text:
\begin{document}
\author{Riccardo Benedetti}
\author{Paolo Lisca}
\address{Dipartimento di Matematica, Largo Bruno Pontecorvo 5, 56127 Pisa, Italy}
\email{riccardo.benedetti@unipi.it,\  paolo.lisca@unipi.it}

\subjclass[2010]{57R25 (57M99)}

\date{\today}

\begin{abstract}
After surveying existing proofs that every closed, orientable $3$-manifold is parallelizable,
we give three proofs using minimal background. In particular, our proofs use neither spin structures
nor the theory of Stiefel-Whitney classes.
\end{abstract}

\maketitle

\section{Introduction}\label{s:intro}

The aim of this note is to provide three proofs ``with bare hands'' of the
following primary result in $3$-dimensional differential topology,
originally attributed to Stiefel \cite{Sti} (1936):

\begin{thm}\label{t:main}
Every orientable,  closed $3$-manifold is parallelizable.
\end{thm}

We realized by searching the literature that there are  at least four
modern proofs of the above result, collected  in~\cite{Ge, FM} in a very clean way.
Each of those proofs requires a somewhat robust mathematical background, so
we asked ourselves whether there might be a proof which uses minimal background.\footnote
{Essentially the same question was asked in the Mathematics StackExchange Forum, see\\
{\tt https://math.stackexchange.com/questions/1107682/elementary-proof-of-the-fact-that-\\
any-orientable-3-manifold-is-parallelizable}, but the answers given there until July 18, 2018
use the same tools employed in the  proofs mentioned above.}
By asking the use of `minimal background' we meant that such
a proof should (i) satisfy the qualitative constraint of adopting a minimal toolbox (the simplest properties of
cohomology and homotopy groups, the basic tools of differential topology and transversality theory
such as given  e.g.~in~\cite{M} or~\cite{GP} and a few well-known facts about vector bundles and their
Euler classes) and (ii) be as self-contained as possible. Eventually we found three such proofs
which, contrary to some of the proofs present nowadays in the literature,
make neither explicit reference to Stiefel-Whitney classes nor to spin structures.

Throughout the paper, $M$ denotes an orientable, closed (i.e.~compact without boundary)
smooth $3$-manifold. It is not restrictive to assume that $M$ is  connected as well.
Recall that a {\em  combing} of $M$ is a nowhere vanishing tangent vector
field on $M$. Moreover, $M$ is {\em parallelizable} if it admits
a {\em framing}, that is a triple $\Ff=(w,z,v)$ of pointwise linearly independent combings.
The existence of a framing is equivalent to the existence of a {\em trivialization}
\[
\tau_\Ff: M\times \R^3 \to TM
\]
of the tangent bundle of $M$. A
framing incorporates an {\em orientation} of $M$ and, {\em vice versa},
if $M$ is oriented and parallelizable, then there are framings inducing the given orientation.
We will always assume that $M$ is {\it oriented}, with a fixed auxiliary orientation.

The paper is organized as follows. In Section~\ref{s:3proofs} we briefly recall the four proofs collected in~\cite{Ge, FM} and
we point out why they do not satisfy our minimal background requirements.
In Section~\ref{s:results} we fix some notation and we recall a few well-known bare hands results.
In each one of Sections~\ref{s:barehands}, \ref{s:barehands2} and~\ref{s:barehands3}  we give
a different bare hands proof of Theorem~\ref{t:main}. The proof of Section~\ref{s:barehands}
is purely $3$-dimensional and could be regarded as a minimalistic version of the available modern
proof based on Stiefel-Whitney classes.
The proofs provided in Sections~\ref{s:barehands2} and~\ref{s:barehands3} could also be regarded as
minimalistic versions of available modern proofs based on even surgery presentations and, respectively,
$4$-dimensional and purely $3$-dimensional considerations. In particular, the proof of Section~\ref{s:barehands3} could
be viewed as a simplification of the available modern proof mainly based on spin structures.

\noindent
{\bf Acknowledgements.} The authors heartfully thank Chuck Livingston for pointing out, after the appearance of
a previous version of this note, the existence of a proof of Theorem~\ref{t:main} based on Kaplan's
results~\cite{Ka}. We are also grateful to Alexis Marin for an interesting e-mail exchange on the topics at hand.

\section{Available modern proofs of Theorem~\ref{t:main}}\label{s:3proofs}

Each of the four modern proofs we survey in this section
argues that $M$ admits a {\em quasi-framing},
that is a framing $\Ff_0$ of a submanifold $M_0$ of $M$ of the form
\[
M_0 = M \setminus {\rm Int}(B),
\]
where $B$ is a smooth $3$-disk
embedded in $M$. The quasi-framing $\Ff_0$ can be extended with bare hands to a framing of
the whole of $M$ as follows. By the uniqueness of disks up to ambient
isotopy, the choice of $B$ is immaterial. Hence, we can
assume that $B$ is contained in a chart of $M$ and looks standard therein.
Upon fixing an auxiliary metric on $M$ and a trivialization of $TM$ over $B$,
the restriction of $\Ff_0$ to $S^2=\partial B$ is encoded by a smooth map
\[
\rho: S^2 \to SO(3).
\]
Since the universal covering space of $SO(3)\cong\PP^3(\R)$ is $S^3$, we have
$\pi_2(SO(3))=\pi_2(S^3)=0$, therefore $\rho$ can be extended over $B$ and
$\Ff_0$ to $M$.

\subsection{The three proofs presented in~\cite{Ge}}\label{ss:Ge-proofs}

We refer the reader to~\cite[\S~4.2]{Ge} for details. The first and third proofs presented in~\cite{Ge}
use a certain mixture of the theory of Stiefel-Whitney classes and spin structures to establish
the existence of a quasi-framing as follows. The first Stiefel-Whitney class $w_1(M)$ vanishes
because $M$ is orientable, and the key point in both proofs is to show
that $w_2(M)$ vanishes as well.
Using obstruction theory to define Stiefel-Whitney classes one
can argue that $w_2(M)=0$ implies the existence of a spin structure on $M$,
and therefore that $M$ admits a quasi-framing. The first and third proofs differ in
the way they establish the vanishing of $w_2(M)$.

The first proof, resting on several properties of Stiefel-Whitney classes,
is perhaps the one requiring the most sophisticated background.
The so-called {\em Wu classes} $v_i\in H^i(M;\Z/2\Z)$ can be characterized
by the property that, for every $x\in H^{3-i}(M;\Z/2\Z)$,
\[
\langle {\rm Sq}^i(x), [M]\rangle = \langle v_i\cup x, [M]\rangle,
\]
where $[M]$ denotes the fundamental class of $M$ in $H_3(M;\Z/2\Z)$
and ${\rm Sq}^i$ is the $i$-th {\em Steenrod square} operation. It follows that
$v_0= 1$ and, for dimensional reasons, $v_i=0$ if $i> 3-i$. Hence,
the only potentially nonzero Wu classes are $v_0$ and $v_1$. Moreover,
Wu classes and Stiefel-Whitney classes are related through {\em Wu's formula}:
\[
w_q(M)= \sum_{i+j=q} {\rm Sq}^i (v_j).
\]
Since ${\rm Sq}^0$ is the identity map and ${\rm Sq}^i (x)=0$ when $i>\deg(x)$,
by Wu's formula we have
\[
0=w_1(M)={\rm Sq}^0 (v_1) + {\rm Sq}^1 (v_0) = v_1.
\]
By Wu's formula again, the vanishing of $v_1$ implies $w_2(M)=0$.

The third proof given in~\cite[\S~4.2]{Ge} goes as follows: first one shows~\cite[Lemma~4.2.2]{Ge}
that if $\Si$ is a closed, possibly non orientable surface embedded in $M$,
then $w_2(E)=0$ where $E=E(\Si)$ is a tubular neighborhood of $\Si$ in $M$.
The proof is elementary modulo the use of the basic {\it Whitney sum formula}
for Stiefel-Whitney classes of vector bundles.
The conclusion is entirely based on the theory of spin structures combined with
some bare hands reasoning. It is a slight simplification of the proof proposed by R. Kirby in
\cite{Ki2}. The argument is by contradiction: if $w_2(M)\neq 0$ then its
Poincar\'e dual in $H_1(M;\Z/2\Z)$ is represented by a knot $K$
embedded in $M$. Then, the assumption implies that:
\begin{enumerate}
\item[(a)]
$M\setminus K$ carries a spin structure $\s$ which cannot be
 extended over any embedded $2$-disk transverse to $K$;
\item[(b)]
there is a compact, closed surface $\Si$ embedded in $M$
intersecting $K$ transversely in a single point $x_0$.
\end{enumerate}
By the general theory of spin structures, the vanishing of $w_2(E)$ implies that
the set of spin structures on $E=E(\Si)$ is non-empty, and in fact
it is an affine space on
\[
H^1(E;\Z/2\Z) \cong H^1(\Si;\Z/2\Z)\cong H^1(\Si\setminus\{x_0\};\Z/2\Z).
\]
It follows that the restriction of $\s$ to $E\setminus K$ extends to the
whole of $E$, contradicting (b).

