TITLE: Atoms in motion - It's known that heat is related only with the movement
QUESTION [0 upvotes]: It's known that heat is related only with the movement of particles inside the body. What is the difference between hot, stationary ball baseball and the ball moving quickly?

REPLY [0 votes]: Basically, de difference is in the direction of the atoms. The atoms in a hot, stationary ball move around randomly (the kinetic energy of the atoms is in their vibrational and rotational motion), whereas all atoms in a moving ball move in the same direction.


TITLE: Uniform convergence of the derivative function $h_n'(x)$
QUESTION [2 upvotes]: I am self-learning Real Analysis from the text Understanding Analysis, by Stephen Abbott. I would like for someone to
(1) Verify my proof for part (a) of this exercise problem.
(2) Do you have any clues for part (b) without giving away the entire solution/proof?

[Abbott 6.3.2] Consider the sequence of functions :

\begin{equation*}
h_{n}( x) =\sqrt{x^{2} +\frac{1}{n}}
\end{equation*}

(a) Compute the pointwise limit of $\displaystyle ( h_{n})$ and then prove that the convergence is uniform on $\displaystyle \mathbf{R}$.

Proof.
Fix $\displaystyle x\in \mathbf{R}$. We know that, if $\displaystyle \lim a_{n} =a$, then $\displaystyle \lim \sqrt{a_{n}} =\sqrt{\lim a_{n}} =\sqrt{a}$. Thus:
\begin{equation*}
\begin{array}{ c l }
\lim _{n\rightarrow \infty } h_{n}( x) & =\lim _{n\rightarrow \infty }\sqrt{x^{2} +\frac{1}{n}}\\
 & =\sqrt{\lim _{n\rightarrow \infty }\left( x^{2} +\frac{1}{n}\right)}\\
 & =\left[\lim _{n\rightarrow \infty } x^{2} +\lim _{n\rightarrow \infty }\frac{1}{n}\right]^{( 1/2)}\\
 & =\sqrt{x^{2}}\\
 & =|x|
\end{array}
\end{equation*}
Consider the expression:
\begin{align*}
|h_{n}( x) -h_{m}( x) | & =\left| \sqrt{x^{2} +\frac{1}{n}} -\sqrt{x^{2} +\frac{1}{m}}\right|  & \\
 & =\frac{\left| \left( x^{2} +\frac{1}{n}\right) -\left( x^{2} +\frac{1}{m}\right)\right| }{\left| \sqrt{x^{2} +\frac{1}{n}} +\sqrt{x^{2} +\frac{1}{m}}\right| } & \\
 & =\frac{\left| \frac{1}{n} -\frac{1}{m}\right| }{\sqrt{x^{2} +\frac{1}{n}} +\sqrt{x^{2} +\frac{1}{m}}} \\
 & \leq \frac{\left| \frac{1}{n} -\frac{1}{m}\right| }{\frac{1}{\sqrt{n}} +\frac{1}{\sqrt{m}}} & \left\{\because x^{2} \geq 0\right\}\\
 & =\frac{\left| \frac{1}{\sqrt{n}} -\frac{1}{\sqrt{m}}\right| \left(\frac{1}{\sqrt{n}} +\frac{1}{\sqrt{m}}\right)}{\left(\frac{1}{\sqrt{n}} +\frac{1}{\sqrt{m}}\right)} & \\
 & =\left| \frac{1}{\sqrt{n}} -\frac{1}{\sqrt{m}}\right|  &
\end{align*}
Pick an arbitrary $\displaystyle \epsilon  >0$. Since $\displaystyle \frac{1}{\sqrt{n}}\rightarrow 0$, and convergent sequences are Cauchy, there exists $\displaystyle N( \epsilon )  >0$, such that for all $\displaystyle n >m\geq N$,
\begin{equation*}
\left| \frac{1}{\sqrt{n}} -\frac{1}{\sqrt{m}}\right| < \epsilon
\end{equation*}
Consequently, by Cauchy criterion for uniform convergence of a sequence of functions, $\displaystyle ( h_{n})$ converges uniformly on $\displaystyle \mathbf{R}$ to $\displaystyle h$.

(b) Note that each $\displaystyle h_{n}$ is differentiable. Show that $\displaystyle g( x) =\lim h_{n} '( x)$ exists for all $\displaystyle x$ and explain how we can be certain that the convergence is not uniform on any neighbourhood of zero.

Proof.
By Chain rule of differentiation, we have:
\begin{equation*}
h_{n} '( x) =\frac{x}{\sqrt{x^{2} +\frac{1}{n}}}
\end{equation*}
Moreover,
\begin{equation*}
\lim h_{n} '( x) =\lim _{n\rightarrow \infty }\frac{x}{\sqrt{x^{2} +\frac{1}{n}}} =\frac{x}{|x|}
\end{equation*}

REPLY [2 votes]: Hint: Uniform Convergence preserves certain properties from the sequence of functions to the limit function.


TITLE: Graded Vector Spaces (definition)
QUESTION [4 upvotes]: I am studying Algebraic Operads with the book Algebraic Operads, by Jean-Louis Loday and Bruno Vallette and I'm having a little problem with the definition of graded vector space. My advisor and I disagree on the definition. The book defines it this way:

my advisor thinks it is equivalent to saying that $V = \bigoplus_{n\in \mathbb{Z}} V_n$. I believe that they are different. In nLab (Graded Vector Spaces) I found the following definition

which strengthens my argument.
Finally, who is correct? Can anyone suggest me any other reference about graded vector spaces?

