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\centerline{\Large{\bf Cotton Solitons on Almost Kenmotsu 3-$h$-Manifolds}}
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\centerline{\large{\textbf{Dibakar Dey and Pradip Majhi}}}
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\centerline{Department of Pure Mathematics}
\centerline{ University of Calcutta,
35, Ballygunge Circular Road}
\centerline{Kolkata-700019, West Bengal, India }
\centerline{E-mail: deydibakar3@gmail.com and Pradip Majhi}

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\textbf{Abstract:} In this paper, we consider the notion of Cotton soliton within the framework of almost Kenmotsu 3-$h$-manifolds. First we consider that the potential vector field is pointwise collinear with the Reeb vector field and prove a non-existence of such Cotton soliton. Next we assume that the potential vector field is orthogonal to the Reeb vector field. It is proved that such a Cotton soliton on a non-Kenmotsu almost Kenmotsu 3-$h$-manifold such that the Reeb vector field is an eigen vector of the Ricci operator is steady and the manifold is locally isometric to $\mathbb{H}^2(-4) \times \mathbb{R}$.\\

\textbf{Mathematics Subject Classification 2010:} Primary 53D15; Secondary 35Q51.\\

\textbf{Keywords:} Almost Kenmotsu manifolds, Cotton solitons, Non-unimodular Lie group.\\

\section{Introduction}

An almost contact metric manifold is an odd dimensional differentiable manifold $M^{2n+1}$ together with a structure $(\varphi,\xi,\eta,g)$ satisfying (\cite{bl}, \cite{bll})
\be
\varphi^{2}X = - X + \eta(X)\xi,\;\; \eta(\xi)=1, \label{1.1}
\ee
\be
 g(\varphi X,\varphi Y) = g(X,Y) - \eta(X)\eta(Y) \label{1.2}
\ee
for all vector fields $X$, $Y$ on $M^{2n+1}$, where $g$ is the Riemannian metric, $\varphi$ is a $(1,1)$-tensor field, $\xi$ is a unit vector field called the Reeb vector field and $\eta$ is a 1-form defined by $\eta(X) = g(X,\xi)$. Here also $\phi\xi=0$ and $\eta\circ\phi=0$; both can be derived from \eqref{1.1} easily. The fundamental 2-form $\Phi$ on an almost contact metric manifold is defined by $\Phi(X,Y)=g(X,\varphi Y)$ for all vector fields $X$, $Y$ on $M^{2n+1}$. The condition for an almost contact metric manifold being normal is equivalent to vanishing of the $(1,2)$-type torsion tensor $N_{\varphi}$, defined by $N_{\varphi}=[\varphi,\varphi]+2d\eta\otimes\xi$, where $[\varphi,\varphi]$ is the Nijenhuis tensor of $\varphi$ \cite{bl}. Almost contact metric manifold such that $\eta$ is closed and $d\Phi=2\eta\wedge\Phi$  are called almost Kenmotsu manifolds (see \cite{dp},\cite{dp2}). Obviously, a normal almost Kenmotsu manifold is a Kenmotsu manifold. Also Kenmotsu manifolds can be characterized by $(\nabla_{X}\varphi)Y = g(\varphi X,Y)\xi-\eta(Y)\varphi X$, for any vector fields $X$, $Y$ on $M^{2n+1}$.\\

The $(0,3)$-Cotton tensor of a 3-dimensional Riemannian manifold $(M^3,g)$ is defined as
\bea
\nonumber C(X,Y,Z) &=& (\nabla_X S)(Y,Z) - (\nabla_Y S)(X,Z) \\ && - \frac{1}{4}[(X(r)g(Y,Z) - Y(r)g(X,Z)], \label{1.3}
\eea
where $S$ is the Ricci tensor and $r$ is the scalar curvature of $M^3$. The Cotton tensor is skew-symmetric in first two indices and totally trace free. It is well known that for $n \geq 4$, an $n$-dimensional Riemannian manifold is conformally flat if the Weyl tensor vanishes. For $n = 3$, the Weyl tensor always vanishes but the Cotton tensor does not vanish in general.\\

In 2008, Kicisel, Sario\u{g}lu and Tekin \cite{tekin} introduced the notion of Cotton flow as a analogy of the Ricci flow. The Cotton flow is based on the conformally invariant Cotton tensor and defined exclusively foe 3-simension as
\bea
\nonumber \frac{\partial g}{\partial t} = C,
\eea
where $C$ is the $(0,2)$-Cotton tensor of $g$. From the Cotton flow, they defined the notion of Cotton soliton as follows:
\begin{definition}
A Cotton soliton is a metric $g$ defined on 3-dimensional smooth manifold $M^3$ such that the following equation
\bea
(\mathcal{L}_V g)(X,Y) + C(X,Y) - \sigma g(X,Y) = 0, \label{1.4}
\eea
holds for a constant $\sigma$ and a vector field $V$, called the potential vector field, where $\mathcal{L}_V$ denotes the Lie derivative along $V$ and $C$ is the $(0,2)$-Cotton tensor defined by
\bea
C_{ij} = \frac{1}{2\sqrt{g}}C_{nmi}\epsilon^{nml}g_{lj} \label{1.5}
\eea
in a local frame of $M^3$, where $g = \operatorname{det}(g_{ij})$, $C_{ijk}$ is the $(0,3)$-Cotton tensor and $\epsilon$ is a tensor density.
\end{definition}
In an orthonormal frame, $\epsilon^{123} = 1$. Also exchange of any two indices will give rise to minus sign and it will be zero if there has two same indices. For example, $\epsilon^{231} = - \epsilon^{213}$ and $\epsilon^{112} = \epsilon^{122} = \epsilon^{223} = 0$. Cotton solitons are fixed points of the Cotton flow upto diffeomorphisms and rescaling. The Cotton soliton is said to be shrinking, steady or expanding according as $\sigma$ is positive, zero or negative respectively. As far as we know, the Cotton soliton was studied by Chen \cite{chen} on certain almost contact metric manifold, precisely on almost coK\"{a}hler 3-manifolds. Motivated by the study of Chen \cite{chen}, we consider the notion of Cotton soliton on an almost Kenmotsu 3-$h$-manifold and prove some related results.

