TITLE: Lie algebras over non-algebraically closed fields
QUESTION [13 upvotes]: I am independently studying Lie algebras (in preparation for grad school) from James Humphrey's text "Introduction to Lie Algebras and Representation Theory. Fairly early on in the development (Chapter 2), the assumption that the underlying field is algebraically closed sneaks in and remains in place all the way through to the classification of semi-simple Lie algebras in Chapter 3. I understand where we need the closure in Lie's theorem etc., but I couldn't help wondering

How far can you get in the development of Lie algebras without assuming algebraic closure of the underlying field?

Are there classifications, or partial results etc.? Where can I find them in a format suitable for a beginner? Thanks for your help!

REPLY [12 votes]: Hmmm...   I guess I should try to say something in self-defense, beyond the reference I made to Jacobson's 1962 book Lie Algebras (since republished in paperback by Dover).   It's conventional in many algebraic classifications to start with a splitting field, then descend to smaller fields by something like Galois cohomology.   In the case of semisimple Lie algebras, which emerged from Elie Cartan's work a century ago on Lie groups and differential geometry, most people simplify by working first over an algebraically closed field of characteristic 0 (often specified arbitrarily as $\mathbb{C}$).  Classically the objective was to understand simple Lie algebras (and Lie groups) over $\mathbb{R}$, but this gets quite complicated even though the field extension involved is quadratic.
The notion of splitting field here is fairly straightforward: you need all eigenvalues of ad-operators to be in your field in order to get a clean structure and classification theory.    This mostly comes down to looking at Cartan subalegras (or "maximal toral subalgebras") in semisimple Lie algebras.    But it's easier just to think about a familiar large field.    Here the classification is relatively elementary, thanks to Killing and Cartan and their successors.   But working over an arbitrary ground field gets quite intricate.   Here some motivation comes from the parallel but more difficult study of semisimple algebraic groups.
Books and lecture notes abound, but usually at a fairly high level, and some books like those in the defunct Dekker series are hard to find as well as being photocopied from typescript.   Reviews I wrote for the AMS Bulletin long ago of a couple of these are free online from AMS at emath.org, written by Goto-Grosshans and by my graduate advisor George Seligman from a different viewpoint.
I didn't mention in my various reference lists the standard old book by Helgason now republished by AMS: Differential Geometry, Lie Groups, and Symmetric Spaces.   These and more recent books are pretty formidable, as are the much older research papers on structure and classification.    It's all there, but are there readable and accessible modern references?    Good luck.

REPLY [9 votes]: Well, certainly things get more complicated when the field is not algebraically closed, as you can see from the classification of finite-dimensional simple Lie algebras over $\mathbb{R}$.  But there are many cases where one just needs to be more careful with hypotheses.

In the proof of Lie's Theorem: for a solvable Lie algebra $\mathfrak{g}$ and a finite-dimensional representation $\pi : \mathfrak{g} \to \mathfrak{gl}(V)$, you can get the theorem as long as you assume that all of the eigenvalues of all of the endomorphisms $\pi(X)$ lie in the field you are working over.  I don't know how often this happens in practice when the field is not algebraically closed, but the proof does go through.
Engel's theorem goes through without a problem over any field, I believe.
The Poincare-Birkhoff-Witt Theorem works over any field (and in fact, for Lie algebras over any commutative ring $R$ where the underlying $R$-module of the Lie algebra is free).
Cartan's Criterion for Semisimplicity works over any subfield of $\mathbb{C}$, essentially because $\mathfrak{g}$ is semisimple if and only if $\mathfrak{g} \otimes \mathbb{K}$ is semisimple for any extension field $\mathbb{K}$ (note that this is not true for simplicity; if $\mathfrak{g}$ is a simple real Lie algebra which is not obtained from a simple complex Lie algebra by restriction of scalars, then the complexification $\mathfrak{g} \otimes \mathbb{C}$ is not simple.)

Hmmm... I'm sure there's a lot more to say.  Let's just wait until Professor Humphreys shows up.


TITLE: Proof complement of sets
QUESTION [0 upvotes]: $P(A \cap  B) \geq 1- P(A^c) - P(B^c)$

I want to prove this. My attempt:
$P(A \cap  B) \geq 1- P(A^c) - P(B^c) \iff P(A \cap  B) \geq P(A) - P(B^c) \iff P(A \cap B) + P(B^c) \geq P(A)$
So I know that visually, $A$ is subsumed by $[(A \cap B )\cup B^c]$ because that expression spans the whole of $A$. But mathematically, using probabilities, how would I finish my proof here?

REPLY [0 votes]: If we split $B^c$ into a union of two disjoint sets (inside $A$ and outside $A$), we get
$\ldots \iff P(A \cap B) + P(B^c) \geq P(A) \iff P(A \cap B) + \left( P(B^c\cap A) + P((A\cup B)^c) \right) \geq P(A)$.
Then we re-group the parts of the sum:
$\ldots \iff \left( P(A \cap B) +  P(B^c\cap A) \right) + P((A\cup B)^c) \geq P(A) \iff P(A) + P((A\cup B)^c) \geq P(A)$
This concludes the proof because $P((A\cup B)^c) \geq 0$.

REPLY [0 votes]: Your attempt is fine. You could finish by stating that:
$$\Pr(A\cap B)+\Pr(B^c)=\Pr((A\cap B)\cup B^c)\geq\Pr(A)$$ Here $=$ because the sets $A\cap B$ and $B^c$ are disjoint and $\geq$ because set $A$ is a subset of set $(A\cap B)\cup B^c$.
In general if $U\subseteq V$ then $U=V\cup (U-V)$ (a disjoint union) so that: $$\Pr(U)=\Pr(V\cup (U-V)=\Pr(V)+\Pr(U-V)\geq\Pr(V)$$


text:
{\bf Problem.} Let $\mathcal P$ be a parabola, and let $V_1$ and $F_1$ be its vertex and focus, respectively. Let $A$ and $B$ be points on $\mathcal P$ so that $\angle AV_1 B = 90^\circ$. Let $\mathcal Q$ be the locus of the midpoint of $\overline{AB}$. It turns out that $\mathcal Q$ is also a parabola, and let $V_2$ and $F_2$ denote its vertex and focus, respectively. Determine the ratio $\frac{F_1F_2}{V_1V_2}$.
{\bf Level.} Level 5
{\bf Type.} Intermediate Algebra
{\bf Solution.} Since all parabolas are similar, we may assume that $\mathcal P$ is the curve $y = x^2,$ so $V_1 = (0,0).$  Then, if $A = (a, a^2)$ and $B = (b, b^2)$, the slope of line $AV_1$ is $a,$ and the slope of line $BV_1$ is $b.$  Since $\angle AV_1 B = 90^\circ,$ $ab = -1$. Then, the midpoint of $\overline{AB}$ is \[
\left( \frac{a+b}{2}, \frac{a^2 + b^2}{2} \right) = \left( \frac{a+b}{2}, \frac{(a+b)^2 - 2ab}{2} \right)
= \left( \frac{a+b}{2}, \frac{(a+b)^2}{2} + 1 \right).
\](Note that $a+b$ can range over all real numbers under the constraint $ab = - 1$.) It follows that the locus of the midpoint of $\overline{AB}$ is the curve $y = 2x^2 + 1$.

Recall that the focus of $y = ax^2$ is $\left(0, \frac{1}{4a} \right)$. We find that $V_1 = (0,0)$, $V_2 = (0,1)$, $F_1 = \left( 0, \frac 14 \right)$, $F_2 = \left( 0, 1 + \frac18 \right)$. Therefore, $\frac{F_1F_2}{V_1V_2} = \boxed{\frac78}$.

set_name: MATH


TITLE: Classification of principal $G$-bundles over closed oriented surfaces with $G$ connected.
QUESTION [4 upvotes]: Let $\Sigma$ be a closed oriented surface and $G$ a connected Lie group. Then, it is known that

principal $G$-bundles $P\to\Sigma$ are topologically classified by a characteristic class $c(P)\in  H^2(\Sigma,\pi_1 G)\cong\pi_1 G$.

So, why is this true? Any reference? Thanks.

REPLY [5 votes]: I don't know of a reference, but this follows from classifying space theory and a bit of homotopy theory. (Having written the answer, I now realise there is probably an easier answer using obstruction theory.)
Let $G$ be a topological group, then there is a topological space $BG$ and a principal $G$-bundle $EG \to BG$, called the universal principal $G$-bundle. It is so called because for any principal $G$-bundle over a paracompact space $X$, there is a map (unique up to homotopy) $f : X \to BG$ such that $f^*EG \to X$ is isomorphic to the given principal $G$-bundle. The universal bundle is uniquely determined (up to isomorphism) by the fact that $EG$ is weakly contractible.
It follows that the isomorphism classes of principal $G$-bundles, denoted $\operatorname{Prin}_G(X)$, are in one-to-one correspondence with homotopy classes of maps $X \to BG$, denoted $[X, BG]$. In particular, for the case you asked about, we have $\operatorname{Prin}_G(\Sigma) = [\Sigma, BG]$.
The long exact sequence in homotopy applied to the universal principal $G$-bundle $EG \to BG$ gives
$$\dots \pi_{n+1}(EG) \to \pi_{n+1}(BG) \to \pi_n(G) \to \pi_n(EG) \to \dots$$
As $EG$ is weakly contractible, we see that $\pi_{n+1}(BG) = \pi_n(G)$.
If $X$ is an $n$-dimensional CW complex and $Y$ is a CW complex, then it follows from the cellular approximation theorem that $[X, Y] = [X, Y^{(n+1)}]$ where $Y^{(n+1)}$ denotes the $(n+1)$-skeleton of $Y$. If $X$ and $Y$ are pointed, then the same is true of basepoint-preserving homotopy classes. In particular, $\pi_n(Y) = \pi_n(Y^{(n+1)})$.
You can attach cells on to $BG$ of dimension at least four so that the resulting space, call it $Z$, has $\pi_n(Z) = 0$ for $n \geq 3$ (attach an $(n + 1)$-cell for each generator of $\pi_n(BG)$)*. Moreover, because the cells that were added were of dimension at least five, $Z$ has the same $3$-skeleton as $BG$. Again by cellular approximation, $\pi_2(Z) = \pi_2(Z^{(3)}) = \pi_2(BG^{(3)}) = \pi_2(BG) = \pi_1(G)$ (note, this implies $\pi_1(G)$ is abelian), and likewise $\pi_1(Z) = \pi_1(BG) = \pi_0(G) = 0$ as $G$ is connected. Therefore $Z$ is an Eilenberg-MacLane space, namely $Z = K(\pi_1(G), 2)$.
The identity map $\operatorname{id} : \pi_1(G) \cong \pi_2(K(\pi_1(G), 2)) \cong H_2(K(\pi_1(G), 2); \mathbb{Z}) \to \pi_1(G)$ gives rise to an element of $\operatorname{Hom}(H_2(K(\pi_1(G), 2); \mathbb{Z}), \pi_1(G))$ and hence** to an element $\alpha \in H^2(K(\pi_1(G), 2); \pi_1(G))$. For any $[g] \in [\Sigma, K(\pi_1(G), 2)]$, $[g] \mapsto g^*\alpha \in H^2(\Sigma; \pi_1(G))$ defines a bijection $[\Sigma, K(\pi_1(G), 2)] \to H^2(\Sigma; \pi_1(G))$, so we see that
$$\operatorname{Prin}_G(\Sigma) = [\Sigma, BG] = [\Sigma, BG^{(3)}] = [\Sigma, Z^{(3)}] = [\Sigma, Z] = [\Sigma, K(\pi_1(G), 2)] = H^2(\Sigma; \pi_1(G)).$$
We have an inclusion $\iota : BG \to K(\pi_1(G), 2)$. The overall isomorphism $[\Sigma, BG] \to [X, K(\pi_1(G), 2)]$ is given by $[f] \mapsto [\iota\circ f]$. Therefore the isomorphism $[\Sigma, BG] \to H^2(\Sigma; \pi_1(G))$ is given by $[f] \mapsto (\iota\circ f)^*\alpha = f^*(\iota^*\alpha) = f^*\beta$ where $\beta := \iota^*\alpha \in H^2(BG; \pi_1(G))$.
If $P \to \Sigma$ is a principal $G$-bundle, then $c(P) \in H^2(\Sigma; \pi_1(G))$ is given by $c(P) = f^*\beta$ where $f : \Sigma \to BG$ is a classifying map for $P$.
As $\Sigma$ is closed and oriented, $H^2(\Sigma; \pi_1(G)) \cong \pi_1(G)$; in fact, you only need $\Sigma$ to be orientable for this to hold. For certain groups $\pi_1(G)$ (e.g. $\mathbb{Z}_2^k$), the isomorphism still holds even in the non-orientable case. If $\Sigma$ is not closed, then $\Sigma$ is homotopy equivalent to a bouquet of circles, so $H^2(\Sigma; \pi_1(G)) = 0$ and hence all principal $G$-bundles over $\Sigma$ are trivial.