REPLY [6 votes]: There are two different categories:

The category $\mathsf{gVec}$ where objects are vector spaces $V$ equipped with a decomposition $V \cong \bigoplus_{n \in \mathbb{Z}} V_n$, and morphisms are linear maps respecting the decomposition;
The category $\mathsf{Vec}^\mathbb{Z}$ of families of vector spaces $\{ V_n \}_{n \in \mathbb{Z}}$ and morphisms $f : \{V_n\} \to \{W_n\}$ are families of linear maps $\{f_n : V_n \to W_n\}_{n \in \mathbb{Z}}$.

These two categories are equivalent; you can construct an explicit equivalence $\mathsf{Vec}^\mathbb{Z} \to \mathsf{gVec}$ given by $\{V_n\} \mapsto V_\bullet = \bigoplus_{n \in \mathbb{Z}} V_n$. So it makes no difference to consider graded vector spaces from the first point of view or from the second point of view.


TITLE: Covering a genus g surface with n disks, and conversely
QUESTION [3 upvotes]: Suppose $X$ is a closed compact surface (i.e. a compact two-dimensional manifold without boundary).

If $X$ has genus $g$, what is the minimum number of (homeomorphic images of) open balls required to cover $X$?
If $X$ can be covered by $n$ (homeomorphic images of) open balls, is there an upper bound for the genus of $X$? If so, what's the upper bound?

One can show that if $X$ can be covered by $2$ open balls, then $X$ has to be a sphere (it's a fun exercise so I won't spoil it with further details). Furthermore, it's not too hard to show that a torus can be covered by $3$ open balls. I'm not sure what the situation is for more balls/higher genus though, so I'm content to leave this one with the internet!

REPLY [1 votes]: I believe that you only need $3$ open balls to cover any closed oriented compact surface $X$ of $g\geq 1$.
Consider the standard fundamental polygons of $X$, it is a CW complex by identifying the boundary of a $2$-cell with the wedge sum of $2g$ circles. Suppose the intersection point of the $2g$ circles is $p$. Let $B_1$ be the $2$-cell of $X$, $B_2$ be a open neighborhood of $p$ which is homeomorphic to an open $2$-ball, then $X-(B_2\cup B_2)$ is a disjoint union of $2g$ paths, each path is homeomorphic to the closed interval. Finally, choose $B_3$ to cover these $2g$ paths, we are done.


TITLE: How does surface tension allow the surface of a liquid to exert an upward force on an object?
QUESTION [0 upvotes]: I've seen partial answers to my question on Wikipedia and several other questions on this site, but I'm not quite grasping how intermolecular forces lead to the creation of a membranous boundary between, for example, water and air.
I can grasp that with no balancing forces at the surface of a water droplet, surface molecules tend to be attracted to the interior of the droplet, but fail to see how that leads to a stronger attraction between the molecules at the surface. Wouldn't the perpendicular forces (and their imbalance) be irrelevant to the effect surface molecules have on each other?

REPLY [0 votes]: here is one way to look at it which might help:
every water molecule that is completely surrounded by other water molecules exhibits a certain amount of affinity for its nearest neighbors. But at a free surface, the water molecules occupying it have only half as many nearest neighbors with which to commune. as a result, the strength of the intermolecular attraction between molecules at the free surface is to first order about twice what it is in the interior of the body of water, which makes the free surface act like a stretched membrane that wants to snap back.


text:
\begin{document}\frenchspacing

\title{\large\bf INVOLUTIONS OF 3-DIMENSIONAL HANDLEBODIES
       \label{Version 0.5 / \today}}

\author{
\fsc Andrea Pantaleoni\\
\fsl Dipartimento di Matematica\\[-3pt]
\fsl Universit\`a di Camerino -- Italy\\
\ftt andrea.pantaleoni@unicam.it
\and
\fsc Riccardo Piergallini\\
\fsl Dipartimento di Matematica\\[-3pt]
\fsl Universit\`a di Camerino -- Italy\\
\ftt riccardo.piergallini@unicam.it
}

\date{}

\maketitle

\begin{abstract}
\baselineskip13.5pt
\smallskip

\noindent
We study the orientation preserving involutions of the orientable 3-dimensional handlebody
$H_g$, for any genus $g$. A complete classification of such involutions is given in terms
of their fixed points.

\medskip\smallskip\noindent
{\sl Keywords}\/: handlebody, involution, fixed point set, double branched covering.

\smallskip\noindent
{\sl AMS Classification}\/: 57M60, 57S17, 55M35.

\end{abstract}

\section*{Introduction}

Involutions of the 3-dimensional orientable handlebody $H_g$ of genus $g$ has been already
classified in \cite{Bar}, \cite{Liv}, \cite{Wal} and \cite{Prz} for $g \leq 2$. Moreover,
a classification of the orientation reversing involutions of $H_g$ was given in \cite{KMG}
(theorem 3.6).

In this paper, we complete the study of the subject, by providing a classification of the
orientation preserving involutions of $H_g$ for any $g \geq 0$. Our argument is direct and
elementary. The same result can also be derived from the general theory of actions on
handlebodies developed in \cite{CMZ}.

