text:
\begin{document}
    \title{
\begin{flushright}
 \mbox{\normalsize ITEP/TH-97/09}
\end{flushright}
\vskip 20pt
On the Homology of Certain Smooth Covers of Moduli Spaces
of Algebraic Curves}
    \author{P.Dunin-Barkowski\footnote{ITEP, Moscow, Russia and KdVI, UvA, Amsterdam, The Netherlands; barkovs@itep.ru}, A.Popolitov\footnote{ITEP, Moscow, Russia; popolit@itep.ru}, G.Shabat\footnote{ITEP and RSHU, Moscow, Russia; george.shabat@gmail.com}, A.Sleptsov\footnote{ITEP, Moscow, Russia; sleptsov@itep.ru}}
    \maketitle
\begin{abstract}
We suggest a general method of computation of the homology of certain smooth covers $\widehat{\mathcal{M}}_{g,1}(\mathbb{C})$ of moduli spaces $\mathcal{M}_{g,1}\br{\mathbb{C}}$ of pointed curves of genus $g$. Namely, we consider moduli spaces of algebraic curves with level $m$ structures. The method is based on the lifting of the Strebel-Penner stratification $\mathcal{M}_{g,1}\br{\mathbb{C}}$. We apply this method for $g\leq 2$ and obtain Betti numbers; these results are consistent with Penner and Harer-Zagier results on Euler characteristics.
\end{abstract}

\tableofcontents

    \section {Introduction}
The homology of moduli spaces of curves deserves much attention
during the last decades ( see, e.g., \cite{F1, F2}). However, the values
of \it all \rm the Betti numbers $b_q(\mathcal{M}_g)$ are far from
being known. There exist some  indications of the existence of the
\it beautiful \rm answer for these numbers; the values of orbifold
Euler characteristics \cite{HZ} and \cite{P} and the generating functions for the intersection numbers \cite{K} and \cite{W} are among the best known.

Most results on the homology of moduli spaces were obtained by
some indirect methods; the papers by Looijenga \cite{L1,L2} is among the few
counterexamples known to the authors.

In the present paper we take the direct approach to the calculation
of $\mathrm{H}_*(\mathcal{M}_g,\mathbb{Z}/2\mathbb{Z})$ based on the stratification of
the moduli spaces labeled by \it dessins d'enfants. \rm
The two
versions of these stratifications were introduced by Penner in \cite{P2} (see also \cite{P3})
and by Strebel (see Looijenga \cite{L1}); they are different set-theoretically but equivalent
combinatorially.

The similar approach was undertaken in \cite{PMil}.

In order to avoid difficulties related to singularities of moduli space we work with smooth cover instead and lift the Strebel-Penner decomposition there.
Then we construct a simplicial complex on which this cover retracts, and to which the standard definition of simplicial homology can be applied.

Another motivation of our work comes from the fact, that moduli spaces $\mathcal{M}_{g,\gamma}$ play important role in string theory. Namely, in order to calculate string scattering amplitude of $\gamma$ (closed) strings, one should perform an integration over $\mathcal{M}_{g,\gamma}$ (this is analogous to the integration over momenta of virtual particles in Feynmann diagram technique for particle physics).
There are many open problems in this field, even the measure of integration in supersymmetric case is known only up to genus 4 (for developement in this field see \cite{DP,DBMS}).

Our paper is organized as follows. In section 2 we show that using dessins d'enfants one defines a structure of cellular complex on $\mathcal{M}_{g,\gamma}\times\mathbb{R}_{>0}^\gamma$. For this reason we first define metrized ribbon graphs as pairs of dessins d'enfant and real positive numbers. Then we homeomorphically map metrised ribbon graphs to the moduli space that provides us a decomposition of moduli space into cells. Then in section 3 we construct a smooth cover $\widehat{\mathcal{M}}_{g,1}$ of $\mathcal{M}_{g,1}$ in order to resolve the singularities of moduli spaces. It is done by introducing an additional structure: a basis in homology of Riemann surfaces. In section 4 we construct a spine $\mathcal{S}$ on $\widehat{\mathcal{M}}_{g,1}$ and prove that it is a simplicial complex. Then we conjecture that the simplicial complex can be represented as a cellular complex $\mathbb{D}$ by combining groups of simplicies into cells. In section 5 we describe our method for calculating the Betti numbers of $\widehat{\mathcal{M}}_{g,1}$. For simplicity it is divided into practical steps. In sections 6 and 7 our method is applied to genus 1 and 2. Explicit answers are given.

\section{Metrized ribbon graphs and moduli spaces}
\subsection{ Dessins d'enfants}
We briefly introduce the main concepts and terminology of the
theory (see \cite{VS} and \cite{LZ} for more details). \\
\\
A triple
$$
X_0\subset X_1\subset X_2=X
$$
is called a \it dessin d'enfant \rm if\\
\\
$\bullet X_0$ is a finite set (of \it vertices \rm);\\
$\bullet X_1$ is a graph, i.e.
$$
X_1\setminus X_0\text{ is homeomorphic to a disjoint union of finite
number of open real intervals (\it{edges})};
$$
 $\bullet X_2$ is a compact connected oriented surface and
 $$
X_2\setminus X_1\text{ is homeomorphic to a disjoint union of finite
number of open discs (\it{cells})}.
$$\\
Throughout the paper we denote by $g$ the genus of a surface $X$ and
by $\gamma$ the number of the cells; we are going mostly to consider
the case $\gamma=1$. Throughout the paper we assume
$(g,\gamma)\notin\{(0,1),(0,2)\}$.\\
\\
Following \cite{Gr}, introduce the \it oriented cartographic group \rm$\mathcal{C}_2^+$; as
an abstract group it is generated by three elements $\rho_{0,1,2}$ satisfying the relations
$$
\rho_2\rho_1\rho_0=1
$$
and
$$
\rho_1^2=0.
$$
(Of course, using the first relation we can delete one of the
generators $\rho_0$ or $\rho_2$ but it is convenient to use both).\\
\\
The group \rm$\mathcal{C}_2^+$ should be thought of as acting on the
\it directed edges \rm of any dessin d'enfant; $\rho_0$ rotates a
directed edge contrary-clockwise around its origin (sending it to
the \it next \rm one), $\rho_1$ changes the direction and $\rho_2$
moves an edge contrary-clockwise around the cell that lies \it to
the left \rm of the directed edge. \\
\\
So the set $\vec{E}(D)$ of the directed edges of a dessin $D$ is a
homogeneous finite $\mathcal{C}_2^+$-set; it is well-known that a
dessin can be restored by this set.\\
\\
By a 0-\it valency \rm of a directed edge $\epsilon\in\vec{E}(D)$ we mean its
$\rho_0$-order, i.e. the minimal positive integer $k$ such that
$\rho_0^k\cdot\epsilon=\epsilon$; obviously this number equals the number of germs of
edges incident to the origin of $\epsilon$.\\
\\
For any pair $(g,\gamma)$  our main tool is the \it finite \rm set
$\mathrm{Dess}_{g;*,*,\gamma[v_0\ge3]}$ of the isomorphic classes of
dessins of genus $g$ with $\gamma$ cells and \it all \rm the
0-valencies exceeding 3. It can be thought of as finite number of
isomorphic classes of objects in the category of $\mathcal{C}_2^+\mathfrak{sets}$.\\
 \\
 For $\vec{E}\in\mathrm{Dess}_{g;*,*,\gamma}$ denote
 $$
 X_{\vec{E}}
 $$
 the corresponding compact surface of genus $g$ with $\gamma$ cells.
 \\
 \\

\subsection{ Metrised ribbon graphs}

 Given $\vec{E}\in\mathrm{Dess}_{g;*,*,\gamma[v_0\ge3]}$, consider the
 real octant
 $$
 \mathrm{Metr}_{\vec{E}}:=\{\ell\in\mathbb{R}_{>0}^{\vec{E}}|
 \rho_1^*\ell=\ell\}.
 $$
 \\
  A pair $(\vec{E},\ell)$ with
  $\vec{E}\in\mathrm{Dess}_{g;*,*,\gamma[v_0\ge3]}$ and
$\ell\in\mathrm{Metr}_{\vec{E}}$
  will be called a \emph{metrised ribbon graph}.\\
  \\
  To each
  $(g,\gamma)\in(\mathbb{N}\times\dot{\mathbb{N}})\setminus\{(0,1),(0,2)\}$
  we associate the finite disjoint union of real octants
  $$
  \mathrm{MRG}_{g,\gamma}:=
\coprod_{\vec{E}\in\mathrm{Dess}_{g;*,*,\gamma[v_0\ge3]}}\mathrm{Metr}_{\vec{E}}
   \cong\{(\vec{E},\ell)|\vec{E}\in\mathrm{Dess}_{g;*,*,\gamma[v_0\ge3]},
    \ell\in\mathrm{Metr}_{\vec{E}}\}.
  $$
  \\
  The most important are the $\rho_2$-invariant \emph{circumferences}
  $$
  c(\epsilon):=\sum_{\epsilon'\in<\rho_2>\epsilon}\ell(\epsilon').
  $$

  \subsection{Strebel atlases}
To a metrised ribbon graph
  $(\vec{E},\ell)\in \mathrm{MRG}_{(g,\gamma)}$ we associate a
  covering
  $$
  X_{\vec{E}}=:\cup_{\epsilon\in\vec{E}}(\mathcal{Z}_\epsilon\cup
  \mathcal{U}_\epsilon\cup\mathcal{W}_\epsilon)
  $$
together with the local coordinates
$$
z_\epsilon:\mathcal{Z}_\epsilon\hookrightarrow\mathbb{C},\quad
u_\epsilon:\mathcal{U}_\epsilon\hookrightarrow\mathbb{C},\quad
w_\epsilon:\mathcal{W}_\epsilon\hookrightarrow\mathbb{C},
$$
\begin{figure}[H]
\centering\leavevmode
\includegraphics[width=15 cm]{pics/patches.png}
\caption{Patches}
\label{basis}
\end{figure}
satisfying
 $$
 z_{\rho_1\cdot\epsilon}=-z_\epsilon,
 $$

 $$
 z_{\rho_2\cdot\epsilon}=z_\epsilon+
 \frac{\ell(\epsilon)+\ell(\rho_2\cdot\epsilon)}{2};
$$

 $$
 u_\epsilon=e^{\frac{2\pi\mathrm{i}}{c(\epsilon)}z_\epsilon};
 $$

$$
w_\epsilon=[z_\epsilon+\frac{\ell(\epsilon)}{2}]^{\frac{2}{v_0(\epsilon)}}.
$$
\\
We have just defined a holomorphic \emph{atlas}
$$
\mathfrak{A}_{(\vec{E},\ell)}:=\{(\mathcal{Z}_\epsilon,z_\epsilon),
(\mathcal{U}_\epsilon,u_\epsilon),
(\mathcal{W}_\epsilon,w_\epsilon)|\epsilon\in\vec{E}\}
$$
on the topological surface $X_{\vec{E}}$.
\\

\subsection{From metrized ribbon graphs to moduli spaces}
The above construction defines a map
$$
\mathrm{MRG}_{g,\gamma}\longrightarrow\mathcal{M}_{g,\gamma}(\mathbb{C}).
$$
Compare the dimensions in the case of trivalent dessins $D$
($v_0\equiv3$):
$$
\dim_{\mathbb{R}}(\mathrm{MRG}_{g,\gamma})=\#\{\text{edges of }D\}=
 \frac{1}{2}\#\vec{E}(D);
$$
now, denoting $\alpha:=\#\{\text{vertices of }D\}$, taking into
account $\alpha-\frac{1}{2}\#\vec{E}(D)+\gamma=2-2g$ and
$\#\vec{E}(D)=3\alpha$ (that follows from our assumption
$v_0\equiv3$) and solving
$\alpha-\frac{1}{2}\#\vec{E}(D)+\gamma=2-2g$, where we substitute
$\alpha=\frac{1}{3}\#\vec{E}(D)$ we arrive at
$$
\dim_{\mathbb{R}}(\mathrm{MRG}_{g,\gamma})=3(2g-2+\gamma)=6g-6+3\gamma,
$$
while
$$
\dim_{\mathbb{R}}\mathcal{M}_{g,\gamma}(\mathbb{C})=
 2\dim_{\mathbb{R}}\mathcal{M}_{g,\gamma}(\mathbb{C})=2(3g-3+\gamma)=6g-6+2\gamma,
$$
so our map reduces real dimensions by $\gamma$. However, the points
of the  space $\mathrm{MRG}_{g,\gamma}$ define exactly $\gamma$
extra (positive) real parameters: the circumferences of the
boundaries of cells.\\
\\
The fantastic theorem of Strebel \cite{S} claims that
$$
\mathrm{MRG}_{g,\gamma}\longrightarrow\mathcal{M}_{g,\gamma}(\mathbb{C})
 \times\mathbb{R}_{>0}^\gamma,
$$
sending a metrized ribbon graph to the Riemann surface defined by
the Strebel atlas and a tuple of circumferences, actually
is a homeomorphism!

In what follows we set $\gamma=1$ and require the sum of the legths of all edges to be equal to $1$, thus dropping $\mathbb{R}_{>0}^\gamma$ factor.

Now we can define strata on the MRG side and map them to the moduli space via our homeomorphism.
Namely, a stratum is an equivalence class, where two MRG's belong to the same class iff they can be transformed
to one another by continiously changing lengths of the edges without letting any edge's length to go zero.
\begin{equation}
 \label{mstrat}
\mathcal{M}_{g,1}=:\mathop{\bigsqcup}_{D\in Dess}C_D
\end{equation}

\section{Covers $\widehat{\mathcal{M}}_{g,1}\longrightarrow
\mathcal{M}_{g,1}$}
\label{cover}
In this section we define $\widehat{\mathcal{M}}_{g,1}$, the so-called \textit{level $m$ covers of $\mathcal{M}_{g,1}$} \cite{ACG}.

\subsection{Definition}
Fix a finite abelian group $\mathbb{A}$.
Then by definition,
$$
\widehat{\mathcal{M}}_{g,\gamma}:=\{(\mathcal{X},\iota)\}
$$
with a fixed isomorphism
$$
\iota:\mathrm{H}^1(\mathcal{X},\mathbb{A})\mathop{\longrightarrow}^{\sim}
\mathbb { A } ^ { 2g } ;
$$
the map $\widehat{\mathcal{M}}_{g,\gamma}\longrightarrow
\mathcal{M}_{g,\gamma}$ forgets $\iota$.

