text:
\begin{document}

\title{Nilpotent groups of class two which underly a unique regular dessin}
\author[1,2]{Kan Hu\thanks{hukan@zjou.edu.cn}}
\author[3,4]{Roman Nedela\thanks{nedela@savbb.sk}}
\author[1,2]{Na-Er Wang\thanks{wangnaer@zjou.edu.cn}}
\affil[1]{School of Mathematics, Physics and Information Science, Zhejiang Ocean University, Zhoushan, Zhejiang 316022, People's Republic of China}
\affil[2]{Key Laboratory of Oceanographic Big Data Mining \& Application of Zhejiang Province, Zhoushan, Zhejiang 316022, People's Republic of China}
\affil[3]{Faculty of Natural Sciences, Matej Bel University, Tajovsk\'eho 40, 974 01, Bansk\'a Bystrica, Slovak Republic}
\affil[4]{Institute of Mathematics and Computer Science, Slovak Academy of Sciences, Bansk\'a Bystrica, Slovak Republic}

\maketitle
\begin{abstract}
A dessin is an embedding of connected bipartite graph into an oriented closed surface. A dessin is regular if its group of colour- and orientation-preserving automorphisms acts transitively on the edges. In the present paper regular dessins with a nilpotent automorphism group are investigated, and attention are paid on those with the highest level of external symmetry. Depending on the algebraic theory of dessins and using group-theoretical methods, we present a classification of nilpotent groups of class two which underly a unique regular dessin.\\[2mm]
\noindent{\bf Keywords} regular dessin, nilpotent group, dessin operation, external symmetry\\
\noindent{\bf MSC(2010)} Primary 14H57; Secondary 14H37, 20B25, 30F10.
\end{abstract}

\section{Introduction}
  A  \textit{Bely\v{\i} function} is a non-constant meromorphic function $\beta\colon S\to\Sigma$ defined over a Riemann surface $S$ with at most three critical values  $0,$ $1$ and $\infty$ on the Riemann sphere $\Sigma$. Every Bely\v{\i} function $\beta$ determines a $2$-cell embedding of a $2$-colored bipartite graph on $S$ called a \textit{dessin}: The embedded bipartite graph is the preimage of the closed interval $[0,1]$ where the black vertices are $\beta^{-1}(0)$ and the white vertices are $\beta^{-1}(1)$; the faces are the components of $S\setminus\beta^{-1}[0,1]$. By Bely\v{\i}'s theorem~\cite{Belyi1979} a compact Riemann surface admits a Bely\v{\i} function (and hence a dessin) if and only if the surface, regarded as an algebraic curve,  is defined over the field $\mathbb{\bar Q}$ of algebraic numbers. The absolute Galois group $\Gal (\mathbb{\bar Q}/\mathbb{Q})$ acts on the coefficients of polynomials and rational functions defining the curves and Bely\v{\i} functions, and hence on the dessins. This provides a combinatorial approach to the absolute Galois group, as first observed by Grothendieck~\cite{Grothendieck1997}.

An \textit{automorphism} of a dessin $D$ is a permutation of its edges which preserves the graph incidence and vertex-colorings, and extends to a self-conformal homeomorphism of its supporting Riemann surface. The set of automorphisms of $D$ form the automorphism group $\Aut(D)$ of $D$ under composition. It is well known that the group $\Aut(D)$ acts semi-regularly on the edges. If this action is transitive, and hence regular, the dessin is called \textit{regular}.

Because the absolute Galois group acts faithfully on regular dessins~\cite{GJ2013} it is important to investigate regular dessins. The classification of regular dessins has been investigated by imposing certain conditions on the supporting surfaces, the embedded graphs or the underlying automorphism groups~\cite{Conder2012,CJSW2013,Hidalgo2013, HNW2014,Jones2010,Jones2013, MNS2012}.
In the present paper we focus on the third direction and consider regular dessins with nilpotent automorphism groups. Our main result is a classification of nilpotent groups of class two which underly a \textit{unique} regular dessin. The uniqueness implies that such dessins possess the highest level of external symmetry. The importance of such dessins lies in  the fact that they play the role of universal covers of other nilpotent regular dessins of the same nilpotence class~\cite[Section 6]{Jones2013}.

\section{External symmetries of dessins}
In this section we briefly outline the algebraic theory of dessins; see \cite{CS1988,JS1996} for more details.

Let $\Omega$ be the set of edges of a dessin $D$. Following the global orientation of the supporting surface of $D$ we obtain two permutations $\rho$ and $\lambda$ which successively permute the edges around the black and white vertices. The black  and white vertices correspond to the cycles of $\rho$ and $\lambda$ respectively, and the connectivity of the underlying graph of $D$ implies that the monodromy group $\Mon(D)$ of $D$ generated by $\rho$ and $\lambda$ is transitive on $\Omega$. It follows that each dessin $D$ determines a transitive permutation representation $\theta\colon F_2\to \Mon(D)$ of $F_2=\la X,Y\mid -\ra$, the free group of rank two. It is isomorphic to the action of $F_2$ (by right multiplication) on the cosets $Ng$ of a subgroup $N\leq F_2$. This subgroup $N$, the \textit{dessin-subgroup} associated with $D$, is the stabiliser in $F_2$ of an element in $\Omega$, and is uniquely determined up to conjugacy. The coverings $D_1\to D_2$ between dessins correspond to group inclusions $N_1\leq N_2$, and the automorphism group $\Aut(D)$ corresponds to the action of $N_{F_2}(N)/N$ on the cosets of $N$ where $N_{F_2}(N)$ is the normaliser of $N$ in $F_2$, acting by left multiplication.  In particular, regular dessins $D$ correspond to normal subgroups $N$ of finite index in $F_2$, in which case $\Aut(D)\cong F_2/N$.

An automorphism $\sigma$ of $F_2$ transforms a dessin $D$ to a dessin $D^{\sigma}$ by sending the dessin subgroup $N$ to $N^{\sigma}$. In particular, if $\sigma$ is an inner automorphism induced by an element $g\in F_2$, then $N^{\sigma}=g^{-1}Ng$, and hence $D\cong D^{\sigma}$. In this way the outer automorphism group $\Out(F_2):=\Aut(F_2)/\Inn(F_2)$,  the group of  \textit{dessin operations}, acts on the isomorphism classes of dessins. A dessin $D$ is said to \textit{possess an external symmetry $\sigma$} if $D\cong D^{\sigma}$ where $ \sigma\in\Aut(F_2) \backslash{\rm Inn}(F_2)$. This is equivalent to that the dessin subgroup $N$ of $D$ is $\sigma$-invariant, that is, $N^\sigma$ is conjugate to $N$. For instance, let $\sigma_1:X\to Y, Y\to X$ be the automorphism of $F_2$ transposing the generators, and let $\iota\colon X\to X^{-1}, Y\mapsto Y^{-1}$ be the automorphism of $F_2$ inverting the generators. If $D\cong D^{\sigma_1}$, then $D$ is called \textit{symmetric}, corresponding to an external symmetry transposing the vertex colors; if $D\cong D^{\iota}$, then $D$ is called \textit{reflexible}, corresponding to an external symmetry reversing the orientation of the supporting surface. A regular dessin which is invariant under all dessin operations will be called \textit{totally symmetric}.

In \cite{James1988} James showed that $\Out(F_2)\cong GL(2,\Z)$; in \cite{JP2010} Jones and Pinto proved that $\Out(F_2)=\la  \sigma_1\Inn(F_2),\sigma_2\Inn(F_2),\sigma_3\Inn(F_2)\ra$ where $\sigma_1$, $\sigma_2$ and $\sigma_3$ are automorphisms of $F_2$ defined by
\begin{align}
&\sigma_1\colon X\mapsto Y,\, Y\mapsto X,\label{Oper1}\\
&\sigma_2\colon X\mapsto Y, \,Y\mapsto X^{-1},\label{Oper2}\\
& \sigma_3\colon X\mapsto YX, Y\mapsto X^{-1}.\label{Oper3}
\end{align}
Hence we have
\begin{proposition}\label{TOTAL}
A regular dessin $D$ is totally symmetric if and only if $D^{\sigma_i}\cong D$ $(i=1,2,3)$.
\end{proposition}

Every regular dessin $D$ can be identified with a triple $(G,x,y)$ where $G=\Aut(D)$ and $x$ and $y$ are automorphisms of $D$ which generate, respectively, the stabilisers of a white vertex and an adjacent black vertex. The monodromy group and the automorphism group of $D$ are identified with the left and the right regular representations of $G$. In such an identification if $D$ possesses an external symmetry $\sigma$, then the assignment $\bar\sigma\colon x\mapsto \bar\sigma(x), y\mapsto\bar\sigma(y)$ extends to an automorphism of $G$.

\section{The universal cover}
In this section, following Jones~\cite{Jones2013} we give a precise construction of a totally symmetric dessin from a given group.

Let $D_i=(G_i,x_i,y_i)$ $(i=1,2)$  be two regular dessins, and let $N_i$ be the associated dessin subgroups in $F_2$. The subgroup $N_1\cap N_2$ of $F_2$, being the intersection of two normal subgroups of finite index in $F_2$, is also a normal subgroup of finite index in $F_2$. The regular dessin corresponding to $N_1\cap N_2$ is called the \textit{parallel product} of $D_1$ and $D_2$ \cite{Wilson1994} and is denoted by $D_1\vee D_2$.
The automorphism group of $D_1\vee D_2$ is isomorphic to a subgroup of $G_1\times G_2$  generated by $(x_1,x_2)$ and $(y_1,y_2)$~\cite{BN2001}.

Note that for a finite two-generated group $G$ there are finitely many non-isomorphic regular dessins $D$ with $\Aut(D)\cong G$. The set $\mathcal{R}(G)$ of isomorphism classes of regular dessins $D$ with $\Aut(D)\cong G$ corresponds bijectively to the set $\mathcal{N}(G)$ of normal subgroups $N$ of $F_2$ such that $F_2/N\cong G$, or to the set of orbits of $\Aut(G)$ acting on the generating pairs of $G$. Define
\begin{align}
K(G)=\bigcap_{N\in \mathcal{N}(G)} N.
\end{align}
Since $K(G)$ is the intersection of finitely many normal subgroups of finite index in $F_2$,  $K(G)$ is also a normal subgroup of finite index in $F_2$. Define
\begin{align}\label{UNIVERSAL}
U(G)=\bigvee_{D\in\mathcal{R}(G)} D\quad\text{and}\quad \bar G= F_2/K(G).
 \end{align}
Then $\Aut(U(G))\cong \bar G$.
By the construction, $K(G)$ is the unique normal subgroup of $F_2$ with quotient isomorphic to $\bar G$, and hence $U(G)$ is the unique regular dessin with automorphism group isomorphic to $\bar G$. In other words, the group $\Aut(\bar G)$ acts transitively on the generating pairs of $\bar G$. Since Galois conjugations preserve the automorphism group, the uniqueness implies that $U(G)$ is invariant under the action of the absolute Galois group, and is therefore defined over $\mathbb{Q}$.

\begin{example}\cite[Example 5.2]{Jones2013} Let $G=C_n$ be  the cyclic group of order $n$. Then there are precisely $\psi(n)$ regular dessins $D$ with $\Aut(D)\cong C_n$ where $\psi(n)$ is the Dedekind's totient function~\cite[Theorem 24]{HNW2014}; see also \cite[Example  3.1]{Jones2013}. Since $G$ is abelian of exponent $n$, $F_2' F_2^n\leq N$ for each $N\in \mathcal{N}(G)$ where $F_2'F_2^n$ is the group generated by the commutators and $n$th powers of elements of $F_2$. The minimality of $K(G)$ implies that $ K(G)=F_2'F_2^n$, so
\[
\bar G=F_2/F'F^n\cong C_n\times C_n.
\]
The regular dessin $U(G)$ is the $n$th degree Fermat dessin, corresponding to the standard embedding of $K_{n,n}$~\cite{Jones2010}.
\end{example}

\begin{example}
Let $G$ be a metacyclic $2$-group defined by the presentation
\[
G=\la g,h\mid g^8=h^8=1, h^g=h^5\ra.
\]
Let $P_1=(g,h)$, $P_2=(h,g)$ and $P_3=(g,gh)$. Then $P_i$ $(i=1,2,3)$ are generating pairs of $G$, giving us all three isomorphism classes of regular dessins with an automorphism group isomorphic to $G$~\cite[Example 2]{HNW2014}. Define
\begin{align*}
&S_1=\{X^8,Y^8, [X,Y]^2,[X,[X,Y]],[Y,[X,Y]], [X,Y]Y^4\}\subseteq F_2,\\
&S_2=\{X^8,Y^8,[X,Y]^2,[X,[X,Y]],[Y,[X,Y]],[X,Y]X^4\}\subseteq F_2,\\
&S_3=\{X^8,Y^8,[X,Y]^2,[X,[X,Y]],[Y,[X,Y]],[X,Y](X^{-1}Y)^4\}\subseteq F_2.
\end{align*}
Then the above dessins have dessin subgroups $S_i^{F_2}$ $(i=1,2,3)$, the normal closures of the sets $S_i$ in $F_2$. Moreover, let
\[T=\{X^8,Y^8,[X,Y]^2,[X,[X,Y]],[Y,[X,Y]]\}.\] Then $K(G)=T^{F_2}$, corresponding to a totally symmetric dessin $U(G)$ of type $(8,8,8)$ and genus $41$. The group $\bar G=F_2/K(G)$ has a presentation
\begin{align*}
\la x,y\mid x^8=y^8=z^2= [x,z]=[y,z]=1, z:=[x,y]\ra.
\end{align*}
Clearly, both $G$ and $\bar G$ are $2$-groups of class 2. It can be directly verified that the automorphism group of $\bar G$ acts transitively on the generating pairs of $\bar G$, and hence $\bar G$ underlies a unique regular dessin.
\end{example}