\smallskip

Namely, we prove the following theorem.

\begin{statement}{Theorem}
Let $h: H_g \to H_g$ be an orientation preserving involution. If $h$ is free, then $g =
2n+1$ for some $n \geq 0$ and $h$ is equivalent to the involution $I_g$ depicted in Figure
\ref{involution1/fig}. If $h$ is not free, then there exist $n,m,l \geq 0$ with $1 \leq n
+ 2m \leq n + 2m + 2l = g + 1$, such that $h$ is equivalent to the involution $L_g^{n,m}$
depicted in Figure \ref{involution2/fig}.
\end{statement}

\begin{Figure}[b]{involution1/fig}
\fig{}{involution1.eps}{}{The free involution $I_g$ for $g = 2n + 1$}\vskip-6pt
\end{Figure}

The free involution $I_g: H_g \to H_g$ with $g = 2n +1$ can be realized by embedding $H_g$
in $R^3$ as in Figure \ref{involution1/fig} and rotating of $\pi$ radians around the axis
orthogonal to the plane of the picture at the dot.

\begin{Figure}[htb]{involution2/fig}
\fig{}{involution2.eps}{}{The non-free involution $L^{n,m}_g$}
\end{Figure}

The description of the involution $L_g^{n,m}: H_g \to H_g$, with $g = n + 2m + 2l -1$, is
a little bit more involved. The fixed point set $\Fix L_g^{n,m}$ consists of $n$ arcs and
$m$ loops, all dashed in Figure \ref{involution2/fig}. We think of the $H_g$ as $H_{n + m
+ 2l - 1}$ with $m$ extra handles attached to it. The handlebody $H_{n + m + 2l - 1}$ is
imbedded in $R^3$ in such a way that it is symmetric with respect to the median horizontal
line and meets it in $n + m$ arcs, while the $m$ extra handles are the non-symmetric ones.
Then, the restriction of $L_g^{n,m}$ to $H_{n + m + 2l - 1}$ is given by the rotation of
$\pi$ radians around this axis. Of course, the fixed point set of this involution of $H_{n
+ m + 2l - 1}$ consists of $n + m$ arcs. Now, we attach each one of the $m$ extra handles
at two disks centered at the end points of a fixed arc. Finally, we extend the rotation to
the such extra handle as the rotation of $\pi$ radians around its core. Hence, the fixed
arc close up to give a fixed loop.

We remark that $L^{g+1,0}_{g}$ coincides the hyperelliptic involution of $H_g$.

\smallskip

As a consequence of our classification, we see that any orientation preserving involution
of $H_g$ is uniquely determined, up to equivalence, by its restriction to the boundary
$T_g = \Bd H_g$. However, it is worth observing that the restrictions to $T_g$ of
non-equivalent involutions of $H_g$ can be equivalent as involutions of $T_g$, by a PL
homeomorphism of $T_g$ which does not extend to $H_g$. Actually, two involutions of $T_g$
are equivalent if and only if they have the same number of fixed points and they give
raise to quotient surfaces of the same genus $g'$, as it follows from the Hurwitz
classification of branched covering between surfaces (\cite{Hur}, see also \cite{BE}).

\smallskip

From a different point of view, we see that the quotient of $H_g$ under the action of any
orientation preserving involution turns out to be a handlebody $H_{g'}$. Namely, $g' = (g
+ 1)/2 = n + 1$ for $H_{g = 2n + 1} / I_g$ and $g' = (g - n + 1)/2 = m + l$ for $H_{g = n
+ 2m + 2l - 1} / L_g^{n,m}$. Therefore, our result could also be reformulated in terms of
double branched coverings $H_g \to H_{g'}$ between handlebodies.

\section{Preliminaries\label{prelim/sec}}

An {\sl involution} of a PL manifold $X$ is any PL homeomorphism $h: X \to X$ such that
$h \neq \id_X$ and $h^2 = \id_X$. We denote by $\Fix h = \{x \in X \;|\; h(x) = x\}$ the
{\sl fixed point set} of $h$. The involution $h$ is called {\sl free} if $\Fix h =
\emptyset$.

If $h': X' \to X'$ is another involution of the PL manifold $X'$, then we say that $h$ and
$h'$ are {\sl equivalent} if there exists a PL homeomorphism $\eta: X \to X'$ such that
$h' = \eta \circ h \circ \eta^{-1}$.

Here, we focus on orientation preserving involutions the 3-dimensional handlebody $H_g$,
consists of one 0-handle and $g$ orientable 1-handles attached to it, for any $g \geq 0$.
If $h: H_g \to H_g$ is such an orientation preserving involution, then $\Fix h$ is a
(possibly empty) proper PL 1-submanifold of $H_g$. Moreover, the canonical projection
$\pi: H_g \to H_g/h$ turns out to be a double branched covering.

\medskip

In particular, we want to prove the theorem stated in the introduction, providing a
complete classification, up to equivalence, of the orientation preserving involutions of
$H_g$ for any $g \geq 0$.

The proof proceeds by induction on the number $g$ of the 1-handles, starting from the
trivial case of $g = 0$. In this case, we have $H_0 \cong B^3 \subset R^3$, whose only
orientation preserving involution, up to equivalence, is the symmetry $(x,y,z) \mapsto
(x,-y,-z)$ with respect to the $x$-axis (cf. \cite{Liv} and \cite{Wal}), which coincides
with $L_0^{1,0}$.

\medskip

The following lemma concerning involutions of 1-handles, tells us how a given orientation
preserving involution of a disjoint union of orientable handlebodies can be extended to
some extra 1-handles equivariantly attached to it. As an immediate consequence, such
extension is uniquely determined by the equivalence class of the involution induced on the
pairs of attaching disks. This fact will used when performing the inductive step.

\begin{lemma}\label{handle/thm}
The 3-dimensional 1-handle $B^1 \times B^2 \subset R^3$ has only two involutions
preserving the attaching disks $\{-1,1\} \times B^2$, up to equivalence preserving such
disks. Namely, they are the symmetries $(x,y,z) \mapsto (x,-y,-z)$ and $ (x,y,z) \mapsto
(-x,y,-z)$. The first one fixes the core $B^1 \times \{0\}$ of the handle and sends each
attaching disk onto itself, while the second one fixes the diameter $\{0\} \times B^1$ of
the co-core of the handle and swaps the attaching disks.
\end{lemma}

\begin{proof}
Taking into account what we have said about involutions of $B^3$, the lemma can be easily
derived just by considering the possible positions of the arc fixed by the involution with
respect to the attaching disks.
\end{proof}

The other main tools for the inductive step is the next lemma, which allows us to split
any orientation preserving involution of $H_g$ as boundary connected sum of involutions of
simpler handlebodies.

