text:
\begin{document}
\title[Euler equations for Cosserat media]{Euler equations for Cosserat media}

\author[Valentin Lychagin]{Valentin Lychagin}
\thanks{The author was partially supported by the Russian Foundation for Basic Research (project 18-29-10013).}
\address{Valentin Lychagin: V. A. Trapeznikov Institute of Control Sciences of Russian Academy of Sciences, Moscow, 117997, Russia.}
\email{valentin.lychagin@uit.no}

\subjclass[2010] {Primary 35Q31; Secondary 76A02, 80A17, 53B20}
\date{Date of Submission   August 15, 2020;  Date of Acceptance  October 10, 2020, Communicated by Yuri E. Gliklikh }
\keywords{Euler equation, SO(3) group, Cosserat medium}

\begin{abstract}
We consider Cosserat media as $\mathbf{SO}(3)$-structures
over a domain $\mathbf{D}\subset\mathbb{R}^{3}$.
Motions of such media are given by
infinitesimal automorphisms of the $\mathbf{SO}(3)$-bundle.
We present Euler-type equations for such media and discuss
their structure.
\end{abstract}

\maketitle

\section{Introduction}

This paper is a realization of approach (\cite{DLT}) for the
case of media formed by `rigid microelements' (\cite{MGC}),
or Cosserat' media.

We show that the geometry hidden behind of the such type of
media is a $\mathbf{SO}(3)$-structure over a spacial domain
$\mathbf{D}$.

Namely, the dynamics in such media are given by vector
fields on the bundle $\pi\colon\Phi \rightarrow \mathbf{D}$
of $\mathbf{SO}(3)$-frames over $\mathbf{D}$. These fields
are solutions of the system of differential equations
(\cite{DLT}) that generalize the well known Navier-Stocks
equations.

To construct these equations we need two additional
geometric objects: (1) connection in the bundle $\pi$, we
call it media connection, and (2) left
$\mathbf{SO}(3)$-invariant metric on fibres of $\pi$. The
last determines the mechanics of `rigid microelements' and
the media connection allows us to compare `rigid
microelements' at different points of $\mathbf{D}$. The
third ingredient that we needed to construct the system of
differential equations is the thermodynamics of media. We
apply here the point of view (\cite{LY}), where
thermodynamics were considered as measurement of extensive
quantities.

Using these remarks and observations we present here  the
Euler type equations that govern the motion. To make the
presentation as compact as possible we restrict ourselves by
this type equations only although the difference between
Euler and Navier-Stocks type equations consists only in the
description of the stress tensor.

The final form of the Euler equations is given by
\eqref{EulerEq}, \eqref{MassConEq}, \eqref{ErgyConserv}.

\section{Geometry of $\mathbf{SO}(3)$}

Here we collect the main properties of the orthogonal
group $\mathbf{SO}(3)$.

Let $(\mathbf{T}, g)$ be a an Euclidian vector space, $\dim
\mathbf{T}=3$, where $g$ is a metric tensor. Let $A\in
\limfunc{End}\left( \mathbf{T}\right)$ be a linear operator
in $\mathbf{T}$ and let $A^{\prime}\in \limfunc{End}\left(
\mathbf{T}\right)$ be the $g$-adjoint operator, i.e
$g\left( AX,Y\right) =g\left(X,AY\right)$ for all $X,Y\in
\mathbf{T}$. Remind that
\begin{equation*}
\mathbf{SO}(3)=
\left\{A\in\limfunc{End}
\left( \mathbf{T}\right)\, \vert\,
AA^{\prime}=\mathbf{1},\,\det A=1\right\} .
\end{equation*}
Geometrically, elements of the group are counterclockwise
rotations $R\left(\phi,\overline{n}\right)$
on angle $\phi$ about the axis through unit
vector $\overline{n}\in \mathbf{T}$. One has $R\left( \phi ,n\right)
=R\left( -\phi ,-n\right) $ and
$R\left( \pi ,n\right) =R\left( \pi,-n\right)$.
Therefore, as a smooth manifold, $\mathbf{SO}(3)$ is
diffeomorphic to the projective space $\mathbb{R}\mathbf{P}^{3}$.

The Lie algebra of the group,
\begin{equation*}
\mathfrak{so}\left( 3\right) =\left\{ A\in \limfunc{End}\left(
\mathbf{T}\right) \,\vert\, A+A^{\prime}=\mathbf{0}\right\},
\end{equation*}
consists of skew symmetric operators.

There is the \textit{hat isomorphism} of the Lie algebras
\begin{equation*}
\wedge\colon\left( \mathbf{T,\times }\right)
\rightarrow \mathfrak{so}\left(3\right),
\end{equation*}
where $\left(\mathbf{T,\times}\right)$ is the Lie algebra
of vectors in $\mathbf{T}$ with respect
to the cross product $\times$.

In an orthonormal basis $\left( e_{1},e_{2},e_{3}\right)$
this isomorphism has the following form
\[
\wedge\colon
w=\left(w_{1},w_{2},w_{3}\right) \mapsto \widehat{w}=w_{1}
\widehat{e}_{1}+w_{2}\widehat{e_{2}}+w_{3}\widehat{e_{3}}=
\begin{bmatrix}
0 & -w_{3} & w_{2} \\
w_{3} & 0 & -w_{1} \\
-w_{2} & w_{1} & 0
\end{bmatrix}.
\]

\subsection{Exponent and logarithm}

In the case of Lie algebra $\mathfrak{so}\left( 3\right)$
the exponential map
\begin{equation*}
\exp\colon\mathfrak{so}\left( 3\right) \rightarrow \mathbf{SO}(3),
\end{equation*}
has the following Rodrigues' form
(see, for example, \cite{Rod}, \cite{Mar})
\begin{equation*}
\exp \left( \phi \widehat{n}\right) =\mathbf{1+}\sin \left( \phi \right)
\widehat{n}+(1-\cos \left( \phi \right) )\widehat{n}^{2},
\end{equation*}
and $\exp \left( \phi \widehat{n}\right) =R\left( \phi ,n\right)$.

This formula gives us the following description of the logarithm map
\begin{equation*}
\ln \left( R\right) =\frac{\arcsin \left( \phi \right) }{2\phi }\left(
R-R^{\prime }\right),
\end{equation*}
where $R\in \mathbf{SO}(3)$ and
\begin{equation*}
\phi =\frac{3+\sqrt{-\limfunc{Tr}R^{2}}}{2}.
\end{equation*}

\subsection{Left invariant tensors on $\mathbf{SO}(3)$.}

The Baker--Campbell--Hausdorff formula in
$\mathfrak{so}\left(3\right)$ has very concrete form.

Namely, let $X,Y\in\mathfrak{so}\left(3\right)$, then
$\exp \left(X\right) \cdot \exp \left( Y\right) \in \mathbf{SO}(3)$ and
therefore has the form $\exp\left(Z(X,Y)\right)$,
for some element $Z(X,Y)\in \mathfrak{so}\left( 3\right)$.

Then (see, for example, \cite{Eng}),
\begin{equation*}
Z\left( X,Y\right) =\alpha\,X+\beta\,Y+\gamma\,\lbrack X,Y\rbrack,
\end{equation*}
where
\begin{equation*}
\alpha =\frac{a\arcsin \left( d\right) }{d\theta },\quad
\beta =\frac{b\arcsin\left( d\right) }{d\phi },\quad
\gamma =\frac{\arcsin \left( d\right) }{d\phi\theta },
\end{equation*}
$a$, $b$, $c$, $d$ are defined as follows
\begin{align*}
&a =\sin \left( \theta \right) \cos ^{2}\left( \frac{\phi }{2}\right)
-\omega \sin \left( \phi \right) \sin ^{2}\left( \frac{\theta }{2}\right),\\
&b=\sin \left( \phi \right) \cos ^{2}\left( \frac{\theta }{2}\right) -\omega
\sin \left( \theta \right) \sin ^{2}\left( \frac{\phi }{2}\right) \\
&c =\frac{1}{2}\sin \left( \theta \right) \sin \left( \phi \right) -2\omega
\sin ^{2}\left( \frac{\theta }{2}\right) \sin ^{2}\left( \frac{\phi }{2}
\right),\\
&d=\sqrt{a^{2}+b^{2}+2\omega ab+\left( 1-\omega ^{2}\right) c^{2}},
\end{align*}
and
\begin{equation*}
\theta =\sqrt{\frac{-\limfunc{Tr}\,X^{2}}{2}},\quad
\phi =\sqrt{\frac{-\limfunc{Tr}\,X^{2}}{2}},\quad
\omega =\theta^{-1}\phi^{-1}\sqrt{\frac{-\limfunc{Tr}\,XY}{2}}.
\end{equation*}
Applying these formulae for the case $X=t\widehat{n},Y$,
where $n$ is a unit vector, we get
\begin{equation*}
\alpha =\frac{\phi }{2}\cot \left( \frac{\phi }{2}\right) +O\left( t\right)
,~\beta =1+t\left( \frac{\omega }{2}\cot \left( \frac{\phi }{2}\right) +
\frac{\omega }{\phi }\right) +O\left( t^{2}\right) ,\ \gamma =\frac{1}{2}
+O\left( t\right).
\end{equation*}

Denote by $E_{1}$, $E_{2}$, $E_{3}$ the left invariant vector fields on
$\mathbf{SO}(3)$ that correspond to the basis $e_{1}$, $e_{2}$, $e_{3}$
in $\mathfrak{so}\left( 3\right)$.

Then the above formulae give us the following expressions for
$E_{1}$, $E_{2}$, $E_{3}$ in the canonical coordinates
$\left(x_{1},x_{2},x_{3}\right)$ of the first kind:
\begin{align*}
E_{1}=&\frac{\phi }{2}\cot \left( \frac{\phi }{2}\right) \partial _{1}+
\frac{1}{2}\left( x_{2}\partial _{3}-x_{3}\partial _{2}\right) +\\
&x_{1}\left(
\frac{1}{2}\cot \left( \frac{\phi }{2}\right) +\frac{1}{\phi }\right) \left(
x_{1}\partial _{1}+x_{2}\partial _{2}+x_{3}\partial _{3}\right),\\
E_{2}=&\frac{\phi }{2}\cot \left( \frac{\phi }{2}\right) \partial _{2}+
\frac{1}{2}\left( x_{3}\partial _{1}-x_{1}\partial _{3}\right) +\\
&x_{2}\left(
\frac{1}{2}\cot \left( \frac{\phi }{2}\right) +\frac{1}{\phi }\right) \left(
x_{1}\partial _{1}+x_{2}\partial _{2}+x_{3}\partial _{3}\right),\\
E_{3}=&\frac{\phi }{2}\cot \left( \frac{\phi }{2}\right) \partial _{3}+
\frac{1}{2}\left( x_{1}\partial _{2}-x_{2}\partial _{1}\right) +\\
&x_{3}\left(
\frac{1}{2}\cot \left( \frac{\phi }{2}\right) +\frac{1}{\phi }\right) \left(
x_{1}\partial _{1}+x_{2}\partial _{2}+x_{3}\partial _{3}\right),
\end{align*}
where
\begin{equation*}
\phi =\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}},\quad \text{and}\quad
\partial_{i}=\frac{\partial }{\partial x_{i}}.
\end{equation*}

Remark that basis vectors $e_{1},e_{2},e_{3}$ have the following commutation
relations:
$\lbrack e_{\sigma \left( 1\right) },e_{\sigma \left( 2\right) }\rbrack=
\limfunc{sign}\left( \sigma \right) e_{\sigma \left( 3\right) }$, for any
permutation $\sigma $ of three letters and accordingly vector fields
$E_{1}$, $E_{2}$, $E_{3}$ inherit the same commutation relations
\begin{equation*}
\lbrack E_{\sigma \left( 1\right) },E_{\sigma \left( 2\right) }]=\limfunc{
sign}\left( \sigma \right) E_{\sigma \left( 3\right) }.
\end{equation*}
Let us denote by $\Omega _{1},\,\Omega_{2},\,\Omega_{3}\in
\Omega^{1}\left(
\mathbf{SO}(3)\right)$ differential $1$-forms on Lie group
$\mathbf{SO}(3)$ such that $\Omega _{i}\left( E_{j}\right) =\delta _{ij},$
then
\begin{equation*}
d\Omega _{\sigma \left( 3\right) }+\limfunc{sign}\left( \sigma \right) \
\Omega _{\sigma \left( 1\right) }\wedge \Omega _{\sigma \left( 2\right) }.
\end{equation*}
Vector fields $E_{i}$ and differential $1$-forms $\Omega _{i}$ give us the
bases (over $\mathbb{R}$) in the space of left invariant vector fields and
correspondingly invariant differential $1$-forms on $\mathbf{SO}(3)$.

