text:
\begin{document}

\begin{abstract}
Ehrhart's famous theorem states
that the number of integral points in a rational polytope is a quasi-polynomial in the integral dilation
factor. We study the case of rational dilation factors. It turns out that the number of integral points
can still be written as a rational quasi-polynomial. Furthermore, the coefficients of this rational
quasi-polynomial are piecewise polynomial functions and related to each other by derivation.
\end{abstract}

\maketitle
\section{Introduction}
Let $\R^n$ be the $n$-dimensional Euclidean space and let $\Z^n$ be the integral lattice.
For a set $M\subset\R^n$, we denote by $\inner(M)$ its interior, by $\conv(M)$ its convex hull, by
$\aff(M)$ its affine hull, by $\vol(M)$ its volume, which is the usual Lebesgue measure of $M$, and by
$\dim(M)$ its dimension, which is defined as the dimension of its affine hull.
By $\vol_{\dim(M)}(M)$ we denote the $\dim(M)$-dimensional volume of $M$.
A polytope
is called integral (rational), if all its vertices have integral (rational) coordinates. For a rational polytope $P$,
we denote by $\d(P)$ the denominator of $P$, that is,
the smallest number $k\in\Z_{>0}$ such that $kP$ is an integral polytope. In other words
$\d(P)$ is the lowest common multiple of the denominators of all coordinates of the vertices of $P$.
Furthermore, let the $i$-index $\ind_i(P)$ of a rational polytope $P$ be the smallest number $k\in\Z_{\geq 0}$
such that for each $i$-face $F$ of $P$ the affine space $k\aff(F)$ contains integral points.\\
A function $p:\Z\to\Z$ is called a \emph{quasi-polynomial with period $d$} if there exist periodic functions
$p_i:\Z_{\geq 0}\to\Z$ with period $d$ such that $p(k)=\sum_{i=0}^np_i(k)k^i$.
Ehrhart's Theorem states the following:

\begin{theorem}[Ehrhart, 1962, \cite{Ehrhart1962}, McMullen, 1978, \cite{McMullen1978}]\label{ehr_quasipoly}
  Let $P\subset\R^n$ be a rational polytope. Then
  \[\G(P,k):=\#(kP\cap\Z^n)=\sum_{i=0}^{\dim(P)}\G_i(P,k)k^i\ \text{ for }k\in\Z_{\geq 0}.\]
  $\G(P,\cdot):\Z_{\geq 0}\to\Z_{\geq 0}$ is called the \emph{Ehrhart quasi-polynomial} of $P$.
  For every $i$, the coefficient $\G_i(P,k)$ depends only on the congruence class of $k$ modulo $\ind_i(P)$.

\end{theorem}

Here, $\G_i(P,\cdot):\Z_{\geq 0}\to\Q$ is a periodic function and $\ind_i(P)$ is a period of $\G_i(P,\cdot)$.
The second part of this statement is due to McMullen. Furthermore, $\G_0(P,0)=1$, and $\G_{\dim(P)}(P,k)=\vol_{\dim(P)}(P)$ for all
$k\in\Z_{\geq 0}$ such that $k\aff(P)$ contains integer points.

For an introduction into Ehrhart theory we refer to
Beck and Robins \cite{Beck&Robins2006}.
Unfortunately, $\ind_i(P)$ is not necessarily the minimal period of $\G_i(P,\cdot)$, that is, there is
possibly a smaller integer number $p<\ind_i(P)$ such that $p$ is a period of $\G_i(P,\cdot)$.
This phenomenon is called period collapse and has been subject to active research in the last years.
McAllister and Woods \cite{McAllister&Woods} studied the $1$- and $2$-dimensional case with the result that period collapse
does not occur in dimension $1$, and they gave a characterization of those rational polygons in dimension $2$
whose Ehrhart quasi-polynomial is a polynomial. They also showed that the minimal periods of $\G_i(P,\cdot)$ are not
necessarily decreasing with $i$.
Woods \cite{Woods2004} gave a polynomial-time algorithm in fixed dimension
which decides whether a given integer $p$ is a period of all $\G_i(P,\cdot)$. In \cite{Beck&Sam&Woods}, Beck, Sam and Woods constructed
polytopes with no period collapse at all. Furthermore, they showed that period collapse never occurs for $\G_{\dim(P)-1}(P,\cdot)$.
Haase and McAllister \cite{Haase&McAllister} gave a conjectural explanation of period collapse involving splitting the polytope into
pieces and applying unimodular transformations onto these pieces.

We show that the Ehrhart quasi-polynomial can be generalized to a rational quasi-polynomial by allowing rational dilation factors,
where a \emph{rational quasi-polynomial with period $d$} is a function $p:\Q\to\Q$ of the
form $p(r)=\sum_{i=0}^np_i(r)r^i$ where $p_i:\Q\to\Q$ is periodic with period $d$.

\begin{theorem}\label{thm:rat_dilations}
  Let $P\subset\R^n$ be a rational polytope. Then for $r\in\Q_{\geq 0}$
  \[\Grat(P,r):=\#(rP\cap\Z^n)=\sum_{i=0}^{\dim(P)}\Grat_i(P,r)r^i.\]
  Here, $\Grat_i(P,\cdot):\Q_{\geq 0}\to\Q$ is a periodic function, and $\ind_i(P)$ is a period of $\Grat_i(P,\cdot)$.
	We call $\Grat(P,\cdot):\Q_{\geq 0}\to\Q$ the \emph{rational Ehrhart quasi-polynomial} of $P$.

\end{theorem}

We remark that $\Grat_i(P,\cdot)$ is an extension of $\G_i(P,\cdot)$ to rational numbers. Hence,
$\Grat_0(P,0)=\G_0(P,0)=1$. Furthermore, we have that $\Grat_{\dim(P)}(P,r)=\vol_{\dim(P)}(P)$ for all $r$ such that
$r\aff(P)$ contains integer points.

We define a rational analogue of the $i$-index. These rational indices allow us to show a refined result on the periods of rational Ehrhart quasi-polynomials.

\begin{definition}
Let the \emph{rational denominator} $\rd(P)$ of $P$ be the smallest positive rational number $r$
such that $rP$ is an integral polytope:
\begin{equation*}
\rd(P)=\min\{r\in\Q_{>0}: rP\text{ is an integral polytope}\},
\end{equation*}
and let the \emph{rational $i$-index} $\rind_i(P)$ of $P$
be the smallest positive rational number $r$ such that for each $i$-face $F$ of $P$ the affine space $r\aff(F)$
contains integral points:
\begin{equation*}
\rind_i(P)=\min\{r\in\Q_{>0}: r\,\aff(F)\cap\Z^n\neq\emptyset\emph{, for all }i\emph{-faces } F\}.
\end{equation*}

\end{definition}

Then we get the following result:

\begin{corollary}\label{cor:periods}
Let $P$ be a rational polytope. Then $\rind_i(P)$ is a period of $\Grat_i(P,\cdot)$.
Furthermore, $\Grat(P,\cdot):\Q_{\geq 0}\to\Z$ is a rational quasi-polynomial with period $\rind_0(P)=\rd(P)$.

\end{corollary}

Ehrhart's reciprocity law is also true for rational Ehrhart quasi-polynomials:

\begin{corollary}\label{cor:reci}
Let $P$ be a rational polytope and let $\Grat(P,r)=\sum_{i=0}^{\dim(P)}\Grat_i(P,r)r^i$ be its rational Ehrhart quasi-polynomial.
Then for $r\in\Q_{\geq 0}$,
\begin{equation*}
\#(r\,\inner(P)\cap\Z^n)=(-1)^{\dim(P)}\Grat(P,-r).
\end{equation*}

\end{corollary}

We show further that the coefficients $\Grat_i(P,\cdot)$ of the rational Ehrhart quasi-polynomial are piecewise polynomials.
Here, we assume $P$ to be full-di\-men\-sional. This assumption is necessary since, if $P$ is contained in an affine hyper\-plane
not containing $0$, we have $\#(rP\cap\Z^m)=0$ for all $r\in\Q_{\geq 0}$ such that $r\aff(P)$ does not contain integral points. Thus, in that case $\Grat_i(P,r)=0$
for nearly all points $r\in\Q_{\geq 0}$. On the other hand, if $P$ is contained in a hyperplane containing $0$, it behaves like
a full-dimensional polytope.

\begin{theorem}\label{thm:polycoeffs}
Let $P\subset\R^n$ be an $n$-dimensional rational polytope and let $\Grat(P,r)=\sum_{i=0}^n\Grat_i(P,r)r^i$ be its rational Ehrhart
quasi-polynomial. Then $\Grat_i(P,\cdot)$ is a piecewise polynomial of degree $n-i$.

\end{theorem}

By $\Grat_i'(P,r)$ we denote the first derivative of $\Grat_i(P,r)$ in $r$ if it exists. Using this
we deduce the following relation between the coefficients of the rational Ehrhart quasi-polynomial of
a polytope $P$:

\begin{theorem}\label{thm:derivatives}
Let $P\subset\R^n$ be an $n$-dimensional rational polytope. Then
\begin{equation*}
\Grat_i'(P,r)=-(i+1)\Grat_{i+1}(P,r),\quad i=0,\ldots,n-1,
\end{equation*}
for all $r\geq 0$ where the derivative exists.
\end{theorem}
This theorem can be seen as a first step towards investigations on period collapses; it implies that,
in contrast to the integral case, the minimal periods of $\Grat_i(P,r)$ are decreasing with $i$
if $0\in P$.

In general nothing is known about extremal values of $G_i(P,k)$ or $\Grat_i(P,r)$.
As a first result in this direction we have in the 2-dimensional case:

\begin{theorem}\label{thm:dim2}
Let $P$ be an arbitrary $2$-dimensional rational polygon. Then $|\Grat_1(P,r)|\leq \Grat_1(P,0)$ for all $0\leq r<\rd(P)$.

\end{theorem}
To this end, we work out an explicit example using the approach presented by Sam and Woods in \cite{Sam&Woods2009}.
An analogous statement for $\Grat_0(P,r)$ is not true.

The paper is organized as follows: In Section \ref{sec:rat} we present all tools used for the proofs of
the results presented in this introduction. The proofs of Theorems \ref{thm:rat_dilations}, \ref{thm:polycoeffs} and \ref{thm:derivatives}
and their Corollaries are given in this section as well. In Section \ref{sec:dim2} we work out a detailed example
(see Theorem~\ref{thm_2d_triangle})
and deduce Theorem \ref{thm:dim2} from the explicit formulas of the rational Ehrhart quasi-polynomial of this example.

For further information on Ehrhart (quasi-)polynomials and similar problems as considered in this work, we refer to
\cite{Barvinok1992,Beck&Haase2008,Brion&Vergne,Chen&Li&Sam2010}.

\section{Rational dilations}\label{sec:rat}
To prove Theorem \ref{thm:rat_dilations}, we need a well-known property of the $\G_i(\cdot,\cdot)$, which we present here:

\begin{lemma}\label{lem:int_dilations}
Let $\G(P,k)=\sum_{i=0}^{\dim(P)}\G_i(P,k)k^i$ be the Ehrhart quasi-polynomial of a rational polytope $P$. Then
$\G_i(mP,k)=\G_i(P,mk)m^i$ for $m,k\in\Z_{\geq 0}$.

\end{lemma}
\begin{proof}
We have $\#(mkP\cap\Z^n)=\G(mP,k)=\sum_{i=0}^{\dim(P)}\G_i(mP,k)k^i$ and
$\#(mkP\cap\Z^n)=\G(P,mk)=\sum_{i=0}^{\dim(P)}\G_i(P,mk)(mk)^i$.
Comparing coefficients yields $\G_i(mP,k)=\G_i(P,mk)m^i$, $\forall m,k\in\Z_{\geq 0}$ (see Barvinok
\cite[Section 4.3]{Barvinok2006} for details on equality of quasi-polynomials).

\end{proof}
\begin{proof}[Proof of Theorem \ref{thm:rat_dilations}]
Let $\G(P,k)=\sum_{i=0}^{\dim(P)}\G_i(P,k)k^i$ for $k\in\Z_{\geq 0}$ be the Ehrhart quasi-polynomial of $P$.
We define
\begin{equation*}
\Grat_i\left(P,\frac{a}{b}\right):=\G_i\left(\frac{1}{b}P,a\right)b^i.
\end{equation*}
$\Grat_i\left(P,\frac{a}{b}\right)$ is well-defined, since for $\frac{a}{b}=\frac{ka}{kb}$
we get $\Grat_i\left(P,\frac{ka}{kb}\right)=\G_i(\frac{1}{kb}P,ka)k^ib^i=\G_i(\frac{1}{b}P,a)b^i=\Grat_i\left(P,\frac{a}{b}\right)$ by Lemma \ref{lem:int_dilations}.
Then
\begin{equation*}
\Grat\left(P,\frac{a}{b}\right)
=\G\left(\frac{1}{b}P,a\right)
=\sum_{i=0}^{\dim(P)}\G_i\left(\frac{1}{b}P,a\right)a^i=\sum_{i=0}^{\dim(P)}\Grat_i\left(P,\frac{a}{b}\right)\left(\frac{a}{b}\right)^i.
\end{equation*}
It remains to show that $\Grat_i\left(P,\frac{a}{b}\right)$ is periodic with period $\ind_i(P)$.
Since $b\,\ind_i(P)$ is a multiple of $\ind_i\left(\frac{1}{b}P\right)$, we get
\begin{equation*}
\Grat_i\left(P,\frac{a}{b}+\ind_i(P)\right)
=\G_i\left(\frac{1}{b}P,a+b\,\ind_i(P)\right)b^i\\
=\G_i\left(\frac{1}{b}P,a\right)b^i
=\Grat_i\left(P,\frac{a}{b}\right).\qedhere
\end{equation*}

\end{proof}

This proof implies that knowing the classical Ehrhart quasi-polynomial of $\frac{1}{b}P$
for all positive integers $b$ is equivalent to knowing the rational Ehrhart quasi-polynomial of $P$.
However, as the next remark shows, it is not enough to know the Ehrhart quasi-polynomial of a polytope
to recover the rational version:

\begin{remark}
$\Grat(P,\cdot):\Q_{\geq 0}\to\Z$ is not invariant under translations of $P$ with respect to integral
vectors. Furthermore, $\Grat(P,\cdot):\Q_{\geq 0}\to\Z$ is not necessarily monotonically increasing if $0\not\in P$.
For instance, let
\begin{equation*}\begin{split}
T_1&=\conv\left(\binom{1/2}{-1/2},\binom{-1/2}{-1/2},\binom{0}{3/2}\right),\\
T_2&=\conv\left(\binom{1/2}{1/2},\binom{-1/2}{1/2},\binom{0}{5/2}\right)
\end{split}\end{equation*}
(see Figure \ref{fig:rem}).Then $T_2=T_1+\binom{0}{1}$.
Nevertheless, we have $\Grat(T_1,2/3)=2$ and $\Grat(T_2,2/3)=1$. Moreover, $\Grat(T_2,2)=7$ and $\Grat(T_2,11/5)=4$.
\begin{figure}[ht]\centering
\includegraphics{figure1.eps}
\caption{Triangles $T_1$ and $T_2$}\label{fig:rem}
\end{figure}

\end{remark}

Furthermore, as for the integral case, there are examples of polytopes with different combinatorial type
that have the same rational Ehrhart quasi-polynomial. Stanley \cite{STANLEY1986} constructed two classes of
polytopes, \emph{order polytopes} and \emph{chain polytopes}. In general these are polytopes of different combinatorial type
but with the same Ehrhart polynomials. Since his consideration is independent of the integrality of dilation factors, these polytopes
have also the same rational Ehrhart polynomials.

\begin{lemma}\label{lem:rat_dilations}
For all $r,s\in\Q$ we have $\Grat_i(sP,r)=\Grat_i(P,sr)s^i$.

\end{lemma}
\begin{proof}
Let $r=\frac{a}{b}$, $s=\frac{c}{d}$. By the definition of $\Grat_i(P,r)$ in the proof of Theorem \ref{thm:rat_dilations},
we get, together with Lemma \ref{lem:int_dilations},
\begin{equation*}\begin{split}
\Grat_i(sP,r)&=\Grat_i\left(\frac{c}{d}P,\frac{a}{b}\right)=\G_i\left(\frac{c}{db}P,a\right)b^i\quad \text{and}\\
\Grat_i(P,sr)s^i&=\Grat_i\left(P,\frac{ac}{bd}\right)\frac{c^i}{d^i}
=\G_i\left(\frac{c}{bd}P,a\right)b^i.\qedhere
\end{split}\end{equation*}

\end{proof}
As in the integral case, the rational indices are divisors of each other.
\begin{lemma}\label{lem:rat_divisors}
Let $P$ be a rational polytope. Then $\rind_{i-1}(P)/\rind_{i}(P)\in\Z$ for $i=1,\ldots,n$.