We recall that a properly embedded PL 2-disk $D$ in a bounded 3-manifold $M$ is called
{\sl boundary parallel\/} if there exists a 2-disk $E \subset \Bd M$ such that $\Bd D =
\Bd E$ and $D \cup E$ bounds a 3-cell in $M$. Moreover, if $D'$ is another properly
embedded PL 2-disk in $M$, then $D$ and $D'$ are called {\sl parallel\/} if they are
disjoint and there exists an annulus $A \subset \Bd M$ such that $\Bd A = \Bd D \cup \Bd
D'$ and the 2-sphere $D \cup A \cup D'$ bounds a 3-cell in $M$.

\begin{lemma}\label{split/thm}
Let $h: H_g \to H_g$ be an orientation preserving involution with $g \geq 1$. Then there
exists a properly embedded PL 2-disk $D$ in $H_g$ which is not boundary parallel and such
that either $h(D) \cap D = \emptyset$ or $h(D) = D$ and this meets $\Fix h$ transversally
at one point. In the first case, denoting by $N$ a regular neighborhood of $h(D) \cup D$,
we can assume that $\Cl(H_g - N)$ is PL homeomorphic to $H_{g-2}$ or $H_{g_1} \sqcup
H_{g_2}$ with $g_1 + g_2 = g-1$. In the second case, denoting by $N$ a regular
neighborhood of $h(D) = D$, we have that $\Cl(H_g - N)$ is PL homeomorphic to $H_{g-1}$ or
$H_{g_1} \sqcup H_{g_2}$ with $g_1 + g_2 = g$.
\end{lemma}

\begin{proof}
The first part of the statement follows from Theorem 3 of \cite{GL}. Concerning the second
part, we first observe that $\Cl(H_g - N)$ is a disjoint union of handlebodies (cf.
\cite{Gro}) and $H_g$ can be thought as $\Cl(H_g - N)$ with one (when $h(D) = D$) or two
(when $h(D) \cap D = \emptyset$) 1-handles attached to it. Hence, the only non-trivial
fact to be proved is that $\Cl(H_g - N)$ can be assumed to have at most two components. In
fact, if $h(D) \cap D = \emptyset$ then $\Cl(H_g - N)$ could also have three components,
say $C_1$, $C_2$ and $C_3$. It is not difficult to see that in this case $h$ swaps two of
them, say $C_1$ and $C_2$, and sends the other one onto itself. Since $D$ is not boundary
parallel, $C_1 \cong C_2 \cong H_{g'}$ with $g' \geq 1$. Hence, we can replace the disk
$D$ by a non-separating disk in $C_1$. After that, we have $h(D) \cap D = \emptyset$ and
$\Cl(H_g - N)$ turns out to be connected.
\end{proof}

By previous lemmas, one can easily determine the orientation preserving involutions of
$H_1 \cong S^1 \times B^2 \subset \Bbb C^2$. Since these are known (cf. \cite{Prz} or
\cite{Bar}), we limit ourselves to list them without proof. Up to equivalence, they are
$I_1: (x,y) \mapsto (-x,y)$, $L_1^{0,1}: (x,y) \mapsto (-x,y)$ and $L_1^{2,0}: (x,y)
\mapsto (\widebar x, \widebar y)$, where the bar denotes the complex conjugation, for any
$(x,y) \in S^1 \times B^2$. The first involution is free, while the fixed point sets of
the last two are respectively $S^1 \times \{0\}$ and $\{-1,1\} \times [-1,1]$.

\medskip

We conclude this section by a characterization of the hyperelliptic involutions of $H_g$
for $g \geq 2$. This will be useful in order to simplify the induction argument for
non-free case in the next section.

\begin{lemma}\label{hyperelliptic/thm}
Let $h$ be a non-free orientation preserving involution of $H_g$ with $g \geq 1$. If for
any 2-disk $D$ in $H_g$ given by Lemma \ref{split/thm} the union $h(D) \cup D$ (possibly
coinciding with $D$ itself) disconnect $H_g$, then $h$ is equivalent to $L_g^{g+1,0}$.
\end{lemma}

\begin{proof}
We proceed by induction on $g$. For $g = 0,1$ the statement follows from the above
classification of the orientation preserving involutions of $H_0$ and $H_1$.

Now, assume $g > 1$. Given a disk $D \subset H_g$ as in Lemma \ref{split/thm}, we denote
by $N$ a regular neighborhood of $D \cup h(D)$. Then, $\Cl(H_g - N)$ is disconnected by
hypothesis, and the second part of that lemma implies that $\Cl(H_g - N) = C_1 \sqcup
C_2$, where $C_i \cong H_{g_i}$ for $i=1,2$, with $g_1 + g_2 = g - 1$ if $h(D) = D$ and
$g_1 + g_2 = g - 2$ if $h(D) \cap D = \emptyset$.

Since $h$ is non-free, we have that each of $C_1$ and $C_2$ is sent onto itself by $h$.
Actually, $h$ could in principle swap $C_1$ and $C_2$ (with $g_1 = g_2$) when $h(D) \cap D
= \emptyset$, but in this case it would be free. Moreover, both the restrictions $h_i =
h_{|C_i}: C_i \to C_i$ obviously satisfy the condition of the lemma. Therefore, by the
inductive hypothesis we have $C_i \cong L_{g_i}^{g_i + 1,0}$ for $i = 1,2$.