Moreover, any left invariant tensor on $\mathbf{SO}(3)$ is a linear
combination of tensor products $E_{i}$ and $\Omega_{j}$ with constant
coefficients.

Thus any left invariant metric $g$ on $\mathbf{SO}(3)$ is defined
by a positive self adjoint operator $\Lambda$ on
$\mathfrak{so}\left(3\right)$, so-called inertia tensor.

We will take basis $e_{1}$, $e_{2}$, $e_{3}$
to be eigenvectors of the operator $\Lambda$.

Thus we get:
\begin{equation}\label{Metr}
g_{\lambda }=\frac{1}{2}\left( \lambda _{1}\Omega _{1}^{2}+\lambda
_{2}\Omega _{2}^{2}+\lambda _{3}\Omega _{3}^{2}\right),
\end{equation}
where $\lambda_{i}$, $\lambda _{i}>0$, are eigenvalues of
the operator and $\Omega _{i}^{2}$ are the symmetric squares
of the $1$-forms.

\subsection{Levi-Civita connections on $\mathbf{SO}(3)$}

Let $\nabla$ be the Levi-Civita connection associated with
left invariant metric $g_{\lambda}$. We denote by
$\nabla_{i}$ the covariant derivative along vector field
$E_{i}$.

Then we have
\begin{equation*}
\nabla _{i}\left( E_{j}\right) =\sum_{k}\Gamma _{ij}^{k}E_{k},
\end{equation*}
where $\Gamma _{ij}^{k}$ are the Christoffel symbols.

This connection preserves the metric and therefore
\begin{equation*}
g\left( \nabla _{i}\left( E_{j}\right) ,E_{k}\right) +g\left( E_{j},\nabla
_{i}\left( E_{k}\right) \right) =0,
\end{equation*}
or
\begin{equation}\label{Christ1}
\lambda _{k}\Gamma _{ij}^{k}+\lambda _{j}\Gamma _{ik}^{j}=0,
\end{equation}
for all $i,j,k=1,2,3.$

The condition for the connection to be torsion-free gives us
the following relations:
\begin{equation*}
\nabla _{\sigma \left( 1\right) }\left( E_{\sigma \left( 2\right) }\right)
-\nabla _{\sigma \left( 2\right) }\left( E_{\sigma \left( 1\right) }\right)
=E_{\sigma \left( 3\right) },
\end{equation*}
for all permutations $\sigma$, or
\begin{equation}\label{Christ2}
\Gamma _{\sigma \left( 1\right) ,\sigma \left( 2\right) }^{k}-\Gamma
_{\sigma \left( 2\right) ,\sigma \left( 1\right) }^{k}=\delta _{k,\sigma
\left( 3\right) }.
\end{equation}

The solution of these equations is the following
\begin{eqnarray}\label{Gamma}
\Gamma _{12}^{3} &=&\frac{\lambda -\lambda _{1}}{\lambda _{3}},\Gamma
_{23}^{1}=\frac{\lambda -\lambda _{2}}{\lambda _{1}},\Gamma _{31}^{2}=\frac{
\lambda -\lambda _{3}}{\lambda _{2}},  \\
\Gamma _{21}^{3} &=&\Gamma _{12}^{3}-1,\Gamma _{32}^{1}=\Gamma
_{23}^{1}-1,\Gamma _{13}^{2}=\Gamma _{31}^{2}-1,  \notag
\end{eqnarray}
where $\lambda =\lambda _{1}+\lambda _{2}+\lambda _{3},$
and all other Christoffel symbols are trivial.

Thus we have the only non-trivial relations:
\begin{align*}
&\nabla_{1}\left( E_{2}\right)=\frac{\lambda -\lambda _{1}}{\lambda _{3}}
[E_{1},E_{2}],\nabla _{2}\left( E_{3}\right) =\frac{\lambda -\lambda _{2}}{
\lambda _{1}}[E_{2},E_{3}],\nabla _{3}\left( E_{1}\right) =\frac{\lambda
-\lambda _{3}}{\lambda _{2}}[E_{3},E_{1}],\\
&\nabla_{2}\left( E_{1}\right)=\frac{\lambda -\lambda _{2}}{\lambda _{3}}
[E_{2},E_{1}],\nabla _{3}\left( E_{2}\right) =\frac{\lambda -\lambda _{3}}{
\lambda _{1}}[E_{3},E_{2}],\nabla _{1}\left( E_{3}\right) =\frac{\lambda
-\lambda _{1}}{\lambda _{2}}[E_{1},E_{3}],
\end{align*}

\begin{theorem}
The Levi-Civita connection for left invariant the metric
$g_{\lambda}$ has the form:
\begin{align*}
\nabla_{\sigma \left( 1\right) }\left( E_{\sigma \left( 2\right) }\right)
&=\limfunc{sign}\left( \sigma \right) \frac{\lambda -\lambda _{\sigma
\left( 1\right) }}{\lambda _{\sigma \left( 3\right) }}E_{\sigma \left(
3\right)},\\
\nabla_{i}\left( E_{i}\right)&=0,
\end{align*}
for all permutations $\sigma \in S_{3}$ and $i=1,2,3$.
\end{theorem}

\section{Cosserat media and $\mathbf{SO}\left(3\right)$-structures}

Let $\mathbf{D}$ be a domain in $\mathbb{R}^{3}$ considered
as Riemannian manifold equipped with the standard metric
$g_{0}$. Then by \textit{Cosserat medium} we mean a medium
composed with solids or having `rigid microstructure',
(\cite{MGC}). We assume that on the microlevel this media is
formed by rigid elements, which we represent as orthonormal
frames $f_{a}\colon\mathbb{R}^{3}\rightarrow
\mathbf{T}_{a}B$.

Thus configuration space of these media is the principal
$\mathbf{SO}\left(3\right)$-bundle
$\pi\colon\Phi\rightarrow\mathbf{D}$ of orthonormal frames
on $\mathbf{D}$, (see, for example, \cite{Ch}).
The projection $\pi$ assign
the centre mass $a$ of element $f_{a}$.

\subsection{Metrics and connections, associated with Cosserat medium}

The group $\mathbf{SO}\left( 3\right)$ acts in the natural
way on fibres of projection $\pi $  and we will continue to
use notations $E_{1}$, $E_{2}$, $E_{3}$ for the induced
vertical vector fields on $\Phi$.

These fields form the basis in the module of
vertical vector fields on $\Phi$ and accordingly
differential $1$-forms
$\Omega_{1}, \Omega_{2}, \Omega_{3}$
define the dual basis in the space of vertical differential
forms.

We assume also that the media is characterized by
a $\mathbf{SO}\left(3\right)$-connection in the bundle $\pi$.
This connection, we call it
\textit{media connection} and denote by $\nabla^{\mu}$, allows us to
compare microelements at different points of $\mathbf{D}$.

To define the connection, we consider a microelement as
orthonormal frame $b=\left( b_{1},b_{2},b_{3}\right)$,
formed by vector fields $b_{i}$ on $\mathbf{D}$.
Then the covariant derivatives $\nabla_{X}$ along vector field
$X$ is defined by the connection form
$\omega\in\Omega^{1}\left( \mathbf{D,}\mathfrak{so}\left( 3\right) \right)$,
i.e. differential $1$-form on $\mathbf{D}$ with values
in the Lie algebra $\mathfrak{so}\left( 3\right)$,
and such that
\begin{equation}
\nabla _{X}^{\mu }\left( b\right) =\omega \left( X\right) b.
\label{MediaConnection}
\end{equation}
By using the hat morphism, we represent the form as before
\begin{equation*}
\omega =
\begin{Vmatrix}
0 & -\omega _{3} & \omega _{2} \\
\omega _{3} & 0 & -\omega _{1} \\
-\omega _{2} & \omega _{1} & 0
\end{Vmatrix},
\end{equation*}
where $\omega_{i}$ are differential $1$-forms on $\mathbf{D}$.

Then formula \eqref{MediaConnection} shows us that
a microelement subject to rotation along vector
$\left(\omega_{1}(X),\,\omega_{2}(X),\,\omega_{3}(X) \right)$
on the angle
$\phi \left(X\right) =\sqrt{\omega_{1}\left( X\right) ^{2}+\omega _{2}\left( X\right)
^{2}+\omega _{3}\left( X\right) ^{2}}$, when we transport it on vector $X$.

Let $\left( x_{1},x_{2},x_{3}\right) $ be the standard Euclidian coordinates
on $\mathbf{D}$ and $\partial =\left( \partial _{1},\partial _{2},\partial
_{3}\right) $ be the corresponding frame. Here $\partial _{i}=\partial
/\partial x_{i}$.

In these coordinates we have
\begin{equation*}
\omega _{i}=\sum_{j=1}^{3}\omega _{ij}d_{j},
\end{equation*}
where $d_{j}=dx_{j}$.

The connection form $\omega$ will be the following
\begin{equation*}
\omega =
\begin{Vmatrix}
0 & -\omega _{31} & \omega _{21} \\
\omega _{31} & 0 & -\omega _{11} \\
-\omega _{21} & \omega _{11} & 0
\end{Vmatrix}
d_{1}+
\begin{Vmatrix}
0 & -\omega _{32} & \omega _{22} \\
\omega _{32} & 0 & -\omega _{12} \\
-\omega _{22} & \omega _{12} & 0
\end{Vmatrix}
d_{2}+
\begin{Vmatrix}
0 & -\omega _{33} & \omega _{23} \\
\omega _{33} & 0 & -\omega _{13} \\
-\omega _{23} & \omega _{13} & 0
\end{Vmatrix}
d_{3}.
\end{equation*}

This connection allows us to split tangent spaces
$T_{b}\Phi$ into direct sum
\begin{equation*}
T_{b}\Phi =V_{b}+H_{b},
\end{equation*}
where $V_{b}$ is the vertical part with basis $E_{1,b}$,
$E_{2,b}$, $E_{3,b}$, and the horizontal space $H_{b}$ is
formed by vectors
\begin{equation*}
X_{a}-\omega_{a}\left(X_{a}\right),
\end{equation*}
where $a=\pi\left(b\right)$.