The second proof presented in~\cite[\S~4.2]{Ge} is less standard. It is based
on the following non-trivial fact due to Hilden, Montesinos and Thickstun~\cite{HMT}:
{\em there exists a branched covering map $\pi: M\to S^3$ such that the branching locus
bounds an embedded $2$-disk in $M$}. Using this fact, it is relatively easy
to lift a framing of $S^3$, which can be constructed directly, to a quasi-framing of $M$. Although this proof is of a
geometric-topological nature, clearly it does not use minimal background.

\begin{remark}
The three proofs described above are quite demanding from
our ``bare hands'' point of view. The first proof, in particular, comes out of a relatively
obscure algebraic machinery -- we would have a hard time deducing from such a machinery
a heuristic justification for the existence of framings on closed 3-manifolds.
\end{remark}

\subsection{The proof presented in~\cite{FM}}\label{ss:FM-proof}
We refer the reader to~\cite[\S 9]{FM} for details. The starting point is
the {\it Lickorish-Wallace  theorem}~\cite{Li, Wa}, stating that the $3$-manifold $M$ can be obtained by
surgery along a framed link $L\subset S^3$. Equivalently, the statement says that $M$ is the boundary
of a $4$-manifold $W$ constructed by attaching $4$-dimensional $2$-handles to the $4$-ball.
Then, an argument essentially due to Kaplan~\cite{Ka} shows that by applying {\it Kirby moves} to $L$,
it is not restrictive to assume that all the framings of $L$ are even. By using this fact one shows
that the $4$-manifold $W$ is parallelizable, hence that $M=\partial W$ is {\it stably-parallelizable}
and eventually admits a quasi framing.

\begin{remark}\label{discussion} The proof presented in~\cite{FM} satisfies to a large extent the first
minimal background requirement from Section~\ref{s:intro}. In fact: the final portion of the argument,
which will be recalled in Section \ref {s:barehands2}, is ``bare hands"; Rourke's proof \cite{Ro} of the Lickorish-Wallace theorem
is completely elementary and constructive provided one allows the use of  {\it Smale's theorem}~\cite{Sm}
so that, for example, one can take for granted that the operation of cutting and re-gluing a $3$-ball does not change a $3$-manifold;
although Kaplan's argument requires the introduction of Kirby calculus, it does {\it not} use the hard part of Kirby's theorem~\cite{Ki1}
on the completeness of the calculus. Everything  considered, we think that the proof
presented in~\cite{FM} is not as self-contained as possible and therefore it does not satisfy the second
minimal background requirement from Section~\ref{s:intro}.
\end{remark}

\section{Some notation and bare hands results}\label{s:results}

In this section we collect some notation and a few well-known facts that we
allow in our minimal toolbox. Let $N$ be a closed, connected manifold of dimension $n$, and let
\[
\xi: B\to N
\]
be a vector bundle of rank $k$, considered up to bundle isomorphisms.
According to our bare hands constraints, in this generality the only allowable
``characteristic'' class of $\xi$ is
\[
\ww(\xi) \in H^k(N;\Z/2\Z)\cong H_{n-k}(N;\Z/2\Z),
\]
defined as the class carried by the transverse self-intersection of $N$ viewed as
the {\em zero section} of $\xi$ inside $B$. The class $\ww(\xi)$ actually coincides with the $k$-th
Stiefel-Whitney class $ w_k(\xi)$, but we shall not need this fact. Moreover, we
will not make use of any other Stiefel-Whitney class. If both $N$ and $\xi$ are
oriented, the same construction defines an integral class
\[
\ee(\xi)\in H^k(N;\Z),
\]
sent to $\ww(\xi)$ by the natural map $H^k(N;\Z)\to H^k(N;\Z/2\Z)$.
In both cases we talk about the {\it Euler class} of $\xi$, referring to
either $\ww(\xi)$ or $\ee(\xi)$ depending on the context.

We will feel free to use the following facts:
\begin{itemize}
\item
if $\xi=\xi_1 \oplus \xi_2$ is the Whitney sum of two vector bundles then
$\ww(\xi) = \ww(\xi_1)\cup \ww(\xi_2)$;
\item
a line bundle $\la$ on $N$ has a nowhere vanishing section if and only if $\ww(\la)=0$;
\item If $N$ is oriented, a rank-2 oriented vector bundle $\xi$ on $N$ has a nowhere vanishing
section if and only if $\ee(\xi)=0$;
\item
if $\xi=\la_1 \oplus \la_2$ is the Whitney sum of two line bundles then
\[
\ww(\det \xi)= \ww(\la_0)+\ww(\la_1);
\]
\item
$\ww(TN)\in H^n(N;\Z/2\Z)$ and
\[
\langle\ww(TN), [N]\rangle = \chi(N)\bmod (2) \in \Z/2\Z;
\]
\item
$N$ is orientable if and only if $\ww(\det TN)=0$;
\item
If $N$ is oriented then
\[
\langle\ee(TN),[N]\rangle=\chi (N) \in \Z;
\]
\item
let $M$ be a closed, oriented $3$-manifold and $\be\in H^2(M;\Z)$. Then, there is an oriented, connected, closed
$1$-submanifold $C\subset M$ which represents the Poincar\'e dual of $\be$. If $\be\in H^j(M;\Z/2\Z)$
with $0\leq j\leq 3$, there is a possibly non orientable, connected and  closed $(3-j)$-submanifold
of $M$ which represents the Poincar\'e dual of $\be$. Moreover, the cup product of two cohomology
classes $\be_1$ and $\be_2$ can be represented by a transverse intersection of submanifolds
representing the Poincar\'e duals of $\be_1$ and $\be_2$;
\item
any closed $3$-manifold $M$ carries a combing.~\footnote{This fact follows from $\chi(M)=0$ using the
Poincar\'e-Hopf index theorem, clearly an allowable tool, together with the  fact that maps $S^2\to S^2$
are classified up to homotopy by their $\Z$-degree.}
\end{itemize}

Given an auxiliary Riemannian metric $g$ on a closed $3$-manifold $M$, by normalization any
combing of $M$ can be made of unitary norm, and by the Gram-Schmidt process any framing of $M$ can be turned into a point-wise $g$-orthonormal framing.
A unitary combing $v$ on $M$ determines an oriented distribution of tangent $2$-planes
\[
F_v = \{F_v(x)\}_{x\in M}\subset TM,
\]
where $F_v(x)\subset T_x M$ is the subspace $g(x)$-orthogonal to $v(x)$. We assume that, for each $x\in M$,
$v(x)$ followed by an oriented basis of $F_v (x)$ gives an oriented basis of $T_x M$.
The restriction of the projection $TM\to M$ gives rise to an oriented rank-2 real vector bundle
$F_v \to M$ whose isomorphism type is independent of the choice of $g$ and depends on $v$
only up to homotopy. We denote by
\[
\ee(F_v)\in H^2(M;\Z)
\]
the Euler class of $F_v$.

\section{First bare hands proof of Theorem~\ref{t:main}}\label{s:barehands}

In this section we provide the first bare hands proof of Theorem~\ref{t:main},
resting neither on the theory of spin structures nor on properties of Stiefel-Whitney classes.
Our tools consist of basic properties of cohomology groups, transversality theory, and
the facts collected in Section~\ref{s:results}. We will also use the notation introduced in
Section~\ref{s:results}.

The section is organized as follows. In Subsection~\ref{ss:comb-fram} we give a bare
hands proof of the following proposition.
\begin{prop}\label{p:wFv=0} $M$ is parallelizable if and only if
there is a combing $v$ of $M$ such that $\ww(F_v)=0$,
in which case $\ww(F_v)=0$  for every combing $v$.
\end{prop}

Proposition~\ref{p:wFv=0} reduces the proof of Theorem~\ref{t:main}
to showing that $M$ carries a combing $v$ such that  $\ww(F_v)=0$. Observe that,
for a combing $v$ on $M$, the property $\ww(F_v)=0$ is equivalent to the fact that,
for every closed, connected, embedded surface $\Si\subset M$, we have
\begin{equation}\label{e:w2FS=0}
\langle\ww(F_v), [\Si]\rangle = \langle \ww(F_v|_\Si), [\Si]\rangle = 0 \in \Z/2\Z.
\end{equation}
We claim that Equation~\eqref{e:w2FS=0} is a consequence of the equation
\begin{equation}\label{e:WSbh}
\ww(F_v|_\Si) = \ww(T\Si) + \ww(\det T\Si)\cup \ww(\nu_\Si),
\end{equation}
where $\nu_\Si\subset TM$ denotes the normal line bundle of $\Si$.
In fact, if $\Si$ is orientable then $\ww(\det T\Si)=0$ and by~\eqref{e:WSbh}
we have $\ww(F_v|_\Si)=\ww(T\Si)$. Therefore,
\[
\langle \ww(F_v|_\Si), [\Si]\rangle =
\langle \ww(T\Si), [\Si]\rangle = \chi(\Si) \bmod (2) = 0\in\Z/2\Z.
\]
If $\Si$ is non-orientable then $\Si$ is homeomorphic to a connected sum $\#^h\PP^2(\R)$ of
$h$ copies of the projective plane.  Since $M$ is orientable, the normal line bundle $\nu_\Si$
is isomorphic to the determinant line bundle $\det T\Si$, and by~\eqref{e:WSbh}
we have
\[
\langle \ww(F_v|_\Si), [\Si]\rangle = \chi(\Si)\bmod (2) + \langle \ww(\nu_\Si)\cup \ww(\nu_\Si), [\Si]\rangle
= 2- h+h \bmod (2) = 0 \in\Z/2\Z.
\]
Proposition~\ref{p:WSF} of Subsection~\ref{ss:WSF} below contains a bare hands proof of~\eqref{e:WSbh},
thus concluding our bare hands proof of Theorem~\ref{t:main}.

Before embarking in the bare hands proofs of Proposition~\ref{p:wFv=0} and~\ref{p:WSF}
it seems worth to point out the existence of a short argument to prove $w_2(M)=0$
without spin structures, yielding a simplification of the first and third proofs from ~\cite[\S~4.2]{Ge}.
More precisely, we prove Proposition~\ref{p:w2=0} below using only the existence of
Stiefel-Whitney classes and the basic Whitney sum formula.
\begin{prop}\label{p:w2=0}
Let $M$ be a closed, oriented 3-manifold. Then, $w_2(M) = 0$.
\end{prop}

\begin{proof}
Let $v$ be a combing on $M$ and $\Si\subset M$ a closed, connected, embedded surface.
Then, we have the Whitney sum decompositions
\[
TM|_\Si = F_v |_\Si \oplus \epsilon = T\Si \oplus \nu_\Si,
\]
where $\epsilon$ is the trivial line bundle generated by $v$.
 By the Whitney sum formula for Stiefel-Whitney classes,  the first decomposition
gives $\langle w_2(M), [\Si]\rangle = \langle w_2(F_v), [\Si]\rangle$, hence
$w_2(M)=0$ if and only if $w_2(F_v)=0$. The second decomposition yields
\begin{equation*}\label{e:WSF}
w_2(F_v|_\Si) = w_2(\Si) + w_1(\Si)\cup w_1(\nu_\Si),
\end{equation*}
which is analogous to Equation~\eqref{e:WSbh}. An argument similar to the one above showing~\eqref{e:WSbh}
$\Rightarrow$ \eqref{e:w2FS=0} gives $\langle w_2(F_v), [\Si]\rangle = 0\in\Z/2\Z$, therefore
we conclude $w_2(F_v)=0$.
\end{proof}

\subsection{Combing and framing $3$-manifolds}\label{ss:comb-fram}
Our purpose in this subsection is to achieve a bare hands proof of Proposition~\ref{p:wFv=0} above.

\begin{lem}\label{e=0}
$M$ is parallelizable if and only if $\ee(F_v)=0$ for some unitary combing $v$.
\end{lem}

\begin{proof}
Let $v$ be a unitary combing of $M$ such that $\ee(F_v)=0$.
Any nowhere vanishing section of $F_v$ can be normalized with respect to $g$ to a unitary
section $w$ of $F_v$, extended to an oriented orthonormal framing $(w, z)$ of $F_v$ and
finally to an oriented orthonormal framing $(w,z,v)$ of $M$. Conversely, for any
orthonormal framing $(w,z,v)$ of $M$ we may view $v$ as a combing of $M$ and $w$ as
a section of $F_v$.
\end{proof}

\subsection*{The comparison class}

We can associate to an ordered pair of unitary combings $(v, v')$ of $M$ a
smooth section $v\x v'$ of $F_v$ as follows.
At a point $x\in M$ where $v(x) \neq\pm v'(x)$, $v\x v'(x) \in F_v(x)\subset T_x M$ is
the ``vector product'' of $v(x)$ and $v'(x)$, i.e.~the only tangent vector such that
\begin{itemize}
\item
$\|v\x v'(x)\|_{g(x)}^2 = 1 - g(v,v')^2$;
\item
$v\x v'(x)$ is $g(x)$-orthogonal to $v(x)$ and $v'(x)$;
\item
$(v(x), v'(x), v\x v'(x))$ is an oriented basis of $T_ x M$.
\end{itemize}
At a point $x\in M$ where $v(x) = \pm v'(x)$, we set $v\x v'(x) = 0$.