At this point, can easily conclude that $h \cong L_g^{g+1,0}$ by Lemma \ref{handle/thm},
after observing that $N$ consists of one (if $h(D) = D$) or two (if $h(D) \cap D =
\emptyset$) 1-handles attached to $C_1 \sqcup C_2$ to give $H_g$.
\end{proof}

\section{Proof of the theorem\label{proof/sec}}

Assume first that $h$ is free. Since the Euler characteristic $\chi(H_g) = 1 - g$ is even,
$g = 2n + 1$ for some $n \geq 0$. We will prove that $h \cong I_g$ by induction on $n$,
based on the case $n = 0$, which follows from the above classification of the involutions
of $H_1$.

So, suppose $n > 0$. Let $D \subset H_g$ be a disk as in Lemma \ref{split/thm}. Then $h(D)
\cap D = \emptyset$, being $h$ free. Now, denoting by $N$ a regular neighborhood of $h(D)
\cup D$ and putting $H' = \Cl(H_g - N)$, we have three cases.

\smallskip
{\sl Case 1. $H' \cong H_{g-2}$}. By the inductive hypothesis, $h' = h_{|H'} \cong
I_{g-2}$. Moreover, $N$ consists of a pair of 1-handles equivariantly attached to $H'$,
which are swapped by $h$. Then, up to equivalence, $h$ is the unique possible extension of
$h'$ to $H_g$. Since, up to equivariant PL homeomorphisms, $I_g$ can be obtained in the
same way from $I_{g-2}$, for example by considering as $D$ the leftmost meridian disk in
Figure \ref{involution1/fig}, we have $h \cong I_g$.

\smallskip
{\sl Case 2. $H' = C_1 \sqcup C_2$, with $C_i \cong H_{g_i}$ and $h(C_i) = C_i$ for $i =
1,2$}. Since $g_1 < g$, by the inductive hypothesis $h_{|C_1} \cong I_{g_1}$. Now, if $g_1
> 1$ we know that there exists a disk $D' \subset C_1 \cong H_{g_1}$, such that $C_1 -
(h(D') \cup D')$ is connected. Then, by replacing $D$ with $D'$ thought as a disk in
$H_g$, we are reduced to case 1. On the other hand, if $g_1 = 1$, for any disk $D'$ in
$C_1$, we have that $C_1 - (h(D') \cup D')$ has two components and these are swapped by
$h_{|C_1}$. Then, since also the two attaching disks of the 1-handles given by $N$ are
swapped by $h_{|C_1}$, we can easily conclude that $H_g - (h(D') \cup D')$ is connected.
So, we can once again reduce ourselves to case 1.

\smallskip
{\sl Case 3. $H' = C_1 \sqcup C_2$, with $C_i \cong H_{g_i}$ and $h(C_i) = C_{3-i}$ for $i
= 1,2$}. In this case we have $1 \leq g_1 = g_2 < g$. Then, there exists a disk $D'
\subset C_1$ such that $C_1 - D'$ is connected. Since $h(D') \subset C_2$ and also $C_2 -
h(D')$ is connected (being PL homeomorphic to $C_1 - D'$), we have that $H_g - (h(D') \cup
D')$ is connected too. This allows the reduction to case 1 as above.

\medskip

Now, we assume that $h$ is non-free. We will prove that $h \cong L_g^{n,m}$ by induction
on $g$, based on the cases $g = 0,1$, which follow from the above classification of the
involutions of $H_0$ and $H_1$, and on the cases considered in Lemma
\ref{hyperelliptic/thm}.

So, suppose $g > 1$. Let $D \subset H_g$ be a disk as in Lemma \ref{split/thm}. If for any
such a disk $D$ the union $h(D) \cup D$ disconnect $H_g$, we are done by Lemma
\ref{hyperelliptic/thm}. Hence, we can assume that $H_g - (h(D) \cup D)$ is connected.
Then, denoting by $N$ a regular neighborhood of $h(D) \cup D$ and putting $H' = \Cl(H_g -
N)$, we have $H' \cong H_{g-1}$ if $h(D) = D$ and $H' \cong H_{g-2}$ if $h(D) \cap D =
\emptyset$. We consider this two cases separately.

\smallskip
{\sl Case 1. $h(D) = D$.} By the inductive hypothesis, $h' = h_{|H'} \cong L_{g-1}^{n,m}$
for some $n$ and $m$ such that $1 \leq n + 2m \leq g$. Moreover, $N$ consists of one
1-handle attached to $H'$, whose attaching disks $D_1,D_2 \subset \Bd H'$ are such that
$h'(D_i) = D_i$ and $D_i \cap \Fix h' = \{p_i\} \subset \Int D_i$, for $i = 1,2$. We have
the following two subcases.

\smallskip
{\sl Case 1.1. $p_1$ and $p_2$ are end points of the same arc $A \subset \Fix h'$.} In
this case, when attaching $N$ to $H'$, the arc $A$ closes up to give a fixed loop for $h$.
Now, if $A$ is the rightmost fixed arc in Figure \ref{involution2/fig}, then clearly $h
\cong L_g^{n-1,m+1}$. On the other hand, the half-twists on the disks $E$ and $E' = h'(E)$
in right side of Figure \ref{twists/fig} allows us to equivariantly exchange to
consecutive arcs in $\Fix h'$, hence all the arcs in $\Fix h'$ are equivalent by an
equivariant PL homeomorphisms. Therefore, the final result is the same for any fixed arc
$A \subset \Fix h'$.