\end{lemma}
\begin{proof}
Let $H^i_1,\ldots,H^i_{m(i)}$ be the respective affine hulls of the $m(i)$ $i$-faces of $P$ and let
$r^i_j$ be the smallest positive rational number such that $H^i_j$ contains integral points.
Then $rH^i_j$ contains integral points if and only if $r$ is an integral multiple of $r^i_j$, for $j=1,\ldots,m(j)$. Thus,
$\rind_i(P)$ is the smallest positive rational number that is an integral multiple of all $r^i_j$.
Furthermore, since $H^{i-1}_j\subset H^{i}_{\tilde{\jmath}}$ for some $\tilde{\jmath}$, we have that $r^{i-1}_{j}$ is an
integral multiple of $r^{i}_{\tilde{\jmath}}$, and thus $\rind_{i-1}(P)$ is an integral multiple of $\rind_{i}(P)$.

\end{proof}
Now we are able to prove that the rational indices are periods of the coefficients of the rational Ehrhart quasi-polynomials.
\begin{proof}[Proof of Corollary \ref{cor:periods}]
Since $\ind_i(\rind_i(P)P)=1$ for all $i$, we know that
\[\Grat_i(\rind_i(P)P,r+k)=\Grat_i(\rind_i(P)P,r),\quad\forall r\in\Q_{\geq_0},k\in\Z_{\geq_0}.\]
This implies, together with Lemma \ref{lem:rat_dilations},
\[\Grat_i(P,r\,\rind_i(P)+k\,\rind_i(P))=\Grat_i(P,r\,\rind_i(P)),\quad\forall r\in\Q_{\geq_0},k\in\Z_{\geq_0},\]
and thus
\[\Grat_i(P,\tilde{r}+k\,\rind_i(P))=\Grat_i(P,\tilde{r}),\quad\forall \tilde{r}\in\Q_{\geq_0},k\in\Z_{\geq_0}.\]
Furthermore, together with Lemma \ref{lem:rat_divisors}, we get that $\rind_0(P)/\rind_{i}(P)\in\Z$ for $i=1,\ldots,n$,
and thus $\rind_0(P)=\rd(P)$ is a period of $\Grat(P,\cdot)$.

\end{proof}

Concerning the minimal periods of Ehrhart quasi-polynomials, Beck, Sam and Woods \cite{Beck&Sam&Woods} showed that
$\ind_{\dim(P)-1}(P)$ is in fact the minimal period of $\G_{\dim(P)-1}(P,\cdot)$. An analogous result is also true in the rational case:

\begin{corollary}
Let $P\subset\R^n$ be a rational polytope and let $\Grat(P,r)=\sum_{i=0}^{\dim(P)}\Grat_i(P,r)r^i$ be its rational Ehrhart quasi-polynomial.
Then $\rind_{\dim(P)-1}(P)$ is the minimal period of $\Grat_{\dim(P)-1}(P,\cdot)$.

\end{corollary}
\begin{proof}
By Lemma \ref{lem:rat_dilations} and since $\rind_{\dim(P)-1}$ is homogeneous, it suffices to show the statement for all $P$
with $\rind_{\dim(P)-1}(P)=1$. Thus we assume that $\frac{s}{t}<1$ is a period of $\Grat_{\dim(P)-1}(P,\cdot)$
with $s,t\in\Z_{>0}$, that is $\Grat_{\dim(P)-1}(P,r)=\Grat_{\dim(P)-1}\left(P,r+\frac{s}{t}\right)$ for all $r\in\Q_{\geq 0}.$
Again by Lemma \ref{lem:rat_dilations} we get
\begin{equation*}
\Grat_{\dim(P)-1}\left(\frac{1}{t}P,rt\right)=\Grat_{\dim(P)-1}\left(\frac{1}{t}P,rt+s\right)\quad\text{for all }r\in\Q_{\geq 0}.
\end{equation*}
In particular this is true for all $rt\in\Z_{\geq 0}$, and
thus, $s$ is a period of $\G_{\dim(P)-1}\left(\frac{1}{t}P,\cdot\right)$, which is a contradiction, since
\begin{equation*}
\ind_{\dim(P)-1}\left(\frac{1}{t}P\right)\geq\rind_{\dim(P)-1}\left(\frac{1}{t}P\right)=t\,\rind_{\dim(P)-1}(P)=t>s.\qedhere
\end{equation*}

\end{proof}

We can also prove the Ehrhart reciprocity law for rational Ehrhart quasi-polynomials. We refer to Beck and Robins \cite[Chapter 4]{Beck&Robins2006}
for details on Ehrhart reciprocity law.

\begin{proof}[Proof of Corollary~\ref{cor:reci}]
Let $r=\frac{a}{b}$ with $a,b\in\Z_{\geq 0}$. Then, by Ehrhart reciprocity law and Lemma \ref{lem:rat_dilations}, we get
\begin{equation*}\begin{split}
\#\left(\frac{a}{b}\,\inner(P)\cap\Z^n\right)
&=(-1)^{\dim(P)}\sum_{i=0}^{\dim(P)}\G_i\left(\frac{1}{b}P,-a\right)(-a)^i\\
&=(-1)^{\dim(P)}\sum_{i=0}^{\dim(P)}\Grat_i\left(P,-\frac{a}{b}\right)\left(\frac{1}{b}\right)^i(-a)^i\\
&=(-1)^{\dim(P)}\Grat\left(P,-\frac{a}{b}\right).\qedhere
\end{split}\end{equation*}

\end{proof}

For the proof of Theorem \ref{thm:polycoeffs} we need the following Lemma:

\begin{lemma}\label{qp_const_lemma}
Let $p:\Q\to\Q$ be a rational quasi-polynomial of degree $n\in\Z_{>0}$ with period $d\in\Q_{>0}$ and constant leading coefficient, that is,
\[p(r)=p_nr^n+p_{n-1}(r)r^{n-1}+p_{n-2}(r)r^{n-2}+\ldots+p_1(r)r+p_0(r),\]
where $0\neq p_n\in\Q$ and $p_i:\Q\to\Q$ are periodic functions with period $d$ for $i=0,\ldots,n-1$. Furthermore,
suppose there exist an interval $(r_1,r_2)\subset \Q$ and $c_k\in\Q$ for $k\in\Z_{\geq 0}$ such that
\[p(r+kd)=c_k,\quad\forall r\in (r_1,r_2), \forall k\in\Z_{\geq 0}.\]
Then $p_i:(r_1,r_2)\to\Q$ is a polynomial of degree $n-i$. Furthermore, if $p_n>0$ then $p_{n-1}$ has negative leading
coefficient.

\end{lemma}
\begin{proof}
We prove this result by induction with respect to $n$. For $n=1$ we have $c_0=p(r)=p_1r+p_0(r)$ for all $r\in(r_1,r_2)$.
Thus, $p_0(r)=c_0-p_1r$ for $r\in(r_1,r_2)$, which is a polynomial of degree $n-0=1$ with negative leading coefficient.\\
Now let $n>1$. We have
\begin{equation*}
c_k=p_n\cdot(r+kd)^n+\sum_{i=0}^{n-1}p_i(r)(r+kd)^i\quad\forall r\in(r_1,r_2), \forall k\in\Z_{\geq 0}.
\end{equation*}
Then $q:\Q\to\Q$ with $q(r):=p_n\cdot\left((r+d)^n-r^n\right)+\sum_{i=0}^{n-1}p_i(r)\left((r+d)^i-r^i\right)$ is a
quasi-polynomial of degree $n-1$ with period $d$ and constant leading coefficient, and
\begin{equation*}\begin{split}
q(r+md)&=p_n\cdot\left((r+(m+1)d)^n-(r+md)^n\right)\\
&\hspace{1cm}+\sum_{i=0}^{n-1}p_i(r)\left((r+(m+1)d)^i-(r+md)^i\right)\\
&=c_{m+1}-c_m\quad\forall r\in(r_1,r_2), \forall m\in\Z_{\geq 0}.
\end{split}\end{equation*}
Thus, we can use the induction hypothesis for $q$, and together with
\begin{equation*}\begin{split}
q(r)&=p_n\cdot\left((r+d)^n-r^n\right)+\sum_{i=0}^{n-1}p_i(r)\left((r+d)^i-r^i\right)\\
&=p_nndr^{n-1} + \sum_{j=0 }^{n-2}\left(p_n\binom{n}{j}d^{n-j}+\sum_{i=j+1 }^{n-1}p_i(r)\binom{i}{j}d^{i-j}\right)r^j\\
\end{split}\end{equation*}
we get that
\begin{equation*}
q_j(r):=p_n\binom{n}{j}d^{n-j}+\sum_{i=j+1 }^{n-1}p_i(r)\binom{i}{j}d^{i-j}
\end{equation*}
is a polynomial of degree $n-1-j$
for $r\in(r_1,r_2)$, for all $j=0,\ldots,n-2$. Since $p_nnd>0$ we get, also by induction, that
$q_{n-2}(r)=p_n\binom{n}{2}d^2+p_{n-1}(r)\binom{n-1}{n-2}d$ has a negative leading coefficient. \\
Now we use induction again to show
that $p_{j+1}$ is a polynomial of degree $n-j-1$
for $r\in(r_1,r_2)$.
For $j=n-2$ we have that $q_{n-2}(r)=p_n\binom{n}{2}d^2+p_{n-1}(r)\binom{n-1}{n-2}d$ is a polynomial of degree $1$ with negative leading coefficient,
hence the same is true for $p_{n-1}$.
For $j<n-2$ write
\begin{equation*}
q_j(r)=p_n\binom{n}{j}d^{n-j}+\sum_{i=j+1 }^{n-1}p_i(r)\binom{i}{j}d^{i-j}=\sum_{i=0}^{n-j-1}\alpha_ir^i\quad \forall r\in(r_1,r_2).
\end{equation*}
Then for $r\in(r_1,r_2)$,
\begin{equation*}
p_{j+1}(r)(j+1)d=\sum_{i=0}^{n-j-1}\alpha_ir^i-\sum_{i=j+2 }^{n-1}p_i(r)\binom{i}{j}d^{i-j}-p_n\binom{n}{j}d^{n-j}
\end{equation*}
which is a polynomial of degree $n-j-1$ since $p_i(r)$ is a polynomial of degree $n-i$ for $i\geq j+2$ by induction
hypothesis.\newline
We conclude that $p_i(r)$ is a polynomial of degree $n-i$ for $r\in(r_1,r_2)$ and $i=1\ldots,n-1$. That $p_0(r)$ is
a polynomial follows immediately from $p_0(r)=c_0-p_nr^n-\sum_{i=1}^{n-1}p_i(r)r^i$.

\end{proof}

Next we show that Ehrhart quasi-polynomials of full-dimensional rational polytopes satisfy the setting in Lemma \ref{qp_const_lemma}.
\begin{proof}[Proof of Theorem \ref{thm:polycoeffs}]
By Theorem \ref{thm:rat_dilations}, $\Grat(P,r)$ is a rational quasi-polynomial of degree $n$ with
period $\rd(P)$ and constant, nonzero leading coefficient. To apply Lemma \ref{qp_const_lemma} it remains to show that there exist
$0=r_0<r_1<\ldots<r_l=\rd(P)$ such that $\Grat(P,r)$ is constant for $r\in (r_i+k\rd(P),r_{i+1}+k\rd(P))$ and $i=0,\ldots,l-1$, $k\in\Z_{\geq 0}$.
To this end we consider $\Grat(P,r)$ as a function $\Q_{\geq 0}\to \Q$. $\Grat(P,r)$ is certainly piecewise constant, and it
jumps whenever integral points leave or enter $rP$, which can only happen if one of the facets of $rP$ lies
in a hyperplane containing integral points. Thus, for every facet $F$ of $P$ let $\alpha_F$ be the smallest positive rational number
such that $\alpha_FF$ lies in a hyperplane containing integral points. Then $\{k\alpha_F: F\text{ facet of }P, k\in\Z_{\geq 0}\}$
are the only possible jump discontinuities of $\Grat(P,r)$. By the definition of $\rd(P)$, for a facet $F$ of $P$ there exists a $k_F\in\Z$
such that $k_F\alpha_F=\rd(P)$. Thus for $\{r_0,\ldots,r_l\}=\{k\alpha_F: k=0,\ldots,k_F, F\text{ facet of }P\}$ we can apply
Lemma~\ref{qp_const_lemma}.

\end{proof}

We refer to Figure \ref{g01pic} in Section \ref{sec:dim2} for a visualization of the $\Grat_i(P,\cdot)$.

\begin{remark}
The rational Ehrhart quasi-polynomials can be extended to real quasi-polynomials
$\Grat(P,\cdot):\R_{\geq 0}\to\Z_{\geq 0}$.
To do that, for $r_0\in\R_{\geq 0}\setminus\Q$ let $r_j\in\Q_{\geq 0}$, $j\geq 1$ with $r_0=\lim_{j\to\infty}r_j\in\R$. We define
$\Grat_i(P,r_0):=\lim_{j\to\infty}\Grat_i(P,r_j)$. This limit exists since $\Grat_i(P,\cdot)$ is piecewise continuous and,
for $j$ large enough, all $r_j$ are contained in the same continuous part of $\Grat_i(P,\cdot)$. Then, since $\Grat(P,\cdot):\R_{\geq 0}\to\Z_{\geq 0}$
only jumps for rational points, we get
\begin{equation*}\begin{split}
\#(r_0P\cap\Z^n)&=\lim_{j\to\infty}\#(r_jP\cap\Z^n)=\lim_{j\to\infty}\sum_{i=0}^n\Grat_i(P,r_j)r_j^i\\
&=\sum_{i=0}^n\lim_{j\to\infty}\Grat_i(P,r_j)r_j^i=\sum_{i=0}^n\Grat_i(P,r_0)r_0^i.
\end{split}\end{equation*}
Baldoni et al.~\cite{KOEPPEetAl2011} generalized this statement to intermediate sums of rational polytopes
and gave an efficient algorithm to compute these real quasi-polynomials.

\end{remark}
Now we show that the coefficients $\Grat_i(P,\cdot)$
are derivatives of each other.
\begin{proof}[Proof of Theorem \ref{thm:derivatives}]
Let $0=r_0<r_1<\ldots<r_l=\rd(P)$ be as in the proof of Theorem \ref{thm:polycoeffs}, and for $m=1,\ldots,l$, $k\in\Z_{\geq 0}$ let
\[c_{m,k}=\Grat(P,r)=\sum_{i=0}^n\Grat_i(P,r)r^i\quad \text{for }r\in(r_{m-1}+k\,\rd(P),r_{m}+k\,\rd(P)).\]
Since $\Grat_i(P,r)$ is a polynomial of degree $n-i$ in $r$, we can write it as
\begin{equation*}
\Grat_i(P,r)=\sum_{j=0}^{n-i}\Grat_{i,j}r^j.
\end{equation*}
Since $\Grat_i(P,r)$ are periodic with period $\rd(P)$, we can write $r=\tilde{r}+k\,\rd(P)$ with $k\in\Z_{\geq 0}$ and
$r_{m-1}\leq \tilde{r}<r_{m}$ for some $m=1,\ldots,l$ and get
\begin{equation*}\begin{split}
c_{m,k}&=\sum_{i=0}^n\Grat_i(P,\tilde{r})(\tilde{r}+k\,\rd(P))^i
=\sum_{i=0}^n\sum_{j=0}^{n-i}\Grat_{i,j}\tilde{r}^j(\tilde{r}+k\,\rd(P))^i\\
&=\sum_{h=0}^n\sum_{i=0}^{n}\sum_{j=\max(0,h-i)}^{\min(h,n-i)}\binom{i}{h-j}\Grat_{i,j}(k\,\rd(P))^{i-h+j}\tilde{r}^{h},\\
\end{split}
\end{equation*}
which is a constant polynomial in $\tilde{r}$. Thus, for $h\neq 0$,
\begin{equation*}
\sum_{i=0}^{n}\sum_{j=\max(0,h-i)}^{\min(h,n-i)}\binom{i}{h-j}\Grat_{i,j}(k\,\rd(P))^{i-h+j}=0
\end{equation*}
and therefore
\begin{equation*}
\Grat(P,\tilde{r}+k\,\rd(P))
=\sum_{i=0}^{n}\Grat_{i,0}(k\,\rd(P))^{i}=\sum_{i=0}^{n}\Grat_{i,0}(r-\tilde{r})^{i}.
\end{equation*}
Expanding to the quasi-polynomial form yields
\begin{equation*}
\begin{split}
\Grat(P,\tilde{r}+k\,\rd(P))&=\sum_{i=0}^{n}\Grat_{i,0}\sum_{j=0}^i\binom{i}{j}r^j\tilde{r}^{i-j}(-1)^{i-j}\\
&=\sum_{j=0}^{n}\left(\sum_{i=0}^{n-j}\binom{i+j}{j}\Grat_{i+j,0}(-1)^{i}\tilde{r}^{i}\right)r^j.
\end{split}
\end{equation*}
This implies that for all $\tilde{r}\in (r_{m-1},r_m)$, $m=1,\ldots,l$,
\begin{equation*}
\Grat_j(P,\tilde{r})=\sum_{i=0}^{n-j}\binom{i+j}{j}\Grat_{i+j,0}(-1)^{i}\tilde{r}^{i}
\end{equation*}
and the claim follows by differentiation.