If the two unitary combings $v$ and $v'$ are generic,
the section $v\x v'$ of $F_v$ is transverse to the zero section and
the zero locus
\[
C := \{x\in M\ |\ v\x v'(x) = 0\} \subset M
\]
is a disjoint collection of simple closed curves. Moreover,  $C = C_+\cup C_-$, where
\[
C_+ = \{x\in M\ |\ v(x) = v'(x)\}\quad\text{and}\quad
C_- = \{x\in M\ |\ v(x) = -v'(x)\}.
\]
By the very definition of $\ee(F_v)$, $C$ can be oriented to represent the Euler class of $F_v$.
Indeed, let $E(F_v)$ denote the total space of $F_v$, $M_0\subset E(F_v)$ the zero-section and
$M_1 = v\x v'(M)\subset E(F_v)$. Under the natural identification of $M$ with $M_0$ the submanifold $C$ is identified
with $M_0\cap M_1$. By transversality, for each $x\in M_0\cap M_1$ the natural projection
$p_x : T_x E(F_v)\to F_v(x)$ maps isomorphically the image under $(v\x v)'_*$ of the fiber $N_x(C)$
of the normal bundle of $TC\subset TM|_C$ onto $F_v(x)$. Therefore, the given orientation on $F_v(x)$
can be pulled-back to $N_x(C)$ and, together with the orientation of $T_x M$, it induces an orientation on
$T_x C$ in a standard way.

\begin{definitions}
An ordered pair of unitary combings $(v,v')$ of $M$ such that $v\x v'$ is a section of $F_v$
transverse to the zero section will be called a {\em generic pair} of unitary combings.
We define the {\em comparison class} $\al(v,v')\in H^2(M;\ZZ)$ of a generic pair of unitary combings
as the Poincar\'e dual of the homology class $[C_-]$ carried by the collection of curves $C_-$
with the orientation induced a a subset of the zero locus $C$
 of $v\x v' : M\to F_v$, where $C$ is oriented as described above, so that the resulting homology class
 $[C]$ represents the Poincar\'e dual of  $\ee(F_v)$.
\end{definitions}

\theoremstyle{plain}

\begin{lem}\label{l:comparison-properties}
Let $(v,v')$ be a generic pair of unitary combings of $M$. Then,
\[
\al(v,v') = -\al(v',v)\quad\text{and}\quad \al(v,-v') = \al(v', -v).
\]
\end{lem}

\begin{proof}
For each $x\in C$ the equality $F_v(x)=F_{v'}(x)$ holds, with the orientations of $F_v(x)$ and $F_{v'}(x)$
being the same or different according to, respectively, whether $x\in C_+$ or $x\in C_-$. We may choose
a tubular neighborhood $U = U(C)$ such that the restrictions of the tangent plane fields $F_v |_U$ and $F_{v'} |_U$
are so close that there is a vector bundle isomorphism $\varphi : F_v |_U\stackrel{\cong}{\to} F_{v'} |_U$
which is the identity map on the intersections $F_v (x)\cap F_{v'} (x)$, $x\in U$, is orientation-preserving near
$C_+ = \{x\in M\ |\ v(x) = v'(x)\}$ and orientation-reversing near
$C_- = \{x\in M\ |\ v(x) = -v'(x)\}$. Since $\varphi\circ (v\x v') = v\x v' = - v'\x v$ and $- v'\x v$ is obtained
by composing the section $v'\x v$ with the orientation-preserving automorphism of $F_{v'}$ given by minus
the identity on each fiber, the orientation on $C_-$ as part of the zero locus of $v\x v' : M\to F_v$ is the opposite of its
orientation as part of the zero locus of $v'\x v = -v\x v': M\to F_{v'}$.  This implies $\al(v,v') = -\al(v',v)$.
Similarly, the orientation on $C_+$ as part of the zero locus of $v\x (-v') : M\to F_v$ coincides with its
orientation as part of the zero locus of $(-v')\x v = v'\x (-v): M\to F_{v'}$, which implies $\al(v,-v') = \al(v',-v)$.
\end{proof}

\begin{lem}\label{l:compare}
Let $(v,v')$ be a generic pair of unitary combings of $M$. Then,
\[
\ee(F_v)-\ee(F_{v'})= 2\alpha(v,v').
\]
\end{lem}

\begin{proof}
According to the definitions we have
\[
\ee(F_v) = \al(v,v') + \al(v,-v')\quad\text{and}\quad
\ee(F_{v'}) = \al(v',v) + \al(v',-v).
\]
The statement follows applying Lemma~\ref{l:comparison-properties} after taking the difference of the two equations.
\end{proof}

\subsection*{Pontryagin surgery}
Let $v$ be a unitary combing of $M$ and $C\subset M$ an oriented, simple closed curve such that
the positive, unit tangent field along $C$ is equal to $v|_C$ and there is a trivialization
\[
j:D^2\times S^1 \stackrel{\cong}{\to} U(C)
\]
of a tubular neighborhood of $C$ in $M$ such that
\[
v \circ j= j_*(\partial /\partial \phi),
\]
where $\phi$ is a periodic coordinate on the $S^1$-factor
of $D^2\x S^1$. Let  $(\rho, \th)$ be polar coordinates on the $D^2$-factor. Following
terminology from~\cite{BP}, we say that a unitary combing $v'$ is obtained from $v$ by
{\it Pontryagin surgery} along $C$ if, up to homotopy, $v'$ coincides with $v$ on $M\setminus U(C)$ and
\[
v' \circ j = j_*\left (-\cos(\pi\rho)\frac{\partial}{\del\phi} -\sin(\pi\rho)\frac{\partial}{\del\rho}\right)
\]
on $U(C)$.

\begin{remark}
A basic fact not used in this paper is that any two combings of $M$ are obtained
from each other, up to homotopy, by Pontryagin surgery~\cite{BP}.
\end{remark}

\begin{lem}\label{pont-comb-surg} Let $v$ be a unitary combing of $M$ and
$\beta \in \ H^2(M;\ZZ)$. Then, possibly after a homotopy of $v$, there is a unitary
combing $v'$ such that $(v,v')$ is a generic pair of unitary combings and
\[
\alpha(v, v')= \beta.
\]
\end{lem}

\begin{proof}
Let $C\subset M$ be an oriented simple closed curve representing the Poincar\'e dual of $\beta$
and let $j : D^2\x S^1 \to U(C)$ be a trivialization of a neighborhood of $C$.
Without loss of generality we may assume that
the pull-back $j^*(g)$ of the auxiliary metric $g$ on $M$ is the standard product metric on $D^2\x S^1$.
After a suitable homotopy of $v$ the assumptions to perform Pontryagin surgery on $v$
along $C$ are satisfied.
Consider a normal disc $D_{\phi_0} = j(D^2\x\{\phi_0\})$ and let $p = D_{\phi_0}\cap C$. Then,
$T_p D_{\phi_0}$ coincides, as an oriented $2$-plane,  with $F_v(p)$  as well as with the $g(p)$-orthogonal subspace
of $T_p C$ inside $T_p M$. Let $v'$ be a unitary combing obtained from
$v$ by first performing a Pontryagin surgery on $U(C)$ and
then applying a small generic perturbation supported on a small neighborhood of $M\setminus U(C)$.
Then, $(v,v')$ is a generic pair of unitary combings and $C = \{x\in M\ |\ v(x) = -v'(x)\}$.
By the definition of $\al(v,v')$, to prove the statement it suffices
to show that the given orientation of $C$ coincides with its orientation as part of the zero set of $v\x v' : M\to F_v$.
Near $C$ we have
\[
(v\x v')\circ j = j_*\left(-\sin(\pi\rho) \frac{\del}{\del\th}\right) =
j_*\left(\frac{\sin(\pi\rho)}{\rho}\left(y\frac{\del}{\del x} - x\frac{\del}{\del y}\right)\right),
\]
where $x = \rho\cos\th$ and $y=\rho\sin\th$ are rectangular coordinates on the $D^2$-factor. Observe that
$j_*$ sends the pair $(\del/\del x, \del/\del y)$ to an oriented framing of $F_v$.
Using the resulting trivialization of $F_v$ we can write locally the restriction of $v\x v'$ to
to the disc $D_{\phi_0}$ followed by projection onto $F_v$ as follows:
\[
v\x v'|_{D_{\phi_0}} : (x,y)\mapsto  \frac{\sin(\pi\rho)}{\rho} (y, -x) = \pi (y, -x)\ +\ \text{higher order terms}.
\]
It is easy to compute that $(v\x v')_* \circ j_*$ sends $\del/\del x$ to $-\pi\del/\del y$ and
$\del/\del y$ to $\pi\del/\del x$, and since the matrix
$\left(\begin{smallmatrix}0&\pi\\-\pi&0\end{smallmatrix}\right)$ has
determinant $\pi^2>0$ this shows that the restriction of $(v\x v')_*$ to the normal bundle to $C$
composed with the projection onto $F_v$ is orientation-preserving along $C$, concluding the proof.
\end{proof}

