text:
\begin{document}
\title[Knot projections with reductivity two]{Knot projections with reductivity two}
\author{Noboru Ito}
\address{Waseda Institute for Advanced Study, 1-6-1 Nishi Waseda Shinjuku-ku Tokyo 169-8050, Japan}
\email{noboru@moegi.waseda.jp}
\author{Yusuke Takimura}
\address{Gakushuin Boys' Junior High School, 1-5-1 Mejiro Toshima-ku Tokyo 171-0031, Japan}
\email{Yusuke.Takimura@gakushuin.ac.jp}

\thanks{2010 {\it{Mathematics Subject Classification.}}  57M25, 57Q35.}

\keywords{reductivity; knot projection; spherical curve}
\date{July 15, 2015}
\maketitle
\begin{abstract}
Reductivity of knot projections refers to the minimum number of splices of double points needed to obtain reducible knot projections.  Considering the type and method of splicing (Seifert type splice or non-Seifert type splice, recursively or simultaneously), we can obtain four reductivities containing Shimizu's reductivity, three of which are new.  In this paper, we determine knot projections with reductivity two for all four of the definitions.  We also provide easily calculated lower bounds for some reductivities.  Further, we detail properties of each reductivity, and describe relationships among the four reductivities with examples.
\end{abstract}
\section{Introduction}\label{intro}

A {\it{knot projection}} is the image of a generic immersion of an unoriented circle into the sphere $S^2$.  By this definition, a knot projection is an unoriented generic immersed curve on $S^2$, also called a {\it{spherical curve}}.  In this study, we assume that every knot projection has at least one double point.
As previously proven, every knot projection without $1$- and $2$-gons has $3$-gons.

As shown in Fig.~\ref{rf0}, there are four types of $3$-gons.
\begin{figure}[h!]
\includegraphics[width=7cm]{rf0.pdf}
\caption{All types of $3$-gons.  For each $3$-gon, three dotted arcs show the connections of branches of three double points.}\label{rf0}
\end{figure}

Of interest is the still-unanswered question below (\cite[Question 3.2]{S}).
\begin{question}[Shimizu \cite{S}]\label{q1}
Is it true that every knot projection without $1$- and $2$-gons has at least one $3$-gon of type $A$, $B$, or $C$, as shown in Fig.~\ref{rf0}?
\end{question}
If it is true, we can more easily prove the triple chord's theorem \cite{ITtriple}:
\begin{fact}[Ito-Takimura \cite{ITtriple}]
Every chord diagram of a knot projection without $1$- and $2$-gons contains \tr.
\end{fact}
Here, a chord diagram is a circle, on which paired points are placed such that each pair of two points, corresponding to two pre-images of a double point of a knot projection, is connected by a chord.

Shimizu's reductivity $r(P)$ \cite{S}, closely related to Question~\ref{q1}, is defined as follows:
Shimizu's reductivity $r(P)$ of a knot projection $P$ is the minimum number of non-Seifert splices defined by Fig.~\ref{rf2}, applied recursively, to obtain a {\it{reducible knot projection}} from $P$.
\begin{figure}[h!]
\includegraphics[width=3cm]{rf2.pdf}
\caption{Non-Seifert splice $A^{-1}$.}\label{rf2}
\end{figure}
Here, a reducible knot projection $P$ is a knot projection having a {\it{nugatory crossing}}, as shown in Fig.~\ref{rf1}.  A non-reducible knot projection is called a {\it{reduced knot projection}}.  Traditionally, an arbitrary non-Seifert splice is often denoted by $A^{-1}$ \cite{ItSh}.  Note that $A^{-1}$ does not depend on the selection of the orientation of a knot projection.
\begin{figure}[h!]
\includegraphics[width=3cm]{rf1.pdf}
\caption{A nugatory crossing between two (1, 1)-tangles, $T$ and $T'$.}\label{rf1}
\end{figure}

Shimizu's Question~\ref{q2} remains unanswered, as well.
\begin{question}[Shimizu \cite{S}]\label{q2}
Is it true that $r(P) \le 3$ for every knot projection $P$?
\end{question}
As Shimizu points out, if the answer to Question~\ref{q1} is {\it{yes}}, the answer to Question~\ref{q2} is also {\it{yes}} \cite{S}.

In general, it is not easy to compute Shimizu's reductivity $r$.  However, this paper provides the lower bounds that can be easily calculated for an infinite family of knot projections (Theorem~\ref{thm_lower} and Examples~\ref{ex1} and \ref{ex2}).  Moreover, we explicitly obtain the necessary and sufficient condition for a knot projection to satisfy $r(P)=2$ (Theorem~\ref{r2}).

As a first step, \cite{ITtriple} establishes the necessary and sufficient condition that a knot projection has $r(P)=1$ (Fact~\ref{f3}).
\begin{fact}\label{f3}
Let $P$ be a reduced knot projection $P$.  There exists a circle intersecting $P$ at just two double points of $P$, as shown in Fig.~\ref{rf6}, if and only if $r(P)=1$.
\end{fact}
As a corollary,
\begin{fact}[Ito-Takimura \cite{ITtriple}]
Every chord diagram of a knot projection $P$ with $r(P)=1$ has a sub-chord diagram \tr.
\end{fact}
In this paper, we take this one step further, proving the following theorem and its corollary.

\begin{theorem}\label{r2}
Let $P$ be a reduced knot projection.  There is no possibility of existence of a simple closed curve, such as the one in Fig.~\ref{rf6}, and there exists a circle intersecting $P$ at just three double points of $P$, as shown in Fig.~\ref{rf5}, if and only if $r(P)=2$.

\begin{figure}[h!]
\includegraphics[width=2cm]{kiyakudo1.pdf}
\caption{A simple closed curve and a type of knot projection.  Dotted arcs show the connections of branches between two double points.}\label{rf6}
\end{figure}
\begin{figure}[h!]
\includegraphics[width=10cm]{kiyakudo2.pdf}
\caption{All types of knot projections with Shimizu's reductivity two.}\label{rf5}
\end{figure}
\end{theorem}
\begin{corollary}
If $r(P)=2$ of a knot projection $P$, its chord diagram has a sub-chord diagram \h.
\end{corollary}
We also introduce three new types of reductivities, and determine knot projections up to reductivity two for each of all the four reductivities.

Here, we state that it is worth considering these three reductivities.  As for Question~\ref{q1}, there is no doubt that it is geometrically important.  We regard Shimizu's formulation (Question~\ref{q2}) as capturing an aspect of Question~\ref{q1}.  If the answer to Question~\ref{q1} is yes, then Question~\ref{q2} obtains one of the necessary conditions.

\noindent$\bullet$ (1) If the answer to Question~\ref{q1} is yes, the answer to Question~\ref{q2} yes ($r \le 3$).

In other words, if the answer to Question~\ref{q2} is no, disproving the conjecture for $r \le 3$ would be worth, i.e., (1)' is the contraposition of (1).

\noindent$\bullet$ (1)'  If there exists a knot projection $P$ such that $r(P) \ge 4$, the answer to Question~\ref{q1} is no.

In this paper, we add another reductivity $t \le r$, giving another necessary condition.  The reductivity $t$ was introduced by Taniyama during the discussions with the authors (private communication).

\noindent$\bullet$ (2) If the answer to Question~\ref{q1} is yes, $t \le 3$.

Obviously, because (1) implies (2), (2) is easier to prove than (1) and is more accessible than Question~\ref{q2}.  Further, (2) is as important as (1) to obtain a negative answer to Question~\ref{q1}.

To the best of our knowledge, an example with $t \ge 4$ and even $t \ge 3$, has not been found yet.  Here, we present our question (cf.~Question~\ref{q2}).

\begin{question}\label{q3}
Is it true that $t (P) \le 2$ for every knot projection $P$?
\end{question}

Thus, it is important to explicitly obtain $P$ that satisfies $t (P) \le 2$.  Question~\ref{q3} is still open, but this study determines $P$ with $t(P) \le 2$.  To determine $t(P) \le 2$, we should determine $y(P) \le 2$ or $i(P) \le 2$, where $y$ and $i$ are two additional reductivities that are closely related to $r(P)$ and $t(P)$.  These reductivities have explicit relations with $t(P)$ (Proposition~\ref{prop1}), and thus their properties should be studied.

This paper consists of the following sections:
Sec.~\ref{sec_def} contains all four definitions of reductivities;
Sec.~\ref{sec_r1} describes knot projections with reductivity one for all four reductivities $t(P)$, $r(P)$, $y(P)$, and $i(P)$ for a knot projection $P$; Sec.~\ref{sec_lower} introduces easily calculated lower bounds; Sec.~\ref{sec_r2} contains a proof of Theorem~\ref{r2};

Sec.~\ref{sec_t2} determines knot projections with reductivity two for each of the three cases $t(P)$, $y(P)$, and $i(P)$.   Finally, Sec.~\ref{sec_table} obtains a list of reductivities for prime reduced knot projections up to seven double points.

\section{Four definitions of reductivities}\label{sec_def}

In Definition~\ref{def_four}, reductivity $r(P)$ is introduced by Shimizu \cite{S}.  The others, $t(P)$, $y(P)$, and $i(P)$, are new.
\begin{definition}\label{def_four}
Let $P$ be a knot projection.
A splice indicates either a Seifert splice (Fig.~\ref{rf4}) or a non-Seifert splice (Fig.~\ref{rf2}).
\begin{itemize}
\item $t(P)$ $=$ min $\{$ number of splices, applied simultaneously, to obtain a reducible knot projection from $P$ $\}$,
\item $r(P)$ $=$ min $\{$ number of non-Seifert splices, applied recursively, to obtain a reducible knot projection from $P$ $\}$ (\cite{S}),
\item $y(P)$ $=$ min $\{$ number of non-Seifert splices, applied simultaneously, to obtain a reducible knot projection from $P$ $\}$,
\item $i(P)$ $=$ min $\{$ number of Seifert splices, applied simultaneously, to obtain a reducible knot projection from $P$ $\}$.
\end{itemize}
\begin{figure}[h!]
\includegraphics[width=3cm]{rf4.pdf}
\caption{A Seifert splice.  The splice preserves the orientation of a curve.}\label{rf4}
\end{figure}
\end{definition}
By the definition, we have the basic properties given in Proposition~\ref{prop1} below.  Note that the coherent $2$-gon referenced in the properties is $2$-gon shown in Fig.~\ref{coh}.
\begin{figure}[h!]
\includegraphics[width=2cm]{cohe.pdf}
\caption{A coherent $2$-gon.  Dotted arcs show the connections of branches of two double points of $2$-gons.}\label{coh}
\end{figure}
\begin{proposition}\label{prop1}
Let $P$ be a reduced knot projection.  Then $P$ has the following properties.
\begin{enumerate}
\item $t(P) \le r(P)$.
\item $t(P) \le y(P)$.
\item $t(P) \le i(P)$.
\item $i(P)$ is an even nonnegative integer.
\item If a reduced knot projection $P$ has at least one coherent $2$-gon, $i(P)=2$.
\item If a reduced knot projection $P$ has at least one $C$-type $3$-gon, $i(P)=2$.
\item If a reduced knot projection $P$ has at least one $A$-type $3$-gon, $1 \le y(P) \le 2$.
\item For any even positive integer $2m$, there exists $P$ such that $i(P)=2m$.
\end{enumerate}
\end{proposition}
\begin{proof}
It is sufficient to verify properties (4) and (8).