\smallskip
{\sl Case 1.2. $p_1$ and $p_2$ are end points of different arcs $A_1, A_2 \subset \Fix
h'$.} In this case, when attaching $N$ to $H'$, the arcs $A_1$ and $A_2$ are joined to
give one fixed arc in $\Fix h$. Now, if $A_1$ and $A_2$ are the rightmost fixed arcs in
Figure \ref{involution2/fig} and the points $p_1$ and $p_2$ are closest the endpoints of
them, then it is not difficult to see that $h \cong L_g^{n-1,m}$. Now, the half-twists on
the disks $E$ and $E' = h'(E)$ in left side of Figure \ref{twists/fig} allows us to
equivariantly exchange the two end points of the same arc in $\Fix h'$. Then, using this
PL homeomorphism, together with that used in the previous case to exchange two consecutive
arcs in $\Fix h'$, we can always equivariantly move the points $p_1$ and $p_2$ in the
preferred position described above. Hence, $h \cong L_g^{n-1,m}$ whatever $p_1$ and $p_2$
are.

\begin{Figure}[htb]{twists/fig}\vskip3pt
\fig{}{twists.eps}{}{Equivariantly inverting a fixed arc
       and exchanging two fixed arcs.}\vskip-3pt
\end{Figure}

{\sl Case 2. $h(D) \cap D = \emptyset$.} By the inductive hypothesis, $h' = h_{|H'} \cong
L_{g-2}^{n,m}$ for some $n$ and $m$ such that $1 \leq n + 2m \leq g-1$. Moreover, $N$
consists of a pair of 1-handles equivariantly attached to $H'$, which are swapped by $h$.
Then, up to equivalence, $h$ is the unique possible extension of $h'$ to $H_g$. Since, up
to equivariant PL homeomorphisms, $L_g^{n,m}$ can be obtained in the same way from
$L_{g-2}^{n,m}$, for example by considering as $D$ and $h(D)$ the rightmost meridian disks
in Figure \ref{involution2/fig}, we have $h \cong L_g^{n,m}$.

config: arxiv


TITLE: Are there first-order statements that second order PA proves that first order PA does not?
QUESTION [5 upvotes]: Are there first-order statements that second order PA proves that first order PA does not? Is this known one way or the other? Could you share an example? (edit: to clarify, by 'second order PA' I don't mean any first order theory)

REPLY [8 votes]: This question more or less reduces to asking for natural arithmetic statements unprovable in PA.  A very interesting but lesser-known example that I learned about recently concerns fusible numbers. The term "fusible" may be unfamiliar but it arises as a natural generalization of a well-known puzzle involving measuring lengths of time by burning fuses (finite pieces of string) of different lengths.  The puzzle is an old one and was not specifically constructed to produce unprovable statements, so the unprovable statements involving fusible numbers are arguably the most "natural" PA-unprovable statements around.


TITLE: An example where GCD depends on the domain
QUESTION [4 upvotes]: First some notation.  Given a domain $R$ and $x,a,b \in R$, I write $x=gcd(a,b)_R$ to mean that $x$ is one gcd of $a$ and $b$ in $R$.
I want to find an example of an GCD-domain $R$, a subdomain $S \subseteq R$, and two elements $a, b \in S$ such that there isn't any $x \in S$ such that $x=gcd(a,b)_R$ and $x=gcd(a,b)_S$.  Notice that it is not enough to find one element $x \in S$ such that $x=gcd(a,b)_R$ but $x \neq gcd(a,b)_S$.
I can prove that this is impossible in as little as a Bezout domain, but I cannot prove that this is impossible in a mere GCD-domain.  I do not know that many examples of GCD-domains which are not Bezout domains in the first place.
ETA:  As suggested below, I also wanted $S$ to be a GCD-domain.

REPLY [1 votes]: Take $S=\mathbb{Z}[X]$ and $R=\mathbb{Z}+X\mathbb{Q}[X]$.  Then $gcd_S (2,X)=1$  and $gcd_R(2,X)=2$, where both $R$ and $S$ are GCD-domains.


TITLE: Question on de Broglie hypothesis
QUESTION [0 upvotes]: I have read about de Broglie hypothesis that matter like light can also exhibit wave particle duality. This gave rise to the following questions in my mind.

The wavelength for electrons is significant. So does that mean that electrons are both particles and waves at the same time or they show different behaviour in different experiments. Or they are only waves.
I cannot understand why only moving objects will possess a wavelength.
What kind of waves will electrons be like light is an electromagnetic wave.
If electrons behave like waves then will the electric current flowing in wires also be considered as waves.

Please provide me the explanation. Thank you very much

REPLY [1 votes]: de Broglie hypothesis is a historically significant development that has very little bearing on what is happening now. IMHO it is only relevant for historians.

Electrons behave like quantum particles with wavefunction (or a more abstract state vector in Hilbert space) governed by Schrodinger equation (other equivalent formalisms exist). They are neither waves, nor particles. Historically, people talked about wave-particle duality, but this concept does not help you to solve the propblem, neither does it help conceptually. Most of the time it simply confuses.
Not all particles in QM can be given a definite wavelength. Definite wavelength is only possible if the particle is strongly de-localized. Lets say you have a particle described by wavefunction $\psi\left(\mathbf{r},\,t\right)$. One extreme is then that location of particle is precise, i.e. point in space and time. Another extreme is that location of particle is nearly arbitrary, so probability of finding the particle is $\left|\psi\left(\mathbf{r},\, t\right)\right|^2=const$ for all positions $\mathbf{r}$. Basic maths, coupled with translational invariance of free space then tell you that $\psi\left(\mathbf{r},\, t\right)\propto\exp\left(i\mathbf{k}.\mathbf{r}\right)$, for some vector $\mathbf{k}$. Wavelength is then $\lambda_{Broglie}=2\pi/\left|\mathbf{k}\right|$. Wether particle is (inertially) moving or not does not change this logic (but does have stipulations for the value of $\mathbf{k}$)
Makes no sense. I think you left out some words.
Sort of. There are a lot of complications there due to impurities and thermal noise. The closest to this simple picture are superconductors. There, indeed, within Ginzburg-Landau treatment, one can describe the current flow with a collective wave-function for all electrons (current then is an operator that you apply to the wavefunction).


