text:
\begin{document}

\begin{abstract}
We show that the stable commutator length vanishes for certain groups defined as infinite unions of smaller
groups. The argument uses a group-theoretic analogue of the Mazur swindle, and goes back to the works
of Anderson, Fisher, and Mather on homeomorphism groups.
\end{abstract}

\maketitle

\section{Introduction}

In this paper we show that the stable commutator length vanishes for certain groups defined as unions of
subgroups that have many conjugate embeddings that commute element-wise. The argument used is a
group-theoretic analogue of the Mazur swindle. Informally, it can be paraphrased by saying that as the size of a table grows,
it becomes easier to sort objects on its surface, so that commutation can be done very efficiently on a table of
infinite size\footnote{I owe this description to N.~A'Campo.}. In the words of Po\'enaru~\cite{Po}: ``The infinite
comes with magic and power.''

Let $\Gamma$ be a group. The commutator length $c(g)$ of an element $g$ in the commutator subgroup
$[\Gamma,\Gamma]\subset\Gamma$ is the minimal number of factors needed to write $g$ as a product of elements that
can be expressed as single commutators. The stable commutator length of $g$ is defined to be
$$
\vert\vert g\vert\vert =\lim_{n\rightarrow\infty}\frac{c(g^n)}{n} \ .
$$
Direct proofs of the vanishing of the stable commutator length sometimes proceed by showing a lot more, namely
that the commutator length itself is bounded. A prototypical argument for this in the context of homeomorphism groups
goes back to Anderson~\cite{A} and Fisher~\cite{F}, and was later used and refined by Mather~\cite{M1} and by
Matsumoto--Morita~\cite{MM}. This argument uses an infinite iteration that leads to complicated behaviour near
one point, and is therefore not suitable for the study of diffeomorphism groups. The argument we employ here is
a variation on this classical one. It works for diffeomorphism groups because instead of an infinite iteration we do
only a finite iteration, however it is important that the finite number of iterations can be taken to be arbitrarily large.
This argument does not prove that the
commutator length is bounded, it only proves the weaker conclusion that the stable commutator length vanishes.

Bavard~\cite{B}, using a Hahn--Banach argument influenced by that of Matsumoto and Morita~\cite{MM},
showed that the vanishing of the stable commutator length on the commutator
subgroup is equivalent to the injectivity of the comparison map $H^2_b(\Gamma;\R)\rightarrow H^2(\Gamma;\R)$,
where $H^*_b(\Gamma)$ denotes bounded group cohomology in the sense of Gromov. Another way to express
this condition is to say that every homogeneous quasi-homomorphism $\Gamma\rightarrow\R$ is in fact a
homomorphism. If $\Gamma$ is a perfect group, then the stable commutator length is defined on the whole
of $\Gamma$, and vanishes identically if and only if there are no non-trivial homogeneous quasi-homomorphisms
from $\Gamma$ to $\R$.

Bavard's result suggests that the commutator calculus arguments discussed above should have a dual formulation
in terms of quasi-ho\-mo\-mor\-phisms. It is this dual formulation that we shall discuss, although we could equally well
argue directly on the commutator side. Thus, what we prove for certain groups is that every homogeneous quasi-homomorphism
is a homomorphism. Using Bavard's result this is equivalent to the vanishing of the stable commutator length.

In Section~\ref{s:alg} we give the algebraic mechanism behind the vanishing results we shall prove.
Section~\ref{s:discrete} applies this mechanism to the stable mapping class group, the braid group on
infinitely many strands, and the stable automorphism groups of free groups. In Section~\ref{s:diff} we
give applications to diffeomorphism groups, and in Section~\ref{s:BIP} we compare our methods and results
to those of Burago, Ivanov and Polterovich~\cite{BIP}.

We refer the reader to~\cite{B,qmorph} for background on quasi-ho\-mo\-mor\-phisms. In fact, the proofs of the vanishing
results below are somewhat reminiscent of the discussion of weak bounded generation in~\cite{qmorph}.

\section{The algebraic vanishing result}\label{s:alg}

A map $f\colon\Gamma\rightarrow\R$ is called a quasi-homomorphism
if its deviation from being a homomorphism is bounded; in other words,
there exists a constant $D(f)$, called the defect of $f$, such that
$$
\vert f(xy)-f(x)-f(y)\vert\leq D(f)
$$
for all $x,y\in\Gamma$. We will always take $D(f)$ to be the smallest number
with this property, i.e.~it is the supremum of the left hand sides over all $x$
and $y\in\Gamma$.

Every quasi-homomorphism can be homogenized by defining
$$
\varphi (g) = \lim_{n\rightarrow\infty}\frac{f(g^{n})}{n} \ .
$$
Then $\varphi$ is again a quasi-homomorphism, is homogeneous in the
sense that $\varphi(g^{n})=n\varphi(g)$, and is constant on conjugacy
classes. (Compare~\cite{B}, Proposition~3.3.1.) Throughout this paper we shall only consider
homogeneous quasi-homomorphisms.

We begin with the following preliminary result.
\begin{lemma}\label{l:pre}
Let $\varphi\colon\Gamma\rightarrow\R$ be a homogeneous quasi-homomorphism. Then the following two
properties hold:
\begin{enumerate}
\item If $x$, $y\in\Gamma$ commute, then $\varphi (xy)=\varphi (x)+\varphi (y)$.
\item If $\varphi$ vanishes on every single commutator, then $\varphi$ is a homomorphism.
\end{enumerate}
\end{lemma}
\begin{proof}
By homogeneity we have the following:
\begin{alignat*}{1}
\vert \varphi(xy)-\varphi(x)-\varphi(y)\vert &= \lim_{n\rightarrow\infty}\frac{1}{n}\vert\varphi ((xy)^n)+\varphi (x^{-n})+
\varphi (y^{-n})\vert\\
&\leq \lim_{n\rightarrow\infty}\frac{1}{n}\left(\vert \varphi((xy)^nx^{-n}y^{-n})\vert+2D(\varphi)\right)\\
&=\lim_{n\rightarrow\infty}\frac{1}{n}\vert \varphi((xy)^nx^{-n}y^{-n})\vert \ .
\end{alignat*}
If $x$ and $y$ commute, then $\varphi((xy)^nx^{-n}y^{-n})=0$ for every $n$, giving the first statement.

Assume now that $\varphi$ vanishes on single commutators. As $(xy)^nx^{-n}y^{-n}$ can be expressed as the product of
$\frac{n}{2}+c$ commutators, see~\cite{B}, the right hand side of the formula is bounded above by $\frac{1}{2}D(\varphi )$.
However, taking the supremum of the left hand side over all $x$ and $y$, we get the defect $D(\varphi )$. Thus
$D(\varphi )\leq \frac{1}{2}D(\varphi )$, showing that the defect vanishes, and $\varphi$ is a homomorphism.
\end{proof}

Here is the main mechanism for the vanishing theorems.
\begin{proposition}\label{p:hom}
Let $\Lambda\subset\Gamma$ be a subgroup with the property that there is an arbitrarily large number of conjugate embeddings
$\Lambda_i\subset\Gamma$ of $\Lambda$ in $\Gamma$ with the property that elements of $\Lambda_i$ and of $\Lambda_j$ commute
with each other in $\Gamma$ whenever $i\neq j$. Then every homogeneous quasi-homomorphism on $\Gamma$ restricts to $\Lambda$
as a homomorphism.
\end{proposition}
Note that homogeneous quasi-homomorphisms are constant on conjugacy classes, so that the restriction of $\varphi$ to $\Lambda_i$
is independent of $i$.
\begin{proof}
By the second part of Lemma~\ref{l:pre} we only have to prove that $\varphi([x,y])=0$ for any $x,y\in\Lambda$. Let $x_i$, $y_i\in\Gamma$
be the images of $x$ and $y$ under the embedding $\Lambda_i\subset\Gamma$.

We have the following equalities:
\begin{alignat*}{1}
n\varphi([x,y]) &=\varphi([x_1,y_1])+\ldots +\varphi([x_n,y_n]) \\
&=\varphi([x_1,y_1]\ldots [x_n,y_n]) \\
&=\varphi([x_1\ldots x_n,y_1\ldots y_n]) \ ,
\end{alignat*}
where the first one comes from the constancy of $\varphi$ on conjugacy classes, the second is the first part of Lemma~\ref{l:pre} applied to the
commuting embeddings, and the third follows directly from the commuting property of the embeddings $\Lambda_i\subset\Gamma$.
On the right hand side $\varphi$ is applied to a single commutator in $\Gamma$, and therefore the right hand side is bounded in absolute
value by the defect of $\varphi$ on $\Gamma$. However, if $\varphi([x,y])\neq 0$, then the left hand side is unbounded because by
assumption we can make $n$ arbitrarily large. Thus $\varphi([x,y])=0$ for all $x,y\in\Lambda$.
\end{proof}

There are several ways in which this mechanism can be  applied to obtain the vanishing of the  stable commutator length in
various groups. We try to give a general statement, in the hope of unifying several different applications of the argument.
\begin{theorem}\label{t:alg}
Let $\Gamma$ be a group in which every element can be decomposed as a product of some fixed number $k$ of elements contained
in distinguished subgroups $\Lambda\subset\Gamma$. If each $\Lambda$ is perfect and has the property in Proposition~\ref{p:hom}, then
the stable commutator length of $\Gamma$ vanishes.
\end{theorem}
\begin{proof}
As the subgroups $\Lambda$ are assumed perfect, the restrictions of quasi-homomorphisms on $\Gamma$ to $\Lambda$ vanish,
because by Proposition~\ref{p:hom} they are homomorphisms. Therefore the value of a quasi-homomorphism on every element of $\Gamma$
is bounded by $k-1$ times the defect. But every bounded homogeneous quasi-homomorphism is trivial.
\end{proof}

\section{Applications to discrete groups}\label{s:discrete}

\subsection{The stable mapping class group}

Let $\Gamma_g^1$ be the group of isotopy classes of diffeomorphisms with compact support in the interior of a compact surface $\Sigma_g^1$
of genus $g$ with one boundary component. Attaching a two-holed torus along the boundary defines the stabilization homomorphism
$\Gamma_g^1\rightarrow\Gamma_{g+1}^1$. The stable mapping class group $\Gamma_{\infty}$ is defined as the limit
$$
\Gamma_{\infty} = \lim_{\stackrel{\longrightarrow}{g}}\Gamma_g^1 \ .
$$
For $g\geq 3$ the groups $\Gamma_g^1$ are perfect, see~\cite{Powell}, hence $\Gamma_{\infty}$ is also perfect.
Recall that in~\cite{EK} it was proved that the stable commutator length is non-trivial on every $\Gamma_g^1$ with $g\geq 2$,
see also~\cite{BF,BK,CF,qmorph}. In contrast with this we have:
\begin{theorem}\label{t:sMCG}
The stable commutator length for $\Gamma_{\infty}$ vanishes identically.
\end{theorem}
\begin{proof}
We apply Theorem~\ref{t:alg} with $k=1$. The subgroups $\Lambda$ are the images of the $\Gamma_g^1$. In detail, every
element of $\Gamma_{\infty}$ is in the image of some $\Gamma_g^1=\Lambda$. This has arbitrarily large numbers of commuting
conjugate embeddings in $\Gamma_{\infty}$ given by taking the boundary connected sum of an arbitrarily large number of copies of
the surface of genus $g$ with one boundary component. Any homogeneous quasi-homomorphism on $\Gamma_{\infty}$
restricts to (the image of) $\Gamma_g^1$ as a homomorphism. However, for $g\geq 3$ this group is perfect, and so the homomorphism
vanishes.
\end{proof}
\begin{remark}
Theorem~\ref{t:sMCG} shows that the second bounded cohomology of mapping class groups does not stabilize.
This contrasts sharply with the Harer stability theorem~\cite{Harer} for the ordinary group cohomology.
\end{remark}
\begin{remark}
Theorem~\ref{t:sMCG} fits in nicely with the form of the estimates for the stable commutator length obtained in~\cite{EK,BK,qmorph}.
The lower bounds given there for the stable commutator length of specific elements in $\Gamma_g^1$ go to zero for $g\rightarrow\infty$.
A similar phenomenon seems to appear in the work of Calegari and Fujiwara~\cite{CF}.
\end{remark}
\begin{remark}
The discussion in this subsection also applies to the stable mapping class groups for surfaces with several boundary components.
\end{remark}

\subsection{The braid group on infinitely many strands}

Let $B_n$ be the Artin braid group on $n$ strands. Adding strands defines injective stabilization homomorphisms
$B_n\longrightarrow B_{n+1}$. The braid  group on infinitely many strands is
$$
B_{\infty} = \bigcup_n B_n \ .
$$
\begin{theorem}\label{t:braids}
Any homogeneous quasi-homomorphism on the infinite braid group $B_{\infty}$ is a homomorphism.
\end{theorem}
\begin{proof}
We apply Proposition~\ref{p:hom}. The subgroups $\Lambda$ are the $B_n$. Any two $x$, $y\in B_{\infty}$
are contained in some finite $B_n=\Lambda$. This has arbitrarily large numbers of commuting conjugate embeddings in
$B_{\infty}$ given by considering braids on $n$ strands placed side by side. Any homogeneous quasi-homomorphism
$\varphi$ on $B_{\infty}$ therefore restricts to $B_n$ as a homomorphism. Thus $\varphi (xy)=\varphi (x) +\varphi (y)$,
showing that $\varphi$ is a homomorphism on $B_{\infty}$.
\end{proof}
We have chosen this formulation of the result because the braid groups have infinite Abelianizations, so that it does
not make sense to speak of the stable commutator lengths of elements. The stabilization is compatible with Abelianization,
so that $B_{\infty}$, like $B_n$ for finite $n$, has infinite cyclic Abelianization. The conclusion of Theorem~\ref{t:braids}
is that every homogeneous quasi-homomorphism is a constant multiple of the Abelianization homomorphism.

For finite $n$, the  braid groups do admit non-trivial homogeneous quasi-homomorphisms that are not proportional to
the Abelianization map $B_n\longrightarrow\Z$, see~\cite{Baader,GG}. Thus Theorem~\ref{t:braids} shows that the bounded
cohomology of braid groups does not stabilize. The quotients $B_n/C$ of finite braid groups modulo their centers have
finite Abelianizations, so that all elements have powers that are products of commutators. Baader~\cite{Baader} proves
some lower bounds for the stable commutator length of certain elements in $B_n/C$. These lower bounds tend to zero as $n\rightarrow\infty$.
This of course fits with Theorem~\ref{t:braids}. The corresponding homogeneous quasi-homomorphisms defined on $B_n/C$,
and, by composition, on $B_n$, do not extend to the braid group on infinitely many strands.

