i-1}$ $\displaystyle\left<Q_{i+1},P_{i}\right>$
$\displaystyle=Q_{i+1}Y_{i}^{-1}P_{i}Y_{i}$ $\displaystyle\hskip
28.45274pt\left<P_{i},Q_{i+2}\right>$
$\displaystyle=Y_{i}^{-1}P_{i}Y_{i}Q_{i+2}$

From the formulas given in the previous section for the pentagram map in the
$X,Y$ variables, we get that the pentagram map transorms the $P,Q$ variables
by

$\displaystyle P_{i}$ $\displaystyle\mapsto
X_{i+1}^{-1}(1+P_{i+1}^{-1})^{-1}X_{i+1}\cdot(1+P_{i+3})Q_{i+2}(1+P_{i+2})\cdot
Y_{i}^{-1}(1+P_{i})Y_{i}$ $\displaystyle Q_{i}$ $\displaystyle\mapsto
X_{i+1}^{-1}P_{i+1}^{-1}X_{i+1}$

##  7 Invariance, Invariants, and Integrability

###  7.1 Invariance of the Induced Bracket

We will show in this section that the induced bracket $\left<-,-\right>$ on
$\mathscr{L}_{\bullet}^{\natural}$ is invariant under the pentagram map. To do
so, we will consider step-by-step how the weights and double bracket change
under the Postnikov moves.

Recall that, starting with the $X,Y$ variables, the sequence of Postnikov
moves which gives the pentagram map is as follows:

  1. 1.

Perform the square move at each square face

  2. 2.

Perform the white-swap move at each edge connecting two white vertices

  3. 3.

Perform the black-swap move at each edge connecting two black vertices

  4. 4.

Perform gauge transformations so all weights are 1 except the bottom and left
of each square face

We will actually start with the $a,b,c,d$ weights instead of the $X,Y$
weights. The result will be the same, as we will show below. Application of a
square move gives

$a$$b$$c$$d$$\widetilde{a}$$\widetilde{b}$$\widetilde{c}$$\widetilde{d}$

The new edge weights are given by

$\displaystyle\widetilde{b}$ $\displaystyle=b+adc$ $\displaystyle\hskip
28.45274pt\widetilde{a}$ $\displaystyle=dc\widetilde{b}^{-1}$
$\displaystyle\widetilde{c}$ $\displaystyle=\widetilde{b}^{-1}ad$
$\displaystyle\hskip 28.45274pt\widetilde{d}$
$\displaystyle=\widetilde{a}bc^{-1}$

The double brackets of these new weights are

$\begin{array}[]{ccc|ccc}\left\\{\\!\\!\left\\{\widetilde{b},\widetilde{a}\right\\}\\!\\!\right\\}&=&\frac{1}{2}\,\left(\widetilde{a}\widetilde{b}\otimes
1\right)&\left\\{\\!\\!\left\\{\widetilde{a},\widetilde{d}\right\\}\\!\\!\right\\}&=&\frac{1}{2}\,\left(\widetilde{a}\otimes\widetilde{d}\right)\\\\[5.16663pt]
\left\\{\\!\\!\left\\{\widetilde{b},\widetilde{c}\right\\}\\!\\!\right\\}&=&\frac{1}{2}\,\left(1\otimes\widetilde{b}\widetilde{c}\right)&\left\\{\\!\\!\left\\{\widetilde{c},\widetilde{d}\right\\}\\!\\!\right\\}&=&\frac{1}{2}\,\left(\widetilde{d}\otimes\widetilde{c}\right)\end{array}$

Next, applying the white-swap and black-swap moves interchanges the square and
octagonal faces. The resulting weights around the square faces is pictured in
Figure 12.

Figure 12: Edge weights after white-swap and black-swap
moves$\widetilde{c}_{i-2}$$\widetilde{d}_{i-1}$$\widetilde{a}_{i+1}$$\widetilde{b}_{i}$

Finally, we perform gauge transformations, as we did to get the original $X,Y$
weights, so that all weights become 1 except those on the bottom and left
edges of the square faces. We call these resulting weights $\widetilde{X}$ and
$\widetilde{Y}$. If we define the “staircase” monomials
$\xi_{k}=\widetilde{a}_{1}\widetilde{b}_{1}\cdots\widetilde{a}_{k}\widetilde{b}_{k}$,
then we can express the new weights as

$\displaystyle\widetilde{X}_{i}$
$\displaystyle=\xi_{i}\,\widetilde{c}_{i}\widetilde{a}_{i+2}^{-1}\widetilde{b}_{i+1}^{-1}\widetilde{a}_{i+1}^{-1}\,\xi_{i}^{-1}$
$\displaystyle\hskip 28.45274pt\widetilde{Y}_{i}$
$\displaystyle=\xi_{i}\,\widetilde{d}_{i+1}\widetilde{a}_{i+3}^{-1}\widetilde{b}_{i+2}^{-1}\widetilde{a}_{i+2}^{-1}\widetilde{b}_{i+1}^{-1}\widetilde{a}_{i+1}^{-1}\,\xi_{i}^{-1}$

These weights are the images of $X_{i}$ and $Y_{i}$ under the pentagram map.
That is, $\widetilde{X}_{i}=T(X_{i})$ and $\widetilde{Y}_{i}=T(Y_{i})$. As
noted in the previous sections, these are given in terms of the original $X,Y$
variables as $\widetilde{X}_{i}=\sigma_{i}^{-1}X_{i}\sigma_{i+2}$ and
$\widetilde{Y}_{i}=\sigma_{i+1}^{-1}Y_{i+1}\sigma_{i+3}$, where
$\sigma_{i}=X_{i}+Y_{i}$. We are now ready to prove the main theorem of this
section, but first we introduce some notation.

Define the map $S\colon\mathscr{L}_{\bullet}\to\mathscr{L}_{\bullet}$ which
shifts all the indices. That is, $S(X_{i})=X_{i-1}$ and $S(Y_{i})=Y_{i-1}$,
where the indices are read cyclically.

######  Theorem 12.

$(a)$ |  In $\mathscr{L}_{\bullet}^{(1)}$, the induced bracket $\left<-,-\right>$ is invariant under the pentagram map.
---|---
$(b)$ |  In $\mathscr{L}_{\bullet}$, the induced bracket is invariant under $S^{2}\circ T$.

###### Proof.

We want to show that the induced brackets of $\widetilde{X}_{i}$ and
$\widetilde{Y}_{i}$ have exactly the same form as the induced brackets of the
$X_{i}$ and $Y_{i}$. Note that for $a,b,c\in\mathscr{L}_{\bullet}$, we have
$\left<a,c\right>=\left<bab^{-1},c\right>$, since
$a^{\natural}=(bab^{-1})^{\natural}$. So we will instead compute the brackets
of the conjugate, but equivalent, elements:

$\displaystyle\widetilde{x}_{i}$
$\displaystyle=\xi_{i}^{-1}\widetilde{X}_{i}\xi_{i}=\widetilde{c}_{i}\widetilde{a}_{i+2}^{-1}\widetilde{b}_{i+1}^{-1}\widetilde{a}_{i+1}^{-1}$
$\displaystyle\hskip 28.45274pt\widetilde{y}_{i}$
$\displaystyle=\xi_{i}^{-1}\widetilde{Y}_{i}\xi_{i}=\widetilde{d}_{i+1}\widetilde{a}_{i+3}^{-1}\widetilde{b}_{i+2}^{-1}\widetilde{a}_{i+2}^{-1}\widetilde{b}_{i+1}^{-1}\widetilde{a}_{i+1}^{-1}$

We will compute the three possible combinations
$\left\\{\\!\\!\left\\{x_{i},x_{j}\right\\}\\!\\!\right\\}$,
$\left\\{\\!\\!\left\\{y_{i},y_{j}\right\\}\\!\\!\right\\}$, and
$\left\\{\\!\\!\left\\{x_{i},y_{j}\right\\}\\!\\!\right\\}$. For
$\left\\{\\!\\!\left\\{x_{i},x_{j}\right\\}\\!\\!\right\\}$, a simple
calculation using the Leibniz rules for double brackets gives:

$\displaystyle\left\\{\\!\\!\left\\{\widetilde{x}_{i},\widetilde{x}_{j}\right\\}\\!\\!\right\\}$
$\displaystyle=\frac{1}{2}\delta_{i,j+1}(\widetilde{c}_{j}\widetilde{a}_{j+2}^{-1}\widetilde{x}_{i}\otimes\widetilde{a}_{j+2}\widetilde{c}_{j}^{-1}\widetilde{x}_{j}+\widetilde{x}_{j}\widetilde{a}_{i}\widetilde{b}_{i}\otimes\widetilde{x}_{i}\widetilde{b}_{i}^{-1}\widetilde{a}_{i}^{-1})$
$\displaystyle\phantom{=}-\frac{1}{2}\delta_{i,j-1}(\widetilde{x}_{j}\widetilde{b}_{j}^{-1}\widetilde{a}_{j}^{-1}\otimes\widetilde{x}_{i}\widetilde{a}_{j}\widetilde{b}_{j}+\widetilde{b}_{j}^{-1}\widetilde{a}_{j}^{-1}\otimes\widetilde{x}_{i}\widetilde{a}_{j}\widetilde{b}_{j}\tilde{x}_{j})$

Therefore their induced brackets are

$\left<\widetilde{x}_{i},\widetilde{x}_{j}\right>=\delta_{i,j+1}(\widetilde{c}_{j}\widetilde{a}_{j+2}^{-1})\,\widetilde{x}_{i}\,(\widetilde{a}_{j+2}\widetilde{c}_{j}^{-1})\,\widetilde{x}_{j}-\delta_{i,j-1}(\widetilde{b}_{j}^{-1}\widetilde{a}_{j}^{-1})\,\widetilde{x}_{i}\,(\widetilde{a}_{j}\widetilde{b}_{j})\,\tilde{x}_{j}$

Conjugating the right hand side by $\xi_{j}$ gives the equivalent formula in
terms of the $\widetilde{X}_{i}$:

$\left<\widetilde{X}_{i},\widetilde{X}_{j}\right>=(\delta_{i,j+1}-\delta_{i,j-1})\widetilde{X}_{i}\widetilde{X}_{j}$

This is exactly the same as the bracket formula for the original $X_{i}$.
There are, however, some exceptions. When $i=n$, we instead have

$\left<\widetilde{X}_{1},\widetilde{X}_{n}\right>=\widetilde{X}_{1}Z^{-1}\widetilde{X}_{n}Z$

The calculations for $\left<\widetilde{Y}_{i},\widetilde{Y}_{j}\right>$ and
$\left<\widetilde{X}_{i},\widetilde{Y}_{j}\right>$ are similar, with
exceptions when $i=n$ and $i=n-1$. So the exact form of the bracket relations
are not invariant, since the exceptional cases (which are conjugated by $Z$)
do not occur at the same indices. But if we set $Z=1$, then the bracket is
invariant in $\mathscr{L}_{\bullet}^{(1)}$. Also, if we shift the indices by 2
and define $\widehat{X}_{i}:=S^{2}(\widetilde{X}_{i})=S^{2}\circ T(X_{i})$,
then the bracket is invariant under $X_{i}\mapsto\widehat{X}_{i}$. ∎

######  Remark.

Theorem 12 suggests that a more appropriate definition of the pentagram map
should include a “rotation”, which shifts the labels/indices of the vertices
by 2. This shift of indices is necessary to make the Poisson structure
invariant. From now on, we will use the notation
$\widehat{T}\colon\mathcal{GP}_{n,N}\to\mathcal{GP}_{n,N}$ for this modified
version of the map. It is worth noting that earlier definitions of the
pentagram map in [OST10] [Sch92] [GSTV16] do not include this shift of
indices, and this phenomenon only appears by considering this noncommutative
generalization.

###  7.2 The Invariants

Let $Q=Q_{n}$ be the quiver/network for the pentagram map, with the $X_{i}$
and $Y_{i}$ weights as before, and $\mathscr{B}=\mathscr{B}_{Q}(\lambda)$ its
boundary measurement matrix, whose entries are elements of the Laurent
polynomial ring $\mathscr{L}_{\bullet}[\lambda^{\pm}]$. For each $i\geq 1$,
denote the coefficients of $\lambda^{k}$ in $\mathrm{tr}(\mathscr{B}^{i})$ by
$t_{ik}$:

$\mathrm{tr}(\mathscr{B}^{i})=\sum_{k}t_{ik}\lambda^{k}$

We will spend the remainder of this section proving the following theorem.

