text:
\begin{document}

\maketitle

\begin{abstract}
\textsf{A recently proposed version of mimetic gravity incorporates a limiting curvature into general relativity by means of a specific potential depending on the d'Alembertian of the scalar field. In the homogeneous and isotropic setting, the resulting theory agrees with the simplest incarnation of effective loop quantum cosmology (LQC) once the limiting curvature is identified with a multiple of the Planck scale. In this paper, we answer the question of whether such a relation can hold in the context of Bianchi I models. Our result is negative: it turns out to be impossible to view the Hamiltonian of limiting curvature mimetic gravity as an effective LQC Hamiltonian due to the appearance of terms that cannot be supported on the polymer Hilbert space underlying LQC. The present analysis complements a related result in the context of spherical symmetry. }
\end{abstract}

\section{Introduction}

Mimetic gravity \cite{ChamseddineMimeticDarkMatter} has been proposed as a modified version of general relativity incorporating cold dark matter, see \cite{SebastianiMimeticGravityA} for an overview. A specifically interesting variant \cite{ChamseddineResolvingCosmologicalSingularities} incorporates a limiting curvature by means of a specific potential depending on the d'Alembertian of a scalar field, the only field entering the action beyond the metric. While the form of the action was argued to be motivated by non-commutative geometry in \cite{ChamseddineResolvingCosmologicalSingularities}, the potential was chosen by hand. However, by relating the resulting theory to homogeneous and isotropic loop quantum cosmology, see \cite{BodendorferCanonicalStructureOf, NouiEffectiveLoopQuantum}, one can make a simplicity argument \cite{BodendorferCanonicalStructureOf} to obtain the potential from quantum mechanics.

Given this situation, it is interesting to ask whether such a relation can hold more generally, i.e. by dropping the symmetry assumptions of homogeneity or isotropy. In particular, it was stressed in \cite{BodendorferCanonicalStructureOf, NouiEffectiveLoopQuantum} that one could then view limiting curvature mimetic gravity or generalised versions of it as effective actions for loop quantum gravity. This is on the one hand particularly interesting for applications of loop quantum gravity to more complicated systems such as black holes, where working directly within the full theory might be too involved computationally. On the other hand, it may shed light on the ongoing debate on possible modifications of general covariance in loop quantum gravity, see e.g. \cite{BojowaldEffectiveLineElements}, as mimetic gravity is manifestly covariant.

The most convenient setting to check the desired relation are Bianchi I cosmological models which are homogeneous, but non-isotropic. Here, the cosmological solutions in \cite{ChamseddineResolvingCosmologicalSingularities} match those of loop quantum cosmology \cite{GuptQuantumGravitationalKasner} qualitatively and thus suggest that equivalence between the two theories may hold for a specific (polymerisation) choice in the loop quantum cosmology dynamics. It turns out that we are able to establish a no-go result based on an analytic computation as follows: we first bring limiting curvature mimetic gravity in a form suitable to be compared with loop quantum cosmology via a variable transformation. By expressing the mimetic gravity Hamiltonian in terms of those variables, we can check whether the resulting expression can be supported as an operator on the Hilbert space underlying loop quantum cosmology. This turns out to be impossible for the potential chosen in \cite{ChamseddineResolvingCosmologicalSingularities}, but also for general potentials by a simple argument.
Our analysis complements a related computation that has been performed in the context of spherical symmetry \cite{AchourNonSingularBlack}.

This paper is organised as follows:\\
Section \ref{sec:Review} reviews previous work and introduces the relevant concepts. The main computation is done in section \ref{sec:Computation}. A brief conclusion is given in section \ref{sec:Conclusion} and a computation omitted in the main part is detailed in the appendix.

\section{Review} \label{sec:Review}

\subsection{Conventions}

In what follows, we will use the Einstein sum convention for repeated indices when covariant and contravariant indices are contracted, while no sum is meant if the repeated indices are all up or down. The signature of the metric is $(-, +, +, +)$. We set $8 \pi G = 1$. For simplicity, we choose an orientation of the spatial slice so that all scale factors are positive. We will restrict here to a 3-torus topology of the spatial slices with unit coordinate volume for simplicity, but all our arguments apply more generally by taking into account fiducial cells. We denote spatial tensor indices by $a,b$ when no symmetry assumptions are present and by $i,j$ in the context of Bianchi I.

\subsection{Limiting curvature mimetic gravity: Canonical structure}

The class of models leading to limiting curvature mimetic gravity is defined by the action \cite{ChamseddineResolvingCosmologicalSingularities}
\be
	S = \int d^4x \, \sqrt{-g} \left( \frac{1}{2} R + \frac{1}{2} \lambda \left(1+ g^{\mu \nu} \partial_\mu \phi \partial_\nu \phi \right) + f (\Box \phi)\right) \label{eq:MCAction} \text{,}
\ee
where we note that $\lambda$ enforces the constraint $g^{\mu \nu} \partial_\mu \phi \partial_\nu \phi = -1$. The potential $f (\Box \phi)$ is a priori arbitrary. A detailed canonical analysis of this action was performed in \cite{BodendorferCanonicalStructureOf}, see also \cite{KlusonCanonicalAnalysisOf} for previous work. It is convenient to first bring the action into the more suitable form
\begin{align}
	S &= \int d^4x \, \sqrt{-g} \left( \frac{1}{2} R + \frac{1}{2} \lambda \left(1+ g^{\mu \nu} \partial_\mu \phi \partial_\nu \phi \right) + f (\chi) + \beta (\chi - \Box \phi)\right)
\end{align}
by introducing the auxiliary fields $\beta$ and $\chi$. The resulting canonical theory is rather cumbersome and lengthy to state explicitly, which is why we will only display the most relevant parts here.

The Hamiltonian is, as usual, a sum of constraints, most notably the Hamiltonian constraint
\begin{align}\label{CMH}
\mathcal{H} =&\; \frac{2}{\sqrt{q}} P^{a b} P^{c d} \left( q_{ac} q_{bd} - \frac{1}{2} q_{a b} q_{cd} \right) - \frac{1}{2} \sqrt{q} R^{(3)} - \frac{1}{2} \lambda \sqrt{q} \left( 1 + q^{a b} \partial_a \phi \partial_b \phi - \frac{P^2_\beta}{q} \right)
\notag
\\
&-\sqrt{q} \left(f(\chi) + \beta \chi\right)-\sqrt{q} q^{a b} \partial_a \beta \partial_b \phi - \frac{1}{\sqrt{q}} P_\phi P_\beta \text{,}
\end{align}
where $P_\beta$ and $P_\chi$ are the conjugate momenta of $\beta$ and $\chi$. Due to the appearance of second class constraints related to the auxiliary fields (see below), the resulting Dirac brackets are quite involved. For this paper, it will be sufficient to only consider
\begin{align}
&\;\left\{q_{ab}(x),q_{cd}(y)\right\}_*=0\;,\label{DB2}\\
&\left\{q_{ab}(x),P^{cd}(y)\right\}_*\overset{\partial_a\phi=0}{=}\delta^{(3)}(x,y)\left(\delta_{(a}^c\delta_{b)}^d-q_{ab}q^{cd}\,\frac{f''(\chi)}{3f''(\chi)-2}\right)\;,\label{DB}\\
&\left\{P^{ab}(x),P^{cd}(y)\right\}_*\overset{\partial_a\phi=0}{=}-\delta^{(3)}(x,y)\left(q^{ab}P^{cd}-q^{cd}P^{ab}\right)\frac{f''(\chi)}{3f''(\chi)-2}\;\label{DB1}\text{.}
\end{align}
where we use the notation $\{\cdot,\cdot\}_*$ for the Dirac bracket and by $\overset{\partial_a\phi=0}{=}$ we mean equality up to terms containing spatial gradients of $\phi$. This restriction was chosen in \cite{BodendorferCanonicalStructureOf} for convenience, as otherwise the Dirac bracket becomes too intricate. In the Bianchi I models considered in this paper, it is automatically satisfied.

An interesting choice for $f(\chi)$ is given by
\be\label{fCM}
f(\chi)=\chi_m^2\left(1+\frac{1}{3}\frac{\chi^2}{\chi_m^2}-\sqrt{\frac{2}{3}}\frac{\chi}{\chi_m}\arcsin\left(\sqrt{\frac{2}{3}}\frac{\chi}{\chi_m}\right)-\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}\,\right)\;.
\ee
which results in singularity-free cosmological models \cite{ChamseddineResolvingCosmologicalSingularities} and Schwarzschild black holes \cite{ChamseddineNonsingularBlackHole} with limiting curvature $\sim \chi_m^2$. This choice can also be extracted from the simplest choice of Hamiltonian in loop quantum cosmology by an inverse Legendre transform \cite{DateEffectiveActionsFrom}. We will adopt this choice in the paper for concreteness, but our argument will in the end generalise to arbitrary $f$.

\subsection{Limiting curvature mimetic gravity: Bianchi I sector}

Bianchi I models are the simplest homogeneous but non-isotropic cosmological models. In a suitable coordinate system $(t,x^i)$, the metric reads
\be\label{3metric}
q_{i j} = q_{i i} \delta_{i j}, \quad q_{i i} = a_{i}^2\;,
\ee
where the $a_i$ are the three scale factors. Similarly, we have
\be\label{PCM}
P^{i j} = P^{i i} \delta^{i j}\;.
\ee
This metric is spatially flat, i.e. $R^{(3)} = 0$. Scalar fields consistent with this symmetry only have a time dependence, i.e. $\partial_a\phi = 0$. For unit lapse, the above constraint $g^{\mu \nu} \partial_\mu \phi \partial_\nu \phi = -1$ leads to $\phi = \pm t + const$.

It is convenient to solve the second class constraints appearing in the above canonical analysis for $\beta, P_\beta$, and $\chi$ in our symmetry reduced context. Without stating their precise general form (see \cite{BodendorferCanonicalStructureOf}), we obtain
\begin{align}
&C_\lambda = 0\qquad \Rightarrow \qquad P^2_\beta = q\;,
\\
&C_\chi = 0 \qquad \Rightarrow \qquad \beta = - f'(\chi)\;,
\\
&D_\lambda = 0 \qquad \Rightarrow \qquad \chi = - \frac{P}{\sqrt{q}} = - \frac{1}{\sqrt{q}}q_{ii} P^{ii}\;.
\label{chi}
\end{align}
Note that in the Bianchi I case $P= \frac{\sqrt{q}}{2}q_{ab} (K^{ab} - q^{ab} K)=-\stackrel{\textbf{\;\;\;\,.}}{\sqrt{q}}$, where $K_{ab}$ is the extrinsic curvature and the dot denotes derivatives w.r.t. the time coordinate (with unit lapse) to avoid confusion with primes, which instead denote generic derivatives w.r.t. the argument of the function. Hence, Eq. \eqref{chi} relates $\chi$ to the expansion of spacetime, namely $\chi=\stackrel{\textbf{\;\;\;\,.}}{\sqrt{q}}/\sqrt{q}$. Inserting all this in the Hamiltonian \eqref{CMH} and performing the trivial spatial integrations gives
\begin{align}
\mathcal{H} = \frac{2}{\sqrt{q}} P^{a b} P^{c d} \left( q_{ac} q_{bd} - \frac{1}{2} q_{a b} q_{cd} \right) -\sqrt{q} \left(f(\chi) - f'(\chi) \chi\right) \pm P_\phi\;.
\end{align}
To compare with loop quantum cosmology, we focus on the gravitational part (dropping $P_\phi$)
\begin{align}
\mathcal{H}_{\text{grav}} =&\; \sqrt{q} \left(\chi^2 -f(\chi) + f'(\chi) \chi \right) -\frac{4}{\sqrt{q}}\left(q_{11} P^{11} q_{22} P^{22} + q_{11} P^{11} q_{33} P^{33}+ q_{22} P^{22} q_{33} P^{33}\right)\;.
\label{HCM3}
\end{align}

\subsection{Loop quantum cosmology}

There are several ways to arrive at loop quantum cosmology, either by quantising a symmetry reduced theory as, e.g., in \cite{AshtekarMathematicalStructureOf, AshtekarQuantumNatureOf} or by considering reduction of full loop quantum gravity \cite{BIII, BVI, OritiBouncingCosmologiesFrom, AlesciImprovedRegularizationFrom}. Here, we will take the first route and present a simplified derivation by direct polymerisation of the symmetry reduced classical Hamiltonian.

