TITLE: How to explain the commutativity of multiplication to middle school students?
QUESTION [8 upvotes]: It's easy for natural numbers: $3\times 5=5\times 3$

*****
*****
*****
but how do you explain that $x.y=y.x$ for any real numbers $x$ and $y$.
Moreover, in $\Bbb{N}$, do you prefer to define $n\times m=\underbrace{n+n+\cdots+n}_{m\text{ times}}$ or $n\times m=\underbrace{m+m+\cdots+m}_{n\text{ times}}$.

REPLY [2 votes]: One option, which my teacher did with us (8th grade, so may not work so well with 6th graders), is to show the class a system where multiplication (or perhaps even addition) is not commutative, like matricies or 3D vectors.  After showing them that that type of multiplication is not commutative, it becomes much easier to understand that there is something special about multiplication on the reals.
The entire rest of the field axioms can be handled the same way, if you can just show systems that don't have all the properties that they take for granted.

commutativity of addition
distributivity of multiplication over addition
associativity of addition/multiplication
existence of identity/zero
existence of an equivalence relation
existence of an order relation
trichotomy (for any $a,b$, exactly one of $a < b$, $a = b$, or $a> b$)
transience of equality
substitution ($a=b \oplus c \wedge c=d \to a=b \oplus d$)

Although most middle schoolers might not be willing to learn this sort of stuff.


TITLE: Proof of sequence limit, using epsilon-delta method.
QUESTION [1 upvotes]: ${\displaystyle (a_n)}$ is a sequence with ${\displaystyle a_n = \frac{1}{\sqrt{n}}}$. Proove that ${ \displaystyle \lim\limits_{n\to\infty}{a_n} = 0 }$, using epsilon-delta method.

First of all, I assume that ${ \displaystyle n \in [1;+\infty) }$.
For all ${n}$, which are larger or equal to ${1}$, ${n^{th}}$ term of sequence is strictly larger than zero and less or equal to one.
${ \forall n \geq 1 : 0 < a_n \leq 1  }$
Assume some ${ \displaystyle \epsilon > 0 }$, such that ${ \displaystyle 1 - \frac{1}{\sqrt{n}} < \epsilon \Longleftrightarrow \left(\frac{1}{1-\epsilon}\right)^2 > n }$
Now let ${ \displaystyle N = 1 + \left\lceil \left(\frac{1}{1-\epsilon}\right)^2 \right\rceil }$, then ${ \displaystyle \left| \frac{1}{\sqrt{n}} - 1 \right| < \epsilon }$ ${ \displaystyle \forall n \geq N }$

Have I understood the task? Is my solution correct?

REPLY [4 votes]: As mentioned in comments, you should be investigating the inequality $\frac{1}{\sqrt{n}}<\epsilon$, not $1-\frac{1}{\sqrt{n}}<\epsilon$.
It's OK to do preliminary side work that establishes $$\frac{1}{\sqrt{n}}<\epsilon\iff n>\frac{1}{\epsilon^2}$$ and then use that to decide that a good choice for $N$ is $\lceil\frac{1}{\epsilon^2}\rceil$. But the logic of an epsilon-N argument doesn't start until:

Let $\epsilon$ be someone's arbitrary idea of what "small" is. That is, let $\epsilon$ be some positive number that is handed to me by someone else.
Turn around back to them and say "based on your $\epsilon$, here is an $N$-value:  $\lceil\frac{1}{\epsilon^2}\rceil$. And I can demonstrate that '$n\ge N\implies \frac{1}{\sqrt{n}}<\epsilon$'."

That last directional implication must be in that direction for the logic of this argument to work.

REPLY [3 votes]: There are a few errors in what you have written. First you want to prove that the limit is $0$. In which case, you want to prove the following.
Given any $\epsilon >0$, there exists a $N(\epsilon) \in \mathbb{N}$,
$$\text{ such that for all $n > N(\epsilon)$, we have }\left \vert a_n - 0 \right \vert < \epsilon \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (\star)$$
When you are proving a limit from first principles, you cannot choose $\epsilon$; $\epsilon$ is given to you. The challenge for you is to find $N(\epsilon)$. So whatever be the '$\epsilon$' I give you, you should be able to find a $N(\epsilon)$ such that $(\star)$ is satisfied.
In your problem, given $\epsilon > 0$, you need to find $N(\epsilon) \in \mathbb{N}$,
$$\text{ such that for all $n > N(\epsilon)$, we have }\left \vert \dfrac1{\sqrt{n}} - 0 \right \vert < \epsilon \text{ i.e. } \dfrac1{\sqrt{n}} < \epsilon $$
Now note that $a_n = \dfrac1{\sqrt{n}}$ is a decreasing sequence and hence if we find $m$ satisfying $(\star)$, for any $n > m$, $(\star)$ will be satisfied since for $n>m$, we have $a_n < a_m$. This motivates us to choose $N(\epsilon)$ as $$N(\epsilon) = \left \lceil\dfrac1{\epsilon^2} \right \rceil$$ Note that $$N(\epsilon) = \left \lceil\dfrac1{\epsilon^2} \right \rceil \geq \dfrac1{\epsilon^2} \implies \sqrt{N} \geq \dfrac1{\epsilon} \implies \dfrac1{\sqrt{N}} \leq \epsilon$$
Hence, for all $n > N$, we have that
$$\dfrac1{\sqrt{n}} < \dfrac1{\sqrt{N}} \leq \epsilon$$


TITLE: How to prove that there is at least one student at the round table whose two neighbours both cheated on the exam
QUESTION [1 upvotes]: A English exam was taken by $50$ students, the teacher found out that $25$ students had cheated on the exam, if the teacher was to place all of them in a round table show that there is at least one student two neighbours (at the table) of whom have cheated on the exam.
-I'm not sure if this problem should have a solution using combinatorics or graph theory. I was thinking if we use graphs to make a bipartite graph and then using the color theorem show that these students can be placed in a manner similar to the Cyclic Graph, but am still not sure if that may be the right answer since there may be a simpler solution.

REPLY [2 votes]: Assume the conclusion does not hold.
Number your students from $1$ to $50$ consistently with the circle.
Consider the smaller circle formed by the students numbered $1$, $3$, $\ldots$, $49$. It contains $25$ students, and no two neighbors cheated, so it contains at most $12$ cheaters.
Same for the circle $2,4, \ldots,50$. In total you have at most $12+12 =24$ cheaters, a contradiction.


TITLE: Generating function of multiset cycles?
QUESTION [3 upvotes]: I've been tossing the idea of multiset cycles around in my head for the past day or two. In Stanley's Enumerative Combinatorics, he defines a multiset cycle to be a sequence $(i_1,i_2,\dots,i_k)$ of positive integers similar to cycles from algebra, but with repetitions allowed. For a cycle $C=(i_1,\dots,i_k)$, the weight of $C$ is defined as $w(C)=x_{i_1}\cdots x_{i_k}$, where $x_1,x_2,\dots$ are indeterminates. Then a multiset permuation is just a multiset of multiset cycles.
I'm curious as to why
$$\prod_C (1-w(C))^{-1}=\sum_\pi w(\pi)$$
($C$ ranges over all multiset cycles and $\pi$ over all multiset permutations). Why is this true?
My thoughts — I try to rewrite the left as
$$
\prod_C\frac{1}{1-w(C)}=(1+w(C)+w(C)^2+\cdots)(1+w(C')+w(C')^2+\cdots)\cdots
$$
for $C,C\,',\dots$ multiset cycles. Now the weight of a permutation $\pi=C_1C_2\cdots C_j$ is then defined by the product $w(\pi)=w(C_1)\cdots w(C_j)$. So the summands of $\sum_\pi w(\pi)$ seem to "factor" into multiset cycles decompositions. Is it fair to then say that the equality holds since for each factors of $w(\pi)$ for some $\pi$ is picked up as a term in the product on the left hand side? (I believe the decomposition is unique up to order of factors, much like in algebra.)
I'm not fully convinced myself. What is the correct and more rigorous way to justify this equality?
Thank you,

REPLY [3 votes]: By definition, a multiset permutation is a multiset of multiset cycles (so not necessarily distinct)
$$\pi=\{C_1,C_2,\dots,C_m\}$$
and the associated weight is $w(\pi)=w(C_1)\cdots w(C_m)$. Suppose we have an enumeration of all of the multiset cycles $\{C_i\}_{i=1}^\infty$ and we consider the partial products given by
$$\ell_n=\prod_{i=1}^n\frac{1}{1-w(C_i)}=\prod_{i=1}^n\left(1+w(C_i)+w(C_i)^2+\cdots\right).$$
There is a formula
$$\left(\sum_{a\in A} a\right)\left(\sum_{b\in B}b\right)=\sum_{(a,b)\in A\times B}ab,$$
which holds true any time certain convergence requirements are met, which generalizes to $n$-ary products via induction, and all of this carries over to the formal power series setting. Hence we may use the notation $\Gamma_n$ to stand for all products of the form $w(C_1)^{a_1}\cdots w(C_n)^{a_n}$ with $a_i=0,1,2,3,\dots$, which of course will be equal to a corresponding $w(\pi)$ for the appropriate multiset permutation $\pi$; with this we see that every summand of the partial product expansions is of the form $w(\pi)$.
Now every multiset of multiset cycles decomposes uniquely into a union of those very multiset cycles, counted with multiplicity obviously, so our final task is to see that for every permutation $\pi$, there is an $n$ such that $w(\pi)\in\Gamma_m$ for all $m\ge n$, and hence $w(\pi)$ shows up in the expansion of the partial product of $\ell(x_1,x_2,\dots)$ at some point. We can see this by letting $n$ be the maximum subscript of all of the $C_i$'s present in the multiset cycle decomposition of $\pi$.
What we've shown then is that every term in the expansion of the partial products of $\ell$ corresponds to a $w(\pi)$, and every $w(\pi)$ is eventually in all of the partial products of $\ell$. Finally, if you're worried about non-uniqueness of the $w(\pi)$, what we can do is consider this argument first where all of the terms $w(C_i)$'s are unknown but distinct variables - in which case the line of reasoning obtains - and then specialize (plug in for) the $w(C_i)$ however we wish afterwards.
Finally, let me point out that this is exactly analogous to the Euler product for the Riemann zeta function (and there are many other generalizations and analogues of this product, in fact). Here the multiset cycles $C_i$ correspond to prime numbers and the $\pi$'s to all the natural numbers.


TITLE: How to find Var($\bar{X}-\bar{Y}$) , the variance of the difference between the sample means?
QUESTION [2 upvotes]: So I tried to do this my own way but I'm not sure if it's correct. I used the equation for variance to get this answer, but I'm not sure if it matches up with what the answer is. Also to be honest, I'm not sure why I'm dividing by m+n? I just sort of guessed but I don't really get why. I thought I'd have to multiply by m and n, not divide. I'm guessing that's because you take out the constant when you calculate variance?

REPLY [2 votes]: I hope a few extra sentences can clear up some of your confusion.
In the notation of the printed problem, the variances of the sample means are $V(\bar X) = \sigma_1^2/n$ and $V(\bar Y) = \sigma_2^2/m$.
Without knowing the context of the chapter you took this from, I'd say it is
reasonable to assume these are two independent samples. That implies that
$\bar X$ and $\bar Y$ are independent random variables.
In general, if $X$ and $Y$ are independent random variables, then
$Var(aX + bY) = a^2Var(X) + b^2Var(y),$ so that $Var(X - Y) = Var(X)+Var(Y),$
letting $a = 1$ and $b = -1.$
Then
$$Var(\bar X - \bar Y) = \frac{\sigma_1^2}{n} + \frac{\sigma_2^2}{m},$$
Note: The covariance plays a role when the two sample means are
not independent. That is the reason for the last sentence in part b.
I'm guessing that this exercise is to get you ready to find a confidence
interval or do a test on the difference between two population means.
The estimate of $\mu_1 - \mu_2$ is $\bar X - \bar Y.$ Part of the rationale
for that statement is that $E(\bar X - \bar Y) = \mu_1 - \mu_2.$ The variance
of this estimate is what we just derived in the displayed equation.

A little more background--because you wondered about $n$'s in denominators.
$$E(\bar X) = E\left(\frac{X_1 + X_2 + \cdots + X_n}{n}\right)
= \frac{1}{n}E(X_1 + X_2 + \cdots + X_n) = \frac{1}{n}[E(X_1)+E(X_2)+\cdots + E(X_n)] = \frac{1}{n}[\mu_1 + \mu_1 + \cdots + \mu_1] \\
 = \frac{1}{n}(n\mu_1) = \mu_1.$$
Also, because $Var(aX) = a^2 Var(X),$ and taking $a = 1/n,$ we have
$$Var(\bar X) = Var\left(\frac{X_1 + X_2 + \cdots + X_n}{n}\right)
= \left(\frac{1}{n}\right)^2 Var(X_1 + X_2 + \cdots + X_n) = \frac{1}{n^2}[Var(X_1)+Var(X_2)+\cdots +  Var(X_n)] = \frac{1}{n^2}[\sigma_1^2 + \sigma_1^2 + \cdots + \sigma_1^2] \\
 = \frac{1}{n^2}(n\sigma_1^2) = \sigma_1^2/n.$$
This says that $\bar X$ is a less-variable (more stable) random variable
than any one of the $X_i$'s. And the variance gets smaller as the
sample size gets larger. That idea is really important in inferential
statistics. More information (carefully collected) is better than less.


text:
\begin{document}
\title[Representation dimension]
{The representation dimension of a class of tame algebras}

\author[Assem]{Ibrahim Assem}
\address{D\'epartement de math\'ematiques, Universit\'e de Sherbrooke,
Sherbrooke, Qu\'ebec, Canada, J1K 2R1.}
\email{ibrahim.assem@usherbrooke.ca}

\author[Coelho]{Fl\'avio U. Coelho}
\address{Departamento de Matem\'atica-IME, Universidade de S\~ao Paulo,
CP 66281, S\~ao Paulo, SP, 05315-970, Brazil}
\email{fucoelho@ime.usp.br}

\author[Trepode]
{Sonia Trepode}
\address{Depto de Matem\'atica, FCEyN, Universidad Nacional de Mar del Plata, 7600,
Mar del Plata, Argentina} \email{strepode@mdp.edu.ar}

\keywords{representation dimension, multicoil algebras, torsionless-finite}
\subjclass{16G70, 16G20, 16E10}
\thanks{ This paper was finished when the second  author was
visiting the Universit\'e de Sherbrooke under the support of NSERC, to whom he is very
grateful.  The first author gratefully acknowledges partial support from the NSERC
of Canada and the Universit\'e de Sherbrooke. The second author thanks the CNPq.
The third author is a researcher of CONICET. The authors thank Grzegorz Bobinski
for pointing out an inacurrancy in a previous version.}

\maketitle

\begin{abstract}
We prove that, if $A$  is a strongly simply
connected algebra of polynomial growth then $A$ is torsionless-finite. In particular,
its representation  dimension is at most three.
\end{abstract}
\vspace{.3 cm}

Among the most useful algebraic invariants are the homological dimensions, which are meant to
measure how much an algebra or a module deviates from a situation considered to be ideal. Introduced
by Auslander in the early seventies \cite{Au}, the representation dimension of an Artin algebra
was long left aside from the mainstream of the theory, until a marked renewal of interest about
ten years ago. It measures the least global dimension of all endomorphism rings  of those finitely
generated modules which are both generators and cogenerators of the module category. Part of the
reason for this new interest comes from the fact that Igusa and Todorov have shown that, if the
representation dimension of an Artin algebra is at most three, then its finitistic dimension is
finite \cite{IT}. Also, Iyama has proved that, for any Artin algebra $A$, the representation
dimension rep.dim.$A$ is finite \cite{Iy}. Since then, there have been several attempts to
understand this invariant and to calculate it for classes of algebras, see for instance \cite{APT,CHU,CP,Op}
or the survey \cite{Ri3}. It was shown by Auslander that an Artin algebra is representation-finite
if and only if its representation dimension is at most two \cite{Au}. Since Auslander's expectation
was that this dimension would measure how far an algebra is from being representation-finite,
it is natural to ask whether tame algebras have representation dimension at most three. The answer
to this question is known to be positive for some classes of tame algebras, such as special
biserial algebras \cite{EHIS} and domestic self-injective algebras socle equivalent to a weakly
symmetric algebra of euclidean type \cite{BHS}.