\section{Almost Kenmotsu 3-$h$-Manifolds}

Let $(M^3,\varphi,\xi,\eta,g)$ be a 3-dimensional almost Kenmotsu manifold. We denote by $l = R(\cdot,\xi)\xi$, $h = \frac{1}{2}\mathcal{L}_\xi \varphi$ and $h' = h \circ \varphi$ on $M^3$, where $R$ is the Riemannian curvature tensor. The tensor fields $l$ and $h$ are symmetric operators and satisfy the following relations (\cite{dp}, \cite{dp2}):
 \bea
h\xi = 0,\;l\xi = 0,\;tr(h) = 0,\;tr(h\varphi) = 0,\;h\varphi+\varphi h = 0, \label{2.1}
 \eea
\bea
 \nabla_{X}\xi = X - \eta(X)\xi - \varphi hX(\Rightarrow \nabla_{\xi}\xi=0), \label{2.2}
\eea
\bea
\nabla_\xi h = - \varphi - 2h - \varphi h^2 - \varphi l. \label{2.3}
\eea
\begin{definition} $\cite{ww}$
A 3-dimensional almost Kenmotsu manifold is called an almost Kenmotsu 3-$h$-manifold if it satisfies $\nabla_\xi h = 0$.
\end{definition}

Let $\mathcal{U}_1$ be the maximal open subset of a 3-dimensional almost Kenmotsu manifold $M^3$ such that $h \neq 0$ and $\mathcal{U}_2$ be the maximal open subset on which $h = 0$. Then $\mathcal{U}_1 \cup \mathcal{U}_2$ is an open and dense subset of $M^3$. Then $\mathcal{U}_1$ is non-empty and there is a local orthonormal basis $\{e_1 = \xi, e_2 = e, e_3 = \varphi e\}$ on $\mathcal{U}_1$ such that $he = \lambda e$ and $h\varphi e = - \lambda \varphi e$ for some positive function $\lambda$. Since $\nabla_\xi h = 0$ for an almost Kenmotsu 3-$h$-manifold, then using Lemma 6 of \cite{cho} and \eqref{2.3}, a direct calculation gives $\xi(\lambda) = a = 0$. Therefore, Lemma 6 of \cite{cho} can be rewritten for an almost Kenmotsu 3-$h$-manifold as
\begin{lemma} \label{lem2.2}
On $\mathcal{U}_1$, the coefficients of the Riemannian connection $\nabla$ of an almost Kenmotsu 3-$h$-manifold with respect to a local orthonormal basis $\{\xi,e,\varphi e\}$ is given by
$$\nabla_\xi \xi = 0,\;\; \nabla_\xi e = 0,\;\; \nabla_\xi \phi e = 0,$$
$$\nabla_e \xi = e - \lambda \varphi e,\;\; \nabla_e e = - \xi - b\varphi e,\;\; \nabla_e \varphi e = \lambda \xi + be,$$
$$\nabla_{\varphi e} \xi = - \lambda e + \varphi e,\;\; \nabla_{\varphi e} e = \lambda \xi + c\varphi e,\;\; \nabla_{\varphi e} \varphi e = - \xi - ce,$$
where $b$ and $c$ are smooth functions.
\end{lemma}
From Lemma \ref{lem2.2}, the Lie brackets can be calculated as follows:
\bea
[e,\xi] = e - \lambda \varphi e,\;\; [e,\varphi e] = be - c\varphi e\;\; \mathrm{and}\;\; [\varphi e,\xi] = - \lambda e + \varphi e. \label{2.4}
\eea
From Lemma 3.2 of \cite{wang2}, the Ricci operator $Q$ of an almost Kenmotsu 3-$h$-manifold can be obtained as
\begin{lemma} \label{lem2.3}
On $\mathcal{U}_1$, the Ricci operator of an almost Kenmotsu 3-$h$-manifold with respect to a local orthonormal basis $\{\xi,e,\varphi e\}$ is given by
$$Q\xi = -2(\lambda^2 + 1)\xi - [\varphi e(\lambda) + 2\lambda b]e - [e(\lambda) + 2\lambda c]\varphi e,$$
$$Qe = - [\varphi e(\lambda) + 2\lambda b]\xi - fe + 2\lambda \varphi e,$$
$$Q\varphi e = - [e(\lambda) + 2\lambda c]\xi + 2\lambda e - f\varphi e,$$
where $f = e(c) + \varphi e(b) + b^2 + c^2 + 2$.
\end{lemma}
The scalar curvature $r$ of an almost Kenmotsu 3-$h$-manifold is given by
\bea
r = g(Qe_i,e_i) = - 2(\lambda^2 + 1) - 2f. \label{2.5}
\eea
Using Lemma \ref{lem2.3}, we obtain
\bea \label{2.6}
\begin{cases}
  S(\xi,\xi) = -2(\lambda^2 + 1),\;  S(\xi,e) = - [\varphi e(\lambda) + 2\lambda b], \\
  S(\xi,\varphi e) = - [e(\lambda) + 2\lambda c],\;  S(e,e) = -f, \\
   S(e,\varphi e) = 2\lambda, \; S(\varphi e,\varphi e) = - f.
 \end{cases}
\eea
It is well known that an almost Kenmotsu 3-manifold is Kenmotsu if and only if $h = 0$. Thus a Kenmotsu metric always admits a almost Kenmotsu 3-$h$-metric structure. We now give an example of a non-Kenmotsu almost Kenmotsu 3-$h$-manifold.
\begin{example} $\cite{wa2}$
Let $M^3$ be a 3-dimensional non-unimodular Lie group with a left invariant local orthonormal frame $\{e_1,e_2,e_3\}$ satisfying
$$[e_1,e_2] = \alpha e_2 + \beta e_3,\; [e_2,e_3] = 0 \; \mathrm{and} \; [e_1,e_3] = \beta e_2 + (2 - \alpha)e_3$$
for $\alpha, \beta \in \mathbb{R}$. If either $\alpha \neq 1$ or $\beta \neq 0$, then $M^3$ admits a non-Kenmotsu almost Kenmotsu 3-$h$-metric struicture.
\end{example}