\end{proof}

\section{Dimension 2}\label{sec:dim2}
In what follows, we denote by $\gauss{.}$ the floor function, that is, $\gauss{x}$ is the largest integer not greater than $x$,
by $\lceil.\rceil$ the ceiling function, that is, $\lceil x\rceil$ is the smallest integer not smaller than $x$,
and by $\rat{.}$ the fractional part, that is, $\rat{x}=x-\gauss{x}$.
For the following calculations, we mention the following fact: Let $n,m,t,r\in\Z$, $m>0$ and $t\equiv r\bmod m$. Then
$\gauss{\frac{nt}{m}}=\frac{nt}{m}-\rat{\frac{nr}{m}}$ and
$\left\lceil\frac{nt}{m}\right\rceil=\frac{nt}{m}+\rat{-\frac{nr}{m}}$.

First, we consider $2$-dimensional triangles of the form
$T=\conv\left(\binom{0}{0},\binom{x_1}{y},\binom{x_2}{y}\right)$,
where $x_1<x_2\in\Q$ and $y\in\Q_{>0}$.
\begin{figure}[ht]\centering
\includegraphics{figure2.eps}
\caption{Triangle $T$}\label{triangle2d}
\end{figure}
\begin{theorem}\label{thm_2d_triangle}
Let $T=\conv\left(\mybinom{0}{0},\mybinom{\frac{s_1}{t_1}\frac{a}{b}}{\frac{a}{b}},\mybinom{\frac{s_2}{t_2}\frac{a}{b}}{\frac{a}{b}}\right)$
 with $a,b,t_1,t_2\in\Z_{>0}$, $s_1,s_2\in\Z$, $\frac{s_2}{t_2}>\frac{s_1}{t_1}$, and $\gcd(a,b)=\gcd(s_1,t_1)=\gcd(s_2,t_2)=1$.
Then for $r\in\Q_{\geq 0}$ the following hold:
\begin{equation*}\begin{split}
\mathrm{(i)}\quad\Grat_2(T,r)&=\frac12\frac{a^2}{b^2}\left(\frac{s_2}{t_2}-\frac{s_1}{t_1}\right)\\
\mathrm{(ii)}\quad\Grat_1(T,r)&=\frac{a}{b}\left(\frac{t_1+t_2}{2t_1t_2}-\left(\rat{\frac{ar}{b}}-\frac12\right)\left(\frac{s_2}{t_2}-\frac{s_1}{t_1}\right)\right) \\
\mathrm{(iii)}\quad\Grat_0(T,r)&=1
-\frac{1}{2}\rat{\frac{ar}{b}}\left(\frac{s_2}{t_2}-\frac{s_1}{t_1}+2\right)
+\frac{1}{2}\rat{\frac{ar}{b}}^2\left(\frac{s_2}{t_2}-\frac{s_1}{t_1}\right)\\
&\quad+\rat{\frac{ar}{b\lcm{t_1,t_2}} }\lcm{t_1,t_2}\left(\frac{t_2-1}{2t_2}+\frac{t_1-1}{2t_1}\right)\\
&\quad-\sum_{i=0}^{\scriptscriptstyle\rat{\gauss{\frac{ar}{b}} /\lcm{t_1,t_2} }\lcm{t_1,t_2}}
\left(\frac{s_2i}{t_2}-\gauss{\frac{s_2i}{t_2}}+\left\lceil\frac{s_1i}{t_1}\right\rceil-\frac{s_1i}{t_1}\right).
\end{split}\end{equation*}

\end{theorem}

\begin{proof}
In what follows, we determine $\Grat(T,t)=\sum_{i=0}^{\gauss{at/b}}\Grat(Q,i)$, $t\in\Q_{\geq 0}$,
$Q=\conv\left(\mybinom{\frac{s_1}{t_1}}{1},\mybinom{\frac{s_2}{t_2}}{1}\right)$, (see Figure \ref{triangle2d},
\cite{Sam&Woods2009}).
For abbreviation we write $l$ instead of $\lcm{t_1,t_2}$ and $r$ for an arbitrary integer number with $r\equiv t\mod lb$.
It is
\begin{equation*}\begin{split}
\Grat(Q,t)&=\#(tQ\cap\Z^2)=\gauss{\frac{s_2t}{t_2}}-\left\lceil\frac{s_1t}{t_1}\right\rceil+1\\
&=\left(\frac{s_2}{t_2}-\frac{s_1}{t_1}\right)t-\left(\rat{\frac{s_2r}{t_2}}+\rat{-\frac{s_1r}{t_1}}\right)+1.
\end{split}\end{equation*}
This implies
\begin{equation}\label{eq:2d-theorem}
\Grat(T,t)
=\sum_{i=0}^{\gauss{at/b}}\left(\frac{s_2}{t_2}-\frac{s_1}{t_1}\right)i-\left(\rat{\frac{s_2i}{t_2}}
+\rat{1-\frac{s_1i}{t_1}}\right)+1.
\end{equation}
Since $\gauss{at/b}=\frac{at}{b}-\rat{\frac{ar}{b}}$, the first part can be written as
\begin{equation*}
\sum_{i=0}^{\gauss{at/b}}i=t^2\frac{a^2}{2b^2}
+t\frac{a}{b}\left(\frac12-\rat{\frac{ar}{b}}\right)
+\frac12\rat{\frac{ar}{b}}^2-\frac12\rat{\frac{ar}{b}}.
\end{equation*}
For the second part, we remark that $\rat{\frac{s_2i}{t_2}}$ is periodic with period $t_2$ and
\begin{equation*}
\sum_{i=0}^{l-1}\rat{\frac{s_2i}{t_2}}=\frac{l}{t_2}\sum_{i=0}^{t_2-1}\rat{\frac{s_2i}{t_2}}
=\frac{l(t_2-1)}{2t_2}.
\end{equation*}
Thus, we get
\begin{equation*}\begin{split}
\sum_{i=0}^{\gauss{at/b}}\rat{\frac{s_2i}{t_2}}
&=\gauss{\frac{ \gauss{\frac{at}{b}} } {l}}\sum_{i=0}^{l-1}\rat{\frac{s_2i}{t_2}}
+\sum_{i=0}^{\scriptscriptstyle\rat{\gauss{\frac{ar}{b}} /l }l}\rat{\frac{s_2i}{t_2}}\\
&=\left(\frac{at}{bl}-\rat{\frac{ar}{bl}}\right)\frac{l(t_2-1)}{2t_2}
+\sum_{i=0}^{\scriptscriptstyle\rat{\gauss{\frac{ar}{b}} /l }l}\rat{\frac{s_2i}{t_2}},
\end{split}\end{equation*}
and similarly
\begin{equation*}
\sum_{i=0}^{\gauss{at/b}}\rat{-\frac{s_1i}{t_1}}
=\left(\frac{at}{bl}-\rat{\frac{ar}{bl}}\right)\frac{l(t_1-1)}{2t_1}
+\sum_{i=0}^{\scriptscriptstyle\rat{\gauss{\frac{ar}{b}} /l }l}\rat{1-\frac{s_1i}{t_1}}.
\end{equation*}
After some elementary algebra, \eqref{eq:2d-theorem} expands to the claim.

\end{proof}

In particular, the theorem shows that (as shown in Section \ref{sec:rat}) $\frac{b}{a}$ is a period of $\Grat_1(T,\cdot)$,
$\Grat_1(T,\cdot)$ is piecewise linear,
and that $\frac{b\lcm{t_1,t_2}}{a}$ is a period of $\Grat_0(T,r)$. Furthermore, $\Grat_0(T,r)$ is piecewise quadratic, and
the pieces are equal up to a constant depending only on $k$ (see Figure~\ref{g01pic}).
\begin{figure}[ht]\centering
\mbox{}\hfill\includegraphics{figure3.eps}\hfill\includegraphics{figure4.eps}\hfill\mbox{}
\caption{Continuous $\Grat_1(P,r)$ and $\Grat_0(P,r)$}\label{g01pic}
\end{figure}

\begin{corollary}\label{g1_corollary}
Let $T$ be as in Theorem \ref{thm_2d_triangle}, $g_i=\conv\left(
\left(\begin{matrix}0\\0\end{matrix}\right),
\left(\begin{matrix}\frac{a}{b}\frac{s_i}{t_i}\\ \frac{a}{b}\end{matrix}\right)\right)$,
for $i=1,2$.
Then $|\Grat_1(T,r)|\leq \Grat_1(T,0)$ for all $0\leq r<b\lcm{t_1,t_2}$.
More precisely, $\Grat_1(T,r)\geq -\Grat_1(T,0)+\Grat_1(g_1,r)+\Grat_1(g_2,r)$.

\end{corollary}
\begin{proof}
It is $\Grat_2(g_i,r)=0$, $\Grat_1(g_i,r)= \frac{a}{bt_i}$, and $\Grat_0(g_i,r)=1-\left(\frac{ar}{bt_i}-\gauss{\frac{ar}{bt_i}}\right)$
for all $r\in\Q$, $i=1,2$.
Since $\frac{s_2}{t_2}-\frac{s_1}{t_1}>0$ we get
\begin{equation*}\begin{split}
\Grat_1(T,r)&=\frac{a}{b}\left(\frac{t_1+t_2}{2t_1t_2}-\left(\rat{\frac{ar}{b}}-\frac12\right)
\left(\frac{s_2}{t_2}-\frac{s_1}{t_1}\right)\right)\\
&\leq \frac{a}{b}\left(\frac{t_1+t_2}{2t_1t_2}+\frac12\left(\frac{s_2}{t_2}-\frac{s_1}{t_1}\right)\right)=\Grat_1(T,0).
\end{split}\end{equation*}
Furthermore
\begin{equation*}\begin{split}
\Grat_1(T,r)&=\frac{a}{b}\left(\frac{t_1+t_2}{2t_1t_2}-\rat{\frac{ar}{b}}
\left(\frac{s_2}{t_2}-\frac{s_1}{t_1}\right)
+\frac12\left(\frac{s_2}{t_2}-\frac{s_1}{t_1}\right)\right)\\
&\geq\frac{a}{b}\left(-\frac{t_1+t_2}{2t_1t_2}
-\frac12\left(\frac{s_2}{t_2}-\frac{s_1}{t_1}\right)\right)+\frac{a}{b}\left(\frac{t_1+t_2}{t_1t_2}\right)\\
&=-\Grat_1(T,0)+\Grat_1(g_1,r)+\Grat_1(g_2,r).\qedhere
\end{split}\end{equation*}

\end{proof}

\begin{remark}
An analogous statement of Corollary \ref{g1_corollary} for $\Grat_0(T,\cdot)$ is not true. To see this, we
consider the triangles
\[T_\alpha=\conv\left(\binom{0}{0},\binom{\frac{\alpha-1}{\alpha}}{1},\binom{\frac{\alpha+1}{\alpha}}{1}\right),\quad \alpha\in\Z.\]
Together with Theorem \ref{thm_2d_triangle}, we get, for $m\in\Z_{\geq 0}$, $0\leq k<\alpha$ and $0\leq \tilde{r}<1$, that
\[\Grat_0(T_\alpha,m\alpha+k+\tilde{r})=\frac{1}{\alpha}\left(k(\alpha-k-2)+\tilde{r}^2-2\tilde{r}\right)+1.\]
Then for $k\sim \alpha/2$,
\[\Grat_0(T_\alpha,m\alpha+k+\tilde{r})\sim\frac{\alpha}{4}+\frac{1}{\alpha}\left(\tilde{r}^2-2\tilde{r}\right)\geq\frac{\alpha}{4}-\frac{1}{\alpha}\]
which tends to infinity, when $\alpha$ goes to infinity, but $\Grat_0(T,0)=1$.

\end{remark}
Now we can deduce the inequality $|G_1(P,r)|\leq G_1(P,0)$ for arbitrary rational polygons:
\begin{proof}[Proof of Theorem \ref{thm:dim2}]
We first consider only integral dilation factors, that is, we show that
\begin{equation}\label{eq::g1}
|\G_1(P,k)|\leq \G_1(P,0)\quad\text{for all }k\in\Z.
\end{equation}
For this, we use several steps:\\
\emph{1. An integral version of Corollary \ref{g1_corollary} holds true for $G_1(P,k)$ for arbitrary triangles $P$ with at least one integral vertex.}
This is true since $G_1(P,k)$ is invariant under translations and unimodular transformations.\\
\emph{2. \eqref{eq::g1} is true for every rational polygon $P$ with one integral point in its interior.}
Let $P$ be an arbitrary $2$-dimensional polygon with $m$ vertices and let $z$ be an integral point in the interior
of $P$. We consider the triangulation of $P$ as given in Figure \ref{6gon}.
\begin{figure}[ht]\centering
\includegraphics{figure5.eps}
\caption{Triangulation of $P$}\label{6gon}
\end{figure}

Here let $g_0:=g_m$. Then for all $k\in\Z$,
\begin{equation*}
\G(P,k)=\sum_{i=1}^m\G(T_i,k)-\sum_{i=1}^m\G(g_i,k)+1.
\end{equation*}
Thus, expanding all Ehrhart polynomials yields
\begin{equation*}
\G_1(P,k)
=\sum_{i=1}^m\left(\G_1(T_i,k)-\frac12\G_1(g_{i-1},k)-\frac12\G_1(g_i,k)\right).
\end{equation*}
Thus, step 1 implies that $\G_1(P,k)\leq \G_1(P,0)$, and
\begin{equation*}\begin{split}
\G_1(P,k)&=\sum_{i=1}^m\left(\G_1(T_i,k)-\frac12\G_1(g_{i-1},k)-\frac12\G_1(g_i,k)\right)\\
&\geq\sum_{i=1}^m\left(-\G_1(T_i,0)+\frac12\G_1(g_{i-1},k)+\frac12\G_1(g_i,k)\right)\\
&= -\G_1(P,0).
\end{split}\end{equation*}
\emph{3. \eqref{eq::g1} is true for arbitrary rational polygons $P$.}
Let $l\in\Z_{\geq 0}$ such that $(l\d(P)+1)P$ contains at least one integral point in its interior.
Then for all $k\in\Z_{\geq 0}$,
\begin{equation*}
\G_1((l\d(P)+1)P,k)=\G_1(P,k)(l\d(P)+1).
\end{equation*}
Thus, from step c2 it follows that $|\G_i(P,k)|\leq \G_i(P,0)$.\\[6pt]
Finally we consider arbitrary rational dilation factors. Let $P$ be an arbitrary rational polygon and let
$r=\frac{p}{q}\in\Q_{\geq 0}$ with $r\leq d(P)$, where $p\in\Z_{\geq  0}$ and $q\in\Z_{>0}$. Then,
by Lemma \ref{lem:rat_dilations},
$\Grat_i(P,r)=\G_i\left(\frac1qP,p\right)q^i$. Hence, \eqref{eq::g1} implies that
$|\Grat_i(P,r)|\leq \Grat_i(P,0)$.
\end{proof}

\section*{Acknowledgements}
\sloppypar
The author would like to thank Martin Henk for helpful discussions and Matthias Beck
and the reviewers for helpful comments improving the manuscript.

