text:
\begin{document}

\title{Dehn surgery and Seifert surface system}

\author{Makoto Ozawa}
\address{Department of Natural Sciences, Faculty of Arts and Sciences, Komazawa University, 1-23-1 Komazawa, Setagaya-ku, Tokyo, 154-8525, Japan}
\email{w3c@komazawa-u.ac.jp}
\thanks{The first author is partially supported by Grant-in-Aid for Scientific Research (C) (No. 23540105 and No. 26400097), The Ministry of Education, Culture, Sports, Science and Technology, Japan}
\author{Koya Shimokawa}
\address{Department of Mathematics, Saitama University, Saitama 338-8570, Japan}
\email{kshimoka@rimath.saitama-u.ac.jp}
\thanks{The second author is partially supported by Grant-in-Aid for Scientific Research (C) (No. 25400080), The Ministry of Education, Culture, Sports, Science and Technology, Japan}

\subjclass[2010]{57M25}

\keywords{link, Dehn surgery, Seifert surface}

\begin{abstract}
For a compact connected 3-submanifold with connected boundary in the 3-sphere, we relate the existence of a Seifert surface system for a surface with a Dehn surgery along a null-homologous link.
As its corollary, we obtain a refinement of the Fox's re-embedding theorem.
\end{abstract}

\maketitle

\section{Introduction}

\begin{definition}
Let $M$ be a compact connected 3-manifold with connected boundary of genus $g$.
A {\em spanning surface system} $\{F_i\}$ for $M$ is a set satisfying the following.
\begin{enumerate}
\item $\{F_i\}$ is a set of disjoint orientable surfaces properly embedded in $M$.
\item $\{\partial F_i\}$ is a set of $g$ disjoint loops $C_1, \ldots, C_g$ which do not separate $\partial M$.
\end{enumerate}
A spanning surface system $\{F_i\}$ for $M$ is {\em completely disjoint} if $\{F_i\}$ is a set of $g$ disjoint orientable surfaces.
\end{definition}

\begin{remark}\label{Lei}
We remark by \cite[Corollary 1.4]{L} that if $M$ is a handlebody and $\{F_i\}$ is a completely disjoint spanning surface system for $M$, then there exists a meridian disk system $\{D_i\}$ for $M$ such that $\partial F_i=\partial D_i$ for $i=1,\ldots,g$.
\end{remark}

By a homological argument, we have the following proposition.

\begin{proposition}\label{existence}
Any compact connected 3-submanifold with connected boundary in $S^3$ admits a spanning surface system.
\end{proposition}

However, a compact connected 3-submanifold with connected boundary in $S^3$ does not always admit a completely disjoint spanning surface system.

\begin{remark}\label{Lambert}
We remark by \cite{Lam} that there exists a compact connected 3-submanifold with connected boundary of genus 2 in $S^3$ which does not admit a completely disjoint spanning surface system.
\end{remark}

\begin{definition}
Let $S$ be a genus $g>0$ closed surface in $S^3$, and put $S^3=V\cup_S V'$.
A {\em Seifert surface system} $(\{F_i\}, \{F_i'\})$ for $S$ is a pair of sets satisfying the following.
\begin{enumerate}
\item $\{F_i\}$ (resp. $\{F_i'\}$) is a spanning surface system for $V$ (resp. $V'$).

\item $|C_i\cap C_j'|=\delta_{ij}$ for $i, j=1,\ldots, g$, where $\{\partial F_i\}=\{C_1, \ldots, C_g\}$ and $\{\partial F_i'\}=\{C_1', \ldots, C_g'\}$
\end{enumerate}
A Seifert surface system $(\{F_i\},\ \{F_i'\})$ for $S$ is {\em completely disjoint} if $\{F_i\}$ and $\{F_i'\}$ are completely disjoint.
\end{definition}

\begin{definition}
Let $L=L_1\cup \cdots \cup L_n$ be a link in $S^3$.
Following \cite{MOS}, we say that $L$ is a {\em reflexive link} if the 3-sphere can be obtained by a non-trivial Dehn surgery along $L$.
In particular, if the surgery slope $\gamma_i$ for $L_i$ is $1/n_i$ for some integer $n_i$ $(i=1,\ldots, n)$, then we call the Dehn surgery a {\em $1/\Bbb{Z}$-Dehn surgery}.