We shall say that the Euler class $\ee(F_v)$ is {\em even} if there exists $\beta \in H^2(M;\Z)$
such that $\ee(F_v)=2\beta$.

\begin{lem}\label{l:even} $M$ is parallelizable if and only
$\ee(F_v)$ is even for every unitary combing $v$.
\end{lem}

\begin{proof}
If $M$ is parallelizable, then $M$ has a unitary framing $(w,z,v)$ and the class
$\ee(F_v)=0$ is obviously even. Let $v'$ be an arbitrary unitary combing of $M$.
After possibly small perturbations of $v'$ and $v$ which do not change
$\ee(F_{v'})$ nor $\ee(F_v)$, the pair $(v',v)$ becomes a generic
pair of unitary combings and by Lemma~\ref{l:compare} we have
\[
\ee(F_{v'}) =\ee(F_{v'}) - \ee(F_v) = 2\alpha(v',v).
\]
Therefore, $e(F_{v'})$ is even as well.
Conversely, suppose that $v$ is a unitary framing with $\ee(F_v)=2\beta\in H^2(M;\Z)$.
By Lemma~\ref{pont-comb-surg}, possibly after a homotopy of $v$ --
which does not change $\ee(F_v)$ --
there is a unitary framing $v'$ such that $(v,v')$ is a generic pair and
$\alpha(v,v')=\beta$. Hence, by Lemma~\ref{l:compare} we have
\[
\ee(F_v) - \ee(F_{v'})= 2\alpha(v,v')=2\beta,
\]
which implies $\ee(F_{v'})=0$, therefore $M$ is parallelizable by Lemma~\ref{e=0}.
\end{proof}

\begin{lem}\label{l:even2} Let $v$ be a unitary combing of $M$. Then,
$\ee(F_v)$ is even if and only if $\ww(F_v)=0$.
\end{lem}

\begin{proof}
The implication $\ee(F_v)=0$ $\Rightarrow$ $\ww(F_v)=0$ is trivial. We give two arguments for the other implication.
The first argument uses a little bit of homological algebra. The short exact sequence of coefficients
$$0\to \Z \stackrel{2\cdot}{\to} \Z \to \Z/2\Z \to 0$$
induces a long exact sequence in cohomology including the segment
\[
\cdots \to  H^2(M;\Z)\stackrel{2\cdot}{\to} H^2(M;\Z)\stackrel{\varphi}{\to} H^2(M;\Z/2\Z) \to \cdots
\]
where the map $\varphi$ is reduction mod $2$. Exactness yields the statement.

The second argument is more geometric. The Poincar\'e dual of $\ee(F_v)$ can be represented
by an oriented  knot $K\subset M$. If $\ww(F_v)=0$ then $K$ bounds an embedded surface $\Si\subset M$.
If $\Si$ is orientable then $[K]=0$, hence $\ee(F_v)=0$, which is obviously even. If $\Si$ is non-orientable
then there is a collection $C$ of simple closed curves in the interior of $\Si$ such that
$\Si\smallsetminus C$ is orientable and a tubular neighborhood $U$ of $C$ in $\Si$ is a union
of M\"obius bands. Orient $\Si\smallsetminus{\stackrel{\circ}{U}}$ so that $K$ is an oriented boundary component
and give $\del U$ the resulting boundary orientation. Orient the cores of $U$ so that the natural projection
$\partial U \to C$ has positive degree. Then, $[K]=[\partial U]= 2[C]$, therefore $\ee(F_v)$ is even.
\end{proof}

\begin{proof}[Proof of Proposition~\ref {p:wFv=0}]
The statement is an immediate consequence of Lemmas~\ref{l:even} and~\ref{l:even2}.
\end{proof}

\subsection{Proof of Equation~\eqref{e:WSbh}}\label{ss:WSF}
The purpose of this subsection is to give a bare hands proof of Proposition~\ref{p:WSF} below,
which establishes Equation~\eqref{e:WSbh}. As explained at the beginning of the present section,
this concludes our bare hands proof of Theorem~\ref{t:main}.

Let $v$ be a unitary combing of $M$ and $\Si\subset M$ a closed, embedded surface.
At each point $x\in\Si$ we have the splittings
\begin{equation}\label{e:splittings}
T_x M = F_v (x)\oplus \ep(x) = T_x \Si\oplus \nu_\Si (x),
\end{equation}
where $\ep(x)$ is the (oriented) line spanned by $v(x)$, while $\nu_\Si (x)$
is the (unoriented) line orthogonal to $T_x \Si$.

\begin{prop}\label{p:WSF}
Let $v$ be a unitary combing of $M$ and $\Si\subset M$ a closed, embedded surface. Then,
\begin{equation}\label{e:bhWSF}
\ww(F_v|_\Si) = \ww(T\Si) + \ww(\det T\Si)\cup \ww(\nu_\Si).
\end{equation}
\end{prop}

\begin{proof}
Let $s:\Si\to F_v|_\Si$ be a generic section of the restriction $F_v$ to $\Si$.
For each $x\in\Si$, the second splitting from~\eqref{e:splittings} induces decompositions
\[
s(x) = s_\Si(x)+ s_\nu(x), \quad v(x)= v_\Si(x) + v_\nu(x).
\]
By transversality we may assume that:
\begin{enumerate}
\item[(i)]
the zero set $\{s=0\}\subset\Si$ consists of a finite number of points representing $\ww(F_v|_\Si)$;
\item[(ii)]
$s_\nu$ and $v_\nu$ are generic sections of $\nu_\Si$, so that both their zero sets
$\{s_\nu = 0\}$ and $\{v_\nu = 0\}$ consist of smooth curves in $\Si$ representing $\ww(\nu_\Si)$.
Moreover, $\{s_\nu = 0\}$ and $\{v_\nu = 0\}$ intersect transversely in $\Si$,
so that the finite set $\{v_\nu = 0\}\cap \{s_\nu = 0\}$ represents
$\ww(\nu)\cup \ww(\nu) = \ww(\det T\Si)\cup \ww(\nu)$;
\item[(iii)]
$\{s=0\}$ and $\{v_\nu = 0\}$ are disjoint subsets of $\Si$;
\item[(iv)]
$s_\Si$ is a generic section of $T\Si$, so that $\{s_\Si = 0\}$ consists of a finite number
of points representing $\ww(T\Si)$.
\end{enumerate}

Given a finite set $X$, denote by $|X|_2\in\Z/2\Z$ the cardinality of $X$ modulo $2$.
Then, we have
\[
\langle\ww(F_v|_\Si), [\Si]\rangle = |\{s=0\}|_2,\quad \langle\ww(T\Si), [\Si]\rangle = |\{s_\Si = 0\}|_2,
\]
\[
\langle\ww(\det T\Si)\cup \ww(\nu_\Si), [\Si]\rangle = |\{v_\nu = 0\}\cap \{s_\nu = 0\}|_2.
\]
Therefore Equation~\eqref{e:bhWSF} is equivalent to the following equality:
\begin{equation}\label{e:equal}
|\{s=0\}|_2 = |\{s_\Si = 0\}|_2 + |\{v_\nu = 0\}\cap \{s_\nu = 0\}|_2.
\end{equation}
The finite set $\{s_\Si = 0\}$ can be tautologically decomposed as a disjoint union:
\[
\{s_\Si = 0\} = (\{v_\nu = 0\}\cap \{s_\Si = 0\})\amalg (\{v_\nu \neq 0\}\cap \{s_\Si = 0\}).
\]
We claim that
\[
\{v_\nu \neq 0\}\cap \{s_\Si = 0\} = \{s=0\}.
\]
In fact, by Assumption~(iii) above we have
\[
\{s=0\} = \{v_\nu \neq 0\}\cap \{s=0\},
\]
and clearly
\[
\{v_\nu \neq 0\}\cap \{s=0\}\subset  \{v_\nu \neq 0\}\cap \{s_\Si = 0\}.
\]
On the other hand, if $x\in \{v_\nu \neq 0\}\cap \{s_\Si = 0\}$ then $s(x)=0$
because, since $v_\nu(x)\neq 0$,
the projection $F_v(x) \to T_x\Si$ is an isomorphism. Thus, the claim is proved.
In order to establish Equality~\eqref{e:equal} it is now enough to check that
\begin{equation}\label{e:equal2}
|\{v_\nu = 0\}\cap \{s_\Si = 0\}|_2 = |\{v_\nu = 0\}\cap \{s_\nu = 0\}|_2.
\end{equation}
Let $C$ be the collection of smooth curves $\{v_\nu = 0\}\subset\Si$. At each $x\in C$
we have a splitting
\[
F_v(x) = (F_v(x)\cap T_x\Si)\oplus \nu_\Si(x),
\]
therefore the restriction $F_v|_C$ splits as a sum of line bundles
\[
F_v|_C = \lambda \oplus  \nu_\Si |_C,
\]
where $\lambda = \{F_v(x)\cap T_x\Si\}_{x\in C}$. We claim that the line bundles $\lambda$ and $\nu_\Si |_C$ are isomorphic. In fact, along each component
of $C$ the bundle $F_v$ is trivial because it is oriented, so the two line bundles are
either both trivial or both non-trivial. Thus,
$\langle\ww(\lambda), [C]\rangle = \langle\ww(\nu_\Si|_C), [C]\rangle$, and
Equality~\eqref{e:equal2} follows from the observation that the restriction of $s_F$ and $s_\nu$ to $C$ are
generic sections of, respectively, $\lambda$ and $\nu_\Si |_C$.
\end{proof}

\section{Second bare hands proof of Theorem~\ref{t:main}}\label{s:barehands2}
The aim of this section is to provide a genuine proof of Theorem~\ref{t:main} using minimal background,
employing some of the ideas we summarized in Section~\ref{ss:FM-proof}.
Let us first outline an elementary proof of the last portion of the proof presented in~\cite{FM}.

