TITLE: tangent plane to surface
QUESTION [1 upvotes]: I was wondering if somebody could check this question from schuam's outline for "Vector analysis" and tell me if this is a valid solution or has a typo...
Execise 3.27
Find an equation of the tangent plane to the surface $x^2+2xy^2-3z^3 = 6$ at the point P(1,2,1).
Solution:
We have
$F_x = 2x + 2y^2$, $F_y = 2x$, $F_z = 3z^2.$
(At this point i'm wondering: is this Schuam's outline problem correct?)
Why isn't
$$\frac{\partial F}{\partial y}=F_y = 4xy,\,\frac{\partial F}{\partial z}=F_z = -9z^2$$
Is there's something i'm missing or misunderstanding?
Here's the rest of the solution for reference.

Thus, at the point, the normal to the surface (And the tangent plane) is $N(P)=[10,2,3]$
The tangent plane E at P has the form $10x+2y+3z=b$. Substituting $P$ in the equation gives $b=10+4+3=17$.  Thus $10x_+2y+3z=17$ is an equation for the tangent plane at $P$.

REPLY [2 votes]: Yes, it is wrong. Note that
$$
F_x=2x+2y^2,\\
F_y=4xy\neq 2x,\\
F_z=-9z^2\neq 3z^2.
$$
Then, at the point $P=(1,2,1)$, you get the normal of the plane $(10,8,-9)$.


TITLE: A characterization of semisimple module related to anihilators
QUESTION [1 upvotes]: I'm trying to solve the following question:

Let $R$ be a ring and let $M$ be an $R$-module. Prove that $M$ is semisimple iff $ann(m)$ is the intersection of finitely many maximal left ideals of $R$ for all $m\in M\backslash\{0\}$. Hint: Try it first for $M$ cyclic.

For the $\Rightarrow$ part, I argued in this way:

Since $M$ is semisimiple, then $M=\oplus_{i\in I}M_i$ where $\{M_i\}$ is a family of simple modules. Choose $m\in M\backslash\{0\}$ and suppose $m\in M_k, k\in I$. As $M_k$ is simple, $M_k\simeq R/ann(m)$. So, $ann(m)$ is a maximal left ideal of $R$. Thus, $ann(m)$ is the intersection (created with only element) of maximal left ideals of $R$.

But I couldn't go that long in the $\Leftarrow$ part and I'd like to seek some hint. Thanks in advance!

REPLY [1 votes]: To complete your argument of $\implies$, this is what you need:
Let $m\in M$ be nonzero. Then in $\oplus_{i\in I}M_i$, $m$ is only nonzero on finitely many coordinates, say in the finite subset $F\subseteq I$.
For each term $0\neq m_i\in M_i$, you argue as above that $ann(m_i)$ is a maximal left ideal. Then we contend that $ann(m)=\bigcap_{i\in F} ann(m_i)$.
The containment $\supseteq$ is obvious, and the other containment $\subseteq$ is almost as obvious due to the definition of the direct sum:
If $rm=0$, then $r(\sum_{i\in F} m_i)=\sum_{i\in F} rm_i=0$, but by the uniqueness of representation of elements in the direct sum, this means $rm_i=0$ for all $i\in F$.
G.S. Zhou has already handled the other direction nicely and I won't bother making another solution to it.


TITLE: Riemann Integrals proof
QUESTION [2 upvotes]: I'm working on a question involving Riemann Sums, and I'm not quite sure how to take the information I've gotten and prove an integration. The question prompts me this:
Let $f(x) = 2x +1$ on $[0,1]$
Let p be the partition $0 < 1 / n < 2 / n < · · · < n / n$.
Compute limn→∞ S¯(p) and limn→∞ S_(p)
For this I got these answers: S¯(p) = $(2j / n) + 1$ and S_(p) = $(2(j-1) / n) + 1 $.
I am then asked to use this information to prove that
$\int_{0}^{1} 2x + 1 dx = 2$.
However I am not exactly sure how to use this information to prove this. Most of the proofs I've done so far has been proving whether a function is integrable or not, not to actually prove the area of an integral. Any help is greatly appreciated!

REPLY [1 votes]: Let

$\displaystyle\mathscr{P} = \left\{ 0,\frac{1}{n},\frac{2}{n},\cdots,\frac{n}{n}\right\}$ be a partiton of $[0,1]$.
$M(f;I_{k})= \sup \left\{\: f(x) \ : \ x \in I_{k}\ \right\}$
$|I_{k}| =\text{Length of the interval $I_{k}$}$

The Upper Riemann Sum corresponding to $\mathscr{P}$ is defined as
\begin{align*}
U\left(f;\mathscr{P}\right) &= \sum_{k=1}^{n} M(f;I_{k})\cdot |I_{k}|
\\&= \left( \frac{2}{n}+1\right) \cdot \frac{1}{n} + \left(\frac{4}{n}+1\right)\cdot \frac{1}{n} + \cdots + \left(\frac{2n}{n}+1 \right)\cdot \frac{1}{n}
\\ &= \frac{2}{n^{2}} \cdot (1+2+ \cdots +n) +1
\\ &= \frac{2}{n^{2}} \cdot \frac{n^{2}+n}{2} + 1
\end{align*}
Now take $\displaystyle\lim_{n \to\infty} U\left(f,\mathscr{P}\right)$ to get the answer.


TITLE: Simplify $\tan^{-1} ( \frac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}} )$ with trigonometric substitution
QUESTION [7 upvotes]: I will explain my approach, help me with the last step please!
$$ \tan^{-1} {\left(\frac {x - \sqrt {1-x^2}}{x + \sqrt {1-x^2}}\right)}$$
substituting x = $\sin \theta$ (as learnt from book) and solving 1-$\sin^2 \theta$ = $\cos^2 \theta$
$$ \tan^{-1} {\left(\frac {\sin \theta - |\cos \theta|}{\sin \theta + |\cos \theta| }\right)}$$
For solving modulus, it was important to determine range of $\theta$ , therefore  I defined it (as it is my variable,i can define it my way) for [-$\pi$/2 , $\pi$/2] so that sine covers all values from $-1$ to $1$ (as ,  $ -1 \le x \le 1 \,$ , from domain  ) and $\cos \theta$ is positive , and hence $|\cos \theta| = \cos \theta$.
$$ \tan^{-1} {\left(\frac {\sin \theta - \cos \theta}{\sin \theta + \cos \theta }\right)}$$
= dividing by $\cos \theta$ $$ \tan^{-1} {\left(\frac {\tan \theta - 1}{\tan\theta + 1 }\right)}$$
= by formula of $\tan (\theta - \pi/4)$ $$ \tan^{-1}( \tan{\left(\theta - \pi/4\right)})$$
That's where I am stuck ,as according to the identity,$\quad$ $tan^{-1} ( \tan \alpha) = \alpha$ $\quad$ only when $\, -\pi/2 <\alpha < \pi/2$ . But here $$ -3\pi/4 \le \,(\theta-\pi/4) \, \le \pi/4 $$
Therefore, I am not going to get ($ \,\theta - \pi/4 $) out of the expression. What  i get will be based on that graph of $\bf {\tan^{-1} (\tan x)}$ .
$$ (\theta - \pi/4) +\pi \,$$  for  $\,-3\pi/4 \le \, (\theta -\pi/4) \, < -\pi/2 \,\,$ and
$$\theta -\pi/4$$
for $\,-\pi/2 < \, (\theta -\pi/4) \, \le \pi/4 \,\,$
My teacher just cancelled arctan and tan and wrote $\theta - \pi/4$ and he didn't even include that modulus function over $\cos \theta$.
So what will be the exact answer because if everyone decide $\theta$ as per they like then there will not be a finite answer. Everyone will have their own answers and in each answer they have multiple cases as I just discussed above.
So please help me, very hopefully I signed up in stackexchange!
Found Solution :-
I was confused because  I was thinking that there can be many solutions differing person to person, but even if you choose any value of $\theta$ , you are going to get two solutions which are in the asked question above. The problem resolves when we write $\theta$ in terms of $sin^{-1} x$  as then we would not simply write like $$ \theta = \sin^{-1} x $$
we would write an equation,$$ \sin^{-1} x = \sin^{-1} (\sin \theta)$$,
now if $\theta$ is not in range of $-\pi/2$ and $\,\pi/2$ , then there would be some constant in $\pi$ (like , $\pi/4 , 2\pi$ etc. we would have to add or subtract according to the graph of 'sin inverse sin' and when we would put that value of $\theta$ , we would end with the solutions as answered by people.
(I write the answer in this edit  to help anyone who will reach here after searching  web , thanks to everyone for answers)

REPLY [1 votes]: The domain of the function is $x\in [-1,-\frac1{\sqrt2})\cup (-\frac1{\sqrt2},1]$. Then, with the substitution $\sin \theta =x$, we have $\theta \in[-\frac\pi2,-\frac\pi4)\cup (-\frac\pi4,\frac\pi2]$ and correspondingly
$$\tan^{-1} {\left(\frac {x - \sqrt {1-x^2}}{x + \sqrt {1-x^2}}\right)}
=\tan^{-1}\left[\tan{\left(\theta - \frac\pi4\right)}\right]$$
$$=
\begin{cases}
 \theta+\frac{3\pi}4 = \sin^{-1}x +\frac{3\pi}4  & x\in [-1,-\frac1{\sqrt2})\\
\theta -\frac\pi4 = \sin^{-1}x -\frac\pi4 & x\in (-\frac1{\sqrt2},1] \end{cases}
$$


TITLE: Easier way to show $\cos(20^\circ)\cos(30^\circ)\cos(40^\circ)=\cos^2(10^\circ)\cos(50^\circ)$
QUESTION [6 upvotes]: The only way I can show
$$\cos(20^\circ)\cos(30^\circ)\cos(40^\circ)=\cos^2(10^\circ)\cos(50^\circ)$$
stinks of sweat and brute force:
If $\cos(10^\circ)=x$ then using $\cos(a+b) = \cos a \cos b - \sin a \sin b$ and $\cos^2 + \sin^2 = 1$ you get
$$
\cos(20^\circ) = 2x^2-1\\
\cos(30^\circ) = 4x^3-3x \\
\cos(40^\circ) = 8x^4-8x^2+1\\
\cos(80^\circ) = 128x^8-256x^6+160x^4-32x^2+1
$$
Then
$$
\sin(10^\circ) = 128x^8-256x^6+160x^4-32x^2+1
$$
and using $\sin (a+b) = \sin(a) \cos(b) + \cos a \sin b$ and $\sin(30^\circ)=\frac12$ you get
$$
\sin(20^\circ)= 256x^9 -512x^7+320x^5-64x^3+2x \\
\cos(50^\circ)=(2x^2-1)(4x^x-3x) - \frac12(256x^9 -512x^7+320x^5-64x^3+2x)\\
\cos(50^\circ) = -128x^9+256x^7-152x^5+22x^3+2x
$$
Now that we have all the needed values of cosines, we have to prove that
$$
(2x^2-1)(4x^x-3x)(8x^4-8x^2+1)=x^2(-128x^9+256x^7-152x^5+22x^3+2x)\\
64x^9-144x^7+112x^5-34x^3+3x=-128x^{11}+256x^9-152x^7+22x^5+2x^3
$$
The tool we have is that $\cos(30^\circ) = \frac{\sqrt{3}}{2}$ so
$$x^3=\frac{\sqrt{3}/2+3x}{4}
$$
We now have to use that replace the highest power of $x$ by 2 powers lower; for example, using
$$x^{11} = \frac{3x^9}{4} + \frac{\sqrt{3}}{8}x^8$$
I had to do this replacement trick thirteen times (seven on the right side and six on the left) before I came to
$$
-x^3 + \frac{\sqrt{3}x^2}{2} + \frac{3x}{4} = -x^3 + \frac{\sqrt{3}x^2}{2} + \frac{3x}{4}
$$
And we come to the original question -- there has got to be an easier way to show this!

REPLY [0 votes]: There is a formal, purely multiplicative proof.
All the information needed is:

$\cos(20) \cos(40) \cos(80)$ and $\sin(20) \sin(40) \sin(80)$ have known values (in fact we only need the product of $\tan(20) \tan(40)\tan(80)$)
$\cos (30)$ is known
basic identities $\sin(x) = \cos(90 - x)$ and $\sin(2x) = 2 \sin x \cos x$

Multiply the alleged formula by $\cos 80 = \sin 10$ , using ( ) to denote known quantities.
c20 c30 c40 c80 ?= s10 c10 c10 c50
(c20 c40 c80) (c30) ?= s10 c10 c10 c50
(c20 c40 c80) (c30) ?= s10 c10 s80 s40
(c20 c40 c80) (c30) ?= (s20 s40 s80) / 2
which is true knowing the values of the (20 40 80) products. The identity is equivalent to the product $\tan(20) \tan(40) \tan(80) = \sqrt{3}$ and does not need separate knowledge of the sine and cosine products.
Until a better reference comes along for the 20-40-80 products, here is a misnamed one, http://en.wikipedia.org/wiki/Morrie%27s_law


TITLE: Project Euler problem #3 (how to do it by hand?)
QUESTION [2 upvotes]: Problem #3 in Project Euler:

What is the largest prime factor of the number $600851475143$?

I want to solve this by hand. (I am doing this with all problems.) What techniques would allow me to figure this out? (Yes, I checked the prime factors up to $53$, but I got quite tired.)