Suppose that $L$ is contained in a compact 3-submanifold $M$ in $S^3$.
We say that $L$ is {\em null-homologous} in $M$ if $[L]=0$ in $H_1(M;\Bbb{Z})$, and that $L$ is {\em completely null-homologous} in $M$ if $[L_i]=0$ in $H_1(M;\Bbb{Z})$ for $i=1,\ldots,n$.
\end{definition}

\begin{theorem}\label{main}
Let $M$ be a compact connected 3-submanifold with connected boundary in $S^3$.
Then the followings hold.
\begin{enumerate}
\item There exists a null-homologous link $L$ in $M$, which is reflexive in $S^3$, such that a handlebody can be obtained from $M$ by a $1/\Bbb{Z}$-Dehn surgery along $L$.
\item $M$ admits a completely disjoint spanning surface system if and only if there exists a completely null-homologous link $L$ in $M$, which is reflexive in $S^3$, such that a handlebody can be obtained from $M$ by a $1/\Bbb{Z}$-Dehn surgery along $L$.
\end{enumerate}
\end{theorem}

We remark that in Theorem \ref{main} (2), we can take a completely null-homologous reflexive link $L$ so that it is disjoint from the completely disjoint spanning surface system.

\begin{remark}\label{equivalent}
Let $M$ be a compact connected 3-submanifold of $S^3$ with connected boundary of genus $g$.
Let $f:M\to X$ be a map onto a genus $g$ handlebody $X$.
We say that $f$ is a {\em boundary preserving map} of $M$ onto $X$ if $f$ is continuous and $f|_{\partial M}$ is a homeomorphism onto $\partial X$.
We say that $M$ is {\em retractable} if $M$ can be retracted onto a wedge of $g$ simple closed curves.
If such a wedge can be chosen to be in $\partial M$, then $M$ is called {\em boundary retractable}.
Set $G=\pi_1(M)$ and define $G_1=[G,G]$, $G_{n+1}=[G_n,G]$, $G_{\omega}=\bigcap_n G_n$.
Then the following conditions are equivalent.
\begin{enumerate}
\item $M$ admits a completely disjoint spanning surface system.
\item There exists a boundary preserving map from $M$ onto a handlebody.
\item $M$ is boundary retractable.
\item The natural map $\pi_1(\partial M)\to G/G_{\omega}$ is an epimorphism.
\end{enumerate}

The equivalence between (1) and (2) was shown in \cite[Theorem 2]{Lam}.
The equivalence between (2) and (3) was shown in \cite[Theorem 3]{JM}.
The equivalence between (3) and (4) was shown in \cite[Theorem 2, 3]{J}.
\end{remark}

Let $M$ be a compact connected 3-submanifold of $S^3$.
By Proposition \ref{existence}, each component of the exterior of $M$ admits a spanning surface system.
If we adapt Theorem \ref{main} (1) to the exterior of $M$, then we obtain a refinement of the Fox's re-embedding theorem as the following corollary.

\begin{corollary}[\cite{F,ST,N}]\label{Fox}
Every compact connected 3-submanifold $M$ of $S^3$ can be re-embedded in $S^3$ so that the exterior of the image of $M$ is a union of handlebodies.
\end{corollary}

\begin{remark}
In relation with Remark \ref{equivalent}, there is an another equivalence condition.
Let $M$ be a compact connected 3-submanifold of $S^3$ with connected boundary of genus $2$.
By Corollary \ref{Fox}, there exists a re-embedding of $M$ so that its exterior is a genus $2$ handlebody $V$.
A handcuff graph shaped spine $\Gamma$ of $V$ is a {\em boundary spine} if its constituent link $L_{\Gamma}$ is a boundary link that admits a pair of disjoint Seifert surfaces whose interiors are contained in $S^3-\Gamma$.
A handlebody $V$ is {\em $(3)_S$-knotted} if it does not admit any boundary spine.
Then it was shown in \cite[Theorem 3.10]{BF} that $M$ admits a completely disjoint spanning surface system if and only if $H$ is not $(3)_S$-knotted.
\end{remark}

Theorem \ref{main} also deduces the following corollary.
It follows from (2) of the next corollary that every closed surfaces with completely disjoint Seifert surface systems can be related by $1/\Bbb{Z}$-Dehn surgeries along completely disjoint null-homologous reflexive links.

\begin{corollary}\label{closed_surface}
Let $S$ be a closed surface in $S^3$ which separates $S^3$ into 3-submanifolds $M$ and $M'$
Then the followings hold.
\begin{enumerate}
\item There exist null-homologous links $L$ and $L'$ in $M$ and $M'$, which are reflexive in $S^3$, such that handlebodies can be obtained from $M$ and $M'$ by $1/\Bbb{Z}$-Dehn surgeries along $L$ and $L'$.
\item $S$ admits a completely disjoint Seifert surface system if and only if there exist completely null-homologous links $L$ and $L'$ in $M$, which are reflexive in $S^3$, and $M'$ such that handlebodies can be obtained from $M$ and $M'$ by $1/\Bbb{Z}$-Dehn surgeries along $L$ and $L'$.
\end{enumerate}
\end{corollary}