In other words, $\widehat{\mathcal{M}}_{g,1}$ is a quotient of Teichmuller space $\mathcal{T}_{g,1}$ by a certain subgroup $\widehat{\Gamma}_{g,1}$ of finite index of modular group $\Gamma_{g,1}$, defined by the short exact sequence
\eq{
1 \longrightarrow \widehat{\Gamma}_{g,1}  \longrightarrow  \Gamma_{g,1}  \longrightarrow \mathrm{Sp}_{2g}\br{\mathbb{A}} \longrightarrow 1
}

\subsection{Smoothness}

For $\mathbb{A} = \mathbb{Z}/3\mathbb{Z}$ (as well as for all $\mathbb{A} = \mathbb{Z}/m\mathbb{Z}$, $m\geq 3$) the above defined $\htM_{g,1}$ are smooth (see Chapter XVI of \cite{ACG}). For $g=1$ the cover of level 2 is also smooth, so we consider this cover when we describe $g=1$ case in detail in Section \ref{gen1}.

\subsection{Stratification}

The stratification (\ref{mstrat}) can be naturally lifted to $\htM_{g,1}$. The strata are then enumerated by dessins d'enfants with a chosen symplectic basis in the first homology of the curve as an additional
structure, $\widehat{D} = (D,\iota)$.

In a particular such stratum, corresponding to a dessin with $k$ edges, all
points are uniquely described by $k$
numbers, sum of which is equal to $1$, i.e. the highest dimension of a stratum
is $6g-4$, because the maximal number of edges in our dessins is $6g-3$ (while
the minimal is $2g$).

It is evident that every stratum is in fact an (open) simplex. The stratification,
however, is \textbf{not} a simplicial complex, because not all faces of each
simplex lie in this stratification. This happens because for some dessins one
cannot retract some of their edges without pinching a handle or a throat. Let us call this stratification a $quasisimplicial complex$, meaning a difference of two simplicial complexes. Denote it as $\mathcal{Q}_g$, i.e.
\eq{
\htM_{g,1} \cong \mathcal{Q}_g = \mathop{\bigsqcup}_{\widehat{D}} C_{\widehat{D}}
}

\section{Spine}

\subsection{Definition}
\label{spine}

Consider a \textit{flag} $F$ of dessins d'enfants with additional structures,
\eq{
F : \widehat{D}_0 \prec \dots \prec \widehat{D}_k,
}
where $\widehat{D} \prec \widehat{D}'$ means that $\widehat{D}'$ can be obtained from $\widehat{D}$ by contraction of some of its edges.

Denote the set of all flags by $\mathcal{F}$.

Recall that to each dessin with additional structure $\widehat{D}$ corresponds an open simplex $S_{\widehat{D}}$ in quasisimplical complex $\mathcal{Q}_{g}$. Note that for all $i$ the stratum $S_{\widehat{D}_i}$ lies in the closure of $S_{\widehat{D}_k}$.

Denote by $S_F$ the open convex hull of all centers of $S_{\widehat{D}_i}$ in the coordinates of the closure of $S_{\widehat{D}_k}$.

Here by ``the center of a stratum'' we mean the point which corresponds to all edges of the dessin being of equal length. Note that $S_F$ is a $k$-simplex due to all of its vertices being in general position by construction.
\defn{
Define the spine of $\hM$ as
\eq{
\S_{g,1} := \bigsqcup_{F\in \Fl} S_F
}
}
\prop{
$\S_{g,1} $ is a simplicial complex.
}

By construction, $\mathcal{F}$ is an abstract simplicial complex with dessins with additional structures as its vertices and flags as faces. Since simplices of our quasisimplicial complex $\mathcal{Q}_g$ are glued together according to this very partial ordering on dessins with additional structures on which flags are based, the geometric realization of $\mathcal{F}$ with simplices $S_F$ is a simplical complex.

\subsection{Retraction}

\prop{

$\S_{g,1}$
is a deformation retract of $\widehat{\mathcal{M}}_{g,1}$.
}

To prove this proposition we will construct a continuous function $f: \hM
\times [0,1] \rightarrow \hM$, such that $f_0$ is identical on $\hM$ and
$f_1(\hM)=\S_{g,1}$.

First, we will define $f$ on a certain dense subset of $\hM \times [0,1]$ and then
we will prove that it can be extended to all $\hM \times [0,1]$ by continuity.

Let $S_{\widehat{D}_0}$ be a stratum of the highest dimension (recall that it is actually
an open simplex, as all strata are). Introduce a particular ordering on the edges $\widehat{D}_0$.
Then we have coordinates $(l_1,\dots,l_{N}),\; \su{i}
l_i = 1$, on the stratum. Consider a subset of the stratum with
$l_1>l_2>\dots>l_{N}$. Denote it by $\A$. We will now define the
function $f$ on $\A\times [0,1]$. Note that by choosing all other possible
orderings of edges and considering analogous subsets, one covers all stratum.
Thus, if we define $f$ on $\A$ and then define it in analogious way on other
such subsets corresponding to other orderings and on other strata of the highest
dimension, we will have $f$ defined on a dense subset of $\hM \times [0,1]$.

Let
\eq{L = (l_1,\dots,l_{N})}
be a point in $\A$. Let us retract the edge of the
dessin which corresponds to the last coordinate (which is the smallest one by
construction of $\A$). Denote the resulting dessin as $\widehat{D}_1$. Then let us try to retract the edge which corresponds to
the next to last coordinate and so on. We will finally face a situation when we
try to retract an edge, retraction of which inevitably leads to pinching a
handle. Let $k+1$ be the index of the coordinate for which we faced such a
situation for the first time. Then denote by $O$ the following point:
\eq{
O = \dfrac{1}{l_1+\dots+l_{k} -
k l_{k+1}}\br{l_1-l_{k+1},l_2-l_{k+1},\dots,l_{k}-l_{k+1},0,\dots,0}
}
This point does not belong to $S_{\widehat{D}}$. Actually, it does not belong to $\hM$.
It lies on a boundary of $S_{\widehat{D}}$ not included in $\hM$.
Consider then the ray $OL$ in $S_{\widehat{D}}$. Note that it intersects one of
simplices of $\S_{g,1}$, namely the simplex corresponding to flag
$F: \widehat{D}_0 \prec \dots \prec \widehat{D}_k$.

To see it, recall the construction
of this simplex. Since this simplex is the convex hull of centers of $S_{\widehat{D}_0},\dots,S_{\widehat{D}_0}$, in this simplex lie all points of $S_{\widehat{D}_0}$ with first $k$ coordinates equal to each other and last $N-k$ ones strictly decreasing and less then the first ones.

It is straightforward to show that ray $OL$ passes through the point
\eq{
P(L) =
\dfrac{1}{1+k l_k-l_1-\dots-l_{k}}\br{l_{k+1},\,l_{k+1},\dots,l_{k+1},\,l_{k+2},\dots,
l_N},
}
where $l_{k+1}$ stand on first $k=1$ places and $N=6g-3$ is the number of edges of $\widehat{D}_0$. Obviously, this point lies on $OL$ further from $O$ then $L$ does.

Because $l_k>l_{k-1}>\dots>l_{N}$ it is evident that this point lies in $S_F$ due to all mentioned requirements being satisfied. It is also evident that this is the only point of intersection of $OL$ with $S_F$, because this is the only point on $OL$, where first $k$ coordinates are equal to each other (there cannot be more such points because $OL$ contains a point, namely point $L$, which has all coordinates mutually \textit{inequal}).

Let then $f_t$ map the point $L$ to the point on $LP$ which divides this segment as $t:\br{1-t}$, i.e.
\eq{
f:\br{L,t} \mapsto L-\br{L-P(L)}t
}
Because this map is linear in $t$ and all $l_i$, it is evidently continuous on $\A$. Also it is evident that $f_0$ is identical and that $f_1$ maps $\A$ on $S_F$ (one only needs to check that something is mapped to \textit{every} point of $S_F$; this can be done by just considering a point in $S_F$ and a point on the corresponding boundary and than taking any point on the connecting interval).

Thus we have defined a continuous map on a dense subset of $\hM\times [0,1]$.
It is easy to see that this map can be continued to entire $\hM$.

\subsection{``Dual'' cell complex}
\label{cells}
All information about homology of $\widehat{\mathcal{M}}_{g,1}$ is encoded in its spine $\S_{g,1}$, a pure simplicial complex of dimension $4g-3$, defined and studied above. However, already for genus 2 the number of simplices in this simplicial complex is far too huge for its homologies being computable on current hardware. Thus we prove that actually it can be represented as a \textit{cellular complex} by combining collections of simplices into cells, with eventual number of cells being far smaller than initial number of simplices.

This cell complex is constructed in the following way.

Let $\T_0$ be the set of centers of all strata of codimension $4g-3$. Evidently, $\T_0 \subset \S_{(0)}$, i.e. $\T_{0}$ is a certain subset of the set of vertices of the simplicial complex. For $T \in \T_{0}$ consider its star $A_T$ in $\S_{g,1}$ (recall that the star of a vertex of a simplicial complex is the union of all simplices of this complex which have this vertex as one of their vertices). Denote the set of all such stars as $\As_{(4g-3)}$. Exclude all simplices that formed these stars from $\S$. The remaining set of simplices will still be a simplicial complex.

Consider then the set $\T_1$, centers of all strata of codimension $4g-4$. Consider stars of elements of this set in the mentioned remaining simplicial complex. They form the set $\As_{(4g-4)}$ and so on.

\begin{lem}
\label{celllemma}
$\As = \mathop{\cup}_{i=0}^{4g-3}\As_{(i)}$ is a cell complex.
\end{lem}
In order to prove this, let us first prove the following proposition.

\prop{
Stars $A_{(4g-3)}(T),\ T \in \T_{0}$ as simplicial complexes are isomorphic to barycentric subdivision of \textit{associahedron $\mathcal{K}^{(4g-3)}$}.
}
For the definition of $\mathcal{K}^{(4g-3)}$ refer to e.g. \cite{Loday}.

Let $T$ be an element of $\T_{0}$ and let $\hD_0$ be the dessin with structure corresponding to the stratum of $\hM$ where $T$ belongs. Note that simplices of its star $A_{(4g-3)}(T)$ correspond to all possible flags of dessins with structures such that $\hD_0$ can be obtained from them by contraction of some number of edges. Let $\mathcal{D}_T$ be the set of all such dessins with structures and let $\mathcal{F}_T$ be the set of all such flags ($A_T$ is then a realization of $\mathcal{F}_T$ as an abstract simplicial complex).

Note the following
\prop{ \label{propdes} For a given $\hD$ and two sets $E_1$ and $E_2$ of its edges, if $E_1$ differs from $E_2$, then dessins with structures $\hD_1$ and $\hD_2$ obtained from $\hD$ by contraction of all edges in $E_1$ and  $E_2$ respectively are different.}
This proposition follows from the fact that a curve with an additional structure of this type does not have any nontrivial automorphisms \cite{ACG}. Let us stress that this proposition is not valid for dessins without additional structures, but it \textit{is} true for dessins with chosen symplectic bases.

This fact leads to the following statement: $\mathcal{D}_T$ is in one-to-one correspondance with the set of all planar trees with $4g-1$ leaves. If one cuts all of the edges of the original dessin $\hD_0$, one will obtain a planar tree. All other dessins in $\mathcal{D}_T$ again turn into trees if one cuts these $4g$ edges in them. Then the previous fact implies that it is a one-to-one correspondance.

These trees label the cells of associahedron $\mathcal{K}^{4g-3}$ (which is a cell complex). A tree with $k$ internal vertices corresponds to a $(4g-2-k)$-cell. The barycentric subdivision of $\mathcal{K}^{4g-3}$ is a realization of an abstract simplicial complex formed by flags of trees. These flags are actually the same as the flags in $\mathcal{F}_T$, via the mentioned correspondance. This implies that $A_T$ is isomorphic to the barycentric subdivision of $\mathcal{K}^{4g-3}$.

This proposition implies that all stars $A_T$ for $T \in T_0$ are homeomorphic to balls of corresponding dimension, due to the associahedron being homeomorphic to a ball.

Analogously, all defined above stars of lower dimesion are homeomorphic to balls as well, which implies the statement of the lemma.

Note that resulting cell complex is ``dual'' to quasisimplicial complex $\mathcal{Q}_g$ in the sense that simplices of the highest dimension of $\mathcal{Q}_g$ are in one-to-one correspondance with vertices of cell complex, simplices of codimension one are in one-to-one correspondance with edges of cell complex and so on.

Also note that the incidence number arising in computation of cellular homology are trivial in the case of complex $\mathcal{A}$. This again follows from Proposition \ref{propdes}.

\section {Description of the computational method}
Let us briefly repeat the practical steps needed to be taken in order to actually compute Betti numbers of the moduli spaces of level $m$ of algebraic curves of genus $g$, where $m=2$ for $g=1$ and $m=3$ otherwise. In the two sections that follow this one we present the details of computation for $g=1$ and $g=2$.

In this practical approach we actually compute the homology of the cell complex $\mathcal{A}_g$ which is a deformation retract of $\hM$. We use the fact that this complex is ``dual'' to $\mathcal{Q}_g$.

The practical steps are the following ones:
        \begin{enumerate}
            \item Describe all possible dessins d'enfants with at least trivalent vertices that can be drawn on a surface
            of a given genus , which give exactly one disc when cut along. It is equivalent to enumerating
            all inequivalent ways of glueing of a $(12g-6)$-gon.
            \item For each glueing scheme find all symplectic bases in 1-homology of the curve. For
            this one needs to define an intersection form on the glueing scheme (see Section \ref{intersect} for details).
            \item Construct the structure of the cell complex by labeling cells with gluing scheme with basis.
            \item Define the boundary operator corresponding to contraction of edges step by step.
            \item Compute dimensions of kernels and images of the boundary operator.
        \end{enumerate}

        \section{Genus 1 case}
\label{gen1}
In this section we compute Betti numbers of $\htM_{1,1}$, where $\htM_{1,1}$ is a six-sheeted covering of $\Mt_{1,1}$.

First, let us enumerate all needed graphs.  For this we calculate Euler's characteristic and impose condition, that graph is trivalent:
\eq{
\chi \equiv V - E + F = 2 - 2g = 0
}
\eq{\left\{
\begin{aligned}
& V - E + 1 = 0 \hspace{0.4cm} \Leftrightarrow \hspace{0.3cm} \\
& 3V = 2E \\
\end{aligned}
\right.
\left\{
\begin{aligned}
V =2 \\
E=3 \\
\end{aligned}
\right.}

There is a unique graph, that satisfies these requirements, which is called $\theta$.

\begin{figure}[!ht]
\begin{center}
\includegraphics[scale=0.3]{pics/g=1.png}
\end{center}
\caption{$\theta$-graph}
\label{pic}
\end{figure}

Second, we define functions on the edges of the graph over $\mathbb{Z}/2\mathbb{Z}$, i.e. we associate $0$ or $1$ with each edge, so that the sum at each vertex is equal to zero. Then there are $4$ functions on this graph:
\eq {
\lb \begin{array}{c}
0  \\
0  \\
0
\end{array} \rb, \
e_1:=\lb \begin{array}{c}
1  \\
1 \\
0
\end{array} \rb, \
e_2:=\lb \begin{array}{c}
0  \\
1 \\
1
\end{array} \rb, \
e_3:=\lb \begin{array}{c}
1  \\
0  \\
1
\end{array} \rb.
}
We consider these functions as vectors generating $2$-dimensional vector space. To come to agreement with canonical intersection form of cycles the basis of the vector space has to be symplectic. Let us enumerate all symplectic bases:
\eq{
(e_1,e_2), \ (e_1,e_3), \ (e_2,e_3), \ (e_2,e_1), \ (e_3,e_1), \ (e_3,e_2).
}

Third, glueing schemes with symlectic bases form cell. Since in $g=1$ case there is the only glueing scheme, then there are $6$ cells as well as a number of symplectic bases.