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text:
\begin{document}
\maketitle
\begin{abstract}
In this survey paper we give an overview of a generalization, introduced by R. Bautista and the author, of the theory of mutation of quivers with potential developed in 2007 by Derksen-Weyman-Zelevinsky. This new construction allows us to consider finite dimensional semisimple $F$-algebras, where $F$ is any field. We give a brief account of the results concerning this generalization and its main consequences.
\end{abstract}
\section{Introduction}
Since the development of the theory of quivers with potentials created by Derksen-Weyman-Zelevinsky in \cite{5}, the search for a general concept of \emph{mutation of a quiver with potential} has drawn a lot of attention. The theory of quivers with potentials has proven useful in many subjects of mathematics such as cluster algebras, Teichm\"{u}ller theory, KP solitons, mirror symmetry, Poisson geometry, among many others. There have been different generalizations of the notion of a quiver with potential and mutation where the underlying $F$-algebra, $F$ being a field, is replaced by more general algebras, see \cite{4,8,10}.
This paper is organized as follows. In Section \ref{section2}, we review the preliminaries taken from \cite{1} and \cite{9}. Instead of working with an usual quiver, we consider the completion of the tensor algebra of $M$ over $S$, where $M$ is an $S$-bimodule and $S$ is a finite dimensional semisimple $F$-algebra. We will then see how to construct a cyclic derivation, in the sense of Rota-Sagan-Stein \cite{12}, on the completion of the tensor algebra of $M$. Then we introduce a natural generalization of the concepts of potential, right-equivalence and cyclical equivalence as defined in \cite{5}.
In Section \ref{section3}, we describe a generalization of the so-called \emph{Splitting theorem} (\cite[Theorem 4.6]{5}) and see how this theorem allows us to lift the notion of mutation of a quiver with potential to this more general setting.
Finally, in Section \ref{section4}, we recall the notion of species realizations and describe how the generalization given in \cite{1} allows us to give a partial affirmative answer to a question raised by J. Geuenich and D. Labardini-Fragoso in \cite{7}.

\pagebreak
\section{Preliminaries} \label{section2}
The following material is taken from \cite{1} and \cite{9}.

\begin{definition} Let $F$ be a field and let $D_{1},\ldots,D_{n}$ be division rings, each containing $F$ in its center and of finite dimension over $F$. Let $S=\displaystyle \prod_{i=1}^{n} D_{i}$ and let $M$ be an $S$-bimodule of finite dimension over $F$. Define the algebra of formal power series over $M$ as the set:
\begin{center}
$\mathcal{F}_{S}(M)=\left\{\displaystyle \sum_{i=0}^{\infty} a(i): a(i) \in M^{\otimes i}\right\}$
\end{center}
where $M^{0}=S$. Note that $\mathcal{F}_{S}(M)$ is an associative unital $F$-algebra where the product is the one obtained by extending the product of the tensor algebra $T_{S}(M)=\displaystyle \bigoplus_{i=0}^{\infty} M^{\otimes i}$.
\end{definition}

Let $\{e_{1},\ldots,e_{n}\}$ be a complete set of primitive orthogonal idempotents of $S$.

\begin{definition} An element $m \in M$ is legible if $m=e_{i}me_{j}$ for some idempotents $e_{i},e_{j}$ of $S$.
\end{definition}

\begin{definition} Let $Z=\displaystyle \sum_{i=1}^{n} Fe_{i} \subseteq S$. We say that $M$ is $Z$-freely generated by a $Z$-subbimodule $M_{0}$ of $M$ if the map $\mu_{M}: S \otimes_{Z} M_{0} \otimes_{Z} S \rightarrow M$ given by $\mu_{M}(s_{1} \otimes m \otimes s_{2})=s_{1}ms_{2}$ is an isomorphism of $S$-bimodules. In this case we say that $M$ is an $S$-bimodule which is $Z$-freely generated.
\end{definition}

Throughout this paper we will assume that $M$ is $Z$-freely generated by $M_{0}$.

\begin{definition} Let $A$ be an associative unital $F$-algebra. A cyclic derivation, in the sense of Rota-Sagan-Stein \cite{12}, is an $F$-linear function $\mathfrak{h}: A \rightarrow \operatorname{End}_{F}(A)$ such that:
\begin{equation}
\mathfrak{h}(f_{1}f_{2})(f)=\mathfrak{h}(f_{1})(f_{2}f)+\mathfrak{h}(f_{2})(ff_{1})
\end{equation}
for all $f,f_{1},f_{2} \in A$. Given a cyclic derivation $\mathfrak{h}$, we define the associated cyclic derivative $\delta: A \rightarrow A$ as $\delta(f)=\mathfrak{h}(f)(1)$.
\end{definition}

We now construct a cyclic derivative on $\mathcal{F}_{S}(M)$. First, we define a cyclic derivation on the tensor algebra $A=T_{S}(M)$ as follows. Consider the map:
\begin{center}
$\hat{u}: A \times A \rightarrow A$
\end{center}
given by $\hat{u}(f,g)=\displaystyle \sum_{i=1}^{n} e_{i}gfe_{i}$ for every $f,g \in A$. This is an $F$-bilinear map which is $Z$-balanced. By the universal property of the tensor product, there exists a linear map $u: A \otimes_{Z} A \rightarrow A$ such that $u(a \otimes b)=\hat{u}(a,b)$.
Now we define an $F$-derivation $\Delta: A \rightarrow A \otimes_{Z} A$ as follows. For $s \in S$, we define $\Delta(s)=1 \otimes s - s \otimes 1$; for $m \in M_{0}$, we set $\Delta(m)=1 \otimes m$. Then we define $\Delta: M \rightarrow T_{S}(M)$ by:
\begin{center}
$\Delta(s_{1}ms_{2})=\Delta(s_{1})ms_{2}+s_{1}\Delta(m)s_{2}+s_{1}m\Delta(s_{2})$
\end{center}
for $s_{1},s_{2} \in S$ and $m \in M_{0}$.
Note that the above map is well-defined since $M \cong S \otimes_{Z} M_{0} \otimes_{Z} S$ via the multiplication map $\mu_{M}$. Once we have defined $\Delta$ on $M$, we can extend it to an $F$-derivation on $A$. Now we define $\mathfrak{h}: A \rightarrow \operatorname{End}_{F}(A)$ as follows:
\begin{center}
$\mathfrak{h}(f)(g)=u(\Delta(f)g)$
\end{center}
We have
\begin{align*}
\mathfrak{h}(f_{1}f_{2})(f)&=u(\Delta(f_{1}f_{2})f) \\
&=u(\Delta(f_{1})f_{2}f)+u(f_{1}\Delta(f_{2})f) \\
&=u(\Delta(f_{1})f_{2}f)+u(\Delta(f_{2})ff_{1}) \\
&=\mathfrak{h}(f_{1})(f_{2}f)+\mathfrak{h}(f_{2})(ff_{1}).
\end{align*}
It follows that $\mathfrak{h}$ is a cyclic derivation on $T_{S}(M)$. We now extend $\mathfrak{h}$ to $\mathcal{F}_{S}(M)$. Let $f,g \in \mathcal{F}_{S}(M)$, then $\mathfrak{h}(f(i))(g(j)) \in M^{\otimes(i+j)}$; thus we define $\mathfrak{h}(f)(g)(l)=\displaystyle \sum_{i+j=l} \mathfrak{h}(f(i))(g(j))$ for every $non$-negative integer $l$.

In \cite[Proposition 2.6]{9}, it is shown that the $F$-linear map $\mathfrak{h}: \mathcal{F}_{S}(M) \rightarrow \operatorname{End}_{F}(\mathcal{F}_{S}(M))$ is a cyclic derivation. Using this fact we obtain a cyclic derivative $\delta$ on $\mathcal{F}_{S}(M)$ given by
\begin{center}
$\delta(f)=\mathfrak{h}(f)(1)$.
\end{center}

\begin{definition} Let $\mathcal{C}$ be a subset of $M$. We say that $\mathcal{C}$ is a right $S$-local basis of $M$ if every element of $C$ is legible and if for each pair of idempotents $e_{i},e_{j}$ of $S$, we have that $C \cap e_{i}Me_{j}$ is a $D_{j}$-basis for $e_{i}Me_{j}$.
\end{definition}
We note that a right $S$-local basis $\mathcal{C}$ induces a dual basis $\{u,u^{\ast}\}_{u \in \mathcal{C}}$, where $u^{\ast}: M_{S} \rightarrow S_{S}$ is the morphism of right $S$-modules defined by $u^{\ast}(v)=0$ if $v \in \mathcal{C} \setminus \{u\}$; and $u^{\ast}(u)=e_{j}$ if $u=e_{i}ue_{j}$.

Let $T$ be a $Z$-local basis of $M_{0}$ and let $L$ be a $Z$-local basis of $S$. The former means that for each pair of distinct idempotents $e_{i}$,$e_{j}$ of $S$, $T \cap e_{i}Me_{j}$ is an $F$-basis of $e_{i}M_{0}e_{j}$; the latter means that $L(i)=L \cap e_{i}S$ is an $F$-basis of the division algebra $e_{i}S=D_{i}$. It follows that the non-zero elements of the set $\{sa: s \in L, a \in T\}$ form a right $S$-local basis of $M$. Therefore, for every $s \in L$ and $a \in T$, we have the map $(sa)^{\ast} \in \operatorname{Hom}_{S}(M_{S},S_{S})$ induced by the dual basis.

\begin{definition} Let $\mathcal{D}$ be a subset of $M$. We say that $\mathcal{D}$ is a left $S$-local basis of $M$ if every element of $\mathcal{D}$ is legible and if for each pair of idempotents $e_{i},e_{j}$ of $S$, we have that $\mathcal{D} \cap e_{i}Me_{j}$ is a $D_{i}$-basis for $e_{i}Me_{j}$.
\end{definition}

Let $\psi$ be any element of $\mathrm{Hom}_{S}(M_{S},S_{S})$. We will extend $\psi$ to an $F$-linear endomorphism of $\mathcal{F}_{S}(M)$, which we will denote by $\psi_{\ast}$.

First, we define $\psi_{\ast}(s)=0$ for $s\in S$; and for $M^{\otimes l}$, where $l \geq 1$, we define $\psi_{\ast}(m_{1}\otimes \dotsm \otimes m_{l})=\psi (m_{1})m_{2} \otimes \cdots \otimes m_{l}\in M^{\otimes (l-1)}$ for $m_{1},\dots,m_{l}\in M$. Finally, for $f\in \mathcal{F}_{S}(M)$ we define $\psi_{\ast} (f)\in \mathcal{F}_{S}(M)$ by setting
$\psi_{\ast} (f)(l-1)=\psi_{\ast}(f(l))$ for each integer $l>1$. Then we define
\begin{center}
$\psi_{\ast} (f)=\displaystyle \sum _{l=0}^{\infty }\psi_{\ast} (f(l)).$
\end{center}

\begin{definition} Let $\psi \in M^{*}=\mathrm{Hom}_{S}(M_{S},S_{S})$ and $f\in \mathcal{F}_{S}(M)$. We define $\delta _{\psi }:\mathcal{F}_{S}(M)\rightarrow \mathcal{F}_{S}(M)$ as
\begin{center}
$\delta _{\psi }(f)=\psi_{\ast}(\delta (f))=\displaystyle \sum _{l=0}^{\infty }\psi_{\ast}(\delta (f(l))).$
\end{center}
\end{definition}

\begin{definition} Given an $S$-bimodule $N$ we define the \emph{cyclic part} of $N$ as $N_{cyc}:=\displaystyle \sum_{j=1}^{n} e_{j}Ne_{j}$.
\end{definition}
\begin{definition} A \emph{potential}  $P$ is an element of $\mathcal{F}_{S}(M)_{cyc}$.
\end{definition}

Motivated by the \emph{Jacobian ideal} introduced in \cite{5}, we define an analogous two-sided ideal of $\mathcal{F}_{S}(M)$.

For each legible element $a$ of $e_{i}Me_{j}$, we let $\sigma(a)=i$ and $\tau(a)=j$.

\begin{definition} Let $P$ be a potential in $\mathcal{F}_{S}(M)$, we define a two-sided ideal $R(P)$ as the closure of the two-sided ideal of $\mathcal{F}_{S}(M)$ generated by all the elements $X_{a^{\ast}}(P)=\displaystyle \sum_{s \in L(\sigma(a))} \delta_{(sa)^{\ast}}(P)s$, $a \in T$.
\end{definition}

In \cite[Theorem 5.3]{1}, it is shown that $R(P)$ is invariant under algebra isomorphisms that fix pointwise $S$. Furthermore, one can show that $R(P)$ is independent of the choice of the $Z$-subbimodule $M_{0}$ and also independent of the choice of $Z$-local bases for $S$ and $M_{0}$.

\begin{definition} An algebra with potential is a pair $(\mathcal{F}_{S}(M),P)$ where $P$ is a potential in $\mathcal{F}_{S}(M)$ and $M_{cyc}=0$.
\end{definition}

We denote by $[\mathcal{F}_{S}(M),\mathcal{F}_{S}(M)]$ the closure in $\mathcal{F}_{S}(M)$ of the $F$-subspace generated by all the elements of the form $[f,g]=fg-gf$ with $f,g\in \mathcal{F}_{S}(M).$

\begin{definition} Two potentials $P$ and $P'$ are called cyclically equivalent if $P-P' \in  [\mathcal{F}_{S}(M),\mathcal{F}_{S}(M)]$.
\end{definition}

\begin{definition} We say that two algebras with potential $(\mathcal{F}_{S}(M),P)$ and $(\mathcal{F}_{S}(M'),Q)$ are right-equivalent if there exists an algebra isomorphism $\varphi: \mathcal{F}_{S}(M) \rightarrow \mathcal{F}_{S}(M')$, with $\varphi|_{S}=id_{S}$, such that $\varphi(P)$ is cyclically equivalent to $Q$.
\end{definition}

The following construction follows the one given in \cite[p.20]{5}. Let $k$ be an integer in $[1,n]$ and $\bar{e}_{k}=1-e_{k}$. Using the $S$-bimodule $M$, we define a new $S$-bimodule $\mu_{k}M=\widetilde{M}$ as:
\begin{center}
$\widetilde{M}:=\bar{e}_{k}M\bar{e}_{k} \oplus Me_{k}M \oplus (e_{k}M)^{\ast} \oplus ^{\ast}(Me_{k})$
\end{center}
where $(e_{k}M)^{\ast}=\operatorname{Hom}_{S}((e_{k}M)_{S},S_{S})$, and $^{\ast}(Me_{k})=\operatorname{Hom}_{S}(_{S}(Me_{k}),_{S}S)$. One can show (see \cite[Lemma 8.7]{1}) that $\mu_{k}M$ is $Z$-freely generated.

\begin{definition} Let $P$ be a potential in $\mathcal{F}_{S}(M)$ such that $e_{k}Pe_{k}=0$. Following \cite{5}, we define
\begin{center}
$\mu_{k}P:=[P]+\displaystyle \sum_{sa \in _{k}\hat{T},bt \in \tilde{T}_{k}}[btsa]((sa)^{\ast})(^{\ast}(bt))$
\end{center}
\end{definition}
where:
\begin{align*}
_{k}\hat{T}&=\{sa: s \in L(k),a \in T \cap e_{k}M\} \\
\tilde{T}_{k}&=\{bt: b \in T \cap Me_{k}, t \in L(k)\}.
\end{align*}

\section{Mutations and potentials} \label{section3}
Let $P=\displaystyle \sum_{i=1}^{N} a_{i}b_{i}+P'$ be a potential in $\mathcal{F}_{S}(M)$ where $A=\{a_{1},b_{1},\ldots,a_{N},b_{N}\}$ is contained in a $Z$-local basis $T$ of $M_{0}$ and $P' \in \mathcal{F}_{S}(M)^{\geq 3}$. Let $L_{1}$ denote the complement of $A$ in $T$, $N_{1}$ be the $F$-vector subspace of $M$ generated by $A$ and $N_{2}$ be the $F$-vector subspace of $M$ generated by $L_{1}$; then $M=M_{1} \oplus M_{2}$ as $S$-bimodules where $M_{1}=SN_{1}S$ and $M_{2}=SN_{S}S$.

