TITLE: substitution on multiple integrals
QUESTION [0 upvotes]: Let $Q$ be the region in $\mathbb{R}^2$ enclosed by the quadrilateral with vertices $(2, 4), (6, 3), (8, 4)$, and  $(4, 8)$. Evaluate the double integral
$$\iint_Q \frac{5y-x}{y^2(y-2)^2}dxdy$$
I was trying to do a substitution but it did not work how can I tackle this question?

REPLY [1 votes]: So it turns out that you just kind of bash your head into this integral with partial fractions and bravery. Split the quadrilateral into two triangles on the line $y=4$, and call the upper one $U$ and the lower one $L$. Then your integral becomes:
$$\iint_U \frac{5y-x}{y^2(y-2)^2}dxdy ~~+~~ \iint_L \frac{5y-x}{y^2(y-2)^2}dxdy.$$
$$=\int_{4}^{8}\!\!\int_{y/2}^{12-y} \frac{5y-x}{y^2(y-2)^2}dxdy ~~+~~ \int_{3}^{4}\!\!\int_{18-4y}^{2y} \frac{5y-x}{y^2(y-2)^2}dxdy.$$
Despite the scariness of these integrands, they are actually just linear polynomials in $x$, i.e. $a+bx$ where $a$ and $b$ are allowed to depend on $y$. So the inner integration is very easy:
$$=\int_{4}^{8} -\frac{9(7y^2-64y+64)}{8y^2(y-2)^2}dy ~~+~~ \int_{3}^{4} \frac{18(2y^2-9y+9)}{y^2(y-2)^2}dy.$$
At this step we apply partial fractions. A little bit of thought or a lot algebra will convince you that the "linear-denominator" terms will vanish, and indeed
$$=\frac{9}{8}\int_{4}^{8} -\frac{16}{y^2}+\frac{9}{(y-2)^2}dy ~~+~~ 18\int_{3}^{4}\frac{9/4}{y^2}-\frac{1/4}{(y-2)^2}dy.$$
And again, we now just have four integrals, each of which is computed with the power rule.
$$=\frac{9}{8} ~~+~~ \frac{9}{8}$$
Therefore, the integral is $\frac{9}{4}$.

Having seen the solution long-hand, it's interesting to me that (1) the answer is nice, (2) because each of the two halves have nice answers, instead of there being some gross terms which cancel between the two, and (3) each half has the same nice answer.
Therefore, I wouldn't be surprised if there is a nice change of variables $(x,y)\mapsto(u,v)$ which both

simplifies the integrands somewhat and also
makes the two triangles congruent in the $(u,v)$-plane.

But, as a general principle, there are lots of changes of variables out there, and I don't think it's wise to sit around hoping that you're going to come up with some magic one for any particular integral. Better just to put in the elbow grease :)


TITLE: Vanishing right-derived functor implies right-exact
QUESTION [1 upvotes]: Let $\mathbf{f}\colon \mathbf{A}\rightarrow \mathbf{B}$ be an additive functor between Abelian categories (not necessarily left-exact), $\mathbf{A}$ with enough injectives.  Suppose that $\operatorname{RDer}^m\mathbf{f}\cong 0$ for all $m\geq 1$.  Is it necessarily the case that $\mathbf{f}$ is right-exact?  If not, what additional assumptions does one need to make?

REPLY [1 votes]: Assuming that by $\text{RDer}^n\mathbf{f}$ you mean the functor obtained by applying $\mathbf{f}$ to an injective resolution and taking degree $n$ cohomology, then the answer is no.
For example, take $\mathbf{f}$ to be a non-zero functor that vanishes on injectives, such as $\text{Ext}^1(X,-)$ for some object $X$.


TITLE: Question about integration
QUESTION [2 upvotes]: I am currently revising some calculus notes, and I got a question I hope somebody could help me with. I seen the following done various places (not in my own notes, but in other proofs)
$\int \text{some-expression-in-terms-of-y  }\frac{dy}{dx} dx = \int \text{some-expression-in-terms-of-y  } dy$
I was told never to think of $\frac{dy}{dx}$ as a fraction, but rather as a symbol. So what exactly is happening there? Its like part of the integral is evaluated, changing the integral so its now with respect to $dy$.

REPLY [3 votes]: Let denote some-expression-in-terms-of-$y $ by $f(y)$ where $y$ is a function that depend to $x$. Now since $dy=y'(x) dx=\frac{dy}{dx}dx$ then  we have
$$\int f(y) y' dx=\int f(y) dy$$


TITLE: Is an automorphism of the field of real numbers without ordering the identity map?
QUESTION [3 upvotes]: We know an automorphism on $\mathbb{R}$ must fix $\mathbb{Q}$. If we assume the usual order structure and topology on $\mathbb{R}$, then we can use the density of $\mathbb{Q}$ to show an automorphism must be the identity map, similar to the answers in this thread.
If we do not assume the usual ordering on $\mathbb{R}$, is it still true that any automorphism is the identity map?

REPLY [6 votes]: Great question - surprisingly, the ordering on $\mathbb{R}$ is definable purely algebraically! We have $$x\le y\quad\iff\quad \exists z(z^2+x=y).$$ This, in combination with the fact that any field automorphism fixes $\mathbb{Q}$ pointwise, shows that the only field automorphism of $\mathbb{R}$ is the identity.
(Note that this suggests a way to get subfields of $\mathbb{R}$ which do have nontrivial automorphisms: whip up a subfield with some crucial elements missing square roots. And this does work: e.g. in $\mathbb{Q}(\pi)$ we can get a nontrivial automorphism by swapping $\pi$ and $-\pi$, roughly because the fact that $\sqrt{\pi}\not\in\mathbb{Q}(\pi)$ prevents us from distinguishing $\pi$ and $-\pi$)

Note that this breaks down when we move to $\mathbb{C}$, since every element has a square root. Indeed, $\mathbb{C}$ has an obvious nice nontrivial automorphism (conjugation) and assuming the axiom of choice has lots of wild nontrivial automorphisms. (If we drop choice it is consistent that the only automorphisms are the identity and conjugation.)


TITLE: if $\|A - B\| > \|A - C\|$, is it then true that $\|A - \frac12(B+C)\| > \|A - C\|$
QUESTION [0 upvotes]: Given that $\|A-B\| > \|A-C\|$ I can show that $\|A - B\| > \|A - 1/2(B+C)\|$:
Using the triangle inequality:
$$1/2(\|A - B\| + \|A - C\|) \geq \|A - 1/2(B+C)\|$$
and since $\|A-B\| > \|A-C\|$, we get:
$$\|A- B\| \geq \|A - 1/2(B+C)\|$$
However I cant prove the statement $\|A - 1/2(B+C)\| \geq \|A - C\|$. Intuitively it seems true and if I draw a few vectors it is true ...
Any help would be appreciated.

REPLY [0 votes]: No, it's not true. A counter example is $A=0$, $B=2$ and $C=-1$ which results in $\frac{1}{2}\left(B+C\right)=\frac{1}{2}$, and thus, $\|A-C\|=1>\|A-\frac{1}{2}\left(B+C\right)\|=\frac{1}{2}$.


TITLE: Derivative of double integral with respect to symmetric limits in square roots
QUESTION [0 upvotes]: Can anyone help me to get the following integral?
$$\frac{d}{dx} \int_{-\sqrt{x}}^{\sqrt{x}} \int_{-x}^x f(y,z) \;dy\; dz$$
Thanks.

REPLY [0 votes]: $\displaystyle{\large{{\rm d} \over {\rm d}x}\int_{-\sqrt{x\,}}^{\sqrt{x,}}\int_{-x}^{x}
{\rm f}\left(y,z\right)\,{\rm d}y\,{\rm d}z:\ ?}$

$$
\int_{-x}^{x}{\rm f}\left(y,z\right)\,{\rm d}y
=
{\rm sgn}\left(x\right)\int_{0}^{\left\vert x\right\vert}
\sum_{\sigma = \pm}{\rm f}\left(\sigma\,y,z\right)\,{\rm d}y
$$

\begin{align}
\int_{-\sqrt{x\,}}^{\sqrt{x\,}}\int_{-x}^{x}{\rm f}\left(y,z\right)
\,{\rm d}y\,{\rm d}z
&=
\sum_{\sigma,\sigma' = \pm}\int_{0}^{\sqrt{x\,}}
{\rm sgn}\left(x\right)\int_{0}^{\left\vert x\right\vert}
{\rm f}\left(\sigma\,y,\sigma'z\right)\,{\rm d}y\,{\rm d}z
\\[3mm]&=
\sum_{\sigma,\sigma' = \pm}\int_{0}^{\infty}
\int_{0}^{\infty}{\rm sgn}\left(x\right)
\Theta\left(\left\vert x\right\vert - y\right)
\Theta\left(\sqrt{x\,} - z\right)
{\rm f}\left(\sigma\,y,\sigma'z\right)\,{\rm d}y\,{\rm d}z
\end{align}

\begin{align}
&{\partial\left[{\rm sgn}\left(x\right)
\Theta\left(\left\vert x\right\vert - y\right)
\Theta\left(\sqrt{x\,} - z\right)\right] \over \partial x}
\\[3mm]&=
2\delta\left(x\right)
\Theta\left(\left\vert x\right\vert - y\right)
\Theta\left(\sqrt{x\,} - z\right)
+
{\rm sgn}\left(x\right)
\delta\left(\left\vert x\right\vert - y\right){\rm sgn}\left(x\right)
\Theta\left(\sqrt{x\,} - z\right)
\\[3mm]&\mbox{}+
{\rm sgn}\left(x\right)
\Theta\left(\left\vert x\right\vert - y\right)
\delta\left(\sqrt{x\,} - z\right)\,{1 \over 2\,\sqrt{x\,}}
\\[3mm]&=
2\delta\left(x\right)\Theta\left(-y\right)\Theta\left(-z\right)
+
\delta\left(x - y\right)\Theta\left(y\right)\Theta\left(\sqrt{x\,} - z\right)
\\[3mm]&\mbox{}+
{\rm sgn}\left(x\right)
\Theta\left(\left\vert x\right\vert - y\right)
\delta\left(\sqrt{x\,} - z\right)\,{1 \over 2\,\sqrt{x\,}}
\end{align}

\begin{align}
&{{\rm d} \over {\rm d}x}\int_{-\sqrt{x\,}}^{\sqrt{x,}}\int_{-x}^{x}
{\rm f}\left(y,z\right)\,{\rm d}y\,{\rm d}z
=
\sum_{\sigma,\sigma' = \pm}
\int_{0}^{\sqrt{x\,}}{\rm f}\left(\sigma\,x,\sigma'z\right)\,{\rm d}z
+
{1 \over 2\,\sqrt{x\,}}\sum_{\sigma,\sigma' = \pm}\int_{0}^{\left\vert x\right\vert}
{\rm f}\left(\sigma\,y,\sigma'\sqrt{x\,}\right)\,{\rm d}y
\end{align}


TITLE: Proof $p \vdash q \Rightarrow p$
QUESTION [2 upvotes]: Firstly, how do I read it? Is below right? With braces
$$p \vdash (q \Rightarrow p)$$
The given proof is:

${p}$, premise
${q}$, assumption

$p$ by (1) // what???