The objective of our paper is to prove that the representation dimension of a strongly simply
connected algebra of polynomial growth (over an algebraically closed field) is at most three.
These algebras form a nice class which has been extensively studied by Skowro\'nski, de la Pe\~na
and others, see, for instance, the survey \cite{AS4}. In particular, it is shown in \cite{Sk2} that
strongly simply connected algebras of polynomial growth are multicoil algebras. We recall also
that, as pointed out in \cite{Ri3}, an algebra which is torsionless-finite (that is, such that
any indecomposable projective module has only finitely many isomorphism classes of indecomposable
submodules) has its representation dimension at most three. Our main result is the following theorem.
\vs\\
{\sc Theorem A.} {\it Let $A$ be a strongly simply connected algebra of polynomial growth. Then $A$ is torsionless-finite. In particular, rep.dim$.A \leq 3$.}
\vs

In the course of the proof, we found that the following result, of independent interest, was necessary.
\vs\\
{\sc Theorem B.} {\it Let $C$ be a tame concealed algebra of type distinct of ${\widetilde{\A}}$, and $(H_{\lambda})_\lambda$ be an infinite family
of pairwise non-isomorphic simple homogeneous $C$-modules. Suppose $M$ is a $C$-module such that
$H_{\lambda} \subset M$ for all $\lambda$. Then $C[M]$ is wild. }
\vs

The paper is organized as follows. After a short preliminary section, in which we fix the notations
and recall facts about multicoil algebras, we prove our Theorem(B) in section 2 and Theorem(A) in section 3.

\section{Multicoil algebras}

\subsection{Notation.}
In this paper, $k$ denotes a fixed algebraically closed field. By an algebra $A$ is meant a basic,
connected, associative finite dimensional $k$-algebra with an identity. Thus, there exists a connected
bound quiver $(Q_A, I)$ and an isomorphism $A \simeq kQ_A/I$. Equivalently, $A$ may be viewed as a
$k$-category, of which the object class $A_0$ is the set of points of $Q_A$ and the set of morphisms
from $x$ to $y$ is the quotient of the $k$-vector space $kQ_A(x,y)$ of linear combinations of paths
in $Q_A$ from $x$ to $y$ by $I(x,y) = I\cap kQ_A(x,y)$, see \cite{BG}. A full subcategory $C$ of $A$ is
{\it convex} (in $A$) if, for any path $x_0 \ra x_1 \ra \cdots \ra x_t$ in $A$ with $x_0, x_t \in C_0$,
we have $x_i \in C_0$ for all $i$. The algebra $A$ is {\it triangular} if $Q_A$ is acyclic.

By $A$-module is meant a finitely generated right $A$-module. We denote by mod$A$ the category of $A$-modules
and by ind$A$ a full subcategory consisting of a complete set of representatives of the isomorphism
classes of indecomposable $A$-modules. We recall that, if $A \simeq kQ/I$, then an $A$-module $M$ is
identified to a corresponding representation $(M(x)_{x\in Q_0}, M(\alpha)_{\alpha\in Q_1})$ of the
bound quiver $(Q,I)$, see \cite{ASS}. For a point $x\in Q_0$, we denote by $P_x$ (or $I_x$, or $S_x$) the
indecomposable projective (or injective, or simple, respectively) $A$-module corresponding to $x$. The
{\it support} of an $A$-module $M$ is the full subcategory Supp$M$ of $A$ with objects those $x\in A_0$
such that $M(x) \neq 0$. For a full subcategory ${\mathcal C}$ of mod$A$, we denote by add${\mathcal C}$
the additive full subcategory with objects the direct sums of direct summands of objects in ${\mathcal C}$.
If ${\mathcal C}$ contains a single module $M$, we write add${\mathcal C} =$ add$M$. For two full
subcategories ${\mathcal C}, {\mathcal C}'$ of ind$A$, the notation Hom$_A({\mathcal C}, {\mathcal C}') =0$
means that Hom$_A(M, M')= 0$ for all $M\in {\mathcal C}, M\in  {\mathcal C}'$. We then denote by
${\mathcal C}'\vee {\mathcal C}$ the full subcategory of ind$A$ having as objects those of
${\mathcal C}_0' \cup {\mathcal C}_0$.

A {\it path} in ind$A$ from $M$ to $N$ (sometimes denoted as $M \leadsto N$) is a sequence of non-zero
morphisms
$$  M = M_0 \lra M_1 \lra \cdots \lra M_t = N \leqno(*) $$
with all $M_i$ in ind$A$. A path $(*)$ is a {\it cycle} if $M=N$ and at least one of the morphisms is not
an isomorphism. An indecomposable module is {\it directed} if it lies on no cycle.

We use freely properties of the Auslander-Reiten translations $\tau_A =$ DTr and $\tau^{-1}_A = $ TrD
and the Auslander-Reiten quiver $\Gamma($mod$A$) of $A$ for which we refer to \cite{ASS,Ri2}. We identify points
in $\Gamma($mod$A$) with the corresponding indecomposable $A$-modules, and components with the
corresponding full subcategories of ind$A$. A component $\Gamma$ is {\it standard} if the category
$\Gamma$ is equivalent to the mesh category $k(\Gamma)$. For tubes, tubular extensions and
coextensions, we refer to \cite{Ri2} and for tame algebras, we refer to \cite{Sk1}.

\subsection{One-point extensions.}
The {\it one-point extension} of an algebra $A$ by an $A$-module $M$ is the matrix algebra
$$ A[M] \ = \ \bema{cc} A & 0 \\ M & k \ema $$
with the usual addition and multiplication of matrices. The quiver of $A[M]$ contains $Q_A$ as a full
convex subquiver and there is an additional (extension) point which is a source. The $A[M]$-modules
are identified with triples $(V, X, \varphi)$, where $V$ is a $k$-vector space, $X$ an $A$-module and
$\varphi \colon V \lra $ Hom$_A(M,X)$ is a $k$-linear map. An $A[M]$-linear map
$(V,X, \varphi) \lra (V', X', \varphi ')$ is a pair $(f,g)$ where $f \colon V \lra V'$ is $k$-linear
and $g \colon X \lra X'$ is $A$-linear such that $\varphi'f = $ Hom$_A(M,g)\varphi$. The dual notion
is that of {\it one-point coextension}.

A {\it vector space category} \cite{Ri1,Ri2} ${\mathcal K}$ is a $k$-category together with a faithful
$k$-linear functor $|\cdot | \colon {\mathcal K} \lra $ mod$k$. The {\it subspace category}
${\mathcal U}({\mathcal K})$ of ${\mathcal K}$ has as objects the triples $(V, X, \varphi)$ where $V$ is a
$k$-vector space, $X$ an object in ${\mathcal K}$ and $\varphi \colon V \lra |X|$ is a $k$-linear
monomorphism. A morphism $(V,X, \varphi) \lra (V', X', \varphi ')$ is a pair $(f,g)$ where
$f \colon V \lra V'$ is $k$-linear and $g \colon X \lra X'$ is a morphism in ${\mathcal K}$
such that $\varphi'f = |g|\varphi$.

If $A$ is an algebra and $M$ an $A$-module, one considers the vector space category Hom$_A(M,$ mod$A$)
whose objects are of the form Hom$_A(M,X)$ with $X$ an $A$-module, and morphisms are of the form
$$ \mbox{Hom}_A(M,f) \colon \mbox{ Hom}_A(M,X) \lra \mbox{ Hom}_A(M',X'), $$
where $f \colon X \lra X'$ is $A$-linear. Then
$|$Hom$_A(M,X)|$ is just the underlying\break
$k$-vector space of Hom$_A(M,X)$. It is shown in \cite{Ri1} that
${\mathcal U}($Hom$_A(M,$mod$A$)) is equivalent to the full subcategory of mod$A[M]$ consisting of
the triples $(V,X, \varphi)$ without non-zero direct summands of the form $(k,0,0)$ or $(0, Y,0)$
where\break
Hom$_A(M,Y) = 0$. We need essentially the following lemma.
\vs\\
{\sc Lemma.} {\it Let $A$ be a tame algebra, and $M$ be an $A$-module. If $L$ is a submodule of
$M$ such that $A[L]$ is wild, then $A[M]$ is wild.}
\begin{proof}
Since $A$ is tame, while $A[L]$ is wild, then the vector space category\break
${\mathcal U}($Hom$_A(L,$mod$A$))
is wild. Now the inclusion $L \hookrightarrow M$ induces an epifunctor
$$ {\mathcal U}(\mbox{Hom}_A(M,\mbox{mod}A)) \lra {\mathcal U}(\mbox{Hom}_A(L,\mbox{mod}A)).  $$
Since the latter is wild, then so is ${\mathcal U}(\mbox{Hom}_A(M,\mbox{mod}A))$. Therefore,
$A]M]$ is wild.
\end{proof}

\subsection{Coils.}
Let $A$ be an algebra, and $\Gamma$ a standard component of $\Gamma($mod$A$). Given a module
$X \in \Gamma$, called the {\it pivot}, the {\it support} Supp$(X, -)|_\Gamma$ of the functor
Hom$_A(X, -)|_\Gamma$ is defined as follows. Let ${\mathcal H}_X$ be the full subcategory of
ind$A$ consisting of the $Y \in \Gamma$ such that Hom$_A(X,Y) \neq 0$, and ${\mathcal I}_X$
be the ideal of ${\mathcal H}_X$ consisting of the morphisms $f \colon Y \lra Y'$ (with
$Y, Y'$ in ${\mathcal H}_X$) such that Hom$_A(X, f) \neq 0$. Then
Supp$(X, -)|_\Gamma = {\mathcal H}_X/ {\mathcal I}_X$. We define three admissible operations:
\benu
\item[(ad1)] Assume Supp$(X,-)|_\Gamma$ consists of an infinite sectional path starting at $X$
$$ X = X_0 \lra X_1 \lra X_2 \lra \cdots $$
Let $t\geq1$, $D = \T_t (k)$ be the full $t\times t$ lower triangular matrix algebra, and $Y$
be the unique indecomposable projective-injective $D$-module. The {\it modified algebra} of
$A$ is $A' = (A\times D)[X \oplus Y]$. If $t=0$, it is simply $A' = A[X]$.
\item[(ad2)] Assume Supp$(X,-)|_\Gamma$ consists of two sectional paths starting at $X$, the first
infinite and the second finite with at least one arrow
$$ Y_t \longleftarrow \cdots \longleftarrow Y_2 \longleftarrow Y_1 \longleftarrow X = X_0
\lra X_1 \lra X_2 \lra \cdots $$
with $t\geq 1$. The {\it modified algebra} is $A' = A[X]$.
\item[(ad3)] Assume Supp$(X,-)|_\Gamma$ consists of two parallel sectional paths, the first
infinite and starting at $X$, and the second finite with at least one arrow
\begin{picture}(60,19)

\put(21,0){$X = X_0$}
\put(37,0){$\lra$}
\put(45,0){$X_1$}
\put(52,0){$\lra$}
\put(63,0){$\cdots$}
\put(71,0){$\lra$}
\put(79,0){$X_{t-1}$}
\put(87,0){$\lra$}
\put(95,0){$X_{t}$}
\put(102,0){$\lra$}
\put(112,0){$\cdots$}
\put(32,6){$\uparrow$}
\put(46,6){$\uparrow$}
\put(80,6){$\uparrow$}
\put(31,11){$Y_1$}
\put(37,11){$\lra$}
\put(45,11){$Y_2$}
\put(52,11){$\lra$}
\put(63,11){$\cdots$}
\put(71,11){$\lra$}
\put(79,11){$Y_t$}

\end{picture}
\vspace{.5 cm}\\
where $t\geq 2$. In particular, $X_{t-1}$ is injective. The {\it modified algebra}
is $A' = A[X]$.
\enu

In each case, $t$ is the {\it parameter} of the operation, and the component $\Gamma'$ of $\Gamma($mod$A'$)
containing $X$ is the {\it modified component}. We also consider the duals of the above operations,
denoted by (ad1*), (ad2*) and (ad3*), respectively. These six operations are the {\it admissible operations}.

An Auslander-Reiten component $\Gamma$ is a {\it coil} if there exists a sequence\break
$\Gamma_0, \Gamma_1, \cdots, \Gamma_m = \Gamma$ where $\Gamma_0$ is a stable tube and, for each $i$,
$\Gamma_{i+1}$ is obtained from $\Gamma_i$ by an admissible operation.