REPLY [2 votes]: Since you're bored, you might try this:  Multiply the first few primes together and call it $M$.  Then compute the gcd of $M$ and your number by Euclidean algorithm.  Since your number has about 12 digits, use enough primes so that $M$ is 14 or 15 digits.  By Lame's theorem, this should take at most 5 times 12 = 60 steps.  If the gcd turns out to be different from 1, you've found a factor.  If not, then make a new $M$ of 14 or 15 digits by multiplying the next few primes together and repeat.
If I were really doing this by hand, I think I'd make $M$ as large as I could stand (maybe 50 digits), because after the first step in Euclidean algorithm, both numbers would be less than 12 digits.  So I could eliminate a lot of candidates at once, at the price of one horrendous long division.


TITLE: Formula for the average of the lowest remaining value in a sequence of multisets
QUESTION [1 upvotes]: $B_1=(b_1,b_2,\ldots,b_n)$ is a finite multiset of positive real numbers $b_i$.
To obtain $B_{i+1}$ we let
$$m_i=\min\{B_i\}\quad\quad\quad\quad k_i=\min\{j\in\mathbb{N}:(B_i)_j=m_i\}\\
B_{i+1}=\left(\underbrace{(B_{i})_1-m_i,\ \ldots\ ,(B_{i})_{{k_i}-1}-m_i}\ ,\ b_{k_i}\ ,\ \underbrace{(B_{i})_{{k_i}+1}-m_i,\ \ldots\ ,(B_{i})_n-m_i}\right)$$
In words you take the minimum of the current $B_i$, subtract it from all entries of $B_i$, and replace the leftmost zero by the value at that entry in $B_1$. I'm interested in the average of the values getting subtracted:
$$\underset{n\to\infty}{\lim}\frac{1}{n}\sum_{i=1}^n m_i$$
If all $b_i$ are integers then all $B_i$ are in a finite space, so the sequence of $B_i$s is periodic. By iterating through one period I found some averages and noticed that all of them were:
$$\frac{b_1b_2\ldots b_n}{b_1b_2\ldots b_{n-1}+b_1b_2\ldots b_{n-2}b_n+\ldots+b_2\ldots b_n}$$
Does anyone know where I can find a proof for this?

REPLY [1 votes]: Define $a_N = \sum_{i=1}^N m_i$, this sequence is just $\bigcup_i b_i \mathbb{Z}_+$ (after ordering from small to large and count multiplicity).
For a positive real number $x$, there are $\sum_i \lfloor \frac{x}{b_i} \rfloor =f(x)$ terms of $a_N$ that is $\le x$, hence $a_{f(x)} \le x < a_{f(x)+1}$.
The average waiting time is
\begin{equation}
\text{lim}_N \frac{a_N}{N} = \text{lim}_x \frac{x}{f(x)} = \frac{1}{\sum_i\frac{1}{b_i}}.
\end{equation}


TITLE: How to differentiate e to a function?
QUESTION [0 upvotes]: I am trying to find the derivative of $$\large e^{2 \pi i t \sin(\pi/(2t))}.$$
I know that I am to take the derivative of the exponent, and then multiply it by the beginning problem - the piece that is giving me trouble is the $i$, I think.
Help is appreciated!

REPLY [1 votes]: Differentiate complex functions the same way you would with real functions.  Treat $i$ as a constant.

REPLY [1 votes]: By the chain rule, $$\frac{d}{dt}e^{u}=\frac{du}{dt}e^{u}$$
You have $u=2\pi it \sin(\pi/2t)$, so by the product rule $$\frac{du}{dt}=2\pi i \sin(\pi/2t)+2\pi i t \frac{d}{dt}\sin(\pi/2t)$$
Now for this derivative, let $v=\frac{\pi}{2t}$ so $\frac{dv}{dt}=-\frac{\pi}{2t^{2}}$. Use the chain rule to get $$\frac{d}{dt}\sin(v)=\cos(v)\frac{dv}{dt}=\cos(\pi/2t)\cdot \frac{-\pi}{2t^{2}}$$
Substituting back in gives
$$\frac{du}{dt}=2\pi i \sin(\pi/2t)-2\pi i t \cos(\pi/2t)\cdot \frac{\pi}{2t^{2}}$$
Therefore
$$\frac{d}{dt}e^{2\pi i t \sin(\pi/2t)}=2\pi i \left(\sin(\pi/2t)-\frac{\pi}{2t}\cos(\pi/2t)\right)e^{2\pi i t \sin(\pi/2t)}$$


TITLE: Clasifying singularities.
QUESTION [1 upvotes]: I would like to clasify the singularities of the function
$$
f(z) = \frac{1}{z^2+2z+1}
$$
I tried to make the Laurent expansion on $ z = -1$, but at this point the function is not holomorphic and I do not know how to calculate the coefficents because it is not posible to use the Cauchy's Integral Formula for the derivatives on points where the function is not holomorphic. The formula says that if you have a function $f$, holomorphic on an open set $\Omega$, and $\bar{D}(c,R)\subset \Omega$, for all $z_0\in \Omega$ you have
$$
f^{(k} (z_0) = \frac{k!}{2\pi i}\int_{\partial D(c;r)} \frac{f(\zeta)}{(\zeta-z_0)^{k+1}}d\zeta
$$
And the coefficients are given by
$$
a_n = \frac{1}{2\pi i}\int_{\gamma_r} \frac{f(\zeta)}{(\zeta+1)^{k+1}}d\zeta
$$
where $\gamma_r$ is the disc $\gamma_r(t) = c + r\exp{it}$ with $ r>0$. Note that $f$ is holomorphic on $\mathbb C\backslash{\{-1\}}$.
Thanks.

REPLY [0 votes]: Note $z^2+2z+1=(z+1)^2$. So
$$f(z)=\frac{1}{(z+1)^2}$$ This is the Laurent series expansion about $z=-1$. I.e. if $f(z)=\sum_{-\infty}^\infty a_n (z+1)^n$, then $$a_n=\begin{cases}1&n=-2\\0&\text{else}\end{cases}$$
Can you now classify the singularity?


text:
\section{Realisation as $\omega$-categories} \label{sec:omegacat}

In this section, we look at how we can interpret constructible polygraphs as strict $\omega$\nbd categories. To this aim, we will use Steiner's theory of directed complexes \cite{steiner1993algebra} as a bridge. As we mentioned in Remark \ref{rmk:precomplex}, directed complexes have an underlying structure --- a directed precomplex --- which is only slightly more general than an oriented graded poset, and, similarly to constructible directed complexes, satisfy a property relative to a class of closed subsets: the \emph{molecules}.

The characteristic property of the molecules is that they are ``globular composites'' of other molecules; compare this with constructible molecules, which can be decomposed as ``mergers''. We will show that constructible molecules are molecules, and derive that constructible directed complexes are directed complexes, which \emph{a posteriori} justifies their name. This will allow us to realise constructible atoms as $\omega$\nbd categories, which we can then extend to realise constructible polygraphs as $\omega$\nbd categories.

First of all, we need to fix the notation for $\omega$\nbd categories; we will use a mixture of Steiner's notation and the one prevalent in the more recent literature on polygraphs \cite{metayer2008cofibrant,lafont2010folk}.

\begin{dfn}
Let $X$ be an $\omega$\nbd graph, and for all $x \in X_n$ and $k < n$, let
\begin{equation*}
	\bord{k}{\alpha}x = \underbrace{\bord{}{\alpha}(\ldots(\bord{}{\alpha}}_{n-k} x)).
\end{equation*}
We call the elements $x \in X_n$ the \emph{$n$\nbd cells} of $X$. Given two $n$\nbd cells $x$, $y$ of $X$, and $k < n$, we say that $x$ and $y$ are \emph{$k$\nbd composable}, and write $x \comp{k} y$, if $\bord{k}{+}x = \bord{k}{-}y$.

We write $X_n \comp{k} X_n \subseteq X_n \times X_n$ for the set of pairs of $k$\nbd composable $n$\nbd cells of $X$.
\end{dfn}

\begin{dfn}
A \emph{partial $\omega$\nbd category} is an $\omega$\nbd graph $X$ together with \emph{unit} and \emph{$k$\nbd composition} operations
\begin{equation*}
	\idd{}: X_n \to X_{n+1}, \qquad \cp{k}: X_n \comp{k} X_n \pfun X_n
\end{equation*}
for all $n \in \mathbb{N},$ and $k < n$, where $\idd{}$ is a total function and the $\cp{k}$ are partial functions. For all $k$\nbd cells $x$, and $n > k$, let
\begin{equation*}
	\idd{n}x := \underbrace{(\idd{}\ldots\idd{})}_{n-k} x,
\end{equation*}
an $n$\nbd cell of $X$. The operations are required to satisfy the following conditions:
\begin{enumerate}
	\item for all $n$\nbd cells $x$, and all $k < n$,
	\begin{align*}
	 	\bord{}{\alpha}(\varepsilon x) & = x, \\
		x \cp{k} \idd{n}(\bord{k}{+}x) & = x = \idd{n}(\bord{k}{-}x) \cp{k} x,
	\end{align*}
	where the two $k$\nbd compositions are always defined;
	\item for all $(n+1)$\nbd cells $x, y$, and all $k < n$, whenever the left-hand side is defined,
	\begin{align*}
		\bord{}{-}(x \cp{n} y) & = \bord{}{-} x, \\
		\bord{}{+}(x \cp{n} y) & = \bord{}{+} y, \\
		\bord{}{\alpha}(x \cp{k} y) & = \bord{}{\alpha} x \cp{k} \bord{}{\alpha} y;
	\end{align*}
	\item for all cells $x, y, x', y'$, and all $n$ and $k < n$, whenever the left-hand side is defined,
	\begin{align*}
		\idd{}(x \cp{n} y) & = \idd{}x \cp{n} \idd{}y; \\
		(x \cp{n} x') \cp{k} (y \cp{n} y') & = (x \cp{k} y) \cp{n} (x' \cp{k} y');
	\end{align*}
	\item for all cells $x, y, z$, and all $n$, whenever either side is defined,
	\begin{equation*}
		(x \cp{n} y) \cp{n} z = x \cp{n} (y \cp{n} z).
	\end{equation*}
\end{enumerate}
A partial $\omega$\nbd category is an \emph{$\omega$\nbd category} if the $\cp{k}$ are total functions.

A \emph{functor} of partial $\omega$\nbd categories is a morphism of the underlying $\omega$\nbd graphs that commutes with units and compositions. A functor is an \emph{inclusion} if it is injective on cells of each dimension. Partial $\omega$\nbd categories and functors form a category $\pomegacat$, with a full subcategory $\omegacat$ on $\omega$\nbd categories.

The inclusion of $\omegacat$ into $\pomegacat$ has a left adjoint $(-)^*: \pomegacat \to \omegacat$. Given a partial $\omega$\nbd category $X$, we call $X^*$ the $\omega$\nbd category \emph{generated} by $X$.
\end{dfn}

\begin{dfn}
Let $X$ be a partial $\omega$\nbd category, $n \in \mathbb{N}$, and $x \in X_n$. We define a number $\dmn{x}$, the \emph{dimension} of $x$, by induction on $n$:
\begin{itemize}
	\item if $x \in X_0$, $\dmn{x} := 0$;
	\item if $x \in X_n$, if $x = \idd{}y$ for some $y \in X_{n-1}$, then $\dmn{x} := \dmn{y}$, else $\dmn{x} := n$.
\end{itemize}
\end{dfn}

\begin{remark}
Given a cell $x$ in a partial $\omega$\nbd category, when we say that ``$x$ is an $\emph{n}$\nbd cell'' we mean $x \in X_n$, and when we say that ``$x$ is $n$\nbd dimensional'' we mean $\dmn{x} = n$.
\end{remark}

\begin{remark}
Steiner defines a partial $\omega$\nbd category as a single set $X$ together with boundary operations $\bord{n}{\alpha}: X \to X$ and partial composition operations $\cp{n}: X \times X \pfun X$ for all $n$.

We recover this picture from our definition by taking $X := (\coprod_n X_n)/\!\!\sim$, where $\sim$ is the equivalence relation generated by $x \sim \idd{n}x$ for all $n$; notice that $X$ is isomorphic to $\coprod_n \{x \in X_n \,|\, \dmn{x} = n\}$ as a set.

For all cells $x$ and $k \in \mathbb{N}$, there is some $n > k$ such that $\bord{k}{\alpha}(\idd{n}x)$ is defined, and the result belongs to the same equivalence class for all such $n$, so the boundary operations descend to $X$. Similarly, given two cells $x$ and $y$ belonging to equivalence classes $[x]$, $[y]$, we let $[x] \cp{k} [y]$ be defined when $\idd{n}x \cp{k} \idd{n}y$ is defined for some $n > k$, and in that case be equal to the latter's equivalence class.
\end{remark}

\begin{cons}
Let $X$ be a partial $\omega$\nbd category. The \emph{$J$\nbd dual} $\oppn{J}{X}$ of $X$ is the partial $\omega$\nbd category with an $n$\nbd cell $\oppn{J}{x}$ for each $n$\nbd cell $x$ of $X$, and
\begin{equation*}
	\bord{}{\alpha}(\oppn{J}{x}) := \begin{cases} \oppn{J}{\bord{}{-\alpha}x} & \text{if } \dmn{x} \in J, \\
	\oppn{J}{\bord{}{\alpha}x} & \text{if } \dmn{x} \notin J, \end{cases}
\end{equation*}
\begin{equation*}
	\idd{}(\oppn{J}{x}) := \oppn{J}{\idd{}x}, \quad \quad \oppn{J}{x} \cp{k} \oppn{J}{y} := \begin{cases} \oppn{J}{y \cp{k} x} & \text{if } k+1 \in J, \\
	\oppn{J}{x \cp{k} y} & \text{if } k+1 \notin J, \end{cases}
\end{equation*}
where $\oppn{J}{x} \cp{k} \oppn{J}{y}$ is defined whenever $y \cp{k} x$ is defined and $k+1 \in J$, or $x \cp{k} y$ is defined and $k+1 \notin J$.
\end{cons}

\begin{dfn}
Let $U_1, U_2 \subseteq P$ be closed subsets of an oriented graded poset. If $U_1 \cap U_2 = \bord{n}{+}U_1 = \bord{n}{-}U_2$, let
\begin{equation*}
	U_1 \cp{n} U_2 := U_1 \cup U_2;
\end{equation*}
this defines partial $n$\nbd composition operations on the closed subsets of $P$, for all $n$.