In what follows, we study the group $\bar G$ and the associated regular dessin in more detail. The following technical lemma will be useful.
\begin{lemma}\label{LIFT}
Let $m$ and $n$ be positive integers where $m|n$. Then for each number $s$, $1\leq s< m$, such that $\gcd(s,m)=1$, there is a number $s'$, $1\leq s'< n$, such that $\gcd(s',n)=1$ and $s'\equiv s\pmod{m}$.
\end{lemma}
\begin{proof}
If $n$ and $m$ contain the same prime factors, then since $\gcd(s,m)=1$, we also have $\gcd(s,n)=1$. In this case we can take $s':=s$. Otherwise, let $n'$ be the maximal factor of $n$ which is coprime to $m$. By the Chinese Remainder Theorem, there is a number $s'$, $1\leq s'<n$, which satisfies the following congruences:
\begin{equation*}
\begin{cases}
x\equiv1\pmod{n'}\\
x\equiv s\pmod{m}
\end{cases}
\end{equation*}
Clearly, $\gcd(s',n)=1$. Hence $s'$ is the desired number.
\end{proof}

\begin{proposition}\label{CHAR}
If a finite group underlies a unique regular dessin, then the dessin is totally symmetric with an underlying graph of multiplicity at most two.
\end{proposition}
\begin{proof}Assume that $G$ is a finite group which underlies a unique regular dessin $D=(G,x,y)$. Recall that dessin operations preserve the automorphism group. The uniqueness implies that the dessin $D$ is invariant under all dessin operations, and is therefore totally symmetric. In particular, $G$ has an automorphism $\bar\sigma_1$ transposing $x$ and $y$. So $o(x)=o(y)$. Assume $o(x)=n$. Recall that the multiplicity $m$ of the underlying graph is equal to the order of $\la x\ra\cap\la y\ra$. We have $m\mid n$ and $\la x\ra\cap\la y\ra=\la x^{n/m}\ra=\la y^{n/m}\ra$. Define $q=n/m$. Then there is a number $r$, $\gcd(r,m)=1$, such that $y^{q}=x^{qr}$. If $m\geq3$, then there is a number $s$, $1<s<m$, such that $\gcd(s,m)=1$. By Lemma~\ref{LIFT}, the number $s$ lifts to a number $s'$, $1<s'<n$, such that $\gcd(s',n)=1$ and $s'\equiv s\pmod{m}$. It follows that $G=\la x,y\ra=\la x^{s'},y\ra$. It follows from the uniqueness again that the generating pairs $(x,y)$ and $(x^{s'},y)$ belong to the single orbit under the action of $\Aut(G)$. Hence,  $x^{s'}$ and $y$ also satisfy the relation $y^q=x^{s'rq}$. So we get $x^{rq}=x^{s'rq}$, that is, $x^{qr(s'-1)}=1$. Since $\gcd(r,m)=1$ and $o(x)=mq$, we have $s'\equiv1\pmod{m}$.  Recall that $s'\equiv s\pmod{m}$, we have $s\equiv1\pmod{m}$. This is a contradiction to our choice of $s$.\end{proof}

The following example shows that the quaternion group underlies a unique regular dessin of multiplicity two.
\begin{example}\label{QUAT}
The quaternion group $Q_8$ has 24 distinct generating pairs. Recall that $\Aut(Q_8)\cong\mathrm{Sym}_4$, the symmetry group of degree 4. Since $\Aut(Q_8)$ acts semiregularly on the generating pairs of $Q_8$, the number of regular dessins with automorphism group isomorphic to $Q_8$ is equal to $24/|\Aut(Q_8)|=1$. The dessin is totally symmetric of type $(4,4,4)$ and genus $2$, corresponding to the $8$-gonal regular embedding of $K_{2,2}^{(2)}$ into the double torus.
\end{example}

The following result summarizes some properties of $\bar G$ when $G$ is solvable or nilpotent.

\begin{proposition}\cite[Section 5]{Jones2013}
Let $G$ be a $2$-generated group, and $\bar G$ defined by \eqref{UNIVERSAL}.
\begin{itemize}

\item[\rm(i)] If $G$ is solvable of derived length ${\rm dl(G)}$, then so is $\bar G$;
\item[\rm(ii)] If $G$ is nilpotent of class ${\rm c}(G)$, then so is $\bar G$.
\end{itemize}
\end{proposition}
\begin{proof}
The result follows from the fact that $\bar G$ is isomorphic to a subgroup of the direct product $G^r$ where $r=|\mathcal{R}(G)|$.
\end{proof}

\section{Classification }
In this section, we classify nilpotent groups of class 2 which underly a unique regular dessin.

The following decomposition theorem reduces the classification of nilpotent regular dessins to the classification of regular $p$-dessins.
\begin{proposition}\cite[Theorem 13]{HNW2014}\label{DECOM}
Every regular dessin with a nilpotent automorphism group $G$ is uniquely decomposed into a parallel product of regular dessins whose automorphism groups are the Sylow subgroups of $G$.
\end{proposition}

Recall that groups of class 1 are abelian. It is shown that an abelian $p$-group which underlies a unique regular dessin is isomorphic to $C_{p^a}\times C_{p^a}$ for some integer $a\geq0$~\cite[Theorem 24]{HNW2014}. In the remainder of the paper, we classify $p$-groups of class 2 which underly a unique regular dessin.

  The following prerequisites are assumed.
\begin{lemma}\label{BINO}\cite[Chapter III, Lemma 1.3]{Huppert1967}
Let $G$ be a nilpotent group of class 2, $x,y\in G$. Then
\[
[x^n,y]=[x,y^n]=[x,y]^n\quad\text{ and  }\quad (xy)^n=x^ny^n[y,x]^{n\choose 2},
\]
where $n\geq1$ is a positive integer.
\end{lemma}

\begin{lemma}\label{DERIVED}\cite[Chapter III, Lemma 1.11]{Huppert1967}
Let $G=\la x,y\ra$ be a group. Then $G'=\la [x,y]^g\mid g\in G\ra$.
\end{lemma}

\begin{lemma}\cite[Chapter III, Theorem 3.15]{Huppert1967}\label{BURNSIDE}
Let $G$ be a $p$-group and $\Phi(G)$ the Frattini subgroup of $G$. If $|G/\Phi(G)|=p^d,$ then every minimal generating set of $G$ contains exactly $d$ elements. In particular, $G=\la x_i\mid i=1,2,\ldots, d\ra$ if and only if $G/\Phi(G)=\la x_i\Phi(G)\mid i=1,2,\ldots, d\ra$.
\end{lemma}

The following theorem is the main result of the paper.

\begin{theorem}\label{MAIN}
A finite $p$-group $G$ of class two which underlies a unique regular dessin is isomorphic to one of the following groups:
\begin{itemize}
\item[\rm(i)] $p$ is odd and $1\leq b\leq a$:
\begin{align}
G=\la x,y\mid x^{p^a}=y^{p^a}=z^{p^b}=[x,z]=[y,z]=1,z:=[x,y]\ra.\label{GP1}
\end{align}
\item[\rm(ii)] $p=2$ and $1\leq b\leq a-1$:
\begin{align}
G=\la x,y\mid x^{2^a}=y^{2^a}=z^{2^b}=[x,z]=[y,z]=1, z:=[x,y]\ra.\label{GP2}\end{align}
\item[\rm(iii)] $p=2$ and $a\geq 2:$
\begin{align}
G=\la x,y\mid x^{2^a}=[x,z]=[y,z]=1, x^{2^{a-1}}=y^{2^{a-1}}=z^{2^{a-2}}, z:=[x,y]\ra.\label{GP3}
\end{align}
\end{itemize}
Moreover, the groups from distinct families, or from the same family but with distinct parameters, are pairwise non-isomorphic.
\end{theorem}
\begin{proof}
Assume that the group $G$ underlies a unique regular dessin $D=(G,x,y)$, so $\Aut(G)$ acts transitively on the generating pairs of $G$. By Proposition~\ref{CHAR}, $D$ is totally symmetric. It follows from the discussion at the end of Section 2 that the automorphisms $\sigma_i$ $(i=1,2,3)$ of $F_2$ defined by \eqref{Oper1}, \eqref{Oper2} and \eqref{Oper3} all induce automorphisms $\bar\sigma_i$ of $G$ given by
\begin{align*}
&\bar\sigma_1:x\mapsto y, y\mapsto x,\\
&\bar\sigma_2:x\mapsto y, y\mapsto x^{-1},\\
&\bar\sigma_3: x\mapsto yx, y\mapsto x^{-1}.
\end{align*}
 Using these properties, we first construct the group $G$ in the following two steps:

\begin{step}[1] Determination of the presentation of $G$.\par
Assume that $o(x)=p^a$ and $o(y)=p^{a'}$, $a,a'\geq 0$.  Define $z=[x,y]$ where $o(z)=p^b$. By Lemma~\ref{DERIVED}, $G'=\la z^g\mid g\in G\ra$. Since ${\rm c}(G)=2$, $1<G'\leq Z(G)$, and hence $G'=\la z\ra\cong C_{p^b}$ where $b\geq 1$.
 Since  $y=\bar\sigma_1(x)$, we have $o(x)=o(y)$, and hence $a=a'$. By induction we have $z^{p^a}=[x^{p^a},y]=1$, so $b\leq a$.

Let $N=\la z,x\ra$. Then $N\normal G$ is abelian and $G/N=\la yN\ra$ is cyclic. Assume $\la z\ra\cap\la x\ra= \la x^{p^c}\ra$ and $\la y\ra\cap N= \la y^{p^d}\ra$ where $0\leq c\leq a$ and $0\leq d\leq a$. Then there exist integers $r$, $s$ and $t$, where
\[0\leq r\leq p^a-1,\quad 0\leq s\leq p^c-1, \quad 0\leq t\leq p^b-1,\]
 such that the following identities hold:
\begin{align}\label{EQUA}
x^{p^{c}}=z^{r}\quad\text{and}\quad y^{p^d}=x^{s}z^{t}
\end{align}
 Apply $\bar\sigma_1$, $\bar\sigma_2$ and $\bar\sigma_3$ to \eqref{EQUA} (use Lemma~\ref{BINO} if necessary) we get
\begin{align}
&y^{p^{c}}=z^{-r},\quad x^{p^d}=y^sz^{-t};&  \text{apply $\bar\sigma_1$ to \eqref{EQUA}} \label{EQUA1}\\
&y^{p^{c}}=z^{r},\qquad x^{-p^d}=y^sz^{t};&  \text{apply $\bar\sigma_2$ to \eqref{EQUA}} \label{EQUA2}\\
&y^{p^c}x^{p^c}=z^{r-{p^c\choose 2}},\quad y^sx^{s+p^d}=z^{-t-{s\choose2}}.& \text{apply $\bar\sigma_3$ to \eqref{EQUA}} \label{EQUA3}
\end{align}
By assumption $d$ is the smallest nonnegative integer $i$ such that $y^{p^i}\in N$, so the first relation in \eqref{EQUA1} implies that $d\leq c$. By induction we have
\[z^{p^d}=[x,y]^{p^d}=[x,y^{p^d}]\stackrel{\eqref{EQUA}}=[x,x^sz^t]=1.\]
Since $o(z)=p^b$, we get $b\leq d$. Summarizing the above inequalities we obtain
\begin{align}\label{INEQ}
1\leq b\leq d\leq c\leq a.
\end{align}

Moreover, by \eqref{EQUA} and \eqref{EQUA1}, substituting $z^{-r}$ and $z^{r}$ for $y^{p^c}$ and $x^{p^c}$ in the first relation of \eqref{EQUA3} we obtain that $z^{r-p^c(p^c-1)/2}=1$, which implies that
\begin{align}\label{CONG1}
r-p^c(p^c-1)/2\equiv0\pmod{p^b}.
\end{align}
Similarly, by \eqref{EQUA2} substituting $x^{-p^d}z^{-t}$ for $y^s$ in the second relation of \eqref{EQUA3}  we obtain that
\begin{align}\label{CONG2}
x^s=z^{-{s\choose2}}.
\end{align}
 Further, we obtain from the first relations in \eqref{EQUA1} and \eqref{EQUA2} that $z^{2r}=1$. So $p^b\mid 2r$. Therefore, if $p>2$, then $p^b\mid r$ ; if $p=2$, then either $2^b\mid r$ or $2^{b-1}\parallel r$. We distinguish 3 cases.

\begin{case}[i] $p>2$ and $p^b| r$.\par
Note that $\la x\ra\cap\la z\ra=z^r$. We have $\la x\ra\cap\la z\ra=1$. Since $\bar\sigma_1(\la x\ra\cap\la z\ra)=\la y\ra\cap\la z\ra$, we also have $\la y\ra\cap\la z\ra=1$.  So by \eqref{EQUA} we have $c=a$, and by \eqref{CONG2} we have $s\equiv0\pmod{p^a}$. In particular, the second relation of \eqref{EQUA} is reduced to $y^{p^d}=z^t$. Since $\la y\ra\cap\la z\ra=1$, we have $d=a$ and $t\equiv0\pmod{p^b}$. Consequently, $G$ has the presentation \eqref{GP1}.