One of the main results proved in \cite{5} is the so-called \emph{Splitting theorem} (Theorem 4.6). Inspired by this result, the following theorem is proved in \cite{1}.

\begin{theorem} (\cite[Theorem 7.15]{1}) \label{split} There exists an algebra automorphism $\varphi: \mathcal{F}_{S}(M) \rightarrow \mathcal{F}_{S}(M)$ such that $\varphi(P)$ is cyclically equivalent to a potential of the form $\displaystyle \sum_{i=1}^{N} a_{i}b_{i}+P''$ where $P''$ is a reduced potential contained in the closure of the algebra generated by $M_{2}$ and $\displaystyle \sum_{i=1}^{N} a_{i}b_{i}$ is a trivial potential in $\mathcal{F}_{S}(M_{1})$.
\end{theorem}

\begin{definition} Let $P\in \mathcal{F}(M)$ be a potential and $k$ an integer in $\{1,\ldots,n\}$. Suppose that there are no two-cycles passing through $k$. Using Theorem \ref{split}, one can see that $\mu _{k}P$ is right-equivalent to the direct sum of a trivial potential $W$ and a potential $Q$ in $\mathcal{F}_{S}(M)^{\geq 3}$. Following \cite{5}, we define the mutation of $P$ in the direction $k$, as $\overline{\mu }_{k}(P)=Q$.
\end{definition}

One of the main results of \cite{5} is that mutation at an arbitrary vertex is a well-defined involution on the set of right-equivalence classes of reduced quivers with potentials. In \cite{1}, the following analogous result is proved.

\begin{theorem} (\cite[Theorem 8.21]{1}) Let $P$ be a reduced potential such that the mutation $\overline{\mu}_{k}P$ is defined. Then $\overline{\mu}_{k}\overline{\mu}_{k}P$ is defined and it is right-equivalent to $P$.
\end{theorem}

\begin{definition} Let $k_{1}, \hdots, k_{l}$ be a finite sequence of elements of $\{1,\hdots,n\}$ such that $k_{p} \neq k_{p+1}$ for $p=1,\hdots, l-1$. We say that an algebra with potential $(\mathcal{F}_{S}(M),P)$ is $(k_{l},\hdots,k_{1})$-nondegenerate if all the iterated mutations $\bar{\mu}_{k_{1}}P$, $\bar{\mu}_{k_{2}}\bar{\mu}_{k_{1}}P, \hdots, \bar{\mu}_{k_{l}} \cdots \bar{\mu}_{k_{1}}P$ are $2$-acyclic. We say that $(\mathcal{F}_{S}(M),P)$ is nondegenerate if it is $(k_{l},\hdots, k_{1})$-nondegenerate for every sequence of integers as above.
\end{definition}
In \cite{5}, it is shown that if the underlying base field $F$ is uncountable then a nondegenerate quiver with potential exists for every underlying quiver. Motivated by this result, the following theorem is proved in \cite{9}.

\begin{theorem} (\cite[Theorem 3.5]{9}) \label{teouncount} Suppose that the underlying field $F$ is uncountable, then $\mathcal{F}_{S}(M)$ admits a nondegenerate potential.
\end{theorem}

\section{Species realizations} \label{section4}
We begin this Section by recalling the definition of \emph{species realization} of a skew-symmetrizable integer matrix, in the sense of \cite{7} (Definition 2.22).

\begin{definition} \label{especies} Let $B=(b_{ij}) \in \mathbb{Z}^{n \times n}$ be a skew-symmetrizable matrix, and let $I=\{1,\hdots, n\}$. A species realization of $B$ is a pair $(\mathbf{S},\mathbf{M})$ such that:
\begin{enumerate}
\item $\mathbf{S}=(F_{i})_{i \in I}$ is a tuple of division rings;
\item $\mathbf{M}$ is a tuple consisting of an $F_{i}-F_{j}$ bimodule $M_{ij}$ for each pair $(i,j) \in I^{2}$ such that $b_{ij}>0$;
\item for every pair $(i,j) \in I^{2}$ such that $b_{ij}>0$, there are $F_{j}-F_{i}$-bimodule isomorphisms
\begin{center}
$\operatorname{Hom}_{F_{i}}(M_{ij},F_{i}) \cong \operatorname{Hom}_{F_{j}}(M_{ij},F_{j})$;
\end{center}
\item for every pair $(i,j) \in I^{2}$ such that $b_{ij}>0$ we have $\operatorname{dim}_{F_{i}}(M_{ij})=b_{ij}$ and $\operatorname{dim}_{F_{j}}(M_{ij})=-b_{ji}$.
\end{enumerate}
\end{definition}

In \cite[p.29]{1}, we impose the following condition on each of the bases $L(i)$. For each $s,t \in L(i)$:
\begin{equation} \label{eq2}
e_{i}^{\ast}(st^{-1}) \neq 0 \ \text{implies} \ s=t \ \text{and} \ e_{i}^{\ast}(s^{-1}t) \neq 0 \ \text{implies} \ s=t
\end{equation}
where $e_{i}^{\ast}: D_{i} \rightarrow F$ denotes the standard dual map corresponding to the basis element $e_{i} \in L(i)$.

In \cite[p.14]{7}, motivated by the seminal paper \cite{5}, J. Geuenich and D. Labardini-Fragoso raise the following question: \\

\textbf{Question} \cite[Question 2.23]{7} Can a mutation theory of species with potential be defined so that every skew-symmetrizable matrix $B$ have a species realization which admit a nondegenerate potential? \\

In \cite[Corollary 3.6]{9} a partially affirmative answer to Question $2.23$ is given by proving the following: let $B=(b_{ij}) \in \mathbb{Z}^{n \times n}$ be a skew-symmetrizable matrix with skew-symmetrizer $D=\operatorname{diag}(d_{1},\ldots,d_{n})$. If $d_{j}$ divides $b_{ij}$ for every $j$ and every $i$, then the matrix $B$ can be realized by a species that admits a nondegenerate potential.

We now give an example (\cite[p.8]{9}) of a class of skew-symmetrizable $4 \times 4$ integer matrices, which are not globally unfoldable nor strongly primitive, and that have a species realization admitting a nondegenerate potential. This gives an example of a class of skew-symmetrizable $4 \times 4$ integer matrices which are not covered by \cite{8}.

Let
\begin{equation} \label{eq3}
B
=\begin{bmatrix}
0 & -a & 0 & b \\
1 & 0 & -1 & 0 \\
0 & a & 0 & -b \\
-1 & 0 & 1 & 0
\end{bmatrix}
\end{equation}
where $a,b$ are positive integers such that $a<b$, $a$ does not divide $b$ and $\gcd(a,b) \neq 1$. \\

Note that there are infinitely many such pairs $(a,b)$. For example, let $p$ and $q$ be primes such that $p<q$. For any $n \geq 2$, define $a=p^{n}$ and $b=p^{n-1}q$. Then $a<b$, $a$ does not divide $b$ and $\gcd(a,b)=p^{n-1} \neq 1$. Note that $B$ is skew-symmetrizable since it admits $D=\operatorname{diag}(1,a,1,b)$ as a skew-symmetrizer. \\

\begin{remark} By \cite[Example 14.4]{8} we know that the class of all matrices given by \eqref{eq3} does \emph{not} admit a global unfolding. Moreover, since we are not assuming that $a$ and $b$ are coprime, then such matrices are not strongly primitive; hence they are not covered by \cite{8}.
\end{remark}

We have the following
\begin{proposition} (\cite[Proposition 5.2]{9})
The class of all matrices given by \eqref{eq3} are not globally unfoldable nor strongly primitive, yet they can be realized by a species admitting a nondegenerate potential.
\end{proposition}

By Theorem \ref{teouncount}, we know that a nondegenerate potential exists provided the underlying field $F$ is uncountable. If $F$ is infinite (but not necessarily uncountable) one can show that $\mathcal{F}_{S}(M)$ admits ``locally'' nondegenerate potentials. More precisely, we have

\begin{proposition} (\cite[Proposition 12.5]{1}) \label{infinite} Let $F$ be an infinite field and let $k_{1},\ldots,k_{l}$ be an arbitrary sequence of elements of $\{1,\ldots,n\}$. Then there exists a potential $P \in \mathcal{F}_{S}(M)$ such that the mutation $\overline{\mu}_{k_{l}} \cdots \overline{\mu}_{k_{1}}P$ exists.
\end{proposition}

We conclude the paper by giving an example of a class of skew-symmetrizable $4 \times 4$ integer matrices that have a species realization via field extensions of the rational numbers. Although in this case we cannot guarantee the existence of a nondegenerate potential, we can guarantee (by Proposition \ref{infinite}) the existence of ``locally'' nondegenerate potentials.

First we require some definitions.

\begin{definition} Let $E/F$ be a finite field extension. An $F$-basis of $E$, as a vector space, is said to be semi-multiplicative if the product of any two elements of the basis is an $F$-multiple of another basis element.
\end{definition}

It can be shown that every extension $E/F$ which has a semi-multiplicative basis satisfies (\ref{eq2}).

\begin{definition} A field extension $E/F$ is called a simple radical extension if $E=F(a)$ for some $a \in E$, with $a^n \in F$ and $n \geq 2$.
\end{definition}

Note that if $E/F$ is a simple radical extension then $E$ has a semi-multiplicative $F$-basis.

\begin{definition} A field extension $E/F$ is a radical extension if there exists a tower of fields $F=F_{0} \subseteq F_{1} \ldots \subseteq F_{l}=E$ such that $F_{i}/F_{i-1}$ is a simple radical extension for $i=1,\ldots, l$.
\end{definition}

As before, let
\begin{equation}
B
=\begin{bmatrix}
0 & -a & 0 & b \\
1 & 0 & -1 & 0 \\
0 & a & 0 & -b \\
-1 & 0 & 1 & 0
\end{bmatrix}
\end{equation}
but without imposing additional conditions on $a$ or $b$.

\begin{proposition} \label{rationals} Let $n,m \geq 2$. The matrix $B$ admits a species realization $(\mathbf{S},\mathbf{M})$ where $\mathbf{M}$ is a $Z$-freely generated $S$-bimodule and $S$ satisfies (\ref{eq2}).
\end{proposition}

\begin{proof} To prove this we will require the following result (cf. \cite[Theorem 14.3.2]{11}).

\begin{lemma} \label{lemroots} Let $n \geq 2$, $p_{1},\ldots,p_{m}$ be distinct primes and let $\mathbf{Q}$ denote the set of all rational numbers. Let $\zeta_{n}$ be a primitive $nth$-root of unity. Then
\begin{center}
$[\mathbf{Q}(\zeta_{n})(\sqrt[n]{p_{1}},\ldots,\sqrt[n]{p_{m}}): \mathbf{Q}(\zeta_{n})]=n^{m}$
\end{center}
\end{lemma}
Now we continue with the proof of Proposition \ref{rationals}. Let $F=\mathbf{Q}(\zeta_{n})$ be the base field and let $p_{1}$ be an arbitrary prime. By Lemma \ref{lemroots}, $F_{2}=F(\sqrt[n]{p_{1}})/F$ has degree $n$. Now choose $m-1$ distinct primes $p_{2},p_{3},\ldots,p_{m}$ and also distinct from $p_{1}$. Define $F_{4}=F(\sqrt[n]{p_{1}},\sqrt[n]{p_{2}},\ldots,\sqrt[n]{p_{m}})/F$, then by Lemma \ref{lemroots}, $F_{4}$ has degree $n^{m}$.
Let $S=F \oplus F_{2} \oplus F \oplus F_{4}$ and $Z=F \oplus F \oplus F \oplus F$. Since $F/\mathbf{Q}$ is a simple radical extension then it has a semi-multiplicative basis; thus it satisfies (\ref{eq2}). On the other hand, note that $F_{2}/\mathbf{Q}(\zeta_{n})$ and $F_{4}/\mathbf{Q}(\zeta_{n})$ are radical extensions. Using \cite[Remark 6, p.29]{1} we get that both $F_{2}$ and $F_{4}$ satisfy (\ref{eq2}); hence, it is always possible to choose a $Z$-local basis of $S$ satisfying (\ref{eq2}). Finally, for each $b_{ij}>0$, define $e_{i}Me_{j}=(F_{i} \otimes_{F} F_{j})^{\frac{b_{ij}}{d_{j}}}=F_{i} \otimes_{F} F_{j}$. It follows that $(\mathbf{S},\mathbf{M})$ is a species realization of $B$.
\end{proof}

config: arxiv


TITLE: How can I calculate the number of nodes on a plucked string?
QUESTION [0 upvotes]: Given the length of the guitar string, can you predict the number of nodes that would exist on that string?
My textbook says that there are an infinite number of harmonics that can be created when a string is plucked, however upon observing guitar strings I find that there are a distinct number of nodes upon plucking. So how would I predict how many nodes would exist on a guitar string when it would be plucked?

REPLY [0 votes]: For a single frequency, there is a finite number of nodes, which you can find by fitting an integer number of half-waves along the string.
But since there are infinite frequencies, there are an infinite number of possible nodes.
Let's say the fundamental frequency is $n_0$ such that the length of the string fits 1/2 wave
$$ n_0 = \frac{c}{2 \ell} $$ where $c$ is the wave speed on the string.
A string vibrating with an integer multiply of the fundamental frequency $n = i\, n_0$ has a total of $i$ nodes along its length. You can check the half-wavelength $\lambda = c/(2 n)$ fits $i$ times in the length $\ell$.
So for each frequency $n = i\,n_0$ there are $i$ nodes, but since $i$ can be anything integer positive $i >0$, there are inifnite possibliities.


text:
\section{General results on the mutual information}\label{app:matrix}

In this appendix we give a more general form of theorem \ref{thm:ws} in section~\ref{sec:matrixWinfotheorresults}. Our analysis by the adaptive interpolation method works for any regime where the sequences $\lambda_n$ and $\rho_n$ verify:
\begin{align}
C\le \lambda_n\rho_n = O(n^{\gamma})\, \quad \text{for some constants} \quad \gamma \in [0,1/2) \quad \text{and} \quad C>0\,.
\label{app-scalingRegime}
\end{align}
Of course this contains the regime \eqref{mainregime} as a special case. Our general result is a statement on the smallness of
\begin{align*}
\Delta I_n \equiv \frac{1}{\rho_n|\ln\rho_n|}\Big|\frac{1}{n}I(\bX;\bW) -\inf_{q\in [0,\rho_n]} i^{\rm pot}_n(q,\lambda_n,\rho_n)\Big|\,.
\end{align*}
The analysis of section \ref{sec:adapInterp_XX} leads to the following general theorem.

\begin{thm}[Sparse spiked Wigner model]\label{thm:wsgeneral}
Let the sequences $\lambda_n$ and $\rho_n$ verify \eqref{app-scalingRegime} and let $\alpha>0$.
There exists a constant $C>0$ independent of $n$, such that the mutual information for the Wigner spike model verifies
\begin{align*}
\Delta I_n\le \frac{C}{|\ln\rho_n|}\max\Big\{\frac{1}{n^\alpha}, \, \frac{\lambda_n}{n\rho_n}, \, \Big(\frac{\lambda_n^4}{n^{1-4\alpha}\rho_n^2}\big(1+\lambda_n\rho_n^2\big)\Big)^{1/3}\Big\}\,.
\end{align*}
In particular, choosing $\lambda_n= \Theta(|\ln\rho_n|/\rho_n)$ (which is the appropriate scaling to observe a phase transition),

\begin{align*}
\Delta I_n \le C\max\Big\{\frac{1}{n^{\alpha}|\ln\rho_n|}, \, \frac{1}{n\rho_n^2}, \, \Big(\frac{|\ln \rho_n|}{n^{1-4\alpha}\rho_n^6}\Big)^{1/3}\Big\}\,.
\end{align*}
If, in addition, we set $\rho_n=\Omega(n^{-\beta})$ for $\beta\ge 0$ (which is the regime in \eqref{mainregime}), then we have
\begin{align*}
\Delta I_n \le C\max\Big\{\frac{1}{n^{\alpha}\ln n}, \, \frac{1}{n^{1-2\beta}}, \, \Big(\frac{\ln n}{n^{1-4\alpha-6\beta}}\Big)^{1/3}\Big\}\,.
\end{align*}
This bound vanishes as $n$ grows if $\beta\in[0,1/6)$ and $\alpha\in(0, (1-6\beta)/4]$. The final bound is optimized (up to polylog factors)  by setting $\alpha=(1-6\beta)/7$. In this case (again, when $\lambda_n= \Theta(|\ln\rho_n|/\rho_n)$ and $\rho_n=\Omega(n^{-\beta})$),
\begin{align*}
\Delta I_n \le C\frac{(\ln n)^{1/3}}{n^{(1-6\beta)/7}}\,.
\end{align*}
\end{thm}


TITLE: Question about the uniqueness of solutions?
QUESTION [0 upvotes]: I was working on a few practice problems for my applied ordinary differential equations class and I was stumped on a really simple intro problem.
If we are told that a DFQ has the following form:
$$\frac{du}{dx} = f(u,x)$$
Is there any way to infer the number of solutions it has? Can $u(x) = x^2$ and $u(x) = x^3$ both be solutions? I'm thinking not, since usually a DFQ with multiple solutions has all solutions differ by a constant, not something with the dependent variable.
Are there standard rules for this?
Any help is extremely appreciated!

