TITLE: Algebraic varieties : How to calculate explicitly the ramification index?
QUESTION [3 upvotes]: Again a question about algebraic varieties ! Actually, I followed to book of Silverman "The Arithmetic of elliptic curve", and I have several questions about ramification index. For $\phi : C_1 \to C_2$ a non constant map of smooth curves, and $P \in C_1$, he's defining the ramification index of $\phi$ at $P$ as : $e_{\phi}(P) = \operatorname{ord}_{P}(\phi^*(t_{\phi(P)}))$ where $t_{\phi(P)}$ is a uniformizer at $\phi(P)$. Then, we have a proposition among which we have the formula : $\forall Q \in C_2 \; \operatorname{deg}(\phi) = \sum_{P \in \phi^{-1}(Q)} e_{\phi}(P)$.
My problem is : how to explicitly calculate $\operatorname{deg}(\phi)$ using this formula. I mean, the author gives then an example :
$\phi : \mathbb{P}^1 \to \mathbb{P}^1 \; [X:Y] \mapsto [X^3(X-Y)^2:Y^5]$, and he says that $\phi$ is unramified everywhere except on $[0:1]$ and $[1:1]$ where we find : $e_{\phi}([0:1]) = 3, e_{\phi}([1:1]) = 2$. So, I tried to understand this example, but I'm stuck.
Actually, I didn't first saw why it is unramified everywhere except on $[0:1]$ and $[1:1]$, so I tried to understand what happens in $[0:1]$ and $[1:1]$ for example.

If $Q=[a:b], \; b \neq 0$, we have : $\mathcal{O}_{\mathbb{P}^1, Q}= k[\frac{X}{Y}]_{(\frac{X}{Y}-\frac{a}{b})}$ with maximal ideal : $\mathcal{m}_q = (\frac{X}{Y}-\frac{a}{b})k[\frac{X}{Y}]_{(\frac{X}{Y}-\frac{a}{b})}$. So, for $Q=[0:1]$ for instance, we have :  $\mathcal{O}_{\mathbb{P}^1, Q}= k[\frac{X}{Y}]_{(\frac{X}{Y})}$ with maximal ideal : $(\frac{X}{Y})k[\frac{X}{Y}]_{(\frac{X}{Y})}$, and a uniformizer is then given by $\frac{X}{Y}$. But we have : $$ e_{\phi}([0:1]) = \operatorname{ord}_{[0:1]}(\frac{X}{Y} \circ [X^3(X-Y)^2 : Y^5]) = \operatorname{ord}_{[0:1]}(\frac{X^3}{Y^3}(X-Y)^2.\frac{1}{Y^2}) $$ which is not in the maximal ideal, but : $$ (\frac{X^3}{Y^5}(X-Y)^2)^2 = \frac{X^6}{Y^6}.\frac{(X-Y)^4}{Y^4} $$ seems to be on the ideal cause it's a quotient of polynomial of same degree of the form : $\frac{X}{Y} \times (*)$. So, we find : $e_{\phi}([0:1]) = 2$ ? So, obviously I'm wrong somewhere, or there is something I didn't understood.

And it's the same for the other one. So my first question is : where I'm wrong, and how to explictly determine the ramification index ?
My second question is : if we know that $\phi : C_1 \to C_2$ is given by $[x:y] \mapsto [f_1(x) : 1]$ for example, and we know in some point $P$ $f_1$ as a pole of order $n$ fixed, and otherwise $f_1$ as neither a pole nor a zero. Can we conclude that : $deg(\phi)= - \operatorname{ord}_P(f_1)) = n$ and the same if we replace the pole by a zero ? Put an other way : is there a link between the pole and zeros of the rational functions defining the map and the degree of the map ?
Sorry for the long post, and thank you by advance for enlighten me !

REPLY [0 votes]: We have $\phi[0,1]=\phi[1,1]=[0,1]$, so by definition
$$e_{\phi}[0,1] = ord_{[0:1]}(\phi^{*}t_{\phi[0,1]}) =  ord_{[0:1]}(\phi^{*}t_{[0,1]}) = ord_{[0:1]}(t_{[0,1]}\circ \phi)$$
Analogously
$$e_{\phi}[1,1] = ord_{[1:1]}(t_{[0,1]}\circ \phi)$$
A uniformizer in [0,1] is $t_{[0,1]}:= (x/y)$ because $ord_{[0,1]}(x/y)=1$, indeed, we will see $ord_{[0,1]}(x)=1$ and $ord_{[0,1]}(y)=0$:
First $y\in\mathcal{O}_{\mathbb{P}^1,[0,1]}$, so $ord_{[0,1]}(y)\geq 0$.
Moreover $y[0,1]=1\neq 0 \Rightarrow ord_{[0,1]}(y)\leq 0$, thus $ord_{[0,1]}(y)=0$.
Second, $x[0,1]=0 \Rightarrow  x\in\mathfrak{m}_{[0,1]} \Rightarrow (x)\subseteq \mathfrak{m}_{[0,1]}$ We have to check $(x)=\mathfrak{m}_{[0,1]}$. We will suppose the opposite and find a contradiction:
If $(x)\subsetneq \mathfrak{m}_{[0,1]}$ then we have a chain of prime ideals of length equal to 2 of the ring $\mathcal{O}_{\mathbb{P}^1,[0,1]}$, this is a contradiction because $dim(\mathcal{O}_{\mathbb{P}^1,[0,1]})=1$.
So we have
$$ord_{[0,1]}(x/y) =  ord_{[0,1]}(x) - ord_{[0,1]}(y)=1-0=1 \Rightarrow  t_{[0,1]}=x/y$$
Making a traslation we wil have
$$t_{[1,1]} = \frac{x}{y}-1$$
Now we can compute the ramification index:
$$e_{\phi}[0,1] = ord_{[0:1]}(t_{[0,1]}\circ \phi) = ord_{[0:1]}(\frac{x}{y}\circ [x^3(x-y)^2,y^5])= ord_{[0:1]}(\frac{x^3(x-y)^2}{y^5}) = ord_{[0:1]}((\frac{x}{y})^3\frac{(x-y)^2}{y^2}) = ord_{[0:1]}((\frac{x}{y})^3)+ord_{[0:1]}((\frac{x}{y}-1)^2)) = 3+0=3$$
Analogously, remembering that $t_{[1,1]} = x/y -1$:
$$e_{\phi}[0,1]) =  ord_{[1:1]}(t_{[0,1]}\circ \phi) = ord_{[0:1]}((\frac{x}{y})^3)+ord_{[0:1]}((\frac{x}{y}-1)^2) = 0+2=2$$


TITLE: Limit of function which is product of $x^x $ primitives
QUESTION [3 upvotes]: I'm trying to analytically derive the following limit: $$ \lim_{x\to \infty}\left(\frac{1}{8}\right)\left(e^{1-\frac{\sqrt{x}}{2}}\right)\left(\frac{x}{x-\frac{\sqrt{x}}{2}+1}\right)^{x-\frac{\sqrt{x}}{2}+1} $$
I've found it extremely difficult because, as you apply L'Hospital's rule, you need to differentiate $x$ times, thus getting an infinite chain of differentiations (each one invoking the product rule).
WorlframAlpha says the limit is equal to $\frac{1}{8e^{1/8}}$ which seems to match perfectly the numerical estimates I've made with my computer up to $x=2\times10^4$, but it gives no explanation for how it derives this result. How can I analytically evaluate this limit? On it's face, it seems intractable. Yet there appears to be an analytical solution.
The purpose of evaluating this limit is to calculate the following limit of a combinatorial probability:
$$\lim_{N\to\infty}\left(\frac{M^2(N!)}{2N^M(N-M+1)!}\right)$$
Where $M=\frac{N^{1/2}}{2}$.
The original limit above can be found by applying Sterling's Approximation on the second limit. If anyone is aware of any alternative techniques for analytically evaluating this second limit, that would be just as helpful.

REPLY [3 votes]: Define $f(x) = 1 - \frac{\sqrt x}{2}$, and thus your limit becomes:
\begin{align}
\lim_{x\to\infty} \frac 18e^{f(x)}\left(\frac x{x+f(x)}\right)^{x+f(x)}&=\frac 18\lim_{x\to\infty} e^{f(x)}\left(1-\frac {f(x)}{x+f(x)}\right)^{x+f(x)}\\
&=\frac 18\exp\lim_{x\to\infty}\left( f(x)+(x+f(x))\ln\left(1-\frac {f(x)}{x+f(x)}\right) \right)\end{align}
Note that $\lim_{x\to\infty}\frac {f(x)}{x+f(x)}=0$ so we can use Taylor expansion of $\ln(1+x)$ around $0$ and your limit becomes:
$$ \frac 18\exp\lim_{x\to\infty}\left[f(x) + (x+f(x))\left(  -\frac {f(x)}{x+f(x)} - \frac {f(x)^2}{2(x+f(x))^2} + O\left(\frac {f(x)^3}{(x+f(x))^3}\right)\right)\right]\\ = \frac 18\exp\lim_{x\to\infty}\left(-\frac {f(x)^2}{2(x+f(x))} + O\left(\frac {f(x)^3}{(x+f(x))^2}\right)\right) \stackrel{(*)}{=} \frac 1{8e^{1/8}}$$

$(*)$
$$\lim_{x\to\infty}\frac {f(x)^2}{2(x+f(x))} = \lim_{x\to\infty}\frac {1-\sqrt x + \frac x4}{2(x+1-\frac{\sqrt x}2)} = \frac 18$$
$$\lim_{x\to\infty}\frac {f(x)^3}{(x+f(x))^2} = \lim_{x\to\infty}\frac {x^{3/2}+\small\text{lower order terms}}{x^2 + \small\text{lower order terms}} = 0$$


TITLE: How do you factor the following cubic polynomial?
QUESTION [1 upvotes]: How do you factor the following polynomial?
$$1-x-x^2-x^3$$
So far, I've got numerical solutions, for example, I know $(0.543689 - x)$ is a factor , but I would really like to have an exact form for the factors, namely a form like the following:
$$(1-ax)(1-bx)(1-cx)$$
with $a,b,c \in \mathbb{C}$.

REPLY [2 votes]: Deflate the polynomial by setting $y:=x+\frac13$ and you get
$$y^3+\frac23y+\frac{34}{27},$$ also, with $z=3y$,
$$z^3+6z+34=0.$$
Then let $z=u^{1/3}-2u^{-1/3}$. Plugging in the equation, and after simplification,
$$u+34-\frac8u=0$$ or
$$u^2+34u-8=0.$$
The solutions of the quadratic equation are
$$\pm3\sqrt{33}-17.$$
Finally,
$$x=\frac13\left(\sqrt[3]{3\sqrt{33}-17}-\dfrac2{\sqrt[3]{3\sqrt{33}-17}}-1\right)$$ which matches the answer given by WA.


TITLE: Extension of unipotent algebraic groups
QUESTION [1 upvotes]: Let $G$ be an algebraic group with closed normal subgroup $N$. Suppose that $N$ and $G/N$ are both unipotent. Does it imply that $G$ is also unipotent?

REPLY [2 votes]: By definition, G is unipotent if and only if every nonzero representation has nonzero fixed vectors. Consider a representation V of G. As N is unipotent, $V^N$ is nonempty. Because N is normal, $V^N$ is stable under G, hence under G/N, and hence has nonzero fixed vectors (because G/N is unipotent). Therefore G is unipotent.


TITLE: Adjacency-Preserving 2D Grid Embedding
QUESTION [4 upvotes]: Consider a 2D grid, and a given planar graph $G$ with $\Delta<4$ (max node degree) and without odd cycles. What conditions should $G$ satisfy so that when it is mapped (or embedded) into the 2D grid, the adjacency of the nodes is maintained (i.e., all adjacent nodes in $G$ remain adjacent in the 2D grid). Accordingly, after embedding of $G$ in the 2D grid, the shortest path distance between adjacent nodes is still 1.
The alternative question is what is the condition for a given planar graph (with $\Delta<4$ and w/o odd cycles) to be a 2D grid?
Thanks!