Let $P$ be an oriented graded poset. For each $n \in \mathbb{N}$, we define a family $\molec{n}{P}$ of closed subsets of $P$, the \emph{$n$\nbd molecules} of $P$, together with a partial order $\submol$ on each $\molec{n}{P}$, to be read ``is a submolecule of''.

Let $U \subseteq P$ be closed. Then $U \in \molec{n}{P}$ if and only if $\dmn{U} \leq n$, and, inductively on proper subsets of $U$, either
\begin{itemize}
	\item $U$ has a greatest element, in which case we call it an \emph{atom}, or
	\item there exist $n$\nbd molecules $U_1, U_2$ properly contained in $U$, and $k < n$ such that $U_1 \cap U_2 = \bord{k}{+}U_1 = \bord{k}{-}U_2$, and $U = U_1 \cp{k} U_2$.
\end{itemize}
We define $\submol$ to be the smallest partial order relation on $\molec{n}{P}$ such that $U_1, U_2 \submol U$ for all triples $U, U_1, U_2$ in the latter situation.
\end{dfn}

We will call a decomposition $U = U_1 \cp{k} U_2$ \emph{proper} if $U_1$ and $U_2$ are properly contained in $U$.

\begin{exm}
All $0$\nbd molecules are atoms: $\molec{0}{P} = \{\{x\} \,|\, \dmn{x} = 0\}$. If $P$ is an oriented thin poset, 0-molecules and 1-molecules are constructible: in fact, $\molec{1}{P}$ is equal to $\glob{0}{P} + \glob{1}{P}$. However, there are 2-molecules that are not constructible, such as the ``whiskering'' of a 2-globe with a 1-globe:
\begin{equation*}
\input{img/s4_whiskering.tex}
\end{equation*}
We will prove that, on the other hand, all constructible molecules are molecules.
\end{exm}

\begin{remark} \label{rmk:molec_ndim}
Clearly, $\molec{n-1}{P} \subseteq \molec{n}{P}$ for all $n > 0$, and the inclusion is order-preserving. Moreover, if $U$ is $n$\nbd dimensional and a molecule, then $U$ is an $n$\nbd molecule. This is obvious if $U$ is an atom. Otherwise, if $U$ decomposes as $U_1 \cp{k} U_2$, where $U_1$ and $U_2$ are $n$\nbd molecules, if $k < n$ then $U$ is an $n$\nbd molecule, and if $k \geq n$, then $U_1 = \bord{k}{+}U_1 = \bord{k}{-}U_2 = U_2$, so $U = U_1 = U_2$, and again $U$ is an $n$\nbd molecule. Therefore $\molec{n}{P} \setminus \molec{n-1}{P}$ only contains $n$\nbd dimensional subsets.
\end{remark}

\begin{dfn} \label{dfn:directedcomplex}
A \emph{directed complex} is an oriented graded poset $P$ such that, for all $x \in P$, with $\dmn{x} = n > 0$, and all $\alpha, \beta$,
\begin{enumerate}
	\item $\bord{}{\alpha}x$ is a molecule, and
	\item $\bord{}{\alpha}(\bord{}{\beta}x) = \bord{n-2}{\alpha}x$.
\end{enumerate}
\end{dfn}

\begin{remark}
Compared to Steiner's definition based on directed precomplexes, ours has in addition the built-in constraint that $\sbord{}{+}x$ and $\sbord{}{-}x$ are disjoint for all $x$.
\end{remark}

\begin{dfn}
A closed $n$\nbd dimensional subset $U$ of an oriented graded poset is \emph{globelike} if $\bord{j}{\alpha}(\bord{k}{\beta}U) = \bord{j}{\alpha}U$ holds for all $\alpha$ and $j < k < n$.
\end{dfn}

\begin{prop} \label{prop:moleculeglobelike}
In a constructible directed complex, all molecules are globelike.
\end{prop}
\begin{proof}
Let $U$ be a molecule in a constructible directed complex. By induction on the definition of a molecule: if $U$ is an atom, then it is constructible, hence globelike by Proposition \ref{prop:globelike}. If $U$ decomposes as $U_1 \cp{k} U_2$, assuming that $U_1$ and $U_2$ are globelike, we obtain that $U$ is globelike as in \cite[Lemma 3.4]{steiner1993algebra}.
\end{proof}

\begin{remark}
The statement of \cite[Lemma 3.4]{steiner1993algebra} assumes that we are in a directed complex, but only the properties of directed precomplexes (or, \emph{a fortiori}, oriented graded posets) are actually used.
\end{remark}

\begin{cor} \label{cor:globelikecor}
Let $U_1, U_2$ be two $n$\nbd molecules in a constructible directed complex, and suppose $U_1 \cap U_2 = \bord{k}{+}U_1 = \bord{k}{-}U_2$.
\begin{enumerate}[label=(\alph*)]
	\item If $k = n-1$, then $\bord{n-1}{-}(U_1 \cp{k} U_2) = \bord{n-1}{-}U_1$ and $\bord{n-1}{+}(U_1 \cp{k} U_2) = \bord{n-1}{+}U_2$.
	\item If $k < n-1$, then $\bord{n-1}{\alpha}U_1 \cp{k} \bord{n-1}{\alpha}U_2$ is defined and equal to $\bord{n-1}{\alpha}(U_1 \cp{k} U_2)$.
\end{enumerate}
\end{cor}
\begin{proof}
The first point holds more generally for closed subsets of an oriented graded poset, see \cite[Proposition 3.1.(vi)]{steiner1993algebra}.

The second point is also proved in [Lemma 3.4, \emph{ibid.}] under the assumption that $U_1$ and $U_2$ be globelike, which follows for us from Proposition \ref{prop:moleculeglobelike}.
\end{proof}

We have the following fact about $n$\nbd dimensional molecules in a constructible directed complex: their $k$\nbd boundaries, for $k \leq n$, are always of the ``right dimension''.
\begin{prop} \label{prop:molecdim}
Let $U$ be an $n$\nbd dimensional molecule in a constructible directed complex. For all $\alpha \in \{+,-\}$ and $k < n$, $\bord{k}{\alpha}U$ is $k$\nbd dimensional, and there exists $x \in \sbord{k}{\alpha}U$ such that $x \notin \sbord{k}{-\alpha}U$.
\end{prop}
\begin{proof}
We proceed by induction on the dimension and proper subsets of $U$. If $U$ is 0-dimensional, there is nothing to prove. If $U$ is an atom of any dimension, then $\bord{k}{\alpha}U$ is a constructible $k$\nbd molecule, and by Lemma \ref{lem:basic} applied to the constructible $(k+1)$\nbd molecule $\bord{k+1}{\beta}U$, we have that $\sbord{k}{-}U$ and $\sbord{k}{+}U$ are disjoint and both inhabited. So any $x \in \sbord{k}{\alpha}U$ satisfies the statement.

Suppose $U$ is $n$\nbd dimensional and has a proper decomposition $U_1 \cp{m} U_2$, where at least one of $U_1$ and $U_2$ is $n$\nbd dimensional. If $m = n-1$, then $\bord{n-1}{-}U = \bord{n-1}{-}U_1$ and $\bord{n-1}{+}U = \bord{n-1}{+}U_2$. In this case, both $U_1$ and $U_2$ are $n$\nbd dimensional (otherwise the decomposition is not proper), so by the inductive hypothesis $\bord{n-1}{\alpha}U$ is $(n-1)$\nbd dimensional. Picking any $x \in \sbord{n-1}{-}U = \sbord{n-1}{-}U_1$ such that $x \notin \sbord{n-1}{+}U_1 = U_1 \cap U_2$, necessarily $x \notin \sbord{n-1}{+}U_2 = \sbord{n-1}{+}U$; similarly we find $y \in \sbord{n-1}{+}U$ such that $y \notin \sbord{n-1}{-}U$. For $k < n-1$, we have $\sbord{k}{\alpha}U = \sbord{k}{\alpha}(\bord{n-1}{-}U) = \sbord{k}{\alpha}U_1$, so we can use the inductive hypothesis on $U_1$ to find suitable $k$\nbd dimensional elements.

If $m < n -1$, then $\sbord{k}{\alpha}U = \sbord{k}{\alpha}(\bord{n-1}{\alpha}U_1\cp{m}\bord{n-1}{\alpha}U_2)$. By the inductive hypothesis, either $\bord{n-1}{\alpha}U_1$ or $\bord{n-1}{\alpha}U_2$ is $(n-1)$\nbd dimensional; therefore, $\bord{n-1}{\alpha}U_1\cp{m}\bord{n-1}{\alpha}U_2$ is an $(n-1)$\nbd dimensional molecule, and again we can apply the inductive hypothesis.
\end{proof}

The key step in our proof is provided by the following technical lemma.

\begin{lem} \label{lem:moleculesubs}
Let $U$ be an $n$\nbd molecule and $V$ a constructible $(n+1)$\nbd molecule in a constructible directed complex. Suppose that $V$ and $\bord{}{\alpha}V$ are molecules, and that $U \cap V = \bord{}{\alpha}V \submol U$. Then $U \cup V$ is an $(n+1)$\nbd molecule and $V \submol U \cup V$.
\end{lem}
\begin{proof}
By definition of the submolecule relation, either $U = \bord{}{\alpha}V$, or $U = U_1 \cp{m} U_2$ for some $n$\nbd molecules $U_1, U_2 \subset V$ and $m < n$, and $\bord{}{\alpha}V \submol U_i$ for some $i \in \{1,2\}$. In the first case, $U \cup V = V$, which is an $(n+1)$\nbd molecule by assumption. Moreover, we trivially have $V \submol U \cup V$ and $\bord{k}{\alpha}V = \bord{k}{\alpha}(U \cup V)$ for all $k < n$.

In the second case, suppose without loss of generality that $\bord{}{\alpha}V \submol U_2$. We can assume, inductively, that $U_2 \cup V$ is an $(n+1)$\nbd molecule, that $V \submol U_2 \cup V$, and that $\bord{k}{\alpha}V = \bord{k}{\alpha}(U_2 \cup V)$ for all $k < n$. Then
\begin{equation*}
	U_1 \cap (U_2 \cup V) = U_1 \cap U_2 = \bord{m}{+}U_1 = \bord{m}{-}U_2 = \bord{m}{-}(U_2 \cup V).
\end{equation*}
Hence, $U \cup V = U_1 \cp{m} (U_2 \cup V)$ is a molecule, $V \submol U_2 \cup V \submol U \cup V$, and by Corollary \ref{cor:globelikecor}, $\bord{n-1}{\alpha}(U \cup V)$ depends only on $\bord{n-1}{\alpha}U_1$ and $\bord{n-1}{\alpha}(U_2 \cup V) = \bord{n-1}{\alpha}U_2$. Because molecules are globelike, it follows that $\bord{k}{\alpha}V = \bord{k}{\alpha}(U \cup V)$ for all $k < n$, which completes the inductive step.
\end{proof}

\begin{thm} \label{thm:globmolec}
Let $U$ be a constructible $n$\nbd molecule in an oriented thin poset. Then $U$ is an $n$\nbd molecule. If $V$ is a constructible submolecule of $U$, it is also a submolecule of $U$.
\end{thm}
\begin{remark}
In other words, if $P$ is an oriented thin poset, there are inclusions of posets $(\glob{n}{P}, \sqsubseteq) \hookrightarrow (\molec{n}{P}, \submol)$ for all $n$.
\end{remark}
\begin{proof}
We proceed by induction on the dimension and number of maximal elements of $U$. If $U$ is an atom of any dimension, it is obviously a molecule. Moreover, $U$ has no proper constructible submolecules, and it is trivially a submolecule of itself.

Suppose $U$ is a non-atomic constructible $(n+1)$\nbd molecule, splitting as $U_1 \cup U_2$; by the inductive hypothesis, $U_1$ and $U_2$ are $(n+1)$\nbd molecules. Let
\begin{equation*}
	\tilde{U}_1 := \bord{}{-}U \cup U_1, \quad \quad \quad \tilde{U}_2 := \bord{}{+}U \cup U_2;
\end{equation*}
then $\tilde{U}_1 \cap \tilde{U}_2 = \bord{}{+}U_1 \cup \bord{}{-}U_2 = \bord{n}{+}\tilde{U}_1 = \bord{n}{-}\tilde{U}_2$, so $U = \tilde{U}_1 \cp{n} \tilde{U}_2$.

By the inductive hypothesis, both $\bord{}{-}U$ and $\bord{}{-}U_1$ are $n$\nbd molecules, and because $\bord{}{-}U_1 \sqsubseteq \bord{}{-}U$, also $\bord{}{-}U_1 \submol \bord{}{-}U$. Since $U_1 \cap \bord{}{-}U = \bord{}{-}U_1$, we fall under the hypotheses of Lemma \ref{lem:moleculesubs}: we deduce that $\bord{}{-}U \cup U_1 = \tilde{U}_1$ is an $(n+1)$\nbd molecule, and that $U_1 \submol \tilde{U}_1$.