TITLE: Condition that a homeomorphism is an isometry
QUESTION [0 upvotes]: The statement that every isometry is a homeomorphism is true (correct me if I am wrong). What is the condition for any homeomorphism is an isometry? In particular, is it true that any homeomorphism from a 2-sphere is an isometry?
Thanks

REPLY [1 votes]: Homeomorphisms wrk with general topological spaces, isometries require metrics. Especially, even if $f\colon (X,d)\to (Y,d')$ is an isometry, we could replace $d'$ with $\frac 12d'$ and with respect to that metric $f\colon (X,d)\to (Y,\frac12d')$ would still be a homeomorhism, but not an isometry (if $Y$ has at least two points).
The trick with simlpy changing one metric may fail if we have $X=Y$ and require the same metric to be used on both sides.
But here's a homeomorphism $S^2\to S^2$ that is not an isometry with respect to any metric on $S^2$ inducing the usual topology:
$$f(x,y,z) = \frac1{\sqrt{x^2+y^2+(z-\frac12)^2}}\cdot(x,y,z-\frac12). $$


TITLE: probability that you pick three cards from the top and they are from different suits given that the first two you picked are?
QUESTION [1 upvotes]: You shuffle a normal deck of 52 playing cards.What is the probability that you pick three cards from the top and they are from different suits given that the first two you picked are?
Let us say A is the probability that 3 cards are from different suits.
B: is the probability that first two cards are from different suits.
The answer to this question is as follows:
P(A given B)=P(A and B)/P(B)
I know P(B) is:

 $$p(\text{two different suits}) = \frac{\dbinom{4}{2}\dbinom{13}{1}^2}{\dbinom{52}{2}}$$

But my problem is that I have no idea how to come up with P(A and B).
Thank you guys for your help

REPLY [2 votes]: Assuming the first two cards are from distinct suits, there are $50$ cards left in the deck, of which $26$ of them qualify to yield $3$ distinct suits, so the desired conditional probability is $26/50$.

Alternatively, using the formula for conditional probability . . .
\begin{align*}
P(B) &= \frac{\bigl((4)(13)\bigr)\bigl((3)(13)\bigr)}{(52)(51)}\\[8pt]
P(A) &= \frac{\bigl((4)(13)\bigr)\bigl((3)(13)\bigr)\bigl((2)(13)\bigr)}{(52)(51)(50)}\\[8pt]
P(A\mid B) &= \frac{P(A \cap B)}{P(B)}\\[4pt]
&= \frac{P(A)}{P(B)}\qquad\text{[since $A \subset B$]}\\[4pt]
&= \frac{(2)(13)}{50}\\[4pt]
&= \frac{26}{50}\\[4pt]
\end{align*}


TITLE: Spinor inner products
QUESTION [2 upvotes]: The spinor inner product in particle physics is given by $\overline{\psi} \psi = \psi^{\dagger} \gamma_0 \psi $, where I take the convention that the zeroth gamma matrix is hermitian while the rest are anti-hermitian. This is invariant under spin group transformations, $\psi \rightarrow e^{\omega_{ab} S^{ab} }\psi$, with $\omega_{ab}$ real parameters and $S^{ab} = \frac{1}{4}[\gamma^a, \gamma^b]$.
However, there is a second invariant inner product given by $\psi^{\dagger} \gamma_0 \gamma_5 \psi$, with $\gamma_5 = i\gamma_0 \gamma_1 \gamma_2 \gamma_3$. My question is, why not the other one? Is it down to the fact that there are observable differences between the two, and one is favoured by experiment?

REPLY [2 votes]: Well the $\gamma^5$ term does appear in the standard model.
As you duly note, both $\bar{\psi}\psi$ and $\bar{\psi} \gamma^5 \psi$ are Lorentz invariant, so the question to ask is, how are they different?
It turns out that the difference lies in their transformations under a Parity transformation. Without proof, I claim that under $P:(x,y,z) \to (-x,-y,-z)$, $\bar{\psi}\psi \to \bar{\psi}\psi$, while $\bar{\psi}\gamma^5\psi \to -\bar{\psi}\gamma^5\psi$.
So the first object is a true scalar while the second object is a pseudoscalar (i.e. it picks up a minus sign under $P$). Thus, if we have the second term in the Lagrangian, it breaks Parity symmetry.
It turns out (experimentally) that $P$ is indeed broken in the weak interaction. So we do have a weak interaction vertex (for a electron + neutrino $ \to W^-$ interaction, for example) that looks like
\begin{align}
\sim -\frac{-i g_{\mu\nu}}{2\sqrt{2}}\gamma^\mu(1-\gamma^5).
\end{align}
This is an axial vector coupling, which is not parity invariant.
However the pseudoscalar $\bar{\psi}\gamma^5\psi$ doesn't appear in the SM because the object $(1-\gamma^5)/2 = \mathbb{P}_L$ is the projector onto the left-handed component of a spinor, so it can only appear if you want a theory in which the mass term of the right-handed and left-handed spinors are different, which we don't observe experimentally.