For (4), if a single Seifert splice is applied to a knot projection, the result obtained is not a knot projection.   Property (8) can be verified if we consider $(2, 2m+1)$-torus knot projections.  Hence, the proposition is verified.
\end{proof}
\begin{example}\label{ex1}
There is an example such that $t(P) \lneqq r(P)$.
\begin{figure}[h!]
\includegraphics[width=3cm]{senya1.pdf}
\caption{Example with $t(P)=2$, $r(P)=3$, $y(P)=2$, and $i(P)=2$.}\label{exa1}
\end{figure}
\end{example}
\begin{proof}
\begin{figure}[h!]
\includegraphics[width=10cm]{f9.pdf}
\caption{Knot projections (1), (2) and (3).}\label{expr}
\end{figure}
From Fig.~\ref{expr} (1), it can be seen that due to its symmetry, $P$ has exactly two types of double points.  Thus, it can be easily seen that any single splice cannot produce a reducible knot projection.  Thus, $2 \le t(P), r(P), y(P),$ and $i(P)$. (More simply, because $|\tau(P)|=1$, we have $2 \le t(P), r(P), y(P), i(P)$ (Theorem~\ref{thm_lower}).)   From Fig.~\ref{expr} (2) and (3), we can easily notice that $t(P), y(P),$ and $i(P) = 2$.  However, from Fig.~\ref{expr} (2) and (3), we can see that $r(P) = 3$.
\end{proof}
Intuitively, as in this case, we can see that the calculations for $t(P)$, $y(P)$, and $i(P)$ are simpler than those for $r(P)$.  This is one of the advantages of considering the reductivities $t(P)$, $y(P)$, and $i(P)$.  Similarly to Example~\ref{ex1}, we have Example~\ref{ex2}.
\begin{example}\label{ex2}
There are examples such that $t(P) \lneqq y(P)$ and $r(P) \lneqq y(P)$.
\begin{figure}[h!]
\includegraphics[width=8cm]{senya2.pdf}
\caption{Examples (infinite family) with $t(P)=2$, $r(P)=2$, and $y(P) \gneqq 2$ ($i(P)=2$).}\label{exa2}
\end{figure}
\end{example}
\section{Reductivity one}\label{sec_r1}
\begin{theorem}\label{r1}
For every reduced knot projection $P$, there exists a circle intersecting $P$ at just two double points of $P$, as shown in Fig.~\ref{rf6}, if and only if $t(P)=1$.

\end{theorem}

\begin{proof}

\begin{itemize}
\item (If part) Consider the case $t(P)=1$.  Let $P'$ be a reduced knot projection obtained from $P$ by applying any splice at a double point, say $a$, of $P$, and let $b$ be a reducible crossing of $P'$.  Then, there exists a simple closed curve that intersects $P$ with $a$ and $b$ only.  There are exactly four cases with respect to the connectivity among the arcs at $a$ and $b$ as shown in Fig.~\ref{rf7}.
\begin{figure}[h!]
\includegraphics[width=8cm]{kiyaku1.pdf}
\caption{All the possibilities for the distributions of arcs and the circle splitting the sphere into two disks before (upper) and after (lower) applying $A^{-1}$.}\label{rf7}
\end{figure}
Since $P$ is one component, the second and third cases remain.  However, the second and third cases are the same on $S^2$.
\item (Only if part) Recall Fact~\ref{f3}.  By the definition of $t(P)$, we immediately observe that $t(P) \le r(P)$ (Proposition~\ref{prop1}).  If there exists a circle with two double points, as shown in Fig.~\ref{rf6}, this reduced knot projection $P$ satisfies $1 \le t(P) \le r(P)=1$.  Therefore, $t(P)=1$.

\end{itemize}

\end{proof}

\begin{theorem}
For every reduced knot projection $P$, there exists a circle intersecting $P$ at just two double points of $P$, as shown in Fig.~\ref{rf6}, if and only if $y(P)=1$.
\end{theorem}
\begin{proof}
\begin{itemize}
\item (If part) Consider a reduced knot projection $P$ with $y(P)=1$.  By the definition, $t(P) \le y(P)$ and $P$ satisfies $t(P) = 1$.  By Theorem~\ref{r1}, for $P$, there exists a circle with two double points, as shown in Fig.~\ref{rf6}.
\item (Only if part) Obvious.

\end{itemize}
\end{proof}
\begin{corollary}\label{try_cor}
Let $P$ be a knot projection.   The following conditions are mutually equivalent.

\noindent $(1)$ $t(P)=1$.

\noindent $(2)$ $r(P)=1$.

\noindent $(3)$ $y(P)=1$.

\end{corollary}
\begin{corollary}
Let $P$ be a reduced knot projection.

\noindent $(1)$ If $r(P)=2$, then $t(P)=2$.

\noindent $(2)$ If $y(P)=2$, then $t(P)=2$.
\end{corollary}
\begin{remark}
There is no knot projection with $i(P)=1$ (see Proposition~\ref{prop1}~(4)).
\end{remark}
\section{Lower bounds for reductivities.}\label{sec_lower}
In this short section we show that {\it{circle numbers}}, first introduced in \cite{ITcircle}, are useful to obtain a lower bound of $r(P)$, $t(P)$, and $y(P)$.
\begin{definition}[circle number \cite{ITcircle}]
A circle number $|\tau(P)|$ of a knot projection $P$ is the number of circles that result from applying splice $A^{-1}$ to every double point of $P$ simultaneously.
\end{definition}
\begin{theorem}\label{thm_lower}
Let $P$ be a reduced knot projection.  If $|\tau(P)|=1$, then
\[2 \le t(P), 2 \le r(P),~{\text{and}}~2 \le y(P).\]
\end{theorem}
\begin{proof}
From Proposition~\ref{prop1} (1) and (2), it is clear that it is sufficient to show that $2 \le t(P)$ if $|\tau(P)|=1$.  Fig.~\ref{rf6} implies that $2 \le |\tau(P)|$ if $t(P)=1$.
\end{proof}
\begin{example}
See Figs.~\ref{exa1} and \ref{exa2}.
\end{example}
\begin{remark}
Known upper bounds are shown in Table \ref{table0} (see \cite{S} or Proposition~\ref{prop1}).  For example, if a reduced knot projection $P$ has at least one coherent $2$-gon, $t(P) \le 2$.
\begin{table}[h!]
\includegraphics[width=5cm]{hyou0.pdf}
\caption{Known upper bounds.  Blank cells denote unknown information.}\label{table0}
\end{table}
\end{remark}
\section{Shimizu's reductivity two}\label{sec_r2}
In this section, we prove Theorem~\ref{r2}.  The inverse operation $A$ of the non-Seifert splice $A^{-1}$ is the operation shown in Fig.~\ref{rf9}.

\noindent{\it{Proof of Theorem~\ref{r2}.}}

\begin{figure}[h!]
\includegraphics[width=5cm]{move.pdf}
\caption{Inverse replacements of a non-Seifert splice $A^{-1}$ when an arbitrary orientation is obtained.}\label{rf9}
\end{figure}
Applying an $A$ to the knot projection in Fig.~\ref{rf8} on the dotted two arcs produces the following three cases.   (Note that in Fig.~\ref{rf10}, we chose an orientation of a knot projection, but another choice of the orientation implies the same conclusion for unoriented knot projections.)
\begin{figure}[h!]
\includegraphics[width=2cm]{kiyakudo1a.pdf}
\caption{Dotted arcs $A$, $B$, $C$, and $D$.}\label{rf8}
\end{figure}
\begin{figure}[h!]
\begin{tabular}{|c|c|c|c|} \hline
Case 1 & Case 2-(i) & Case 2-(ii) & Case 3 \\ \hline
\includegraphics[width=2.5cm]{case1.pdf}&
\includegraphics[width=2.5cm]{case2a.pdf}&
\includegraphics[width=2.5cm]{case2b.pdf}&
\includegraphics[width=2.5cm]{case3.pdf}\\ \hline
\end{tabular}
\caption{Cases 1--3.}\label{rf10}
\end{figure}
\begin{itemize}
\item Case 1: Connect dotted arcs A and B. In this case, the second numbered knot projections are obtained (Fig.~\ref{rf10}).
\item Case 2: Connect dotted arcs A and C.  Pairs (A, D), (B, C), and (B, D) return the case (A, C).  (Note that each mirror image of every knot projection in Fig.~\ref{rf5} is the same as itself).  In this case, the the first, fourth, or fifth numbered knot projections are obtained (Fig.~\ref{rf10}).
\item Case 3: Connect dotted arcs C and D.  In this case, the third numbered knot projections are obtained (Fig.~\ref{rf10}). \end{itemize}
\hfill$\Box$

\section{Other reductivities two}\label{sec_t2}
In this section, we determine knot projections with $i(P)$, $y(P)$, or $t(P)$ $\le 2$.  The inverse operation of a non-Seifert splice (resp.~Seifert splice) is denoted by $A$ (resp.~$B$).

When proving claims in this section, we used an orientation of every knot projection for presenting an unoriented knot projection.  This orientation eliminates confusion.  Since a knot projection is a one-component curve, it is easy to find an unoriented knot projection by simply omitting the orientation.  Thus, references to the orientations of figures may be unnecessary in the proofs of Theorems~\ref{i_kiyakudo2_thm}, \ref{y_kiyakudo2_thm}, and \ref{t_kiyakudo2_thm}.
Note that we do not distinguish between a knot projection and its mirror image in this section.

Based on Fig.~\ref{rf11}, we consider all possible connections of two local places.
\begin{figure}[h!]
\includegraphics[width=5cm]{kiyaku2a.pdf}
\caption{Inverse moves $A$ or $B$ to consider all possibilities of producing a reducible knot projection after applying Seifert splice $A^{-1}$ or non-Seifert splice $B^{-1}$.}\label{rf11}
\end{figure}
When we choose $B$ from these possibilities, and we consider Fig.~\ref{rf12b}, we have Theorem~\ref{i_kiyakudo2_thm}.
\begin{theorem}\label{i_kiyakudo2_thm}
Let $P$ be a reduced knot projection.  There exists a circle intersecting $P$ at just two or three double points of $P$, as shown in Fig.~\ref{ito}, if and only if $i(P)=2$.
\begin{figure}[h!]
\includegraphics[width=8cm]{ito.pdf}
\caption{Knot projections with $i(P) = 2$.

}\label{ito}
\end{figure}
\end{theorem}
\begin{figure}
\includegraphics[width=10cm]{itokiyaku2.pdf}
\caption{Knot projections with $i(P)=2$.}\label{rf12}
\end{figure}
\begin{figure}[h!]

\includegraphics[width=5cm]{itokiyaku2aex.pdf}
\caption{Knot projections with $i(P)=2$.

}\label{rf12b}
\end{figure}
\begin{proof}
One Seifert splice changes the number of components of a spherical curve, and, therefore, $2 \le i(P)$ for a reduced knot projection $P$.  As shown in Fig.~\ref{rf12} and Fig.~\ref{rf12b}, we obtain all possible ways to apply $B$ at the two places of a reducible knot projection.
\end{proof}

Next, we consider the necessary and sufficient condition $y(P)=2$.  Similar to the above, we consider any possibilities of connections at two places, as shown in Fig.~\ref{rf11}.  It is clear that the case in the first line of Fig.~\ref{rf13} is impossible.  All the possible cases appear in the second line of Fig.~\ref{rf13}.
\begin{figure}[htbp]
\includegraphics[width=12cm]{yusukekiyaku2.pdf}
\caption{Knot projections with $y(P) \le 2$.}\label{rf13}
\end{figure}
\begin{theorem}\label{y_kiyakudo2_thm}
Let $P$ be a knot projection.  There is no possibility of existence of a simple circle, as in Fig.~\ref{rf6}, and there exits a circle intersecting $P$ at just three double points of $P$, as shown in Fig.~\ref{yusuke},

if and only if $y(P)=2$.
\begin{figure}[h!]
\includegraphics[width=8cm]{yusuke1.pdf}
\caption{Knot projections with $y(P) \le 2$.}\label{yusuke}
\end{figure}
\end{theorem}
\begin{proof}
Fig.~\ref{rf13} shows all possible ways to apply $A$ at the two places of a reducible knot projection.
\end{proof}
\begin{corollary}
If $y(P)=2$ for a knot projection $P$, its chord diagram has a sub-chord diagram \h.
\end{corollary}

Finally, we determine knot projections with $t(P)=2$ using Theorems~\ref{i_kiyakudo2_thm} and \ref{y_kiyakudo2_thm}.  Based on Fig.~\ref{rf11}, we have considered all possibilities of connections consisting of one kind of operation, either (B, B) or (A, A), at two places.  Therefore, we add all the cases of mixed type (A, B).  There are four such cases. (See Fig.~\ref{rf14}.)
\begin{figure}[h!]