Let $A$ be an algebra $A$. A family ${\mathcal R} = ( {\mathcal R}_{\lambda})_{\lambda \in \Lambda}$ of
components of $\Gamma($mod$A$) is {\it weakly separating} if the indecomposable $A$-modules not in
${\mathcal R}$ split into two classes ${\mathcal P}$ and ${\mathcal Q}$ such that
\benu
\item[(1)] The ${\mathcal R}_\lambda$ are standard and pairwise orthogonal.
\item[(2)] Hom$_A({\mathcal Q}, {\mathcal P}) $ = Hom$_A({\mathcal Q}, {\mathcal R}) $ =
Hom$_A ({\mathcal R},{\mathcal P}) = 0$.
\item[(3)] Any morphism from ${\mathcal P}$ to ${\mathcal Q}$ factors through add${\mathcal R}$.
\enu

If ${\mathcal R}$ is a weakly separating family in $\Gamma$(mod$A$) consisting of stable tubes, then
an algebra $B$ is a coil enlargement of $A$ using modules from ${\mathcal R}$ if there exists a finite sequence
of algebras $A = A_0, A_1, \cdots , A_m = B$ such that, for each $i$, $A_{i+1}$ is obtained from $A_i$
by an admissible operation with pivot either on a stable tube of ${\mathcal R}$ or on a coil of
$\Gamma($mod$A_i)$ obtained from a stable tube of ${\mathcal R}$ by means of the admissible operations
done so far. The {\it coil type} $c_B = ( c^-_B, c^+_B)$ of $B$ is a pair of functors $\Lambda \lra \N$
defined by induction on $i$, for each $\lambda \in \Lambda$, as follows.
\benu
\item[(a)] $c_A = (c^-_0, c^+_0)$ is such that $c^-_0(\lambda) = c^+_0(\lambda)$ is the rank of the
stable tube ${\mathcal R}_\lambda$.
\item[(b)] If $c_{A_{i-1}} = (c^-_{i-1}, c^+_{i-1})$ is known and $t_i$ is the parameter of the
operation from $A_{i-1}$ to $A_i$, then $c_{A_i} = (c^-_i, c^+_i)$ is defined by:
$$ c^-_i(\lambda) \ = \ \left\{ \begin{array}{ll} c^-_{i-1}(\lambda) + t_i + 1 &
\mbox{if the operations is (ad1*), (ad2*), (ad3*)}\\
& \mbox{with pivot in the coil of } \Gamma(\mbox{mod}A_{i-1})\\
&  \mbox{arising from } {\mathcal R}_\lambda .\\
c^-_{i-1}(\lambda) & \mbox{otherwise} \end{array} \right. $$
and
$$ c^+_i(\lambda) \ = \ \left\{ \begin{array}{ll} c^+_{i-1}(\lambda) + t_i + 1 &
\mbox{if the operations is (ad1), (ad2), (ad3)}\\
& \mbox{with pivot in the coil of } \Gamma(\mbox{mod}A_{i-1})\\
&  \mbox{arising from } {\mathcal R}_\lambda .\\
c^+_{i-1}(\lambda) & \mbox{otherwise} \end{array} \right. $$
\enu

If all but at most finitely many values of each of $c^-_B, c^+_B$ equal 1, we replace each
sequence by a finite sequence containing at least two terms, and including all those exceeding 1.
\vsd\\
{\sc Theorem.} \cite{AST} {\it Let $A$ be an algebra having a weakly separating family of stable
tubes ${\mathcal R}$, and $B$ be a coil enlargement of $A$ using modules from ${\mathcal R}$. Then
\benu
\item[(a)] There is a unique maximal branch coextension $B^-$ (or extension $B^+$) of $A$ which is
a full convex subcategory of $B$, and having $c^-_B$ as coextension type (or $c^+_B$ as extension
type, respectively).
\item[(b)] ind $B = {\mathcal P}' \vee {\mathcal R}' \vee {\mathcal Q}'$, where ${\mathcal R}'$ is a
weakly separating family of ind$B$ obtained from ${\mathcal R}$ by the sequence of admissible operations,
and separating ${\mathcal P}'$ from ${\mathcal Q}'$. Moreover, ${\mathcal P}'$ consists of $B^-$-modules,
while ${\mathcal Q}'$ consists of $B^+$-modules.
\enu }

\subsection{Tame coil enlargements.}
Let $B$ be a coil enlargement of a tame concealed algebra. Its {\it coil type} $c_B = (c^-_B, c_B^+)$ is
tame if each of the sequences $c_B^-$, $c_B^+$ is one of the following: $(p,q)$ with $1 \leq p\leq q$,
$(2,2,r)$ with $r \geq 2$, (2,3,3), (2,3,4), (2,3,5), (2,4,4), (2,3,6) or(2,2,2,2). We have the following
result.
\vsd\\
{\sc Corollary.} \cite{AST}(4.3) {\it Let $B$ a coil enlargement of a tame concealed algebra. The
following conditions are equivalent:
\benu
\item[(a)] $B$ is tame.
\item[(b)] $B$ is of polynomial growth.
\item[(c)] $c_B$ is tame.
\item[(d)] $B^-$ and $B^+$ are tame.
\enu
Moreover, $B$ is domestic if and only if $B^-$ and $B^+$ are tilted of euclidean type. }

\subsection{Multicoil algebras.}
An Auslander-Reiten component $\Gamma$ is a {\it multicoil} if it contains a full translation
subquiver $\Gamma'$ which is a disjoint union of coils such that no point in $\Gamma \setminus \Gamma'$
belongs to a cyclical path.

An algebra $A$ is a {\it multicoil algebra} if, for any cycle $M_0 \ra M_1 \ra \cdots \ra M_t= M_0$
in ind$A$, all the $M_i$ lie in one standard coil of a multicoil of $\Gamma($mod$A$). The first part of
the following theorem  is \cite{AS2} (4.6), the second part is \cite{Sk2}(4.1).
\vsd\\
{\sc Theorem.} \cite{AS2,Sk2} {\it Let $A$ be a multicoil algebra, then $A$ is of polynomial growth. If $A$ is strongly
simply connected, then the converse also holds. }
\vs

If $A$ is a multicoil algebra, then it is triangular (hence of finite global dimension) \cite{AS3}(3.5).
Also, any full convex subcategory of $A$ is a multicoil algebra \cite{AS3}(5.6).

\subsection{Supports.} Supports of indecomposable modules over multicoil algebras are characterised
in the following lemma.
\vsd\\
{\sc Lemma.} Let $M$ be an indecomposable module over a multicoil algebra $A$, and let $B =$ Supp$M$.
Then:
\benu
\item[(a)] If $M$ is directed, then $B$ is tame and tilted.
\item[(b)] If $M$ is not directed, then $B$ is a tame coil enlargement of a tame concealed algebra.
\enu
\begin{proof} Since $B$ is a full subcategory of $A$, which is tame, then $B$ is also tame. Then (a)
follows from \cite{ASS}(IX.2.8, p. 366) and (b) follows from \cite{AS3}(5.9).
\end{proof}

Note that, if $L$ is an $A$-submodule of $M$, and $x \in A_0$ is such that $L(x) \neq 0$,
then $M(x) \neq 0$. Hence, $L$ is also a $B$-submodule of $M$.

\section{One-point extension of tame concealed algebras}

\subsection{}
In this section, we prove that, if $C$ is a tame concealed algebra and $M$ is a $C$-module containing an
infinity of pairwise non-isomorphic simple homogeneous submodules, then $C[M]$ is wild. We start
with reduction lemmata.
\vsd\\
{\sc Lemma.} {\it Let $C$ be a tame concealed algebra, let $M$ be a preinjective $C$-module,
and $H$ a simple homogeneous submodule of $M$. Then:
\benu
\item[(a)] $M$ is sincere.
\item[(b)] If $C[M]$ is not wild, then dim$_k $(top$M) \leq 2$. Then, either $M$ is
indecomposable, or $M \simeq M_1 \oplus M_2$, with $M_1, M_2$ indecomposables with simple top.
\enu }
\begin{proof}
(a) This follows from the sincerity of $H$.

(b) If dim$_k($top$M) =d  >2$, then $C[M]$ contains a wild full subcategory of the form\\

\begin{picture}(60,10)

\multiput(61,0)(22,0){2}{\circle*{1}}

\multiput(80,3)(0,-6){2}{\vector(-1,0){16}}
\multiput(72,2)(0,-1){5}{\circle*{.1}}

\put(70,5){$\alpha_1$}
\put(70,-8){$\alpha_d$}
\end{picture}
\vspace{1 cm}\\
a contradiction. The second statement follows from Nakayama's lemma, and (1.2).
\end{proof}

\subsection{}
Let $A = kQ/I$ be a bound quiver algebra. It is well-known (see, for instance,
\cite{ASS}(III.2.2, p.77)) that, if $M$ is an $A$-module, then (top$M)(x) = M(x)$ is $x$ is a source,
and (top$M)(y) = \sum \{ $Coker$(M(\alpha)) \colon \alpha \colon x \lra y \}$, if $y$ is not a
source.

If $M$ satisfies the hypothesis of (2.1)(b), and $x$ is a source, we then have dim$_kM(x) \leq 2$.
\vsd\\
{\sc Corollary.}{\it Let $C$ be a tame concealed algebra, let $M$ be a preinjective $C$-module,
and $H$ a simple homogeneous submodule of $M$. If $C[M]$ is not wild, then:
\benu
\item[(a)] $C$ has at most two sources.
\item[(b)] If $C$ has just one source $s$ and dim$_k H(s) = d \geq 2$, then $d=2$
and top$H$ = top$M$.
\item[(c)] If $C$ has two sources $s_1, s_2$ then dim$_kH(s_1) = $ dim$_k H(s_2) = 1$ and
top$H = $ top$M$.
\enu }
\begin{proof} (a) If $C$ has at least three sources $s_1, s_2, s_3$, the remark above
implies that top$M$ has $S_{s_1} \oplus S_{s_2} \oplus S_{s_3}$ as a summand, and
dim$_k($top$M) \geq 3$, a contradiction.

(b) Since $H(s) \subset M(s)$, which is at most two-dimensionsl, then $d=2$ and top$H = $top$M = S^2$.

(c) Clearly, $S_{s_1} \oplus S_{s_2}$ is a summand of both top$H$ and top$M$. Since
dim$_k($top$M) \leq 2$, then top$M = S_{s_1} \oplus S_{s_2}$. We claim that top$H =  S_{s_1} \oplus S_{s_2}$.
If this is not the case, then there exists $a \notin \{ s_1, s_2 \}$ such that $S_a$ is a
summand of top$H$. Hence (top$H)(a) \neq 0$. By the remark above, there exists an arrow
$ \alpha \colon b \lra a$ in $C$ such that $H(\alpha)$ is not surjective. On the other hand, since $S_a$ is
not a summand of top$M$, then $M(\alpha)$ is surjective. Now there is a path from one of the
$s_i$ (with $i \in \{ 1,2 \}$) to $a$ passing through $\alpha$, which we may assume, without loss of
generality, to be of minimal length:
$$s_i = b_t \lra b_{t-1} \lra \cdots \lra b_1 = b \stackrel{\alpha}{\lra} a \leqno(*)$$
We claim that dim$_kM(b) \geq 2$. Indeed, if this in not the case, then the inclusion
$j \colon H \hookrightarrow M$ yields a commutative square\\
\begin{picture}(60,16)

\multiput(60,8)(0,-16){2}{\vector(1,0){24}}
\multiput(51.5,6)(38.5,0){2}{\vector(0,-1){10}}

\put(48,7){$H(b)$}   \put(87,7){$H(a)$}
\put(48,-9){$M(b)$}   \put(87,-9){$M(a)$}
\put(68,10){$H(\alpha)$}   \put(68,-6){$M(\alpha)$}
\put(46,-1){$j_b$}   \put(92,-1){$j_a$}
\end{picture}
\vspace{1.4 cm}\\
and $M(b) \simeq H(b) \simeq k$ gives that $j_b$ is an isomorphism. Also, since $H$ is sincere and $M(\alpha)$
is surjective, we have $M(a) \simeq H(a) \simeq k$. So $H(\alpha)$ is an isomorphism, a contradiction
to its non-surjectivity. Since dim$_k M(s_i) = 1$, there exists an arrow $\beta \colon c \lra d$ on the
path $(*)$ above such that 1 = dim$_k M(c) <$ dim$_k M(d)$. Therefore Coker$M(\beta) \neq 0$, hence
$S_d$ is a summand of top$M$, a contradiction to top$M = S_{s_1} \oplus S_{s_2}$ and $ d \notin \{ s_1, s_2 \}$.
\end{proof}

\subsection{}
{\sc Lemma. } {\it Let $C$ be a tame concealed algebra of type ${\widetilde{\D}_4}$
and $(H_{\lambda})_{\lambda}$ be an infinite family of pairwise non-isomorphic simple homogeneous modules.
If $M$ is a module such that $H_{\lambda} \subset M$ for all $\lambda$, then $C[M]$ is wild. }
\begin{proof}
Assume first $C$ to be non-schurian, then $C$ is given by the quiver \\
\begin{picture}(60,15)

\multiput(56,0)(16,0){3}{\circle*{1}}
\multiput(72,10)(0,-20){2}{\circle*{1}}

\multiput(70,0)(16,0){2}{\vector(-1,0){12}}
\multiput(70,9)(16,-10){2}{\vector(-3,-2){12}}
\multiput(70,-9)(16,10){2}{\vector(-3,2){12}}

\put(52,-1){$p$} \put(90,-1){$s$}
\put(71,12){$a$} \put(71,2){$b$} \put(71,-8){$c$}
\put(80,7){\small$\alpha$}   \put(77.5,1){\small$\beta$}   \put(79.5,-8){\small$\gamma$}
\put(60.5,6.5){\small$\alpha '$}   \put(64,1){\small$\beta '$}   \put(59,-8){\small$\gamma '$}
\end{picture}
\vspace{1.4 cm}\\
bound by $\alpha \alpha' + \beta \beta' + \gamma \gamma' = 0$. For $\lambda \in k$, $H_\lambda$ is
given by \\
\begin{picture}(60,17)

\multiput(51,-1)(20,0){3}{$k$}
\multiput(71,10)(0,-22){2}{$k$}

\multiput(68,0)(20,0){2}{\vector(-1,0){12}}
\multiput(68,10)(20,-12){2}{\vector(-3,-2){12}}
\multiput(68,-10)(20,12){2}{\vector(-3,2){12}}

\put(81.5,8){\small 1}   \put(81.5,1){\small 1}   \put(81.5,-10){\small 1}
\put(60,7){\small$\lambda$}   \put(62,1){1}   \put(50,-9.5){\small$-\lambda -1$}
\end{picture}
\vspace{1.6 cm}\\
Since $H_\lambda \subset M$ for any $\lambda$, then $M$ is preinjective and we may further assume
that dim$_k($top$M) \leq 2$. Also, soc$H_\lambda = S_p$ is a summand of soc$M$ while $S_s$ is a
summand of top$M$. A straightforward examination of the preinjective component of $C$ shows that
$M$ is indecomposable and is one of the four modules $I_p, \tau_C^2 I_a, \tau_C^2 I_b$ or
$\tau_C^2 I_c$. In the first case, $C[I_p]$ has quiver  \\
\begin{picture}(60,15)

\multiput(48,0)(16,0){4}{\circle*{1}}
\multiput(64,10)(0,-20){2}{\circle*{1}}

\multiput(62,0)(16,0){2}{\vector(-1,0){12}}
\multiput(94,1)(0,-2){2}{\vector(-1,0){12}}
\multiput(62,9)(16,-10){2}{\vector(-3,-2){12}}
\multiput(62,-9)(16,10){2}{\vector(-3,2){12}}