Similarly, we obtain that $\tilde{U}_2$ is an $(n+1)$\nbd molecule, and $U_2 \submol \tilde{U}_2$. It follows that $U$ is an $(n+1)$\nbd molecule and that $U_1, U_2 \submol U$. This completes the proof.
\end{proof}

\begin{cor}
Constructible directed complexes are directed complexes.
\end{cor}
\begin{proof}
Let $P$ be a constructible directed complex, and $x \in P^{(n)}$, with $n > 0$. Then $\bord{}{\alpha}x$ is a constructible molecule, hence a molecule by Theorem \ref{thm:globmolec}, and $\bord{}{\alpha}(\bord{}{\beta}x) = \bord{n-2}{\alpha}x$ is an instance of Proposition \ref{prop:globelike}.
\end{proof}

Before moving on, we briefly discuss the loop-freeness conditions that Steiner considered for directed complexes, in order to show that constructible directed complexes do not fall into one of the loop-free classes.

\begin{dfn} \label{dfn:loopfree}
Given a constructible directed complex $P$, for each $n \in \mathbb{N}$, let $\loopd{n}{P}$ be the bipartite directed graph with $P$ as set of vertices, and an edge $y \to x$ if and only if
\begin{itemize}
	\item $\dmn{y} \leq n$, $\dmn{x} > n$, and $y \in \bord{n}{-}x \setminus \bord{n-1}{}x$, or
	\item $\dmn{y} > n$, $\dmn{x} \leq n$, and $x \in \bord{n}{+}y \setminus \bord{n-1}{}y$.
\end{itemize}
We say that $P$ is \emph{loop-free} if $\loopd{n}{P}$ is acyclic for all $n$. We say that $P$ is \emph{totally loop-free} if $\hasso{P}$ is acyclic.
\end{dfn}

Being totally loop-free is a quite natural condition: it applies to globes, cubes and oriented simplices, and it is preserved by suspensions, lax Gray products, and joins \cite[Theorem 2.19]{steiner1993algebra}. Totally loop-free constructible directed complex are loop-free [Proposition 2.15, \emph{ibid.}], which is a strictly weaker condition (and seemingly less natural, as it is not preserved by lax Gray products or joins), that nevertheless is sufficient to prove a number of results.

\begin{remark}
This characterisation of total loop-freeness is inspired by acyclic matchings in discrete Morse theory, see \cite[Chapter 11]{kozlov2008combinatorial}. In a totally loop-free constructible directed complex, the ``flow'' on cells described in Remark \ref{exm:standard} never returns to a cell once it has left it.
\end{remark}

\begin{exm} \label{exm:nonloopfree}
The following is an example of an constructible 3\nbd atom $U$ that is neither loop-free nor totally loop-free, based on \cite[Example 3.12]{power1991pasting}:
\begin{equation*}
\input{img/s4_loop.tex}
\end{equation*}
Indeed, $x_1 \to y_1 \to x_2 \to y_2 \to x_1$ is a loop in $\loopd{0}{U}$ and in $\hasso{U}$. Thus, not all constructible directed complex are loop-free.
\end{exm}

\begin{remark}
Example \ref{exm:nonloopfree} is also a counterexample to the restriction of Proposition \ref{prop:atomicnglobe} to loop-free or totally loop-free constructible molecules: $\bord{}{-}U$ and $\bord{}{+}U$ are individually totally loop-free. This is one reason why we do not consider loop-free atoms to be an adequate shape category for rewriting theory: given two composable diagrams with the same boundary, we want to be able to rewrite one into the other, which is modelled by a higher-dimensional cell.
\end{remark}

On the other hand, constructible directed complexes do satisfy a restricted loop-freeness condition, in that there can be no looping via top-dimensional elements in a molecule; this makes them less general than directed complexes, where looping is possible at every level.

\begin{prop} \label{prop:loopd-acyclic}
Let $U$ be an $(n+1)$\nbd dimensional molecule in a constructible directed complex. Then $\loopd{n}{U}$ is acyclic.
\end{prop}
\begin{proof}
We proceed by induction on proper subsets of $U$. If $U$ is an atom, the statement is true because $\sbord{}{-}U$ and $\sbord{}{+}U$ are disjoint. Otherwise, suppose $U$ has a proper decomposition $U_1 \cp{k} U_2$, where necessarily $k \leq n$; it is easy to check that $\loopd{n}{U}$ contains both $\loopd{n}{U_1}$ and $\loopd{n}{U_2}$.

If $k < n$, then $U_1 \cap U_2 = \bord{k}{+}U_1 = \bord{k}{-}U_2$ is contained in the union of the $\bord{n-1}{}x$ for $x \in U$, so it cannot contain any element of $\bord{n}{\alpha}x\setminus\bord{n-1}{}x$. It follows that there can be no edges connecting vertices of $\loopd{n}{U_1}$ to vertices of $\loopd{n}{U_2}$, and any path stays within one or the other.

If $k = n$, finite paths in $\loopd{n}{U}$ are either contained in $\loopd{n}{U_1}$ or in $\loopd{n}{U_2}$, or they reach an element $x \in \bord{}{+}U_1 = \bord{}{-}U_2 = U_1 \cap U_2$ from an $(n+1)$\nbd dimensional element $y \in U_1$, before entering an $(n+1)$\nbd dimensional element $y' \in U_2$. But $y$ can only be reached from an element of $\bord{}{-}y\setminus\bord{n-1}{}y$, which does not belong to $\bord{}{+}U_1$, and from $y'$ the path can only reach an element of $\bord{}{+}y'\setminus\bord{n-1}{}y'$, which does not belong to $\bord{}{-}U_2$. Hence, any path in $\loopd{n}{U}$ consists of a path in $\loopd{n}{U_1}$, followed by a path in $\loopd{n}{U_2}$; by the inductive hypothesis, $\loopd{n}{U}$ is acyclic.
\end{proof}

\begin{cons}
Let $P$ be a constructible directed complex. By Proposition \ref{prop:moleculeglobelike}, the diagram
\begin{equation*}
\begin{tikzpicture}
	\node[scale=1.25] (0) at (0,0) {$\molec{0}{P}$};
	\node[scale=1.25] (1) at (2.5,0) {$\molec{1}{P}$};
	\node[scale=1.25] (2) at (5,0) {$\ldots$};
	\node[scale=1.25] (3) at (7.5,0) {$\molec{n}{P}$};
	\node[scale=1.25] (4) at (10,0) {$\ldots$};
	\draw[1c] (1.west |- 0,.15) to node[auto,swap] {$\bord{}{+}$} (0.east |- 0,.15);
	\draw[1c] (1.west |- 0,-.15) to node[auto] {$\bord{}{-}$} (0.east |- 0,-.15);
	\draw[1c] (2.west |- 0,.15) to node[auto,swap] {$\bord{}{+}$} (1.east |- 0,.15);
	\draw[1c] (2.west |- 0,-.15) to node[auto] {$\bord{}{-}$} (1.east |- 0,-.15);
	\draw[1c] (3.west |- 0,.15) to node[auto,swap] {$\bord{}{+}$} (2.east |- 0,.15);
	\draw[1c] (3.west |- 0,-.15) to node[auto] {$\bord{}{-}$} (2.east |- 0,-.15);
	\draw[1c] (4.west |- 0,.15) to node[auto,swap] {$\bord{}{+}$} (3.east |- 0,.15);
	\draw[1c] (4.west |- 0,-.15) to node[auto] {$\bord{}{-}$} (3.east |- 0,-.15);
\end{tikzpicture}
\end{equation*}
is an $\omega$\nbd graph $\molec{}{P}$.

For any $n$\nbd molecule $U$, let $\idd{}(U) := U$ as an $(n+1)$\nbd molecule, and for any pair $U_1, U_2$ of $n$\nbd molecules, let $U_1 \cp{k} U_2$ be defined if and only if $U_1 \cap U_2 = \bord{k}{+}U_1 = \bord{k}{-}U_2$, and in that case be equal to $U_1 \cup U_2$. By \cite[Proposition 2.9]{steiner1993algebra}, this makes $\molec{}{P}$ a partial $\omega$\nbd category.
\end{cons}

For any partial $\omega$\nbd category $X$, if $X^*$ is the $\omega$\nbd category generated by $X$, we obtain a functor $X \to X^*$ from the unit of the adjunction between $\pomegacat$ and $\omegacat$.

\begin{prop}\emph{\cite[Theorem 2.13]{steiner1993algebra}} \label{thm:steiner213} Let $P$ be a directed complex. Then the unit $\molec{}P \to (\molec{}P)^*$ is an inclusion of partial $\omega$\nbd categories.
\end{prop}

The fact that $\molec{}P \to (\molec{}P)^*$ is an inclusion allows us to identify the cells in its image with unique molecules of $P$.

\begin{prop}
The assignment $P \mapsto (\molec{}{P})^*$ for constructible directed complexes extends to a faithful functor $(\molec{}{-})^*: \globpos \to \omegacat$.
\end{prop}
\begin{proof}
The $\omega$\nbd category $(\molec{}{P})^*$ is composition-generated by the atoms of $P$, and an inclusion $\imath: P \hookrightarrow Q$  determines a sequence of functions $\clos{\{x\}} \mapsto \clos{\{\imath(x)\}}$ sending generators of $(\molec{}{P})^*$ to generators of $(\molec{}{Q})^*$. These are compatible with boundaries and compositions (because they are between the partial $\omega$\nbd categories $\molec{}{P}$ and $\molec{}{Q}$), so they determine a unique map of $\omega$\nbd categories. Faithfulness is a consequence of Proposition \ref{thm:steiner213}.
\end{proof}

Restricting to the subcategory $\globe$, we obtain a functor $\hatto{-}: \globe \to \omegacat$. Now, $\globe$ is a small category (there are countably many constructible atoms, and only finitely many morphisms between each two), $\cpol$ is locally small, and $\omegacat$ is cocomplete \cite{batanin1998monoidal}. By \cite[Theorem X.4.1-2]{maclane1971cats}, the left Kan extension of $(\molec{}{-})^*$ along the Yoneda embedding $\globe \hookrightarrow \cpol$ exists, and is given, for each constructible polygraph $X$, by the coend
\begin{equation*}
	X \mapsto \hatto{X} := \int^{U \in \globe} \hatto{U} \times X(U).
\end{equation*}
This functor has a right adjoint $P: \omegacat \to \cpol$, defined by
\begin{equation*}
	PX(-) := \mathrm{Hom}_{\omegacat}(\hatto{-},X)
\end{equation*}
for each $\omega$\nbd category $X$.

The lax Gray product of $\omega$\nbd categories was defined by Al-Agl, Brown, and Steiner \cite{al2002multiple}, and later studied by Crans \cite{crans1995pasting}. A definition based on augmented directed complexes was given by Steiner in \cite{steiner2004omega}; the proof of its correctness was completed by Ara and Maltsiniotis \cite{ara2016joint}, who also defined the join of $\omega$\nbd categories.

Let $(\omegacat, \tensor, 1)$ and $(\omegacat, \join, \emptyset)$ denote the monoidal structures on $\omegacat$ corresponding, respectively, to the lax Gray product and the join. It seems almost inevitable that the following should hold.

\begin{conj} \label{conj:monoidality}
The functor $\hatto{-}: \cpol \to \omegacat$ is monoidal both from $(\cpol, \tensor, 1)$ to $(\omegacat, \tensor, 1)$ and from $(\cpol, \join, \emptyset)$ to $(\omegacat, \join, \emptyset)$.
\end{conj}
Proving this, however, seems to require developing the relation between our theory and that of Steiner, Ara, and Maltsiniotis further than we intended. We give a few partial results, and leave a full proof to future work.

\begin{lem} \label{lem:restrictmonoidal}
The following hold:
\begin{enumerate}[label=(\alph*)]
	\item $\hatto{-}: (\cpol, \tensor, 1) \to (\omegacat, \tensor, 1)$ is a monoidal functor if and only if its restriction to $\globe$ is monoidal;
	\item $\hatto{-}: (\cpol, \join, \emptyset) \to (\omegacat, \join, \emptyset)$ is a monoidal functor if and only if its restriction to $\globe_+$ is monoidal.
\end{enumerate}
\end{lem}
\begin{proof}
For the first point: $\globe$ is a dense subcategory of $\cpol$, and the lax Gray product, being part of a biclosed structure, preserves colimits separately in each argument. Thus, for any pair of constructible polygraphs $X, Y$,
\begin{equation*}
	X \tensor Y \simeq \colim{U \to X}U \tensor \colim{V \to Y}V \simeq \colim{U \to X, V \to Y}(U \tensor V),
\end{equation*}
where $U, V$ range over atoms. Moreover, $\hatto{-}$, being a left adjoint, preserves colimits, and by \cite[Th\'eor\`eme A.15]{ara2016joint} the lax Gray product of $\omega$\nbd categories also preserves colimits separately in each variable: therefore, if $\hatto{-}: (\globe, \tensor, 1) \to (\omegacat, \tensor, 1)$ is monoidal,
\begin{equation*}
	\hatto{(X \tensor Y)} \simeq \colim{\hatto{(U \to X)}, \hatto{(V \to Y)}}\hatto{(U \tensor V)} \simeq \colim{\hatto{(U \to X)}, \hatto{(V \to Y)}}(\hatto{U} \tensor \hatto{V}) \simeq \hatto{X} \tensor \hatto{Y}.
\end{equation*}
and all isomorphisms are natural in $X$ and $Y$. The other direction is obvious.