\end{case}
\begin{case}[ii] $p=2$ and $2^b| r$.\par
In this case we also have $\la x\ra\cap\la z\ra=\la y\ra\cap\la z\ra=1$. So by \eqref{EQUA} we have $c=a$ and by \eqref{CONG2} we get $s\equiv0\pmod{2^a}$. It follows that the second relation of \eqref{EQUA} is reduced to $y^{p^d}=z^t$, so $d=a$ and $t\equiv0\pmod{2^b}$. Note that \eqref{CONG1} is reduced to $2^{a-1}(2^a-1)\equiv0\pmod{2^b}$, so $b\leq a-1$. Therefore, $G$ is defined by \eqref{GP2}.
\end{case}

\begin{case}[iii] $p=2$ and $2^{b-1}\parallel r$.\par
We have $\la x\ra\cap\la z\ra\cong\la y\ra\cap\la z\ra\cong C_2$. By the first relation of \eqref{EQUA} we have $c=a-1$. By \eqref{CONG1} we have $2^{b-1}(r'-2^{c-b}(2^c-1))\equiv0\pmod{2^b}$ where $r=2^{b-1}r'$, $r'$ is odd. This implies that $r'-2^{c-b}(2^c-1)$ is an even number. Hence $b=c$. By \eqref{INEQ} $b=d=c=a-1.$
So $y^{2^{a-1}}=z^{2^{a-2}}$ and $y^{2^{a-1}}=x^{s}z^t$. By \eqref{CONG2}, $s\equiv0\pmod{2^{a-1}}$. Recall that $0\leq s\leq 2^{a-1}-1$. Then we have $s=0$, implying $y^{2^{a-1}}=z^t$. Since $\la y\ra\cap\la z\ra\cong C_2$ and $o(z)=2^{a-1}$, we have $t=2^{a-2}$. Consequently we obtain the presentation \eqref{GP3}.
\end{case}
\end{step}

\begin{step}[2] Proof that $G$ underlies a unique regular dessin.\par
Equivalently, we shall prove that $\Aut(G)$ acts transitively on the generating pairs of $G$. Note that every element of $G$ can be written as the form $x^iy^jz^k$. Let $x_1=x^iy^jz^k$ and $y_1=x^ry^sz^t$. By Lemma~\ref{BURNSIDE}, $G=\la x_1,y_1\ra$ if and only if $G/\Phi(G)=\la \bar x^i\bar y^j, \bar x^r\bar y^s\ra$. Since $G/\Phi(G)\cong C_p\times C_p$ is an elementary $p$-group of rank 2, this is equivalent to that the matrix $\begin{pmatrix} i&j\\r &s\end{pmatrix}$ is invertible in $\Z_p$, that is,
\begin{align}\label{GEN}
is-jr\not\equiv0\pmod{p}.
\end{align}
It is sufficient to show that $x_1$ and $y_1$ satisfy the stated presentations in the respective case. Define $z_1=[x_1,y_1]$. We have
\[z_1=[x^iy^jz^k,x^ry^sz^t]=[x^iy^j,x^ry^s]=[x^i,x^ry^s][y^j,x^ry^s]=[x^i,y^s][y^j,x^r]=z^{is-jr}.\] By \eqref{GEN}, $p\nmid is-jr$, so $o(z_1)=o(z)$.  Clearly, $[x_1,z_1]=[y_1,z_1]=1$. Note that by Lemma~\ref{BINO} we have
\begin{align}\label{POWER}
(x^iy^jz^k)^n=(x^iy^j)^nz^{kn}=x^{in}y^{jn}[y^j,x^i]^{n\choose2}z^{kn}=x^{in}y^{jn}z^{kn-ij{n\choose2}},
\end{align}
where $n$ is a positive integer.  We distinguish 3 cases as before.

\begin{case}[i]By \eqref{POWER} we have  $x_1^{p^{a}}=x^{ip^{a}}y^{jp^{a}}z^{kp^{a}-ij{p^{a}\choose2}}=1.$ Since $\la x\ra\cap\la z\ra=1$ and $\la y\ra\cap\la x,z\ra=1$, by \eqref{GEN} and \eqref{POWER} we have
\begin{align*}
&x_1^{p^{a-1}}=(x^iy^j)^{p^{a-1}}z^{kp^{a-1}}=x^{ip^{a-1}}y^{jp^{a-1}}z^{kp^{a-1}-ij{p^{a-1}\choose2}}\neq1.
\end{align*}
 Therefore $o(x_1)=p^a$. Similarly, $o(y_1)=p^a$. So $x_1$ and $y_1$ satisfy the presentation~\eqref{GP1}.

\end{case}

\begin{case}[ii] By \eqref{POWER} we have $x_1^{2^a}=x^{i2^a}y^{j2^a}z^{k2^a-ij{2^a\choose2}}=1$. Since $\la x\ra\cap\la z\ra=1$ and $\la y\ra\cap\la x,z\ra=1$, by \eqref{GEN} and \eqref{POWER} we have
\begin{align*}
&x_1^{2^{a-1}}=(x^iy^j)^{2^{a-1}}z^{k2^{a-1}}=x^{i2^{a-1}}y^{j2^{a-1}}z^{k2^{a-1}-ij{2^{a-1}\choose2}}\neq1.
\end{align*}
 Therefore $o(x_1)=2^a$. Similarly, $o(y_1)=2^a$. It follows that $x_1$ and $y_1$ satisfy the presentation~\eqref{GP2}.

\end{case}

\begin{case}[iii] By \eqref{POWER} we have $x_1^{2^a}=x^{i2^a}y^{j2^a}z^{k2^a-ij{2^a\choose2}}=1$. Since $x^{2^{a-1}}=y^{2^{a-1}}=z^{2^{a-2}}$, we have
\begin{align*}
&x_1^{2^{a-1}}=x^{i2^{a-1}}y^{j2^{a-1}}z^{k2^{a-1}-ij{2^{a-1}\choose2}}=z^{(i+j+ij)2^{a-2}},\\
&y_1^{2^{a-1}}=x^{r2^{a-1}}y^{s2^{a-1}}z^{t2^{a-1}-rs{2^{a-1}\choose2}}=z^{(r+s+rs)2^{a-2}}.
\end{align*}
By \eqref{GEN}, $i+j+ij\equiv 1\pmod{2}$ and $r+s+rs\equiv1\pmod{2}$, so $o(x_1)=2^{a-1}o(z^{(i+j+ij)2^{a-2}})=2^a$ and $x_1^{2^{a-1}}=y_1^{2^{a-1}}=z_1^{2^{a-2}}$. Therefore $x_1$ and $y_1$ satisfy the presentation \eqref{GP2}.
\end{case}
\end{step}

 To finish the proof we need to show that the groups are uniquely determined by the parameters. This is easily seen from the following table which summarises the invariant types of the derived subgroups $G'$ and the abelianisations $G^{\mathrm{ab}}=G/G'$.
\[
\begin{array}{llll}
\text{Case} & G' &  G^{\mathrm{ab}} & \text{Condition}\\
\hline
{\rm(i)} &C_{p^a} & C_{p^a}\times C_{p^a} & 1\leq b\leq a\\
{\rm(ii)} &C_{2^{b+1}} & C_{2^{a-1}}\times C_{2^a} & 1\leq b\leq a-1\\
{\rm(iii)}&C_{2^{a-1}} & C_{2^{a-1}}\times C_{2^{a-1}} &a\geq 2\\
\end{array}
\]
\end{proof}

As a natural consequence of Theorem~\ref{MAIN} we have
\begin{corollary}Let $G$ be the groups from Theorem~\ref{MAIN}. Then the sizes of the group $G$ and its automorphism group $\Aut(G)$, and the type and genus of the associated regular dessin $U$ with $\Aut(U)\cong G$ are summarized as follows.
\[
\begin{array}{lllll}\label{TT}
Family & |G| &  |\Aut(G)| & \text{Type of $U$}&\text{Genus of $U$} \\
\hline
{\rm (i)} &p^{2a+b}& (p+1)(p-1)^2p^{4a+2b-3} & (p^a,p^a,p^a) &p^{a+b}(p^a-3)/2+1\\
{\rm(ii)} &2^{2a+b} & 3\cdot 2^{4a+2b-3} & (2^a,2^a,2^a)&2^{a+b-1}(2^a-3)+1\\
{\rm(iii)} & 2^{3a-4}& 3\cdot 2^{6a-9} &(2^a,2^a,2^a)&2^{2a-5}(2^a-3)+1\\
\end{array}
\]
\end{corollary}

\begin{proof}
From the proof of Theorem~\ref{MAIN}, we see that $|G|=p^d|N|=p^{b+c+d}$. The order $|G|$ of $G$ follows from substitution for $a$ and $c$ in the respective case. Since $\Aut(G)$ acts freely and transitively on the generating pairs of $G$, the order $|\Aut(G)|$ is equal to the number of generating pairs of $G$. The type and genus of the associated dessin $U$ are straightforward.\end{proof}

\section*{Acknowledgement}
The authors are grateful to the anonymous referee(s) for the helpful suggestions which have simplified the proof of Lemma~\ref{LIFT} and improved the presentation of the paper. The first and third author are supported by the following grants: Scientific Research Foundation(SRF) of Zhejiang Ocean Univesity and National Natural Science Foundation (NNSF: 60673096). The second author is supported by the following grants: APVV-0223-10, and the grant APVV-ESF-EC-0009-10 within the EUROCORES Programme EUROGIGA (Project GReGAS) of the European Science Foundation and the Slovak-Chinese bilateral grant APVV-SK-CN-0009-12.


TITLE: Let a, b, c, d be rational numbers...
QUESTION [2 upvotes]: Let $a, b, c, d$ be rational numbers, where $\sqrt{b}$ and $\sqrt{d}$ exist and are irrational. If $a + \sqrt{b} = c + \sqrt{d}$, prove that $a=c$ and $b=d$.

REPLY [0 votes]: Hint $\,\ 0\ne  \alpha = \sqrt b\! -\!\sqrt d\in\Bbb Q\,\Rightarrow\ \alpha' = \sqrt b\!+\!\sqrt d = \dfrac{\ b-d}{\sqrt b \!-\! \sqrt d}\in\Bbb Q\,\Rightarrow\,\alpha\!+\!\alpha' = 2\sqrt b\in \Bbb Q\,\Rightarrow\Leftarrow$


TITLE: Second Sylow theorem's proof
QUESTION [1 upvotes]: In the proof of the 2nd Sylow theorem, I don't understand why the fact tha $O(S)| m$ implies that there exists a suborbit with one element. This is part of the proof

Second Sylow Theorem. Let $G$ be a finite group with order
$p^a \cdot m$ and $p$ a prime such that $m$ is not divisible by $p$.
Then any $p$-subgroup $H$ of $G$ is contained in a $p$-Sylow subgroup.

Proof. Let $P$ be the set of $p$-Sylow subgroups of $G$, and
let $H$ be a $p$-subgroup of $G$, and let $|H| = p^r$, $r \le; a$. Let
$S$ a fixed $p$-Sylow subgroup and $O(S)$ its orbit under the
conjugation action. Let $H$ act on $O(S)$: $hT = hTh^{-1} \, \forall T
> \in O(S)$ : it's an action on on $O(S)$ because the conjugate of an
element is still in $O(S)$. $O(S)$ is thus partitioned into suborbits,
each one having a cardinality which divides $p^r$. The cardinality
$O(S)$ is the sum of the cardinalities of all these suborbits.
Since the cardinality $O(S)$ divides $m$ and is relatevely
prime to $p$, there exists at least one suborbit having just one
element; that is $\exists T \in O(S)$ such that $hTh^{-1} = T \quad
> \forall; h \in H$.

REPLY [1 votes]: (i) $|O(S)|$ is the number of distinct conjugates of the Sylow-$p$ subgroup $S$.
(ii) We ask: when two memebers of $G$ give same conjugate of $S$?
$$
\begin{align*}
\mbox{ $g,h\in G$ give same conjugate of $S$} & \Longleftrightarrow  gSg^{-1}=hSh^{-1}\\
& \Longleftrightarrow h^{-1}gS(h^{-1}g)^{-1}=S\\
& \Longleftrightarrow h^{-1}g \mbox{ normalizes } S\end{align*}
$$
The normalizer $N(S)$ of $S$ in $G$ is a subgroup of $G$; it contains $S$. Thus,
$$
\mbox{$g,h$ give same conjugate of $S$ } \Longleftrightarrow h^{-1}g\in N(S) \Longleftrightarrow gN(S)=hN(S).
$$
In other words, the number of distinct conjugates of $S$ is equal to index of normalizer of $S$ in $G$. Now,
$$S\subseteq N(S)\subseteq G$$
and index of $S$ in $G$ is $m$, so index of $N(S)$ in $G$ divides $m$; i.e. $|O(S)|$ divides $m$.

The bold-faced conclusion is true even if $S$ is not Sylow-$p$ subgroup; the arguments show that we not used any special thing about $S$.


TITLE: The quadratic equation $x^2=c$ in a division ring
QUESTION [1 upvotes]: Let $D$ be a division ring. We denote $D'$ by the derived subgroup of the multiplicative group $D\setminus\{0\}$, that is, the subgroup generated by all the commutators of $D\setminus\{0\}$. For $c\in D',$ does there exist an element $a$ of $D$ such that $a^2=c$? In other words, is the quadratic equation $x^2=c$ solvable for each $c\in D'$?
It is true for $D=\mathbb{H},$ which is the real quaternion division ring. Thanks for all your support.

REPLY [1 votes]: Let us analyze the problem in a particular case.
Assume that $D$ has prime degree $p$ over its center $F$. Then it is known that $D'$ is the set of elements of reduced norm $1$.
Assume that $c$ has reduced norm $1$, and $c\notin F$. Since $F[c]$ is a commutative subfield of $D$, its degree over $F$ divides $p$. It cannot be $1$ since $c\notin F$. Hence $F[c]/F$ has degree $p$, and $[F[c]:F]=p$. Assume that there exists $a\in D$ such that $a^2=c$. Once again, $F[a]/F$ has degree $p$ (otherwise $c=a^2\in F$ !). Since $F[c]\subset F[a]$, we get $F[a]=F[c]$. Hence $a\in F[c]$.
So we have to solve the equation $a^2=c$ in $F[c]$. Since $F[c]/F$ is a MAXIMAL commutative subfield of $D$, $Nrd_D(x)=N_{F[c]/F}(x)$ for all $x\in F[c]$.
Hence, we need to solve the equation $a^2=c$ in the field $F[c]$, where $N_{F[c]/F}(c)=1$. However, this equation does not have always solutions.
We can now construct explicit counter examples. Take $D=(d,-1)_\mathbb{Q}$, where $d$ is not a sum of two squares (for example $d$ is a prime congruent to $3$ mod $4$, or negative), so $D$ is division. Let $j\in D$ the basis element such that $j^2=-1$.
Then $Nrd_D(j)=-j^2=1$, so $j\in D'$. Now the above analysis shows that, if $a^2=j$, then $a\in \mathbb{Q}[j]$.
Write $a=x+yj,x,y\in \mathbb{Q}$. Then $x^2-y^2+2xyj=j$, so $y=\pm x$ and $2xy=1$. We then get $\pm 2x^2=1\in\mathbb{Q}$, which has no solution.
Of course, you can avoid the general analysis if you only work with this particular example, and I am pretty sure you can prove by hand that $j\in D'$, but I think it is still interesting to know how counterexamples are found...


TITLE: Uniform convergence of a sequences of functions to a complete metric space
QUESTION [3 upvotes]: I would appreciate clarification regarding Theorem 2.5 (page 10) in Gamelin and Greene "Intro to Topology."
The theorem states:

Let $S$ be a set and let $X$ be a complete metric space. If $\{f_n\}_{n=1}^{\infty}$ is a Cauchy sequences from $S$ to $X$, then there exists a function $f$ from $S$ to $X$ such that $\{f_n\}$ converges uniformly to $f$.

The proof goes as follows:

For each fixed $s \in S, $ $\{f_n\}_{n=1}^{\infty}$ is a Cauchy sequence in $X$. Since $X$ is complete, $\{f_n\}$ converges to some point of $X$, which we define to be $f(s)$.
Let $\epsilon>0$. Choose an integer $N$ such that $d(f_n(s),f_m(s))< \epsilon$ for all $s \in S$ and all $n,m \geq N$. Then
$$d(f_n(s),f(s)) \leq d(f_n(s),f_m(s)) + d(f_m(s),f(s)) $$
and $$d(f_n(s),f(s)) < \epsilon + d(f_m(s),f(s))$$
whenever $n,m \geq N$.