Let us introduce the following notation: if the edge is labeled by $\br{0,1}$, we denote this edge by $\textbf{a}$;  if the edge is labeled by $\br{1,0}$, we denote it by $\textbf{b}$, otherwise $\textbf{c}$. All zeroes situation is not considered since it is not presented in any of abovewritten bases superficially. So, there are 6 cells of highest dimension in our complex:

\begin{figure}[H]
\centering\leavevmode
\includegraphics[width=12 cm]{pics/full_basis.png}
\caption{FULL BASIS}
\label{Fbasis}
\end{figure}

Note that some glueing schemes with bases are equivalent with respect to rotation. Thus, there are only 2 nonequivalent glueing schemes with bases, i.e. we have $2$ cells of highest dimensions:
\begin{figure}[H]
\centering\leavevmode
\includegraphics[width=9 cm]{pics/reduced_basis.png}
\caption{REDUCED BASIS}
\label{rbasis}
\end{figure}

Fourth, to find out how these cells are connected to each other, we contract each edge of $\theta$-graph by turn and again factorize by rotation. Thus, we have 3 nonequivalent glueing schemes:
\begin{figure}[H]
\centering\leavevmode
\includegraphics[width=12 cm]{pics/1_cells.png}
\caption{1-CELLS}
\label{1cells}
\end{figure}

If we contract one more edge, than we will contract torus handle and genus will get down. Thus, three 0-cells $\textbf{a},\textbf{b},\textbf{c}$ are not included in our cellular complex. So, our cellular complex consists of two $2$-dimensional cells $\textbf{abc}$ and $\textbf{acb}$, three $1$-dimensional cells $\textbf{ab}$, $\textbf{bc}$ and $\textbf{ac}$. Geometrically, we can illustrate the complex as follows:
\begin{figure}[H]
\centering\leavevmode
\includegraphics[width=5 cm]{pics/complex.png}
\caption{CELLULAR COMPLEX}
\label{complex}
\end{figure}

Fifth, we construct dual complex by the following way $$ {\rm 2-}cell \ \longmapsto \ {\rm 0-}cell \hspace{2cm} {\rm1-}cell \ \longmapsto \ {\rm 1-}cell $$
\begin{figure}[H]
\centering\leavevmode
\includegraphics[width=8 cm]{pics/dual_complex.png}
\caption{DUAL CELLULAR COMPLEX}
\label{ducomplex}
\end{figure}

Thus, clearly, Betty numbers of moduli space for genus 1 with 1 marked point are
\eq{
\begin{array}{|c|}
\hline\\
dim\mathbb{H}^0(\Mt_{1,1},\mathbb{Z}/2\mathbb{Z})=1\\
dim\mathbb{H}^1(\Mt_{1,1},\mathbb{Z}/2\mathbb{Z})=2\\
dim\mathbb{H}^k(\Mt_{1,1},\mathbb{Z}/2\mathbb{Z})=0, k \geq 2\\ \\
\hline
\end{array}
}

\section {Genus 2 case}

First of all, for this $g=2$ case we computed the homology of stars and it turned out to be trivial, thus allowing us to compute the homology of the moduli space as cellular homology of the quotient of spine by star structure rather than just as simplicial homology of the spine itself.

\par
Also, the case of genus two is rather special,
due to existence of hyperelliptic involution the covering of $\Mm_{2,1}$
obtained with the help of symplectic bases in homologies of the surface over field
$\mathbb{Z}/2\mathbb{Z}$ is still not smooth and one needs to consider symplectic bases in homologies
over $\mathbb{Z}/3\mathbb{Z}$.

The results on ribbon graphs on genus-2 surfaces are summarised in the following table:

\begin{tabular}{|c||c@{\hspace{7mm}}|c@{\hspace{7mm}}|c@{\hspace{7mm}}|c@{\hspace{7mm}}|c@{\hspace{7mm}}|c@{\hspace{7mm}}|c@{\hspace{7mm}}|c@{\hspace{7mm}}|c|}
 \hline
 number &  \multicolumn{8}{|c|}{number of schemes with symmetry of order  } &  total number \\

\cline{2-9}
 of edges & 1  & 2 & 3 & 4 & 5 & 6 & 8 & 10 & of schemes \\
\hline
\hline
9 & 3 & 5 & 1 &   &   &   &   &    & 9 \\
\hline
8 & 24 & 4 &   & 1 &   &   &   &    & 29 \\
\hline
7 & 41 & 11  &   &  &   &   &   &    & 52 \\
\hline
6 & 37 & 5 & 1  & 1 &   & 1  &   &    & 45 \\
\hline
5 & 14 & 5 &   & 4 & 1  &   &   & 1  & 21 \\
\hline
4 & 2 & 1 &   &  &   &   & 1  &    & 4 \\
\hline
\end{tabular}

\

On an asymmetric graph there are 51840 different symplectic bases in homologies over $\mathbb{Z}/3\mathbb{Z}$ on the curve. On a graph with a symmetry of order $r$ there are exactly $51840/r$ different bases. Graphs with chosen bases enumerate cells in our complex, with graphs with 9 edges corresponding to 0-cells in (dual) complex, with 8 edges -- to 1-cells, and so on. Thus, for the numbers $n_j$ of cells of dimension $j$ we have
\eqna{
n_0 = 51840\cdot & \displaystyle\br{\frac{1}{3}+\frac{5}{2}+3} &= 302400\\
n_1 = 51840\cdot & \displaystyle\br{\frac{1}{4}+\frac{4}{2}+24} &= 1360800\\
n_2 = 51840\cdot & \displaystyle\br{\frac{11}{2}+41} &= 2410560\\
n_3 = 51840\cdot & \displaystyle\br{\frac{1}{6}+\frac{1}{4}+\frac{1}{3}+\frac{5}{2}+37} &= 2086560\\
n_4 = 51840\cdot & \displaystyle\br{\frac{1}{10}+\frac{1}{5}+\frac{5}{2}+14} &= 870912\\
n_5 = 51840\cdot & \displaystyle\br{\frac{1}{8}+\frac{1}{2}+2} &= 136080\\
}
Euler characteristic of this complex is thus
\eq{
\chi_{\widehat{\Mm}_{2,1}} = 302400 - 1360800 + 2410560 - 2086560 + 870912 - 136080 = 432
}
Since Euler characteristic of a covering is equal to Euler characteristic of the base times the degree, we have
\eq{
\chi_{{\Mm}_{2,1}} = \displaystyle\frac{432}{51840} = \frac{1}{120}= \zeta\br{-3}
}
i.e. Euler characteristic of original noncompactified moduli space ${\Mm}_{2,1}$ is precisely equal to the value of Riemann zeta function at $-3$, which is in perfect agreement with results of Harer and Zagier.

As mentioned above, to compute Betti numbers of this covering of the moduli space it is suffecient to consider homologies of this cell complex over $\mathbb{Z}/2\mathbb{Z}$. We thus have 5 matrices filled with zeros and ones as matrices of boundary operators. Their ranks are
\eqna{
r_{1\rightarrow 0} &=& 302399\\
r_{2\rightarrow 1} &=& 1058377\\
r_{3\rightarrow 2} &=& 1352011\\
r_{4\rightarrow 3} &=& 733978\\
r_{5\rightarrow 4} &=& 135716\\
}
Thus, for Betti numbers we have
\eqna{
b_0 &=& 1\\
b_1 &=& 24\\
b_2 &=& 172\\
b_3 &=& 571\\
b_4 &=& 1218\\
b_5 &=& 364\\
}

\section{Conclusion}

In the present paper we constructed a direct method of computation of homology of certain smooth covers of moduli spaces of pointed curves via construction of the spine, a simplicial complex, on which the cover of moduli space retracts. Then we conjectured the equivalence of this spine to certain cell complex, which allows direct computation for the case of genus 2. We carried out this calculation and obtained Betti numbers for this case.

The proof of the mentioned conjecture remains a work in progress.

Since in our work we computed Betti numbers, then, recalling the mentioned in introduction connection with string theory, it seems natural to ask if there exist some generalized Penner model, i.e. a matrix model (or, perhaps, $\beta$-ensemble), that generates Betti numbers and not just Euler characterstics. It is very interesting if this model is an integral representation of some important quantity (such as Nekrasov function) on the string theory side. We are currently working in these directions.

\appendix
    \section {Details of computation}
        \subsection {Intersection form}
\label{intersect}

        \par
        In order to find possible symplectic bases on a given curve (parametrized by a glueing scheme),
        one needs to know an intersection form on cycles, that is
        bilinear map, which for any two cycles $a$ and $b$ gives their intersection number.
        Intersection form for $g = 1$ is very simple and is described in the corresponding section,
        so here we are focusing on the form for genus two.
	Note, that while we consider curve homology with coefficients in $\mathbb{Z}/3\mathbb{Z}$ for $g=1$, for $g\geq 2$ we use homology with coefficients in $\mathbb{Z}/3\mathbb{Z}$.

        Note, that once we have found sympletic bases for glueing schemes of highest dimension, possible
        bases on cells of lower dimension are induced by contraction of some edges of cells of highest dimension.
        So, in what follows we concentrate on obtaining the intersection form on cells of highest dimension.

        \par Each one cycle, drawn on a curve, can be deformed in such a way, that it will run
        along edges of the glueing scheme, so each cycle is a map from edges of a glueing scheme to $\mathbb{Z}/3\mathbb{Z}$.
        Now, the answer is as follows: for any two cycles $a$ and $b$ their intersection number is equal to
        \eq {
            Int_{ab} = \lb \sum_{i} \Omega (a(i) \times a(i-1), b(i) \times b(i-1)) \rb \mod 3,
		\label {int_number}
	  }
        where sum is taken over vertices of a glueing scheme, $a(i)$ denotes the value
        of $a$ on the i-th edge. Note, that $\Omega$ defined below is neither \textit{antisymmetric}, nor
        \textit{polylinear}. Nevertheless, the intersection number, computed with help of it,
        is antisymmetric and polylinear function of $a$ and $b$.
        \par $\Omega$ is equal to
        \eq {
            \Omega = \lb \begin{array}{cccccccccc}
                          & 0\times0 & 0\times1 & 0\times-1 & 1\times0 & 1\times1 &
                          1\times-1 & -1\times0 & -1\times1 & -1\times-1 \\
                          0\times0: & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
                          0\times1: & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
                          0\times-1: & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
                          1\times0: & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
                          1\times1: & 0 & -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 0 & 1 & -\frac{1}{2} & -1 & 0 \\
                          1\times-1: & 0 & \frac{1}{3} & -\frac{1}{3} & \frac{1}{3} & -\frac{1}{3} & -\frac{1}{2} & -\frac{1}{3} & \frac{1}{2} & \frac{1}{3} \\
                          -1\times0: & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
                          -1\times1: & 0 & -\frac{1}{3} & \frac{1}{3} & -\frac{1}{3} & \frac{1}{3} & \frac{1}{2} & \frac{1}{3} & -\frac{1}{2} & -\frac{1}{3} \\
                          -1\times-1: & 0 & \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & 0 & -1 & \frac{1}{2} & 1 & 0
                     \end{array}
                    \rb
        }
        Let us turn to the derivation of this formula. Each cycle over $\mathbb{Z}/3\mathbb{Z}$ can
        be thought of as oriented ribbon, where $+1$ means that orientation of the ribbon coincides
        with that of an edge and $-1$ means that it is opposite. Also, since $1+1+1 = 0$, trivalent glueings
        of the ribbon are allowed, such that orientation of all glued ribbons are incoming (outgoing),
        respectively.
        \par Overall sign of the intersection number is a matter of convention.
        We use the following convention: left picture counts as $+1$ while right picture counts
        as $-1$.

\begin{figure}[!ht]
\begin{center}
\includegraphics[scale=0.5]{pics/positive_negative_intersection.png}
\end{center}
\caption{Positive and negative intersections}
\label{pos_neg_int}
\end{figure}

        If one looks at some vertex for a given cycle, one can see one of the following pictures.
\begin{figure}[!ht]
\begin{center}
\includegraphics[scale=0.25]{pics/trivalent_vertex_examples.png}
\end{center}
\caption{Possible behavior of a cycles at the vertex}
\label{tri_vert_exampl}
\end{figure}
        The idea is, that for each vertex, for each pair of possible pictures, an intersection
        number can be explicitly written.
        \par
        To succeed, without loss of generality assume, that whenever values of $a$ and $b$ on a given edge
        attached to some vertex are both non-zero $a$ is on the ``left'' side of the edge, while
        $b$ is on the ``right'' side:
\begin{figure}[!ht]
\begin{center}
\includegraphics[scale=0.4]{pics/a_b_order.png}
\end{center}
\caption{ordering of cycles}
\label{cycle_order}
\end{figure}

\par One further distinguishes two kinds of situations: when both cycles are of type $L$ or $R$
at a given vertex and follow the same road (see pic), and all other situations.

\begin{figure}[!ht]
\begin{center}
\includegraphics[scale=0.4]{pics/trivial_intersection.png}
\end{center}
\caption{Example of the trivial behavior at the vertex}
\label{trivial_vert_ex}
\end{figure}
Let us call situations (and vertices) of the first kind \textit {trivial}, others - \textit {nontrivial}.
It is easy to see that trivial vertices unify in sequences, in which $a$ and $b$ run along eash other,
and there are nontrivial vertices on both ends of such a sequence. Clearly, due to our ordering
convention, one should attribute either $+1$ or $-1$ intersection number to the whole sequence
of trivial vertices. Or, alternatively, one may prescribe an additional $\pm \frac{1}{2}$ to
nontrivial vertices, and do not consider trivial vertices at all.
\begin{figure}[!ht]
\begin{center}
\includegraphics[scale=0.4]{pics/trivial_taken_into_account.png}
\end{center}
\caption{Sequence of trivial vertices adds $+1$ or $-1$ to the intersection number}
\label{trivial_account}
\end{figure}

\par The next step is to consider each essential situation and prescribe intersection number
to it, which is fairly straightforward.
\par Next, one should notice, that type of the situation that actually takes place at given
vertex of a graph can be unambiguously reconstructed from values of cycles on just two of three
incoming edges (that is, by looking on vertex of a glueing scheme and edges, attached to it).
\par This way, one can rewrite sum over vertices of a graph on a curve as sum over vertices of
a glueing scheme. The only thing that one should take into account is, that risks to count
some vertices more than once, so symmetry factors like $\frac{1}{2}$ and ${1}{3}$ should be
introduced.
\par If one proceeds through all these steps, one arrives at the formula (\ref{int_number}).