REPLY [1 votes]: You can get both $u=x^2$ and $u=x^3$ as solutions if the exponent is the integration constant, that is, the general solution is $u=x^C$. Isolating $C$ and taking the derivative results in
$$
C=\frac{\ln u}{\ln x}\implies 0=\frac{u'}{u\ln x}-\frac{\ln u}{x(\ln x)^2}
$$
so that the corresponding differential equation is
$$
u'=\frac{u\ln u}{x\ln x}.
$$

You can also take other forms of interpolation like $u=Cx^2+(1-C)x^3$ so that
$$
C=\frac{u-x^3}{x^2-x^3}\implies 0=\frac{u'-3x^2}{x^2-x^3}-\frac{(u-x^3)(2x-3x^2)}{(x^2-x^3)^2}
$$
leading to the differential equation
$$
u'=3x^2+\frac{(2-3x)(u-x^3)}{x(1-x)}
$$
etc.


TITLE: What are models of inconsistent theory?
QUESTION [2 upvotes]: Consider arbitrary language $\mathcal{L}$, inconsistent theory $T$ (e.g. $T=\{ \forall x P(x) \land \forall x \neg P(x) \}$ for $\mathcal{L}=\{P\}$) and $A = \{\ \mathcal{M} \mid \mathcal{M} \models T \}$. How does the set $A$ look like?

REPLY [1 votes]: An inconsistent theory has no models, so $A=\varnothing$.


TITLE: Infinite domain with finite number of prime ideals(elements)
QUESTION [2 upvotes]: While trying to prove one property of commutative rings with units I can't prove one fact without assuming existence of infinitely many different prime ideals or elements. I tried to test if it was the neccesary assumptions, but I failed, since I don't know any "toy"-examples of such rings.
I know only one example of this kind ($\mathbf{Q} [x]/ (x^2) $ ) but it's not a domain.
So,
Are there any infinite domains with finite number of prime ideals?
If no, then are there any infinite domains with finite (but nontrivial) number of prime elements?
I am interested in noncommutative examples as well.
Sorry if this question is too elementary.

REPLY [1 votes]: For a noncommutative example you could take the first Weyl algebra $A_1(k)$ with generated over a field $k$ by elements $x$ and $y$ subject to the relation $xy-yx-1$. This is a Noetherian domain which is simple, that is the only two-sided ideals are 0 and the ring itself.


TITLE: $\int_0^{2\pi} \text{arctan} \Big[\frac{\text{sin}\ \theta}{\text{cos}\ \theta +3}\Big] d\theta$
QUESTION [0 upvotes]: I've been asked to solve this definite integral:
$$ \int_0^{2\pi} \text{arctan} \Big[\frac{\text{sin}\ \theta}{\text{cos}\ \theta +3}\Big] d\theta$$
It's obvious that it is finite and a bad approximation is $<\pi^2$. One can say more because it can be proven that the integrand is odd, so the result is $0$.
However I'm not interested in this solution because I need a way to see it using Complex Analysis.

REPLY [3 votes]: Clearly, the integrand equals
$$\Im \ln (\cos \theta+3+i\sin\theta)=\Im\ln(3+e^{i\theta})$$
By the substitution $z=e^{i\theta}$, the integral becomes
$$\Im\oint_{|z|=1}\ln(3+z)\frac{dz}{iz}=\Im \, 2\pi i\cdot\frac{\ln 3}{i}=0$$
by residue theorem.


TITLE: How to show that $\cos(x)=\sum\limits_{n=0}^\infty (-1)^n \frac {x^{2n}}{ (2n)!} $
QUESTION [0 upvotes]: We know that cos(x) is infinitely differentiable and the Lagrange remainder $\rightarrow 0$ for all $x$, so the Taylor series indeed produces the function. We also know that
$\cos^{(4k)}(x)=\cos(x)$ at $x=0$ its $1$
$\cos^{(4k+1)}(x)=-\sin(x)$ at $x=0$ its $0$
$\cos^{(4k+2)}(x)=-\cos(x)$ at $x=0$ its $-1$
$\cos^{(4k+3)}(x)=\sin(x)$ at $x=0$ its $0$
Now the question is how do i get that sum from these informations?

REPLY [0 votes]: By Taylor's expansion (with $f^n(a) = \left.\frac{d^n}{dx^n}f(x)\right|_{x=a}$)
$$
f(0+x) = \sum_{n=0}^{\infty} \frac{x^nf^n(0)}{n!}
$$
$\cos(x)$ becomes:
\begin{align*}
\cos(x) &= \sum_{n=0}^{\infty} \left(\frac{x^{4n}\cdot1}{(4n)!}+\frac{x^{4n+1}\cdot 0}{(4n+1)!}+\frac{x^{4n+2}\cdot(-1)}{(4n+2)!}+\frac{x^{4n+3}\cdot 0}{(4n+3)!} \right)
\\
&= \sum_{n=0}^{\infty} \left(\frac{x^{4n}\cdot 1}{(4n)!}+\frac{x^{4n+2}\cdot(-1)}{(4n+2)!}\right)\\
&= \sum_{n=0}^{\infty} \left(\frac{x^{2(2n)}\cdot(-1)^{2n}}{(2(2n))!}+\frac{x^{2(2n+1)}\cdot (-1)^{(2n+1)}}{(2(2n+1))!}\right)
\end{align*}
Now, put $2n=m$, to get
$$
\cos(x) = \sum_{m=0,2,4}^{\infty} \left(\frac{x^{2m}.(-1)^{m}}{(2m)!}+\frac{x^{2(m+1)}.(-1)^{(m+1)}}{(2(m+1))!}\right)
$$
Now, the first term uses even powers for $-1$ and second term uses odd. So, this can be rewritten as the required form.
PS: It might be easier to prove using the expansion of $e^{jx}$ and collecting the real part of that.


TITLE: Calculate the number of paths found on the basis of probabilities
QUESTION [2 upvotes]: Let G=<V,E> - weighted directed graph
$w(e_{ij})$ - transition probability from node $v_i$ to node $v_j$ ($w \in[0;1]$)
So my first question is: how calculate the probability of path $u \rightarrow v $ ?
My answer (that I would like to check on the truth): $p(u \rightarrow v) = \prod p(e_j)$, where $e_j$ are edges in this path.
Second question is: how calculate the probability of reaching from $u\in V$ to $v\in V$ (e.g. using information about the probabilities of transitions or probability of paths $u \rightarrow v $)?
Third question is: Let $P(u,v)$ - probability of reaching from $u$ to $v$. $p(u \rightarrow v)$ - probability of path from $u$ to $v$. Can I count the numbers of paths that still need to be found in addition to the found $u→v$ one?
For example, let:
$P(u,v)=0.5$
$p(u \rightarrow v) = 0.2$
$Result = \frac{p(u \rightarrow v)}{P(u,v)}=\frac{0.2}{0.5}*100$% $= 40$% path found.
I'll be glad to answer any of these questions

REPLY [1 votes]: First question: You are basically right, but let's lay some notation.
I assume your graph is finite (finite edges and finite vertices). Let a path be a set $\{e_i\}$ for $i=1,2,...,k$ where each $e_i$ is an edge and the $e_i$ lands where $e_{i+1}$ starts, for all $i=1,2,...,k-1$.
Now, given this path, call the probability of following it $p\big(\{e_i\}\big)$. This will be equal to the product of the weights of each edge: $$p\big(\{e_i\}\big) =\prod_{i=1}^kw(e_i)$$
Second question: The probability of reaching a vertex $v$ starting from $u$ will be the sum of the probabilities of following a path that starts at $u$ and ends at $v$. Notice that these paths from $u$ to $v$ can in principle include paths that come back to $u$, but the last edge will always be the only one to land at $v$.
Third question: Short answer is no. You can see it from your example: using your method you conclude that you have found $40%$ of the paths from $u$ to $v$. But if $1=\frac{40}{100}N$, then $N=2.5$ which is not an integer!
The reason you cannot do this in general, is that knowing the probability of one path only fixes yo sum of the probabilities of the other paths (in your example this would be $0.5-0.2=0.3$). You could have many other paths, with low probabilities, or just one more path.
In general, to answer the question of how many paths from $u$ to $v$ exist, you will need to look at other properties of the graph.


TITLE: Differentiable in real analysis problem
QUESTION [0 upvotes]: Let $f$ be a real value function that is differentiable in $[a,b]$, with .
Show that there is exists $x_0$ in $(a,b)$ such that:

I don't know how to prove it. Please, could anyone give me a hint. Thanks.

REPLY [1 votes]: First, it's almost certainly "show there's an $x_0 \in [a, b]$ rather than $[a. v]$.
But the hypotheses seem to have a problem. If $f(a) = f(b) = 0$ and $f$ is everywhere differentiable, then $f'(c) = 0$ for some $c \in (a, b)$ by Rolle's Theorem. Yet you've got a hypothesis that says $f'(x) \ne 0$ for all $x \in [a, b]$.
I'd carefully check that your problem is stated correctly.
As stated, the problem's easy: a false hypothesis implies any conclusion.


TITLE: Construction of invariants of 4-manifolds with the Kirby calculus
QUESTION [14 upvotes]: I'm an undergraduate student, interested in the low dimensional topology, in particular, the 4-manifold theory.
I have a question.
In the knot theory, the Reidemeister moves play fundamental roles.
For instance, to prove the fact that the Jones polynomial is an invariant of knots, we can use the Reidemeister moves.
On the other hand, in the 4-manifold theory, there is the Kirby calculus, which play roles similar to the Reidemeister moves.
So, are there some studies about construction of an invariant of 4-manifolds by using the Kirby calculus?
Thanks for your help.

REPLY [11 votes]: Disclaimer: Shameless self-advertising.

Yes, it can be done, and it's really beautiful! You can define the Crane-Yetter invariant with Kirby calculus, and possibly other TQFTs ("dichromatic models"). I've written this down in this article:
https://doi.org/10.1007/s00220-017-3012-9
If you want a version with more impressions and pictures, and less text, look at the talk slides:
https://www.manuelbaerenz.de/article/understanding-crane-yetter-model
The Broda invariant is a special case of the dichromatic framework, which was developed by Jerome Petit (and probably Alain Bruguières).
The dichromatic and Crane-Yetter invariants are stronger than signature and Euler characteristic. They are sensitive to the fundamental group (not just homology), but still fail to distinguish $S^2 \times S^2$ and $\mathbb{CP}^2 \# \overline{\mathbb{CP}}^2$. They are probably still homotopy invariants, but this is a conjecture.


TITLE: A smooth $Q \in \mathbb R^N \to \mathbb R$ close to, but strictly below min
QUESTION [1 upvotes]: So, I've noticed that in many realworld applications, strict bounds are a requirement. I'll use a factory with $N$ inputs as an example. Suppose the inputs are organised into lots, and one output requires exactly one lot of each input. The factory simply can't run if even one of its inputs runs out, so the factory's maximum production rate (in widgets/second) is
$$P_{max}(I) = \min(I)$$
where $I \in \mathbb R^N$ is a vector of all the input rates.
We'll start with what we're not doing. We aren't worrying about whether the unused inputs will accumulate without bound. In other words, we are supposing that the actual production rate $P$ can be set to any value below or equal to $P_{max}$ without causing issues in the input end. Perhaps the extra is thrown away, or more likely, it is used to inform the future value of $I$, with a low value of $P$ causing it to lower. We're also supposing $I$ is differentiable with respect to time.
What I'm looking at is the problems that arise when you actually do set $P$ to $P_{max}$. Even if the input rates are differentiable with respect to time, $P_{max}$ will not be. Every time the lowest two inputs cross each other, a discontinuity in $P_{max}$'s derivative arises. This produces a discontinuity in the acceleration a.k.a. a jerk of the moving components. In a fake factory like Factorio or Satisfactory, the machines switch on and off instantly, so there isn't even a problem with a discontinuity in velocity—never mind acceleration. In a real factory, the jerk creates vibrations and causes wear, not to mention a lot of noise.
The naïve solution to this problem is to try to set $P$ to some moving average of $P_{max}$, but this is not going to work! Such a moving average carries no guarantee that $P < P_{max}$, which it absolutely must!
A solution that will work is to find some function $Q \in \mathbb R^N \to \mathbb R$ that is smooth and satisfies
$$\forall I, \quad 0 < Q(I) < \min(I)$$
While being as close to $\min(I)$ as possible within the limits of the maximum mechanical jerk.
I've been taking some stabs at some candidates, and there are some very good solutions for the $N = 2$ case, based off of smoothings of the absolute or sign functions, but nothing immediately good and simple comes to mind for the general case. Do any of you know a function like this?

REPLY [1 votes]: While researching this problem I actually found an answer! In real analysis, there's a much-used result:
$$\min(I) = \lim_{k \to \infty} \left( \sum_{a = 0}^{<N} \frac 1 {I_a^k} \right)^{- \frac 1 k}$$
To get the $Q$ I'm looking for, all I have to do is get rid of the limit:
$$Q(I) = \left( \sum_{a = 0}^{<N} \frac 1 {I_a^k} \right)^{- \frac 1 k}$$
This function is smooth and strictly smaller than the minimum.
To choose $k$, the easiest thing to do is substitute in the "even keel" state, where all input rates are equal. Say $\forall a, I_a = v$. If you require that in this state, $Q(I) = \alpha v$ for some $0 < \alpha < 1$, then
$$k = -\frac {\log N} {\log \alpha}$$
Thus to maintain $\alpha$, $k$ only needs slight adjustment, even for large increases in $N$.


TITLE: $R$ is a ring. Prove that $R/(0_R)\cong R$
QUESTION [2 upvotes]: $R$ is a ring. Prove that $R/(0_R)\cong R$.

I don't quite understand what $R/(0_R)$ looks like. By definition of quotient ring, it should have cosets $a+(0_R)$ where $a\in R$. So $R/(0_R)$ and $R$ itself really are the same thing?

REPLY [1 votes]: We can said that this two sets are the same in the following sense: if you remember the definition of the quotient ring, the elements of $R/(0_R)$ are, like you are saying, of the form $$a+(0_R)=\{a\},$$ i.e., are sets consisting only by an element $a\in R$. So the function $\varphi : R\to R/(0_R)$ by $$\varphi(a) =a+ (0_R)=\{a\}$$ is clearly an isomorphism.


TITLE: One-dimensional Sheaves and Pushforwards of Vector Bundles
QUESTION [1 upvotes]: Throughout, I want my ambient space $X$ to be a smooth projective variety of dimension at least two.
I believe we can at least say the following.  If $\mathcal{E}$ is a pure one-dimensional coherent sheaf on $X$, and the support of $\mathcal{E}$ is a smooth projective curve $C \subset X$, then $\mathcal{E}$ is the pushforward of a vector bundle on $C$.  I believe this is true because pure sheaves are torsion-free restricted to their support, and torsion-free sheaves on a smooth projective curve are locally-free.
Now, removing the assumption that the support be smooth, I am wondering to what extent (if any) pure one-dimensional coherent sheaves on smooth projective varieties are pushforwards of vector bundles.  Given a pure one-dimensional sheaf $\mathcal{E}$ supported on $C$, if we compose the normalization with the inclusion
$$\widetilde{C} \rightarrow C \hookrightarrow X,$$
can we always realize $\mathcal{E}$ as the pushforward of a locally-free sheaf on $\widetilde{C}$ to $X$?  What if $\mathcal{E}$ has scheme-theoretic support on a non-reduced curve $C$?

REPLY [1 votes]: The pushforward of a sheaf is always scheme-theoretically supported on the corresponding subscheme. So, if the scheme-theoretical support of a sheaf is not reduced, it is not the pushforward from the corresponding reduced subscheme.


TITLE: Is every regular (excellent) scheme separated?
QUESTION [2 upvotes]: Sorry for one more stupid AG question. I need schemes that are regular, excellent and separated. Are these three conditions independent?