Remark that geometrically spaces $H_{b}$ represent `constant
frames' due to \eqref{MediaConnection} and the basis in this
space is
\begin{eqnarray*}
&&\partial _{1}-\omega _{11}E_{1}-\omega _{21}E_{2}-\omega _{31}E_{3}, \\
&&\partial _{2}-\omega _{12}E_{1}-\omega _{22}E_{2}-\omega _{32}E_{3}, \\
&&\partial _{3}-\omega _{13}E_{1}-\omega _{23}E_{2}-\omega _{33}E_{3}.
\end{eqnarray*}
The horizontal distribution
$H\colon b\in \Phi \rightarrow H_{b}\subset T_{b}\Phi$
could be also defined as the kernel of the following
differential $1$-forms on
$\Phi$:
\begin{align*}
&\theta _{1} =\Omega _{1}+\omega _{11}d_{1}+\omega _{12}d_{2}+\omega
_{13}d_{3}, \\
&\theta _{2} =\Omega _{2}+\omega _{21}d_{1}+\omega _{22}d_{2}+\omega
_{23}d_{3}, \\
&\theta _{3} =\Omega _{2}+\omega _{31}d_{1}+\omega _{32}d_{2}+\omega
_{33}d_{3}.
\end{align*}

By metric associated with the left invariant metric
$g_{\lambda}$ and standard metric $g_{0}$ and media
connection form $\omega$ we mean direct sum of metric
$g_{\lambda}$ on the vertical space $V$ and the standard
metric $g_{0}$ on the horizontal space $H$.

We also call the media \textit{homogeneous} if the
connection form $\omega$ as well as the inertia tensor
$\Lambda$ are constants. Moreover, for the case of
homogeneous media the Euclidian coordinates
$\left(x_{1},x_{2},x_{3}\right)$ in the domain $\mathbf{D}$
are chosen in a such way that operators
$\widehat{\partial_{i}}\in \mathfrak{so}\left( 3\right)$
are eigenvectors of the inertia tensor $\Lambda$.

Summarizing we get the following
\begin{proposition}
The metric $g^{\mu}$ associated with the triple $\left( g_{\lambda
},g_{0},\omega \right)$ has the following form
\begin{equation*}
g^{\mu }=\frac{1}{2}\sum_{i=1}^{3}\left( \lambda _{i}\Omega
_{1}^{2}+d_{i}^{2}\right).
\end{equation*}
\end{proposition}

\begin{remark}
Frame $\left( E_{1},E_{2},E_{3},\partial _{1}-\omega \left( \partial
_{1}\right) ,\partial _{2}-\omega \left( \partial _{2}\right) ,\partial
_{2}-\omega \left( \partial _{2}\right) \right)$ and coframe $\left( \Omega
_{1},\Omega _{2},\Omega _{3},d_{1},d_{2},d_{3}\right)$
are dual and $g^{\mu}$-orthogonal.
\end{remark}

\subsection{Levi-Civita connection on Cosserat media}

Assume that the media is homogeneous and let $\nabla$ be the
Levi-Civita connection on the configuration space $\Phi$
associated with metric $g^{\mu}$. Then for basic vector
fields $\partial_{i},E_{j}$, where $i,j=1,2,3$, we have the
following commutation relations
\begin{equation*}
\lbrack\partial _{i},\partial_{j}\rbrack=
\lbrack\partial _{i},E_{j}\rbrack=0,\quad
\lbrack E_{\sigma(1)}, E_{\sigma(2)}\rbrack=
\limfunc{sign}\left( \sigma\right)E_{(3)}.
\end{equation*}

Moreover, it is easy to see that the equations for
Christoffel symbols do not depend on the connection form
$\omega$. On the other hand, it is clear that the Levi-Civita
connection $\nabla$, for the case $\omega =0$, is the direct
sum of the trivial connection on $\mathbf{D}$ and the left
invariant Levi-Chivita connection on $\mathbf{SO}\left(
3\right)$. Thus we get the following result.

\begin{theorem}
The Levi-Civita connection $\nabla^{c}$ on the configuration
space $\Phi$ associated with metric $g^{\mu}$ and
homogeneous media has the form, where the only non trivial
covariant derivatives are
\begin{equation*}
\nabla _{E_{\sigma \left( 1\right) }}\left( E_{\sigma \left( 2\right)
}\right) =\limfunc{sign}\left( \sigma \right) \frac{\lambda -\lambda
_{\sigma \left( 1\right) }}{\lambda _{\sigma \left( 3\right) }}E_{\sigma
\left( 3\right) },
\end{equation*}
for all permutations $\sigma\in S_{3}$.
\end{theorem}

\subsection{Deformation tensor}

Dynamics we describe by $\pi$-projectable vector fields
on $\Phi$ (see \cite{DLT}, for more details):
\begin{equation*}
U:=\sum_{i=1}^{3}\left( X_{i}\left( x\right) \partial _{i}+Y_{i}\left(
x,y\right) E_{i}\right).
\end{equation*}
Here $x=\left( x_{1},x_{2},x_{3}\right) $ are the Euclidian
coordinates on $\mathbf{D}$ and
$y=\left(y_{1},y_{2},y_{3}\right)$ are the canonical
coordinates on $\mathbf{SO}\left( 3\right)$.

Due to the above theorem and relations \eqref{Gamma} the covariant
differential $\mathcal{D}$ with respect to the Levi-Civita connection $
\nabla ^{c}$ acts in the following way:
\begin{eqnarray}
\mathcal{D}\left( \partial _{i}\right)  &=&0,\quad i=1,2,3,  \label{CovDif} \\
\mathcal{D}\left( E_{1}\right)  &=&\left( \alpha _{2}+1\right)\Omega
_{3}\otimes E_{2}+\left( \alpha _{3}-1\right) ~\Omega _{2}\otimes E_{3},
\notag \\
\mathcal{D}\left( E_{2}\right)  &=&\left( \alpha _{1}-1\right) \Omega
_{3}\otimes E_{1}+\alpha _{3}~\Omega _{1}\otimes E_{3},  \notag \\
\mathcal{D}\left( E_{3}\right)  &=&\alpha _{1}\Omega _{2}\otimes
E_{1}+\alpha _{2}\Omega _{1}\otimes E_{2},  \notag
\end{eqnarray}
where
\begin{align*}
\Gamma _{32}^{1}&=\alpha _{1}-1,\quad
& \Gamma _{23}^{1}&=\alpha _{1}=\frac{\lambda-\lambda _{2}}{\lambda _{1}}, \quad
& \Gamma _{31}^{2}&=\alpha _{2}+1,\\
\Gamma _{13}^{2}&=\alpha _{2}=\frac{\lambda_{1}}{\lambda _{2}}, \quad
& \Gamma _{21}^{3}&=\alpha _{3}-1,\quad
& \Gamma _{12}^{3}&=\alpha _{3}=\frac{\lambda-\lambda _{1}}{\lambda _{3}}.
\end{align*}
Respectively, for the dual frame $\left( \Omega _{1},\Omega _{2},\Omega
_{3}\right) ,$ we have
\begin{align}
\mathcal{D}\left( \Omega _{1}\right)  &=-\alpha _{1}\Omega _{2}\otimes
\Omega _{3}-\left( \alpha _{1}-1\right) \Omega _{3}\otimes \Omega _{2},
\notag \\
\mathcal{D}\left( \Omega _{2}\right)  &=-\alpha _{2}\Omega _{1}\otimes
\Omega _{3}-\left( \alpha _{2}+1\right) \Omega _{3}\otimes \Omega _{1},
\label{CovDifW} \\
\mathcal{D}\left( \Omega _{3}\right)  &=-\alpha _{3}\Omega _{1}\otimes
\Omega _{2}-\left( \alpha _{3}-1\right) \Omega _{2}\otimes \Omega _{1}.
\notag
\end{align}

By the \textit{rate of deformation tensor}
$\Delta\left(U\right)$ we mean tensor
\begin{multline*}
\Delta \left( U\right) =\mathcal{D}\left( U\right) =\sum_{i,j=1}^{3}\left(
\partial _{i}X_{i}\,d_{j}\otimes \partial _{i}+\partial
_{i}Y_{i}\,d_{j}\otimes E_{i}+E_{j}\left( Y_{i}\right) \Omega _{j}\otimes
E_{i}\right)+\sum_{i=1}^{3}Y_{i}\mathcal{D}\left( E_{i}\right) .
\end{multline*}

\section{Thermodynamics of Cosserat media}

The thermodynamics of the Cosserat media is based on
measurement (see, \cite{LY}, \cite{DLT}) of
\textit{extensive quantities:} \textit{inner energy}
$\mathit{E}$, \textit{volume} $V$, mass $m$, and deformation
$D$. The corresponding dual, or \textit{intensive
quantities} are the temperature $T$, preasure $p$, chemical
potential $\xi$ and the stress tensor $\sigma$.

The first law of thermodynamics requires that on
thermodynamical states the following differential $1$-form
\begin{equation}\label{thermo form1}
dE-(T\,dS-p\,dV+\limfunc{Tr}(\sigma^{\ast}dD)+\xi\,dm)
\end{equation}
should be zero.

In other words, the thermodynamical state is a maximal
integral manifold $L$ of differential form \eqref{thermo form1}.

It is more convenient to use another but proportional
differential $1$-form
\begin{equation}\label{thermo form2}
dS-T^{-1}\left( dE+p\,dV-\limfunc{Tr}(\sigma ^{\ast }dD)-\xi\,dm\right).
\end{equation}

Here differential forms: $dE$, $T\,dS$,
$-p\,dV+\limfunc{Tr}(\sigma^{\ast }dD)$
and $\xi\,dm$ represent change of inner energy,heat,work and mass
respectively.

Extensivity of quantities $E,V,D,m,S$ means that their simultaneous
rescaling does not change the intensives as well as the thermodynamic state.

In other words, if we represent $\left(S, p, \sigma, \xi\right)$ as
functions of the extensive variables $\left(E,V,D,m\right)$,
then $S=S\left(E,V,D,m\right)$ is a homogeneous function of degree one.

Let $\left(s, \varepsilon, \rho, \Delta \right)$
be the densities of $\left(S, E, m, D\right)$.

Then substituting expression
$S=Vs(\varepsilon, \rho, \Delta)$ into \eqref{thermo form2}
we get that $1$-form
\begin{equation*}
\psi=ds-T^{-1}\left( d\varepsilon -\limfunc{Tr}\,(\sigma ^{\ast }d\Delta
)-\xi\,d\rho \right)
\end{equation*}
should be zero and
\begin{equation*}
\varepsilon -Ts=\limfunc{Tr}\,(\sigma ^{\ast }\Delta )-p+\xi \rho.
\end{equation*}
The last condition shows that the density of Gibbs
free energy $\varepsilon-Ts$ equals sum of density
of deformation $\limfunc{Tr}\,(\sigma ^{\ast}\Delta )$
and mechanical $-p$ and chemical $\xi\,\rho$ works.

All these observations could be formulated as follows.