The proof of the second point is analogous, using the fact that any constructible polygraph is the colimit of a \emph{connected} diagram in $\globe_+$ (because the latter has an initial object), and that joins of $\omega$\nbd categories preserve connected colimits separately in each argument \cite[Th\'eor\`eme 7.28]{ara2016joint}.
\end{proof}

Let $\globe^\mathrm{tlf}$ and $\globe^\mathrm{tlf}_+$ be the full subcategories of $\globe$ and $\globe_+$ on totally loop-free constructible atoms (Definition \ref{dfn:loopfree}). By \cite[Theorem 2.19]{steiner1993algebra}, they form monoidal subcategories of $(\globe, \tensor, 1)$ and of $(\globe_+, \join, \emptyset)$, respectively.

\begin{lem}
The restriction of $\hatto{-}$ to $\globe^\mathrm{tlf}$ is monoidal from $(\globe^\mathrm{tlf}_+, \tensor, 1)$ to $(\omegacat, \tensor, 1)$ and from $(\globe^\mathrm{tlf}_+, \join, \emptyset)$ to $(\omegacat, \join, \emptyset)$.
\end{lem}
\begin{proof}[Sketch of the proof]
If $U$ is a totally loop-free constructible atom, then the augmented directed complex $KU$ (Definition \ref{cons:adc}) is a strong Steiner complex (\emph{complexe de Steiner fort}) with basis $U$, in the sense of \cite[2.10]{ara2016joint}. Strong Steiner complexes form a full monoidal subcategory $\cat{St}_\mathrm{f}$ of $\adc$ both with the tensor product and the join.

There is a functor $\nu: \adc \to \omegacat$, described in \cite[Definition 2.8]{steiner2004omega}, which restricted to $\cat{St}_\mathrm{f}$ is full and faithful, and exhibits $(\cat{St}_\mathrm{f}, \otimes, I)$ and $(\cat{St}_\mathrm{f}, \join, 0)$ as full monoidal subcategories of $(\omegacat, \tensor, 1)$ and of $(\omegacat, \join, \emptyset)$, respectively \cite[Th\'eor\`eme A.15 and Th\'eor\`eme 7.28]{ara2016joint}.

Moreover, by \cite[Theorem 6.1]{steiner2004omega}, if $K$ is a strong Steiner complex with basis $B$, the $\omega$\nbd category $\nu K$ is a polygraph, whose generators correspond to elements of $B$. We can use this and Proposition \ref{prop:molec_hatto_iso} to establish an isomorphism between $(\molec{}{U})^*$ and $\nu K(U)$, natural in $U$, for all totally loop-free molecules $U$.

It follows that $\hatto{-}$ restricted to $\globe^\mathrm{tlf}$ or $\globe^\mathrm{tlf}_+$ factors as $\nu K(-)$ through $\adc$, and we conclude by Proposition \ref{prop:adcmonoidal}.
\end{proof}

By the two lemmas combined, we have reduced the task of proving Conjecture \ref{conj:monoidality} to proving that the definitions of lax Gray products and joins coincide on non-totally-loop-free molecules. Unfortunately, the monoidality of $K: \globpos \to \adc$ does not seem directly useful for this extension, as $\nu$ is in general only lax monoidal: we would need to prove both that $\nu K(P)$ is naturally isomorphic to $(\molec{}{P})^*$ for all constructible directed complexes $P$, and that $\nu$ is monoidal on the image of $K$.

\begin{remark}
Something we can do is to restrict $K$ to $\globe$, and extend it along colimits to obtain a monoidal functor $K: (\cpol,\tensor,1) \to (\adc, \otimes, I)$. Post-composed with the forgetful functor from augmented directed complexes to chain complexes of abelian groups, this coincides, on constructible polygraphs, with the linearisation functor of \cite[Subsection 3.3]{metayer2003resolutions}.
\end{remark}

We conclude this section with the proof that $\hatto{-}$ is compatible with $J$\nbd duals.

\begin{prop}
Let $J \subseteq \mathbb{N}^+$. There exist isomorphisms $\hatto{(\oppn{J}{X})} \simeq \oppn{J}{\hatto{X}}$, natural in the constructible polygraph $X$.
\end{prop}
\begin{proof}
It suffices to construct isomorphisms $\hatto{(\oppn{J}{U})} \simeq \oppn{J}{\hatto{U}}$ natural in atoms $U$ and inclusions, because $\oppn{J}{-}$ is an invertible endofunctor both in $\cpol$ and in $\omegacat$, so in particular it preserves colimits; we will then have
\begin{align*}
	\hatto{(\oppn{J}{X})} & \simeq \hatto{(\colim{U \to X}\oppn{J}{U})} \simeq \colim{\hatto{(U \to X)}}\hatto{(\oppn{J}{U})} \simeq \colim{\hatto{(U \to X)}}\oppn{J}{\hatto{U}} \simeq \\
	& \simeq \oppn{J}{\colim{\hatto{(U \to X)}} \hatto{U}} \simeq \oppn{J}{\hatto{X}}.
\end{align*}
Let $P$ be a constructible directed complex. We will prove that $U$ is an $n$\nbd molecule in $P$ if and only if $\oppn{J}{U}$ is an $n$\nbd molecule in $\oppn{J}{P}$, and $\bord{k}{\alpha}(\oppn{J}{U}) = \oppn{J}{\bord{k}{\pm \alpha}U}$, the sign depending on whether $k+1 \in J$ or not.

If $U$ is an atom, this is immediate from Proposition \ref{prop:oppglobpos} and the characterisation of $k$\nbd boundaries of constructible molecules in Proposition \ref{prop:globelike}. If $U$ has a proper decomposition $U_1 \cp{k} U_2$, then $U_1 \cap U_2 = \bord{k}{+}U_1 = \bord{k}{-}U_2$, and, equivalently by the inductive hypothesis, $\oppn{J}{U_1} \cap \oppn{J}{U_2} = \bord{k}{\pm}\oppn{J}{U_1} = \bord{k}{\mp}\oppn{J}{U_2}$, which means that either $\oppn{J}{U_1} \cp{k} \oppn{J}{U_2}$ or $\oppn{J}{U_2} \cp{k} \oppn{J}{U_1}$ is defined and equal to $\oppn{J}{U}$. The relation between the $k$\nbd boundaries of $U$ and of $\oppn{J}{U}$ is then a consequence of Corollary \ref{cor:globelikecor}.

This determines an isomorphism of the underlying $\omega$\nbd graphs of $\molec{}{(\oppn{J}{P})}$ and $\oppn{J}{\molec{}{P}}$; naturality in $P$ and compatibility with units and compositions are obvious. Restricting to atoms and whiskering with the functor $(-)^*$, we obtain a natural isomorphism between $\hatto{(\oppn{J}{-})}$ and $(\oppn{J}{\molec{}{-}})^*$;  to conclude, it suffices to show that $\oppn{J}{-}^*$ and $\oppn{J}{-^*}$ are naturally isomorphic, which is a simple exercise.
\end{proof}

config: arxiv


TITLE: Domain and Range with functions in brackets
QUESTION [1 upvotes]: For the function
$f(x, y) = \sqrt{(x-1)^2+(y-2)^2 -1}$
Would it be sufficient to write the domain as
$(x-1)^2 + (y-2)^2 ≤ 1$
or is that too vague? Then, the range would be $z ≥ 0$ but how could that be expressed in terms of $x$ if the value has to be greater than or equal to $1$?

REPLY [2 votes]: Your idea is pretty fine, but be careful about the direction of the inequality. If you have $\sqrt{z}$, then you want $z \ge 0$, not $z \le 0$. Since $f : \Bbb R^2 \to \Bbb R$, you could write the domain by
$$\mathrm{dom}(f) = \left\{ (x,y) \in \Bbb R^2 \; \middle| (x-1)^2 + (y-2)^2 \ge 1 \right\}$$
as a result. The range of $f$ could simply be given by $\Bbb R_{\ge 0}$ if you want a symbol for it. You just want the points it is possible for $f$ to map to, regardless of the constraints that need be placed on $x,y$ in order for that to happen.
For instance, you could have $x=y-1$. Then the radicand is given by
$$(x-1)^2 + (y-2)^2 - 1 \longrightarrow 2(y-2)^2 - 1$$
Clearly the function $g(y) = \sqrt{2(y-2)^2 - 1}$, for wherever $y$ ensures $g(y)$ is defined, has range $\Bbb R_{\ge 0}$.


TITLE: Connection between separable measure spaces and $\sigma$-finite measure spaces
QUESTION [12 upvotes]: I recently came across a theorem which makes a hypothesis that a certain measure space is separable (the definition can be found here). In order to avoid confusion, I'll add the definition here:
We have a space $X$ with measure. Then we define a distance on measurable sets $A$ and $B$ as $d(A,B)=meas(A\triangle B) = meas ((A\setminus B)\cup(B\setminus A))$. The space is called separable measure space iff the space of measurable subsets is separable with respect to distance $d$. In other words, there exists a countable sequence $S$ of measurable subsets of $X$, such that for all measurable $A\subset X$  and $\forall \delta>0$ there exists $B\in S$ such that $meas(A\triangle B)<\delta $.
On the other hand, it's established that in non-$\sigma$-finite spaces some weird things can happen (Fubini's theorem no longer valid, etc. This wiki article briefly touches this subject). Therefore, I'd like to have $\sigma$-finiteness.
We know that $L^p$ spaces for finite $p$ over separable measure spaces are separable; on the other hand, we can build a nonseparable $L^2$ space over a $\sigma$-finite measure space (see this answer), therefore $\sigma$-finiteness does not imply separability (in the sense of measures).
My intuition suggests that separability of the measure space implies $\sigma$-finiteness, however, I'm not able to prove or disprove it (this doesn't immediately follow from the definition).
I'd be grateful if you could provide me with references where such things are discussed. And hints on how to prove/disprove the hypothesis above are welcome, too.
Edit
One can establish that a semifinite separable measure is $\sigma$-finite, as was indicated in now deleted answer by Norbert. Moreover, the accepted answer by Michael Greinecker shows that a non-semifinite separable measure space can be non-$\sigma$-finite.

REPLY [10 votes]: Separable measure spaces do not have to be $\sigma$-finite. Let $X=\{0\}$, $\Sigma=\{X,\emptyset\}$, and let $\mu(X)=\infty$. Then $\Sigma$ is also a countable dense subset.


TITLE: In which finite groups is there a non-central g such that, for all irreducible characters, Chi(i)(g) <> zero?
QUESTION [2 upvotes]: What is the character of Pi(G), the tensor product of all inequivalent irreducible representations of G?

REPLY [6 votes]: I'm refering to the question in the title. Elements such that $\chi(g)\neq 0$ for all $\chi\in \operatorname{Irr}G$ are called nonvanishing elements in the literature. They were studied by Isaacs, Navarro and Wolf (Group elements were no irreducible character vanishes, J. Algebra 222 (1999), 413-423, MR1733678, Doi) and others (search for papers that cite the paper just mentioned). Isaacs, Navarro and Wolf prove (among other things) that in a nilpotent group, all nonvanishing elements are central. On the other hand, Miyamoto (J. Algebra 364 (2012), 88-89, MR2927049, doi) showed that any nontrivial solvable group has nonvanishing elements different from $1$. Thus at least every solvable groups with trivial center is a group as in the question title. A paper of Dolfi et. al. (J. Algebra 323 (2010), 540-545, MR2564856, doi) contains a number of nonsolvable examples.


TITLE: If a function is continuous and converges finitely, then it is bounded
QUESTION [1 upvotes]: Suppose that $f$ is a continuous function on $[a,+\infty)$ and that there exists a finite limit $\lim \limits_{x \to +\infty} f(x)$, then $f$ is bounded.
I know that from the assumption I can conclude that $f$ is uniformly continuous on that interval, but how can I use this to show that $f$ is bounded.

REPLY [1 votes]: Definition. $\lim _{ x\rightarrow \infty  }{ f(x)=k\Longleftrightarrow  } \forall \epsilon >0\;, \exists K:x>K\Rightarrow \left| f(x)-k \right| <\epsilon$

Taking $\epsilon = 1$, we have
$\left| f(x)-k \right| < 1\Rightarrow -1+k<f(x)<1+k\;  \text{(This is the upper )} $
for all $x\in (K ,\infty )$.
Now take $M = \max\limits_{x\in[a,K]}f(x)$ (It exists because $f$ is continuous on a compact)
Finally, we take $H=\max { \left\{ M,k+1 \right\} } $ and we have $f(x)\le H\; \forall x\in [a,\infty)$


TITLE: $\nabla \cdot f + w \cdot f = 0$
QUESTION [7 upvotes]: Let $w(x,y,z)$ be a fixed vector field on $\mathbb{R}^3$. What are the solutions of the equation
$$
\nabla \cdot f + w \cdot f = 0 \, ?
$$
Note that if  $w = \nabla \phi $, then the above equation is equivalent to
$$
\nabla \cdot (e^\phi f) = 0,
$$
for which the solutions are of the form $f = e^{-\phi} \nabla \times g$ for some arbitrary $g$.