\includegraphics[width=10cm]{taniyamakiyaku2a.pdf}

\caption{Four cases obtaining knot projections with $t(P) \le 2$.}\label{rf14}
\end{figure}
This leads to Theorem~\ref{t_kiyakudo2_thm}.
\begin{theorem}\label{t_kiyakudo2_thm}
Let $P$ be a reduced knot projection.
There is no possibility of existence of a simple closed curve, as shown in Fig.~\ref{rf6}, and there exists a circle intersecting $P$ at just two or just three double points of $P$, as shown in Fig.~\ref{taniyama},

if and only if $t(P)=2$.
\begin{figure}[h!]
\includegraphics[width=8cm]{taniyama.pdf}
\caption{Knot projections with $t(P) \le 2$.

}\label{taniyama}
\end{figure}
\end{theorem}

\section{Tables}\label{sec_table}
In this section, we provide tables of prime reduced knot projections with the four types of reductivities.  A {\it{prime knot projection}} is defined as a knot projection that can not be a connected sum of two non-trivial knot projections.  A knot projection is called a {\it{prime reduced}} knot projection if the knot projection is prime and reduced (cf.~Sec.~\ref{intro}).

\begin{table}[htbp]
\includegraphics[width=11cm]{hyou1a.pdf}
\caption{Prime reduced knot projections and reductivities ($3_1$--$7_1$).}
\end{table}
\begin{table}[htbp]
\includegraphics[width=11cm]{hyou2a.pdf}
\caption{Prime reduced knot projections and reductivities ($7_2$--$7_C$).}
\end{table}

\section*{Acknowledgements}
The authors would like to thank Professor Kouki Taniyama for his useful comments.  The authors also thank Ms.~Senja Barthel for fruitful discussion and providing us with her examples.  The part of this work contributed by N.~Ito was supported by a Waseda University Grant for Special Research Projects (Project number: 2014K-6292) and the JSPS Japanese-German Graduate Externship.

config: arxiv


TITLE: Understand the properties of this function
QUESTION [3 upvotes]: We define a function
$f(t):=\sum_{n=0}^{\infty}e^{-nt}= \frac{1}{1-e^{-t}}= \frac{e^{\frac{t}{2}}}{e^{\frac{t}{2}}-e^{-\frac{t}{2}}}=\frac{2e^{\frac{t}{2}}}{\sinh\left(\frac{t}{2} \right)}$
observe that although the series is only defined for $t >0$ the right-hand side is defined everywhere. Thus, we can extend $f$ to all of $\mathbb{R}\backslash \{0\}.$ Observe that this function though has poles on the imaginary-axis ($2 \pi i n$ in the complex plane)
Now, we define a function $g(t):=\sum_{n=0}^{\infty} e^{-\sqrt{n}t}.$ In that case the trick with the geometric series does no longer work. What I would like to understand though is whether there is a continuation of this function as above which has roots roughly like $n^2$ on the imaginary axis? Can this conjecture be true?

REPLY [1 votes]: There is a meromorphic continuation of $g(t)$ with only pole of order $2$ at $t=0$. To prove this, lets consider the Mellin transform $G(s)$ of $g(t)-1$. It is easy to see that $g(t)-1$ decreases exponentially when $t$ goes to infinity and that $g(t)\ll \frac{1}{t^2}$ when $t\to 0$. So, the integral
$$G(s)=\int\limits_0^{+\infty} (g(t)-1)t^{s-1}dt$$
converges absolutely when $\mathrm{Re}\,s>2$. Now, we have
$$\int\limits_0^{+\infty} e^{-t\sqrt n}t^{s-1}dt=n^{-s/2}\Gamma(s),$$
thus,
$$G(s)=\zeta(s/2)\Gamma(s).$$
For fixed $\sigma$ and $|T|\to \infty$, we have $\zeta((\sigma+iT)/2)\Gamma(\sigma+iT)\ll e^{-\frac{\pi |T|}{2}}|T|^{O(1)}$, so we can apply Mellin inversion formula to obtain the following:
$$g(t)-1=\frac{1}{2\pi i}\int\limits_{3-i\infty}^{3+i\infty} G(s)t^{-s}ds.$$
As $G(s)$ decays exponentially when $\mathrm{Im}\,s$ is large, we can move the contour of integration to the line $\mathrm{Re}\,s=-N-\frac12$. $G(s)$ has poles in the points $s=2$ and $s=-n$ for $n \in \mathbb Z_{\geq 0}$. Thus, using Cauchy's integral formula and estimating the integral over the line $\sigma=-N-\frac12$, we get
$$g(t)-1=\mathrm{Res}_{s=2}\,G(s)t^{-s}+\sum\limits_{n=0}^{N} \mathrm{Res}_{s=-n}G(s)t^{-s}+O(t^{N+1/2})=\frac{2}{t^2}+\sum\limits_{n=0}^N \frac{(-1)^n\zeta(-n/2)t^n}{n!}+O(t^{N+1/2}).$$
By the functional equation for the Riemann zeta function,
$$\zeta(-n/2)=-2^{-n/2}\pi^{-n/2-1}\sin\left(\frac{\pi n}{4}\right)\Gamma(1+n/2)\zeta(1+n/2)\ll n^{n/2}.$$
Thus, the asymptotic expansion above is convergent and letting $N \to +\infty$ we prove (for $0<t<1$) that
$$g(t)-1=\frac{2}{t^2}+\sum\limits_{n=0}^{+\infty} \frac{a_nt^n}{n!}$$
with $|a_n|\ll n^{n/2}$. So, $\lim\limits_{n\to \infty} \left(\frac{|a_n|}{n!}\right)^{1/n}=0$ and the series for $g(t)-1-\frac{2}{t^2}$ have infinite radius of convergence. Thus, by analytic continuation, the function $g(t)-\frac{2}{t^2}$ is entire, as needed.


text:
\begin{definition}[Definition:Intersection (Geometry)]
The '''intersection''' of two [[Definition:Line|lines]] $AB$ and $CD$ is denoted by $AB \cap CD$.
The '''intersection''' of two [[Definition:Geometric Figure|geometric figures]] is the set of [[Definition:Point|points]] shared by both [[Definition:Geometric Figure|figures]].
Note that this use of $\cap$ is consistent with that of its more usual context of [[Definition:Set Intersection|set intersection]].
When two [[Definition:Line|lines]] '''intersect''', they are said to '''cut''' each other.
\end{definition}

config: wiki


TITLE: Limit of ratio of incomplete gamma function
QUESTION [1 upvotes]: In order to derive Sterling's approximation, I need to show that the following integral decays quicker than at least $\mathcal{O}(n^2)$:
$\lim_{n\to\infty}\dfrac{\int_{2n}^\infty x^ne^{-x}dx}{\int_{0}^{2n} x^ne^{-x}dx}$ is at most $\mathcal{O}(n^2)$, this integral can be written as $\lim_{n\to\infty}\dfrac{\Gamma(n+1,2n)}{\gamma(n+1,2n)}$ in terms of incomplete gamma functions. I have been trying analytical methods since a month but to no use. I tried to plot the ratio of this ratio to $\frac{1}{n^6}$ i.e. $\dfrac{n^6\int_{2n}^\infty x^ne^{-x}dx}{\int_{0}^{2n} x^ne^{-x}dx}$ vs n as shown below and so I am sure that it decays at least as quick as $\mathcal{O}(n^6)$, maybe it decays exponentially, however I need to produce an analytical upper bound for ratio.
Is there some light? or some identity on incomplete gamma functions?

REPLY [2 votes]: We can use some fairly brutal estimates. On the one hand,
\begin{align}
\int_0^{2n} x^n e^{-x}\,dx
&> \int_n^{2n} x^n e^{-x}\,dx \\
&> n^n \int_n^{2n} e^{-x}\,dx \\
&= n^ne^{-n}(1-e^{-n}) \\
& > \frac{n^n}{2e^n}
\end{align}
for $n \geqslant 1$.
On the other hand, with $g(x) = x^n e^{-x/2}$ we have
$$g'(x) = \biggl(\frac{n}{x} - \frac{1}{2}\biggr)g(x) \leqslant 0$$
for $x \geqslant 2n$, whence
\begin{align}
\int_{2n}^{\infty} x^ne^{-x}\,dx
&= \int_{2n}^{\infty} g(x) e^{-x/2}\,dx \\
&\leqslant g(2n) \int_{2n}^{\infty} e^{-x/2}\,dx \\
&= 2g(2n)e^{-n} \\
&= 4\cdot \frac{n^n}{2e^n}\cdot \biggl(\frac{2}{e}\biggr)^n\,.
\end{align}
Hence
$$\frac{\Gamma(n+1,2n)}{\gamma(n+1,2n)} \leqslant 4\biggl(\frac{2}{e}\biggr)^n\,,$$
i.e. we have exponential decay.


text:
\begin{document}

	\title{Euler Calculus on spaces homeomorphic to definable sets and some applications}

	\author[E. Mac\'{\i}as-Virg\'os]{E. Mac\'{\i}as-Virg\'os}
	\address{Department of Mathematics, University of Santiago de Compostela, Spain.}
	\email{quique.macias@usc.es}

	\author[D. Mosquera-Lois]{D. Mosquera-Lois}
	\address{Institute of Mathematics, University of Santiago de Compostela, Spain.}
	\email{david.mosquera.lois@usc.es}

	\thanks{The first author was partially supported by MINECO-FEDER research project MTM2016-78647-P. The second author was partially supported by an IMAT research grant}

	\date{\today}

	\subjclass[2010]{03C64,
		28A25,
		55R05,
		}
	\keywords{ Euler calculus, Euler-Poincar\'e characteristic, Euler integration,  fiber bundles, sensor networks}

\begin{abstract}
We show that integration with respect to the Euler-Poincar\'e characteristic can be extended from the setting of definable sets to the setting of topological spaces homeomorphic to definable sets. We use that extension to generalize a result regarding sensor networks due to Ghrist and Baryshnikov, in order to make it more flexible in applications. Finally, we obtain both an extension and a combinatorial proof of a classical result  about the  Euler-Poincar\'e characteristic of fiber bundles.
\end{abstract}

\maketitle

\section{Introduction}

For topological spaces,  the Euler-Poincar\'e characteristic can be defined as the alternating sum of the ranks of the singular homology groups. For finite simplicial complexes this coincides with the classical combinatorial definition. As a consequence, the  Euler-Poincar\'e characteristic is a topological invariant. In the context of o-minimal structures and definable sets \cite{Viro}, a combinatorial definition of the characteristic is used \cite{Dries} in order to define an integration with respect to Euler-Poincar\'e characteristic. However, it has the drawback of no longer being a homotopy invariant. Moreover,  definable sets can be quite restrictive when studying topological questions or using topological arguments.

In the work of Curry, Baryshnikov, Ghrist and Robinson \cite{Ghristsurvey} it is implicitly suggested that the combinatorial Euler-Poincar\'e characteristic is a topological invariant. Furthermore, spaces which are homeomorphic to definable sets are implicitly used. However, neither this is stated explicitly nor an integration theory is developed in this broader context. We aim to fill this gap by formalizing these ideas. The original formulation of Euler Calculus used subanalytic sets \cite{Schapira1991,Schapira1995}. More recent work uses definable sets. The present paper applies a theorem from Beke (if two definable sets are homeomorphic, where the homeomorphism does not need to be definable, then they have equal characteristic) to extend Euler calculus to spaces homeomorphic to definable sets. We use that extension to provide generalizations of some results regarding enumeration of objectives. Moreover, by means of that extension and the heuristics of Euler calculus, we provide a combinatorial proof of the multiplicative behavior of the Euler-Poincar\'e characteristic for fiber bundles where the total space, the base and the fiber are homeomorphic to definable sets, by discretizing Fubini's theorem and avoiding the usage of spectral sequences. \\

\noindent{\em Acknowledgement} The second author would like to thank Robert Ghrist for enlightening discussions on the topic of this work.