Let us introduce the thermodynamic phase space of Cosserat
medium (\cite{DLT}) as the contact space
$\widetilde{\Psi}=\mathbb{R}^{5}\times \limfunc{End}
\left( \mathbf{T}^{\ast }\right) \times \limfunc{End}
\left( \mathbf{T}\right)$, $\dim \widetilde{\Psi }=23$
with points $(s,T,\varepsilon,$ $\xi, \rho, \sigma,$
$\Delta) \in \Psi$,
where $\sigma \in \limfunc{End}\left( \mathbf{T}^{\ast }\right)$,
$\Delta \in \limfunc{End}\left( \mathbf{T}\right)$
and equipped with contact form $\psi$.
Then by thermodynamic
states we mean Legendrian manifolds
$\widetilde{L}\subset \widetilde{\Psi}$,
$\dim \widetilde{L}=11$, of the differential $\psi$, or respectively,
after eliminating entropy from the consideration,
their Lagrangian
projections $L\subset\Psi$ into
$\Psi=\mathbb{R}^{4}\times \limfunc{End}
\left( \mathbf{T}^{\ast }\right) \times
\limfunc{End}\left( \mathbf{T}\right)$, where the symplectic structure on $\Psi$ given by the
differential $2$-form
\begin{equation*}
d\psi =\tau ^{-2}\left( dT\wedge d\varepsilon -\xi dT\wedge d\rho -dT\wedge
\limfunc{Tr}\left( \sigma ^{\ast }d\Delta \right) \right) -T^{-1}\left( d\xi
\wedge d\rho -\limfunc{Tr}\left( d\sigma ^{\ast }d\Delta \right) \right).
\end{equation*}

As it was shown in (\cite{LY},\cite{DLT}) we should require, in addition,
that the differential symmetric form $\kappa $ shall define the Riemannian
structure on $L.$

Also, similar to (\cite{DLT}) we consider only such Legendrian manifolds $
\widetilde{L}$ where $\left( T,\rho ,\Delta \right) $ are coordinates. In
this case we will write down form $T^{-1}\psi$ as
\begin{equation*}
d(\varepsilon -Ts)-\left(s\,dT+\limfunc{Tr}\,\left( \sigma ^{\ast }d\Delta
\right) +\xi\,d\rho \right).
\end{equation*}
Therefore, manifold $\widetilde{L}$ is Legendrian if and only if
\begin{equation}\label{state eq}
s=h_{T},\quad\sigma =h_{\Delta },\quad\xi=h_{\rho},
\end{equation}
where $h$ is the Gibbs free energy
\begin{equation*}
h=\varepsilon -Ts.
\end{equation*}
In the case of Newton-Cosserat media we have in addition
$\mathbf{SO}\left(3\right) \times \mathbf{SO}\left(3\right)$
symmetry and  Gibbs free energy $h$ is a function of
$\mathbf{SO}\left(3\right) \times \mathbf{SO}\left(3\right)$-invariants
(see \cite{DLT}, for more details).

Here we consider the Euler case (\cite{DLT}), when
\begin{equation*}
h\left(T,\rho,\Delta\right)=
p_{1}\left(T,\rho \right)\limfunc{Tr}\left( \Delta \right)+
p_{2}\limfunc{Tr}\left(\Delta\Pi_{V}\right),
\end{equation*}
where $\Pi_{V}$ is the projector on the vertical part of $T\Phi$.

In this case the stress tensor $\sigma$ equals
\begin{equation*}
\sigma=
p_{1}\left(T,\rho\right)+
p_{2}\left(T,\rho\right)\Pi_{V},
\end{equation*}
or
\begin{align}\label{EulerSigma}
\sigma=&\left( p_{1}\left( T,\rho \right) +p_{2}\left( T,\rho \right)
\right) ~\left( E_{1}\otimes \Omega _{1}+E_{2}\otimes \Omega
_{2}+E_{3}\otimes \Omega _{3}\right)+\\
&p_{2}\left( T,\rho \right) ~\left( \partial _{1}\otimes d_{1}+\partial
_{2}\otimes d_{2}+\partial _{3}\otimes d_{3}\right)
+p_{2}\sum_{i,j=1}^{3}\omega _{ij}E_{i}\otimes d_{j},\notag
\end{align}
and energy density equals
\begin{equation*}
\varepsilon =\left( p_{1}-T p_{1,T}\right) \limfunc{Tr}\left( \Delta \right)
+\left(p_{2}-T p_{2,T}\right) \limfunc{Tr}\left(\Delta\Pi_{V}\right)
\end{equation*}

\section{Euler equations for Cosserat media}

The general form of the Navier-Stocks equations and Euler equations as well,
for media with inner structures have the form (\cite{DLT}):

\begin{enumerate}
\item Moment conservation, or Navier-Stocks equation:
\begin{equation}\label{NSeq}
\rho \left( \frac{\partial U}{\partial t}+
\nabla_{U}^{c}\left( U\right)
\right) =\func{div}^{\flat }\sigma,
\end{equation}
where $\nabla^{c}$ is the Levi-Civita connection,
$\func{div}^{\flat}\sigma$ is a vector field dual to
the differential form $\func{div}\sigma\in \Omega^{1}
\left(\Phi\right)$, with respect to the canonical metric $g$.

\item Conservation of mass:
\begin{equation}\label{MassEq}
\frac{\partial \rho }{\partial t}+U\left( \rho \right) +\func{div}\left(
U\right) \rho =0.
\end{equation}

\item Conservation of energy:
\begin{equation*}
\frac{\partial \varepsilon }{\partial t}+
\varepsilon\,\func{div}\left( U\right)-
\func{div}\left( \zeta\,\func{grad}\left( T\right) \right) +
\limfunc{Tr}\left( \sigma ^{\prime }~\mathcal{D}U\right) =0,
\end{equation*}
where $\zeta $ is the thermal conductivity.

\item State equations:
\begin{equation*}
\sigma =h_{\Delta},\quad\varepsilon=h-T\,h_{T}.
\end{equation*}

\item In addition, we require that vector field
$U$ preserves the bundle
$\pi\colon\Phi\rightarrow\mathbf{D}$, or that $U$ is
a $\pi$-projectable vector field.
\end{enumerate}

In the case of Euler equations we have relation
\eqref{EulerSigma}, and using the property (see \cite{DLT})
\begin{equation*}
\func{div}\left( X\otimes \omega \right) =\func{div}\left( X\right) \omega
+\nabla _{X}\left( \omega \right),
\end{equation*}
where $X$ is a vector field and $\omega$ is
a differential $1$-form, we get, due to \eqref{CovDifW},
\begin{align*}
&\func{div}\left( E_{i}\otimes \Omega_{i}\right)=\nabla _{E_{i}}\left(
\Omega _{i}\right) =0,\\
&\func{div}\left( \partial _{i}\otimes d_{i}\right)=\nabla _{\partial
_{i}}\left( d_{i}\right) =0,
\end{align*}
and
\begin{equation*}
\func{div}\,\sigma =\sum_{i=1}^{3}\left( E_{i}\left( p_{1}+p_{2}\right) \Omega
_{i}+(\partial _{i}-\sum_{j=1}^{3}\omega _{ji}E_{j})\left( p_{2}\right)
~d_{i}\right).
\end{equation*}
Therefore,
\begin{equation*}
\func{div}^{\flat}\,\sigma =\sum_{i=1}^{3}\left( \lambda _{i}^{-1}E_{i}\left(
p_{1}+p_{2}\right) E_{i}+(\partial _{i}-\sum_{j=1}^{3}\omega
_{ji}E_{j})\left( p_{2}\right) \partial_{i}\right).
\end{equation*}

Assume now that vector field $U$ has the form
\begin{equation*}
U=\sum_{i=1}^{3}\left( X_{i}\left( x\right) \partial _{i}+Y_{i}\left(
x,y\right) E_{i}\right).
\end{equation*}
Then,
\begin{align*}
\nabla _{\partial _{1}}\left( U\right)  &=\sum_{i=1}^{3}\left( \partial
_{1}\left( X_{i}\right) \partial _{i}+\partial _{1}\left( Y_{i}\right)
E_{i}\right) ,\\
\nabla _{\partial _{2}}\left( U\right) &=\sum_{i=1}^{3}\left(
\partial _{2}\left( X_{i}\right) \partial _{i}+\partial _{2}\left(
Y_{i}\right) E_{i}\right) , \\
\nabla _{\partial _{3}}\left( U\right) &=\sum_{i=1}^{3}\left( \partial
_{3}\left( X_{i}\right) \partial _{i}+\partial _{3}\left( Y_{i}\right)
E_{i}\right) , \\
\nabla _{E_{1}}\left( U\right)  &=\sum_{i=1}^{3}\left( E_{1}\left(
Y_{i}\right) E_{i}+\lambda _{1i}Y_{i}[E_{1},E_{i}]\right)
=\sum_{i=1}^{3}E_{1}\left( Y_{i}\right) E_{i}+\lambda
_{12}Y_{2}E_{3}-\lambda _{13}Y_{3}E_{2}, \\
\nabla _{E_{2}}\left( U\right)  &=\sum_{i=1}^{3}\left( E_{2}\left(
Y_{i}\right) E_{i}+\lambda _{2i}Y_{i}[E_{2},E_{i}]\right)
=\sum_{i=1}^{3}E_{2}\left( Y_{i}\right) E_{i}-\lambda
_{21}Y_{1}E_{3}+\lambda _{23}Y_{3}E_{1}, \\
\nabla _{E_{3}}\left( U\right)  &=\sum_{i=1}^{3}\left( E_{3}\left(
Y_{i}\right) E_{i}+\lambda _{3i}Y_{i}[E_{3},E_{i}]\right)
=\sum_{i=1}^{3}E_{3}\left( Y_{i}\right) E_{i}+\lambda
_{31}Y_{1}E_{2}-\lambda _{32}Y_{2}E_{1},
\end{align*}
where
\begin{align*}
&\lambda_{12}=\frac{\lambda -\lambda_{1}}{\lambda _{3}},\quad
\lambda_{13}=\frac{\lambda -\lambda_{1}}{\lambda _{2}},\quad
\lambda_{21}=\frac{\lambda -\lambda_{2}}{\lambda _{3}},\\
&\lambda _{23}=\frac{\lambda -\lambda _{2}}{\lambda _{1}},\quad
\lambda _{31}=\frac{\lambda -\lambda _{3}}{\lambda _{2}},\quad
\lambda _{32}=\frac{\lambda -\lambda _{3}}{\lambda _{1}}.
\end{align*}

Therefore,
\begin{eqnarray*}
\nabla _{U}\left( U\right)  &=&\sum_{j,i=1}^{3}X_{j}\partial _{j}\left(
X_{i}\right) \partial _{i}+\sum_{j,i=1}^{3}X_{j}\partial _{j}\left(
Y_{i}\right) E_{i}+\sum_{j,i=1}^{3}Y_{j}E_{j}\left( Y_{i}\right) E_{i}+ \\
&&\left( \lambda _{23}-\lambda _{32}\right) Y_{2}Y_{3}E_{1}+\left( \lambda
_{31}-\lambda _{13}\right) Y_{1}Y_{3}E_{2}+\left( \lambda _{12}-\lambda
_{21}\right) Y_{1}Y_{2}E_{3}.
\end{eqnarray*}

Summarizing, we get the following system of Euler equations:
\begin{eqnarray}
&&\rho \left( \partial _{t}X_{i}+\sum_{j=1}^{3}X_{j}\partial _{j}\left(
X_{i}\right) \right) =(\partial _{i}-\sum_{j=1}^{3}\omega _{ji}E_{j})\left(
p_{2}\right) ,\ \ i=1,2,3,  \label{EulerEq} \\
&&\rho \left( \partial _{t}Y_{1}+\sum_{j=1}^{3}X_{j}\partial _{j}\left(
Y_{1}\right) +\sum_{j=1}^{3}Y_{j}E_{j}\left( Y_{1}\right) \right) +\left(
\lambda _{23}-\lambda _{32}\right) Y_{2}Y_{3}=\lambda _{1}^{-1}E_{1}\left(
p_{1}+p_{2}\right) ,  \notag \\
&&\rho \left( \partial _{t}Y_{2}+\sum_{j=1}^{3}X_{j}\partial _{j}\left(
Y_{2}\right) +\sum_{j=1}^{3}Y_{j}E_{j}\left( Y_{2}\right) \right) +\left(
\lambda _{31}-\lambda _{13}\right) Y_{1}Y_{3}=\lambda _{2}^{-1}E_{2}\left(
p_{1}+p_{2}\right) ,  \notag \\
&&\rho \left( \partial _{t}Y_{3}+\sum_{j=1}^{3}X_{j}\partial _{j}\left(
Y_{3}\right) +\sum_{j=1}^{3}Y_{j}E_{j}\left( Y_{3}\right) \right) +\left(
\lambda _{12}-\lambda _{21}\right) Y_{1}Y_{2}=\lambda _{3}^{-1}E_{3}\left(
p_{1}+p_{2}\right).  \notag
\end{eqnarray}
Remark that the first three equations in \eqref{EulerEq}
are very close to the classical Euler equations and the
second three equations are the Euler type equations on
the Lie group $\mathbf{SO}\left(3\right)$.