So far so good, now comes where my question lies. The proof then says:

Letting $m$ tend to $\infty$ we obtain $d(f_n(s),f(s)) < \epsilon$ for all $s \in S$. Consequently $\{f_n\}$ converges uniformly to $f$ on $S$.

-- What is the difference between $d(f_n(s),f(s))$ on the LHS and $d(f_m(s),f(s))$ on the RHS? i.e., if it's valid on the RHS, just let $n$ go to $\infty$ on the LHS and not need a proof.
-- Also where in the proof does the specificity "for each $s \in S$" which looks like pointwise convergence turn to uniform convergence?
Thanks
EDIT Maybe I should restate my question. The first term on the right in the long inequality $d(f_n(s),f_m(s))$ holds for all $s$ based on the definition of a Cauchy sequence of functions. So it comes down to the other two terms. In that there doesn't seem to be any inherent difference between them and the proof claims the one on the right becomes negligible for all $s$, then why could you not just start and end with that, using term $d(f_n(s),f(s))$ and not need the inequality at all.
And my second question is where exactly in the proof does it change from pointwise convergence? It was always the case for $d(f_n(s),f_m(s))$. But where does it change, e.g., for $d(f_m(s),f(s))$

REPLY [1 votes]: Maybe it would help if you choose $N_\epsilon$ so that $d(f_n(s), f_m(s)) < \epsilon/2$ for $n,m > N$. Then use the fact that $f_m \to f$ pointwise to choose $M_{\epsilon,s} > N$ so that $d(f_m(s), f(s)) <\epsilon/2$ for $m> M_{\epsilon, s}$. So $d(f_n(s), f(s) ) < \epsilon $ for $n > N_\epsilon$. This is uniform convergence since the final inequality depended on $\epsilon$ (not $s$).
The uniformity of the convergence is hidden in the definition of cauchy-sequence of the functions. We were able to choose $N_\epsilon$ such that $d(f_n(s), f_m(s))<\epsilon$ independent of/for all $s\in S$.


TITLE: Why is this infinite series _intuitionistically_ Cauchy?
QUESTION [0 upvotes]: I'm currently writing a short paper on Intuitionism for uni. The subject of this paper is the decay of the intermediate value theorem under intuitionism. I have found a proof for this but I have a question about a step the author makes in this proof. It is abbout the following:

The problem I have with using this in my paper is that in the introduction I proclaim that under intuitionism you abandon the law of the excluded middle and then proceed to give an implicit proof for this sequence being Cauchy where (at least in my head) you give the two possibilities where the Goldbach conjecture is true (the limit is 0)  or false (the limit is 2^{-k0}) It could be that this is nothing to worry about, but the reasoning seems a little crooked to me.
If anyone could maybe clarify this to me it would be greatly appreciated!

REPLY [1 votes]: There is no need to use the law of excluded middle here.
In general, consider some sequence $\{s_n \in \{0, 1\}\}_{n \in \mathbb{N}}$. We can define a corresponding Cauchy sequence
$$a_n = \begin{cases}
   2^{-k_0} & k_0 \leq k \text{ is the smallest }k_0 \text{ such that } s_{k_0} = 1 \\
   2^{-k} & otherwise
\end{cases}$$
We can prove by induction that for all $k \in \mathbb{N}$, there either is a least $k_0 \leq k$ such that $s_{k_0} = 1$, or there is no $k_0 \leq k$ such that $s_{k_0}$. Thus, $a$ is well-defined.
I claim that for all $i, j \geq k$, we have $|a_i - a_j| < 2^{-k}$. This means that $a$ is a Cauchy sequence.
To prove this, we suppose WLOG that $i \leq j$. We consider three cases.
Case 1: there is no $k_0 \leq j$ such that $s_{k_0} = 1$. If this is the case, then there is also no $k_0 \leq i$ such that $s_{k_0} \leq i$. So we have $|a_i - a_j| = 2^{-i} - 2^{-j} < 2^{-i} \leq 2^{-k}$.
Case 2: there is a least $k_0 \leq k$ such that $s_{k_0} = 1$. Then we have two sub cases:
Case 2a: $k_0 \leq i$. In this case, $a_i = a_j = 2^{-k_0}$, so $|a_i - a_j| = 0 < 2^{-k}$.
Case 2b: $i < k_0$. In this case, we have $|a_i - a_j| = 2^{-i} - 2^{-k_0} < 2^{-i} \leq 2^{-k}$.
This completes the proof that $a$ is Cauchy.
Your Goldbach situation is a special case of this general situation. We have
$$s_k = \begin{cases}
  1 & k \text{ is a counterexample to the Goldbach conjecture}\\
  0 & otherwise
\end{cases}$$
Finally, note that $\lim\limits_{n \to \infty} s_n > 0$ if and only if there exists some $k$ such that $s_k = 1$.


TITLE: Really Confusing Question on Wronskian and ODEs
QUESTION [1 upvotes]: The following question was given to as in a recent assignment, namely
Let $$y''+p(t)y'+q(t)y=0~.$$ It is given that $(1+t)^2$ is a solution to this differential equation, and that the wronskian of any two solutions is constant. Find the coefficients of $p$ and $q$.
Say for example the wronskian is a constant, say $C$. Then if $u$ is another solution, then $$u'+ \frac{2u}{(1+t)^2} = C~,$$ from which I get $$u = \frac{Ct+A}{(1+t)^2}~,$$ where $A$ is another constant.
How do I obtain the coefficients $p$ and $q$ ? I could solve simultaneous equations, but then they are messy, as the second derivative of $q$ would involve many terms.
How can Abel's identity for differential equations help me? Another problem is that I don't know if the wronskian is $0$ or not.
But $p$ and $q$ are analytic, so if their wronskian is zero then the two functions are linearly dependent; just putting some facts together.
Thanks,
Ben

REPLY [4 votes]: As DJC pointed out, the differential equation you get by setting the Wronskian to a constant is
$$u'(t) = {2 \over 1 + t} u(t) + {C \over (1 + t)^2}$$
This is a first-order linear equation which is solved by the usual methods to
$$u(t) = -{C' \over 1 + t} + D(1 + t)^2$$
We can choose any of these to be the second solution, other than a multiple of the first solution $(1 + t)^2$. So we take $u(t) = {1 \over 1 + t}$. To find out what $p(t)$ and $q(t)$ are, you can just plug your two solutions into the original equation and solve for $p(t)$ and $q(t)$. The first one gives
$$2(1 + t)p(t) + (1 + t)^2 q(t) = -2$$
The second one gives
$$-{1 \over (1 + t)^2}p(t) + {1 \over 1 + t}q(t) = -2{1 \over (1 + t)^3}$$
or equivalently
$$-(1 + t)p(t) + (1 + t)^2q(t) = -2$$
Subtract these equations to get $p(t) = 0$. Then solve for $q(t) = -2{1 \over (1 + t)^2}$.

Edit: The Wronskian is actually of the form $Ce^{-\int p(t)}$ by something known as Abel's identity. So if you are willing to use that, you can shorten your work by getting $p(t) = 0$ immediately this way. Then you can just plug in your first solution and solve for $q(t)$.


TITLE: A simple osmosis problem
QUESTION [1 upvotes]: The following is adapted from a problematic question* asked on the Bio site. I would like to ask it free of those distractions here. If there is anything unduly artificial about the problem please advise and I'll try to revise the question. I think it belongs to physics at least as much as to chemistry.
Question: A bag whose walls form a semi-permeable membrane contains a mixture of 5% of solute A and 10% of solute B (both in water). The membrane allows passage of B particles but not A particles.
The bag is lowered into a solution containing a 10% solution of solute B. Instead of asking a multiple choice question let me keep it open-ended. What would we expect in terms of the movement of water/solutes in this situation? Can we predict, in rough terms, a final equilibrium state?
Thanks for any clarification!
*The problem is that one of the solutes was poorly chosen.

REPLY [1 votes]: What would we expect in terms of the movement of water/solutes in this situation? Can we predict, in rough terms, a final equilibrium state?

Some amount of particles B will exit the bag, so as to equalized the osmotic pressure on both sides of the membrane.
The final state will be $<10$% B inside the bag and $>10$% B outside the bag.
The question does not specify whether the percentages are weight percent or something else.  Without knowing more information such as molecular weight or molarity one can not make a quantitative determination of the final state.
Also, the question doesn't explicitly state that water can pass through the membrane, although such is usually the case.  Water would initially enter the bag stretching the bag and raising the hydraulic pressure inside the bag.  If the bag is elastic and not permanently deformed, extra water will eventually leave the bag, as osmotic pressure balances by B exiting the bag.


text:
\begin{document}
\maketitle

\begin{center}Dipartimento di Fisica ``G. Galilei'' and Istituto
Nazionale di Fisica Nucleare \\
Universit\`a di Padova, Via Marzolo, 8 -- 35131 Padova,
Italy\end{center}

\bigskip

\begin{abstract} We show that a recently conjectured
expression for the superstring three-point amplitude, in the
framework of the Cacciatori, Dalla Piazza, van Geemen -- Grushevsky
ansatz for the chiral measure, fails to vanish at three-loop, in
contrast with expectations from non-renormalization theorems. Based
on analogous two-loop computations, we discuss the possibility of a
non-trivial correction to the amplitude and propose a natural
candidate for such a contribution. Thanks to a new remarkable
identity, it is reasonable to expect that the corrected three-point
amplitude vanishes at three-loop, recovering the agreement with
non-renormalization theorems.
\end{abstract}

\bigskip

In the last years there has been a considerable progress in the
conceptual understanding and in derivation of explicit formulas for
multiloop superstring amplitudes. In a series of papers
\cite{D'Hoker:2001zp,D'Hoker:2004xh,D'Hoker:2004ce,D'Hoker:2005jc,D'Hoker:2005ia,D'Hoker:2007ui},
D'Hoker and Phong have derived from first principles an explicitly
gauge independent expression for the $2$-loop chiral superstring
measure on the moduli space of Riemann surfaces, given by
\begin{equation}\label{twoloopmeas}d\mu[\delta]=\theta[\delta](0)^4\Xi_6[\delta]d\mu_{Bos}\ ,
\end{equation}
where $\delta\in\ZZ_2^2$ is an even spin structure, $\Xi_6[\delta]$
is a modular form of weight $6$ for the subgroup
$\Gamma(2)\subset\Sp(4,\ZZ)$ leaving theta characteristics invariant
and $d\mu_{Bos}$ is the (genus $2$) bosonic string measure.
Analogous procedures also led them to prove the
non-renor\-ma\-liza\-tion of the cosmological constant and of the
$n$-point functions, $n\le 3$, up to $2$-loops, as expected by
space-time supersymmetry arguments \cite{Martinec:1986wa}.
Furthermore, the $4$-point amplitude has been computed and checked
against the constraints coming from S-duality \cite{D'Hoker:2005ht}.

Direct computations of higher loop corrections to superstring
amplitudes appear, at the moment, out of reach. However, the strong
constraints coming from modular invariance and from factorization
under degeneration limits, together with the explicit $2$-loop
expressions, can lead to reliable conjectures on such corrections.
This is the point of view adopted, for example, in
\cite{Matone:2005vm}, where a general formula for the higher loop
contributions to the $4$-point function has been proposed.

In \cite{D'Hoker:2004xh,D'Hoker:2004ce}, D'Hoker and Phong
conjectured that the $3$-loop chiral superstring measure may be
expressed in the same form as \eqref{twoloopmeas} for a suitable
modular form $\Xi_6[\delta]$ of weight $6$. Such a form is required
to fulfill a series of constraints related to holomorphicity,
modular invariance and factorization; however, no form has been
found satisfying all such conditions. In \cite{Cacciatori:2008ay}
Cacciatori, Dalla Piazza and van Geemen (CDvG) showed that a series
of equivalent constraints can be solved by assuming the more general
expression
\begin{equation}\label{threeloopmeas}
d\mu[\delta]=\Xi_8[\delta]d\mu_{Bos}\ ,
\end{equation}
for the $3$-loop chiral measure, where $\Xi_8[\delta]$ is a modular
form of weight $8$ (not necessarily divisible by
$\theta[\delta](0)^4$). In \cite{DallaPiazza:2008}, it has been
proved that such constraints admit a unique solution for genus $3$,
thus ruling out the more restrictive assumption by D'Hoker and
Phong. Hereafter, we will drop the subscript in $\Xi_8[\delta]$.

The CDvG ansatz for the $3$-loop measure has been generalized by
Grushevsky \cite{Grushevsky:2008zm} to a formula defined for every
loop. In particular, it has been shown that the D'Hoker and Phong
formula for genus $2$ and the CDvG ansatz can be re-expressed in
terms of modular forms associated to isotropic spaces of theta
characteristics, and this leads to a straightforward generalization
of $\Xi$, solving the constraints for all genera. A possible problem
in Grushevsky construction concerns the existence of holomorphic
roots of modular forms appearing in the definition of $\Xi$ for
genus $g\ge 5$. Proving that such roots are well-defined seems
highly non-trivial and, up to now, it has been shown only for genus
$5$ \cite{SalvatiManni:2008qa}.

\medskip

Further consistency checks for chiral measure ans\"atze may be
provided by the non-renormalization of the cosmological constant and
of the $1$,$2$,$3$-point functions. In this respect, it is useful to
consider more closely the general form of the $g$-loop contribution
to the $n$-point function in superstring theories. Using the
notation of \cite{D'Hoker:2007ui}, the general $n$-point amplitude
can be expressed as
$$A(k_1,\ldots,k_n,\epsilon_1,\ldots,\epsilon_n)=\int\prod
dp^\mu_j\int_{\M_g}\int_{C^n}\Bigl|\sum_{\delta\text{
even}}\B[\delta](k_i,\epsilon_i,z_i,p^\mu_j)\Bigr|^2\ ,
$$
where $k_i,\epsilon_i$, $i=1,\ldots,n$, are the space-time momenta
and polarizations of the external states, $p_j^\mu$, $j=1,\ldots,g$,
are the internal loop momenta, $\M_g$ is the moduli space of Riemann
surfaces of genus $g$, $C$ is the Riemann surface at the
corresponding point of $\M_g$ and $z_i$, $i=1,\ldots,n$ are the
insertion position of the vertex operators on $C$. The sum is over
all the even spin structures $\delta$ and
 $\B[\delta](k_i,\epsilon_i,z_i,p^\mu_j)$ are the so-called chiral amplitudes.
In principle, upon applying a consistent gauge fixing procedure, the
chiral amplitudes $\B[\delta]$ can be obtained by computing suitable
CFT correlators of products of vertex operators, supercurrents and
stress-energy tensor operators \cite{D'Hoker:2005jc}. It is often
useful to split the chiral amplitudes into a sum
$$\B[\delta]:=\B^c[\delta]+\B^d[\delta]\ .$$ Here, the so-called
disconnected part $\B^d[\delta]$ includes all the terms where the
Wick contractions between the vertex operators are disconnected from
the contractions of the supercurrents and of the stress-energy
tensor; the connected part $\B^c[\delta]$ includes all the other
terms \cite{D'Hoker:2005jc}.