REPLY [16 votes]: - Separated, excellent, regular: Spec$(k)$.
- Separated, excellent, not regular: Spec$(k[\epsilon]/\epsilon^2)$.
- Separated, not excellent, regular: See http://en.wikipedia.org/wiki/Excellent_ring
- Separated, not excellent, not regular: Spec$(k[\epsilon_1,\epsilon_2,\ldots]/\langle\epsilon_1^2,\epsilon_2^2,\ldots\rangle$.
- Not separated, excellent, regular: Glue Spec$(\mathbb{Z})$ to itself along the complement of a closed point.
To get the other three, take the disjoint union of the fifth example with any of the second, third, or fourth examples.


text:
\section{Definitions}
\label{Sec: Fractional Calculus}

Let $ \xi \in [-1,1]$. The left- and right-sided fractional derivative of order $\sigma$, $n-1 < \sigma \leq n$, $n \in \mathbb{N}$, are defined as (see e.g., \cite{Miller93, Podlubny99})

\begin{align}
\label{Eq: left RL derivative}

(\prescript{RL}{-1}{\mathcal{D}}_{\xi}^{\sigma}) u(\xi) = \frac{1}{\Gamma(n-\sigma)}  \frac{d^{n}}{d\xi^n} \int_{-1}^{\xi} \frac{u(s) ds}{(\xi - s)^{\sigma+1-n} },\quad \xi >-1 ,

\\
\label{Eq: right RL derivative}

(\prescript{RL}{\xi}{\mathcal{D}}_{1}^{\sigma}) u(\xi) = \frac{1}{\Gamma(n-\sigma)}  \frac{(-d)^{n}}{d\xi^n} \int_{\xi}^{1} \frac{u(s) ds}{(s - \xi)^{\sigma+1-n} },\quad \xi < 1 ,

\end{align}

respectively.

An alternative approach in defining the fractional derivatives is the left- and right-sided Caputo derivatives of order $\sigma$, defined, as

\begin{align}
\label{Eq: left Caputo derivative}

&
(\prescript{C}{-1}{\mathcal{D}}_{\xi}^{\sigma} u) (\xi) = \frac{1}{\Gamma(n-\sigma)}  \int_{-1}^{\xi} \frac{u^{(n)}(s) ds}{(\xi - s)^{\sigma+1-n} },\quad \xi>-1,
\\
\label{Eq: right Caputo derivative}

&
(\prescript{C}{\xi}{\mathcal{D}}_{1}^{\sigma} u) (\xi) =  \frac{1}{\Gamma(n-\sigma)}  \int_{\xi}^{1} \frac{u^{(n)}(s) ds}{(s-\xi)^{\sigma+1-n} },\quad \xi<1.

\end{align}

By performing an affine mapping from the standard domain $[-1,1]$ to the interval $t \in [a,b]$, we obtain

\begin{eqnarray}
\label{Eq: RL in xL-xR}

\prescript{RL}{a}{\mathcal{D}}_{t}^{\sigma} u  &=&  (\frac{2}{b-a})^\sigma (\prescript{RL}{-1}{\mathcal{D}}_{\xi}^{\sigma} \, u )(\xi),
\\
\label{Eq: Caputo in xL-xR}
\prescript{C}{a}{\mathcal{D}}_{t}^{\sigma} u  &=&  (\frac{2}{b-a})^\sigma (\prescript{C}{-1}{\mathcal{D}}_{\xi}^{\sigma} \, u) (\xi).

\end{eqnarray}

Hence, we can perform the operations in the standard domain only once for any given $\sigma$ and efficiently utilize them on any arbitrary interval without resorting to repeating the calculations. Moreover, the corresponding relationship between the Riemann-Liouville and Caputo fractional derivatives in $[a,b]$ for any $\sigma \in (0,1)$ is given by

\begin{equation}
\label{Eq:  Caputo vs. Riemann}
(\prescript{RL}{a}{\mathcal{D}}_{t}^{\sigma} \, u) (t)  =  \frac{ u(a)}{\Gamma(1-\sigma) (t-a)^{\sigma}}  +   (\prescript{C}{a}{\mathcal{D}}_{t}^{\sigma} \, u) (t).
\end{equation}

\begin{define}
\textup{We define the following left- and right-sided integro-differential operator with logarithmic-power law kernel, namely \textit{Log-Pow integro-differential operator}, given as,}
\end{define}

\begin{align}

\label{Eq: left sided LP kernel derivative}

&
\prescript{RL-LP}{a}{\mathcal{D}}_{x}^{\sigma} u (x) =
\frac{1}{\Gamma(n-\sigma)}  \frac{d^n}{dx^n} \int_{a}^{x} \frac{ \log(x-s) \, u(s)}{(x - s)^{\sigma -n +1}} ds,

\\
\label{Eq: right sided LP kernel derivative}
&
\prescript{RL-LP}{x}{\mathcal{D}}_{b}^{\sigma} u (x) =
\frac{1}{\Gamma(n-\sigma)}  \frac{(-d)^n}{dx^n} \int_{x}^{b} \frac{ \log(s-x) \, u(s)}{(s - x)^{\sigma-n+1}} ds,

\\

\label{Eq: left sided LP kernel derivative Caputo}

&
\prescript{C-LP}{a}{\mathcal{D}}_{x}^{\sigma} u (x) =
\frac{1}{\Gamma(n-\sigma)}  \, \int_{a}^{x} \frac{ \log(x-s) \, u^{(n)}(s)}{(x - s)^{\sigma -n +1}} ds,

\\
\label{Eq: right sided LP kernel derivative Caputo}
&
\prescript{C-LP}{x}{\mathcal{D}}_{b}^{\sigma} u (x) =
\frac{1}{\Gamma(n-\sigma)}  \, \int_{x}^{b} \frac{ \log(s-x) \, u^{(n)}(s)}{(s - x)^{\sigma-n+1}} ds,

\end{align}

where $RL-LP$ and $C-LP$ stand for Log-Pow integro-differential operator, which partially resemble the fractional derivative in Riemann-Liouville and Caputo sense, respectively. The following lemma shows a useful relation between the two aforementioned operators.

\begin{lem}

	\label{lem: Caputo vs. Riemann PowLog}

	Let $x\in [a,b]$. Then, the following relation holds.

	\noindent Part A: $\sigma \in (0,1)$

	\begin{align}
	\label{Eq:  Caputo vs. Riemann PowLog 1}

	\prescript{RL-LP}{a}{\mathcal{D}}_{x}^{\sigma} \, u (x)  =  \frac{ u(a)}{\Gamma(1-\sigma)}\frac{\log(x-a)}{(x-a)^{\sigma}}  +   \prescript{C-LP}{a}{\mathcal{D}}_{x}^{\sigma} \, u (x).

	\end{align}

	\noindent Part B: $\sigma \in (1,2)$

	\begin{align}
	\label{Eq:  Caputo vs. Riemann PowLog 2}

	\prescript{RL-LP}{a}{\mathcal{D}}_{x}^{\sigma} \, u (x)  =  \frac{u(a)}{\Gamma(2-\sigma)} \frac{1+(1-\sigma)\log(x-a)}{(x-a)^{\sigma}} + \frac{u'(a)}{\Gamma(2-\sigma)} \frac{\log(x-a)}{(x-a)^{\sigma-1}}  +   \prescript{C-LP}{a}{\mathcal{D}}_{x}^{\sigma} \, u (x).

	\end{align}

\end{lem}

\begin{proof}

	See Appendix \ref{Sec: App. proof of RL Caputo PL derivative} for proof.

\end{proof}

\subsection{\textbf{Fractional Sobolev Spaces}}
\label{Sec: Sol Test Space FPDE}

We define some functional spaces and their associated norms \cite{kharazmi2017petrov,Li2009}. By $H^\sigma(\mathbb{R}) = \big{\{} u(t) \vert u \in L^{2}(\mathbb{R});\, (1+\vert \omega \vert^2)^{\frac{\sigma}{2}} F(u)(\omega) \in L^{2}(\mathbb{R}) \big{\}}$, $\sigma \geq 0$, we denote the fractional Sobolev space on $\mathbb{R}$, endowed with norm $\Vert u \Vert_{H^{\sigma}_{\mathbb{R}}}=\Vert (1+\vert \omega \vert^2)^{\frac{\sigma}{2}} F(u)(\omega) \Vert_{L^{2}(\mathbb{R})}$, where $\mathcal{F}(u)$ represents the Fourier transform of $u$. Subsequently, we denote by $H^\sigma(\Lambda) =\big{\{} u\in L^{2}(\Lambda)\, \vert \,\exists \tilde{u} \in H^{\sigma}(\mathbb{R})\, \, s.t. \, \,\tilde{u}\vert_{\Lambda}=u \big{\}}$, $\sigma \geq 0$, the fractional Sobolev space on any finite closed interval, e.g. $\Lambda = (a,b)$, with norm $ \Vert u \Vert_{H^{\sigma}(\Lambda)}= \underset{\tilde{u}\in H^{\sigma}_{\mathbb{R}},\, \tilde{u}\vert_{\Lambda}=u }{\inf} \, \Vert \tilde{u} \Vert_{{}H^{\sigma}(\mathbb{R})}$. We define the following useful norms as:

\begin{align*}

&\Vert \cdot \Vert_{{^l}H^{\sigma}_{}(\Lambda)} = \Big(\Vert \prescript{}{a}{\mathcal{D}}_{x}^{\sigma}\, (\cdot)\Vert_{L^2(\Lambda)}^2+\Vert \cdot \Vert_{L^2(\Lambda)}^2 \Big)^{\frac{1}{2}},
\\
&\Vert \cdot \Vert_{{^r}H^{\sigma}_{}(\Lambda)} = \Big(\Vert \prescript{}{x}{\mathcal{D}}_{b}^{\sigma}\, (\cdot)\Vert_{L^2(\Lambda)}^2+\Vert \cdot \Vert_{L^2(\Lambda)}^2 \Big)^{\frac{1}{2}},
\\
&\Vert \cdot \Vert_{{^c}H^{\sigma}_{}(\Lambda)} = \Big(\Vert \prescript{}{x}{\mathcal{D}}_{b}^{\sigma}\, (\cdot)\Vert_{L^2(\Lambda)}^2+\Vert \prescript{}{a}{\mathcal{D}}_{x}^{\sigma}\, (\cdot)\Vert_{L^2(\Lambda)}^2+\Vert \cdot \Vert_{L^2(\Lambda)}^2 \Big)^{\frac{1}{2}},

\end{align*}

where the equivalence of $\Vert \cdot \Vert_{{^l}H^{\sigma}_{}(\Lambda)}$ and $\Vert \cdot \Vert_{{^r}H^{\sigma}_{}(\Lambda)}$ are shown in \cite{Li2009,ervin2007variational,Li2010}. We show the equivalence of these two norms with $\Vert \cdot \Vert_{{^c}H^{\sigma}_{}(\Lambda)}$ in the following lemma.

\begin{lem}

	\label{Lem: norm equivalence 1}

	Let $\sigma \geq 0$ and $\sigma \neq n-\frac{1}{2}$. Then, the norms $\Vert \cdot \Vert_{{^l}H^{\sigma}_{}(\Lambda)}$ and $\Vert \cdot \Vert_{{^r}H^{\sigma}_{}(\Lambda)}$ are equivalent to $\Vert \cdot \Vert_{{^c}H^{\sigma}_{}(\Lambda)}$.

\end{lem}

\begin{proof}

	See Appendix \ref{Sec: App. proof of norm equivalence} for proof.

\end{proof}

We also define $C^{\infty}_{0}(\Lambda)$ as the space of smooth functions with compact support in $(a,b)$. We denote by $\prescript{l}{}H^{\sigma}_0(\Lambda)$, $\prescript{r}{}H^{\sigma}_0(\Lambda)$, and $\prescript{c}{}H^{\sigma}_0(\Lambda)$ as the closure of $C^{\infty}_{0}(\Lambda)$ with respect to the norms $\Vert \cdot \Vert_{{^l}H^{\sigma}_{}(\Lambda)}$, $\Vert \cdot \Vert_{{^r}H^{\sigma}_{}(\Lambda)}$, and $\Vert \cdot \Vert_{{^c}H^{\sigma}_{}(\Lambda)}$.

\begin{lem}[\cite{ervin2007variational,Li2010}]

	\label{Lem: sobolev equivalence}
	The Sobolev spaces $\prescript{l}{}H^{\sigma}_0(\Lambda)$, $\prescript{r}{}H^{\sigma}_0(\Lambda)$, and $\prescript{c}{}H^{\sigma}_0(\Lambda)$ are equal and their seminorms are equivalent to $\vert \cdot \vert_{{}H^{\sigma}_{}(\Lambda)}^{*} = \big\vert \left( \prescript{}{a}{\mathcal{D}}_{x}^{\sigma}\, (\cdot), \prescript{}{x}{\mathcal{D}}_{b}^{\sigma}\, (\cdot) \right) \big\vert_{\Lambda}^{\frac{1}{2}}$

\end{lem}

Based on Lemma \eqref{Lem: sobolev equivalence}, and assuming that $\big{\vert}(\prescript{}{a}{\mathcal{D}}_{x}^{\sigma}\, u, \prescript{}{x}{\mathcal{D}}_{b}^{\sigma}\, v )_{\Lambda}^{} \big{\vert} > 0$ and $\big{\vert}(\prescript{}{x}{\mathcal{D}}_{b}^{\sigma}\, u, \prescript{}{a}{\mathcal{D}}_{x}^{\sigma}\, v )_{\Lambda}^{} \big{\vert} > 0$, we can prove that $\big{\vert}(\prescript{}{a}{\mathcal{D}}_{x}^{\sigma}\, u, \prescript{}{x}{\mathcal{D}}_{b}^{\sigma}\, v )_{\Lambda}^{} \big{\vert} \geq \beta_1 \, \vert u \vert_{{^l}H^{\sigma}_{}(\Lambda)}\, \vert  v \vert_{{^r}H^{\sigma}_{}(\Lambda)}$ and $\big{\vert}(\prescript{}{x}{\mathcal{D}}_{b}^{\sigma}\, u, \prescript{}{a}{\mathcal{D}}_{x}^{\sigma}\, v )_{\Lambda}^{} \big{\vert} \geq \beta_2 \, \vert u \vert_{{^r}H^{\sigma}_{}(\Lambda)}\, \vert  v \vert_{{^l}H^{\sigma}_{}(\Lambda)}$, where $\beta_1$ and $\beta_2$ are positive constants. Following \cite{Samiee2016unified}, we define the corresponding solution and test spaces of our problem. Thus, by letting $\Lambda_1 = (a_1,b_1)$, $\Lambda_j = (a_j,b_j) \times \Lambda_{j-1}$ for $j=2,\cdots,d$, we define $\mathcal{X}_1 = H^{\frac{\beta_1}{2}}_{0}(\Lambda_1)$, which is associated with the norm $ \Vert \cdot \Vert_{{^c}H^{\frac{\beta_1}{2}}_{}(\Lambda_1)}$, and accordingly, $\mathcal{X}_j, \, j=2,\cdots,d$ as

\begin{eqnarray}

\mathcal{X}_2 &=& H^{\frac{\beta_2}{2}}_0 \Big((a_2,b_2); L^2(\Lambda_1) \Big) \cap L^2((a_2,b_2); \mathcal{X}_1),
\\
&\vdots&
\nonumber
\\
\mathcal{X}_d &=& H^{\frac{\beta_d}{2}}_0 \Big((a_d,b_d); L^2(\Lambda_{d-1}) \Big) \cap L^2((a_d,b_d); \mathcal{X}_{d-1}),

\end{eqnarray}

associated with norms $\Vert \cdot \Vert_{\mathcal{X}_j} = \bigg{\{} \Vert \cdot \Vert_{H^{\frac{\beta_j}{2}}_0 \Big((a_j,b_j); L^2(\Lambda_{j-1}) \Big)}^2 + \Vert \cdot \Vert_{ L^2\Big((a_j,b_j); \mathcal{X}_{j-1}\Big)}^2 \bigg{\}}^{\frac{1}{2}}, \,\, j=2,3,\cdots,d$.

\vspace{0.1 in}

\begin{lem}

	\label{space norm 1}
	Let $\sigma \geq 0$ and $\sigma \neq n-\frac{1}{2}$. Then, for $j=1,2,\cdots,d$

	\begin{align*}

	\Vert \cdot \Vert^{2}_{\mathcal{X}_j} \equiv  \sum_{i=1}^{j} \Big(\Vert \prescript{}{x_i}{\mathcal{D}}_{b_i}^{\beta_i/2}\, (\cdot)\Vert_{L^2(\Lambda_j)}^2+\Vert \prescript{}{a_i}{\mathcal{D}}_{x_i}^{\beta_i/2}\, (\cdot)\Vert_{L^2(\Lambda_j)}^2 \Big) + \Vert  \cdot \Vert_{L^2(\Lambda_j)}^2.

	\end{align*}

\end{lem}

\begin{proof}
	See Appendix \ref{Sec: App. proof of norm Xd} for proof.
\end{proof}

\vspace{0.1 in}

Moreover, by letting ${_0}C^{\infty}(I)$ and $C^{\infty}_{0}(I)$ be the space of smooth functions with compact support in $(0,T]$ and $[0,T)$, respectively, we define $\prescript{l}{}H^{s}(I)$ and $\prescript{r}{}H^{s}(I)$ as the closure of ${_0}C^{\infty}(I)$ and $C^{\infty}_{0}(I)$ with respect to the norms $\Vert \cdot \Vert_{\prescript{l}{}H^{s}(I)}$ and $\Vert \cdot \Vert_{\prescript{r}{}H^{s}(I)}$.