If one assumes that $G$ is also simply connected, then the same argument can be used to show that principal $G$-bundles $P \to M$ where $M$ is a manifold (or even just a CW complex) of dimension up to four are classified by a characteristic class $c(P) \in H^4(M; \pi_3(G))$. For example, in this answer the above argument is applied when $G = SU(2)$ and the characteristic class $c(P)$ is identified as $c_2(P)$ (or possibly $-c_2(P)$).

*In fact, $\pi_3(BG) = \pi_2(G) = 0$ as the second homotopy group of a Lie group is always zero, so we don't even need to add any four-dimensional cells and hence $Z$ and $BG$ have the same $4$-skeleton (not that we need this fact).
**Here we are using the fact that $\pi_1(G)$ is abelian, and therefore a $\mathbb{Z}$-module, so we can use the Universal Coefficient Theorem for cohomology.


text:
\begin{document}
\maketitle

\begin{abstract}
Let $F$ be an $n$-point set in $\mathbb{K}^d$ with $\mathbb{K}\in\{\mathbb{R},\mathbb{Z}\}$ and $d\geq 2$. A (discrete) X-ray of $F$ in direction $s$ gives the number of points of $F$ on each line parallel to $s$. We define $\psi_{\mathbb{K}^d}(m)$ as the minimum number $n$ for which there exist $m$ directions $s_1,\dots,s_m$ (pairwise linearly independent and spanning $\mathbb{R}^d$) such that two $n$-point sets in $\mathbb{K}^d$ exist that have the same X-rays in these directions. The bound $\psi_{\mathbb{Z}^d}(m)\leq 2^{m-1}$ has been observed many times in the literature. In this note we show $\psi_{\mathbb{K}^d}(m)=O(m^{d+1+\varepsilon})$ for $\varepsilon>0$. For the cases $\mathbb{K}^d=\mathbb{Z}^d$ and $\mathbb{K}^d=\mathbb{R}^d$, $d>2$, this represents the first upper bound on $\psi_{\mathbb{K}^d}(m)$ that is polynomial in~$m$. As a corollary we derive bounds on the sizes of solutions to both the classical and two-dimensional Prouhet-Tarry-Escott problem. Additionally, we establish lower bounds on $\psi_{\mathbb{K}^d}$ that enable us to prove a strengthened version of R{\'e}nyi's theorem for points in $\mathbb{Z}^2$.
\end{abstract}

\section{Introduction}
The problem of reconstructing point sets from their X-rays has a long history; perhaps the 1952 paper~\cite{renyi52} by R{\'e}nyi represents one of the first works in this field. Of special interest are questions of uniqueness. Two sets with the same X-rays are said to be \emph{tomographically equivalent} \cite{gritzmannlangfeld}, \cite{hajdutijdeman01}; the sets are also commonly referred to as \emph{switching components} \cite{kongherman98}, \cite{shilferstein} or \emph{ghosts} \cite[Sect.~15.4]{herman09}. In \cite{matousek08} Matou{\v{s}}ek, P{\v{r}}{\'{\i}}v{\v{e}}tiv{\'y}, and {\v{S}}kovro{\v{n}} show that almost all sets of $m$~directions (in the sense of measure) allow for a unique reconstruction of $2^{Cm/\log(m)}$-point sets in the real plane (here $C>0$ is a constant and the result holds for large~$m$). For almost all choices of $m$ directions there thus exist only superpolynomial size switching components. By a careful selection of directions, however, we can reduce them to a polynomial size.

To make this precise, let $F$ be an $n$-point set in $\mathbb{K}^d$ with $\mathbb{K}\in\{\mathbb{R},\mathbb{Z}\}$ and $d\geq 2$. A (discrete) X-ray of~$F$ in direction $s$ gives the number of points of $F$ on each line parallel to $s$. We define $\psi_{\mathbb{K}^d}(m)$ as the minimum number $n$ for which there exist $m$ directions $s_1,\dots,s_m$ (pairwise linearly independent and spanning $\mathbb{R}^d$) such that two different $n$-point sets in $\mathbb{K}^d$ exist that have the same X-rays in these directions. We derive lower and upper bounds on $\psi_{\mathbb{K}^d}$.

Two constructions are known to yield upper bounds on $\psi_{\mathbb{K}^d}$.
The first construction is based on regular polygons. The two disjoint $m$-point sets of alternate vertices of a regular $2m$-gon in $\mathbb{R}^2$ yield $\psi_{\mathbb{R}^2}(m)\leq m$. This cannot be transfered to $\mathbb{Z}^d$ as any (planar) regular polygon with integer vertices must have $3$, $4$ or $6$ vertices~\cite{scherrer,chrestenson63}. The functions $\psi_{\mathbb{R}^2}$ and $\psi_{\mathbb{Z}^2}$ are, in fact, different functions as we show $\psi_{\mathbb{Z}^2}(m)\geq m+1$ if $m=5$ or $m>6$ (see Thm.~\ref{thm:lowerbound}). From this we derive a strengthened version of R{\'e}nyi's theorem (see Thm.~\ref{thm:thm1} and Cor.~\ref{cor:cor1} in Sect.~\ref{sect:lowerbounds}).

The second well-known construction for upper bounds on $\psi_{\mathbb{K}^d}$ is based on two-colorings of the unit cube $[0,1]^m$ in $\mathbb{Z}^m$. More precisely, two different sets with equal X-rays in coordinate directions are obtained as the two disjoint sets of $2^{m-1}$ alternate vertices of  $[0,1]^m$. By projecting into~$\mathbb{Z}^d$, the bound $\psi_{\mathbb{Z}^d}(m)\leq 2^{m-1}$ is obtained. This construction seems to be due to Lorentz~\cite{lorentz49}; see also \cite{bianchilonginetti}, \cite[Lem.~2.3.2]{gardner}, and \cite[Thm.~4.3.1]{gardnergritzmannschapter4}. As $\mathbb{Z}^d\subseteq \mathbb{R}^d$, this, of course, yields also $\psi_{\mathbb{R}^d}(m)\leq 2^{m-1}$.

Our main observation is contained in the statement of Thm.~\ref{thm:maintheorem}, where we prove $\psi_{\mathbb{Z}^d}(m)=O(m^{d+1+\varepsilon})$ for $\varepsilon>0$. This is, to our knowledge, the first upper bound on $\psi_{\mathbb{K}^d}(m)$ that is polynomial in~$m$. Our proof is non-constructive.

We conclude in Sect.~\ref{sect:consequences} by stating some remarks and consequences that relate our bounds to the \emph{Prouhet-Tarry-Escott} problem from number theory (see, e.g., \cite[Sect.~21.9]{hardywright08}).

Throughout the paper, $\zeta$ is the Riemann zeta function, $m$ and $n$ denote natural numbers, and $\mathbb{Z}$, $\mathbb{R}$, $\mathbb{N}=\{1,2,\dots\}$ are, respectively, the sets of integers, reals, and natural numbers. We use the notation $\mathbb{N}_0=\mathbb{N}\cup\{0\}$, $[n]=\{1,\dots,n\}$, and $2\mathbb{Z}=\{2z\::\:z\in \mathbb{Z} \}$. With $G_n^d=[n]^d$ we denote the set of $d$-tuples of positive integers less than or equal to $n$. If $\xi\in\mathbb{R}$, then $\lceil \xi \rceil$ denotes the smallest integer greater than or equal to $\xi$. The symbol $O$ has the usual meaning: $f(m)=O(g(m))$ means that $f(m)/g(m)$ is bounded as $m \to \infty$. A property is said to hold for large $m$ if that property holds for all $m$ larger than some $m_0$.

\section{Lower bounds}\label{sect:lowerbounds}
In this section we derive lower bounds on $\psi_{\mathbb{K}^d}$. The key ideas are not new, but appear scattered and isolated in different contexts in the literature (see \cite{renyi52} and the proof of Thm.~\ref{thm:lowerbound} in \cite{alpers-tijdeman-07}).

\begin{thm}\label{thm:thm1}
For every $d\geq2$ we have $\psi_{\mathbb{K}^d}(m)\geq m$.
\end{thm}
\begin{proof}
This is a reformulation of R{\'e}nyi's theorem (proved in \cite{renyi52} and generalized to arbitrary dimensions by Heppes~\cite{heppes}), which states that any $n$-point set in $\mathbb{K}^d$ is uniquely determined by its X-rays from $n+1$ different directions.
For completeness, we reproduce a short proof. Suppose there are two sets $F, F'$ with equal X-rays in $m+1$ directions, each set containing at most $m$ points. Without loss of generality there exists a point $p \in F\setminus F'$. Since $F$ and $F'$ have equal X-rays, there needs to be a point of $F'$ on each of the $m+1$ lines through $p$. This implies that $F'$ contains at least $m+1$ points, a contradiction.
\end{proof}

The bound is tight for $m\in\{1,2,3,4,6\}$ and $\mathbb{K}^d=\mathbb{Z}^2$; examples showing this for $m=1,2,3,4,6$ are respectively provided by any two $1$-point sets in $\mathbb{Z}^2$,  two-colorings of the unit cube in $\mathbb{Z}^2$, the sets $F=\{(0,0),(1,2),(2,1)\}$, $F'=\{(1,0),(0,1),(2,2)\}$, and the examples shown in Fig.~4.3 and Fig.~4.5 of~\cite{gardnergritzmannschapter4}. For the remaining cases, however, we can improve the bound as stated in the following result.
\begin{thm} \label{thm:lowerbound}
If $m=5$ or $m>6$ then $\psi_{\mathbb{Z}^2}(m)\geq m+1$.
\end{thm}
\begin{proof}
Let $m=5$ or $m>6$, and suppose there exist different $n$-point sets $F,F' \subseteq \mathbb{Z}^2$ with equal X-rays in $m\geq n$ directions. Without loss of generality we can assume that $F \cap F'=\emptyset$. The convex hull $P$ of $F\cup F'$ is a non-degenerate polygon with at most $2n$ vertices.  Parallel to each of the $m$ directions there are two lines that support $P$ with each line containing a single point from $F$ and $F'$, respectively (since otherwise one of $F$ and $F'$ contains more than $n$ points). Since this implies that $P$ has at least $2m$ edges, we conclude that at least $2m$ of the elements of $F\cup F'$ are vertices of $P$ (i.e, $n=m$), proving that $F\cup F'$ is the set of vertices of the non-degenerate convex $2m$-gon $P$. Since $F$ and $F'$ have the same X-rays, $P$ has the property that any line through a vertex of $P$ in any of the $m$ directions meets another vertex of~$P$. Such polygons are known as lattice $U$-gons with $U$ denoting the set of $m$ directions. They, however, do not exist for $m >6$ (see Thm.~4.5 in \cite{gardnergritzmann97}). As is shown in the proof of Thm.~4.5 of \cite{gardnergritzmann97} or (more simply) in Thm.~6 of \cite{alpers-tijdeman-07}, there are also no lattice $U$-gons for exactly $5$ directions. In other words, we have $\psi_{\mathbb{Z}^2}(m)> m$ for $m=5$ or $m>6$.
\end{proof}

The bound is tight for $m=5$.  For this consider the $6$-point sets \[F=\{(0,2),(1,4),(2,2),(3,0),(4,3),(5,1)\} \textnormal{ and } F'=\{(0,3),(1,1),(2,4),(3,2),(4,0),(5,2)\}.\] It is easily verified that $F$ and $F'$ have the same X-rays in the $5$ directions \[S=\{(1,0),(0,1),(1,1),(1,-1), (-2,1)\}.\]