It is worth noticing that $\B^c[\delta]$ is a sum of a huge number
of terms and that, in general, each of these terms strongly depends
on the details of the gauge fixing; only the whole sum is gauge
independent. For the $2$-loop $n$-point functions with $n\le 3$
(but, for example, not for $n=4$), all the different contributions
to $\sum_\delta\B^c[\delta]$ simply cancel each other, just giving
$\sum_\delta\B^c[\delta]=0$. It follows that the non-renormalization
theorems at $2$-loops are implemented by $\sum_\delta\B^c[\delta]$
and $\sum_\delta\B^d[\delta]$ vanishing separately. For arbitrary
genus $g>2$, a general consistent gauge fixing procedure is not
known (even though several steps of the genus $2$ derivation
generalize to genus $3$). As a consequence, the precise mechanism
for the cancelation of the gauge fixing ambiguities among such terms
is not understood in full detail. It is reasonable, therefore, to
assume that, in analogy with the genus $2$ case, all the ambiguous
terms exactly cancel giving
\begin{equation}\label{connvanish}\sum_{\delta\text{ even}}\B^c[\delta]=0\ ,\end{equation}
for $n\le 3$ and for arbitrary genus.

Under the assumption \eqref{connvanish}, the non-renormalization
theorems are equivalent to the vanishing of the disconnected part
$\B^d[\delta]$ of the chiral amplitude, which can be easily
expressed in terms of the chiral superstring measure. In fact, given
the ansatz \eqref{threeloopmeas}, it can be proved that the
vanishing of the $n$-point function is equivalent to
\begin{equation}\label{Npoint}A_n(z_1,\ldots,z_n):=
\sum_{\delta\text{ even}} \Xi[\delta]\A[\delta](z_1,\ldots,z_n)=0\ ,
\end{equation}
where $\A[\delta](z_1,\ldots,z_n)$ can be derived from the Wick
contractions of the $n$ vertex operators, inserted at the points
$z_1,\ldots,z_n$ of the Riemann surface. More precisely, the
$0$-point function is obtained by setting $\A[\delta]=1$, so that
the non-renormalization theorem for the cosmological constant is
just equivalent to
\begin{equation}\label{zeropvanish}A_0\equiv\sum_{\delta\text{ even}}\Xi[\delta]=0\ .
\end{equation} Notice that \eqref{connvanish}
trivially holds for $n=0$; the vanishing of the cosmological
constant is, therefore, the most reliable check of the correctness
of the chiral superstring measure ans\"atze. The $1$-point function
vanishes automatically under the assumption \eqref{connvanish},
whereas the non-renormalization of the $2$- and $3$-point amplitudes
corresponds to
\begin{align}
A_2(a,b)&\equiv \sum_{\delta\text{ even}}\Xi[\delta]S_\delta(a,b)^2=0\ ,\label{twopvanish}\\
A_3(a,b,c)&\equiv \sum_{\delta\text{
even}}\Xi[\delta]S_\delta(a,b)S_\delta(b,c)S_\delta(c,a)=0\label{threepvanish}\
,
\end{align}
respectively, where $a,b,c$ are arbitrary points of the genus $g$
Riemann surface $C$ and $S_\delta(a,b)$ is the Szeg\"o kernel. The
identity \eqref{zeropvanish} has been proved for the CDvG-Grushevsky
(CDvG-G) ansatz at genus $3$ \cite{Cacciatori:2008ay} and $4$
\cite{Grushevsky:2008zm}. Remarkably, for genus $4$, $A_0$
corresponds to a non-zero Siegel modular form of weight $8$ (the
Schottky form), which vanishes only on the locus of Jacobians of
Riemann surfaces. A strong argument for the identities
\eqref{twopvanish} and \eqref{threepvanish} to hold on the
hyperelliptic locus for any genus has been given by Morozov in
\cite{Morozov:2008xd}, whereas in \cite{Grushevsky:2008qp}
Grushevsky and Salvati Manni proved \eqref{twopvanish} for genus
$3$.

\medskip

 In this paper, we will prove that
\eqref{threepvanish} does not hold for any non-hyperelliptic Riemann
surface of genus $3$. More precisely, one of the main results of the
paper is the following theorem
\medskip
\begin{theorem}\label{th:main} Let $C$ be a Riemann surface of genus
three. Then, $A_3(a,b,c)=0$ for all $a,b,c\in C$ if and only if $C$
is hyperelliptic.\end{theorem}
\medskip
In particular, we will prove that the remarkable identity
\begin{equation}\label{A3formula}A_3(p_1,p_2,p_3)d\mu_{Bos}=\frac{c}{c_3}\det\omega_i(p_j)\prod_{i\le
j}d\tau_{ij}\ ,
\end{equation}
where $c_3:=2^6\pi^{18}$ and $c\in\CC$ is a non-vanishing constant,
holds for genus $3$, so giving a close expression for $A_3$.
Moreover, we discuss the possibility of a further non-vanishing
contribution to the three-point function coming from the connected
part of the chiral amplitude and give some arguments suggesting that
the natural candidate for such a term should exhibit the same
structure of the right-hand side of \eqref{A3formula}. In this
respect, \eqref{A3formula} can be interpreted as the statement of
the non-renormalization of the three-point function at three-loop.

The paper is organized as follows. In section \ref{s:mathback} we
introduce the mathematical background on Riemann surfaces and theta
functions needed for the later construction. In section
\ref{s:vanish} we give a necessary and sufficient condition for
$A_3(a,b,c)$ to vanish identically and we find that such a condition
is trivially satisfied for $g=2$, whereas, for genus $3$, it is
fulfilled only for hyperelliptic surfaces. In section
\ref{s:goodform}, a strikingly simple formula for $A(a,b,c)$ is
provided for genus $3$, which does not include any summation over
spin characteristics; equation \eqref{A3formula} is an immediate
consequence of such a formula. In section \ref{s:discussion}, we
discuss how the non-renormalization theorems could be implemented in
the framework of the CDvG-G ans\"atze for the chiral superstring
measures in view of our results.

\section{Theta functions and Riemann surfaces}\label{s:mathback}

In this section, we provide the basic background on theta functions
and Riemann surfaces necessary for the subsequent derivations. We
refer to \cite{Mumford:1983,Fay:1973,Farkas:1992} for proofs and
further details.

\medskip

Let $\Sieg_g$ denote the Siegel upper half-space, i.e. the space of
$g\times g$ complex symmetric matrices with positive definite
imaginary part
$$\Sieg_g:=\{\tau\in M_{g\times g}(\CC)\mid \tp\tau=\tau\,
,\im\tau>0 \}\ .
$$
Let $\Sp(2g,\ZZ)$ be the symplectic modular group, i.e. the group of
$2g\times 2g$ complex matrices $M:=\bigl(\begin{smallmatrix}A &B\\
C & D\end{smallmatrix}\bigr)$, where $A,B,C,D$ are $g\times g$
blocks satisfying
$$\tp A C=\tp C A\ ,\quad \tp BD=\tp D B\ ,\quad \tp
D A-\tp B C=\II_g\ .
$$
Let us define the action of $\Sp(2g,\ZZ)$ on $\CC^g\times\Sieg_g$ by
\begin{equation}\label{modull}
(M\cdot z,M\cdot\tau):=\bigl(\tp(C\tau+D)^{-1}z,(A\tau+B)(C\tau+D)^{-1}\bigl)\ ,\end{equation} where $M\equiv\bigl(\begin{smallmatrix}A &B\\
C & D\end{smallmatrix}\bigr)\in\Sp(2g,\ZZ)$ and
$(z,\tau)\in\CC^g\times\Sieg_g$.

\medskip

Let $\ZZ_2:=\ZZ/(2\ZZ)$ be the additive group with elements
$\{0,1\}$. For each $\delta',\delta'' \in\ZZ_2^{g}$, the theta
function $\theta[\delta]\equiv\theta\spin{\delta'}{\delta''}\colon
\CC^g\times\Sieg_g\to \CC$ with characteristics $[\delta]
\equiv\spin{\delta'}{\delta''}$
 is defined by
$$\theta[\delta](z,\tau):=\sum_{k\in\ZZ^g}
\exp{\pi
i\Bigl[\tp\Bigl(k+\frac{\delta'}{2}\Bigr)\tau\Bigl(k+\frac{\delta'}{2}\Bigr)+
2\tp\Bigl(k+\frac{\delta'}{2}\Bigr)\Bigl(z+\frac{\delta''}{2}
\Bigr)\Bigr]\ ,}
$$
where $(z,\tau)\in\CC^g\times\Sieg_g$. For each fixed $\tau$,
$\theta[\delta](z,\tau)$ is an even or odd function on $\CC^g$
depending whether $(-1)^{\delta'\cdot\delta''}$ is $+1$ or $-1$,
respectively. Correspondingly, there are $2^{g-1}(2^g+1)$ even and
$2^{g-1}(2^g-1)$ odd theta characteristics. Under translations
$z\mapsto z+\lambda$, $z\in\CC^g$, $\lambda
\in\ZZ^g+\tau\ZZ^g\subset\CC^g$, theta functions get multiplied by a
nowhere vanishing factor
$$\theta \spin{\delta'}{\delta''}\left(z+n+\tau m,\tau\right)=
e^{-\pi i \tp{m}\tau m-2\pi i\tp{m}z+\pi
i(\tp{\delta'}n-\tp{\delta''}m)}\theta\spin{\delta'}{\delta''}\left(z,\tau\right)
\ ,
$$
$m,n\in\ZZ^g$. It follows that, for any fixed $\tau$, the theta
functions can be seen as sections of line bundles on the complex
torus $A_\tau:=\CC^g/(\ZZ^g+\tau\ZZ^g)$, with a well defined divisor
on $A_\tau$. We denote by $\Theta$ the divisor of
$\theta(z)\equiv\theta[0](z,\tau)\equiv\theta\spin{0}{0}(z,\tau)$
and by $\Sing\Theta$ its singular locus, i.e. the locus of points at
which $\theta(z)$ and all its first partial derivatives vanish.

The second order theta functions are defined by
$$\Theta[\epsilon](z,\tau):=\theta\spin{\epsilon }{0}(2z,2\tau)\ ,
$$
for all $\epsilon\in\ZZ^g$. They are a basis for
$H^0(A_\tau,\OO(2\Theta))$ and are related to the first order theta
functions by the Riemann bilinear identities
\begin{equation}\label{RiemBil}\theta\spin{\epsilon}{\delta
}(z_1+z_2,\tau)\,\theta\spin{\epsilon}{\delta
}(z_1-z_2,\tau)=\sum_{\sigma\in\ZZ^{g}}(-1)^{\delta\cdot\sigma}\Theta[\sigma](z_1,\tau)\,\Theta[\sigma+\epsilon
](z_2,\tau)\ ,
\end{equation}
for all $z_1,z_2\in\CC^g$, $\epsilon ,\delta \in\ZZ_2^g$.

\medskip

Let us define the action of $\Sp(2g,\ZZ)$ on $\ZZ_2^{2g}$ by
 \begin{equation}
\label{modulcharac}M\cdot\delta\equiv M\cdot
\Biggl(\begin{matrix}\delta' \\
\delta''\end{matrix}\Biggr):=\Biggl(\begin{matrix}D& -C\\ -B
&A\end{matrix}\Biggr)\Biggl(\begin{matrix}\delta' \\
\delta''\end{matrix}\Biggr)+\Biggl(\begin{matrix}\diag (C\tp
D)\\\diag (A\tp B)\end{matrix}\Biggr)\mod 2\ .
\end{equation}
Theta characteristics are invariant under the action of the subgroup
$\Gamma(2)\subset\Sp(2g,\ZZ)$, where
$$\Gamma(n):=\{M\in\Sp(2g,\ZZ)\mid M=\II_{2g}\mod n\}\ ,
$$ is the subgroup of elements of $\Sp(2g,\ZZ)$ congruent to the $2g\times 2g$
identity matrix mod $n$. The action of $\Sp(2g,\ZZ)$ on $\ZZ_2^{2g}$
factorizes through the action of
$\Sp(2g,\ZZ_2)\equiv\Sp(2g,\ZZ)/\Gamma(2)$. Symplectic
transformations preserve the parity of the characteristics and, for
any two $\delta,\epsilon\in\ZZ_2^{2g}$ of the same parity, there
exists $M\in\Sp(2g,\ZZ_2)$ such that $\epsilon=M\cdot\delta$.

A (Siegel) modular form $f$ of weight $k\in\ZZ$ for a subgroup
$\Gamma\in\Sp(2g,\ZZ)$ is a holomorphic function on $\Sieg_g$ such
that
$$f(M\cdot \tau)=\det(C\tau+D)^kf(\tau)\ ,
$$
for all $M\in\Gamma$. A condition of regularity is also required for
$g=1$, but it is automatically satisfied for $g>1$.

\medskip
 The modular transformation of the theta function is given
by
\begin{equation}\label{modultheta}
\theta[M\cdot\delta](M\cdot z,M\cdot \tau)=\kappa(M)\det
(C\tau+D)^\frac{1}{2} e^{\pi i[\phi[\delta](M)+\tp
z(C\tau+D)^{-1}Cz]}\theta[\delta](z,\tau)\ ,\end{equation} where
$\kappa(M)$ is an eighth root of $1$ depending on $M$ and
$$4\phi\spin{\delta'}{\delta''}(M):=(\tp \delta' \ \tp \delta'')\Biggl(\begin{matrix} -\tp BD & \tp BC\\ \tp BC & -\tp
AC\end{matrix}\Biggr)\Biggl(\begin{matrix}\delta'\\ \delta''
\end{matrix}\Biggr)+2\diag(A\tp B)\cdot (D\delta'-C\delta'')\ .$$
Powers of theta constants
$\theta[\delta](\tau)\equiv\theta[\delta](0,\tau)$ are the basic
building blocks for modular forms, at least for low genera.