\subsection{Automorphism groups of free groups}

The canonical homomorphisms $F_n\longrightarrow F_n\star\Z=F_{n+1}$ give rise to injective homomorphisms
 $\Aut(F_n)\longrightarrow \Aut(F_{n+1})$. Thus we define
$$
\Aut_{\infty}(F) = \bigcup_{n}\Aut(F_n) \ .
$$
Note that this is smaller than $\Aut(F_{\infty})$.
\begin{theorem}\label{t:sAut}
The stable commutator length for $\Aut_{\infty}(F)$ vanishes identically.
\end{theorem}
\begin{proof}
The Abelianization of $\Aut(F_n)$ is of order $2$, and is stable. Thus, up to taking squares, all elements in $\Aut(F_n)$
and in $\Aut_{\infty}(F)$ are products of commutators.
We again apply Proposition~\ref{p:hom}. The subgroups $\Lambda$ are the $\Aut(F_n)$. In detail, every
element of $\Aut_{\infty}(F)$ is contained in some $\Aut(F_n)=\Lambda$. This has arbitrarily large numbers of commuting
conjugate embeddings in $\Aut_{\infty}(F)$ given by embedding $F_n\star F_n\star\ldots\star F_n$ in some large $F_N$.
Although the elements in the different copies of $F_n$ do not commute, the induced embeddings of $\Aut(F_n)$ in
$\Aut_{\infty}(F)$ do commute and are conjugate to each other. Any homogeneous quasi-homomorphism on $\Aut_{\infty}(F)$
restricts to $\Aut(F_n)$ as a homomorphism. As the Abelianization of $\Aut(F_n)$ is finite, the homomorphism is trivial.
\end{proof}

One could also consider the outer automorphism groups $\Out(F_n)$, and of course quasi-ho\-mo\-mor\-phisms
on these induce quasi-homomorphisms on $\Aut(F_n)$ by composition with the projection. However, there is no natural
way  of stabilizing the outer automorphism groups.

Unlike for mapping class groups, no unbounded quasi-ho\-mo\-mor\-phisms are known on the automorphism groups of
free groups. The groups $\Aut(F_n)$ are analogous to the so-called extended mapping class groups, consisting of the
isotopy classes of all diffeomorphisms of surfaces. The usual mapping class groups we considered in Theorem~\ref{t:sMCG}
are index $2$ subgroups of the extended mapping class groups. A single Dehn twist has non-zero stable commutator length
in a mapping class group~\cite{EK,BK,qmorph}, but is conjugate to its inverse in the extended mapping class group~\cite{NAC},
so that in this latter group its stable commutator length must vanish. By analogy with this phenomenon, it may be more promising
to look for quasi-homomorphisms on the special automorphism group $S\Aut(F_n)$, rather than
on $\Aut (F_n)$, where by  $S\Aut(F_n)$ we denote the automorphisms preserving an orientation on the Abelianization $\Z^n$
of $F_n$. Theorem~\ref{t:sAut} also applies to the  stabilized special automorphism group $S\Aut_{\infty}(F)$.

\section{Applications to diffeomorphism groups}\label{s:diff}

In this section we apply the algebraic vanishing result to some groups of diffeomorphisms. If $M$ is a smooth closed
manifold, we consider the identity component $G=\Diff_0(M)$ of the full diffeomorphism group. If $M$ is open, we consider
the identity component $G=\Diff_0^c(M)$ of the group of compactly supported diffeomorphisms. (Sometimes this notation
is redundant because the group of compactly supported diffeomorphisms may be connected.) In both cases we assume
that the diffeomorphisms are of class $C^r$ with $1\leq r\leq\infty$ and $r\neq 1+\dim (M)$. The classical results of Herman,
Thurston, Epstein and Mather then ensure that $G$ is a perfect group; see~\cite{Ban} and the references quoted there.

\subsection{The full diffeomorphism groups}

We would like to prove that the stable commutator length vanishes on the diffeomorphism groups. Unfortunately, we can only
achieve this goal in the following special cases:
\begin{theorem}\label{t:easy}
The stable commutator length vanishes for the following diffeomorphism groups:
\begin{enumerate}
\item $\Diff_0^c(D)$ for a ball $D$,
\item $\Diff_0^c(D\times M)$ for any $M$ and a ball $D$ of positive dimension,
\item $\Diff_0(S^n)$ for any sphere,
\item $\Diff_0(\Sigma^n)$ for any exotic sphere of dimension $\neq 4$.
\end{enumerate}
\end{theorem}
\begin{proof}
Let $D$ be an $n$-dimensional ball, and $G=\Diff_0^c(D)$ the group of diffeomorphisms of $D$ with compact support in the interior of $D$.
Fix an exhaustion of $D$ by smaller nested balls $D_i$ so that each $D_i$ has closure contained in the interior of $D_{i+1}$.
If $G_i$ is the group of diffeomorphisms of $D_i$ with compact support in the interior of $D_i$, then we have injective homomorphisms
$G_i\rightarrow G_{i+1}$ induced by the  inclusion $D_i\subset D_{i+1}$, and
$$
G = \bigcup_{i} G_i \ .
$$
We now apply Theorem~\ref{t:alg} with $k=1$. The subgroups $\Lambda$ are the $G_i$. In detail, every element of
$G$ is contained in some $G_i$. This has arbitrarily large numbers of commuting conjugate embeddings in $G$,
because we can embed arbitrarily large numbers of disjoint (smaller) balls in the annulus $D\setminus D_i$. Any
homogeneous quasi-homomorphism on $G$ then restricts to $G_i$ as a homomorphism. But this group is perfect,
and so the homomorphism vanishes.
This completes the proof of the first claim.
The same argument works for $D\times M$ using the exhaustion by $D_i\times M$.

Now let $G=\Diff_0(S^n)$.
We can compose any $g\in G$ with a suitable $f_1^{-1}\in G$ supported in an open ball to achieve that $f_1^{-1} g$ has a fixed point.
It follows that we can write $f_1^{-1} g=f_2 f_3$, with $f_3$ having support in a ball around the fixed point, and $f_2$ having support
in the complement of a (smaller) ball around the fixed point. Because the manifold is a sphere, this complement is again a ball.
Using $g=f_1 f_2 f_3$, we apply Theorem~\ref{t:alg} with $k=3$. Let $\Lambda=\Diff_c(D)$ be the group of compactly supported
diffeomorphisms of a ball. This is perfect and admits infinitely many conjugate commuting embeddings in $G$.

The argument given for spheres also works for exotic spheres in dimensions $n\neq 4$, because they are all twisted spheres
obtained from two standard balls by gluing along the boundary~\cite{Kos}.
\end{proof}

After this paper had been submitted, Tsuboi~\cite{T} proved a much stronger result. He proved that $\Diff_0(M)$ is uniformly
perfect for every closed manifold, with the possible exception of even-dimensional manifolds having no handle decomposition
without handles of middle dimension. The simplest $M$ for which the question is unresolved, is $T^2$. Because of Tsuboi's
results, I have removed some material from the original version of this paper, which was concerned with the continuity, with
respect to the $C^0$ topology, of potential homogeneous quasi-homomorphisms on diffeomorphism groups. That discussion is
rather delicate technically, and I hope to return to it elsewhere, unless it is rendered completely empty by a generalization of
Tsuboi's work.

\subsection{Diffeomorphism groups preserving a symplectic or volume form}

The first statement in Theorem~\ref{t:easy} can be phrased for the compactly supported diffeomorphism group of Euclidean
space $\R^n$ instead of a ball $D$. However, the two cases are rather different if we consider diffeomorphism groups
preserving a symplectic or volume form $\omega$, and we fix our conventions so that $\R^n$ has infinite (symplectic) volume
and a ball has finite volume.

Let $G=\Diff_0^c(\R^n,\omega)$ be the identity component of the group of compactly supported diffeomorphisms
preserving $\omega$. This is not perfect because the Calabi invariant
\begin{equation}\label{eq:Cal}
\Cal\colon \Diff_0^c(\R^n,\omega)\longrightarrow\R
\end{equation}
is a non-trivial homomorphism. (See for example~\cite{Ban} for the definition of $\Cal$.)
By results of Thurston and Banyaga, see~\cite{Ban}, its kernel is a perfect group.
It follows that $\Cal$ is the Abelianization homomorphism. So far there is no difference between the finite and
infinite volume cases, and we have the same statements if we replace $(\R^n,\omega)$ by a ball of finite volume.
\begin{theorem}\label{t:Ham}
The stable commutator length vanishes on the kernel of the Calabi homomorphism in $\Diff_0^c(\R^n,\omega)$.
On $\Diff_0^c(\R^n,\omega)$ every homogeneous quasi-homomorphism is a constant multiple of the Calabi invariant.
\end{theorem}
\begin{proof}
As in the previous proof we can write the kernel of $\Cal$ as
$$
G = \bigcup_{i} G_i \ ,
$$
with each $G_i$ consisting of those elements of $G$ supported in a ball of radius $i$, say. Then each element of $G$
is contained in some $G_i$, but this $G_i$ has arbitrarily large numbers of commuting conjugate embeddings given by
disjoint Hamiltonian displacements of $D_i$ in $\R^n$. Therefore we can apply Theorem~\ref{t:alg} to conclude that every
homogeneous quasi-homomorphism vanishes on $\Ker\Cal$.

Applying the same argument to $G=\Diff_0^c(\R^n,\omega)$ with each $G_i$ equal to $\Diff_0^c(D_i,\omega)$, we conclude that the
restriction to $G_i$ of any homogeneous quasi-ho\-mo\-mor\-phism $\varphi$ on $G$ is a homomorphism, and is therefore a constant multiple
of the Calabi invariant. But any two balls in $\R^n$ are both contained in some larger ball, and therefore the restriction
of $\varphi$ is the same multiple of $\Cal$ on all balls.
\end{proof}
This argument clearly does not work in the finite volume case because the number of commuting conjugate embeddings
one can construct is bounded in terms of the available volume. The abstract isomorphism type of the group of symplectic
or volume-preserving diffeomorphisms of a ball is always the same. In particular it does not depend on the size or volume
of the ball. Therefore one can always find arbitrarily large numbers of commuting embeddings of such a group in the corresponding
group of a larger ball, or in itself, but these embeddings are conjugate only in the full diffeomorphism group, and not
in the diffeomorphism group preserving $\omega$.

\section{Comparison with results on quasi-norms}\label{s:BIP}

In this section we compare our results on the non-existence of homogeneous quasi-ho\-mo\-mor\-phisms on various
stable groups to results about existence and non-existence of quasi-norms. The following definitions are due to
Burago, Ivanov and Polterovich~\cite{BIP}.
\begin{definition}
A function $q\colon G\longrightarrow\R$ on a group is called a quasi-norm if it is almost subadditive, almost
invariant under conjugacy, and unbounded.

A group $G$ is called unbounded if it admits a quasi-norm, and is called bounded otherwise.
\end{definition}
Here almost subadditive and almost invariant under conjugation means that there is a constant $c$ such that
$$
q(xy)\leq q(x)+q(y)+c
$$
and
$$
\vert q(xyx^{-1})-q(y)\vert\leq c
$$
for all $x$ and $y\in G$.

If $\varphi$ is a non-trivial homogeneous quasi-homomorphism, then its absolute value is a quasi-norm.
Therefore boundedness of a group is an even stronger property than the non-existence of homogeneous
quasi-homomorphisms. A bounded group has finite Abelianization and vanishing stable commutator length.
If the Abelianization is trivial, then the group is uniformely perfect~\cite{BIP}.

Burago, Ivanov and Polterovich~\cite{BIP} prove boundedness of many diffeomorphism groups by an
argument that is rather stronger than ours. In particular, they prove boundedness for the groups we considered
in Theorem~\ref{t:easy}. An easy example that illustrates the difference between the vanishing mechanism
in~\cite{BIP} and the one considered here is the following.
\begin{example}
Consider the group $G=\Diff_0^c(\R^n,\omega)$ from Theorem~\ref{t:Ham}. Then the function $q$ which
assigns to every $g\in G$ the (symplectic) volume of its support is a quasi-norm which is non-trivial on
the kernel of the Calabi homomorphism. Therefore this kernel is an unbounded group which nevertheless
has vanishing stable commutator length.

This also brings out another aspect of the difference between the finite and infinite volume cases mentioned earlier.
If we take $G=\Diff_0^c(D,\omega)$ with $D$ of finite (symplectic) volume, then $q$ is bounded and therefore not
a quasi-norm.
\end{example}

This idea of a quasi-norm defined by support size actually applies to all the discrete groups we considered in Section~\ref{s:discrete}:
\begin{theorem}
The following groups are unbounded:
\begin{enumerate}
\item the stable mapping class group $\Gamma_{\infty}$,
\item the braid group $B_{\infty}$ on infinitely many strands,
\item the stable automorphism group of free groups $Aut_{\infty}(F)$, and
\item the stable special automorphism group of free groups $S\Aut_{\infty}(F)$.
\end{enumerate}
\end{theorem}
\begin{proof}
Perhaps the  case of the braid group is easiest to visualize. Define $q\colon B_{\infty}\longrightarrow\Z$
by sending a braid $x$ to the smallest number $k$ of strands needed to express it. These need not be the
first $k$ strands, but can be any $k$ strands. The function $q$ is clearly invariant under conjugation and
is unbounded. It is also subadditive, because a set of strands used to express $x$ and a set of strands
used to express $y$ can together be used to  express $xy$. Thus $q$ is a quasi-norm.

On the stable mapping class group define $q\colon\Gamma_{\infty}\longrightarrow\Z$ by mapping an element
$x$ to the smallest genus $g$ of a compact surface with one boundary component on which a diffeomorphism representing
the isotopy class $x$ can be supported. Again this does not have to be the embedding of $\Sigma^1_g$ from
the definition of the stabilization procedure, but can be any such subsurface of the infinite genus surface.
By definition, this function is conjugacy-invariant. It is also not hard to see
that it is unbounded, for example by considering the action of mapping classes on homology. To check (almost)
subadditivity, one has to consider several cases depending on how compact surfaces with one boundary
component supporting representatives for $x$ and $y$ sit in the infinite genus surface. In all cases the union
of these two subsurfaces can be enlarged slightly, without increasing the genus, to obtain a compact surface
with one boundary component supporting a representative for $xy$. The genus of this surface is at most
$q(x)+q(y)$. Therefore $q$ is in fact subadditive and a quasi-norm.