REPLY [8 votes]: Embedding planar graphs (with max degree four) in an adjacency-preserving way onto a grid is NP-complete, meaning that there's unlikely to be simple necessary and sufficient conditions. Actually that's still true even for embedding trees into a grid. See:
S. Bhatt and S. Cosmodakis. The complexity of minimizing wire lengths in VLSI layouts.
Inform. Proc. Lett. 25:263–267, 1987.


TITLE: Prove that real multiplication distributes over addition
QUESTION [0 upvotes]: The distributive property of real numbers states that $“$for all $a, b, c \in \mathbb{R}$, we've $a⋅(b + c) = a⋅b + a⋅c$ and $(b + c)⋅a = b⋅a + c⋅a”$. How to prove this field property of real numbers? Is there any rigorous proof of this? Why was this property accepted as an axiom? It doesn't seem trivial to me like other axioms of real numbers. Was the multiplication between two negative real numbers or a negative real number & a positive real number defined before setting up the distributive property of real numbers as an axiomatic property of real numbers?
Basically, I want to know the proof of the following property of $\mathbb{R}$: $“(-a)⋅(-b + c) = (-a)⋅(-b) + (-a)⋅c$ and $(-b + c)⋅(-a) = (-b)⋅(-a) + c⋅(-a)$ where $-a$ and $-b$ are any negative real number and $c$ is any positive real number$”$.

REPLY [2 votes]: This depends on how you define $\mathbb{R}$ and addition. In the most formal of treatments, you start by formalizing integer arithmetic, then rational arithmetic, and then real arithmetic. To go from rationals to reals, read about Dedekind cuts--any introductory analysis book (ie, Baby Rudin) will have some information about them. A set theory book will teach you how to go from sets to defining integers/rationals and there arithmetic.


TITLE: Who has winning strategy? Modified from $X + Y$ movie
QUESTION [1 upvotes]: Recently I watch a youtube video, a scene from $X + Y$ movie, which shows a boy named Nathan solve a problem. Here is the link to the video.
Original Problem: There are 21 cards facing down put in a row. We choose one card that facing down and one card to the right then turn both card. Show that, no matter which card you choose, this move will terminate.
I found this problem interesting and wonder what if there are two people playing this, and who will have the winning strategies.
Question: With the same rule as above, two person, say $A$ and $B$ , moving alternately. Who will have winning strategy(if any)? The person start first or the second one? Or is there any other condition? The loser is the one that can no more move(no more card to choose).
more generaly,
Problem: If there are $n$ cards facing down, who has the winning strategy?
I don't know how to solve this, but I'm just curious about this one.

REPLY [4 votes]: The initial state of the game is
D D D D D D D D D D D D D D D D D D D D D

The ending state is always
U U U U U U U U U U U U U U U U U U U U D

Now for each state let's count how many face-down cards there are in even-numbered positions on the table
D D D D D D D D D D D D D D D D D D D D D
  ^   ^   ^   ^   ^   ^   ^   ^   ^   ^

This number starts at $10$ and ends at $0$. In each move exactly one of these cards will be flipped so the number changes by $\pm 1$.
Therefore there is always an even number of moves in the game, no matter which strategy anyone follows.

This analysis generalizes easily to other numbers of cards. We see that who wins depends only on how many cards there are modulo 4.


text:
\begin{document}
\title{The Volume of a Local Nodal Domain}
\author{Dan Mangoubi}
\date{}
\maketitle

\begin{abstract}
Let $M$ either be a closed real analytic Riemannian manifold or a
closed $C^\infty$-Riemannian surface. We estimate from below the
volume of a nodal domain component in an arbitrary ball, provided
that this component enters the ball deeply enough.
\end{abstract}

\section{Introduction}
\subsection{Main Results}
Let $(M, g)$ be a closed $C^\infty$-Riemannian manifold of dimension $n$.
Let $\Delta=-\DIV\circ\grad$ be the Laplace--Beltrami operator on $M$.
We consider the eigenvalue equation
\begin{equation}
  \Delta \vphi_\lambda = \lambda \vphi_\lambda
\end{equation}
For any $\lambda$-eigenfunction $\vphi_\lambda$
the null set $\{\vphi_\lambda=0\}$ is called the
$\vphi_\lambda$-\emph{nodal set}
and any connected component of the set $\{\vphi_\lambda\neq 0\}$
is called a $\lambda$-\emph{nodal domain}.

The Faber-Krahn Inequality (\cite{EgoKon}) shows that the volume of any
$\lambda$-nodal domain $\Al_\lambda$ is $\geq
C/(\sqrt{\lambda})^n$.

Donnelly and Fefferman initiated in~\cite{DonFef90} the study of a
local version of the Faber-Krahn inequality. Namely, they gave
a lower bound on the volume of \emph{local} nodal domains:
\begin{thm}[\cite{DonFef90, ChaMuc91, Lu93}]
 \label{thm:local-courant}
   Let $\vphi_\lambda$ be a $\lambda$-eigenfunction.
   Let $B$ be an arbitrary metric ball in $M$,
   and let $\Omega_\lambda$ be a connected component
   of $\{\vphi_\lambda\neq 0\}\cap B$.
   If $\Omega_\lambda\cap\Bhf\neq\emptyset$ then
     $$\frac{|\Omega_\lambda|}{|B|}\geq
     \frac{C_1}{(\sqrt{\lambda})^{\alpha(n)}(\log\lambda)^{4n}}\ ,$$
   where $\alpha(n)=4n^2+n/2$.
  \end{thm}
\noindent\textbf{Notations.} In the above theorem and throughout
the paper $rB$ denotes a concentric ball whose radius is $r$ times
that of $B$. $C_1, C_2, \ldots$ denote constants which depend only
on the metric $g$. The enumeration of constants is different in
each section.\vspace{1Ex}

Donnelly and Fefferman note that the estimates above are clearly
non-sharp and they conjecture sharp estimates. The aim of the
present paper is to give practically sharp estimates in the case
of real analytic metrics and in the case of $C^{\infty}$-surfaces.
We prove:
\begin{thm}
 \label{thm:local-courant-analytic}
   Let $(M, g)$ be a closed real analytic Riemannian manifold.
   Let $\vphi_\lambda$ be as above. Let $B\subseteq M$ be an
   arbitrary metric ball of radius $R$,
   and let $\Omega_\lambda$ be a connected component of
   $\{\vphi_\lambda\neq 0\}\cap B$. If
$\Omega_\lambda\cap\Bhf\neq\emptyset$ then
     $$\frac{|\Omega_\lambda|}{|B|}\geq
     \frac{C_2}{(\sqrt{\lambda})^{2n-2}R'(\log\lambda)^{n-1}}\ ,$$
   where $R' =\max\{R, 1/\sqrt{\lambda}\}$.
\end{thm}

In particular, for $R<1/\sqrt{\lambda}$, we have
\begin{equation}
\label{local-courant-small} |\Omega_\lambda|/|B| \geq
C_2(\sqrt{\lambda})^{-(n-1)} (\log\lambda)^{-(n-1)}\ .
\end{equation}

To better understand Theorem~\ref{thm:local-courant-analytic} consider a harmonic function
$\vphi$ defined in the ball of radius $2$, $B_2\subseteq\Rb^n$.
Assume $\vphi(0)=0$, and suppose also that its growth $\beta$ is given
by $$\beta:= \log\frac{\sup_{B_1}|\vphi|}{\sup_{B_{1/2}}|\vphi|}\ .$$
We consider the positive and negative components of $\vphi$ in the unit ball
$B_1$. Take one positive component $\Omega$ of $\{\vphi>0\}\cap B_1$.
We show that if $\Omega\cap B_{1/2}\neq\emptyset$ then
\begin{equation}
\label{harmonic-vol}
  \Vol(\Omega)\geq \frac{C}{\beta^{n-1}}\ .
\end{equation}
In fact we prove a similar estimate for a small perturbation of harmonic functions,
and use a well known scaling argument to pass to the estimate for eigenfunctions
on the wavelegth $(1/\sqrt{\lambda})$ scale.

One of our main motivation to prove the estimate~(\ref{harmonic-vol})
besides the interest raised by Donnelly and Fefferman is the following
result in~\cite{NazPolSod05}:
\begin{thm}
\label{thm:naz-pol-sod05}
Under the assumptions above in dimension $n=2$
$$\Vol(\vphi>0) \geq \frac{C}{\log \beta}\ .$$
\end{thm}
Theorem~\ref{thm:naz-pol-sod05} was proved by using complex-analysis methods,
and we hope that estimate~(\ref{harmonic-vol}) on each positive component separately
may lead to a real analysis proof of Theorem~\ref{thm:naz-pol-sod05}.

We show by a series of examples on the $n$-dimensional round
sphere $\Sb^n$, that estimate~(\ref{local-courant-small}) is sharp
up to the $(\log\lambda)^{n-1}$ factor:
\begin{thm}
Consider the standard round sphere $\Sb^n$.
For every eigenvalue $\lambda$ and $R<1/\sqrt{\lambda}$
there exists an eigenfunction $\vphi_\lambda$ on $\Sb^n$,
a nodal domain $\Al_\lambda$ and a ball $B$ of radius $R$
such that
$$\frac{|\Al_\lambda\cap B|}{|B|} \leq
\frac{C_3(n)}{(\sqrt{\lambda})^{n-1}}\ ,$$
and $\Al_\lambda\cap \Bhf\neq\emptyset$.
\end{thm}
For $R \sim 1$,
we have
\begin{equation}
\label{frm-large-balls}
|\Omega_\lambda|/|B| \geq C_2(\sqrt{\lambda})^{-(2n-2)}
(\log\lambda)^{-(n-1)} \ .
\end{equation}
An example (see Section~\ref{sec:example-tn})
on a flat torus suggests that the power $2n-2$
in~(\ref{frm-large-balls}) could be improved to $n-1$. However, we
believe that the large ball behavior should depend on the dynamics
of the manifold, and it would be interesting to prove a result
in this direction.

We also consider in this paper $C^{\infty}$-surfaces. In this case, we use the
methods from~\cite{NazPolSod05} in order to reduce the analysis to
the case of flat metrics. Then, we can apply complex analysis. We
prove
\begin{thm}
 \label{thm:local-courant-dim-2}
   Let $(\Sigma, g)$ be a closed $C^{\infty}$-Riemannian surface.
     Then
     $$\frac{|\Omega_\lambda|}{|B|}\geq
     \frac{C_4}
{\lambda R' (\log\lambda)^{1/2}}\ ,$$ where $\lambda,
\vphi_\lambda, B, R, R', \Omega_\lambda$ are as in
Theorem~\ref{thm:local-courant-analytic}.
\end{thm}
\begin{remark}
Comparing the last theorem with
Theorem~\ref{thm:local-courant-analytic} in the case $n=2$, we see
that we gain a $(\log\lambda)^{1/2}$ factor. This is due to
complex analysis methods.
\end{remark}

\vspace{1Ex}
 One can interpret
Theorem~\ref{thm:local-courant-dim-2} as an estimate on the size
of the so called ``avoided crossings'' discussed in the physics
literature (\cite{smilansky}). Roughly speaking, two nodal lines
cannot approach each other too much for a long period of time.
Theorem~\ref{thm:local-courant-dim-2} shows that two nodal lines
cannot approach each other much less than a distance of
$1/\lambda$ (interestingly, no square root here) along a line of
length $\geq C/\sqrt{\lambda}$. On the other hand, we showed
in~\cite{Man-IRND} that two adjacent nodal lines cannot stay much
closer than $C/\sqrt{\lambda}$ at all times.