\begin{lem}\label{l:even-4par} Let  $N=\chi(S^3, L)$ be a $3$-manifold obtained by surgery along a framed link
$L\subset S^3$ such that all framings are {\rm even}. Let $W$ be the corresponding $4$-manifold
obtained by attaching $4$-dimensional $2$-handles to the $4$-ball, so that $N=\partial W$.
Then, $W$ is parallelizzable.
\end{lem}

\begin{proof} We refer the reader to~\cite{FM} for further details. For simplicity, assume that $L$ is a one-component
link with even framing $n$. As we can assume that the attaching tubes of the $2$-handles are pairwise
disjoint, this is not really restrictive.  Let $N(L)\subset \partial D^4$ be the attaching tube of the corresponding $2$-handle
attached to $D^4$. Both $D^4$ and $D^2\times D^2$ are parallelizable,
so we have to show that they carry some framings which match on $N(L)$.
Fix a reference framing $\Ff_0$ on $TD^4$. Then, the restriction to $N(L)$ of any framing $\Ff$ on the $2$-handle
is encoded by a map $\rho: N(L)\to SO(4)$.  Viewing $S^3$ as the group of unit quaternions
one can construct a 2-fold covering map $S^3\x S^2\to SO(4)$ showing that $\pi_1(SO(4))= \Z/2\Z$.
As the solid torus $N(L)$ retracts onto $L\cong S^1$, $\rho$ determines an element $\overline\rho\in\Z/2Z$ which
vanishes if and only if the two framings coincide on $N(L)$. It is easy to see that $\overline\rho$ is equal to
$n\bmod 2$.
\end{proof}

\begin{cor}\label{c:stab-par} If a $4$-manifold $W$ is parallelizable, then $\partial W$ is {\rm stably-parallelizable}.
In fact, the Whitney sum of the tangent bundle $T\partial W$ with a trivial line bundle $\epsilon$ is a product bundle.
\end{cor}

\begin{proof}
By the existence of a collar of $\partial W$ in $W$ it is immediate that $T\partial W\oplus\epsilon\cong TW|_{\partial W}$.
\end{proof}

\begin{lem}\label{l:stably-almost} If a closed, connected, orientable $3$-manifold $N$
is stably-parallizable, then it admits a quasi framing, hence it is parallelizable.
\end{lem}

\begin{proof}
We reproduce the short bare hands  argument  of ~\cite[Lemma~3.4]{KM}. With the usual notation, let
$N_0 = N\setminus {\rm Int}(B)$.  Since $TN_0$ is oriented, a bundle isomorphism
$TN_0\oplus\ep\cong\ep^4$ gives rise to a map from $N_0$ to the Grassmannian $Gr(3,4)$ of
oriented $3$-planes in $\R^4$. Since $Gr(3,4)\cong S^3$ and  $N_0$ has a $2$-dimensional spine, by transversality
any such map is not surjective up to homotopy, hence it is homotopically trivial, therefore $TN_0$ is trivial.
\end{proof}

The following lemma is trivial.

\begin{lem}\label{l:trivial}
Let $M$ and $M'$ be closed, connected, oriented $3$-manifolds.
If $M\# M'$ is parallelizable, then both $M$ and $M'$ admit a quasi framing, hence
they are parallelizable.
\end{lem}

\begin{proof}
Let $N = M\# M'$. Obviously $M_0$ embeds into $N$ and $TM_0$ is the restriction of $TN$ to $M_0$.
The same holds for $M'$.
\end{proof}

Combining Corollary~\ref{c:stab-par} with Lemmas~\ref{l:stably-almost} and~\ref{l:trivial},
to complete our second bare hands proof of Theorem~\ref{t:main} we are
reduced to providing a proof using minimal background of the following proposition.

\begin{prop}\label{p:mainprop} For every connected, closed, oriented $3$-manifold $M$,
there exists another such $3$-manifold $M'$ such that  $N=M\# M'$ is of the form $N=\chi(S^3,L)$ for some
framed link $L\subset S^3$ such that all framings are even.
\end{prop}

\begin{proof}
We use some basic facts about Heegaard splittings of $3$-manifolds.
Let us start with any Heegaard splitting of $M$ of some genus $g$. Up to diffeomorphisms,
$M_0$ can be relized as follows. Given a handlebody $\HG_g$ of genus $g$, the orientable
surface $\Sigma_g = \partial \HG_g$ contains a non separating system $C=\{c_1,\dots, c_g\}$ of
$g$ pairwise disjoint smooth circles. A tubular neighbourhood $N(C)$ in $\Sigma_g$ is formed by a system of
pairwise disjoint attaching tubes for $3$-dimensional $2$-handles, which, when attached to $\HG_g$
give $3$-manifold $M_0$. The closed $3$-manifold $M$ is obtained by attaching a further final $3$-handle.
The union of the above $2$- and $3$-handles gives the second handlebody $\HG'_g$ of the Heegaard
splitting, glued to $\HG_g$ along the common boundary $\Sigma_g$. Fix any standard embedding
of $\HG_g$ into $S^3$, so that the closure of $S^3\setminus \HG_g$ is a handlebody as well.
This embedding realizes a genus-$g$ Heegaard splitting of $S^3$. The collection of curves
$C\subset \partial \HG_g \subset S^3$ becomes a link $L$ in $S^3$, with each component of $L$ framed
by a parallel curve in $\partial \HG_g$. Now we can apply the key basic Lemma 1 of \cite{Ro},
which has a bare hands proof. In our situation, the lemma implies that $$\chi(S^3,L)=M\cs M'$$
for some $3$-manifold $M'$. It is an immediate consequence of the above description of $M_0$ and of the definition
of surgery along a framed link that $\chi(S^3,L)$ is obtained by gluing $M_0$ and $M'_0$ along
their spherical boundaries. Applying Smale's theorem~\cite{Sm} we can conclude that
$\chi(S^3,L)=M\cs M'$.

Now fix a complete system $\Mm =\{m_1,\dots, m_g\} $ of meridians of $\HG_g$. The curves $m_i$
bound a system of disjoint $2$-disks properly embedded into $(\HG_g,\Sigma_g)$. Denote by $\tau_i$
the Dehn twist on $\Sigma_g$ along $m_i$. Since every $\tau_i$ extends to a diffeomorphism $\overline\tau_i$
of the whole $\HG_g$, we can modify a given embedding of $\HG_g$ into $S^3$ by applying
any finite sequence of such $\overline\tau_i$'s. So we are reduced to show that in this way we can
obtain an embedding such that the framing of each $c_i$ determined as above by
the embedding in $\Sigma_g$ is even.  This is the content of~\cite[Lemma~8.4.1]{BP}
(proved  therein to have a treatment with bare hands of Kaplan's result for the double
$D(M)=-M \cs M$ of $M$). The proof of~\cite[Lemma~8.4.1]{BP} boils down to solving
a certain $\Z/2\Z$-linear system.
\end{proof}

\begin{remarks}
(1) Smale's theorem~\cite{Sm} is an essential ingredient of Rourke's clever proof of the Lickorish-Wallace theorem.
We refer to it also in the proof of Proposition~\ref{p:mainprop}. However, this is not really necessary.
In fact, the description of $\chi(S^3,L)$ as obtained by gluing together $M_0$ and $M'_0$ along their spherical
boundaries suffices. Thus,  {\em Smale's theorem can be discarded from the background of our second bare hands
proof of Theorem~\ref{t:main}}.
(2) By \cite{Wa}, the Lickorish-Wallace theorem is bare hands equivalent to $``\Omega_3=0"$.
Modulo Smale's theorem, Rourke's proof is the simplest one. However, a more basic proof that
$\Omega_3=0$ could probably be concocted by combining a bare hands proof that $M$ is parallelizable
with a specialization of the elementary proof of a theorem by Thom given in~\cite{BH}.
\end{remarks}

\section{Third bare hands proof of Theorem~\ref{t:main}}\label{s:barehands3}

We shall make use of Lemma~\ref{l:seiferttwist} below, which could be viewed as a `ground zero'
fact about spin structures. Let $N$ be an oriented $3$-manifold, $K\subset N$ an oriented knot and
$n:K\to TN|_K$ a unitary normal vector field along $K$.
The orientation of $K$ determines the unitary tangent vector field $t:K\to TN|_K$ and an orthonormal oriented
framing $\Ff_n = (t,n,b)$ of $TN|_K$. Let $F\subset N$ be a smoothly embedded, oriented surface with
$\del F=K$. Since $F$ retracts onto a one-dimensional CW-complex, $TF$ is trivial. Let $(a,b)$ be any
oriented, orthonormal framing of $TF$ and $(a,b,c)$ the orthonormal framing of $TN|_F$ obtained by adding the oriented
unit normal vector field $c$ to $F$. From now on, we shall implicitely use the framing $(a,b,c)$ to identify, at any point of $F$,
the set of orthonormal framings of $TN$ with $SO(3)$ and the set of unit vectors of $TN$ with $S^2$.
Define the map $\varphi_n: K\to S^1$ by $\varphi_n(x) = e^{i\theta(x)}$, where $\theta(x)$ is the
counterclockwise angle between $c(x)$ and $n(x)$ measured in the oriented normal plane to $K$ at $x$.

\begin{lem}\label{l:seiferttwist}
The framing $\Ff_n$ of $TN|_K$ extends to a framing of $TN|_F$ if and only if $\deg(\varphi_n)$ is odd.
\end{lem}

\begin{proof}
Let $\psi_n:K\to SO(3)$ be the map given by $\psi_n(x)=\Ff_n(x)$. Clearly $\Ff_n$ extends to a framing of $TN|_F$
if and only if $\psi_n$ extends to a map $F\to SO(3)$, which happens if and only if the image of $[K]\in H_1(K;\Z/2\Z)$
under the induced map $(\psi_n)_*:H_1(K;\Z/2\Z)\to H_1(SO(3);\Z/2\Z)\cong \Z/2\Z$ is trivial.
Consider the $S^1$-fibrations $\pi_1,\pi_2:SO(3)\to S^2$ given by
$\pi_1(a',b',c')= a'$ and $\pi_2(a',b',c')= b'$ and homotope $\Ff_n$ until there are two disjoint intervals
$A,B\subset K$ such that $n=c$ on $K\setminus A$
and $t=a$ on $K\setminus B$. Then, setting $C := \pi_1^{-1}(\sm{1\\0\\0})$ and $C':=\pi_2^{-1}(\sm{0\\1\\0})$,
it is easy to check that
\[
(\psi_n)_*([K]) = \chi(F)[C]+\deg(\varphi_n)[C']\in H_1(SO(3);\Z/2\Z).
\]
Since $[C]=[C']$ is a generator of $H_1(SO(3);\Z/2\Z)$, we deduce that $\psi_n$
extends to $F$ if and only if $\chi(F) + \deg(\varphi_n)$ is even. But $\chi(F)$
is always odd, therefore the statement holds.
\end{proof}