The argument for $Aut_{\infty}(F)$ and $S\Aut_{\infty}(F)$ is completely analogous.
\end{proof}

\bigskip
\noindent
{\bf Acknowledgements:}
I am very grateful to all the colleagues with whom I discussed the subject matter of this paper, including in particular
N.~A'Campo, D.~Burago, D.~Calegari, K.~Fujiwara, V.~Markovic, S.~Matsumoto, S.~Morita, L.~Polterovich, P.~Py and
T.~Tsuboi. Special thanks are due to P.~Py for an inspiring conversation
about the work of Burago, Ivanov and Polterovich, which got me started in the first place.

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\begin{document}

\title{Geometry of syzygies via Poncelet varieties }
\author{Giovanna Ilardi, Paola Supino, Jean Vall\`es}

\address{Dip. Matematica e Applicazioni, Univ. di Napoli,
via Cinthia, 80126 Napoli (Italy);
 {email:  giovanna.ilardi@dma.unina.it } }

\address{Dip. Matematica, Univ. RomaTre,
Largo L. Murialdo, 00146 Roma (Italy);
 {email: psupino@uniroma3.it } }

\address{Laboratoire de Mathematiques appliques
de Pau et de Pays de l'Adour,
 Avenue de l'Universit\'e
 64000 Pau(France);
 {email:  jean.valles@univ-pau.fr } }

\maketitle

\begin{abstract}
 We consider the  Grassmannian $\mathbb{G}r(k,n)$ of $(k+1)$-dimensional linear
subspaces of
 $V_n=H^0({\P^1},\O_{\P^1}(n))$. We define
  $\frak{X}_{k,r,d}$
as the classifying  space of the
$k$-dimensional linear systems  of degree $n$ on $\P^1$ whose
basis realize a fixed number of polynomial relations of fixed
degree, say a fixed number of syzygies of a certain degree.
The first result of this paper is the computation of the dimension of   $\frak{X}_{k,r,d}$.
In the second part we make a link between   $\frak{X}_{k,r,d}$ and the  Poncelet varieties.
 In particular, we prove that the existence of linear syzygies
implies  the existence of singularities on the Poncelet varieties.
\end{abstract}

\pagestyle{myheadings} \thispagestyle{plain} \markboth{\today}{}

\section{Introduction and set up}
In this paper we are interested in linear systems $\Lambda$ on $\P^1$ of degree
 $n$ and  projective dimension $k$ (so, from now we assume that $n>k$), more particulary in those
having an  algebraic limitation, namely   the syzygies.
A  {\em syzygy} of degree $d$ for $\Lambda$ is a
$(k+1)$-uple of homogeneous forms of degree $d$, $(g_0,..,g_k)$,
 such that
$\sum_{i=0}^k g_if_i=0$.
We say that $\Lambda$ has $r$ syzygies of degree $d$ if there
exist $r$ linearly independent $k+1$-uples $(g_{0,j},..,
g_{k,j})$, where $0\le  j \le r$.

  The locus
$\frak{X}_{k,r,d}$ lives inside $\mathbb{G}r(k,n)$ in a natural way:
\begin{align}\label{}
\frak{X}_{k,r,d}\colon &= \big\{ \Lambda\in \G r(k, n) \text { having }
 \notag\\
 & \text {at least $r$ syzygies of degree $d$} \big\}. \notag
\end{align}

The first result of this paper is  the computation of the dimension of
$\frak{X}_{k,r,d}$.
The subvarieties
$\frak{X}_{k,r,d}$
turn out to be determinantal varieties for a suitable map of
vector bundles on the Grassmannian. This  extends  the main result (corollary 4.4) in \cite{IS}, where the computation was only proved
in the  $k=1$ case.
In a second part we  give a geometric interpretation of the varieties
$\frak{X}_{k,r,d}$ in terms of Poncelet varieties. These varieties were
introduced by Trautmann
in \cite{T}, but, except for the case of curve, they have not been actually studied.
We prove that the existence of linear syzygies
implies  the existence of singularities on the Poncelet varieties.

\section{The dimension of the varieties $\frak{X}_{k,r,d}$}

Let  $\Lambda$ be a linear system on $\P^1$ of degree
 $n$ and dimension $k$. We choose
$u, v$ a system of coordinates on $\P^1$, and denote by $V_n$
the  $n+1$-dimensional vector space $H^0({\P^1},\O_{\P^1}(n))$ of
binary homogeneous forms of degree $n$, otherwise said binary {\em
quantics} of degree $n$. A base for $V_n$ is $x_0,\dots,x_n$ where
$x_i=u^iv^{n-i}$. \\
Choose a linear subspace $\Lambda$ of $\P V_n$, and let
$\{f_0,\dots,f_k\}$ be
 a basis for $\Lambda$.
 It   defines  a  morphism of vector bundles on
$\P^1$
\begin{equation}\label{phiLambda}
 \begin{CD}
 \phi_{\Lambda} :\, \Lambda\otimes \O_{\P^1} @>>> \O_{\P^1}(n)
\end{CD}
\end{equation}
which is surjective when $\Lambda$ has no base points.
The system $\Lambda$ gives a  map from $\P^1$ to $\P^k$
\begin{equation}\label{fLambda}
f_{\Lambda} :\, \P^1 \stackrel{\Lambda} \longrightarrow \P^k.
\end{equation}
Its image is a rational curve of degree
$n$ when $ \phi_{\Lambda}$ is surjective, and less than $n$ when $\Lambda$ has base points.
For general $\Lambda$, $\phi_{\Lambda}$  is a surjective morphism
of vector bundles on $\P^1$, thus there is  an exact sequence
\begin{equation}\label{ELambda}
 \begin{CD}
0 @>>> E_{\Lambda} @>>>\Lambda\otimes \O_{\P^1} @>>>
\O_{\P^1}(n)@>>> 0,
\end{CD}
\end{equation}
where the kernel is a vector bundle $E_{\Lambda}$ on  $\P^1$ of
rank $k$ and degree $-n$.\\

The short exact sequence
(\ref{ELambda}) twisted by $\O_{\P^1}(d)$
\[
 \begin{CD}
0@>>>E_{\Lambda}(d)@>>>\Lambda\otimes \O_{\P^1}(d) @>>>
\O_{\P^1}(n+d)@>>> 0
\end{CD}
\]
suggests that $\Lambda$ has exactly $r$ independent syzygies of
degree $d$ if and only if $h^0(E_{\Lambda}(d))=r$.
\\
Since any $k$-bundle of degree $-n$ splits as
$\O_{\P^1}(-a_0)\oplus\dots\oplus
\O_{\P^1}(-a_k)$ for suitable positive $a_0\leq\dots\leq a_k$ such
that $a_0+\dots+a_k= n$, one can stratify the varieties $\frak{X}_{k,r,d}$ by all possible
splitting  of the integer $n$ in $k$ pieces. This point of view is developed by Ramella  in \cite{R} to sudy  the stratification of the Hilbert scheme of rational curves $C$ embedded in projective space, by the splitting of the restriction of the tangent bundle to $C$, and by the splitting of the normal bundle. We will use this point of view
(in thm \ref{dime}) in order to prove that the dimension is the expected one.
\begin{exmp}
When $n=5$ and $k=3$
the only possible cases for $E_{\Lambda}$ are:
\[
E_{\Lambda}=\O_{\P^1}(-1)\oplus \O_{\P^1}(-2)\oplus
\O_{\P^1}(-2),
\] which is the general case, and
\[
E_{\Lambda}=\O_{\P^1}(-1)\oplus \O_{\P^1}(-1)\oplus
\O_{\P^1}(-3).
\]
In the general case $h^0(E_{\Lambda}(1))=1$, in the other $h^0(E_{\Lambda}(1))=2$, thus the general stratum is
$\frak{X}_{3,1,1}=\G r(3,5)$, and the stratum $\frak{X}_{3,2,1}$ is strictly contained in $\G r(3,5)$.
\end{exmp}
Note that if  $n<2k$ the general splitting has the first term $a_0=1$, thus $h^0(E_{\Lambda}(1))\geq 1$,
which implies that  $\frak{X}_{k,1,1}^n= \G r(k,n)$, that
is: there always exists a linear syzygy.\\

\begin{thm}\label{dime}  $\mathrm{codim} (\frak{X}_{k,r,d},\G r(k,n))=(dk+k-n+r)r.$
Moreover the varieties $\frak{X}_{k,r,d}$ are Cohen-Macaulay with singular locus
$\frak{X}_{k+1,r,d}$.
\end{thm}
\begin{proof}

Consider the
  universal vector bundle $\mathcal{U}=\mathcal{U}_{\G r(k,n)}$ on the Grassmannian
\[
\mathcal{U}=\{ (f,\Lambda)\in V_n\times \G r(k,n) \mid f\in\Lambda
\},
 \]
and the canonical map of vector bundles
$$
 \mathcal{U} \hookrightarrow V_n\otimes \O_{\G r(k,n)}
 $$
On  the product variety $\G r(k,n)\times \P^1$ let  $p$ and $q$ be
the two projections
\[
 \begin{CD}
\G r(k,n) @<p<<  \G r(k,n)\times \P^1 @>q>> \P^1.
\end{CD}
\]
 The  map  $p$ composed
with the evaluation map gives on $\G r(k,n)\times \P^1$
\[
 \begin{CD}
 p^{*}\mathcal{U}
  @>>> V_n\otimes \O_{\G r(k,n)\times \P^1}@>>> q^{*}
  \O_{\P^1}(n).
\end{CD}
\]
For all $d$ we also have a morphism
\[
 \begin{CD}
 p^{*}\mathcal{U}\otimes  q^{*} \O_{\P^1}(d)
  @>>> q^{*} \O_{\P^1}(n+d).
\end{CD}
\]
If we take now the direct image $p_*$ on the Grassmannian,we
obtain
\[
 \begin{CD}
 \mathcal{U} \otimes V_d
  @>>> V_{n+d}
\end{CD}
\]
which is just the relative version of our map $ \phi_{\Lambda}$ in (\ref{phiLambda})
twisted by $H^0(\O_{\P^1}(d))$
\[
\Lambda \otimes H^0(\O_{\P^1}(d)) \rightarrow H^0(\O_{\P^1}(n+d)).
\]
 Therefore
\[
\frak{X}_{k,r,d}=\lbrace \Lambda \in \G r(k,n) \mid
rk(\Phi_{\Lambda})\le (k+1)(d+1)-r\rbrace.
\]

 Applying  Thom-Porteous formula we compute
the expected codimension for
$\frak{X}_{k,r,d}$ as
$r(n+r-(d+1)k)$.
\\
It is a classical fact that when the codimension is exactly the expected one than the Chow class is
$det(c_{n-k(d+1)+r+j-i}(\mathbb{C}^{(d+1)(n+d+1)}\otimes\mathcal{U}^*))$.
\\
We  compute now the codimension of the tangent space in a generic
point ${\Lambda}$. Consider its associated bundle $E_{\Lambda}$, by  genericity we have
\[
E_{\Lambda}= \O_{\P^1}^r(-d) \oplus \O_{\P^1}^{k-r-B}(-A)
\oplus \O_{\P^1}^{B}(-A-1)
\]
 where $A$ and $B$ are uniquely
defined by
\[
(n-dr)=A(k-r)+B, \,\, 0\le B<k-r
\]
and, by
hypothesis on the syzygy, $A>d$. The
codimension of the tangent space in the point ${\Lambda}$ is then
$h^1(E_{\Lambda}\otimes E_{\Lambda}^{\vee})$.
 Since
\[
 E_{\Lambda}\otimes E_{\Lambda}^{\vee}=
\O_{\P^1}^{r(k-r-B)}(d-A) \oplus \O_{\P^1}^{rB}(d-A-1) \oplus R,
\]
where $R$ is a suitable bundle  with $h^1(R)=0$, we have $h^1(E_{\Lambda}\otimes E_{\Lambda}^{\vee})=r(n+r-(d+1)k)$   which is the expected codimension.
\end{proof}

\section{Geometric description as Poncelet  varieties}
At the end
of  his paper  \cite{T}, Trautmann has introduced a generalization of Poncelet curves, namely the Poncelet varieties, in higher dimension. Those are in bijective correspondence with the points of the Grassmannian.
The aim of this part is to describe the points of $\frak{X}_{k,r,d}$  as Poncelet varieties. In particular we will show that Poncelet varieties corresponding to  $\frak{X}_{k,1,1}$  are singular (see theorem \ref{teorema}).
Following \cite{V}, we define the Poncelet varieties  as determinant of sections of
Schwarzenberger bundles, therefore we start by  recalling the definition of Schwarzenberger bundle and we
describe the zero locus of their section ( see proposition \ref{prozero}).
\subsection{Schwarzenberger bundles}
We denote by     $(x_i=u^{i}v^{k+1-i})$,
$(y_j=u^{j}v^{n-k-1-j})$, and $(z_{l}=u^{l}v^{n-l})$ the basis of
$V_{k+1}, V_{n-k-1}
\,\, \textrm{and} \,\,  V_{n}$ respectively.
The multiplication of homogeneous polynomials in two variables
\[
 \begin{array}{c}
 V_{k+1}\otimes V_{n-k-1} \stackrel{\phi_{\times}} \longrightarrow V_{n},\\
(u^{i}v^{k+1-i}, u^{n-k-1-j}v^{j})\longmapsto   u^{n-j+i}v^{n-i+j}
\end{array}
\]
induces the following bundle on $\P V_n$:
\begin{equation}\label{SCH}
\begin{CD}
0 @>>> V_{n-k-1}\otimes \O_{\P V_{k+1}}(-1)@>A>>  V_{n}
\otimes \O_{\P V_{k+1}} @>B>>
E_{n}@>>> 0
\end{CD}
\end{equation}
where
$$A^T=
\left (
             \begin{array}{cccccccc}
              x_0 & x_1 & \cdots  & x_{k+1} & &   & &\\
               & x_0 & x_1 &\cdots      &x_{k+1} &  & &  \\
               &    &  x_0 & x_1 &\cdots      &x_{k+1} &  &   \\
               &   &  &\ddots & \ddots&  &\ddots&\\
                & & &   &x_0 & x_1 & \cdots  & x_{k+1}
             \end{array}
      \right )$$
and
$B=(z_0,\dots,z_{n})$. These bundles are called {\em Schwarzenberger bundles} because for $n=2$ they were first introduced by Schwarzenberger in \cite{S}.