Following Arnowitt, Deser, and Misner \cite{ArnowittTheDynamicsOf}, classical canonical general relativity is usually written using the spatial metric $q_{ab}$ and its conjugate $P^{ab} = \frac{\sqrt{q}}{2} (K^{ab} - q^{ab} K)$, where $K_{ab}$ is the extrinsic curvature. In the context of a Bianchi I model as above, $q_{i j} = \text{diag}(a_1^2, a_2^2, a_3^2)_{i j}$ and one may consider the variables
\be
	p_1= a_2a_3, ~~ p_2= a_1a_3, ~~ p_3= a_1a_2, \label{eq:ps}
\ee
as well as
\be
	c^1 = K_{11} / a_1, ~~ c^2 = K_{22} / a_2, ~~ c^3 = K_{33} / a_3
\ee
with the Poisson brackets $\{c^i, p_j\} = \delta_i^j$. The Hamiltonian constraint can be written as
\be
\mathcal{H}_{\text{grav}} =-\sqrt{p_1 p_2 p_3} \left(\frac{ c^2 c^3}{p_1} +\frac{ c^3 c^1}{p_2} +\frac{ c^1 c^2}{p_3}   \right) \text{.} \label{eq:BianchiIH}
\ee
Another way to arrive at this formulation is to start with the SU$(2)$ Ashtekar-Barbero variables of $3+1$-dimensional canonical general relativity \cite{AshtekarNewVariablesFor, BarberoRealAshtekarVariables}, an SU$(2)$ connection $A^i_a$ and its conjugate, the densitised triad $E_i^a$. $c^i$ and $p_j$ are related to them as
\be\label{LQCvar}
A_a^i=c^i(L_i)^{-1}{\,}^{o}{\omega}_a^i,\qquad E_i^a=p_iL_iV_o^{-1}\sqrt{{}^oq}{\,}^oe_i^a\;,
\ee
where $L_i$, $i=1,2,3$, denote the lengths of the edges of the elementary cell in the three spatial directions, $V_o=L_1L_2L_3$ its volume measured by using the fiducial flat metric ${}^oq_{ab}$, and ${}^oe_i^a$, ${}^o\omega^i_a$ the fiducial triads and co-triads dual to them.

The mathematical structure of the Hilbert space underlying loop quantum cosmology is inspired by loop quantum gravity \cite{RovelliQuantumGravity, ThiemannModernCanonicalQuantum}, which in turn can be described as a diffeomorphism invariant extension of lattice gauge theory, where the (dynamical) lattice encodes gravity. As in lattice gauge theory, the scalar product of the theory is obtained by integrating an exponentiated connection against the Haar measure of the underlying group. In the cosmological setting, this translates into considering point-holonomies of $c^i$. More precisely, the Hilbert space is the completion of the functions
\be
	\psi(c) = \sum_{\rho \in \mathbb R} \sum_{j=1}^3 \alpha^j_\rho e^{i \rho c^j}, ~~~~ \alpha_{\rho}^j \in \mathbb C
\ee
w.r.t. to the norm induced by the scalar product
\be
	\braket{\psi}{\psi'} = \sum_{\rho \in \mathbb R} \sum_{j=1}^3 \overline{\alpha^j_\rho} \, \alpha'{}^{j}_\rho \text{.}
\ee
satisfying the square integrability condition $\braket{\psi}{\psi}<\infty$, which enforces that at most countably many $\alpha_\rho^j$ can be non-zero. These functions (for a single $j$) are known as (square integrable) almost periodic functions, see e.g. \cite{SubinDifferentialAndPseudodifferential}, and the underlying compact group is the Bohr compactification of the real line, which, roughly speaking, generalises U$(1)$ to the case of non-integer representation labels $\rho$.

The crucial point is now that while linear combinations of $e^{i \rho c^j}$ can be promoted to (multiplication) operators on this space, functions such as $c^j$ are not well defined as operators. When quantising a given system based on this Hilbert space, one thus has to approximate all phase space functions depending on $c^j$ by linear combinations of exponentials as above, a process known as polymerisation\footnote{The choice of approximant is guided by physical principles as no naive lattice-like continuum limit is available in the background-independent setting. In the case of cosmology, one chooses $\rho$ as a function of the $p_i$ so that quantum gravity effects become important near the Planck curvature while the classical theory is recovered for small curvatures \cite{AshtekarQuantumNatureOf, AshtekarLoopQuantumCosmologyBianchi}.} \cite{CorichiPolymerQuantumMechanics}. We note that this is the direct analogue to writing the lattice Hamiltonian in terms of parallel transports.

It turns out that for large volumes, the quantum evolution of the expectation value of $\hat p_i$ and $\hat c^j$ is well approximated by solving effective classical Hamiltonian equations of motion where all operators are approximated by their classical analogues \cite{RovelliWhyAreThe}. The original classical theory is then deformed by corrections to the equations of motion in powers of $\hbar$ stemming from substituting $c^j$ by a polymerisation, e.g. $\frac{\sin(\bar \mu_j(p) c^j)}{\bar \mu_j(p)}$, for a suitable choice of $\bar \mu_j(p)$ \cite{AshtekarLoopQuantumCosmologyBianchi}.

It is crucial to note that due to this change of Hamiltonian, the original relation of $c^j$ to time derivatives of the spatial metric is obscured, since such a relation is dictated by the Hamiltonian. Therefore, while $\{c^i, p_j\} = \delta^i_j$ still holds for the effective theory, $c^i$ is a complicated function of $\dot a_j$ depending on the chosen polymerisation scheme.

\section{Comparison with loop quantum gravity} \label{sec:Computation}

\subsection{Strategy}

Comparing the Hamiltonian formulations of loop quantum cosmology and limiting curvature mimetic gravity consists of two steps:
\begin{enumerate}
\item We look for a canonical transformation $(q_{ii},P^{ii})\mapsto(c^i(q,P),p_i(q,P))$\footnote{Here $(q,P)$ is meant just as a shorthand notation for the dependence from all the $q_{ii}$ and the $P^{ii}$ and it must not be confused with their determinants.} such that the canonical structures of the two theories agree. In other words, we demand the Dirac brackets between $c^i(q,P)$ and $p_j(q,P)$ in limiting curvature mimetic gravity to be equal to the canonical Poisson brackets of effective loop quantum cosmology
\be\label{PBDB1}
\left\{c^i(q,P),p_j(q,P)\right\}_*\overset{!}{=}\left\{c^i,p_j\right\}=\delta^i_j\;,
\ee
which, together with
\be\label{PBDB2}
\left\{c^i(q,P),c^j(q,P)\right\}_*\overset{!}{=}\left\{c^i,c^j\right\}=0\;,
\ee
provide a set of first-order partial differential equations (PDEs) in the unknowns $c^i(q,P)$, $p_j(q,P)$.
\item Once this dependence is known, we can invert these functions and insert the resulting expressions into the Hamiltonian \eqref{HCM3}. We then check whether the resulting Hamiltonian can be written as a linear combination of operators that are well defined on the corresponding loop quantum gravity Hilbert space.
\end{enumerate}

A simplification in the problem occurs since $p_j$ as a configuration variable is unchanged by the modification in the LQC Hamiltonian and should thus only depend on $q_{ij}$ as in \eqref{eq:ps}.

\subsection{Canonical transformation}

With the above simplification, the Dirac brackets on the l.h.s. of \eqref{PBDB1} read
\begin{align}\label{cpDB}
\left\{c^i(q,P),p_j(q)\right\}_*&=\frac{\partial c^i}{\partial q_{kl}}\{q_{kl},q_{mn}\}_*\frac{\partial p_j}{\partial q_{mn}} +\frac{\partial c^i}{\partial P^{kl}}\{P^{kl},q_{mn}\}_*\frac{\partial p_j}{\partial q_{mn}} \notag\\
&=-\frac{\partial c^i}{\partial P^{kl}}\left(\delta_{(m}^k\delta_{n)}^l-q_{mn}q^{kl}\,\frac{f''(\chi)}{3f''(\chi)-2}\right)\frac{\partial p_j}{\partial q_{mn}}\;,
\end{align}
where in the last line we have used the Dirac bracket \eqref{DB} together with the Dirac bracket \eqref{DB2} \cite{BodendorferCanonicalStructureOf}. Recalling now that $P^{kl}=P^{kk}\delta^{kl}$ and $q_{mn}=q_{mm}\delta_{mn}$, \eqref{cpDB} gives