\medskip

Let $C$ be a Riemann surface of genus $g>1$. The choice of a marking
for $C$ provides a set of generators
$\{\alpha_1,\ldots,\alpha_g,\beta_1,\ldots,\beta_g\}$ for the first
homology group $H_1(C,\ZZ)$ on $C$, with symplectic intersection
matrix, that is
\begin{equation}\label{sympl}\alpha_i\cdot\alpha_j=0=\beta_i\cdot\beta_j\ ,\qquad
\alpha_i\cdot\beta_j=\delta_{ij}\ ,
\end{equation}
for all $i,j=1,\ldots,g$. The choice of such generators canonically
determines a basis $\{\omega_1,\ldots,\omega_g\}$ for the space
$H^0(K_C)$ of holomorphic $1$-differentials on $C$, with normalized
$\alpha$-periods
$$\oint_{\alpha_i}\omega_j=\delta_{ij}\ ,$$
for all $i,j=1,\ldots,g$. The $\beta$-periods define the Riemann
period matrix of the Riemann surface $C$
$$\tau_{ij}\equiv\oint_{\beta_i}\omega_j\ ,
$$
which is symmetric and with positive-definite imaginary part, so
that $\tau\in\Sieg_g$. By Torelli's theorem, the complex structure
of $C$ is completely determined by giving its Riemann period matrix.

By the conditions \eqref{sympl}, a general change of marking of $C$
corresponds to a symplectic transformation on the set of generators
of $H_1(C,\ZZ)$
\begin{equation}\label{modulhomol}
\Biggl(\begin{matrix}\alpha\\
\beta\end{matrix}\Biggr)\mapsto

\Biggl(\begin{matrix}\tilde\alpha\\
\tilde\beta\end{matrix}\Biggr):=\Biggl(\begin{matrix}D & C \\ B &
A\end{matrix}\Biggr) \Biggl(\begin{matrix}\alpha\\
\beta\end{matrix}\Biggr)\ ,\qquad\qquad
M\equiv\Biggl(\begin{matrix}A & B
\\ C & D\end{matrix}\Biggr)\in \Sp(2g,\ZZ) \ , \end{equation} under which
\begin{equation}\label{modulbasis}
(\omega_1,\ldots,\omega_g)\mapsto(\tilde\omega_1,\ldots,\tilde\omega_g):=(\omega_1,\ldots,\omega_g)(C\tau+D)^{-1}\
,
\end{equation}
whereas $\tau\mapsto\tilde\tau:=M\cdot\tau$ transforms as in
\eqref{modull}.

\medskip

 The
complex torus $J_C:=\CC^g/(\ZZ^g+\tau\ZZ^g)$ associated to the
Riemann period matrix of $C$ is called the Jacobian torus of $C$.
For a fixed base-point $p_0\in C$, let $I\colon C\to J_C$ denote the
Abel-Jacobi map, defined by
$$p\mapsto
I(p):=\tp\Bigl(\int_{p_0}^p\omega_1,\ldots,\int_{p_0}^p\omega_g\Bigr)\in
J_C\ .
$$
Note that different choices of the path of integration from $p_0$ to
$p$ correspond, by the formula above, to points in $\CC^g$ differing
by elements in the lattice $\ZZ^g+\tau\ZZ^g$, so that $I$ is
well-defined only on $\CC^g/(\ZZ^g+\tau\ZZ^g)$. The Abel-Jacobi map
extends to a map from the Abelian group of divisors on $C$ to $J_C$
by
$$I\bigl(\smsum\nolimits_i p_i-\smsum\nolimits_i q_i\bigl):=\smsum\nolimits_i I(p_i)-\smsum\nolimits_i I(q_i)\ .$$
Such a map is independent of the base point $p_0$ when restricted to
zero degree divisors. In the following, when no confusion is
possible, we will identify such zero degree divisors with their
image in $J_C$ through $I$. In particular, we will omit $I$ when
considering the theta functions on the Jacobian evaluated at (the
image of) some zero degree divisor on $C$. Furthermore, the argument
$\tau$ for theta functions associated to a marked Riemann surface
will be understood.

\medskip

Given a Riemann surface $C$ with marking, one can canonically
associate to each theta characteristic $\delta\in\ZZ_2^{2g}$ a spin
structure, correspondent to a line bundle $L_\delta$ on $C$ such
that $L_\delta^2\cong K_C$, with $K_C$ the canonical line bundle on
$C$. Such a correspondence can be defined as follows. Let $\delta$
be a non-singular theta characteristic (that is, such that at least
one among $\theta[\delta](z)$ and its first partial derivatives does
not vanish at $z=0$) and, for an arbitrary $y\in C$, set
$f_{\delta,y}(x):=\theta[\delta](x-y)$. By the Riemann vanishing
theorem \cite{Farkas:1992}, the divisor $2(f_{\delta,y})-2y$ is
linearly equivalent to the canonical divisor, so that
$(f_{\delta,y})-y$ defines the divisor class of a spin bundle, that
we denote by $L_\delta$. It can be proved that such a divisor class
$[(f_{\delta,y})-y]$, and thus also $L_\delta$, is independent of
$y\in C$, so that for each marked Riemann surface we have a
correspondence $\delta\mapsto L_\delta$. By \eqref{modultheta},
under a change of marking \eqref{modulhomol}, such a correspondence
transforms into the map $\delta\mapsto \tilde L_\delta$, where
\begin{equation}\label{modulspin}\tilde L_{M\cdot\,\delta}=L_{\delta}\
.\end{equation}

Fix a non-singular odd spin structure $\nu\in\ZZ_2^{2g}$ and
consider the holomorphic $1$-differential
$$\sum_{i=1}^g\left.\frac{\partial\theta[\nu](z)}{\partial
z_i}\right\rvert_{z=0}\omega_i\ .
$$
It can be proved that such a $1$-differential has $g-1$ double
zeroes and corresponds to the square $h_\nu^2$ of a holomorphic
section of the line bundle $L_\nu$.

Let us define the prime form by
$$E(a,b):=\frac{\theta[\nu](b-a)}{h_\nu(a)h_\nu(b)}\ ,
$$
$a,b\in C$, for an arbitrary non-singular odd spin-structure $\nu$.
The prime form is a section of a line bundle on $C\times C$, it is
antisymmetric in its arguments and vanishes only on the diagonal
$a=b$. Furthermore, it does not depend on the choice of $\nu$.

For each non-singular even characteristic $\delta\in\ZZ^{2g}$, the
Szeg\"o kernel is defined by
$$S_\delta(a,b)\equiv S(a,b;L_\delta):=\frac{\theta[\delta](a-b)}{\theta[\delta](0)\,E(a,b)}\
.
$$ For each fixed $b\in C$, $S_\delta(a,b)$ is the unique meromorphic
section of $L_\delta$ with a single pole of residue $-1$ at $b$ and
holomorphic elsewhere. Such a characterization implies that, for a
fixed spin bundle $L$, $S(a,b;L)$ is independent of the marking. It
follows that, under a change of marking corresponding to
\eqref{modulhomol}, by \eqref{modulspin} we have
\begin{equation}\label{modulSzego}
\tilde S_{M\cdot\,\delta}(a,b)\equiv S(a,b;\tilde
L_{M\cdot\,\delta})= S(a,b;L_{\delta}) =S_{\delta}(a,b)\
,\end{equation} with $M\cdot\delta$ given by \eqref{modulcharac}.

Finally, we denote by $$\omega_{a-b}(x):=\frac{\partial}{\partial
x}\log\frac{E(x,a)}{E(x,b)}\ ,$$ $a,b,x\in C$, the Abelian
$1$-differential of the second kind with single poles on $a$ and $b$
with residue $+1$ and $-1$, respectively, holomorphic on
$C\setminus\{a,b\}$, and with vanishing $\alpha$-periods.

\section{Proof of theorem \ref{th:main}}\label{s:vanish}

Let $\Xi[\delta](\tau)$ be the modular form of weight $8$ for
$\Gamma(2)\subset\Sp(2g,\ZZ)$ defined in \cite{Cacciatori:2008ay}
for genus $3$ and in \cite{Grushevsky:2008zm} for arbitrary genus.
It satisfies the property
\begin{equation}\label{modulXi}\Xi[M\cdot\delta](M\cdot\tau)=\det
(C\tau+D)^8\,\Xi[\delta](\tau)\ ,
\end{equation}
for an arbitrary $M\in\Sp(2g,\ZZ)$.

In terms of $\Xi[\delta]$ and the Szeg\"o kernel, one can define the
sections $A_2(a,b)$ on $C\times C$ and $A_3(a,b,c)$ on $C\times
C\times C$ by \eqref{twopvanish} and \eqref{threepvanish},
respectively. In \cite{Grushevsky:2008qp}, it has been proved that
$A_2(a,b)=0$ for all $a,b\in C$, where $C$ is an arbitrary Riemann
surface of genus $2$ or $3$. It is useful to recall the main points
of such a derivation.

Let $C$ be a marked Riemann surface of genus $g>1$ and let $\tau$ be
its Riemann period matrix. Define the function $X\colon\CC^g\to\CC$
by
\begin{equation}\label{Xsect}X(z):=\sum_{\delta\text{ even}}\Xi[\delta]\frac{\theta[\delta
](z,\tau)^2}{\theta[\delta ](0,\tau)^2}\ ,
\end{equation}
$z\in \CC^g$, corresponding to a section of $|2\Theta|$ on the
Jacobian $J_C$. The restriction of such a section to
$$C-C:=\{(a-b)\in J_C, a,b\in C\}\subset J_C\ ,
$$
is related to $A_2(a,b)$ by
$$A_2(a,b)=\frac{X(a-b)}{E(a,b)^2}\ .
$$
Thus, $A_2(a,b)=0$ for all $a,b\in C$ if and only if the restriction
of $X$ to $C-C$ vanishes identically. On the other hand, as
conjectured in \cite{VanGeemen:1986} and proved in
\cite{Welters:1986}, the space of sections of $|2\Theta|$ vanishing
on $C-C$ is
$$\Gamma_{00}:=\{f\in H^0(J_C,\OO(2\Theta))\mid
f(0)=0=\partial_i\partial_j f(0),\ i,j=1,\ldots,g\}\ ,
$$
so that the following theorem follows.
\medskip
\begin{theorem}[Grushevsky, Salvati Manni \cite{Grushevsky:2008qp}]\label{th:GrSalMan} The function $A_2(a,b)=0$ for all $a,b\in C$
if and only if $X\in\Gamma_{00}$.
\end{theorem}
\medskip
For genus $g>1$, the dimension of $\Gamma_{00}$ is $2^g-1-g(g+1)/2$;
in particular, $\dim\Gamma_{00}=0$ for $g=2$ and $\dim\Gamma_{00}=1$
for $g=3$. By applying the Riemann bilinear relations, $X(z)$ can be
expressed in terms of second order theta functions. In
\cite{Grushevsky:2008qp}, such an expression is used to show that
the function $X(z)$ vanishes identically on $\CC^g$ for $g=2$,
 whereas for $g=3$ $X(z)$ is a generator of
$\Gamma_{00}$. By theorem \ref{th:GrSalMan}, this proves that
$A_2(a,b)\equiv 0$ for $g=2,3$.

\medskip

Let us show that, for an arbitrary marked Riemann surface $C$ of
genus $g>1$, also $A_3(a,b,c)$ can be expressed in terms of $X(z)$.
The starting point is the following identity
\begin{equation}\label{FaySzego}\frac{S_\delta(c,a)S_\delta(b,c)}{S_\delta(a,b)}=\omega_{a-b}(c)+\sum_{i=1}^g
\left.\frac{\partial\log\theta[\delta](z)}{\partial z_i}\right\vert
_{z=a-b}\,\omega_i(c)\ , \end{equation} $a,b,c\in C$, which is an
immediate consequence of \cite{Fay:1973}, formula 38 page 25. It is
instructive to derive such an identity from the famous Fay's
trisecant identity \cite{Fay:1973}, written in the form
\begin{equation}\label{Fayid}\frac{\theta[\delta](a+c-b-d)E(a,c)E(b,d)}
{\theta[\delta](0)E(a,b)E(a,d)E(c,b)E(c,d)}=S_\delta(a,d)S_\delta(c,b)-
S_\delta(a,b)S_\delta(c,d) \ ,
\end{equation}
which holds for arbitrary $a,b,c,d\in C$ and for each non-singular
even spin structure $\delta$. By comparing the Laurent expansion of
both sides of \eqref{Fayid} in the limit $d\to c$, with respect to
some local coordinate centered in $c$, we obtain an infinite tower
of (possibly trivial) identities, one for each order in $(d-c)$.
Using $E(c,d)^{-1}=(d-c)^{-1}(1+O(d-c)^2)$, it is easy to check that
the first non-trivial identity is obtained at $O(1)$
$$-\frac{\theta[\delta](a-b)}{\theta[\delta](0)E(a,b)}\left.\frac{d}{d
x}
\Bigl(\log\theta[\delta](a+c-b-x)+\log\frac{E(b,x)}{E(a,x)}\Bigr)\right|_{x=c}\!\!\!=S_\delta(a,c)S_\delta(c,b)\
,
$$
and \eqref{FaySzego} follows immediately. (Note that second term on
the RHS of \eqref{Fayid} is odd in
$(d-c)$, so that it does not contribute to $O(1)$.)