\section{Cotton Soliton}

In this section, we consider the notion of Cotton soliton within the framework of almost Kenmotsu 3-$h$-manifolds. To study the notion of Cotton soliton, we need to compute the components of the $(0,2)$-Cotton tensor. In this regard, we prove the following Lemma:
\begin{lemma} \label{lem3.1}
The components of the $(0,2)$-Cotton tensor $C$ with respect to an orthonormal frame $\{\xi,e,\varphi e\}$ of a non-Kenmotsu almost Kenmotsu 3-$h$-manifold $M^3$ can be expressed as follows:
\bea
\nonumber  C_{11} = C(\xi,\xi) &=& b[\varphi e(\lambda) + 2\lambda b] - c[e(\lambda) + 2\lambda c] \\ && - e(e(\lambda) + 2\lambda c) + \varphi e(\varphi e(\lambda) + 2\lambda b), \label{3.1}
\eea
\bea
\nonumber  C_{12} = C(\xi,e) &=& 2[e(\lambda) -3\lambda \varphi e(\lambda) + 2\lambda c - 2\lambda^2 b] \\ && + \xi(e(\lambda) + 2\lambda c) - \frac{1}{4}\varphi e(r), \label{3.2}
\eea
\bea
\nonumber  C_{13} = C(\xi,\varphi e) &=& -2[\varphi e(\lambda) -3\lambda e(\lambda) + 2\lambda b - 2\lambda^2 c] \\ && -\xi(\varphi e(\lambda) + 2\lambda b) + \frac{1}{4}e(r), \label{3.3}
\eea
\bea
  C_{22} = C(e,e) = 2\lambda^3 - f\lambda +c[e(\lambda) + 2\lambda c]  - \varphi e(\varphi e(\lambda) + 2\lambda b), \label{3.4}
\eea
\be
C_{23} = C(e,\varphi e) = - \xi(f) - f + 2 + e(\varphi e(\lambda) + 2\lambda b) + b[e(\lambda) + 2\lambda c] - \frac{1}{4}\xi(r), \label{3.5}
\ee
\be
C_{33} = C(\varphi e,\varphi e) = - 2\lambda^3 + f\lambda - b[\varphi e(\lambda) + 2\lambda b] + e(e(\lambda) + 2\lambda c). \label{3.6}
\ee
\end{lemma}
\begin{proof}
The components of the metric tensor $g$ with respect to an orthonormal frame $\{\xi,e,\varphi e\}$ of a non-Kenmotsu almost Kenmotsu 3-$h$-manifold $M^3$ is given by
\bea
\nonumber (g_{ij}) = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)
\eea
and hence $\operatorname{det}(g_{ij}) = 1$. Therefore, equation \eqref{1.5} reduces to
\bea
\nonumber C_{ij} = \frac{1}{2}C_{nmi}\epsilon^{nmj},\;\; i,j = 1,2,3,
\eea
where $C_{ijk} = C(e_i,e_j,e_k)$. Also, $C_{ijk} = - C_{jik}$ and $C_{iik} = 0$ for all $i,j,k = 1,2,3$. It can be easily obtained that (see \cite{chen})
\bea
\nonumber C_{11} = C_{231},\; C_{12} = C_{311},\; C_{13} = C_{121},\; C_{22} = C_{312},\; C_{23} = C_{122},\; C_{33} = C_{123}.
\eea
Making use of \eqref{1.3}, we get the followings:
\bea
 C_{11} = C_{231} = C(e,\varphi e,\xi) = (\nabla_e S)(\varphi e,\xi) - (\nabla_{\varphi e} S)(e,\xi), \label{3.7}
\eea
\bea
C_{12} = C(\varphi e,\xi,\xi) = (\nabla_{\varphi e} S)(\xi,\xi) - (\nabla_\xi S)(\varphi e,\xi) - \frac{1}{4}\varphi e(r), \label{3.8}
\eea
\bea
C_{13} = C(\xi,e,\xi) = (\nabla_\xi S)(e,\xi) - (\nabla_e S)(\xi,\xi) + \frac{1}{4}e(r), \label{3.9}
\eea
\bea
C_{22} = C(\varphi e,\xi,e) = (\nabla_{\varphi e} S)(\xi,e) - (\nabla_\xi S)(\varphi e,e), \label{3.10}
\eea
\bea
C_{23} = C(\xi,e,e) = (\nabla_\xi S)(e,e) - (\nabla_e S)(\xi,e) - \frac{1}{4}\xi(r), \label{3.11}
\eea
\bea
 C_{33} = C(\xi,e,\varphi e) = (\nabla_\xi S)(e,\varphi e) - (\nabla_e S)(\xi,\varphi e). \label{3.12}
\eea
Using \eqref{2.6}, Lemma \ref{lem2.2} and $\xi(\lambda) = 0$, we now obtain the followings:
\bea
\begin{cases}
(\nabla_e S)(\varphi e,\xi) = 2\lambda^3 - f\lambda + b[\varphi e(\lambda) + 2\lambda b] - e(e(\lambda) + 2\lambda c),\\
(\nabla_{\varphi e} S)(e,\xi) = 2\lambda^3 - f\lambda + c[e(\lambda) + 2\lambda c] -\varphi e(\varphi e(\lambda) + 2\lambda b). \label{3.13}
\end{cases}
\eea
\bea
\begin{cases}
(\nabla_{\varphi e} S)(\xi,\xi) = 2[e(\lambda) -3\lambda \varphi e(\lambda) + 2\lambda c - 2\lambda^2 b],\\
(\nabla_\xi S)(\varphi e,\xi) = - \xi(e(\lambda) + 2\lambda c). \label{3.14}
\end{cases}
\eea
\bea
\begin{cases}
(\nabla_\xi S)(e,\xi) = -\xi(\varphi e(\lambda) + 2\lambda b),\\
(\nabla_e S)(\xi,\xi) = 2[\varphi e(\lambda) -3\lambda e(\lambda) + 2\lambda b - 2\lambda^2 c]. \label{3.15}
\end{cases}
\eea
\bea
\begin{cases}
(\nabla_{\varphi e} S)(\xi,e) = 2\lambda^3 - f\lambda + c[e(\lambda) + 2\lambda c] -\varphi e(\varphi e(\lambda) + 2\lambda b),\\
(\nabla_\xi S)(\varphi e,e) = 0. \label{3.16}
\end{cases}
\eea
\bea
\begin{cases}
(\nabla_\xi S)(e,e) = -\xi(f),\\
(\nabla_e S)(\xi,e) = f - 2 - e(\varphi e(\lambda) + 2\lambda b) - b[e(\lambda) + 2\lambda c]. \label{3.17}
\end{cases}
\eea
\bea
\begin{cases}
(\nabla_\xi S)(e,\varphi e) = 0,\\
(\nabla_e S)(\xi,\varphi e) = 2\lambda^3 - f\lambda + b[\varphi e(\lambda) + 2\lambda b] - e(e(\lambda) + 2\lambda c). \label{3.18}
\end{cases}
\eea
We now complete the proof by substituting the equations \eqref{3.13}-\eqref{3.18}
in the equations \eqref{3.7}-\eqref{3.12} respectively.
\end{proof}