######  Theorem 13.

For each $i$ and $j$, the classes $t_{ij}^{\natural}$ are invariant under both
$T$ and $\widehat{T}$.

To begin proving this, we first make the following simple observation:

######  Lemma 5.

Let $R$ be an associative ring, and $p,q\in R[\lambda^{\pm}]$. Then every
coefficient of $[p,q]$ is in $[R,R]$.

###### Proof.

Let $p=\sum_{i}p_{i}\lambda^{i}$ and $q=\sum_{j}q_{j}\lambda^{j}$. Then
$[p,q]=\sum_{i,j}[p_{i},q_{j}]\lambda^{i+j}$. ∎

This implies the following

######  Corollary 1.

Let $f=\sum_{i}f_{i}\lambda^{i}$ and $g=\sum_{j}g_{j}\lambda^{j}$ be Laurent
polynomials in $A:=R[\lambda^{\pm}]$. If $f\equiv g~{}\mathrm{mod}~{}[A,A]$,
then $f_{i}\equiv g_{i}~{}\mathrm{mod}~{}[R,R]$ for each $i$.

We now consider traces of powers of matrices over general rings:

######  Lemma 6.

Let $R$ be an associative ring, and $A,B\in\mathrm{Mat}_{n}(R)$. Then
$\mathrm{tr}((AB)^{k})\equiv\mathrm{tr}((BA)^{k})~{}\mathrm{mod}~{}[R,R]$ for
all $k$.

###### Proof.

If the matrices are given by $A=(a_{ij})$ and $B=(b_{ij})$ then

$\displaystyle\mathrm{tr}(AB)^{k}$
$\displaystyle=\sum_{i,j_{1},\dots,j_{k-1}}(AB)_{ij_{1}}(AB)_{j_{1}j_{2}}\cdots(AB)_{j_{k-1}i}$
$\displaystyle=\sum_{i,j_{1},\dots,j_{k-1}}\left(\sum_{\ell_{1}}a_{i\ell_{1}}b_{\ell_{1}j_{1}}\right)\cdots\left(\sum_{\ell_{k}}a_{j_{k-1}\ell_{k}}b_{\ell_{k}i}\right)$
$\displaystyle=\sum_{i,j_{1},\dots,j_{k-1}}\sum_{\ell_{1},\dots,\ell_{k}}a_{i\ell_{1}}b_{\ell_{1}j_{1}}\cdots
a_{j_{k-1}\ell_{k}}b_{\ell_{k}i}$

Similarly, using $BA$ instead, we get

$\mathrm{tr}(BA)^{k}=\sum_{i,j_{1},\dots,j_{k-1}}\sum_{\ell_{1},\dots,\ell_{k}}b_{i\ell_{1}}a_{\ell_{1}j_{1}}\cdots
b_{j_{k-1}\ell_{k}}a_{\ell_{k}i}$

After re-indexing by $\ell_{t}\mapsto j_{t}$ (for $1\leq t\leq k-1$),
$\ell_{k}\mapsto i$, $i\mapsto\ell_{1}$, $j_{t}\mapsto\ell_{t+1}$, we get

$\mathrm{tr}(BA)^{k}=\sum_{i,j_{1},\dots,j_{k-1}}\sum_{\ell_{1},\dots,\ell_{k}}b_{\ell_{1}j_{1}}a_{j_{1}\ell_{2}}\cdots
b_{\ell_{k}i}a_{i\ell_{1}}$

Clearly, this is the same as the expression for $\mathrm{tr}((AB)^{k})$ mod
$[R,R]$, since each term with corresponding indices is the same after cyclicly
shifting the last $a_{i\ell_{1}}$ to the beginning. ∎

Now we may prove Theorem 13:

###### Proof.

As discussed earlier, the map $T$ is given, in the $X,Y$ variables, by a
sequence of Postnikov moves and gauge transformations. The square move and
gauge transformations do not change the boundary measurement matrix at all,
and the “white-swap” and “black-swap” moves only change the boundary
measurement matrix up to conjugation, since it amounts to a refactorization
$\mathscr{B}=\mathscr{B}_{1}\mathscr{B_{2}}\mapsto\widetilde{\mathscr{B}}=\mathscr{B}_{2}\mathscr{B}_{1}$.
If we let $R=\mathscr{L}_{\bullet}$ and
$A=\mathscr{L}_{\bullet}[\lambda^{\pm}]$, then by Corollary 1 and Lemma 6,
$\mathrm{tr}(\widetilde{\mathscr{B}}^{k})$ and $\mathrm{tr}(\mathscr{B}^{k})$
differ by an element of $[A,A]$, and so $t_{ij}^{\natural}$ is invariant under
$T$. The map $\widehat{T}$ is just $T$ followed by a shift of indices, which
is equivalent to cutting part of the network off one end, and glueing it to
the opposite end. Again, this amounts to a re-factorization of the boundary
measurement measurement matrix, which changes it only up to conjugation. ∎

######  Remark.

In [Izo18], Izosimov interprets the pentagram map, as well as some
generalizations, in terms of re-factorizations in Poisson-Lie groups. At the
end of the paper, he poses the question of whether his techniques, when
applied to matrix-valued coefficients, give rise to the Grassman pentagram map
of Mari Beffa and Felipe. It seems that the proof of Theorem 13, which
realizes the pentagram map as a re-factorization of the boundary measurement
matrix, suggests a positive answer to Izosimov’s question.

###  7.3 Involutivity of the Invariants

In this section, we will prove that the invariants from the previous section
are an involutive family with respect to the induced bracket. More
specifically:

######  Theorem 14.

Let $Q_{n}$ be the network for the pentagram map on $\mathcal{GP}_{n,N}$, and
$\mathscr{B}=\mathscr{B}_{Q}(\lambda)$ its boundary measurement matrix, with
$t_{ik}$ the homogeneous components of $\mathrm{tr}(\mathscr{B}^{i})$ as
defined before. Then for all $i,j,k,\ell$:

$\left<t_{ik},t_{j\ell}\right>=0$

Recall that we use as the cut the identified top/bottom edge of the rectangle
on which we draw the quiver. So the element $t_{ik}\in\mathscr{L}$ is the sum
over all loops in $Q$ which are homologous to $(i,k)$ cycles on the torus.
That is, if we lift these paths to the universal cover, then they cross
fundamental domains $i$ times horizontally and $k$ times vertically (with
sign). Let $A_{ik}$ be the set of all such paths, so that

$t_{ik}=\sum_{p\in A_{ik}}p$

Then by bilinearity and the formula from Theorem 7,

$\left<t_{ik},t_{j\ell}\right>=\sum_{f\in A_{ik}}\sum_{g\in
A_{j\ell}}\left<f,g\right>=\sum_{f\in A_{ik}}\sum_{g\in
A_{j\ell}}\sum_{\bullet\in f\cap
g}\varepsilon_{\bullet}(f,g)f_{\bullet}g_{\bullet}$

We want to prove that this expression is zero. To do so, we will define a
sign-reversing, fixed-point-free permutation on the set of terms appearing in
the sum. That is, each term which appears can be paired with another term with
opposite coefficient and the same monomial in $A^{\natural}$.

We start by choosing an arbitrary non-zero term in the sum, of the form
$\varepsilon_{\bullet}(f,g)\,f_{\bullet}g_{\bullet}$. To define a permutation
as suggested above, we need to know what are the other terms in the sum that
are equivalent to $f_{\bullet}g_{\bullet}$ in $A^{\natural}$. The answer is
given by the following lemmas:

######  Lemma 7.

Suppose $f_{\bullet}g_{\bullet}\equiv f^{\prime}_{\ast}g^{\prime}_{\ast}$ mod
$[\mathscr{L}_{\bullet},\mathscr{L}_{\bullet}]$, and that
$\varepsilon_{\ast}(f^{\prime},g^{\prime})\neq 0$. Then $\ast$ is in either
$f\cap g$, $f\cap f$, or $g\cap g$.

###### Proof.

Certainly $f^{\prime}$ and $g^{\prime}$ together have the same combined set of
edges as $f$ and $g$ since
$f_{\bullet}g_{\bullet}=f^{\prime}_{\ast}g^{\prime}_{\ast}$ cyclically. Then
since two paths come together at $\ast$ (since
$\varepsilon_{\ast}(f^{\prime},g^{\prime})\neq 0$), it must be that either $f$
and $g$ meet at $\ast$, or $f$ or $g$ meets itself. ∎

######  Lemma 8.

Let $\bullet$ and $\ast$ be as in the previous lemma, with
$f_{\bullet}g_{\bullet}=f^{\prime}_{\ast}g^{\prime}_{\ast}$. Additionally
assume that $(f,g)$ and $(f^{\prime},g^{\prime})$ are both in $A_{ik}\times
A_{j\ell}$. Then
$(a)$ If $\ast\in f\cap g$, then $(f^{\prime},g^{\prime})$ are obtained from
$(f,g)$ by swapping the segments of $f$ and $g$ between $\bullet$ and $\ast$.
$(b)$ If $\ast\in f\cap f$, then $(f^{\prime},g^{\prime})$ are obtained from
$(f,g)$ by swapping the entire loop $g$ with the subloop of $f$ based at
$\ast$.

###### Proof.

$(a)$ Denote by $a$ and $b$ the segments of $f$ between $\bullet$ and $\ast$,
and similarly let $x$ and $y$ be the segments of $g$ between $\bullet$ and
$\ast$, so we may write $f=\bullet a\ast b$ and $g=\bullet x\ast y$. Then

$f_{\bullet}g_{\bullet}=\bullet a\ast b\bullet x\ast y$

Since we assumed that
$f^{\prime}_{\ast}g^{\prime}_{\ast}=f_{\bullet}g_{\bullet}$ up to cyclic
permutation, then we can write it starting at $\ast$ as

$f^{\prime}_{\ast}g^{\prime}_{\ast}=\ast b\bullet x\ast y\bullet a$

This means $f^{\prime}=\ast b\bullet x$ and $g^{\prime}=\ast y\bullet a$, or
the other way around: $f^{\prime}=\ast y\bullet a$ and $g^{\prime}=\ast
b\bullet x$. The two possibilities differ by switching the roles of
$f^{\prime}$ and $g^{\prime}$. But since we assume $(f^{\prime},g^{\prime})\in
A_{ik}\times A_{j\ell}$, and since $A_{ik}\neq A_{j\ell}$, only one of the
possibilities will be correct.

After cyclically permuting, we have $f^{\prime}=\bullet x\ast b$ and
$g^{\prime}=\bullet a\ast y$ (or the other way around). Thus we see that
$f^{\prime}$ and $g^{\prime}$ are obtained from $f$ and $g$ by swapping either
$x$ with $a$ or $y$ with $b$, which are the portions of the paths in between
their common subpaths $\bullet$ and $\ast$. This is illustrated in Figure 13.
Note that since $(f^{\prime},g^{\prime})\in A_{ik}\times A_{j\ell}$, the
swapped segments must have the same intersection index with the cut.

Figure 13: A “type I”
swap$f$$g$$\bullet$$\ast$$\longleftrightarrow$$f^{\prime}$$g^{\prime}$$\bullet$$\ast$

$(b)$ Now suppose $\ast\in f\cap f$. Let $a,b,c$ be the segments of $f$
between $\bullet$ and $\ast$, so that $f=\bullet a\ast b\ast c$. Also let $x$
be the rest of $g$ after $\bullet$, so that $g=\bullet x$. Then

$f_{\bullet}g_{\bullet}=\bullet a\ast b\ast c\bullet x$

Cyclically permuting, and using the assumption that
$f_{\bullet}g_{\bullet}=f^{\prime}_{\ast}g^{\prime}_{\ast}$, we get

$f^{\prime}_{\ast}g^{\prime}_{\ast}=\ast c\bullet x\bullet a\ast b$

Again there are two possibilities. Either $f^{\prime}=\ast c\bullet x\bullet
a$ and $g^{\prime}=\ast b$, or $f^{\prime}$ and $g^{\prime}$ are flipped. But
the condition that $f^{\prime}\in A_{ik}$ and $g^{\prime}\in A_{j\ell}$
ensures that only one of the two choices is correct. Suppose it is the first
case, and $f^{\prime}=\ast c\bullet x\bullet a$ and $g^{\prime}=\ast b$. Then
$f^{\prime}$ is obtained from $f$ by removing the loop based at $\ast$ and
adding the loop $g$, and $g^{\prime}$ is simply the subloop of $f$ based at
$\ast$. This is illustrated in Figure 14. Note that since
$(f^{\prime},g^{\prime})\in A_{ik}\times A_{j\ell}$, the loops $\bullet x$ and
$\ast b$ must cross the cut and rim the same number of times.