Fix a Heegaard splitting $M=\HG_g\cup\HG'_g$ and let $C=\{c_1,\dots, c_g\}\subset \del\HG_g=\del\HG'_g$
be a complete system of meridians for $\HG'_g$.  Consider a standard embedding of $\HG_g$ in $\R^3$ and unit vector
field $n_i$ along the curves $c_i\subset \partial \HG_g$, normal to $\del\HG_g$ and pointing towards $\HG_g$.
As in the proof of Proposition~\ref{p:mainprop}, using~\cite[Lemma~8.4.1]{BP} we can choose the embedding
so that each $n_i$ defines an even
framing of $c_i$ with respect to the Seifert framing in $\R^3$. Note that, by Lemma~\ref{l:seiferttwist},
this is equivalent to saying that the induced framing $\Ff_{n_i}$ of $T\R^3|_{c_i}$ does not extend to
a framing of $T\R^3$ over a Seifert surface. The vector fields $n_i$ coincide with the unit normal vector fields
determined by collars of each curve $c_i$ in the corresponding $2$-disk $D_i$ properly embedded into $(\HG_g', \partial \HG'_g)$.
Let if $B_i\subset M$ be a $3$-disk containing $D_i$. By Lemma~\ref{l:seiferttwist} the framings $\Ff_{n_i}$,
regarded as framings of $TB_i|_{c_i}$, do not extend to framings of $TB_i|_{D_i}$. On the other hand,
the restriction of the standard framing $\Ff$ of $\R^3$ to each $c_i$ is homotopic to a framing $\Ff_{m_i}$ determined
by a unit vector field $m_i$ normal to $c_i$ and defining an odd framing with respect to the Seifert framing.
Again by Lemma~\ref{l:seiferttwist}, this means that $\Ff$ can be extended along each $D_i$, yielding a quasi-framing of $M$.
This concludes the third bare hands proof of Theorem~\ref{t:main}.


TITLE: Applying Time Dilation Twice
QUESTION [1 upvotes]: I have recently been thinking about the equations of special relativity and have run across something interesting. I was thinking about the following scenario: one person is standing stationary on a road, another person is driving at 0.125c, and a third person is driving at 0.25c.
For the rest of this post, I will use $\gamma_v$ to denote the relativistic factor for a speed v and $t_v$ to represent the time measured by the person moving at speed v.
So, I thought of two ways of analyzing the time dilation when comparing the stationary person and the person moving at 0.25c. The simple way is:
$t_0= \gamma_{0.25c}t_{0.25c}$ (1)
I then thought about using the time measured by the person moving at 0.125c to determine the time dilation. According to the person moving at 0.125c, the person moving at 0.25c is only moving at 0.125c, so:
$t_{0.125c}=\gamma_{0.125c}t_{0.25c}$ (2)
Additionally, the following should be true:
$t_{0}=\gamma_{0.125c}t_{0.125c}$ (3)
Rearranging equation 3 gives:
$t_{0.125c}=t_{0} / \gamma_{0.125c}$ (4)
Plugging equation 4 into equation 2 gives:
$t_{0} / \gamma_{0.125c} = \gamma_{0.125c}t_{0.25c} $ (5)
Rearranging equation 5 gives:
$t_{0} = {\gamma_{0.125c}}^2 t_{0.25c}$ (6)
Finally, plugging equation 6 into equation 1 gives:
${\gamma_{0.125c}}^2 t_{0.25c} = \gamma_{0.25c}t_{0.25c}$
${\gamma_{0.125c}}^2 = \gamma_{0.25c}$ (7)
However, in carrying out the calculations, equation 7 appears to be false. It appears as if I have made a mistake, but I am not sure where. Any help would be very much appreciated. Thank you!

REPLY [1 votes]: There are two errors:

Relative-velocity is not subtractive--instead, it's $$v_{BA}=\frac{v_{BO}-v_{AO}}{1-v_{BO}v_{AO}}.$$
Time-dilation factors are not multiplicative--there is an extra factor $$\gamma_{BA}=\gamma_{BO}\gamma_{AO}(1-v_{BO}v_{AO}).$$

These are easier to recognize if you work with rapidities (Minkowski angles), where $v_{BO}=\tanh\theta_{BO}$, $\gamma_{BO}=\cosh\theta_{BO}$, etc...

Relative-rapidity is $\theta_{BA}=\theta_{BO}-\theta_{AO}$.

Relative-velocity is $v_{BA}=\tanh(\theta_{BA})=\tanh(\theta_{BO}-\theta_{AO})=\frac{\tanh\theta_{BO}-\tanh\theta_{AO}}{1-\tanh\theta_{BO}\tanh\theta_{AO}}$.

Relative-time-dilation-factor is $\begin{align}\gamma_{BA}=\cosh(\theta_{BA})=\cosh(\theta_{BO}-\theta_{AO})&=\cosh\theta_{BO}\cosh\theta_{AO}-\sinh\theta_{BO}\sinh\theta_{AO}\\&= \cosh\theta_{BO}\cosh\theta_{AO}(1-\tanh\theta_{BO}\tanh\theta_{AO})\end{align}$.

Here's an example with a spacetime diagram on rotated graph paper.
Here $v_{CA}=-3/5$ and $v_{BA}=3/5$--so $\gamma_{CA}=5/4$ and $\gamma_{BA}=5/4$.
So, $v_{BC}=\frac{v_{BA}-v_{CA}}{1-v_{BA}v_{CA}}=15/17$.
Note that $\gamma_{BC}=17/8$ (graphically)
and $\gamma_{BA}\gamma_{CA}(1-v_{BA}v_{CA})=(\frac{5}{4})(\frac{5}{4})(1-(\frac{3}{5})(\frac{-3}{5}))=17/8$.

postscript: This diagram shows that although Alice says that events N and Q are simultaneous, they are not simultaneous according to Bob or Carol.


TITLE: If $a, b$ are positive integers, does $\;b\mid(a^2 + 1)\implies b\mid (a^4 + 1)\quad?$
QUESTION [3 upvotes]: If $a, b$ are positive integers, does $\;b\mid(a^2 + 1)\implies b\mid (a^4 + 1)\;$?
Explain if this is true or not. If no, give a counterexample.

REPLY [0 votes]: If positive integer $b$ divides $a^2+1$ and $a^4+1$ for some integer $a$
$b$ must divide $(a^4+1)-(a^2-1)(a^2+1)=2$
So, $b=1$ or $2$


TITLE: Existence Of Electric Field Lines
QUESTION [1 upvotes]: Can an Electric Field  with field lines Like So Exist:
http://puu.sh/tWkJ
One Of my friends said it couldn't as the field lines here are not conservative ; so it cannot exist ; Is he right?
Or can it be made to exist

REPLY [4 votes]: Yes, your friend is right. Within electrostatics, an electric field $\vec{E}$ should be curl-free $\vec{\nabla} \times\vec{E}= \vec{0}$. The drawn electric field lines looks like the electric field is of the form
$$ E_x=E_x(y), \qquad E_y=0, \qquad E_z=0, $$
cf. the rule that to depict the magnitude $|\vec{E}|$, a selection of field lines is drawn such that the density of field lines (number of field lines per unit perpendicular area) at any location is proportional to $|\vec{E}|$ at that point. Here the $x$-axis is horizontal, the $y$-axis is vertical, and the $z$-axis perpendicular to the plane.
This is only curl-free if $E_x=E_x(y)$ is independent of $y$, which it isn't on the figure.


text:
\begin{document}
\subsection{Low-rank method}
\label{subsection:low_rank_method}

We now make use of these low-rank tensor formats to approximate the marginal distribution $\pPM$ as the solution of~\cref{eq:linear_system}.
Since the exact solution $\pPM$ is a probability distribution, its entries sum up to one, i.e., it fulfills
\begin{align}
\label{eq:probability_condition}
	\langle \e , \pPM \rangle = 1
\end{align}
where $\e \in \R^S$ is the tensor of all ones.
To allow for a probabilistic interpretation of an approximation of $\pPM$, we need to treat~\cref{eq:probability_condition} as an additional constraint.
To this end, we now present an iterative method based on the Neumann series~\cite{Dubois1979} and the uniformization method~\cite{Grassmann1977}.