TITLE: Integer partitions with subset sums "not divisible" by p
QUESTION [1 upvotes]: I have the following questions: Let $N \in \mathbb{N}$ and
\begin{equation}
\sum_{i=1}^k n_i = N,
\end{equation}
with $n_i \in \mathbb{N}$ for $1 \le i \le k$ and some $k \in \mathbb{N}$, be an integer partition of $N$, such that $n_i \le p$ and for all non-empty subsets $I \subset \{1,\ldots,k\}$ we either have
\begin{equation}
\sum_{i \in I} n_i \notin p \mathbb{N}
\end{equation}
 or
\begin{equation}
n_i = p, \;\;\;\;\;\;\;\; \forall i \in I.
\end{equation}
(I.e., in words, any subset sum is not a multiple of $p$ except it consists of the number $p$ only). How can I generate these partitions efficiently? Is there a closed expression for the number of such partitions?
For example, for $N=16$ and $p = 5$ I count the following 5 partitions:
\begin{equation}
\begin{split}
5+5+5+1 &= 16, \\
5+5+4+2 &= 16, \\
5+5+3+3 &= 16, \\
5+4+4+3 &= 16, \\
4+4+4+4 &= 16. \\
\end{split}
\end{equation}

REPLY [1 votes]: I do not know the literature on this problem.  Hopefully someone here can provide a reference.  Erdos, Turan, and others considered various problems related to your problem. Here is one result that shows k is less than p.
Take some sequence b_i of positive integers and look at s_j, the sum of the first j integers. When s_k - s_j is a multiple of p, then there is a subsum which is a multiple of p. To avoid this, all the s_j must be distinct and nonzero mod p. But then there have to be fewer than p of the s_j. So k is less than p.
I suspect dynamic programming is going to help here. Note that if j is a number in the partition, then p-j cannot be a member of the same partition.  So start by creating a structure which is a bag of numbers, followed by a list of permitted elements allowed to be added to the bag, as well as a list of sums mod p. For each bag, and for each permitted element, make a new bag which has the contents of the old bag, and the new element, and compute the superset of sums mod p and use this to trim the permitted list of elements to add. In practice, most bags will not grow beyond sqrt(p) in size, and the number of distinct elements will often be below log p in size.  For p less than 100, this should be a feasible computation.  Use the information to guide your partition generation.
Gerhard "Is Talkin' 'Bout Our Generation" Paseman, 2018.07.19.


TITLE: Can we measure kenetic energy on an absolute scale, or does it have to be measured with reference to something else?
QUESTION [2 upvotes]: I recently read the question Kinetic energy of an object at rest where it was suggested that the kinetic energy of an object on Earth is measured in the reference frame of Earth.
Can we measure kinetic energy free of a frame of reference? Does the concept even makes sense? And if kinetic energy needs to be measured in a frame of reference, then does potential energy need to as well?

REPLY [1 votes]: Kinetic energy is the energy associated with motion. Therefore, we can no more measure kinetic energy free of a reference frame than we can say something is moving free of a reference frame. It must always be moving with respect to something.
Energy is conserved because the total energy of a system is the same before and after a process in the same frame of reference. If you change the reference frame, it has no meaning to compare it to another frame in terms of whether energy is conserved or not


TITLE: How to compute $\frac{\Gamma(n-\alpha)}{\Gamma(\alpha)}?$
QUESTION [2 upvotes]: How to simplify the gamma function:
$$\frac{\Gamma(n-\alpha)}{\Gamma(\alpha)}?$$
where $n\in \mathbb{Z}$ and $\alpha$ some real number.

REPLY [2 votes]: Hint: If you assume $0<\alpha<1$, you should be able to express this in terms of $\Gamma(\alpha/2)$ and $\Gamma(1-\alpha/2)$ using the fact that $n!= n(n-1)!$. Once you have that, you may be able to use the reflection formula
$$\Gamma(z)\Gamma(1-z) =\frac {\pi}{\sin \pi z}$$This is only speculation on my part, but I think it is a path worth exploring.
Addendum: The reflection formula means you could write $$\frac{1}{\Gamma(1-z)}= \frac{1}{\pi}\Gamma(z)\sin \pi z$$ so $$\frac{\Gamma(z)}{\Gamma(1-z)}= \frac {1}{\pi}\Gamma(z)^2\sin \pi z$$


TITLE: Why is the Continuum Hypothesis (not) true?
QUESTION [61 upvotes]: I'm making my way through Thomas W Hungerfords's seminal text "Abstract Algebra 2nd Edition w/ Sets, Logics and Categories" where he makes the statement that the Continuum Hypothesis (There does not exist a set with a cardinality less than the reals and no set strictly greater than the natural numbers.) is neither true or false.
This is utterly baffling to me, If it's possible to construct a set between $\mathbb{N}$ and $\mathbb{R}$ then this statement is demonstrably false, but if not then the statement is true.
This seems to be a straitforward deduction, but many with a more advanced understanding of the topic matter believe CH to be neither.
How can this be?

REPLY [0 votes]: (1)To make your wording more accurate, I assume you mean "there exists" when you state it as "It's possible to construct."
(2)Truth and Falsity in logic might have a different meaning than you think. Truth and Falsity here can only be talked
within a scope, which is given by ZFC axiom system. Under the consistency assumption of ZFC, the system is incomplete,
meaning that there are statements there not having a T/F answer.
(3) When people say CH "is neither true or false" in this context, they really just mean that such Truth or Falsity cannot be
 deduced from ZFC system.  More precisely, they mean that under the consistency assumption of ZFC, if you add CH or
 its negation to ZFC, the system remains consistent, and therefore CH and its negation cannot be deduced from ZFC.
(4) In a larger system, it is possible to give a definitive answer to CH. As a trivial example, if we add CH to ZFC, then
  CH would be true.