\section{Preliminaries}

In this section we summarize some definitions and results taken from \cite{Beke,Ghristsurvey,Dries}.

\begin{definition}
	An {\em open $n$-simplex} is the interior of an $n$-simplex $[v_0,\ldots, v_n]$, i.e., the set
	$$\{t_0v_0+ \cdots+ t_nv_n \in \mathbb{R}^m \; \colon \; \sum_i{t_i}=1 , \; t_i> 0 \; \forall i\},$$
	where $v_0, \ldots, v_n$  are affine independent points in $\mathbb{R}^m$, $m\geq n$.
\end{definition}

\begin{definition}
	A {\em generalized simplicial complex} is a finite collection $\K $ of open simplices in $\R^n$, satisfying that given two open simplices in the complex, the intersection of their closures is the empty set or the closure of an open simplex in the complex.
\end{definition}

\begin{definition}
	We define the {\em Euler-Poincar\'e  characteristic} of the generalized simplicial complex $\K$ as the alternating sum of the number of open simplices in each dimension, i.e.,
	$$\chi(\K)=\sum_{i=0}^{n}(-1)^{i}c_i,$$ where
	$c_i$ is the number of open $i$-simplices.
\end{definition}

\begin{definition}
	Given a topological space $X$, a {\em triangulation} is a homeomorphism $h \colon |\K | \rightarrow X$ where $|\K |$ is the geometric realization of the generalized simplicial complex $\K $.
\end{definition}

\begin{remark}
	The Euler-Poincar\'e  characteristic of an open $n$-simplex is $(-1)^n$, hence, it is not homotopy invariant. However, it is homotopy invariant on compact sets.
\end{remark}

\begin{definition}
	An {\em o-minimal structure} (over $\R$) is a colection $\A =\{\A _n\}_{n\in \N}$, where each $\A _n$ is an algebra of subsets of $\R^n$,
	satisfying:
	\begin{enumerate}

		\item $A\in \A _n$, $B\in \A _m \Rightarrow A\times B\in \A _{n+m}$.
		\item $A\in \A _n \Rightarrow \pi(A)\in \A _{n-1}$, where $\pi$ is the projection into the first $n-1$ coordinates.
		\item $\Delta_{i,j}=\{(x_1,\ldots,x_n)\in \R^n \colon x_i=x_j\} \in \A _n$ for all $1\leq i<j\leq n$.
		\item $\{(x,y)\in \R^2\colon x<y\} \in \A _2$.
		\item $\A _1=\{\text{finite unions of points and open intervals of $\R$}\}.$
	\end{enumerate}

	Given $A\subset \R^n$, we say that $A$ is  {\em definable in $\A $} if $A \in \A _n$. A map between definable sets is a {\em definable map} if its graph is definable.
\end{definition}

\begin{theorem}[{Definable triangulation theorem \cite{Dries}}]
	Assume that $X\subset \R^n$ is a definable set and $\{X_i\}_{i=1}^{m}$ is a finite family of definable subsets of $X$. Then there exists a definable triangulation of $X$ compatible with the collection of subsets.
\end{theorem}

So, the {\em Euler-Poincar\'e characteristic of a definable set} is defined as the characteristic of the associated generalized simplicial complex.

\begin{theorem}[{\cite{Dries}}]
	Assume $X\subset \R^n$ and $Y\subset \R^m$
	are definable sets. Then:
	\begin{enumerate}
		\item $\chi(X \cup Y)=\chi(X)+\chi(Y)-\chi(X \cap Y).$

		\item $\chi(X\times Y)=\chi(X)\cdot \chi(Y).$
	\end{enumerate}
\end{theorem}

\begin{definition}
	If $X$ is a definable subset of $\R^n$, we define the $\mathbb{Z}$-module of {\em constructible functions} $CF(X)$ as follows: a function $h\colon  X \rightarrow \Z$ belongs to $CF(X)$ iff it is bounded, has compact support and the level sets are definable.
\end{definition}

\begin{theorem}[Deconstruction of constructible functions] \label{tma:deconstruction:functions} If $X$ is a definable set and $h \in CF(X)$, then $h=\sum_{\alpha}c_{\alpha}\mathbb{1}_{\sigma_{\alpha}}$, where  $c_{\alpha}\in \Z$, $\sigma_{\alpha}$ are the open simplices of a triangulation and $\mathbb{1}_{\sigma_{\alpha}}$ are the characteristic functions of $\sigma_{\alpha}$.
\end{theorem}

\begin{definition}
	The \emph{Euler integral} is the homomorphism
	$$\int\nolimits_X (\bullet) \, d\chi \colon CF(X) \rightarrow \Z$$ given by
	$$\int\nolimits_X \bigg(\sum_\alpha c_\alpha\mathbb{1}_{\sigma_\alpha}\bigg) \, d\chi=\sum_\alpha c_\alpha\chi(\sigma_\alpha).$$
\end{definition}

\begin{theorem}[{\cite{Ghristsurvey}}] The Euler integral satisfies the following properties:
	\begin{enumerate}
		\item Additivity with respect to the integrand:
		$$\int\nolimits_X\bigg(\sum_ih_i\bigg) \, d\chi =\sum_i\int\nolimits_Xh_i \, d\chi.$$
		\item Additivity with respect to the domain:
		$$\int\nolimits_{A\cup B}h \, d\chi =\int\nolimits_A h \, d\chi + \int\nolimits_B h \, d\chi - \int\nolimits_{A\cap B}h \, d\chi.$$
		\item Fubini's theorem:
		If $f\colon  X \rightarrow Y$ is a continuous definable map and $h\in CF(X)$, then
		$$
		\int\nolimits_X h \, d\chi=\int\nolimits_Y\bigg(\int\nolimits_{f^{-1}(y)}h \, d\chi(x)\bigg) \, d\chi(y).
		$$
	\end{enumerate}
\end{theorem}

\section{Spaces homeomorphic to definable sets and Euler Calculus}

We begin by recalling the following theorem of Beke \cite{Beke}.

\begin{theorem}[\cite{Beke}] \label{theorem:Beke}
	Let $X$ and $Y$ be definable sets. If they are homeomorphic (not necessarily a definable homeomorphism), then $\chi(X)=\chi(Y)$.
\end{theorem}

\begin{definition}
	Given a topological space $X$ homeomorphic to a definable set $A$, we define its {\em Euler-Poincar\'e characteristic} as $\chi(X)\coloneqq\chi(A)$.
\end{definition}

Theorem \ref{theorem:Beke} guarantees that the Euler-Poincar\'e characteristic of   spaces homeomorphic to  definable sets is well-defined.

\begin{example}
	Examples of spaces homeomorphic to definable sets are:
	\begin{itemize}
		\item Subcomplexes of regular finite CW-complexes.
		\item Smooth compact manifolds (with or without boundary).
		\item Topological compact manifolds (with or without boundary) of dimension less or equal to three.
	\end{itemize}
\end{example}

\begin{theorem}
	\label{tma:propiedades:conjuntos:homeo:a:definibles}
	Let $X$ and $Y$ be topological spaces   homeomorphic to definable sets. Then:
	\begin{enumerate}
		\item given a finite family $\{X_i\}_{i=1}^m$ of subsets of $X$ such that each member $X_i$ is homeomorphic to a definable set, there exists a triangulation of $X$ compatible with the collection of subsets.
		\item If $X$ is compact, then the simplicial complex associated with the triangulation of $X$ is compact (a finite simplicial complex).
		\item If $X$ is compact, then its Euler-Poincar\'e characteristic defined counting open simplices equals its Euler-Poincar\'e characteristic using singular (co)\-ho\-mo\-lo\-gy, i.e., $$\chi(X)=\sum_n(-1)^n \rank H_n(X,\Z).$$
		\item Inclusion-exclusion principle:
		\[\chi(X \cup Y)=\chi(X)+\chi(Y)-\chi(X \cap Y).\]
		\item Multiplicativity of the characteristic: $$\chi(X\times Y)=\chi(X) \cdot \chi(Y).$$
	\end{enumerate}
\end{theorem}

Moreover, given a space $X$ homeomorphic to a definable set $A$, the existence of the homeomorphism $\beta\colon  X \rightarrow A$ guarantees that the theorem about deconstruction of constructible functions (Theorem \ref{tma:deconstruction:functions}) extends to functions defined on $X$ such that their level sets are mapped by $\beta$ to definable subsets of $A$.

\begin{definition}
	Assume that $X$ is a  space homeomorphic to a definable set $A$ by the homeomorphism $\beta \colon  X \rightarrow A$. We define the $\Z$-module $HCF(X)$ of {\em homeo-integrable functions}, that is, functions $h\colon  X \rightarrow \Z$ which are bounded, have compact support and their level sets are homeomorphic to definable subsets of $A$ by restrictions of the homeomorphism $\beta$.
\end{definition}

\begin{theorem} \label{tma:deconstruc_funciones:homeo:construibles}
	If $X$ is a space homeomorphic to a definable set, and $h \in HCF(X)$, then $h$ can be decomposed uniquely as a finite linear combination of characteristic functions of the open simplices of a triangulation compatible with the support of $h$, i.e.,  $h=\sum_\alpha c_\alpha\mathbb{1}_{\sigma_\alpha}$, where  $c_\alpha\in \Z$ and $\sigma_\alpha$
	are the open simplices of the triangulation. The function $h$ can also be written as a linear combination of characteristic functions on the level sets.
\end{theorem}

Now, we can extend Euler integration to functions in $HCF(X)$, where $X$ is a space homeomorphic to a definable set.

\begin{definition}
	Given a space $X$  homeomorphic to a definable set, we define the {\em integral with respect to the Euler-Poincar\'e characteristic} of the function $h=\sum_\alpha c_\alpha\mathbb{1}_{\sigma_\alpha} \in HCF(X)$ as  $$\int\nolimits_X h \, d\chi\coloneqq\sum_\alpha c_\alpha\chi(\sigma_\alpha).$$
\end{definition}

Theorem \ref{tma:deconstruc_funciones:homeo:construibles} guarantees that this integral  is well-defined. As it happens in classical Euler Calculus, if $U$ is a subspace of $X$ homeomorphic to a definable set, then $\mathbb{1}_U\in HCF(X)$ and $\int\nolimits_X\mathbb{1}_U \, d\chi=\chi(U)$.  The results regarding the additivity of the integral also extend without major modifications.

\begin{theorem}[Additivity of the integral with respect to the integrand]
	If $X$ is a space homeomorphic to a definable set, and $\{h_i\}_{i=1}^m$ is a finite set of functions in $HCF(X)$, then
	$$
	\int\nolimits_X{\bigg(\sum_{i=1}^m h_i\bigg)} \, d\chi=\sum_{i=1}^m{\int\nolimits_X h_i} \, d\chi.
	$$
\end{theorem}

\begin{theorem}[Additivity of the integral with respect to the domain] \label{theorem:aditivity:domain}
	Let $X$ be a space homeomorphic to a definable set, and let $h\colon  X \rightarrow \Z$ be a function in $HCF(X)$. If $\{X_i\}_{i=1}^m$ is a finite cover of $X$ by subspaces homeomorphic to definable subsets, then
	\begin{align*}
	\int\nolimits_{X}h \, d\chi=&\sum_i  \int\nolimits_{X_i}h \, d\chi - \sum_{i> j} \int\nolimits_{X_i\cap X_j}h \, d\chi
	\\
	&\quad +\sum_{i>j> k} \int\nolimits_{X_i\cap X_j \cap X_k}h \, d\chi - \cdots
	+ (-1)^{m+1} \int\nolimits_{X_1 \cap  \cdots \cap X_m}h \, d\chi.
	\end{align*}
\end{theorem}

\section{Consequences}

In this section we present generalizations of Ghrist's results and two theorems regarding the Euler-Poincar\'e characteristic. First, we obtain a generalization of \cite[Theorem 3.2]{Bary2}.

\begin{theorem}
	Assume that $X$ is a space homeomorphic to a definable set and $h\colon  X \rightarrow \N$ is a function in $HCF(X)$, $h=\sum_{\alpha\in I}\mathbb{1}_{U_\alpha}$, where the supports $\{U_\alpha\}_{\alpha \in I}$ are subspaces homeomorphic to definable subsets. If the Euler-Poincar\'e characteristic of all the supports  $U_\alpha$ is $N$ and different from zero, then the number of observables is
	$$\vert I\vert=\frac{1}{N} \int\nolimits_X h \, d\chi.$$
\end{theorem}

\begin{remark}
	Another results about sensor networks presented in \cite{Bary2} can be generalized to spaces homeomorphic to definable sets by discretizing time.
\end{remark}