\be
\left\{c^i(q,P),p_j(q,P)\right\}_*=-\frac{\partial c^i}{\partial P^{kk}}\left(\delta_m^k-q_{mm}q^{kk}\,\frac{f''(\chi)}{3f''(\chi)-2}\right)\frac{\partial p_j}{\partial q_{mm}}\;,
\ee
from which, using the expressions \eqref{eq:ps} for $p_j(q)$ and \eqref{fCM} for the function $f(\chi)$, it follows that
\be\label{cDB}
\begin{cases}
\left\{c^i(q,P),p_1(q)\right\}_*=-\frac{1}{2}\sqrt{\frac{q_{33}}{q_{22}}}\frac{\partial c^i}{\partial P^{22}}-\frac{1}{2}\sqrt{\frac{q_{22}}{q_{33}}}\frac{\partial c^i}{\partial P^{33}}+\frac{\sqrt{q_{22}q_{33}}}{3}\left(1-\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}\right)\sum_{k=1}^3\frac{1}{q_{kk}}\frac{\partial c^i}{\partial P^{kk}}\\
\left\{c^i(q,P),p_2(q)\right\}_*=-\frac{1}{2}\sqrt{\frac{q_{33}}{q_{11}}}\frac{\partial c^i}{\partial P^{11}}-\frac{1}{2}\sqrt{\frac{q_{11}}{q_{33}}}\frac{\partial c^i}{\partial P^{33}}+\frac{\sqrt{q_{11}q_{33}}}{3}\left(1-\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}\right)\sum_{k=1}^3\frac{1}{q_{kk}}\frac{\partial c^i}{\partial P^{kk}}\\
\left\{c^i(q,P),p_3(q)\right\}_*=-\frac{1}{2}\sqrt{\frac{q_{22}}{q_{11}}}\frac{\partial c^i}{\partial P^{11}}-\frac{1}{2}\sqrt{\frac{q_{11}}{q_{22}}}\frac{\partial c^i}{\partial P^{22}}+\frac{\sqrt{q_{11}q_{22}}}{3}\left(1-\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}\right)\sum_{k=1}^3\frac{1}{q_{kk}}\frac{\partial c^i}{\partial P^{kk}}
\end{cases}
\ee
Let us focus on the case $i=1$. Imposing the condition \eqref{PBDB1} for the Dirac brackets above, it yields the following system of first-order PDEs for the unknown function $c^1(q,P)$
\be\label{PDE1}
\begin{pmatrix}
A_1 & A_2 & A_3\\
B_1 & B_2 & B_3\\
C_1 & C_2 & C_3
\end{pmatrix}\begin{pmatrix}
\frac{\partial c^1}{\partial P^{11}}\\
\frac{\partial c^1}{\partial P^{22}}\\
\frac{\partial c^1}{\partial P^{33}}
\end{pmatrix}=\begin{pmatrix}
1\\
0\\
0
\end{pmatrix}\;,
\ee
where, according to \eqref{cDB}, the matrix of coefficients is given by
\be\label{coeffPDE1}
\begin{pmatrix}
\frac{1}{3}\frac{\sqrt{q_{22}q_{33}}}{q_{11}}\left(1-\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}\right) & -\frac{1}{3}\sqrt{\frac{q_{33}}{q_{22}}}\left(\frac{1}{2}+\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}\right) & -\frac{1}{3}\sqrt{\frac{q_{22}}{q_{33}}}\left(\frac{1}{2}+\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}\right)\\
-\frac{1}{3}\sqrt{\frac{q_{33}}{q_{11}}}\left(\frac{1}{2}+\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}\right) & \frac{1}{3}\frac{\sqrt{q_{11}q_{33}}}{q_{22}}\left(1-\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}\right) & -\frac{1}{3}\sqrt{\frac{q_{11}}{q_{33}}}\left(\frac{1}{2}+\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}\right)\\
-\frac{1}{3}\sqrt{\frac{q_{22}}{q_{11}}}\left(\frac{1}{2}+\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}\right) & -\frac{1}{3}\sqrt{\frac{q_{11}}{q_{22}}}\left(\frac{1}{2}+\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}\right) & \frac{1}{3}\frac{\sqrt{q_{11}q_{22}}}{q_{33}}\left(1-\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}\right)
\end{pmatrix}\;.
\ee
Solving \eqref{PDE1} for the derivatives $\frac{\partial c^1}{\partial P^{kk}}$, we get
\begin{align}
&\frac{\partial c^1}{\partial P^{11}}=\frac{1}{3}\frac{q_{11}}{\sqrt{q_{22}q_{33}}}\left(4-\frac{1}{\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}}\right)\;,\label{dc1dP11}\\
&\frac{\partial c^1}{\partial P^{22}}=-\frac{2}{3}\sqrt{\frac{q_{22}}{q_{33}}}\left(1+\frac{1}{2}\frac{1}{\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}}\right)\label{dc1dP22}\;,\\
&\frac{\partial c^1}{\partial P^{33}}=-\frac{2}{3}\sqrt{\frac{q_{33}}{q_{22}}}\left(1+\frac{1}{2}\frac{1}{\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}}\right)\;,\label{dc1dP33}
\end{align}
which, recalling the expression \eqref{chi} for $\chi$, can be integrated to
\be\label{c1}
c^1(q,P) = G_1(q) + \frac{4}{3} \frac{q_{11}}{\sqrt{q_{22} q_{33}}} P^{11} - \frac{2}{3} \sqrt{\frac{q_{33}}{q_{22}}} P^{33} - \frac{2}{3} \sqrt{\frac{q_{22}}{q_{33}}} P^{22} + \frac{\chi_m}{\sqrt{6}} \sqrt{q_{11}} \arcsin\left(\sqrt{\frac{2}{3}} \frac{\chi}{\chi_m}\right)\;.
\ee
Similar expressions are obtained for $c^2$ and $c^3$:
\be\label{c2}
c^2(q,P) = G_2(q) + \frac{4}{3} \frac{q_{22}}{\sqrt{q_{11} q_{33}}} P^{22} - \frac{2}{3} \sqrt{\frac{q_{33}}{q_{11}}} P^{33} - \frac{2}{3} \sqrt{\frac{q_{11}}{q_{33}}} P^{11} + \frac{\chi_m}{\sqrt{6}} \sqrt{q_{22}} \arcsin\left( \sqrt{\frac{2}{3}} \frac{\chi}{\chi_m}\right)\;,
\ee
\be\label{c3}
c^3(q,P) = G_3(q) + \frac{4}{3} \frac{q_{33}}{\sqrt{q_{11} q_{22}}} P^{33} - \frac{2}{3} \sqrt{\frac{q_{11}}{q_{22}}} P^{11} - \frac{2}{3} \sqrt{\frac{q_{22}}{q_{11}}} P^{22} + \frac{\chi_m}{\sqrt{6}} \sqrt{q_{33}} \arcsin\left( \sqrt{\frac{2}{3}} \frac{\chi}{\chi_m} \right)\;.
\ee

It is possible to leave the functions $G_i(q)$ free and to determine a differential equation for them by demanding $\left\{c^i(q,P),c^j(q,P)\right\}_*=0$. A comparison to loop quantum cosmology on the other hand tells us to set $G_i=0$, as $c^i \rightarrow 0$ for $\dot p_i \rightarrow 0$ for generic polymerisation functions consistent with the classical limit\footnote{One needs to demand that $f(c) \sim c$ as $c \rightarrow 0$ for any good polymerisation function $f$, as otherwise the classical dynamics is changed already for low curvatures.}. We will thus drop $G_i$ in the following.

As shown in the appendix, the condition \eqref{PBDB2} for the Dirac brackets between the $c^i$ given in \eqref{c1}-\eqref{c3} is satisfied and \eqref{c1}-\eqref{c3} with $G_i=0$ thus provide the desired solution for $c^i(q,P)$.

\subsection{Hamiltonian in LQC-variables}

The relations $p_i(q)$ and $c^i(q,P)$ found in the previous section now have to be inverted in order to insert the resulting expressions for the $q_{ii}$ and $P^{ii}$ in terms of the $c^i,p_i$ into the Hamiltonian \eqref{HCM3}. Inverting the functions $p_i(q)$ is straight forward:
\be\label{qofp}
q_{11}=\frac{p_2p_3}{p_1},\quad q_{22}=\frac{p_1p_3}{p_2},\quad q_{33}=\frac{p_1p_2}{p_3}\;.
\ee
On the other hand, inverting $c^i(q,P)$ is not trivial since also $\chi$ depends on $P^{ii}$. Therefore, let us first consider
\be
\frac{\eqref{c1}}{\sqrt{q_{11}}} + \frac{\eqref{c2}}{\sqrt{q_{22}}} + \frac{\eqref{c3}}{\sqrt{q_{33}}} = \sum_{i=1 }^3 \frac{c^i}{\sqrt{q_{ii}}} = \sqrt{\frac{3}{2}} \chi_m \arcsin\left( \sqrt{\frac{2}{3}} \frac{\chi}{\chi_m} \right)\;,
\ee
from which, we find
\be
\chi = \sqrt{\frac{3}{2}} \chi_m \sin\left(\sqrt{\frac{2}{3}} \frac{1}{\chi_m} \sum_{i=1}^3 \frac{c^i}{\sqrt{q_{ii}}} \right) \stackrel{\eqref{qofp}}{=} \sqrt{\frac{3}{2}} \chi_m \sin\left(\sqrt{\frac{2}{3}} \frac{1}{\chi_m} \frac{p_i c^i}{\sqrt{p_1p_2p_3}} \right) \;.
\ee
Inserting this back into \eqref{c1}-\eqref{c3} and using also \eqref{qofp}, we get a system of linear equations in the $P^{ii}$:
\begin{equation}\label{eqsys}
\begin{pmatrix}
4 \frac{p_2 p_3}{p_1^2} & -2 \frac{p_3}{p_2} & -2 \frac{p_2}{p_3} \\
-2 \frac{p_3}{p_1} & 4 \frac{p_1 p_3}{p_2^2} & -2 \frac{p_1}{p_3} \\
-2 \frac{p_2}{p_1} & -2 \frac{p_1}{p_2} & 4 \frac{p_1 p_2}{p_3^2}
\end{pmatrix}
\begin{pmatrix}
P^{11} \\
P^{22} \\
P^{33}
\end{pmatrix}
=
\begin{pmatrix}
2 c^1 - \frac{p_2}{p_1} c^2 - \frac{p_3}{p_1} c^3 \\
2 c^2 - \frac{p_3}{p_2} c^3 - \frac{p_1}{p_2} c^1 \\
2 c^3 - \frac{p_2}{p_3} c^2 - \frac{p_1}{p_3} c^1
\end{pmatrix}\;.
\end{equation}
This system of equations is underdetermined since the determinant of the left-most matrix vanishes. Indeed, solving the second equation of \eqref{eqsys} for $P^{22}$, we have
\begin{equation}\label{P2}
P^{22} = \frac{1}{4} \frac{p_2^2}{p_1 p_3} \left( 2c^2 - \frac{p_3}{p_2} c^3 - \frac{p_1}{p_2} c^1 + 2 \frac{p_3}{p_1} P^{11} + 2 \frac{p_1}{p_3} P^{33} \right)\;,
\end{equation}
and, inserting it into the third equation of \eqref{eqsys}, it yields
\begin{equation}\label{P33}
P^{33} = - \frac{p_3}{2p_2}c^1 + \frac{p_3^2}{2 p_1 p_2}c^3 + \frac{p_3^2}{p_1^2} P^{11}\; .
\end{equation}
Inserting this back into \eqref{P2}, we get
\begin{equation}\label{P22}
P^{22} = - \frac{p_2}{2p_3}c^1 + \frac{p_2^2}{2 p_1 p_3}c^2 + \frac{p_2^2}{p_1^2} P^{11}\; .
\end{equation}
Inserting both back into the first equation of \eqref{eqsys} solves it identically, thus showing that the system is underdetermined. Nevertheless, the condition
\begin{align}
&-\sqrt{\frac{3}{2}} \chi_m \sqrt{p_1 p_2 p_3} \sin\left(\sqrt{\frac{2}{3}} \frac{1}{\chi_m} \frac{p_i c^i}{\sqrt{p_1p_2p_3}} \right)  \nonumber \\ &= -\sqrt{q} \chi \overset{\eqref{chi}}{=} q_{ii} P^{ii} = 3 \frac{p_2 p_3}{p_1} P^{11} - p_1 c^1 + \frac{1}{2} p_2 c^2 + \frac{1}{2} p_3 c^3
\end{align}
fixes $P^{11}$ and hence $P^{22}$ and $P^{33}$ given in \eqref{P33}, \eqref{P22} to be
\begin{align}
P^{11} =& - \frac{\chi_m}{\sqrt{6}} \frac{p_1}{p_2 p_3} \sqrt{p_1 p_2 p_3} \sin\left( \sqrt{\frac{2}{3}} \frac{1}{\chi_m} \frac{p_i c^i}{\sqrt{p_1p_2p_3}} \right) + \frac{1}{3} \frac{p_1^2}{p_2 p_3} c^1 - \frac{1}{6} \frac{p_1}{p_3} c^2 - \frac{1}{6} \frac{p_1}{p_2} c^3\;, \\
P^{22} =& - \frac{\chi_m}{\sqrt{6}} \frac{p_2}{p_1 p_3} \sqrt{p_1 p_2 p_3} \sin\left( \sqrt{\frac{2}{3}} \frac{1}{\chi_m} \frac{p_i c^i}{\sqrt{p_1p_2p_3}} \right) + \frac{1}{3} \frac{p_2^2}{p_1 p_3} c^2 - \frac{1}{6} \frac{p_2}{p_3} c^1 - \frac{1}{6} \frac{p_2}{p_1} c^3\;, \\
P^{33} =& - \frac{\chi_m}{\sqrt{6}} \frac{p_3}{p_1 p_2} \sqrt{p_1 p_2 p_3} \sin\left( \sqrt{\frac{2}{3}} \frac{1}{\chi_m} \frac{p_i c^i}{\sqrt{p_1p_2p_3}} \right) + \frac{1}{3} \frac{p_3^2}{p_1 p_2} c^3 - \frac{1}{6} \frac{p_3}{p_2} c^1 - \frac{1}{6} \frac{p_3}{p_1} c^2\;.
\end{align}
These expressions can now be inserted into the Hamiltonian \eqref{HCM3}, giving
\begin{align}
\mathcal{H}_{grav} =&-\frac{1}{\sqrt{p_1p_2p_3}} \left( p_1c^1 p_2c^2 + p_1c^1 c^3p_3 + p_2c^2p_3c^3\right)
\notag
\\&-2\sqrt{p_1p_2p_3} \chi_m^2 \sin^2\left(\frac{1}{\sqrt{6} \chi_m} \frac{p_i c^i}{\sqrt{p_1p_2p_3}} \right)  +\frac{1}{3}\sqrt{p_1p_2p_3}\left(\frac{p_i c^i}{\sqrt{p_1p_2p_3}}\right)^2\;, \label{HCMcp}
\end{align}
or equivalently
\begin{align} \label{eq:FinalH}
\mathcal{H}_{grav}=&-\sqrt{p_1 p_2 p_3} \left(\frac{ c^2 c^3}{p_1} +\frac{ c^3 c^1}{p_2} +\frac{ c^1 c^2}{p_3} \right) - 2 \sqrt{p_1p_2p_3} \chi_m^2 \left( \sin^2(x) - x^2 \right) , \; x = \frac{1}{\sqrt{6} \chi_m} \frac{p_i c^i}{\sqrt{p_1p_2p_3}}.
\end{align}
Taking the last expression, it is easy to check that the classical theory is obtained in the limit $\chi_m \rightarrow \infty$, hence $x\rightarrow 0$, where the $c^i$ also assume their classical meaning.