Figure 14: A “type II”
swap$f$$g$$\bullet$$\ast$$\longleftrightarrow$$f$$g$$\bullet$$\ast$

∎

The next lemma is a converse to the previous one, so in fact the conditions
given above completely characterize the set of terms in the expansion of
$\left<t_{ik},t_{j\ell}\right>$ with a given term $f_{\bullet}g_{\bullet}$.

######  Lemma 9.

Let $(f,g)\in A_{ik}\times A_{j\ell}$ and $\bullet\in f\cap g$ such that
$\varepsilon_{\bullet}(f,g)\neq 0$.
$(a)$ If there exists an $\ast\in f\cap g$ such that the segments of $f$ and
$g$ between $\bullet$ and $\ast$ (or between $\ast$ and $\bullet$) cross the
cut and rim the same number of times, then swapping those segments gives
$(f^{\prime},g^{\prime})\in A_{ik}\times A_{j\ell}$ such that
$f^{\prime}_{\ast}g^{\prime}_{\ast}=f_{\bullet}g_{\bullet}$. $(b)$ If there
exists an $\ast\in f\cap f$ such that $g$ and the subloop of $f$ based at
$\ast$ cross the cut and rim the same number of times, then swapping $g$ and
this subloop gives $(f^{\prime},g^{\prime})\in A_{ik}\times A_{j\ell}$ such
that $f^{\prime}_{\ast}g^{\prime}_{\ast}=f_{\bullet}g_{\bullet}$.

We will call the swaps from part $(a)$ of the previous two lemmas “type I”
swaps, and the swaps from part $(b)$ will be called “type II” swaps. We are
now ready to define our sign-reversing map, which we will denote $\sigma$,
which acts on the set of terms appearing in the sum.

######  Lemma 10.

There exists a fixed-point-free permutation $\sigma$ of the non-zero terms
appearing in the expansion of $\left<t_{ik},t_{j\ell}\right>$ such that
$\sigma(x)=-x$ for each term $x$.

###### Proof.

Recall that $f$ is a term in $t_{ik}$ and $g$ a term in $t_{j\ell}$. Assume
that $i\geq j$, so that $f$ goes around the torus more times “horizontally”.
Starting at $\bullet$, walk along $f$ until we reach the first admissible
swap. If the first admissible swap is of type I, and occurs at $\ast$, then we
will define the image of the term
$\varepsilon_{\bullet}(f,g)f_{\bullet}g_{\bullet}$ under $\sigma$ to be the
term
$\varepsilon_{\ast}(f^{\prime},g^{\prime})f^{\prime}_{\ast}g^{\prime}_{\ast}$,
where $f^{\prime}$ and $g^{\prime}$ are obtained by performing the type I swap
between $\bullet$ and $\ast$. In order for
$\varepsilon_{\ast}(f^{\prime},g^{\prime})\neq 0$, it must be that $f$ and $g$
cross at $\ast$ (rather than touch). This guarantees that
$\varepsilon_{\bullet}(f,g)f_{\bullet}g_{\bullet}$ is not a fixed point of
$\sigma$. To see that
$\varepsilon_{\ast}(f^{\prime},g^{\prime})=-\varepsilon_{\bullet}(f,g)$, note
that by the preceding lemmas, the segments of $f$ and $g$ from $\bullet$ to
$\ast$ must cross the cut and rim the same number of times. This means that on
the universal cover, the segments of $f$ and $g$ between $\bullet$ and $\ast$
bound a contractible disc. Thus after performing the swap, $f^{\prime}$ and
$g^{\prime}$ touch with opposite orientation – i.e. if $f$ was to the left,
and $g$ to the right, then $f^{\prime}$ is on the right, and $g^{\prime}$ on
the left. This can be seen in Figure 13.

If, on the other hand, the first admissible swap is of type II, we consider
two separate cases. Recall that a type II swap means $f$ intersects itself at
$\ast$, so we may write $f_{\ast}=f^{\prime}_{\ast}f^{\prime\prime}_{\ast}$,
and that $f^{\prime\prime}_{\ast}\in A_{j\ell}$ (the same as $g$). First we
consider the case that $f^{\prime\prime}_{\ast}$ does not intersect $g$. In
this case, we define the action of $\sigma$ on
$\varepsilon_{\bullet}(f,g)f_{\bullet}g_{\bullet}$ to be the term
corresponding to performing the type II swap. Now, consider the second case,
in which $f^{\prime\prime}_{\ast}$ does intersect $g$. In this case, let
$\star$ be the first intersection point after $\ast$. Then we define the
action of $\sigma$ on $\varepsilon_{\bullet}(f,g)f_{\bullet}g_{\bullet}$ to be
the term corresponding to the type I swap from $\bullet$ to $\star$. ∎

Theorem 14 now follows immediately as a corollary, since the lemma implies
that there are an even number of terms equivalent to any given
$f_{\bullet}g_{\bullet}$, half with coefficient $+1$ and half with coefficient
$-1$.

##  8 Recovering the Lax Representation

In [FMB15], the authors gave a Lax representation of the Grassmann pentagram
map, which gives a family of invariants. However, they did not formulate
Liouville integrability, since there was no associated Poisson structure. In
this section, we explain how to apply the results of the present paper to
recover the invariants from [FMB15], and additionally give a Poisson structure
in which these invariants commute.

As described earlier, we may lift the vertices of a twisted Grassmann $n$-gon
to $3N$-by-$N$ matrices $V_{i}$, and we get a relation of the form

$V_{i+3}=V_{i}A_{i}+V_{i+1}B_{i}+V_{i+2}C_{i}$

As before, we define the $3N$-by-$3N$ matrices
$\mathbf{V}_{i}=(V_{i}V_{i+1}V_{i+2})$. Then the formula above can be re-
phrased by saying $\mathbf{V}_{i+1}=\mathbf{V}_{i}L_{i}$, where $L_{i}$ is the
block matrix

$L_{i}=\left(\begin{array}[]{ccc}0&0&A_{i}\\\ \mathrm{Id}_{N}&0&B_{i}\\\
0&\mathrm{Id}_{N}&C_{i}\end{array}\right)$

Assuming the lift was twisted, then the monodromy matrix of the polygon is
related to the $L_{i}$ by

$M\,\mathbf{V}_{i}=\mathbf{V}_{i}\,L_{i}L_{i+1}\cdots L_{i+n-1}$

In that paper, the authors proved the following

######  Theorem 15.

[FMB15] The Grassmann pentagram map is invariant under the scaling
$B_{i}\mapsto\lambda^{-1}B_{i}$, $C_{i}\mapsto\lambda C_{i}$.

In light of this theorem, we may define the modified matrix with parameter
$\lambda$:

$L_{i}(\lambda)=\left(\begin{array}[]{ccc}0&0&A_{i}\\\
\mathrm{Id}_{N}&0&\lambda^{-1}B_{i}\\\ 0&\mathrm{Id}_{N}&\lambda
C_{i}\end{array}\right)$

In [FMB15], the authors proved that the conjugacy class of
$L(\lambda):=L_{1}(\lambda)\cdots L_{n}(\lambda)$ is preserved under the
pentagram map, so the spectral invariants of $L(\lambda)$ are invariants of
the map. It is not hard to see that the pentagram map is also invariant under
the scaling $A_{i}\mapsto\lambda A_{i}$, $B_{i}\mapsto\lambda B_{i}$. We will
call the corresponding matrix $\widetilde{L}_{i}(\lambda)$:

$\widetilde{L}_{i}(\lambda)=\left(\begin{array}[]{ccc}0&0&\lambda A_{i}\\\
\mathrm{Id}_{N}&0&\lambda B_{i}\\\ 0&\mathrm{Id}_{N}&C_{i}\end{array}\right)$

In [GSTV16], the authors present a very similar Lax representation for the
pentagram map on $\mathbb{P}^{2}$, and explicitly describe the connection with
the boundary measurement matrix used in the combinatorial proof of
integrability. We now mimic this approach to connect the earlier results from
this paper to the Lax representation described above from [FMB15]. In order to
more easily generalize this approach, we change some of our notations and
conventions from earlier to more closely resemble those from [GSTV16]. In
particular, we re-index the $Y_{i}$’s by 1, so that our relation reads

$V_{i+3}=V_{i}Y_{i-1}+V_{i+1}X_{i}+V_{i+2}$

Also, recall we use an oriented curve called the “cut” on the cylinder to
determine the powers of $\lambda$ in the boundary measurements. Earlier in the
paper, we chose the convention that when we draw the cylinder as a rectangle
(identifying the top/bottom edges), we take the cut to be the the top/bottom
edge. We change this convention now so that the cut goes diagonally down and
to the right, crossing each edge that connects two white vertices. With this
convention, the quiver can be realized as the concatenation of the elementary
networks pictured in Figure 15.

Figure 15: Elementary Networks$X_{i}$$Y_{i}$$1$$3$$2$$2$$1$$3$

The boundary measurement matrix of the $i^{\mathrm{th}}$ elementary network is
given by

$\mathscr{B}_{i}(\lambda)=\left(\begin{array}[]{ccc}0&X_{i}&X_{i}+Y_{i}\\\
\lambda&0&0\\\ 0&1&1\end{array}\right)$

The boundary measurement matrix of the entire network $Q_{n}$ is then the
product
$\mathscr{B}(\lambda):=\mathscr{B}_{1}(\lambda)\cdots\mathscr{B}_{n}(\lambda)$.
Although our matrices have entries in the non-commutative ring
$\mathscr{L}_{\bullet}[\lambda^{\pm}]$, the following result from [GSTV16] is
true in this more general context:

######  Proposition 13.

[GSTV16]
$\mathscr{B}_{i}(\lambda)=\mathscr{A}_{i}(\lambda)\widetilde{L}_{i+1}(\lambda)\mathscr{A}_{i+1}(\lambda)^{-1}$,
where the matrices $\mathscr{A}_{i}(\lambda)$ are given by

$\mathscr{A}_{i}(\lambda):=\left(\begin{array}[]{ccc}\lambda^{-1}&0&X_{i}\\\
0&1&0\\\ 0&0&1\end{array}\right)$

In particular, $\mathscr{B}(\lambda)$ is conjugate to
$\widetilde{L}(\lambda):=\widetilde{L}_{1}(\lambda)\cdots\widetilde{L}_{n}(\lambda)$.

Therefore, the spectral invariants of $\mathscr{B}(\lambda)$ are the same as
those of $\widetilde{L}(\lambda)$. The coefficients of $\lambda$ in the
expansions of $\mathrm{tr}(\mathscr{B}(\lambda)^{k})$ can be written as non-
commutative polynomials in the $X_{i}$ and $Y_{i}$. Thus the (traces of the)
spectral invariants can be interpreted as functions on $\mathcal{GP}_{n,N}$,
after identifying it with
$\mathrm{GL}_{N}^{2n+1}/\mathrm{Ad}\,\mathrm{GL}_{N}$, as in Theorem 4. The
Poisson bracket is given by taking the trace of the induced bracket on
$\mathscr{L}_{\bullet}^{\natural}$:

$\\{\mathrm{tr}(X_{i}),\,\mathrm{tr}(Y_{j})\\}=\mathrm{tr}\left<X_{i},Y_{j}\right>$

As before, we let $t_{ij}$ denote the components of the spectral invariants,
so that $\mathrm{tr}(\mathscr{B}(\lambda)^{j})=\sum_{i}t_{ij}\lambda^{i}$. The
discussion above tells us that $\mathrm{tr}(t_{ij})$ may be interpreted as
functions on $\mathcal{GP}_{n,N}$. Although in the present paper we chose a
different normalization convention (namely the $X_{i}$, $Y_{i}$, $Z$ matrices)
than Mari Beffa and Felipe, these invariant functions are still essentially
the same as those mentioned in [FMB15]. Theorem 14 then implies that these
functions form an involutive family with respect to the induced Poisson
structure.