\put(44,-1){$p$} \put(79,2){$s$}
\put(63,12){$a$} \put(63,2){$b$} \put(63,-8){$c$}
\put(72,7){\small$\alpha$}   \put(69.5,1){\small$\beta$}   \put(71.5,-8){\small$\gamma$}
\put(52.5,6.5){\small$\alpha '$}   \put(56,1){\small$\beta '$}   \put(51,-8){\small$\gamma '$}
\put(87,3){\small$\lambda$}  \put(87,-5){\small$\mu$}
\end{picture}
\vspace{1.4 cm}\\
bound by $\lambda \alpha = \mu \alpha$, $\lambda \beta = \mu \beta$, $\lambda \gamma = \mu \gamma$,
$\alpha \alpha' + \beta \beta' + \gamma \gamma' = 0$. Changing the presentation (replacing $\lambda$
by $\lambda - \mu$) gives the same quiver bound by $\lambda \alpha = 0, \lambda \beta = 0, \lambda \gamma = 0,
\alpha \alpha' + \beta \beta' + \gamma \gamma' = 0$. This is a split extension \cite{ACT} of the
algebra given by the quiver \\

\begin{picture}(60,17)
\multiput(48,0)(16,0){4}{\circle*{1}}
\multiput(64,10)(0,-20){2}{\circle*{1}}

\multiput(62,0)(16,0){2}{\vector(-1,0){12}}
\multiput(94,0)(0,-2){1}{\vector(-1,0){12}}
\multiput(62,9)(16,-10){2}{\vector(-3,-2){12}}
\multiput(62,-9)(16,10){2}{\vector(-3,2){12}}

\put(44,-1){$p$} \put(79,2){$s$}
\put(63,12){$a$} \put(63,2){$b$} \put(63,-8){$c$}
\put(72,7){\small$\alpha$}   \put(69.5,1){\small$\beta$}   \put(71.5,-8){\small$\gamma$}
\put(52.5,6.5){\small$\alpha '$}   \put(56,1){\small$\beta '$}   \put(51,-8){\small$\gamma '$}
\put(87,2){\small$\mu$}
\end{picture}
\vspace{1.4 cm}\\
bound by $\alpha \alpha' + \beta \beta' + \gamma \gamma' = 0$, which is evidently wild.

In the second case, $C[\tau_C^2 I_a]$ has quiver \\
\begin{picture}(60,15)

\multiput(48,0)(16,0){4}{\circle*{1}}
\multiput(64,10)(0,-20){2}{\circle*{1}}

\multiput(62,0)(16,0){2}{\vector(-1,0){12}}
\multiput(94,1)(0,-2){2}{\vector(-1,0){12}}
\multiput(62,9)(16,-10){2}{\vector(-3,-2){12}}
\multiput(62,-9)(16,10){2}{\vector(-3,2){12}}

\put(44,-1){$p$} \put(79,2){$s$}
\put(63,12){$a$} \put(63,2){$b$} \put(63,-8){$c$}
\put(72,7){\small$\alpha$}   \put(69.5,1){\small$\beta$}   \put(71.5,-8){\small$\gamma$}
\put(52.5,6.5){\small$\alpha '$}   \put(56,1){\small$\beta '$}   \put(51,-8){\small$\gamma '$}
\put(87,3){\small$\lambda$}  \put(87,-5){\small$\mu$}
\end{picture}
\vspace{1.4 cm}\\
bound by $\lambda \beta = \mu \beta$, $\lambda \gamma = \mu \gamma$,
$\alpha \alpha' + \beta \beta' + \gamma \gamma' = 0$. It contains as full convex subcategory the wild
hereditary algebra with quiver \\
\begin{picture}(60,10)

\multiput(70,0)(0,-2){1}{\vector(-1,0){14}}
\multiput(88,1)(0,-2){2}{\vector(-1,0){14}}

\multiput(54,0)(18,0){3}{\circle*{1}}
\put(63,2){$\alpha$}
\put(81,3){$\lambda$}  \put(81,-5){$\mu$}

\put(50,-1){$a$}  \put(71,-4){$s$}
\end{picture}
\vspace{1 cm}\\
The remaining cases follow by symmetry.

We may thus assume that $C$ is schurian. If $C[M]$ is not wild, then $C$ is hereditary and has one of the
quivers\\

\begin{picture}(60,17)

\multiput(12,9)(27.4,0){2}{\circle*{1}}
\multiput(12,-9)(27.4,0){2}{\circle*{1}}
\put(25.7,0){\circle*{1}}
\multiput(38,8)(-14,-9){2}{\vector(-3,-2){11}}
\multiput(38,-8)(-14,9){2}{\vector(-3,2){11}}
\put(6,9){$p_1$}   \put(6,-9.5){$p_2$}  \put(42,9){$s_1$}   \put(42,-9.5){$s_2$}  \put(24.5,2){$a$}

\multiput(65,0)(14,0){3}{\circle*{1}}
\multiput(65,9)(0,-18){2}{\circle*{1}}
\multiput(77,0)(14,0){2}{\vector(-1,0){10}}
\put(77,1){\vector(-3,2){10}}   \put(77,-1){\vector(-3,-2){10}}
\put(59,9){$p_1$}   \put(59,-0.5){$p_2$}  \put(59,-9.5){$p_3$}   \put(92,2){$s$}  \put(78,2){$a$}

\multiput(118,9)(0,-18){2}{\circle*{1}} \multiput(118,3.5)(0,-7){2}{\circle*{1}} \put(132,0){\circle*{1}}
\put(131,1){\vector(-3,2){11}}  \put(131,0.5){\vector(-4,1){11}}
\put(131,-0.5){\vector(-4,-1){11}}  \put(131,-1){\vector(-3,-2){11}}
\put(112,9){$p_1$}   \put(112,3){$p_2$}  \put(112,-4){$p_3$}   \put(112,-10){$p_4$}  \put(132,2){$s$}
\end{picture}
\vspace{1.6 cm}\\
Recall that soc$H_\lambda$ is a summand of soc$M$ and dim$_k($top$M) \leq 2$. It is easily seen that
in the first and third cases, no preinjective $C$-module satisfies these conditions. In the second case,
however, we have the possibility $M = \tau^2_C I_s$, but then $C[M]$ is given by the quiver\\

\begin{picture}(60,17)

\multiput(48,0)(16,0){4}{\circle*{1}}
\multiput(48,10)(0,-20){2}{\circle*{1}}

\multiput(62,0)(16,0){2}{\vector(-1,0){12}}
\multiput(94,1)(0,-2){2}{\vector(-1,0){12}}
\put(62,1){\vector(-3,2){12}}
\put(62,-1){\vector(-3,-2){12}}

\put(42,10){$p_1$}    \put(42,0){$p_2$}     \put(42,-10){$p_3$}
\put(79,2){$s$}       \put(63,2){$a$}
\put(71,1){\small$\alpha$}
\put(56,7){\small$\beta$}   \put(53.5,1){\small$\gamma$}   \put(55.5,-8.5){\small$\delta$}
\put(87,3){\small$\lambda$}  \put(87,-5){\small$\mu$}
\end{picture}
\vspace{1.6 cm}\\
bound by $\lambda \alpha \beta = \mu \alpha \beta$, $\lambda \alpha \gamma = \mu \alpha \gamma,
\lambda \alpha \delta = \mu \alpha \delta$. It contains the wild hereditary full subcategory \\
\begin{picture}(60,10)

\multiput(70,0)(0,-2){1}{\vector(-1,0){14}}
\multiput(88,1)(0,-2){2}{\vector(-1,0){14}}

\multiput(54,0)(18,0){3}{\circle*{1}}
\put(63,2){$\alpha$}
\put(81,3){$\lambda$}  \put(81,-5){$\mu$}

\put(50,-1){$a$}  \put(71,-4){$s$}
\end{picture}
\vspace{1 cm}\\
This establishes the lemma.
\end{proof}

\pn {\sc Example.} Before proving our next result, we observe that the above lemma does not hold true
for tame concealed algebras of type ${\widetilde{\A}}$. Indeed, let $C$ to be the Kronecker algebra, that is, the
hereditary algebra given by the quiver \\
\begin{picture}(60,7)
\multiput(79,1)(0,-2){2}{\vector(-1,0){14}}

\multiput(63,0)(18,0){2}{\circle*{1}}

\put(59,-1){$p$}  \put(83,-1){$s$}
\end{picture}
\vspace{.7 cm}\\
Observe that the indecomposable injective $C$-module $I_p$ at the point $p$ contains all the
indecomposable regular modules $H_\lambda$ given by \\
\begin{picture}(60,7)
\multiput(79,1)(0,-2){2}{\vector(-1,0){14}}

\multiput(63,0)(18,0){2}{\circle*{1}}

\put(59,-1){$k$}  \put(83,-1){$k$}
\put(72,3){$1$}  \put(72,-5){$\lambda$}
\end{picture}
\vspace{.7 cm}\\
Now, the one point extension $C[I_p]$ is the radical square zero algebra given by the
quiver \\
\begin{picture}(60,7)
\multiput(70,1)(0,-2){2}{\vector(-1,0){14}}
\multiput(88,1)(0,-2){2}{\vector(-1,0){14}}

\multiput(54,0)(18,0){3}{\circle*{1}}
\put(63,3){$\beta$}  \put(63,-5){$\delta$}
\put(81,3){$\alpha$}  \put(81,-5){$\gamma$}
\put(50,-1){$p$}  \put(71,-4){$s$}
\end{picture}
\vspace{.7 cm}\\
bounded by $\alpha \delta = 0$, $ \gamma \beta = 0 $ and $\alpha \beta = \gamma \delta$,
which is clearly tame.
\vs

\pn {\sc Remark.} Notice, however, that if $C$ is tame concealed of type ${\widetilde{\A}}$ and schurian
and $M$ satisfies the conditions of (2.3), then $C[M]$ is wild. Indeed, if this is the case then,
because of (2.2), $C$ has at most 2 sources, hence is hereditary with quiver \\

\begin{picture}(60,25)

\multiput(48,12.5)(16,0){4}{\circle*{1}}
\multiput(64,3.5)(16,0){2}{\circle*{1}}
\multiput(48,-14)(16,0){4}{\circle*{1}}
\multiput(64,-5)(16,0){2}{\circle*{1}}

\multiput(62,12.5)(32,0){2}{\vector(-1,0){12}}
\put(62,4.5){\vector(-2,1){12}}
\put(94,10.5){\vector(-2,-1){12}}
\multiput(62,-14)(32,0){2}{\vector(-1,0){12}}
\put(62,-6){\vector(-2,-1){12}}
\put(94,-12){\vector(-2,1){12}}

\multiput(66,12.5)(3,0){5}{\circle*{.1}}
\multiput(66,-14)(3,0){5}{\circle*{.1}}
\multiput(66,-4)(3,1.5){5}{\circle*{.1}}
\multiput(78,-4)(-3,1.5){5}{\circle*{.1}}

\put(41.5,11.5){$p_1$}
\put(98.5,11.5){$s_1$}
\put(41.5,-15){$p_2$}
\put(98.5,-15){$s_2$}
\end{picture}
\vspace{2.5 cm}\\
Here, soc$H_\lambda = S_{p_1} \oplus S_{p_2}$ is a summand of soc$M$ while top$M = $ top$H_\lambda = S_{s_1}\oplus S_{s_2}$. It is easily seen that no preinjective $C$-module satisfies these conditions.

\subsection{}
{\sc Lemma.} {\it Let $C$ be a tame concealed algebra of type ${\widetilde{\D}_n}$, with $n \geq 5$,
or ${\widetilde{\E}}$ and $H$ be a simple homogeneous $C$-module. If $M$ is preinjective such that
$H \subset M$, then $C[M]$ is wild. }
\begin{proof}
We claim that $C[M]$ is wild. Indeed, the tubular type of $C$ is of the form $(a,b,c)$ with at
least one of $a, b, c$ larger than or equal to 3. Taking then a one-point extension by a simple
homogeneous module yields a wild algebra. This establishes our claim. Applying now (1.2), we get
that $C[M]$ is wild, as required.
\end{proof}

\subsection{}
We are now able to prove the main result of this section.
\vs\\
{\sc Theorem.} {\it Let $C$ be a tame concealed algebra of type distinct of ${\widetilde{\A}}$, and
$(H_\lambda)_\lambda$ be an infinite
family of pairwise non-isomorphic modules. Suppose $M$ is a $C$-module such that $H_\lambda \subset M$
for all $\lambda$. Then $C[M]$ is wild. }
\begin{proof} This follows immediately from (2.3) and (2.4).
\end{proof}

\section{Torsionless finiteness and representation dimension.}

\subsection{}
Let $A$ be an Artin algebra. An $A$-module $M$ is called a {\it generator} (of mod$A$) if $A_A \in$
add$M$ and a {\it cogenerator} if (D$A)_A = $ Hom$_k(_AA,k) \in $ add$M$. Let $A$ be a non-semisimple
algebra. The {\it representation dimension} rep.dim$.A$ of $A$ is the infimum of the global dimensions
of the algebras End$_AM$, where $M$ is a generator and a cogenerator of mod$A$, see \cite{Au}.

An Artin algebra $A$ is called {\it torsionless-finite} if every indecomposable projective $A$-module
has only finitely many isomorphism classes of indecomposable submodules. One defines {\it cotorsionless-finite}
dually.  An Artin algebra is torsionless-finite if and only if it is cotorsionless-finite \cite{AB}.
We need essentially the following result (see \cite{Ri3}).
\vsd\\
{\sc Theorem.} {\it If $A$ is torsionless-finite, then rep.dim.$A \leq 3$. }

\subsection{}
{\sc Lemma.} {\it Let $A$ be a (possibly wild) branch enlargement of a tame concealed algebra. Then
$A$ is torsionless-finite. In particular, rep.dim.$A \leq 3$. }
\begin{proof}
Using the description of mod$A$ in \cite{LS}, we see that, if $P_A$ is an indecomposable projective
$A$-module, then either $P_A$ is postprojective (in which case it clearly has only finitely many
isomorphism classes of indecomposable modules) or it lies in an inserted tube $\Gamma$. But in this
latter case, $P$ has only finitely many indecomposable submodules lying in $\Gamma$ and, using \cite{AR},
there are only finitely many postprojective modules $X$ such that dim$_k X \leq $ dim$_k P$. Since
the other tubes in the same family as $\Gamma$ are orthogonal to $\Gamma$, the proof is complete.
\end{proof}

\subsection{}
{\sc Corollary.} {\it Let $A$ be a tame quasi-tilted algebra, then $A$ is torsionless-finite.
In particular, rep.dim.$A \leq 3$. }
\begin{proof}
This follows from (3.2) and \cite{Sk3}
\end{proof}

\vsd\pn
{\sc Remark.} This corollary is a particular case of the main result of \cite{Op}.
Further, the same proof as in (3.2) and \cite{AS1} give that if $A$ is iterated tilted of euclidean
type, then $A$ is torsionless-finite, and so rep.dim.$A \leq 3$. This is a particular case of the main
result of \cite{CHU}.