By Corollary \ref{closed_surface} (1), we can obtain a Seifert surface system from a meridian-longitude disk system for the handlebodies by tubing along the null-homologous links.

\begin{corollary}\label{existence2}
Any closed surface in $S^3$ admits a Seifert surface system.
\end{corollary}

Let $M$ be a 3-manifold.
Let $L\subset M$ be a submanifold with or without boundary.
When $L$ is 1 or 2-dimensional, we write $E(L)=M-int\,N(L)$.
When $L$ is of 3-dimension, we write $E(L)=M-int\,N(L)$.

\section{Proof}

Let $V$ be a genus $g$ handlebody in $S^3$, and $\{D_i\}$ be a meridian disk system for $V$.
Since $V-\bigcup_i int N(D_i)$ is a 3-ball, there exists a spine $\Gamma$ of $V$ such that:
\begin{enumerate}
\item $\Gamma$ consists of $g$ loops $l_1,\ldots,l_g$ and $g$ arcs $\gamma_1,\ldots,\gamma_g$ connecting $l_i$ to a point $x$.
\item The point $x$ is the center of the 3-ball $V-\bigcup_i int N(D_i)$, and which is homeomorphic to $N(x\cup \gamma_1\cup\cdots\cup\gamma_g)$.
\item Each loop $l_i$ is dual to $D_i$.
\end{enumerate}
We call this spine $\Gamma$ a {\em $g$-handcuff graph shaped spine} for $V$ with respect to $\{D_i\}$.

\begin{figure}[htbp]
	\begin{center}
	\includegraphics[trim=0mm 0mm 0mm 0mm, width=.6\linewidth]{spine.eps}\\

	\end{center}
	\caption{A $g$-handcuff graph shaped spine for $V$ with respect to $\{D_i\}$}
	\label{spine}
\end{figure}

Next, let $\{F_i\}$ be a set of orientable surfaces with boundary and without closed component.
We say that $\{F_i\}$ is a {\em Seifert surface system} for $\Gamma$ if $(\bigcup_i F_i) \cap \Gamma=\bigcup_i \partial F_i=\bigcup_i l_i$.

\begin{lemma}\label{handcuff_bound}
Any $g$-handcuff graph shaped spine in $S^3$ admits a Seifert surface system.
\end{lemma}

\begin{proof}
We take a regular diagram of $\Gamma$ such that $x\cup \gamma_1\cup\cdots\cup\gamma_g$ has no crossing.
Then we apply the Seifert's algorithm (\cite{S}) to loops $l_1\cup\cdots\cup l_g$ with arbitrary orientations, and obtain a Seifert surfaces $\{F_i'\}$ for the loops.
\end{proof}

The following lemma states that from any meridian disk system for a handlebody, we can obtain a Seifert surface system for the boundary of the bandlebody.

\begin{lemma}\label{extension}
Let $V$ be a genus $g$ handlebody in $S^3$ with a meridian disk system $\{D_i\}$.
Then there exists a spanning surface system $\{F_i\}$ for $E(V)$ such that $(\{D_i\},\{F_i\})$ is a Seifert surface system for $\partial V$.
\end{lemma}

\begin{proof}
Let $\Gamma$ be a $g$-handcuff graph shaped spine $\Gamma$ for $V$ with respect to $\{D_i\}$.
By Lemma \ref{handcuff_bound}, $\Gamma$ admits a Seifert surface system $\{F_i'\}$.
The restriction $\{F_i'\}$ to $E(V)$ gives a spanning surface system, say $\{F_i\}$, for $E(V)$ such that $(\{D_i\},\{F_i\})$ is a Seifert surface system for $\partial V$.
\end{proof}

Let $\Gamma$ be a $g$-handcuff graph shaped spine with a Seifert surface system $\{F_i\}$.
We call the operation of (1) in Figure \ref{crossing} a {\em band-crossing change} of $\{F_i\}$, and the operation of (2) in Figure \ref{crossing} a {\em full-twist} of $\{F_i\}$.
We remark that these operations can be obtained by a $1/\Bbb{Z}$-Dehn surgery along trivial links in the complement of $\{F_i\}$.