Using the generalization of Euler Calculus to spaces homeomorphic to definable sets we prove now the relationship between the Euler-Poincar\'e characteristics of the spaces involved in a locally trivial fiber bundle. The proof is mainly of combinatorial nature, avoiding the use of spectral sequences as in \cite[Theorem 1, p. 481]{Spanier} or \cite{Serre}, and improves \cite{Spanierjoven2}. Moreover, the theorem below generalizes related results since it works for all spaces homeomorphic to definable sets and we only need the compactness assumption on the base space.

\begin{theorem}
	Let $p \colon  E \rightarrow B$ be a locally trivial fiber bundle where
	\begin{enumerate}
		\item[(a)] the total space $E$, the base $B$ and the fiber $F$ are homeomorphic to definable sets;
		\item[(b)] $B$ is compact.
	\end{enumerate}
	Then $\chi(E)=\chi(B)\cdot \chi(F)$.
\end{theorem}

\begin{proof}
	It follows from the Lebesgue number lemma, Theorem \ref{theorem:aditivity:domain} and the multiplicativity of the characteristic. Being more precise, since $B$ is compact we can cover it with a finite trivialization by open sets $\{U_j\}_{j\in J}$. Consider a triangulation of $B$ and subdivide it until we have a finite refinement $\{B_i\}_{i\in I}$ by subspaces  homeomorphic to definable subsets. Assume $I=\{1,\ldots,m\}$. Then we have:
	\begin{align*}
	\chi(E)= &\int\nolimits_X\mathbb{1}_E \, d\chi\\
	= &\sum_i \int\nolimits_{p^{-1}(B_i)}\mathbb{1}_E \, d\chi - \sum_{i\neq j} \int\nolimits_{p^{-1}(B_i)\cap p^{-1}(B_j)}\mathbb{1}_E \, d\chi
	\\
	&\quad+\sum_{i> j > k} \int\nolimits_{p^{-1}(B_i)\cap p^{-1}(B_j) \cap p^{-1}(B_k)}\mathbb{1}_E \, d\chi \\
	&\quad - \cdots + (-1)^{m+1} \int\nolimits_{p^{-1}(B_1) \cap  \cdots \cap p^{-1}(B_m)}\mathbb{1}_E \, d\chi\\
	& \text{}\\
	= &\sum_i \int\nolimits_{p^{-1}(B_i)}\mathbb{1}_E \, d\chi - \sum_{i> j} \int\nolimits_{p^{-1}(B_i\cap B_j)}\mathbb{1}_E \, d\chi \\
	&\quad+\sum_{i> j > k} \int\nolimits_{p^{-1}(B_i\cap B_j \cap B_k)}\mathbb{1}_E  \, d\chi \\
	&\quad - \cdots
	+ (-1)^{m+1} \int\nolimits_{p^{-1}(B_1 \cap  \cdots \cap B_m)}\mathbb{1}_E  \, d\chi\\
	& \text{}\\
	= &\sum_i  \chi(p^{-1}(B_i)) - \sum_{i> j} \chi(p^{-1}(B_i\cap B_j))\\
	&\quad+\sum_{i> j > k} \chi(p^{-1}(B_i\cap B_j \cap B_k))\\
	&\quad - \cdots + (-1)^{m+1} \chi(p^{-1}(B_1 \cap  \cdots \cap B_m))\\
	& \text{}\\
	= &\sum_i  \chi(B_i \times F) - \sum_{i> j} \chi((B_i\cap B_j)\times F)\\
	&\quad+\sum_{i> j > k } \chi((B_i\cap B_j \cap B_k) \times F )\\
	&\quad - \cdots + (-1)^{m+1} \chi((B_1 \cap  \cdots \cap B_m)\times F)\\
	& \text{}\\
	= &\sum_i  \chi(B_i) \cdot \chi(F) - \sum_{i> j} \chi(B_i\cap B_j) \cdot \chi(F) \\
	&\quad +\sum_{i> j > k} \chi(B_i\cap B_j \cap B_k)\cdot \chi(F)\\
	&\quad  - \cdots + (-1)^{m+1} \chi(B_1 \cap  \cdots \cap B_m) \cdot \chi(F)\\
	& \text{}\\
	= &\chi(F) \cdot \bigg( \sum_i  \chi(B_i) - \sum_{i> j} \chi(B_i\cap B_j) \\
	&\quad+\sum_{i> j > k} \chi(B_i\cap B_j \cap B_k)\\
	&\quad - \cdots + (-1)^{m+1} \chi(B_1 \cap  \cdots \cap B_m) \bigg)\\
	& \text{}\\
	= & \chi(F) \cdot \chi(B).
	\end{align*}
\end{proof}

Notice that this result can be thought as being essentially Fubini's theorem, where the characteristic $\chi(F)$ is a constant.

By using the same ideas, a purely combinatorial proof of the Riemann-Hurwitz Formula, alternative to the one presented in \cite{Viro},  can also be given.

\begin{theorem}[Riemann-Hurwitz]
	If $f\colon  \widetilde{S} \rightarrow S$ is a ramified covering between two compact Riemann surfaces without boundary, i.e., a non constant holomorphic map, then
	\begin{equation*}
	\chi(\widetilde{S})=n\chi(S) -\sum_{x\in S}(e_f(x)-1),
	\end{equation*}
	where $e_f(x)$ is the ramification index.
\end{theorem}

\bibliographystyle{plain}
\bibliography{biblio_EC}

\end{document}


TITLE: A simple criterion for semi-simplicity
QUESTION [1 upvotes]: I made a simple observation about semi-simplicity of rings:

A ring $R$ is semi-simple iff for  every $x\in R$ there is an element $y\in R$ s.t. $xyx=x$.

It seems very interesting to have such a simple criterion for semi-simplicity. It should be known but I couldn't find a reference. Does there exists good references and other analogs?
Edit. As mentioned by Dietrich-Burde in the answers the above criterion is wrong. But one can prove that if all of endomorphism rings in an abelian category are Von Neumann  regular then every object is semisimple.

REPLY [3 votes]: Such rings are called von Neumann regular rings. Every semisimple ring is von Neumann regular, but conversely not every von Neumann regular ring is semisimple. This is true, though, if we assume left (or right) Noetherian in addition.


TITLE: Prove that central limit theorem Is applicable to a new sequence
QUESTION [0 upvotes]: Let $\xi_1, \xi_2...$ be i.i.d. random variables. $\mathbb{E}\xi_i = 0, \mathbb{E}\xi^2_i = 1 \ \forall i $. Let $\lambda_1, \lambda_2, ...$ be a sequence of real number such that
$$
\frac{\max_{k \in \{1, ..., n\}} \lambda_k^2}{\sum\limits_{i=1}^n \lambda_i^2} \longrightarrow 0
$$
Prove that CLT is applicable for sequence of random variables $\lambda_1 \xi_1, \cdots \lambda_2 \xi_n\cdots$.

REPLY [1 votes]: That's my solution.
Let $\phi(t)$ be a characteristic function of $\xi$. Then remembering, that $\mathbb{E}\xi = 0, \mathbb{E}\xi^2 = 1$, one can see
$$
\phi(t) = 1 - \frac{t^2}{2} + r(t),
$$
where $r(t) = o(t^2)$.
Let $\zeta_n$ be equal to $\frac{\lambda_1 \xi_1 + ... + \lambda_n \xi_n}{\sqrt{\lambda_1^2 + ... + \lambda_n^2}}$. Then, using independence of random variable, we get
$$
\phi_{\zeta_n}(t) = \prod_{i=1}^n\phi\left(\frac{\lambda_i}{\sqrt{\lambda_1^2 + ... + \lambda_n^2}}t\right)
$$
Taking $\log $ of both sides one gets
$$
\log{\phi_{\zeta_n}(t)} = \sum\limits_{i=1}^n \log{\left(1 - \frac{\lambda_i^2}{\lambda_1^2 + ... + \lambda_n^2}\frac{t^2}{2} + r\left(\frac{\lambda_i}{\sqrt{\lambda_1^2 + ... + \lambda_n^2}}t\right)\right)}
$$
Considering Taylor's series, let's recall $\log(1 + x) = \sum\limits_{k=1}^{\infty} \frac{(-1)^{k-1} x^k}{k!}$.
Therefore if $\log(1+x) = x + \alpha(x)$ and $|x|\leq \frac12$, then
$$
|\alpha(x)| = |\sum\limits_{k=2}^{\infty} \frac{(-1)^{k-1} x^k}{k!}| \leq \frac 12 \sum\limits_{k = 2}^{\infty} |x|^k \leq |x|^2.
$$
For characteristic functions it can be proved that (considering $\mathbb{E} \xi = 0)$
$$
|\phi_{\xi}(t) - 1| \leq \frac{t^2}{2} \cdot \mathbb{E}\xi^2 = \frac{t^2}{2}
$$
Let's fix $T > 0$ and choose $n_0$ so that $\forall n \geq n_0$ $\max \frac{\lambda_i^2}{\lambda_1^2 + \cdots \lambda_n^2} \frac{T^2}{2} \leq \frac 12$. Then $\forall t \in [-T, T]$
$$
|\phi_{\zeta_n} + \frac{t^2}{2}| \leq \sum\limits_{i=1}^n \left|r\left(\frac{\lambda_i}{\sqrt{\lambda_1^2 + ... + \lambda_n^2}}t\right)\right| + \sum\limits_{i=1}^n \left|\frac{\lambda_i^2}{{\lambda_1^2 + ... + \lambda_n^2}} \frac{t^2}{2}\right|^2 \leq
$$
$$
\leq \sum\limits_{i=1}^n \left|r\left(\frac{\lambda_i}{\sqrt{\lambda_1^2 + ... + \lambda_n^2}}t\right)\right| + \frac{T^4}{4} \max_i \frac{\lambda_i^2}{\lambda_1^2 + \cdots \lambda_n^2} \cdot \sum\limits_{i=1}^n \frac{\lambda_i^2}{\lambda_1^2 + ... + \lambda_n^2} =
$$
$$
= \sum\limits_{i=1}^n \left|r\left(\frac{\lambda_i}{\sqrt{\lambda_1^2 + ... + \lambda_n^2}}t\right)\right| + \frac{T^4}{4} \max_i \frac{\lambda_i^2}{\lambda_1^2 + \cdots \lambda_n^2}
$$
Last step is to evaluate first sum. Let's recall that $r(t) = o(t^2)$. So $\forall t_0 \in (-T, T) \ \exists C_{t_0}: \ |r(t)|\leq C_{t_0} \cdot t^2$ on $(-t_0, t_0)$, and $C_{t_0}$ converges to $0$ while $t_0$ converges to $0$.
$t':= \max \frac{\lambda_i}{\sqrt{\lambda_1^2 + \cdots + \lambda_n^2}} T$
Then (it's an only place where I use that all random variables are identically distributed)
$$
\sum\limits_{i=1}^n \left|r\left(\frac{\lambda_i}{\sqrt{\lambda_1^2 + ... + \lambda_n^2}}t\right)\right| \leq C_{t'} T^2 \sum\limits_{i=1}^n \left(\frac{\lambda_i}{\sqrt{\lambda_1^2 + ... + \lambda_n^2}}\right)^2 = C_{t'}T^2
$$
Finally, we get
$$
\left|\phi_{\zeta_n}(t) + \frac{t^2}{2}\right| \leq C_{t'}T^2 + \frac{T^4}{4} \max_i \frac{\lambda_i^2}{\lambda_1^2 + \cdots \lambda_n^2}
$$
The right side converges to $0$ on $[-T, T]$ while $n \longrightarrow \infty$.


TITLE: The abstract definition of commutative monoids
QUESTION [4 upvotes]: In trying to begin to understand the idea of a $k$-tuply monoidal $n$-category, I'm already a bit stuck on the idea (Baez, nLab) that a commutative monoid can be defined as a monoid object in the category Mon of monoids. So what I have in a monoid object in Mon is a normal old monoid $M$ equipped with a multiplication morphism $\mu: M\times M \rightarrow M$ such that $\mu\circ(id_M\times \mu):M\times M\times M\rightarrow M=
\mu\circ(\mu\times id_M): M\times M\times M \rightarrow M$
and $\mu: 1\times M\rightarrow M=\mu: M\times 1\rightarrow M=\pi_1(M\times 1)=\pi_2(1\times M)$.
In the last part the trivial monoid $1$ is identified with its image in $M$, while the formula before just says $\mu$ defines an associative binary operation on $M$. I've also identified $M\otimes M$, the monoidal category tensor, with $M\times M$. That works here, right?
All of this is separate from the multiplication on $M$ that I get just from being in Mon. Now I need somehow to connect this monoid-object-in-Mon multiplication with the original multiplication $\cdot$ on $M$ to get $a\cdot b=b\cdot a$ for each $a,b \in M$. I don't know how to do this, in part because I'm not sure how to think about that last statement categorically. Do I just define $\cdot$ as another monoidal structure on $M$ that follows the same rules as $\mu$, and try to relate them?
I appreciate any help.