\subsection{}
{\sc Lemma.} {\it Let $A$ be a strongly simply connected tame coil enlargement of a tame concealed algebra,
then $A$ is torsionless-finite. In particular, rep.dim.$\leq 3$. }
\begin{proof}
By (1.3), $A$ contains as full convex subcategories a unique maximal branch coextension $A^-$ and a
unique maximal branch extension $A^+$ of a tame concealed algebra, and mod$A$ contains a weakly separating
family of coils ${\mathcal T}'$ such that ind$A = {\mathcal P}' \vee {\mathcal T}'\vee {\mathcal Q}'$
where ${\mathcal P}' \subset $ ind$A^-$ and ${\mathcal Q}' \subset $ ind$A^+$. Let $P_A$ be an
indecomposable projective module. We have three cases:

(a) Assume $P \in {\mathcal P}'$, then $P$ is an indecomposable projective $A^-$-module. Because of
(3.2), and the fact that $A^-$ is a branch enlargement of a tame concealed algebra, $P$ has only
finitely many isomorphism classes of indecomposable submodules in mod$A^-$. Now, the indecomposable
submodules of $P$ in mod$A$ and mod$A^-$ coincide.

(b) Assume $P \in {\mathcal Q}'$. For the same reason, $P$ has only finitely many isomorphism classes
of indecomposable submodules in mod$A^+$, hence in mod$A$.

(c) Assume $P \in {\mathcal T}'$, and let $M$= rad$P$. There exists a sequence of full convex
subcategories of $A$
$$ A^- \ = \ A_0 \ \subset\ A_1 \ \subset \ \cdots \ \subset \ A_t = A$$
which are iterated one-point extensions of $A^-$, and an index $i$ such that $A_{i+1} = A_i[M]$, that is,
$P$ is the unique indecomposable projective in mod$A_{i+1}$ which is not in mod$A_i$. Also, $M$ is the
pivot of an admissible operation and then is indecomposable except perhaps in the case (ad1). Then,
$M = M' \oplus M''$, where $M''$ is a directed indecomposable, while $M'$ is an indecomposable lying in
a coil. Since $M''$ has only finitely many isomorphism classes of indecomposable submodules and $A_{i+1}$
is of the form $(A_i'\times D)[M'\oplus M'']$, where $D$ is a triangular matrix algebra, and $A_i'$ is the
full subcategory of $A_i$ with objects $(A_i')_0 = (A_i)_0 \setminus D_0$, then it suffices to consider the
submodules of $M'$. We may thus for simplicity assume that $M = M'$. \\
Let $M^-$ be the largest $A^-$-submodule of $M$. Assume that $M^-$ has infinitely many isomorphim classes
of indecomposable $A^-$-submodules. Because of (3.2), $M^-$ has to be a preinjective $A^-$-module. By (2.5),
$A^-[M^-]$ is wild. Since $A^-$ is a full convex subcategory of $A_i$, then $A_i[M^-]$ is wild. By (1.2),
and the tameness of $A_i$, we get that $A_{i+1} = A_i[M]$ is wild. This is absurd, because $A_{i+1}$ is a
full convex subcategory of $A$ which is tame. This shows that $M^-$ has only finitely many non-isomorphic
indecomposable submodules. Since a submodule of $M$ is either a submodule of $M^-$, or lies in the coil
containing $M$, we are done.
\end{proof}

\subsection{}
{\sc Corollary.} {\it Let $A$ be a multicoil algebra, and $P$ be an indecomposable projective $A$-module
lying in a coil $\Gamma$ of $\Gamma($mod$A$), then $P$ has only finitely many isomorphism classes of
indecomposable submodules. }
\begin{proof}
By \cite{AS3}(5.9), we can assume that $\Gamma$ is obtained from a stable tube over a tame concealed algebra
$C$ by a sequence of admissible operations and the support algebra $B$ of $\Gamma$ is obtained from $C$ by the
corresponding sequence of one-point extensions and coextensions. Since the $A$-submodules and the $B$-submodules
of $P$ coincide, the result follows from (3.4).
\end{proof}

\subsection{}
We are now able to state and prove the main result of this paper.
\vsd\\
{\sc Theorem.} {\it Let $A$ be a strongly simply connected algebra of polynomial growth, then
$A$ is torsionless-finite. In particular, rep.dim.$A \leq 3$.}
\begin{proof}
We may clearly assume that $A$ is representation-infinite. Let $P$ be an indecomposable projective
$A$-module. By (1.6), if $P$ is not directed, then the support algebra $B$ of $P$ is a tame coil
enlargement of a tame concealed algebra. By (3.4), $P$ has only finitely many isomorphism classes of
indecomposable $B$-submodules, hence $A$-submodules. On the other hand, if $P$ is directed, then, by
(1.6) again, $B$ is tame and tilted. Since, clearly, $B$ has a sincere directed indecomposable module
(namely $P$), then, by \cite{Pe}, $B$ is domestic in at most two one-parameters. Moreover, by \cite{Bo},
$B$ is a full convex subcategory of $A$. \\
Let $\Gamma$ denote the component of $\Gamma$(mod$B$) containing $P$, and $\Sigma$ denote the full
subquiver of $\Gamma$ consisting of the indecomposable modules $X \in \Gamma$ such that there is a
path $X \leadsto P$, and every such path is sectional. By \cite{ASS}(IX.2.6, p. 364), $\Sigma$ is a
complete slice in $\Gamma$(mod$B$). Note also that in $\Gamma($mod$B$), there are finitely many directed
components preceding $\Gamma$ (actually, by \cite{Pe}, at most two). Since indecomposable modules lying
in a directed component are uniquely determined by their composition factors \cite{AR}, then $P$ has
infinitely many isomorphism classes of indecomposable submodules if and only if there exists an
infinite family $({\mathcal T}_\lambda)_\lambda$ of homogeneous tubes over a tame concealed algebra $C$,
and an infinite family of homogeneous modules $H_\lambda \in {\mathcal T}_\lambda$ such that
$H_\lambda \subset $ rad$P$, for each $\lambda$. Clearly, then, all the $H_\lambda$ are contained in the
largest $C$-submodule $R$ of rad$P$. Moreover, by the structure of homogeneous tubes, we can assume
the $H_\lambda$ to be simple homogeneous. By (2.5), if this is the case, then $C[R]$ is wild, hence
so is $B$. This shows that $P$ has only finitely many isomorphism classes of indecomposable $B$-submodules,
hence $A$-submodules.
\end{proof}

\subsection{}
{\sc Remark.} As is seen from the example in Section 2, it is not true in general that multicoil algebras
are torsionless-finite. However, notice that the algebra \\
\begin{picture}(60,7)
\multiput(70,1)(0,-2){2}{\vector(-1,0){14}}
\multiput(88,1)(0,-2){2}{\vector(-1,0){14}}

\multiput(54,0)(18,0){3}{\circle*{1}}
\put(63,3){$\beta$}  \put(63,-5){$\delta$}
\put(81,3){$\alpha$}  \put(81,-5){$\gamma$}

\end{picture}
\vspace{.7 cm}\\
bounded by $\alpha \delta = 0$, $ \gamma \beta = 0 $ and $\alpha \beta = \gamma \delta$
is tilted and so, by \cite{APT}, has representation dimension at most 3.

\makeatletter
\renewcommand{\@biblabel}[1]{\hfill#1.}\makeatother

config: arxiv


TITLE: Product of gradients of x=0 and y=0
QUESTION [1 upvotes]: A friend asked me this question:
The product of the gradient of any two lines perpendicular to each other is $-1$.
Now, the lines $x=0$ and $y=0$ are perpendicular to each other. If you take the product of both their gradients, you dont get $-1$? Or do you? The question is missing on something fairly fundamental here as far as I think. Could you point it out and clear this problem? I have some ideas but I'm not writing them down because they might be too stupid.

REPLY [1 votes]: By "gradient", I suppose you mean the slope of the line $y = f(x)$. If so, then notice that the line $x = 0$ cannot be properly written as a function of $x$, and so its gradient is not defined.  That is, we don't have a change in $y$ over a change in $x$ because between any two points, the change in $x$ is $0$.


TITLE: Prove $f(z)=\int_{[0,1]}\frac{1}{1-wz}dw$ is holomorphic in the open unit disk.
QUESTION [3 upvotes]: Define $f:D[0,1] \rightarrow \mathbb C$ through
$$f(z)=\int_{[0,1]}\frac{1}{1-wz}dw$$
The integration path is from 0 to 1 along the real line.
Prove that $f$ is holomorphic in the open unit disk $D[0,1]$.
I was trying to use the Cauchy's Integral Formula
$$\begin{aligned}
f'(w) & =\frac{1}{2\pi i}\int_{\gamma}\frac{f(z)}{(z-w)^2}dz\\\
& =\frac{1}{2\pi i}\int_{\gamma}\frac{1}{(z-w)^2}\left[\int^1_0\frac{1}{1-tz}dt\right]dz\\
& =\frac{1}{2\pi i}\int^1_0\left[\int_{\gamma} \frac{\frac{1}{1-tz}}{(z-w)^2}dz\right]dt\\
& =\frac{1}{2\pi i}\int^1_0 \left[\frac{\partial}{\partial z}\frac{1}{1-tz}\left.\right|_{z=w}\right]dt\\
& =\frac{1}{2\pi i}\int^1_0 \frac{t}{(1-tw)^2}dt\\
& =\cdots
\end{aligned}$$
But the Cauchy's Integral formula need $f$ to be holomorphic at the first place...Any help with this?
Many Thanks!

REPLY [1 votes]: No need to justify switching integrals. Just obtain a closed form of $f(z)$, which allows to prove continuity and that $\int_{\gamma} f =0$:
$$f(z) := \int_{[0,1]} \frac{dw}{1-wz} = -\frac1z\operatorname{Ln}(1-z)1_{z\ne0} + 1_{z=0},$$

$f(z)$ is continuous on $D[0,1]$ because $$\lim_{z \to 0}-\frac1z\operatorname{Ln}(1-z) = 1 = f(0)$$
$$\int_{\gamma} f(z) dz = \int_{\gamma} -\frac1z\operatorname{Ln}(1-z)1_{z\ne0} + 1_{z=0} dz$$

$$ = \int_{\gamma} -\frac1z\operatorname{Ln}(1-z)1_{z\ne0} dz + \int_{\gamma} 1_{z=0} dz = 0$$
Now all conditions of Morera's Thm 5.6 are satisfied without (explicit) reference to antiderivatives or Fubini's theorem, measure spaces, uniform continuity, etc


TITLE: Faraday boxes (cages) are nice and all but what if you need wires or tubes to enter / exit your box?
QUESTION [1 upvotes]: Then is there still a way to EMP-protect a device?

REPLY [3 votes]: EMP is typically a high frequency signal. It is possible to allow low frequency signals to enter your Faraday cage by decoupling the signal - a choke in series and a capacitor in parallel. You need to make sure that the choke does not saturate at the current spikes expected - and that the voltage rating of the capacitor is sufficient to absorb the energy.
A technical paper on the subject can be found at http://www.kwset.fi/kW_Consulting/pdf/Yleis_PDF_GeneralTechnicalInformation.pdf
Another solution can be to use optical interfaces to permit the transmission of high frequency signals while protecting from unwanted RF. In that case you let the optical signal cross a small hole (small compared to the wavelength of RF you want to stop) after which you can safely convert to electrical. This is done, for example, with electronics that runs in certain MRI machines - allowing digital electronic circuits to live in the vicinity of the MRI scanner without interfering. "Wanted" signals travel across the boundary, but "unwanted" signals are stopped.


TITLE: Average electric field calculated from average potential
QUESTION [4 upvotes]: In a Griffiths problem (3.4) I'm asked to find the average electric field over the surface of a sphere due to point charges inside/outside the surface. The calculation is done in his solutions starting with some point charge a distance $z$ away from the origin, and then integrating the electric field over the surface of the sphere and dividing by the total area:
$\mathbf{E}_{ave} = \frac{1}{A} \int_A \mathbf{E} dA$.
This calculation is all well and good to me, but I'm asked by my job to make a solution video to this problem, and the solution they provide me first finds the average potential over the surface, then takes the gradient to find the average electric field:
$V_{ave} = \frac{q}{4\pi\epsilon_0z}, \implies \mathbf{E}_{ave} = -\partial_z V_{ave} \hat{\mathbf{z}} = \frac{q}{4\pi\epsilon_0 z^2} \hat{\mathbf{z}}.$
This doesn't make much sense to me. In the end you get the average potential as a function of $z$, but it's not the potential at $z$, it's just the average over the sphere given some fixed $z$. By taking the gradient you get the right answer, but I feel that this is a coincidence, since I don't think a derivative with respect to $z$ means what it should here. Is this correct?
Edit: Griffiths Introduction to Electrodynamics, 4e.

REPLY [3 votes]: In electrostatics we have
\begin{equation}
\mathbf{E}\left(\mathbf{r}\right)=\boldsymbol{-\nabla_{\mathbf{r}}} V\left(\mathbf{r}\right)
\tag{01}\label{eq01}
\end{equation}
In a region free of charges the electrostatic potential $\;V\left(\mathbf{r}\right)\;$ at a point $\;\mathrm P\;$ with position vector $\;\mathbf{r}\;$ is equal to the average potential on a spherical surface $\,\mathrm S(\mathbf{r},R)\,$with center at $\;\mathrm P\;$ and radius $\;R$, see Figure-01 above
\begin{equation}
V\left(\mathbf{r}\right)=\dfrac{1}{4\pi R^2}\!\!\oint\limits_{\mathrm S(\mathbf{r},R)}\!V\left(\mathbf{r'}\right)\mathrm{dS}=V_{\bf ave}^{\mathrm S}
\tag{02}\label{eq02}
\end{equation}
Note that this value is independent of the radius $\;R\;$  of the sphere as long as it does not contain charges. Inserting expression \eqref{eq02} in equation \eqref{eq01} we have
\begin{equation}
\mathbf{E}\left(\mathbf{r}\right)=\boldsymbol{-\nabla_{\mathbf{r}}}\left[\dfrac{1}{4\pi R^2}\!\!\oint\limits_{\mathrm S(\mathbf{r},R)}\!\!V\left(\mathbf{r'}\right)\mathrm{dS}\right]
\tag{03}\label{eq03}
\end{equation}
Now we must examine if in the rhs of equation \eqref{eq03} it's permissible to insert into the integral the gradient $\;\boldsymbol{\nabla_{\mathbf{r}}}\;$ with respect to $\;\mathbf{r}\;$  as gradient $\;\boldsymbol{\nabla_{\mathbf{r'}}}\;$ with respect to $\;\mathbf{r'}$. In this case
\begin{equation}
\mathbf{E}\left(\mathbf{r}\right)=\dfrac{1}{4\pi R^2}\!\!\oint\limits_{\mathrm S(\mathbf{r},R)}\!\!\bigl[\boldsymbol{-\nabla_{\mathbf{r'}}}V\left(\mathbf{r'}\right)\bigr]\mathrm{dS}=\dfrac{1}{4\pi R^2}\!\!\oint\limits_{\mathrm S(\mathbf{r},R)}\!\!\mathbf{E}\left(\mathbf{r'}\right)\mathrm{dS}=\mathbf{E}_{\bf ave}^{\mathrm S}
\tag{04}\label{eq04}
\end{equation}
That from \eqref{eq03} we could have \eqref{eq04} is proved below.