\begin{figure}[htbp]
	\begin{center}
	\begin{tabular}{cc}
	\includegraphics[trim=0mm 0mm 0mm 0mm, width=.5\linewidth]{crossing.eps}&
	\includegraphics[trim=0mm 0mm 0mm 0mm, width=.15\linewidth]{crossing2.eps}\\
	(1) a band-crossing change & (2) a full-twist\\
	\end{tabular}
	\end{center}
	\caption{A band-crossing change and a full-twist of $\{F_i\}$}
	\label{crossing}
\end{figure}

\begin{lemma}\label{unknotted}
Any $g$-handcuff graph shaped spine $\Gamma$ with a Seifert surface system $\{F_i\}$ can be unknotted by band-crossing changes and full-twists of $\{F_i\}$.
\end{lemma}

\begin{figure}[htbp]
	\begin{center}
	\includegraphics[trim=0mm 0mm 0mm 0mm, width=.6\linewidth]{seifert.eps}\\

	\end{center}
	\caption{A $g$-handcuff graph shaped spine $\Gamma$ with a Seifert surface system $\{F_i\}$, which is a ``standard planar form''}
	\label{standard}
\end{figure}

\begin{proof}
It is observed that $\Gamma$ with $\{F_i\}$ can be transformed to be a ``standard planar form'' (cf. \cite{K}) by the following operations.
\begin{enumerate}
\item a band-crossing change of $\{F_i\}$
\item a full-twists of $\{F_i\}$
\item a crossing change between $\{F_i\}$ and $\{\gamma_i\}$
\item a crossing change among $\{\gamma_i\}$
\end{enumerate}
However, the operations (3) and (4) can be exchanged for the operation (1).
If $\Gamma$ with $\{F_i\}$ has a standard planar form, then $\Gamma$ is unknotted and this completes the proof.
We remark that in a standard planar form, $l_1\cup\cdots\cup l_g$ is the trivial link.
\end{proof}

\begin{lemma}\label{core}
Let $L$ be a reflexive link in $S^3$ which is contained in a compact 3-submanifold $M$ in $S^3$.
Suppose that $L$ in null-homologous (resp. completely null-homologous) in $M$.
Then the core link $L^*$ in the 3-submanifold $M'$ obtained by a $1/\Bbb{Z}$-Dehn surgery along $L$ is also null-homologous (resp. completely
null-homologous) in $M'$.
\end{lemma}

\begin{proof}
Suppose that $L$ in null-homologous (resp. completely null-homologous) in $M$.
Then $L$ bounds a Seifert surface $F$ (resp. completely disjoint Seifert surface) in $M$.
Put $F^*=F\cap E(L)$.
By a $1/\Bbb{Z}$-Dehn surgery, the meridian of the core link $L^*$ intersects each component of $\partial F^*$ in one point.
This shows that $F^*$ can be extended to a a Seifert surface (resp. completely disjoint Seifert surface) for $L^*$ in $M'$.
Thus $L^*$ is also null-homologous (resp. completely
null-homologous) in $M'$.
\end{proof}

\begin{lemma}\label{handlebody}
Let $V$ be a handlebody in $S^3$.
Then the followings hold.
\begin{enumerate}
\item $\partial V$ admits a Seifert surface system if and only if there exists a null-homologous link $L$ in $E(V)$, which is reflexive in $S^3$, such that a handlebody can be obtained from $E(V)$ by a $1/\Bbb{Z}$-Dehn surgery along $L$.
\item $\partial V$ admits a completely disjoint Seifert surface system if and only if there exists a completely null-homologous link $L$ in $E(V)$, which is reflexive in $S^3$, such that a handlebody can be obtained from $E(V)$ by a $1/\Bbb{Z}$-Dehn surgery along $L$.
\end{enumerate}
\end{lemma}

\begin{proof}
(1) Suppose that there exists a null-homologous reflexive link $L$ in $E(V)$ such that a handlebody, say $W$, can be obtained from $E(V)$ by a $1/\Bbb{Z}$-Dehn surgery along $L$.
There exists a meridian disk system $\{D_i\}$ for $W$.
Since $L$ is null-homologous in $E(V)$, by Lemma \ref{core}, the core link $L^*$ is also null-homologous in $W$.
Therefore, we can obtain a Seifert surface system for $E(V)$ by tubing $\{D_i\}$ along $L^*$.

Conversely, suppose that $\partial V$ admits a Seifert surface system $(\{F_i\}, \{F_i'\})$, where $\{F_i\}$ and $\{F_i'\}$ are spanning surface systems for $V$ and $E(V)$.
By Remark \ref{Lei}, we may assume that each $F_i$ is a disk.
Then there exists a $g$-handcuff graph shaped spine $\Gamma$ and $F_i'$ can be extended to a Seifert surface system for $\Gamma$.
By Lemma \ref{unknotted}, $\Gamma$ with $\{F_i'\}$ can be unknotted, hence $E(V)$ is a handlebody, by band-crossing changes and full-twists of $\{F_i'\}$.
This operations can be obtained by a $1/\Bbb{Z}$-Dehn surgery along a trivial link $L$ in $\Gamma\cup \bigcup_i F_i'$.
Since $L$ is contained in $E(V)-\bigcup_i F_i'$, $L$ is a null-homologous link in $E(V)$.
Hence we obtain a null-homologous reflexive link $L$ in $E(V)$ such that a handlebody can be obtained from $E(V)$ by a $1/\Bbb{Z}$-Dehn surgery along $L$.