REPLY [2 votes]: You can try to proceed along the following lines:
\begin{equation*}
(\partial _{\mathbf{x}}+\mathbf{w})\cdot \mathbf{f}=0
\end{equation*}
Special case $\mathbf{w}$ is constant. Then
\begin{equation*}
\exp [-\mathbf{w\cdot x}]\partial _{\mathbf{x}}\exp [+\mathbf{w\cdot x}
]=\partial _{\mathbf{x}}+\mathbf{w}
\end{equation*}
so
\begin{eqnarray*}
\exp [-\mathbf{w\cdot x}]\partial _{\mathbf{x}}\exp [+\mathbf{w\cdot x}%
]\cdot \mathbf{f} &=&0 \\
\partial _{\mathbf{x}}\cdot \{\exp [+\mathbf{w\cdot x}]\mathbf{f}\} &=&0 \\
\exp [+\mathbf{w\cdot x}]\mathbf{f} &\mathbf{=a}&+\partial _{\mathbf{x}%
}\times \mathbf{b(x)} \\
\mathbf{f(x)} &=&\exp [-\mathbf{w\cdot x}]\{\mathbf{a}+\partial _{\mathbf{x}%
}\times \mathbf{b(x)}\}
\end{eqnarray*}
In general you have to find $\mathbf{U}(\mathbf{x})$ ($3\times 3$ matrix)
such that
\begin{equation*}
\mathbf{U}^{-1}(\mathbf{x})\cdot \partial _{\mathbf{x}}\cdot \mathbf{U}(%
\mathbf{x})=\partial _{\mathbf{x}}+\mathbf{w(x})
\end{equation*}
Let
\begin{eqnarray*}
\mathbf{U}(\mathbf{x}) &=&\exp [\mathbf{A(x)}] \\
\partial _{\mathbf{x}}\cdot \mathbf{U}(\mathbf{x}) &=&\mathbf{U}(\mathbf{x}%
)[\partial _{\mathbf{x}}+\{\partial _{\mathbf{x}}\cdot \mathbf{A(x)\}]} \\
\partial _{\mathbf{x}}\cdot \mathbf{A(x)} &=&\mathbf{w(x})
\end{eqnarray*}
Then
\begin{eqnarray*}
(\partial _{\mathbf{x}}+\mathbf{w(x)})\cdot \mathbf{f} &=&\mathbf{U}^{-1}(%
\mathbf{x})\cdot \partial _{\mathbf{x}}\cdot \{\mathbf{U}(\mathbf{x})\cdot
\mathbf{f}\}=0 \\
\partial _{\mathbf{x}}\cdot \{\mathbf{U}(\mathbf{x})\cdot \mathbf{f}\} &=&0
\\
\mathbf{U}(\mathbf{x})\cdot \mathbf{f(x)} &=&\{\mathbf{a}+\partial _{\mathbf{
x}}\times \mathbf{b(x)}\} \\
\mathbf{f(x)} &=&\mathbf{U}^{-1}(\mathbf{x})\cdot \{\mathbf{a}+\partial _{
\mathbf{x}}\times \mathbf{b(x)}\}
\end{eqnarray*}


TITLE: When does equality hold in the Minkowski's inequality $\|f+g\|_p\leq\|f\|_p+\|g\|_p$?
QUESTION [19 upvotes]: I would like to see a proof of when equality holds in Minkowski's inequality.

Minkowski's inequality. If $1\le p<\infty$ and $f,g\in L^p$, then $$\|f+g\|_p \le \|f\|_p + \|g\|_p.$$

The proof is quite different for when $p=1$ and when $1<p<\infty$. Could someone provide a reference? Thanks!

REPLY [7 votes]: Let $q$ be a conjugate exponent of $p$, meaning $\frac{1}{q} + \frac{1}{p} = 1$. Now $\|f + g\|_p = \|f\|_p + \|g\|_p$ iff $\|f+ g\|_p^p = \|f + g\|_p^{p - 1}(\|f\|_p + \|g\|_p)$.
So since
$$
\begin{split}
\|f + g\|_p^p  & = \int_X |f + g|^p d\mu = \int_X (|f + g|)|f + g|^{p - 1} d\mu \\
&\leq \int_X \big(|f| + |g|\big)|f + g|^{p - 1} d\mu = \int_X |f||f + g|^{p-1} d\mu + \int_X |g||f + g|^{p-1} d\mu  \\
& = \|f(f + g)^{p - 1}\|_{L^1} + \|g(f + g)^{p - 1}\|_{L^1} \\
&\leq \|f\|_p\|(f + g)^{p - 1}\|_q + \|g\|_p \|(f + g)^{p - 1}\|_q \\
& = \big(\|f\|_p + \|g\|_p\big)\|(f + g)^{p - 1}\|_q \\
& = \big(\|f\|_p + \|g\|_p\big)\left(\int_X \big(|f(x) + g(x)|^{p - 1}\big)^q d\mu\right)^{1/q} \\
& = \big(\|f\|_p + \|g\|_p\big)\left(\int_X |f(x) + g(x)|^{pq - q} d\mu\right)^{1/q} \\
& = \big(\|f\|_p + \|g\|_p\big)\left(\int_X |f(x) + g(x)|^{p} d\mu\right)^{1/q} \\
& = \big(\|f\|_p + \|g\|_p\big)\left(\int_X |f(x) + g(x)|^{p} d\mu\right)^{1 - 1/p} \\
& = \big(\|f\|_p + \|g\|_p\big)\left(\int_X |f(x) + g(x)|^{p} d\mu\right)^{\frac{p - 1}{p}} = \|f + g\|_p^{p - 1}\big(\|f\|_p + \|g\|_p\big)
\end{split}
$$
where the $3^{rd}$ inequality follows from the triangular inequality and the $6^{th}$ inequality holds from Hölder's inequality.
Therefore we have that Minkowski's inequality gives us equality iff the triangular inequality is an equality almost everywhere and Hölder's inequality is an equality. Now the triangular inequality is an equality almost everywhere, iff $g(x)$ and $f(x)$ have the same sign almost everywhere, or if on every set of positive measure where the two diﬀer by sign at least one of them is zero. We have equality for Hölder’s if $|f|^p$ is a constant multiple of $|g|^q$ almost everywhere.


text:
\begin{document}
\title[on Kirchhoff equations]{Nondegeneracy of positive solutions to a Kirchhoff problem with critical Sobolev growth}

     \author[G. Li and C.-L. Xiang]{Gongbao Li and Chang-Lin Xiang}

\address[Gongbao Li]{School of Mathematics and Statistics and Hubei Key Laboratory of Mathematical Sciences,  Central China Normal University, Wuhan,  430079, P.R. China}
\email[]{ligb@mail.ccnu.edu.cn}

\address[Chang-Lin Xiang]{School of Information and Mathematics, Yangtze University, Jingzhou 434023, P.R. China.}
\email[]{changlin.xiang@yangtzeu.edu.cn}
\thanks{Corresponding author: Chang-Lin Xiang}
\thanks{{\bf 2010 Mathematics Subject Classification:} 35A02, 35B09, 35J61}
\thanks{Li is supported by  NSFC (No. 11771166), and Program for Changjiang Scholars and Innovative Research Team in University \# IRT13066. Peng is financially supported by NSFC (No. 11571130). The corresponding author Xiang is supported by NSFC (No. 11701045) and   the Yangtze Youth Fund (No. 2016cqn56).}

\begin{abstract}
In this paper, we prove uniqueness and nondegeneracy of positive solutions to the following Kirchhoff equations with critical growth \begin{eqnarray*} -\left(a+b\int_{\mathbb{R}^{3}}|\nabla u|^{2}\right)\Delta u=u^{5}, & u>0 & \text{in }\mathbb{R}^{3},\end{eqnarray*} where $a,b>0$ are positive constants. This result has potential applications in singular perturbation problems concerning Kirchhoff equaitons.
\end{abstract}

\maketitle

{\small
\keywords {\noindent {\bf Keywords:} Kirchhoff equations; Positive solutions; Uniqueness; Nondegeneracy}
\smallskip
\newline

\bigskip

\section{Introduction and main result}

In this paper, we are concerned about the nonlocal Kirchhoff type
problem
\begin{eqnarray}
-\left(a+b\int_{\R^{3}}|\na u|^{2}\right)\De u=u^{5}, & u>0 & \text{in }\R^{3},\label{eq: limiting Kirchhoff}
\end{eqnarray}
where $a,b>0$ are constants, $\De=\sum_{i=1}^{3}\pa_{x_{i}x_{i}}$
is the usual Laplacian operator in $\R^{3}$.

Denote by $D=D^{1,2}(\R^{3})$ the completion of $C_{0}^{\wq}(\R^{3})$
under the seminorm
\[
\|\var\|_{D}^{2}\equiv\int_{\R^{3}}|\na\var|^{2}.
\]
A (weak) solution to Eq. (\ref{eq: limiting Kirchhoff}) is a function
$u\in D$ satisfying
\[
\left(a+b\int_{\R^{3}}|\na u|^{2}\right)\int_{\R^{3}}\na u\cdot\na\var=\int_{\R^{3}}u^{5}\var
\]
for all $\var\in D$. By the Sobolev embedding $D\subset L^{6}(\R^{3})$,
all the integrals in the above equation are well defined.

Problem (\ref{eq: limiting Kirchhoff}) and its variants have been
studied extensively in the literature. Physician Kirchhoff \cite{Kirchhoff-1883}
proposed the following time dependent wave equation
\[
\rho\frac{\pa^{2}u}{\pa t^{2}}-\left(\frac{P_{0}}{h}+\frac{E}{2L}\int_{0}^{L}\left|\frac{\pa u}{\pa x}\right|^{2}\right)\frac{\pa^{2}u}{\pa x^{2}}=0
\]
for the first time, in order to extend the classical D'Alembert's
wave equations for free vibration of elastic strings.
\cite{Bernstein-1940} and Pohozaev \cite{Pohozaev-1975} contributed
some early research on the study of Kirchhoff equations. Much attention
was received until the work \cite{Lions-1978} of J.L. Lions. For
more interesting results in this respect, we refer to e.g. \cite{Arosio-Panizzi-1996,DAncona-Spagnolo-1992}
and the references therein. From a mathematical point of view, the
interest of studying Kirchhoff equations comes from the nonlocality
of Kirchhoff type equations. For instance, the consideration of the
stationary analogue of Kirchhoff's wave equation leads to problem
of the type
\begin{eqnarray}
-\left(a+b\int_{\R^{3}}|\na u|^{2}\right)\De u=f(x,u) &  & \text{in }\Om\label{eq: general Kirchhoff eq.}
\end{eqnarray}
where $\Om\subset\R^{3}$ is a smooth domain. Note that the term $\left(\int|\na u|^{2}\D x\right)\De u$
depends not only on the pointwise value of $\De u$, but also on the
integral of $|\na u|^{2}$ over the whole domain. In this sense, Eqs.
(\ref{eq: limiting Kirchhoff}) and (\ref{eq: general Kirchhoff eq.})
are no longer the usual pointwise equalities. This new feature brings
new mathematical difficulties that make the study of Kirchhoff type
equations particularly interesting. We refer to e.g. \cite{Deng-Peng-Shuai-2015,Figueiredo et al-2014,Guo-2015,He-2016-JDE,He-Li-2015,He-Zou-2012,Li-Li-Shi-2012,LLPWX-2017,Li-Ye-2014,Perera-Zhang-2006,Wang-Tian-Xu-Zhang-2012}
and the references therein for mathematical researches on the existence
of solutions and many other problems.

In this paper, we study the uniqueness and nondegeneracy of positive
solutions to problem (\ref{eq: limiting Kirchhoff}). These quantitative
properties play a fundamental role in singular perturbation problems.
Let us briefly recall some results in this respect. Kwong \cite{Kwong-1989}
established uniqueness and nondegeneracy of positive solutions to
the Schr\"odinger equations
\begin{eqnarray*}
-\De w+w=w^{q}, & w>0 & \text{in }\R^{N},
\end{eqnarray*}
see also Chang et al. \cite{Chang et al-2007}; For quasilinear Schr\"odinger
equations such as
\begin{eqnarray*}
-\Delta u-u\Delta|u|^{2}+\omega u-|u|^{q-1}u=0 &  & \text{in }\R^{N},
\end{eqnarray*}
where $\omega>0$ is a constant, $q$ is an index denoting subcritical
growth of the nonlinearity and $N\ge1$, see e.g. Selvitella \cite{Selvitella-2015},
Xiang \cite{Xiang-2016} and Adachi et al. \cite{Shinji-Masataka-Tatsuya-2016};
For fractional Schr\"odinger equations such as
\begin{eqnarray*}
\left(-\De\right)^{s}w+w=w^{q}, & w>0 & \text{in }\R^{N},
\end{eqnarray*}
where $0<s<1\le N$ and $q$ is an index denoting subcritical growth
of the nonlinearity, see e.g. Frank and Lenzmann \cite{Frank-Lenzmann-2013}
and Frank, Lenzmann and Silvestre \cite{Frank-Lenzman-Silvestre-2016},
Fall and Valdinoci \cite{Fall-Valdinoci-2014}. For elliptic equations
with critical growth
\begin{eqnarray}
(-\De)^{s}w=w^{\frac{N+2s}{N-2s}}, & w>0 & \text{in }\R^{N},\label{eq: Yamabe problems}
\end{eqnarray}
see Ambrosetti et al. \cite{Ambrosetti-GA-Peral-1999} and D\'avila
et al. \cite{Davila-Pino-Sire-2013} for $s=1$ and $0<s<1$, respectively.
In particular, Ambrosetti and Malchiodi \cite{Ambrosetti-Malchiodi-Book}
provides a systematical research on nondegeneracy of ground states
to various types of elliptic problems together with applications in
singular perturbation problems.