\subsection {Calculation of ranks}
Matrices of the boundary operator, that appear in our calculation, are pretty large, so it is challenging to calculate their ranks. Luckily, there is a wonderful collection of linear algebra tools called\\
\textit{linbox} (http://www.linalg.org/linbox-html/index.html). So, we used the implementation of the Wiedemann algorithm, presented there, to calculate ranks of our matrices.

\section*{Acknowledgements}
Authors are very grateful to Sergey Shadrin for the idea of proof of Lemma \ref{celllemma} and to Elena Kreines for helpful discussions.
It is a special pleasure to thank Robert Penner for stimulating discussions and very helpful remarks. Authors are also grateful to Jim Milgram for a useful reference and to Jean-Guillaume Dumas for the help with \textit{linbox} software. This work is partly supported by NWO grant 613.001.021 (P.D.-B.); by RFBR grants 10-01-00709 (G.Sh.), 10-02-00509 (A.P., A.S.) and 10-02-00499 (P.D.-B.); by Federal Agency for Science and Innovations of Russian Federation (contract 02.740.11.0608); by Russian Federal Nuclear Energy Agency, by the joint grants  10-02-92109-Yaf-a (P.D.-B., A.P.), 09-02-91005-ANF (P.D.-B., A.P.), 09-02-93105-CNRSL (A.P.) and 09-01-92440-CE (P.D.-B., A.S.).


TITLE: How does the Earth rotate, given that the torque acting on it while revolving is zero?
QUESTION [0 upvotes]: I've come to understand that the torque acting on the Earth while revolving the Earth is zero. Torque is the force responsible for rotation of a body. So how does the Earth rotate?

REPLY [5 votes]: Moving bodies have inertia which means that they will continue to move at a constant velocity unless acted upon by an external force (this is Newton's first law of motion).
Similarly, rotating bodies have a moment of inertia, meaning that they will continue to rotate unless acted upon by an external force (torque).
Therefore, torque is only required to either accelerate (or decelerate) a rotating body, or to act against a resistive force so as to maintain a constant rotational velocity.
Since there is only a very tiny resistance to the rotation of the earth, no torque is required. Note, however, that since there is resistance (primarily from tidal forces) then the earth is actually slowing down.


TITLE: Let S be the set of all reals where every other decimal place, starting with the first one, contains a 1.
QUESTION [0 upvotes]: (so, for instance, S contains 23.101816191... but not 0.123419...)
(a) Show that the cardinality of positive integers is less than or equal to the cardinality of the set S
(b) Is the cardinality of set S equal to the cardinality of the real numbers?
For (a) I have been playing around with the concept that the cardinality of X is less than or equal to the cardinality of Y if there is an injection from X to Y
For (b) I know the answer is no because the reals are uncountable, but I'm not sure how to explain this/prove this.
Thanks in advance for your help!

REPLY [1 votes]: You are correct with your idea for a.  Can you find an explicit injection from $\Bbb N$ into $S$?  Hint:  $345 \to 0.1314151\overline{01}$ For b, you are not correct that $S$ is countable.  Can you find an injection from $(0,1)$ into $S$ using the same idea?  As all the members of $S$ are reals, that shows that $S$ has the same cardinality as $\Bbb R$


TITLE: Heat equation: Show $\int\limits_{\mathbb{R}^n} u(x,t)\,dx=\int\limits_{\mathbb{R}^n} g(x)\,dx$ for $t>0$
QUESTION [2 upvotes]: Assume $g\in C(\mathbb{R}^n)$, $g\in L^1(\mathbb{R}^n)$, $|g|\leq M$. Let $u$ be the bounded solution to
\begin{gather*}
\Delta u -u_t=0 \text{ for } t>0, x\in\mathbb{R}^n, \\
u(x,0)=g(x) \text{ for } x\in\mathbb{R}^n
\end{gather*}
Show

$\lim\limits_{t\to\infty} \sup\limits_{x\in\mathbb{R}^n} |u(x,t)|=0$
$\int\limits_{\mathbb{R}^n} u(x,t)\,dx=\int\limits_{\mathbb{R}^n}
    g(x)\,dx$ for $t>0$

Proof of (a): Assume $g\in C(\mathbb{R}^n)$, $g\in L^1(\mathbb{R}^n)$, $|g|\leq M$. Let $u$ be a solution to the PDE above. Recall the solution to this PDE is
$$u(x,t)=\dfrac{1}{(4\pi t)^{n/2}} \int_{\mathbb{R}^n} \exp\left[ -\dfrac{|x-y|^2}{4t}\right] g(y)\,dy,$$
where $x\in \mathbb{R}^n$ and $t>0$. So,
\begin{equation*}
\begin{aligned}
|u(x,t)|
& =\left|\dfrac{1}{(4\pi t)^{n/2}} \int_{\mathbb{R}^n} \exp\left[ -\dfrac{|x-y|^2}{4t}\right] g(y)\,dy\right| \\
& \leq \dfrac{1}{(4\pi t)^{n/2}} \int_{\mathbb{R}^n} \left|\exp\left[ -\dfrac{|x-y|^2}{4t}\right]\right| \cdot \left|g(y)\right|\,dy \\
& \leq \dfrac{1}{(4\pi t)^{n/2}} \int_{\mathbb{R}^n} \exp\left[ -\dfrac{|x-y|^2}{4t}\right]\cdot M\,dy
\end{aligned}
\end{equation*}
Recall also the formula for the integral of a Gaussian function
$$\int_{\mathbb{R}} a\exp\left[-\dfrac{(x-b)^2}{2c^2}\right]\,dx=\sqrt{2a} \cdot |c|\cdot \sqrt{\pi}.$$
We can extend this result to $\mathbb{R}^n$ and use it in our inequality above.
\begin{equation*}
\begin{aligned}
|u(x,t)|
& \leq \dfrac{M}{(4\pi t)^{n/2}} \int_{\mathbb{R}^n} \exp\left[ -\dfrac{|y-x|^2}{2(\sqrt{2t})^2}\right]\,dy \\
& = \dfrac{M}{(4\pi t)^{n/2}} \cdot \sqrt{2\pi} \cdot|\sqrt{2t}|
= \dfrac{M}{(4\pi t)^{n/2}} \cdot \sqrt{4\pi t}
= \dfrac{M}{(4\pi t)^{(n-1)/2}}
\end{aligned}
\end{equation*}
Hence
$$\lim\limits_{t\to\infty} \sup\limits_{x\in\mathbb{R}^n} |u(x,t)| \leq
\lim\limits_{t\to\infty} \sup\limits_{x\in\mathbb{R}^n} \dfrac{M}{(4\pi t)^{(n-1)/2}}
= \lim\limits_{t\to\infty} \dfrac{M}{(4\pi t)^{(n-1)/2}} =0
$$
i.e.
$$\lim\limits_{t\to\infty} \sup\limits_{x\in\mathbb{R}^n} |u(x,t)| =0
$$
Proof of (b):
\begin{equation*}
\begin{aligned}
\int\limits_{\mathbb{R}^n} u(x,t)\,dx
& = \int\limits_{\mathbb{R}^n}\left[\dfrac{1}{(4\pi t)^{n/2}} \int_{\mathbb{R}^n} \exp\left[ -\dfrac{|x-y|^2}{4t}\right] g(y)\,dy \right]\,dx \\
& = \dfrac{1}{(4\pi t)^{n/2}} \int\limits_{\mathbb{R}^n}\left[ \int_{\mathbb{R}^n} \exp\left[ -\dfrac{|x-y|^2}{4t}\right] \,dx \right]g(y)\,dy  \\
& = \dfrac{1}{(4\pi t)^{n/2}} \int\limits_{\mathbb{R}^n}\left[ \sqrt{2\pi} \cdot|\sqrt{2t}|\right]g(y)\,dy  \\
& = \dfrac{1}{(4\pi t)^{(n-1)/2}} \int\limits_{\mathbb{R}^n}
g(y)\,dy  \\
& = \dfrac{1}{(4\pi t)^{(n-1)/2}} \int\limits_{\mathbb{R}^n}
g(x)\,dx  \\
\end{aligned}
\end{equation*}
Questions:

Is my proof of (a) correct?
I know I am very close to finishing the proof of b. What am I missing?

REPLY [1 votes]: If $g(x) = M$ then you can easily check that the solution is $u(x,t) \equiv M$ for which $u\not\to 0$ as $t\to\infty$ which contradicts your result. This does not contradict the original claim since $g\not\in L^1$, but this assumption is not used in your derivation. Your mistake is in computing the integral, and if you do the calculation right you will find
$$|u| \leq \int K(x,y,t)|g(y)|\,{\rm d}y \leq M\int K(x,y,t)\,{\rm d}y = M$$
since the integral of the Heat kernel is unity, i.e. $\int K(x,y,t){\rm d}y = 1$. Thus this approach is not good enough to show $u \to 0$, only that $u$ remains bounded.
Since both $g\in L^1$ and the properties of the heat kernel is critical for the claim to hold true it's natural to apply some integral inequality to try to bound the integral. Applying Cauchy-Schwarz we find
$$|u|^2 = \left(\int K(x,y,t)g(y)\,{\rm d}y\right)^2\leq \left(\int K^2(x,y,t)\,{\rm d}y\right)\left(\int g^2(y)\,{\rm d}y\right)$$
Now you can check that $K^2(x,y,t)= \frac{1}{\sqrt{8\pi t}^{n/2}}K(x,y,t/2)$ which gives you the first term and for the second term note that $g^2(y) \leq M|g(y)|$ and use the fact that $g\in L^1$.


text:
\begin{document}

\title{Application of Mean Curvature Flow for surface parametrizations}

\author{Ka Wai Wong}
\curraddr{Department of Mathematics, University of California -- Davis}
\email{ucdwong@ucdavis.edu}

\keywords{Conformalized Mean Curvature Flow, Discrete Conformal Map, Conformal Parametrization, Shape Analysis}

\begin{abstract}
This is an expository article describing the conformalized mean curvature flow, originally introduced by Kazhdan, Solomon, and Ben-Chen. We are interested in applying mean curvature flow to surface parametrizations. We discuss our own implementation of their algorithm and some limitations.
\end{abstract}

\maketitle

\section{Introduction}
We would like to apply extrinsic geometric flows on the parametrization of a closed genus-zero suface to obtain a conformal map from the surface onto a unit sphere. Conformal parametrizations have been widely studied in shape analysis and have found applications in anthropology, neurobiology, and image processing. Given a closed surface of genus zero, the uniformization theorem guarantees the existence of a conformal map from this surface onto a unit sphere, but such a map is not necessarily unique. In practice, surfaces are represented and visualized by discrete meshes in real-world applications for the sake of computation. As a result, different notions of discrete conformality between surfaces and their corresponding discrete conformal maps have been introduced and studied.

\bigbreak

We provide an overview on different methods in conformal parametrization of two-dimensional surfaces in recent years.
\begin{enumerate}
\item[(i)] A method introduced by Bobenko, Pinkall, Schr\"{o}der, and Springborn is to flatten a mesh onto a plane by minimizing a convex energy functional. Their energy functional describes a precise notion of discrete conformal equivalence that captures the conformal equivalence of surfaces in the continuous case \cite{bps}, \cite{ssp}. However, if the input mesh has a region full of ``flat" triangles, i.e. triangles with one angle close to $\pi$, the triangle inequality might fail during the energy minimization, then the output set of edge lengths cannot be embedded in $\mathbb{R}^3$. Edge flipping or subdivision of triangles can be used to fix this problem.
\item[(ii)] A method introduced by Chow, Luo, and Gu is to use Ricci flow to evolve a genus-zero surface to a sphere \cite{cl}, \cite{jklg}, \cite{zzglg}. More precisely, their idea is to distribute evenly the total curvature  ($4\pi$ by Gauss-Bonnet Theorem) over the discretized surface through assigning a discrete conformal factor on each vertex. These factors can be obtained by minimizing another convex energy functional which is different from the energy functional in (i).
\item[(iii)] A method introduced by Aigerman and Lipman is to generalize Tutte's embedding to a bijective map from a given surface onto an Euclidean orbifold by solving a sparse linear system \cite{al2}, \cite{al0}, \cite{al1}. Their subsequent work extended this bijective parametrization onto spherical and hyperbolic cone-surfaces. Their algorithm uses a large set of landmark points on the given surface and the corresponding parametrization domain as input.
\end{enumerate}
\vskip 5pt
Methods (i) and (ii) are derived from Ricci flow which is an intrinsic flow but with different notions for discrete conformal equivalence of surfaces.

\bigbreak

In this article, we are interested in the conformalized mean curvature flow (cMCF), originally introduced by Kazhdan, Solomon, and Ben-Chen \cite{ksbc}. In the following, we will briefly describe the algorithm they have developed, as well as our own implementation of this algorithm. Our interest in this article is to illustrate its applications as well as highlight some of its limitation.

\section{Conformalized Mean Curvature Flow}
In this section, we present the cMCF as it was introduced in \cite{ksbc}. \\
We start with a finite element discretization of the mean curvature flow (MCF) on a 2 dimensional surface $M$. Let $\Phi_t \colon M \rightarrow \mathbb{R}^3$ be a smooth family of immersions with time $t \geq 0$ and the induced metric $g_t$ at time $t$. MCF is
\begin{align}
\dfrac{\partial \Phi_t}{\partial t} = \varDelta_{g_t} \Phi_t =  2H_t \hat{N_{t}}
\end{align}
where $H_t(p)$ is the scalar mean curvature and $\hat{N_t} (p)$ is the inward unit surface normal. We approximate $\Phi_t$ by $\sum_{i=1}^{N} x_i(t) B_i(p)$ with a finite set of function basis $B_i \colon M \rightarrow \mathbb{R}$ for $1 \leq i \leq n$ and a set of coefficient vectors denoted by $ \vec{X}(t) = \left\{ x_1(t), \cdots, x_N(t) \right\} \subset \mathbb{R}^3$. Using the weak formulation and applying backward Euler method to discretize the time derivative of $\vec{X}(t)$, we solve for $\vec{X}(t+ \tau)$ in $x,y,z-$ directions in the following linear system
\begin{align}
 (D^t - \tau L^t) \vec{X}(t+\tau) = D^t \vec{X}(t)
\end{align}
The mass matrix entries $D^t_{ij} = \int_M B_i \cdot B_j \, dA_t$ are the inner products of the basis on $M$ and the stiffness matrix entries $L^t_{ij} =  - \int_M g_t(\nabla_{t}B_i, \nabla_{t} B_j ) dA_t$ are the inner products of the basis gradients on $M$. $dA_t$ denotes the volume form on $M$ at time $t$. We use the hat basis in practice. Formulas for the mass and stiffness matrices using the hat basis can be found in \cite{ksbc}.\\
MCF cannot always provide a spherical parametrization since a singularity sometimes emerges. Below is the mesh ``spot" with $2,930$ vertices and $5,856$ faces. \textit{Surface area is kept constant at every step}, with step size $ \tau =0.05$. The spot's head evolves into a spike where singularity forms and the flow stops within $4$ steps due to the formation of the spike.

\begin{figure}[H]
\begin{center}
\begin{minipage}[!h]{.28\textwidth}
\centering
\includegraphics[width=\textwidth]{spot}
\end{minipage}
\hfill
\begin{minipage}[!h]{0.28\textwidth}
\centering
\includegraphics[width=\textwidth]{spotMCF1}
\end{minipage}
\hfill
\begin{minipage}[!h]{0.28\textwidth}
\centering
\includegraphics[width=\textwidth]{spotMCF5}
\end{minipage}
\end{center}
\end{figure}

The main idea of cMCF is to replace the metric $g_t$ with a metric $\tilde{g}_t = \sqrt{|g_t||g_0|^{-1}}g_0$ that is conformal to the initial metric $g_0$. The flow becomes
\begin{align}
&\dfrac{\partial \Phi_t}{\partial t} = \sqrt{|g_0||g_t|^{-1}} \varDelta_{g_0} \Phi_t\\
 &(D^t - \tau L^0) \vec{X}(t+\tau) = D^t \vec{X}(t)
\end{align}
While the mass matrix $D^t$ is updated at every step, we use the initial stiffness matrix $L^0$ and do not update it.