(2) This can be proved by the argument similar to (1).
\end{proof}

\begin{lemma}\label{heegaard}
Let $S$ be a Heegaard surface in $S^3$ which decomposes $S^3$ into two handlebodies $V$ and $V'$.
Let $\{D_i\}$ be a meridian disk system for $V$.
Then there exist a null-homologous reflexive link $L'$ in $V'$ which yields a handlebody $V''$ by a $1/\Bbb{Z}$-Dehn surgery on $L'$ and a meridian disk system $\{D_i''\}$ for $V''$ such that $(\{D_i\},\{D_i''\})$ is a completely disjoint Seifert surface system for $S$ in $V\cup V''$.

\end{lemma}

\begin{proof}[Proof of Lemma \ref{heegaard}]
We take a $g$-handcuff graph shaped spine $\Gamma$ of $V$ with respect to $D_i$.
Since $\Gamma$ can be unknotted by crossing changes, there exists a null-homologous reflexive link $L'$ in $V'$ such that after a $1/\Bbb{Z}$-Dehn surgery along $L'$, every loops of $\Gamma$ bound mutually disjoint disks.
Therefore, a handlebody $V''$ obtained from $V'$ by a $1/\Bbb{Z}$-Dehn surgery along $L'$ admits a meridian disk system $\{D_i''\}$ so that $(\{D_i\},\{D_i''\})$ is a completely disjoint Seifert surface system for $S$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{main}]
(1) We prove by an induction on the genus $g(\partial M)$.
Since the 3-sphere does not contain an incompressible closed surface, there exists a compressing disk $D$ for $\partial M$ in $S^3$.
We divide the proof into two cases.

\begin{description}
\item[Case 1] $D\subset M$
\item[Case 2] $D\subset E(M)$
\end{description}

In Case 1, put $M'=M-{int}\,N(D)$.
By the assumption of the induction, there exists a null-homologous reflexive link $L'$ in $M'$ such that handlebodies can be obtained from $M'$ by a $1/\Bbb{Z}$-Dehn surgery along $L'$.
This proves the theorem since $M$ is obtained by adding 1-handle $N(D)$ to $M'$.

In Case 2, we take a maximal compression body $W$ for $\partial M$ in $E(M)$ \cite{B}.
If $W$ is a handlebody (i.e. $W=E(M)$), then the theorem follows Lemma \ref{extension} and Lemma \ref{handlebody} (1).
Otherwise as $g(\partial W)<g(\partial M)$, by the assumption of the induction, there exists a null-homologous reflexive link $L'$ in each component of $E(M)-{int}\,W$ such that handlebodies can be obtained from the component by a $1/\Bbb{Z}$-Dehn surgery along $L'$.
After these $1/\Bbb{Z}$-Dehn surgery, $E(M)$ is a handlebody.
Therefore, again by Lemma \ref{extension} and Lemma \ref{handlebody} (1), there exists a null-homologous reflexive link $L$ in $M$ such that a handlebody can be obtained from $M$ by a $1/\Bbb{Z}$-Dehn surgery along $L$.
Finally, we recover the previous $1/\Bbb{Z}$-Dehn surgery on each component of $E(M)-{int}\,W$ to obtain the original $E(M)$.

(2) Suppose that there exists a completely null-homologous reflexive link $L$ in $M$ such that a handlebody can be obtained from $M$ by a $1/\Bbb{Z}$-Dehn surgery along $L$.
There exists a meridian disk system $\{D_i\}$ for the resultant handlebody.
Since $L$ is completely null-homologous in $M$, we can obtain a completely disjoint spanning surface system for $M$ by tubing $\{D_i\}$ along $L$.

Conversely, suppose that $M$ admits a completely disjoint spanning surface system $\{F_i\}$.
In the following 3 steps, we convert $M$ and $E(M)$ into two handlebodies $V$ and $V''$ so that $(V,V'')$ admits a meridian disk system $(\{D_i\},\{D_i''\})$ with $\partial F_i=\partial D_i$.

Step 1: By (1) of this theorem, there exists a null-homologous link reflexive $L$ in $E(M)$ such that a handlebody can be obtained from $E(M)$ by a $1/\Bbb{Z}$-Dehn surgery along $L$.
Let $V'$ the resultant handlebody obtained from $E(M)$ and note that $M\cup V'$ is again the 3-sphere.