Let $\gamma \geq \max_{x \in \Space} \vert \QPM[x,x] \vert > 0$ be a bound on the diagonal entries of the infinitesimal generator, then
\begin{align*}
	\pPM = \left(\Id - \QPM\right)^{-1} \pN = \frac{1}{1 + \gamma} \left(\Id - \left(\frac{\gamma}{1 + \gamma} \Id + \frac{1}{1 + \gamma} \QPM\right)\right)^{-1} \pN.
\end{align*}
Owing to the ordering assumption~\ref{ass:(iii)} and the fact that all diagonal entries of $\QPM$ are non-positive, the spectral radius is smaller than one, i.e.,
$\rho\left(\frac{\gamma}{1 + \gamma} \Id + \frac{1}{1 + \gamma} \QPM\right) = \frac{\gamma}{1 + \gamma} < 1$, and so the Neumann series converges.
Hence, we can represent the marginal distribution as
\begin{align}
\label{eq:Neumann_representation}
	\pPM = \frac{1}{1 + \gamma} ~ \sum\limits_{m = 0}^{\infty} ~ \left(\frac{\gamma}{1 + \gamma}\right)^m \left(\Id + \frac{1}{\gamma} \QPM\right)^m \pN.
\end{align}
Alternatively we can derive~\cref{eq:Neumann_representation} using the uniformization method.
Here, the idea is to describe a continuous-time Markov chain by a discrete-time Markov chain with a time increment that is exponentially distributed.
With this interpretation, we write the time-dependent probability distribution in~\cref{eq:transient_distribution_with_time} as
\begin{align*}
	\pPM(t) = \sum\limits_{m = 0}^{\infty} \frac{\left(\gamma t\right)^m}{m!} \exp(-\gamma t) \left(\Pgamma\right)^m \pN \qquad \text{with} \qquad \Pgamma = \Id + \frac{1}{\gamma} \QPM
\end{align*}
for a time $t \geq 0$.
Note that $\Pgamma $ is the transition probability matrix of a discrete-time Markov chain, since $\gamma$ is a bound on the diagonal entries of $\QPM$.
Marginalization of time similar to~\cref{eq:marginal_distribution_integral} and substitution leads to~\cref{eq:Neumann_representation}:
	\begin{align*}
		\pPM
		&= \int\limits_{0}^{\infty} \exp(-t) ~ \pPM(t) ~\mathrm{d}t
		~=~ \sum\limits_{m = 0}^{\infty} ~ \frac{\gamma^m}{m!} ~\int\limits_{0}^{\infty} t^m ~ \exp(- (\gamma + 1 ) t) ~\mathrm{d}t ~ \left(\Pgamma\right)^m \pN
		\\
		&= \sum\limits_{m = 0}^{\infty} ~ \frac{\gamma^m ~ \Gamma(m + 1 )}{ m! \left(1 + \gamma\right)^{m+1}} ~  \left(\Pgamma \right)^m \pN
		~=~ \frac{1}{1 + \gamma} ~ \sum\limits_{m = 0}^{\infty} ~ \left(\frac{\gamma}{1 + \gamma} ~\Pgamma \right)^m \pN
	\end{align*}
where $\Gamma$ denotes the gamma function. A natural approximation based on~\cref{eq:Neumann_representation} would be
\begin{align*}
	\pktilde= \frac{1}{1 + \gamma} ~ \sum\limits_{m = 0}^{k} ~ \left(\frac{\gamma}{1 + \gamma} ~\Pgamma \right)^m \pN
\end{align*}
for $k \in \N$. Based on the properties of the Neumann series this sequence converges linearly to $\pPM$, but $\tilde{\mathbf{p}}^{(k)}$ does not satisfy the normalization condition~\cref{eq:probability_condition} for any $k \in \N$.
As $\Pgamma $ is a transition probability matrix, its application to a probability distribution leads to a probability distribution again satisfying~\cref{eq:probability_condition}, and thus
\begin{align*}
	\langle \e, \sum\limits_{m = 0}^{k} ~ \left(\frac{\gamma}{1 + \gamma} ~ \Pgamma\right)^m \pN \rangle=  \sum\limits_{m = 0}^{k} ~ \left(\frac{\gamma}{1 + \gamma}\right)^m = \frac{\left(1 + \gamma\right)^{k + 1} - \gamma^{k+1}}{\left(1 + \gamma\right)^k}
\end{align*}
for all $k \in \N$.
By scaling each element of the sequence, we obtain
\begin{align}
\label{eq:pk}
	\pk =  \frac{\left(1 + \gamma\right)^k}{\left(1 + \gamma\right)^{k + 1} - \gamma^{k+1}} ~ \sum\limits_{m = 0}^{k} ~ \left(\frac{\gamma}{1 + \gamma}~ \Pgamma\right)^m \pN
\end{align}
for $k \in \N$, which is now an approximation to $\pPM$ that satisfies~\cref{eq:probability_condition}. We prove its linear convergence to $\pPM$ in the following theorem.
\begin{theorem}
\label{lemma:convergence_pk}
	Let $\PM$ be some given parameters, $\pPM$ be the solution of the corresponding linear system~\cref{eq:linear_system}, and $\pk$ be defined by~\cref{eq:pk} for all $k \in \N$. Then $\pk$ converges linearly to $\pPM$ as $k$ approaches infinity, i.e.,
	\begin{align*}
		\lim\limits_{k \to \infty} \pk = \pPM \qquad \text{with} \qquad \big\Vert \pk - \pPM \big\Vert \leq c \cdot \biggl(\frac{\gamma}{1 + \gamma}\biggr)^k \quad \text{for all } k \in \N
	\end{align*}
	where $c =\big \Vert \frac{1}{1 + \gamma} \pN - \pPM \big\Vert + \gamma \Vert \pPM \Vert$.
\end{theorem}
\begin{proof}
	For $\alpha_k = \frac{(1 + \gamma)^k}{(1 + \gamma)^{k + 1} - \gamma^{k+1}} $, $\alpha = \frac{1}{1+\gamma}$ and any $k \in \N_0$ we have
	\begin{align*}
		\vert \alpha_k - \alpha \vert &= \bigg\vert \frac{ \gamma^{k+1}}{(1+\gamma) ((1+\gamma)^{k+1} - \gamma^{k+1})}  \bigg\vert < \frac{\gamma}{1+\gamma} ~ \vert \alpha_{k-1} - \alpha \vert
		\\
		&< \biggl(\frac{\gamma}{1 + \gamma}\biggr)^k ~ \vert \alpha_0 - \alpha \vert = \biggl(\frac{\gamma}{1 + \gamma}\biggr)^{k+1}.
	\end{align*}
	Furthermore we obtain
	\begin{align*}
		\big\Vert \pk - \pPM \big\Vert &\leq \big\Vert \pk - \pktilde \big\Vert + \big\Vert \pktilde - \pPM \big\Vert
		\\
		&\leq \vert \alpha_k - \alpha \vert \bigg\Vert \sum\limits_{m = 0}^{k} ~ \biggl(\frac{\gamma}{1 + \gamma}~\Pgamma\biggr)^m \pN \bigg\Vert + \biggl( \frac{\gamma}{1+\gamma}\biggl)^k \Vert \alpha \pN - \pPM \Vert
		\\
		&\leq \bigl( \Vert \alpha \pN - \pPM\Vert + \gamma \Vert \pPM \Vert \bigr) \cdot \biggl(\frac{\gamma}{1 + \gamma}\biggr)^k.
	\end{align*}
	Since $\frac{\gamma}{1 + \gamma} < 1$, the linear convergence follows.
\end{proof}

In order to employ low-rank tensor formats, we extend the iterative method corresponding to~\cref{eq:pk} by truncation.
As shown in~\cite{Hackbusch2008}, a convergent iterative method combined with truncation still converges if the truncation error is sufficiently small.
However, the truncation could violate condition~\cref{eq:probability_condition}.
Therefore, we have to extend our method to ensure that
\begin{align*}
	\langle \e, \tau\left( \left(\Pgamma\right)^m \pN \right) \rangle = 1 \qquad \text{and} \qquad \langle \e, \tau( \pk )\rangle = 1
\end{align*}
for all $m, k \in \N$ where $\tau$ denotes the truncation operator. We implement these conditions by rescaling after truncation as shown in~\cref{alg:uniformtization}, see lines $7$ and $11$.
The procedure should be continued until the norm of the relative residual is smaller than a given tolerance $\tol >0$.

\begin{algorithm}[tbhp]
	\caption{Low-rank uniformization$\left(\PM, \gamma, \tol \right)$}
	\label{alg:uniformtization}
	\begin{algorithmic}[1]
		\STATE $\Pgamma= \Id + \frac{1}{\gamma} \QPM$
		\STATE $k = 0$
		\STATE $s = 1, \csum = 1$
		\STATE $\pk = \pN, \psum = \pN$
		\WHILE{ $\frac{\Vert \left(\Id - \QPM\right) \pk / \csum -  \pN \Vert}{\Vert \pN\Vert } \geq \tol$ }
		\STATE{ $ \psum = \tau\left(\Pgamma \psum\right)$ }
		\STATE{ $ \psum = \frac{ \psum }{ \langle e, \psum \rangle}$ }
		\STATE{$s = \frac{ \gamma}{ \left( 1 + \gamma\right)} \cdot s$}
		\STATE{$\csum = \csum + s$}
		\STATE{$\mathbf{p}^{(k+1)} = \tau\left(\pk + s~ \psum\right)$}
		\STATE{$\mathbf{p}^{(k+1)} =  \frac{\csum }{ \langle e, \pk \rangle} \cdot \mathbf{p}^{(k+1)}$}
		\STATE{$k = k + 1$}
		\ENDWHILE
		\STATE{$\pk = \frac{\pk}{ \csum}$}
		\RETURN $\pk$
	\end{algorithmic}
\end{algorithm}
In~\cref{alg:uniformtization} we have to choose a bound $\gamma$ on the diagonal entries of $\QPM$.
The cost for computing all diagonal entries to find the maximum is in $\mathcal{O}(n^d)$.
In case of parameters $\PM$ that vary little,
\begin{align}
\label{eq:upper_bound}
	\gamma  = \sum\limits_{\nu = 1}^d ~ \max\limits_{i_{\nu}} \left( \sum\limits_{\left(i_{\nu}, j_{\nu}\right) \in T_{\nu}} ~ \prod\limits_{\mu = 1}^d ~ \max\limits_{i_{\mu}} ~ \PM_{\left(i_{\nu}, j_{\nu}\right), i_{\mu}} \right)
\end{align}
can be used as an inexpensive bound.
For the computation of~\cref{eq:upper_bound} we need $\mathcal{O}\left(d^2 n^2 T\right)$ comparisons and evaluations\footnote{Strictly speaking, the computational cost of the method is therefore quadratic, rather than linear, in $d$. However, $\gamma$ only needs to be precomputed once, the computational cost of which is negligible.} of the parameter $\PM$ where $T = \max_{\nu} \vert T_{\nu} \vert$ is the maximum number of possible transitions in an automaton.
In the case of strongly varying parameters $\PM$,~\cref{eq:upper_bound} may be a gross overestimation for the diagonal entries of $\QPM$.
The question of how to determine a tighter bound with polynomial effort using low-rank tensor formats will be dealt with in future work.

\end{document}

config: arxiv


TITLE: What's wrong with my calculations?
QUESTION [0 upvotes]: I am trying to solve for theta. I get the wrong result returned, when plugging in values. I am trying to solve for theta and cannot seem to find what's wrong with my calculations.

REPLY [0 votes]: Hint.
Calling $z = \sqrt{v^2u^2+2gh}$ we have
$$
\frac{v^2}{g}(1-2u^2)+\frac{v^3}{g}u(1-u^2)\frac 1z-\frac vg uz=\frac{v}{g z}\left(u v+z\right)((1-u^2)v-u z)=0
$$


TITLE: Double Integral Limits of Integration Without Graphing
QUESTION [0 upvotes]: Usually I change the order & limits of integration in double integrals by graphing the curves in the domain & working it out from the geometry.
However in this double integral:
$$ \smallint_0^{\alpha} \tfrac{T(E)dE}{\sqrt{\alpha - E}} = \sqrt{2m} \int_0^\alpha \int_0^E [\tfrac{dx_2}{dU} - \tfrac{dx_1}{dU}] \tfrac{dUdE}{\sqrt{(\alpha - E)(E - U)}}$$
I have no idea how the following happens:
$$ \sqrt{2m} \int_0^\alpha \int_0^E [\tfrac{dx_2}{dU} - \tfrac{dx_1}{dU}] \tfrac{dUdE}{\sqrt{(\alpha - E)(E - U)}} = \sqrt{2m} \int_0^\alpha \int_U^\alpha [\tfrac{dx_2}{dU} - \tfrac{dx_1}{dU}] \tfrac{dUdE}{\sqrt{(\alpha - E)(E - U)}}$$
as is done in section 12 of Landau.
So I guess the question is how does one do this in general without graphing anything, i.e. change the limits of integration, in a way that makes sense of the above integral? Thanks!