\subsection{Methods of Proof}

The main tool which we use in the proof of
Theorem~\ref{thm:local-courant-analytic}
is a generalization of Hadamard's 3-circles theorem due to Nadirashvili.
This lets us eliminate difficult Carleman type estimates
which were used in~\cite{DonFef90}.
In more details we exploit the following three properties
of eigenfunctions:

\bigskip
\noindent\textbf{``Reduction'' to Harmonic Functions.}
We follow the principle that on balls of small
radius with respect to the wavelength $1/\sqrt{\lambda}$ a
$\lambda$-eigenfunction is almost harmonic.
This principle was developed in \cite{DonFef88}, \cite{DonFef90}, \cite{Nad91}
and~\cite{NazPolSod05}.
After rescaling an eigenfunction $\vphi_\lambda$
in a ball of radius $\sim 1/\sqrt{\lambda}$ to the unit ball $B_1$,
one arrives at a solution $\vphi$ of a second order
self adjoint elliptic operator $L$ in the
unit ball $B_1\subseteq\Rb^n$,
where $L$ has coefficients bounded independently of $\lambda$.
$\vphi$ is close to a harmonic function in a sense to be clarified below.
Moreover, the growth (defined below) of $\vphi$ in the unit ball is bounded
in terms of $\lambda$ (\cite{DonFef88}).

\bigskip
\noindent\textbf{Rapid Growth in Narrow Domains.}
If a harmonic function $\vphi$ vanishes on the boundary of a domain
which is long and narrow $\vphi$ must grow exponentially fast
along the direction in which the domain is long.
A corresponding property is true for eigenfunctions on
any $C^\infty$-manifold.

This was extensively developed and investigated
by Landis (\cite{Lan63}) for a certain class of solutions
of second order elliptic equations. The version we found in~\cite{Lan63}
cannot be directly applied to eigenfunctions.
A version for eigenfunctions but
with slightly weaker estimates than in the present paper
 was proved in~(\cite{DonFef90}).

In Section~\ref{sec:properties}
we formulate a sharp version of this property
which can be applied for eigenfunctions. We prove it
in Section~\ref{sec:rapid-growth-narrow-domains}.
The proof combines the ideas from~\cite{Lan63} and~\cite{DonFef90}.
We replace some arguments from~\cite{DonFef90} by more elementary ones.

\bigskip
\noindent\textbf{Nadirashvili's-Hadamard's 3-Circles Theorem.}
This is a ``propagation of smallness'' principle:
Let $\vphi$ be a harmonic function in the unit ball,
and suppose $|\vphi|\leq 1$.
If a harmonic function $\vphi$ is small on
a subset $E\subseteq B_r$, where $r<1$ and $|E|/|B_r|$ is large
then $|\vphi|$ can be estimated from above in any
concentric ball $B_R$ containing $B_r$.
When $E$ is a ball centered at $0$, this reduces to the classical
Hadamard's 3-Circles Theorem. Nadirashvili replaced the innermost
circle by an arbitrary measurable subset $E$.
When this principle is adapted to eigenfunctions,
we are restricted to consider real analytic metrics.
The sharp estimate in the generalized Hadamard Theorem is the main source
from which we get the improvement in Theorem~\ref{thm:local-courant-analytic}
relative to Theorem~\ref{thm:local-courant}.

\subsection{Organization of the paper}
In Section~\ref{sec:rescaling} we rescale the problem on balls of
small radius compared with the wavelength to a problem on the unit
ball in $\Rb^n$. Thus, we arrive to consider a problem on almost
harmonic functions whose growth is controlled. In
Section~\ref{sec:properties} we explain Rapid Growth in Narrow
Domains and Propagation of Smallness. In Section~\ref{sec:proof}
we prove Theorem~\ref{thm:local-courant-analytic}. In
Section~\ref{sec:rapid-growth-narrow-domains} we prove Rapid
Growth in Narrow Domains. In Section~\ref{sec:dimension2} we
consider the case of smooth surfaces. We give a new estimate for
harmonic functions in dimension two, and show how
Theorem~\ref{thm:local-courant-dim-2} follows.
 In Section~\ref{sec:examples} we give two examples.
The first one is a sequence of spherical harmonics which
demonstrate that Theorem~\ref{thm:local-courant-analytic} is sharp
up to a logarithmic factor for balls of radius smaller than the
wavelength. The second one is a sequence of eigenfunctions on the
standard flat torus concerning the sharp bound in
Theorem~\ref{thm:local-courant-analytic} for balls of radius $\sim
1$.

\bigskip
\noindent\textbf{Acknowledgements:} I would like to heartily thank
Kolya Nadirashvili for indicating to me that the Generalized
Hadamard Theorem may be fruitful. I owe many thanks to Leonid
Polterovich and Misha Sodin for their continuous encouragement and
helpful discussions. I am grateful to Alexander Eremenko for
removing the extra log factor in our estimate in dimension two.
Finally, I would like to thank Alexander Borichev and
Dima Jakobson for fruitful discussions.
This paper was written in the IHES, MPIM-Bonn while the author
was an EPDI fellow, and the CRM, Montreal.
The support of the EPDI, IHES, MPIM-Bonn and the CRM
is gratefully acknowledged.

\section{Passing to the Wavelength Scale}
\label{sec:rescaling}

Consider an eigenfunction $\vphi_\lambda$ on a ball $B\subseteq M$
of radius small compared to the wavelength. Namely,
consider $\vphi_\lambda$ as a function defined on a Euclidean ball of radius
$\sqrt{\eps/\lambda}$ contained in $\Rb^n$ as a coordinate neighbourhood. Now,
rescale $\vphi_\lambda$ to a function $\vphi$
on the unit ball $B_1\subseteq\Rb^n$. $\vphi$ satisfies
\begin{equation}
\label{solution}
      L\vphi=0\ ,
\end{equation}

where $L$ is a second order elliptic operator with $C^{\infty}$ coefficients.
Equation~(\ref{solution}) with real analytic coefficients will be denoted by~(\ref{solution}.RA).
$L$ is of the form
\begin{equation}
\label{L-definition}
   Lu := -\partial_i (a^{ij}\partial_j u) -\eps qu \ .
\end{equation}
$a^{ij}$ is symmetric and satisfies ellipticity bounds:
\begin{equation}
\label{ellipticity}
   \kappa_1 |\xi|^2\leq a^{ij}\xi_i\xi_j \leq \kappa_2 |\xi|^2.
\end{equation}

$a^{ij}, q$ are bounded:
\begin{equation}
\label{bounds}
  \|a^{ij}\|_{C^1(\overline{B_1})} \leq K,\ |q| \leq K\ ,
 \end{equation}
and we will assume $\eps<\eps_0$, and $\eps_0$ is small.

It is a fundamental fact due to Donnelly and Fefferman
that the eigenvalue $\lambda$
controls the growth of the eigenfunction $\vphi_\lambda$
and hence also of $\vphi$:
For a Euclidean ball $B\subseteq B_1$, we define the $r$-growth exponent as:
\begin{eqnarray}
\label{def:beta}
\beta_r(\vphi; B)
:=\log\frac{\sup_{B}|\vphi|}{\sup_{rB}|\vphi|}
 &,&\beta_r(\vphi) :=\sup_{B\subseteq B_1} \beta_r(\vphi; B)\ .
\end{eqnarray}
Donnelly and Fefferman proved~(\cite{DonFef88})
that for the rescaled eigenfunction $\vphi$ we have
\begin{equation}
\label{DF-growth} \frac{\beta_r(\vphi)}{\log(1/r)}\leq
C\sqrt{\lambda} \ .
\end{equation}

\section{Rapid Growth and Propagation of Smallness}
\label{sec:properties}

In this section we give precise formulations of the properties
mentioned in the introduction. We use them in Section~\ref{sec:proof}.

\begin{itemize}
\item \textbf{Rapid Growth in Narrow Domains}:
This property tells that if  a solution $\vphi$ of~(\ref{solution})
has a deep and narrow positivity component $\Omega$,
then $\vphi$ grows rapidly in $\Omega$.

A first version of it
is given by
\begin{thm}
\label{thm:rapid-growth-1st-version}
Let $\vphi$ satisfy~(\ref{solution}).
Let $\Omega$ be a connected component
of \mbox{$\{\vphi>0\}$} which intersects $B_{1/2}$.
Let $\eta>0$ be small enough. If
$$\frac{|\Omega|}{|B_1|} \leq \eta^{n-1} \ ,$$
then
   $$\frac{\sup_{\Omega}{\vphi}}{\sup_{\Omega\cap B_{1/2}} \vphi} \geq
     \me^{C_1/\eta}\ .$$
\end{thm}
We remark that $\eta$ can be considered as a bound on the cross sections
of $\Omega$.
We will use an iterated version of the above property:
\begin{thm}
\label{thm:rapid-growth}
Let $\vphi$ satisfy~(\ref{solution}).
Let $0<r_0\leq 1/2$. Let $\Omega$ be a connected component of
\mbox{$\{\vphi>0\}$} which intersects $B_{r_0}$.
Let $\eta>0$ be small enough. If $|\Omega\cap B_r|/|B_r| \leq \eta^{n-1}$
for all $r_0<r < 1$, then
$$\frac{\sup_{\Omega} \vphi}
{\sup_{\Omega\cap B_{r_0}} \vphi} \geq
\left(\frac{1}{r_0}\right)^{C_1/\eta}\ . $$
\end{thm}

\item \textbf{Nadirashvili's-Hadamard's 3-Circles Theorem:}

\begin{thm}[\cite{Nad76}]
\label{thm:nad-had}
Let $\vphi$ satisfy~(\ref{solution}.RA),

Let $E\subseteq B_R$, where $R<1$.
\begin{equation*}
 \mbox{If } \frac{\sup_E |\vphi|}{\sup_{B_1} |\vphi|}
\leq \left(\frac{|E|}{|B_1|}\right)^{\gamma/n},\  \mbox{then }
\frac{\sup_{B_R} |\vphi|}{\sup_{B_1}|\vphi|}
\leq (c_0 R)^{c_1\gamma},
\end{equation*}
whenever $\gamma>\gamma_0$, and where
$\gamma_0, c_0, c_1$ depend on $\kappa_1, \kappa_2, K, \eps_0, n$.
\end{thm}

When $L$ is the Euclidean Laplacian, $n=2$ and $E$ is a ball centered
at $0$ we get the classical Hadamard Theorem which says that
the maximal function $\max_{B_r} |\vphi|$ is a logarithmic convex function of
$\log r$. When $E=B_R$ and $n>2$,
this theorem was proved by Gerasimov in~\cite{Ger66}.

\begin{observation} Theorem~\ref{thm:nad-had} is equivalent to
\begin{equation}
\label{propagation}
  \sup_{B_1} |\vphi| \leq \sup_E |\vphi| \cdot
  \left(\frac{|B_1|}{|E|}\right)^{C\beta_R(\vphi; B_1)/\log(1/R)}\ ,
\end{equation}
for all $R$ small enough and where $E\subseteq B_R$.
One should compare~(\ref{propagation}) with inequality~(D)
in~\cite[p. 636]{DonFef90}.
\end{observation}

\begin{remark}
The proof in~\cite{Nad76} is for harmonic functions. When one goes
through the proof, one sees that the only point where the
harmonicity of $\vphi$ is used is an interior elliptic regularity
estimate which is true for solutions of any second order elliptic
operator with real analytic coefficients (See~\cite[Theorem
7.5.1]{Hor64}).
\end{remark}

\end{itemize}

\section{Proof of Theorem~\ref{thm:local-courant-analytic}}
\label{sec:proof}

From Section~\ref{sec:rescaling} it follows that on balls of
radius $R<\sqrt{\eps_0/\lambda}$
Theorem~\ref{thm:local-courant-analytic} is equivalent to
Theorem~\ref{thm:local-courant-analytic}' below. In section~\ref{sec:bigballs}
we complete the argument for larger balls.