Step 2: We note that there exists a degree one map from $M$ to a handlebody $V$ which sends each $F_i$ to a meridian disk $D_i$ of $V$ and preserves the boundary of $M$ (cf. {\cite[Theorem 2]{Lam}}, {\cite[Theorem 5]{Hem}}).
We naturally extend this degree one map to a degree one map $\phi:S^3=M\cup V' \to X=V\cup V'$ as follows.
\begin{enumerate}
\item $V'$ is contained in $X$ by an inclusion.

\item Each $F_i$ is sent to a meridian disk $D_i$ of the handlebody $\phi(M)=V$.
\item The remnant $M-\bigcup int\,N(F_i)$ is sent to the 3-ball $V-\bigcup int\,N(D_i)$.
\end{enumerate}
Since $\phi_*:\pi_1(S^3)\to \pi_1(X)$ is surjective \cite[Lemma 15.12]{H}, $X$ is homeomorphic to $S^3$ \cite{P1,P2,MT}.

Step 3: By Lemma \ref{heegaard}, there exists a null-homologous reflexive link $L'$ in $V'$ and a meridian disk system $\{D_i''\}$ for a handlebody $V''$ obtained from $V'$ by a $1/\Bbb{Z}$-Dehn surgery along $L'$ such that $(\{D_i\}, \{D_i''\})$ is a completely disjoint Seifert surface system for $(V, V'')$.

Since the degree one map $\phi$ is a boundary preserving map by the condition (1), $(\{F_i\}, \{D_i''\})$ is a completely disjoint Seifert surface system for $(M, V'')$.
By Lemma \ref{handlebody} (2), there exists a completely null-homologous reflexive link $L_0$ in $M$ such that a handlebody can be obtained from $M$ by a $1/\Bbb{Z}$-Dehn surgery along $L_0$.
Moreover, by the proof of Lemma \ref{handlebody} (2), we can take $L_0$ so that $L_0\cap \bigcup F_i=\emptyset$.
Thus the completely disjoint spanning surface system $\{F_i\}$ is contained in the resultant handlebody $V_0$ obtained from $M$ by a $1/\Bbb{Z}$-Dehn surgery along $L_0$.
\end{proof}

\bigskip

\bibliographystyle{amsplain}


TITLE: Did I get hit by a car this morning?
QUESTION [27 upvotes]: Quantum mechanics allows us to describe a particle as a wave, and also a collection of particles, which a car happens to be. What separates a typical wave from a classical particle is that the position is not well defined for a wave.
I think that it is hard to tell exactly how big an atom is, because the electron of the atom may be anywhere, but the probability of finding the electron gets very small as you look far away from the nuclei. I think in some sense the atom is as large as we want, depending on the definition.
When a car drives past me, is the position of the car separate from my position in a quantum mechanical way, or is it correct to say that "part" of the car hit "part" of me? I understand that no matter the answer the force from the hit is so small that it is negligible in every way, but that is not the core of the question.
In addition, if I made any wrong claims, please correct me, thanks in advance.

REPLY [52 votes]: Asher's comment written above is simply wrong, and the reason is rather fundamental in quantum mechanics. "The car slightly hits you" isn't how it works in quantum mechanics.
The reason is that weak effects – such as very small but nonzero values of the wave function of an electron very far from the nucleus – do not cause tiny but observable effects like they do in classical physics.
Instead, the wave function has a probabilistic interpretation. So its being nonzero very far from the nucleus means that there is a small probability that a finite effect takes place.
So in quantum mechanics, you can ask whether a car hit you – whether some reaction took place. In quantum mechanics, questions about physical systems (including this one) have to be answered by measurements, otherwise they are physically and scientifically meaningless.
And quantum mechanics predicts some probabilities for one outcome of the measurement or another. If only 0.0001% of the integrated $\int |\psi|^2$ of the electron was located at the distance $R$ equal to the distance of your body from the car, then it means that there was a 99.9999% probability that no interaction has taken place at all.
On the other hand, there is a 0.0001% probability that some interaction did take place, but it wouldn't be a small interaction. It would be a rather large one, such as the ionization of one atom – of the car or your body.
Again, in quantum mechanics, small wave functions don't mean small forces. They mean finite (not so small) jumps happening with tiny probabilities.
Again, it's very likely that you didn't get hit by a car in the morning at all.
One can give as many serious examples of that as we want. For example, when a radioactive nucleus decays, the wave function for an alpha particle gradually grows outside the nucleus. It doesn't mean that the alpha particle is "weakly felt" at all times. Instead, it means that there is some probability that the nucleus has decayed up to some moment, and some probability that it remained intact.
Also, one may consider collisions of particles, such as those at the LHC. Various quantum processes may take place; they are described by Feynman diagrams. The two colliding protons have an infinite cross section to interact through the electrostatic repulsion. So yes, there was always some interaction of this electrostatic kind. It's equivalently interpreted as the certainty that some very soft photons are being emitted. (One could say that the car-body contact above was electrostatic elastic between two electrons as well, in that case, it would always occur.)
But there are also other, short-range forces, that predict finite cross sections. In some reactions, some finite process like that – accompanied by the production of pions etc. – can take place. But in others, it strictly doesn't take place. The atoms effectively interact e.g. with van der Waals forces that decrease as a faster power law of the distance and they have a finite cross section. So when we talk about similar interactions between the whole atoms, then the interaction probably didn't occur at all – strictly zero at a probability that is almost 100%.