By multiplying both sides of \eqref{FaySzego} by
$\Xi[\delta]S_{\delta}(a,b)^2$ and summing over all the even spin
structures, we obtain
\begin{multline*}\sum_{\delta\text{ even}}\Xi[\delta]S_\delta(c,a)S_\delta(b,c)S_\delta(a,b)=
\omega_{a-b}(c)\sum_{\delta\text{ even}}\Xi[\delta]S_\delta(a,b)^2\\
+ \frac{1}{E(a,b)^2}\sum_{\delta\text{ even}}\frac{\Xi[\delta]}
{\theta[\delta](0)^2}\theta[\delta](a-b)
\sum_{i=1}^g\left.\frac{\partial\theta[\delta](z)}{\partial
z_i}\right\vert_{z=a-b}\omega_i(c)\ .
\end{multline*}
By \eqref{Xsect}, such an identity can be written as
\begin{equation}\label{threepoint}A_3(a,b,c)=
\frac{1}{E(a,b)^2}\Bigl[\omega_{a-b}(c)X(a-b)+
\frac{1}{2}\sum_{i=1}^g\left.\frac{\partial X(z)}{\partial
z_i}\right\vert_{z=a-b}\omega_i(c)\Bigr]\ .\end{equation} Since
$\omega_{a-b},\omega_1,\ldots,\omega_g$ are linearly independent,
the condition that $A_3(a,b,c)=0$ for all $a,b,c\in C$, is
equivalent to the $g+1$ conditions
\begin{align*}&X(a-b)=0\ ,\\ &\left.\frac{\partial
X(z)}{\partial z_i}\right\vert_{z=a-b}=0\ ,\qquad\qquad
i=1,\ldots,g\ ,
\end{align*} to hold for all $a,b\in C$. Note that the first
condition is equivalent to $A_2(a,b)=0$. It is natural to define the
following subspace of $\Gamma_{00}$
$$\Gamma^{(2)}_{00}:=\{f\in \Gamma_{00}\mid \mult_{a-b}(f)\ge 2,\ \forall a,b\in C\}\
.
$$
We have proved the following theorem.
\medskip
\begin{theorem}\label{th:threevanish} For an arbitrary Riemann surface $C$ of genus $g>1$,
$A_3(a,b,c)=0$ for all $a,b,c\in C$  if and only if
$X\in\Gamma^{(2)}_{00}$.
\end{theorem}
\medskip
The space $\Gamma_{00}^{(2)}$ and other remarkable subspaces of
$\Gamma_{00}$ have been extensively studied in the last few years,
in particular because of their relationship with the geometry of the
moduli space $\mathcal{S}U_C(2,K)$ of semi-stable bundles of rank 2
with fixed canonical determinant on a smooth projective curve $C$
\cite{Oxbury:1998,Pauly:2001,Izadi:2001}. There is a simple
procedure to construct elements of $\Gamma_{00}^{(2)}$. Let
$e\in\Sing\Theta$ be a point of the singular locus of the theta
function, i.e. such that $\theta(z)$ and all its first derivatives
vanish at $z=e$; note that, by the parity of the theta function,
also $-e\in\Sing\Theta$. By the Riemann singularity theorem
\cite{Farkas:1992}, $\theta(a-b+e)=0$ for all $a,b\in C$. It follows
immediately that
$$F_e(z):=\theta(z+e)\theta(z-e)=\sum_{\sigma\in\ZZ_2^g}\Theta[\sigma](e)\,\Theta[\sigma](z)\ ,
$$
is an element of $\Gamma^{(2)}_{00}$. In \cite{Pauly:2001} (Theorem
1.1), it has been proved that, for an arbitrary non-hyperelliptic
Riemann surface $C$, $\Gamma^{(2)}_{00}$ is generated by the
sections $F_e(z)$ as $e$ varies in $\Sing\Theta$, that is
$\Gamma^{(2)}_{00}=\langle F_e\rangle_{e\in \Sing\Theta}$.

\medskip

Let us consider the consequences of such results for low genera. For
genus $2$, the identity $X\equiv 0$ proved in
\cite{Grushevsky:2008qp}, together with theorem
\ref{th:threevanish}, implies that $A_3(a,b,c)\equiv0$, thus
reobtaining the result of \cite{D'Hoker:2005jc}.

\medskip
 For genus $3$, in the
non-hyperelliptic case, $\Sing\Theta$ is empty, so that
$\dim\Gamma_{00}^{(2)}=0$. On the other hand, in
\cite{Grushevsky:2008qp} it has been proved that, in this case,
$X\neq 0$, so that we conclude that $A_3(a,b,c)$ does not vanish
identically on $C\times C\times C$.

The hyperelliptic curves are not considered in \cite{Pauly:2001}. In
this case, one has just the weaker inclusion $\langle
F_e\rangle_{e\in \Sing\Theta}\subseteq \Gamma^{(2)}_{00}$. On the
other hand, if $C$ is hyperelliptic of genus $3$,
$\Sing\Theta\subset J_C$ consists of a unique point of order $2$,
corresponding to a singular even spin-structure $\delta_{sing}$. It
follows that, in this case, $\Gamma_{00}^{(2)}$ has at least one
non-trivial section, namely
$$F_{\delta_{sing}}(z):=\theta[\delta_{sing}](z)^2\ .
$$
Since $\Gamma_{00}$ is $1$-dimensional and contains
$\Gamma_{00}^{(2)}$ as a subspace, we get
$\Gamma_{00}=\Gamma_{00}^{(2)}$ so that $X\in\Gamma_{00}^{(2)}$ and
$A_3(a,b,c)=0$ for all $a,b,c\in C$, as suggested by the arguments
in \cite{Morozov:2008xd}. This concludes the proof of theorem
\ref{th:main}.

\medskip
More generally, for genus $g\ge 3$, the space $\langle
F_e\rangle_{e\in \Sing\Theta}$ has dimension
$2^g-\sum_{i=0}^3\binom{g}{i}$, which, for non-hyperelliptic
surfaces, corresponds to the dimension of $\Gamma_{00}^{(2)}$ . In
particular, for a non-hyperelliptic surface of genus $4$,
$\dim\Gamma_{00}^{(2)}=1$. In this case, $\Sing\Theta$ has only two
(possibly coincident) points $\pm e$ and the generator of
$\Gamma_{00}^{(2)}$ is $\theta(z+e)\theta(z-e)$. It would be
interesting (but probably highly non-trivial) to check whether $X$
is proportional to such a section in this case.

\section{A simple expression for $A_3(a,b,c)$}\label{s:goodform}

Let us show that $A_3(a,b,c)$ admits the  alternative expression
\eqref{A3formula} that will be useful in the following.

First of all, note that, if $A_2(a,b)$ vanishes identically on a
Riemann surface $C$ of genus $g>1$, then $A_3(a,b,c)$ is a
holomorphic $1$-differential in each variable. Furthermore, it is
anti-symmetric under permutation of such variables and, in
particular, it must vanish on the diagonals of $C\times C\times C$.
For $g=3$, this is enough to conclude that
$$A_3(p_1,p_2,p_3)\equiv\sum_{\delta\text{ even}}\Xi[\delta]S_\delta(p_1,p_2)S_\delta(p_2,p_3)S_\delta(p_3,p_1)
=f\det\omega_i(p_j)\ ,
$$
$p_1,p_2,p_3\in C$, for some holomorphic function $f$, independent
of $p_1,p_2,p_3$, on the Teichm\"uller space of genus $3$ Riemann
surfaces with marking. Under a change of marking, corresponding to a
transformation \eqref{modulhomol} for some $M\equiv\bigl(\begin{smallmatrix}A & B\\
C & D\end{smallmatrix}\bigr)\in\Sp(2g,\ZZ)$, by
\eqref{modulbasis}\eqref{modulSzego}\eqref{modulXi} we get
\begin{align*}
A_3(p_1,p_2,p_3)&\mapsto\det(C\tau+D)^8A_3(p_1,p_2,p_3)\ ,\\
\det \omega_i(p_j)&\mapsto\det(C\tau+D)^{-1}\det\omega_i(p_j)\
.\end{align*} It follows that $f$ transforms as
$$f\mapsto\det(C\tau+D)^9f\ ,
$$
and thus it corresponds to a Teichm\"uller modular form of weight
$9$ and degree $3$ \cite{Ichikawa:1995}. In general, a Teichm\"uller
modular form of weight $d\ge 0$ and degree $g$ is defined as a
holomorphic section of $\lambda_1^{\otimes d}$ on the moduli space
$\M_g$ of genus $g$ Riemann surfaces; here, $\lambda_1$ is the line
bundle whose fiber at the point corresponding to the surface $C$ is
$\wedge^g H^0(K_C)$. By \cite{Ichikawa:1995}, there is a unique (up
to a constant), holomorphic section $\mu_{3,9}$ of
$\lambda_1^{\otimes 9}$ on $\M_3$, so that
$$f=c\,\mu_{3,9}\ ,
$$
for some $c\in\CC$; theorem \ref{th:main} implies that $c\neq 0$.
The section $\mu_{3,9}$ vanishes only on the hyperelliptic locus,
consistently with theorem \ref{th:main}, and its square
$(\mu_{3,9})^2$ corresponds to the (Siegel) modular form
$$\Psi_{18}(\tau):=\prod_{\delta\text{ even}}\theta[\delta](0,\tau)\
,
$$
of weight $18$. (More precisely, $(\mu_{3,9})^2$ is the image of
$\Psi_{18}$ under the homomorphism, induced by the Torelli map,
mapping Siegel modular forms to Teichm\"uller modular forms
\cite{Ichikawa:1995}; for such a reason, $\mu_{3,9}$ itself is often
identified with the parabolic form $\Psi_9$, that is a holomorphic
square root of $\Psi_{18}$).

Furthermore, $\mu_{3,9}$ also appears in the explicit formula for
the chiral bosonic string measure for genus $3$
\cite{Belavin:1986tv}
$$d\mu_{Bos}=\frac{1}{c_3}\frac{\prod_{i\le j}d\tau_{ij}}{\mu_{3,9}}\ ,
$$
with $c_3=2^6\pi^{18}$ \cite{D'Hoker:2004ce}, and the identity
\eqref{A3formula} follows.

It would be interesting to compute exactly the constant $c$. This
may be done, for example, by using the factorization properties of
string amplitudes under degeneration limits.

\section{Non-renormalization theorems and chiral measure ans\"atze}\label{s:discussion}

Theorem \ref{th:main} shows that the CDvG-G ansatz and the
assumption \eqref{connvanish} are not compatible with the
non-renormalization theorems at $3$-loops. It is necessary,
therefore, to discuss the validity of the assumptions leading to
such a result.

For genus $3$, upon assuming the form \eqref{threeloopmeas}, the
chiral superstring measure is completely determined by the
constraints related to holomorphicity, modular invariance and
factorization. Even though there are no strong first-principle
arguments suggesting that \eqref{threeloopmeas} holds for genus
higher than $2$, a posteriori there are several hints that the
ansatz should be correct, at least for low genera: the uniqueness of
the genus $3$ solution for the constraints, the existence of general
well-defined solutions at least up to genus $5$, the
non-renormalization of the cosmological constant at $3$ and $4$
loops (a result that, as noted in the introduction, is independent
of the assumption \eqref{connvanish}).

The evidence for the assumption \eqref{connvanish} is much weaker.
In fact, it does not hold, for example, for the $2$-loop
contribution to the $4$-point function \cite{D'Hoker:2005jc}.
Therefore, it is reasonable to consider the possibility of a
non-vanishing contribution from the connected part of the chiral
amplitude. Some hints on the possible structure of such a
contribution can be obtained analyzing the $4$-point function at $2$
loops. In this case, the connected part of the chiral amplitude
gives two non-vanishing terms, both corresponding to Wick
contractions between the vertex operators and the stress-energy
tensor. One of such contributions exactly cancels the disconnected
part of the amplitude. The other one has, very schematically, the
following structure
\begin{equation}\label{A4formula}A_4(p_1,p_2,p_3,p_4)d\mu_{Bos}=
\sum_{I_1\sqcup I_2=\{1,2,3,4\}}\!
K_{I_1}\,\det_{\substack{i=1,2\\j\in I_1}}\omega_i(p_j)
\det_{\substack{i=1,2\\j\in I_2}}\omega_i(p_j)\prod_{i\le
j}d\tau_{ij}\ ,
\end{equation}
where $K_{I_1}$ is a kinematical factor and the sum is over all the
possible ways of splitting the set $\{1,2,3,4\}$ into the disjoint
union of sets $I_1$ and $I_2$ of two elements. Hence, it seems
reasonable for a non-vanishing $3$-loop contribution corresponding
to Wick contractions between vertex operators and stress tensor to
exhibit a structure analogous to the right-hand side of
\eqref{A4formula}.

On the other hand, the most natural generalization of
\eqref{A4formula} to the case of a $3$-loop contribution to the
$3$-point function, satisfying the fundamental consistency
constraints (modular weight $5$ and conformal weight $1$ in each
variable), is precisely the right-hand side of \eqref{A3formula}
(times a kinematical factor). In this respect, note that the
requirements we are imposing on the possible structure of such a
contribution are very restrictive. In fact, no consistent
generalization of \eqref{A4formula} can be defined for the $2$-loop
contribution to the $3$-point function or for the $2$- and $3$-loop
contributions to the $2$-point function. The fact that, for such
amplitudes, the disconnected part vanishes separately enforces the
validity of our analysis. Remarkably, a structure similar to
\eqref{A3formula} has been proposed in \cite{Matone:2005vm} for the
higher loop contributions to the $4$-point function.

Thus, by the identity \eqref{A3formula} and by the arguments above,
it is reasonable to conjecture that the contributions from the
connected chiral amplitude could exactly cancel the disconnected
part, thus giving the expected non-renormalization theorems. It
would be very interesting to check such a conjecture by explicitly
computing some of the terms coming from the relevant Wick
contractions. Unfortunately, even though such a computation should
be considerably simpler than a complete first-principles derivation
of the amplitudes (for example, we are neglecting the Wick
contractions between supercurrents and vertex operators, together
with a huge number of subtleties already emerging at genus $2$), it
is not clear, at the moment, how it should be performed.