\eqref{eq:FinalH} puts us into a position to answer the central question of the paper. While the $\sin^2(x)$ term can be written as a linear combination of point holonomies and therefore arise as a part of an effective loop quantum cosmology Hamiltonian, the other terms, being linear or quadratic in $c^i$, cannot. We thus conclude that already in the context of Bianchi I models, limiting curvature mimetic gravity cannot be written as an effective loop quantum cosmology type model.

It also becomes clear at this point that changing the function $f$ given in \eqref{fCM} cannot change this conclusion: since $f$ depends only on $\chi$ and thus a certain combination of the $c^i$, we cannot turn all individual terms containing $c^i$ in \eqref{eq:BianchiIH} into almost-periodic functions by a modification of the Hamiltonian as in \eqref{HCM3}.

To reconcile this conclusion with the previous work, let us consider the isotropic limit in which $p_1 = p_2 = p_3=p$ and $c^1=c^2=c^3=c$. The Hamiltonian \eqref{HCMcp} then reduces to
\be
\mathcal{H}_{grav} = -2\sqrt{p^3} \chi_m^2 \sin^2\left(\sqrt{\frac{3}{2}}\frac{1}{\chi_m} \frac{c}{\sqrt{p}}\right).
\ee
For the isotropic case, the relations
\be
c = -\frac{\sqrt{p}}{3} b \, , \qquad v = \sqrt{p^3}
\ee
hold. Thus, the Hamiltonian can be written as
\begin{equation}
\mathcal{H}_{grav} = -2 v \chi_m^2 \sin^2\left(\frac{1}{\sqrt{6} \chi_m} b\right) = -\frac{1}{3} v \frac{\sin^2\left(\lambda b\right)}{\lambda^2} \label{eq:IsotropicH}
\end{equation}
with polymerisation scale $\lambda = \frac{1}{\sqrt{6} \chi_m}$, in agreement with \cite{BodendorferCanonicalStructureOf, NouiEffectiveLoopQuantum}. The problematic terms in \eqref{eq:FinalH} drop and the above argument fails since \eqref{eq:IsotropicH} depends only on one gravitational variable.

\section{Conclusion} \label{sec:Conclusion}

In this paper, we have investigated whether limiting curvature mimetic gravity can be interpreted as effective loop quantum cosmology beyond the homogeneous and isotropic setting. Our result was negative, owing to the appearance of terms in the Hamiltonian constraint that cannot be implemented as well-defined operators on the loop quantum cosmology Hilbert space. While the computation was based on a specific choice of potential leading to the limiting curvature feature, our conclusion should hold for general potentials as explained above. Despite the differences encountered between effective loop quantum cosmology and limiting curvature mimetic gravity, it is still interesting to consider such models and generalisations thereof, see e.g. \cite{NouiEffectiveLoopQuantum}, as toy models for loop quantum gravity. This is due to a similar qualitative behaviour in the homogeneous, but non-isotropic context such as the presence of the bounce interpolating between two classical Bianchi I spacetimes as well as the same rules relating the Kasner exponents before and after the bounce, cfr. Eq. (45) in \cite{ChamseddineResolvingCosmologicalSingularities} and Eq. (40) in \cite{WilsonEwingTheLoopQuantum} (see however \cite{BrahmaOnSingularityResolution} for some subtleties). It would be interesting to continue this comparison by dropping the homogeneity assumption and proceed to compare black hole solutions \cite{ChamseddineNonsingularBlackHole, CorichiLoopQuantizationOf} as initiated in \cite{AchourNonSingularBlack}.

As a byproduct, the variable transformation derived in this paper brought the Dirac bracket for Bianchi I models in limiting curvature mimetic gravity into canonical form. Independently of a relation to a polymer quantisation, we may now take this choice of variable as a starting point for a treatment \`a la Wheeler-de Witt.

\section*{Acknowledgments}

The authors were supported by an International Junior Research Group grant of the Elite Network of Bavaria. Discussions with Andreas Sch\"afer and John Schliemann are gratefully acknowledged.

\begin{appendix}

\section{Dirac bracket $\left\{c^i(q,P),c^j(q,P)\right\}_*$}

We have to impose the condition \eqref{PBDB2} for the Dirac brackets between the $c^i$ given in \eqref{c1}-\eqref{c3}. Therefore, by using the Dirac brackets \eqref{DB}, \eqref{DB1} and the expression \eqref{fCM} for $f(\chi)$, the l.h.s. of \eqref{PBDB2} gives
\be\label{cDB2}
\begin{split}
\{c^i&(q,P),c^j(q,P)\}_*=\frac{\partial c^i}{\partial q_{kl}}\{q_{kl},P^{mn}\}_*\frac{\partial c^j}{\partial P^{mn}}+\frac{\partial c^i}{\partial P^{mn}}\{P^{mn},q_{kl}\}_*\frac{\partial c^j}{\partial q_{kl}}+\frac{\partial c^i}{\partial P^{kl}}\{P^{kl},P^{mn}\}_*\frac{\partial c^j}{\partial P^{mn}}\\
&=\left(\frac{\partial c^i}{\partial q_{kl}}\frac{\partial c^j}{\partial P^{mn}}-\frac{\partial c^j}{\partial q_{kl}}\frac{\partial c^i}{\partial P^{mn}}\right)\{q_{kl},P^{mn}\}_*+\frac{\partial c^i}{\partial P^{kl}}\frac{\partial c^j}{\partial P^{mn}}\{P^{kl},P^{mn}\}_*\\
&=\left(\frac{\partial c^i}{\partial q_{kl}}\frac{\partial c^j}{\partial P^{mn}}-\frac{\partial c^j}{\partial q_{kl}}\frac{\partial c^i}{\partial P^{mn}}\right)\left(\delta_{(k}^m\delta_{l)}^n-\frac{1}{3}q_{kl}q^{mn}\left(1-\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}\right)\right)\\
&\quad-\frac{\partial c^i}{\partial P^{kl}}\frac{\partial c^j}{\partial P^{mn}}\left(q^{kl}P^{mn}-q^{mn}P^{kl}\right)\frac{1}{3}\left(1-\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}\right)\\
&=\left(\frac{\partial c^i}{\partial q_{kk}}\frac{\partial c^j}{\partial P^{kk}}-\frac{\partial c^j}{\partial q_{kk}}\frac{\partial c^i}{\partial P^{kk}}\right)-\frac{1}{3}\left(1-\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}\right)\\
&\quad\cdot\left(\left(\frac{\partial c^i}{\partial q_{kk}}\frac{\partial c^j}{\partial P^{mm}}-\frac{\partial c^j}{\partial q_{kk}}\frac{\partial c^i}{\partial P^{mm}}\right)q_{kk}q^{mm}+\left(\frac{\partial c^i}{\partial P^{kk}}\frac{\partial c^j}{\partial P^{mm}}-\frac{\partial c^j}{\partial P^{kk}}\frac{\partial c^i}{\partial P^{mm}}\right)q^{kk}P^{mm}\right)\;.
\end{split}
\ee
Let us compute the three sums involving the derivatives of the $c^i$ separately. For instance for $i=1$ and $j=2$, using the expressions \eqref{c1}, \eqref{c2} for $c^1$, $c^2$ and \eqref{chi} for $\chi$, we have