Recall the definition~(\ref{def:beta}) of $\beta_r$. We set
\begin{equation*}
\beta_r'(\vphi; B) := \max\{\beta_r(\vphi; B), 3\},\quad
\beta_r'(\vphi):=\max(\beta_r(\vphi), 3)\ .
\end{equation*}
\begin{thm1.3p}
\label{thm:rescaled}
Let $\vphi$ satisfy~(\ref{solution}.RA).
Let $\Omega$ be a
 connected component of $\{\vphi\neq 0\}$. Suppose $\Omega\cap B_{1/2}\neq\emptyset$.
Then, $$\frac{|\Omega|}{|B_1|}
\geq \frac{C_1}{(\beta_{\rho_0}'(\vphi)\log\beta_{\rho_0}'(\vphi))^{n-1}}\ ,$$
where $0<\rho_0<1$ depends only on $K, \kappa_1, \kappa_2, \eps_0$ and $n$.
\end{thm1.3p}

Theorem~\ref{thm:local-courant-analytic}' immediately follows from
\begin{thm1.3pp}
Let $\vphi$ satisfy~(\ref{solution}.RA).
Suppose $\vphi(0)>0$. Let $\Omega$ be the connected component of
$\{\vphi>0\}$ which contains $0$. Then
$$\frac{|\Omega|}{|B_1|}
\geq \left(\frac{C_1\eps}
{\beta_{\rho_0}'(\vphi; B_1)}\right)^{(n-1)/(1-\eps)}$$
for all $0<\eps<1/2$, and where $0<\rho_0<1$ depends only on
$K, \kappa_1, \kappa_2, \eps_0, n$.
\end{thm1.3pp}
\begin{proof}[Theorem~\ref{thm:local-courant-analytic}''
implies Theorem~\ref{thm:local-courant-analytic}']
One may use an affine transformation in order
to move the center of the ball to $\Omega\cap B_{1/2}$.
Also, one can check that
$$\sup_{0<\eps<1/2} (\eps/A)^{1/(1-\eps)} \geq C/A\log A\ .$$
Indeed,
   with $\eps= 1/\log A$, one has
   $$\left(\frac{\eps}{A}\right)^{1/(1-\eps)} \geq
  \frac{\eps}{A}\left(\frac{\eps}{A}\right)^{2\eps} =
  \frac{\me^{2\eps\log\eps - 2}}{A\log A} \geq \frac{\me^{-3}}{A\log A}\ .$$
\end{proof}

We now give the heart of the argument which proves
Theorem~\ref{thm:local-courant-analytic}''. It is close to the argument
in~\cite{DonFef90}. The main new tool we introduce in the argument
is the propagation of smallness principle in the form of
Nadirashvili's-Hadamard's 3-circles Theorem.
\begin{proof}[Idea of Proof of Theorem~\ref{thm:local-courant-analytic}'']
If $|\vphi|\leq 1$ in the ball $B_1$
and $|\Omega|$ is very small, then by Rapid Growth in Narrow Domains
$\vphi$ should be very small on $E_{1/2}:=|\Omega\cap B_{1/2}|$.
Then by the propagation of smallness
from $E_{1/2}$ to $B_{1/2}$, we know that $|\vphi|$ must be small also
on the ball $B_{1/2}$, but if it is too small we will get
that the growth of $\vphi$ on $B_1$ is bigger than $\beta_{1/2}(\vphi; B_1)$.
The argument above has to be modified a little bit,
since we can apply propagation of smallness only if we know that
the relative size of $E_{1/2}$ in $B_{1/2}$ is large.
All one has to do is to replace $E_{1/2}$ by $E_r$ for
a suitable $r<1/2$.
\end{proof}

We now give the complete proof.
\begin{proof}[Proof of Theorem~\ref{thm:local-courant-analytic}'']
We may assume
\begin{equation}
\label{assumptioneta}
  |\Omega|/|B_1| < (\eps/\gamma_1)^{(n-1)/(1-\eps)},
\end{equation}
where $\gamma_1$ is a large enough constant.
Otherwise, the theorem becomes trivial.

Let $c_0>1$ be as in Theorem~\ref{thm:nad-had}, and let
$\rho_0 = 1/(2c_0)$.
Define
\begin{equation}
\label{r0-definition}
r_0 := \sup\left\{r:\, \frac{|\Omega\cap B_r|}{|B_r|}
\geq \frac{1}{\rho_0^n}\left(\frac{|\Omega|}{|B_1|}\right)^{1-\eps}\right\}\ .
\end{equation}
Let $E:=\Omega\cap B_{r_0}$.
\begin{claim}
$$r_0 <\rho_0 (|E|/|B_1|)^{\eps/n}\ .$$
\end{claim}
\begin{proof}[Proof of Claim]
By definition,
$$(r_0/\rho_0)^n\leq(|\Omega|/|B_1|)^{\eps}
(|E|/|\Omega|)\leq
(|\Omega|/|B_1|)^{\eps}(|E|/|\Omega|)^{\eps} .$$
\end{proof}
In particular,
\begin{equation}
\label{r0-romegacapbr0} r_0 \leq \min\{\rho_0,
(|E|/|B_1|)^{\eps/n}\}\ .
\end{equation}
Theorem~\ref{thm:rapid-growth} and
Inequality~(\ref{r0-romegacapbr0}) give together
\begin{multline}
\label{small-in-omegar0}
  \log\frac{\sup_{E} |\vphi|}{\sup_{B_1} |\vphi|}\leq \log\frac{\sup_{E} |\vphi|}{\sup_{\Omega} |\vphi|}
\leq
C_1(\log r_0)\rho_0^{n/(n-1)}/(|\Omega|/|B_1|)^{(1-\eps)/(n-1)}\leq \\
 C_2\eps(\log(|E|/|B_1|)^{1/n})/(|\Omega|/|B_1|)^{(1-\eps)/(n-1)} \ .
\end{multline}
Due to assumption~(\ref{assumptioneta}) and to $E\subseteq B_{\rho_0}$
we can apply Theorem~\ref{thm:nad-had}:
$$-\beta_{\rho_0}(\vphi; B_1)
=\log\frac{\sup_{B_{\rho_0}} |\vphi|}{\sup_{B_1} |\vphi|}
\leq {C_3\eps (\log c_0 \rho_0)/(|\Omega|/|B_1|)^{(1-\eps)/(n-1)}}\ .$$
The last inequality can be rewritten as
 $$\frac{|\Omega|}{|B_1|} \geq
\left(\frac{\eps C_4}{\beta_{\rho_0}(\vphi; B_1)}\right)^{(n-1)/(1-\eps)} \ .$$
\end{proof}

\subsection{Handling Large Balls}
\label{sec:bigballs}

So far, we proved Theorem~\ref{thm:local-courant-analytic} for small balls.
We now treat the case of balls $B$ whose radius
radius $R> \sqrt{\eps_0/\lambda}$. There are
two cases: \begin{itemize}
\item[1)] $\Omega_\lambda=\Al_\lambda\subseteq B$. By the Faber-Krahn
Inequality, we know that
  $|\Omega_\lambda| \geq C_2/(\sqrt{\lambda})^n$.
Hence,
$$\frac{|\Omega_\lambda|}{|B|} \geq
\frac{C_3}{(R\sqrt{\lambda})^n}\ ,$$ which gives the estimate in
Theorem~\ref{thm:local-courant-analytic}.

\item[2)] $\Omega_\lambda$ touches $\partial B$. We
decompose $B\setminus \Bhf$ into spherical layers, each of width
$\sqrt{\eps_0/\lambda}$. In each spherical layer we can find a
ball $B'$ of radius $(\sqrt{\eps_0/\lambda})/2$ such that
$\Omega_\lambda$ cuts $\Bhf'$. The number of such layers is
$\sim R\sqrt{\lambda/\eps_0}$. By Theorem~\ref{thm:local-courant-analytic}
for small balls the total
volume of $\Omega_\lambda$ is
$$|\Omega_\lambda|\geq \sum_{B'} |\Omega_\lambda\cap B'|\geq
C_5
R\sqrt{\lambda/\eps_0}|B'|/(\sqrt{\lambda}\,\log\lambda)^{n-1}\
.$$

The last inequality gives
$$\frac{|\Omega_\lambda|}{|B|}\geq
\frac{C_6}{(R\sqrt{\lambda})^{n-1}(\sqrt{\lambda}\,\log\lambda)^{n-1}}\ .$$
\end{itemize}
This completes the proof of Theorem~\ref{thm:local-courant-analytic}.

\section{Rapid Growth in Narrow Domains}
\label{sec:rapid-growth-narrow-domains}

In this section we prove Theorems~\ref{thm:rapid-growth-1st-version}
and~\ref{thm:rapid-growth}.
They follow from the classical growth Lemma:

Let $\vphi$ satisfy~(\ref{solution}).
Let $B^y_R=B(y, R)\subseteq B_1$.
Suppose $\vphi(y)>0$, and let
$\Omega_y$ be the connected component of $\{\vphi>0\}\cap B^y_R$
which contains $y$.
The Growth Lemma is
\begin{lem}[{\cite[Lemma~3.1]{Lan63}, \cite[p.~651]{DonFef90}}]
\label{lem:growth-lemma}
For all $A>1$ there exists $\gamma(A)>0$ such that if
$$\frac{|\Omega_y|}{|B^y_R|} \leq \gamma(A)\ ,$$
then
$$\frac{\sup_{\Omega_y} |\vphi|}
{\sup_{\Omega_y \cap B^y_{R/2}} |\vphi|} \geq A \ . $$
In particular, $\gamma$ does not depend on $R$, neither on $y$.
\end{lem}
We give a proof of this lemma in Section~\ref{sec:growth-lemma}.
As a corollary we obtain
Theorem~\ref{thm:rapid-growth-1st-version}:

\begin{proof}[Proof of Theorem~\ref{thm:rapid-growth-1st-version}]
We notice that $\Omega$ touches $\partial B_1$. Otherwise, since $\eps<\eps_0$
the maximum principle (\cite[Cor. 3.8]{GilTru83}) tells us that $\vphi$ is
identically $0$ in $\Omega$, which contradicts the definition of $\Omega$.

We decompose $B_1\setminus B_{1/2}$ into $N$ equally distanced
spherical layers, where $N$ will be chosen below.
Let $t_k = (1/2+k/(2N))$, $k=0\ldots N$.
Let $A_0 = B_{1/2}$, and  $A_k = B_{t_k}\setminus B_{t_{k-1}}$
for $k=1\ldots N$.
Set  $\wtld{A_0} = A_0\cup A_1$, $\wtld{A_N} = A_{N-1}\cup A_N$,
and
  $$\wtld{A_k} = A_{k-1}\cup A_k\cup A_{k+1}$$
for $1\leq k\leq N-1$.