${q} \Rightarrow p$ by implication introduction with 2 and 2.1
QED...

I can't really get the link between $p$ and $q$ in 2.1 ...

REPLY [4 votes]: Many/most systems of natural deduction for classical (non-relevantist) logic allow (i) reiteration, and also (ii) unrestricted discharge of assumptions -- so we are allowed to write

$p\quad\quad\quad$   Premiss
$\quad|\quad q\quad$ Supposition
$\quad|\quad p\quad$ From (1), by reiteration
$q \to p\quad\ $  Conditional Proof, by proof from (2) to (3)

There is indeed no 'link' between $p$ and $q$ at step (3). But that isn't needed at step (4), in most systems. The CP rule is: given a sub-proof starting from $A$ and concluding $B$ we can discharge the assumption $A$ and infer $A \to B$ (on the remaining assumptions/premisses). We don't, in typical classical systems, have to check that the assumption $A$ is actually invoked in getting to $B$.
Does that mean we shouldn't like reiteration and/or should restrict discharge? Well, actually that wouldn't much affect things in the presence of other standard rules. Thus consider the proof

$p\quad\quad\quad\quad$   Premiss
$\quad|\quad q\quad\quad$ Supposition
$\quad|\quad p \land q\quad$ From (1), (2)
$\quad|\quad p\quad\quad$ From (3)
$q \to p\quad\quad\ $  Conditional Proof, by proof from (2) to (4)

And now $q$ is invoked en route to getting to line (4).


TITLE: If there must be at least one person in each table, in how many ways can 6 people be seated in 3 tables?
QUESTION [4 upvotes]: If there must be at least one person in each table, in how many ways can 6 people be seated in 3 tables?

I know there are three possible ways to split the set of people P into three distinct groups ($1+1+4,1+2+3,2+2+2$). Consider only the case $2+2+2$. At first I thought that the number of ways to choose the first pair of people to sit on a table is ${6\choose2}$ but for some reason it is $\frac{1}{3}\times {6\choose2}$. Can anyone explain why this is the case?

REPLY [3 votes]: We consider the 3 cases mentioned above:
1) If they are split 4/1/1, there are $\dbinom{6}{4}$ ways to select the 4 people sitting together, and $3!$ ways to arrange them at their table, so this gives $\dbinom{6}{4}(3!)=15(6)=90$ possibilities.
2) If they are split 3/2/1, there are $\dbinom{6}{3}$ ways to select the 3 people sitting together, $2!$ ways to arrange them at their table, and $\dbinom{3}{2}$ ways to select the 2 people sitting together, so this gives $\hspace{.2 in}\dbinom{6}{3}\cdot2\cdot\dbinom{3}{2}=20(2)(3)=120$ possibilities.
3) If they are split 2/2/2, then there are $\displaystyle\frac{\binom{6}{2}\binom{4}{2}\binom{2}{2}}{3!}$ or $\;5\cdot3\cdot1=15$ ways to pair up the 6 people.
Therefore there are a total of $90+120+15=$225 possible ways to do this.

This is s(6,3), a Stirling number of the first kind.


TITLE: Ring homomorphism $\mathbb{Z}/8\mathbb{Z} \to \mathbb{Z}/5\mathbb{Z}$ implies $1=0$
QUESTION [2 upvotes]: Let $f$ be an ring homomorphism from $R_1$  to $R_2$ and define $f^*$ as the homomorphism from the group of units of $R_1$ to the group of units of $R_2$. Suppose $f^*$ is surjective, the question is to decide if this implies that $f$ is surjective.
After several attempts I got to this example. Take $R_1 =\mathbb{Z}/8\mathbb{Z}$ (so integers mod 8), $R_2 = \mathbb{Z}/5\mathbb{Z}$, and define $f(x) = x \mod 5$ for every $x \in Z/8Z$.
Now I noticed something weird:
$$1 = f(1) = f(5\cdot 5 \mod 8) = f(5)\cdot f(5) = (5 \mod 5)\cdot (5 \mod 5) = 0\cdot 0 = 0$$
What is wrong with my map? I don't get it, everything seems to be correct. I've checked whether $f$ is a ring homomorphism and it is (altough I'm not so sure anymore). I can't think of other mistake, but there has to be one of course.

REPLY [1 votes]: Notice that if $\mathcal{O}^{(*)}(a) < \infty$ then there is an homomorphism $f: \langle a\rangle \to G$ such that $f(a) = b$ if, and only if, $\mathcal{O}(b)$ divides $\mathcal {O}(a)$.
$(*)$ here $\mathcal{O}(a)$ means the order of the element $a$.
So if $G = \mathbb{Z}/5\mathbb{Z} = \langle \overline{1}\rangle$ and $\mathcal{G} = \mathbb{Z}/8\mathbb{Z} = \{\overline{1},\overline{2},\ldots,\overline{7}\}$ the only element $b$ in $\mathcal{G}$  whose order divides $\mathcal{O}(a) = 5$ is $\overline{0}$ which means that
$$\begin{align}f: \mathbb{Z}/5\mathbb{Z} &\to \mathbb{Z}/8\mathbb{Z}\\\overline{n} &\mapsto \overline{0}\end{align} $$
and the trivial homomorphism is the only homomorphism.


TITLE: How to prove that elementary row operations preserves the solution set
QUESTION [0 upvotes]: I would like to understand how to proceed, if want to prove that elementary row operations ($R_i \rightarrow R_i + \lambda R_j$, with $\lambda \in \mathbb{R}$, preserves the solution set (does not add or loose solutions).

REPLY [2 votes]: Hint: An elementary row operation corresponds to an invertible linear transformation, $R$, that's an isomorphism of the domain $V$.  It follows that they preserve linear relations between the columns.
If $c_i$ is the $i$-th column of $A$, then $Ax=b\iff x_1c_1+\dots +c_nx_n=b\iff x_1Rc_1+\dots+x_nRc_n=Rb$, where $x=(x_1,\dots,x_n)$.


text:
\section{Invariance of the positive Rabinowitz Floer homology}
The goal of this section is to show the following invariance of $RFH^+(H)$:
\begin{theorem} \label{thm:posRFHtent}
For a strongly tentacular Hamiltonian $H$ the positive Rabinowitz Floer homology $RFH^+(H)$ does not depend on the almost complex structure $J$ or on the Morse-Smale pair $(f,g)$ on $\operatorname{Crit} (\mathcal{A}^H )$. Moreover if $\{H_s\}$ is a $1$-parameter family of Hamiltonians in the affine space of compactly supported perturbations of $H$, then $RFH^+(H_s)$ is constant along $\{H_s\}$.
\end{theorem}
As seen above, the positive Rabinowitz Floer homology $RFH^+(H)$ is generated by non-constant periodic orbits and thus vanishes in the absence of closed characteristics.