\bigskip

\section*{Acknowledgements}
We are grateful to Riccardo Salvati Manni for several interesting
discussions, to Emma Previato for clarifying comments about the
space $\Gamma_{00}^{(2)}$ and to Duong Phong and Samuel Grushevsky
for helpful discussions on superstring amplitudes.

config: arxiv


TITLE: How can I differentiate $f(x,z(x,y))$ w. r. to x
QUESTION [0 upvotes]: How can I differentiate $f(x,z(x,y))$ w. r. to x

If $z(x,y)=c=\text{constant}$ and $\hat{y}=f(x,c)$ then what is $d\hat y/dx$
If I just differentiate $f$ w.r. to $x$ without knowing whether $z$ is constant or not $f_x(x,z(x,y))=f_x(z(x,y))=f_z(x,z(x,y))\cdot z_x(x,y)=\frac{\partial \hat y}{\partial z}z_x(x,y)=\frac{1}{\frac{\partial z}{\partial \hat y}}\cdot z_x(x,y)=\frac{z_x(x,y)}{z_y(x,y)}$
but the result should be negative
It is here on page 10 (18) Question 2.4 https://docs.google.com/file/d/0B21HoBq6u9TsWGtxWnNqVGtCSnM/edit?pref=2&pli=1&embedded=true

REPLY [0 votes]: Engrave this method to memory. I find that I never forget it this way;
$$\frac{d}{dt} \  f(\vec{c}(t))  = \nabla f|_{\vec{c}(t)} \cdot \vec{c}'(t)$$
Just let $\vec{c}(x,y) = \langle x,z(x,y) \rangle$ then differentiating with respect to $x$, allows you to view $\vec{c}(x,y)$ as just a function of $x$ say $c(x)$. Then we have;
$$\frac{d}{dt} \ f(\vec{c}(x)) = \langle f_x,f_y \rangle|_{\vec{c}(x)} \cdot \langle 1,z_x(x,y) \rangle$$


TITLE: A confusion about why can't a statistical mixture be modelled as a superposition of pure states?
QUESTION [0 upvotes]: I have read Cohen's book, and various posts in this site; however, I'm still not convinced why we can't model a statistical mixture as a superpositions of pure states ?
For example, consider the Stern-Gerlach experiment; the book claims that when the atoms go outfrom the oven, we don't know their states; however, even when they pass from the apparatus, we still do now know the states of the particles coming from the $+z$ direction of the apparatus if we were to measure their $x$ component next. Of course, one can argue that, no we do know the state; it is $| +z \rangle$, but that state is composed of $|+x\rangle $ and $|-x\rangle$, so if were to measure their $x$ component, practically we just know that %50 percent of the atom will come out from the $+x$ direction.
Similarly, when we first pass the incoming beam coming out from the oven, we, statistically, knew that %50 percent of the beam was going to come out from $+z$ direction, so, in practice, I don't see any reason why can't we model the statistical mixture as a superposition of pure states.
For example, why don't we say that the state of the atoms in the beam coming out from the oven is $$\psi = 0.5 |+z \rangle + 0.5 |-z\rangle~?$$

REPLY [1 votes]: The core difference between a mixture and a superposition is that, if we use orthogonal states, different parts of a mixture cannot interfere whereas different parts of a superposition will interfere.
In other word, if you define for instance
$$
\vert \pm\rangle =\frac{1}{\sqrt{2}}
\left(\vert\psi_1\rangle \pm \vert\psi_2\rangle\right)  \tag{1}
$$
with $\langle \psi_k\vert\psi_j\rangle =\delta_{kj}$ then an average value for a superposition
$$
\langle +\vert \hat A\vert +\rangle
= \frac{1}{2}
\left(  \langle \psi_1\vert \hat A\vert\psi_1\rangle
+\langle \psi_1\vert \hat A\vert\psi_2\rangle
+\langle \psi_2\vert \hat A\vert\psi_1\rangle
+\langle \psi_2\vert \hat A\vert\psi_2\rangle
\right)
$$
and likewise for $\langle -\vert \hat A\vert -\rangle $
with cross terms occurring between different states $\vert \psi_1\rangle$
and $\vert \psi_2\rangle$ in the superposition,  whereas the average value for a mixture
$$
\hat \rho =\alpha \vert +\rangle \langle +\vert
+ (1-\alpha) \vert -\rangle \langle -\vert
$$would be
$$
\hbox{Tr}(\hat \rho\hat A)
=\alpha \langle +\vert \hat A\vert +\rangle
+ (1-\alpha) \langle -\vert \hat A\vert -\rangle   \tag{2}
$$
with no cross-terms occurring between different states $\vert +\rangle$ and $\vert -\rangle$ of the mixture.  Notice also that, in Eq.(2), the average value is the weight sum of $\langle \pm \vert \hat A\vert \pm \rangle$ with the coefficients $\alpha$ and $(1-\alpha)$, which are necessarily real, functioning as statistical weight and thus sum to 1.
On the other hand, there is no restriction on the possible coefficients of a linear combination of states beyond the normalization condition: these coefficients can be complex or real.  The normalization condition involves a sum of their modulus squared and not a sum of the coefficients themselves.


TITLE: Eigenvalues for fingerprint classification
QUESTION [0 upvotes]: i acquired waveshare's (uart capacitive) fingerprint scanner which provides me with eigenvalues of a finger. these values are used to compare/classify fingerprints as per the scanners documentation.
Fingerprint algorithms generally use PCA (principle component analysis) to get eigenvectors which are used for comparisons as per :
Principal Component Analysis for Minutiae Verification on Fingerprint
Image (HSIEH,SHYU) &
Reconstruction of Fingerprints from Minutiae
Points (Jidnya A. Shah)
Eigenvectors can be obtained from the eigenvalues if the original matrix A is given which we dont have as per the following equation.
(A - $\lambda$I)x = 0   where x is the eigenvector and $\lambda$ is the eigenvalue
And we know from studying the scanner that eigenvalues can either be used to directly match samples of fingerprint or there is a way to convert eigenvalues to eigenvector.
We already used Euclidean distance to directly compare eigenvalues of sample fingerprints. this resulted in results which were incorrect i.e. false samples getting matched etc.
What is the way therefore to use eigenvalues for classification?

REPLY [1 votes]: there is a way to convert eigenvalues to eigenvector.

I'm not sure how this is possible. One is a scalar, the other is a vector. An eigenvalue can correspond to a particular eigenvector via depending on the system matrix.

Fingerprint algorithms generally use PCA (principle component analysis) to get eigenvectors

Usually PCA is used for dimensionality reduction, not really to "get" eigenvectors. You find principal directions, which are eignvectors of the covariance matrix of your data, and then you project the data down onto your principal vector directions.
The links you provided simply show how to find some optimal representation in a lower dimension to best represent your fingerprint data.
As for classification, I dont' know enough of your domain, but based on some references  just read, it seems you would need more than just a Euclidean distance between eigenvalues for a consistently good classifier.


TITLE: Bessel type function?
QUESTION [7 upvotes]: The matrix below
$\left(
\begin{array}{ccccc}
1 & 0 & 1 & 0 & 1 & 0 & 1 & \dots\\
\frac{1}{2} & \frac{1}{2} & 0 & 0 & \frac{1}{2} & \frac{1}{2} & 0 & \dots\\
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} & 0 & 0 & 0 & \frac{1}{3} & \dots\\
\frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & 0 & 0 & 0 & \dots\\
\frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 & \dots\\
\frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & 0 & \dots\\
\frac{1}{7} & \frac{1}{7} & \frac{1}{7} & \frac{1}{7} & \frac{1}{7} & \frac{1}{7} & \frac{1}{7} & \dots\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots\\
\end{array}
\right)$
follows a pattern whereby the first line begins $\{1,0,1,0,1,0,1,0,1,0,1,0\dots\}$
the second: $\{\frac{1}{2},\frac{1}{2},0,0,\frac{1}{2},\frac{1}{2},0,0,\frac{1}{2},\frac{1}{2},0,0\dots\}$
the third: $\{\frac{1}{3},\frac{1}{3},\frac{1}{3},0,0,0,\frac{1}{3},\frac{1}{3},\frac{1}{3},0,0,0\dots\}$
and so on.
Summing the first $m$ rows together and subtracting $\frac{1}{2}\sum_{k=1}^{m}\frac{1}{k}$ from each element from the resultant array gives a sequence of numbers that fluctuates between positive and negative values. The resultant cummulative plot looks like this:

y = 3000; m = 500;
ListPlot[Accumulate[Total[Take[Flatten[ConstantArray
[#,Ceiling[(y)/Length@#]]], y] & /@ Table[Join[ConstantArray[1/row, row],
ConstantArray[0, row]], {row, 1, m}]] - Sum[1/k, {k, 1, m}]/2]]

where $y$ is the number of columns, and $m$ the number of rows, and is apparently scalable by dividing by $\dfrac{m}{4}$:

Is it true that this is related to the family of Bessel functions (inclusive of trigonometric integrals, which are related to spherical Bessel functions)?
Below is a plot of a Bessel function of the second kind, for illustrative purposes:

given by
$$
Y_0(z)=\dfrac{2}{\pi}\bigg(\bigg(\log(z/2)+\gamma\bigg)\sum_{m=0}^{\infty}\dfrac{(-1)^m}{m!\ \Gamma(m+1)}(z/2)^{2m}+\sum_{k=1}^{\infty}(-1)^{k+1}\sum_{i=1}^{\infty}\dfrac{1}{i}\bigg(\dfrac{(z^2/4)^k}{(k!)^2}\bigg)\bigg)
$$
Whether it is the case or not, I would be interested to know why it follows this pattern. The periodicity up to $y$ follows a particular type of randomness, seen also in the differences between zeta zeros, for example. I am curious to find out whether there is any connection.
Update

plotted with this (ridiculously long) piecewise code, thanks to Ruslan's answer below.

REPLY [6 votes]: This is unlikely to be related to Bessel functions or Fresnel integral or sine/cosine integral. Let's take your code, assigning the ListPlot argument to variable data:
y = 3000; m = 500;
data = Accumulate[
   Total[Take[Flatten[ConstantArray[#, Ceiling[(y)/Length@#]]], y] & /@
       Table[Join[ConstantArray[1/row, row],
        ConstantArray[0, row]], {row, 1, m}]] -
    Sum[1/k, {k, 1, m}]/2];
ListPlot[data]

Now we'd need to smooth this somewhat. Let's use a moving average with e.g. 30 points:
smdata = ListConvolve[#/Total[#] &@Table[1., {i, 1, 30}], data];
ListPlot[smdata, PlotRange -> All]

Now this looks somewhat more tractable by differentiating, although it does have some noise, we can still see some structure in the derivative. For this, make an interpolation:
di = Interpolation[smdata];

And plot the derivative of the interpolant:
Plot[di'[x], {x, 1, First@Dimensions@data - 500}, PlotRange -> All]

We can see that the derivative may even have some cusps at its extrema, so it doesn't look like a smooth function, which would be if it were a Bessel function or Fresnel integral or anything supposed in the comments.
The plot, however, gives some ideas. What if it's just "broken" and (quasi)periodically reflected? Let's try adding -0.45-di'[x] to the plot:
Plot[{di'[x], -0.45 - di'[x]}, {x, 1, First@Dimensions@data - 500}, PlotRange -> All]

This seems to smoothly fit the guess. I think we could do the similar thing with other cusps and recover something like a logarithmic function. Here're some more added plots:
Plot[{di'[x], -0.45 - di'[x], di'[x] - 0.70, -0.85 - di'[x], di'[x] - 1}, {x, 1,
  First@Dimensions@data - 500}, PlotRange -> All]

Defining a piecewise function, we can plot the single curve:
pw[x_] = Piecewise[{
    {di'[x], x < 500},
    {-0.45 - di'[x], x < 1000},
    {di'[x] - 0.70, x < 1500},
    {-0.85 - di'[x], x < 2000},
    {di'[x] - 1, True}}];
Plot[pw[x], {x, 1, First@Dimensions@data - 500}, PlotRange -> All]

Now, more interesting. Let's define analytically what we get. First, we can notice that our matrix can be generated by the following function:
$$f(x,y)=\frac1{2y}\left\{1+\text{sgn}\left[\sin\left(\frac{\pi x}y-\varepsilon\right)\right]\right\},$$
where to get the matrix we should take $\lim_{\varepsilon\to+0}$ and use $x,y=1,2,...$.
Now, we aren't interested in any pixellation noise, so we can switch from taking sums to taking integrals, and from matrix to the generator function itself. Then we don't have to bother with $\varepsilon$, and can set $\varepsilon=0$. Now the sum over column index $y$ and subtracting the sum over $\frac1k$ will give us a function of row index $x$:
$$s(x)=\int_1^m\frac1{2y}\text{sgn}\left(\sin\frac{\pi x}y\right)\text{d} y,$$
where $m$ is the same as m in your code.
Now, taking $y=300$, $m=50$, I get the resulting $s(x)$:

Here, at, e.g. $x=30$, we have:
$$s(30)=-\frac12\left(\log\frac35+\sum_{n=2}^{30}(-1)^n \log\frac n{n-1}\right).$$
Similar expressions in terms of logarithms are for other $x$ values.
So, the correct description of the function is in terms of simple logarithms. Then the function which you thought as Bessel-related, being an integral of this function, is also expressible in terms of logarithms.
Note however, that the function isn't really smooth. See it magnified:


TITLE: Optional Stopping Theorem and Uniform Integrability
QUESTION [2 upvotes]: Here's a problem related to the Optional Stopping Theorem:
Let $S_n$ be a symmetric simple random walk on $\mathbb{Z}$ (as in, at each step, the random walk moves a distance of 1 in a uniformly random direction) starting at 0, and let $T=\min \{n:S_n = M$ or $S_n = -M\}$ where $M$ is an integer.
Find constants $b$ and $c$ so that $Y_n = S_n^4 - 6nS_n^2 +bn^2 +cn$ is a martingale, and compute $\mathbb{E}[T^2]$.
I understand how to get that $b=3$, $c=2$. The problem intends for one to then apply the Optional Stopping Theorem to $Y_{n\wedge T}$ (along with the fact that $E[T] = M^2$ as follows:
$0=\mathbb{E}[M^4-(6T)M^2+3T^2+2T] = M^4+(2-6M^2)\mathbb{E}[T]+3\mathbb{E}[T^2]=M^4+(2-6M^2)M^2+3\mathbb{E}[T^2] \Rightarrow \mathbb{E}[T^2] = \frac{5M^4-2M^2}{3}$.
Per multiple online sources, the following three conditions are sufficient for the OST to apply to a martingale:

The stopping time has to be bounded a.s.
The value of the martingale is bounded a.s.
The expectation of the stopping time is finite, and the difference of the martingale's increments are bounded a.s.

As far as I can tell, $Y_{n\wedge T}$ does not meet any of these conditions: the stopping time is not bounded a.s., and, if $n$ becomes arbitrarily large, so do $Y_{n}$ and the martingale's increments.
However, my professor said that the optional stopping theorem still applies here because $Y_{n\wedge T}$ is uniformly integrable.
So, in general, for a discrete-time martingale, is uniform integrability a sufficient condition for the Optional Stopping Theorem?  If so, I understand that the three conditions above would be useful as sufficient (but not necessary) conditions for uniform integrability.
If not, why would the Optional Stopping Theorem apply here?

REPLY [1 votes]: Yes, uniform integrability of the martingale is sufficient to apply the Optional Stopping Theorem to any stopping time. In fact, your condition 2. is a special case since bounded $\Rightarrow$ uniformly integrable.
Here, you don't need uniform integrability however and you can simply apply condition 1.
Fix $k \geqslant 0$ and apply the OST to the martingale $(Y_n)_{n \geqslant 0}$ and the stopping time $k \wedge T$ which is bounded a.s. by $k$. We get
$$0 = \mathbb{E}\left[S_{k \wedge T}^4 - 6(k \wedge T)S_{k \wedge T}^2+ 3 (k\wedge T)^2 +2 k\wedge T \right].$$
Then, by the dominated convergence theorem, letting $k \to \infty$ yields the result.