We also define

\begin{align*}

\prescript{l}{0}H^{\frac{\alpha}{2}} \Big(I; L^2(\Lambda_d) \Big) =
\Big{\{} u \, \big|\, \Vert u(t,\cdot) \Vert_{L^2(\Lambda_d)} \in H^{\frac{\alpha}{2}}(I), u\vert_{t=0}=u\vert_{x=a_j}=u\vert_{x=b_j}=0,\, j=1,2,\cdots,d  \Big{\}},

\\
\prescript{r}{0}H^{\frac{\alpha}{2}} \Big(I; L^2(\Lambda_d) \Big) =
\Big{\{} v \,\big|\, \Vert v(t,\cdot) \Vert_{L^2(\Lambda_d)} \in H^{\frac{\alpha}{2}}(I), v\vert_{t=T}=v\vert_{x=a_j}=v\vert_{x=b_j}=0,\, j =1,2,\cdots,d  \Big{\}},

\end{align*}

equipped with norms $\Vert u \Vert_{\prescript{l}{}H^{\frac{\alpha}{2}}(I; L^2(\Lambda_d))}$ and $\Vert u \Vert_{\prescript{r}{}H^{\frac{\alpha}{2}}(I; L^2(\Lambda_d))}$, respectively, which take the following forms

\begin{align}

\label{space norm 2}
&
\Vert u \Vert_{\prescript{l}{}H^{\frac{\alpha}{2}}(I; L^2(\Lambda_d))} = \Big{\Vert} \, \Vert u(t,\cdot) \Vert_{L^2(\Lambda_d)}\, \Big{\Vert}_{{^l}H^{\frac{\alpha}{2}}(I)}
=\Big(\Vert \prescript{}{0}{\mathcal{D}}_{t}^{\frac{\alpha}{2}}\, (u)\Vert_{L^2(\Omega)}^2 + \Vert u\Vert_{L^2(\Omega)}^2 \Big)^{\frac{1}{2}},

\\
&\Vert u \Vert_{\prescript{r}{}H^{\frac{\alpha}{2}}(I; L^2(\Lambda_d))} = \Big{\Vert} \, \Vert u(t,\cdot) \Vert_{L^2(\Lambda_d)}\, \Big{\Vert}_{{^r}H^{\frac{\alpha}{2}}(I)}
= \Big(\Vert \prescript{}{t}{\mathcal{D}}_{T}^{\frac{\alpha}{2}}\, (u)\Vert_{L^2(\Omega)}^2+\Vert u\Vert_{L^2(\Omega)}^2\Big)^{\frac{1}{2}}.

\end{align}

\subsection*{\textbf{Solution and Test Spaces}}

We define the solution space $U$ and test space $V$, respectively, as

\begin{align}

\label{Eq: solution test space}
U  = \prescript{l}{0}H^{\frac{\alpha}{2}}\Big(I; L^2(\Lambda_d) \Big) \cap L^2(I; \mathcal{X}_d),

\quad
V  = \prescript{r}{0}H^{\frac{\alpha}{2}}\Big(I; L^2(\Lambda_d)\Big) \cap L^2(I; \mathcal{X}_d),

\end{align}

endowed with norms

\begin{align}

\label{Eq: solution test space norm}
\Vert u \Vert_{U} &=
\Big{\{}\Vert u \Vert_{\prescript{l}{}H^{\frac{\alpha}{2}}(I; L^2(\Lambda_d))}^2 + \Vert u \Vert_{L^2(I; \mathcal{X}_d)}^2 \Big{\}}^{\frac{1}{2}},

\quad

\Vert v \Vert_{V} &=
\Big{\{}\Vert v \Vert_{\prescript{r}{}H^{\tau}(I; L^2(\Lambda_d))}^2  + \Vert v \Vert_{ L^2(I; \mathcal{X}_d)}^2 \Big{\}}^{\frac{1}{2}},

\end{align}

Using Lemma \ref{space norm 1}, we can show that

\begin{align}

\label{space norm 3}
\Vert u \Vert_{L^2(I; \mathcal{X}_d)}
=
\Big{\Vert} \, \Vert u(t,.) \Vert_{\mathcal{X}_d}\,\Big{\Vert}_{L^2(I)}
=
\Big{\{}  \Vert u \Vert_{L^2(\Omega)}^2 + \sum_{j=1}^{d} \big( \Vert \prescript{}{x_j}{\mathcal{D}}_{b_j}^{\frac{\beta_j}{2}}\, u  \,\Vert_{L^2(\Omega)}^2
+ \Vert \prescript{}{a_j}{\mathcal{D}}_{x_j}^{\frac{\beta_j}{2}}\, u \,\Vert_{L^2(\Omega)}^2 \big)
\Big{\}}^{\frac{1}{2}}.

\end{align}

Therefore, by \eqref{space norm 2} we write \eqref{Eq: solution test space norm} as

\begin{align}

\label{Eq: solution space norm}
\Vert u \Vert_{U}
&=
\Big{\{}  \Vert u \Vert_{L^2(\Omega)}^2 + \Vert \prescript{}{0}{\mathcal{D}}_{t}^{\frac{\alpha}{2}}\, u\, \Vert_{L^2(\Omega)}^2
+ \sum_{j=1}^{d} \big( \Vert \prescript{}{x_j}{\mathcal{D}}_{b_j}^{\frac{\beta_j}{2}}\, u \, \Vert_{L^2(\Omega)}^2+\Vert \prescript{}{a_j}{\mathcal{D}}_{x_j}^{\frac{\beta_j}{2}}\, u \, \Vert_{L^2(\Omega)}^2 \big) \Big{\}}^{\frac{1}{2}},

\\
\label{Eq: test space norm}

\Vert v \Vert_{V}
& =
\Big{\{}  \Vert v \Vert_{L^2(\Omega)}^2 + \Vert \prescript{}{t}{\mathcal{D}}_{T}^{\frac{\alpha}{2}}\, v \, \Vert_{L^2(\Omega)}^2
+ \sum_{j=1}^{d} \big( \Vert \prescript{}{x_j}{\mathcal{D}}_{b_j}^{\frac{\beta_j}{2}}\, v \, \Vert_{L^2(\Omega)}^2
+\Vert \prescript{}{a_j}{\mathcal{D}}_{x_j}^{\frac{\beta_j}{2}}\, v \, \Vert_{L^2(\Omega)}^2 \big) \Big{\}}^{\frac{1}{2}}.

\end{align}

The following lemmas help us obtain the weak formulation of our problem, construct the numerical scheme and further prove the stability of our method.

\begin{lem}
	\label{Lem: left frac proj}
	\cite{Li2009}: For all $\alpha \in  (0,1)$, if $u \in H^1([0,T])$ such that $u(0)=0$, and $v \in H^{\alpha/2}([0,T])$, then $( \prescript{}{0}{ \mathcal{D}}_{t}^{\,\,\alpha} u, v )_{\Omega} =  (\, \prescript{}{0}{ \mathcal{D}}_{t}^{\,\,\alpha/2} u \,,\, \prescript{}{t}{ \mathcal{D}}_{T}^{\,\,\alpha/2} v\, )_{\Omega}$, where $(\cdot , \cdot)_{\Omega}$ represents the standard inner product in $\Omega=[0,T]$.

\end{lem}

\begin{lem}
	\label{Lem: fractional integ-by-parts 1 and 2}

	\cite{kharazmi2017petrov}: Let $1 < \beta < 2$, $a$ and $b$ be arbitrary finite or infinite real numbers. Assume $u \in H^{\beta}(a,b)$ such that $u(a)=0$, also $\prescript{}{x}{\mathcal{D}}_{b}^{\beta/2}v$ is integrable in $\Omega = (a,b)$ such that $v(b) = 0$. Then, $( \prescript{}{a}{\mathcal{D}}_{x}^{\beta} u \,,\,v )_{\Omega} = ( \prescript{}{a}{\mathcal{D}}_{x}^{\beta/2} u \,,\,\prescript{}{x}{\mathcal{D}}_{b}^{\beta/2} v )_{\Omega}$.

\end{lem}

We generalize Lemma \ref{Lem: fractional integ-by-parts 1 and 2} to the two-sided $(1+d)$-dimensional case (see Appendix \ref{Sec: App. proof of gen int by part} for proof).

\begin{lem}

	\label{lem_generalize}

	Let $1<\beta_j<2$ for $j=1,2,\cdots,d$, and $u,v \in  \mathcal{X}_d$. Then,

	\begin{align*}

	\big(\prescript{}{a_j}{\mathcal{D}}_{x_j}^{\beta_j} u, v\big)_{\Lambda_d}=\big(\prescript{}{a_j}{\mathcal{D}}_{x_j}^{\frac{\beta_j}{2}} u, \prescript{}{x_j}{\mathcal{D}}_{b_j}^{\frac{\beta_j}{2}} v\big)_{\Lambda_d},
	\qquad
	\big(\prescript{}{x_j}{\mathcal{D}}_{b_j}^{\beta_j} u, v\big)_{\Lambda_d}=\big(\prescript{}{x_j}{\mathcal{D}}_{b_j}^{\frac{\beta_j}{2}} u, \prescript{}{a_j}{\mathcal{D}}_{x_j}^{\frac{\beta_j}{2}} v\big)_{\Lambda_d}.

	\end{align*}

\end{lem}


TITLE: System of parameters for a local ring
QUESTION [1 upvotes]: I need some help to solve this problem. This is the kind of problem that makes me stuck at the very beginning.

Let $K$ be an algebraically closed field, $X = \{(x,y)\in\mathbb{A}^2_K: \ y^2-x^3=0\}$ an affine variety and $\mathfrak{m}$ the maximal ideal in $X$ corresponding to the point $(0,0)\in X$. I want to localize the coordinate ring $\mathcal{O}(X)$ in $\mathfrak{m}$, so I will have a system of parameters for the local ring $\mathcal{O}_\mathfrak{m}$.

All I know about system of parameters is the following:
Proposition: let $R$ be a Noetherian local ring with maximal ideal $\mathfrak{m}$. Then the dimension of $R$ is the minimum $n\in\mathbb{N}$ such that there is $r_1,\ldots,r_n\in R$ with $\sqrt{(r_1,\ldots,r_n)} = \mathfrak{m}$. In this case, the $r_i$'s a system of parameters for $R$.
But this is not helping much. Thank you very much for the help.

REPLY [3 votes]: Thought isn't clear, the question seems to ask for finding a sop of the local ring $$K[X,Y]_{(X,Y)}/(X^3-Y^2).$$
This ring is $1$-dimensional, so a sop has only one element. There are two clear contenders. Prove that each one of them does the job.


TITLE: Is there a local ring but not Noetherian?
QUESTION [2 upvotes]: I was looking for an example of a non-Noetherian local commutative ring with $1$.

I would appreciate if anyone can point to a reference.

REPLY [4 votes]: Take a field $k$ and the ring $k[X_1,\ldots ,X_n,\ldots ]$. Localize it at the maximal ideal $(X_1,\ldots , X_n,\ldots )$. Then the ideals $(X_1)\subset (X_1,X_2)\subset ... \subset (X_1,\ldots , X_n) \subset \ldots$ still form a strictly increasing union of ideals, and so the localization is local but not noetherian
(To see that the inclusions are strict, for instance kill everyone but $X_n$ and then include in power series to get a map $k[(X_i)] \to k[[X_n]]$ that sends anyone $\notin$ the maximal ideal (that is, anyone with a constant term) to an invertible element, and so that factors through our ring, and sends $(X_1,\ldots , X_{n-1})$ to $0$ but not $X_n$).


TITLE: Transversal sections of open and closed sets in the product topology.
QUESTION [1 upvotes]: Problem: Let be $X$ and $Y$ topological spaces and $O\subseteq X\times Y$.
$(i)$ Suppose that $O$ is an open set, where $X\times Y$ have the product topology. Then $p_X\left( X\times\left\lbrace y\right\rbrace\cap O \right)$ is open in $X$, where $p_X:X\times Y\ni (x,y)\mapsto x\in X$ is the projection in the first component.
$(ii)$ Supose that $O$ is closed. Then $p_Y\left(\left\lbrace x\right\rbrace\times Y\cap O \right)$ is closed in $Y$, where $p_Y:X\times Y\ni (x,y)\mapsto y\in Y$ is the projection in the second component.
Is known that, the projection functions are open and, in general aren´t closed, but this fact It does not seem to help much here, because  $X\times\left\lbrace y\right\rbrace\cap O$ is not necessarily open.
I do not find counterexamples of these statements either.

REPLY [3 votes]: Let $x \in p_X[(X \times \{y\}) \cap O]$, which is true iff $(x,y) \in O$. Then let $U \times V$ (where $U \subseteq X, V \subseteq Y$ are open) such that $(x,y) \in U \times V \subseteq O$. Note that this implies that for any $p \in U$, we have that $(p,y) \in U \times V$, hence $(p,y) \in O$ so $p \in p_X[(O \times \{y\}) \cap O]$. So $U$ witnesses that $x$ is an interior point of $p_X[(X \times \{y\}) \cap O]$. As $x$ was arbitrary, $p_X[(X \times \{y\}) \cap O]$ is open.
Suppose $C$ is closed (I prefer not to call my closed sets $O$) and let $D = p_Y[(\{x\} \times Y) \cap C]$ and try to show $D$ is closed.
So let $q \in Y$ be such that $q \notin D$. This means that $(x,q) \notin C$.
As $C$ is closed we find a basic open set $U \times V \ni (x,q)$ that is disjoint from $C$. Then $V$ contains $q$ and is disjoint from $D$ (otherwise $q' \in D \cap V$ exists and $(x,q') \in C$, but $(x,q') \in (U \times V) \cap C$ which is empty by construction). So every point not in $D$ has an open neighbourhood that is disjoint from $D$, so $D$ is closed.


TITLE: Preservation of the cross product by parametrization
QUESTION [0 upvotes]: Let $S$ be a regular surface and $X:U \subset \mathbb{R}^2\longrightarrow X(U)\subset S\subset \mathbb{R}^3$ a local parametrization. Does the following hold?
If $e_1, e_2$ are two linearly independent vectors in $\mathbb{R}^2$, and we assume $\mathbb{R}^2$ is embedded in $\mathbb{R}^3$ (we assume $X:\mathbb{R}^3\longrightarrow\mathbb{R}^3$)
¿$dX(e_1\times e_2)$ parallel to $dX(e_1)\times dX(e_2)$?

REPLY [1 votes]: Note that the cross product $e_{1} \times e_{2}$ depends on whether you're interpreting these vectors as elements of the plane (in which case the product is customarily viewed as the scalar $1$) or as elements of space (in which case $e_{1} \times e_{2} = e_{3}$).
The first line of your question suggests $X$ is defined (only) on $U$, in which case $dX(e_{1} \times e_{2})$ doesn't make sense.
If instead you mean that $X$ is defined on some neighborhood of the plane in space (and your surface is obtained by restricting $X$ to an open set in the plane), then "no, $dX(e_{1} \times e_{2})$ is not generally parallel to $dX(e_{1}) \times dX(e_{2})$". Indeed, this isn't even true for linear transformations $X$. If we write $X$ as a $3 \times 3$ real matrix, then
$$
dX(e_{1} \times e_{2}) = dX(e_{3})
$$
is the third column of $X$, while $dX(e_{1}) \times dX(e_{2})$ is the cross product of the first two columns of $X$.
There's no reason for the cross product of the first two columns to be proportional to the third column.


TITLE: Find exponential function given two points
QUESTION [0 upvotes]: $f(x) = ar^x $ given that $r > 0$.
I'm given two points ($3$, $\frac{8}{9}$) and ($4$, $\frac{16}{27}$).
My textbook then says  $$r = \frac{\frac{16}{27}}{\frac{8}{9}}$$
Why does this work? I can't figure out why $y_2/y_1 = r$.
There's some other similar questions asked on this site but the answers are too complicated, using integrals and natural logarithms and such to explain something this simple.

REPLY [2 votes]: Like Mick said, all you're doing is solving the system
$$ar^3=\frac89$$
$$ar^4=\frac {16}{27}$$
If you divide the second equation by the first, you get
$$\frac{ar^4}{ar^3}=\frac{\frac{16}{27}}{\frac{8}{9}}$$
$$\Rightarrow r=\frac{\frac{16}{27}}{\frac{8}{9}}$$