There exist $\geq N/2$ values of $1 \leq k\leq N$ for which
$$|\Omega\cap \wtld{A_k}|\leq 6|\Omega|/ N\ .$$
Let $l$ be any one of these values.
Let $y\in \Omega\cap A_l$, and let $R=1/(2N)$.
Consider the ball $B^y_R=B(y, R)$.
$B^y_R\subseteq \wtld{A_l}$, and we check that
\begin{equation}
\label{growth-lemma-start}
\frac{|\Omega\cap B^y_R|}{|B^y_R|} \leq
\frac{|\Omega\cap \wtld{A_l}|}{|B_1|(1/(2N))^{n}}
\leq \frac{6|\Omega|(2N)^n}{N|B_1|}\leq 12(2\eta N)^{n-1} \ .
\end{equation}

Set $A=10\me/9$ and take
$N =\lfloor((\gamma(A)/12)^{1/(n-1)}/(2\eta))\rfloor$.
Inequality~(\ref{growth-lemma-start}) and Lemma~\ref{lem:growth-lemma}
applied in $B^y_R$ show that
$$\sup_{\Omega\cap \wtld{A_l}} \vphi \geq
\sup_{\Omega\cap B^y_{R}} \vphi \geq A\vphi(y)\ .$$
Since this is true for all $y\in \Omega\cap A_l$, we get
\begin{equation}
\label{almost-recursion}
  \sup_{\Omega\cap B_{t_{l+1}}}\vphi
\geq \sup_{\Omega\cap \wtld{A_l}} \vphi \geq
A\sup_{\Omega\cap A_l}\vphi\ .
\end{equation}

Now we apply the following maximum principle:
\begin{thm}[{\cite[Corollary~3.8]{GilTru83}}]
Let $\vphi$ satisfy~(\ref{solution}). Then
$$\sup_{\Omega\cap A_k} \vphi \geq 0.9\sup_{\Omega\cap B_{t_k}}\vphi\ ,$$
whenever $\eps<\eps_0$.
\end{thm}
Hence, from~(\ref{almost-recursion}) we obtain

 $$ \sup_{\Omega\cap B_{t_{l+1}}}\vphi
    \geq 0.9 A\sup_{\Omega\cap B_{t_l}}\vphi =
    \me \sup_{\Omega\cap B_{t_l}}\vphi\ .$$
And since this is true for $\geq N/2$ values of $k$ we finally have
$$ \frac{\sup_{\Omega} \vphi}{\sup_{\Omega\cap B_{1/2}}\vphi} \geq
\me^{N/2}
\geq \me^{C_1/\eta}\ .$$
\end{proof}

An iteration of Theorem~\ref{thm:rapid-growth-1st-version} gives
Theorem~\ref{thm:rapid-growth}:
\begin{proof}[Proof of Theorem~\ref{thm:rapid-growth}]
Let $N$ be a positive integer for which $(1/2)^{N+1}<r_0 \leq (1/2)^N$.
$N=\lfloor\log(1/r_0)/\log 2\rfloor$.
Set $t_k = (1/2)^k$. It follows by scaling from
Theorem~\ref{thm:rapid-growth-1st-version} that
$$\frac{\sup_{\Omega\cap B_{t_k}} |\vphi|}
{\sup_{\Omega\cap B_{t_{k+1}}} |\vphi|} \geq
\me^{C_1/\eta}\ ,$$
for all $0\leq k \leq N-1$.
The point is that the bounds~(\ref{ellipticity})-(\ref{bounds})
on the operator $L$ remain true
after rescaling.
Hence,
\begin{equation*}
  \frac{\sup_{\Omega} |\vphi|}{\sup_{\Omega\cap B_{r_0}} |\vphi|}
\geq
\frac{\sup_{\Omega} |\vphi|}{\sup_{\Omega\cap B_{t_N}} |\vphi|} \geq
\me^{C_1 N/\eta}
\geq \me^{C_1\log(1/r_0)/(2\eta\log 2)} =
(1/r_0)^{C_2/\eta}\ .
\end{equation*}
\end{proof}

\subsection{Proof of the Growth Lemma}
\label{sec:growth-lemma}

Its proof is based on ideas
from the proof in~\cite{DonFef90},
where we replaced several arguments by more elementary ones.

\begin{proof}[Proof of Lemma~\ref{lem:growth-lemma}]
Let $g(t)$ be a smooth function defined on $\Rb$
with the following properties
\begin{itemize}
\item $g(t) = 0$ for $t\leq 1$,
\item $g(t) = t-2$ for $t\geq 3$,
\item $g''(t) \geq 0$.
\end{itemize}
Let $\delta>0$ be small and
let $g_\delta(t) = \delta g(t/\delta)$.
Let $\vphi_\delta = (g_\delta\circ \vphi) \cdot \chi_{\Omega_y}$,
where $\chi_{\Omega_y}$ is the characteristic function of $\Omega_y$.
$\vphi_\delta$ is a smooth function in $B^y_R$ with
compact support in $\Omega_y$.
We notice that $0\leq\vphi\chi_{\Omega_y}-\vphi_\delta\leq 2\delta$.
In particular, $\vphi_\delta\to\vphi\chi_{\Omega_y}$
uniformly as $\delta\to 0$.

We now calculate $L\vphi_\delta$:
\begin{multline}
\label{almost-subharmonic}
  L\vphi_\delta = \chi_{\Omega_y}\eps q \vphi\cdot(g_\delta'\circ\vphi)
-\eps q \vphi_\delta-\chi_{\Omega_y}a^{ij}\partial_i\vphi\partial_j
\vphi\cdot(g_\delta''\circ\vphi) \\ \leq
  \chi_{\Omega_y}\eps q\vphi\cdot(g_\delta'\circ\vphi)
-\eps q \vphi_\delta\ .
\end{multline}
The last inequality is true due
to the fact that $a^{ij}$ is positive definite and
$g$ is convex. Let us denote the last expression in~(\ref{almost-subharmonic})
by $f$.
\begin{lem}
\label{lem:ln-norm}
  $$\|f\|_{L^n(B^y_R)} \leq 3\eps_0 \delta K |\Omega_y|^{1/n}$$
\end{lem}
We postpone the proof of this lemma to the end of this section.
Recall the
following local maximum principle:
\begin{thm}[{\cite[Theorem 9.20]{GilTru83}}]
Suppose $Lu\leq f$ in $B^y_R\subseteq B_1$. Then,
$$\sup_{B^y_{R/2}} u
 \leq \frac{C_1}{|B^y_R|}\int_{B^y_R} u^+ \dx + C_2\|f\|_{L^n(B^y_R)}\ ,$$
where $C_1, C_2$ depend only on $\kappa_1, \kappa_2$ and $K$.
\end{thm}
Applying the local maximum principle to $\vphi_\delta$ gives
$$\sup_{B^y_{R/2}} \vphi_\delta \leq \frac{C_1}{|B^y_R|}\int_{B^y_R}
\vphi_\delta\dx +
3C_2\delta\eps_0 K |\Omega_y|^{1/n} \ .$$
Letting $\delta\to0$ we obtain that
$$\sup_{\Omega_y\cap B^y_{R/2}}\vphi\leq
\frac{C_1}{|B^y_R|}\int_{\Omega_y} \vphi \dx
\leq \frac{C_1|\Omega_y|}{|B^y_R|} \sup_{\Omega_y} \vphi\ . $$
Thus, we may take $\gamma(A) = 1/(C_1 A)$.
\end{proof}
To complete the proof of the Growth Lemma~\ref{lem:growth-lemma}
it remains to prove Lemma~\ref{lem:ln-norm}.
\begin{proof}[Proof of Lemma~\ref{lem:ln-norm}]
We notice that $t-2\delta\leq g_\delta(t)\leq t$
and $0\leq g_\delta'(t) \leq 1$.
Hence,
$f\leq \eps q\vphi\chi_{\Omega_y}
- \eps q (\vphi-2\delta)\chi_{\Omega_y} = 2 \eps \delta q\chi_{\Omega_y}$.
When $\vphi \geq 3\delta$, $f=2\eps\delta q\chi_{\Omega_y}$,
and when $\vphi\leq 3\delta$, $f\geq -\eps q \vphi \chi_{\Omega_y}
\geq -3\eps q \delta\chi_{\Omega_y}$.
We have shown that $|f| \leq 3\eps_0 \delta q\chi_{\Omega_y}$.
Integration gives
$$\|f\|_{L^n(B^y_R)} \leq 3 \eps_0 \delta K |\Omega_y|^{1/n}.$$
\end{proof}

\section{Dimension Two - $C^{\infty}$ case}
\label{sec:dimension2}

In dimension two we can use the techniques in~\cite{NazPolSod05}
in order to reduce the analysis to the case of harmonic functions.

\subsection{An estimate for harmonic functions}
Let $\vphi$ be a harmonic function in the unit ball $B_1\subseteq\Rb^2$.
Let $\Omega$ be a connected component of $\{\vphi\neq 0\}$.
Suppose $\Omega\cap B_{1/2}\neq\emptyset$.
Theorem~\ref{thm:local-courant-analytic}' shows
$$|\Omega|/|B_1|>C/\beta_{\rho_0}'(\vphi)\log\beta_{\rho_0}'(\vphi)$$
for some $0<\rho_0<1$.
We will show that in fact,
\begin{thm}
\label{thm:dim2-harmonic}
$$|\Omega|/|B_1| \geq C/\beta_{1/2}'(\vphi; B_1)\ .$$
\end{thm}

In order to see this we define
$h=\vphi$ in $\Omega$ and $h=0$ elsewhere in $B_1$. $h$
is a subharmonic function. The key is the
following proposition:
\begin{thm}[Eremenko]
\label{thm:eremenko}
$$\beta_{3/4}(h; B_1) \leq C_1\beta_{1/2}(\vphi; B_1)+ C_2\ .$$
\end{thm}
\begin{remark}
Nadirashvili's-Hadamard's Theorem is equivalent to a related inequality with
 $C_1=C_1(\Omega)$. Namely,
\begin{equation}
\label{nad-had-equiv}
\frac{\beta_r(h; B_1)}{\log(1/(|\Omega\cap B_r|/|B_1|)^{1/n})}
\leq C_1 \frac{\beta_r(\vphi; B_1)}{\log1/(C_2 r)}\ .
\end{equation}
The important point in Theorem~\ref{thm:eremenko} is that by
allowing a slightly bigger $r$ in the nominator of the LHS of~(\ref{nad-had-equiv})
we obtain a constant
$C_1$ which is independent of $\Omega$.
\end{remark}
\begin{proof}[Proof of Theorem~\ref{thm:eremenko}]
The proof is based on the harmonic majorant principle and on
the Beurling-Nevannlina projection theorem.
For a function $u$ defined in $B_1$,
let $M_u(r):=\max_{|z|\leq r} |u|$. We normalize $\vphi$
so that $M_{\vphi}(1)=1$.
It is well known (\cite{hor-convexity})
that $M_h(r)$ is a convex function of $\log r$,
that is, $t M_h'(t)$ is a monotonically increasing function of $t$.
Thus, we have for all $1/2\leq t\leq2/3$

\begin{multline*}
  \me^{-\beta_{3/4}(h; B_1)}\geq M_h(3/4)-M_h(t) = \int_{t}^{3/4}
\frac{rM_h'(r)}{r}\,\dif r \geq \\
tM_h'(t)4(3/4-t)/3  \geq tM_h'(t) / 9 \geq M_h'(t)/18 \ .
\end{multline*}
Hence, $M_h'(t) < 20 \me^{-\beta_{3/4}(h; B_1)}$ for all $1/2\leq t\leq 2/3$.