REPLY [4 votes]: Since you are interested in higher categories, you should see the Eckmann-HIlton argument as a  special case of the use of the interchange law, or exchange law. A general formulation of this is for double categories: this is a set, or class, $C$, with two category structures $\circ_1, \circ_2$ each of which is a morphism for the other. This amounts to the rule
$$(x \circ_1 y)\circ_2(z\circ_1 w)= (x \circ_2 z) \circ_1 (y \circ_2 w)$$whenver both sides are defined, which can also be represented in matrix terms as there is only one way of evaluating $$\left[ \matrix{ x&z \\y&w }\right]
 $$
($\circ_2$ is horizontal, $\circ_1 $ is vertical, like matrices). One then proves that while such a structure is not  itself  abelian, it does contain a family of abelian monoids.  (The fact that such morphisms preserve identities is a crucial part of the argument.)  It was this fact that raised the possibility  of non abelian higher homotopy groupoids, and indeed of higher groupoids having a potential in wide areas of mathematics and physics, giving a kind of "higher dimensional group theory".


TITLE: Golden Ratio plus 1
QUESTION [3 upvotes]: There was an interesting puzzle by Presh Talwalker in 'MindYourDecisions' about finding the radius of a circle that was cotangent to two larger circles.
https://www.youtube.com/watch?v=i0dZukEw1JY
I extended the problem by considering adding a second small circle adjacent to the circle of radius 2, (now 16 in my diagram) and wondered what the radius of the two large circles would have to be to make the two small circles have equal radii. The attached diagram will make this more clear.

I was surprised to find that the ratio of the two circles must be equal to 2.61803398875 which is, of course, the Golden Ratio + 1 The straight lines tangent to the larger circles subtend an angle of 53.130102 degrees. The series of circles can be extended to the right indefinitely and will give pairs of circles of equal radii. See second image.

Has this been found before? I have looked on the Internet without finding anything.

REPLY [3 votes]: I don't know if this has been seen before, although I suspect it's known. Here's a quick proof.

Let $\bigcirc P$, $\bigcirc Q$, $\bigcirc R$ (of radii $p$, $q$, $r$) be successively-tangent circles with common external tangent lines. Note that similarity guarantees a common proportionality factor, $k$, such that
$$\frac{q}{p} = k = \frac{r}{q}$$
Now, suppose that $\bigcirc P^\prime$, tangent to $\bigcirc Q$, $\bigcirc R$, and one of those lines, has radius $p$. By Descartes' Theorem for "kissing circles" (one of which is a line, hence a circle of zero curvature), we have
$$\left(\;\frac1{p}+\frac1{q}+\frac1{r}\;\right)^2 = 2\left(\;\frac1{p^2}+\frac1{q^2}+\frac1{r^2}\;\right)
\quad\to\quad
\left(\;1+\frac1k+\frac1{k^2}\;\right)^2= 2\left(\;1+\frac1{k^2}+\frac1{k^4}\;\right)$$
so that
$$(k^2 - 3 k + 1) (k^2 + k + 1) = 0$$
The latter factor has no real roots, and the former gives $k = \frac12\left(3\pm\sqrt{5}\right) = 1 + \phi^{\pm 1}$, where $\phi$ is the Golden Ratio. The smaller of these turns out to be extraneous, so we conclude $k = 1 + \phi$. $\square$
There's probably a less-algebraic "picture proof" demonstration of the relation, but I'm not seeing it at the moment.

The fact that "[t]he series of circles can be extended to the right indefinitely and will give circles of equal radii" is a simple consequence of similarity.


TITLE: Show that for vectors $\bf u$ and $\bf v$ in $ℝ^3$, $\bf u \times  v = (-v) \times u$
QUESTION [2 upvotes]: How would I use the properties of determinants to show that for any two vectors $\bf u$ and $\bf v$ in $ℝ^3$
$$\bf u \times  v = (-v) \times u$$

REPLY [0 votes]: Several nice answers have been given. I would like to explain it from another point of view.
The fact $u\times v= (-v)\times u$ is a very basic property for cross product. There are of course several ways to prove it. But I think the best way is to explain and prove it using the definition of cross product. Otherwise, if we need to calculate a lot to prove it, I don't think the calculation can help you to understand it better, and the fundamental thing may be concealed by the calculation.
Hint: In order to see $u\times v= (-v)\times u$, we need first show $u\times v= -v\times u$ and $-v\times u= (-v)\times u$. The best way is to use the definition (see wiki)
$u\times v=\|u\|\|v\|\sin\theta\mathbf{n}$.


TITLE: Finding the least squares solution for the system of equations $y=Ax^2+B$
QUESTION [1 upvotes]: Find the least squares solution for the system of equations $y = Ax^2 + B$ where
$(x, y)$ belongs to the set {$(0, 1),(1, 5),(−1, 3)$}. What is the geometric (graphical) interpretation
of the solution?
Is this done the same way as a least squares solution for $Ax=B$? If not then how do I start this?

REPLY [0 votes]: Constant + quadratic
The trial function is
$$y(x) = \alpha + \gamma x^2.
$$
The input vectors are
$$
x = \left[
    \begin{array}{r}
      0\\
      1\\
     -1
    \end{array}
\right], \quad
y = \left[
    \begin{array}{r}
      1\\
      5\\
      3
    \end{array}
\right]
$$
They imply the linear system of column vectors:
$$
\begin{align}
 \mathbf{A} c &= y \\
  \left[
    \begin{array}{rr}
      1 & x^2
    \end{array}
    \right]
    \left[
    \begin{array}{r}
      \alpha \\
      \gamma
    \end{array}
    \right]
      & =
    \left[
    \begin{array}{rr}
      y
    \end{array}
    \right]
\end{align}
$$
The system matrix $\mathbf{A}$ is built with the $x$ data, the data vector is constructed with the $y$ data. The task is to find the coefficient vector $c$.
Using the input data, we have
$$
\begin{align}
 \mathbf{A} c &= y \\
\left[
    \begin{array}{rr}
      1 & 0\\
      1 & 1\\
      1 & 1
    \end{array}
\right]
\left[
    \begin{array}{r}
      \alpha \\
      \gamma
    \end{array}
\right]
&  =
\left[
    \begin{array}{r}
      1\\
      5\\
      3
    \end{array}
\right]
\end{align}
$$
Use the normal equations
$$
\mathbf{A}^{*}\mathbf{A} c =\mathbf{A}^{*}y
$$
 to find the least squares solution:
$$
c = \left( \mathbf{A}^{*}\mathbf{A} \right)^{-1}\mathbf{A}^{*}y
$$
The product matrix and its inverse are
$$
\mathbf{A}^{*}\mathbf{A} =
%
    \left[
    \begin{array}{rr}
      3 & 2\\
      2 & 2
    \end{array}
    \right], \qquad
%
\left( \mathbf{A}^{*}\mathbf{A} \right)^{-1} =
%
\left[
    \begin{array}{rr}
      1 & -1\\
     -1 & \frac{3}{2}
    \end{array}
\right].
$$
The solution vector is
$$
c = \left[
    \begin{array}{r}
      \alpha\\
      \gamma
    \end{array}
\right]
 =
\left[
    \begin{array}{r}
      1\\
      3
    \end{array}
\right],
$$
that is, the solution function is $y(x) = 1 + 3x^2$. The error vector is
$$
  r = \mathbf{A} c - y =
\left[
    \begin{array}{r}
       0 \\
      -1 \\
       1
    \end{array}
\right], \qquad
r \cdot r = 2.
$$
The solution curve is plotted against the data points below.

Constant + linear
With the quadratic fit complete, turn to the linear fit. The trial function becomes
$$
y(x) = \alpha + \beta x
$$
The $x$ and $y$ vectors are unchanged. The linear system of column vectors is
$$
\begin{align}
 \mathbf{A} c &= y \\
\left[
    \begin{array}{rr}
      1 & 0\\
      1 & 1\\
      1 & -1
    \end{array}
\right]
\left[
    \begin{array}{r}
      \alpha \\
      \beta
    \end{array}
\right]
&  =
\left[
    \begin{array}{r}
      1\\
      5\\
      3
    \end{array}
\right]
\end{align}
$$
With the input data, we have
$$ \left[
    \begin{array}{rr}
      1 & 0\\
      1 & 1\\
      1 & -1
    \end{array}
\right]
\left[
    \begin{array}{r}
      \alpha \\
      \beta
    \end{array}
\right]
  =
\left[
    \begin{array}{r}
      1\\
      5\\
      3
    \end{array}
\right]
$$
The product matrix and its inverse are
$$
\mathbf{A}^{*}\mathbf{A} =
%
    \left[
    \begin{array}{rr}
      3 & 0 \\
      0 & 2
    \end{array}
    \right], \qquad
\left( \mathbf{A}^{*}\mathbf{A} \right)^{-1} = \frac{1}{6}
\left[
    \begin{array}{rr}
      2 & 0 \\
      0 & 3
    \end{array}
\right].
$$
The solution vector here is
$$c =
\left[
    \begin{array}{r}
      3\\
      1
    \end{array}
\right],
$$
so the solution function is $y(x) = 3 + x$. The error vector is
$$
  r = \mathbf{A} c - y =
\left[
    \begin{array}{r}
       2 \\
      -1 \\
       1
    \end{array}
\right], \qquad
r \cdot r = 6.
$$
The first solution represented a quadratic equation, the second a linear equation (see plot below).

To emphasize the difference between the quadratic solution $y_{q}$ and the linear solution $y_{l}$, they are written with terms in increasing order:
$$
        \begin{matrix}
        y_{q}(x) & = & \alpha_q &   &         & + & \gamma x^2\\
        y_{l}(x) & = & \alpha_l & + & \beta x
        \end{matrix}
$$
Constant + linear + quadratic
The final case is
$$
  y(x) = \alpha + \beta x + \gamma x^{2}.
$$
The linear system is
$$
\begin{align}
\mathbf{A} c &= y \\
 \left[
    \begin{array}{rr}
      1 &  0 & 0\\
      1 &  1 & 1\\
      1 & -1 & 1
    \end{array}
\right]
\left[
    \begin{array}{r}
      \alpha \\
      \beta \\
      \gamma
    \end{array}
\right]
 & =
\left[
    \begin{array}{r}
      1\\
      5\\
      3
    \end{array}
\right]
\end{align}
$$
The product matrix and its inverse are
$$
\mathbf{A}^{*}\mathbf{A} =
%
\left[
\begin{array}{ccc}
 3 & 0 & 2 \\
 0 & 2 & 0 \\
 2 & 0 & 2 \\
\end{array}
\right], \qquad
\left( \mathbf{A}^{*}\mathbf{A} \right)^{-1} = \frac{1}{2}
\left[
\begin{array}{rrr}
 2 & 0 & -2 \\
 0 & 1 & 0 \\
 -2 & 0 & 3 \\
\end{array}
\right].
$$
The solution vector is:
$$c =
\left[
    \begin{array}{r}
      1\\
      1\\
      3
    \end{array}
\right].
$$
The solution vector is $y(x)  = 1 + x + 3x^{2}$. The error vector is
$$
  r = \mathbf{A} c - y =
\left[
    \begin{array}{r}
       0 \\
       0 \\
       0
    \end{array}
\right], \qquad
r \cdot r = 0.
$$


TITLE: Property function counting prime numbers
QUESTION [2 upvotes]: Let $\Pi$ be the usual function counting prime numbers. Show that for any $n\in \mathbb{Z}$, exists $k\in \mathbb{Z}$ such that
$$\Pi((k+1)^2)-\Pi(k^2) \geq n$$
I tried to use the Prime Numbers Theorem to estimate how many prime numbers should exist between $(k+1)^2$ and $k^2$ in the form
$$\lim_{n\to \infty} \frac{p_n}{n\log n} =1 $$
But that was not good to estimate the difference of the function between consecutive squares

REPLY [3 votes]: If each prime less than $x$ is put into a group according to which squares it is surrounded by, there are $\lfloor \sqrt{x}\rfloor$ groups.  Therefore at least one of those groups has more than $\frac{\pi(x)}{\sqrt{x}}$ primes, and by the prime number theorem this value tends to infinity.