Into the product $\P V_{k+1}\times\P V_{n}$ the projective bundle $\P E_{n}$ is defined by the equations
$$
BA^T = ( \sum_{i=0}^{i=k+1}x_i z_{i+j}=0)_{j=0,\cdots,n-k-1}
$$
\\
Let $s\in H^0(E_{n})=V_{n}$ be a non zero section.
We describe now the zero locus $Z(s)$ geometrically.
\\
We call $C_{k+1}^{\vee}\in \P (V_{k+1}^*)$ the rational normal curve of the $k$-osculating planes to $C_{k+1}$.

\begin{prop}
\label{prozero}
Let   $s\in H^0(E_{n})$ be a non zero section and  $D_{n}(s)$
be the corresponding effective
divisor of degree $n$ on  $C_{k+1}^{\vee}$. We denote by  $Z(s)$ the
zero-scheme of $s$. Then,
$$a\in Z(s)   \Leftrightarrow a^{\vee}\cap C_{k+1}^{\vee} \subset
D_{n}(s).$$

More generally we have

$${\mathcal I}_{Z(s)}\subset {\frak m}_a^{r+1} \Leftrightarrow (a^{\vee})^{r+1}\cap
C_{k+1}^{\vee}\subset D_{n}(s).$$
\end{prop}

\begin{proof}
The section
$s$ corresponds to an
 hyperplane $H_s \subset \P(V_{n})$ or to an effective divisor of degree
$n$ on the
rational curve $C_{n}$, but also on $C_{k+1}^{\vee}$. The section $s$
$$
\begin{CD}
(E_{n})^{*} @>{s}>> {\O}_{\P V_{k+1}} @>>>
{\O}_{Z(s)}@>>>  0
\end{CD}
$$
induces a rational map
$
\P(V_{k+1}) \longrightarrow \P((E_{n})^{*})
$
which is not defined over the zero-scheme $Z(s)$.

We call $\pi$ the canonical projection $\pi:\P(E_{n})
\longrightarrow \P(V_{k+1})$.

Over a point
$a\in \P(V_{k+1})$ the fiber is
$$\pi^{-1}(a)=\left\{(\sum_{i=0}^{i=k+1}a_iz_{i+j}=0)_{j=0,\dots,n-k-1}\right\}=
\P(E_{n}(a
))=\P^{k}.$$
The rational map $
\P(V_{k+1}) \longrightarrow \P((E_{n})^{*})
$
sends a point $a\in \P(V_{k+1}) $ onto $H_s\cap \pi^{-1}(a)$ which is, in
general, a $\P^{k-1}$, i.e a point in
$\P((E_{n})^{*}(a))=\P^{k\vee}$. This map is not defined when
$\pi^{-1}(a) \subset H_s $. The hyperplane
$H_s$ cuts the rational curve $C_{n}$ along an effective divisor
$D_{n}(s)$ of length $n$.  When $D_{n}(s)$ is smooth, there are $\binom{n}{k+1}$
subschemes $D_{k+1} \subset D_{n}(s)$ of length $k+1$; they generate
the $k$-planes $k+1$ secant to $C_{n}$ which are contained in
$H_s$.  Since $\P V_1 \simeq C_{n}\simeq C_{k+1}^{\vee}$ it is clear that  $D_{n}(s)$ corresponds to degree $n$ divisor on
$C_{k+1}^{\vee}$. We still denote it by $D_{n}(s)$.

Then the zero-scheme $Z(s)$ is the set of points $a\in \P(V_{k+1})$
such that the divisor $a^{\vee}\cap
C_{k+1}^{\vee}$
 of degree $n$ belongs to $D_{n}(s)$. When $D_{n}(s)$ is smooth, we get $n$
osculating hyperplanes of
$C_{k+1}$ in $\P(V_{k+1})$. Every subset of $k+1$-osculating hyperplanes gives
a point in $\P(V_{k+1})$. These points
are   the zero-scheme of the section $s$. We have proved the first part of the
proposition.
\\
Assume  ${\mathcal I}_{Z(s)}\subset {\frak
m}_a^{r+1}$. Let $H_s$ the hyperplane corresponding to the
section $s$ and $D_{k+1}$ the divisor on $C_{k+1}^{\vee}$ corresponding  to
$a$. Then $\mid(r+1)D_n\mid^{\vee}\subset H_s$. This proves
$(a^{\vee})^{r+1}\cap C_{k+1}^{\vee}\subset D_{n}(s)$.
\\
On the other hand, the inclusion  $\mid(r+1)D_{k+1}\mid^{\vee}\subset H_s$  proves that
the exceptional divisor
 $\mid D_{k+1} \mid^{\vee}$ of  $\P({\mathcal I}_{Z(s)})$ appears with  multiplicity
$(r+1)$, it means that  ${\mathcal I}_{Z(s)}\subset {\frak m}_a^{r+1}$.
\end{proof}
\subsection{Poncelet varieties}
The group  $SL(2, {\mathbb{C}})$   acts on $\G r(k,n)$ and we have an equivariant morphism $\label{Plu}\G r(k,n)\hookrightarrow \P (\bigwedge ^{k+1} V_n)$.
The $SL(2)-$modules $\bigwedge ^{k+1} V_n$ and  $S^{k+1}V_{n-k}$     are isomorphic (see \cite {FH}, p. 160).
Moreover by Hermite reciprocity formula (see \cite {FH}, p. 82 and p.160), we have $S^{k+1}V_{n-k}\cong S^{n-k}V_{k+1}$. So the Plucker embedding  becomes

\begin{equation}\label{Tmap}
T\colon \G r(k,n)\hookrightarrow \P(S^{n-k}V_{k+1}).
\end{equation}
It  associates to $\Lambda=\langle f_0,\dots,f_k\rangle$ the hypersurface of degree $n-k$ in $\P V_{k+1}$ with equation $f_0\wedge\dots\wedge f_k=0$.
Indeed, since from \eqref{SCH}
\[
H^0(E_{n})\cong H^0(\O_{\P^1}(n)),\]
we consider $f_i$ as section of $E_n$.  We summarize these facts in
 the following commutative diagram:
\begin{equation}\label{comdia}
 \begin{CD}
@.@. \Lambda\otimes\O_{\P^{k+1}} @ =  \Lambda\otimes\O_{\P^{k+1}}\\
@. @. @VVV   @ VVV\\
 0@>>> V_{n-k-1}\otimes \O_{\P^{k+1}}(-1)@>>>V_n\otimes \O_{\P^{k+1}}  @>>> E_{n}@>>>0\\
  @.                           @V=VV              @ VVV         @VVV                         \\
0 @ >>> V_{n-k-1}\otimes \O_{\P^{k+1}}(-1)@>>>V_n/\Lambda\otimes \O_{\P^{k+1}}  @>>> {\mathcal L}@>>>0.
\end{CD}
\end{equation}
where the support of the sheaf ${\mathcal L}$ is given by $f_0\wedge\dots\wedge f_k=0$.
\begin{defn} The
  varieties defined as determinant of $k+1$ sections of  Schwarzenberger bundles
on $\P^{k+1}$ are called {\em Poncelet varieties} of dimension $k$.
\end{defn}
We point out that a section of $E_n$ corresponds to $n$ points on the rational normal curve $C_{k+1}$, and it vanishes along $\binom{n}{k+1}$
points in $\P^{k+1}$ which are the intersection points of $(k+1)$-osculating hyperplanes of $C_{k+1}$ in ${k+1}$ points chosen among the previous $n$ points. Hence these varieties are characterized by the nice following geometric fact: they contain the vertices of $k+1$ dimensional linear space of osculating polytopes. The case of curves is well-known since Darboux, but in higher dimension quite nothing exists in literature.

\begin{prop}
\label{Tpenc}
Let $\Lambda$  be a linear system in $\G r(k,n)$  with $d$ base points. Then $T(\Lambda)$
is the union of $d$ osculating hyperplanes to $C_{k+1}\subset \P V_{k+1}$ and a degree $(n-k-d)$ Poncelet variety of dimension $k$.
  \end{prop}
\begin{rem}In particular when $\Lambda$ is a pencil, a base point corresponds to a linear syzygy. The morphism $T$ sends  $
\frak{X}_{1,1,d}$ on the locus of Poncelet curves which are union
of a Poncelet curve of degree $d-1$ with
  $n-d$  tangent lines to  $C_2$.
\end{rem}
\begin{proof}
In the case of curves it is proved by Trautmann (see proposition 1.11 of \cite{T}).
In general assume that $\Lambda$ has $d<n$ base points and let $f=0$ be one equation for this base locus.
Then   we have a factorization:
\[
 \begin{CD}
\O_{\P^{1}}^{k+1}@>{\frac{\Lambda}{f}}>> \O_{\P^{1}}(n-d)\stackrel{f}\hookrightarrow  \O_{\P^{1}}(n).
\end{CD}
\]
The first arrow  gives the following vector bundle map:
\[
 \begin{CD}
 0@>>> \O^{k+1}_{\P(V_{k+1})}@>{\frac{\Lambda}{f}}>>E_{n-d}  @>>>K@>>>0
\end{CD},
\]
where $K$ is supported on a Poncelet variety of degree $n-k-d$.
The second arrow gives
\[
 \begin{CD}
 0@>>> E_{n-d}@>>>E_n  @>>> \oplus_{i=1}^{d}\O_{H_i}@>>>0
\end{CD}
\]
where $H_i$ is the osculating hyperplane to $C_{k+1}$ which corresponds to a base point on $C_{k+1}^{\vee}$.
Now the result follows from the following commutative diagram:
\[
 \begin{CD}
@. \O_{\P^{k+1}} @ =  \O_{\P^{k+1}}\\
@.  @V{\frac{\Lambda}{f}}VV   @ V{\Lambda}VV\\
 0@>>> E_{n-d}@>>>E_n  @>>> \oplus_ {i=1}^{d}\O_{H_i}@>>>0\\
  @.                           @VVV              @ VVV         @V=VV                         \\
0 @ >>> K@>>>{\mathcal L}  @>>> \oplus_{i=1}^{d} \O_{H_i}@>>>0,
\end{CD}
\]
where ${\mathcal L}$  is a sheaf supported by the Poncelet variety corresponding to $\Lambda$.
 \end{proof}

The following examples are done as suggested by the  commutative diagram  in  (\eqref{comdia}).

\begin{exmp}  In $\G r(2,4)$, $\frak{X}_{2,1,2}= \G r(2,4)$ and ${\mbox{\rm codim\,}} \frak{X}_{2,1,1}= 1$.

a) The net $\Lambda= \langle u^4,u^2v^2,v^4\rangle \in \frak{X}_{2,1,2}\setminus \frak{X}_{2,1,1}$ corresponds to the smooth Poncelet quadric
$$ \mathrm{det} \left (
             \begin{array}{cc}
              x_1 & x_3   \\
              x_0&x_2
             \end{array}
      \right )=x_1x_2-x_0x_3 = 0.
$$

b) The net $\Lambda= \langle u^4,u^3v,v^4\rangle \in \frak{X}_{2,1,1}$  has associated Poncelet cone
$$ \mathrm{det} \left (
             \begin{array}{cc}
              x_2 & x_3   \\
              x_1&x_2
             \end{array}
      \right )=x_2^2-x_1x_3=0.
$$
\end{exmp}
\begin{exmp}\label{fixpt}
The net $ \Lambda= \langle u^3v,uv^3,v^4\rangle$ has a base point, its associated Poncelet quadric is
$$ \mathrm{det} \left (
             \begin{array}{cc}
              x_0 & x_2   \\
              0&x_1
             \end{array}
      \right )=x_1x_0=0.
$$
This quadric consists in  two planes, one of which osculating the rational normal curve $C_3$.
\end{exmp}

\begin{thm}\label{teorema}
The Poncelet
variety associated to any element of $\frak{X}_{k,1,1}$  is singular. Moreover the Poncelet variety associated to a general element of
$\frak{X}_{k,1,1}$ contains $\binom{n-1}{k}$ lines and is singular in the  $\binom{n-1}{k+1}$ vertices of this configuration.
\end{thm}
\begin{proof}
Let $\Lambda$ have a syzygy of degree one : we can say that $\Lambda =\langle uf, vf,f_2,\dots,f_k\rangle$, where $f\in V_{n-1}$.
The curve $\Gamma$ in $\P^{k+1}$ defined by the determinant of the two sections $uf, vf$ of $E_n$ is obtained as follows.
\\
The pencil $\langle uf,vf\rangle$  defines $n-1$ fixed points and a moving point $p$ on the rational normal curve $C_{k+1}^{\vee}$.
Therefore, the curve $\Gamma$ in $\P^{k+1}$ consists in $\binom{n-1}{k}$ lines. They are the
lines of $k$-planes in $\P^{(k+1)\vee}$ passing through $p$ and each subset of $k$ points chosen among the fixed ones.
\\
Each $(k+1)$-uple of points on $C_{k+1}^{\vee}$, chosen among the $(n-1)$ fixed one, gives a point on $\Gamma\subset \P^{(k+1)}$. It is the  intersection point
of $(k+1)$ lines corresponding to each choice of $k$ points among the $(k+1)$. Since the $(k+1)$ points are distinct on the rational normal curve $C_{k+1}^{\vee}$, the $\P^{k-1}$ generated by each choice of $k$ points do not have a common point, dually,  the configuration of lines is not contained in an hyperplane.
Then, since the hypersurface defined by $\Lambda$ contains the curve $\Gamma$, it has singularities in the $\binom{n-1}{k+1}$ vertices of the configuration of lines $\Gamma$.
\end{proof}

\begin{exmp}
The net $\Lambda=\langle u^5+v^5,u^5-u^4v+u^3v^2-u^2v^3+uv^4,u^5-v^5\rangle$ has  a syzygy of degree $1$. It corresponds to
the Poncelet cubic surface
$$ \mathrm{det} \left (
             \begin{array}{ccc}
              x_1+x_2 & x_2+x_3 &x_3  \\
              x_0+x_1&x_1+x_2&x_3+x_2\\
              x_0&x_1+x_0&x_1+x_2\\
             \end{array}
      \right )=0.
$$
This surface has four singular points, which is the maximum number of ordinary double points. Thus it is a Cayley cubic.
\end{exmp}
It could be interesting to explore the link between the sygyzies of higher degrees and the singularities
of the associated Poncelet varieties.


TITLE: Why doesn't spaghetti knot when stirred?
QUESTION [3 upvotes]: When I cook spaghetti, they never tangle themselves so much to form knots, neither when in the boiling water nor when they're mixed with sauce.
Why is this so?
I experience this behaviour both with durum wheat flour spaghetti (stiffer and higher diameter, 2 mm when cooked) and rice noodles (thinner, 1mm when cooked, and more flexible). The length is the same, about 25 cm.