The mass conservation equation takes the form
\begin{equation}\label{MassConEq}
\partial _{t}\rho +\sum_{i=1}^{3}\left( X_{i}\partial _{i}\rho
+Y_{i}E_{i}\rho \right) +\sum_{i=1}^{3}\left( \partial
_{i}X_{i}+E_{i}Y_{i}\right) \rho =0.
\end{equation}

Finally, the energy conservation equation takes the form
\begin{equation}
\frac{\partial \varepsilon }{\partial t}+\sum_{i=1}^{3}\left( \partial
_{i}X_{i}+E_{i}Y_{i}\right) ~\varepsilon +\sum_{i=1}^{3}E_{i}\left(
Y_{i}\right) =\func{div}\left(\zeta\,\func{grad}\left( T\right) \right) ,
\label{ErgyConserv}
\end{equation}
because
\begin{equation*}
\limfunc{Tr}\left(\sigma^{\ast}\,\mathcal{D}U\right)
=\sum_{i=1}^{3}E_{i}\left( Y_{i}\right),
\end{equation*}
and here
\begin{equation*}
\varepsilon =\left( p_{2}-T\ p_{2,T}\right) \left( E_{1}\left( Y_{1}\right)
+E_{2}\left( Y_{2}\right) +E_{3}\left( Y_{3}\right) \right) +\left(
p_{1}-T~p_{1,T}\right) \func{div}\,U.
\end{equation*}


TITLE: Transitive vs Acyclic Relations
QUESTION [1 upvotes]: My question is that if a relation is transitive does it have to be acyclic?
My first thought is yes, because if aRb, bRc, cRd then by transitivity aRc, bRd and applying transitivity once more implies aRd, thus R is acyclic because c cannot be related to a.

REPLY [1 votes]: Let's take the simplest relation you know : the equality relation ($x \mathcal{R} y \Leftrightarrow x = y$).
Obviously this relation is transitive and is not acyclic.
More generally, if your relation is an equivalence relation, it is then transitive and symetric, so it cannot be acyclic.


TITLE: Group extension
QUESTION [2 upvotes]: I am getting blurred about group extensions. Let $A,B,C$ be groups.

If $G=(A{:}B).C$ and $A$ is characteristic in $G$, then $G=A.(B.C)$. But is it also true that $G=A{:}(B.C)$ ?
If $G=A.(B{:}C)$, then $G=(A.B).C$. But is it true that $G=(A.B){:}C$ ?

Note that $G=A.B$ means $G/A\cong B$; and $G=A{:}B$ means semidirect product of $A$ and $B$ with $A\lhd G$.

REPLY [3 votes]: As to (1), no. Take $G$ to be the direct product of a cyclic group of order $9$ by a group of order $2$. Let $A$ be the unique subgroup of order $3$, and $B$ the unique subgroup of order $2$, so $C$ is cyclic of order $3$. Then $AB$ is the unique subgroup of order $6$, so $A$ cannot have a complement $B.C$ of order $6$.
The very same example show that the answer to (2) is also negative, as there's no subgroup $C$ of order $3$ besides $A$.


TITLE: What do we need to define a category?
QUESTION [6 upvotes]: I used some results from the category theory without thinking about its foundations. However, after reading a few topics on MSE, this subject haunts me.
My question is:

What do we need to define a category?

According to some books, a category consists of a class $\text{Obj}$ of objects and a set $\text{Hom}$ of morphisms which satisfy some axioms. For me it means, that to define a category we need some set theory. But there are many different set theories. Do they raise different category theories?
Also, as I understand, when we are talking about specific categories, like $\text{Set}$, $\text{Grp}$,... we mean models (interpretations) of the axioms of a category. Is it correct?

REPLY [2 votes]: First, one needs to adopt a foundation of mathematics to define a set, category, and other mathematical objects!
Different foundations give different category theories. Some 'categories', which are called big in one foundation, do not exists in another, e.g. the functor category between two large categories and the localisation of a category with respect to a proper class (large set) of its morphisms. The meanings of the terms (small) set and (proper) class, and the operations you can perform on them, depend on the adopted foundation. Shulman's Set theory for category theory and Mac Lane's One universe as a foundation for category theory discuss the effect of the foundation on the resulting category theory, although the latter is more focused on the advantages of a specific foundation.


TITLE: Simple discrete math question
QUESTION [0 upvotes]: It's been a long time since I took discrete math in college and I was never good at it, so I'm asking for some help with something I need to populate a chart in C#.
X = 2, the output for 2 is 0, and with each iteration the input increases by 3 and the output increases by 1.
input | output
______|_______
  2   |   0
______|_______
  5   |   1
______|_______
  8   |   2

and so on. I tried finding a value for a for input % a = output but didn't come up with anything. I feel like this is a simple problem but it just isn't my thing. Is there a function that given an input will give me the proper output? I need this to work, like if I put in 23 I need to get 7 without knowing that 20 was 6.

REPLY [1 votes]: Try $$f(x) = \frac{x-2}{3}$$ and plug input for $x$, then $f(x)$ will return output.


TITLE: What is the relationship between the speed of the end of the thread and the angular velocity of the spool which was pulled by the winding thread?
QUESTION [1 upvotes]: On a rough ground, there is a cylindrical spool of inelastic thin thread (as the picture following), the inner and outer radii are $r$ and $R$ respectively, the thin thread is wound in its inner cylinder, its outer cylinder and the ground contact.
Now a force is pulled horizontally on the end of the thread and the thread is tangent to the lower end of the inner cylinder (the thread is always taut!), so that the spool of thread on the ground to do pure rolling (let angular velocity be $\omega$), and at the same time the speed of the end of the thread $v$ is also uniform, seeking the relationship between the speed of the end of the thread and angular velocity of the spool of thread?

The following image shows a simplified model.

I have a thought that $v=\omega(R-2r)$, but I'm not sure about its correctness.

REPLY [1 votes]: Let the centre of the spool have a linear velocity of $u$.
Because the bottom of the spool is not slipping, $u=R\omega$.
Relative to the centre, the point at which the string meets the inner disc has a speed $r\omega$ in the opposite direction to $v$.
Therefore $v=u-r\omega$.
So eliminating $u$ gives $$v=\omega(R-r)$$


TITLE: Embedding of the cotangent of the n sphere in R^2n
QUESTION [3 upvotes]: It is true that $T^* S^1$ embeds in $\mathbb{R}^2$ as the cotangent of $S^1$ is trivial and $S^1 \times \mathbb{R}$ is diffeomorphic to the punctured plane. Is it true in general that $T^* S^n$ embeds in $\mathbb{R}^{2n}$ ?

REPLY [2 votes]: This is false for $n = 2$. If $T^\ast S^2$ could be embedded in $\Bbb R^4$, then we would also have an embedding of the unit cotangent bundle $ST^\ast S^2 \cong \Bbb R P^3$ in $\Bbb R^4$. But it is known that $\Bbb R P^3$ does not embed in $\Bbb R^4$ (see for example here).


TITLE: In how many ways 3 persons can solve N problems.
QUESTION [4 upvotes]: There are $3$ friends $(A,B,C)$ preparing for math exam. There are $N$ problems to solve in $N$ minutes.
It is given that:

Each problem will take $1$ minute to solve. So all $N$ problems will be solved in exactly $N$ minutes.
Only $1$ person will solve a problem, that is if a problem $xyz$ is solved by $A$ , then $B$ and $C$ need not to solve it.
Only one person will solve a problem in any given time. That is if $A$ is solving a problem at any given time, then $B$ and $C$ should remain idle(This will ensure that $N$ problems will be solved in exactly $N$ minutes).

Now there are some constraints:

$A$ being a lover of the number $k$ has decided that the total number of problems solved by him will be a multiple of $k$. (If $k=2$ then $A$ will solve $0$ or $2$ or $4$... problems)
$B$ is a lazy guy. So he will not solve any problem in consecutive minutes. He needs rest after solving $1$ problem.
$C$ will solve atleast $1$ problem.

Determine the number of ways they can solve the problems.
Example: if $N=3$ and $k=0$ , then $A$ will not solve any problem, so there are $4$ ways:($B$ solve $1$st, $C$ solve $2$nd and $3$rd)  or ($B$ solve $2$nd, $C$ solve $1$st and $3$rd) or ($B$ solve $3$rd and $C$ solve $1$st and $2$nd) or ($B$ solve $1$st and $3$rd and $C$ solve $2$nd)
Given: $0\le k\le10$

REPLY [0 votes]: I don't think any formula will be able to capture those rules, for non trivial values of N and k.
What is certainly doable is to write an algorithm to determine that number; or even better, the infinite sequence of ways for each natural n, given a fixed k.
Such algorithm will basically "inductively" compute each element of the series based on the previous element. Such element will be internally represented into a split form into a few useful partitions, one for each of the possible state of the current problem solving situation.
Allow me to further explain.
Finite state machine
One instance of your sequence of N problems solutions can be imagined as a sequence of N+1 states of progress into the problem solving. The things you want to remember, during that process, are just
$$
a \in [0..k[ = \text{how many problems A solved so far (in modulo k)}\\
b \in [0,1] = \text{1 iff the last problem was solved by B, else 0}\\
c \in [0,1] = \text{1 iff C already solved any problem, else 0}
$$
By only knowing these facts (or variables) you are able to determine which are all possible further states, and if one of them is an ending state (i.e. if the current sequence of problem solvers will be one of the ways counted in $w(N)$).
For example, if $a=0, b=0, c=0$ I know that I can go to, depending on the next person who will solve the problem,

$A -> a=1, b=0, c=1$ (not an ending state)
$B -> a=0, b=1, c=1$ (an ending state)
$C -> a=0, b=0, c=0$ (an ending state)