A reformulation of Thm.~\ref{thm:lowerbound} provides a strengthened version of R{\'e}nyi's theorem for $\mathbb{Z}^2$.
\begin{cor}\label{cor:cor1}
Any $n$-point set in $\mathbb{Z}^2$ with $n=5$ or $n>6$ is uniquely determined by its X-rays taken from at least~$n$ different directions.
\end{cor}

\section{Upper bounds}\label{sect:upperbounds}
In this section we prove a polynomial upper bound on $\psi_{\mathbb{K}^d}$. As a prelude, we prove an upper bound on the number of lines parallel to a given direction that intersect points of $G_n^d$. This is followed by a lemma that asserts the existence of certain coverings of a specified finite part of the integer lattice by $m$ families of parallel lines.

\begin{lem}\label{lem:lemq1}
For any relatively prime $d$-tuple $s=(\sigma_1,\dots,\sigma_d)\in\mathbb{N}_0^d\setminus\{0\}$ with $d\geq2$ there are at most $dn^{d-1}\cdot\max\{\sigma_1,\dots,\sigma_d\}$ lines parallel to $s$ that intersect~$G_n^d$.
\end{lem}
\begin{proof}
For each line $\ell$ parallel to $s=(\sigma_1,\dots,\sigma_d)$ that intersects $G_n^d$, there is a unique point $p \in \ell\cap G_n^d$ for which $p-s \not\in G_n^d$. The point $p-s$ needs to have a non-positive component, i.e.,
\[p\in V_i=\{(\xi_1,\dots,\xi_d)\in G_n^d\::\: 1\leq \xi_i\leq \sigma_i \}\] for an $i \in [d]$. As the number of points in $\bigcup_{i=1}^dV_i$ is clearly bounded by $dn^{d-1}\cdot\max\{\sigma_1,\dots,\sigma_d\}$, we obtain the claimed result. (Tight bounds can be obtained similary via the inclusion-exclusion principle, but they are not needed in the present context.)
\end{proof}

\begin{lem} \label{lem:lemT1}
Let $\varepsilon>0$, $m \in \mathbb{N}$, $d\geq 2$, and $n\in\left\{\left\lceil m^{1+(1+\varepsilon)/d}\right\rceil,\left\lceil m^{1+(1+\varepsilon)/d}\right\rceil+1\right\}$. Then, for large $m$ there is a set $S=\{s_1,\dots,s_m\}\subseteq \mathbb{Z}^d$ with the property that
\begin{enumerate}[(i)]
\item the elements of $S$ are pairwise linearly independent spanning $\mathbb{R}^d$;
\item the total number $l$ of lines that are parallel to a direction in $S$ and intersect $G_{n}^d$ is bounded from above by

$2^{1+1/d} d {n}^{d-1} m^{1+1/d}.$

\end{enumerate}
\end{lem}
\begin{proof}
For the number $R(p,d)$ of relatively prime $d$-tuples in $G_p^d$, $p\in\mathbb{N}$, it holds by \cite{nymann72} that \[
\lim_{p \to \infty}\frac{R(p,d)}{p^d}=\frac{1}{\zeta(d)}.
\] As $\zeta$ decreases for values larger than $1$ and since $\zeta(2)=\pi^2/6<2$, we have \[ R(p,d)> p^d/2 \] for large $p$.

Setting $q=\left\lceil (2m)^{1/d}\right\rceil$, we note that $q \leq 2(2m)^{1/d}$ and $q \leq n$ for $m\geq2$. For large $m$ we have
\[ R(q,d)> q^d/2\geq m, \] so for our set $S$ we can select $m$ elements from $G_{q}^d\subseteq G_{n}^d$. We can assume that the elements of $S$ span~$\mathbb{R}^d$ since otherwise we replace~$d$ of the directions by the standard unit vectors. Property~(i) is thus fulfilled (note that the elements of $S$ are relatively prime $d$-tuples).

The entries of the elements in $S$ are bounded by $q$, so by Lem.~\ref{lem:lemq1} we have at most
\[
mdn^{d-1}q\leq 2^{1+1/d} d {n}^{d-1} m^{1+1/d}
\]
lines parallel to a direction in $S$ that intersect $G_{n}^d$.
\end{proof}

\begin{thm} \label{thm:maintheorem}
For every $\varepsilon>0$ and $d\geq2$ it holds that $\psi_{\mathbb{Z}^d}(m)=O(m^{d+1+\varepsilon})$.
\end{thm}
\begin{proof}
We assume that $m$ is large enough that the set $S$ from Lem.~\ref{lem:lemT1} exists. We set \[n=\left\{\left\lceil m^{1+(1+\varepsilon)/d}\right\rceil,\left\lceil m^{1+(1+\varepsilon)/d}\right\rceil+1\right\}\cap 2\mathbb{Z}\] and $k=\frac{1}{2}n^d$. Note that $k \in \mathbb{N}$,  $k=O(m^{d+1+\varepsilon})$, and that we can assume that $n \geq 4$.

Let $l_i$, $i\in[m]$, denote the number of lines parallel to $s_i$ that intersect $G_n^d$.
The X-ray in direction $s_i$ of a set in $G_{n}^d$ with cardinality $k$ gives a \emph{weak $k$-composition of $l_i$}, i.e., a  solution to $\xi_1+\cdots+\xi_{l_i}=k$ in nonnegative integers \cite[p.~15]{stanley}. (The converse is generally false, because the corresponding X-ray lines may intersect~$G_{n}^d$ in fewer points than provided by a weak $k$-composition of $l_i$.)
The number of weak $k$-compositions of $l_i$ is given by \[N(k,l_i)= {k+l_i-1 \choose l_i-1}\] and thus represents an upper bound for the number of different X-rays of $k$-point subsets of $G_n^d$ in the direction $s_i$.

With $l=l_1+\cdots+l_m$ we thus obtain the following upper bound on the number of different X-rays (for the directions in $S$) that can originate from a subset of $G_{n}^d$ with cardinality $k$:
\[
 \prod_{i=1}^mN(k,l_i)\leq\prod_{i=1}^m{n^d/2+l_i\choose l_i} \leq \prod_{i=1}^m\left(\frac{(n^d/2+l_i)e}{l_i}\right)^{l_i}
=\prod_{i=1}^m\left( \frac{n^de}{2l_i}+e\right)^{l_i} \leq (ne+e)^{l}\leq n^{2l};
\]
here the inequalities (from left to right) follow from $N(k,l_i)\leq N(k,l_i+1)$, a standard inequality for binomial coefficients (see, e.g., \cite[Eq.~(4.9)]{odlyzko}), $l_i \geq n^{d-1}$, and  $n\geq4$, respectively.

There are
\[
{n^d\choose n^d/2} \geq 2^{n^d/2}\label{eq:FF1}
\]
subsets of cardinality $k$ in $G_{n}^d$. We claim that
\[
n^{2l} < 2^{n^d/2}
\] holds for large $m$, which, by the pigeonhole principle, concludes the proof as it implies the existence of two sets in $G_{n}^d$ with cardinality $k$ and equal X-rays in the directions in $S$.

For the claim we first note that
\begin{equation}m^{1+(1+\varepsilon)/d} \leq n\leq 3m^{1+(1+\varepsilon)/d}   \label{eq:eqww1} \end{equation}
holds as $m^{1+(1+\varepsilon)/d}\geq 1$. It is easy to see that $\lim_{x \to \infty}x^{a}/2^{x^{b}}=0$ for $a,b>0$.
Thus for large~$m$ and $C=2^{3+1/d}d$
we have
\[ 3^{C}m^{C(1+(1+\varepsilon)/d)}<2^{m^{\varepsilon/d}},\] which, by~(\ref{eq:eqww1}) and Property~(ii) of Lem.~\ref{lem:lemT1}, gives
\[ n^{C}< 2^{m^{\varepsilon/d}}
\Rightarrow  n^{Cm^{1+1/d}}<2^n
\Rightarrow  n^{Cn^{d-1}m^{1+1/d}}<2^{n^d}
\Rightarrow  n^{4l}<2^{n^d},
\]
proving the claim.
\end{proof}

\section{Remarks and consequences}\label{sect:consequences}
The previously mentioned regular $2m$-gon construction in $\mathbb{R}^2$, together with the inequality $\psi_{\mathbb{R}^d}(m)\leq \psi_{\mathbb{Z}^d}(m)$ for $d\geq 2$, yields the following corollary to Thm.~\ref{thm:maintheorem}.
\begin{cor}
 For every $\varepsilon>0$ and $d\in\mathbb{N}$, it holds that
 \[ \psi_{\mathbb{R}^d}(m)=\left\{ \begin{array}{lll} m &\textnormal{if}& d=2,\\ O(m^{d+1+\varepsilon}) &\textnormal{if}& d>2.\end{array}\right.\]
\end{cor}

In \cite{alpers-tijdeman-07} the  \emph{general Prouhet-Tarry-Escott problem} (PTE$_r$) was introduced: Given $k$, $n$, $r \in \mathbb{N}$, find two different multi-sets $\{x_1,\dots,x_n\}\!$, $\{y_1,\dots,y_n \}\subseteq\mathbb{Z}^r$
where $x_i=(\xi_{i1}, \dots, \xi_{ir})$, $y_i=(\eta_{i1}, \dots, \eta_{ir})$ for $i\in[n]$  such that
$$  \sum_{i=1}^n \xi_{i1}^{j_1}\xi_{i2}^{j_2} \cdots \xi_{ir}^{j_r}=\sum_{i=1}^n \eta_{i1}^{j_1}\eta_{i2}^{j_2} \cdots \eta_{ir}^{j_r} $$
for all nonnegative integers  $j_1, \dots, j_r$ with $j_1+\cdots+ j_r \leq k.$ The parameter $k$ is called the \emph{degree} and~$n$ the \emph{size} of the solution.
Tracing back to works of Euler and Goldbach \cite[p.~705]{dickson}, the Prouhet-Tarry-Escott problem (PTE$_1$) is an old and largely unsolved problem in Diophantine analysis. The following corollary sharpens the bound of \cite[Thm.~12]{alpers-tijdeman-07} on the size of solutions, which for (PTE$_1$) is due to Prouhet~\cite{prouhet51}.

\begin{cor} \label{cor:cor2}
For every $\varepsilon>0$ there exists a constant $C>0$ such that there are solutions of (PTE$_2$) of degree $k$ and size bounded by $Ck^{3+\varepsilon}$.
\end{cor}
\begin{proof}
In Thm.~8 of \cite{alpers-tijdeman-07} it was shown that tomographically equivalent sets in $\mathbb{Z}^2$ for $m$ directions yield (PTE$_2$) solutions of degree $m-1$. This and Thm.~\ref{thm:maintheorem} for $d=2$ imply the statement of this corollary.
\end{proof}

\begin{rema}
As the products cancel, it is evident that solutions of (PTE$_1$) can be obtained by applying to (PTE$_2$) solutions a suitable linear functional that maps $(\xi_1,\xi_2)\in\mathbb{Z}^2$ to $\alpha_1\xi_1+\alpha_2\xi_2$
where $\alpha_1,\alpha_2 \in \mathbb{Z}$ are suitably chosen. The current best bounds for (PTE$_1$) are quadratic in~$k$ (see \cite{mezak61}, \cite{wright35}); the bound from Thm.~\ref{thm:maintheorem} is in this case weaker.
\end{rema}

config: arxiv


TITLE: Zero volume at zero Kelvin
QUESTION [1 upvotes]: Why does the volume of a gas become zero at 0 Kelvin? Can a Bose Einstein condensate be considered as matter? (I mean the volume becomes zero)