Now, let $z_t$ be a point where $h(z_t)=M_h(t)$. Then,
$$0\leq \frac{\partial h}{\partial r}(z_t)\leq M_h'(t),\quad
\frac{\partial h}{\partial\theta}(z_t)=0\ .$$

Thus, we get that on $z_t$
\begin{equation*}
|\nabla \vphi(z_t)|\leq 20\me^{-\beta_{3/4}(h; B_1)}\ .
\end{equation*}
The function $v:= \log|\nabla \vphi|$ is subharmonic in $B_1$,
and on $z_t$,
\begin{equation}
\label{v-zt}
v(z_t)\leq -\beta_{3/4}(h; B_1) +C_1\ .
\end{equation}
Since on $|z|\leq 1$, $|\vphi|\leq 1$,
on the circle $|z| = 2/3$, $|\nabla\vphi| \leq C_2$. So,
\begin{equation}
\label{v-23}
v|_{|z|=2/3} \leq C_3\ .
\end{equation}
Now we apply the harmonic majorant principle:
Let $\omega(z, \gamma, B_{2/3})$ be the harmonic measure of
$\gamma =\{z_t : 1/2\leq t\leq 2/3\}$ with respect
to $B_{2/3}$.
$\omega$ is a harmonic function on $B_{2/3}\setminus\gamma$,
which tends to $1$ on $z_t$
and to $0$ on $|z|=2/3$.
Thus, in light of~(\ref{v-zt}) and~(\ref{v-23}), we get
\begin{equation}
\label{majorant}
v\leq (-\beta_{3/4}(h; B_1)+C_1) \omega + C_3 \leq
-\beta_{3/4}(h; B_1)\omega +C_4\ .
\end{equation}
Also, since $M_\vphi(1/2)\geq \me^{-\beta_{1/2}(\vphi; B_1)}$,
there exists a point $z_0$, $|z_0|\leq 1/2$, where
\begin{equation}
\label{gradient}
v(z_0) \geq
-\beta_{1/2}(\vphi; B_1) + C_5\ .
\end{equation}
By the Beurling-Nevannlina Projection Theorem (\cite{ahlfors-conf-inv})
\begin{equation}
\label{beurling-nevannlina}
\omega(z_0, \gamma, D) \geq \omega(-|z_0|; [1/2, 2/3], B_{2/3}) = C_6 \ .
\end{equation}

Combining (\ref{beurling-nevannlina}) with~(\ref{majorant})
and~(\ref{gradient}),
we obtain

$$-\beta_{1/2}(\vphi; B_1)+ C_5\leq v(z_0)
\leq -C_6 \beta_{3/4}(h; B_1) + C_4\ .$$
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thm:dim2-harmonic}]
By Theorem~\ref{thm:rapid-growth-1st-version}
$$\beta_{3/4}(h; B_1) \geq C/(|\Omega|/|B_1|)\ .$$
On the other hand, Theorem~\ref{thm:eremenko} tells us
$$\beta_{3/4}(h; B_1) \leq C_1\beta_{1/2}(\vphi; B_1)+C_2\ .$$
Combining these two inequalities we get
$$|\Omega|/|B_1| \geq C_1/(\beta_{1/2}(\vphi; B_1) + C_2)
\geq C_3/\beta_{1/2}'(\vphi; B_1)\ .$$
\end{proof}

\subsection{Proof of Theorem~\ref{thm:local-courant-dim-2}}

The technique developed in~\cite{NazPolSod05}
shows that one can use a quasiconformal mapping in order
to pass to estimates for harmonic functions.

From Section~\ref{sec:rescaling} and by using conformal coordinates
we see that
Theorem~\ref{thm:local-courant-dim-2} is equivalent
for small balls to

\begin{thm}
\label{thm:quasiconformal-transfer}
Let $\Delta\vphi - \eps q \vphi = 0$ in $B_1\subseteq\CC$.
Suppose $\vphi(0)>0$ and let $\Omega\subseteq B_1$ be the connected
component of $\{\vphi>0\}$ which contains $0$. Then,
$$\frac{|\Omega|}{|B_1|} \geq
\frac{C_3}{\beta_{1/2}'(\vphi)
(\log\beta_{1/2}'(\vphi))^{1/2}} \ .$$
\end{thm}
\begin{proof}
The theorem follows from Theorem~\ref{thm:dim2-harmonic}
in exactly the same way which is explained in~\cite{NazPolSod05}.
So, we omit the proof.
\end{proof}
Then, for large balls we proceed in the same way as in
Section~\ref{sec:bigballs}.

\section{Examples}
\label{sec:examples}

\subsection{An Example on $\Sb^n$}
\label{sec:example-sn}

In this section we show that Theorem~\ref{thm:local-courant-analytic}
for balls of radius $R<1/\sqrt{\lambda}$
is sharp up to the $(\log\lambda)^{n-1}$ factor.
The example we give will be a sequence of
spherical harmonics on the standard sphere $\Sb^n$.
Let us denote by $\Hl^n_k$ the
space of spherical harmonics on $\Sb^n$ of degree $k$.
\begin{prop}
\label{prop:max-nodal-domains}
There exists a sequence
 $(Y_k^n)_{k\geq 1} \in \Hl^n_k$
with the following properties:
\begin{enumerate}
\item The number of nodal domains of $Y_k^n$ is
$\geq c_{1,n} k^n$.
\item \label{prop:2} There exist $\geq c_{2,n} k^{n-1}$ nodal
domains of $Y_k^n$ which have the north pole on their boundary.
\end{enumerate}
\end{prop}

\begin{cor}
For every eigenvalue $\lambda$ and $r<1/\sqrt{\lambda}$
there exists an eigenfunction $\vphi_\lambda$,
a nodal domain $\Al_\lambda$ and a ball $B$ of radius $r$
such that
$$\frac{\Vol(\Al_\lambda\cap B)}{\Vol(B)} \leq
\frac{C_3(n)}{(\sqrt{\lambda})^{n-1}}\ ,$$
and $\Al_\lambda\cap \Bhf\neq\emptyset$.
\end{cor}
\begin{proof}
$\lambda = k(k+n-1)$ for some integer $k\geq 0$.
Let $Y^n_k$ be as in Proposition~\ref{prop:max-nodal-domains}.
Let $B$ be a ball of radius $r<1/k$ centered at the north pole.
By Proposition~\ref{prop:max-nodal-domains}
there exists a nodal domain $\Al_\lambda$
for which
$$\frac{\Vol(\Al_\lambda\cap B)}{\Vol(B)} \leq
\frac{C_4(n)}{k^{n-1}}\ .$$
The result follows since $\lambda\sim k^2$.

\end{proof}

We now prove Proposition~\ref{prop:max-nodal-domains}.
First, we introduce spherical coordinates and we
review elementary facts about spherical harmonics.
\begin{lemma}
A point on the sphere $\Sb^n$ is parametrized by
$(\theta_1, \ldots, \theta_{n-1}, \vphi)$,
where $0<\theta_l<\pi$, $0\leq \vphi\leq 2\pi$,
and
$$
\begin{array}{lcl}
x_1 &=& \cos \theta_1\ ,\\
\vdots & & \\
x_{n-1} &=& \sin\theta_1\ldots\sin\theta_{n-2}\cos\theta_{n-1} \ ,\\
x_{n} &=& \sin\theta_1\ldots\sin\theta_{n-1}\cos\vphi\ , \\
x_{n+1} &=& \sin\theta_1\ldots\sin\theta_{n-1}\sin\vphi\ .
\end{array}
$$
\end{lemma}

We recall the definition of the zonal spherical harmonics and
Legendre Polynomials. Details can be found in chapter~$3$ of~\cite{groemer}.
Consider the natural action of the orthogonal group
$O(n+1)$ on $\Sb^n$. It induces a representation of $O(n+1)$ on $\Hl_k^n$.
The \emph{zonal} spherical harmonic $Z^n_{k, p}$
of degree $k$ with pole $p\in \Sb^n$
is defined as the unique spherical harmonic in $\Hl_k^n$, which
is fixed by the stabilizer of the point $p$ in $O(n+1)$, and
admits the value~$1$ at~$p$.

The Legendre polynomial $P^{n+1}_k(t)$
is defined to be the polynomial on $[-1, 1]$, for which

$$Z^{n}_{k, p_0}(\theta_1, \ldots, \theta_{n-1}, \phi)
= P^{n+1}_k(\cos\theta_1)\ ,$$
where $p_0$ is the north pole.

It is easy to see that for any $p\in\Sb^n$
$$Z^n_{k, p}(x) = P^{n+1}_k(\langle p, x \rangle)\ .$$

\begin{lemma}[{\cite[Proposition 3.3.7]{groemer}}]
\label{lem:explicit-legendre}
$P^{n}_k$ is given by
   $$P^{n}_k(t)=\alpha_n (-1)^k (1-t^2)^{-(n-3)/2}
\frac{\partial^k}{\partial t^k} (1-t^2)^{k+\frac{n-3}{2}}\ ,$$
where $\alpha_n$ are some constants which depend on $n$.
\end{lemma}

We define also the associated Legendre functions:
 $$E^{n}_{k, j}(t) = (1-t^2)^{j/2}(\partial_t^j P^{n}_k)(t)\ .$$

The next lemma is an inductive construction of spherical harmonics:
\begin{lemma}[{\cite[Lemma~3.5.3]{groemer}}]
\label{lem:standard-basis}
  Given $G\in\Hl^{n-1}_j$, let
  $$H(\theta_1, \ldots\theta_{n-1},\vphi):=
E^{n+1}_{k, j}(\cos\theta_1) G(\theta_2, \ldots \theta_{n-1}, \vphi)\ . $$
Then, $H\in\Hl^n_k$.
\end{lemma}

\begin{proof}[Proof of Proposition~\ref{prop:max-nodal-domains}]
We prove it by induction on $n$.
For $n=1$, we take $Y_k^1(\vphi) = \sin k\vphi$.
Suppose the result is true for $n-1$.

Set
$$H^n_{k, j}(\theta_1,\ldots, \theta_{n-1},\vphi)
:= E^{n+1}_{k, j }(\cos\theta_1)
Y^{n-1}_{j}(\theta_2,\ldots, \theta_{n-1},\vphi)\ .$$

By Lemma~\ref{lem:standard-basis} $H^n_{k, j} \in \Hl^n_k$.
From Lemma~\ref{lem:explicit-legendre} one can see that
$E^{n+1}_{k, j}$ has exactly $k-j$ distinct zeroes in the interval $(-1, 1)$
it follows that
the number of nodal domains of $H^k_n$ is $\geq c_{1,n-1} (k-j) j^{n-1}$,
of which $c_{1,n-1}j^{n-1}$ touch the north pole.

We define
$Y^{n}_k := H^n_{k, \lfloor k/2\rfloor}$.

\end{proof}

\subsection{An Example on $\Tb^n$}
\label{sec:example-tn}

\begin{thm}
Consider the standard flat torus $\Tb^n$.
For every eigenvalue $\lambda$
there exists an eigenfunction $\vphi_\lambda$ on $\Tb^n$,
a nodal domain $\Al_\lambda$ and a ball $B$ of radius $\sim 1$
such that
$$\frac{\Vol(\Al_\lambda\cap B)}{\Vol(B)} \leq
\frac{C_3(n)}{(\sqrt{\lambda})^{n-1}}\ ,$$ where
$\Al_\lambda\not\subseteq B$ and
$\Al_\lambda\cap\Bhf\neq\emptyset$.
\end{thm}
\begin{proof}
Let $\Tb^n = \Rb^n/\ZZ^n$ be the standard flat torus parametrized
by the standard coordinates $(x_1, x_2, \ldots x_n)$, where $0\leq
x_j<1$. Let $k$ be an integer, and let
$\vphi_k = \Pi_{j=1}^{n-1}\sin 2\pi k x_j$. $\vphi_k$ is an
eigenfunction corresponding to the eigenvalue $\lambda_k=
4(n-1)k^2\pi^2$. Each nodal domain has cross sections in normal direction
to the $x_n$-axis of area $<c/k^{n-1}$.
Hence, if we take a ball~$B\subseteq\Tb^n$ of
radius~$1$ and we let $\Al_\lambda$ be a nodal domain which
contains the center of the ball, we have
$$\frac{\Vol(\Al_\lambda\cap B)}{\Vol(B)}\leq
   \frac{C(n)}{(\sqrt{\lambda})^{n-1}}\ .$$
\end{proof}

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config: arxiv


TITLE: How can I apply Newton's sums to solve this problem?
QUESTION [4 upvotes]: Given $x_1,x_2,x_3,x_4$ real numbers such that $x_1+x_2+x_3+x_4 = 0$ and $x_1^7+x_2^7+x_3^7+x_4^7 = 0,$ how can I use symmetric functions and Newton's sums to prove that $x_1(x_1+x_2)(x_1+x_3)(x_1+x_4)=0$?
This is what I have so far:
Note that $x_1 = -(x_2+x_3+x_4)$ and the relation between Newton's sums, elementary symmetric polynomials, and power sums gives us that $P_7-S_1 P_6+S_2 P_5-S_3 P_4+S_4P_3=0,$ where $P_k$ denotes the $kth$ power sum and $S_k$ denotes the $kth$ elementary symmetric polynomial of $x_1,x_2,x_3,x_4.$
We want to show that $(x_2+x_3+x_4)(x_2+x_3)(x_2+x_4)(x_3+x_4) = 0.$
But how?