##  9 Conclusions and Suggestions for Future Work

We have shown in Theorem 13 and Theorem 14 that the coefficients $t_{ij}$ of
the traces of powers of the boundary measurement matrix are non-commutative
invariants of the pentagram map, and that they form an involutive family under
the induced Lie bracket $\left<-,-\right>$ on $\mathscr{L}^{\natural}$. In
this sense, we have established a form of non-commutative integrability for
the Grassmann pentagram map, in terms of the $X_{i},Y_{i}$ variables/matrices.
Furthermore, this induces a Poisson structure on the moduli space
$\mathcal{GP}_{n,N}$ in which Mari-Beffa and Felipe’s invariants Poisson-
commute.

However, in the usual commutative setting, the definition of Liouville
integrability requires not only a Poisson-commuting family of invariants, but
also that these invariants are independent, and that they are a maximal family
of such independent commuting invariants. These last two aspects —
independence and maximality — were not addressed in the present paper. It
would be nice to have a proof that among this infinite family of commuting
invariants, one can find a maximal independent subset, so that we have a
Liouville integrable system.

The proof of Theorem 14 is very combinatorial. It would be interesting to give
an alternate proof using $R$-matrices, which would more closely resemble the
method of proof in the classical case [GSTV16]. This would involve formulating
an $R$-matrix double bracket on the noncommutative space of matrices, and
formulating some result saying that the spectral invariants are in involution
with respect to the induced $H_{0}$-Poisson structure.

##  10 Acknowledgments

The author would like to give thanks to Michael Shapiro for suggesting the
problem, and for helpful feedback and conversations; and to Semeon Artamonov
and Leonid Chekhov for helpful discussions. The author was partially supported
by the NSF grant DMS-1702115.

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# Equivalence and Duality of Polycyclic Codes Associated with Trinomials over
Finite Fields††thanks: This research is supported by the National Natural
Science Foundation of China (12071001) and the Excellent Youth Foundation of
Natural Science Foundation of Anhui Province (1808085J20).

Minjia Shi, Haodong Lu, Shuang Zhou, Jiarui Xu, Yuhang Zhu
smjwcl.good@163.comhdlu818@163.comshzhou0815@163.comjrxu39<EMAIL_ADDRESS>Key Laboratory of Intelligent Computing and Signal Processing, Ministry of
Education, School of Mathematical Sciences, Anhui University, Hefei 230601,
China.

###### Abstract

In this paper, several conjectures proposed in [2] are studied, involving the
equivalence and duality of polycyclic codes associated with trinomials.
According to the results, we give methods to construct isodual and self-dual
polycyclic codes, and study the self-orthogonal and dual-containing polycyclic
codes over $\mathbb{F}_{2}$.

Keywords: polycyclic codes, trinomials, equivalence, duality.
MSC 94B05, 05E30

##  1 Introduction

Polycyclic codes is a generalization of cyclic codes since the concept of
associate polynomials is proposed. Algebraically, a polycylic code is an ideal
of $\mathbb{F}_{q}/\langle f(x)\rangle$ for some $f(x)\in\mathbb{F}_{q}[x]$,
and $f(x)$ is called the associate polynomial of this polycyclic code.
Polycyclic codes include cyclic codes, constacyclic codes and etc, as
$x^{n}-1$ can be associated with cyclic codes and $x^{n}-\lambda$ can be
associated with constacyclic codes of length $n$ over $\mathbb{F}_{q}$.
Polycylic codes have been known since $1972$ in [8]. As is well known,
polycyclic codes are shortened cyclic codes, and conversely shortened cyclic
codes are polycyclic (P. 241, [8]). However, polycyclic codes were never
received the same level of attention as cyclic codes and some of their
generalizations. In recent years, polycyclic codes has also been studied
widely in [2, 4, 6, 10, 11, 12], which shows that polycyclic codes begun to
receive more and more attention.

In this paper, we mainly solve several conjectures proposed in [2], which
mainly around the equivalence and duality of polycyclic codes associated with
trinomials. Some kinds of duality of polycyclic codes have been studied, such
as the results in [1, 3], and we will also give some information and results
about the duality of polycyclic codes associated with trinomials in this
paper.

In Section 2, we mainly give the basic concepts and results of polycyclic
codes, they are the basis of this paper and will appear frequently in the
following sections. In Section 3, a mapping $\varphi$ will be given to
illustrate the relationship between two kinds of polycyclic codes
corresponding to $f(x)$ and $f^{*}(x)$ respectively, where $f(x)$ is trinomial
with $f(0)\neq 0$. This particular relationship cannot be obtained simply by
choosing polynomials of the same order, which shows that Conjecture 3.5 in [2]
is wrong. A counter example of Conjecture 3.5 would be given in this section.
In Section 4, a conjecture about isodual in [2] (i.e. Conjecture 4.5) will be
proven, which gives a simple way to construct some isodual polycyclic codes
associated with trinomials. However, we will give a counter-example to
illustrate conjecture 4.2 in [2] is wrong, and point out that this method is
not the only way to construct isodual polycyclic codes associated with
trinomials over $\mathbb{F}_{2}$. In Section 5, we study the properties of
generator polynomials for self-dual polycyclic codes associated with
trinomials, and it turns out that the generators have a unified form
$x^{n}-a$, where $a^{2}=-1$ in $\mathbb{F}_{q}$. And these results show that
Conjecture 4.7 is correct in [2]. In Section 6, we discussed Conjecture 4.10
and Conjecture 4.11 in [2]. The results of the discussion show that it is
impossible to find self-orthogonal or nontrivial dual-containing polycyclic
codes over $\mathbb{F}_{2}$, and a dual-containing polycyclic code with prime
length will be given to show that Conjecture 4.11 is wrong. Section 7
concludes this paper.

##  2 Preliminaries

In this part, we recall the basic definitions and theorems of polycyclic
codes.

######  Definition 2.1.

[2] A linear code $C$ is said to be right polycyclic with respect to
$v=(v_{0},v_{1},\cdots,v_{n-1})$
$\in\mathbb{F}_{q}^{n}$, if for any codeword $(c_{0},c_{1},\cdots,c_{n-1})\in
C$, its right polycyclic shift, i.e.
$(0,c_{0},c_{1},\cdots,c_{n-2})+c_{n-1}(v_{0},v_{1},\cdots,v_{n-1})$ is also a
codeword of $C$.

Similarly, we can define $C$ is left polycyclic. If $C$ is both left and right
polycyclic, then it is bi-polycyclic. In our paper, when referring to
polycyclic codes, we suppose it to be right polycyclic codes.

On the premise that there is no difference between vector $v$ and polynomial
$v(x)$, each polycyclic code $C$ of length $n$ is associated with a polynomial
$v(x)$ of degree less than $n$, and we say that $C$ is a polycyclic code
associated with $x^{n}-v(x)$. Note that an associate polynomial of a
polycyclic code may not be unique, i.e. a polycylic code can have multiple
associate polynomials. Furthermore, polycyclic codes associated with
$f(x)=x^{n}-v(x)$ are ideals of the factor ring $\mathbb{F}_{q}[x]/\langle
f(x)\rangle$.

######  Definition 2.2.

[6] A polycyclic code $C$ over $\mathbb{F}_{q}$ of length $n$ and dimension
$k$ with associate polynomial $v(x)$ has a monic polynomial $g(x)$ of minimal
degree $n-k$ that belongs to $C$, which divides $x^{n}-v(x)$ and is called the
generator polynomial of $C$.

For $\lambda\in\mathbb{F}_{q}$ and any
$v=(v_{0},v_{1},\cdots,v_{n-1})\in\mathbb{F}_{q}^{n}$, the
$\lambda$-constacyclic shift $\tau_{\lambda}$ on $\mathbb{F}_{q}^{n}$ is the
shift $\tau_{\lambda}(v_{0},v_{1},\cdots,v_{n-1})=(\lambda
v_{n-1},v_{0},v_{1},\cdots,v_{n-2})$. A linear code $C$ is said to be a
$\lambda$-constacyclic code if $\tau_{\lambda}(C)\subseteq C$. Note that the
associate polynomials of constacyclic codes have the form $x^{n}-\lambda$,
where the length of constacyclic codes is $n$. Since polycyclic codes are a
generalization of cyclic codes, so in the same way as cyclic codes, we can
construct a generator matrix for $C$ from its generator polynomial.

######  Theorem 2.1.

[6] A code $C\in\mathbb{F}_{q}^{n}$ is right polycyclic associated with
polynomial $f(x)$ if and only if it has a $k\times n$ generator matrix of the
form

$G=\begin{pmatrix}g_{0}&g_{1}&\cdots&g_{n-k}&0&0&\cdots&0\\\
0&g_{0}&g_{1}&\cdots&g_{n-k}&0&\cdots&0\\\
\vdots&\ddots&\ddots&\ddots&\ddots&\ddots&\ddots&\vdots\\\
0&\cdots&0&g_{0}&g_{1}&\cdots&g_{n-k}&0\\\
0&\cdots&0&0&g_{0}&g_{1}&\cdots&g_{n-k}\\\ \end{pmatrix}_{k\times n},$

with $g_{n-k}\neq 0$. In this case $\langle
g_{0}+g_{1}x+\cdots+g_{n-k}x^{n-k}\rangle$ is an ideal of
$\mathbb{F}[x]/\langle f(x)\rangle$.

In this article, If we say that a polycyclic code $C$ is generated by
$g(x)\mid f(x)$, which also means $f(x)$ is the associate polynomial of $C$
and $g(x)$ is the generator polynomial of $C$.

Since we are concerned with the duality of polycyclic codes, it is necessary
to mention some definitions of it.

######  Definition 2.3.

Two codes $C_{1},C_{2}$ over $\mathbb{F}_{q}$ with generator matrices $G_{1}$
and $G_{2}$ respectively are equivalent if there exists a monomial matrix over
$\mathbb{F}_{q}$ such that $G_{1}M=G_{2}$. A monomial matrix is a square
matrix with exactly one non-zero entry in each row and each column.

######  Definition 2.4.

The dual of $C$, denoted by $C^{\perp}$, is the set of vectors orthogonal to
every codeword of $C$ under the Euclidean inner product. A code $C$ is self-
dual if $C=C^{\perp}$; $C$ is iso-dual if $C$ is equivalent to $C^{\perp}$;
$C$ is self-orthogonal if $C\subseteq C^{\perp}$ and $C$ is dual-containing if
$C^{\perp}\subseteq C$.

Finally, here are some notations and theorems that will be used in this paper.

######  Definition 2.5.

[9] Let $f\in\mathbb{F}_{q}[x]$ be a nonzero polynomial. If $f(0)\neq 0$, then
the least positive integer $e$ such that $f(x)$ divides $x^{e}-1$ is called
the order of $f$ and denoted by $\operatorname{\text{Ord}}(f)$. If $f(0)=0$,
then $f(x)=x^{h}g(x)$, where $h\in\mathbb{N}$ and $g\in\mathbb{F}_{q}[x]$ with
$g(0)\neq 0$ are uniquely determined, thus we can define
$\operatorname{\text{Ord}}(f)$ as $\operatorname{\text{Ord}}(g)$.

######  Definition 2.6.

[9] Let $g(x)=\sum_{i=0}^{k}g_{i}x^{i}$ be a polynomial of degree $k$
($g_{k}\neq 0$) over $\mathbb{F}_{q}$. Define the reciprocal polynomial
$g^{*}(x)$ of $g(x)$ by $g^{*}(x):=x^{k}\,g(1/x)=\sum_{i=0}^{k}g_{k-i}x^{i}$.

######  Definition 2.7.

Let $C$ be a code, its reversed code is defined as $C^{\prime}$ if
$(c_{n-1},c_{n-2},\cdots,c_{0})\in C^{\prime}$, where
$(c_{0},\cdots,c_{n-2},c_{n-1})\in C$.