Throughout this section we will denote by $(M, \omega)$ any exact symplectic manifold. In order to prove Theorem \ref{thm:posRFHtent}, we need compactness results for homotopies of Hamiltonians and almost complex structures, which are stronger than the ones proved in \cite{pasquotto2017}. To obtain these results, we recall the notion of \emph{uniform continuity of \emph{(PO)}}, as introduced in \cite{pasquotto2018}:
\begin{definition}
We say that $H$ satisfies property (PO) if
for any fixed action window, all non-degenerate periodic orbits are contained in a compact subset of $M$.
Moreover, we say that property (PO) is \emph{uniformly continuous} at $H$ if there exists an open neighbourhood $\mathcal{O}(H)$ of $0$ in $C^{\infty}_{c}(M)$ and an exhaustion of $M$ by compact sets $\{K_{n}\}_{n\in\mathbb{N}}$, such that for every $n\in \mathbb{N}$ and every $h\in \mathcal{O}(K_{n})=\mathcal{O}(H)\cap C^{\infty}_{0}(K_{n})$, whenever
\[
(v,\eta)\in \Crit(\cA^{H+h})\quad\textrm{and}\quad 0<\Big|\cA^{H+h}(v,\eta)\Big|\leq n,
\]
then $v(S^{1})\subseteq K_{n} \cap (H{+}h)^{-1}(0)$.
\end{definition}
\begin{remark}
By \cite[Lem.\ 8.4]{pasquotto2018} every strongly tentacular Hamiltonian satisfies the axiom of uniform continuity of (PO).
\end{remark}
Define: $\displaystyle\qquad \Crit^+(\cA^H) \coloneqq \Crit(\cA^H )\cap \left(\cA^H\right)^{-1}((0,+\infty)).$
\begin{remark}\label{rem:C2small}
Let $H \colon T^*\R^n\to \R$ be a Hamiltonian on $(T^*\R^n,\omega_n)$. If $(v,\eta)\in \Crit^+(\cA^H)$, then the following adaptation of \cite[Lem.\ 2.2]{laudenbach2004} to the Rabinowitz Floer framework holds:
\[
|\eta| \sup_{v(S^1)}\|DX_H\|\geq 2\pi.
\]
\end{remark}
\begin{lem}\label{lem:delta1}
Let $H$ be a Hamiltonian on an exact symplectic manifold $(M,\omega)$, such that $\Sigma \coloneqq H^{-1}(0)$ is of contact type.
If property (PO) is uniformly continuous at $H$ then there exists an open neighbourhood $\mathcal{O}(H)$ of $0$ in $C^\infty_c(M)$, such that for every compact set $K\subseteq M$ there exists a constant $\delta>0$, such that for all $h\in \mathcal{O}(H)\cap C_c^\infty(K)$
\[
\inf \left\lbrace \cA^{H+h}(v,\eta)\ \Big| \ (v,\eta)\in \Crit^+(\cA^{H+h})\right\rbrace>\delta.
\]
\end{lem}
\begin{proof}
We argue by contradiction. Let $\widetilde{\mathcal{O}}(H)$ be the open neighbourhood from property (PO). Without loss of generality, we can assume that $K=K_n$ for some $K_n$ in the exhaustion of $M$. Suppose that there exists a sequence $h_n\in \widetilde{\mathcal{O}}(H)\cap C_c^\infty(K)$ and a sequence $(v_n,\eta_n)\in \Crit^+(\cA^{H+h_n})$, such that $\lim_{n\to\infty}h_n= 0$ and
\begin{equation}\label{A->0}
\lim_{n\to\infty}\cA^{H+h_n}(v_n,\eta_n)= 0.
\end{equation}
Property (PO) is uniformly continuous at $H$, hence all the periodic orbits $v_n$ are contained in the compact set $K$. By requiring certain bounds on the derivatives for $h\in\widetilde{\mathcal{O}}(H)$, we can find an open subset $\mathcal{O}(H)\subset\widetilde{\mathcal{O}}(H)$ where the following uniform bounds are satisfied:
\begin{equation}\label{DX_(H+h)}
\sup_{h\in \mathcal{O}(H)\cap C_c^\infty(K)}\sup_{K}\|DX_{H+h}\|<+\infty.
\end{equation}
As $\Sigma$ is of contact type there exists a Liouville vector field $Y$, such that $dH(Y)>0$ on $\Sigma$. In particular, we can assume the existence of an open neighbourhood $\mathcal{U}$ of $\Sigma$ and of a constant $\delta'>0$, such that
\[
\inf_{U\cap K} dH(Y)>\delta'>0.
\]
By possibly shrinking $\mathcal{O}(H)$, we can assume that for all $h\in \mathcal{O}(H)\cap C_c^\infty(K)$ we have $(H{+}h)^{-1}(0)\cap K \,\subseteq\, U\cap K$ and
\[
\inf_{U\cap K} d(H{+}h)(Y)>\frac{\delta'}{2}.
\]
Thus we obtain $\displaystyle \qquad\cA^{H+h_n}(v_n,\eta_n)=\eta_n\int_{S^1} d_{v_n}(H{+}h_n)(Y)>\frac{\eta_n \delta'}{2}$.\medskip\\
With (\ref{A->0}), we conclude that $\lim_{n\to \infty}\eta_n=0$. On the other hand, as $(v_n,\eta_n)\in \Crit^+(\cA^{H+h_n})$, we get uniform bounds on the derivative of $v_n$
\[
v_n(t)\in K, \qquad \|\partial_tv_n(t)\|\leq |\eta_n|\sup_{K}\|X_{H+h_n}\|\quad \forall\ n\in \N,\ t\in S^1.
\]
This allows us to use the Arzel{\'a}-Ascoli theorem, which yields a convergent subsequence (which we denote the same)
\[
\lim_{n\to\infty}(v_n,\eta_n)=(x,0)\in \Sigma{\times}\{0\}.
\]
Let $x\in\mathcal{V} \subseteq M,\; \varphi \,\colon \mathcal{V}\to\R^{2m}$ be a coordinate chart around $x$. For $n$ big enough we can assume $v_n(S^1)\subseteq \mathcal{V}$. Then $x_n \coloneqq \varphi \circ v_n$ satisfies the respective Hamiltonian equation $\partial_tx_n=\eta_n D\varphi^{-1}(X_{H+h_n})(x_n)$. Hence $\partial_{tt}x_n=\eta_n D\big(D\varphi^{-1}(X_{H+h_n})\big)(\partial_tx_n)$. However, from Remark \ref{rem:C2small} we can infer that
\[
\eta_n \big\|D\left(D\varphi^{-1}(X_{H+h_n})\right)\big\|\geq 2\pi,
\]
which together with \eqref{DX_(H+h)} contradicts $\lim_{n\to \infty}\eta_n= 0$.
\end{proof}
Fix a strongly tentacular Hamiltonian $H \colon T^*\mathbb{R}^n\to \mathbb{R}$ and let $\mathcal{O}(H)$ be an open neighbourhood of $0$ in $C_c^\infty(T^*\R^n)$, such that all Hamiltonians from $H+\mathcal{O}(H)$ are strongly tentacular and $\mathcal{O}(H)$ is as in Lemma \ref{lem:delta1}. Fix sets $K\subseteq \mathcal{V}\subseteq T^*\R^n$, such that $K\neq \emptyset$ is compact and $\mathcal{V}$ is open and precompact. Let $\delta>0$ be a constant as in Lemma \ref{lem:delta1}; let $\y\in (0,\delta)$ and let $0<\varepsilon_{0},\, \tilde{c}<\infty$ be constants as in \cite[Lem.\ 2.1]{pasquotto2017} depending only on $H$.

Finally, fix $h_\pm\in \mathcal{O}(H)\cap C^\infty_c(K)$ and let
$$
\mathbb{R}\ni s\longmapsto (H_s,J_s)\in \big(H+\mathcal{O}(H)\big)\times \mathcal{J}(\mathcal{V}{\times}(\R \setminus [-\y,\y]), \mathbb{J}),
$$
 be a smooth homotopy of Hamiltonians and $\omega$-compatible almost complex structures, constant outside of $[0,1]$, such that $H_s=H+h_-$ for $s\leq 0$ and $H_s=H+h_+$ for $s\geq 1$. In this setting we formulate the following lemma:

\begin{lem}\label{lem:delta2}
Let $(v^\pm,\eta^\pm)$ be a pair of critical points of $\cA^{H+h_\pm}$.
If the homotopy $\{H_s,J_s\}_{s\in\R}$ satisfies
\begin{equation}
\|\partial_{s}H_{s}\|_{L^{\infty}}<\min \left\lbrace \frac{\varepsilon_{0}}{8\big(\tilde{c}\varepsilon_0+\|J\|_{L^{\infty}}^{3/2}\big)},\frac{7\delta}{16\tilde{c}(1+\delta)}\right\rbrace,
\label{Hs}
\end{equation}
and if $u$ is a solution to the equation $\partial_s u=\nabla_{J_s}\cA^{H_s}(u)$ with $\displaystyle\lim_{s\to\pm\infty}u(s)=(v^\pm,\eta^\pm)$, then $\cA^{H+h_-}(v^-,\eta^-)>\delta$ implies
$\cA^{H+h_+}(v^+,\eta^+)>\delta$.
\end{lem}
\begin{proof}
The following proof is an adjustment of the proof of \cite[Cor.\ 3.8]{CieliebakFrauenfelder2009} to the setting of strongly tentacular Hamiltonians and is based on the results proven in \cite[Prop.\ 3.3]{pasquotto2017}.
Abbreviate
\[
a \coloneqq  \cA^{H+h_-}(v^-,\eta^-),\qquad b \coloneqq \cA^{H+h_+}(v^+,\eta^+).
\]
By assumption $a\geq \delta$. Let $u=(v,\eta)\in C^{\infty}\big(\mathbb{R}, C^{\infty}(S^{1}, T^*\R^n) {\times} \mathbb{R}\big)$ be a solution to the equation $\partial_s u=\nabla_{J_s}\cA^{H_s}(u)$ with $\lim_{s\to\pm\infty}u(s)=(v^\pm,\eta^\pm)$. Then our setting satisfies the assumptions of \cite[Prop. 3.3]{pasquotto2017}.

First assume additionally $|b|\leq a$. Then $b-a\leq 0$ and \cite[Prop.\ 3.3, eq.\ (3.7)]{pasquotto2017} gives
\begin{align}
\|\eta\|_{L^{\infty}(\mathbb{R})} & \leq \frac{8}{7} \left(\tilde{c}\Big(\max\{|a|, |b|\}+1\Big)+\frac{b-a}{\varepsilon_{0}}\|J\|_{L^{\infty}}^{\frac{3}{2}}\right)\nonumber\\
& \leq \frac{8}{7} \tilde{c}(a+1)\leq a \tilde{c}\frac{8}{7}\left(1+\frac{1}{\delta}\right)\label{eta}
\end{align}
On the other hand, by equation (3.5) from the proof of \cite[Prop. 3.3]{pasquotto2017}, we have that
\begin{equation}
\|\partial_{s}u\|_{L^{2}(\mathbb{R}\times S^{1})}^{2} \leq \|J\|_{L^{\infty}} (b-a + \|\eta\|_{L^{\infty}}\|\partial_{s}H_{s}\|_{L^{\infty}}).
\label{partialU}
\end{equation}
Combined with \eqref{eta} and \eqref{Hs}, we obtain the following estimate:
\[
b\geq a - \|\eta\|_{L^{\infty}(\mathbb{R})}\|\partial_{s}H_{s}\|_{L^{\infty}}\geq a \left(1- \tilde{c}\frac{8}{7}\Big(1+\frac{1}{\delta}\Big)\|\partial_{s}H_{s}\|_{L^{\infty}}\right)\geq \frac{1}{2}a.
\]
In particular, $\cA^{H+h_-}(v^-,\eta^-)=b>0$. By assumption $h_-\in \mathcal{O}(H)\cap C_c^\infty(K)$, hence by Lemma \ref{lem:delta1} we can conclude that $\cA^{H+h_-}(v^-,\eta^-)\geq \delta$. This implies the result provided $|b|\leq a$.