REPLY [3 votes]: Here is another useful application of the AM-GM inequality(didn't think of it, did you)
Tricky factorization:
$$0=-(a^7+b^7+c^7+d^7)$$$$=7(b+c)(c+d)(d+a)E$$ where $E=(b^2+c^2+d^2+bc+cd+da)^2+bcd(b+c+d)$. However, $$4E=((b+c)^2+(c+d)^2+(b+d)^2)^2)-4abcd$$$$=((b+a)^2+(c+a)^2+(a+d)^2)^2)-4abcd$$$$=(3a^2+2a(b+c+d)+b^2+c^2+d^2)^2-4abcd$$ The last expression reduces to $$(a^2+b^2+c^2+d^2)^2-4abcd$$ However we see that $$(a^2+b^2+c^2+d^2)^2-4abcd \geq 12|abcd| \geq 0$$ from the AM-GM inequality. We conclude that $4E \geq 0$ with equality if and only if $|a|=|b|=|c|=|d|=0$. Otherwise one of $b+c,b+d,c+d$ must be $0$. Hence, the result follows.


TITLE: How to understand the laws of physics correctly?
QUESTION [0 upvotes]: Are the laws of physics something that exists separately from the universe or is it a description of the physical properties of the universe and objects in it?

REPLY [1 votes]: The laws that actually govern the universe are of course a part of the universe. What you are getting confused about is that most people who use the phrase "laws of physics" are referring only to their hypothesized laws that they believe or have empirically verified to be accurate approximations of the actual laws. Nobody knows what the actual laws are, but we can be very sure that some models of certain physical phenomena that we consider as a part of solid mainstream science are extremely accurate in the domain of experiments that we have performed so far, such as Einstein's theory of general relativity. It may be that the universe deviates slightly from what Einstein's theory predicts (such as at the quantum level). Even if we later come to a general consensus on a more accurate theory of gravity, it can never be provably correct.
So it is true that physicists' hypotheses about the laws of physics are more or less descriptive in nature, but that has nothing to do with the real laws (which we will never have access to).
I also want to emphasize that (contrary to another answer) we definitely cannot claim that the real laws governing the universe can be expressed as statements in modern mathematics. If you know just a little about foundations of mathematics, you will know that modern mathematics is based on ZFC but there is no physically meaningful ontological justification for ZFC, so it is unclear whether every statement over ZFC has physical meaning or not, or whether real-world phenomena can be meaningfully expressed over ZFC or not. Mathematics, after all, is a sociohistorical product of the human race, not something intrinsic to the universe.
In short, not only do we not know what the real laws of reality are, but we also do not know what foundational system may be compatible with them. In fact, it appears that there is no real-world embedding of the standard mathematical conception of natural numbers even though PA yields accurate real-world predictions. I shall leave you to draw your own (philosophical) conclusions.


TITLE: Can you compare curvatures in space, spacetime and hidden space?
QUESTION [0 upvotes]: Spacetime curvature is given by the cosmological constant, that produces a De-Sitter spacetime. It is non-zero.
But space curvature is nearly zero (how close to zero, compared to the cosmological constant?)
If we accept Kaluza-Klein derived theories (I like to believe that that encompasses string theory and M-theory as well, but one is never sure these days), then one would expect that curvatures of hidden dimensions are huge to be able to stay hidden (by the way, one can conceive high curvature manifolds that are not bounded, hence is not clear how high curvature helps hidden dimensions to stay hidden in all cases)
I'm wondering if one can think of individual dimensional curvatures as cosmological variables, and time was the last one to expand exponentially, while space dimensions where older than time (in some mathematically yet unrealised sense), and electromagnetic gauge $U(1)$ is just the higher inner dimension
Can we compare the curvature of a typical Kaluza-Klein theory compactification with the curvature of the cosmological constant? what about the curvature of the U(1) electromagnetic gauge dimension? can one express the 1/137 coupling constant of electromagnetism as a curvature?
I'm aware that there is no evidence for the electromagnetic dimension to budge or expand in spectroscopic analysis of light from the farthest observed galaxies, but I think that's still an interesting scenario to think about

REPLY [1 votes]: Can you compare curvatures in space, spacetime

Yes, you can compare spatial curvature with spacetime curvature. Have a look at this Baez article: "Similarly, in general relativity gravity is not really a 'force', but just a manifestation of the curvature of spacetime. Note: not the curvature of space, but of spacetime. The distinction is crucial". There's sometimes some confusion concerning curved spacetime and curved space. Curved spacetime is not curved space and curved time. It's a curvature of "the metric". A curvature in your plot of measurements, as it were.

and hidden space?

I don't know anything about hidden space I'm afraid.

Spacetime curvature is given by the cosmological constant, that produces a De-Sitter spacetime. It is non-zero.

Spacetime curvature usually relates to the tidal force and the second derivative of gravitational potential. Have a google on that. Then check out the De Sitter universe on Wikipedia where you can read that it "models the universe as spatially flat". There's room for confusion here. I'll try to clarify...

But space curvature is nearly zero (how close to zero, compared to the cosmological constant?)

See Einstein talking about a gravitational field as space that's "neither homogeneous nor isotropic". Then see this paper which says curved spacetime is the same thing as inhomogeneous space. Then note how the FLRW metric starts with "the assumption of homogeneity and isotropy of space". That's a reasonable assumption wherein we might claim there's no overall gravitational field in the universe, because it didn't collapse when it was small and dense. Note that spacetime is curved because the universe is expanding, but that this isn't the same as the spacetime curvature of a gravitational field. In the former situation space is inhomogeneous over time, in the latter it's inhomogeneous across space.

If we accept Kaluza-Klein derived theories (I like to believe that that encompasses string theory and M-theory as well, but one is never sure these days), then one would expect that curvatures of hidden dimensions are huge to be able to stay hidden

There's no supporting evidence for string theory or M theory or any hidden dimensions.

Can we compare the curvature of a typical Kaluza-Klein theory compactification with the curvature of the cosmological constant?

I would say no, because they're chalk and cheese.

what about the curvature of the U(1) electromagnetic gauge dimension?

I would encourage you to pursue this. Check out electromagnetic geometry and things like this by Percy Hammond: "We conclude that the field describes the curvature that characterizes the electromagnetic interaction". For an analogy, imagine you're standing on a headland overlooking a flat calm sea near an estuary. You see a single wave, and you notice that its path curves a little. That's because there's a salinity gradient in the water, which is inhomogeneous. The wave is standing in for a photon, and the inhomogeneous water is standing in for a gravitational field. Now look at the surface of the sea where the wave is. It's curved.

can one express the 1/137 coupling constant of electromagnetism as a curvature?

Not directly, because it's running constant. See NIST. At an energy corresponding to the mass of the W boson : "α(mW) is approximately 1/128 compared with its zero-energy value of approximately 1/137. Thus the famous number 1/137 is not unique or especially fundamental".


TITLE: Why is using small signal models of nonlinear systems for controller design effective?
QUESTION [0 upvotes]: my question does not contain any specific maths, but rather a general question about controller design using small signal models.
I'm familiar with the fact that it is often times hard to synthesize controllers for nonlinear systems, even when the system is described well from a mathematical viewpoint. The most common way of dealing with that problem is then to linearize the nonlinear system around an operating point (i.e. equilibrium point) and use the resulting linear model to derive a controller. This whole procedure makes sense to me - as long as the behaviour of the linearized system is approximately the same for the behaviour of that system being driven into this operating point.
As an example, imagine a simplified DC-DC converter with a constant load R and control input D (D for duty cycle of some switch in the circuit). My goal is now e.g. to drive the output of the converter from initially 0V / 0A to the equilibrium of 100V / 10mA. Say we'd need a duty cycle of D = 0.7 to reach that goal. We now need to design a controller that tries to do exactly that transition 0V / 0A -> 100V / 10mA using as little time as possible. Now, why does it make sense to linearize the plant about the operating point 100V / 10mA and use that as a basis for the controller design, when actually the resulting linear system dynamics covers the real, non-linear system dynamics only in close proximity of 100V / 10mA and nowhere near the whole transition 0V / 0A -> 100V / 10mA. I.e. the dynamics of the whole transition could be completely different from that in close to the equilibrium point 100V / 10mA. That means the controller would only work well if it e.g. should control the transition 100V / 10mA -> 110V / 10mA but behave completely different for much greater changes in equilibria.
So, how can the procedure be effective for the mentioned bigger transitions? Does one just assume that the system dynamics is approximately the same?
Greetings
David

REPLY [0 votes]: It would be very inappropriate to use a controller synthesized using a linearized model near the target operating point to drive the "plant" there when its initial output is very far away from such a target point. Following this approach the controller will soon bring the plant to its limits (in your case, D=1; in other case the plant can be broken or its protection can intervene limiting the input signal) that means it will be working in open loop. For such high output changes, you should try other approaches (for instance, bang-bang control) and put on line the small signal controller only near the point about which the plant was linearized.


TITLE: Gravitational lensing from an extended body Vs. a point mass
QUESTION [5 upvotes]: I am interested in gravitational lensing caused by a cluster of galaxies (say it has a diameter of 1 Mpc and mass of $10^{12}$ solar masses). How close must a light travel as it passes by to be notably bent - if compared to a case where there would be much smaller object in a spot where is now a center of cluster, like one isolated galaxy?
I am trying to determine angle of deflection for a special case of gravitational potential and I have no feeling for the distances at which lens still has an effect on light.

REPLY [2 votes]: The bending of light in GR is by an angle which is twice the ratio of the Schwartzschild radius of the mass divided by the distance to the light. It falls off as 1/distance, so it is a very slow falloff, and essentially if you get noticible bending at one distance, you get noticible bending at any distance.
A solar mass has a Schwarzschild radius of 3km, so 1012 solar masses is 3 x 1012 km, or in term of light-seconds 104 light-seconds, or three light-hours, 1/3000 light-year, so the deflection of light passing 105 light years out will be deflected by 2 x 10-8 turns, or about two hundredth of an arcsecond.
Many cases are worked out in this answer: How does gravitational lensing account for Einstein's Cross?


text:
\shrink
\section{Robust \PERCEPTION as Polynomial Optimization}
\label{sec:pop}
\shrink

In this paper we develop certifiable algorithms to solve~\eqref{eq:generalPerception}
 for the case when the cost $\rho$ is a  \emph{truncated least squares} (\TLS) cost:

\begin{equation} \label{eq:generalTLS}
f^\star = \min_{\vxx \in \calX} \; \; \sumallpoints \min \left\{ \frac{r^2(\vxx,\measured_i)}{\beta_i^2}, \barcsq \right\}, \tag{\TLS}
\end{equation}
where $\min \{\cdot,\cdot\}$ denotes the minimum between two scalars, $\beta_i$ is a known constant that can be
used to model the inlier standard deviation (potentially different for each measurement $i$),
and $\barc$ is the maximum admissible residual for a measurement to be considered an inlier.
Intuitively, problem~\eqref{eq:generalTLS} implements a nonlinear least squares where measurements with large residuals
 (\ie outliers) do not influence the estimate (\ie lead to a constant cost of $\barcsq$).
Problem~\eqref{eq:generalTLS}
is known to be robust to large amounts of outliers~\cite{Yang10NIPS-relaxedClipping,Maronna19book-robustStats}.
However, its global minimum

$f^\star$ is hard to compute due to the non-convexity and non-smoothness of the cost (which adds to the typical non-convexity of the
domain $\calX$).
In the following, we briefly review a few instantiations of robust \perception.

\begin{example}[Single Rotation Averaging~\cite{Hartley13ijcv}] \label{eg:singleRotationAveraging}
Given $N$ measurements of an unknown 3D rotation: $\MR_i, i=1,\dots,N$, single rotation averaging seeks to find the best \emph{average} rotation $\MR$. In this case, $\vxx = \MR \in \SOthree$, $\measured_i = \MR_i$, and the residual function can be chosen as $r(\vxx,\measured_i) = \left\| \MR - \MR_i \right\|_F$ (the {chordal distance} between two rotations~\cite{Hartley13ijcv}), where $\| \cdot \|_F$ denotes the Frobenius norm.
\end{example}

\begin{example}[Shape Alignment~\cite{Yang20ral-GNC}] \label{eg:shapeAlignment}
Given a set of 3D points $\MB_i \in \Real{3}$ and a set of 2D pixels $\vb_i \in \Real{2}$ ($i=1,\dots,N$), with putative correspondences $\vb_i \leftrightarrow \MB_i$, shape alignment seeks to find the best scale $s \in [0,\sub]$ (where $\sub$ is a given upper bound for the scale)

and 3D rotation $\MR \in \SOthree$ of the point set,
such that the 3D points project onto the corresponding pixels.