\section{Implementation}
We programmed in \texttt{C++} with an open source library \texttt{OpenMesh} \cite{om}. Snapshots were taken using the software \texttt{MeshLab}. Our implementation does not deviate from the original algorithm as it was introduced in \cite{ksbc}. Kazhdan has made his source code for the cMCF available at \cite{kazhdan}. \\
We decided to have our implementation as part of a larger software platform which is to assess the efficiency and limitations of different parametrization methods. The platform is still under construction and is our ongoing work.

\section{Results}
We present results from our implementation of cMCF. Here are two criteria to measure conformality:
\begin{enumerate}
\item[(I).] Angular distortion associated to each triangle, i.e. $\underset{\substack{\theta = \alpha, \beta, \gamma}}{\text{max}}\left(\frac{|\theta - \theta'|}{\theta}\right)$ where $ \alpha, \beta, \gamma$ are the three angles in a triangle and $ \alpha', \beta', \gamma'$ are the transformed angles. The value $0$ implies no angular distortion.
\item[(II).] Deviation of the length-cross-ratio (lcr.) associated to each edge, i.e. $\frac{\mathfrak{c}'_{ij}}{\mathfrak{c}_{ij}} $ where the lcr. of an edge $e_{ij}$ is defined to be $\mathfrak{c}_{ij} = \frac{\ell_{im} \cdot \ell_{jk}}{ \ell_{mj}\cdot \ell_{ki}}$ (as seen in  the figure below) whereas $\mathfrak{c}'_{ij}$ denotes the lcr. of the transformed edge. Two meshes are conformally equivalent if and only if $\mathfrak{c}_{ij} = \mathfrak{c}'_{ij}$ for all edges $e_{ij}$ \cite{ssp}, i.e.  $\frac{\mathfrak{c}'_{ij}}{\mathfrak{c}_{ij}} $ is equal to $1$.
\end{enumerate}
\begin{figure}
\centering
\includegraphics[width =0.3\textwidth]{crosslengthratio}
\end{figure}
In addition, we use the parameter $s = \frac{\left( 36 \pi V^2 \right)^{\frac{1}{3}}}{A}$ to measure the sphericity, where $V$ and $A$ are the volume and the total surface area of the mesh respectively. Clearly, $0 \leq s \leq 1$ and $s = 1$ implies that the mesh is a sphere.\\
We applied the cMCF on ``spot", a human brain mesh (with $65,538$ vertices and $131,072$ faces), and a Christmas deer mesh (with $113,780$ vertices and $227,556$ faces).

\begin{figure}[H]
\begin{center}
\begin{minipage}[!h]{.27\textwidth}
\centering
\includegraphics[width=\textwidth]{spot}
\caption*{$s = 0.679$}
\end{minipage}
\hfill
\begin{minipage}[!h]{0.27\textwidth}
\centering
\includegraphics[width=\textwidth]{spotCMCF5}
\caption*{$s = 0.839$} \label{fig:spotcmcf1}
\end{minipage}
\hfill
\begin{minipage}[!h]{0.27\textwidth}
\centering
\includegraphics[width=\textwidth]{spotCMCF25}
\caption*{$s = 0.999$} \label{fig:spotcmcf5}
\end{minipage}
\end{center}
\end{figure}

\begin{figure}[H]
\begin{center}
\begin{minipage}[!h]{.27\textwidth}
\centering
\includegraphics[width=\textwidth]{subject1}
\caption*{$s = 0.5877$}
\end{minipage}
\hfill
\begin{minipage}[!h]{0.27\textwidth}
\centering
\includegraphics[width=\textwidth]{subject1CMCF1}
\caption*{$s = 0.9290$} \label{fig:subjectcmcf1}
\end{minipage}
\hfill
\begin{minipage}[!h]{0.27\textwidth}
\centering
\includegraphics[width=\textwidth]{subject1CMCF14}
\caption*{$s = 0.9999$} \label{fig:subjectcmcf14}
\end{minipage}
\end{center}
\end{figure}

\begin{figure}[H]
\begin{center}
\begin{minipage}[!h]{.33\textwidth}
\centering
\includegraphics[width=\textwidth]{christmasdeer}
\caption*{$s = 0.4055$} \label{fig:christmasdeer}
\end{minipage}
\hfill
\begin{minipage}[!h]{0.33\textwidth}
\centering
\includegraphics[width=\textwidth]{christmasdeerCMCF2}
\caption*{$s = 0.5678$} \label{fig:christmasdeercmcf2}
\end{minipage}
\hfill
\begin{minipage}[!h]{0.33\textwidth}
\centering
\includegraphics[width=\textwidth]{christmasdeerCMCF22b}
\caption*{$s = 0.9995$} \label{fig:christmasdeercmcf22}
\end{minipage}
\end{center}
\end{figure}

Below are the distributions of the angular distortion (ang.) and that of the deviation of the lcr. for the ``spot" and brain models. \textit{Conformality measurements on the deer are not valid since many triangles either shrink to a point or collapse into a line.} Their area becomes zero. Our future work is to improve the cMCF to avoid mesh degeneracy.
\begin{table}[H]
\begin{center}
\begin{minipage}[!h]{.4\textwidth}
\centering
\begin{tabular}{c|c|c}
\texttt{spot}& mean $\mu$ & std $\sigma$ \\ \hline
ang. & 0.0910 & 0.0767 \\ \hline
clr. & 1.0010 & 0.0441 \\
\end{tabular}
\end{minipage}
\hfill
\begin{minipage}[!h]{0.4\textwidth}
\centering
\begin{tabular}{c|c|c}
\texttt{brain}& mean  $\mu$ & std $\sigma$ \\ \hline
ang. & 0.0326 & 0.0207 \\ \hline
clr. & 1.0004 & 0.0294 \\
\end{tabular}
\end{minipage}
\end{center}
\end{table}

\section{Acknowledgement}
The author would like to thank Joel Hass and Patrice Koehl for their help. The author is partly supported by their NSF grant (NSF DMS 1719582). Thanks to Yanwen Luo and Maria Trnkova for discussion and feedback. Thanks to Michael Kazhdan for very useful discussion on the cMCF.\\
The ``spot" model is from Keenan Crane 3D Model Repository \cite{keenan}. The brain model is from Joel Hass and Patrice Koehl \cite{hk1}. The Christmas deer model is from the dataset of 3D-Printing Models \texttt{Thingi10K} \cite{thingi10k}.

\bibliographystyle{acm}
\bibliography{bibliography}

\end{document}

config: arxiv


TITLE: Tensor product of two irreducible $\mathfrak sl_2$ modules
QUESTION [1 upvotes]: I know that given an irreducible $\mathfrak sl_2$ module $V$, it is $V\cong V(m), m\in \mathbb Z_+$, whose basis $(v_0,\cdots,v_m$) is such that:
$$H\cdot v_j = (m-2j)v_j, Y\cdot v_j = (j+1)v_{j+1}, X\cdot v_j = (m-j+1)v_{j-1},
\mbox{for } j\geq 0 \mbox{ and }v_{-1} = 0 $$
and $V(m) = \bigoplus^m_{k= -m} V_{m-2k}$.

Now given $V=V(m), W=W(n)$ irreducible $\mathfrak sl_2$ modules, why is that $V\otimes W \cong V({m+n})\oplus \cdots \oplus V({|n-m|})$?

Well, we can choose such a basis for $V$ and $W$, respectively: $(v_0,\cdots, v_m)$ and $(w_0,\cdots, w_n$). Then the set $v_i\otimes w_j, 0\leq i \leq n, 0 \leq j \leq m ,$ is a basis for the tensor product $V\otimes W$. Ok, now applying an element of the cartan subalgebra give us:
$H\cdot v_i\otimes w_j = (m+n-2(i+j))v_i\otimes w_j.$
Since $i+j$ lies in the set $\{1,2,\cdots, n+m\}$, it follows that the eigenvalue for $v_i\otimes w_j$ also lies in the set $\{n+m, n+m-2, \cdots, -n-m+2,-n-m\}$. Now how to proceed from here? Looking for what we got to prove, we see that there are no negative eigenvalue anymore; why is that? any hint on how to proceed?

REPLY [1 votes]: Define the character $\text{ch}_V$ of a (finite-dimensional)
representation of $\frak sl_2$ as
$\sum_k \dim E_k(V) t^k$ where $E_k(V)$ is the eigenspace of
$H$ acting on $V$ for the eigenvalue $k$. Then $\text{ch}(V\oplus W)
=\text{ch}(V)+\text{ch}(W)$ and $\text{ch}(V\otimes W)
=\text{ch}(V)\text{ch}(W)$. Also
$$\text{ch}(V(n))=t^n+t^{n-2}+t^{n-4}+\cdots +t^{-n}=\frac{t^{n+1}-t^{-n-1}}
{t-t^{-1}}.$$
The characters for the $V(n)$ are linearly independent, and the
character determines a representation up to isomorphism.
Let $n\ge m$. Then
\begin{align}
(t-t^{-1})\text{ch}(V(n))\text{ch}(V(m))&=(t^{n+1}-t^{-n-1})
(t^m+t^{m-2}+\cdots+t^{-m})\\
&=(t^{m+n+1}-t^{-m-n-1})+(t^{m+n-1}-t^{-m-n+1})+\cdots
+(t^{m-n+1}-t^{-m+n-1})\\
&=(t-t^{-1})(\text{ch}(V(m+n))+\text{ch}(V(m+n-2))+\cdots+\text{ch}(V(m-n)))
\end{align}
and so
$$V(n)\otimes V(n)\cong V(m+n)\oplus V(m+n-2)\oplus\cdots\oplus V(m-n).$$


TITLE: Rolling with slipping
QUESTION [0 upvotes]: What do we mean by the statement "Rolling with slipping"? I couldn't imagine this situation as a real life scenario.. I mean to say that if the body (say a solid sphere) is slipping , then how can it rotate at the same time ? If it will be rotating it won't be slipping . How can the two happen together?

REPLY [1 votes]: It is when there is relative movement between a rotating object and the ground at the point of contact so the condition
translational velocity  $=$ radius of object $\times$ angular speed of object
is not  satisfied.
Push down too hard on the accelerator pedal and the drive wheels of your car will spin with the car and produce a lot of smoke due to the tyres getting very hot because there is relative movement between the tyres and the road - slipping.


text:
\section{Detecting isomorphisms of Howe curves}\label{sec:IsomTest}
In this section, we give an efficient criterion for determining
whether two Howe curves are isomorphic or not.
This criterion will be used in both the first and the second approach
to enumerating superspecial Howe curves over a finite field.

We continue to work over an algebraically closed field of characteristic $p\neq 2$.
Recall from Section \ref{sec:Howe} that a Howe curve is the desingularization of
the fiber product of two genus-$1$ double covers of $\bbP^1$, where the
ramification loci of the two covers overlap in exactly one point. This means
that a Howe curve is precisely a genus-$4$ curve $H$ that fits into a $V_4$-diagram
of the following form, where $C$ is a curve of genus~$2$ and $E_1$ and $E_2$ are curves of genus $1$:
\[
\xymatrix@=3ex{
&&H\ar[lld]\ar[rrd]\ar[d]&&\\
E_1\ar[rrd] & &C\ar[d] & &E_2\ar[lld]\\
&&\bbP^1\rlap{.}&&
}
\]

If $E_1\to\bbP^1$ ramifies at points $P$, $Q_1$, $Q_2$, and $Q_3$, and
if $E_2\to\bbP^1$ ramifies at $P$, $R_1$, $R_2$, and $R_3$, then the Weierstrass
points of $C$ are the points lying over $Q_1$, $Q_2$, $Q_3$, $R_1$, $R_2$, and $R_3$.
On the other hand, the point $P$ splits in the cover $C\to\bbP^1$,
and we let $P_1$ and $P_2$ be the points of $C$ lying over~$P$.

Thus, to specify a Howe curve, it is enough to provide three pieces of information:
\begin{enumerate}
\item A genus-$2$ curve $C$.
\item An unordered pair of disjoint sets $\{W_1, W_2\}$, each consisting of
      three Weierstrass points of $C$.
\item An unordered pair of distinct points $\{P_1, P_2\}$ on $C$ that are mapped to
      one another by the hyperelliptic involution.
\end{enumerate}

These three things determine the $V_4$-diagram above, and hence also determine
the double cover $\eta\colon H\to C$, which we call the \emph{structure map} for the given data.
Of course, if $\alpha$ is an automorphism of $C$ then $\{\alpha(W_1), \alpha(W_2)\}$ and
$\{\alpha(P_1), \alpha(P_2)\}$ will give us a double cover $H\to C$ that is isomorphic to $\eta$,
namely $\alpha \eta$.

\begin{lemma}
\label{L:data}
The data specifying a Howe curve is recoverable \textup(up to automorphisms of $C$\textup)
just from the structure map $\eta\colon H\to C$.
\end{lemma}

\begin{proof}
The map $C\to\bbP^1$ is unique (up to automorphism of $\bbP^1$), so we recover the entire map
$H\to C\to \bbP^1$ from $\eta$. This map is a Galois extension with group~$V_4$,
so we recover the genus-$1$ curves in the extension, and hence the division of
the Weierstrass points of $C$. The pair of points $\{P_1, P_2\}$ is simply the set
of ramification points of $\eta$.
\end{proof}

\begin{theorem}
Two structure maps $\eta_1\colon H \to C_1$ and $\eta_2\colon H\to C_2$ starting from the
same Howe curve $H$ are isomorphic to one another. That is, there is an isomorphism
$\gamma\colon C_1\to C_2$ and an automorphism $\delta\colon H\to H$ such that the
following diagram commutes\textup{:}
\[
\xymatrix@C=4ex{
H\ar[rr]^\delta\ar[d]_(0.4){\eta_1}&&H\ar[d]^(0.4){\eta_2}\\
C_1\ar[rr]^\gamma && C_2\rlap{.}
}
\]

\end{theorem}

\begin{proof}
Let $U_1$ and $U_2$ be the $V_4$-subgroups of $\Aut H$ specified by $\eta_1$ and $\eta_2$,
and let $S$ be the $2$-Sylow subgroup of $\Aut H$ that contains $U_1$.
By conjugating $U_2$ by an automorphism $\delta$ (and thereby replacing $\eta_2$
with $\eta_2 \delta$) we may assume that $U_2$ is also contained in~$S$.
Let $\alpha_1$ and $\alpha_2$ be the involutions of $H$ corresponding to the
double covers $\eta_1$ and $\eta_2$, and for each $i$ let $\beta_i$ and $\gamma_i$
be the other nonzero elements of $U_i$.