TITLE: Calculus Made Easy -Exercises IX, Question 10 (rates of change)
QUESTION [4 upvotes]: I am working with this exercise, which should be pretty easy, as a similar one is exemplified in the book.

A spherical balloon is increasing in volume. If, when its radius is r feet, its volume is increasing at the rate of 4 cubic feet per second, at what rate is its surface then increasing?

This is what i did: the formula for the volume of a sphere is $\ \frac43^3\;$.
Rate of change of volume with respect to time is 4, so $$\frac{dv}{dr}\cdot\frac{dr}{dt} = 4\implies 4r^2\cdot\frac{ dr}{dt} = 4$$, and $$\frac{dr}{dt} = \frac{1}{r^2}.$$
As what is being asked is rate of change of surface area, I need to calculate now $\dfrac{dS}{dr} \cdot\dfrac{ dr}{dt}$. Surface area of a sphere is $4r^2$, its derivative with respect to $r$ is $8r$, and multiplied with the previously found $\dfrac{dr}{dt} = \dfrac{1}{r^2}$, it gives a result of $\dfrac{8}{r}$.
When I check the solutions of the book -rather bare, with no steps- it says the answer should be 8 times the square root of .
What have I done wrong?

REPLY [1 votes]: Your answer is correct.  The answer in the book cannot be correct since it has the wrong units for the rate at which the surface area is changing.
Setting up the problem in a different way:
\begin{align*}
V & = \frac{4}{3}\pi r^3\\
\frac{dV}{dt} & = 4\pi r^2 \frac{dr}{dt}\\
\frac{1}{4\pi r^2} \frac{dV}{dt} & = \frac{dr}{dt}\\
\frac{1}{4\pi r^2} 4~\frac{\text{ft}^3}{\text{s}} & = \frac{dr}{dt}\\
\frac{1}{\pi r^2}~\frac{\text{ft}^3}{\text{s}} & = \frac{dr}{dt}
\end{align*}
We will substitute this result into the rate at which the surface area is changing.
\begin{align*}
S & = 4\pi r^2\\
\dfrac{dS}{dt} & = 8\pi r \frac{dr}{dt}\\
               & = 8\pi r \cdot \frac{1}{\pi r^2}~\frac{\text{ft}^3}{\text{s}}\\
               & = \frac{8}{r}~\frac{\text{ft}^3}{\text{s}}
\end{align*}
The units make sense since $r$ is measured in feet, leaving us with units of $\text{ft}^2/\text{s}$ for the rate at which the surface area is changing.
The units in the book's answer do not make sense since the lack of an $r$ in the denominator would mean that the units for $dS/dt$ would be $\text{ft}^3/\text{s}$.


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In this work, the Kronecker tensor product of matrices and the proofs of some of its properties are formalized. Properties which have been formalized include associativity of the tensor product and the mixed-product property. This formalization of tensor product of matrices relies on the formalization of matrices by Christian Sternagel and Rene Thiemann under the title
 `Executable Matrix Operations on Matrices of Arbitrary Dimensions'.
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TITLE: Finding the angle of two congruent isosceles triangles inscribed in a semi circle.
QUESTION [3 upvotes]: Question
This question comes from Question 10 of the 2020 AIMO

A circle with centre $O$ has diameter $AD$. With $X$ on $AO$ and points $B$ and $C$ on the circle, triangles $ABX$ and $XCO$ are similar isosceles with base angle $\alpha$ as shown. Find, with proof, the value of $\alpha$.