Proof  (see Figure-02 in the bottom)
\begin{equation}
V\left(\mathbf{r}+\mathrm d\mathbf{r} \right)=\dfrac{1}{4\pi R^2}\!\!\oint\limits_{\mathrm \Sigma(\mathbf{r}+\mathrm d\mathbf{r},R)}\!\!\!V\left(\mathbf{r'}+\mathrm d\mathbf{r}\right)\mathrm{d\Sigma}\stackrel{\mathrm d\mathbf{r}\equiv\mathrm d\mathbf{r'}}{=\!=\!=\!=}\dfrac{1}{4\pi R^2}\!\!\oint\limits_{\mathrm \Sigma(\mathbf{r}+\mathrm d\mathbf{r},R)}\!\!\!V\left(\mathbf{r'}+\mathrm d\mathbf{r'}\right)\mathrm{d\Sigma}
\tag{05}\label{eq05}
\end{equation}
\begin{align}
V\left(\mathbf{r}+\mathrm d\mathbf{r} \right)-V\left(\mathbf{r}\right)& =\dfrac{1}{4\pi R^2}\!\!\oint\limits_{\mathrm S(\mathbf{r},R)}\!\bigl[V\left(\mathbf{r'}+\mathrm d\mathbf{r'}\right)-V\left(\mathbf{r'}\right)\bigr]\mathrm{dS}
\nonumber\\
&= \dfrac{1}{4\pi R^2}\!\!\oint\limits_{\mathrm S(\mathbf{r},R)}\!\biggl[\dfrac{V\left(\mathbf{r'}+\mathrm d\mathbf{r'} \right)-V\left(\mathbf{r'}\right)}{\mathrm d\mathbf{r'}}\biggr]\boldsymbol{\cdot}\mathrm d\mathbf{r'}\mathrm{dS}
\nonumber\\
&\stackrel{\mathrm d\mathbf{r'}\equiv\mathrm d\mathbf{r}}{=\!=\!=\!=}\left( \dfrac{1}{4\pi R^2}\!\!\oint\limits_{\mathrm S(\mathbf{r},R)}\!\biggl[\dfrac{V\left(\mathbf{r'}+\mathrm d\mathbf{r'} \right)-V\left(\mathbf{r'}\right)}{\mathrm d\mathbf{r'}}\biggr]\mathrm{dS}\right)\boldsymbol{\cdot}\mathrm d\mathbf{r}
\tag{06}\label{eq06}
\end{align}
\begin{equation}
\dfrac{V\left(\mathbf{r}+\mathrm d\mathbf{r} \right)-V\left(\mathbf{r}\right)}{\mathrm d\mathbf{r}} = \dfrac{1}{4\pi R^2}\!\!\oint\limits_{\mathrm S(\mathbf{r},R)}\!\biggl[\dfrac{V\left(\mathbf{r'}+\mathrm d\mathbf{r'} \right)-V\left(\mathbf{r'}\right)}{\mathrm d\mathbf{r'}}\biggr]\mathrm{dS}
\tag{07}\label{eq07}
\end{equation}
\begin{equation}
\lim_{\mathrm d\mathbf{r}\rightarrow\boldsymbol{0} }\dfrac{V\left(\mathbf{r}+\mathrm d\mathbf{r} \right)-V\left(\mathbf{r}\right)}{\mathrm d\mathbf{r}} = \dfrac{1}{4\pi R^2}\!\!\oint\limits_{\mathrm S(\mathbf{r},R)}\!\biggl[\lim_{\mathrm d\mathbf{r'}\rightarrow\boldsymbol{0}}\dfrac{V\left(\mathbf{r'}+\mathrm d\mathbf{r'} \right)-V\left(\mathbf{r'}\right)}{\mathrm d\mathbf{r'}}\biggr]\mathrm{dS}
\tag{08}\label{eq08}
\end{equation}
\begin{equation}
\dfrac{\mathrm d V\left(\mathbf{r}\right)}{\mathrm d\mathbf{r}} = \dfrac{1}{4\pi R^2}\!\!\oint\limits_{\mathrm S(\mathbf{r},R)}\!\!\!\dfrac{\mathrm d V\left(\mathbf{r'}\right)}{\mathrm d\mathbf{r'}}\mathrm{dS}
\tag{09}\label{eq09}
\end{equation}
\begin{equation}
\boldsymbol{\nabla_{\mathbf{r}}} V\left(\mathbf{r}\right) = \dfrac{1}{4\pi R^2}\!\!\oint\limits_{\mathrm S(\mathbf{r},R)}\!\!\boldsymbol{\nabla_{\mathbf{r'}}}V\left(\mathbf{r'}\right)\mathrm{dS}
\tag{10}\label{eq10}
\end{equation}


TITLE: Integral $\int_0^{\infty}\frac{\ln(x)}{x^2+nx+n^2}dx=\frac{2\pi}{3\sqrt{3}}\frac{\ln(n)}{n}$
QUESTION [4 upvotes]: I was working on some integrals, and I proved the following really beautiful identity:
$$\int_0^{\infty}\frac{\ln(x)}{x^2+nx+n^2}dx=\frac{2\pi}{3\sqrt{3}}\frac{\ln(n)}{n}$$
I found it by solving $\int_0^{\infty}\frac{x^{k-1}}{ax^2+bx+c}dx$, and then differentiating my result with respect to $k$ and evaluating at $k=1$. When I saw that it yielded a nice formula for the above integral, I was surprised, which is why I posted this question.
But my proof is very long and messy, and since the result is so elegant, I was wondering if there was a simple and elegant proof of this identity.

REPLY [5 votes]: Let $I(n)$ be the integral in question. Then, using the substitution
$x=ny$, we get
$$I(n)=\int_0^\infty\frac{\ln(nx)}{n(x^2+x+1)}\,dx=I(1)+
\frac{\ln n}n\int_0^\infty\frac{dy}{y^2+y+1}.$$
That latter integral can be done by a arctangent substitution. But
using the substitution $y=1/x$ gives $I(1)=-I(1)$ so that $I(1)=0$.

REPLY [2 votes]: Call your integral $I$. With $x\mapsto\frac{n^2}{x}$, $I=\int_0^\infty\frac{\ln(n/x)dx}{x^2+nx+n^2}$. Averaging, $I=\frac12\ln n\cdot\int_0^\infty\frac{dx}{x^2+nx+n^2}$.


TITLE: With Choice, is any linearly ordered set well-ordered if no subset has order type $\omega^*$?
QUESTION [9 upvotes]: I've been fumbling around with order types and ordinals these past few days. I read about partial, total, and well-ordered structures, and I'm curious to see if a linearly ordered set has no subset with order type $\omega^*$, then it is in fact well ordered. Here $\omega^*$ is the order type of the negative integers.
My idea was this. Let $X$ be some totally ordered set such with subset with order type $\omega^*$. Take any nonempty subset $Y$ of $X$. If $Y$ is finite, it must have a least element, so assume $Y$ is infinite. By way of contradiction, I assume $Y$ has no least element. My strategy was to then somehow construct a subset of $Y$ which has order type $\omega^*$. I do this by first picking an element $a_0$ from $Y$. Since $Y$ has no least element, and since $Y$ is also totally ordered, there must be some other element $a_1\in Y$ such that $a_1\prec a_0$. Again, $a_1$ is not the least element of $Y$, so we can find an element $a_2\in Y$ such that $a_2\prec a_1\prec a_0$. Continuing along, I would eventually have a set $Z=\{\dots,a_3,a_2,a_1,a_0\}$, where I have written it in increasing order. But then $Z$ has order type $\omega^*$, a contradiction.
Does this argument hold up? If so, I feel it is a little handwavey, and the "continuing along" part needs to be formalized. Is there a way to do so with Choice or maybe (transfinite) recursion? I fear I may have written nonsense, as I've done many times in the past. Thanks for any criticism and insights.

REPLY [7 votes]: Yunone:
Your argument is fine. The axiom of dependent choices (DC) is what you are using to "recursively" pick the members of your sequence. DC is a consequence of the axiom of choice, but it is strictly weaker. It says that if you have a relation $R$ on a set $X$ with the property that for any $a\in X$ there is a $b\in X$ with $a R b$, then there is a sequence $(a_n\mid n\in\omega)$ with $a_n R a_{n+1}$ for all $n$.
If you have a set that is not well-ordered, and DC holds, you can pick a nonemepty $X$ subset of your set without a least element and set $aRb$ iff $b<a$.
DC is strictly weaker than choice, because there are models where DC holds but full choice fails. For example, Shelah built a model of ZF+DC where all sets of reals have the property of Baire. But choice proves that some sets of reals don't have this property.
Actually, DC is an overkill for the result that you are interested in, but some form of choice is needed. The proposition you are considering (a linearly ordered set is well-ordered iff it has no subsets of type $\omega^*$) was first considered by Sierpiński in "L'axiome de M. Zermelo et son rôle dans la théorie des ensembles et l'analyse", Bull. Int. Acad. Sci. Cracovie Cl. Math. Nat., (1918), 97-152.
This is form 77 in the book by Paul Howard and Jean E. Rubin, "Consequences of the axiom of choice", AMS Mathematical surveys and monographs, vol 59, 1998. The book also provides references to models where this form holds while form 43 (DC) fails, and models where form 77 fails. Another place where some of these examples are discussed is Jech's book "The axiom of choice" (recently reprinted by Dover in 2008).
Let me point you to a great resource online for choice-related questions: There is a companion webpage to the Howard-Rubin book, Consequences of the Axiom of Choice Project Homepage,  (I have the Howard-Rubin book, but I usually consult the webpage first to identify the forms that I can then go and look for in the book).

REPLY [6 votes]: Your argument is alright. There is nothing wrong with it. It's very common practice -when the axiom of choice is assumed- to say something along the lines of what you wrote. Having said that, here's a couple of ways to "formalize" your argument:
One thing you can do is use the Principle of Dependent Choices: What this says is that if we have a binary relation $E$ on a (nonempty) set $A$ and for every $a\in A$ there is a $b\in A$ such that $bEa$, then there is a sequence $a_0,\ldots,a_n,\ldots$ of elements of $A$ such that $a_{n+1}Ea_n$ for all $n\in\mathbb{N}$. Of course you can see how your argument can be formalized through this principle. Given a subset $Y\subset X$ with no least element of course for every $a\in Y$ there is a $b\in Y$ such that $b<a$. Then by the principle of dependent choice there exists a sequence that defines a set with order-type $\omega^*$.
But assume you just want to use the axiom of choice. We have that $\mathcal{P}(Y)\setminus\{\varnothing\}$ has a choice function $f$. Let $a_0=f(Y)$. Having defined $a_n$, let $Y_{n+1}=\{a\in Y : a<a_n\}$ (these are not empty since $Y$ doesn't have a least element). Then let $a_{n+1}=f(Y_{n+1})$. Every $a_n$ is well defined since $f$ is fixed. Thus the sequence is well defined, and thus the set with order type $\omega^*$ is well defined (these are all results of the recursion theorem).
As you see both "formalizations" are simple. This is why (as I said) your argument is perfectly valid and I am sure that you will encounter it in many books. I just wrote these in case you were curious as to how the axiom of choice leads to the part of your argument that you considered problematic.


TITLE: P-adic expansion construction
QUESTION [3 upvotes]: Can anyone teach me about p-adic expansion? especially the case where we have to expand a square root.
I need to know how to construct them. for example: the 7-adic expansion of $\sqrt{305}$.
This is a general question and not an assignment or anything, so i cannot post the progress.
That is just a random example, please use any other example to explain if it's easier to understand.
edit: I'm stuck with the cases where i have to expand the square roots, expanding rational numbers is fine though

REPLY [1 votes]: The most naive way:
305=4(mod 7) this gives 2 or 5 for last digit.
305=11(mod 49) and we have 16^2 or 33^2=11(mod 49). this gives 22 or 45 written in 7-adic of last two digit.
continuing this procedure we have $\sqrt{305}=\dots22$ or $...45$
Of course you can use Taylor expansion:
$\sqrt{305} = \sqrt{7*43+4}=2+\frac{1}{4}*(7*43)-\frac{1}{64}*(7*43)^2+\frac{1}{512}*(7*43)^2-\dots$
it's not hard to find it converge w.r.t 7-adic topology


TITLE: Alternating sum of Stirling numbers equals 0
QUESTION [3 upvotes]: How can I prove that the following formula is true for Stirling numbers of first kind.

For any $n \geq 2$, we have
$$\sum_{k=1}^n(-1)^k\left[\begin{matrix} n\\k\end{matrix}\right] =0$$

Actually I want to prove that the number of permutations of $\left\{1,2,\ldots,n\right\}$ with an even number of cycles equals the number of permutations of $\left\{1,2,\ldots,n\right\}$ with an odd number of cycles. This is equivalent to the above identity by the combinatorial interpretation of Stirling numbers.