If $\alpha_1$ and $\alpha_2$ are conjugate to one another in $S$ (or even in $\Aut H$), we are done.
So assume, to get a contradiction, that $\alpha_1$ and $\alpha_2$ lie in different conjugacy classes of~$S$.

We know that the quotient of $H$ by the subgroup $\langle\alpha_i\rangle$ has genus~$2$,
while the quotients of $H$ by $\langle\beta_i\rangle$ and by $\langle\gamma_i\rangle$ have
genus~$1$. The same is true for all of the conjugates of $\alpha_i$, $\beta_i$, and
$\gamma_i$ in $S$. Moreover, if we have any $V_4$-subgroup of $S$, giving rise to
a diagram
\begin{equation}
\begin{split}
\label{EQ:V4}
\xymatrix@=3ex{
&&H\ar[lld]\ar[rrd]\ar[d]&&\\
Y_1\ar[rrd] & &Y_2\ar[d] & &Y_3\ar[lld]\\
&&X\rlap{,}&&
}
\end{split}
\end{equation}
then none of the curves $Y_i$ can have genus~$0$ (by Lemma~\ref{HoweIsCanonical}),
and so the only possibilities are that (a) all of the $Y_i$ have genus~$2$ and $X$ has genus~$1$,
or (b) one of the $Y_i$ has genus~$2$, the other two have genus~$1$, and $X$ has genus~$0$.
(This follows from the fact that in any diagram such as~\eqref{EQ:V4}, the genus of $H$
is the sum of the genera of the $Y_i$ minus twice the genus of $X$; this follows, for
instance, from~\cite[Theorem~B]{KR}.)
Thus, given two commuting involutions in $S$, if we know the genera of the quotients
of $H$ they produce, we can deduce the genus of the quotient of $H$ by their product.

Our strategy, then, will be to enumerate all possible $2$-groups $S$ that occur as the
$2$-Sylow subgroup of the automorphism group of a non-hyperelliptic curve~$H$ of genus~$4$,
along with all possible pairs $U_1$ and $U_2$ of $V_4$-subgroups of $S$ that contain
elements $\alpha_1$ and $\alpha_2$ that are not conjugate in~$S$. We will assume that
$\alpha_1$ and $\alpha_2$ generate genus-$2$ curves, while the other involutions in $U_1$ and $U_2$
generate genus-$1$ curves. Given these assumptions, we deduce, for as many involutions as we can, the
genera of the curves associated to these involutions.

Suppose $\delta$ is an involution in $S$ for which we know that the quotient ${Y\coloneqq H/\langle\delta\rangle}$
has genus~$2$. Let $T$ be the centralizer of $\delta$ in $S$.
Then the quotient $T/\langle\delta\rangle$ is contained in the automorphism group of the genus-$2$ curve $Y$.
Using Igusa's classification of the automorphism groups of genus-$2$ curves~\cite[\S8]{Igusa},
we can show that there are only eight $2$-groups that appear as subgroups of the automorphism
groups of genus-$2$ curves. If $T/\langle\delta\rangle$ is not one of these groups, then we have shown that the values
of $U_1$, $U_2$, $\alpha_1$, and $\alpha_2$ cannot correspond to two different realizations
of $H$ as a Howe curve.

In order to use this strategy, we need a good bound on the sizes of automorphism groups
of non-hyperelliptic curves of genus $4$ in characteristic not~$2$.
A result of Henn~\cite[Satz~1]{Henn} (see also~\cite{GK}) shows
that in characteristic $p>2$, the order of the automorphism group of a curve
of genus~$g$ is strictly less than $8g^3$, except possibly when the curve is of one of
the following types:
\begin{enumerate}
\item $x^n + y^m = 1$, where $n = 1 + p^a$ for some $a>0$ and $m\mid n$, or
\item $y^p - y = x^n$, where $n = 1 + p^a$ for some $a>0$.
\end{enumerate}
The first type of curve has genus $(n-2)(m-1)/2$, and if this is equal to $4$
then either we have $n=10$ and $m=2$ (and $p=3$) or we have $n=6$ and $m=3$ (and $p=5$).
In the first case the curve is hyperelliptic; in the second case, as Henn
notes, the automorphism group has order $360$, which is less than $8g^3$.
The second type of curve has genus $p^a (p-2)/2$, which is never equal to~$4$.
Thus, it will suffice for us to look at every $2$-group $S$ of order less than $8\cdot 4^3 = 512$.

We implemented this computation in Magma; the code is available on the third author's web site.
We ran our code on all $2$-groups of order less than $512$, and the only group not eliminated was
$S\cong(\bbZ/2\bbZ)^3.$

For this $S$, our computation shows that of the seven involutions in $S$, three
give genus-$2$ quotients and four give genus-$1$ quotients, and the three elements
that give genus-$2$ quotients sum to zero. Now consider the seven $V_4$-subgroups $T$ of $S$.
Each such $T$ gives us a diagram like~\eqref{EQ:V4} above.
For the $T$ that contains the three genus-$2$ involutions, the genus of $H/T$ is~$1$,
while for the other six $V_4$-subgroups $T$, the genus of $H/T$ is~$0$.

Let us consider the diagram of subextensions between $H$ and its quotient $H/S\cong \bbP^1$.
We label the elements of $S$ by vectors in $\bbF_2^3$, and we label the $V_4$-subgroups in the
same way, with the convention that a $V_4$-subgroup labeled by $v$ contains the elements with labels $g$
such that the dot product of $v$ and $g$ is~$0$. Then the diagram of subextensions,
with their genera, is as follows:
\[
\xymatrix@=3ex{
&&&\genfrac{}{}{0pt}{}{H}{\textup{genus 4}}\ar@{-}[llld]\ar@{-}[lld]\ar@{-}[ld]\ar@{-}[d]\ar@{-}[rd]\ar@{-}[rrd]\ar@{-}[rrrd]\\
\genfrac{}{}{0pt}{}{001}{\textup{genus 2}}&
\genfrac{}{}{0pt}{}{010}{\textup{genus 2}}&
\genfrac{}{}{0pt}{}{011}{\textup{genus 2}}&
\genfrac{}{}{0pt}{}{100}{\textup{genus 1}}&
\genfrac{}{}{0pt}{}{101}{\textup{genus 1}}&
\genfrac{}{}{0pt}{}{110}{\textup{genus 1}}&
\genfrac{}{}{0pt}{}{111}{\textup{genus 1}}\\
\genfrac{}{}{0pt}{}{001}{\textup{genus 0}}&
\genfrac{}{}{0pt}{}{010}{\textup{genus 0}}&
\genfrac{}{}{0pt}{}{011}{\textup{genus 0}}&
\genfrac{}{}{0pt}{}{100}{\textup{genus 1}}&
\genfrac{}{}{0pt}{}{101}{\textup{genus 0}}&
\genfrac{}{}{0pt}{}{110}{\textup{genus 0}}&
\genfrac{}{}{0pt}{}{111}{\textup{genus 0}}\\
&&&\genfrac{}{}{0pt}{}{\bbP^1}{\textup{genus 0}}\ar@{-}[lllu]\ar@{-}[llu]\ar@{-}[lu]\ar@{-}[u]\ar@{-}[ru]\ar@{-}[rru]\ar@{-}[rrru]&&&\\
}
\]
(For visual clarity, we have left off the heads of the arrows, and omitted the $21$ arrows between the middle layers.)
But this configuration of genera is not possible; consider for example the following subdiagram:
\[
\xymatrix@=3ex{
&\genfrac{}{}{0pt}{}{100}{\textup{genus 1}}\ar@{-}[ld]\ar@{-}[d]\ar@{-}[rd]&\\
\genfrac{}{}{0pt}{}{001}{\textup{genus 0}}&
\genfrac{}{}{0pt}{}{010}{\textup{genus 0}}&
\genfrac{}{}{0pt}{}{011}{\textup{genus 0}}\\
&\genfrac{}{}{0pt}{}{\bbP^1}{\textup{genus 0}}\ar@{-}[lu]\ar@{-}[u]\ar@{-}[ru]&\\
}
\]
This diagram violates the genus property we mentioned below diagram~\eqref{EQ:V4}.

This contradiction shows that the involutions $\alpha_1$ and $\alpha_2$ corresponding
to the structure maps $\eta_1$ and $\eta_2$ lie in the same conjugacy class of $\Aut H$, so
that $\eta_1 = \eta_2\delta$ for an automorphism $\delta$ of $H$.
\end{proof}

\begin{corollary}
\label{C:isomorphism}
Two triples $(C, \{W_1, W_2\}, \{P_1, P_2\})$
and $(C', \{W_1', W_2'\}, \{P_1', P_2'\})$ give isomorphic Howe curves
if and only if there is an isomorphism $C \to C'$ that takes
$\{W_1, W_2\}$ to $\{W_1', W_2'\}$ and $\{P_1, P_2\}$ to $\{P_1', P_2'\}$. \qed
\end{corollary}
This isomorphism test is very fast; it simply requires determining whether there are any automorphisms of $\bbP^1$ that
respect the sets of Weierstrass points and their divisions, and that take the $x$-coordinate of $P_1$ and $P_2$ to that of $P_1'$ and $P_2'$.

\if0
We note that the isomorphism test obtained from this corollary is
very fast: We simply check whether any of the six linear fractional transformations
that take the points of $W_1$ to those of $W_1'$ also take the points of $W_2$ to
those of $W_2'$, and, if so, whether they take the $x$-coordinate of $P_1$ and $P_2$
to that of $P_1'$ and $P_2'$; then we repeat, with the roles of $W_1'$ and $W_2'$
reversed. Even in the worst case, this can be carried out with well under $1000$ multiplications,
and is faster than computing the Igusa invariants of the the curves $C$ and $C'$.
\fi


text:
In this section, we study an extension of the system model in Section \ref{sec:model_2_controllers} to the case where instead of 1 local controller, we have $N$ local controllers, $C^1,C^2, \ldots,C^N$, each associated to a co-located plant as shown in Fig. \ref{fig:SystemModel}. We use $\mathcal{N}$ to denote the set $\{1,2, \ldots, N\}$ and $\overline{\mathcal{N}}$ to denote $\{0,1,\ldots, N\}$.
The linear dynamics of plant $n \in \mathcal{N}$ are given by
\begin{align}
&X_{t+1}^n \!=\! A^{nn} X_t^n + B^{nn}U^{n}_t+ B^{n0} U^0_t + W_t^n, t=0,\dots,T,
 \label{Model:system}
\end{align}
where $X_t^n\in \R^{d_X^n}$ is the state of the plant $n$ at time $t$,
$U^n_t \in \R^{d_U^n}$ is the control action of the controller $C^{n}$,  $U^0_t \in \R^{d_U^0}$ is the control action of the controller $C^{0}$, and $A^{nn}, B^{nn}, B^{n0}$ are matrices with appropriate dimensions.
We assume that $X^n_0 = 0$, and that  $W^n_t$, $n \in \mathcal{N}, t \geq 0$, are i.i.d  random variables with zero mean and $\cov(W_t^n) = \mathbf{I}$. Note that we do not assume that random variables $W_{t}^{n}$, $n \in \mathcal{N}, t \geq 0$, are Gaussian.

The overall system  dynamics can be written as
\begin{align}
X_{t+1} = A X_t + BU_t + W_t,
\label{overall_state_dynamic}
\end{align}
where $X_t = \vecc(X^{1:N}_t), U_t = \vecc(U^{0:N}_t),W_t = \vecc(W^{0:N}_t)$ and $A,B$ are defined as

\begin{align}
A &= \begin{bmatrix}
   A^{11} & & \text{\huge0}\\
          & \ddots & \\
     \text{\huge0} & & A^{NN}
\end{bmatrix},
B=
\begin{bmatrix}
B^{10} &  B^{11} & & \text{\huge0}\\
\vdots     &     & \ddots & \\
B^{N0}  &   \text{\huge0} & & B^{NN}
\end{bmatrix}.
\label{eq:thm_matricesABB}
\end{align}

\begin{singlespace}

\input{system_model}
\end{singlespace}

\subsubsection*{Communication Model}
\label{subs:comm_model}

The communication model is similar to the one described in  Section \ref{subs:comm_model_2C}. In particular, for each $n \in \mathcal{N}$, there is an unreliable link with link failure probability $p^n$ from the local controller $C^{n}$ to the remote controller $C^0$. The local controller $C^{n}$ uses its unreliable  link to send the  state  $X_t^n$ of its co-located plant to the remote controller. The state of this link at time $t$ is described by a Bernoulli random variable $\Gamma_t^n$ and the output of this link at time $t$ is denoted by $Z_t^n$, where $\Gamma_t^n$ and $Z_t^n$ are described by equations  similar to \eqref{eq:gamma} and \eqref{Model:channel_2C}. We assume that $\Gamma_{0:t}^{1:N}$, $t \geq 0$, are independent random variables and that they are independent of $W_{0:t}^{1:N}$, $t \geq 0$.

Unlike the unreliable uplinks, we assume that there exist perfect links from $C^0$ to $C^{n}$, for  each $n \in \mathcal{N}$. Therefore, $C^0$ can share $Z_t^{1:N}$ and $U_{t-1}^0$ with all local controllers $C^{1:N}$.
All controllers select their control actions at time $t$ after observing $Z_t^{1:N}$  and  $U_{t-1}^0$.
We assume that for each $n \in \mathcal{N}$, the links from controllers $C^{n}$ and $C^0$ to  plant $n$ are perfect.