Track all states
Now, notice that all possible states are a finite number, and actually relatively few. The exact amount is $4k$, and in the worst case in which k = 10 (thanks for binding k), you will only have 40 states.
If you know, for each of the $4k$ states, in how many ways you could have got there in exactly n steps, you can determine in how many ways you can get to each of the $4k$ states after step n+1.
If you start with 0 problems, with 1 only in $w_0(0,0,0)$ (you can only start from this state in which A solved 0, B didn't solve the last one, and C didn't solve any), and compute the amount of ways $w_i$ for each state, for each i up to n, you will simply have to sum together the ways $w_n(0,0,1) + w_n(0,1,1)$ to get your ways in which n problems were solved and all your rules were honored.
The code
I coded this in Scala, I hope you don't mind the language choice.
Beware that your number grows pretty quickly, and if its value overflows the maximum Java long value ($2^{63}-1$), you will have to replace long variables with a BigInteger or something similar.
import collection.Map // generic Map interface

object ProblemSolverz {
  def main(args: Array[String]): Unit = {
    // C
    println(new ProblemSolverz(3).ways(1))

    // CC, BC, CB
    println(new ProblemSolverz(3).ways(2))

    // CCC, BCC, CBC, CCB, BCB
    println(new ProblemSolverz(3).ways(3))

    // CCCC
    // BCCC, CBCC, CCBC, CCCB
    // BCBC, BCCB, CBCB
    // AAAC, AACA, ACAA, CAAA
    println(new ProblemSolverz(3).ways(4))

    // a LOT of ways
    println(new ProblemSolverz(3).ways(20))
  }
}

class ProblemSolverz(k: Int) {
  def ways(N: Int) =
    // number of states in which
    // A solved a number of problems equal to 0 in modulo k,
    // C solved at least one problem,
    // B whatever
    states(N - 1)((0, 0, 1)) + states(N - 1)((0, 1, 1))

  val states = makeNextStates(Map((0, 0, 0) -> 1L)) // initial state

  def makeNextStates(previousStates: Map[(Int, Int, Int), Long]): Stream[Map[(Int, Int, Int), Long]] = {
    // initialize all states to 0 (makes for cleaner code later)
    val states = scala.collection.mutable.Map((for {
      a <- 0 until k
      b <- 0 to 1
      c <- 0 to 1
    } yield ((a, b, c), 0L)): _*)

    // for each previous state, compute its successors
    previousStates.foreach({
      case ((a, b, c), ways) => {
        // A could solve this problem
        states(((a + 1) % k, 0, c)) += ways
        // B could solve this problem, if he is rested
        if (b == 0)
          states((a, 1, c)) += ways
        // C could solve this problem
        states((a, 0, 1)) += ways
      }
    })

    // contribute this state to the stream
    Stream.cons(states, makeNextStates(states))
  }
}


TITLE: Absolutely continuous measure which is not equivalent
QUESTION [0 upvotes]: Does there exist a simple example of a probability measure on $[0,1]$ such that ${\rm spt}\,\mu = [0,1]$ and $\mu \ll \mathcal{L}$ but $\mathcal{L} \not\ll \mu$? Here $\mathcal{L}$ denotes the Lebesgue measure.

REPLY [1 votes]: Let $C$ be  a fat  Cantor set and $\mu (E)=\mathcal L (E\setminus C)$. Then $\mu$ has full support (since $C$ has no interior) and it is absolutely continuous w.r.t. Lebesgue measure but $\mu (C)=0$ whereas Lebesgue measure of $C$ is  $>0$.


TITLE: Number $N$ of summands adding up to $5$ whose product is maximal.
QUESTION [1 upvotes]: In how many parts $N$ should the integer $5$ be partitioned so that the product of these parts is maximized?
I tried using A.M$\geq$G.M but could not go further.

REPLY [3 votes]: Suppose you partition a number $n$ into exactly $k$ summands $n_1 + n_2 + \cdots + n_k$. It is a routine calculation to show that the product $n_1 \cdots n_k$ is maximized when all summands are equal. In this case $n_i = \frac{n}{k}$ for all $1 \le i \le k$ and the maximum product is $\left (\dfrac{n}{k} \right)^k$.
So, what value of $k$ maximizes $\left (\dfrac{n}{k} \right)^k$? Clearly $1 \le k \le n$. Define
$$f(x) = \left (\dfrac{n}{x} \right)^x, \quad 1 \le x \le n$$
and apply the first derivative test. Since $f(x) = e^{x(\log n - \log x)}$ you have
$$f'(x) = e^{x(\log n - \log x)} (\log n - \log x - 1) = e^{x(\log n - \log x)} \log\left( \frac{n}{xe}\right)$$ so that $f'(x) > 0$ if $xe < n$ and $f'(x) < 0$ if $xe > n$.
It follows that $f$ attains its maximum value when $x = \dfrac ne$, and that the maximum value that $f(k)$ takes among integer values $k$ is attained either at
$k=\lfloor n/e \rfloor$ or $k = \lceil n/e \rceil$.
In case $n=5$ you have $1 < 5/e < 2$ so the product is maximized using a partition with either $1$ or $2$ summands. When $k=1$ the product is $5$ and when $k=2$ it is $\left(\dfrac 52 \right)^2 = 6.25$.
The maximum product is 6.25 using 2 (equal) parts.


TITLE: Solution of ODE strictly monotone increasing
QUESTION [2 upvotes]: Let $f \in C^1(\mathbb R)$ and look at $y'(t) = f(y(t))$, $y(0) = y_0$ such that $f(y_0) > 0$. Show that every solution is strictly monotone increasing.

My idea:
It is clear, if $f>0$ on $\mathbb R$. So assume that $y'$ is not strictly increasing. So there exists $a$ with $y'(a)=f(y(a))\leq 0$. By intermediate value theorem there exists $\xi$ between $0$ and $a$ so that $y'(\xi)=f(y(\xi))=0$. Now I don't know how to got on.

REPLY [5 votes]: Let $\psi:I\to\mathbb R$, be a solution of the IVP
$$
y'=f(y), \quad y(t_0)=y_0, \qquad (\star)
$$
where $I$ interval, with $t_0\in I$. It suffices to show that
$$
f\big(\psi((t)\big)>0, \quad\text{for all $t\in I$}.
$$
Suppose that $f\big(\psi((t_1)\big)=0$, for some $t_1\in I$. If $y_1=\psi(t_1)$, then $f(y_1)=0$, and hence the constant function $\tilde \psi(t)\equiv y_1$, would be a solution of the IVP
$$
y'=f(y), \quad y(t_1)=y_1. \qquad (\star\star)
$$
But $\psi$ satisfies $(\star\star)$ as well, and due to uniqueness enjoyed by both IVPs (since $f$ is $C^1$), we have that $\psi=\tilde\psi$. Contradiction, since $\psi'(t_0)=f\big(\psi(t_0)\big)=f(y_0)>0$.
If $f\big(\psi((t_1)\big)<0$, for some $t_1\in I$, then due to Intermediate Value Theorem, there would be also a $t_2\in I$, where $f\big(\psi(t_1')\big)=0$, again leading to a contradiction.


TITLE: If a category has both an initial and terminal object, does that mean they are the same object?
QUESTION [0 upvotes]: If a category C has initial object I and terminal object T, then from what I read that means that I = T. From the definitions of initial and terminal objects, every object A has unique morphisms I -> A, and A -> T. But unless I and T are the same (zero) object, then won't I have a distinct morphism from I -> A -> T for every object A through composition?
I haven't found this mentioned anywhere so far so I just wanted to make sure I am reading the definitions correctly.

REPLY [0 votes]: If a category $\mathcal{C}$ has initial object $I$ and terminal object $T$, then from what I read that means that $I = T$.

No, initial and terminal objects need not be the same.
Consider the following examples:

Consider $\mathbf{Set}$.
There exists a unique initial object in $\mathbf{Set}$, namely the empty set.
The terminal objects in $\mathbf{Set}$ are precisely the singleton sets (i.e., those sets that contain precisely one element).
But the empty set is not a singleton set.

Consider the category $\mathbf{Ring}$ of rings.
The initial object of $\mathbf{Ring}$ is $ℤ$, and the terminal object is the zero ring.
These two rings are very much not the same.

Let $P$ be a partially ordered set, considered as a category in the usual way.¹
An initial object of $P$ is the same as a minimum of $P$, and a terminal object of $P$ is the same as a maximum of $P$.
Unless $P$ consists of only a single element, these two will not be the same.

We have in the given arbitrary category $\mathcal{C}$ a unique morphism from $I$ to $T$.
(This hold because $I$ is initial, and it also holds because $T$ is terminal.)
However, the reverse is not necessarily true:
there does not need to exist a morphism from $T$ to $I$.
We actually have the following statement.

Proposition.
Let $\mathcal{C}$ be a category with initial object $I$ and terminal object $T$.
The following conditions on $I$ and $T$ are equivalent:

The object $I$ is also terminal in $\mathcal{C}$.
The object $T$ is also initial in $\mathcal{C}$.
The unique morphism from $I$ to $T$ is an isomorphism.
There exists a morphism from $T$ to $I$.

If one (and thus all) of these equivalent conditions are satisfied, then one calls the initial object of $\mathcal{C}$ (which is then also its terminal object) a zero object.
Some examples of this:

In $\mathbf{Grp}$, the trivial group is a zero object.
Given a ring $R$, the zero module is a zero object in $R\text{-}\mathbf{Mod}$.

From the definitions of initial and terminal objects, every object $A$ has unique morphisms $I \to A$, and $A \to T$.
But unless $I$ and $T$ are the same (zero) object, then won't I have a distinct morphism from $I \to A \to T$ for every object $A$ through composition?

No, the morphisms $I \to A \to T$ don’t need to be distinct.
Let us denote the unique morphism from $I$ to $A$ by $f_A$ and the unique morphism from $A$ to $T$ by $g_A$.
The composite $g_A ∘ f_A$ is a morphism from $I$ to $T$, and there is only one such morphism.
This means that the composite $g_A ∘ f_A$ must be the same for every object $A$.
Let us look at an explicit example:

For every ring $R$, the unique homomorphism of rings $f_R$ from $ℤ$ to $R$ is given by $f_R(n) = n ⋅ 1_R$ for every integer $n$.
The unique homomorphism of rings $g_R$ from $R$ to $0$ (the zero ring) is the constant map $r \mapsto 0$.
The composite $g_R ∘ f_R$ is the map
$$
  ℤ \to 0 \,, \quad n \mapsto 0 \,.
$$
It doesn’t matter which ring $R$ we choose, the resulting homomorphism $ℤ \to R \to 0$ is always the same.

¹ Given a partially ordered set $P$, the corresponding category $\mathcal{P}$ is defined as follows:
The objects of $\mathcal{P}$ are the elements of $P$.
For every two elements $x$ and $y$ of $P$, there exists at most one morphism from $x$ to $y$ in $\mathcal{P}$, and this morphism exists if and only if $x ≤ y$.
The composition of morphisms in $\mathcal{P}$ is defined in the only possible way.


TITLE: Relations between elements in a Geometric Progression
QUESTION [0 upvotes]: What are the different kinds of relations one can obtain between the elements of a finite Geometric progression other than the fact that one element upon the previous element gives the common ratio?