REPLY [2 votes]: One reason you might expect to see few if any 'knots' in a pot of cooking noodles is that without external stirring (and assuming the noodles are dilute), the motion of each noodle strand is determined only by currents of the boiling water. Since noodle strands are relatively massive and aren't very stretchy, the response of the noodle to changes in the flow field can be quite slow. The velocity profile of boiling water tends to form cycles that transport heat from the region nearest to the stove up to the surface. These cycles might cause groups of noodles to twist around each other, but couldn't carry any individual noodles into a knotted configuration. On smaller scales (of order the persistence length), fluctuations in the velocity profile would tend to cancel over extended segments of noodle. However, if you also stir the noodles directly, then you can artificially make knots with long spaghetti strands (and maybe measure how long they persist). For a dilute strand, I would expect the lifetime of simple knots to depend strongly on how long the spaghetti has been cooked. As mentioned in the comments, the stiffness of spaghetti will tend to favor the elimination of curvature over time. Once the total curvature in the spaghetti gets below a certain threshold, it is impossible for knots to exist. A simple threshold value for total curvature is given by the total curvature of a circle (or more accurately, a closed loop with a single fold defect, that acts like a hinge). You can analytically estimate the lifetime of a spaghetti knot by solving for the flow of curvature density along the spaghetti as a function of time, with local sources that represent the effect of the fluid flow. Very roughly, I would expect this to be similar to a diffusion equation with a continuous source, representing the twisting motion of the boiling fluid.
Now that we've discussed the behavior of a single noodle in response to the flow field generated by boiling water, let's consider the effect of many-noodle interactions. At high densities, interactions between noodles become important (and the effect of noodles on the velocity field). When the separation between noodles is of order the persistence length, one might expect the noodles to behave somewhat like a liquid crystal, favoring alignment between adjacent strands. You can then group together clusters of parallel noodles, and think of each cluster as an "effective noodle" on larger scales. In the rescaled system, there is some distribution of binding stiffness among effective noodles, but it's reasonable to expect a longer effective persistence length than before on average. "Effective noodles" can be grouped together again, leading to "effective effective noodles" at even larger scales. If the order persists up to the scale of the typical noodle length, then knotting will knot occur spontaneously.
However, if you tied a spaghetti strand in a knot and placed it in a dense pile, then many-spaghetti interactions will cause the knot to have a longer life-time than it would have in the dilute phase. The rate at which curvature density 'diffuses' along a preknotted noodle is controlled by the friction, inertia, and density of neighboring strands.


TITLE: Barometer and toriclli vaccum
QUESTION [0 upvotes]: What happens when
Reading of barometer when the space over Hg contains gas or vapour does it increase or decrease the height of mercury

REPLY [1 votes]: The space over the mercury column contains only the very low vapor pressure saturated mercury vapor. When you put a gas or vapor into this space, the space will expand due to the higher gas pressure so that the height of the mercury column will decrease.


TITLE: Affine open set of spec$A$ is distinguished?
QUESTION [1 upvotes]: Is any affine open subset of spec$A$ necessarily of the form $D(f)$? Why?
Here $D(f):=\{ \mathfrak p  \text{ prime ideal of } A | f \notin\mathfrak p\}$.

REPLY [4 votes]: Let $E$ be an  elliptic curve, say over $\mathbb C$ , with origin $O$ and let $X=E\setminus \{O\}=\operatorname {Spec} A$ the affine curve obtained by deleting that origin.
Choose $P\in E$ such that for every $n\gt0$ we have  $n[P]-n[O]\neq 0\in \operatorname {Cl}E$, the class group of $E$.
This is equivalent to $P\oplus\cdots\oplus P\neq 0$ ($n$ summands) in the underlying abelian group of $E$, in other words the condition is that $P$  not be a torsion element of that group.
Then I claim that:

The affine open subset $U=X\setminus \{P\}\subset X$ cannot be written $U=D(f)$ for any $f\in A$.

Indeed, if such an $f$ existed it could be seen as a rational function $F\in \operatorname {Rat}E$ with divisor necessarily of the form $\operatorname {Div}(F)=nP-nO\in \operatorname {Div}(E)$.
But this implies for the class $[nP-nO]$ that  $[nP-nO]=n[P]-n[O]=0\in \operatorname {Cl}(E)$, a contradiction with the choice of $P$.


text:
\begin{document}

\begin{center}
\textsc{Cartesian products of the $g$-topologies are a $g$-topology}\\
\medskip
\textsl{Jumaev Davron Ilxomovich}\footnote{\scriptsize Tashkent Institute of Architecture and Civil Engineering, 7, Kichik Xalqa yuli Str.,  Tashkent 100084, Uzbekistan,\\
e-mail: d-a-v-ron@mail.ru},\\
\textsl{Ishniyazov Baxrom Normamatovich}\footnote{\scriptsize Tashkent State Agrarian University, 2, University Str., Kibray District, Tashkent Region 100700, Uzbekistan,\\
e-mail: ishniyazov73@mail.ru},\\
\textsl{Tagaymuratov Abror Olimovich} \footnote{\scriptsize Chirchik State Pedagogical Institute, 104,  Amir Temur Str., Chirchik town, Tashkent Region 111700, Uzbekistan,\\
e-mail: abror.t93@mail.ru}

\end{center}
\begin{abstract}
We show that unlike the usual topologies the $g$-topologies are closed with respect to the Cartesian products. Moreover, we bring much detailed explanations some examples of concepts related the statistical metric spaces.\\

2010 \textit{Mathematics Subject Classification.} 54A08, 54A10.

\textit{Key words and phrases:} Statistical metric space, $g$-topology, Cartesian product.
\end{abstract}

\section{Introduction}
As mentioned in \cite{Menger1942} the three principal applications of statistical metrics are to macroscopic, microscopic and
physiological spatial measurements. Statistical metrics are designed to provide us firstly with a method removing conceptual difficulties
from microscopic physics and transferring them into the underlying geometry, secondly  with a treatment of thresholds of spatial sensation
eliminating the intrinsic paradoxes of the classical theory.

The notion of distance is defined in terms of functions, points and
sets. Indeed, in many situations, it is appropriate to look upon the distance concept as a statistical rather than a determinate one.
More precisely, instead of associating a number to the distance $d(p,q)$ with every pair of points $p$, $q$, one should associate a
distribution function $F_{pq}$ and for any positive number $x$, interpret $F_{pq}(x)$ as the probability that the distance from p to q be
less than $x$.

Using this idea, K.~Menger in \cite{Menger1942} defined a statistical metric space using the probability function in the year
1942. In 1943, shortly after the appearance of Mengers article, Wald published an article \cite{Wald1943} in which he criticized Mengers generalized triangle inequality.

In \cite{Thorp1962} the following questions raised by Thorp in statistical metric spaces:
\begin{itemize}
\item What are the necessary and sufficient conditions that the $g$-topology of type $V$ to be of type $V_{D}$?

\item What are the necessary and sufficient conditions that the $g$-topology of type $V_{\alpha}$ to be the $g$-topology of type $V_{D}$?

\item What conditions are both necessary and sufficient for the $g$-topology of type $V_{\alpha}$ to be a topology?
\end{itemize}

In \cite{RenVad2019} it had given partial answer to the above questions. Also, it was provided the basis for carrying out analysis in statistical metric spaces, in particular for the development of various $g$-topologies, neighborhoods defined in a statistical metric space and also the improvement of $\lambda_{\Omega}$-open sets in a generalized metric space. The authors of \cite{RenVad2019} had given more examples of the neighborhoods defined in a statistical metric space and the special kind of relationship between various $g$-topologies defined by Thorp in a $SM$ space. We seem that the examples have shortcomings. That is why we in the present paper completed the shortcomings and give more detail clarifies.

Further, we give positive answers to the question, expressed in \cite{Zaitov2019NUUz}, which asks: Is the Cartesian product of the $g$-topological spaces a $g$-topological space?

\section{Preliminaries}

A statistical metric space ($SM$ space) is an ordered pair $(S,\, F)$ where $S$ is a non-null set and $F$ is a mapping from $S\times S$ into the set of distribution functions (that is, real-valued functions of a real variable which are everywhere defined, non decreasing,
left-continuous and have infimum $0$ and supremum $1$).

The distribution function $F(p,\, q)$ associated with a pair of points p and q in S is denoted by $F_{pq}$. Moreover, $F_{pq}(x)$ represents the probability that the ``distance'' between $p$ and $q$ is less than $x$.

The functions $F_{pq}$ are assumed to satisfy the following:

\begin{itemize}
\item[$(SM\mbox{-}I)$] 	$F_{pq}(x)=1$ for all $x>0$ if and only if $p=q$.

\item[$(SM\mbox{-}II)$] 	$F_{pq}(0)=0$.

\item[$(SM\mbox{-}III)$] 	$F_{pq}=F_{qp}$.

\item[$(SM\mbox{-}IV)$] 	If $F_{pq}(x)=1$ and $F_{qr}(y)=1$, then $F_{pr}(x+y)=1$.
\end{itemize}

We often find it convenient to work with the tails of the distribution functions rather than with these distribution functions
themselves. Then the tail of $F_{pq}$, denoted by $G_{pq}$, is defined by $G_{pq}(x)=1-F_{pq}(x)$ for all $ x \in\mathbb{R}.$

Let $(S,\, F)$ be a statistical metric space. Then the Menger inequality is,

\begin{itemize}
\item[$(SM\mbox{-}IVm)$] 	$F_{pr}(x+y)\geq T(F_{pq}(x),\, F_{qr} (y))$
\end{itemize}
holds for all points $p$, $q$, $r \in S$ and for all numbers $x$, $y\geq 0$ where $T$ is a $2$-place function on the unit square satisfying:

\begin{itemize}
\item[$(T\mbox{-}I)$] $0 \leq T(a,\, b)\leq 1$ for all $a,\, b\in [0,1]$.

\item[$(T\mbox{-}II)$] $T(c,\, d)\geq T(a,\, b)$ for all $a,\, b,\, c,\, d \in [0,1]$ such that $c\geq a$, $d\geq b$ (monotonicity).

\item[$(T\mbox{-}III)$] $T(a,\, b)=T(b,\, a)$ for all $a, b\in [0,1]$ (commutativity).

\item[$(T\mbox{-}IV)$] $T(1,\, 1)=1$.

\item[$(T\mbox{-}V)$] $T(a,\, 1)>0$ for all $a>0$.
\end{itemize}

Let $(S,\ F)$ be a statistical metric space, $p\in S$ and $u$, $v$ be positive numbers. Then
\begin{gather*}
N_{p}(u,\, v)= \{q\in S:\, F_{pq} (u)>1-v\}=\{q\in S:\, G_{pq} (u)<v\}
\end{gather*}
is called the $(u,\, v)$-sphere with the center $p$.

The following example shows the existence of $(u,\, v)$-sphere in a statistical metric space.

\begin{Ex}\label{Npuv}
{\rm Consider the $SM$ space $(S,\, F)$ where $S$ denotes the possible outcomes of getting a tail when a coin is tossed once. Then $S=\{0,\, 1\}$. Let $F_{pq}(u)$ be the probability that the ``distance'' between $p$ and $q$ is less than u where $u>0$ and $p,\, q\in S$. We have (A.~A.~Zaitov):

$F_{00} (u)=F_{11} (u)=1$ for all $u>0$;
and
\begin{gather*}
F_{01} (u)=F_{10} (u)=
\begin{cases}
0, & \text{if}\,\, 0<u\leq 1,\\
1, & \text{if}\,\, u>1.
\end{cases}
\end{gather*}

Fix $p=0$. Then
\begin{gather*}
N_{0} (u,\, v)=
\begin{cases}
\{0\}, & \text{if}\,\, 0<u\leq 1, 0\leq v\leq 1,\\
\{0,1\}, &\text{if}\,\,   0<u\leq 1, v>1,\\
\{0,1\}, &\text{if}\,\,   u>1, v\geq 0.
\end{cases}
\end{gather*}

Fix $p=1$. Then
\begin{gather*}
N_{1} (u,\, v)=
\begin{cases}
\{1\}, & \text{if}\,\,  0<u\leq 1, 0\leq v\leq 1,\\
\{0,1\}, & \text{if}\,\,  0<u\leq 1, v>1,\\
\{0,1\}, & \text{if}\,\,   u>1, v\geq 0.
\end{cases}
\end{gather*}}
\end{Ex}

For fixed positive numbers $u$ and $v$, define a set
\begin{gather*}
U(u,\, v)=\{(p,q)\in S\times S: G_{pq} (u)<v\}.
\end{gather*}

Let us illustrate such sets in the following example.

\begin{Ex}\label{Uuv}
{\rm Consider the $SM$ space $(S,F)$ where $S$ denotes the possible outcomes of rolling a dice. Then $S=\{1,\, 2,\, 3,\, 4,\, 5,\, 6\}$ and the distribution function $F_{pq}(x)$ is the probability that the ``distance'' between $p$ and $q$ is less than u where $u>0$ and $p,\, q\in S$. Consider the usual metric on $S$ induced on the real line, i.~e. $d(p,\, q)=|q-p|$. We have (A.~A.~Zaitov):

\begin{align*}
&F_{pp}(u)=1\,\, \text{for all}\,\, u>0\,\, \text{and}\,\, p=1,\, 2,\, 3,\, 4,\, 5,\, 6;\\
&F_{p(p+1)}(u)=F_{(p+1)p}(u)=
\begin{cases}
0, &  0<u\leq 1,\\
1, &  u>1,
\end{cases} \qquad\mbox{where}\quad p=1,\, 2,\, 3,\, 4,\, 5;\\
&F_{p(p+2)}(u)=F_{(p+2)p}(u)=
\begin{cases}
0, & 0<u\leq 2,\\
1, & u>2,
\end{cases} \qquad\mbox{where}\quad p=1,\, 2,\, 3,\, 4;\\
&F_{p(p+3)}(u)=F_{(p+3)p}(u)=
\begin{cases}
0, & 0<u\leq 3,\\
1, & u>3,
\end{cases} \qquad\mbox{where}\quad p=1,\, 2,\, 3;\\
&F_{p(p+4)}(u)=F_{(p+4)p}(u)=
\begin{cases}
0, & 0<u\leq 4,\\
1, & u>4,
\end{cases} \qquad\mbox{where}\quad p=1,\, 2;\\
&F_{p(p+5)}(u)=F_{(p+5)p}(u)=
\begin{cases}
0, & 0<u\leq 5,\\
1, & u>5,
\end{cases} \qquad\mbox{where}\quad p=1.
\end{align*}