REPLY [3 votes]: An ideal gas at 0 K doesn't necessarily have zero volume. It depends on the experimental setup; "experimental" in an idealized sense, since the ideal gas is a purely theoretical construct.
Let's consider it from a macroscopic point of view first, that is, within thermostatics. Its pressure $p$, volume $V$, temperature $T$, and mole number $n$ are related by the ideal gas law:
$$pV = nRT,$$
with $R$ the ideal-gas constant.
If $T=0$ we have $pV=0$. This means that we can have non-zero volume, but the pressure will be zero. This equation also says that we could have non-zero pressure, but will have zero volume then.
Let's consider these two situations from a microscopic point of view now. The ideal gas is a collection of point particles: they have mass, momentum, kinetic energy but don't occupy any space; and they don't exert any kind of force on one another. The pressure of the gas on the walls of a container comes from the exchange of momentum between particles and walls. The temperature of the gas is the spatial average of the molecules' kinetic energies (I say more about statistical averages below).
At constant temperature we can control the pressure of an ideal gas by letting one wall be freely moving with a constant force exerted on the other side. For example a piston pressed by a weight, or pushed by a spring. In equilibrium the wall is still and the pressure exerted by the gas must equal the force/area on the piston.
We can control the volume by simply letting the walls be unmovable.
And how do we control the temperature of an ideal gas? One way is to let the walls of the container have local microscopic movements: when a particle of the gas is approaching a wall, we can let the impact region of the wall have an extra velocity – hence kinetic energy and momentum – towards the particle, or receding from it, for a very short time, then restore its position. This will give extra kinetic energy and momentum to the particle, or take some from it. Obviously this mental experiment is mimicking the effect of the molecules that constitute a real wall. Another way to control temperature is possible when the particles have some extra physical property that allow us to exert a force at a distance on them. Electrical or gravitational forces are excluded, however, otherwise the particles would exert forces on one another (this also tells us that the particles' masses must be enough small for gravitational attraction to be negligible).
At 0 K the particles of the ideal gas have no velocity: they stay still. If the gas is within an enclosure with unmovable walls, it will have that volume but will exert no pressure, as we found from the ideal-gas law. It's interesting to note that if we reach 0 K by modifying the particles' kinetic energies in the way described above, all the particles must eventually be in contact with the walls: the wall takes the last amount of kinetic energy from a particle, via a microscopic receding motion when the particles hits, and the particle thus stays still in contact with the wall.
Reaching 0 K at constant pressure, the moving wall eventually reaches the opposite wall of the container, so the final pressure is really just the contact, normal force exerted by the opposite wall, rather than the momentum released by the point particles.
This is just to give a pictorial intuition. The extreme situations implied by the processes envisaged above make clear that this is a purely mental construct. Already gravity alone introduces contradictions between the particles' having mass and yet be non-interacting within a volume shrinking to zero.
References for all this are, for example,

Chapman, S., Cowling, T. G. (1995): The Mathematical Theory of Non-uniform Gases: An Account of the Kinetic Theory of Viscosity, Thermal Conduction and Diffusion in Gases (3rd ed., Cambridge University Press).
Truesdell, C. A., Muncaster, R. G. (1980): Fundamentals of Maxwell's Kinetic Theory of a Simple Monatomic Gas: Treated as a Branch of Rational Mechanics (Academic Press).

Note on the definition of temperature
We can also define temperature as a statistical average if we introduce ensembles (Gibbs's statistical mechanics) or number-density functions (kinetic theory). The results are the same. More about all these kinds of temperatures and the interesting and troublesome relations between them can be found in

Biró, T. S.: Is There a Temperature?: Conceptual Challenges at High Energy, Acceleration and Complexity (Springer 2001).


TITLE: Number of compositions of $n$ such that each term is less than equal to $k.$
QUESTION [6 upvotes]: Let $n$ be an integer $\geq 1.$ Then a partition of $n$ is a sequence of positive integers (greater than equal to $1$) such that their sum equals $n.$ So for instance if $n=4$ then
$$[[4], [1, 3], [1, 1, 2], [1, 1, 1, 1], [1, 2, 1], [2, 2], [2, 1, 1], [3, 1]]$$
is a collection of all partitions of $n$ with order. Clearly for any $n$ the number of such ordered partitions is $2^{n-1}.$ However I want to count the number of paritions of $n$ where each integer in the parition is less than equal to some integer $k.$ So for instance if $n=4$ and $k=2$ then
$$[[1, 1, 2], [1, 1, 1, 1], [1, 2, 1], [2, 2], [2, 1, 1]]$$
is a collection of all the $2$-partitions of $4.$ Is there a general formula for finding this or maybe an asymptotic expression?

REPLY [1 votes]: First a note about terminology:
 - a Partition of a positive integer $n$ is a non-decreasing sequence of positive integers summing to $n$;
 - a Composition of a positive integer $n$ is an unordered sequence of positive integers summing to $n$.
That premised, you are speaking of the number of Compositions of $n$, whose terms (parts) are not-greater than $k$.
Consider the case in which you seek for the number of compositions of the positive number $s+m$
into $m$ parts not exceeding $r+1$
$$
{\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{
  {\rm 1} \le {\rm integer}\;y_{\,j}  \le r + 1 \hfill \cr
  y_{\,1}  + y_{\,2}  + \; \cdots \; + y_{\,m}  = s + m \hfill \cr}  \right.
$$
that's the same as
$$N_{\,b} (s,r,m) = \text{No}\text{. of solutions to}\;\left\{ \begin{gathered}
  0 \leqslant \text{integer  }x_{\,j}  \leqslant r \hfill \\
  x_{\,1}  + x_{\,2}  +  \cdots  + x_{\,m}  = s \hfill \\
\end{gathered}  \right.$$
$N_b$ is given by the following sum
$$
N_b (s,r,m)\quad \left| {\;0 \leqslant \text{integers  }s,m,r} \right.\quad  =
\sum\limits_{\left( {0\, \leqslant } \right)\,\,j\,\,\left( { \leqslant \,\frac{s}{r+1}\, \leqslant \,m} \right)}
{\left( { - 1} \right)^k \binom{m}{j}
 \binom
 { s + m - 1 - j\left( {r + 1} \right) }
 { s - j\left( {r + 1} \right)}\ }
$$
as widely explained in this related post.
Going back to your case, the
number of compositions of $n$ into $m$ parts not greater than $k$
will then be
$$
\eqalign{
  & N_c (n,k,m)\quad \left| {\;1 \le n,m,k} \right.\quad  =   \cr
 & = \sum\limits_{\left( {0\, \le } \right)\,\,j\,\,\left( {\, \le \,m} \right)} {\left( { - 1} \right)^{\,j} \binom{m}{j}
\binom{n - 1 - j\,k}{ n - m - j\,k}} \cr}
$$
while the
overall number of compositions of $n$ into parts not greater than $k$
 will be the sum
of the above for $0 \le m$ : if the sum extends over $n$ it will maintain the value $2^{n-1}$.
$$ \bbox[lightyellow] {
\eqalign{
  & N_{c\,t} (n,k)\quad \left| {\;0 \le n,k \in \mathbb Z} \right.\quad  =   \cr
  &  = \sum\limits_{0\, \le \,\,m} {\;\sum\limits_{\left( {0\, \le } \right)\,\,j\,\,\left( {\, \le \,m} \right)} {\left( { - 1} \right)^{\,j}
    \binom{m}{j} \binom{ n - 1 - j\,k}{n - m - j\,k}
} }  \cr}
}$$
In the example with $n=4$, the above gives for $k=1,2,3,4$
$$
1,5,7,8
$$
which in fact correspond to
$$
\eqalign{
  & \left[ {1,1,1,1} \right]  \cr
  & \left[ {1,1,1,1} \right]\left[ {1,1,2} \right]\left[ {1,2,1} \right]\left[ {2,1,1} \right]\left[ {2,2} \right]  \cr
  & \left[ {1,1,1,1} \right]\left[ {1,1,2} \right]\left[ {1,2,1} \right]\left[ {2,1,1} \right]\left[ {2,2} \right]\left[ {1,3} \right]\left[ {3,1} \right]  \cr
  & \left[ {1,1,1,1} \right]\left[ {1,1,2} \right]\left[ {1,2,1} \right]\left[ {2,1,1} \right]\left[ {2,2} \right]\left[ {1,3} \right]\left[ {3,1} \right]\left[ 4 \right] \cr}
$$
Finally note that:
 - $N_{c\,t}(n,k)$ correctly checks with OEIS seq. A126198, which provides further properties of these numbers;
 - the o.g.f. of $N_{c\,t}$ in $n$ is in fact the $F(x)$ given by @jmerry's answer (re. to the o.g.f. for $N_b$ given in related post);
$$
\sum\limits_{0\, \le \,\,n} {N_{c\,t} (n,k)\,x^{\,n} }  = {{1 - x} \over {1 - 2x + x^{\,k + 1} }}
$$
 - $N_{c\,t}(n,k)$ satisfies the recursion given by @NobleMushtak
$$
N_{c\,t} (n,k) = \sum\limits_{j = 1}^k {N_{c\,t} (n - j,k)}  + \left[ {n = 0} \right]
$$
where $[P]$ denotes the Iverson bracket.


TITLE: When $K[x]$ is a field?
QUESTION [3 upvotes]: If $K$ is a ring consisting only of zero, then $K[x]$ is a field (edit: from the comments below I learned that it's not). Are there another rings with this property?
I think no. If $R$ contains $1 \neq 0$, then $1 \cdot x \in R[x]$ has no inverse because a degree of a product of two polynomials is a sum of degrees of the factors.
How do you think?

REPLY [2 votes]: Yes, you are right.
$$R[X]=\{\sum_{i=1}^na_iX^i| a_i \in R ~~\text{and}~~ n \in \mathbb N\}$$
Let's call the elements in $R[X]$ a polynomial in $X$.
A polynomial in $X$ cannot have negative/ fractional exponents in $X$, by definition.
And, as pointed out in the comments, even if $R=\{0\}$, the ring $R[X]$ is not a field. That amounts to saying, there is no field with a single element. Read the comments above for more enlightening dose of information.


TITLE: Taylor series remainder question
QUESTION [0 upvotes]: Let $f(x)=\frac{\sin(x)}{x}$ when $x\neq 0$ and $f(x)=1$ when $x=0$. Starting with the Taylor polynomial of degree $2n+1$ for $\sin(x)$ and the estimate for the remainder term, show that
$f(x)=(1-\frac{x^2}{3!}+\frac{x^4}{5!}+\cdots +(-1)^n \frac{x^{2n}}{(2n+1)!})+R_{2n,0,f}(x)$ where $|R_{2n,0,f}(x)|\leq \frac{|x|^{2n+1}}{(2n+2)!}$ and use this to conclude that $\int_0^1 f \approx \int_0^1 (1-\frac{x^2}{3!}+\frac{x^4}{5!})dx = \frac{1703}{1800}\approx .946$ with an error of less than $10^{-3}$.
The first part of this seems pretty straightforward unless I'm missing something with the piecewise nature of $f$. The second statement is not immediately clear to me, though.