TITLE: Integration in normed vector spaces
QUESTION [2 upvotes]: Given an interval $I\subset \mathbb{R}$ and a normed vector space $X$, I want to know if I am able to define the Lebesgue space $L^p(I;X)$ of all $p$-integrable functions $f:I\to X$. I know that this can be done if $X$ is a Banach space (which leads to the notion of Bochner spaces), but for general normed vector spaces I don't know. So I would be grateful for some help and references if it is possible. Thanks.

REPLY [2 votes]: No you can’t. Completeness is critical.
The problem is that for a sequence of simple functions $(s_n)$ such that $\lim\limits_{n \to \infty} \int \Vert f -s_n\Vert =0$ the sequence $\int s_n$ may not converge in $X$.


TITLE: Geometric calculations using Grassmann variables
QUESTION [9 upvotes]: Physicists seem to get huge computational value by introducing Grassmann variables and Grassmann integration into differential geometric calculations.
See: http://en.wikipedia.org/wiki/Grassmann_number
Can someone here motivate these techniques mathematically, and include the simplest  pure-math example where their use and value can be illustrated.  I have thought they were invented for computing volumes of constrained moduli spaces, although physicists did not originally describe them this way.  Do any mathematicians use them rigorously to actually solve problems?

REPLY [4 votes]: Grassman variables are a neat and handy way to talk about exterior algebras and the geometric objects built on them, like forms and spinors. The physics notation in particular gets a lot of mileage out of Grassman calculations that look formally like Gaussian integrals and behave remarkably like them.  The best intro to that, and the bet purely mathematical application I know, is Mathai and Quillen's paper where they get an explicit representation of the Thom form that is quite powerful.  They can be a little intimidating because physicists speak about them in language grating to mathematicians and tend to use them in the vicinity of highly nonrigorous thinking, but they themselves can be worked with mathematically without difficulty.


TITLE: Position of points in an isometric image after rotation
QUESTION [0 upvotes]: I'm developing a game using isometric images for the units.
Each units has weapons represented in red on the first image below.
To attach animations to the weapons, I have to be able to calculate the position of each of those points after a rotation of X degrees (units move in 8 directions).
The isometric view is set with an angle of 45 degrees top down :
Points on the image for the main angle
My goal is to find a way to manually spot the red dots on one angle, and then apply a mathematical transformation to get the position of those points for all others angles.
Image after a rotation
Image after an other rotation
I have no clue on how to approach the problematic and I'd be happy to get some help here !
The 3 images are properly scaled if you want to give it a try.
Here is the first point coordinates in pixels on each image, from the top left corner (0,0) :
Point A - 0 degree rotation : (89,522)
Point A - 45 degree rotation : (573,550)
Point A - 90 degree rotation : (609,207)

So the goal is to have (89,522) and deduce (573,550) and (609,207) from that.
Thanks a lot !
G.
Edit :
It seems that a rotation on one frame around a given center point would not work as shown in that picture, probably because of the 45 degree top down angle.

REPLY [0 votes]: You need to know not just the angle of the rotation, but also what axis the rotation is about. A 30 degree rotation about the x-axis is going to give a much different location than a 30 degree rotation about the y-axis. And a rotation about the y-axis passing through the ankles is going to give a different result than the same rotation about the y-axis passing through the shoulders.
Game developers tend to prefer quaternion to matrix notation for these rotations, and I find it hard to believe you don't already such rotations built into your application. If you really haven't worked out how to move the camera around, then I urge you to put aside your immediate problem and start working your way through the wealth of material out there that explains how it's done, because it is far more than I (or anyone else) can adequately explain in a forum post.
I don't have much need for 3D rotations, so I never remember how quaternion rotation works, but if you are interested, this is how it works by vectors:
Suppose you want to rotate the point $\mathbf p$ by $\theta$ degrees about the axis passing through the point $\mathbf q$ and extending in the direction of the vector $\hat n$, where $\|\hat n\| = 1$, and $\theta$ is positive when rotating in the same direction as the fingers of your right-hand curl when your extended thumb points in the direction of $\hat n$.
Let $\mathbf o = \mathbf q + ((\mathbf p - \mathbf q)\cdot \hat n)\hat n$ and let $\vec p = \mathbf p - \mathbf o$ be the vector pointing to $\mathbf p$ from $\mathbf o$. Then $\mathbf o$ is the point on the axis of rotation that is closest to $\mathbf p$, and $\vec p$ is perpendicular to $\hat n$. Let $\vec r = \hat n \times \vec p$. The rotated point that you want is
$$\mathbf p' = \mathbf o + \vec p\cos \theta  + \vec r\sin\theta $$

FYI - the reason for the somewhat unusual mixture of bold and $\to$ notation is that the bold letters represent points in space, while $\vec v$ represents a vector (direction and magnitude only). And the hat notation $\hat n$ is a common way of denoting unit vectors.


TITLE: questions about a proof to a question (about relations)
QUESTION [0 upvotes]: 1) $R,S,T$ are relations on the same set.
Prove that $R(S\cup T)=RS\cup ST$
The proof that I stumbled upon was the following:
$(a,b)\in R(S\cup T)⇒((a,x)\in R)∧((x,b)\in S∨(x,b)\in T)⇒(a,b)\in RS∨(a,b)\in RT$
At this stage I got stuck. How exactly does the last transition make sense? Could someone please explain this?
2) When multiplying relations, let's say the relations $R$ and $S$ - do we say that $(x,y)\in R ∨ (x,y)\in S$ ?
Or perhaps $(x,y)\in R ∧ (x,y)\in S$? ('and' instead of 'or')
And why is it?
Thank you.

REPLY [1 votes]: $(a,b) \in RZ \iff (a,x)\in R \land (x,b) \in Z$
This means: $a$ and $b$ are related by the product $RZ$ if, and only if, there exists an $x$ such that $a$ and $x$ are related by $R$, and $x$ and $b$ are related by $Z$.
$(a,b) \in S\cup T \iff (a,b) \in S \lor (a,b) \in T$
This means: $a$ and $b$ are related by the union of $S$ and $T$ if, and only if, either $a$ and $b$ are related by $S$ or $a$ and $b$ are related by $T$.
Put it together.
$$(a,b) \in R(S\cup T) \\ \iff (a,x)\in R \land (x,b) \in (S\cup T) \\ \iff (a,x)\in R \land \Bigl((x,b) \in S \lor (x,b) \in T\Bigr) \\ \iff \Bigl((a,x)\in R\land (x,b) \in S\Bigr)\lor\Bigl((a,x)\in R \land (x,b) \in T\Bigr) \\ \iff (a,b) \in RS \lor (a,b)\in RT \\ \iff (a,b)\in RS\cup RT \\\therefore R(S\cup T)\equiv RS\cup RT$$


TITLE: Is every G-invariant function on a Lie algebra a trace?
QUESTION [18 upvotes]: I am in the (slow) process of editing my notes on Lie Groups and Quantum Groups (V Serganova, Math 261B, UC Berkeley, Spring 2010.  Mostly I can fill in gaps to arguments, but I have found myself completely stuck in one step of one proof.  One possibility that would get me unstuck is a positive answer to the following (which may be obviously false or trivial, but I'm not thinking well):

Question: Let $\mathfrak g$ be a finite-dimensional Lie algebra over $\mathbb K$, and if necessary you may assume that $\mathbb K = \mathbb C$ and that $\mathfrak g$ is semisimple.  Then $\mathfrak g$ acts on itself by the adjoint action, and on polynomial functions $f : \mathfrak g \to \mathbb K$ via derivations.  A polynomial $f: \mathfrak g \to \mathbb K\,$ is $\mathfrak g$-invariant if $\mathfrak g \cdot f = 0$.  For example, let $\pi: \mathfrak g \to \mathfrak{gl}(V)$ be any finite-dimensional reprensentation.  Then $x \mapsto \operatorname{tr}_V \bigl(\pi(x)^n\bigr)$ is $\mathfrak g$-invariant for any $n\in \mathbb N$.  Is every $\mathfrak g$-invariant function of this form?  Or at least a sum of products of functions on this form?

When $\mathfrak g$ is one of the classical groups $\mathfrak{sl},\mathfrak{so},\mathfrak{sp}$, or the exceptional group $G_2$ the answer is yes, because we did those examples in the aforementioned class notes.  But I have no good grasp for the $E$ series, and I don't know if the statement holds for non-semisimples.
What I'm actually trying to prove is a weaker statement, but I figured I'd ask the stronger question, because to me the answer is not obviously "no".  The weaker statement:

Claim: Let $\mathfrak g$ be a finite-dimensional semisimple Lie algebra over $\mathbb C$.  Then every $\mathfrak g$-invariant function is constant on nilpotent elements of $\mathfrak g$.  (Recall that $x\in \mathfrak g$ is nilpotent if $\operatorname{ad}(x) = [x,] \in \mathfrak{gl}(\mathfrak g)$ is a nilpotent matrix — some power of it vanishes.)

It's clear that the spectrum of any nilpotent matrix is $\{0\}$, and for a semisimple Lie algebra, any nilpotent element acts nilpotently in all representations.  For the classical groups, in the notes we exhibited generators for the rings of $\mathfrak g$-invariant functions as traces of representations, and so we can just check the above claim.  But we did not do the $E$ series or $F_4$.

REPLY [5 votes]: If the representation is fixed as the fundamental representation, then in the case of $\mathfrak{so}(2n)$, you need Pfaffians as well as traces.


TITLE: Setting up Double Integral
QUESTION [0 upvotes]: Problem:Find the Moment of Inertia of a circular disk of uniform density about an axis which passes through the center and makes an angle of $\dfrac{\pi}{6}$ with the plane of the disc.

Moment of Inertia ($I$) is $\int r^2dm$ where $r$ is the perpendicular distance from the chosen axis (and can vary) and $dm$ is an elemental mass.
$$$$
I interpreted the disk as a 2 Dimensional structure, hence it has mass per unit area ($\sigma$)
$$dm=\sigma dA$$
Thus, $$I=\int r^2dm=\int \int_A \sigma dA$$
$$$$
Using a change of variables to Cylindrical Co-ordinates,
$$dA=rdrd\theta$$
$$$$
$$\Longrightarrow I=\int\int \sigma r^3 dr d\theta$$
$$$$
However, this is the Integral for finding the Moment of Inertia for an axis perpendicular to the plane of the body. I cannot understand how to modify it for the given question.
$$$$
I would be grateful if somebody could solve this question without  differential equations.

REPLY [0 votes]: I am doing this in two stages. The disc needs to be divided into elements in the form of infinitesimally thin and concentric hoops, which we then integrate. But first we have to find the MI of a hoop about an axis at angle $\frac {\pi}{6}$ to the plane of the hoop, through the centre of the hoop.
Consider a circle of radius $r$ in the $xy$ plane, centre $O$, and a point on the circumference with coordinates $P(r\cos\theta,r\sin\theta, 0)$.
We will write the axis, as specified in the question, as the line $$\underline{r}=t\left(\begin{matrix}\frac{\sqrt{3}}{2}\\0\\\frac 12\end{matrix}\right)$$
The distance $d$ from $P$ to the line is given by the magnitude of the cross product $$\overrightarrow{OP}\times\underline{\hat{r}}$$
A quick calculation gives $$d^2=\frac 14r^2(1+3\sin^2\theta)$$
Now let the circle be a hoop of mass $m$ and density per unit length $\rho$
The MI of an element of mass $\delta m$ at $P$ is $\delta I=\delta m d^2=\rho r \delta\theta d^2$
Therefore for the whole hoop, $$I=\frac 14\rho r^3\int_0^{2\pi}(1+3\sin^2\theta)d\theta=\frac 58mr^2$$
Now finally we can consider the disc of radius$R$, mass $m$ and density per unit are $\rho$ divided into concentric elements of thickness $\delta x$ and radius $x$
Then $$\delta I=\frac 58\delta mx^2=\frac 54\pi\rho x^3\delta x$$
$$\Rightarrow I=\frac 54\pi\rho\int_0^Rx^3dx$$
$$\Rightarrow I=\frac{5}{16}mR^2$$


TITLE: Is $X_{0}$ a subset of $X_{0}\sqcup X_{1}$?
QUESTION [1 upvotes]: Since the disjoint union of $X_{0}= \{x_{1}, x_{2}, x_{3}\}$ and $X_{1}= \{x_{1}, x_{2}\}$ is: $ X_{0}\sqcup X_{1}= \{(x_{1},0), (x_{2},0), (x_{3},0), (x_{1},1), (x_{2},1)\}$ I have to ask; is $X_{0}$ a subset of $X_{0}\sqcup X_{1}$? I just would like to know because I can't tell if there ought to be an inclusion function.
The inclusion function is the identity map $id: x \mapsto x$ but the elements of the disjoint union are indexed...