Consequently:

for every pair of $u$, $v$ such that $0<u\leq 1$, $0<v\leq 1$ we have:
\begin{gather*}
U(u,\, v)=\{(p,\, q)\in S\times S:\, d(p,\, q)=0\}=\{(p,\, p):\, p\in S\} = \Delta(S);
\end{gather*}

for every pair of $u$, $v$ such that $1<u\leq 2$, $0<v\leq 1$:
\begin{multline*}
U(u,\, v)=\{(p,\, q)\in S\times S:\, d(p,\, q)\leq 1\} = \\
=\Delta (S)\cup \{(p,\, p+1):\, p=1,\, 2,\, 3,\, 4,\, 5\}\cup \{(p+1,\, p):\, p=1,\, 2,\, 3,\, 4,\, 5\}\stackrel{def} = \Delta_{1}(S);
\end{multline*}

for every pair of $u$, $v$ such that $2<u\leq 3$, $0<v\leq 1$:
\begin{multline*}
U(u,\, v) = \{(p,\, q)\in S\times S:\, d(p,\, q)\leq 2\} =\\
=\Delta_{1}(S)\cup \{(p,\, p+2):\, p=1,\, 2,\, 3,\, 4\}\cup \{(p+2,\, p):\, p=1,\, 2,\, 3,\, 4\}\stackrel{def} = \Delta_{2}(S);
\end{multline*}

for every pair of $u$, $v$ such that $3<u\leq 4$, $0<v\leq 1$:
\begin{multline*}
U(u,\, v)=\{(p,\, q)\in S\times S:\, d(p,\, q)\leq 3\}=\\
=\Delta_{2}(S)\cup \{(p,\, p+3):\, p=1,\, 2,\, 3\}\cup \{(p+3,\, p):\, p=1,\, 2,\, 3\}\stackrel{def} = \Delta_{3}(S);
\end{multline*}

for every pair of $u$, $v$ such that $4<u\leq 5, 0<v\leq 1$:
\begin{multline*}
U(u,\, v)=\{(p,\, q)\in S\times S:\, d(p,\, q)\leq 4\} = \\
=\Delta_{3}(S)\cup \{(p,\, p+4):\, p=1,\, 2\}\cup \{(p+4,\, p):\, p=1,\, 2\}\stackrel{def} = \Delta_{4}(S);
\end{multline*}

for every pair of $u$, $v$ such that $u>5$, $0<v\leq 1$:
\begin{multline*}
U(u,\, v)=\{(p,\, q)\in S\times S:\, d(p,\, q)\leq 5\}=\\
=\Delta_{4}(S)\cup \{(p,\, p+5):\, p=1\}\cup\{(p+5,\, p):\, p=1\} = S\times S.
\end{multline*}

Also, it is easy to see that $U(u,\, v)=S\times S$ for every $u$, $v$ such that $u>0$, $v>1$.}
\end{Ex}

For any set $Z$ of ordered pairs of positive numbers, i.~e. $Z\subset (0,+\infty)^{2}$, let
\begin{gather*}
\mathcal{N}(Z)=\{N_{p}(u,\, v):\, (u,\, v)\in Z,\, \, p \in S\}\qquad \text{and}\qquad \mathcal{U}(Z)=\{U(u,\, v):\, (u,\,v)\in Z\}.
\end{gather*}

A non-null collection $\{N_{p}\}$ of subsets $N_{p}$ of a set $S$ associated with a point $p\in S$ is a {\itshape family of neighborhoods} for $p$ if each $N_{p}$ contains $p$. Let the family of neighborhoods be associated with each point $p$ of a set $S$.
\begin{itemize}
\item[$V$]\label{typeV}
In this case the set $S$ and the collection of neighborhoods is called the $g$-topological space of type $V$ \cite{Thorp1962}.
\end{itemize}

The {\itshape closure} of a subset $E$ of $S$, written $\overline{E}$, is the set of points $p$ such that each neighborhood of $p$ intersects $E$. The {\itshape interior} of $E$ is the complement of the closure of the complement of $E$. A $g$-topological space $S$ is {\itshape symmetric} if, for every pair of points $p$ and $q$, $p$ is in $\overline{\{q\}}$ iff $q$ is in $\overline{\{p\}}$.

E.~Thorp introduced the following $g$-topologies in a statistical metric space $(S,\, F)$.
\begin{itemize}
\item[$N_{0}$] is type $V$.
\item[$N_{1}$]. For each point $p$ and each neighborhood $U_{p}$ of $p$, there is a neighborhood $W_{p}$ of $p$ such that for each point $q$ of $W_{p}$, there is a neighborhood $U_{q}$ of $q$ contained in $U_{p}$.

\item[$N_{2}$]. For each point $p$ and each pair of neighborhoods $U_{p}$ and $W_{p}$ of $p$, there is a neighborhood of $p$ contained in the intersection of $U_{p}$ and $W_{p}$.
\end{itemize}

The following are various $g$-topologies in a statistical metric space $(S,F)$ defined by E.~Thorp.
\begin{itemize}
\item[$V_{D}$]. If the conditions $N_{0}$ and $N_{2}$ are satisfied, then the collection of neighborhoods on $S$ is called the {\itshape $g$-topology} of type $V_{D}$.

\item[$V_{\alpha}$]. The collection of neighborhoods on $S$ is called the {\itshape $g$-topology} of type $V_{\alpha}$ if the conditions $N_{0}$ and $N_{1}$ are satisfied.

\item[$Top$]. A $g$-topology is a $topology$ if the conditions $N_{0}$, $N_{1}$ and $N_{2}$ are satisfied.

\end{itemize}

Let $S$ be a set and $(P,\, <)$ be a partially ordered set with least element $0$. A {\itshape generalized \'{e}cart} ({\itshape $g$-\'{e}cart} for short) is a mapping
\begin{gather*}
G\colon\, S\times S \to P.
\end{gather*}
If a $g$-\'{e}cart $G$ satisfies $G(p,\, p)=0$ and the set $S$ consists of more than one point, the {\itshape $g$-\'{e}cart $g$-topology} for $S$ is the $g$-topology determined from $G$, and its partially ordered range set $P$, as follows:

For each $f>0$ in $P$ and each $p\in S$, the $f$-sphere for $p$ is a set of the form
\begin{gather*}
N_{p}(f)=\{q\in S:\, G(p,\, q)<f\}.
\end{gather*}
Then for each $p\in S$, the collection of $f$-spheres
\begin{gather*}
\mathcal{N}_{p}(P)=\{N_{p}(f):\, f>0,\, p\in P\}
\end{gather*}
is a family of neighborhoods for $p$.

The $g$-\'{e}cart associated with a statistical metric space $(S,F)$ is the mapping $G$ defined by $G(p,\, q)=G_{pq}$.

\begin{Ex}\label{Npf}{\rm
Let $S=\mathbb{N}$ and $P=\mathbb{N}\cup\{0\}$ be a partially ordered set with the relation $<$ where $\mathbb{N}$ denote the set of all positive integers. Let $A=\{1,\, 2,\, 3\}$ be a subset of $S$. Define
\begin{gather*}
G(p,\, q)=\begin{cases}
1, &\text{if}\,\, p\notin A,\, q\in S,\\
1, &\text{if}\,\, \in A,\, q\notin S,\\
0, &\text{if}\,\, \notin A,\, q\notin S.
\end{cases}
\end{gather*}
and for $p\in A$, $q\in A$ define $G(p,\, q)$ as follows:
\begin{gather*}
G(1,\, 1)=0,\qquad G(1,\, 2)=2,\qquad G(1,\, 3)=3,\\
G(2,\, 1)=4,\qquad G(2,\, 2)=0,\qquad G(2,\, 3)=6,\\
G(3,\, 1)=1,\qquad G(3,\, 2)=2,\qquad G(3,\, 3)=0.
\end{gather*}

$f$-sphere for each $p\in S$ has the following form (A.~A.~Zaitov):

\textbf{Case} $p=1$:
\begin{gather*}
N_{1}(f)=\begin{cases}
\emptyset, &\text{if}\,\, f=0,\\
\{1\}, &\text{if}\,\, f=1,\\
S\setminus \{2,\, 3\}, &\text{if}\,\, f=2,\\
S\setminus \{3\}, &\text{if}\,\, f=3,\\
S, &\text{if}\,\, f\geq 4.
\end{cases}
\end{gather*}

\textbf{Case} $p=2$:
\begin{gather*}
N_{2}(f)=\begin{cases}
\emptyset, &\text{if}\,\, f=0,\\
\{2\}, &\text{if}\,\, f=1,\\
S\setminus \{1,\, 3\}, &\text{if}\,\, 2\leq f\leq 4,\\
S\setminus \{3\}, &\text{if}\,\, 5\leq f\leq 6,\\
S, &\text{if}\,\, f\geq 7.
\end{cases}
\end{gather*}

\textbf{Case} $p=3$:
\begin{gather*}
N_{3}(f)=\begin{cases}
\emptyset, &\text{if}\,\, f=0,\\
\{3\}, &\text{if}\,\, f=1,\\
S\setminus \{2\}, &\text{if}\,\, f=2,\\
S, &\text{if}\,\, f\geq 3.
\end{cases}
\end{gather*}

\textbf{Case} of arbitrary $p\in S\setminus A$:
\begin{gather*}
N_{p}(f)=\begin{cases}
\emptyset, &\text{if}\,\, f=0,\\
S\setminus A, &\text{if}\,\, f=1,\\
S, &\text{if}\,\, f\geq 2.
\end{cases}
\end{gather*}}
\end{Ex}

Given a statistical metric space $(S,\, F)$, for each pair of points $p$ and $r$ in $S$, number $u>0$, the $r$-sphere with center $p$,
$N_{p} (r;\, u)$ is defined to be the sphere
\begin{gather*}
N_{p}(r;\, u)=N(G_{pr}(u))=\{q:\, G_{pq} (u)<G_{pr} (u)\}.
\end{gather*}
The $R$-$g$-topology for $(S,\, F)$ is the structure whose family of neighborhoods at each point $p$ is the collection
\begin{gather*}
\mathcal{N}_{p}(r)=\{N_p (r;\,u):\, r \in S,\, u>0\}.
\end{gather*}

\begin{Ex}{\rm
Consider the $SM$ space $(S,\, F)$ where $S=\mathbb{N}$ and the distribution function
\begin{gather*}
F_{pq}(x)=\begin{cases}
\frac{x}{d(p,\, q)}, &\text{if}\,\, 0<x<d(p,\, q), \,\, d(p,\, q)\neq 0,\\
1, &\text{if}\,\, x\geq d(p,\, q).
\end{cases}
\end{gather*}
where $d(p,\, q)=|q-p|$, $p,\, q \in S$.

Fix $p=1$ and $r=2$ from $S$. Let $x=\frac{1}{4}$. Then $G_{pr}(x)=G_{12}(\frac{1}{4})=1-F_{12}(\frac{1}{4})=0.75$. We have (A.~A.~Zaitov):
$F_{11}(\frac{1}{4})=1>0.25$ and $F_{1q}(\frac{1}{4})=\frac{1}{4}(q-1)\leq 0.25$ for every $q\geq 2$. That is why
\begin{multline*}
N_{1}\left(2;\, \frac{1}{4}\right)=N\left(G_{12}\left(\frac{1}{4}\right)\right) =\\
= \left\{q\in S:\, G_{1q} \left(\frac{1}{4}\right)<0.75\right\} = \left\{q\in S:\, F_{1q}(\frac{1}{4})>0.25\right\}=\{1\}.
\end{multline*}

Note that $N(G_{pr})  = \varnothing$ if $p=r$. Really, for every $u>0$ and $p\in S$ one has $F_{pp}(u)=1$, consequently,
$G_{pp}(u)=0$. Since $0\leq G_{pq}(u)\leq 1$ for all $u>0$, $p,\, q\in S$, there exists no
$q\in S$ such that $G_{pq}(u)<G_{pp}(u)$. Hence $N(G_{pp})=\varnothing$, $p\in S$.}
\end{Ex}

\begin{Rem}{\rm In a $SM$ space $(S,F)$, we use the following notations:
\begin{itemize}
\item[$(a)$] Let $\tau$ denote the $g$-topology of type $V$.
\item[$(b)$] Let $\tau_{D}$ denote the $g$-topology of type $V_{D}$.
\item[$(c)$] Let $\tau_{\alpha}$ denote the $g$-topology of type $V_{\alpha}$.
\item[$(d)$] Let $\tau_{e}$ denote the $g$-\'{e}cart $g$-topology.
\item[$(e)$] Let $\tau_{R}$ denote the $R$-$g$-topology.
\item[$(f)$] Each element in $\mathcal{N}(X)$ is called a $\tau$-neighborhood.
\item[$(g)$] Each element in $\mathcal{N}_{p}(P)$ is called a $\tau_{e}$-neighborhood.
\item[$(h)$] Each element in $\mathcal{N}_{p}(r)$ is called a $\tau_{R}$-neighborhood.
\end{itemize}}
\end{Rem}

\section{Main part}

Let $(S^{1},\, F^{1})$ and $(S^{2},\, F^{2})$ be statistical metric spaces, $(p^{1},\, p^{2})$, $(q^{1},\, q^{2})$, $(r^{1},\, r^{2})\in S^{1}\times S^{2}$. Put
\begin{gather}\label{p.ofd.f.}
F^{12}_{(p^{1},\, p^{2})(q^{1},\, q^{2})} (x) = F^{1}_{p^{1}q^{1}}(x)\cdot  F^{2}_{p^{2}q^{2}}(x).
\end{gather}
Obviously, conditions $(SM\mbox{-}I)$ -- $(SM\mbox{-}III)$ are true. Let us show $(SM\mbox{-}IV)$ is also satisfied. Assume that $F^{12}_{(p^{1},\, p^{2})(q^{1},\, q^{2})}(x)=1$ and $F^{12}_{(q^{1},\, q^{2})(r^{1},\, r^{2})}(y)=1$. These equalities mean that the ``distance'' between $(p^{1},\, p^{2})$ and $(q^{1},\, q^{2})$ less then $x$, and the ``distance'' between $(q^{1},\, q^{2})$ and $(r^{1},\, r^{2})$ less than $y$. Then clearly, that the ``distance'' between $(p^{1},\, p^{2})$ and $(r^{1},\, r^{2})$ less than $x+y$. Consequently, the probability that the ``distance'' between $(p^{1},\, p^{2})$ and $(r^{1},\, r^{2})$ is less than $x+y$, is 1, i.~e. $F^{12}_{(p^{1},\, p^{2})(r^{1},\, q^{2})}(x+y)=1$.