REPLY [1 votes]: There is a misprint in that formula. The domain of integration is $\{ (U,E) \colon 0 < E < \alpha, 0 < U < E \}$. Now if you start with $U$ then you can see that $U$ is always between $0$ and $\alpha$ and $E$ is larger than $U$. So
$$\int_0^{\alpha} \int_0^E f(U,E)dUdE = \int_0^{\alpha} \int_U^{\alpha} f(U,E)dEdU.$$


TITLE: polynomial, power series, radius of convergnce
QUESTION [2 upvotes]: Let $p(x)$ be a polynomial of degree $N$ then the radius of convergence of the power series $$\sum_{n=0}^{\infty}p(n)x^n$$

depends on $N$
is $1$ for all $N$
is $0$ for all $N$
is $\infty$ for all $N$

Radius of convergence $r=\lim\limits_{n\to\infty}\frac{p(n)}{p(n+1)}\rightarrow1$ ?

REPLY [3 votes]: You have this. For any nonzero polynomial $p$, you have
$$\lim_{n\to\infty} \root{n}\of{|P(n)|} = 1.$$
Now invoke the root test.

REPLY [3 votes]: You (probably) know that if $p$ and $q$ are polynomials with the same degree and leading coefficients $p_0$ and $q_{\,0}$, then
$$\lim_{x\to \infty}\frac{p(x)}{q(x)}=\frac{p_0}{q_{\,0}}$$
Your line of thought is good. Since $p(n)$ and $p(n+1)$ have the same degree and the same leading coefficient, say $p_0$, then
$$\lim_{n\to \infty}\frac{p(n)}{p(n+1)}=\frac{p_0}{p_0}=1$$


TITLE: Understanding the slope of a line as a rate of change
QUESTION [15 upvotes]: I thought I was confident about what is meant by the slope of a line and how it relates to a rate of change, but I'm having doubts and I'm hoping that someone will be able to help me clear them up.
As I understand it, intuitively the slope of a line is a number $m$ that quantifies the amount it inclines from the horizontal, and its direction. Now, if $y$ is a function of $x$, such that $y=mx+c$, then one can quantify the rate of change in $y$ as one changes $x$ via the the ratio of their coordinate differences, such that $$m=\frac{\Delta y}{\Delta x}$$ Heuristically, can one (hopefully correctly) understand this quantity as follows:
The value of the function $y$ changes by an amount of $\Delta y$ units for every $\Delta x$ units change in $x$. Therefore, the value of $y$ changes by an amount $\frac{\Delta y}{\Delta x}$ units per unit change in x, which is exactly the rate of change in $y$ with respect to $x$, since it quantifies the amount the value of $y$ changes per unit change in $x$.
Would this be a correct understanding at all?

REPLY [4 votes]: Yes, your interpretation of slope is correct. In fact, some people take this interpretation as the definition of the slope of a line.


TITLE: Real analysis - Series
QUESTION [0 upvotes]: Hi guys could you help me with this:
proof if $\sum a_{n}^{2}$ convergs then $\sum \frac{a_{n}}{n}$ convergs absolutely.
i tried:
if $\sum a_{n}^{2}$ convergs then exists m $\geq n\geq n_{0}$ ; given $\varepsilon > 0$
so.  $| a_{n+1}^2 + a_{n+2}^2 + ... + a_{n+m}^2| < \varepsilon $
i was goona use triangular inequality but it doesn't helps.
and i can't keep continue. Maybe i've taken the wrong way to solve this problem.

REPLY [0 votes]: If $m,n\in\mathbb N$ and $m\geqslant n$, then$$\sum_{k=n}^m\left|\frac{a_k}k\right|\leqslant\left(\sum_{k=n}^m{a_k}^2\right)\left(\sum_{k=n}^m\frac1{k^2}\right)$$by the Cauchy-Schwarz inequality. Now, apply Cauchy's convergence test.


TITLE: What is a virtual state?
QUESTION [7 upvotes]: In quantum mechanics / Raman spectroscopy, what is a virtual state?  What is the difference between a virtual state and a superposition of states?  Can you simply think of the virtual state as a superposition of eigenstates?
If you have the complete set of eigenstates for a system, then you can represent any configuration of the system as a linear combination of the eigenstates, so it seems that you should be able to represent the virtual state that way.  Is there something more to the virtual state than that?

REPLY [6 votes]: There are differences in the spin that the term "virtual state" gets depending on context. In the context of Raman scattering, one sometimes refers to transitions to virtual states.  Other terminology used is "virtual transition" to a real state.   The latter terminology is a little easier to understand in my way of thinking.  And note that the "virtual state" can be constructed as a superposition of real states, as you point out.  So whatever we say about virtual transitions to real states immediately applies to the virtual state picture.  But the virtual state picture is more complicated.
In either point of view, the incident radiation in Raman scattering is not resonant with any real state, so real transitions are not possible.  We imagine that the system makes a transition to a real state, and a quantum of EM excitation is destroyed.  Because the real state is not resonant with the radiation, energy is not conserved.  This is possible as long as the lifetime of the system in that state is short.  Heisenberg's uncertainty principle allows violations of conservation of energy for short time intervals:  $\Delta E\Delta t \leq \hbar/2$.
Here's a semi-classical take on Raman scattering.
Consider a diatomic molecule in its ground state.  Upon irradiation, it makes a virtual transition to an excited state, and stays there for a short period of time.  But the characteristics of the chemical bond in the excited state are different from what they were in the ground state.  In particular, the equilibrium length of the bond will be different.  The molecule is promoted to the excited state, but it's initial bond length up there is not the equilibrium length for the excited state.  So the atoms feel a force and move towards the new equilibrium length.  But in a short period of time the system returns back to the ground electronic state.   But now the bond length is no longer the ground state equilibrium length; the atoms feel a force and move towards the original bond length.  Now there's nothing to stop them:  the atoms are oscillate back and forth past the original equilibrium length:  it's vibrating.


TITLE: Find the values of $\alpha$ and $\beta$ for which the function $f$ is of bounded variation on $[0,1]$?
QUESTION [1 upvotes]: The question is :

Let $f : [0,1] \longrightarrow \mathbb {R}$ be defined by

$f(x) = x^{\alpha} \sin {\frac {1} {x^{\beta}}}$ , whenever $x \in (0,1]$ and $=0$ , whenever $x = 0$.
Then find the values of $\alpha$ and $\beta$ for which $f$ becomes a function of bounded variation on $[0,1]$.
Please give me a right way to proceed.Thank you in advance.
I have just obtained the result which is
$\alpha > 2$ and $\beta < \alpha - 2$.Is it correct at all?Please verify it.

REPLY [0 votes]: Claim. Let $\alpha>0$, $\beta > 0$ and let $f$ be as given. Then $f \in \operatorname{BV}(\left[0,1\right])$ if and only if $\alpha > \beta$.
Proof. Suppose that $\alpha>\beta$. Let $h:\left[0,1\right]\to\mathbf{R}$ be defined by
$$
h(x)=
\begin{cases}
\displaystyle{x^{\alpha/(\alpha-\beta)}\sin\left(\frac{1}{x^{\beta/(\alpha-\beta)}}\right)}&,\text{ if } 0<x\le 1\\
0&,\text{ if } x = 0.
\end{cases}
$$
Then $h$ is differentiable at every $x\in \left[0,1\right]$. In fact, we have
$$
h^{\prime}(x)=
\begin{cases}
\displaystyle{\frac{\alpha}{\alpha-\beta}x^{\beta/(\alpha-\beta)}\sin\left(\frac{1}{x^{\beta/(\alpha-\beta)}}\right)-\frac{\beta}{\alpha-\beta}\cos\left(\frac{1}{x^{\beta/(\alpha-\beta)}}\right)}&,\text{ if }0<x\le1\\
0&,\text{ if }x=0.
\end{cases}
$$
In particular, $h^{\prime}$ is bounded, hence $h$ is Lipschitzian of order $1$, i.e., there is some $C>0$ such that for all $x\in\left[0,1\right]$ and all $y\in\left[0,1\right]$, we have
$$ \left|h(x)-h(y)\right|\le C\left|x-y\right|. $$
Substituting $x\mapsto x^{\alpha-\beta}$, $y\mapsto y^{\alpha-\beta}$, we finally get
$$ \left|f(x)-f(y)\right|\le C\left|x^{\alpha-\beta}-y^{\alpha-\beta}\right|, $$
therefore the sums appearing in the definition of total variation are bounded above by $C(t_{n}^{\alpha-\beta} - t_{0}^{\alpha-\beta})=C(1-0)=C$, which implies $f\in\operatorname{BV}(\left[0,1\right])$.
Suppose that $\alpha\le\beta$ and consider the following partition
$$ \left\lbrace 0, \left(\frac{2}{\pi(2n+1)}\right)^{1/\beta}, \left(\frac{2}{\pi(2n)}\right)^{1/\beta},\left(\frac{2}{\pi(2n-1)}\right)^{1/\beta},\cdots,\left(\frac{2}{\pi}\right)^{1/\beta},1 \right\rbrace. $$
The following holds:
$$ \begin{aligned}
\sum_{k=1}^{2n+2}\left|f(t_{k})-f(t_{k-1})\right|&\ge\sum_{k=2}^{2n+1}\left|f(t_{k})-f(t_{k-1})\right|\\
&=\left(\frac{2}{\pi}\right)^{\alpha/\beta}\sum_{k=2}^{2n+1}\left|\frac{1}{(2n+2-k)^{\alpha/\beta}}\sin\left(\frac{k\pi}{2}\right)-\frac{1}{(2n+2-(k-1))^{\alpha/\beta}}\sin\left(\frac{(k-1)\pi}{2}\right)\right|
\\
&\ge\left(\frac{2}{\pi}\right)^{\alpha/\beta}\sum_{k=1}^{n}\frac{1}{\left(2(n-k)+1\right)^{\alpha/\beta}}\\
&=\left(\frac{2}{\pi}\right)^{\alpha/\beta}\sum_{k=0}^{n-1}\frac{1}{\left(2k+1\right)^{\alpha/\beta}}.
\end{aligned} $$
Because $0<\alpha\le\beta$, the last sum can be made arbitrarily large. Consequently, $f\notin \operatorname{BV}(\left[0,1\right])$ and the claim is proved.