\subsubsection*{Information structure and cost}
\label{subs:info_cost_2C}
Let $H^{n}_t$ denote the information available to controller $C^{n}$, $n \in {\color{black} \overline{\mathcal{N}}}$,  at time $t$.
Then,
\begin{align}
H^{n}_t&= \{X^n_{0:t}, U^{n}_{0:t-1}, Z_{0:t}^{1:N}, U^0_{0:t-1}\}, \hspace{2mm} n \in \mathcal{N}, \notag \\
H^0_t &= \{Z_{0:t}^{1:N}, U^0_{0:t-1}\}.
\label{Model:info}
\end{align}

Let $\mathcal{H}^{n}_t$ be the space of all possible realizations of $H_t^n$.
Then, $C^{n}$'s actions are selected according to
\begin{align}
U^{n}_t &= g^{n}_t(H^{n}_t), \hspace{2mm}  n \in {\color{black} \overline{\mathcal{N}}},
\label{Model:strategy}
\end{align}
where $g^{n}_t:\mathcal{H}^{n}_t{\color{black} \to } \R^{d_U^n}$ is a Borel measurable mapping.
We use $g:=(g^0_{0},g^0_1,\dots,g^1_{0},g^1_1,\dots, g^N_{0},g^N_1,\dots,)$ to collectively denote the control strategies of all $N+1$ controllers.

The instantaneous cost $c_t(X_t, U_t)$ of the system is a quadratic function similar to the one described in \eqref{Model:cost_2C} where $X_t = \vecc(X^{1:N}_t), U_t = \vecc(U^{0:N}_t)$ and

\begin{align}
Q= \begin{bmatrix}
Q^{11} &\ldots &Q^{1N} \\
\vdots & \ddots & \vdots \\
Q^{N1} & \ldots & Q^{NN}
\end{bmatrix}, R= \begin{bmatrix}
R^{00} &R^{01} &\ldots &R^{0N} \\
R^{10} &R^{11} &\ldots &R^{1N} \\
\vdots &\vdots & \ddots & \vdots \\
R^{N0}  & \ldots & \ldots & R^{NN}
\end{bmatrix}.
\label{matrix_structure}
\end{align}
$Q$ is a symmetric positive semi-definite (PSD) matrix and $R$ is a symmetric positive definite (PD) matrix.

\subsubsection*{Problem Formulation}
\label{subs:prob_formulaiton}

Let $\mathcal{G}$ denote the set of all possible control strategies of controllers $C^{0},\ldots,C^N$.
The performance of control strategies $g$ over a finite horizon $T$ is measured by $J_T(g)$ defined in \eqref{Model:J_T_2C}.

For the decentralized networked control system (DNCS) described above, we consider the problem of strategy optimization  over finite and infinite time horizons. These two problems are formally defined below.
\begin{problem}
\label{problem_finite}
For the DNCS described above,
solve the following strategy optimization problem:
\begin{align}
\inf_{g \in\mathcal{G}}
J_T(g)
\label{Model:obj_finite}
\end{align}
\end{problem}

\begin{problem}
\label{problem_infinite}
For the DNCS described above,
solve the following strategy optimization problem:
\begin{align}
&\inf_{g \in\mathcal{G}}
J_{\infty}(g)
:= \inf_{g \in\mathcal{G}}
\limsup_{T\rightarrow\infty} \frac{1}{T+1} J_{T}(g).
\label{Model:obj_infinite}
\end{align}
\end{problem}

Due to stability issues in the infinite horizon problem, we make the following  assumptions on the system and cost matrices.
\begin{assumption}
\label{assum:det_stb}
$(A,Q^{1/2})$ is detectable and $(A,B)$ is stabilizable.
\end{assumption}

\begin{assumption}
\label{assum:det_stb_2}
$\big(A^{nn},(Q^{nn})^{1/2} \big)$ is detectable for all $n \in \mathcal{N}$.
\end{assumption}

The finite horizon strategy optimization problem (Problem \ref{problem_finite}) has been solved in \cite{asghari_ouyang_nayyar_tac_2018}.  We summarize the finite horizon results below.
\begin{lemma}(\cite[Theorem 2]{asghari_ouyang_nayyar_tac_2018})
\label{lm:opt_strategies}
The optimal control strategies of Problem \ref{problem_finite} are given by

\begin{align}
\bmat{U^{0*}_t \\ U^{1*}_t \\ \vdots \\ U^{N*}_t} = K_t^0 \hat X_t +
\begin{bmatrix}
\mathbf{0} & \ldots & \mathbf{0} \\
&&\\
K_t^{1} &     &  \text{\huge0}  \\
 &\ddots     &   \\
  \text{\huge0} &  & K_t^{N}
\end{bmatrix} \left(X_t - \hat X_t\right),
\label{eq:opt_U}
\end{align}
where $\hat X_t = \vecc(\hat X^{1:N}_t) = \vecc(\ee[X^1_t|H^0_t], \ldots, \ee[X^N_t|H^0_t]) = \ee[X_t|H^0_t]$ is the  estimate (conditional expectation) of $X_t$ based on the common information $H^0_t$.  The estimate $\hat X_t$ can be computed recursively as follows: for $n \in \mathcal{N}$,
\begin{align}
\hat X_0^n = & 0,
\label{eq:estimator_0}
\\
\hat X_{t+1}^n
= &\left\{
\begin{array}{ll}
 \big([A]_{n,:} + [B]_{n,:} K^0_t\big)\hat X_t& \text{ if }Z_{t+1}^n= \emptyset,\\
 X_{t+1}^n & \text{ if }Z_{t+1}^n = X_{t+1}^n.
\end{array}\right.
\label{eq:estimator_t}
\end{align}
The gain matrices $K_t^0$ and $K_t^{n}$, $n \in \mathcal{N}$, are given by
\begin{align}
& K^0_t = \Psi(P_{t+1},R,A,B),
\label{eq:K_finite}
\\
&K_t^n = \Psi \big ((1-p^n)[P_{t+1}^0]_{n,n}+p^n P_{t+1}^{n},R^{nn},A^{nn},B^{nn} \big),
\label{eq:tildeK_finite}
\end{align}
where $P^0_t = \begin{bmatrix}
[P_{t}^0]_{1,1} &\ldots &[P_{t}^0]_{1,N} \\
\vdots & \ddots & \vdots \\
[P_{t}^0]_{N,1} & \ldots & [P_{t}^0]_{N,N}
\end{bmatrix} \in \mathbb{R}^{(\sum_{n=1}^N d^n_X)\times (\sum_{n=1}^N d^n_X)}$ and $P_t^{n} \in \mathbb{R}^{ d^n_X\times d^n_X} $, for $n \in \mathcal{N}$, are PSD matrices obtained recursively as follows:
where $P_t^0$ and $P_t^{n}$,  $n \in \mathcal{N}$, are PSD matrices obtained recursively as follows:
\begin{align}
&P_{T+1}^0 = \mathbf{0}, \quad P_{T+1}^{n} = \mathbf{0}, \label{eq:P_N_init}\\
&P_t^0 =  \Omega(P_{t+1}^0,Q,R,A,B),
\label{eq:P_finite}
\\
&P_t^{n} =  \Omega \big((1-p^n)[P_{t+1}^0]_{n,n}+p^n P_{t+1}^{n},Q^{nn},R^{nn},A^{nn},B^{nn}\big).
\label{eq:tildeP_finite}
\end{align}
Furthermore, the optimal cost is given by
\begin{align}
J^*_{T} = & \sum_{t = 0}^T \sum_{n = 1}^N \tr \Big((1-p^n)[P_{t+1}^0]_{n,n}+p^n P_{t+1}^{n} \Big).
\label{eq:opcost_finite}
\end{align}
\end{lemma}

\begin{remark}
Note that remote controller's action $U^{0*}_t$ in \eqref{eq:opt_U} is a function  of $\hat{X}_t$ only while the local controller $C^n$'s action $U^{n*}_t$ is a function of both $\hat{X}_t$ and $X_t^n$. Further, as per \eqref{eq:estimator_0} and \eqref{eq:estimator_t}, $\hat{X}_t$ is computed recursively based only on the knowledge of $Z_{0:t}^{1:N}$.
\end{remark}

\subsection{Infinite Horizon Optimal Control}
\label{sec:infinite_N_controllers}
As in Section \ref{sec:infinite_2_controllers},  the infinite horizon  problem (Problem \ref{problem_infinite}) can be solved by answering the following three questions:

\begin{enumerate}[label=(\mylabel{Q}{\arabic*})]
\item Do matrices $P_t^{0},\ldots,P_t^N$, defined in \eqref{eq:P_N_init}-\eqref{eq:tildeP_finite} converge as $t \to -\infty$ to $P_*^0,\ldots, P_*^N$, that satisfy the coupled fixed point  equations \eqref{eq:P_finite_NC_fixed} - \eqref{eq:tildeP_finite_NC_fixed} below?
\begin{align}
&P_*^0 =  \Omega \big(P_{*}^0,Q,R,A,B\big),
\label{eq:P_finite_NC_fixed}
\\
& P_*^{n} =  \Omega \big((1-p^n)[P_{*}^0]_{n,n}+p^n P_{*}^{n},Q^{nn},R^{nn},A^{nn},B^{nn}\big).
\label{eq:tildeP_finite_NC_fixed}
\end{align}
\item If matrices $P_t^{0},\ldots,P_t^N$ converge and we define matrices $K_*^0, \ldots, K_*^N$, using matrices $P_*^{0},\ldots,P_*^N$ as follows,
\begin{align}
& K_*^0 = \Psi(P_{*}^0,R,A,B),
\label{eq:K_finite_NC_fixed}
\\
& K_*^n = \Psi \big((1-p^n)[P_{*}^0]_{n,n}+p^n P_{*}^n,R^{nn},A^{nn},B^{nn}\big),
\label{eq:tildeK_finite_NC_fixed}
\end{align}
are the following strategies optimal for Problem \ref{problem_infinite}?
\begin{align}
\bmat{U^{0*}_t \\ U^{1*}_t \\ \vdots \\ U^{N*}_t} = K_*^0 \hat X_t +
\begin{bmatrix}
\mathbf{0} & \ldots & \mathbf{0} \\
&&\\
K_*^{1} &     &  \text{\huge0}  \\
 &\ddots     &   \\
  \text{\huge0} &  & K_*^{N}
\end{bmatrix} \left(X_t - \hat X_t\right),
\label{eq:opt_U_NC_fixed}
\end{align}
where $\hat{X}_t$ can be computed recursively using \eqref{eq:estimator_0} - \eqref{eq:estimator_t}  by replacing $K^0_t$ with $K^0_*$.

\item If matrices $P_t^{0},\ldots,P_t^N$ do not converge, is it still possible to find control strategies with finite cost for Problem \ref{problem_infinite}?

\end{enumerate}

\subsection{Answering Q1}
As in Section \ref{sec:Q1_2C}, we will answer Q1 by establishing a connection between the recursions for matrices $P_t^{0},\ldots,P^N_t$ in our DNCS problem and the recursions for matrices $P_t^{\diamond}(m)$, $m \in \mathcal{M},$ in the MJLS problem.  One obstacle in making this connection is the fact that the matrices $P_t^{0},\ldots,P^N_t$ in our DNCS problem do not have the same dimensions while the matrices $P_t^{\diamond}(m)$, $m \in \mathcal{M},$ in the MJLS problem  all have the same dimensions. This obstacle was not present in Section \ref{sec:infinite_2_controllers}. To get around this difficulty, we first provide a new representation $\bar P_t^{0},\ldots,\bar P_t^{N}$ for the matrices $P_t^{0},\ldots,P^N_t$ in our DNCS problem such that the new  matrices $\bar P_t^{0},\ldots,\bar P_t^{N}$ all have the same dimensions.

\begin{lemma}
\label{lm:opt_strategies_new_rep}
Define matrices $\bar P_t^{n} \in \mathbb{R}^{(\sum_{k=1}^N d^k_X)\times (\sum_{k=1}^N d^k_X)}$, $n =0,1,\ldots,N$, recursively as follows:
\begin{align}
&\bar P_{T+1}^n = \mathbf{0}, \label{eq:barP_init}\\
&\bar P_t^0 = \Omega(\bar P_{t+1}^0,Q,R,A,B),
\label{eq:barP_finite_0}
\\
&\bar P_t^{n} = \Omega \big((1-p^n)\bar P_{t+1}^0+p^n \bar P_{t+1}^n,
 \mathcal{L}_{zero}(Q,Q^{nn},n,n),  \notag \\
& \hspace{1.4cm}  \mathcal{L}_{iden}(R,R^{nn},n+1), \mathcal{L}_{zero}(A,A^{nn},n,n), \notag \\
 & \hspace{1.4cm} \mathcal{L}_{zero}(B,B^{nn},n,n+1) \big),
\label{eq:barP_finite}
\end{align}
where the operators $\mathcal{L}_{zero}$ and $\mathcal{L}_{iden}$ are as defined in Section \ref{sec:operators}.
Then, for $ t \leq T+1$,
\begin{align}
\label{new_rep1}
&\bar P_t^0 =  P_t^0, \\
& \bar P_t^{n} = \mathcal{L}_{zero}(P^0_t, P_t^{n},n,n), ~~n=1,\ldots,N.
\label{new_rep2}
\end{align}
Consequently, matrices $P_t^{0},\ldots,P^N_t$ converge as $t \to -\infty$ if and only if matrices $\bar P_t^{0},\ldots,\bar P_t^{N}$ converge as $t \to -\infty$.
\end{lemma}
\begin{proof}
See Appendix \ref{proof_lm:opt_strategies_new_rep} for a proof.
\end{proof}

We can now proceed with constructing an auxiliary MJLS.
\subsubsection*{\textbf{Step 1: Constructing an auxiliary MJLS}}
Consider an auxiliary MJLS where the set $\mathcal{M}$ of modes is $ \{0,1,\ldots,N\}$.  Then, we have the following $N+1$ sequences of  matrices, $P^{\diamond}_t(0),P^{\diamond}_t(1),\ldots,P^{\diamond}_t(N)$, defined recursively using \eqref{eq:CARE_init} and \eqref{eq:CARE_finite} for this MJLS:

\begin{align}
&P^{\diamond}_{T+1}(m)  =\mathbf{0}, ~\forall m \in \mathcal{M}, \label{eq:P_N_MJ_init}\\
&P^{\diamond}_t(0) =  \notag \\
&\Omega\big(\sum_{k=0}^N \theta^{0k} P^{\diamond}_{t+1}(k),Q^{\diamond}(0),R^{\diamond}(0),A^{\diamond}(0),B^{\diamond}(0) \big),
\label{P_MJ_cmp_NC_0}
\\
&P^{\diamond}_t(n)=   \notag \\
& \Omega\big(
\sum_{k=0}^N \theta^{nk} P^{\diamond}_{t+1}(k),Q^{\diamond}(n),R^{\diamond}(n),A^{\diamond}(n),B^{\diamond}(n) \big).
\label{P_MJ_cmp_NC_1}
\end{align}
Furthermore, we have the recursions of \eqref{eq:barP_init}-\eqref{eq:barP_finite} for matrices $\bar P_t^{0},\ldots, \bar P_t^N$ in our DNCS problem.