REPLY [0 votes]: Basically you estimate $ f (x) $ with the first 3 terms of it's expansion. There will be an error involved, whose value is given in your question. Since we are taking the expansion up to the $ n=2$ term, our remainder is
$ R \leq \frac {x^5}{6!}$
To estimate$ \int_0^1 f (x) dx$ we take the integral of the first three terms of the expansion (which is easily computed). The error involved then is
$\displaystyle\int_0^1  \frac {x^5}{6!} =
\frac {1}{6(6!)} = \frac {1}{4320} \leq 10^{-3}$
So you have your second part


TITLE: A question about measure theory
QUESTION [0 upvotes]: I was studying a paper, I had problem understanding the proof of a proposition, First
notice that T is a measurable mapping from $([0, 1],B([0, 1]), μ)$ into itself, where μ is a T -
invariant probability measure with support $[0, 1]$ and density $f$ with respect to (w.r.t.) the
Lebesgue measure m on $([0, 1],B([0, 1])$.
Throughout the paper, we assume that any non-empty open interval in $[0, 1]$
contains a non-empty open interval on which $f$ is bounded.We also assume that $T$ admits
a finite number of discontinuity points as well as a finite number of local extrema. We
denote by $\mathcal{P}$ the set $\lbrace 0 = a_0 < a_1 < \cdots < a_l < a_{l+1} = 1\rbrace ⊂ [0, 1]$ $(l \geq 0)$ of discontinuity
points and local extrema of $T$ . By the very definition of $\mathcal{P}$, the transformation $T$ is continuous
and monotone on each interval $I_j =]a_j, a_{j+1}[$ for all $j \in \lbrace 0,\cdots, l\rbrace$. As a matter
of fact, $T$ is strictly monotone on each interval $I_j$.
I was trying to understand the proof of a proposition, that is this:

If $λ>0$ $μ-a.s$., then:\begin{align*}
\max_{j\in \lbrace 0 , 1 ,\cdots ,\mathit{l}_{n}\rbrace} \big(diam(I_{j}^{n})\big) \to 0 \quad as \quad n \to \infty
\end{align*}

The proof is this:

Let us assume that
  \begin{align*}
\limsup_{n \to \infty} \max_{j\in \lbrace 0 , 1 ,\cdots ,\mathit{l}_{n}\rbrace}\big(diam(I_{j}^{n})\big) > 0
\end{align*}
  If this is the case, there exists $ε > 0$ and an increasing sequence $(n_p)_{p\geq 0}$ such that for
  all $p \geq 0$, $\exists j_{p} \in \lbrace 0, \cdots, l_{n_{p}} \rbrace$ with diam $  diam(I_{j_{p}}^{n_{p}}) \geq \varepsilon$
  . By compactness of $[0, 1]$, we can assume
  without loss of generality that there exists a non-empty open interval $J$ such that for all
  $p \geq 0$,  $ I_{j_{p}}^{n_{p}} \supset J $ . Now, according to Birkhoff’s ergodic theorem,
  \begin{align*}
\frac{1}{n_{p}} \int_{J} \log \vert (T^{n_{p}})^{\prime} \vert d\mu \to \int_{J} \lambda d\mu \quad as \quad p \to \infty
\end{align*}
  By Jensen’s concave inequality and the fact that $μ(J ) > 0$ (since $J$ is a non-empty open
  interval and the support of $μ$ is $[0, 1]$),
  \begin{align}
\liminf_{p \to \infty} \frac{1}{n_{p}} \log \int_{J} \vert (T^{n_{p}})^{\prime}\vert \frac{d\mu}{\mu(J)}\geq \frac{1}{\mu(J)} \int_{J} \lambda d\mu. (\star)
\end{align}
  By assumption on $f$ , there exists a positive real number $M$ such that $M = \sup_{x \in J} f (x)$.
  Moreover, for all $p \geq 0$, $T^{n_{p}}$ is monotone on $J$ according to a Proposition in the paper.  Consequently,
  \begin{align*}
\int_{J} \vert (T^{n_{p}})^{\prime} \vert \ d\mu &=\int_{J} \vert(T^{n_{p}})^{\prime}(x) \vert f(x) dx \\
&\leq M \int_{J} \vert (T^{n_{p}})^{\prime}(x) \vert dx\\
&= M\vert \int_{J} (T^{n_{p}})^{\prime}(x) dx \vert \\
& = M \vert T^{n_{p}}(s_{2}) - T^{n_{p}}(s_{1}) \vert \quad , \quad  J=(s_{1} , s_{2}) \\
&\leq M
\end{align*}
  With $ \star $, one deduces that
  \begin{align*}
\int_{J} \lambda d\mu \leq 0
\end{align*}
  and, since $μ(J )> 0$, that
  \begin{align*}
 \mu \big( \lbrace x \in [0,1] : \lambda(x) >0 \rbrace \big) < 1
\end{align*}
  This concludes the proof of the proposition.

My questions :
1- How can I deduce $ \int_{J} \lambda d\mu \leq 0 $ ?
2- Is the proof done by contradiction ?

REPLY [0 votes]: Since $\int_J |(T^{n_p})'|\,d\mu \le M$ as argued, we have $\log \int_{J} \vert (T^{n_{p}})^{\prime}\vert \frac{d\mu}{\mu(J)} \le \log \frac{M}{\mu(J)}$.  Hence $(\star)$ implies
$$\frac{1}{\mu(J)} \int_J \lambda\,d\mu \le \liminf_{p \to \infty} \frac{1}{n_p} \log \frac{M}{\mu(J)} = 0$$
because $n_p \to \infty$ as $p \to \infty$.
The overall structure of the proof seems complete to me, though I would call this more of a proof by contrapositive rather than by contradiction.  We start by supposing it is not true that $\max_{j\in \lbrace 0 , 1 ,\cdots ,\mathit{l}_{n}\rbrace} \big(diam(I_{j}^{n})\big) \to 0$, in which case its limsup must be greater than zero (since the liminf is obviously at least zero).  We finish by showing that it is not the case that $\lambda > 0$ $\mu$-a.s. (because there is a set of positive measure, namely $J$, over which its integral is non-positive).


TITLE: Find the integrating factor of form $x^m y^n$ of first order ODE to make it exact
QUESTION [0 upvotes]: For the first-order ODE: $$(2y^2-3xy)dx + (4xy-3x^2)dy = 0,$$
find the integrating factor (of form below) that makes the ODE exact... $$\mu(x,y)=x^my^n$$
My attempt:
$$M_y = 4y - 3x; N_x = 4y - 6x$$
I'm pretty much stuck at this point...any help?
Thanks

REPLY [0 votes]: $$(2y^2-3xy)dx + (4xy-3x^2)dy = 0,$$
Rearrange terms:
$$(-3xydx-3x^2dy)+2(y^2dx+2xydy)=0$$
$$-3xdxy+2(y^2dx+xdy^2)=0$$
$$-3xd(xy)+2d(xy^2)=0$$
Integrating factor is
$$\mu (x,y)=xy^2$$
$$-3(xy)^2d(xy)+2xy^2d(xy^2)=0$$
Integrate
$$(xy)^3-(xy^2)^2=K$$
$$\boxed {x^3y^3-x^2y^4=K}$$


TITLE: Maximum charge storeable on a copper wire
QUESTION [0 upvotes]: What is the maximum charge in Coulombs that can be stored in a copper wire? Lets say I have a thin copper wire (0.1 mm wide) that is 1 meter long. I'm asking about what could I practically achieve, not the theoretical maximum. For example how would I calculate the total charge of the wire after connecting it to a 100kV source, waiting a few seconds and then disconnecting it?
Would the largest charge in Coulombs be positive (by removing as many electrons from the wire as possible) or negative (by cramming as many extra electrons onto the wire as possible?
Additionally, if I wanted to build an object that would hold as much charge as possible for the least mass how would I do so? Would I use a conductor like copper or somehow statically charging an insulator? Would I use a long 1 dimensional shape like a wire, or a thin plate or a solid sphere or something else?
I have read some related questions How much electric charge can a copper sphere, made of N atoms, contain? and What limits the maximum sustainable surface charge density of a sphere in space? , but I am looking for what could be practically achieved, not a theoretical maximum. Even a rough order of magnitude estimation would be much appreciated.

REPLY [3 votes]: The objects which are designed to hold the most charge for the least mass are the capacitors which you can buy very cheap from any electronics store, such as RS Components. They do this by storing an equal amount of +ve and -ve charge in close proximity yet separated by an insulator with a high dielectric constant, which increases the charge-storing storage capacity by orders of magnitude compared with having air between the stored charges. Wires will not be anywhere near as effective.
The specification of particular capacitors will tell you the maximum voltage $V$ which can be applied. This is related to charge stored $Q$ and capacitance $C$ by $Q=CV$.

Conducting wire can be modelled as a coaxial capacitor with the outer plate at infinity. In practice the outer plate will be any of several nearby conducting surfaces, which could be anything between a few mm to a few m away.
The capacitance of a wire of length $1m$ and diameter $0.1mm$ is about $10pF$ (pF=$10^{-12}$ Farads). Typical electrolytic capacitors are between $0.1\mu F$ and $0.1mF$, a factor of $10^4$ to $10^7$ times larger - ie for the same applied voltage they can store around a million times as much charge.
Charge storage capacity is limited by the breakdown strength of the insulating material between the capacitor plates. In the case of bare wire this is air, which breaks down at a field strength of $E=3kV/mm$. (Mica, used in old capacitors, breaks down at $118kV/mm$.) Using the relation $E=\frac{\sigma}{\epsilon_0}$ this requires a surface charge density $\sigma$ of about $30pC/mm^2$. Your $1m$ wire would have an area of about $600mm^2$. Assuming a uniform charge density, it could hold at most $20nC$. With capacitance of $10pF$ the breakdown voltage would be $V=Q/C=20nC/10pF=2kV$.
The weak points for wire, at which air will break down will occur, are the sharp ends, where surface charge density and electric field are highest. So with a more rigorous calculation the maximum voltage and charge stored will be significantly lower.


TITLE: Law of the iterated logarithm for a Standard Brownian Motion (SBM)
QUESTION [2 upvotes]: Let $(B_t)_{t\geq0}$ be a SBM, $a>0, b>0$, and $$\tau:=\inf\{t\geq0:B_t=b\sqrt{a+t}\}, (\inf\emptyset:=0).$$ I want to show that $\mathbb{P}(\tau<\infty)=1$ with the law of iterated logarithm for a SBM, but I don't know how to do it. I have a solution for it but i don't use it here the law of iterated logarithm for a SBM. So
$$ M_t:=\exp \left( \xi B_{\tau \wedge t}-\dfrac{1}{2}\xi^2t\wedge \tau\right), \quad\xi\in \mathbb{R}, $$
then
$$ \mathbb{E}(M_t)=\mathbb{E}(M_0)=\text{e}^{\xi x},$$
and
$$ \lim_{t\to\infty} \exp \left( \xi B_{\tau \wedge t}-\dfrac{1}{2}\xi^2t\wedge \tau\right) = \begin{cases}
     \exp\left(\xi B_\tau-\dfrac{1}{2}\tau\xi^2\right),&\text{if } \tau<\infty,\\
     0, & \text{if } \tau=\infty.
    \end{cases} $$
And if $\xi\to 0 \Longrightarrow   \mathbb{P}(\tau<\infty)=1$, where $t\wedge\tau:=\min\{t,\tau\}.$

REPLY [2 votes]: $\tau=\infty$ implies that $B_t \leq b\sqrt {a+t}$ for all $t$. This implies that $\frac {B_t} {\sqrt {t \log \log t}} \to 0$. But then $\lim \sup \frac {B_t} {\sqrt {t \log \log t}} =0$. LIL for SBM shows that this event has probability $0$. [ $P(\lim \sup \frac {B_t} {\sqrt {t \log \log t}} =1)=1$].


TITLE: Reversing sign of third derivative
QUESTION [2 upvotes]: I don't understand some part of the solution given to this question:

I can understand how the sign of $f''(x)$ can be reversed (i.e. flipping over the x-axis) but I don't get why changing $x$ to $-x$ would change the sign of $f'''(x)$?

REPLY [3 votes]: Let $g(x)=f(-x)$.  Then using the chain rule in each step, we get $$g'(x)=-f'(-x),$$ $$g''(x)=-(-f''(-x))=f''(-x),$$ and finally $$g'''(x)=-f'''(-x).$$  So if $f'''$ is always positive then $g'''$ is always negative, and vice versa.


TITLE: Find (x, y) coordinates from relative position and quaternion angle
QUESTION [0 upvotes]: I'm making a 3d game and I'm running into some mathematical calcualtions that are eluding me.
I have a door with a 3d quaternion rotation and I have a reference point located so many units away on the line perpendicular to the door, and so many units away on the line parallel to the door. How can I calculate the unrotated x, y coordinates relative to the door (using the door as the center of the coordinate system)?
Thanks all...