Now assume $b<-a$. Then $b-a\leq 0$ and by \cite[Prop.\ 3.3, eq.\ (3.7)]{pasquotto2017} we have
\begin{align}
\|\eta\|_{L^{\infty}(\mathbb{R})} & \leq \frac{8}{7} \left(\tilde{c}\Big(\max\{|a|, |b|\}+1\Big)+\frac{b-a}{\varepsilon_{0}}\|J\|_{L^{\infty}}^{\frac{3}{2}}\right)\nonumber\\
& \leq \frac{8}{7} \tilde{c}(1-b) \leq -b \tilde{c}\frac{8}{7}\left(1+\frac{1}{\delta}\right),
\end{align}
where the last inequality comes from the assumption $-b\geq a\geq \delta$. Combining it with \eqref{partialU} and \eqref{Hs}, we obtain the following inequality:
\begin{align*}
a \leq b + \|\eta\|_{L^{\infty}(\mathbb{R})}\|\partial_{s}H_{s}\|_{L^{\infty}} \leq b\left( 1- \tilde{c}\frac{8}{7}\Big(1+\frac{1}{\delta}\Big)\|\partial_{s}H_{s}\|_{L^{\infty}}\right)\leq \frac{1}{2}b< -\frac{1}{2}a,
\end{align*}
which contradicts the assumption $a>\delta>0$. That excludes $b<-a$ and proves the lemma.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:posRFHtent}:]
Let us fix a Hamiltonian $H \in \mathcal{H}$. By \cite[Lem.\ 8.8]{pasquotto2018} and \cite[Lem.\ 8.9]{pasquotto2018} for every two close enough regular couples $(H{+}h_1,J_1)$, $(H{+}h_2,J_2)$ with $h_1,h_2\in C_c^\infty(T^*\R^n),\ J_1,J_2\in \scrJ_\star$ and a homotopy $\Gamma=\{H{+}h_s,J_s\}_{s\in\R}$ of compactly perturbations $h_s$ of Hamiltonian $H$ and almost complex structures $J_s\in \scrJ_\star$ one can construct a homomorphism
\[
\phi^\Gamma \colon  CF_*\left(H+h_1,f_1\right)\to CF_*\left(H+h_2,f_2\right),
\]
which is defined by counting the perturbed flow lines with cascades (cf.\ \cite[Sec.\ 11.1]{Audin2014}, \cite[Lem.\ 7.2]{pasquotto2018}). Moreover, $\phi^\Gamma$ satisfies $\phi^\Gamma\circ \partial_1= \partial_2 \circ \phi^\Gamma$, thus induces a homomorphism on the homology level
\[
\Phi^\Gamma \colon RFH(H+h_1,J_1)\to RFH(H+h_2,J_2).
\]
Moreover, by Lemma \ref{lem:delta2} we know that if the two regular couples $(H{+}h_1,J_1),\linebreak[1](H{+}h_2,J_2)$ are close enough, then the preimage of $CF_*^+(H{+}h_2,f_2)$ under $\phi^\Gamma$ lies in $CF_*^+(H{+}h_1,f_1)$. In other words
\begin{equation}\label{phiGamma}
\phi^\Gamma\left(CF^{\leq 0}\left(H+h_1,f_1\right)\right)\subseteq CF^{\leq 0}\left(H+h_2,f_2\right).
\end{equation}
This together with \eqref{filtration} allows us to infer that the restriction $\phi^\Gamma_+$ of $\phi^\Gamma$ to\linebreak[4] $CF_*^+\left(H{+}h_1,f_1\right)$ constructed by counting the perturbed flow lines with cascades between critical points with positive action, commutes with the respective boundary operators $\partial^{\scriptscriptstyle +}_1$ and $\partial^{\scriptscriptstyle +}_2$ and thus induces a homomorphism $\Phi^\Gamma_+ \colon  RFH^+(H{+}h_1,J_1)\to RFH^+(H{+}h_2,J_2)$.

Now, we show that $\Phi^\Gamma_+$ is an isomorphism, using the fact that $\Phi^\Gamma$ is an isomorphism by \cite[Prop.\ 11.2.9]{Audin2014}. Via the inverse homotopy $\Gamma^{-1}=\{H{+}h_{1-s},J_{1-s}\}_{s\in\R}$, one can construct analogously a homomorphism
\[
\phi^{\Gamma^{-1}} \colon  CF_*\left(H+h_2,f_2\right)\to CF_*\left(H+h_1,f_1\right).
\]
As this homomorphism also satisfies $\phi^{\Gamma^{-1}}\circ \partial_2 = \partial_1 \circ \phi^{\Gamma^{-1}}$ and
\begin{equation}
\phi^{\Gamma^{-1}}\left(CF^{\leq 0}\left(H+h_2,f_2\right)\right) \subseteq CF^{\leq 0}\left(H+h_1,f_1\right),\label{phiGamma-1}
\end{equation}
it induces a homomorphism $\Phi^{\Gamma^{-1}}_+ \colon  RFH^+(H{+}h_2,J_2)\to RFH^+(H{+}h_1,J_1)$.

Finally, by \cite[Prop.\ 11.2.9]{Audin2014} there exists a homomorphism
\begin{fleqn}
\begin{align}
S \colon  CF_*\left(H+h_1,f_1\right) &\to CF_{*+1}\left(H+h_1,f_1\right),\notag\\
\text{satisfying}\mspace{100mu} \phi^{\Gamma^{-1}}\circ \phi^\Gamma- \operatorname{Id} & = S\circ \partial_1+\partial_1\circ S\label{S1}\\
\text{and} \mspace{80 mu} S \left(CF^{\leq 0}\left(H+h_1,f_1\right)\right) & \subseteq CF^{\leq 0}\left(H+h_1,f_1\right),\label{S2}
\end{align}
\end{fleqn}
where \eqref{S2} comes from applying Lemma \ref{lem:delta2} once again this time to the flowlines with cascades coming from a homotopy of homotopies (cf.\ \cite[Thm.\ 11.3.11]{Audin2014}). By combining \eqref{filtration}, \eqref{phiGamma}, \eqref{phiGamma-1} and \eqref{S2} we infer that the restriction of $S$ to $CF^+\left(H{+}h_1,f_1\right)$ also satisfies \eqref{S1} with $\phi^\Gamma_+$ and $\phi^{\Gamma^{-1}}_+$, thus establishing that $\Phi^\Gamma_+$ and $\Phi^{\Gamma^{-1}}_+$ are also isomorphisms on the homology level.

This proves that $RFH^+(H)$ is invariant under small enough\footnote{The invariance for any compact perturbation follows by splitting a given perturbation into a sequence of smaller perturbations (see \cite{CieliebakFrauenfelder2009}, p.\ 275)} compactly supported perturbations of $H$ and (via $h_1{=}h_2{=}0$) is invariant of $J$ or $(f,g)$.
\end{proof}

config: arxiv


TITLE: Intersection of all $p$-Sylow subgroups is normal
QUESTION [1 upvotes]: Let $G$ be a finite group, $p$ a prime number that divides $|G|$ and $O_p(G)=\bigcap_{P \in Syl_p(G)}P$. Prove that
1) $O_p(G) \lhd G$
2) $O_p(G)$ is maximal among the normal $p$-subgroups of $G$.
3) $O_p(G/O_p(G))=1$
I think I could do 2) using the following proposition which I will just enunciate:
Proposition
Let $P$ be a $p$-Sylow and $H \subset G$ a $p$-subgroup. Then there is $x \in G$ such that $xHx^{-1} \subset P$.
Solution to 2):
Suppose $O_p(G) \subset H$ for some $H \lhd G$ p-subgroup. By the proposition we have that for each $P$ p-Sylow group, there is $x: H=xHx^{-1} \subset P$. But then $H \in \bigcap_{P \in Syl_p(G)} P=O_p(G)$, so $O_p(G)=H$, it follows $O_p(G)$ is maximal.
I am pretty lost in 1) and 3), I would appreciate suggestions or hints rather than complete answers. Thanks in advance.

REPLY [4 votes]: Nice facts to add: there is a famous theorem of Jerald Brodkey (see for instance the book of I.M. Isaacs, Finite Group Theory, 1.37), saying that if the Sylow $p$-subgroups are abelian, then the $O_p(G)$ can be realized as the intersection of a certain pair of different Sylow $p$-subgroups. If $G$ has exactly $p+1$ Sylow $p$-subgroups, then $O_p(G)$ can be realized as the intersection of any pair of different Sylow $p$-subgroups!


TITLE: Iterated Reduced Tensor Power of Graded Vector spaces
QUESTION [2 upvotes]: This might be inappropriate for the MO-level. If so I'll delete it...
Suppose $V$ is a $\mathbb{Z}$-graded vector space and
$\overline{T}(V):=V \oplus V\otimes V \oplus \otimes^3 V \ldots$
is the 'reduced tensor power' of $V$ in the graded sense.i.e:
$V\otimes V:=\oplus_{z\in\mathbb{Z}}\oplus_{p+q=z}V_p\otimes V_q$
We don't see $T(V)$ as a (co)algebra, just as a plain graded vector
space.
Then on a first "empirical observation" I would say, that if we apply
the reduced tensor functor again, i.e make $\overline{T}(\overline{T}(V))$
then this is naturally isomorphic to $\overline{T}(V)$ because
"all tensor powers are already there".
So this should be a consequence of the 'freeness' of $\overline{T}(V)$, but I can't
see how this can be proofen.

REPLY [3 votes]: If $V$ is $\mathbb Z$-graded, then
$\overline{T}(V)$ is $\mathbb Z_{\ge1}\times \mathbb Z$-graded with
$\overline{T}(V)_{p,q} = \bigoplus_{z_1+z_2+\dots+z_p = q}V_{z_1}\otimes V_{z_2}\otimes\dots\otimes V_{z_p}$.
$\overline{T}$ is a functor from $\mathbb Z$-graded vector spaces to $\mathbb Z_{\ge1}\times \mathbb Z$-graded algebras which is left adjoint to the forgetful functor, where one has to pay attention to the degrees ...


TITLE: Convergence of $\int_2^\infty \left[\left(0.99 - \frac{1}{x^2}\right)^x\cdot\left(\frac{x-1}{x}\right)\right]dx$
QUESTION [0 upvotes]: I'm taking a calculus course and I was show this improper integral in class. the Professor told us that this integral converges and left us to prove it as a challenge. The thing is, I can't find a series that converges as well that could help me prove this integral converges and the integral itself is a bit of a handful so I couldn't integrate it directly as well.
This is the intgral: $$\int_2^\infty \left[\left(0.99 - \frac{1}{x^2}\right)^x\cdot\left(\frac{x-1}{x}\right)\right]dx$$
I would really aprricate a hint to the way of proving this, so far I have tried comparing it with other known integrals/series, altering the function to $e^{xln(0.99-\frac{1}{x^2}) + ln(\frac{x-1}{x})}$ and then integrating it, also tried deriving $xln(0.99-\frac{1}{x^2}) + ln(\frac{x-1}{x})$ and show that it's limit is $-\infty$ but all these ways led me to a dead end.
Thanks in advance to all who share their wisdom!

REPLY [1 votes]: Since @Kavi Rama Murthy already provided all the elements, let me consider the more genral case of
$$I(\epsilon)=\int_2^\infty \left(1-\epsilon - \frac{1}{x^2}\right)^x\,\left(\frac{x-1}{x}\right)\,dx$$ where $\epsilon \ll 1$.
At the lower bound the integrand is already small $\left(\frac{(3-4 \epsilon )^2}{32}\right)$
Take the logarithm of the integrand and expand it as a series for large values of $x$
$$\log\Bigg[ \left(1-\epsilon - \frac{1}{x^2}\right)^x\,\left(\frac{x-1}{x}\right)\Bigg]=x\log(1-\epsilon)+O\left(\frac{1}{x}\right)$$
$$\left(1-\epsilon - \frac{1}{x^2}\right)^x\,\left(\frac{x-1}{x}\right)\sim (1-\epsilon)^x $$
So,
$$I(\epsilon)\sim\int_2^\infty (1-\epsilon)^x \,dx=-\frac{(1-\epsilon )^2}{\log (1-\epsilon )}=\frac{1}{\epsilon }-\frac{5}{2}+\frac{23 \epsilon }{12}-\frac{3 \epsilon
   ^2}{8}+O\left(\epsilon ^3\right)$$ For $\epsilon=\frac 1{100}$ the truncated series would give, as an approximation,
$$I\left(\frac{1}{100}\right)=\frac{23404591}{240000} \sim 97.5191$$ while the numerical integration would lead to $91.3737$.
For $\epsilon=\frac 1{1000}$, the series would give $997.502$ while the numerical integration would lead to $986.852$.