\begin{align}
&\frac{\partial c^1}{\partial q_{11}}=\frac{4}{3}\frac{P^{11}}{\sqrt{q_{22}q_{33}}}+\frac{\chi_m}{2\sqrt{6}}\frac{1}{\sqrt{q_{11}}}\arcsin\left( \sqrt{\frac{2}{3}} \frac{\chi}{\chi_m}\right)\notag\\
&\qquad\quad+\frac{\sqrt{q_{11}}}{3}\frac{1}{\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}}\left(-\frac{1}{2}\frac{P^{11}}{\sqrt{q_{11}q_{22}q_{33}}}+\frac{1}{2q_{11}}\sqrt{\frac{q_{22}}{q_{11}q_{33}}}P^{22}+\frac{1}{2q_{11}}\sqrt{\frac{q_{33}}{q_{11}q_{22}}}P^{33}\right)\;,\label{dc1dq11}\\
&\frac{\partial c^1}{\partial q_{22}}=-\frac{2}{3}\frac{q_{11}P^{11}}{q_{22}\sqrt{q_{22}q_{33}}}+\frac{1}{3}\frac{1}{q_{22}}\sqrt{\frac{q_{33}}{q_{22}}}P^{33}-\frac{1}{3}\frac{P^{22}}{\sqrt{q_{22}q_{33}}}\notag\\
&\qquad\quad+\frac{\sqrt{q_{11}}}{3}\frac{1}{\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}}\left(-\frac{1}{2}\frac{P^{22}}{\sqrt{q_{11}q_{22}q_{33}}}+\frac{1}{2q_{22}}\sqrt{\frac{q_{11}}{q_{22}q_{33}}}P^{11}+\frac{1}{2q_{22}}\sqrt{\frac{q_{33}}{q_{11}q_{22}}}P^{33}\right)\;,\label{dc1dq22}\\
&\frac{\partial c^1}{\partial q_{33}}=-\frac{2}{3}\frac{q_{11}P^{11}}{q_{33}\sqrt{q_{22}q_{33}}}-\frac{1}{3}\frac{P^{33}}{\sqrt{q_{22}q_{33}}}+\frac{1}{3}\frac{1}{q_{33}}\sqrt{\frac{q_{22}}{q_{33}}}P^{22}\notag\\
&\qquad\quad+\frac{\sqrt{q_{11}}}{3}\frac{1}{\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}}\left(-\frac{1}{2}\frac{P^{33}}{\sqrt{q_{11}q_{22}q_{33}}}+\frac{1}{2q_{33}}\sqrt{\frac{q_{11}}{q_{22}q_{33}}}P^{11}+\frac{1}{2q_{33}}\sqrt{\frac{q_{22}}{q_{11}q_{33}}}P^{22}\right)\label{dc1dq33}\;,\\
&\frac{\partial c^2}{\partial q_{11}}=-\frac{2}{3}\frac{q_{22}P^{22}}{q_{11}\sqrt{q_{11}q_{33}}}+\frac{1}{3}\frac{1}{q_{11}}\sqrt{\frac{q_{33}}{q_{11}}}P^{33}-\frac{1}{3}\frac{P^{11}}{\sqrt{q_{11}q_{33}}}\notag\\
&\qquad\quad+\frac{\sqrt{q_{22}}}{3}\frac{1}{\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}}\left(-\frac{1}{2}\frac{P^{11}}{\sqrt{q_{11}q_{22}q_{33}}}+\frac{1}{2q_{11}}\sqrt{\frac{q_{22}}{q_{11}q_{33}}}P^{22}+\frac{1}{2q_{11}}\sqrt{\frac{q_{33}}{q_{11}q_{22}}}P^{33}\right)\label{dc2dq11}\;,\\
&\frac{\partial c^2}{\partial q_{22}}=\frac{4}{3}\frac{P^{22}}{\sqrt{q_{11}q_{33}}}+\frac{\chi_m}{2\sqrt{6}}\frac{1}{\sqrt{q_{22}}}\arcsin\left( \sqrt{\frac{2}{3}} \frac{\chi}{\chi_m}\right)\notag\\
&\qquad\quad+\frac{\sqrt{q_{22}}}{3}\frac{1}{\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}}\left(-\frac{1}{2}\frac{P^{22}}{\sqrt{q_{11}q_{22}q_{33}}}+\frac{1}{2q_{22}}\sqrt{\frac{q_{11}}{q_{22}q_{33}}}P^{11}+\frac{1}{2q_{22}}\sqrt{\frac{q_{33}}{q_{11}q_{22}}}P^{33}\right)\;,\label{dc2dq22}
\end{align}
\begin{align}
&\frac{\partial c^2}{\partial q_{33}}=-\frac{2}{3}\frac{q_{22}P^{22}}{q_{33}\sqrt{q_{11}q_{33}}}-\frac{1}{3}\frac{P^{33}}{\sqrt{q_{11}q_{33}}}+\frac{1}{3}\frac{1}{q_{33}}\sqrt{\frac{q_{11}}{q_{33}}}P^{11}\notag\\
&\qquad\quad+\frac{\sqrt{q_{22}}}{3}\frac{1}{\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}}\left(-\frac{1}{2}\frac{P^{33}}{\sqrt{q_{11}q_{22}q_{33}}}+\frac{1}{2q_{33}}\sqrt{\frac{q_{22}}{q_{11}q_{33}}}P^{22}+\frac{1}{2q_{33}}\sqrt{\frac{q_{11}}{q_{22}q_{33}}}P^{11}\right)\;,\label{dc2dq33}
\end{align}
\begin{align}
&\frac{\partial c^2}{\partial P^{11}}=-\frac{2}{3}\sqrt{\frac{q_{11}}{q_{33}}}\left(1+\frac{1}{2}\frac{1}{\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}}\right)\;,\label{dc2dP11}\\
&\frac{\partial c^2}{\partial P^{22}}=\frac{1}{3}\frac{q_{22}}{\sqrt{q_{11}q_{33}}}\left(4-\frac{1}{\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}}\right)\;,\label{dc2dP22}\\
&\frac{\partial c^2}{\partial P^{33}}=-\frac{2}{3}\sqrt{\frac{q_{33}}{q_{11}}}\left(1+\frac{1}{2}\frac{1}{\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}}\right)\;,\label{dc2dP33}
\end{align}

\noindent
from which, together with \eqref{dc1dP11}-\eqref{dc1dP33}, it follows that
\begin{align}
&\frac{\partial c^1}{\partial q_{kk}}\frac{\partial c^2}{\partial P^{kk}}-\frac{\partial c^2}{\partial q_{kk}}\frac{\partial c^1}{\partial P^{kk}}=-\frac{2}{3}\frac{1}{q_{33}}\left(1-\frac{1}{\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}}\right)\left(\sqrt{\frac{q_{11}}{q_{22}}}P^{11}-\sqrt{\frac{q_{22}}{q_{11}}}P^{22}\right),\label{first}\\
&\left(\frac{\partial c^1}{\partial q_{kk}}\frac{\partial c^2}{\partial P^{mm}}-\frac{\partial c^2}{\partial q_{kk}}\frac{\partial c^1}{\partial P^{mm}}\right)q_{kk}q^{mm}=0,\label{second}\\
&\left(\frac{\partial c^1}{\partial P^{kk}}\frac{\partial c^2}{\partial P^{mm}}-\frac{\partial c^2}{\partial P^{kk}}\frac{\partial c^1}{\partial P^{mm}}\right)q^{kk}P^{mm}=\frac{1}{q_{33}}\frac{2}{\sqrt{1-\frac{2}{3}\frac{\chi^2}{\chi_m^2}}}\left(\sqrt{\frac{q_{11}}{q_{22}}}P^{11}-\sqrt{\frac{q_{22}}{q_{11}}}P^{22}\right)\;.\label{third}
\end{align}
Using \eqref{first}-\eqref{third} and similar expressions for the other Dirac brackets, we find that \eqref{cDB2} is satisfied and that \eqref{c1}, \eqref{c2}, and \eqref{c3} provide the desired solution for $c^i(q,P)$.

\end{appendix}


text:
{\bf Problem.} Ben rolls 5 fair 12-sided dice.  The 12 faces of each die are numbered from 1 to 12. What is the probability that exactly two of the dice show an even number?
{\bf Level.} Level 3
{\bf Type.} Counting & Probability
{\bf Solution.} There is a $\frac{1}{2}$ probability that a 12-sided die will show an even number and a $\frac{1}{2}$ probability that it will show an odd number. We can choose which dice will show the even numbers in $\binom{5}{2}=10$ ways. For each way, there is a $\left( \frac{1}{2} \right) ^5=\frac{1}{32}$ probability that the chosen dice actually roll even numbers and the other dice roll odd numbers. Therefore, the probability that exactly two of the dice show an even number is $10\cdot \frac{1}{32}=\boxed{\frac{5}{16}}$.

set_name: MATH


TITLE: Norm of a Positive Operator
QUESTION [0 upvotes]: For a positive operator $A\in B(\mathcal{H})$ on a complex Hilbert space $\mathcal{H}$, I want to prove that $\|A\|=\sup\{\lambda: \lambda \in \sigma(A)\}$, where $\sigma(A)$ is the spectrum of $A$.
So, in effect, I need to show that the spectral radius for a positive operator on a complex Hilbert space is equal to it's norm. I began by showing that the spectral radius $\rho(T)$ of any self-adjoint operator is less than or equal to $\|T\|$. I am stuck in the reverse inequality.

REPLY [0 votes]: The spectral radius of $T$ is $\lim_n \|T^{n}\|^{1/n}$. However, $\|T^{2}\|=\|T\|^{2}$ for any selfadjoint $T$. This is because
$$
       \|T\|^{2}=\sup_{\|x\|=1}\|Tx\|^{2}=\sup_{\|x\|=1}(T^{2}x,x) \le \sup_{\|x\|=1}\|T^{2}x\|\|x\|=\|T^{2}\|.
$$
On the other hand,
$$
                  \|T^{2}\|\le \|T\|\|T\|=\|T\|^{2}.
$$
You get the final result by noticing that $\|T^{2^{n}}\|^{1/2^{n}}=\|T\|$.