REPLY [0 votes]: The logarithms follow an arithmetic progression.


text:
\begin{document}

\title{WaveComBox: a Matlab Toolbox for Communications using New Waveforms}

\author{Fran\c{c}ois Rottenberg,
		Mathieu Van Eeckhaute,
        Fran\c{c}ois Horlin,
        and~J\'er\^ome Louveaux

\thanks{Fran\c{c}ois Rottenberg and J\'er\^ome Louveaux are with the Universit\'e catholique de Louvain, 1348 Louvain-la-Neuve, Belgium (e-mail: francois.rottenberg@gmail.com;jerome.louveaux@uclouvain.be).}
\thanks{Fran\c{c}ois Rottenberg, Mathieu Van Eeckhaute and Fran\c{c}ois Horlin are with the Universit\'e libre de Bruxelles, 1050 Brussel, Belgium (e-mail: mveeckha@ulb.ac.be;fhorlin@ulb.ac.be).}
}

\maketitle

\begin{abstract}
Future generations of communications systems will target a very wide range of applications, with possibly very different specifications. This implies the need for a physical layer technology allowing for an ultra flexible allocation of the time-frequency resources. In this regard, the most popular modulation nowadays, namely, the orthogonal frequency division multiplexing (OFDM) modulation, is characterized by a very poor frequency localization, making frequency resource allocation much less flexible. In the light of these limitations, communications using new waveforms, relying on more sophisticated time windowing or filtering, have attracted a lot of attention for the last decade, such as offset-QAM-based filterbank multicarrier (FBMC-OQAM), universal filtered multicarrier (UFMC) or generalized frequency division multiplexing (GFDM).

At the same time, the higher complexity of these new waveforms, not only in terms of implementation but especially conceptually, creates an entrance barrier that slow down their adoption by industries, standardization bodies and more generally in the telecommunication community.

The WaveComBox toolbox, available at \href{www.wavecombox.com}{\texttt{www.wavecombox.com}}, is a user-friendly, open-source and well documented piece of software aiming at simplifying the use and considerably lowering the entrance barrier of recently proposed waveforms. By using an abstract architecture, the toolbox allows for simple implementation of typical physical layer functionalities and for comparison with state-of-the-art technologies.
\end{abstract}

\begin{IEEEkeywords}
New waveforms, modulation, communications, Matlab toolbox, OFDM, FBMC, GFDM, UFMC.
\end{IEEEkeywords}

\IEEEpeerreviewmaketitle

\input{Section/Introduction.tex}

\input{Section/Architecture.tex}

\input{Section/Example.tex}

\section*{Acknowledgment}

The research reported herein was partly funded by Fonds pour la Formation \`a la Recherche dans l'Industrie et dans l'Agriculture (F.R.I.A.). The authors also want to thank Nicolas Boumal, the founder of the Manopt toolbox, for his precious advice.

\ifCLASSOPTIONcaptionsoff
  \newpage
\fi

\bibliographystyle{IEEEtran}

\bibliography{IEEEabrv,refs}

\end{document}

config: arxiv


TITLE: How to correctly compute this limit?
QUESTION [0 upvotes]: I'm trying to solve this limit. But I'm having some problems. Can you tell me how to do this?
I have:
$$\lim_{x\to\infty}\frac{\phi^{2x}\ln(\phi)+\ln(\phi)\cos(\pi x)+\pi\sin(\pi x)}{\phi^{2x}-\cos(\pi x)}$$
Thank you for any help!
EDIT:
I split the limit in three parts:
$$\lim_{x\to\infty}\frac{\phi^{2x}\ln(\phi)}{\phi^{2x}-cos(\pi x)}+\lim_{x\to\infty}\frac{\ln(\phi)cos(\pi x)}{\phi^{2x}-cos(\pi x)}+\lim_{x\to\infty}\frac{\pi\sin(\pi x)}{\phi^{2x}-cos(\pi x)}$$
So I can say that the last two limits converges to 0.
Now I only have to solve the first limit:
$$\lim_{x\to\infty}\frac{\phi^{2x}\ln(\phi)}{\phi^{2x}-cos(\pi x)}$$
I can see on the graph that this converges to $\ln(\phi)$ , so $\lim_{x\to\infty}\frac{\phi^{2x}}{\phi^{2x}-cos(\pi x)}$ should converge to 1. But how can I demonstrate this?

REPLY [2 votes]: $\phi=1,6...> 1$ therefore $|\phi^{-x}\cos{\pi x}|<\phi^{-x}\longrightarrow 0$
The other hand $\phi^x\longrightarrow \infty$. Combining this two results gives the limit
Restarting to your new edit. The main term in the numerator is $\phi^{2x}\log {\phi}$ and the denominator $\phi^{2x}$. The limit is therefore $\log {\phi}$


TITLE: Calculating a norm of an operator
QUESTION [1 upvotes]: Let $T \in (C([a, b]))^*$,
$$ T(u) = \underset{a}{\overset{(a+b)/2}\int} u(x) dx - \underset{(a+b)/2}{\overset{b}\int} u(x) dx. $$
Show that $ || T || = b - a $.
We have that
$$|| T || = \underset{u \in C([a, b]), ||u|| = 1}{sup} | T (u) |$$
I calculated that $$| T(u) | \leq b - a, \forall u \in C([a, b]), ||u|| = 1$$
i.e. $$ ||T|| \leq b - a, $$
but I can't show that  $ b - a \leq ||T|| $.
$$ | T(u) | = \left| \underset{a}{\overset{(a+b)/2}\int} u(x) dx - \underset{(a+b)/2}{\overset{b}\int} u(x) dx \right| \geq \left| \left| \underset{a}{\overset{(a+b)/2}\int} u(x) dx  \right|- \left| \underset{(a+b)/2}{\overset{b}\int} u(x) dx \right| \right|  $$
and then?
Thank you!

REPLY [2 votes]: A sketch:
Look at the expression for $T(u)$ and suppose you were allowed to use $u$ defined on $[a,b]$ by $u(x) = 1$ on $[a,(a+b)/2]$ and $u(x) = -1$ on $[(a+b)/2, b]$.  The expression evaluates to $b-a$ here.  Of course, $u$ is not continuous so this does not work for you.
Instead take a sequence of continuous functions that approximate this function - you can even form a particular approximation by a straight line connecting the graph at a bit before the midpoint to a bit after.  You can do some easy estimates to see the integrals converge.  Then you will have that $T(u_n)$ converges to $b-a$, so $\left\| T \right\| \geq b - a$ .
To add a bit of explanation, this is a normal thing to happen when computing operator or functional norms.  There is an "obvious" upper bound that you can infer, so that $\left\| T \right\| \leq M$, but it is not obvious from algebra and calculus that the reverse inequality holds.
Then the decision tree goes first: is there an element in the space that attains the bound?  A $u$ with $\left\| u\right\| = 1$ so that $|T(u)| = M$?  Or failing that, if you can find an element outside the space where the action of $T$ attains $M$, can you make a sequence in your space that converges to this element?
Here is why a convergent sequence suffices.  You know that $|T(u)| = M$ and that $|T(u_n)| \rightarrow |T(u)| = M$.  So, for any $\epsilon >0$ $|T(u_n)|$ is within $\epsilon$ of $M$ eventually for large enough $n$.  The norm $\|T\|$ is the sup of evaluations at points of unit norm, and therefore for any $\epsilon$, $\| T \| \geq M - \epsilon$ since we have an element that comes that close to $M$.


text:
In \cite{Macdonald72}, Macdonald proved the following remarkable identity for finite root systems $\Phi$:
\begin{align}
  \label{eq:15}
\sum_{w \in W} t^{\ell(w)} = \sum_{w \in W} w \left( \prod_{\alpha \in \Phi^+} \frac{1 - t e ^\alpha}{1 - e^\alpha} \right)
\end{align}
Here $W$ is the Weyl group, $\ell:W \rightarrow \ZZ_{\geq 0}$ is the length function, and $\Phi^+$ is the set of positive roots. Still more remarkably, this formula no longer holds for infinite type Kac-Moody root systems. Instead one has:
\begin{align}
  \label{eq:16}
\sum_{w \in W} t^{\ell(w)} = \m \sum_{w \in W} w \left(\prod_{\alpha \in \Phi_{\real}^+} \frac{1 - t e ^\alpha}{1 - e^\alpha}\right)
\end{align}
where $\Phi^+_{\real}$ is now the set of positive {\em{real}} roots, and the correction factor $\m$ is a $W$-invariant power series supported on the positive imaginary cone of the root lattice. In \cite{Macdonald03}, Macdonald computes $\m$ in the affine Kac-Moody case. An explicit formula for $\m$ can be obtained from the so-called constant term of the Macdonald kernel. In fact, the computation of $\m$ in affine type is equivalent to the Macdonald constant term conjecture \cite{Macdonald82}, proven by Cherednik \cite{Cherednik95}. We will review the formula for $\m$ in affine type in \S \ref{sect:affinecase}.

In this paper, our goal is to initiate the explicit study of the correction factor $\m$ beyond affine type to arbitrary Kac-Moody type. In \S \ref{sect:peterson} we give an algorithm to compute the power series $\m$, based only on knowledge of the real roots in $\Phi$. In \S \ref{sect:GenBermanMoody} we give a closed formula for $\m$. These tools are used to establish properties of $\m$ which hold across all finite, affine, and indefinite Kac-Moody types.

In addition to its appearance in \eqref{eq:16} the power series $\m$ appears as a correction factor in many other formulas. In this sense $\m$ captures much of the complexity that arises when passing from finite to infinite type. Below, we briefly recall some situations where $\m$ appears.

\subsubsection{$p$-adic groups and Macdonald's formula for the spherical function}

Our initial motivation for studying the factor $\m$ comes from the theory of $p$-adic Kac-Moody groups. In \cite{BGKP}, Braverman, Garland, Kazhdan, and Patnaik show that the discrepancy between the affine Gindikin-Karpelevich formula and the naive generalization from finite type is this factor $\m$. The same is true in Macdonald's formula for the spherical function \cite{BKP} and the Casselman-Shalika formula for the spherical Whittaker function \cite{Patnaik}. Work of Patnaik and the second author \cite{patnaik2017metaplectic} extends these formulas to metaplectic $p$-adic Kac-Moody groups, and $\m$ similarly appears.

Gaussent and Rousseau have recently initiated the study of (not necessarily affine) $p$-adic Kac-Moody groups using their theory of {\em {masures}}, previously known as {\em{hovels}}. The work of  Bardy-Panse, Gaussent, Rousseau \cite[Theorem 7.3]{BGR} show that the role of $\m$ in affine type generalizes to arbitrary Kac-Moody type for the Macdonald formula for the spherical function. We expect that this should continue to hold for the Gindikin-Karpelevich and Casselman-Shalika formulas. We mention H\'ebert's \cite{Hebert-17} work on the well-definedness of Gindikin-Karpelevich formula in the general Kac-Moody setting.

The general principle appears to be that formulas from finite dimensional $p$-adic groups directly generalize to the Kac-Moody case, and the sole new phenomenon is the appearance of the factor $\m$.

\subsubsection{Symmetrizers for Kac-Moody Hecke algebras}
Identities relating Hecke symmetrizers, the Weyl-Kac characters formula, and the factor $\m$  appear in the study of the infinite-dimensional (ordinary and metaplectic) generalizations of the Casselman-Shalika formula for spherical Whittaker functions. The classical Casselman-Shalika formula expresses a spherical Whittaker function in terms of the character of a representation. By the Weyl character formula, this character can be produced by acting on a highest-weight monomial by a ``symmetrizer'', which is a certain sum over the Weyl group.

A Hecke symmetrizer is a sum:
\begin{equation}\label{eq:Hecke_symm}
\sum_{w\in W} T_w
\end{equation}
where the $T_w$ are Demazure-Lusztig operators. Each $T_w$ can be expanded in terms of Weyl group elements that are bounded by $w$ in the Bruhat order, giving rise to an identity relating Weyl and Hecke symmetrizers.