It is also known that the uniqueness and nondegeneracy of ground states
are of fundamental importance when one deals with orbital stability
or instability of ground states. It mainly removes the possibility
that directions of instability come from the kernel of the related
linear operator. The uniqueness and nondegeneracy of ground states
also play an important role in blow-up analysis for the corresponding
standing wave solutions in the corresponding time-dependent equations,
see e.g. \cite{Frank-Lenzmann-2013,Frank-Lenzman-Silvestre-2016}
and the references therein.

For Kirchhoff problems, not much is known in this respect. Recently,
Li et al. \cite{LLPWX-2017} established uniqueness and nondegeneracy
for positive solutions to Kirchhoff equations with subcritical growth.
More precisely, they proved that the following Kirchhoff equation
\begin{eqnarray*}
-\left(a+b\int_{\R^{3}}|\na u|^{2}\right)\De u+u=u^{p}, & u>0 & \text{in }\R^{3},
\end{eqnarray*}
where $1<p<5$, has a unique nondegenerate positive radial solution.
As a counterpart to this result, we have the following theorem for
Kirchhoff equations with critical Sobolev growth.

\begin{theorem} \label{thm: uniqueness and nondegeneracy} For any
positive constants $a,b>0$, there exists a unique positive solution
$u\in D$ to equation (\ref{eq: limiting Kirchhoff}) up to scalings
and translations. Moreover, $u$ is nondegenerate in the sense that
\[
\operatorname{Ker}\L_{+}={\rm span}\left\{ u_{x_{1}},u_{x_{2}},u_{x_{3}},u/2+x\cdot\na u\right\} ,
\]
where $\L_{+}:D\to D$ is defined as
\begin{equation}
\L_{+}\var=-\left(a+b\int_{\R^{3}}|\na u|^{2}\right)\De\var-2b\left(\int_{\R^{3}}\na u\cdot\na\var\right)\De u-5u^{4}\var\label{eq: linear operator}
\end{equation}
for all $\var\in D$.\end{theorem}

Remark that our result can also be viewed as an extension of the nondegeneracy
result of Ambrosetti \cite{Ambrosetti-GA-Peral-1999} for equations
(\ref{eq: Yamabe problems}) with $s=1$, since in the case $b=0$,
problem (\ref{eq: limiting Kirchhoff}) is reduced to (\ref{eq: Yamabe problems})
with $s=1$.

Our notations are standard. For simplicity, we write by $\int u$
the integral $\int_{\R^{3}}u\D x$ unless otherwise stated. We also
write $u(x)=u(|x|)$ whenever $u$ is a radially symmetric function.

\section{Proof of Theorem \ref{thm: uniqueness and nondegeneracy}}

In this section we prove Theorem \ref{thm: uniqueness and nondegeneracy}.
Let
\[
Q(x)=\frac{3^{1/4}}{(1+|x|^{2})^{1/2}}
\]
be the unique positive radial solution with $Q(0)=3^{1/4}$ that satisfies
\begin{eqnarray*}
-\De Q=Q^{5} &  & \text{in }\R^{3},
\end{eqnarray*}
 see e.g. Ambrosetti et al. \cite{Ambrosetti-GA-Peral-1999}. For
any $\la>0$ and $x_{0}\in\R^{3}$, it is direct to verify that the
functions $x\mapsto\la^{-1/2}Q((x-x_{0})/\la)$ are also solutions
to the above equation.

\subsection{Proof of uniqueness}

\begin{proof}[Proof of uniqueness] Let $u\in D$ be an arbitrary
positive solution to Eq. (\ref{eq: limiting Kirchhoff}) and let
\[
c=a+b\int|\na u|^{2}.
\]
Then the function $\bar{u}(x)=u(\sqrt{c}x)$ solves
\begin{eqnarray*}
-\De\bar{u}=\bar{u}^{5} &  & \text{in }\R^{3}.
\end{eqnarray*}
Hence, the uniqueness of $Q$ implies that $\bar{u}(x)=\la^{-1/2}Q((x-x_{0})/\la)$
for some $x_{0}\in\R^{3}$ and some $\la>0$. That is,
\[
u(x)=\la^{-1/2}Q\left(\left(\frac{x}{\sqrt{c}}-x_{0}\right)/\la\right)
\]
 This gives $\int|\na u|^{2}=\sqrt{c}\int|\na Q|^{2}.$ Hence
\[
c=a+b\sqrt{c}\int|\na Q|^{2},
\]
which yields
\[
\sqrt{c}=\frac{1}{2}\left(b\|\na Q\|_{2}^{2}+\sqrt{b^{2}\|\na Q\|_{2}^{4}+4a}\right).
\]
Hence,
\begin{equation}
u(x)=\la^{-1/2}Q\left(\left(\frac{x}{b\|\na Q\|_{2}^{2}+\sqrt{b^{2}\|\na Q\|_{2}^{4}+4a}}-x_{0}\right)/\lambda\right)\label{eq: solution expression}
\end{equation}
for some $x_{0}\in\R^{3}$ and some $\la>0$. This proves that (\ref{eq: limiting Kirchhoff})
has a unique positive energy solution up to translations and scalings.\end{proof}

We point out that $c$ depends only on $a,b$ since $Q$ is explicitly
given. In other words, $c$ is independent of the choice of the positive
solution $u$. As a consequence, we conclude that all the positive
energy solutions to (\ref{eq: limiting Kirchhoff}) is given by
\[
S=\left\{ \la^{-1/2}Q\left(\left(\frac{x}{b\|\na Q\|_{2}^{2}+\sqrt{b^{2}\|\na Q\|_{2}^{4}+4a}}-y\right)/\lambda\right):\la>0,y\in\R^{3}\right\} .
\]

\subsection{Proof of nondegeneracy}

With no loss of generality, we assume that $u(x)=u(|x|)$ is the unique
positive radial energy solution to Eq. (\ref{eq: limiting Kirchhoff})
with $\la=1$ in (\ref{eq: solution expression}). Still write $c=a+b\int|\na u|^{2}$.
Keep in mind that $c$ depends only on $a,b$. Define ${\cal A}:D\to D$
by
\[
{\cal A}\var=-c\De\var-5u^{4}\var.
\]
It is straightforward to derive from \cite[Chapter 5]{Ambrosetti-Malchiodi-Book}
that
\begin{equation}
\operatorname{Ker}{\cal A}={\rm span}\left\{ u/2+x\cdot\na u,u_{x_{1}},u_{x_{2}},u_{x_{3}}\right\} .\label{eq: known nondegeneracy results}
\end{equation}
Note that $u/2+x\cdot\na u=u/2+ru^{\prime}(r)$ with $r=|x|$ is a
radial function in $D$. For simplicity, denote $D_{\rad}=\{v\in D:v(x)=v(|x|)\}$.

To prove the nondenegeracy of $\L_{+}$, first we prove

\begin{proposition}\label{prop: radial nondengeracy}Let $\L_{+}\var=0$
and $\var\in D_{\rad}$. Then $\var=\la\left(u/2+x\cdot\na u\right)$
for some $\la\in\R$. \end{proposition}
\begin{proof}
Direct computation shows that $u/2+x\cdot\na u=u/2+ru^{\prime}(r)$
is indeed a radial solution to equation $\L_{+}\var=0$. We have to
prove that $u/2+x\cdot\na u$ is the unique radial solution to equation
$\L_{+}\var=0$ in $D_{\rad}$ up to a constant.

Let $\var\in D_{\rad}$ satisfy $\L_{+}\var=0$. It is equivalent
to
\[
{\cal A}\var=2b\left(\int\na u\cdot\na\var\right)\De u.
\]
Write $e_{0}=u/2+x\cdot\na u$ for simplicity. Since $D_{\rad}$ is
a Hilbert space, denote by $D_{0}$ the orthogonal complement of $\R e_{0}$
in $D_{\rad}$. Then $\var=\la e_{0}+\tilde{\var}$ for some $\la\in\R$
and $\tilde{\var}\in D_{0}$. By a direct computation, we find that
$\int\na u\cdot\na e_{0}=0$. This implies $u\in D_{0}$. Moreover,
note that (\ref{eq: known nondegeneracy results}) implies that ${\cal A}$
is invertible on $D_{0}$. It follows from ${\cal A}e_{0}=0$ and
$\int\na u\cdot\na e_{0}=0$ that $\tilde{\var}$ satisfies
\[
{\cal A}\tilde{\var}=2b\left(\int\na u\cdot\na\tilde{\var}\right)\De u=-\frac{2b}{c}\left(\int\na u\cdot\na\tilde{\var}\right)u^{5}.
\]

Claim that $\int\na u\cdot\na\tilde{\var}=0$. To this end, first
note that ${\cal A}u=-4u^{5}$. Thus, the above identity implies
\[
{\cal A}\left(\tilde{\var}-\frac{b}{2c}\left(\int\na u\cdot\na\tilde{\var}\right)u\right)=0.
\]
We know $u\in D_{0}$. Hence, (\ref{eq: known nondegeneracy results})
yields
\[
\tilde{\var}-\frac{b}{2c}\left(\int\na u\cdot\na\tilde{\var}\right)u=0.
\]
 This yields
\[
\na\tilde{\var}=\frac{b}{2c}\left(\int\na u\cdot\na\tilde{\var}\right)\na u.
\]
It follows
\[
\int\na u\cdot\na\tilde{\var}=\frac{b}{2c}\left(\int\na u\cdot\na\tilde{\var}\right)\int|\na u|^{2}.
\]
Since $c>b\int|\na u|^{2}$, we have $b\int|\na u|^{2}/(2c)<1/2$.
We deduce from the above equation that $\int\na u\cdot\na\tilde{\var}=0$.
This proves the claim.

Therefore, ${\cal A}\tilde{\var}=0$. Recall $\tilde{\var}\in D_{0}$.
Applying (\ref{eq: known nondegeneracy results}) gives $\tilde{\var}=0$.
Thus, we obtain $\var=\la e_{0}$. The proof is complete.
\end{proof}
Now we can prove the nondegeneracy part of Theorem \ref{thm: uniqueness and nondegeneracy}.

\begin{proof}[Proof of nondegeneracy] Since we have proved Proposition
\ref{prop: radial nondengeracy}, the rest proof for the nondegeneracy
of $\L_{+}$ is standard. We refer the readers to \cite[Chapter 5]{Ambrosetti-Malchiodi-Book}
for details. \end{proof}


TITLE: Trivial p-adic measures
QUESTION [3 upvotes]: I am looking at p-adic distributions, and in this case p-adic measures.    To say that $\mu$ is a distribution means that the arguments of $\mu$ are compact open subsets of $\mathbb{Z}_p$, $\mu$ is finitely additive, and the values $\mu$ takes are in $\mathbb{C}_p$.  To say that $\mu$ is a measure means that $\mu$ is a distribution and the values $\mu$ takes are bounded.  Let $\mu$ be a measure.
Suppose that $lim_{n\to\infty}\mu(a+p^n\mathbb{Z}_p)=0$ for all $a\in\mathbb{Z}_p$.  Does this imply that $\mu\equiv 0$?

REPLY [5 votes]: Yes. Suppose that $\mu$ of some set is nonzero. Then $\mu$ of some interval of the form $a+p^n \mathbb Z_p$ is zero. Then it must be nonzero modulo $p^k$ for some $k$. By finite additivity:
$\sum_{t=0}^{p-1} \mu(a+p^{n}t+ p^{n+1} \mathbb Z_p) = \mu(a+p^n \mathbb Z_p) \not\equiv 0$ modulo $p^k$.
So for some $t$, $\mu(a+p^{n}t+ p^{n+1} \mathbb Z_p) \not \equiv 0$ modulo $p^k$.
Let $a_1 = a+p^n t$ then repeat this process using $a_1$ and $n+1$ to produce $a_2$ , and so on. Then $a_i$ is a sequence that converges $p$-adically, so
$\mu( \lim_{i\to \infty} a_i + p^{n+j} \mathbb Z_p) = \mu( a_j + p^{n+j} \mathbb Z_p) \not \equiv 0$ modulo $p^k$.
and a sequence which is never $0$ modulo $p^k$ cannot converge to $0$ $p$-adically.


TITLE: Eigenvectors matrix multiplied by its transpose $\boldsymbol{\chi} \boldsymbol{\chi}^T $
QUESTION [0 upvotes]: Let $V$ be the set of datapoint and assume that each point can be represented by a vertex. Then, given a similarity matrix $ \mathbf{M}$, we define a graph $G = (V, \mathbf{M})$  generated using $k$-nearest neighbors.

Let $\mathbf{D}$ denote the diagonal degree matrix of $\mathbf{M}$. Then, we deﬁne the normalized weight matrix $\mathbf{W}$ using $\mathbf{D}$, so that

\begin{equation*}
 \mathbf{W}= \mathbf{D}^{-\frac{1}{2}} \mathbf{M} \mathbf{D}^{-\frac{1}{2}}.
\end{equation*}

Therefore, we define the normalized Laplacian matrix of $G$ as
\begin{equation*}
\mathbf{L} = \mathbf{I} -  \mathbf{W}= \mathbf{I}- \mathbf{D}^{-\frac{1}{2}} \mathbf{M} \mathbf{D}^{-\frac{1}{2}},
\end{equation*}
where $\mathbf{I}$ is the identity matrix.