\begin{proposition} \label{prop3.2}
If the Reeb vector field of a non-Kenmotsu almost Kenmotsu 3-$h$-manifold $M^3$ is an eigen vector of the Ricci operator, then  $M^3$ is locally isometric to a non-unimodular Lie group equipped with a left invariant non-Kenmotsu almost Kenmotsu structure.
\end{proposition}
\begin{proof}
Since $\xi$ is an eigen vector of $Q$, then Lemma \ref{lem2.3} implies
\bea
\begin{cases}
\varphi e(\lambda) + 2\lambda b = 0,\\
e(\lambda) + 2\lambda c = 0. \label{3.19}
\end{cases}
\eea
It is well known that $$\frac{1}{2}X(r) = (\operatorname{div}Q)X = \sum_{i=1}^3 g((\nabla_{e_i} Q)X,e_i).$$
From the preceding equation, we can write
\bea
\frac{1}{2}X(r) = (\nabla_\xi S)(X,\xi) + (\nabla_e S)(X,e) + (\nabla_{\varphi e} S)(X,\varphi e). \label{3.20}
\eea
Making use of \eqref{2.6}, \eqref{3.19} and $\xi(\lambda) = 0$, we obtain the followings:
\be
(\nabla_\xi S)(\xi,\xi) = 0,\; (\nabla_e S)(\xi,e) = f - 2,\; (\nabla_{\varphi e} S)(\xi,\varphi e) = f - 2, \label{3.21}
\ee
\be
(\nabla_\xi S)(e,\xi) = 0,\; (\nabla_e S)(e,e) = 4\lambda b - e(f),\; (\nabla_{\varphi e} S)(e,\varphi e) = - 4\lambda b, \label{3.22}
\ee
\be
(\nabla_\xi S)(\varphi e,\xi) = 0,\; (\nabla_e S)(\varphi e,e) = - 4\lambda c,\; (\nabla_{\varphi e} S)(\varphi e,\varphi e) = 4\lambda c - \varphi e(f).\label{3.23}
\ee
Now, substituting $X = \xi,\;e\;\mathrm{and}\;\varphi e$ in \eqref{3.20} and then using \eqref{3.21}, \eqref{3.22} and \eqref{3.23} respectively, we obtain
\bea
\xi(r) = 4(f - 2),\;\; e(r) = -2e(f),\;\; \varphi e(r) = -2\varphi e(f). \label{3.24}
\eea
Using \eqref{3.19} and $\xi(\lambda) = 0$, we get from \eqref{2.5}
\bea
\xi(r) = -2\xi(f),\; e(r) = -2e(f) + 8\lambda^2 c,\; \varphi e(r) = -2\varphi e(f) + 8\lambda^2 b. \label{3.25}
\eea
Since $\lambda$ is a positive function, then the second and third equations of \eqref{3.24} and \eqref{3.25} implies $b = c = 0$. From Lemma \ref{lem2.3}, we get $f = 2$. Also from \eqref{3.19}, we get $e(\lambda) = \varphi e(\lambda) = 0$ and therefore $\lambda$ is a constant. Now, the Lie brackets given in \eqref{2.4} reduces to
$$[e,\xi] = e - \lambda \varphi e,\;\; [e,\varphi e] = 0\;\; \mathrm{and}\;\; [\varphi e,\xi] = - \lambda e + \varphi e.$$
Therefore, according to Milnor \cite{milnor}, $M^3$ is locally isometric to a non-unimodular Lie group equipped with a left invariant non-Kenmotsu almost Kenmotsu structure.
\end{proof}
Combining Lemma \ref{lem3.1} and Proposition  \ref{prop3.2}, the components of the Cotton tensor can be written as given in the follwing Corollary.
\begin{corollary} \label{cor3.3}
If the Reeb vector field of a non-Kenmotsu almost Kenmotsu 3-$h$-manifold $M^3$ is an eigen vector of the Ricci operator, then the components of the $(0,2)$-Cotton tensor $C$ with respect to an orthonormal frame $\{\xi,e,\varphi e\}$ on $M^3$ can be expressed as follows:
$$ C_{11} = C(\xi,\xi) = 0,\;  C_{12} = C(\xi,e) = 0,\;  C_{13} = C(\xi,\varphi e) = 0,$$
$$ C_{22} = C(e,e) = 2\lambda^3 - 2\lambda,\; C_{23} = C(e,\varphi e) = 0,\; C_{33} = C(\varphi e,\varphi e) = - 2\lambda^3 + 2\lambda.$$
\end{corollary}