TITLE: Find k $\frac{1}{(3a-b)^2}+\frac{1}{(3b-c)^2}+\frac{1}{(3c-a)^2} \ge \frac{k}{a^2+b^2+c^2}$
QUESTION [1 upvotes]: Find k with
$$\frac{1}{(3a-b)^2}+\frac{1}{(3b-c)^2}+\frac{1}{(3c-a)^2} \ge \frac{k}{a^2+b^2+c^2}$$

REPLY [0 votes]: In other words, you want to minimize $$f(a,b,c) = (a^2 + b^2 + c^2)\left(\frac{1}{(3a-b)^2} + \frac{1}{(3b-c)^2} + \frac{1}{(3c-a)^2}\right)$$
(presumably for $a,b,c$ real numbers).
By homogeneity, we may restrict to the sphere $a^2 + b^2 + c^2 = 1$.
Using a Lagrange multiplier, we look for critical points of $F(a,b,c,\lambda) = f(a,b,c) + \lambda (a^2 + b^2 + c^2-1)$.  The solution is rather unpleasant.  It seems the minimum is approximately $0.808645463814078870$.
EDIT: In fact this number turns out to be the real root of the polynomial
$$2460640\,{z}^{3}-2787824\,{z}^{2}+905866\,z-210681$$


text:
\begin{document}

\title{Good Codes From Generalised Algebraic Geometry Codes}

\author{\IEEEauthorblockN{Mubarak Jibril\IEEEauthorrefmark{1},
Martin Tomlinson\IEEEauthorrefmark{1},
Mohammed Zaki Ahmed\IEEEauthorrefmark{1} and
Cen Tjhai\IEEEauthorrefmark{1}}
\IEEEauthorblockA{\IEEEauthorrefmark{1}School of Computing and Mathematics\\
Faculty of Technology\\ University of Plymouth\\United Kingdom\\ Email: mubarak.jibril@plymouth.ac.uk}}

\maketitle

\begin{abstract}

Algebraic geometry codes or Goppa codes are defined with places of degree one. In constructing generalised algebraic geometry codes places of higher degree are used. In this paper we present $41$ new codes over $\mathbb{F}_{16}$ which improve on the best known codes of the same length and rate. The construction method uses places of small degree with a technique originally published over 10 years ago for the construction of generalised algebraic geometry codes.
\end{abstract}

\section{Introduction}

In coding theory, it is desirable to obtain an error correcting code
with the maximum possible minimum distance $d$,  given a code length $n$
and code dimension $k$. Algebraic geometry (AG) codes have good
properties and some families of these codes have been shown to be
asymptotically superior as they exceed the well-known Gilbert Vashamov
bound \cite{zink} when the defining finite field  $\mathbb{F}_q$ has
size $q \geq 49$ with $q$ always a square. A closer look at tables of
best known codes in \cite{grassl} and \cite{mint}  shows that algebraic
geometry codes feature as the best known linear codes for an appreciable
range of code lengths for different field sizes $q$. Algebraic geometry
codes are codes derived from curves and were first discovered by Goppa
\cite{goppa} in 1981. Goppa's description uses rational places of the
curve to define these codes. Rational places are called places of degree
one. A generalised construction of algebraic geometry codes was
presented by Xing \textit{et al} in \cite{xing3}\cite{xing4} and Ozbudak
\textit{et al} in \cite{ozbudak}. An extension of the method which utilises places of
higher degrees as well as a concatenation concept was introduced in \cite{xing1}. This method was shown in \cite{san}\cite{ding}\cite{xing2} to be effective in constructing codes that are better than the best known codes and many codes were presented for finite fields up to $\mathbb{F}_9$. In this paper we present several new codes over finite field $\mathbb{F}_{16}$. These codes represent improvements on minimum distance compared to some previously best known codes. We first give a description of the codes and an exposition of the construction in the  Section II.  Finally we present our results in Section III. \\

\section{Construction}
A two dimensional affine space $\mathbb{A}^2(\mathbb{F}_q)$ is given by the set of points $\lbrace (\alpha,\beta):\alpha,\beta\in \mathbb{F}_q \rbrace$
while its projective closure $\mathbb{P}^2(\mathbb{F}_q)$ is given by
the set of equivalence points $\lbrace
\lbrace(\alpha:\beta:1)\rbrace\cup \lbrace(\alpha:1:0)\rbrace\cup
\lbrace (1:0:0) \rbrace : \alpha,\beta\in \mathbb{F}_q \rbrace$. Given a
homogeneous polynomial $F(x,y,z)$, a curve $\mathcal{X}$ defined in
$\mathbb{P}^2(\mathbb{F}_q)$ is a set of distinct points $\lbrace P \in \mathbb{P}^2(\mathbb{F}_q) : F(P)=0\rbrace$. We are only interested in the case where $\mathcal{X}$ is irreducible and is non-singular in order to obtain AG codes. Let $\mathbb{F}_{q^\ell}$ be an extension of the field $\mathbb{F}_q$, the Frobenius automorphism is given as
\begin{align*}
\phi_{q,\ell}&:\mathbb{F}_{{q}^\ell} \rightarrow \mathbb{F}_{q^\ell} &\\
\phi_{q,\ell}&(\beta)=\beta^{q}&\beta \in \mathbb{F}_{q^\ell}
\end{align*}
and its action on a projective point $(x:y:z)$ in $\mathbb{F}_{q^\ell}$ is
$$
\phi_{q,\ell}((x:y:z))=(x^q:y^q:z^q).
$$
A place of degree $\ell$ \cite{judy} is a set of $\ell$ points of a
curve defined in the extension field $\mathbb{F}_{q^\ell}$ denoted by
$\lbrace P_0,P_1,\dots,P_{\ell-1}\rbrace $ where each
$P_i=\phi_{q,l}^i(P_0)$. Places of degree one are called rational
places. An example of a place of degree two is a pair of points
$\lbrace P_0,P_1 \rbrace$ such that $P_0=(x,y)$ has coordinates in
$\mathbb{F}_{q^2}$ and $ P_1=\phi_{q,2}(P_0)=(x^q,y^q)$.\\
We will now describe two maps that are useful in the Xing \textit{et al} construction of generalised AG codes. We observe that $\mathbb{F}_q$ is a subfield of $\mathbb{F}_{q^\ell}$ for all $\ell \geq 2$. It is then possible to map $\mathbb{F}_{q^\ell}$ to an $\ell$-dimensional vector space with elements from $\mathbb{F}_q$ using a suitable basis. We define the mapping,
\begin{align*}
\pi_{\ell}&:\mathbb{F}_{{q}^\ell} \rightarrow  \mathbb{F}^\ell_{q} &\\
\pi_{\ell}&(\beta_j)=[c^j_1 \thickspace c^j_2 \dots c^j_l]~~~\beta_j \in \mathbb{F}_{q^\ell}~,~c^j_i \in \mathbb{F}_q.
\end{align*}
Suppose $[\gamma_1 \gamma_2 \dots \gamma_\ell]$ forms a suitable basis of the vector space $\mathbb{F}^\ell_{q}$, then $\beta_j=c^j_1 \gamma_1 + c^j_2 \gamma_2 + \cdots + c^j_{\ell} \gamma_{\ell}$. Finally we use $\sigma_{\ell,n}$ to represent an encoding map from an $\ell$-dimensional message space in $\mathbb{F}_q$ to an $n$-dimensional code space,
\begin{align*}
\sigma_{\ell,n}&:\mathbb{F}^{\ell}_{{q}} \rightarrow  \mathbb{F}^n_{q} &\\
\end{align*}
with $\ell \leq n$.\\
We now give a description of generalised AG codes as presented in
\cite{xing1}\cite{ding}\cite{xing2}. Let $\mathit{F}=F(x,y,z)$ be a
homogeneous polynomial defined in $\mathbb{F}_q$. Let $g$ be the genus
of the curve $\mathcal{X}/\mathbb{F}_q$ corresponding to the polynomial $\mathit{F}$. Also let $P_1,P_2,\dots,P_r$ be $r$ distinct places of
$\mathcal{X}/\mathbb{F}_q$ and $k_i=deg(P_i)$ ($deg$ is degree of).  $W$
is a divisor of the curve $\mathcal{X}/\mathbb{F}_q$ such that $W=P_1+P_2+\cdots+P_r$ and $G$ a divisor so that $supp(W)\cap supp(G)=\varnothing$. More specifically $G=m(Q-R)$ where $deg(Q)=deg(R)+1$. Associated with the divisor $G$ is a Riemann-Roch space $\mathcal{L}(G)$ with $m=deg(G))$ an integer, $m\geq 0$ . From the Riemann-Roch theorem we know that the dimension of $\mathcal{L}(G)$ is given by $l(G)$ and
$$
l(G)\geq m-g+1
$$
with equality when $m\geq 2g-1$. Also associated with each $P_i$ is a $q$-ary code $C_i$ with parameters $[n_i,k_i=deg(P_i),d_i]_{q}$ with the restriction that $d_i \leq k_i$. We denote $\lbrace f_1,f_2,..,f_k:f_l \in \mathcal{L}(G)\rbrace$ as a set of $k$ linearly independent elements of $\mathcal{L}(G)$ that form a basis. We can create a generator matrix for a generalised AG code as such,

$$
M=
\left [ \begin{smallmatrix}
\sigma_{k_1,n_1}(\pi_{k_1}(f_1(P_1))) & \dots\dots &\sigma_{k_r,n_r}(\pi_{k_r}(f_1(P_r))) \\
\sigma_{k_1,n_1}(\pi_{k_1}(f_2(P_1))) &\dots \dots &  \sigma_{k_r,n_r}(\pi_{k_r}(f_2(P_r)))\\
            .                         & . & .\\
            .                         & . & .\\
            .                         & . & .\\
\sigma_{k_1,n_1}(\pi_{k_1}(f_k(P_1))) & \dots \dots &  \sigma_{k_r,n_r}(\pi_{k_r}(f_k(P_r)))
\end{smallmatrix} \right]
$$
 where $f_l(P_i)$ is an evaluation of a polynomial and basis element
 $f_l$ at a point $P_i$, $\pi_{k_i}$ is a mapping from
 $\mathbb{F}_{q^{k_i}}$ to $\mathbb{F}_q$ and $\sigma_{k_i,n_i}$ is the
 encoding of a message vector in $\mathbb{F}^{k_i}_{q}$ to a code vector
 in $\mathbb{F}^{n_i}_{q}$. It is desirable to choose the maximum
 possible minimum distance for all codes $C_i$ so that $d_i=k_i$. The
 same code is used in the map $\sigma_{k_i,n_i}$ for all points of the
 same degree $k_i$ i.e. the code $C_j$ has parameters $[n_j,j,d_j]_q$
 for a place of degree $j$. Let $A_j$ be an integer denoting the number
 of places of degree $j$ and  $B_j$ be an integer such that $0\leq B_j
 \leq A_j$. If $t$ is the maximum degree of any place $P_i$ we choose to
 use in the construction, then the generalised AG code is represented as
 a $C(k;t;B_1,B_2,\dots,B_t;d_1,d_2,\dots,d_t)$. Let $[n,k,d]_q$
 represent a linear code in $\mathbb{F}_q$ with length $n$, dimension
 $k$ and minimum distance $d$, then a generalised AG code is given by
 the parameters \cite{xing1},
\begin{align*}
k&=l(G)\geq m-g+1\\
n&=\sum_{i=1}^{r}  n_i=\sum_{j=1}^t B_jn_j \\
d&\geq\sum_{i=1}^{r} d_i-g-k+1=\sum_{j=1}^t B_jd_j-g-k+1.
\end{align*}

\begin{comment}
\section{Example}
Let $F(x,y,z)=x^3+xyz+xz^2+y^2z$ \cite{xing1} be a polynomial in $\mathbb{F}_2$. The curve $\mathit{F}/\mathbb{F}_2$ has genus $g=1$ and $A_1=4$ places of degree $1$ and $A_2=2$ places of degree $2$.
\begin{table}[!t]
\caption{Places of $\mathit{F}/\mathbb{F}_2$}
\label{example}
\centering
\begin{tabular}{|c|c|}
\hline
$P_i$ & $deg(P_i)$\\
\hline
 $(0:1:0)$ & $1$\\
\hline
$(0:0:1)$ & $1$\\
\hline
$(1:0:1)$ & $1$\\
\hline
$(1:1:1)$ & $1$\\
\hline
$\lbrace(\alpha:1:1),(\alpha^2:1:1)\rbrace$ & $2$ \\
\hline
$\lbrace(\alpha:\alpha+1:1),(\alpha^2:\alpha:1)\rbrace$ & $2$ \\
\hline
\end{tabular}
\end{table}
Table \ref{example} gives the places of $\mathit{F}/\mathbb{F}_2$ up to degree $2$. The field $\mathbb{F}_{2^2}$ is defined by a primitive polynomial $s^2+s+1$ with $\alpha$ as its primitive element. We also choose points $R=(1:a^3+a^2:1)$ as a place of degree $4$ and $Q=(1:b^4+b^3+b^2:1)$ is a place of degree $5$ while $a$ and $b$ are primitive elements of $\mathbb{F}_{2^4}$ (defined by the polynomial $s^4+s^3+s^2+s+1$) and $\mathbb{F}_{2^5}$ (defined by the polynomial $s^5+s^2+1$) respectively. The basis of the Riemann-Roch space $\mathcal{L}(2D)$ with $D=Q-R$ and $m=2$ is obtained with computer algebra software {\small MAGMA} \cite{magma} as,

\begin{equation*}
\begin{split}
f_1&=(x^7 + x^3 + x)/(x^{10} + x^4 + 1)y\\
&\quad + (x^{10} + x^9 + x^7 + x^6 + x^5 + x + 1)/(x^{10} + x^4 + 1)\\
f_2&=(x^8 + x^7 + x^4 + x^3 + x + 1)/(x^{10} + x^4 + 1)y \\
&\quad + (x^8 + x^4 + x^2)/(x^{10} + x^4 + 1)
\end{split}
\end{equation*}