TITLE: Why is a Topological Field Theory equivalent to a Frobenius algebra?
QUESTION [15 upvotes]: How can a physicist understand a 2-dimensional topological field theory as a Frobenius algebra? Are there some explicit examples in order to understand this relation?
The definition (e.g. on Wikipedia) of the Frobenius algebra is quite clear, it is a finite dimensional associative algebra equipped with a special non-degenerate bilinear form. If this algebra is represented by $n\times n$ matrices then the bilinear form is the trace.
Now, I fail to see how this is a TFT.
In specific two very famous topological field theories in 2-dimensions (that physicists work with a lot) are the topological A-model and the topological B-model (a la Witten). These models are related via mirror symmetry. How are they defined as a Frobenius algebra and what are their differences as a Frobenius algebra? And natural question to ask is wether mirror symmetry in the TFT side relates somehow the two Frobenius algebras on the algebra side.
So I would like to understand intuitively why a Frobenius algebra and a TFT are the same thing and the detail I ask in specific about the A-model and the B-model which are of my main interest.

REPLY [15 votes]: Arun Debray's answer is good, with one addendum: 2d TQFT is the same thing as a commutative Frobenius algebra.  But to truly understand what's going on, one also needs some pictures.  A commutative Frobenius algebra can be defined using pictures that just happen to look exactly like the relations holding between 2-dimensional cobordisms.  The best place to learn about this is here:

Joachim Kock, Frobenius Algebras and 2d Topological Quantum Field Theories, Cambridge U. Press, Cambridge, 2004.

But briefly, a Frobenius algebra has a multiplication and unit obeying associativity and the left and right unit laws:

and also a comultiplication and counit obeying coassociativity and the left and right counit laws:

where the multiplication and comultiplication obey the Frobenius laws:

A Frobenius algebra is commutative if switching two things and then multiplying them is the same as multiplying them:

Furthermore, a Frobenius algebra is commutative iff it is cocommutative, meaning this:

The ultimate statement of this fact is that $2\mathrm{Cob}$, the category with compact oriented 2-dimensional cobordisms as morphisms, is the free symmetric monoidal category on a commutative Frobenius monoid object.  The story gets much more interesting in 3 dimensions, and for that I recommend this paper:

Bruce Bartlett, Christopher L. Douglas, Christopher J. Schommer-Pries and Jamie Vicary, Extended 3-dimensional bordism as the theory of modular objects.


TITLE: The value of nested square roots
QUESTION [0 upvotes]: Find the value of $$x=\sqrt{11-\sqrt{11+\sqrt{11-\sqrt{11+\cdots }}}}$$
we can write $x$ as
$$x=\sqrt{11-\sqrt{11+x}}$$ Squaring both sides we get
$$x^2=11-\sqrt{11+x}$$ or
$$11-x^2=\sqrt{11+x}$$ again squaring i got as
$$x^4-22x^2-x+110=0$$ which is fourth degree equation. i could not find any roots by guessing.

REPLY [3 votes]: Suppose the quartic factors into two quadratics:
$$x^4-22x^2-x+110= (x^2-mx+p)(x^2+mx+q)$$
Then equating the coefficients of $x^2$, $x$, and unity, respectively, we obtain
$$
p+q-m^2 = -22 \\ (p-q)m = -1\\ pq = 110$$
Since (barring going all the way to the quartic formula, which is for practical purposes going to be so messy as to be useless here) we woun't find a nice factoring unless $p$ and $q$ are integer factors of $110$, there are only a very few combinations to try.  And inspired by the second equation, we can guess that $m = \pm 1$ which leads us to $p = -11, q = -10, m=1$.
So we have
$$x^4-22x^2-x+110= (x^2-x-11)(x^2+x-10)$$
The two candidate answers are, then, the positive roots of those two quadratic factors, namely $\frac{\sqrt{45}+1}{2}$ and $\frac{\sqrt{41}-1}{2}$.
One of these, however, is the value of
$$x=\sqrt{11+\sqrt{11-\sqrt{11+\sqrt{11-\cdots \infty}}}}$$
which leads to the same quartic for $x$.  That expression is clearly greater than the expression we want, so the answer we want is $x=\frac{\sqrt{41}-1}{2}$.


TITLE: Computing Fourier transform for $L^2$ function
QUESTION [8 upvotes]: For a function $f\in L^1(\mathbb{R})$, its Fourier transform is defined as $$\hat{f}(y)=\int_{-\infty}^\infty f(x)e^{-ixy}dx$$
For a function $f\in L^2(\mathbb{R})$, its Fourier transform is defined as the unique continuous mapping $g:L^2(\mathbb{R})\rightarrow L^2(\mathbb{R})$ that extends the mapping $h:S\rightarrow L^2(\mathbb{R})$, where $S$ is the Schwartz class, and the Fourier transform of a function in the Schwartz class is defined as in the first paragraph. (We may assume that this continuous mapping $g$ exists and is unique.)
Suppose $f\in L^2(\mathbb{R})$, and let $c>0$.  Show that $$\lim_{c\rightarrow\infty}\int_{-c}^cf(x)e^{-ixy}dx$$ exists in the $L^2$ sense and is equal to $\hat{f}$ defined above.

Define $f_c(x)$ to be $f(x)$ when $|x|\leq c$ and $0$ when $|x|>c$. Then the limit in question is $$\lim_{c\rightarrow\infty}\int_{-\infty}^\infty f_c(x)e^{-ixy}dx$$
The questions are:
1) Why does this limit exist?
2) Why does it equal $\hat{f}$ defined as the unique extension from the Schwartz class?
We know by the dominated convergence theorem that $\|f_c-f\|_2\rightarrow 0$ as $c\rightarrow\infty$. Might that help?

REPLY [3 votes]: So $\int_{-c}^c f(x) e^{-ixy} \, dx = \frac1{2i}(H_{-c} - H_c) \hat f(y)$, where $H_c g(x) = e^{icx} Hg (e^{-icx} x)$, and $H$ is the Hilbert transform.  The Hilbert transform is known to be an isometry on $L_2$.
To prove what you need, it is sufficient to show $-i H_c g \to g$ in $L_2$ as $c \to -\infty$.  But
$$ \|g + iH_c g\|_2 = \| 2 \hat g I_{(-\infty,c]} \|_2 .$$
I know I'm invoking some big machinery.  But it is not clear to me what you are allowed to assume and not allowed to assume when answering the question.
http://en.wikipedia.org/wiki/Hilbert_transform


TITLE: Is the Quot-scheme over non-singular curve reduced
QUESTION [6 upvotes]: Let $k$ be an algebraically closed field, $C$ a non-singular projective curve over $k$ of genus at least $2$ and $\mathcal{F}$ a locally free sheaf on $C$. Let $r,d$ be two integers satisfying $\mathrm{gcd}(r,d)=1$ and $r>0$. Denote by $Q$ the Quot-scheme parametrizing quotients of $\mathcal{F}$ of rank $r$ and degree $d$. Is every irreducible component of $Q$ (generically) reduced i.e., for a general point $x$ of the irreducible component of $Q$, the local ring $\mathcal{O}_{Q,x}$ is reduced?

REPLY [3 votes]: This is false, at least for special $C$ (perhaps it is true for $C$ that are sufficiently general in moduli).  The simplest counterexample I know of is a genus $4$ curve $C$ that is non-hyperelliptic and whose canonical image is contained in a singular quadric hypersurface in $\mathbb{P}^3$.  In this case, the scheme $G^1_3$ equals $W^1_3$ inside $\text{Pic}^3_C$, and this is one nonreduced point.  For the invertible sheaf $\mathcal{L}$ on $C$ parameterized by this point, $\mathcal{L}$ is globally generated.  Thus for $\mathcal{F}$ defined to be $H^0(C,\mathcal{L})\otimes_k \mathcal{O}_C$, the natural homomorphism, $$\phi:\mathcal{F} \to \mathcal{L},$$ is surjective.
Since being invertible is an open condition on flat families of coherent $\mathcal{O}_C$-modules, every small deformation of the quotient $\mathcal{L}$ is an invertible $\mathcal{O}_C$-module of degree $3$ that has a $2$-dimensional vector space of global sections, and conversely.  Thus, there is a smooth morphism from a Zariski open neighborhood $U$ of $[\phi]$ in the Quot scheme to $W^1_3$.  Since $W^1_3$ is nonreduced with nilradical a rank $1$ vector space over the residue field, also $U$ is nonreduced with nilradical an invertible sheaf on the reduced scheme of $U$ (which is smooth).


TITLE: Estimating a sum containing a uniformly distributed but deterministic term
QUESTION [1 upvotes]: In a previous question Summing something with random and non-random parts I asked how one could estimate (specifically I am interested in an upper bound).
$$\sum_{k=1}^n e^{-k B_k^2} $$
where $B_k$ is a uniformly distributed sequence on [0,1]. The response was to take the expected value of the summand. But what can I do if my $B_k^2$ are deterministic (and not random variables?). To clarify, by uniformly distributed I mean $\{B_k\}$ satisfies
$$ \lim_{n \rightarrow \infty} \frac{|[a,b] \cap \{B_1,B_2,\ldots B_n\}|}{n} = b-a$$.
Edit: Suppose $B_n = \{ \alpha n \}$ is the fractional part of $\alpha n$  for some irrational number $\alpha$. By Weyl's Criterion I know that $\{B_n\}$ is uniformly distributed. From some numerical calculations it seems that there exists some $K > 0$ that only depends on $\alpha$ such that
$$\sum_{k=1}^n e^{-k B_k^2} < K \sqrt{n}$$
for all sufficiently large $n$. How would I go about demonstrating this?
Edit 2
Here is an idea I had.
$$
\sum_{k=1}^n e^{-k B_k^2} = \sum_{k \leq n \: : e^{-k B_k^2} \leq n^{-1/2} } e^{-k B_k^2} + \sum_{k \leq n \: : e^{-k B_k^2} > n^{-1/2}} e^{-k B_k^2}
$$
which is less than
$$
\sum_{k = 1}^n n^{-1/2} + \sum_{k \leq n \: : e^{-k B_k^2} > n^{-1/2}} 1
$$
Now the left sum is just $n^{1/2}$. Now $ \#\{e^{-k B_k^2} > n^{-1/2} : k \leq n\} = \#\{B_k < \sqrt{\log (n)/(2k)}  : k \leq n\}$.
I believe that
$$\#\{B_k < \sqrt{\log (n)/(2k)}  : k \leq n\} \approx \int_1^n \sqrt{\frac{\log (n)}{2t}} \, dt$$
Or maybe that integral should be a sum, I don't know. But that implies there exists $\epsilon > 0$ such that for sufficiently large $n$,
$$\#\{B_k < \sqrt{\log (n)/(2k)}  : k \leq n\}  < n^{1/2+\epsilon}$$
So finally we have
$$ \sum_{k=1}^n e^{-k B_k^2} < n^{1/2} + n^{1/2+\epsilon} < 2 n^{1/2+\epsilon}$$
Clearly the above argument has quite a few gaps in it. I am hoping that someone can give me some feedback as to whether or not the gaps can be filled.