REPLY [1 votes]: I would rotate the unit basis vectors $i=(1, 0, 0)^\top$, $j=(0, 1, 0)^\top$ and $k=(0, 0, 1)^\top$ to get the basis vectors of your "door" coordinate frame.  To find your x-y points, just scale these new unit basis vectors by how far away you need to go.
I'll use the following notation:

$i_W$ is the $i$ vector in the world frame,
$i_D$ is the $i$ vector in the door frame
$q_W^D$ is the quaternion describing the rotation from the world to the door coordinate frame.
$\otimes$ is the quaternion multiplication operator

I believe all you need is to find $i_D$ and $j_D$.  Just rotate $i_W$ and $j_W$ to get them.
There are two ways to do this:  The most common is to convert your vector to a quaternion by just adding a 0 for the scalar part (sometimes called the $w$ term) and performing:
$$
\begin{bmatrix} 0\\ i_D\end{bmatrix} = q_I^D \otimes \begin{bmatrix} 0 \\ i_W \end{bmatrix} \otimes q_I^{D,-1}
$$
However, there is actually a more efficient way I ran into a while back:  molecular musings
t = 2 * cross(q.xyz, v)
v' = v + q.w * t + cross(q.xyz, t)
I have used this to great effect.
One thing to be really careful about is the direction of your quaternion.  In engineering, a quaternion is usually defined as "passive".  This means it relates two coordinate frames - i.e.  "rotating" a vector by a quaternion results in the same vector object, described in the new frame.  An active rotation results in a new, rotated object described in the original frame.
This is why sometimes rotating a quaternion vectors with quaternions result in the opposite of what you might intuitively expect if you are unclear on what kind of quaternion you have. Luckily, passive and active rotations are the inverse of one another (just flip the sign of the $w$ term).  I would play with some known rotations to convince yourself that you have the right quaternion.  Once you have that, quaternions aren't too bad, and much more efficient than using trigonometry.


text:
\begin{document}

\section{Latent phase space learning}
This second part of the document corresponds to the steps described in the diagram in \ref{problem_scheme_2}.
The methods and the training process described below, are schematically displayed in figure \ref{training scheme}.

\subsection{Methods}

Once we have studied simpler models, in this second part of the work, we will move forward to deal with dynamical systems described by PDEs \eqref{general_dynamical_system}, with a large number of degrees of freedom. These can be reduced by searching for a suitable coordinate change, which better represents the structure of the latent phase space.
The first training stage will consists of an unsupervised learning stage, where we aim to capture the manifold where the trajectory exists,this is finding an initial basis to integrate the latent dynamical system, this step involves the minimization of the reconstruction loss term \ref{latent_probability_fun_to_maximize_pre}.

The probability distribution that we are modelling now is $P(X_{t+k}|X_{t},\theta)$, and we want to factorize it into a projection to a reduced coordinate system $P(h_{t}|X_{t})$ and  a latent integration part $P(h_{t+k}|h_{t})$, this factor would play the role of $P(x_{t+k}|x_{t})$ in last section. Under the assumption of normal errors of constant variance, this factorization leads us to  $P(X_{t+k}|X_{t},\theta)=P(X_{t+k}|h_{t+k})P(h_{t+k}|h_{t \dots t+k-1},f_{t},X_{t},\theta)=\mathcal{N}(\mu_{1}=\hat{X}_{t+k}(h_{t+k},\theta_{encoder}),\sigma)\mathcal{N}(\mu_{2}=\hat{h}_{t+k}(\theta_{integration},h_{t}(X_{t},\theta_{decoder})),\sigma)$. With this probability distribution we can lay out the optimization problem in the same way as in part I, as it is shown in \eqref{latent_probability_fun_to_maximize3}.

\begin{eqnarray}\label{latent_probability_fun_to_maximize_pre}
\displaystyle\underset{\theta}{\text{argmax }}\mathcal{L}(X,\theta) &=&\underset{\theta}{\text{argmax}}\dfrac{1}{N}\sum \log(P(X_{t}| \hat{X_{t}})\nonumber\\ &=& \underset{\theta}{\text{argmin }}\dfrac{1}{N}\sum \left\|X_{t}-\hat{X}_{t}(h_{t}(\theta,X_{t})\right\|^{2}
\end{eqnarray}
\begin{eqnarray}\label{latent_probability_fun_to_maximize3}
\underset{\theta}{\text{argmax }}\mathcal{L}(X,\theta)=
\displaystyle\underset{\theta}{\text{argmax } }\dfrac{1}{N}\sum_{k=1}^{n}log(P(X_{t+k}|h_{t+k})&\nonumber\\P(h_{t+k}| h_{t+1}\dots h_{t+k-1},f_{ht+1}\dots f_{ht+k},X_{t+1}\dots X_{t+k-1}))&
\end{eqnarray}

Inserting the multi-step approximation of the latent derivative in the latent space of the autoencoder to model $P(h_t+k|h_{t})$, we get the cost function \eqref{latent_probability_fun_to_maximize4}. The autoencoder and the derivative parameters are denoted $\theta_{\textit{aut}}$ and $\theta_{\textit{der}}$, respectively.
\begin{eqnarray}\label{latent_probability_fun_to_maximize4}
\mathcal{L}(X,\theta) &=& \sum_{k=1}^{n}\left\|X_{t+k}-\hat{X}_{t+k}(h_{t+k})\right\|^{2}\nonumber\\ &+& \left\|h_{t+k}-\sum_{i=0}^{k-1}\alpha_{i}h_{t+i}(X,\theta_{\mathit{aut}})-\sum_{i=0}^{k}f_{ht+i}(h,\theta _{\mathit{der}})\right\|^{2}
\end{eqnarray}

\textbf{Training}\\

Training and evaluation is carried out by following these steps:
\begin{enumerate}
    \item In a pre-training stage, $X_{t}$ is reconstructed.
    \item In the training stage, we insert the derivative network studied in the first part of this work, that models $P(f_h|h)$.
    \item In the evaluation stage, we unplug the trained derivative network, encode an initial condition $X_{0}\rightarrow h_{0}$ and integrate the derivative network in the latent space. We later decode this integrated solution and recover the $X$ representation of the trajectory.
\end{enumerate}

For the autoencoder, it was used a 100 neuron and 9 hidden layers network with ReLU activations, except for the latent and regressor layers. It was used a width factor between layers of $3/4$, and a $60$ ReLU, $40$ ReLU, $20$ linear network to model the latent derivative.

A denoising autoencoder \cite{vincent2008extracting} architecture was fundamental to learn useful robust features. A corrupted noise of amplitude $10^{-3}$ was injected in the normalized input data.

Adam optimization algorithm \cite{kingma2014adam} was used in the \textit{backpropagation} optimization to train the network, along with a $0.1$ factor decaying learning rate every $200$ epochs and a batch-size of $200$ samples. Smaller networks would be enough to successfully learn most of the problems, but for consistency sake, the architecture was kept fixed.

\begin{figure}[H]
\centering
\includegraphics[width=1\linewidth]{graphics/custom_svg/training_latent_jpg.jpg}
\caption{Autoencoder and latent integration, training and running scheme}
\label{training scheme}
\end{figure}

\section{Results}
The PDE training data was acquired by solving different equations with the MATLAB spectral solver package \texttt{Chebfun}. We used $512$ nodes and periodic boundary conditions in all cases, down-sampled to $64$ to train the network.

When solving equations using a spectral method with this resolution, some numerical artifacts may occur. This is the case of the formation of shock waves in the \textit{Burgers equations}. High frequency modes, which cannot be represented in discretization, give rise to phenomena like the Gibbs one. As will be seen below, this is not a problem for a neural network, since it achieves a more compact and richer representation than the spectral one.

The integration time for each experiment was set in order to clearly manifest the characteristic dynamics given by each equation with every coefficient set to 1. Burger's equation integration time should capture the formation and damping of the shock wave. In the Korteweg-de-Vries equation, the time used captured 3-4 cycles of soliton (semi)periodic movement. In the Kuramoto Sivashinsky equation, the time was set to ensure fully developed turbulent behaviour, where the intermittency characteristics of the turbulent patterns where shown.

Spatial domain bounds in all plots were scaled to  $[-1,1]$. The spatial domain for the Burger's equation was $[-8,8]$,with an initial condition of  $e^{-(x+2)^{2}}$ and integration time of $70$. For the Korteweg-de-Vries equation, in the single soliton case, the spatial domain is $[-10,10]$, with an initial condition $e^{-(x-7)^{2}/10}$ and an integration time of $12$. For the multiple solition case we used a domain of $[-20,20]$ and an initial condition $sin(\frac{\pi x}{20})$. With Kuramoto Sivashinsky equation, the spatial domain was $[50,0]$ and the integration time was $250$, with the initial condition $cos(\frac{x}{8})(1+sin(\frac{x}{8}))$.

The  temporal discretization used is problem dependent. We  observed that, in general, the fitted networks are able to preserve numerical stability with temporal deltas of greater magnitude than in the case of the spectral method integration, so, in every experiment, with the integration time fixed, we sought for the biggest $dt$ that made the spectral-method integration convergent, and then downsampled it by a factor of 2 to feed the network, we saved $25\%$ of the last time samples for testing and used the rest for training.\\

\textbf{Viscid Burgers}\\

Burger's equation \eqref{burger_eq} is a simplification of  the incompressible Navier-Stokes equation. It is useful to understand mathematical problems that arise in the Navier-Stokes equations under certain conditions, such as limit of small viscosities, formation of shock waves, and so on. This is the simplest model that combines non-linear advection and diffusion. Eliminating the diffusion term, the equation exhibits discontinuities as the shock waves are formed. The presence of the diffusion term counteracts the effect of non-linearity, resulting in an equilibrium between the non-linear advection term and linear diffusion.
\begin{equation} \label{burger_eq}
 \dfrac{\partial{u}}{\partial{t}}+u\dfrac{\partial{u}}{\partial{x}}=\mu \dfrac{\partial^{2}{u}}{\partial^{2}{x}}
\end{equation}

In figure \ref{burgers}, both learnt and ground truth dynamics are shown for the Burgers equation. In figure \ref{burgers_moving}, it is clear where the formation of the shock wave happens, as well as how the dissipative dynamics of the underlying system are correctly captured.
\begin{figure}[H]
\centering
\includegraphics[width=0.8\linewidth]{graphics/order_reduction/burgers_moving/latent_full_time_odeint.png}
\caption{x vs t Burgers equation dynamics. Learnt on the left and ground truth on the right.}
\label{burgers}
\end{figure}

As mentioned above, once the training phase has been completed, the integration of the trajectories has been done using an off-the-shelf ODEs integrator (\texttt{lsoda} BDF-4, specifically). The latent derivative of the trained network is passed as the system derivative function, together with the initial condition, $X_{0}$, encoded into $h_{0}$ coordinates through the encoder network.\\

\begin{figure}[H]
\centering
\begin{subfigure}[h]{0.48\linewidth}
\includegraphics[width=\linewidth]{graphics/order_reduction/burgers_moving/3d_train_time_extension_odeint}
\end{subfigure}
\begin{subfigure}[h]{0.48\linewidth}
\includegraphics[width=\linewidth]{graphics/order_reduction/burgers_moving/Net_phase_space}
\end{subfigure}
\caption{Latent phase space and two principal component projection of the learnt latent phase space for the training and test time together.}
\label{burgers_moving}
\end{figure}

\textbf{Korteweg-de-Vries}\\

Korteweg-de-Vries equation \eqref{KdV} was first conceived as model for waves on shallow water surfaces. It describes the behaviour of stable self-reinforcing wave-packets called solitons, that arise from the equilibrium of dispersive and non-linear effects in the medium.
\begin{equation}\label{KdV}
 \dfrac{\partial{u}}{\partial{t}}+ \dfrac{\partial^{3}{u}}{\partial^{3}{x}}-u \dfrac{\partial{u}}{\partial{x}}=0
\end{equation}

This is one of the few PDEs with a known exact solution \eqref{KdV_sol}, corresponding to a functional family of soliton waves of different amplitude and speed.
\begin{equation} \label{KdV_sol}
\phi(x,t)=-\dfrac{1}{2}c\sec^{2}\left(\dfrac{\sqrt{c}}{2}(x-ct-a)\right)
\end{equation}

In figure \ref{k1}, learnt and ground truth dynamics are represented for one soliton.
\begin{figure}[H]
\centering
\includegraphics[width=0.8\linewidth]{graphics/order_reduction/Kdv/latent_full_time_odeint.png}
\caption{x vs t One soliton learnt dynamics in the left side and ground truth dynamics in the right side. The black line means the training-test.}
\label{k1}
\end{figure}

In figure \ref{k2} the extended network integration of the learnt dynamics is shown in the $X$ and $h$ spaces, where the conservative character of the equation has been captured by the network.