For the map $\sigma_{k_i,n_i}$ we use codes; $C_1$ a $[1,1,1]_2$ cyclic code for places of degree 1 and $C_2$ a $[3,2,2]_2$ cyclic code for places of degree $2$. For the map $\pi_2$ which applies to  places of degree $2$ we use a polynomial basis $[\gamma_1,\gamma_2]=[1,\alpha]$. We  use only the first point in the place $P_i$ for $deg(P_i)=2$ in the evaluation of $f_1$ and $f_2$ at $P_i$. The generator matrix $M$ of the resulting $[10,2,6]_2$ generalised AG code over $\mathbb{F}_2$ is,
$$
M=
\begin{bmatrix}
1 & 1 &  0 & 1&  0&  1& 1&  0& 1 &1\\
0 & 0 &  1 & 1&  1&  1& 0&  1& 0 & 1
\end{bmatrix}
$$
\end{comment}

\section{Results}
 We use two polynomials and their associated curves to obtain codes in
 $\mathbb{F}_{16}$ better than the best known codes in \cite{mint}. The
 two polynomials are given in  Table  \ref{poly2} while Table \ref{poly}
 gives a summary of the properties of their associated  curves (with $t=4$). The number of places of degree $j$, $A_j$, is determined by computer algebra system {\small MAGMA}~\cite{magma}. The best known linear codes from \cite{mint} over $\mathbb{F}_{16}$ with $j=d_j$ for $1 \leq j\leq 4$ are
\begin{equation*}
[1,1,1]_{16}~~~[3,2,2]_{16}~~~[5,3,3]_{16}~~~[7,4,4]_{16}
\end{equation*}
which correspond to $C_1$, $C_2$, $C_3$ and $C_4$ respectively. Since $t=4$ for all the codes in this paper and
$$
[d_1,d_2,d_3,d_4]=[1,2,3,4]
$$
we shorten the representation \\
$$
C(k;t;B_1,B_2,\dots,B_t;d_1,d_2,\dots,d_t) \equiv C(k;B_1,B_2,\dots,B_t).
$$
Tables \ref{f1}-\ref{f2} give codes obtained from the two curves
associated with the two polynomials $F_i$ for $1\leq i\leq 2$ that
improve on the best constructible codes in the tables in
\cite{mint}. Table \ref{f3} gives new codes that improve on both
constructible and non-constructible codes in \cite{mint}. It is also worth noting that codes of the form $C(k;N,0,0,0)$ are simply Goppa codes (defined with only rational points). The symbol \# in the Tables \ref{f1}-\ref{f2} denotes the number of new codes from  each generalised AG code $C(k;B_1,B_2,\dots,B_t)$ . The tables in \cite{geer2} contain curves known to have the most number of rational points for a given genus. Over $\mathbb{F}_{16}$ the curve with the highest number of points with genus $g=12$ from \cite{geer2} has $88$ rational points, was constructed using class field theory and is not defined by an explicit polynomial. On the other hand the curve $\mathcal{X}_{1}/\mathbb{F}_{16}$ obtained by Kummer covering of the projective line in \cite{shabat} has $A_1=83$ rational points and genus $g=12$ and is explicitly presented. Codes from this curve represent the best constructive codes in $\mathbb{F}_{16}$ with code length $83$. The curve $\mathcal{X}_{2}/\mathbb{F}_{16}$ is defined by the well-known Hermitian polynomial.
\begin{table}[!t]
\caption{Polynomials in $\mathbb{F}_{16}$}
\label{poly2}
\centering

\begin{tabular}{|p{7.5cm}|}
\hline
$\mathit{F}_1=x^5z^{10} + x^3z^{12} + xz^{14} + y^{15}$\\
\hline
$\mathit{F}_2=x^5+y^4z+yz^4$\\
\hline
\end{tabular}
\end{table}

\begin{table}[!t]
\caption{Properties of $\mathcal{X}_i/\mathbb{F}_{16}$}
\label{poly}
\centering
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
$F(x,y,z)$ & Genus & $A_1$ & $A_2$ & $A_3$ & $A_4$ & Reference\\
\hline
$\mathcal{X}_1$ & $12$ & $83$ & $60$ & $1320$ & $16140$ & \cite{shabat}\\
\hline
$\mathcal{X}_2$ & $6$ & $65$ & $0$ & $1600$ & $15600$ & \\
\hline
\end{tabular}
\end{table}

\begin{table}[!t]
\caption{Best Constructible Codes from $\mathcal{X}_1$}
\label{f1}
\centering
\begin{tabular}{|l|c|l|c|}
\hline
Codes & $k$ Range & Description & \#\\
\hline
$[83,k,d\geq 72-k]_{16}$& $8 \leq k \leq 52$ & $C(k;[83,0,0,0])$& $45$\\
\hline

$[89,k,d\geq 76-k]_{16}$& $9 \leq k \leq 54$ & $C(k;[83,2,0,0])$& $46$\\
\hline
$[94,k,d\geq 79-k]_{16}$& $10 \leq k \leq 57$ & $C(k;[83,2,1,0])$& $48$\\
\hline

$[92,k,d\geq 78-k]_{16}$& $9 \leq k \leq 57$ & $C(k;[83,3,0,0])$ & $49$\\
\hline

$[98,k,d\geq 82-k]_{16}$& $11 \leq k \leq 59$ & $C(k;[83,5,0,0])$ & $49$\\
\hline

\end{tabular}
\end{table}

\begin{table}[!t]
\caption{Best Constructible Codes from $\mathcal{X}_2$}
\label{f2}
\centering
\begin{tabular}{|l|c|l|c|}
\hline
Codes & $k$ Range & Description & \#\\
\hline
$[72,k,d\geq 64-k]_{16}$& $11 \leq k \leq 50$ & $C(k;[65,0,0,1])$ & $40$ \\
\hline
$[79,k,d\geq 68-k]_{16}$& $11 \leq k \leq 48$ & $C(k;[65,0,0,2])$ & $38$ \\
\hline
$[77,k,d\geq 67-k]_{16}$& $10 \leq k \leq 51$ & $C(k;[65,0,1,1])$ & $42$\\
\hline

$[75,k,d\geq 66-k]_{16}$& $9 \leq k \leq 51$ & $C(k;[65,0,2,0])$& $43$\\
\hline

\end{tabular}
\end{table}

\begin{table}[!t]
\caption{New Codes from $\mathcal{X}_1$}
\label{f3}
\centering
\begin{tabular}{|l|c|l|c|}
\hline
Codes & $k$ Range & Description & \#\\
\hline
$[70,k,d\geq 63-k]_{16}$& $10 \leq k \leq 50$ & $C(k;[65,0,1,0])$ & $41$ \\
\hline
\end{tabular}
\end{table}

\begin{table}[!t]
\caption{New Codes in $\mathbb{F}_{16}$}
\label{f4}
\centering
\begin{tabular}{|c|c|c||c|}
\hline
 $N$ & $K$ & $D$ & $D_m$ \\
\hline
$70$ & $10$ & $53$ & $52$\\
\hline
$70$ & $11$ & $52$ & $51$\\
\hline
$70$ & $12$ & $51$ & $50$\\
\hline
$70$ & $13$ & $50$ & $49$\\
\hline
$70$ & $14$ & $49$ & $48$\\
\hline
$70$ & $15$ & $48$ & $47$\\
\hline
$70$ & $16$ & $47$ & $46$\\
\hline
$70$ & $17$ & $46$ & $45$\\
\hline
$70$ & $18$ & $45$ & $44$\\
\hline
$70$ & $19$ & $44$ & $43$\\
\hline
$70$ & $20$ & $43$ & $42$\\
\hline
$70$ & $21$ & $42$ & $41$\\
\hline
$70$ & $22$ & $41$ & $40$\\
\hline
$70$ & $23$ & $40$ & $39$\\
\hline
$70$ & $24$ & $39$ & $38$\\
\hline
$70$ & $25$ & $38$ & $37$\\
\hline
$70$ & $26$ & $37$ & $36$\\
\hline
$70$ & $27$ & $36$ & $35$\\
\hline
$70$ & $28$ & $35$ & $34$\\
\hline
$70$ & $29$ & $34$ & $33$\\
\hline
$70$ & $30$ & $33$ & $32$\\
\hline
$70$ & $31$ & $32$ & $31$\\
\hline
$70$ & $32$ & $31$ & $30$\\
\hline
$70$ & $33$ & $30$ & $29$\\
\hline
$70$ & $34$ & $29$ & $28$\\
\hline
$70$ & $35$ & $28$ & $27$\\
\hline
$70$ & $36$ & $27$ & $26$\\
\hline
$70$ & $37$ & $26$ & $25$\\
\hline
$70$ & $38$ & $25$ & $24$\\
\hline
$70$ & $39$ & $24$ & $23$\\
\hline
$70$ & $40$ & $23$ & $22$\\
\hline
$70$ & $41$ & $22$ & $21$\\
\hline
$70$ & $42$ & $21$ & $20$\\
\hline
$70$ & $43$ & $20$ & $19$\\
\hline
$70$ & $44$ & $19$ & $18$\\
\hline
$70$ & $45$ & $18$ & $17$\\
\hline
$70$ & $46$ & $17$ & $16$\\
\hline
$70$ & $47$ & $16$ & $15$\\
\hline
$70$ & $48$ & $15$ & $14$\\
\hline
$70$ & $49$ & $14$ & $13$\\
\hline
$70$ & $50$ & $13$ & $12$\\
\hline
\end{tabular}
\end{table}

Table \ref{f4} gives the new codes obtained from $\mathcal{X}_2$. The
codes have length $N$, dimension $K$ and minimum distance $D$. $D_m$ is
the lower bound on the minimum distance of codes from \cite{mint} with the same length and
dimension as the constructed generalised AG codes.
\section{Conclusion}

We have presented $41$ new codes  and $400$
improvements on constructible codes in $\mathbb{F}_{16}$  over the codes in
\cite{mint}. The construction method yields many good codes as shown
here,  as long as curves with many places of small degree are used,
however traditional search for good codes has focused primarily on
finding rational curves with many points. In order to obtain more codes
from generalised AG codes, curves with small genera and many places of small degree need to be found.