We first consider the Cotton soliton with potential vector field $V$  pointwise collinear with the Reeb vector field. In this regard, we prove the following non-existing result.

\begin{theorem} \label{thm3.4}
On a non-Kenmotsu almost Kenmotsu 3-$h$-manifold such that the Reeb vector field is an eigen vector of the Ricci operator, there exist no Cotton soliton with potential vector field pointwise collinear with the Reeb vector field.
\end{theorem}
\begin{proof}
Suppose that the potential vector field $V$ is pointwise collinear with the Reeb vector field $\xi$. Then there exist a non-zero smooth function $\alpha$ on $M^3$ such that $V = \alpha \xi$. Now, substituting $X = e$ and $Y = \varphi e$ in \eqref{1.4} and using Lemma \ref{lem2.2} and Corollary \ref{cor3.3}, we get $2\lambda \alpha = 0$. This gives either $\lambda = 0$ or $\alpha = 0$. In either cases, we get a contradiction. This completes the proof.
\end{proof}

From Theorem \ref{thm3.4} and Proposition \ref{prop3.2}, we can say the following:

\begin{corollary} \label{cor3.5}
On a 3-dimensional non-unimodular Lie group equipped with a left invariant non-Kenmotsu almost Kenmotsu structure, there exist no Cotton soliton with potential vector field pointwise collinear with the Reeb vector field.
\end{corollary}

It is now quite tempting to consider the potential vector field $V$ as orthogonal to the Reeb vector field. In this setting, we prove the following:

\begin{theorem} \label{thm3.6}
Let $(M^3,g)$ be a non-Kenmotsu almost Kenmotsu 3-$h$-manifold such that the Reeb vector field is an eigen vector of the Ricci operator. If $g$ is a Cotton soliton with potential vector field orthogonal to the Reeb vector field, then $M^3$ is locally isometric to $\mathbb{H}^2(-4) \times \mathbb{R}$ and the Cotton soliton is steady.
\end{theorem}
\begin{proof}
For a non-Kenmotsu almost Kenmotsu 3-$h$-manifold such that the Reeb vector field is an eigen vector of the Ricci operator, Proposition \ref{prop3.2} gives  $b = c = 0,\; f = 2,\; \lambda = \mathrm{constant}\; \mathrm{and}\; r = \mathrm{constant}$. Since $V$ is orthogonal to $\xi$, then there exist two smooth functions $\alpha_1$ and $\alpha_2$ on $M^3$ such that $V = \alpha_1 e + \alpha_2 \varphi e$. With the help of Lemma \ref{lem2.2}, we now obtain the components of $\mathcal{L}_V g$ as follows:
\bea
\begin{cases}
(\mathcal{L}_V g)(\xi,\xi) = 0,\; (\mathcal{L}_V g)(\xi,e) = \xi(\alpha_1) - \alpha_1 + \lambda \alpha_2,\\
(\mathcal{L}_V g)(\xi,\varphi e) = \xi(\alpha_2) - \alpha_2 + \lambda \alpha_1,\; (\mathcal{L}_V g)(e,e) = 2e(\alpha_1),\\
(\mathcal{L}_V g)(e,\varphi e) = e(\alpha_2) + \varphi e(\alpha_1),\; (\mathcal{L}_V g)(\varphi e,\varphi e) = 2\varphi e(\alpha_2). \label{3.26}
\end{cases}
\eea
We now use Corollary \ref{cor3.3} and the set of equations \eqref{3.26}. Substituting $X = Y = \xi$ in \eqref{1.4}, we get $\sigma = 0$. This shows that the Cotton soliton is steady. Now, substitution of $X = \xi$, $Y = e$ in \eqref{1.4} yields
\bea
\xi(\alpha_1) - \alpha_1 + \lambda \alpha_2 = 0. \label{3.27}
\eea
Replacing $X$ by $\xi$ and $Y$ by $\varphi e$ in \eqref{1.4}, we get
\bea
\xi(\alpha_2) - \alpha_2 + \lambda \alpha_1 = 0. \label{3.28}
\eea
Putting $X = Y = e$ in \eqref{1.4}, we obtain
\bea
2e(\alpha_1) + 2\lambda^3 - 2\lambda = 0. \label{3.29}
\eea
Substitution of $X = e$ and $Y = \varphi e$ in \eqref{1.4} yields
\bea
e(\alpha_2) + \varphi e(\alpha_1) = 0. \label{3.30}
\eea
Putting $X = Y = \varphi e$ in \eqref{1.4}, we infer
\bea
2\varphi e(\alpha_2) - 2\lambda^3 + 2\lambda = 0. \label{3.31}
\eea
Since $b = c = 0$, the Lie brackets given in \eqref{2.4} reduces to
\bea
[e,\xi] = e - \lambda \varphi e,\;\; [e,\varphi e] = 0\;\; \mathrm{and}\;\; [\varphi e,\xi] = - \lambda e + \varphi e. \label{3.32}
\eea
Since $\lambda$ is a positive constant, then from \eqref{3.27} and \eqref{3.29}, we obtain $$e(\xi(\alpha_1)) = e(\alpha_1) - \lambda e(\alpha_2)\;\; \mathrm{and}\;\; \xi(e(\alpha_1)) = 0.$$
Applying the first Lie bracket of \eqref{3.32} in the preceding eqaution, we get $\varphi e(\alpha_1) = e(\alpha_2)$. Hence, equation \eqref{3.30} implies $\varphi e(\alpha_1) = e(\alpha_2) = 0$. Now, from \eqref{3.28}, we get $e(\xi(\alpha_2)) = - \lambda e(\alpha_1)$. Also, we have $\xi(e(\alpha_2)) = 0$. Again, using these two in the first Lie bracket of \eqref{3.32} yields $\varphi e(\alpha_2) = e(\alpha_1)$. Applying \eqref{3.29} and \eqref{3.31} in the preceding relation and using the fact that $\lambda$ is a positive function, we obtain $\lambda = 1$. Now, it is easy to check that $\nabla Q = 0$. Notice that, a Riemannian 3-manifold is Ricci parallel if and only if it is locally symmetric. The rest of the proof follows from Theorem 5 of \cite{cho1} which says that ``A  non-Kenmotsu almost Kenmotsu 3-manifold $M^3$ is locally symmetric if and only if $M^3$ is locally isometric to the product space $\mathbb{H}^2(-4) \times \mathbb{R}$".
\end{proof}

As a combination of Proposition \ref{prop3.2} and Theorem \ref{thm3.6}, we have the following:
\begin{corollary}
If $g$ is a Cotton soliton with potential vector field orthogonal to the Reeb vector field on a 3-dimensional non-unimodular Lie group $M^3$ equipped with a left invariant non-Kenmotsu almost Kenmotsu structure, then $M^3$ is locally isometric to $\mathbb{H}^2(-4) \times \mathbb{R}$ and the Cotton soliton is steady.
\end{corollary}

\subsection*{Acknowledgement} The author Dibakar Dey is thankful to the Council of Scientific and Industrial Research, India (File no: 09/028(1010)/2017-EMR-1) for their assistance.


TITLE: Linear approximation to the infinite product $\prod_{k=0}^\infty \frac{1}{2} (1+x^{1/p^k})$
QUESTION [0 upvotes]: Let's consider the function defined by the infinite product:
$$f_p(x)=\prod_{k=0}^\infty \frac{1}{2} (1+x^{1/p^k})$$
$$p \geq 2$$
The only closed form I know is for $p=2$.
$$f_2(x)=\frac{x^2-1}{2 \ln x}$$
However, numerically, it turns out that this function is very close to linear, and the larger $p$, the better the approximation.
For my purposes, it's enough to consider:
$$0 \leq x \leq 1$$
For small $x$ and small $p$ the linear approximation doesn't fit. See for $p=3$:

However, for $p=15$ linear approximation fits much better:

Thus, my question is:

Can we obtain a highly accurate linear approximation $f_p(x)=A_p x+B_p$ without numerically computing the product for several points and using regression (i.e. least squares)?

Also, how can we prove that:

$$\lim_{p \to \infty} f_p(x)=\frac{1}{2}(1+x)$$

Motivation:
If we introduce a sequence:
$$a_{n+1}^p=a_n \frac{(a_n+b_n)^{p-1}}{2^{p-1}},\qquad b_{n+1}^p=b_n \frac{(a_n+b_n)^{p-1}}{2^{p-1}}$$
$$a_0>b_0, \qquad x=\frac{b_0}{a_0}$$
Then:
$$\lim_{n \to \infty} a_n=\lim_{n \to \infty} b_n=(f_p(x))^{\frac{p-1}{p}} a_0$$

REPLY [2 votes]: As $x \to 1$, $p \to \infty$ or $k \to \infty$, I think the following are right
$$ \begin{align}
x^{1/p^k} &= \exp( p^{-k} \log x)
\\&= \exp\left( p^{-k} (x-1) + O(p^{-k} (x-1)^2 ) \right)
\\&= 1 + p^{-k} (x-1) + O\left(p^{-k} (x-1)^2 \right)
\end{align}$$
$$ \frac{1}{2} \left(1 + x^{1/p^k} \right) = 1 + \frac{x-1}{2p^k} + O(p^{-k} (x-1)^2) $$
$$\begin{align}
 \log \prod_{k=0}^{\infty} \frac{1}{2} \left(1 + x^{1/p^k} \right)
&= \sum_{k=0}^{\infty} \frac{x-1}{2p^k} + O\left(p^{-k}(x-1)^2 \right)
\\&= \frac{p(x-1)}{2(p - 1)} + O((x-1)^2)
\end{align}$$
$$ \prod_{k=0}^{\infty} \frac{1}{2} \left(1 + x^{1/p^k} \right)
= 1 + \frac{p(x-1)}{2(p-1)} + O((x-1)^2)
$$