REPLY [2 votes]: No, formally $X_0$ is not a subset of $X_0\sqcup X_1$. But there is a natural injective function
$$i:X_0\to X_0\sqcup X_1$$
$$i(x)=(x,0)$$
and therefore many authors will treat $X_0$ as a subset of $X_0\sqcup X_1$ via the embedding above, i.e. they will replace $X_0$ with the image $i(X_0)$. This is so natural that most authors will completely forget about indexing.


TITLE: to prove a coordinate transformation V->U
QUESTION [0 upvotes]: Let U be
$U=M=\mathbb{R^2} \backslash \{(x,0):x \leq 0\} $
and $x:U \rightarrow M$ be the parametrization by cartesian coordinates $(x,y)$.
Let also V be $V=\{(r,\varphi) \in \mathbb{R^2}: r >0,\varphi \in (-\pi, \pi)\}.
  $
Now I've got to show with the help of polar coordinates,  that a coordinate transformation $V \rightarrow U$ is defined. Furthermore, I got to determine the metric coefficients of both parametrizations of M. I really don't understand why there is a coordinate transformation! I don't get it.

REPLY [0 votes]: I will change the name of your map for obvious reasons. Consider the map $g: U \to M$ defined by $g(u) = (x(u),y(u))$. You know that $x(u) = r \cos \theta, y(u) = r \sin \theta$, where $r = \sqrt{x(u)^2+y(u)^2}$. Here $\theta \ (= \varphi)$ is just the possible angle that the vector extending from the origin to $x(u)$ can make. Well, you've thrown out the negative $x$-axis, which corresponds to all points $P \in \mathbb{R}^2$ such that the angle between $\vec{OP}$ and the positive x-axis is $\pm \pi$ i.e $\theta \in (-\pi, \pi)$.


TITLE: Classes of sheaves and the Riemann-Roch theorem (Grothendieck 1957)
QUESTION [1 upvotes]: I couldn't find this paper (preferably english translation) on the web, math.stackexchange, or mathoverflow.net; could someone please point me to the document to read? Thanks!

REPLY [3 votes]: As pointed out in the comments, Grothendieck didn't publish these results. They appeared in

Armand Borel and Jean-Pierre Serre, Le théorème de Riemann-Roch, Bulletin de la S. M. F., tome 86 (1958), p. 97-136 doi:10.24033/bsmf.1500

The paper is subtitled (d'après des résultats inédits de A. GROTHENDIECK).


TITLE: About Springs Nature
QUESTION [0 upvotes]: Why when we coil some straight wires (may be metallic or nonmetallic) then it gives a restoring force (some what like spring)?Why when the wire was straight it doesn't give that appreciable amount of restoring force?
Does in this case also Force is proportional to length the coiled wire extended. If yes then consider giving the ratio of Force to length extended (i.e. Effective spring constant) in terms of young's modulus(Y) and other required terms.

REPLY [7 votes]: The deformation mode for a straight wire is stretching, which is a relatively stiff mode of deformation.  The spring stiffness for a straight wire is given by:
$$k=\frac{EA}{L}$$
where $E$ is Young's modulus, $A$ is the cross-sectional area, and $L$ is the length of the wire.  For a coiled spring the deformation is dominated by torsion of the wire, which is a relatively compliant mode of deformation.  For a coiled spring the stiffness is approximately:
$$k=\frac{G a^4}{4 n R^3}$$
where $a$ is the radius of the wire, $R$ is the radius of the coil, $G$ is the shear modulus, and $n$ is the number of coils.
EDIT: To address the question of where the second formula comes from, I think the following explanation is the simplest I can come up with.  Let's say the axial force applied to the spring is $P$ and it is applied at the center of the coils. The displacement of the spring is $\Delta$.  The stiffness of the spring is then $k=P/\Delta$. The torque that is carried at any cross-section of the wire is $T=PR$, where $R$ is the radius of the coils. Now, during the process of loading the spring the force $P$ does work and the coils store strain energy.  If we equate the work done by the force during the loading process to the strain energy stored in the coils, then we should be able to derive the formula.
$$\frac{1}{2} P\Delta=2n\pi R\int_A \frac{\tau^2}{2 G} dA$$
where again $n$ is the number of coils, $G$ is the shear modulus, and $A$ is the cross-section of the wire. Now, for a wire, the shear stress can be related to the torque as,
$$\tau=\frac{Tr}{I_p}$$
where $r$ is the radial position within the cross-section, and $I_p=\pi a^4/2$ is the polar moment of inertia of the cross-section that has a wire radius of $a$. So, our integral then becomes:
$$2n\pi R\left(\frac{8\pi T^2}{2G\pi^2 a^8}\int_{0}^{a} r^3 dr\right)=\frac{2nRT^2}{Ga^4}=\frac{2nR^3P^2}{Ga^4}=\frac{P^2}{2k}$$
Solving for $k$ gives:
$$k=\frac{Ga^4}{4nR^3}$$


TITLE: Let $A$ be a set such that each member of $A$ is a non-empty set. Then $A \times X \neq x$ for all $X \in A, x \in \bigcup A$.
QUESTION [0 upvotes]: I tried so hard but to no avail while proving the statement:

Let $A$ be a set such that each member of $A$ is a non-empty set. Then $A \times X \neq x$ for all $X \in A, x \in \bigcup A$. Here $A \times X$ is the Cartesian product of $A$ and $X$.

Please shed some light! Thank you for your help!

REPLY [0 votes]: $A$ is a collection of sets, $X$ is a set in $A$ and $x$ is an
element. $x = A\times X$ has to violate regularity. Proof.
There is some $K \in A$ with $x \in K$ and some $y \in X$.
Thus $x \in K \in \{K,y\} \in \{\{K\}, \{K,y\}\} = (K,y) \in A \times X = x$.


TITLE: Basic question about using potential divider to limit capacitor charging voltage
QUESTION [1 upvotes]: This is for a DIY project I'm working on. I know I could just monitor the charging process and unplug before the capacitor blows, or get a more suitable power supply, but it just so happens that I have a bunch of capacitors rated at 1000 V and a transformer that gives me 6000 V DC. I'd like to use this rather than spend extra on getting another transformer.
Let's say I put the capacitor in parallel with a resistor, and this in series with another resistor, i.e. I'm using a potential divider to limit the capacitor output, like this:

Initially, the capacitor is going to be like a short circuit, so it'll charger very rapidly, but as it fills up, more charge will go through R1 instead. When the capacitor is charged at $V = V_{in} \times R_1 / (R_1+R_2)$, there should be no current passing through the capacitor.
This seems like a very simple situation to me, but for some reason I don't see anyone doing this or any information about a circuit like this online (or maybe I'm just not using the right search terms...)
Am I missing out something fatal that'll completely screw up my capacitors if I try this?
Also, any tips on choice of resistor size? I'd probably want to use the most resistant resistors I have to keep current flow small.

REPLY [0 votes]: So to get 1000V from the 6000V supply you need a ratio of 1:5
higher resistance will cause slower charging.
Pick a resistance that would give an acceptable result if used with a 1000V supply, then choose your R1,R2 such that their parallel combination is that resistance. and their ratio is 1:5.
can you instead use a step-down tranformer on the input to the 6000V supply?


TITLE: Prove that a product of Borel algebras is a sigma algebra
QUESTION [0 upvotes]: I was working on the following question and the very last step in my proof required the following:

If $X_1, \dots, X_n$ are separable metric spaces, then $\prod\limits_{i=1}^n \mathcal{B}(X_i)$ is a sigma algebra.

How do I show this? I'm asking a new question because this short claim is much simpler than what I originally asked. I no longer need people to answer the original complicated question, just this simpler and weaker statement. At the same time, I want to leave the other question available on its own to help people in the future who are looking for the full problem.

REPLY [2 votes]: You cannot prove this. $\mathbb R^{2} \setminus [(0,1) \times (0,1)]$ cannot be expressed as  a product $A \times B$ for any sets $A$ and $B$. Hence the collection of sets $A \times B$ such that $A$ and $B$ are Borel sets in $\mathbb R$ is not  a sigma algbra.


TITLE: Simple identity for entire function
QUESTION [0 upvotes]: If $f(-\overline{z})=-f(\overline{z})$ for all $z\in\mathbb{C}$, and if $f$ is entire, then $f(-z)=-f(z)$ for all $z\in\mathbb{C}$.
Is it false when the condition that $f$ is entire is removed from the above statement?

REPLY [1 votes]: $f(-z)=-f(z)$ means that the graph of $f$ (for instance in this sense) looks the same if you turn it around $180^\circ$ about the origin (it has a $180^\circ$ rotational symmetry about the origin). $f(-\overline{z})=-f(\overline{z})$ means that if you mirror the graph of $f$ over the real axis, then the resulting graph has a $180^\circ$ rotational symmetry about the origin. It's quite clear intuitively that these two properties are equivalent.


TITLE: Can "integer" be used as an adjective?
QUESTION [1 upvotes]: Although "integer" in Latin means "whole", as far as I know in English it is used as a noun. Personally, I have not seen an integer "number", but only "integers". My question is that is it wrong to say
k is an integer "number"?

REPLY [2 votes]: The adjective form is integral, even though this is overloaded with the word integral as it pertains to integration.
As for saying integer number, it's used as a noun here, in the sense that computer science is valid but contains two nouns.


TITLE: probability - distinguishable vs indistinguishable objects/persons
QUESTION [1 upvotes]: I am solving one simple problem just for fun.
The problem is as follows.
$n$ people have to be seated randomly in a cinema hall which has $(n+k)$ seats.
What is the probability that in this process, some fixed $m$ places ($m \leq n$) get occupied/taken?
The book gives this answer:
${{n} \choose {m}} \cdot {{n-k-m}\choose{n-m}} / {{n+k} \choose {n}}$
There are 2 things I don't like here:
1) I think that ${{n-k-m}\choose{n-m}}$ is obviously a typo and should read ${{n + k-m}\choose{n-m}}$
Why? Well to say the least $n-k-m \leq n-m$ so this binomial coefficient does not make much sense here.
2) This answer seems to treat the $n$ people as indistinguishable which is strange. Since these are $n$ people/persons these are $n$ different people (we can never have identical people, let alone $n$ of them, even twins are different persons). Thus it makes more sense to treat the people as distinguishable. So the order in which the people take the places also matters, right?
But in that case I am getting this answer:
${{n} \choose {m}} \cdot m! \cdot {{n + k-m}\choose{n-m}} \cdot (n-m)! / ( {{n+k} \choose {n}} \cdot n! ) $
What is your opinion of these two items?
Am I correct in both or at least in one of them?

REPLY [1 votes]: The term $\binom{n}{m}$ counts the number of ways $m$ of the $n$ people can occupy the specified $m$ seats.  That leaves $n + k - m$ seats available.  The remaining people can occupy $n - m$ of them.  Hence, the number of favorable cases is
$$\binom{n}{m}\binom{n + k - m}{n - m}$$
as you suspected.  The denominator $\binom{n + k}{n}$ represents the number of ways the $n$ people can select $n$ of the $n + k$ seats in the theater.
The people are distinguishable in the sense that we are selecting which $m$ of the $n$ people sit in the designated seats.  What does not matter is which particular person sits in which particular seat, just which seats are occupied.