My Attempt
I knew we probably have to use the fact that the isosceles triangles have a base on the diameter and vertex on the circle. Then $\angle ACD=\angle ABD=90^{\circ}$. Since $\angle ABX=180-2\alpha=\angle XBC$, then $AB||XC$.
Now, when constructing $CD$ and $BD$, I notice that $CD$ looked very much like the angle bisector of $\angle XCD$ and $BD$ looked very much like the angle bisector of $\angle CDA$. Let $CD$ and $BD$ intersect at $I$. I would need to prove that $I$ is in fact the incenter of $XCD$ to give a rigorous proof. Note that $\angle XCO=180-2\alpha$ and since $CO=OD$ (both a radius), then $\angle ODC=\angle COD=\frac{\alpha}{2}$. If $I$ were to be the incenter of $XCD$, then $\angle XCO=\angle OCD\implies 180-2\alpha=\frac{\alpha}{2}\implies \alpha=72^{\circ}$. This looked to be the right answer (could confirm with geogebra).
However, this is also the step that I am struggling to prove. I tried constructing the altitude of $\triangle XID$, $\triangle CIX$ and $\triangle DIC$ to be $IF,IG$ and $IH$ repsetively. Note that since $AB||XC$, $\angle ABD=\angle XGD=90^{\circ}$ and thus the points $B,G,I,D$ are colinear. Refer to the following figure:

It would suffice to prove that $\triangle CGI\cong\triangle CHI$ and $\triangle XGI\cong\triangle XFI$.
Any hints or solutions would be appreciated. It would be nice if you would be able to show how to prove the last step in my attempt. Otherwise, giving me a hint to a better approach would also be much appreciated. Thank you in advance!

REPLY [2 votes]: B' is mirror of B about AD.
$\angle AB'C=\frac{arc AC}2=\frac{\alpha}2$.
C', B'' are mirrors of C and B about vertical diameter and $\angle COC'=\angle OCB=\angle AB'C$, therefore $arc CC=arc CB=arc AB=arc C'B''=arc B''D$
Finally:
$\alpha =\frac {2}5 (arc AD)=\frac{2}5\times 180=72^0$


TITLE: Determining whether two 2D polynomial curves are everywhere close to each other
QUESTION [2 upvotes]: Let's say we have two curves $P(t), Q(t): [0, 1] \to \mathbb{R}^2$. $P_x(t), P_y(t), Q_x(t), Q_y(t)$ are all polynomials of some degree $n$. We can further restrict this to Bernstein basis polynomials (Bezier curves), if it makes a difference.
It is well known that if two polynomials of degree $n$ are equal at $n+1$ points, then they are equal everywhere.
Is it possible to use that fact to derive a similar statement to the following effect: if $P, Q$ differ by no more than $\delta$ (by "differ" I mean $L_2$ norm of the difference, but $L_1$ or $L_\infty$ would also be OK), at some set of $n+1$ time values in $[0, 1]$, then there is no time value for which they differ by more than $f(n, \delta)$?
A further question would be whether it is possible to generalize such a statement to images of the curves. There are many Bezier curves that have different control points, but their images contain the same points. (For example, all the possible curves that form a straight line between endpoints.) I would be interested in a statement similar to: if for a set of $n+1$ time values in $P$, the closest point in the image of $Q$ is no further than $\delta$, then the images of $P$ and $Q$ are nowhere more distant than some $f(n, \delta)$.

REPLY [0 votes]: Probably not, take $P_1(x)=k x$ and $P_2(x)=-k x$. Since they have the same value at $x=0$ and are continuous, for any $\delta>0$ we can find $2$ (distinct) points such that the difference between the functions is less than $\delta$ (just pick points close enough to zero). The difference between the two functions at $x=1$ is $2k$. Now we can make $k$ arbitrarily large to reach a contradiction.


TITLE: Does gravity exist between quarks in a neutron?
QUESTION [1 upvotes]: Does gravity exist between quarks in a neutron?
I mean, I've learned that gravity exists between objects that have mass/energy. But does this apply between quarks in a neutron?

REPLY [2 votes]: As far as we know the classical (i.e. non-quantum) laws of gravity apply at all length scales. There are theoretical reasons to suppose that the classical description fails at scales approaching a Planck length, but this is far, far smaller than the size of a neutron.
So inside a neutron we would expect the classical laws of gravity to apply, and in particular we would expect Newton's law to be  good description - it doesn't seem likely we'd have to use general relativity. A simple calculation will show you that the gravitational energy is immeasurably small compared to the strong force binding energies, but in principle there will be some gravitational interactions.
But if you have an image of the neutron with the three quarks as little balls orbiting each other then you need to abandon that image. Firstly the quarks are delocalised so they exist as fuzzy objects distributed throughout the nucleus, and secondly 99% (ish) of the stuff inside a neutron is interaction energy not the three valence quarks. Exactly how you would calculate the gravitational interactions in such a system is unclear.