It’s clear that $(g^{*}(x))^{*}=g(x)$ and $(C^{\prime})^{\prime}=C$.

######  Theorem 2.2.

[9] Let $C$ be a linear code and let $H$ be a parity-check matrix for $C$.
Then the following statements are equivalent:
(i) $C$ has distance $d$;
(ii) any $d-1$ columns of $H$ are linearly independent and $H$ has $d$ columns
that are linearly dependent.

##  3 Equivalence of Polycyclic Codes with Special Associate Polynomials

In this section, we are ready to analyze Conjecture 3.5 in [2]. In fact,
Conjecture 3.5 in [2] itself is not correct, which can be seen in the counter-
example Example 3.1. However, if we change the condition
“$\operatorname{\text{Ord}}(t_{1}(x))=\operatorname{\text{Ord}}(t_{2}(x))$" to
“$t_{2}(x)=t_{1}(0)^{-1}t_{1}^{*}(x)$", we can obtain a new conclusion (see
Theorem 3.3 below). First, let us recall some claims from [2].

######  Proposition 3.1 (Proposition 5.2 in [2]).

For any polynomial $f(x)$, $\deg(f(x))\neq\deg(f^{*}(x))$ if and only if
$x\mid f(x)$.

Since we need $g(x)\mid x^{n}-ax^{i}-b$ with $b\neq 0$, using the result of
Proposition 3.1, we have $\deg(g(x))=\deg(g^{*}(x))$.

######  Theorem 3.1 (Theorem 4.9 in [2]).

Let $C$ be a polycyclic code, then its reversed code $C^{\prime}$ is generated
by the reciprocal polynomial of the generator polynomial of $C$. Furthermore,
$C^{\prime}$ is always equivalent to $C$ because of the isomorphism.

######  Theorem 3.2 (Theorem 6.1 in [2]).

$($same order$)$. Let trinomials $f_{1}(x)=x^{n}-ax^{i}-b$ and
$f_{2}(x)=x^{n}-a^{\prime}x^{n-i}-b^{\prime}$, where $a^{\prime}=-ab^{-1}$ and
$b^{\prime}=b^{-1}$, be mutually reciprocal. Then
$\operatorname{\text{Ord}}(f_{1}(x))=\operatorname{\text{Ord}}(f_{2}(x))$.

######  Conjecture 3.1 (Conjecture 3.5 in [2]).

Let $t_{1}(x)=x^{n}-ax^{i}-b$, $t_{2}(x)=x^{n}-a^{\prime}x^{n-i}-b^{\prime}$
be such that
$\operatorname{\text{Ord}}(t_{1}(x))=\operatorname{\text{Ord}}(t_{2}(x))$. Let
$S_{1}$ be the set of all polycyclic codes of length $n$ over $\mathbb{F}_{q}$
associated with $t_{1}(x)$ and $S_{2}$ be the set of all polycyclic codes of
length $n$ over $\mathbb{F}_{q}$ associated with $t_{2}(x)$. Then $S_{1}$ and
$S_{2}$ are in a one-to-one correspondence where corresponding codes are
equivalent to each other.

Now we give a counter example of Conjecture 3.1.

######  Example 3.1.

Consider $t_{1}(x)=x^{10}+x^{8}+1$ and $t_{2}(x)=x^{10}+x^{2}+2$ over
$\mathbb{F}_{3}$. As
$\operatorname{\text{Ord}}(t_{1}(x))=\operatorname{\text{Ord}}(t_{2}(x))=156$,
and

$\displaystyle t_{1}(x)$ $\displaystyle=(x+1)^{2}(x+2)^{2}(x^{6}+2x^{2}+1),$
$\displaystyle t_{2}(x)$
$\displaystyle=(x^{2}+1)^{2}(x^{3}+x^{2}+x+2)(x^{3}+2x^{2}+x+1).$

It is clear that $S_{1}$ contains 18 linear codes, while $S_{2}$ contains 12
linear codes. Therefore, no mapping exists to satisfy the requirement since
the mapping need to be one-to-one. (The data can be obtained by Magma.)

######  Proposition 3.2.

$f(x)g(x)=h(x)$ if and only if $f^{*}(x)g^{*}(x)=h^{*}(x)$, where $f(0)\neq 0$
and $g(0)\neq 0$.

###### Proof.

Let $\deg(f(x))=k,\deg(g(x))=n-k,\deg(h(x))=n$, we have

$\displaystyle f(x)g(x)$ $\displaystyle=h(x),$ $\displaystyle x^{k}f(1/x)\cdot
x^{n-k}g(1/x)$ $\displaystyle=x^{n}h(1/x),$ $\displaystyle f^{*}(x)g^{*}(x)$
$\displaystyle=h^{*}(x).$

Since
$h(x)=(h^{*}(x))^{*}=(f^{*}(x)g^{*}(x))^{*}=(f^{*}(x))^{*}(g^{*}(x))^{*}=f(x)g(x)$.
∎

Note that
$\operatorname{\text{Ord}}(t_{1}(x))=\operatorname{\text{Ord}}(t_{2}(x))$ if
$t_{2}(x)=t_{1}(0)^{-1}t_{1}^{*}(x)$ in Theorem 3.2, thus the following
theorem (Theorem 3.3) is a special case of Conjecture 3.5. By using
Proposition 3.2, if $g(x)$ is a factor of $h(x)$, then $g^{*}(x)$ is a factor
of $h^{*}(x)$, which means $h(x)$ and $h^{*}(x)$ have the same number of
factors, and combined with Proposition 3.1, $h(x)$ and $h^{*}(x)$ have the
same number of factors of the same degree.

######  Theorem 3.3.

Let $t_{1}(x)=x^{n}-ax^{i}-b$, $t_{2}(x)=x^{n}-a^{\prime}x^{n-i}-b^{\prime}$
be such that $t_{2}(x)=t_{1}(0)^{-1}t_{1}^{*}(x)$. Let $S_{1}$ be the set of
all polycyclic codes of length $n$ over $\mathbb{F}_{q}$ associated with
$t_{1}(x)$ and $S_{2}$ be the set of all polycycliccodes of length $n$ over
$\mathbb{F}_{q}$ associated with $t_{2}(x)$. Then $S_{1}$ and $S_{2}$ are in a
one-to-one correspondence where corresponding codes are equivalent to each
other.

###### Proof.

By Proposition 3.2, if $g(x)$ is a factor of $t_{1}(x)$, then $g^{*}(x)$ is a
factor of $t_{2}(x)$, and $\deg(g(x))=\deg(g^{*}(x))$. If the generator of a
polycyclic code $C$ is $g(x)$, then $g(0)^{-1}g^{*}(x)$ is the generator of
$C^{\prime}$. Based on the above facts, define the mapping

$\displaystyle\varphi:\;S_{1}$ $\displaystyle\rightarrow S_{2},$
$\displaystyle C$ $\displaystyle\mapsto C^{\prime}.$

$\varphi$ satisfies: 1) $\varphi$ is an injection. For $\>C_{1},C_{2}\in
S_{1}$ and $C_{1}\neq C_{2}$, let the generators of $C_{1}$, $C_{2}$ be
$f(x)$, $g(x)$, respectively. If $\varphi(C_{1})=\varphi(C_{2})$, then
$f(0)^{-1}f^{*}(x)=g(0)^{-1}g^{*}(x)$, which means
$f(0)^{-1}f(x)=g(0)^{-1}g(x)$ and $C_{1}=C_{2}$, which is a contradiction. 2)
$\varphi$ is surjective. For $\>C\in S_{2}$, let the generator of $C$ be
$g(x)$. Since $g(x)\mid t_{2}(x)$, then $g(0)^{-1}g^{*}(x)\mid t_{1}(x)$,
which means that there exists $C_{0}=C^{\prime}\in S_{1}$ with the generator
$g(0)^{-1}g^{*}(x)$, i.e. $\varphi^{-1}(C)=C^{\prime}$. 3) $C$ is equivalent
to $\varphi(C)=C^{\prime}$, which is the result of Theorem 3.1. ∎

######  Example 3.2.

Let $t_{1}(x)=x^{5}+x^{4}+1$ and $t_{2}(x)=x^{5}+x+1$ over $\mathbb{F}_{2}$,
and

$\displaystyle t_{1}(x)$
$\displaystyle=(x^{2}+x+1)(x^{3}+x+1)=f_{1}(x)f_{2}(x),$ $\displaystyle
t_{2}(x)$ $\displaystyle=(x^{2}+x+1)(x^{3}+x^{2}+1)=f_{1}^{*}(x)f_{2}^{*}(x).$

Therefore, $S_{1}=\\{C_{1},C_{2},C_{3},C_{4}\\}$, where the generator of
$C_{1},C_{2},C_{3},C_{4}$ are $1,f_{1}(x),f_{2}(x),t_{1}(x)$, respectively.
And $S_{2}=\\{C_{1}^{\prime},C_{2}^{\prime},C_{3}^{\prime},C_{4}^{\prime}\\}$.

##  4 A way to Construct Isodual Polycyclic Codes

In this section, Conjecture 4.1, 4.2 and 4.5 in [2] are studied. Before we
discuss these conjectures, here is a useful proposition.

######  Proposition 4.1.

Over $GF(q)$, where $q$ is a power of $2$. If a polycyclic code $C$ associated
with trinomial $x^{2m}+a^{2}x^{2i}+b^{2}$ and generated by
$g(x)=x^{m}+ax^{i}+b$ is isodual, then the polycyclic code $C_{1}$ generated
by $g_{1}(x)=x^{km}+ax^{ki}+b\mid x^{2km}+a^{2}x^{2ki}+b^{2}$ is also isodual,
where $k$ is a positive integer.

###### Proof.

In fact, the generator matrix of $C$ is

$G=\begin{pmatrix}b&\cdots&a&\cdots&1&{}&{}\\\
{}&\ddots&{}&\ddots&{}&\ddots&{}\\\ {}&{}&b&\cdots&a&\cdots&1\\\
\end{pmatrix}_{m\times 2m}.$

By changing $b$ to $bI_{k}$, $a$ to $aI_{k}$, $1$ to $I_{k}$ and $0$ to
$O_{k}$, we get a new matrix $G_{1}$ which is the generator matrix of $C_{1}$.
Here, $O_{k}$ is the all-zero matrix and $I_{k}$ is the identity matrix. Let
$H$ and $H_{1}$ be the parity-check matrices of $C$ and $C_{1}$, respectively,
then if there exists a monomial matrix $M$ satisfing $GM=H$, then the matrix
$M_{1}$ also satisfies $G_{1}M_{1}=H_{1}$, where $M_{1}$ is obtained by
replacing the elements $r$ in $M$ with $rI_{k}$, and $M_{1}$ is also monomial.
∎

Now we discuss Conjecture 4.5 in [2], which consists of Theorem 4.1, 4.2 and
4.3.

######  Theorem 4.1.

Over $GF(q)$, where $q$ is a power of an odd prime. A polycyclic code $C$
associated with trinomial $x^{2m}\pm 2cx^{m}+c^{2}$ and generated by
$g(x)=x^{m}\pm c$ is isodual.

###### Proof.

Since $g(x)=x^{m}\pm c$, and $g(x)$ is the generator polynomial of $C$, we
have a generator matrix $G$ of the form

$G=(\pm cI_{m}\mid I_{m}),$

where $I_{m}$ is an identity matrix. Since $(\pm c^{-1})G=(I_{m}\mid(\pm
c^{-1})I_{m})$, then the parity-check matrix of $C$ is $H=((\mp
c^{-1})I_{m}\mid I_{m})$. We can define

$M=\begin{pmatrix}-c^{2}I_{m}&{}\\\ {}&I_{m}\\\ \end{pmatrix},$

and $M$ is a monomial matrix, which satisfies $HM=G$, and $C$ is isodual. ∎

######  Theorem 4.2.

Over $GF(q)$, where $q$ is a power of 2. A polycyclic code $C$ associated with
trinomial $x^{2m}+a^{2}x^{2i}+b^{2}$ and generated by $g(x)=x^{m}+ax^{i}+b$ is
isodual, where $i\mid m$.