So, (\ref{p.ofd.f.}) is defined correctly. Its tail
\begin{gather*}
G^{12}_{(p^{1},\, q^{1})(p^{2},\, q^{2})}(x) = 1 - F^{12}_{(p^{1},\, q^{1})(p^{2},\, q^{2})}(x)\qquad \mbox{for all}\qquad x \in\mathbb{R}.
\end{gather*}

Now we check the Menger inequality. Let $T(a,\, b) = ab$, $a,\, b\in [0,\, 1]$. Clearly, it satisfies conditions $(T\mbox{-}I)$ -- $(T\mbox{-}V)$. Suppose
\begin{gather*}
F^{1}_{p^{1}r^{1}}(x+y)\geq T(F^{1}_{p^{1}q^{1}}(x),\, F^{1}_{q^{1}r^{1}}(y))\quad \mbox{and}\quad F^{2}_{p^{2}r^{2}}(x+y)\geq T(F^{2}_{p^{2}q^{2}}(x),\, F^{2}_{q^{2}r^{2}}(y)).
\end{gather*}
Then
\begin{align*}
F^{12}_{(p^{1},\, p^{2})(r^{1},\, r^{2})}(x+y) &= F^{1}_{p^{1}r^{1}}(x+y)\cdot F^{2}_{p^{2}r^{2}}(x+y)\geq\\
&\geq T(F^{1}_{p^{1}q^{1}}(x),\, F^{1}_{q^{1}r^{1}}(y))\cdot T(F^{2}_{p^{2}q^{2}}(x),\, F^{2}_{q^{2}r^{2}}(y)) =\\
&=F^{1}_{p^{1}q^{1}}(x)\cdot F^{1}_{q^{1}r^{1}}(y)\cdot F^{2}_{p^{2}q^{2}}(x)\cdot F^{2}_{q^{2}r^{2}}(y)=\\
&=F^{12}_{(p^{1},\, p^{2})(q^{1},\, q^{2})} (x)\cdot F^{12}_{(q^{1},\, q^{2})(r^{1},\, r^{2})} (y) = \\
&= T(F^{12}_{(p^{1},\, p^{2})(q^{1},\, q^{2})}(x),\, F^{12}_{(q^{1},\, q^{2})(r^{1},\, r^{2})} (y)),
\end{align*}
i.~e.
\begin{gather*}
F^{12}_{(p^{1},\, p^{2})(r^{1},\, r^{2})}(x+y)\geq T(F^{12}_{(p^{1},\, p^{2})(q^{1},\, q^{2})}(x),\, F^{12}_{(q^{1},\, q^{2})(r^{1},\, r^{2})} (y)).
\end{gather*}

Now we claim the following statements the proofs each of them consists just directly verification.

\begin{Th}{\rm
Let $(S^{1},\, \tau^{1})$ and $(S^{2},\, \tau^{2})$ be $g$-topological spaces of type $V$. Then $\tau^{1}\times \tau^{2}$ is a $g$-topology of type $V$ on $S_{1}\times S_{2}$.}
\end{Th}

\begin{Th}{\rm
Let $(S^{1},\, \tau^{1}_{D})$ and $(S^{2},\, \tau^{2}_{D})$ be $g$-topological spaces of type $V_{D}$. Then $\tau^{1}_{D}\times \tau^{2}_{D}$ is a $g$-topology of type $V_{D}$ on $S_{1}\times S_{2}$.}
\end{Th}

\begin{Th}{\rm
Let $(S^{1},\, \tau^{1}_{\alpha})$ and $(S^{2},\, \tau^{2}_{\alpha})$ be $g$-topological spaces of type $V_{\alpha}$. Then $\tau^{1}_{\alpha}\times \tau^{2}_{\alpha}$ is a $g$-topology of type $V_{\alpha}$ on $S_{1}\times S_{2}$.}
\end{Th}

\begin{Th}{\rm
Let $(S^{1},\, \tau^{1}_{e})$ and $(S^{2},\, \tau^{2}_{e})$ be $g$-\'{e}cart  $g$-topological spaces. Then $\tau^{1}_{e}\times \tau^{2}_{e}$ is  a $g$-\'{e}cart  $g$-topology on $S_{1}\times S_{2}$.}
\end{Th}

\begin{Th}{\rm
Let $(S^{1},\, \tau^{1}_{e})$ and $(S^{2},\, \tau^{2}_{e})$ be $R$-$g$-topological spaces. Then $\tau^{1}_{e}\times \tau^{2}_{e}$ is  a $R$-$g$-topology on $S_{1}\times S_{2}$.}
\end{Th}

\begin{Rem}{\rm
Note that the Cartesian product of topologies must not be a topology, i.~e. usual topology does not close under product. Unlike usual topology, $g$-topologies are closed with respect to product.

}
\end{Rem}


TITLE: What is the proper method to integrate $x^2e^{-(ax^2)}$?
QUESTION [0 upvotes]: What is the proper method to evaluate the integral $$\int_{-\infty}^{+\infty} x^2e^{-ax^2}dx$$ ?
I know that for $$\int_{-\infty}^{+\infty} e^{-ax^2}dx$$ we can convert to polar coordinates and then use the standard trick.

REPLY [0 votes]: Note that the "standard trick" (using polar coordinates) can be extended to handle this case as well.
$$
\int x^2 e^{-ax^2}dx\cdot\int e^{-ay^2}dy=\iint x^2 e^{-a(x^2+y^2)}dxdy=\frac{1}{2}\iint (x^2+y^2) e^{-a(x^2+y^2)}dxdy,
$$
where the second equality is by symmetry; so
$$
\int x^2 e^{-ax^2}dx=\frac{\iint(x^2+y^2)e^{-a(x^2+y^2)}dxdy}{2\int e^{-ax^2}dx}.
$$
The new double integral can be evaluated in polar coordinates, and the denominator is known.  Indeed, this approach can be used to solve $\int x^k e^{-ax^2}$ for any even $k$.


TITLE: Probability distribution (pdf)
QUESTION [0 upvotes]: I have this problem:
20% of items produced on a machine are outside tolerance. Determine the probability distribution of the number of defectives in a pack of five items.
I have came to the conclusion that sample size = 5 but I don't know how to continue.

REPLY [2 votes]: It is a probability distribution supported on $\{0,1,2,3,4,5\}$. You just have to give the 6 values of the probability mass function; assuming the fact that a given item is faulty is independent of the others, let $X$ be the number of defectives in the set. You get
\begin{align*}
\mathbb{P}\{X=0\} &= \left(1-\frac{20}{100}\right)^5 \tag{All are good} \\
\mathbb{P}\{X=1\} &= \binom{5}{1}\left(1-\frac{20}{100}\right)^4\left(\frac{20}{100}\right)^1 \tag{One is bad out of 5} \\
\vdots
\end{align*}
You can explicitly compute them. Note that this is actually a binomial distribution $\operatorname{Binom}(0.2, 5)$ — can you see intuitively why?

REPLY [1 votes]: The question wether one of the items is okay (or not okay) does not depend on the condition of the others. Secondly the items have all the same probability of being okay (or not okay). This tells you that you deal here with a binomial distribution. You only must find the correct parameters. Can you take it from here?
The pdf of a binomial distribution with parameters $n$ and $p$ is:$$P_{X}\left(k\right)=\binom{n}{k}p^{k}\left(1-p\right)^{n-k}$$


TITLE: Momentum Conservation Principle and elastic collision vs inelastic collision
QUESTION [0 upvotes]: According to the Momentum Conservation Principle:
For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.
Lets say we have a 2 carts of a certain mass. We launch 2 balls of equal mass with equal speed, 1 into each cart, 1st made from plasticine ( inelastic collision), the 2nd made from rubber (elastic collision). Does it mean that a cart hit by rubber ball would have after an impact a greater speed?

REPLY [2 votes]: AFAIK collision is defined by two objects coming into contact, hence the name.
This may be of assistance in terms of determining an inelastic versus elastic collision: https://en.wikipedia.org/wiki/Coefficient_of_restitution
Other than that, here are your two situations that I can think of and can work with -
Case 1 - Elastic
$$m_{\text{cart}}v_{\text{cart}_0} + m_{\text{ball}}v_{\text{ball}_0} = m_{\text{ball}}v_{\text{ball}_0}$$
(Cart is at rest; ball is assumed to not roll.)
Due to conservation of momentum -
$$m_{\text{ball}}v_{\text{ball}_0} = m_{\text{cart}}v_{\text{cart}} + m_{\text{ball}}v_{\text{ball}}$$
Our second equation of use is conservation of kinetic energy -
$$\frac{1}{2}m_{\text{cart}}v^2_{\text{cart}_0} + \frac{1}{2}m_{\text{ball}}v^2_{\text{ball}_0} = \frac{1}{2}m_{\text{ball}}v^2_{\text{ball}_0}$$
$$\frac{1}{2}m_{\text{ball}}v^2_{\text{ball}_0} = \frac{1}{2}m_{\text{cart}}v^2_{\text{cart}} + \frac{1}{2}m_{\text{ball}}v^2_{\text{ball}}$$
You can work your way to a result from there.
Case 2 - Perfectly Inelastic
$$m_{\text{ball}}v_{\text{ball}_0} = \left(m_{\text{cart}} + m_{\text{ball}} \right)v$$
Kinetic energy is not conserved here, but the final velocity can be found quite easily.

REPLY [1 votes]: What exactly you mean by launching a ball "into" a cart is important. If this means that the rubber ball lands on a horizontal surface of the cart and bounces back out, the cart doesn't have to move at all (the momentum exchange can be purely in the vertical direction for a perfectly elastic collision) while it does with the sticky ball.
If what you mean is the balls impact the cart from behind, pushing it forward, then yes, since the rubber ball bounces back, conservation of momentum implies that the car impacted by it moves faster.


TITLE: Stretching vector in a specific direction and calculating its lenght
QUESTION [1 upvotes]: I know this should be very easy but unfortunately I couldn't attend few of my classes and now I don't understand how to calculate.
This is the problem:
I am given two points: $A=(1,-1)$ and $B=(-4,5)$. I calculated $AB$'s length, it is  $\sqrt{61}$.
I should extend $\overrightarrow{AB}$ vector to a point that my let's say $AC$'s length is  $\sqrt{549}$
.
I am really sorry if this is kinda gibberish, English isn't my main language, I know general English but not terms and such.

REPLY [1 votes]: $$\overrightarrow{AB} = B - A= (b_1-a_1, b_2-a_2) = (-5,6)$$
$$\overrightarrow{AC} = k\cdot\overrightarrow{AB}= k\cdot(-5,6)=(-5k, 6k)$$
$$\left|\overrightarrow{AC}\right| = \Big|(5k, 6k)\Big| = \sqrt{(-5k)^2 + (6k)^2} = k\cdot\sqrt{61}$$
Now, you want $\left|\overrightarrow{AC}\right| = \sqrt{549},\ $ i. e.
$$k\cdot\sqrt{61} = \sqrt{549}$$
You will obtain $k = 3,$ so $$\overrightarrow{AC} = 3\cdot\overrightarrow{AB} = 3\cdot(-5,6)=(-15, 18)$$
and
$$\color{red}C = A + \overrightarrow{AC} = (1,-1) + (-15,18) =\color{red}{(-14, 17)}$$


TITLE: coefficient extension for fraction field $K(T) \otimes_K L$
QUESTION [0 upvotes]: Let $L/K$ be an algebraic field extension. Denote by $K(T)= \mathrm{Frac}(K[T])$ the transcendental field extension of $K$. I would like to find out how to show that the equation $$K(T) \otimes_K L = L(T)$$ holds and especially where the requirement that $L$ is algebraic flows in.

That seems to be essential since for transcendent $L:= K(T)$ the formula above fails since $K(T) \otimes_K K(T)$ is not a field.
Thanks in advance!

REPLY [0 votes]: by induction you must show that $K(T)\otimes K[\alpha]=K[\alpha](T)$ write $K[\alpha]=\frac{K[T]}{(f)}$ and prove the morphism $g\otimes (h+(f))\to g\pi(h)$
($\pi$ is the natural morphism from  $K[T]/(f)$ to L)is isomorphism.
(construct the converse by sending $\sum f_i(\alpha)T^i$ to $\sum T^i\otimes (f_i+(f))$)


TITLE: Taylor Series Clarification
QUESTION [0 upvotes]: For $\sin(x)$, $e^x$, $\cos(x)$... When we are building the $n$-th taylor polynomial, why is it that we always evaluate the functions first $k$ derivatives at $x=0$? In my textbook when they were deriving the formula for each of those 3 functions, they would always evaluate all the derivatives at $x=0$ but why not at some other value of $x$ that is more general?

REPLY [2 votes]: There are at least three answers to this question:

For the functions you list, you don't gain a lot by expanding about another point, because you have identities like
$$ e^x = e^{a+(x-a)} = e^a e^{x-a}, $$
so the Taylor series of $e^x$ about the point $x=a$ is
$$ e^x = \sum_{k=0}^{\infty} e^a \frac{(x-a)^k}{k!}, $$
which has much the same properties as the Taylor series of $e^x$. You can do the same thing with $\sin{x}$ and $\cos{x}$ using the addition formulae like
$$ \sin{(a+(x-a))} = \sin{a}\cos{(x-a)} + \cos{a} \sin{(x-a)}, $$
and you know how to expand those about $x=a$ from your results at $0$.
More generally, it is often useful to expand about a point other than $x=0$. A classic example is the logarithm: $\log{x}$ is not differentiable at $x=0$, but at $x=1$ we do have an expansion, which you may not have noticed because we often write it a different way:
$$ \log{(1+x)} = x-\frac{x^2}{2}+\frac{x^3}{3}+\dotsb $$
The most general answer is that the Maclaurin series of the function is the Taylor series of the translate of a function ($f(x-a)$, basically). Often, when looking at a finite number of terms, it's just easier to deal with a polynomial in $x$

And sometimes, expansions about different points can look quite different. Consider the following series:
$$ 1-x^2+x^4-\dotsb = \frac{1}{1+x^2} \quad (|x|<1) \tag{1} $$
Now consider this series:
$$ \tfrac{1}{2}-\tfrac{1}{2}x+\tfrac{1}{4}x^2-\tfrac{1}{8}x^4+\tfrac{1}{8}x^5-\dotsb. \tag{2} $$
What you don't know about this series is that it's actually the Taylor series of
$$ \frac{1}{1+(1+x)^2}, $$
about $x=0$, so it's also the Taylor series of $(1+x^2)^{-1}$ about $x=1$. (2) looks a lot worse than (1) (you and I would struggle to guess what (2)'s $n$th term is, for one thing), but (using techniques you learn in Analysis courses at university), you can show that (2) actually converges for $|x-1|<\sqrt{2}$, so in a sense it's more useful, because it covers a larger range of possible $x$ values.
Therefore I'd say the normal reason people would give when asked is that there are just some points for which the series is easier to calculate, and this series works for the values of $x$ we're interested in, which is good enough. (In the case of the exponential and sine and cosine, all values of $x$, in fact.)