TITLE: Proving that a set is an ultrafilter
QUESTION [1 upvotes]: Let $\mathfrak{A}$ be a finite Boolean algebra, and $A_0$ be an atom of $\mathfrak{A}$. I would like to prove that the set $$U(A_0)=\{A\in\mathfrak{A}:A_0<A\}$$ is the principal ultrafilter on $\mathfrak{A}$ generated by $A_0$.
I have already proved that $U(A_0)$ is the principal filter generated by $A_0$. So, $\mathfrak{A}$ being a Boolean algebra, I just have to show that for every $A\in\mathfrak{A}$ either $A$ or $-A$ is in $U(A_0)$.
If $A=1$, this is obvious because $-1=0$ is covered by every atom, in particular $A_0$. How do I prove the property for an arbitrary $A$ of $\mathfrak{A}$ though?

REPLY [1 votes]: I'm not sure what notation you are using for your two binary boolean operations. I will use $\wedge$ for conjunction and $\vee$ for disjunction. I will also use $\neg$ for negation.
One useful fact is that for all $A,B\in\mathfrak A,$ we have $B<A$ if and only if $A\wedge B=B.$ From this, we can derive yet another useful fact: given $A_0\in\mathfrak A,$ we have that $A_0$ is an atom of $\mathfrak A$ if and only if for all $A\in\mathfrak A$ either $A_0\wedge A=0$ or $A_0\wedge A=A_0.$ (Note: the "either...or" means that we can't have both, which rules out $A_0=0,$ in particular.) It is good practice to prove these facts.
Now, assuming that $A_0\not<A,$ as you did, we have $A_0\wedge A\neq A_0$ by our first fact, so $A_0\wedge A=0$ by our second fact. We want to show that $A_0<\neg A,$ or equivalently that $A_0\wedge\neg A=A_0.$
Since $$A_0\wedge\neg A=0\vee(A_0\wedge\neg A)=(A_0\wedge A)\vee(A_0\wedge\neg A)=A_0\wedge(A\vee\neg A),$$ what can we conclude?


TITLE: Action of Lie derivative on 1-forms
QUESTION [1 upvotes]: In Sean Carroll's spacetime and geometry appendix B he derives the action of the Lie derivative on 1-forms. He finds that $\mathcal{L}_X Y^\mu = [X, Y]^\mu$, which I believe is meant as $\mathcal{L}_X(Y)^\mu$ since other books and wikipedia quote $\mathcal{L}_X(Y) = [X, Y]$ for some vector fields $X$ and $Y$. He then acts on a contraction $Y^\mu \omega_\mu$. Since this is equivalent to simply applying the vector field to the scalar he finds
$$
\mathcal{L}_X(Y^\mu \omega_\mu) = X^\lambda (\partial_\lambda \omega_\mu) Y^\mu + X^\lambda \omega_\mu (\partial_\lambda Y^\mu)
$$
He then compares this to the expression one finds by instead applying the Leibniz rule. Here he then goes on to write
\begin{align*}
\mathcal{L}_X(Y^\mu \omega_\mu) = \mathcal{L}_X(\omega)_\mu Y^\mu + \omega_\mu \mathcal{L}_X(Y)^\mu
\end{align*}
But I don't understand why the indices appear outside all of the sudden. I am unable to justify this step, but without further knowledge I am unable to derive the Lie derivative of a 1-form.
Here's my attempt: starting only with linearity, Leibniz rule, aswell as $\mathcal{L}_X(f) = X(f)$ and $\mathcal{L}_X(Y) = [X, Y]$. By expanding the commutator defintion I find
\begin{align*}
\mathcal{L}_X(Y) &= \mathcal{L}_X(Y^\mu) \partial_\mu + Y^\mu \mathcal{L}_X(\partial_\mu) \\
\left(X^\lambda \partial_\lambda Y^\mu - Y^\lambda \partial_\lambda X^\mu \right)\partial_\mu &= \mathcal{L}_X(Y^\mu)\partial_\mu + Y^\mu \left(-(\partial_\mu x^\lambda) \partial_\lambda\right)
\end{align*}
Where on the right hand side I treat $\partial_\mu$ as a vector field and then compute the commutator according to the defintion of $\mathcal{L}_X(\partial_\mu)$ as is done here. Comparing both sides I then find $\mathcal{L}_X(Y^\mu) = X^\lambda\partial_\lambda Y^\mu$, which makes sense to me since the components of a vector are just functions again. But this is at odds with the derivation of Carroll since clearly $\mathcal{L}_X(Y^\mu) \neq \mathcal{L}_X(Y)^\mu$ in this case. I don't know how to continue then however, I can show again that $\mathcal{L}_X(\omega_\mu) = X^\lambda \partial_\lambda \omega_\mu$, but it seems to me that I need further information such as the action on a basis 1-form which is quoted on wikipedia as $\mathcal{L}_X(\mathrm{d}x^\mu) = (\partial_\lambda X^\mu)\mathrm{d}x^\lambda$. But I do not know enough about differential geometry to understand the derivation of that. This question originates from an old exam of mine so I believe that it should be possible without any further information.