###### Proof.

Let $m=ki$, according to Propsition 4.1, we consider the polycyclic code
generated by $g(x)=x^{k}+ax+b$. Using row operations, we can change $G$ to the
following standard form

$\displaystyle G$
$\displaystyle=\left(\begin{array}[]{cccc|cccc}1&0&\cdots&0&\,b^{-1}+(b^{-1}a)^{k}&b^{-2}a&\cdots&b^{-1}(b^{-1}a)^{k-1}\\\
0&1&\cdots&0&\,(b^{-1}a)^{k-1}&b^{-1}&\cdots&b^{-1}(b^{-1}a)^{k-2}\\\
\vdots&\vdots&\ddots&\vdots&\,\vdots&\vdots&\ddots&\vdots\\\
0&0&\cdots&1&\,b^{-1}a&0&\cdots&b^{-1}\end{array}\right)=(I_{k}\mid A).$

Then the parity-check matrix of $C$ is

$\displaystyle H$
$\displaystyle=\left(\begin{array}[]{cccc|cccc}b^{-1}+(b^{-1}a)^{k}&(b^{-1}a)^{k-1}&\cdots&b^{-1}a&\,1&0&\cdots&0\\\
b^{-2}a&b^{-1}&\cdots&0&\,0&1&\cdots&0\\\
\vdots&\vdots&\ddots&\vdots&\,\vdots&\vdots&\ddots&\vdots\\\
b^{-1}(b^{-1}a)^{k-1}&b^{-1}(b^{-1}a)^{k-2}&\cdots&b^{-1}&\,0&0&\cdots&1\end{array}\right)=(A^{T}\mid
I_{k}).$

We give the following four steps:

Step 1. Reducing the power of $b$ to nonnegative in $A$ and $A^{T}$ by doing
row operations, respectively, we gain $A_{1}$, $B_{1}$.

$A_{1}=\begin{pmatrix}\begin{array}[]{cccccc}b^{k-1}+a^{k}&b^{k-2}a&b^{k-3}a^{2}&\cdots&ba^{k-2}&a^{k-1}\\\
a^{k-1}&b^{k-2}&b^{k-3}a&\cdots&ba^{k-3}&a^{k-2}\\\
a^{k-2}&0&b^{k-3}&\cdots&ba^{k-4}&a^{k-3}\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\ a^{2}&0&0&\cdots&b&a\\\
a&0&0&\cdots&0&1\end{array}\end{pmatrix},$
$B_{1}=\begin{pmatrix}\begin{array}[]{cccccc}b^{k-1}+a^{k}&ba^{k-1}&b^{2}a^{k-2}&\cdots&b^{k-2}a^{2}&b^{k-1}a\\\
a&b&0&\cdots&0&0\\\ a^{2}&ba&b^{2}&\cdots&0&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
a^{k-2}&ba^{k-3}&b^{2}a^{k-4}&\cdots&b^{k-2}&0\\\
a^{k-1}&ba^{k-2}&b^{2}a^{k-3}&\cdots&b^{k-2}a&b^{k-1}\end{array}\end{pmatrix}.$

Step 2. Swaping the $i$-th ($i\geq 2$) row of $A_{1}$ with the $(k-i+2)$-th
row of $A_{1}$, we gain $A_{2}$.

$A_{2}=\begin{pmatrix}\begin{array}[]{cccccc}b^{k-1}+a^{k}&b^{k-2}a&b^{k-3}a^{2}&\cdots&ba^{k-2}&a^{k-1}\\\
a&0&0&\cdots&0&1\\\ a^{2}&0&0&\cdots&b&a\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
a^{k-2}&0&b^{k-3}&\cdots&ba^{k-4}&a^{k-3}\\\
a^{k-1}&b^{k-2}&b^{k-3}a&\cdots&ba^{k-3}&a^{k-2}\end{array}\end{pmatrix}.$

Step 3. Each column of $A_{2}$ and $B_{1}$ is multiplied by the inverse of its
greatest common factor, respectively, we gain $A_{3},B_{2}$.

$A_{3}=\begin{pmatrix}\begin{array}[]{cccccc}b^{k-1}+a^{k}&a&a^{2}&\cdots&a^{k-2}&a^{k-1}\\\
a&0&0&\cdots&0&1\\\ a^{2}&0&0&\cdots&1&a\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
a^{k-2}&0&1&\cdots&a^{k-4}&a^{k-3}\\\
a^{k-1}&1&a&\cdots&a^{k-3}&a^{k-2}\end{array}\end{pmatrix},$
$B_{2}=\begin{pmatrix}\begin{array}[]{cccccc}b^{k-1}+a^{k}&a^{k-1}&a^{k-2}&\cdots&a^{2}&a\\\
a&1&0&\cdots&0&0\\\ a^{2}&a&1&\cdots&0&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
a^{k-2}&a^{k-3}&a^{k-4}&\cdots&1&0\\\
a^{k-1}&a^{k-2}&a^{k-3}&\cdots&a&1\end{array}\end{pmatrix}.$

Step 4. Swaping the $i$-th ($i\geq 2$) column of $A_{3}$ with the $(k-i+2)$-th
column of $A_{3}$, we gain $A_{4}$.

$A_{4}=\begin{pmatrix}\begin{array}[]{cccccc}b^{k-1}+a^{k}&a^{k-1}&a^{k-2}&\cdots&a^{2}&a\\\
a&1&0&\cdots&0&0\\\ a^{2}&a&1&\cdots&0&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
a^{k-2}&a^{k-3}&a^{k-4}&\cdots&1&0\\\
a^{k-1}&a^{k-2}&a^{k-3}&\cdots&a&1\end{array}\end{pmatrix}.$

After the above steps, $A$ becomes $A_{4}$ and $A^{T}$ becomes $B_{2}$, and we
have $A_{4}=B_{2}$, which means there exist $S$, $M$, such that $SGM=H$. Thus
$C$ is isodual. ∎

######  Theorem 4.3.

Over $GF(q)$, where $q$ is a power of $2$. A polycyclic code $C$ associated
with trinomial $x^{2m}+a^{2}x^{2i}+b^{2}$ and generated by
$g(x)=x^{m}+ax^{i}+b$ is isodual, where $i\nmid m$.

###### Proof.

Let $m=ki+t\>(1\leq t<i)$. Similar to the proof of Theorem 4.2, we discuss it
by dividing it into the following four steps.

Step 1. Reducing the power of $b$ to nonnegative in $A$ and $A^{T}$ as we do
in Case 1, respectively, we gain $A_{1},B_{1}$.

Step 2. Each column of $A_{1}$ and $B_{1}$ is multiplied by the inverse of its
greatest common factor as we do in Case 1, respectively, we gain
$A_{2},B_{2}$.

Step 3. Switching the first $i$ rows of $A_{2}$ with the center axis of this
$i$ rows, and do the same for the last $(k-1)i+t$ rows, we gain $A_{3}$.

Step 4. Switching the first $i$ columns of $A_{3}$ with the center axis of
them, and do the same for the last $(k-1)i+t$ columns, we gain $A_{4}$.

Therefore, we also have $A_{4}=B_{2}$, which means there exist $S,M$, such
that $SGM=H$, and $C$ is isodual. ∎

Here, we give an examples to illustrate the steps in Theorem 4.3.

######  Example 4.1.

Considering the polycyclic code $C$ generated by $g(x)=x^{5}+ax^{2}+b\mid
x^{10}+a^{2}x^{4}+b^{2}$ over $\mathbb{F}_{2^{m}}$, where $2\nmid 5$. We have
the generator matrix of the form

$G=\begin{pmatrix}\begin{array}[]{ccccc|ccccc}b&0&a&0&0&1&0&0&0&0\\\
0&b&0&a&0&0&1&0&0&0\\\ 0&0&b&0&a&0&0&1&0&0\\\ 0&0&0&b&0&a&0&0&1&0\\\
0&0&0&0&b&0&a&0&0&1\\\ \end{array}\end{pmatrix},$
$RG=\begin{pmatrix}\begin{array}[]{ccccc|ccccc}1&0&0&0&0&b^{-1}&b^{-3}a^{3}&b^{-2}a&0&b^{-3}a^{2}\\\
0&1&0&0&0&b^{-2}a^{2}&b^{-1}&0&b^{-2}a&0\\\
0&0&1&0&0&0&b^{-2}a^{2}&b^{-1}&0&b^{-2}a\\\ 0&0&0&1&0&b^{-1}a&0&0&b^{-1}&0\\\
0&0&0&0&1&0&b^{-1}a&0&0&b^{-1}\\\ \end{array}\end{pmatrix}=(I_{5}\mid A).$

Let $H$ have the form

$H=\begin{pmatrix}\begin{array}[]{ccccc|ccccc}b^{-1}&b^{-2}a^{2}&0&b^{-1}a&0&1&0&0&0&0\\\
b^{-3}a^{3}&b^{-1}&b^{-2}a^{2}&0&b^{-1}a&0&1&0&0&0\\\
b^{-2}a&0&b^{-1}&0&0&0&0&1&0&0\\\ 0&b^{-2}a&0&b^{-1}&0&0&0&0&1&0\\\
b^{-3}a^{2}&0&b^{-2}a&0&b^{-1}&0&0&0&0&1\\\
\end{array}\end{pmatrix}=(A^{T}\mid I_{5}).$

After Step 1 and Step 2, we have

$A_{2}=\begin{pmatrix}b^{2}&a^{3}&a&0&a^{2}\\\ a^{2}&b&0&a&0\\\
0&a^{2}&1&0&a\\\ a&0&0&1&0\\\ 0&a&0&0&1\\\
\end{pmatrix},\;\;B_{2}=\begin{pmatrix}b&a^{2}&0&a&0\\\
a^{3}&b^{2}&a^{2}&0&a\\\ a&0&1&0&0\\\ 0&a&0&1&0\\\ a^{2}&0&a&0&1\\\
\end{pmatrix}.$

After Step 3, we can get

$A_{3}=\begin{pmatrix}a^{2}&b&0&a&0\\\ b^{2}&a^{3}&a&0&a^{2}\\\ 0&a&0&0&1\\\
a&0&0&1&0\\\ 0&a^{2}&1&0&a\\\
\end{pmatrix},\;\;B_{2}=\begin{pmatrix}b&a^{2}&0&a&0\\\
a^{3}&b^{2}&a^{2}&0&a\\\ a&0&1&0&0\\\ 0&a&0&1&0\\\ a^{2}&0&a&0&1\\\
\end{pmatrix}.$

After Step 4, we can get

$A_{4}=\begin{pmatrix}b&a^{2}&0&a&0\\\ a^{3}&b^{2}&a^{2}&0&a\\\ a&0&1&0&0\\\
0&a&0&1&0\\\ a^{2}&0&a&0&1\\\
\end{pmatrix},\;\;B_{2}=\begin{pmatrix}b&a^{2}&0&a&0\\\
a^{3}&b^{2}&a^{2}&0&a\\\ a&0&1&0&0\\\ 0&a&0&1&0\\\ a^{2}&0&a&0&1\\\
\end{pmatrix}.$

Combining all the above steps, there exist $R^{\prime}$ and $M$ such that
$R^{\prime}GM=H$, which means $g(x)$ would generate an isodual polycyclic
code.

In [2], the authors show that if $g^{2}(x)=x^{n}-ax^{i}-b$, then over
$\mathbb{F}_{q}$, where $q$ is a power of an odd prime, $g(x)=x^{n/2}\pm c$
with $i=n/2$, $-a=\pm 2c$ and $b=-c^{2}$, and over $\mathbb{F}_{q}$, where $q$
is a power of 2, $g(x)=x^{n/2}+a^{\prime}x^{i/2}+b^{\prime}$ with
$(a^{\prime})^{2}=-a$ and $(b^{\prime})^{2}=-b$. Combining the results of
Theorems 4.1, 4.2, and 4.3, we give Theorem 4.4, which is also Conjecture 4.1
in [2]

######  Theorem 4.4.

(Conjecture 4.1 in [2]) If a polycyclic code $C$ is generated by $g(x)\mid
x^{n}-ax^{i}-b$ and $g^{2}(x)=x^{n}-ax^{i}-b$, then $C$ is isodual.