REPLY [2 votes]: Here is an algebraic proof. The combinatorial species of permutations with cycles marked is given by $$\mathfrak{P}(\mathcal{U}\mathfrak{C}(\mathcal{Z}))$$
which yields the generating function
$$G(z,u) = \exp\left(u\log\frac{1}{1-z}\right).$$
This immediately produces
$$\left[ n\atop k\right] = n! [z^n] \frac{1}{k!}\left(\log\frac{1}{1-z}\right)^k.$$
Substitute this into the sum to get
$$n! [z^n] \sum_{k=1}^n \frac{(-1)^k}{k!}\left(\log\frac{1}{1-z}\right)^k.$$
Supposing we have $n\ge 1$ we may extend the lower limit to include zero and the upper limit to infinity because $\log\frac{1}{1-z}$ starts at $z.$ This gives
$$n! [z^n] \sum_{k=0}^\infty \frac{(-1)^k}{k!}\left(\log\frac{1}{1-z}\right)^k.$$
Simplify to obtain
$$n! [z^n] \exp\left(-\log\frac{1}{1-z}\right) = n! [z^n] (1-z).$$
We assumed that $n\ge 1$ so this gives minus one for $n=1$ and zero otherwise. Indeed there is one permutation on $1$ element and it consists of a single odd cycle.


text:
\begin{document}

\title{On Matrix Consequence (Extended Abstract)}

\author{
Alexei Muravitsky
}
\maketitle
\begin{center}
 Louisiana Scholars' College \\
Northwestern State University\\
Natchitoches, LA 71497, U.S.A. \\
\emph{Email}: alexeim@nsula.edu
\end{center}

\begin{abstract}
These results are a contribution to the model theory of matrix consequence. We give a semantic characterization of uniform and couniform consequence relations. The first concept was introduced by {\L}o\'{s} and Suszko (1958), and the second by W\'{o}jcicki (1970) to characterize structural consequence relations having single adequate matrices. In different variants of the single-adequate-matrix theorem of {\L}o\'{s} and Suszko (corrected by W\'{o}jcicki), the uniformity of consequence is used in conjunction with other properties (e.g. finitariness), and in the W\'{o}jcicki theorem uniformity and couniformity of consequence are used together. These properties have never been treated individually, at least in a semantic manner. We consider these notions from a purely semantic point of view and separately, introducing the notion of a uniform bundle/atlas and that of a couniform class of logical matrices. Then, we show that any uniform bundle defines a uniform consequence; and if a structural consequence is uniform, then its Lindenbaum atlas is uniform. Thus, any structural consequence is uniform if, and only if, it is determined by a uniform bundle/atlas. On the other hand, any couniform set of matrices defines a couniform structural consequence.  Also, the Lindenbaum atlas of a couniform structural consequence is couniform. Thus, any structural consequence is couniform if, and only if, it is determined by a couniform bundle/atlas. We then apply these observations to compare structural consequence relations that are defined in different languages when one language is a primitive extension of another. We obtain that for any structural consequence defined in a language having (at least) a denumerable set of sentential variables, if this consequence is uniform and couniform, then it and the \emph{ W\'{o}jcicki's consequence} corresponding to it, which is defined in any primitive extension of the given language, are determined by one and the same atlas which is both uniform and couniform.
\end{abstract}

\section{Preliminaries}
We consider a structural consequence relation, not necessarily finitary (or compact), in an abstract propositional language $\Lan$. We will be using a unifying term \textit{abstract logic} (usually denoted by $\aLog$) in order to employ in one and the same context the consequence relation $\vdashS$ and the consequence operator $\textbf{Cn}_{\mathcal{S}}$ both associated with $\aLog$.

The set of all $\Lan$-variables is denoted by $\VarL$, the set of all $\Lan$-formulae by $\FormsL$. For any $\alpha\in\FormsL$,
$\Var(\alpha)$ is the set of all variables occurring in $\alpha$. Further, if $\Var\subseteq\VarL$, we denote:
\[
\FormsL[\Var]:=\set{\alpha\in\FormsL}{\Var(\alpha)\subseteq\Var}.
\]
$\FormAl$ stands for a formula algebra in $\Lan$. Given a set $X\subseteq\FormsL$ and an abstract logic $\aLog$, $X$ is an $\aLog$-\textbf{\textit{theory}} if $\ConS{X}=X$.

It is well known (see, e.g.,~\cite{wojcicki1988}) that any structural consequence is determined by a class of (logical) matrices and vice versa. We remind that, given a logical matrix $\mat{M}=\langle\alg{A},D\rangle$, where $\alg{A}$ is an algebra of the same signature as the signature of $\Lan$ and $D\subseteq|\alg{A}|$ (we call such a matrix an $\Lan$-\textit{\textbf{matrix}} and $D$ its \textit{\textbf{logical filter}}), a \textit{\textbf{matrix consequence relative to}} \mat{M} (or \mat{M}-\textit{\textbf{consequence}} for short) is the following relation:
\[
X\models_{\mat{M}}\alpha~\define~(v[X]\subseteq D\Longrightarrow v[\alpha]\in D,~\text{for any valuation $v$ in \alg{A}}),
\]
where $X\cup\lbrace\alpha\rbrace\subseteq\FormsL$.

The last definition is extended to a (nonempty) class $\cM=\lbrace\mat{M}_i\rbrace_{i\in I}$ of $\Lan$-matrices as follows:
\begin{equation}\label{E:matrix-consequence}
X\vdash_{\mathcal{M}}\alpha~\define~(X\models_{\mat{M}_i}\alpha,~\textit{for each $i\in I$}).
\end{equation}
We call the last relation a \textit{\textbf{matrix consequence relative to}} $\cM$ (or $\cM$-\textit{\textbf{consequence}} for short).

Given an abstract logic $\aLog$, a logical matrix $\mat{M}$ is an $\aLog$-\textit{\textbf{model}} if
\[
X\vdashS\alpha~\Longrightarrow X\models_{\mat{M}}\alpha.
\]
The set of all $\aLog$-models is denoted by $\Mod{\aLog}$.\\

In fact, if a consequence relation is determined by a class of $\Lan$-matrices, it is also determined by a bundle or atlas.
\begin{defn}\label{D:bundle/atlas}
	A $($nonempty$)$ family {\em$\mathcal{B}=\lbrace\langle\alg{A}_{i},D_i\rangle\rbrace_{i\in I}$} of $\Lan$-matrices is called a \textbf{bundle} if for any $i,j\in I$, the algebras {\em$\alg{A}_i$} and {\em$\alg{A}_j$} are isomorphic. Ignoring the differences between the algebras {\em$\alg{A}_i$}, we write {\em$\mathcal{B}=\lbrace\langle\alg{A},D_i\rangle\rbrace_{i\in I}$}.  A pair {\em$\langle\alg{A},\lbrace D_i\rbrace_{i\in I}\rangle$}, where each $D_i$ is a logical filter in $($or of$)$ {\em$\alg{A}$}, is called an \textbf{atlas}. By definition, the \textbf{matrix consequence relative to an atlas}
	{\em$\langle\alg{A},\lbrace D_i\rbrace_{i\in I}\rangle$} is the same as the matrix consequence related the bundle {\em$\lbrace\langle\alg{A},D_i\rangle\rbrace_{i\in I}$}, which in turn is defined in the sense of~{\em \eqref{E:matrix-consequence}}. An atlas {\em$\LinS[\theory]=\langle\FormsL,\Sigma_{\mathcal{S}}\rangle$}, where $\Sigma_{\mathcal{S}}$ is the set of all $\aLog$-theories, is called a \textbf{Lindenbaum atlas} $($relative to $\aLog$$)$.
\end{defn}

The transition from a class $\cM$ of matrices to a bundle (or atlas), which would define the $\cM$-consequence, can be done (at least) in two ways.

An indirect way for such a transition, which is based on the fact that  $\cM$-consequence is structural, uses the Lindenbaum theorem; cf., e.g., W\'{o}jcicki~\cite{wojcicki1970}, theorem 2.4,~\cite{wojcicki1988} or~\cite{citkin-muravitsky2019}, corollary 3.3.5.1.

According to a direct method, given $\mathcal{M}=\lbrace\langle\alg{A}_i,D_i\rangle\rbrace_{i\in I}$, we define an atlas  $\mathcal{M}^{\ast}=\langle\alg{A},\lbrace D_{i}^{\ast}\rbrace_{i\in I}\rangle$ as follows:
\[
\begin{array}{rll}
\alg{A}:=\prod_{i\in I}\alg{A}_i
&\text{and, given $i\in I$,}~D_{i}^{\ast}:=\prod_{j\in I} H_j,
&\text{where}~H_j:=\begin{cases}
\begin{array}{cl}
D_i &\text{if $j=i$}\\
|\alg{A}_j| &\text{if $j\neq i$}.
\end{array}
\end{cases}
\end{array}
\]

We claim that
\[
X\vdash_{\mathcal{M}}\alpha~\Longleftrightarrow~X\vdash_{\mathcal{M}^{\ast}}\alpha.
\]

In~\cite{los-suszko1958}, the authors used the notion of uniform consequence (a \textit{\textbf{uniform abstract logic}} in our terminology) in an attempt to prove a criterion for a \textit{\textbf{single-matrix consequence}}, that is when a structural consequence can be determined by a single matrix. As W\'{o}jcicki (see, e.g.,~\cite{wojcicki1988}) has shown, in addition to the notion of uniformity, the notion of a \textit{\textbf{couniform abstract logic}} (in our terminology) is needed for such a criterion.

Originally, the concepts of uniformity and couniformity were defined syntactically. Below we treat them semantically.

\section{Uniform consequence}
For an arbitrary $\Lan$-matrix $\mat{M}=\langle\alg{A},D\rangle$, we define
an operator $\mathfrak{S}:\mat{M}\mapsto\fS{\mat{M}}$ where
\begin{equation}\label{E:S-operator-in-L}
\begin{array}{r}
X\in\mathfrak{S}(\mat{M})~\define~\text{there is an $\Lan$-valuation $v$ in \alg{A} such that}\\
X=\set{\alpha\in\FormsL}{v[\alpha]\in D}.
\end{array}
\end{equation}

We notice that
\[
X\models_{\mat{M}}\alpha~\Longleftrightarrow~\forall Y\in\fS{\mat{M}}.~X\subseteq Y~\Longrightarrow~\alpha\in Y.
\]

In case that we have a set $\cM=\lbrace\mat{M}_i\rbrace_{i\in I}$ of $\Lan$-matrices, the corresponding $\cM$-consequence can be reformulated in terms of $\mathfrak{S}$-operator as follows:
\begin{equation}\label{E:M-con-via-S-operator}
X\vdash_{\cM}\alpha~\Longleftrightarrow~\forall i\in I\,\forall Y\in\fS{\mat{M}_i}.~X\subseteq Y~\Longrightarrow~\alpha\in Y.
\end{equation}

We continue with the following definition.
\begin{defn}[uniform bundle/atlas]\label{D:uniform-bundle}
	Let {\em$\cM=\lbrace\mat{M}_i\rbrace_{i\in I}$} be a bundle.
	We say that $\cM$ is \textit{\textbf{uniform}} if for any {\em$X\cup Y\subseteq\FormsL$} with {\em$X\subseteq\FormsL[\VarL\setminus\Var(Y)]$}, the following condition is fulfilled:
	{\em\[
		\begin{array}{l}
		\forall i,j\in I\,\forall Z_i\in\fS{\mat{M}_i}\forall Z_j\in\fS{\mat{M}_j}\exists k\in I\exists Z_k\in\fS{\mat{M}_k}.\,X\subseteq Z_i\,\&\,Y\subseteq Z_j\subset\FormsL\\
		\Longrightarrow
		Z_i\cap\FormsL[\VarL\setminus\Var(Y)]=Z_k\cap\FormsL[\VarL\setminus\Var(Y)]\,\&\,
		Z_j\cap\FormsL[\Var(Y)]\subseteq Z_k.
		\end{array}
		\]}
	An atlas is \textbf{uniform} if the corresponding bundle is uniform.
\end{defn}

The above definition is justified by the following two propositions.
\begin{prop}\label{P:M-uniform-con-is-uniform}
	Let {\em$\cM$} be a uniform bundle.
	Then the $\cM$-consequence is uniform.
\end{prop}
\begin{prop}\label{P:LinS-uniform}
	If a structural abstract logic $\aLog$ is uniform, then its Lindenbaum atlas {\em$\LinS[\theory]$} is uniform.
\end{prop}

\begin{cor}\label{P:uniformity-by-uniform-bundle}
	A structural abstract logic is uniform if, and only if, it is determined by a uniform bundle/atlas.
\end{cor}

\section{Couniform consequence}
We start with the following observation.
\begin{prop}\label{P:couniformity-criterion}
	Let $\aLog$ be a structural abstract logic that is determined by
	a set {\em$\cM=\lbrace\mat{M}_j\rbrace_{j\in J}$} of $\Lan$-matrices. The logic $\aLog$ is couniform if, and only if, for any collection $\lbrace X_i\rbrace_{i\in I}$ of formula sets with $\Var(X_i)\cap\Var(X_j)=\varnothing$, providing that $i\neq j$, and $\Var(\bigcup_{i\in I}\lbrace X_i\rbrace) \neq\Var_{\Lan}$, the following condition is satisfied:
	{\em\begin{equation}\label{E:couniformity-2}
		\begin{array}{r}
		\forall i\in I\exists j\in J\exists Z_j\in\mathfrak{S}(\mat{M}_j)(X_i\subseteq Z_j\subset\FormsL)~\Longrightarrow\\
		\exists k\in J\exists Z_k\in\mathfrak{S}(\mat{M}_k)(\bigcup_{i\in I}\lbrace X_i\rbrace\subseteq Z_k\subset\FormsL).
		\end{array}
		\end{equation}}
\end{prop}

The last proposition leads to the following definition.
\begin{defn}\label{D:couniform-set-matrices}
	A nonempty set {\em$\cM=\lbrace\mat{M}_j\rbrace_{j\in J}$} of $\Lan$-matrices is called \textbf{couniform} if for any collection $\lbrace X_i\rbrace_{i\in I}$ of formula sets with $\Var(X_i)\cap\Var(X_j)=\varnothing$, providing that $i\neq j$, and $\Var(\bigcup_{i\in I}\lbrace X_i\rbrace) \neq\Var_{\Lan}$, the condition~\eqref{E:couniformity-2} is satisfied. An atlas {\em$\langle\alg{A},\lbrace D_i\rbrace_{i\in I}\rangle$} is \textbf{couniform} if the corresponding bundle {\em$\lbrace\langle\alg{A},D_i\rangle\rbrace_{i\in I}$} is couniform.
\end{defn}

Now we obtain
\begin{prop}\label{P:LinS-couniform}
	Let $\aLog$ be a structural couniform logic. Then its Lindenbaum atlas {\em$\LinS[\theory]$} is couniform.
\end{prop}

\begin{cor}
A structural logic is couniform if, and only if, it is determined by a couniform bundle/atlas.
\end{cor}

Using W\'{o}jcicki's criterion (see, e.g.,~\cite{wojcicki1988}, theorem 3.2.7, or~\cite{citkin-muravitsky2019}, proposition 4.1.2), we also obtain the following.
\begin{cor}
A structural abstract logic is determined by a single matrix if, and only if, it can be determined by a uniform bundle/atlas and a couniform bundle/atlas.
\end{cor}