REPLY [1 votes]: Gold's answer is perfectly fine, but I want to spell out some of the formulas and point out what confused me so much.
For me there's two ways to interpret the Leibniz rule:

Treating it just like a normal derivative $\mathcal{L}_X(Y^\mu\omega_\mu) = \mathcal{L}_X(Y^\mu)\omega_\mu + Y^\mu \mathcal{L}_X(\omega_\mu)$.
Product rule with respect to the interior product i.e. $Y(\omega) = Y^\mu\omega_\mu$ which becomes $\mathcal{L}_X(Y(\omega)) = \mathcal{L}_X(Y)(\omega) + Y(\mathcal{L}_X(\omega)) = \mathcal{L}_X(Y)^\mu\omega_\mu + Y^\mu \mathcal{L}_X(\omega)_\mu$.

Notice in the first case we have the index inside the derivative which means that the Lie derivative simply treats it as a scalar $\mathcal{L}_X(Y^\mu) = X^\lambda \partial_\lambda Y^\mu$. In the second case the index is outside of the derivative, so the derivative acts on the 1-form as a whole instead of the 'scalar' components.  The point that tripped me up now is that $\mathcal{L}_X(Y^\mu) \neq \mathcal{L}_X(Y)^\mu$ since the LHS is simply $X^\lambda\partial_\lambda Y^\mu$, whereas the RHS is the $\mu$-th component of the commutator $X^\lambda\partial_\lambda Y^\mu - Y^\lambda\partial_\lambda X^\mu$. However the derivative of the 1-form contains a term that exactly cancels this contribution such that in sum they are equal.
\begin{align*}
\mathcal{L}_X(Y^\mu\omega_\mu) &= \mathcal{L}_X(Y^\mu)\omega_\mu + Y^\mu\mathcal{L}_X(\omega_\mu) = X^\lambda(\partial_\lambda Y^\mu)\omega_\mu + Y^\mu X^\lambda(\partial_\lambda \omega_\mu) \\
\mathcal{L}_X(Y(\omega)) &= \mathcal{L}_X(Y)^\mu\omega_\mu + Y^\mu \mathcal{L}_X(\omega)_\mu = \left(X^\lambda\partial_\lambda Y^\mu - Y^\lambda \partial_\lambda X^\mu\right)\omega_\mu + Y^\mu\left(X^\lambda\partial_\lambda \omega_\mu + \omega_\lambda\partial_\mu X^\lambda\right)
\end{align*}
From this second interpretation it is much easier to find the action of the Lie derivative on a 1-form as derived in Gold's answer. Namely $\mathcal{L}_X(\omega(Y)) = \mathcal{L}_X(\omega)(Y) + \omega(\mathcal{L}_X(Y))$, so that
\begin{align*}
\mathcal{L}_X(\omega)(Y) = X(\omega(Y)) - \omega([X, Y]) &= X^\mu\partial_\mu (\omega_\lambda Y^\lambda) - \omega_\mu \left(X^\lambda\partial_\lambda Y^\mu - Y^\lambda \partial_\lambda X^\mu\right) \\
&= \left(X^\lambda\partial_\lambda \omega_\mu + \omega_\lambda \partial_\mu X^\lambda \right) Y^\mu
\end{align*}
To generalize the action to higher order tensors I think it's useful to think of the Lie derivative acting differently on the three distinct objects that are out our disposal:

components of vectors/1-forms or any other scalar $\mathcal{L}_X(Y^\mu) = X^\lambda\partial_\lambda Y^\mu$ or $\mathcal{L}_X(\omega_\mu) = X^\lambda\partial_\lambda \omega_\mu$
coordinate basis $\mathcal{L}_X(\partial_\mu) = - (\partial_\mu X^\lambda)\partial_\lambda$
basis 1-forms $\mathcal{L}_X(\mathrm{d}x^\mu) = (\partial_\lambda X^\mu) \mathrm{d}x^\lambda$

Now simply apply the product rule on the whole tensor $T = T^{\mu\dots\nu}_{\qquad\ \lambda\dots\sigma}\partial_\mu\dots\partial_\nu\ \mathrm{d}x^\lambda\dots\mathrm{d}x^\sigma$.


TITLE: same average with different medians
QUESTION [0 upvotes]: The following is an exercise in statistics:

Let $A$ and $B$ be two finite sets of datas with possible values $0,1,2,3,4$. The median of $A$ is 2 while the median of $B$ is $3$. Also, $A$ and $B$ have the same average. Can we say that the numbers in $B$ is generally bigger than the numbers in $A$?

I know nothing but the definitions of median and average in this exercise. Also, I'm confused with the meaning of "generally bigger". Could anyone help?

[Added:]
This question is motivated by the following problem from a practice exam of one of my friends:

REPLY [0 votes]: This is a very sloppily written question. You can already tell from the confusing spelling mistakes – "students'" is misspelled as "student's" three times in as many sentences, and this is not just an irrelevant formal glitch but makes it more difficult to understand the question (as perhaps evidenced by your reference to "Manuel's score" in a comment, which makes no sense since the question is about the scores of Manuel's students, not Manuel's score).
I cannot think of any reasonable definition of "generally higher" (much less an obvious one that could be presupposed without providing it) that would make any of the first three answers correct. You can extend your example by an arbitrary number of scores of $1$ and $4$, e.g. $(1,1,1,2,4,4,4)$ and $(1,1,1,3,3,4,4)$, and it makes no sense by any stretch of the expression "generally higher" to apply it to that case (and the expression "no discernable difference" is even worse – what would prevent a difference from being discerned?).