Theorem 4.4 gives a simple way to construct isodual polycyclic codes, but it’s
not the only way. Example 4.2 illustrates this, which is also a counter-
example of Conjecture 4.2 in [2]. Here is the content of Conjecture 4.2.
(denotes it as Conjecture 4.1.)

######  Conjecture 4.1.

(Conjecture 4.2 in [2]) Over $\mathbb{F}_{2}$, a polycyclic code $C$ generated
by $g(x)\mid x^{n}-ax^{i}-b$ is isodual if and only if
$g^{2}(x)=x^{n}-ax^{i}-b$.

The sufficiency of Conjecture 4.1 can be regarded as a direct corollary
(denoted as Corollary 4.1) of Theorem 4.4. As for the necessity of Conjecture
4.1, it is not correct as the following counter example shows.

######  Corollary 4.1.

Over $\mathbb{F}_{2}$, a polycyclic code $C$ generated by $g(x)\mid
x^{n}-x^{i}-1$ is isodual if $g^{2}(x)=x^{n}-x^{i}-1$.

######  Example 4.2.

Let $C$ be a polycyclic code generated by $g(x)\mid x^{20}+x^{10}+1$ over
$\mathbb{F}_{2}$, where

$\displaystyle x^{20}+x^{10}+1$
$\displaystyle=(x^{2}+x+1)^{2}(x^{4}+x+1)^{2}(x^{4}+x^{3}+1)^{2},$
$\displaystyle g(x)$ $\displaystyle=(x^{2}+x+1)(x^{4}+x+1)^{2},$
$\displaystyle g^{2}(x)$ $\displaystyle\neq x^{20}+x^{10}+1.$

Using Magma, $g(x)$ can also generate an isodual polycyclic code.

##  5 The Structure of Self-dual Polycyclic Codes

This section is devoted to solving Conjecture 4.7 in [2] (see Theorem 5.1 in
this section). Let us first give the following lemma and proposition.

######  Lemma 5.1 (Lemma 4.6 in [2]).

Let $g(x)=a_{0}+a_{1}x+\cdots+a_{k}x^{k}\mid x^{n}-ax^{i}-b$ be the generator
polynomial of a polycyclic code $C$ of length $n$, and let
$e=Ord(x^{n}-ax^{i}-b)$ and
$h(x)=b_{0}+b_{1}x+\cdots+b_{e-k}x^{e-k}=\dfrac{x^{e}-1}{g(x)}$. Then a
parity-check matrix $H$ of $C$ is given by

$H=\begin{pmatrix}\begin{array}[]{ccccccc}b_{e-k}&b_{e-k-1}&b_{e-k-2}&\cdots&\cdots&\cdots&\cdots\\\
0&b_{e-k}&b_{e-k-1}&b_{e-k-2}&\cdots&\cdots&\cdots\\\
\vdots&\ddots&\ddots&\ddots&\ddots&\ddots&\vdots\\\
0&\cdots&0&b_{e-k}&b_{e-k-1}&b_{e-k-2}&\cdots\\\
\end{array}\end{pmatrix}_{k\times n},$

which is a submatrix of the parity check matrix of the cyclic code of length
$e$ generated by $g(x)\mid x^{e}-1$.

######  Proposition 5.1.

Let the monic polynomial $g(x)\in\mathbb{F}_{q}[x]$ with $\deg(g(x))=k>0$. If
both $g(x)\mid x^{2k}+1$ and $g(x)\mid x^{2k}-bx^{i}-c$, where $b,c\neq 0$,
then $k=i$, $c+1\neq 0$ and $g(x)=x^{k}-a$, where $a\neq 0$.

###### Proof.

If $c+1=0$, then $x^{2k}-bx^{i}+1=(x^{2k}+1)-bx^{i}$, which means
$\gcd(x^{2k}+1,x^{2k}-bx^{i}+1)=\gcd(x^{2k}+1,x^{i})=1$. Since $g(x)\mid 1$,
which means $\deg(g(x))=0$, which is a contradiction. Hence $c+1\neq 0$, we
have

$\displaystyle x^{2k}-bx^{i}-c$ $\displaystyle\equiv-
bx^{i}+\alpha\;(\bmod\;x^{2k}+1),$

where $\alpha=-(c+1)\neq 0$. Since $g(x)\mid\gcd(x^{2k}-bx^{i}-c,x^{2k}+1)$,
we have $g(x)\mid\gcd(x^{2k}+1,-bx^{i}+\alpha)$, i.e. $i\geq k$. If $2k>i>k$,
let $2k=i+t$ ($0<t<k$), then

$x^{2k}+1\equiv\alpha b^{-1}x^{t}+1\;(\bmod\;-bx^{i}+\alpha).$

We have $\gcd(x^{2k}+1,-bx^{i}+\alpha)=\gcd(-bx^{i}+\alpha,\alpha
b^{-1}x^{t}+1)$, then $g(x)\mid\gcd(-bx^{i}+\alpha,\alpha b^{-1}x^{t}+1)$,
which means $\deg(\alpha b^{-1}x^{t}+1)=t\geq k$, which is a contradiction.
Therefore, $k=i$. If $-bx^{k}+\alpha\mid x^{2k}+1$, since $\deg(g(x))=k$, then
$g(x)=\gcd(x^{2k}+1,-bx^{k}+\alpha)=x^{k}-b^{-1}\alpha$. If
$-bx^{k}+\alpha\nmid x^{2k}+1$, then

$x^{2k}+1\equiv\gamma\;(\bmod\;-bx^{k}+\alpha),$

where $\gamma\neq 0$. Therefore,
$\gcd(x^{2k}-bx^{i}-c,x^{2k}+1)=\gcd(-bx^{k}+\alpha,\gamma)=1$, which means
$g(x)\mid 1$ and $\deg(g(x))=0$. Therefore, $g(x)=x^{k}-a$, where
$a=b^{-1}\alpha=-(c+1)b^{-1}\neq 0$. ∎

The generators of self-dual polycyclic codes have the same form, which can be
easily seen in the following theorem.

######  Theorem 5.1.

(Conjecture 4.7 in [2]) A polycyclic code $C$ generated by the monic
polynomial $g(x)\mid x^{n}-bx^{i}-c$ is self-dual if and only if
$g(x)=x^{k}-a\mid x^{n}-bx^{k}-c$, where $a^{2}=-1$,
$\operatorname{\text{Ord}}(x^{n}-ax^{i}-b)\geq 3k$ and $n=2k$. Also, the
minimum distance of all self-dual polycyclic codes is $2$ and they are all
actually constacyclic codes.

###### Proof.

We first prove the necessity. Let
$\operatorname{\text{Ord}}(x^{n}-bx^{i}-c)=e$ and
$g(x)=\sum_{i=0}^{k}g_{i}x^{i}$, where $\deg(g(x))=k$ and $g_{k}=1$. Since $C$
is self-dual, we need $n-k=k$, i.e. $n=2k$. As $G$ has the form in Theorem
2.1, and let the parity-check matrix of $C$ be $H$, which has the form in
Lemma 5.1. Here, let $h(x)=\sum_{i=0}^{e-k}b_{i}x^{i}$, then
$h(x)g(x)=x^{e}-1$, which means $b_{e-k}=1$ and $b_{0}g_{0}=-1$. Let
$G_{1}=g_{0}^{-1}G$. In fact, $G_{1}$ and $H$ has the form

$G_{1}=\begin{pmatrix}\begin{array}[]{ccccc|ccccc}1&g_{0}^{-1}g_{1}&g_{0}^{-1}g_{2}&\cdots&g_{0}^{-1}g_{k-1}&g_{0}^{-1}&{}\hfil&{}\hfil&{}\hfil&{}\hfil\\\
{}\hfil&1&g_{0}^{-1}g_{1}&\cdots&g_{0}^{-1}g_{k-2}&g_{0}^{-1}g_{k-1}&g_{0}^{-1}&{}\hfil&{}\hfil&{}\hfil\\\
{}\hfil&{}\hfil&\ddots&\vdots&\vdots&\vdots&\vdots&\ddots&{}\hfil&{}\hfil\\\
{}\hfil&{}\hfil&{}\hfil&\ddots&\vdots&\vdots&\vdots&\ddots&\ddots&{}\hfil\\\
{}\hfil&{}\hfil&{}\hfil&{}\hfil&1&g_{0}^{-1}g_{1}&g_{0}^{-1}g_{2}&\cdots&\cdots&g_{0}^{-1}\\\
\end{array}\end{pmatrix}_{k\times 2k},$
$H=\begin{pmatrix}\begin{array}[]{cccc|cccc}1&b_{e-k-1}&b_{e-k-2}&\cdots&\cdots&\cdots&\cdots&\cdots\\\
0&1&b_{e-k-1}&b_{e-k-2}&\cdots&\cdots&\cdots&\cdots\\\
\vdots&\ddots&\ddots&\ddots&\ddots&\ddots&\ddots&\vdots\\\
0&\cdots&0&1&b_{e-k-1}&b_{e-k-2}&\cdots&b_{e-2k}\\\
\end{array}\end{pmatrix}_{k\times 2k}.$

Assume that $W_{1}G_{1}=(I_{k}\mid G_{2})$ and $W_{2}H=(I_{k}\mid H_{1})$,
where $W_{1},W_{2}$ are the product of elementary matrices, respectively. In
fact, let $g_{j}^{\prime}=g_{0}^{-1}g_{j}$, we have

$W_{1}=\prod_{r=2}^{k}A_{r},\;\;W_{2}=\prod_{r=2}^{k}B_{r}.$

Here, change the $r$-th column of $I_{k}$ to
$(-g_{r-1}^{\prime},-g_{r-2}^{\prime},\cdots,-g_{1}^{\prime},1,0,\cdots,0)^{T}$,
we get $A_{r}$, and change the $r$-th column of $I_{k}$ to
$(-b_{e-k-r+1},-b_{e-k-r+2},\cdots,-b_{e-k-1},1,0,\cdots,0)^{T}$, we get
$B_{r}$, where $2\leq r\leq k$. Note that $C=C^{\perp}$ if and only if
$W_{1}G_{1}=W_{2}H$. Checking the $k$-th row of $W_{1}G_{1}$ and $W_{2}H$, we
would have

$b_{e-k-r}=g_{r}^{\prime},{\rm where}\>1\leq r\leq k.$

which means $W_{1}=W_{2}$, then $C$ is self-dual if and only if $G_{1}=H$. Let
$h(x)=h_{H}(x)+h_{J}(x)$, where $h_{H}(x)=g_{0}^{-1}x^{e-2k}g^{*}(x)$ and
$\deg(h_{J}(x))\leq e-3k$. If $e-3k<0$, since $G_{1}=H$, we have $h(0)=0$,
which contradicts $h(x)g(x)=x^{e}-1$. Therefore, $e\geq 3k$. As
$g(x)h(x)=x^{e}-1$, we have

$x^{e}-1=h(x)g(x)=g_{0}^{-1}x^{e-2k}g^{*}(x)g(x)+h_{J}(x)g(x).$

Here, $e-2k\geq\deg(h_{J}(x)g(x))$. If $\deg(h_{J}(x)g(x))\neq e-2k$, which
means there will leave the term $x^{e-2k}$. Hence, $\deg(h_{J}(x)g(x))=e-2k$,
i.e. $h_{J}(x)g(x)=-x^{e-2k}-1$. Therefore, we have

$x^{2k}+1=g_{0}^{-1}g^{*}(x)g(x).$

Since $g(x)\mid x^{2k}-bx^{i}-c$, by Proposition 5.1, we have $i=k$ and
$g(x)=x^{k}-a$. Then

$-a(x^{2k}+1)=g(x)g^{*}(x)=(x^{k}-a)(-ax^{k}+1)=-a(x^{2k}+1)+(1+a^{2})x^{k},$

i.e. $a^{2}=-1$. Then $g(x)$ has the form $g(x)=x^{k}-a$, where $a^{2}=-1$.