\section{Extensions}
We start with the following two definitions.
\begin{defn}[extension and primitive extension of a language]
	Let $\Lan$ and $\Lan^{\prime}$ be two propositional languages with $\VarL\cup\mathcal{C}_{\mathcal{L}}\cup\mathcal{F}_{\mathcal{L}}\subseteq\Var_{\mathcal{L}^{\prime}}\cup\mathcal{C}_{\mathcal{L}^{\prime}}\cup\mathcal{F}_{\mathcal{L}^{\prime}}$, then $\Lan^{\prime}$ is called an \textbf{extension} of $\Lan$. If $\mathcal{C}_{\mathcal{L}}\cup\mathcal{F}_{\mathcal{L}}=\mathcal{C}_{\mathcal{L}^{\prime}}\cup\mathcal{F}_{\mathcal{L}^{\prime}}$ and $\VarL\subseteq\Var_{\mathcal{L}^{\prime}}$, then we say that $\Lan^{\prime}$ is a \textbf{primitive extension} of $\Lan$, symbolically {\em$\Lan\ext\LanP$}.
\end{defn}

\begin{defn}
	Let $\aLog$ be an abstract logic in a language $\Lan$ and $\aLog^{\prime}$ be an abstract logic in a language $\Lan^{\prime}$ which is an extension of $\Lan$. Then $\aLog^{\prime}$ is called a \textbf{conservative extension of}
	$\aLog$, symbolically $\aLog\prec_{c}\aLog^{\prime}$, if for any set {\em$X\cup\lbrace\alpha\rbrace\subseteq\FormsL$},
	\[
	X\vdashS\alpha~\Longleftrightarrow~X\vdash_{\mathcal{S}^{\prime}}\alpha.
	\]
\end{defn}

\begin{prop}\label{P:equality-of-Mod-classes}
	Let $\aLog$ and $\aLog^{\prime}$ be structural abstract logics in languages $\Lan$ and $\Lan^{\prime}$, respectively, with $\Lan\prec_{p}\Lan^{\prime}$ and
	$\card{\Var_{\mathcal{L}^{\prime}}}=\card{\VarL}\ge\aleph_{0}$. If $\aLog\prec_{c}\aLog^{\prime}$, then {\em$\Mod{\aLog}=\Mod{\aLog^{\prime}}$}.
\end{prop}

\begin{defn}[consequence relation $\vdash_{(\mathcal{M})^{+}}$]
	Let $\mathcal{M}$ be a set of logical matrices of type $\Lan$. Assume that $\Lan\prec_{p}\LanP$. A consequence relation in $\LanP$ determined by the set $\mathcal{M}$, is denoted by $\vdash_{(\mathcal{M})^{+}}$.
\end{defn}

\begin{prop}
	Let $\mathcal{M}$ be a set of logical matrices of type $\Lan$. Assume that $\Lan\prec_{p}\LanP$. The consequence relation $\vdash_{(\mathcal{M})^{+}}$ if finitary if, and only if,
	$\vdash_{\mathcal{M}}$ is finitary.
\end{prop}

\begin{defn}[relation $\vdash_{(\mathcal{S})^{+}}$]\label{D:(S)^+}
	Let a language $\Lan^{+}$ be an extension of a language $\Lan$. Also, let $\mathcal{S}$ be an abstract logic in $\Lan$. We define a relation $\vdash_{(\mathcal{S})^{+}}$ in {\em$\mathcal{P}(\Forms_{\Lan^{+}})\times\Forms_{\Lan^{+}}$} as follows:
	{\em\[
		\begin{array}{rl}
		X\vdash_{(\mathcal{S})^{+}}\alpha\stackrel{\text{df}}{\Longleftrightarrow}
		\!\!\!&\textit{there is a set {\em$Y\cup\lbrace\beta\rbrace\subseteq\Forms_{\Lan}$} and an $\Lan^{+}$\!-substitution}\\
		&\textit{$\sigma$ such that $\sigma(Y)\subseteq X$, $\alpha=\sigma(\beta)$ and $Y\vdash_{\mathcal{S}}\beta$}.
		\end{array}
		\]}
\end{defn}

We borrow some results from~\cite{citkin-muravitsky2019}, collecting them in one lemma.
\begin{lem}[\cite{citkin-muravitsky2019}, Proposition3.5.4, Proposition 3.5.5, Lemma 4.1.2, Lemma 4.1.4]\label{L:lemma-in-Extensions}
	Let $\aLog$ be a structural logic which is both uniform and couniform in a language $\Lan$ with $\card{\Var_{\Lan}}\ge\aleph_{0}$ and $\Lan^{+}$ be a primitive extension of $\Lan$. Then $(\mathcal{S})^{+}$ is a structural abstract logic which is both uniform and couniform.
\end{lem}

Based on Lemma~\ref{L:lemma-in-Extensions}, given an abstract logic $\aLog$,
we call $(\mathcal{S})^{+}$ a \textit{\textbf{W\'{o}jcicki logic}},
and the consequence related to the latter a \textit{\textbf{W\'{o}jcicki consequence}}.\\

Using the last lemma, we obtain the following.
\begin{prop}
	Let $\aLog$ be a structural logic which is both uniform and couniform in a language $\Lan$ with $\card{\Var_{\Lan}}\ge\aleph_{0}$ and $\Lan^{+}$ be a primitive extension of $\Lan$. Then $\aLog$ and $(\mathcal{S})^{+}$ are determined by one and the same atlas which is both uniform and couniform.
\end{prop}

\paragraph{\textbf{Acknowledgment}}
I am grateful to Alex Citkin for numerous discussions regarding the topic of this article.


TITLE: Tournaments of order 7 are there that are not strongly connected?
QUESTION [0 upvotes]: How many tournaments of order 7 are there that are not strongly connected? Also, how many tournaments of order 7 exist? I know there are 56 tournaments of order 6, 12 of order 5, and 4 of order 4, but I am confused as to how this is determined.

REPLY [0 votes]: You can look at some of the links from sequence A000568 in the OEIS to learn about the general methods; the formula on that page comes from applying Burnside's lemma (Wikipedia link) to reduce it to a problem about counting labeled tournaments that are left unchanged by some given permutation of the vertex set. But this requires some group theory to implement.
For very small $n$, it's easier to approach the problem by a structural observation about tournaments: we can divide the vertex set of any tournament into blocks $B_1, B_2, \dots, B_k$ such that each $B_i$ induces a strongly connected tournament while the edges between the blocks are completely acyclic: if $v \in B_i$ and $w \in B_j$ and $i>j$, then the edge between $v$ and $w$ is oriented $v \to w$.
This is still tedious for $n=7$, but in your case, if you can assume that there are $56$ tournaments of order $6$, $12$ tournaments of order $5$, $4$ tournaments of order $4$, and $2$ tournaments of order $3$, there's a simple argument by inclusion-exclusion.
All tournaments of order $7$ that are not strongly connected must fall into one of four types (with some overlap):

Tournaments where the first block has $|B_1| = 1$.
Tournaments where the first block has $|B_1| = 3$.
Tournaments where the last block has $|B_k| = 1$.
Tournaments where the last block has $|B_k| = 3$.

This is because we can't have a block of size $2$ (there are no strongly connected tournaments on $2$ vertices), and we must have at least two blocks if the tournament is not strongly connected, so at most one block can have size $>3$.
These are easy to individually count. There are $56$ tournaments of type 1: removing the first block gives a tournament on $6$ vertices. There are $4$ tournaments of type 2: the first block can only be a cycle on $3$ vertices, and removing it gives a tournament on $4$ vertices. Similarly, there are $56$ tournaments of type 3, and $4$ tournaments of type $4$.
But there is some overlap. There are $12$ tournaments that are both type 1 and type 3: removing the first and last block leaves a $5$-vertex tournament. By the same argument, there are $2$ tournaments that are type 1 and 4, $2$ tournaments that are type 2 and 3, and $1$ tournament that is type 2 and 4.
So the final answer is
$$(56 + 4 + 56 + 4) - (12 + 2 + 2 + 1) = 120 - 17 = 103.$$


TITLE: Subgroups of free products of cyclic groups
QUESTION [6 upvotes]: Consider the free product $\mathbb{Z}_{3} \star \mathbb{Z}_{3}$. How would one determine the number of subgroups of this product up to isomorphism? It is routine for the case of the product $\mathbb{Z}_{2} \star \mathbb{Z}_{2}$ either by considering the combinations of elements from each group separately or the number of covering spaces of the wedge product of projective spaces. I feel that the easiest way would be to consider the subgroups from the group directly without resorting to topology, but how would one do this?

REPLY [4 votes]: There is a generalization, and it is called "Bass-Serre theory". What you learn from that theory is that $\mathbb{Z}_3 * \mathbb{Z}_3$ is a considerably more complicated group than $\mathbb{Z}_2 * \mathbb{Z}_2$. Every infinite subgroup of $\mathbb{Z}_2 * \mathbb{Z}_2$ is isomorphic to either itself or to the infinite cyclic group, whereas $\mathbb{Z}_3 * \mathbb{Z}_3$ contains subgroups isomorphic to free groups of any countable rank, as well as many different kinds of free products whose free factors are mixtures of free groups and cyclic groups of order 3.


TITLE: Using the identity theorem to prove existence of non-identically zero function
QUESTION [2 upvotes]: If we consider the region $U = \{z\in\mathbb{C} : Im(z) \ne 0 \}$ and the sequence $z_n = (1+n^{-1})i$ can we find a holomorphic function that is not identically zero, but is zero at $z_n$?
Now I've answered this question before when $im(z) = 0$ wasn't excluded, in which case the answer is no by the identity theorem. However now that this problem point has been removed, I'm guessing the answer is probably yes, but I am struggling to construct such a function, any tips would be appreciated.
Thanks

REPLY [3 votes]: The sequence $z_n = (1+n^{-1})i$ converges to $i\in U$, so the identity theorem tells us that every holomorphic function vanishing in all $z_n$ must vanish identically on the connected component of $U$ that contains $i$.
Thus an $f\in \mathscr{O}(U)$ vanishing in all $z_n$ is identically zero on the upper half-plane.
But since the upper and lower half-plane are separated by the real axis (and all $z_n$ lie in the upper half plane), the behaviour of $f$ on the lower half-plane is not influenced at all by the behaviour on the upper half-plane. So we can take any function $g$ holomorphic in the lower half-plane, and
$$f(z) = \begin{cases} \; 0 &, \operatorname{Im} z > 0\\ g(z) &, \operatorname{Im} z < 0\end{cases}$$
is a function holomorphic on $U$ that vanishes in all $z_n$, and $f \equiv 0$ if and only if $g \equiv 0$. Choosing $g(z) = 1$ or $g(z) = z$ (to list two simple explicit examples), we obtain a function $f\in\mathscr{O}(U)$ vanishing in all $z_n$ without being identically $0$.


TITLE: Determine whether the set $\{v_1 + v_2 - v_3, 2v_1 + 2v_3, -v_1 + v_2 - 3v_3\}$ is linearly dependent or independent.
QUESTION [1 upvotes]: We had a question on our last test that was very similar to this and I only got $2$ points of $6$ and I want to make sure I do it right this time. Here's my solution to that one:
Let $v_1, v_2,$ and $v_3$ be three linearly independent vectors.
My teacher told me that to qualify for full credit every detail of my suloution must be presented and the logical steps that lead to my conclusion must be clear, justified, and readable.
Here's what my solution would be to the question, but i don't think it's enough.
Here's my updated solution, this should be enough right?
Thanks guys

REPLY [1 votes]: i think you can explain what you are doing in words. you have not made clear that you understand what linear dependence of vectors mean. true, you are showing some row reduction, but why and how does it relate to the question.
one way to do is to explain how you determine linear independence by supposing $$a(v_1+v_2-v_3) + b(2v_1+2v_3) + c(-v_1 + v_2 - 3v_3) = 0 $$ rewrite as $$v_1(a+2b-c)+v_2(a+2c)+v_3(-a+2b-3c)= 0 $$ now, use the linear independence of $v_1, v_2, v_3$ to get three equations.
do you see the point?


TITLE: $A ^3$ is congruent to $A$ for a singular symmetric real matrix $A$
QUESTION [2 upvotes]: I am asked to show that $A ^3$ is congruent to $A$ for all symmetric real matrices $A$.
If $A$ is invertible, then -
$A^3 = A * A * A = A ^t * (A) * A$
and they are congruent by definition, because $A$ is non-singular and symmetric as stated.
I am stuck on the case where $A$ is singular, and I believe I shouldn't make that distinction of cases from the first place.
Thanks for any suggestions.

REPLY [1 votes]: Since $A$ is a symmetric matrix, there exists an orthogonal matrix $P$ such that $P^TAP$- say $D$- is a diagonal matrix. We can assume WLOG that $$D=\begin{pmatrix} D_* &0\\0&0\end{pmatrix}$$ where $D_*$ is a diagonal invertible square matrix of order $k=\mathrm{rank}(A)$. We have$$A^3=(PDP^T)^3=PD^3P^T=P\begin{pmatrix} D_*^3 &0\\0&0\end{pmatrix}P^T=P\begin{pmatrix} D_* &0\\0&I\end{pmatrix}\begin{pmatrix} D_* &0\\0&0\end{pmatrix}\begin{pmatrix} D_* &0\\0&I\end{pmatrix}P^T=P\begin{pmatrix} D_* &0\\0&I\end{pmatrix}P^TPDP^TP\begin{pmatrix} D_* &0\\0&I\end{pmatrix}P^T=P\begin{pmatrix} D_* &0\\0&I\end{pmatrix}P^TAP\begin{pmatrix} D_* &0\\0&I\end{pmatrix}P^T=QAQ$$ Since $Q$ is a symmetric invertible matrix, we are done!


TITLE: How many different kinds of fruit does it take to mix 10 different fruit bags with 4 fruits in each?
QUESTION [0 upvotes]: I have difficulties even with how to classify the problem. Am I looking on a permutation or a combination? I think it is a combination because it's in a bag and therefore it has no order.
Also, which theorems, definitions etc.  can I use?

REPLY [3 votes]: Assuming that each bag contains $4$ different fruits, your question boils down to find the smallest $n$ such that
$$\binom n4 \geq 10$$
Since, $\binom 54 = 5$, you may try $\binom 64 = \binom 62 = 15$. So $n=6$ is the solution in this case.
If the fruits can be repeated, you are looking for the smallest $n$ such that $x_1 + x_2 + \cdots +x_n = 4$ has at least $10$ different solutions.
$$\binom{n+4-1}{4} = \binom {n+3}{4} \geq 10$$
Here, $n=3$ is the solution.


TITLE: What are the physics definitions for time and energy?
QUESTION [0 upvotes]: I'm a student of physics. Please could someone give me the explanations of the observed physical phenomena of time and energy. Especially, the differences between the two and how they relate to oneanother. Many thanks.

REPLY [2 votes]: It's recommended that you learn about energy to start with, looking up the definition of the unit of energy, the Joule and learning about mechanical energy  or 'work done'.  You could then move on to find out about other types of energy.
Time is different and you probably won't find a simple explanation of what it actually is.  The physics unit of time is the second.
Power is also a physics quantity that you could look into, in the equation for power, time and energy are both there.
$Power = \frac{energy}{time}$
Depending on how deeply you are thinking of looking into all this, you might want to see another question here Intuitive explanation for why time symmetry implies conservation of energy?