REPLY [2 votes]: There are some uniformly distributed sequences $B_k$ for which $\Sigma_k=0^n e^{-kB_k^2}$ grows faster than $\sqrt n$ :
Simply pick $B_k = 0$ about $n^{2/3}$ times out of $n$ and pad the rest of the values with any uniformly distributed sequence. You will still get a uniformly distributed sequence because the density $n^{2/3} / n$ is still convergent to $0$.
But with such a sequence, you have $\Sigma_{k=0}^n e^{-kB_k^2} > n^{2/3}$. (and we could pick any function $g(n)$ provided g(n)/n converges to $0$)
Truly random sequences have a lot of better properties than "uniformly distributed".
Here, a property that would work and that random sequences have, is that for any interval $I$,
$$ \lim_{n \rightarrow \infty} \frac{\#\{k \in \{0\ldots n\}, \sqrt{k}B_k \in I\}}{\sqrt{n} \text{length}(I)} = 2$$.
Then you can show that the sum $\Sigma_{k=0}^n e^{-kB_k^2}$, behaves the sames as random sequences, like:
$$\begin{array}{rcl}
\Sigma_{k=0}^n \int_{t=0}^1 e^{-kt^2} dt &<& \Sigma_{k=0}^n \int_{t=0}^\infty e^{-kt^2} dt
 = \Sigma_{k=0}^n \int_{t=0}^\infty \frac{e^{-t^2}}{\sqrt{k}} dt \\
&=& \left(\int_{t=0}^\infty e^{-t^2} dt \right) \left(\Sigma_{k=0}^n \frac{1}{\sqrt{k}} \right)
\sim \sqrt{\pi n} \end{array}
$$
The property described above implies that in an interval $I = [x;x+dx]$, there are about $2\sqrt{n} dx$ integers $k$ such that $\sqrt{k}B_k$ is in $I$. By dividing $\mathbb{R}^+$ in such intervals and regrouping the terms according to which interval they land into, you obtain bounds for $\Sigma_{k=0}^n e^{-k B_k^2}$. The finer you divide, the closer you get to the value of the integral, until you arrive at $\sqrt{\pi n}$


TITLE: Action of $SO_n$ on $\mathbb{S}^{n-1}$ induces fibre bundle.
QUESTION [3 upvotes]: Real compact Lie group $SO_n$ acts smoothly and transitively on $\mathbb{S}^{n-1} \subseteq \mathbb{R}^n$ with obvious action. Isotropy subgroup of each point in $\mathbb{S}^{n-1}$ is isomoprhic to $SO_{n-1}$. Let $\{e_1,\ldots,e_n\}$ be standard orthonormal base for $\mathbb{R}^n$. I want to see that map
$$p \colon SO_n \to \mathbb{S}^{n-1}$$
$$A \mapsto A e_n$$
is fibre bundle.
Map $p$ is obviously continuous surjection. I suppose one could take open cover $\{U^+,U^-\}$ of $\mathbb{S}^{n-1}$, where $U^\pm := \mathbb{S}^{n-1} \setminus \{\pm e_n\}$. Now, I should construct two homeomorphisms $\alpha_1$, $\alpha_2$, such that the following diagrams commute:

(source: presheaf.com)

(source: presheaf.com)
I cannot make this happen. How to define $\alpha_1$ and $\alpha_2$? Or should I use some other cover? Any help would be appreciated.
The reason I want this is that I would like to use homotopy lifting property (or exact sequence of homotopy groups) to calculate fundamental groups of $SO_n$ by induction.

REPLY [2 votes]: If $G$ is a compact Lie group and $H$ is a closed subgroup, then $G\rightarrow G/H$ is a principal $H$-bundle. In particular, it is a fibre bundle with fibre $H$.
So, $SO(n)\rightarrow SO(n)/SO(n-1)$ is a fibre bundle with fibre $SO(n-1)$. We then compose this quotient map with the diffeomorphism $SO(n)/SO(n-1)\rightarrow S^{n-1}$ to get a fibre bundle $SO(n)\rightarrow S^{n-1}$. It then just remains to compute this map. It is precisely $A\mapsto[A]\mapsto Ae_n$, as desired.


text:
\begin{document}

\title{On Some Low Distortion Metric Ramsey Problems}

\date{}

\author{Yair Bartal\thanks{Supported in part by a grant from the
Israeli National Science Foundation.} \and Nathan Linial\thanks{Supported in
part by a grant from the Israeli National Science Foundation.} \and Manor
Mendel\thanks{Supported in part by the Landau Center.} \and Assaf Naor }

\maketitle

\begin{abstract}
In this note, we consider the metric Ramsey problem for the normed
spaces $\ell_p$. Namely, given some $1\le p \le \infty$ and
$\alpha \ge 1$, and an integer $n$, we ask for the largest $m$
such that every $n$-point metric space contains an $m$-point
subspace which embeds into $\ell_p$ with distortion at most $
\alpha$. In \cite{blmn0} it is shown that in the case of $\ell_2$,
the dependence of $m$ on $\alpha$ undergoes a phase transition at
$\alpha =2$. Here we consider this problem for other $\ell_p$, and
specifically the occurrence of a phase transition for $p\neq 2$.
It is shown that a phase transition does occur at $\alpha=2$ for every $p\in
[1,2]$. For $p>2$ we are unable to determine the answer, but
estimates are provided for the possible location of such a phase
transition. We also study the analogous problem for isometric
embedding and show that for every $1<p<\infty$ there are
arbitrarily large metric spaces, no four points of which embed
isometrically in $\ell_p$ .
\end{abstract}

\section{Introduction}\label{introduction}

A Ramsey-type theorem states that large systems necessarily contain large,
highly structured sub-systems. Here we consider Ramsey-type problems for
finite metric spaces, interpreting  ``highly structured" as having low
distortion embedding in $\ell_p$.

A mapping between two metric spaces $f:M
\rightarrow X$, is called an embedding of $M$ in $X$.
The \emph{distortion} of the embedding is defined as
\[
\mathrm{dist}(f)=\sup_{\substack{x,y\in M\\x\neq
y}}\frac{d_X(f(x),f(y))}{d_M(x,y)}\cdot \sup_{\substack{x,y\in M\\x\neq
y}}\frac{d_M(x,y)}{d_X(f(x),f(y))}.
\]
The least distortion required to embed $M$ in $X$ is denoted by
$c_X(M)$. When $c_X(M)\leq \alpha$ we say that $M$ $\alpha$-embeds
in $X$. In  this note we study the following notion.

\begin{definition}[Metric Ramsey function]
We denote by
$R_{X}(\alpha,n)$ the largest integer $m$ such
that every $n$-point metric space has a subspace of size $m$ that
$\alpha$-embeds into $X$.
\end{definition}
 When $X=\ell_p$ we use the notations $c_p$ and $R_{p}$. Note that for $p\in [1,\infty]$, it is always true that
$R_p(\alpha,n)\geq R_2(\alpha,n)$. When $\alpha=1$ we drop it from the
notation, i.e., $R_X(n)=R_X(1,n)$.

Bourgain, Figiel, and Milman~\cite{bfm}  study this function for
$X=\ell_2$, as a metric analog of Dvoretzky's theorem~\cite{dvo}.
They prove

\begin{theorem}[\cite{bfm}] \label{thm:bfm86}
For any $\alpha>1$ there exists $C(\alpha)>0$ such that $R_2(\alpha,n) \geq
C(\alpha) \log n$. Furthermore, there exists $\alpha_0>1$ such that
$R_2(\alpha_0,n)=O(\log n)$.
\end{theorem}

In~\cite{blmn0} the metric Ramsey problem is studied
comprehensively. In particular, the following
phase transition is established in the case of $X=\ell_2$.

\begin{theorem}[\cite{blmn0}]
\label{thm:phase} Let  $n\in \mathbb{N}$. Then:
\begin{enumerate}
\item\label{item:upper} For every $1<\alpha<2$: \( c(\alpha)\log
n\le R_2(\alpha,n)\le 2\log n+C(\alpha), \) where
$c(\alpha),C(\alpha)$ may depend only on $\alpha$.

\item\label{item:lower} For every $\alpha>2$: \( n^{c'(\alpha)}\le
R_2(\alpha,n)\le n^{C'(\alpha)}, \) where $c'(\alpha),C'(\alpha)$
depend only on $\alpha$ and $0<c'(\alpha)\leq C'(\alpha)<1$.
Moreover, $c'(\alpha)$ tends to $1$ as $\alpha$ tends to $\infty$.
\end{enumerate}
\end{theorem}

By Dvoretzky's theorem, the lower bound in part~\ref{item:lower}
of Theorem~\ref{thm:phase} implies in particular that if
$\alpha>2$, and $X$ is any infinite dimensional normed space, then
$R_X(\alpha,n)\ge n^{c'(\alpha)}$. Therefore, in our search
for a possible phase transition for $R_p(\cdot,n)$, $p\neq
2$, it is natural to extend the upper bound in
part~\ref{item:upper} of Theorem~\ref{thm:phase} to this range.
The main result proved in this note is the following:

\begin{theorem}\label{thm:<2 a} There is an absolute constant $c>0$ such that
for every $0<\delta<1$,
\begin{enumerate}
\item For $1\le p< 2$, \( R_p({2-\delta},n)\le
e^{\frac{c}{\delta^2}}\log n. \) \item For $2<p<\infty$, $
R_p({2^{2/p}-\delta},n)\le e^{\frac{c}{p^2\delta^2}}\log n. $
\end{enumerate}
\end{theorem}

Thus we extend the result of \cite{blmn0} to show that a phase
transition occurs in the metric Ramsey problem for $\ell_p$,
$p\in[1,2)$, at $\alpha=2$. The asymptotic behavior of
$R_p(\alpha,n)$ for $p>2$, and $\alpha \in [2^{2/p},2]$, is left
as an open problem. In particular, we do not know whether or not
this function undergoes a similar phase transition. We find this
problem potentially significant: if there is a phase transition at
$2$ also in the range $2<p<\infty$, then this result will
certainly be of great interest. On the other hand, if it is
possible to improve the lower bound in part~\ref{item:lower} of
Theorem~\ref{thm:phase} for $p>2$ and certain distortions strictly
less than $2$, then this would involve an embedding technique that
is different from the method used in~\cite{blmn0}, which doesn't
distinguish between the various $\ell_p$ spaces.