Now we prove the sufficiency. Since $g(x)=x^{k}-a$ and $a^{2}=-1$, the
generator matrix $G$ of $C$ has the form $G=(-aE_{k}\mid E_{k})$, and
$aG=(E_{k}\mid aE_{k})$, then the parity-check matrix of $C$ is
$H=(-aE_{k}\mid E_{k})$, which means $G=H$, then $C$ is self-dual.
Furthermore, since the first and $(k+1)$-th columns of $H$ are linearly
dependent, by Theorem 2.2, the minimum distance of $C$ is $2$. Since
$g(x)g^{*}(x)=g(0)(x^{2k}+1)$, i.e. $g(x)\mid x^{2k}+1$, which means $C$ is
actually a constacyclic code since $x^{2k}+1$ is also an associate polynomial
of $C$. ∎

######  Example 5.1.

Over $GF(5)$, a polycyclic code of length $4$ generated by $g(x)=x^{2}+2\mid
x^{4}+4x^{2}+4$ is self-dual.

######  Example 5.2.

Over $GF(5)$, a polycyclic code of length $4$ generated by $g(x)=x^{2}-2\mid
x^{4}-4x^{2}+4$ is self-dual.

##  6 Self-orthogonal and Dual-containing Polycyclic Codes

In [2], the authors provided two cases in which self-orthogonal or dual-
containing polycyclic codes do not exist. This section is devoted to
discussing Conjecture 4.10 and Conjecture 4.11 in [2]. We will prove that
Conjecture 4.10 in [2] is correct (see Theorem 6.1), while we will prove that
Conjecture 4.11 is wrong (see Conjecture 6.1) by giving a counter example
(i.e. Example 6.1).

######  Proposition 6.1.

Over $\mathbb{F}_{2}$, $\gcd(x^{n}+1,x^{n}+x^{i}+1)=1$.

###### Proof.

In fact, using the Euclidean algorithm, we have

$\displaystyle x^{n}+x^{i}+1$ $\displaystyle=(x^{n}+1)\cdot 1+x^{i},$
$\displaystyle x^{n}+1$ $\displaystyle=x^{i}\cdot x^{n-i}+1,$ $\displaystyle
x^{i}$ $\displaystyle=1\cdot x^{i}.$

Therefore, $\gcd(x^{n}+1,x^{n}+x^{i}+1)=\gcd(x^{n}+1,x^{i})=\gcd(x^{i},1)=1$.
∎

######  Theorem 6.1.

(Conjecture $4.10$ in [2]) There are no self-dual or self-orthogonal
polycyclic codes or nontrivial dual-containing polycyclic codes over the
binary field, where these codes are associated with trinomials.

###### Proof.

(i) For a self-dual polycylic code $C$, it has been proved in [2], so we will
not repeat it.

(ii) For a self-orthogonal polycyclic code $C$, let $C$ be generated by
$g(x)=\sum_{i=0}^{n-k}g_{i}x^{i}\mid x^{n}+x^{i}+1$ with $g_{n-k}=1$. Let $G$
be the generator matrix of $C$, which has the form in Theorem 2.1. If we
denote the $t$-th row of $G$ as $\bm{\delta}_{t}$ for $\>1\leq t\leq k$, since
$C$ is self-orthogonal, then we have

$\bm{\delta}_{1}\cdot\bm{\delta}_{1}^{T}=0\Leftrightarrow
g_{0}^{2}+g_{1}^{2}+\cdots+g_{n-k}^{2}=0,$

which means that $g(x)$ has even terms. Since $g(x)\mid x^{n}+x^{i}+1$, let
$x^{n}+x^{i}+1=g(x)f(x)$, hence $g(x)f(x)$ also has even terms, even if there
are some terms will cancel each other out as cancelling terms will come in
pairs over $\mathbb{F}_{2}$. Then it leads to a contradiction because
$x^{n}+x^{i}+1$ has odd terms. So there are no self-orthogonal polycyclic
codes associated with trinomials.

(iii) For a dual-containing polycyclic code $C$, let $C$ be generated by
$g(x)=\sum_{r=0}^{s}g_{r}x^{r}\mid x^{n}-x^{i}-1$, where $g_{s}=1$. We denote
the $r$-th row of the generator matrix $G$ by $\bm{\delta}_{r}$ as we do in
(ii). Here, $g(x)$ has odd terms since $x^{n}-ax^{i}-b$ has odd terms. Let
$h(x)g(x)=x^{e}-1$, where $e=\operatorname{\text{Ord}}(x^{n}-ax^{i}-b)$ and
$h(x)=\sum_{r=0}^{e-s}h_{r}x^{r}$, thus the parity-check matrix $H$ of $C$ has
the form in Lemma 5.1. Let $\bm{\beta}_{r}$ be the $r$-th row of $H$, since
$C$ is dual-containing, then we have $\bm{\beta}_{r}\cdot\bm{\beta}_{r}^{T}=0$
for $\>1\leq r\leq s$. Checking $\bm{\beta}_{d}$ and $\bm{\beta}_{d+1}$, we
find that

$\bm{\beta}_{d+1}\cdot\bm{\beta}_{d+1}^{T}+h_{e-s-n+d}^{2}=\bm{\beta}_{d}\cdot\bm{\beta}_{d}^{T}\Rightarrow
h_{e-s-n+d}=0,\;{\rm where}\>1\leq d\leq s-1.$

Then we can simplify $H$ to the following form

$H=\begin{pmatrix}\begin{array}[]{ccccccccc}h_{e-s}&h_{e-s-1}&\cdots&\cdots&h_{e-n}&0&\cdots&\cdots&0\\\
0&h_{e-s}&h_{e-s-1}&\cdots&\cdots&h_{e-n}&0&\cdots&0\\\
\vdots&\ddots&\ddots&\ddots&\ddots&\ddots&\ddots&\ddots&\vdots\\\
0&\cdots&0&h_{e-s}&h_{e-s-1}&\cdots&\cdots&\cdots&h_{e-n}\\\
\end{array}\end{pmatrix}_{s\times n}.$

Here, $n-s\geq s$. Let the first non-zero element of the consequence
$h_{e-n},\cdots,h_{e-s}$ are recorded as $h_{u}$. Since
$g_{0}=g_{s}=h_{e-s}=1$ and $C^{\perp}\subseteq C$. There are three cases to
be discussed.

Case 1. If $u>e-2s$, then $\bm{\beta}_{1}$ cannot be represented as a linear
combination of $\bm{\delta}_{1},\cdots,\bm{\delta}_{n-s}$.

Case 2. If $u=e-2s$, then $\bm{\delta}_{1}=\bm{\beta}_{1}$, then
$\bm{\delta}_{1}\bm{\delta}_{1}^{T}=0$, which means $g(x)$ has even terms.

Case 3. If $e-n\leq u<e-2s$, let $h(x)=h_{H}(x)+h_{J}(x)$, where
$h_{H}(x)=x^{u}(\sum_{k=u}^{e-s}h_{k}x^{k-u})$ and $deg(h_{J}(x))\leq e-s-n$.
Here we need $e\geq s+n>3s$, or $h(0)=0$, which leads to a contradiction since
$h(x)g(x)=x^{e}-1$. Since we can write
$\bm{\beta}_{1}=\sum_{k=1}^{e-2s-u+1}c_{k}\bm{\delta}_{k}$, where
$c_{1}=c_{e-2s-u+1}=1$, we have
$(x^{-u}h_{H}(x))^{*}=(\sum_{k=1}^{e-2s-u+1}c_{k}x^{k-1})g(x)=(1+\cdots+x^{e-2s-u})g(x)$.
Therefore,

$\displaystyle h(x)$
$\displaystyle=x^{u}\bigg{(}\sum_{k=1}^{e-2s-u+1}c_{k}x^{k-1}\bigg{)}^{*}g^{*}(x)+h_{J}(x),$
$\displaystyle x^{e}-1=h(x)g(x)$
$\displaystyle=x^{u}(1+\cdots+x^{e-2s-u})g^{*}(x)g(x)+h_{J}(x)g(x).\;\;\;\;\;(1)$

Since $\deg(x^{u})=u\geq e-n\geq deg(h_{J}(x)g(x))$, then there will leave the
term $x^{u}$ if $u\neq e-n$ or $e-n\neq\deg(h_{J}(x)g(x))$, which means the
equation $(1)$ is wrong. If $u=e-n=\deg(h_{J}(x)g(x))$, then
$h_{J}(x)g(x)=x^{u}-1$. We have

$\displaystyle x^{e}-1$
$\displaystyle=x^{u}(1+\cdots+x^{e-2s-u})g^{*}(x)g(x)+x^{u}-1,$ $\displaystyle
x^{n}-1$ $\displaystyle=(1+\cdots+x^{e-2s-u})g^{*}(x)g(x).$

Then $g(x)\mid x^{n}-1$. Since $g(x)\mid x^{n}-x^{i}-1$, according to
Proposition 6.1, we have $g(x)=1$, which is the generator polynomial of
$\mathbb{F}_{2}^{n}$, and this is the extreme situation. Thus, there are no
nontrivial dual-containing polycyclic codes associated with trinomials. ∎

######  Conjecture 6.1.

(Conjecture 4.11 in [2]) A polycyclic code is not dual-containing if the
generator polynomial $g(x)\mid x^{n}-ax^{i}-b$, where $n$ is prime.

The following example is a counter example of Conjecture 6.1.

######  Example 6.1.

Over $\mathbb{F}_{3}$, $C$ is a polycyclic code generated by $g(x)=x+1\mid
x^{3}+x-1$, where $x^{3}+x-1=(x+1)(x^{2}-x-1)$. The generator matrix $G$ and
the parity-check matrix $H$ are

$G=\begin{pmatrix}1&1&0\\\ 0&1&1\\\
\end{pmatrix},\;\;H=\begin{pmatrix}1&2&1\\\ \end{pmatrix}.$

Since $110+011=121$, we have $C^{\perp}\subseteq C$, i.e. $C$ is dual-
containing.

##  7 Conclusion

The main contributions in this paper are as follows.

  1. (i)

We studied Conjecture 3.5 in [2] in Section 3, which is incorrect because of
Example 3.1. Example 3.1 illustrates that the polycyclic codes constructed
from polynomials of the same order do not always find a particular mapping
between them, which needs to be one-to-one and the corresponding codes are
equivalent to each other. By modifying the conditions in Conjecture 3.5, we
give a mapping $\varphi$ to connect two kinds of polycyclic codes associated
with $f(x)$ and $f^{*}(x)$ respectively, where $f(0)\neq 0$ in Theorem 3.3.
Based on equivalence, we only need to study polycyclic codes associate with
$f(x)$ to get the structure of polycyclic codes associate with $f^{*}(x)$.

  2. (ii)

We prove Conjecture 4.1 and Conjecture 4.5 in [2], and give Example 4.2 to
show that Conjecture 4.2 in [2] is wrong in Section 4. We study isodual
polycyclic codes associated with trinomials in this section, and give a useful
way to construct isodual polycyclic codes, which are generated by $g(x)$ and
associated with $g^{2}(x)$ in Theorem 4.4 (corresponds to Conjecture 4.1 and
4.5 in [2]). However, in Example 4.2, we give an isodual polycyclic code
generated by $g(x)\mid x^{n}+x^{i}+1$ with $g^{2}(x)\neq x^{n}+x^{i}+1$ over
$\mathbb{F}_{2}$, which means $g^{2}(x)=x^{n}+x^{i}+1$ is just a necessary
condition of isodual polycyclic codes over $\mathbb{F}_{2}$, and over
$\mathbb{F}_{q}$.

  3. (iii)

We prove Conjecture 4.7 in [2] in Section 5. We show the unique form of
generator of self-dual polycyclic codes associated with trinomials, that is
$x^{k}-a$ and $a^{2}=-1$ over $\mathbb{F}_{q}$ in Theorem 5.1 (corresponds to
Conjecture 4.7 in [2]). And the minimum distance of all self-dual polycyclic
codes is $2$. Theorem 5.1 also shows that there is only one way to construct a
polycyclic code associated with trinomials.

  4. (iii)

We discussed Conjecture 4.10 and Conjecture 4.11 in Section 6. Theorem 6.1
(corresponds to Conjecture 4.10 in [2]) illustrates that it is impossible to
find self-dual, self-orthogonal or nontrivial dual-containing polycyclic codes
associated with trinomials over $\mathbb{F}_{2}$. Example 6.1 is a counter
example of Conjecture 4.11 in [2], which means that there exist prime length
polycyclic codes associated with trinomials over $\mathbb{F}_{q}$.

To sum up, we have solved all the conjectures in [2].

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