\begin{figure}[H]
\centering
\begin{subfigure}[h]{0.48\linewidth}
\includegraphics[width=\linewidth]{graphics/order_reduction/Kdv/latent_time_extension_odeint.png}
\end{subfigure}
\begin{subfigure}[h]{0.48\linewidth}
\includegraphics[width=\linewidth]{graphics/order_reduction/Kdv/3d_train_time_extension_odeint.png}
\end{subfigure}
\caption{Time extension of the network integrated trajectory showing stable dynamics. Latent trajectory on the right side, corresponding to a conservative closed orbit, with a loop encoding the periodic boundary condition.}
\label{k2}
\end{figure}

As it was done for one soliton, in the case of several solitons, figures \ref{k11} and \ref{k22} are obtained. In figure \ref{k22} a two times training time latent trajectory is shown, proving as well the conservative character of the learnt solution, that appears in the latent space as a limit cycle.\\

\begin{figure}[H]
\centering
\includegraphics[width=0.9\linewidth]{graphics/order_reduction/Kdv1/latent_full_time_odeint.png}
\caption{x vs t Korteweg-de-Vries solution for an initial condition that splits mainly in three solitons. Learnt dynamics in the left side and real dynamics in the right side. Black line means the training-test.}
\label{k11}
\end{figure}
\begin{figure}[H]
\centering
\begin{subfigure}[h]{0.4\linewidth}
\includegraphics[width=\linewidth]{graphics/order_reduction/Kdv1/latent_time_extension_odeint.png}
\end{subfigure}
\begin{subfigure}[h]{0.5\linewidth}
\includegraphics[width=\linewidth]{graphics/order_reduction/Kdv1/3d_train_time_extension_odeint.png}
\end{subfigure}
\caption{Network time extension of the training trajectory, showing stable conservative dynamics. After the initial gaussian transient reachs the limit cycle.}
\label{k22}
\end{figure}

\begin{figure}[H]
\centering
\makebox[\textwidth][c]{
\begin{subfigure}[h]{0.9\linewidth}
\makebox[\textwidth][c]{\includegraphics[width=\linewidth]{graphics/order_reduction/Kdv1/Test_solutions_snapshots_different1.png}}
\end{subfigure}}
\begin{subfigure}[h]{0.45\linewidth}
\includegraphics[width=\linewidth]{graphics/order_reduction/Kdv1/different_h0_mesh_radious1_principal}
\end{subfigure}
\begin{subfigure}[h]{0.45\linewidth}
\includegraphics[width=\linewidth]{graphics/order_reduction/Kdv1/Different_h0_X_1_radious_principal_axis}
\end{subfigure}
\caption{In the upper side are represented non-seen test solutions, the first row belongs to the training trajectory. Down it is shown the mesh plot and time snapshots of different latent generated trajectories, moving in the first principal component axis of the latent activation space. In orange the training trajectory and the generated one in blue.}
\label{k44}
\end{figure}

\textbf{Kuramoto Sivashinsky}\\

Kuramoto Sivashinsky equation \eqref{KS} is a fourth-order nonlinear PDE, that models laminar flame fronts with diffusive instabilities. The sign of the diffusive term acts as an energy source and it has a destabilising effect. The convective $uu_{x}$ term transfers energy to smaller wavelengths, where the fourth order derivative term dominates and acts as a stabilizing factor, as it is explained with the dispersion relation $-iw=k^{2}-k^{4}$. This richer set of mechanisms to transfer energy, can lead to very complex behaviours, including chaoticity. Kuramoto Sivashinsky is the simplest known PDE that exhibits this phenomena.
\begin{equation}\label{KS}
 \dfrac{\partial{u}}{\partial{t}}+ \dfrac{\partial^{4}u}{\partial^{4}x}+\dfrac{\partial^{2}{u}}{\partial^{2}{x}}+u\dfrac{\partial u}{\partial x}=0
\end{equation}

By using $L=22$ as the domain width and the same initial condition solutions, Kuramoto Sivashinsky has a positive Lyapunov exponent and exhibits chaotic behaviour. The intrinsic Hausdorff dimension of the chaotic atractor is greater than $3$, so encoding it in a 3D space becomes complicated. In figure \ref{ks22} it is shown the long term prediction capability of a $4$-neuron architecture. In figure \ref{ks66} in shown as in chaotic dynamics, variations are expected for slightly different initial conditions.

\begin{figure}[H]
\centering
\begin{subfigure}[h]{0.4\linewidth}
\includegraphics[width=\linewidth]{graphics/order_reduction/kschaos/latent_time_extension_custom.png}
\end{subfigure}
\begin{subfigure}[h]{0.4\linewidth}
\includegraphics[width=\linewidth]{graphics/order_reduction/kschaos/real_extension.png}
\end{subfigure}
\caption{In the upper side,x vs t extended network dynamics. In the lower side, real dynamics. Black line means the training-test.}
\label{ks22}
\end{figure}

\begin{figure}[h]
\centering
\includegraphics[width=0.9\linewidth]{graphics/order_reduction/kschaos/Test_solutions_snapshots_different1ode.png}
\caption{Time snapshots of different test solutions. First row corresponds to the training one. Neural network results in orange and ground truth in blue.}
\label{ks66}
\end{figure}

\end{document}


TITLE: Is the coaction $\delta: A \to H \otimes A$ injective?
QUESTION [1 upvotes]: Let $A$ be an algebra and let $H$ be a bialgebra. Suppose that $A$ is an $H$-comodule. Then we have a coaction $\delta: A \to H \otimes A$. Is the coaction $\delta: A \to H \otimes A$ always injective? Thank you very much.

REPLY [3 votes]: Yes, it is even a split monomorphism: If $(C,\Delta,\eta)$ is a ${\mathbb k}$-coalgebra and $(M,\nabla)$ is a $C$-comodule, then (by definition) you have that $M\xrightarrow{\nabla} M\otimes_{\mathbb k} C\xrightarrow{\text{id}_M\otimes\eta} M\otimes_{\mathbb k} {\mathbb k}\cong M$ is the identity on $M$. In particular, $\nabla: M\to M\otimes_{\mathbb k} C$ is injective. As Tobias said, this is dual to the perhaps more obvious (by habit) argument that the action of any (unitary) ring on any of its modules is surjective: just look at the action of the unit.


text:
\begin{document}

\title{The maximal tree with respect to the exponential of the second Zagreb index\thanks{This work is supported by the National Natural Science Foundation of China (11971164) and the Department of Education of Hunan Province (19A318).}}
\author{Mingyao Zeng, Hanyuan Deng\thanks{Corresponding author: hydeng@hunnu.edu.cn}\\
{\small College of Mathematics and Statistics, Hunan Normal University,}
 \\{\small  Changsha, Hunan 410081, P. R. China.}
}

\date{}
\maketitle

\begin{abstract}
The second Zagreb index is $M_2(G)=\sum_{uv\in E(G)}d_{G}(u)d_{G}(v)$. It was found to occur in certain approximate expressions of the total $\pi$-electron energy of alternant hydrocarbons and used by various researchers in their QSPR and QSAR studies. Recently the exponential of a vertex-degree-based topological index was introduced. It is known that among all trees with $n$ vertices, the exponential of the second Zagreb index $e^{M_2}$ attains its minimum value in the path $P_n$. In this paper, we show that $e^{M_2}$ attains its maximum value in the balanced double star with $n$ vertices and solve an open problem proposed by Cruz and Rada [R. Cruz, J. Rada, The path and the star as extremal values of vertex-degree-based topological indices among trees, MATCH Commun. Math. Comput. Chem. 82 (3) (2019) 715-732].

{\bf Keywords}: Exponential of the second Zagreb index; Maximal tree
\end{abstract}

\maketitle

\makeatletter
\renewcommand\@makefnmark
{\mbox{\textsuperscript{\normalfont\@thefnmark)}}}
\makeatother

\baselineskip=0.25in

\section{Introduction}
\label{1}

The first Zagreb index $M_1$ and the second Zagreb index $M_2$ were introduced 50 years ago \cite{gt}. The Zagreb indices and their variants have been used to study molecular complexity, chirality, ZE-isomerism and heterosystems whilst the overall Zagreb indices exhibited a potential applicability for deriving multilinear regression models. Zagreb indices are also used by various researchers in their QSPR and QSAR studies. Mathematical properties of the Zagreb indices have also been studied. Readers can refer to the paper \cite{nkmt} and the cited literature. The second Zagreb index was found to occur in certain approximate expressions of the total $\pi$-electron energy of alternant hydrocarbons \cite{gt}. We encourage the reader to consult \cite{bkmr,dg,deng,dsa,eg,fs,jjt,ytl} for the historical background, computational techniques, and mathematical properties of the second Zagreb index.

For simple graph $G$ with edge set $E(G)$, the second Zagreb index of $G$ is defined as
$$M_2(G)=\sum_{uv\in E(G)}d_{G}(u)d_{G}(v)$$
where $d_{G}(v)$ is the degree of the vertex $v$ in $G$. It is a vertex-degree-based (VDB for short) topological index, also referred as bond incident degree index.

A formal definition of a VDB topological index is as follows. Let $\mathcal{G}_n$ be the set of graphs with $n$ non-isolated vertices. Consider the set
$$K =\{(i,j)\in \mathbf{N}\times \mathbf{N}: 1\leq i\leq j\leq n-1\}$$
and for a graph $G\in \mathcal{G}_n$, denote by $m_{i,j}(G)$ the number of edges in $G$ joining vertices of degree $i$ and $j$. A VDB topological index over $\mathcal{G}_n$ is a function $\varphi: \mathcal{G}_n\rightarrow \mathbf{R}$ induced by real numbers $\{\varphi_{i,j}\}_{(i,j)\in K}$ defined as
$$\varphi(G)=\sum_{(i,j)\in K}m_{i,j}(G)\varphi_{i,j}$$
for every $G\in \mathcal{G}_n$.

Many important topological indices are obtained from different choices of $\varphi_{i,j}$. For example, the first Zagreb index $M_1$ induced by numbers $\varphi_{i,j}=i+j$; the second Zagreb index $M_2$ induced by $\varphi_{i,j}=ij$; the Randi\'{c} index $\chi$ induced by $\varphi_{i,j}=\frac{1}{\sqrt{ij}}$, et al. For details on VDB topological indices, see \cite{cr,g,rada,tc}.

In order to study of the discrimination ability of topological indices,  Rada \cite{rada} introduced the exponential of a vertex-degree-based topological index. Given a vertex-degree-based topological index $\varphi$, the exponential of $\varphi$, denoted by $e^{\varphi}$, is defined as
$$e^{\varphi}(G)=\sum_{(i,j)\in K}m_{i,j}(G)e^{\varphi_{i,j}}$$
The extremal value problem of $e^{\varphi}$ over the set $\mathcal{T}_n$ of trees with $n$ vertices was initiated in \cite{cr}, and it was shown that $e^{M_1}$, $e^{M_2}$ attain their minimum value in the path $P_n$, $e^{\chi}$, $e^{H}$, $e^{GA}$, $e^{SC}$, $e^{AZ}$ attain their minimum value in the star $S_n$, $e^{M_1}$, $e^{ABC}$ attain their maximum value in the star $S_n$, $e^{H}$, $e^{GA}$, $e^{SC}$ attain their maximum value in the path $P_n$. In \cite{cmr}, it was shown that $e^{\chi}$ attains its maximum value in the path $P_n$. These results are summarized in Table \ref{tab-1}.

\begin{table}
\caption{Results on extremal trees for exponential of well known VDB topological indices.}\label{tab-1}
\begin{center}
\begin{tabular}{ccccccccc}
\hline
\quad& $e^{M_1}$ & $e^{M_2}$ & $e^{\chi}$ & $e^{H}$ & $e^{GA}$ &$e^{SC}$ & $e^{ABC}$ & $e^{AZ}$\\
\hline
min & $P_n$ & $P_n$ & $S_n$ & $S_n$ & $S_n$ & ${S_n}$ & $?$ & ${S_n}$\\
max & $S_n$ & $?$ & $P_n$ & $P_n$ & $P_n$ & ${P_n}$ & ${S_n}$ & $?$\\
\hline
\end{tabular}
\end{center}
\vskip -0.5cm
\end{table}

The maximum value of $e^{M_2}$ over $\mathcal{T}_n$ is still an open problem. In this paper, we prove that the maximum value of $e^{M_2}$ over $\mathcal{T}_n$ is attained in the balanced double star $S_{\lfloor\frac{n-2}{2}\rfloor, \lceil\frac{n-2}{2}\rceil}$, and solve an open problem proposed by Cruz and Rada \cite{cr}.