The proof of the upper bound on $R_2(\alpha,n)$ for $\alpha<2$
stated in Theorem~\ref{thm:phase} uses the Johnson-Lindenstrauss
dimension reduction lemma for $\ell_2$ \cite{jl}. For $\ell_p$,
$p\neq 2$, no such dimension reduction is known to hold. (Recent
work~\cite{br:cha, lee:naor} shows that dimension reduction does
not, in genereal, hold in $\ell_1$.) Our proof is based on a
non-trivial modification of the random construction in~\cite{bfm},
in the spirit of Erd\"os' upper bound on the Ramsey
numbers~\cite{erdos, bollobasrandom}.  In the process we prove
tight bounds on the embeddability of the metrics of complete
bipartite graphs in $\ell_p$. Specifically we show that
\[ c_p(K_{n,n})=\begin{cases} 2-\Theta(n^{-1}) & p\in[1,2] \\
2^{2/p} -\Theta((pn)^{-1}) & p>2. \end{cases}
\]

\medskip
The second part of this note addresses the isometric Ramsey
problem for $p\in (1,\infty)$. It turns out that this problem is
naturally tackled within the class of uniformly convex normed
spaces (see Section~\ref{section:isometric} for the definition).

\begin{comment}
Section~\ref{section:isometric} deals with the isometric Ramsey
problem. Every $3$-point metric is, of course, isometric to a
Euclidean triangle. We show that $R_2(n)=3$ for every $n \ge 3$.
Namely, there exist arbitrarily large metric spaces in which no
4-tuple isometrically embeds in $\ell_2$. In fact, the same
conclusion holds if $\ell_2$ is replaced by any uniformly convex
normed space. In particular we have
\end{comment}

\begin{theorem}[Isometric Ramsey Problem]\label{thm:isometric-lp}
Let $X$ be a uniformly convex normed space with
$\mathrm{dim}(X)\ge 2$. Then $R_X(1,n) = 3$ for $n\ge 3$.
\end{theorem}
Since $\ell_p$ is uniformly convex for $p\in (1,\infty)$, the
conclusion of Theorem~\ref{thm:isometric-lp} holds in these cases.
Note that the theorem does not apply for $\ell_1$ and
$\ell_\infty$ which are not uniformly convex. Specifically, it is
known that $\ell_\infty$ is universal in that it contains an
isometric copy of every finite metric space, whence
$R_\infty(n)=n$. It is known \cite{dezalau} that any 4-point
metric space is isometrically embeddable in $\ell_1$, and
therefore $R_1(n)\geq 4$ for $n \ge 4$. The determination of
$R_1(n)$ is left as an open problem.

\section{An Upper Bound For $\alpha<2$} \label{section:a<2}

In this section we prove that for any $\alpha<\min\{2,2^{2/p}\}$,
$R_p(\alpha,n)=O(\log n)$. Our technique both improves and
simplifies the technique of \cite{bfm}, which is itself in the
spirit of Erd\"os' original upper bound for the Ramsey coloring
numbers. The basic idea is to exploit a
universality property of random graphs $G\in G(n,1/2)$.
Namely, that any fixed graph of constant size appears
as an induced subgraph of every induced subgraph of $G$ of size
$\Omega(\log n)$. More precisely, we define the following notion
of universality.

\begin{definition} Let $H$ be a graph. A graph $G$ is called
$({H},s)$-universal if every set of $s$ vertices in $G$ contains
an induced subgraph isomorphic to $H$.
\end{definition}

\begin{proposition} \label{prop:ramsey-graph}
For every $k$-vertex graph ${H}$ there exists a constant $C>0 $
and an integer $n_0$ such that for any $n>n_0$ there exists a
$({H},C\log n)$-universal graph on $n$ vertices. Furthermore,
\[
C\le O\left(k^2 2^{\binom{k}{2}}\right)\quad \mathrm{and} \quad n_0 \le
O\left(k^3 2^{\binom{k}{2}}\right).
\]
\end{proposition}

Such facts are well-known in random graph theory, and similar
arguments can be found for example in \cite{ly}. We sketch the
standard details for the sake of completeness.

Recall that a family of sets $\mathcal{F}$ is called \emph{almost
disjoint} if $|A \cap B| \leq 1$ for every $A,B\in \mathcal{F}$.
In what follows, given a set $S$ and an integer $k$, we denote by
$\binom{S}{k}$ the set of all $k$-point subsets of $S$.
\begin{lemma} \label{prop:pairwise}
For every integer $k$ and a finite set $S$ of cardinality
$s=|S| > 2k^2$,
there exists an almost disjoint family $K\subset \binom{S}{k}$,
such that $|K|\geq \left\lfloor \tfrac{s}{2k}\right\rfloor ^2$.
\end{lemma}
\begin{proof}
Let $p$ be a prime satisfying $\tfrac{s}{2k}\leq p\leq
\tfrac{s}{k}$, and assume that
 \[ L=\{(i,j); i,j \in \mathbb{Z}_p,\ i\in\{0,\ldots,k-1\} \}
 \subseteq S.\]
For each $a,b\in \mathbb{Z}_p$ (the field of residues modulo $p$), define
\[ A_{a,b}=\{(i,j); \ j\equiv ai+b \pmod{p},\ i\in\{0,\ldots,k-1\}\} ,\]
and take $K=\{A_{a,b} | a,b\in \mathbb{Z}_p\}$. The set $K$ is
easily checked to satisfy the requirements.
\end{proof}

As usual $G(n,1/2)$ denotes the model of random graphs in
which each edge on $n$ vertices is chosen independently with
probability $1/2$.

\begin{lemma} \label{lem:induced}
Let $H$ be a $k$-vertex graph and let $s>2k^2$. The probability that
a random graph $G \in G(s,1/2)$ does not contain an
induced subgraph isomorphic to $H$, is at most
$(1-2^{-\binom{k}{2}})^{\left\lfloor\tfrac{s}{2k}\right\rfloor^2} $.
\end{lemma}
\begin{proof}
Construct, as in Lemma~\ref{prop:pairwise}, an almost disjoint
family $\mathcal{F}$ of $\left\lfloor\tfrac{s}{2k}
\right\rfloor^2$ subsets of $\{1,\ldots,s\}$, the vertex set of
$G$. If $F_1 \neq F_2 \in \mathcal{F}$, then the event that the
restriction of $G$ to ${F_1}$ (resp. $F_2$) is isomorphic to $H$
are independent. Hence, the probability that none of the sets $F
\in \mathcal{F}$ spans a subgraph isomorphic to $H$ is at most
$(1-2^{-\binom{k}{2}})^{\left\lfloor\tfrac{s}{2k}\right\rfloor^2}
$.
\end{proof}

\begin{proof} [Proof of Proposition~\ref{prop:ramsey-graph}]
Let $G$ be a random graph in $G(n,1/2)$. By the previous lemma,
the expected number of sets of $s$ vertices which contain no
induced isomorphic copy of $H$ is at most $\binom{n}{s}
\left(1-2^{-\binom{k}{2}}\right)^{\left\lfloor\tfrac{s}{2k}\right\rfloor^2}$.
If this number is $< 1$, then there is an $(H,s)$-universal graph,
as claimed. It is an easy matter to check that this holds with the
parameters as stated.
\end{proof}

\begin{comment}
\begin{remark}\label{remark:dependence}
For further applications we record here that in
Proposition~\ref{prop:ramsey-graph}, as long as $\eta\ge
\frac{1}{2}$, we have the following estimates:
$$
C\le \frac{8k^2}{\log\left(\frac{1}{\eta}\right)}\quad
\mathrm{and} \quad \frac{n_0}{\log n_0}\le \frac{8k^2}{\log
\left(\frac{1}{\eta}\right)}.
$$
\end{remark}
\end{comment}

A class $\cal C$ of finite metric spaces is called a \emph{metric
class} if it is closed under isometries. $\cal C$ is said to be
\emph{hereditary}, if $M\in \cal C$ and $N\subset M$ imply $N\in
\cal C$. We call a metric space $(X,d)$ a $\{0,1,2\}$ metric space
if for all $x,y\in X$, $d(x,y)\in \{0,1,2\}$. There is a simple
1:1 correspondence between graphs and $\{0,1,2\}$ metrics. Namely,
associated with a $\{0,1,2\}$ metric space $M=(X,d)$ is the graph
$G=(X,E)$ where $\{x,y\} \in E$ iff $d_M(x,y)=1$.

\begin{lemma} \label{lem:p3}
Let $\cal C$ be a hereditary metric class of finite metric spaces,
and suppose that there exists some finite $\{0,1,2\}$ metric space
${M}_0$ which is not in $P$. Then there exist metric spaces
$M=M_n$ of arbitrarily large size $n$ such that every subspace
$S\subset M_n$ with at least $C \log n$ points is not in $\cal C$.
The constant $C$ depends only on the cardinality of $M_0$.
\end{lemma}
\begin{proof}
Let ${H}_0$ be the graph corresponding to the metric space
${M}_0$.