By comparing \eqref{P_MJ_cmp_NC_0}-\eqref{P_MJ_cmp_NC_1} with \eqref{eq:barP_finite_0}-\eqref{eq:barP_finite}, we find that the following definitions would make the two sets of equations identical:
\begin{align}
&A^{\diamond}(0) = A, A^{\diamond}(n) = \mathcal{L}_{zero}(A,A^{nn},n,n), \hspace{1mm} n \in \mathcal{N},
\label{A_mj}
\end{align}
\begin{align}
&B^{\diamond}(0) = B, B^{\diamond}(n) =  \mathcal{L}_{zero}(B,B^{nn},n,n+1), n \in \mathcal{N},
\label{B_mj}
\end{align}
\begin{align}
&Q^{\diamond}(0) = Q, Q^{\diamond}(n) =   \mathcal{L}_{zero}(Q,Q^{nn},n,n), \hspace{1mm} n \in \mathcal{N},
\label{Q_mj}
\\
&R^{\diamond}(0) = R, R^{\diamond}(n) =  \mathcal{L}_{iden}(R,R^{nn},n+1), \hspace{1mm} n \in \mathcal{N},
\label{R_mj}
\\
&\Theta =
\begin{blockarray}{rccccc}
&0 &1 &2  &\ldots &N  \\
\begin{block}{r[ccccc]}
                     0 & 1 & 0 & \ldots &\ldots & 0 \\
                     1 & 1-p^1 & p^1 &\ddots  & & \vdots \\
                     2 & 1-p^2 &0  &p^2  & \ddots& \vdots \\
                     \vdots & \vdots & \vdots  &\ddots  &\ddots & 0 \\
                     N & 1-p^N & 0  &\ldots  &0 &p^N \\
\end{block}
\end{blockarray}.
\label{transition_prob}
\end{align}
To complete the definition of the auxiliary MJLS, we need to define the initial state and mode probability distributions $\pi_{X_0^{\diamond}}$ and $\pi_{M_0}$. These can be defined arbitrarily and for simplicity we assume that the initial state is fixed to be $X^{\diamond}_0 = 0$ and the initial mode $M_0$ is uniformly distributed over the set $\mathcal{M}$.  The following lemma summarizes the above discussion.

\begin{lemma}
\label{equality_recursions_NC}

For the auxiliary MJLS described by \eqref{A_mj}-\eqref{transition_prob},
 the coupled recursions in \eqref{eq:P_N_MJ_init}-\eqref{P_MJ_cmp_NC_1} are identical to  the coupled recursions in \eqref{eq:barP_init}-\eqref{eq:barP_finite}.

\end{lemma}
\begin{proof}
The lemma can be proved by  straightforward algebraic manipulations.
\end{proof}

\subsubsection*{\textbf{Step 2: Using MJLS results to answer Q1}}
Now that we have constructed an auxiliary MJLS where $P^{\diamond}_t(m) = \bar P_t^{m}$ for $m=0,\ldots,N$, we can use the MJLS results about convergence of  matrices $P^{\diamond}_t(m)$ (that is, Lemmas \ref{lm:MJ_infinite} and \ref{lm:ss})  to answer Q1. The following lemma states this result.

\begin{lemma}
\label{lm:pc_NC}
Suppose Assumptions \ref{assum:det_stb} and \ref{assum:det_stb_2} hold. Then, the matrices $ P^0_t,\ldots,  P^N_t$ defined in \eqref{eq:P_N_init}-\eqref{eq:tildeP_finite} converge as $t \to -\infty$ to matrices  $P_*^0,\ldots, P_*^N$ that satisfy the coupled fixed point  equations \eqref{eq:P_finite_NC_fixed}-\eqref{eq:tildeP_finite_NC_fixed} if and only if  $p^n < p_c^n$ for all $n \in \mathcal{N}$, where the critical thresholds $p_c^n, n \in \mathcal{N}$, are given by
\begin{align}
\frac{1}{\sqrt{p_c^n}} =  \min_{K  \in \mathbb{R}^{d^n_U \times d^n_X}}\rho(A^{nn}+B^{nn}K).
\label{eq:pc}
\end{align}

\end{lemma}
\begin{proof}
See Appendix \ref{proof_lm:pc}.
\end{proof}

\subsection{Answering Q2 and Q3}
Assuming that $P_{t}^n \rightarrow P_*^n$ as $t \rightarrow -\infty$ for $n=0,\ldots,N,$ we want to know whether the control strategies of \eqref{eq:K_finite_NC_fixed}-\eqref{eq:opt_U_NC_fixed} are optimal for Problem \ref{problem_infinite_2C}. The following result shows that these control strategies are indeed optimal.

\begin{lemma}
\label{lm:Q2_NC}
If $P_{t}^n \rightarrow P_*^n$ as $t \rightarrow -\infty$ for $n=0,\ldots,N$, then
\begin{enumerate}
\item Problem \ref{problem_infinite} has finite optimal cost,
\item The strategies described by \eqref{eq:K_finite_NC_fixed}-\eqref{eq:opt_U_NC_fixed} are optimal for Problem \ref{problem_infinite},
\item Under the strategies described by \eqref{eq:K_finite_NC_fixed}-\eqref{eq:opt_U_NC_fixed}, $X_t$ and $(X_t-\hat X_t)$ are mean square stable.
\end{enumerate}
\end{lemma}
\begin{proof}
See Appendix \ref{sec:proof_lm_Q2_NC}.
\end{proof}

The following lemma answers Q3.
\begin{lemma}\label{lm:Q3_NC}
If matrices $P_t^{0},\ldots,P_t^{N}$ do not converge as $t \rightarrow -\infty$, then Problem \ref{problem_infinite} does not have finite optimal cost.
\end{lemma}
\begin{proof}
See Appendix \ref{sec:proof_lm_Q3_NC}.

\end{proof}

Now that we have answered  Q1, Q2 and Q3, we can summarize our results for the infinite horizon DNCS problem (Problem \ref{problem_infinite}).

\subsection{Summary of the Infinite Horizon Results}
Based on the answers to Q1-Q3, the following theorem summarizes our results for Problem   \ref{problem_infinite}.
\begin{theorem}
\label{thm:DC_infinite_NC}
Suppose Assumptions \ref{assum:det_stb} and \ref{assum:det_stb_2} hold. Then,
\begin{enumerate}[(i)]
\item Problem \ref{problem_infinite} has finite optimal cost if and only if
for all $n \in \mathcal{N}$, $p^n < p_c^n$ where the critical threshold $p_c^n$ is given by \eqref{eq:pc}.

\item If $p^n < p_c^n$ for all $n \in \mathcal{N}$, there exist symmetric positive semi-definite matrices $P_*^0,\ldots, P_*^N$ that satisfy \eqref{eq:P_finite_NC_fixed}-\eqref{eq:tildeP_finite_NC_fixed} and the optimal strategies for Problem \ref{problem_infinite} are given by
\begin{align}
\bmat{U^{0*}_t \\ U^{1*}_t \\ \vdots \\ U^{N*}_t} = K_*^0 \hat X_t +
\begin{bmatrix}
\mathbf{0} & \ldots & \mathbf{0} \\
&&\\
K_*^{1} &     &  \text{\huge0}  \\
 &\ddots     &   \\
  \text{\huge0} &  & K_*^{N}
\end{bmatrix} \left(X_t - \hat X_t\right),
\label{eq:opt_U_NC_thm}
\end{align}
where the  estimate $\hat X_t$ can be computed recursively using \eqref{eq:estimator_0}-\eqref{eq:estimator_t} by replacing $K^0_t$ with $K^0_*$ and the gain matrices $K_*^0, \ldots,K_*^N$ are given by
\begin{align}
& K_*^0 = \Psi(P_{*}^0,R,A,B),
\label{eq:K_finite_NC_final}
\\
& K_*^n = \Psi \big((1-p^n)[P_{*}^0]_{n,n}+p^n P_{*}^n,R^{nn},A^{nn},B^{nn}\big).
\label{eq:tildeK_finite_NC_final}
\end{align}
\item If $p^n < p_c^n$ for all $n \in \mathcal{N}$, then under the strategies described in part (ii) above, $X_t$ and $(X_t-\hat X_t)$ are mean square stable.
\end{enumerate}
\end{theorem}

\begin{corollary}
Suppose the local controllers are just  sensors (i.e., $B^{nn} = 0$ for $n=1,\ldots,N$) and the remote controller is the only controller present. Then, if $\rho(A) < 1/\sqrt{p^n}$ for all $n=1,\ldots,N$, the optimal controller of this multi-sensor,  single-controller NCS is given by $U^{0*}_t$ in \eqref{eq:opt_U_NC_thm}, and the corresponding closed-loop system is mean-square stable.
\end{corollary}

\begin{remark}
If Assumption \ref{assum:det_stb_2} is not true, define, for $n=1,\ldots,N$, $p^n_c = \min \{p^n_s, p^n_d\},$  where
\begin{align}
\frac{1}{\sqrt{p_s^n}} &=  \min_{K  \in \mathbb{R}^{d^n_U \times d^n_X}}\rho(A^{nn}+B^{nn}K).
\label{eq:pc_ss}
\\
\frac{1}{\sqrt{p_d^n}} &=  \min_{H \in \mathbb{R}^{d^n_X \times d^n_X}}\rho \big(A^{nn}+H (Q^{nn})^{1/2} \big).
\label{eq:pc_sd}
\end{align}
Then, using arguments similar to those used for proving Theorem \ref{thm:DC_infinite_NC}, we can show that if $p^n < p_c^n$ for all $n \in \mathcal{N}$, the strategies in \eqref{eq:opt_U_NC_thm} are optimal for Problem \ref{problem_infinite}. Moreover, Problem \ref{problem_infinite} has finite optimal cost and the system state is mean square stable under optimal strategies.
\end{remark}


TITLE: Proof of (Strong) Markov Property using sigma-algebras
QUESTION [3 upvotes]: I would like to ask if any of you know of a good resource containing rigorous proof (using sigma-algebras) of Markov Property and Strong Markov Property respectively in terms of Discrete Time Markov Chains.
(All resources I have found dealt with Continuous Time Markov Chains or with discrete MP/SMP in terms of Brownian Motion.)
Thanks,
C.

REPLY [0 votes]: See Theorem 6.3.4. of Probability: Theory and Applications (4th edition) by Richard Durrett.
Also, a very accessible proof is found in Theorem 1.4.2 of Markov Chains by J.R. Norris.


TITLE: Linear Equation in 4 variables- No of solutions
QUESTION [0 upvotes]: If 3a+6b+9c+4d = 100  and a ,b,c and d are natural numbers , then how many values d can take?
How to approach this type of problem?

REPLY [2 votes]: As $3,6,9$ are multipliers of $3$, then $$3|(3a+6b+9c).$$
It is known that $$100\equiv 1 (\bmod 3).$$
So, $d$ must be $$d\equiv 1 (\bmod 3)$$ to fit modular equation $$3a+6b+9c+4d\equiv 1 (\bmod 3).$$
As $a,b,c,d\in\mathbb N$, then $4d\le100-3-6-9=82$, so

$d\equiv 1 (\bmod 3)$
$1\le d \le 20$.

Then there are only these values for $d$:
$$1,4,7,10,13,16,19.$$
Examples:
$d=1$: $(a,b,c,d)=(27,1,1,1)$,
$d=4$: $(a,b,c,d)=(23,1,1,4)$,
$d=7$: $(a,b,c,d)=(19,1,1,7)$,
$d=10$: $(a,b,c,d)=(15,1,1,10)$,
$d=13$: $(a,b,c,d)=(11,1,1,13)$,
$d=16$: $(a,b,c,d)=(7,1,1,16)$,
$d=19$: $(a,b,c,d)=(3,1,1,19)$.

REPLY [1 votes]: I assume that $a,b,c,d\ge 1$.
Since
$$3a+6b+9c+4d=100\tag1$$$$\iff d=3(33-a-2b-3c-d)+1$$
$d$ can be written as $d=3e-2$ where $e$ is a positive integer.
So,
$$(1)\iff 3a+6b+9c+12e-8=100\iff a+2b+3c+4e=36\tag2$$
Here, since $a\ge 1,2b\ge 2,3c\ge 3$,
$$4e=36-a-2b-3c\le 36-1-2-3=30\Rightarrow 1\le e\le 7.$$
For each $e$ such that $1\le e\le 7$, $(a,b,c)=(1,16-2e,1)$ satisfies $(2)$.
Hence, $e$ can take $7$ values. So $d$ can take $7$ values : $d=1,4,7,10,13,16,19$.

Assuming that $a,b,c,d\ge 0$, $d$ can take $9$ values since $(a,b,c,d)=(0,2,0,22),(0,0,0,25)$ satisfy $(1)$.


TITLE: How do these big-oh equalities hold?
QUESTION [1 upvotes]: Chong and Zak, in their book An Introduction to Optimization, 4th ed, write

... we write ... $f(x) = O(g(x))$ to mean that the quotient $\frac{||f(x)||}{|g(x)|} $ is bounded near $0$; that is, there exist numbers $K>0$ and $\delta > 0$ such that if $||x||<\delta$, $x \in \Omega$, then $\frac{||f(x)||}{|g(x)|} \leq K$.

then give following example:

$\begin{bmatrix} x^3 \\ 2x^2+3x^4 \end{bmatrix} = O(x^2)$

Neither the definition nor the example make sense to me.
The definition does not make sense to me, because the quotient should be bounded in the infinity I think, not near $0$?
Can you explain?

REPLY [1 votes]: If $f(x)=\begin{bmatrix} x^3 \\ 2x^2+3x^4 \end{bmatrix} $, then an easy computation gives
$\frac{||f(x)||}{|g(x)|}=\sqrt{4+13x^2+9x^4}$, hence
$\frac{||f(x)||}{|g(x)|} \le \sqrt{4+13+9} \le 6$ for $|x| <1$
This shows that $\begin{bmatrix} x^3 \\ 2x^2+3x^4 \end{bmatrix} = O(x^2)$


TITLE: The volume for truncated pyramid with irregular base
QUESTION [1 upvotes]: According to here, for truncated pyramid with rectangular base:

The volume is given by:
$volume= h/3*(a*b+c*d+(a*d+b*c)/2)$
What if the base is an irregular surface area, defined by $n$ sets of coordinates?
How to extend the above volume formula to cater for irregular surface area?

REPLY [0 votes]: Prove this holds for any pyramid with a triangular base then triangulate your polygon and sum the volumes.
More generally, if you have a closed curve in a plane and a vertex not in that plane the volume of the solid obtained by joining the vertex to the curve is going to be the integral of the sections of area parallel to the curve from it to the vertex: $\int_0^h a(t)dt$ where $a(t)$ is the area of the intersection of the solid and the plane parallel to the one which contains the curve and whose distance to the vertex is t and h is the distance from the vertex to the plane which contains the curve. By a similarity argument we obtain that $a(t)$ is proportional to $t^2$, so let $a(t)=kt^2$ then $a(h)=kh^2\implies k=\frac{a(h)}{h^2}$ so $\int_0^h a(t)dt=\int_0^h kt^2dt=\frac{kh^3}{3}=\frac{a(h)h}{3}$, the familiar formula.
The formula in the question can then be obtained by subtracting a small pyramid from a big pyramid.