\section{Trees with maximum exponential second Zagreb index}

We first show in this section that in a maximal tree with respect to $e^{M_2}$ over $\mathcal{T}_n$, the distance between any pendant vertex and any vertex with the maximum degree in $T$ is at least 2

\begin{figure}[ht]
\begin{center}
  \includegraphics[width=10cm,height=3cm]{fig-2.jpg}
 \end{center}
\vskip -0.5cm
\caption{Trees $T$ and $T'$ in Lemma \ref{l-2}}\label{fig-2}
\end{figure}

\begin{lemma}\label{l-2}
If $T$ is a maximal tree with respect to $e^{M_2}$ in $\mathcal{T}_n$, then the distance between a pendant vertex and a vertex with the maximum degree in $T$ is at most 2.
\end{lemma}
\textbf{Proof}. Otherwise, there is a vertex $u$ with the maximum degree in $T$ and a path $P=uvw$ such that $w$ is not a pendent vertex in $T$. Let $T_u$, $T_v$ and $T_w$ be the components of $T-uv-vw$ containing $u$, $v$ and $w$, respectively, see Figure \ref{fig-2}. Let $d(u)=\triangle$, $d(v)=s$, $d(w)=t$, where $\triangle\geq s\geq 2$, $\triangle\geq t\geq 2$. The set of neighbours of $v$ in $T$ is $N_T(v)=\{u,w,v_1,v_2,\cdots,v_{s-2}\}$ with $d_T(v_i)=x_i$ ($i=1,2,\cdots,s-2$), and the set of neighbours of $w$ is $N_T(w)=\{v,w_1,w_2,\cdots,w_{t-1}\}$ with $d(w_j)=y_j$ ($j=1,2,\cdots,t-1$). Let $T'=T-\{ww_1,ww_2,\cdots,ww_{t-1}\}+\{vw_1,vw_2,\cdots,vw_{t-1}\}$, then $d_{T'}(u)=\triangle$, $d_{T'}(v)=s+t-1$, $d_{T'}(w)=1$ and
\begin{align*}
e^{M_2}(T')-e^{M_2}(T)
&=e^{\triangle(s+t-1)}+\sum_{i=1}^{s-2}e^{x_i(s+t-1)}+\sum_{j=1}^{t-1}e^{y_j(s+t-1)}+e^{s+t-1}-(e^{\triangle s}+\sum_{i=1}^{s-2}e^{x_is}+\sum_{j=1}^{t-1}e^{y_jt}+e^{st})\\
&=(e^{\triangle(s+t-1)}-e^{\triangle s}-e^{st})+\sum_{i=1}^{s-2}(e^{x_i(s+t-1)}-e^{x_is})+\sum_{j=1}^{t-1}(e^{y_j(s+t-1)}-e^{y_jt})+e^{s+t-1}\\
&>0
\end{align*}

So, $e^{M_2}(T')>e^{M_2}(T)$ and $T$ is not a maximal tree with respect to $e^{M_2}$ in $\mathcal{T}_n$.
\hfill$\Box$

\begin{figure}[ht]
\begin{center}
  \includegraphics[width=5cm,height=5cm]{fig-3.jpg}
 \end{center}
\vskip -0.5cm
\caption{A tree $T$}\label{fig-3}
\end{figure}

{\bf Remark}. Let $T\in \mathcal{T}_n$ ($n\geq 4$), in which all distances between any pendent vertex and any vertex with the maximum degree are at most 2, then $T$ has the form shown in Figure \ref{fig-3}, where $u$ is its unique vertex with the maximum degree if $k>1$, or $T$ is a double star if $k=1$.

\begin{figure}[ht]
\begin{center}
  \includegraphics[width=11cm,height=4.5cm]{fig-4.jpg}
 \end{center}
\vskip -0.5cm
\caption{The trees $T$ and $T'$ in Lemma \ref{l-3}.}\label{fig-4}
\end{figure}

\begin{lemma}\label{l-3}
Let $T$ and $T'$ be the trees on $n$ vertices given in Figure \ref{fig-4}, where $u$ is a vertex with the maximum degree and $T_1$ is a subtree of $T$. If $d_1\geq d_2\geq 1$, then $e^{M_2}(T)<e^{M_2}(T')$.
\end{lemma}
\textbf{Proof}.
Let $d_T(u)=\Delta$ be the maximum degree in $T$. The sets of neighbours of $u_1$ and $u_2$ in $T$ are $N_T(u_1)=\{u,v_1,\cdots,v_{d_1}\}$ and $N_T(u_2)=\{u,w_1,\cdots,w_{d_i}\}$, respectively, where $d(v_1)=d(v_2)=\cdots=d(v_{d_1})=1$, $d(w_1)=d(w_2)=\cdots=d(w_{d_2})=1$. Let $T'=T-\{u_2w_1\}+\{u_1w_1\}$, then $d_{T'}(u_1)=d_1+2$, $d_{T'}(u_2)=d_2$, and
\begin{align*}
e^{M_2}(T')-e^{M_2}(T)
=&e^{\Delta(d_1+2)}+(d_1+2)e^{(d_1+2)}+e^{\Delta d_2}+(d_2-1)e^{d_2}\\
&-[e^{\Delta(d_1+1)}+d_1e^{(d_1+1)}+e^{\Delta (d_2+1)}+d_2e^{(d_2+1)}]\\
=&[e^{\Delta(d_1+1)}\cdot e^{\Delta}-e^{\Delta(d_1+1)}]+[d_1e^{(d_1+1)}\cdot e-d_1e^{(d_1+1)}]\\
&+e^{(d_1+2)}+(e^{\Delta d_2}-e^{\Delta d_2}\cdot e^{\Delta})+(d_2e^{d_i}-d_2e^{d_2}\cdot e)-e^{d_2}\\
=&[e^{\Delta(d_1+1)}-e^{\Delta d_2}][e^{\Delta}-1]+[d_1e^{(d_1+1)}-d_2e^{d_2}](e-1)\\
&+[e^{(d_1+2)}-e^{{d_2}}]>0
\end{align*}
So, $e^{M_2}(T)<e^{M_2}(T')$.
\hfill$\Box$

In the following, we consider the double star $S_{x,y}$, it is a tree on $x+y+2$ vertices with exactly two non-pendent vertices of degrees $x+1$ and $y+1$, respectively.

\begin{lemma}\label{l-4}
Let $T=S_{x,y}$ be the double star on $n$ vertices, where $x+y=n-2$. If $T$ is a maximal tree with respect to $e^{M_2}$ among all double stars on $n$ vertices, then $|x-y|\leq 1$, i.e., $T$ is the balanced double star with $n$ vertices. And

$$e^{M_2}(S_{\lfloor\frac{n-2}{2}\rfloor, \lceil\frac{n-2}{2}\rceil})=\left\{
\begin{array}{ll}
  e^{\frac{n^2}{4}}+(n-2)e^{\frac{n}{2}}, & \mbox{if $n$ is even}; \\
  e^{\frac{n^2-1}{4}}+(\frac{n-3}{2})e^{\frac{n-1}{2}}+(\frac{n-1}{2})e^{\frac{n+1}{2}}, & \mbox{if $n$ is odd}.
\end{array}
\right.
$$
\end{lemma}
\textbf{Proof}.
Let $u$ and $v$ be two non-pendent vertices of the double star $T=S_{x,y}$, $d_T(u)=x+1$ and $d_T(v)=y+1$. Without loss of generality, we assume that $x\leq y$.
If $|x-y|>1$, let $T'=T-vv_1+uv_1$, where $v_1\neq u$ is a neighbour of $v$ in $T$, then $T'\simeq S_{x+1,y-1}$ and
\begin{align*}
e^{M_2}(T')-e^{M_2}(T)
&=[e^{(x+2)y}+(x+1)e^{(x+2)}+(y-1)e^{y}]-[e^{(x+1)(y+1)}+xe^{x+1}+ye^{y+1}]\\
&=e^{(x+2)y}-e^{(x+1)(y+1)}-ye^{y+1}+(x+1)e^{(x+2)}-xe^{x+1}+(y-1)e^{y}\\
&>e^{xy+2y}-e^{xy+x+y+1}-ye^{y+1}\\
&=e^{xy+y}[e^y-e^{x+1}]-ye^{y+1}\\
&\geq e^{xy+y}[e^{x+2}-e^{x+1}]-ye^{y+1}\\
&=e^{xy+y+x+1}[e-1]-ye^{y+1}\\
&>e^{xy+y+x+1}-ye^{y+1}\\
&=e^{y+1}[e^{x(y+1)}-y]>0
\end{align*}
So, $e^{M_2}(T')>e^{M_2}(T)$ and $T$ is not maximal with respect to $e^{M_2}$ among all double stars on $n$ vertices.
\hfill$\Box$

\begin{theorem}\label{t-1}
If $T\in \mathcal{T}_n$ and $T\not\simeq S_{\lfloor\frac{n-2}{2}\rfloor, \lceil\frac{n-2}{2}\rceil}$, $n\geq 4$, then $T$ is not maximal with respect to $e^{M_2}$ over $\mathcal{T}_n$.
\end{theorem}
\textbf{Proof}.
In fact, $e^{M_2}(S_{\lfloor\frac{n-2}{2}\rfloor, \lceil\frac{n-2}{2}\rceil})>e^{M_2}(S_n)$ for $n\geq 4$, the star $S_n$ on $n$ vertices is not maximal with respect to $e^{M_2}$ over $\mathcal{T}_n$.

If $T$ is not a double star, then by Lemma \ref{l-2}, we may assume that all distances between a pendant vertex and a vertex with the maximum degree in $T$ are at least 2 and $T$ has the form depicted in Figure \ref{fig-3}, where $k>1$. By Lemma \ref{l-3}, $T$ is not maximal.

So, $T$ is a double star and the result follows from Lemma \ref{l-4}.
\hfill$\Box$

\section*{References}

\bibliographystyle{elsarticle-num}
\bibliography{<your-bib-database>}


TITLE: Second eigenvalue and the last eigenvalue
QUESTION [1 upvotes]: Note : All eigenvalues that I would referring to below would of the adjacency matrix of the graph
My question arises from having read about Expander Graphs from a few different sources. The most common definition of Expander Graphs that have seen is in terms of the spectral gap which is $d - \lambda _2$ where $\lambda _2$ is the second highest eigenvalue. However certain sources also like to define expansion in terms of $\lambda$ which is the second highest eigenvalue in absolute value. Now I am slightly confused about what is a better definition and why these two values are so mixed in literature ?
For instance Cheeger's inequalities are defined in terms of $\lambda _2$ and the Expander Mixing Lemma is defined is terms of $\lambda$ (I am not sure if there exist versions of each in terms of the other quantity). Also I have seen versions of the Alon-Bopanna bound in terms of both $\lambda$ and $\lambda _2$.
I guess my question is that is there a fundamental connection between these two quantities and is one more natural than the other to be used to define expansion ?
Also is there something fundamental that $\lambda _n$ (by $\lambda _n$ I mean the smallest eigenvalue) represent ?
I am not sure if I have been able to express my confusion clearly but I am sure I am missing something fundamental in my understanding.
Thanks

REPLY [5 votes]: Regarding the last eigenvalue:
The last eigenvalue $\lambda_n$ measures (roughly) how close is the graph to be bipartite. For example, $\lambda_n = -d$ if and only if the graph is bipartite (this is a fairly easy exercise). You can read more about it in Luca Trevisan's blog:
http://lucatrevisan.wordpress.com/2008/06/13/max-cut-and-the-smallest-eigenvalue/
http://lucatrevisan.wordpress.com/2008/09/23/max-cut-gain-and-the-smallest-eigenvalue/