Since the normalized Laplacian matrix $\mathbf{L}$ is a positive semi-definite matrix, the matrix $\mathbf{L}$ is decomposed into an orthogonal set of eigenvectors $\mathbf{U}=[u_1,...u_n]$ and eigenvalues $\mathbf{\Lambda}=[\lambda_1,...,\lambda_n]$ represented as follows
\begin{equation*}
    \mathbf{L}= \mathbf{U}\Lambda \mathbf{U}^{T}.
\end{equation*}

In the graph setting, the eigenvalues of $ \mathbf{L}$ can be treated as graph frequencies, and are always situated in the interval $[0, 2]$ for $\mathbf{L}$.
......................

Now, let $\mathbf{H}=\mathbf{U} h(\mathbf{\Lambda}) \mathbf{U}^{T}$ where $h(\mathbf{\Lambda})$=diag$(h(\lambda_1),...,h(\lambda_n))$.

I assume that   $h(\mathbf{\lambda})=1$ if $\lambda \leq \lambda_k$ and $h(\mathbf{\lambda})=0$ if not.

Therefore, $\mathbf{U} h(\mathbf{\Lambda}) \mathbf{U}^{T}= \boldsymbol{\chi} \boldsymbol{\chi}^T $, where $\boldsymbol{\chi} \in \mathbb{R}^{n\times k}$ contains the first $k$ eigenvectors of the normalized Laplacian $\mathbf{L}$.
My questions:

I know that $UU^{T}=\mathbf{I}$ but what about $\boldsymbol{\chi} \boldsymbol{\chi}^T $ that contains the first $k$ eigenvectors of the normalized Laplacian $\mathbf{L}$ ?

Assume that I sum up all elements of the matrix $\boldsymbol{\chi} \boldsymbol{\chi}^T $ (call it $S$). Is there any condition to ensure that the sum $S$ decreases or takes the smallest possible value?

Is there any relation between $S$, $\boldsymbol{\chi} \boldsymbol{\chi}^T $ and the  the topology of the graph?

REPLY [0 votes]: know that $UU^{T}=\mathbf{I}$ but what about $\boldsymbol{\chi} \boldsymbol{\chi}^T $ that contains the first $k$ eigenvectors of the normalized Laplacian $\mathbf{L}$ ?

Because $\chi^T\chi = I_k$, $\chi\chi^T$ is the orthogonal projection onto the span of the columns of $\chi$, i.e. the first $k$ eigenvectors of L.

Assume that I sum up all elements of the matrix $\boldsymbol{\chi} \boldsymbol{\chi}^T $ (call it $S$). Is there any condition to ensure that the sum $S$ decreases or takes the smallest possible value?

Let $\chi_k$ denote the matrix $\chi$ constructed above with $k$ columns. Let $e = (1,1,\dots,1)^T$. The sum of the elements of $\chi_k\chi_k^T$ is equal to
$$
S_k = e^T[\chi_k\chi_k^T]e = [e^T\chi_k][\chi_k^Te] = (\chi_k^Te)^T(\chi_k^Te) = \sum_{j=1}^k (e^Tv_j)^2,
$$
where $v_j$ denotes the $j$th unit eigenvector (i.e. the $j$th column of $\chi_k$). Note that $e^Tv_j$ is the sum of the entries of $v_j$. With that, it is clear that $S$ increases as $k$ increases since each term in the sum is non-negative.

Is there any relation between $S$, $\boldsymbol{\chi} \boldsymbol{\chi}^T $ and the  the topology of the graph?

There is no relation to the topology of the graph that I can see. If you're interested in a more thorough answer, then this question should probably be asked as its own, separate post.


TITLE: Weak topologies in normed spaces
QUESTION [2 upvotes]: Let $(X,\|\cdot\|)$ be a normed space over complex or real field and $\tau_X$ is the topology generated by the norm $\|\cdot\|$, i.e. just norm topology. 1) Is there a locally convex topology $\tau$ such that $\tau$ is weaker than $\tau_X$ and stronger than $\sigma(X,X^*)$, where $X^*$ is Banach adjoint? 2) If $\tau$ is any locally convex topology on $X$, then is it true that $\tau=\sigma(X,Y)$ for some subset $Y$ of $X^+$, where $X^+$ means the algebraic adjoint of $X$, i.e. the set of all linear functionals on $X$?

REPLY [3 votes]: Let's first consider the case that $X$ is finite-dimensional. Since then there is only one Hausdorff vector space topology on $X$, and $\sigma(X,X^{\ast})$ is Hausdorff, the answer to the first question is "no" if we understand "weaker" and "stronger" in the strict sense [in the non-strict sense the answer is trivially "yes"]. The answer to the second question is "yes" for finite-dimensional $X$, even if we consider also non-Hausdorff topologies (let $N$ be the $\tau$-closure of $\{0\}$, then $\tau = \sigma(X, N^{\perp})$).
If $X$ is infinite-dimensional, the answers are different. Every $\sigma(X,Y)$-neighbourhood of $0$ contains a linear subspace of finite codimension, thus no topology having a neighbourhood of $0$ that only contains subspaces of infinite codimension — like for example norm-topologies, that have neighbourhoods of $0$ containing no nontrivial subspace at all — can be a $\sigma(X,Y)$ for any subspace $Y \subset X^{+}$.
This property allows an easy construction of a topology strictly between the norm topology and the weak topology if $X$ contains infinite-dimensional closed subspaces with infinite codimension (all common spaces do): Let $Z$ be such a subspace. Put $U = B + Z$, where $B$ is the unit ball of $X$.
Then the locally convex topology generated by $\sigma(X,X^{\ast}) \cup \{U\}$ is strictly finer than $\sigma(X,X^{\ast})$ — $U$ doesn't contain a subspace of finite codimension — and strictly weaker than the norm topology — every neighbourhood of $0$ contains an infinite-dimensional subspace.
If $X$ is such that every closed subspace has either finite dimension or finite codimension (I don't know whether that's possible), this construction doesn't work.


TITLE: Just finished Spivak. How does real analysis differ?
QUESTION [3 upvotes]: I just finished working through and understanding everything in Calculus by Michael Spivak. Now I'm contemplating reading Rudin, since it's said commonly that real analysis is the next step. I just realized though, I don't really understand what Rudin has that Spivak doesn't. Calculus formally defines limits, convergence, derivatives, integrals and everything. How exactly does Rudin (and real analysis in general) differ? I previously thought they differed roughly by rigor, but Spivak has covered what I previously thought was real analysis.

REPLY [5 votes]: It sounds like you're doing this for self-study. Personally, I would recommend against baby Rudin (Principles of Mathematical Analysis) and even moreso green Rudin (Real & Complex analysis). It's not that Rudin isn't good -- he's great -- but I don't think it's a good progression. I'll suggest a few directions for you to go in.

Spivak's Calculus on Manifolds. This will require you know some basic linear algebra, and you might have to check a few definitions, but this might be an exciting continuation of your course on calculus. You would be familiar with the style of writing, and more importantly, the book is absolutely fantastic. Multivariable calculus really takes on a life of its own, and you can really feel the full force of everything you've just learned with immense applications to real-life phenomena, especially electromagnetism. To this end, Div, grad, curl and all that is an equally charming book that could whet your appetite.
Stein and Shakarchi's Princeton lectures in analysis, Vol. 1 : Fourier Analysis. The text is very readable, and most importantly, doesn't require you to know about Lebesgue integration. Moreover, Fourier analysis motivates a lot of analysis. You might get lost pretty quickly, but then you get start going through an introductory analysis textbook (like Bartle and Sherbert).
Something completely different! Analysis is just one part of mathematics. I think it's a great idea to get an early exposure to group theory. I learned from John & Margaret Maxfield's Abstract algebra and solutions by radicals. It introduces you to groups, permutations, Cayley's theorem, the basics of rings and fields and even field extensions, all in a pretty accessible short book.


TITLE: Counting conjugacy classes in simple groups of Lie type
QUESTION [11 upvotes]: Finite groups of Lie type include those obtained as rational points of a connected simple algebrraic group over a finite field $k = \mathbb{F}_q$ of characteristic $p$: these are split or quasi-split.   There are also several families obtained less directly from algebraic groups: the Suzuki groups $^2\!B_2(q)$ and the Ree groups $^2\!G_2(q),\: ^2\!F_4(q)$, where $q = p^{2n+1}$ with $n \geq 1$ is an odd power of (respectively) $2,3,2$.  [Caution: Sometimes $q$ is written here as $q^2$ to emphasize the similarity of group orders to those of untwisted groups.]
Counting the total number of conjugacy classes in such a group is a natural problem, relative to the determination of ordinary characters.  Case-by-case study has been done for many of the families, especially the exceptional types; but it's unclear how much can be expressed uniformly.   (Older results are surveyed in Chapter 8 of my 1995 AMS book on conjugacy classes.)    It helps to assume the ambient algebraic group is simply connected, in which case Steinberg proved for split and quasi-split types that the number of semisimple classes (whose elements have order prime to $p$) is $p^r$ with $r$ the Lie rank.   For Suzuki or Ree groups, $r$ is the rank of the BN-pair: 1, 1, 2.
For Suzuki groups the total number of classes is $q + 3$ (Deriziotis), while for Ree groups of type $G_2$ the number is $q+8$ (Ward).  The latter groups came up recently here.

How many conjugacy classes do the Ree groups of type $F_4$ have?

This should be computable from known data, though not easily.  Judging from all special cases I'm aware of, the answer should be given by a polynomial $q^2 + aq + b$, where $a, b \in \mathbb{Z}$ are independent of $q$.  [Note however that for most other families of groups of Lie type, there are extra complications related to isogeny type, bad primes, and such.]
The underlying question, of course, is whether one can predict a priori what the polynomials will look like for all groups (coming from simply connected algebraic groups) starting with the highest degree term $q^r$.

REPLY [8 votes]: According to

Fleischmann, Peter; Janiszczak, Ingo.
  On the computation of conjugacy classes of Chevalley groups.
  Appl. Alg. in Eng., Comm. and Comp. 1996, 7(3), 221--234

the class number of ${\rm F}_4(q)$ is

$q^4 + 2q^3 + 6q^2 + 10q + 19$ if $q = 2^n$,
$q^4 + 2q^3 + 7q^2 + 15q + 30$ if $q = 3^n$, and
$q^4 + 2q^3 + 7q^2 + 15q + 31$ if $q = p^n$ where $p > 3$

(see page 233).
According to Frank Lübeck's database on finite groups of Lie type,
the class number of $^2{\rm F}_4(q^2)$ is $q^4+4q^2+17$.
In that database you also find class numbers for many other types of finite
groups of Lie type.


TITLE: Clarification of radius pertaining to shell integration
QUESTION [0 upvotes]: When doing shell integration, I'm unsure just what the radius of a function is, and haven't found any answers in either my text or online examples.
Since $p(x)$ and $h(x)$ is the radius and height of the integral $2\pi\int_a^bp(x)h(x)dx$, and $h(x)$ is represented by the particular line formula, where does the radius come from?
Is the radius based on $f(x)$ or $f(y)$?

REPLY [1 votes]: Since your integral is with respect to $x$, and you’re using the shell method, you must be slicing your region into vertical strips and revolving those strips around a vertical axis. (Here vertical means ‘parallel to the $y$-axis’.) You have one of these strips, of ‘width’ $dx$, at each value of $x$ from $a$ to $b$. Now consider the strip at a particular $x$: the radius of the shell that it generates is the distance from $x$ to the axis of revolution.
Suppose, for example, that $a=2$ and $b=5$. If the axis of revolution is the line $x=1$, then for any $x \in [2,5]$ the distance from $x$ to the axis is $x-1$, and that is the radius of the corresponding shell. If, on the other hand, the axis is $x=7$, the distance from $x$ to the axis is $7-x$, and now that is the radius of the corresponding shell. And if the axis is $x=-7$, the distance from $x$ to the axis (and hence the radius of the shell) is $x-(-7) = x+7$.
Remember, the horizontal distance between any two points is simply the larger $x$-coordinate minus the smaller $x$-coordinate. If you were using the shell method with a horizontal axis of revolution, your slices would be horizontal (parallel to the $x$-axis), and your integral would be with respect to $y$. In that case the radius of the shell generated by the slice at a particular value of $y$ would be the vertical distance from $y$ to the axis of revolution.


TITLE: Correspondences between Borel algebras and topological spaces
QUESTION [36 upvotes]: Though tangentially related to another post on MathOverflow (here), the questions below are mainly out of curiosity. They may be very-well known ones with very well-known answers, but...

Suppose $\Sigma$ is a sigma-algebra over a set, $X$. For any given topology, $\tau$, on $X$ denote by $\mathfrak{B}_X(\tau)$ the Borel algebra over $X$ generated by $\tau$.

Question 1. Does there exist a topology, $\tau$, on $X$ such that $\Sigma = \mathfrak{B}_X(\tau)$?

If the answer to the previous question is affirmative, it makes sense to ask for the following too:

Question 2. Denote by ${\frak{T}}_X(\Sigma)$ the family of all topologies $\tau$ on $X$ such that $\Sigma = \mathfrak{B}_X(\tau)$ and let $\tau_X(\Sigma) := \bigcap_{\tau \in {\frak{T}}_X(\Sigma)} \tau$. Is $\Sigma = \mathfrak{B}_X({\frak{T}}_X(\Sigma))$?

Updates. Q2 was answered in the negative by Mike (here).

REPLY [1 votes]: EDIT: It's best just to ignore this nonsensical post.
The product space mentioned in the answer by Gerald Edgar does not provide a counterexample. The product $\sigma$-algebra is determined by countably many coordinate, the product topology by finitely many coordinates. A set determined by countably many coordiantes is a countable intersection of sets determined by finitely many coordinates.
Remark: This answer contained a flawed "proof" that the answer to the first question is yes. I've edited it.