In this case, $\vxx = (\MR,s)$, $\measured_i = (\MB_i,\vb_i)$ and the residual function is the reprojection error under the weak perspective camera model: $r(\vxx,\measured_i) = \| \vb_i - s\Pi\MR\MB_i \|$, where $\Pi = [1,0,0;0,1,0] \in \Real{2\times 3}$.
\end{example}

\begin{example}[Point Cloud Registration~\cite{Yang19rss-teaser}] \label{eg:pointCloudRegistration}
Given two sets of 3D points $\va_i,\vb_i \in \Real{3}, i=1,\dots,N$, with putative correspondences $\va_i \leftrightarrow \vb_i$, point cloud registration seeks the best 3D rotation $\MR \in \SOthree$ and translation $\vt \in \Real{3}$ to align them.\footnote{\label{footnote1}For mathematical convenience, we assume the translation is bounded by a known value $T$, \ie $\| \vt \| \leq T$.}
In this case, $\vxx = (\MR, \vt)$, $\measured_i = (\va_i,\vb_i)$ and the residual function is the Euclidean distance between registered pairs of points: $r(\vxx, \measured_i) = \|  \vb_i - \MR\va_i - \vt  \|$.
\end{example}

\begin{example}[Mesh Registration~\cite{Briales17cvpr-registration}] \label{eg:meshRegistration}
Consider a 3D mesh $\{\va_i,\vu_i\}_{i=1}^N$ and a 3D point cloud with estimated normals $\{\vb_i,\vv_i\}_{i=1}^N$, where $\va_i \in \Real{3}$ is an arbitrary point on a face of the mesh, and $\vu_i$ is the unit normal of the same face, while $\vb_i \in \Real{3}$ is a 3D point and $\vv_i$ is the estimated unit normal at \edit{$\vb_i$}.
Given putative correspondences $(\va_i,\vu_i) \leftrightarrow (\vb_i,\vv_i)$,
mesh registration seeks the best 3D rotation $\MR \in \SOthree$ and translation $\vt \in \Real{3}$ to align the mesh with the point cloud.\footref{footnote1}

 In this case, $\vxx = (\MR, \vt)$, $\measured_i = (\va_i,\vu_i,\vb_i,\vv_i)$, and the residual function is the weighted sum of the point-to-plane distance and normal-to-normal distance: $r^2(\vxx,\measured_i) = \| (\MR\vu_i)\tran(\vb_i - \MR\va_i - \vt) \|^2 + w_i\| \vv_i - \MR \vu_i \|^2$, where $w_i>0$ is the relative weight between normal-to-normal distance and point-to-plane distance.
\end{example}

The following proposition states that all the four examples above lead \edit{to~\eqref{eq:generalTLS} problems} that can be cast as polynomial optimization problems (POPs).

\begin{proposition}[\PERCEPTION as POP]\label{prop:pop}
Robust \perception~\eqref{eq:generalTLS}, with residual functions as in Examples~\ref{eg:singleRotationAveraging}-\ref{eg:meshRegistration}, is equivalent to the following polynomial optimization (POP):
\bea
f^\star = \min_{\xextend \in \Real{\dimxextend}} & f(\vp)  \label{eq:pop}\\
\subject & \quad \quad h_j(\vp) = 0, j = 1,\dots,\nrEq,  \nonumber \\
& 1 \geq g_k(\vp) \geq 0, k = 1,\dots,\nrIneq, \nonumber
\eea
with $\dimxextend \doteq n + N,$ and $\xextend \doteq [\vxx\tran,\vtheta\tran]\tran \in \Real{\dimxextend}$,
where $\vxx \in \calX$ contains the to-be-estimated geometric model, and the vector of binary variables $\vtheta \in \{\pm 1 \}^N$

is such that $\theta_i = +1$ (resp. $\theta_i = -1$) when the $i$-th measurement $\measured_i$ is estimated to be an inlier (resp. outlier). In this POP, $f$ is a polynomial in $\xextend$ with $\deg{f} \leq 3$, while $h_j,g_k$ are quadratic (degree-2) polynomials
 in $\xextend$ that are used to define the domains $\calX$ and $\{\pm 1 \}^N$.
 The polynomials $f,h_j,g_k$ possess the following structural properties:
\begin{enumerate}[label=(\roman*)]
\item \label{prop:pop-objective} (objective function sparsity) $f$ can be written as a sum of $N$ polynomials $f_i,i=1,\dots,N$, and each $f_i$ is a polynomial in $\vxx$ and $\theta_i$ of degree lower or equal to 3,~\ie~$f = \sumallpoints f_i, f_i \in \polyringin{\vxx,\theta_i},\deg{f_i} \leq 3$;

\item \label{prop:pop-constraint} (constraints sparsity) let $\vh \doteq \{h_j\}_{j=1}^{l_h}$ and $\vg = \{ g_k \}_{k=1}^{l_g}$.

Then, $\vg \subset \polyringin{\vxx}$ are polynomials in $\vxx$ (\ie do not depend on $\vtheta$).
Moreover, $\vh$ can be partitioned into $N+1$ disjoint subsets: $\vh = \vh^{\theta}  \cup \vh^{x}$, with $\vh^{\theta} = \cup_{i=1}^N \vh^{\theta_i}$, where $\vh^{\theta_i} \subset \polyringin{\theta_i}$ are polynomials in $\theta_i$ (\ie do not depend on $\vxx$ and $\theta_j,\forall j \neq i$), $\vh^x \subset \polyringin{\vxx}$ are polynomials in $\vxx$ (\ie do not depend on $\vtheta$);

\item \label{prop:pop-archimedean} (Archimedeanness) the feasible set $\calxextend$ of the POP~\eqref{eq:pop} is \emph{Archimedean}.\footnote{Archimedeanness is a stronger condition than compactness, see~\cite[Definition 3.137, p.~115]{Blekherman12Book-sdpandConvexAlgebraicGeometry}.}

\end{enumerate}
\end{proposition}
\vspace{-2mm}
The \supp provides a proof of Proposition~\ref{prop:pop} and the expressions of $f,h_j,g_k$ for Examples~\ref{eg:singleRotationAveraging}-\ref{eg:meshRegistration}. Proposition~\ref{prop:pop} is based on three insights. \edit{First, each inner minimization $\min \{a,b\}$ ($a,b \in \Real{}$)
can be written as $\substack{\min \\ \theta \in \{\pm 1\}} \frac{1+\theta}{2} a + \frac{1-\theta}{2} b$,
which gives rise to
the
binary variables and leads to the objective sparsity in~\ref{prop:pop-objective}.
Second, the constraint sets of $\vxx$ and each $\theta_i$ are mutually independent, and can be described by quadratic equality and inequality constraints, leading to the constraints sparsity in~\ref{prop:pop-constraint}. Third, the unknown variables, including $\MR \in \SOthree, s \in [0,\sub], \| \vt \| \leq T$, and $\theta_i \in \{\pm 1\}$, live in compact domains described by polynomials, leading to the Archimedeanness property~\ref{prop:pop-archimedean}.}


TITLE: "Expansion" mapping on a compact
QUESTION [0 upvotes]: Let $(K,d)$ be a compact metric space. We consider $f: K -> K$ :
$$\forall x,y \in K: d(x,y) \leq d(f(x), f(y))$$
Show that $$\forall \epsilon >0  :d(f(x), f(y)) \leq d(x, y)+\epsilon$$
What I've tried to do was following:
$$d(f(x), f(y)) \leq d(f(x),x) + d(x,y) + d(f(y),y)$$
Now I need to show that $$\forall \epsilon >0  :d(f(x), x) \leq \epsilon$$
Let $D(x):=d(x,f(x))$ - compact-continuous mapping => uniformly continuous (Heine–Cantor theorem) - how can I use that though? Let $D(x)$ reach its maximum at $x_0$, so$$D(f(x_0))=d(f(f(x_0)),f(x_0)) \geq d(f(x_0),x_0)=D(x_0)=max(D(x))=>$$ $$=>D(x_0)=D(f(x_0))=...=D(f^n(x_0))=...$$
Can I somehow condlude that $$\forall \epsilon >0: D(x_0) \leq \epsilon$$?
What bothers me is that I haven't used much the fact that K is a compact - although I don't see any ways to use it. Perhaps I should consider 2 sequences $$\{x, f(x), f(f(x)),...\}, \{y, f(y), f(f(y)),...\}$$ and show that their subsequential limits are equal to $x, y$ respectively.
Thanks in advance.

REPLY [1 votes]: Your approach cannot work, because that would imply $d(f(x),x)=0$, and hence $f=\text{id}_K$, but clearly if $K=[-1,1]$  say, then $f(x)=-x$ is also an isometry. So, we need another approach. Also, in your assumptions it is not stated that $f$ is continuous, yet you have invoked continuity. I think whenever it comes to questions about contractions/expansions, it is always a good idea to consider iterations of a given point as in your last paragraph.
Here's a proof which I think is low-tech and straight forward, but a little messy because we have to repeatedly take subsequences to make everything work out. First of all we make an observation: given $x\in K$, let's define $x_n:=f^{(n)}(x)$. By compactness, there is a convergent subsequence $\{x_{n_j}\}_{j=1}^{\infty}$. By passing to another subsequence, we may as well assume $n_2-n_1<n_3-n_2<\dots$. In other words, if we define $m_j=n_{j+1}-n_j$, then $\{x_{m_j}\}_{j=1}^{\infty}$ is a subseqence of $\{x_n\}_{n=1}^{\infty}$. Now, the nice thing is for every $j\in \Bbb{N}$,
\begin{align}
d(x_{m_j},x)\leq d(f^{n_j}(x_{m_j}),f^{n_j}(x))=d(x_{n_{j+1}},x_{n_j})
\end{align}
and the RHS approaches $0$ as $j\to \infty$ ($x_{n_j}$ and $x_{n_{j+1}}$ have the same limit; another way of saying it is that $\{x_{n_j}\}_{j=1}^{\infty}$ is a Cauchy sequence).
Now, let $x,y\in K$ be arbitrary. By what we have shown above, $\{x_n=f^{(n)}(x)\}_{n=1}^{\infty}$ has a subsequence $\{x_{m_j}\}_{j=1}^{\infty}$ such that $x_{m_j}\to x$. Now, by similar reasoning, the sequence $\{y_{m_j}=f^{(m_j)}(y)\}_{j=1}^{\infty}$ also has a subsequence $\{y_{m_{j_l}}\}_{l=1}^{\infty}$ converging to $y$. Therefore, for any $l\in\Bbb{N}$, we have by the expansion property,
\begin{align}
d(f(x),f(y))\leq d(x_{m_{j_l}},y_{m_{j_l}}),
\end{align}
and the RHS approaches $d(x,y)$. Thus, we have shown $d(x,y)=d(f(x),f(y))$.
In fact, we can actually go one step further and prove $f$ is bijective (injectivity is obvious from the expansion), for surjectivity, note that the first part of our proof shows $f(K)$ is dense in $K$, but now $f$ being an isometry implies $f$ is continuous, hence $f(K)$ is compact hence closed, hence $K=\overline{f(K)}=f(K)$, which shows surjectivity.