The spherical Whittaker function can be expressed in terms of an operator as in \eqref{eq:Hecke_symm} acting on a highest weight monomial in finite, affine, and indefinite Kac-Moody type. In the finite dimensional setting the sum \eqref{eq:Hecke_symm} satisfies an {\em{operator Casselman-Shalika formula}} \cite[\S 2.12]{Patnaik}, an identity of operators similar to the Demazure character formula, connecting the Whittaker function expressed with \eqref{eq:Hecke_symm} to its usual expression in terms of a Weyl character. In this case the correction factor is just $\m=1.$

The analogous identity between \eqref{eq:Hecke_symm} and the Weyl-Kac character in the affine and indefinite Kac-Moody setting \cite{Patnaik,patnaik2017metaplectic} involves the Weyl invariant correction factor $\m .$ In this case, both symmetrizers are infinite sums. Indeed, the proof that the Hecke symmetrizer \eqref{eq:Hecke_symm} yields a well-defined operator depends on its relation to the Weyl symmetrizer. The proof proceeds by comparing two similar versions of \eqref{eq:Hecke_symm} corresponding to spherical and Whittaker functions respectively. (See \cite[Section 3]{patnaik2017metaplectic} for details.)

\subsubsection{Dolbeault cohomology and the failure of the Hodge theorem}

As Macdonald explains in \cite{Macdonald72}, formula \eqref{eq:15} can be interpreted as a computation of the Betti numbers of a flag variety using Hodge theory. The left hand side is the computation given by counting Schubert cells. The right hand side corresponds to a computation of Dolbeault cohomology using localization at fixed points for the action of the maximal torus. Because the flag variety is smooth and projective, the Hodge theorem establishes that Dolbeault cohomology is equal to Betti cohomology. Equation \eqref{eq:15} results.

The fact that this argument no longer holds in Kac-Moody type is an indication that the Hodge theorem no longer holds for Kac-Moody flag varieties. This is very counterintuitive. It is known that Kac-Moody flag varieties are (ind-)projective. This means that they are not smooth for the purposes of the Hodge theorem. Because they are homogeneous for a group action, Kac-Moody flag varieties are in a sense everywhere singular.

In affine type, the situation is well-understood by the work of Fishel-Grojnowski-Teleman \cite{Fishel-Grojnowski-Teleman08}. They explicitly compute the Dolbeault cohomology of the affine flag variety. In particular, their work shows how the Hodge theorem fails in affine type. One of their results is a proof of the ``strong Macdonald conjecture,'' a conjecture on Lie algebra cohomology which implies the constant term conjecture.

The work of \cite{Fishel-Grojnowski-Teleman08} suggests that the generalization of the strong Macdonald conjecture to general Kac-Moody types should consist of an explicit computation of Dolbeault cohomology for Kac-Moody flag varieties. This should contain strictly more information than the factor $\m$ because one loses the bigrading of Dolbeault cohomology. In finite type only $(p,p)$ classes occur, so no information is lost, but in general other classes must occur.

\subsection{Our work}

Our approach is based on the idea that the problem of computing $\m$ generalizes the problem of computing multiplicities of imaginary roots. In fact, when specialized at $t=0$, a product formula for $\m$ exactly encodes the imaginary root multiplicities. Our main results are two methods for computing this product formula. The methods are $t$-deformations of two existing approaches to computing imaginary root multiplicities: the Peterson algorithm \cite{Peterson83}, \cite{kang2001peterson}, \cite[Exercises 11.11, 11.12]{Kac90} and the Berman-Moody formula \cite{Berman-Moody79}.

Let us write $Q^+_{\imag}$ for the {\it positive imaginary root cone}, which consists of all elements of $Q_+$ (the positive root cone) which remain in $Q_+$ after applying any element of the Weyl group. Our generalized Peterson algorithm is a recursive procedure that allows us to express:
\begin{align}
  \label{eq:2}
\m=\prod_{\lambda \in Q^+_{\imag}} \prod_{n\geq0}(1-t^n e^{\lambda})^{-m(\lambda, n)}
\end{align}
For fixed $\lambda$, the $m(\lambda,n)$ are integers that vanish for $n$ sufficiently large (the bound depends on $\lambda$). The exponent $m(\lambda,0)$ is the multiplicity of $\lambda$. In particular it lies in $\ZZ_{\geq 0}$ and it is non-zero precisely when $\lambda$ is a root.  Our generalized Peterson algorithm is fast, depending only on operations with polynomials. We have implemented it in SageMath, and a small sample of the output is included in the appendix.

It is convenient to package the coefficients $m(\lambda,n)$ into a polynomial $\displaystyle m_\lambda(t) = \sum_{n\geq0} m(\lambda,n) t^n$. The generalized Berman-Moody formula is the following closed expression for the polynomial $m_\lambda(t)$.

\newtheorem*{thm:BermanMoody}{Theorem \ref{thm:BermanMoody}}
\begin{thm:BermanMoody}
For all $\lambda \in Q_{\imag}^+$, we have:
\begin{equation} \label{eq:multformula-intro}
m_\lambda(t) = \sum_{\kappa | \lambda} \mu\left(\lambda/\kappa\right)\left(\frac{\lambda}{\kappa}\right)^{-1} \sum_{\underline{\kappa}\in \text{Par}(\kappa)} (-1)^{|\underline{\kappa}|}\frac{B(\underline{\kappa})}{|\underline{\kappa}|} \prod_{i=1}^{|\underline{\kappa}|} P_{\kappa_i}(t^{\lambda/\kappa})
\end{equation}
\end{thm:BermanMoody}

We refer the reader to \S \ref{sect:GenBermanMoody} for an explanation of the symbols appearing here. Although not as efficient as the generalized Peterson algorithm, this formula is useful for answering theoretical questions about $\m$. For example, we prove the following:

\newtheorem*{thm:roots_only}{Theorem \ref{thm:roots_only}}
\begin{thm:roots_only}
For all $\lambda \in Q^+_{\imag}$, $m_\lambda(t)$ is nonzero if and only if $\lambda$ is a root.
\end{thm:roots_only}

\newtheorem*{thm:order2vanishing}{Theorem \ref{thm:order2vanishing}}
\begin{thm:order2vanishing}
For imaginary roots $\lambda$, the polynomial $m_\lambda(t)$ is divisible by $(1-t)^2$.
\end{thm:order2vanishing}

We note that in affine type Theorem \ref{thm:order2vanishing} is inherent in Macdonald's formatting of the constant term conjecture. Moreover, in affine type, the polynomial $\chi_\lambda(t) = \frac{m_\lambda(t)}{(1-t)^2}$ is more fundamental. It has positive coefficients which are manifestly equal to the cardinalities of certain finite sets. For example, in untwisted affine type, if $\lambda$ is imaginary then $\chi_\lambda(t)$ is the generating polynomial of heights of positive roots of the underlying finite root system (see \S \ref{sect:affinecase} for details).

Beyond affine type the polynomials $\chi_\lambda$ no longer have positive coefficients. However, from computer calculation via the generalized Peterson algorithm, we conjecture the following.
\begin{Conjecture} \label{conj:alternating}
In rank two hyperbolic type, the polynomials $\chi_\lambda$ have alternating sign coefficients.
\end{Conjecture}
Beyond rank two, this does not hold. It is conceivable that the positivity phenomena of affine type is present in the general $\chi_\lambda$ in a more subtle form.

\subsection{Further directions}

The constant term of the polynomial $\chi_\lambda$ is precisely the multiplicity of the root $\lambda.$ It must be positive as it is the dimension of the corresponding root space in the Kac-Moody Lie algebra. It would be illuminating to give a similar interpretation for the other coefficients.

\begin{Problem}
Interpret all coefficients of $\chi_\lambda$ in terms of the Kac-Moody Lie algebra.
\end{Problem}

Computing imaginary root multiplicities is a well-known difficult problem. We propose that one should try to instead study $\chi_\lambda$, which of course contains the information of root multiplicities. Our hope is that this generalized problem might be more tractable. For example, a long-standing problem is Frenkel's conjecture, which provides a conjectural bound for hyperbolic root multiplicities. A natural question is to phrase (and prove) a generalization of this conjecture.

\begin{Problem}
Give upper bounds for the degree and coefficients of $\chi_\lambda(t)$ in terms of $\lambda$.
\end{Problem}

\subsection{Acknowledgments}

We thank Alexander Braverman, Paul E. Gunnells, Kyu-Hwan Lee, Dongwen Liu, Peter McNamara, Manish Patnaik for helpful conversations. At the beginning of this project the first author was partially supported by a PIMS postdoctoral fellowship and the second author was supported through Manish Patnaik's Subbarao Professorship in Number Theory and an NSERC Discovery Grant at the University of Alberta. The project started at the workshop ``Whittaker functions: Number Theory, Geometry and Physics'' at the Banff International Research Station in 2016; we thank the organizers of this workshop.


TITLE: Can gravitational waves escape from inside of black holes?
QUESTION [5 upvotes]: I understand that light cannot escape from inside of an event horizon because the spacetime curvature is too warped for photons to escape.
On the other hand, gravitational waves are ripples of spacetime curvature itself (as opposed to light being IN the spacetime), so if there is some internal dynamical structures inside of blackholes (forget about no hair theorem for the moment) that produce gravitational waves, can those gravitational waves escape from inside of event horizon?

REPLY [2 votes]: The answer to your question is that nothing can travel faster than light, and light can't escape through the event horizon. Therefore gravitational waves can't escape either. I give an algebraic proof that light can't escape in my answer to Why is a black hole black?, and a more visual proof in my answer to Would the inside of a black hole be like a giant mirror? (I'm linking my own answers because I remember them - there must be many similar answers on this site).


TITLE: Prove that if two norms on V have the same unit ball, then the norms are equal. Proof Explanation.
QUESTION [2 upvotes]: This problem has already been solved, the solution is at: Prove that if two norms on V have the same unit ball, then the norms are equal. , my question is in regards to the proof. Particularly where did the idea/motivation to use $\frac{v}{p_{1}(v)}$ come from? What was particular about this random vector from the space that allowed it to work for our purposes? Looking at it, it appears as if it is a vector divided by the norm, but why this? Does it give us something to work with?
I understand the "mechanics" of the solution and how it works out, but the intution behind the choice is passing me.

REPLY [2 votes]: It's the standard way to create a unit vector, i.e. a vector of length $1$.
We are aiming, by contradiction, that the two norms are not equal, meaning there is some vector where they disagree. But for all we know that vector could be far, far away from the origin. There is, a priori, no reason that such a vector should be anywhere close to the unit ball. And if this vector is not close to the unit ball, you can't really exploit the fact that the unit balls are equal to get a contradiction.
You get a similar problem if the vector is really close to the origin.
In order to leverage that the two unit balls are equal, we need to have a vector living right on the edge. This construction ensures that this happens.


TITLE: Is $\sqrt{x}$ concave?
QUESTION [6 upvotes]: I have function $f(x)= \sqrt(x)$. To check is it concave or convex i am checkin $f''(x). $
Which is $ -\frac{1}{4x^{\frac{3}{2}}} < 0$
So the $f(x)$ is concave.  Is it correct ?
And is is the same with $f(x)=arctan(x)$,  $f''(x)= -\frac{2x}{(x^{2}+1)^{2}}<0$. It is concave ?

REPLY [4 votes]: $\sqrt{x}$ is indeed sure concave in its domain, i.e. $[0,+\infty)$, but $\arctan(x)$ in concave only for $x>0$ (the inteval where the last inequality you wrote holds), but for $x<0$ it's convex and in $x=0$ it has an inflection point.