TITLE: How does the levitating water illusion work?
QUESTION [0 upvotes]: After watching a movie in which levitating water was used as an illusion, I was curious to discover how it works. Unfortunately I cannot find any great explanations of the science behind the effect, only that strobe lights are what cause it. I want to know why strobe lights cause this effect.
Here is a video of the illusion.
It is my assumption that there is a steady stream of water, and the strobe lights are flickering on and off at a certain rate. Why then do I not see the entire stream of water being lit up, rather than individual droplets? What causes the droplets to appear to "move?" And why are there dark spaces between the droplets when there is apparently water there?

REPLY [4 votes]: It is an optical illusion that uses stroboscopic light. It is the same principle that allows you to see still pictures into a movie. For instance, when you see the drops floating still, the drop is not the same, the light frequency is synchronized with that of the the falling
 drops, so that every time the light is on it shows you the image of a different drop at the same position (when it is off the drop keeps falling and another replaces it). The stream in not continuous, it a "steam" of drops.
If the next dot is actually slightly higher than the previous drop when the light gets on then your brain interprets that it as the same drop moving up.


TITLE: Excess pressure inside a non spherical drop
QUESTION [2 upvotes]: How can we derive the formula for excess pressure inside a non spherical drop which is $T(1/a_1 + 1/a_2)$ as given here .I was unable to find any derivation on the internet. Also what is $a_1$ and $a_2$ in the formula is unclear to me. It says that $a_1$ and $a_2$ are radius of curvatures of the intersection of perpendicular planes with the surface which raises the question why the intersection gives a circle. Is there any general equation of the surface of drop resting on a plane surface?

REPLY [1 votes]: There are general equations for drops on a surface but they are different equations and require other/more information. When a fluid interacts with a surface, the hydro -phobicity or -philicity also needs to be taken into account, this is usually expressed via a contact angle.
The equation talks about drops that are non-spherical in the sense that they aren't perfect spheres, as in an oval is not a circle, they are usually not talking about completely irregular shapes (which is why this come across as confusing).
a1 or a2, usually called R1 and R2, are the radii of the droplet or surface.
Where there's generally two options, the first one being the clearest example.
Think of it as a deformed spherical droplet, stretched/squeezed along 1 axis (let's say z). That way instead of looking circular from the front the droplet will look oval and you will need two radii to describe it.
As in the image below
https://www.slideserve.com/betty_james/young-laplace-equation-equation-of-capillarity
The second is a bit more complex.
You can describe any surface as a a surface made up of many (to infinity) very small (infinitesimal) surfaces. This is what's usually done when integrating over a more complex surface. Such a small tiny surface does not have to be considered flat but can also be considered(approximated) curved. This can then be used to calculate the integral of the surface tension, to approximate any irregular shape.
Where each small section could be represented as in the image below
https://upload.wikimedia.org/wikipedia/commons/f/f5/Curvature_radii.JPG
If you really want to understand the subject well, there's a derivation for the Laplace-Young equation here:
http://www.ux.uis.no/~s-skj/PetFys04/Notater/Young-Laplace/you-lap.pdf
(warning fairly complex and arduous math)


TITLE: Some questions regarding partial derivatives
QUESTION [1 upvotes]: I know these questions may seem trivial, but I can't understand the following thing:
Say there's a function $F(x,y,y')=x+y+y', where ~y(x)=x^3$.
$\frac{\partial F}{\partial x} = 1+ \frac{\partial (y)}{\partial x}+\frac{\partial (y')}{\partial x}$
here, on the textbook, it states that $\frac{\partial (y)}{\partial x}=0 ~and ~\frac{\partial (y')}{\partial x}=0$
...
At first glance, this seemed obvious because the terms y and y' don't have 'x' in them. However, since $y=x^3$, shouldn't $\frac{\partial (y)}{\partial x}=\frac{\partial(x^3)}{\partial x} =3x^2 $ and so on?
Also, in the case above, what is $\frac{\partial F}{\partial y}$?
Shouldn't the $\partial y'/\partial y$ not equal zero, as y' is related to y?

REPLY [1 votes]: Roughly speaking, taking a partial derivative means that we treat all variables of the function as independent variable and we are taking the derivative with respect to one of those independent variables. Say, for a function $F(x,y,y')$ in general or for your example $F(x,y,y')=x+y+y'$ specifically, we treat it as a function of the three independent variables $x$, $y$, and $y'$, and therefore there are three separate partial derivatives:
$$\frac{\partial F}{\partial x}=1, \quad \frac{\partial F}{\partial y}=1, \quad\frac{\partial F}{\partial y'}=1.$$
And when treating all three of them as independent variables, then of course, the derivative of any one with respect to another is zero.
However, if one of these variables in turn depends on some other variables, then the multivariate Chain Rule kicks in. What you calculated in your question, where $y=x^3$, is not a partial derivative at all, but the total derivative with respect to $x$. It would even be denoted differently:
$$\frac{\mathrm{d}F}{\mathrm{d}x}=\frac{\partial F}{\partial x}\cdot\frac{\mathrm{d}x}{\mathrm{d}x}+\frac{\partial F}{\partial y}\cdot\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{\partial F}{\partial y'}\cdot\frac{\mathrm{d}y'}{\mathrm{d}x}=1\cdot1+1\cdot3x^2+1\cdot6x=3x^2+6x+1.$$


TITLE: How to verify that $U\oplus W$ is a direct sum
QUESTION [0 upvotes]: Suppose
$U = \{(x, y, 0)\in F^3: x, y \in F \}$ and $W = \{(0, 0, z) \in F^3:z \in F\}$.
Then $F^3 = U\oplus W$. But why?
$U + W = \{(x, y, z)\in F^3:x, y, z\in F\}$. But can't I express the sum using $U' = \{(x, 0, 0)\in F^3:x \in F\}$ and $W' = \{(0, y, z) \in F^3: y, z \in F\}$ ? I mean, I have no idea why $U\oplus W$ can be expressed in only one way.

REPLY [1 votes]: When we say that $V = U\oplus W$, this means that every element of $V$ can be uniquely written as a sum of elements from $U$ and $W$. What it does not mean, is that $U$ and $W$ are unique subspaces that sum to $V$. The uniqueness is in the elements of $U$ and $W$, not in $U$ and $W$ themselves. In fact, in most situations, you will have many $U_i$ and $W_i$ such that $U_i \oplus W_i = V$ So, you are only trying to show that the summation is unique here, not that $U,W$ are unique in any sense. As with your example above, they definitely are not unique.
See, when you are trying to show uniqueness of summation, what do you essentially need to show? That if there are two ways of writing the sum, then these ways must be the same.

That is, for $v \in F^3$, you must prove that if $v = u_1 + w_1$ and $v = u_2+w_2$, where $u_1,u_2 \in U$ and $w_1,w_2 \in W$, then it is true that $u_1=u_2$, and $w_1 = w_2$.

So starting from the above, we basically get $u_1 + w_1 = u_2 + w_2$, since both are equal to $v$. Since $u_1,u_2 \in U$, we know how they look. That is, $u_1 = (x_{u_1},y_{u_1},0)$, and $u_2 = (x_{u_2},y_{u_2},0)$ for some elements $x_{u_1},x_{u_2},y_{u_1},y_{u_2} \in F$. Similarly, since $w_1,w_2 \in W$, we get that $w_1 = (0,0,z_{w_1}),w_{2} = (0,0,z_{w_2})$ for some elements $z_{w_1},z_{w_2} \in F$.
Now performing vector addition, we get $u_1 + w_1 = (x_{u_1},y_{u_1},z_{w_1})$, and $u_2  + w_2 = (x_{u_2},y_{u_2},z_{w_2})$.
Now, we are saying that the two vectors above are equal. When does this happen? Only when all their components are equal. That is, above, we must have the first components , which are $x_{u_1}$ and $x_{u_2}$, must be equal. Similarly, $y_{u_1} = y_{u_2},z_{w_1}=z_{w_2}$. From here, it's easy to see that $u_1=u_2,w_1=w_2$. So $V$ is a direct sum of $U$ and $W$.
It's similarly shown that for the $U'$ and $W'$ that you have written above, it is true that $U'\oplus W =V$. So, to reiterate, the subspaces that direct sum to $V$ are in no way unique. It is their elements which are key to the uniqueness element of the direct sum.
You will still have doubts, and that's good, it can only help your understanding of the subject. Do ask, of course.


text:
section \<open>Sets with at Most 2 Elements\<close>

(*<*)
theory Bool_Bounded_Set
imports
  "HOL-Cardinals.Cardinals"
begin
(*>*)

typedef 'a bset =
  "{A :: 'a set. |A| \<le>o ctwo}"
  morphisms set_bset Abs_bset
  by (rule exI[of _ "{}"]) (auto simp: card_of_empty4 ctwo_def)

setup_lifting type_definition_bset

lift_definition map_bset ::
  "('a \<Rightarrow> 'b) \<Rightarrow> 'a bset \<Rightarrow> 'b bset" is image
  using card_of_image ordLeq_transitive by blast

inductive rel_bset :: "('a \<Rightarrow> 'b \<Rightarrow> bool) \<Rightarrow> 'a bset \<Rightarrow> 'b bset \<Rightarrow> bool" for R where
  "set_bset x \<subseteq> {(x, y). R x y} \<Longrightarrow> rel_bset R (map_bset fst x) (map_bset snd x)"

lift_definition bempty :: "'a bset" is "{}"  by (auto simp: card_of_empty4 ctwo_def)
lift_definition bsingleton :: "'a \<Rightarrow> 'a bset" is "\<lambda>x. {x}" by (auto simp: ctwo_def)
lift_definition bdoubleton :: "'a \<Rightarrow> 'a \<Rightarrow> 'a bset" is "\<lambda>x y. {x, y}"
  by (auto simp: ctwo_def finite_card_of_iff_card2 card.insert_remove card_Diff_singleton_if)

lemma map_bset_bempty[simp]: "map_bset f bempty = bempty"
  by transfer simp

lemma map_bset_eq_bempty[simp]:
  "map_bset f x = bempty \<longleftrightarrow> x = bempty"
  "bempty = map_bset f x \<longleftrightarrow> x = bempty"
  by (transfer, simp)+

lemma map_bset_bsingleton[simp]: "map_bset f (bsingleton x) = bsingleton (f x)"
  by transfer simp

lemma map_bset_bdoubleton[simp]: "map_bset f (bdoubleton x y) = bdoubleton (f x) (f y)"
  by transfer simp

lemma bdoubleton_same[simp]: "bdoubleton x x = bsingleton x"
  by transfer simp

lemma bempty_neq_bsingleton[simp]:
  "bempty \<noteq> bsingleton x"
  "bsingleton x \<noteq> bempty"
  by (transfer, simp)+

lemma bempty_neq_bdoubleton[simp]:
  "bempty \<noteq> bdoubleton x y"
  "bdoubleton x y \<noteq> bempty"
  by (transfer, simp)+

lemma bsinleton_eq_bdoubleton[simp]:
  "bsingleton x = bdoubleton y z \<longleftrightarrow> (x = y \<and> y = z)"
  "bdoubleton y z = bsingleton x \<longleftrightarrow> (x = y \<and> y = z)"
  by (transfer, auto)+

lemma bsinleton_inj[simp]:
  "bsingleton x = bsingleton y \<longleftrightarrow> (x = y)"
  by (transfer, simp)

lemma bdoubleton_eq_iff[simp]:
  "bdoubleton x y = bdoubleton z w \<longleftrightarrow> (x = z \<and> y = w \<or> x = w \<and> y = z)"
  by transfer (simp add: doubleton_eq_iff)

lemmas [simp] = bempty.rep_eq bsingleton.rep_eq bdoubleton.rep_eq

lemma bset_cases:
  "\<lbrakk>X = bempty \<Longrightarrow> P; \<And>x. X = bsingleton x \<Longrightarrow> P; \<And>x y. \<lbrakk>x \<noteq> y; X = bdoubleton x y\<rbrakk> \<Longrightarrow> P\<rbrakk> \<Longrightarrow> P"
proof (transfer fixing: P)
  fix Z :: "'a set"
  assume "|Z| \<le>o ctwo"
  moreover then have "finite Z" by (metis card_of_ordLeq_infinite ctwo_def finite_code)
  ultimately have "card Z \<le> 2" unfolding ctwo_def by (subst (asm) finite_card_of_iff_card2) auto
  moreover assume "Z = {} \<Longrightarrow> P" "\<And>x. Z = {x} \<Longrightarrow> P" "\<And>x y. \<lbrakk>x \<noteq> y; Z = {x, y}\<rbrakk> \<Longrightarrow> P"
  ultimately show P using \<open>finite Z\<close>
  proof (induct "card Z")
    case (Suc m) from Suc(2)[symmetric] Suc(2-7) show P by (cases m) (auto simp: card_Suc_eq)
  qed simp
qed

bnf "'k bset"
  map: map_bset
  sets: set_bset
  bd: natLeq
  rel: rel_bset
proof (standard, goal_cases)
  case 1 then show ?case
    by transfer simp
next
  case 2 then show ?case
    apply (rule ext)
    apply transfer
    apply auto
    done
next
  case 3 then show ?case
    apply transfer
    apply (auto simp: image_iff)
    done
next
  case 4 then show ?case
    apply (rule ext)
    apply transfer
    apply simp
    done
next
  case 5 then show ?case by (simp add: card_order_csum natLeq_card_order)
next
  case 6 then show ?case by (simp add: cinfinite_csum natLeq_cinfinite)
next
  case 7 then show ?case
    apply transfer
    apply (erule ordLeq_transitive[OF _ ordLess_imp_ordLeq[OF ctwo_ordLess_natLeq]])
    done
next
  case (8 R S)
  then show ?case
  proof (safe elim!: rel_bset.cases, unfold rel_bset.simps)
    fix z1 z2
    assume "set_bset z1 \<subseteq> {(x, y). R x y}" "set_bset z2 \<subseteq> {(x, y). S x y}" "map_bset snd z1 = map_bset fst z2"
    then show "\<exists>x. map_bset fst z1 = map_bset fst x \<and> map_bset snd z2 = map_bset snd x \<and>
      set_bset x \<subseteq> {(x, y). (R OO S) x y}"
      by (cases z1 z2 rule: bset_cases[case_product bset_cases])
        (fastforce intro: exI[of _ "bsingleton (a, b)" for a b]
        dest: spec[of _ "bdoubleton (a, b) (c, d)" for a b c d])+
  qed
next
  case (9 R)
  then show ?case
    by (auto simp: fun_eq_iff intro: rel_bset.intros elim: rel_bset.cases)
qed

(*<*)
end
(*>*)

subset_name: curated