\newpage

\bibliographystyle{IEEEtran}

\bibliography{IEEEabrv,references}

\end{document}

config: arxiv


TITLE: Showing $\left ( a^{-1} ba\right )^{n}=a^{-1}b^{n}a$
QUESTION [0 upvotes]: Prove that $\left ( a^{-1} ba\right )^{n}=a^{-1}b^{n}a$.

Here's what I have: for positive $n$, $\left ( a^{-1}ba \right )^{n}=a^{n}b^{n}a^{-n}$ by the socks and shoes property.
Any hints for me to go further?

REPLY [3 votes]: Notice that
$$(aba^{-1})^n=aba^{-1}\cdot aba^{-1}\cdot \ldots \cdot aba^{-1} = ab(a^{-1}\cdot a)ba^{-1}\cdot\ldots \cdot aba^{-1}$$

REPLY [2 votes]: Hint: For $n = 2$ we have $(a^{-1}ba)^2 = a^{-1}baa^{-1}ba$. Now simplify.


TITLE: Select $n^2 + 1$ points in the unit square. Show that at least two points are no more than a distance $\sqrt{2}/n$ apart
QUESTION [9 upvotes]: I know I need to use the pigeonhole principle to prove this, but I don't know exactly how.
What I think I could do is divide the unit square into $n^2$ squares.
Using Pythagoras theorem, the maximum distance between 2 points in these squares is $\sqrt{2}/n$ (diagonals).
Then by the pigeonhole principle there must be a square with $2$ points in it since there are $n^2$ squares and $n^2 + 1$ points.
Would this be a correct way to prove the statement?

REPLY [3 votes]: Your argument is correct.
Of course, you have to think about that the squares overlap, so a point could be in more than one square. But the pigeonhole principle still works if some pigeons get duplicated and put into multiple holes.


TITLE: Topological and Metric space sequence convergence
QUESTION [0 upvotes]: Proposition:
In a metric space $(X,d)$, the topological definition of convergence is equivalent to the metric space definition.
My Proof:
Let $(X,d)$ be a metric space. Suppose $\forall \epsilon>0$ $\exists N\in \mathbb{N}$ $:$ $\forall n\in \mathbb{N}$, $n\geq N$ $\implies$ $x_n \in B(x,\epsilon)$. Let $U$ be an arbitrary open set containing $x$, so by definition $\exists r>0$ $:$ $B(x,r) \subseteq U$.
So by assumption, $\exists N\in \mathbb{N}$  such that  for $n\in \mathbb{N}$ whenever $n\geq N$ then $x_n \in B(x,r) \subseteq U$.
For the converse, suppose $\forall U(open)$ $\owns x$ $\exists N\in \mathbb{N}$ $:$ $\forall n\in \mathbb{N}$, $n\geq N$ $\implies$ $x_n \in U$.
Let $\epsilon >0$. Consider $B(x,\epsilon)$. As an open ball is an open set, there exists a natural number $N$ such that for any other natural number n satisfying $n\geq N$ we have $x_n \in B(x,\epsilon)$
Is my proof correct? May I have feedback, please?

REPLY [1 votes]: To have an answer (so the question does not remain forever unanswered):
Yes, this is correct. In short, a metric open set contains a ball around each of its points so we can apply the $\varepsilon$-definition to the used radius and vice versa open balls are open, so the topological version applies to them.
Next on the list: continuity?


TITLE: Area of shaded region circle help
QUESTION [3 upvotes]: Find the area of the shaded region

Area of the sector is $240^\circ$ or $\frac{4\pi}{3}$
Next find $\frac{b\cdot h}{2}$ which is $\frac{2\cdot2}{2}$ which is $2$.
Then subtract the former from the latter: $\frac{4\pi}{3} - 2$
Therefore the answer is $~2.189$?
Is this correct?

REPLY [0 votes]: No, the  area that  corresponds to 120 degrees  or $\frac{2\pi}{3}$ is one third of the total area of the circle $4\pi$ the area of the triangle is $\frac{bh}{2}$ but $h$ is not $2$ you need to project it,  since $\cos (\, 30 \text{ degrees}) = \frac{\sqrt{3}}{2}$ degrees to find $h = \sqrt{3}$ Therefore
$$\text{area} = \frac{4\pi}{3} - \sqrt{3}  = 2.4567 \cdots  $$


TITLE: Show that for all integers $k \ge 2$ every graph $G$ contains a $k$-partite subgraph $H$ with $e(H) \ge \frac{k-1}{k}e(G)$.
QUESTION [1 upvotes]: Show that for all integers $k \ge 2$ every graph $G$  contains a $k$-partite subgraph $H$ with $e(H) \ge \frac{k-1}{k}e(G)$.

REPLY [1 votes]: See my answer to this question for $k=3$. The proof for an arbitrary $k$ is similar.


TITLE: Find the first four nonzero terms in the Maclaurin of $e^x \sin x$
QUESTION [0 upvotes]: Given the following theorem $f$ and $g$
    $$f(x) = \sum_{n=0}^{\infty} a_n (x-x_0)^n, |x-x_0| < R_1 \\ g(x) = \sum_{n=0}^{\infty} b_n (x-x_0)^n, |x-x_0| < R_2$$
    then
    $$f(x)g(x) = \sum_{n=0}^{\infty} c_n (x-x_0)^n, |x-x_0| < R$$
    Where
    $$c_n =  \sum_{r=0}^{n} a_rb_{n-r} = \sum_{r=0}^{n} a_{n-r}b_{r} \\ R \geq \min \{R_1, R_2\}$$

Use the theorem above to find the first four nonzero terms in the Maclaurin of $$e^x \sin x$$

My understanding RE-EDIT based on Answers
$$f(x) \rightarrow 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \frac{1}{4!}x^4 +\frac{1}{5!} x^5 ...
\\ g(x) \rightarrow 0+ x + 0 - \frac{1}{3!} x^3  +    0  + \frac{1}{5!} x^5 ...  $$
$$c_0 = \sum_{r=0}^{0} a_r b_{n-r} = a_0b_0 = 1 \cdot 0 = 0$$
$$c_1 = \sum_{r=0}^{1} a_r b_{n-r} = a_0b_1 +a_1b_0  = 1 \cdot 1 + 1 \cdot 0 =1$$
$$c_2 = \sum_{r=0}^{2} a_r b_{n-r} = a_0b_2 + a_1b_1 + a_2b_0 = 1$$
$$c_3 = ... = \frac{1}{3}$$
$$c_4 = ... = 0$$
$$c_5 = ... = - \frac{1}{30}$$
The first four nonzero terms of $e^x \sin x$:
$$ \implies c_1 x^1+ c_2 x^2  +c_3 x^3 + c_5 x^5 $$
$$ \implies  x+ x^2 +\frac{1}{3} x^3 - \frac{1}{30} x^5$$
Would this be correct?
Thx for your input

REPLY [2 votes]: $$1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \frac{1}{4!}x^4 +\cdots\\\times
\\ 0+ x + 0 - \frac{1}{3!} x^3  +    0  +  \frac{1}{5!} x^5+\cdots$$
Collect the products of coefficients leading to a common power and you get
$$x^0\to1\cdot0\\
x^1\to1\cdot0+1\cdot1\\
x^2\to\frac1{2!}0+1\cdot1+1\cdot0\\
x^3\to\frac1{3!}0+\frac1{2!}1+1\cdot0-1\frac1{3!}\\
x^4\to\frac1{4!}0+\frac1{3!}1+\frac1{2!}0-1\frac1{3!}+1\cdot0\\
x^5\to\frac1{5!}0+\frac1{4!}1+\frac1{3!}0-\frac1{2!}\frac1{3!}-1\cdot0+1\cdot\frac1{5!}$$


TITLE: Find $\tan x$ if $x=\arctan(2 \tan^2x)-\frac{1}{2}\arcsin\left(\frac{3\sin2x}{5+4\cos 2x}\right)$
QUESTION [3 upvotes]: Find $\tan x$   if $$x=\arctan(2 \tan^2x)-\frac{1}{2}\arcsin\left(\frac{3\sin2x}{5+4\cos 2x}\right) \tag{1}$$
First i converted $$\frac{3 \sin 2x}{5+4 \cos 2x}=\frac{6 \tan x}{9+\tan^2 x}$$
So
$$\arcsin\left( \frac{6 \tan x}{9+\tan^2x}\right)=\arctan \left( \frac{6 \tan x}{9-\tan^2x}\right)$$ Now using above result $(1)$ can be written as
$$2x=2 \arctan(2 \tan^2 x)-\arctan \left( \frac{6 \tan x}{9-\tan^2x}\right) \tag{2}$$
But $$2\arctan (\theta)=\arctan \left(\frac{2 \theta}{1-\theta^2}\right)$$
So $(2)$ becomes
$$2x=\arctan \left(\frac{4 \tan^2x}{1-4 \tan^4x}\right)-\arctan \left( \frac{6 \tan x}{9-\tan^2x}\right)$$
Now using $$\arctan(a)-\arctan(b)=\arctan\left(\frac{a-b}{1+ab}\right)$$ i am getting a sixth degree polynomial in $\tan x$.
is there any better approach?

REPLY [1 votes]: Where you have left off,
$$\dfrac{6\tan x}{9-\tan^2x}=\dfrac{2\cdot\dfrac{\tan x}3}{1-\left(\dfrac{\tan x}3\right)^2}$$
Using my answer here, Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,
$$2\arctan\dfrac{\tan x}3=\begin{cases} \arctan\dfrac{6\tan x}{9-\tan^2x} &\mbox{if } \left(\dfrac{\tan x}3\right)^2\le1\iff-3\le\tan x\le3\\ \pi+\arctan\dfrac{6\tan x}{9-\tan^2x} & \mbox{if } \tan x>3\text{ or }\tan x<-3\end{cases}$$
Now the tougher case $\tan x>3$ or $\tan x<-3;$
$$x=\arctan(2\tan^2x)+\dfrac\pi2-\arctan\dfrac{\tan x}3$$
$$\tan x=\tan\left(\dfrac\pi2+\arctan(2\tan^2x)-\arctan\dfrac{\tan x}3\right)$$ $$=-\cot\left(\arctan(2\tan^2x)-\arctan\dfrac{\tan x}3\right)$$
$$\iff\tan x=-\dfrac{1+2\tan^2x\cdot\dfrac{\tan x}3}{2\tan^2x-\dfrac{\tan x}3}$$
On simplification, we should get a cubic equation in $\tan x$
Can you handle the simpler case  $-3\le\tan x\le3$ to reach at $$\tan x(\tan^3x-3\tan x+2)=0$$