We apply
Proposition~\ref{prop:ramsey-graph}, to construct arbitrarily
large graphs $G_n=(V_n,E_n)$ with $|V_n|=n$, in which every
set of $\ge C\log n$ vertices contains an
induced subgraph isomorphic to $H_0$. Let $M_n$ be the $n$-point
metric space corresponding to $G_n$. It follows that
every subspace of $M_n$ of size $\ge C \log n$ contains
a metric subspace that is isometric to $M_0$.
Since $\cal C$ is hereditary, $S\notin \cal C$.
\end{proof}

Note that $\{M; M \text{ is a metric space, } c_p(M)\leq \alpha\}$
is a hereditary metric class. Therefore, in order to show that for
$\alpha<2$, $R_p(\alpha,n)=O(\log n)$, it is enough to find a
$\{0,1,2\}$ metric space whose $\ell_p$ distortion is greater than
$\alpha$. We use the complete bipartite graphs $K_{n,n}$. The
$\ell_p$-distortion of $K_{n,n}$, $1\le p<\infty$, is estimated in
the following proposition.

\begin{comment}
The shortest path metrics on the complete bipartite graphs
$K_{n,n}$ are good candidates: They are $\{0,1,2\}$ metric spaces,
and we show below that:
$$
c_2(K_{n,n})=2\sqrt{\frac{n-1}{n}}.
$$
In fact, we will calculate the asymptotics of the
$\ell_p$-distortion of $K_{n,n}$, $1\le p<\infty$.
\end{comment}

\begin{proposition}\label{prop:knn} For every $1\le p\le 2$,
$$
2\left(\frac{n-1}{n}\right)^{1/p}\le c_p(K_{n,n})\le
2\sqrt{\frac{n-1}{n}}
$$
For every $2\le p<\infty$,
$$
2^{2/p}\left(\frac{n-1}{n}\right)^{1/p}\le c_p(K_{n,n})\le
2^{2/p}\left(1-\frac{1}{2n}\right) ^{1/p}.
$$
\end{proposition}

Before proving Proposition~\ref{prop:knn}, we will deduce the main
result of this section:

\begin{theorem}\label{thm:<2} There is an absolute constant $c>0$ such that
for every $0<\delta<1$, if $1\le p\le 2$ then:
$$
R_p({2-\delta},n)\le e^{\frac{c}{\delta^2}}\log n,
$$
and if $2<p<\infty$ then:
$$
R_p({2^{2/p}-\delta},n)\le e^{\frac{c}{p^2\delta^2}}\log n.
$$
\end{theorem}
\begin{proof}
\begin{comment}
Fix an integer $k$. Let $\eta_k$ be the probability that a random
graph on $2k$ vertices is not isomorphic to $K_{k,k}$. The
probability that

a random graph on $2k$ vertices is isomorphic to $K_{k,k}$ is the
probability that there exists a subset of the vertices of size $k$
such that the edges in the graph are precisely the edges that join
this subset and its complement. Hence:
$$
\eta_k=1-{2k\choose k}2^{-{2k\choose 2}}\le 1-2^{-2k^2}.
$$

When we apply Proposition~\ref{prop:ramsey-graph} with
$G_0=K_{k,k}$ we need to assume that:
$$
\frac{n}{\log n}\ge
\frac{32k^2}{\log\left(\frac{1}{1-2^{-2k^2}}\right)},
$$
and this condition is satisfied when $n\ge 2^{Ck^2}$ for some
absolute constant $C$.

Summarizing, we have proved that there is an absolute constant $C$
such that for every $n\ge 2^{Ck^2}$ there exists a $\{0,1,2\}$
metric space $M_n$ such that any subset $S\subset M_n$ of
cardinality at least $2^{Ck^2}\log n$ contains an isometric copy
of $K_{k,k}$.
\end{comment}

Proposition~\ref{prop:ramsey-graph} implies that
there is an absolute constant $C$ such that for every $n\ge
2^{Ck^3}$ there exists a $\{0,1,2\}$ metric space $M_n$ such that
any subset $S\subset M_n$ of cardinality at least $2^{Ck^2}\log n$
contains an isometric copy of $K_{k,k}$.

We start with $1\le p\le 2$. Let
$k=\left\lfloor\frac{2}{\delta}\right\rfloor+1$.
By Proposition~\ref{prop:knn},
$$
c_p(K_{k,k})\ge 2\left(1-\frac{1}{k}\right)^{1/p}>
2\left(1-\frac{\delta}{2}\right)=2-\delta,
$$
so that for $n$ large enough ($\ge e^{\frac{C'}{\delta^3}}$),
and hence for all $n$ (by proper choice of constants),
$$
R_p(2-\delta,n)\le e^{\frac{C'}{\delta^2}}\log n.
$$

When $p>2$ take $k=2\left\lfloor\frac{4}{p\delta}\right\rfloor$.
In this case one easily verifies that:
$$
c_p(K_{k,k})\ge 2^{2/p}\left(1-\frac{1}{k}\right)^{1/p}\ge
2^{2/p}-\delta,
$$
from which the required result follows as above.
\end{proof}

\medskip

In order to prove Proposition~\ref{prop:knn}, we need some
preparation.

\begin{lemma}\label{lem:matrix}
Let $A=(a_{ij})$ be an $n\times n$ matrix and $2\le p<\infty$.
Then:
$$
\sum_{i=1}^n\sum_{j=1}^n\left(\left|\sum_{k=1}^n
a_{ik}-\sum_{k=1}^n a_{jk}\right|^p+ \left|\sum_{k=1}^n
a_{ki}-\sum_{k=1}^n a_{kj}\right|^p\right)\le
 \frac{(2n)^p}{2}\sum_{i=1}^n\sum_{j=1}^n |a_{ij}|^p.
$$
\end{lemma}
\begin{proof} We identify $\ell_p^{n^2}$ with
the space of all $n\times n$ matrices
$A=(a_{ij})$, equipped with the $\ell_p$ norm:
$$
\|A\|_p=\left(\sum_{i=1}^n\sum_{j=1}^n |a_{ij}|^p\right)^{1/p}.
$$
Define a linear operator $T:\R^{n^2}\to \R^{n^2}\oplus\R^{n^2}$
by:
$$
T(a_{ij})=\left( \sum_{k=1}^n a_{ik}-\sum_{k=1}^n
a_{jk}\right)_{ij}\oplus \left(\sum_{k=1}^n a_{ki}-\sum_{k=1}^n
a_{kj}\right)_{ij}.
$$
For $q\ge 1$ denote $\|T\|_{q\to q}=\max_{A\neq
0}\|T(A)\|_q/\|A\|_q$. Our goal is to show that $\|T\|_{p\to p}\le
2^{1-1/p}n$. By a result from the complex interpolation theory for
linear operators (see \cite{bl}), for $2\le p\le \infty$,
$\|T\|_{p\to p}\le \|T\|_{2\to 2}^{2/p}\cdot \|T\|_{\infty\to
\infty}^{1-2/p}$. It is therefore enough to prove the required
estimate for $p=2$ and $p=\infty$. The case $p=\infty$ is simple:
$$
\|T(A)\|_{\infty}=\max_{1\le i,j\le n}\max\left\{ \left|
\sum_{k=1}^n a_{ik}-\sum_{k=1}^n a_{jk}\right|, \left|\sum_{k=1}^n
a_{ki}-\sum_{k=1}^n a_{kj}\right|\right\}\le 2n\|A\|_{\infty}.
$$
For $p=2$ we have to show that:
$$
\sum_{i=1}^n\sum_{j=1}^n\left(\left|\sum_{k=1}^n
a_{ik}-\sum_{k=1}^n a_{jk}\right|^2+ \left|\sum_{k=1}^n
a_{ki}-\sum_{k=1}^n a_{kj}\right|^2\right)\le
 2n^2\sum_{i=1}^n\sum_{j=1}^n |a_{ij}|^2.
$$
This inequality follows from the following elementary identity:
{\setlength\arraycolsep{1pt}
\begin{eqnarray*}
2n^2\sum_{i=1}^n\sum_{j=1}^na_{ij}^2
&=&\sum_{i=1}^n\sum_{j=1}^n\left[\left(\sum_{k=1}^na_{ik}-\sum_{k=1}^na_{jk}\right)^2+
\left(\sum_{k=1}^na_{ki}-\sum_{k=1}^na_{kj}\right)^2\right]+\\
&\phantom{\le}&+
2\sum_{i=1}^n\sum_{j=1}^n\left(na_{ij}-\sum_{k=1}^na_{ik}-\sum_{k=1}^na_{kj}\right)^2.
\end{eqnarray*}}
\end{proof}

\begin{corollary}\label{corr:roundness} Let $1\le p <\infty$ and $x_1,\ldots,x_n, y_1\ldots
y_n\in \ell_p$. Then if $2\le p<\infty$,
$$
\sum_{i=1}^n\sum_{j=1}^n
\left(\|x_i-x_j\|_p^p+\|y_i-y_j\|_p^p\right)\le
2^{p-1}\sum_{i=1}^n\sum_{j=1}^n \|x_i-y_j\|_p^p.
$$
If $1\le p\le 2$ then:
$$
\sum_{i=1}^n\sum_{j=1}^n
\left(\|x_i-x_j\|_p^p+\|y_i-y_j\|_p^p\right)\le
2\sum_{i=1}^n\sum_{j=1}^n \|x_i-y_j\|_p^p.
$$
\end{corollary}
\begin{proof} By summation it is clearly enough to prove these inequalities
for $x_1,\ldots,x_n,y_1,\ldots,y_n\in \R$. If $2\le p<\infty$ then
the required result follows from an application of
Lemma~\ref{lem:matrix} to the matrix $a_{ij}=x_i-y_j$. If $1\le
p\le 2$ then consider $\ell_p$ equipped with the metric
$d(x,y)=\|x-y\|_p^{p/2}$. It is well known (see \cite{ww}) that
$(\ell_p,d)$ embeds isometrically in $\ell_2$, so that the case
$1\le p \le 2$ follows from the case $p=2$.

\end{proof}

\medskip\noindent{\bf Remark.}
In \cite{enfloroundness} P. Enflo defined the notion on generalized roundness of a metric
space. A metric space $(M,d)$ is said to have generalized
roundness $q\ge 0$ if for every $x_1,\ldots,x_n,y_1,\ldots, y_n\in
M$,
$$
\sum_{i=1}^n\sum_{j=1}^n (d(x_i,x_j)^q+d(y_i,y_j)^q)\le
2\sum_{i=1}^n\sum_{j=1}^nd(x_i,y_j)^q.
$$
Enflo proved that Hilbert space has generalized roundness $2$ and
in \cite{ltw} the concept of generalized roundness was
investigated and was shown to be equivalent to the notion of
negative type (see \cite{dezalau,ww} for the definition).
Particularly, it was proved in \cite{ltw} that for $1\le p<2$,
$\ell_p$ has generalized roundness $p$, which is precisely the
second statement in Corollary \ref{corr:roundness}. For the case $p=1$
simpler, more direct proofs can be given
which do not use reduction to the case $p=2$,
see e.g.~\cite{dezalau}. Observe that
Lemma~\ref{lem:matrix} would follow simply by convexity had it not
been for the additional factor $1/2$ on the right-hand side. This
factor is crucial for our purposes, and this is why the
interpolation argument was needed.

\begin{proof}[Proof of Proposition~\ref{prop:knn}.] We identify $K_{n,n}$
with the metric on $\{u_1,\ldots,u_n,v_1,\ldots,v_n\}$ where
$d(u_i,u_j)=d(v_i,v_j)=2$ for all $i\neq j$, and
$d(u_i,v_j)=1$ for every $1\le i,j\le n$.
Fix some $1\le p<\infty$ and let
$f:\{u_1,\ldots,u_n,v_1,\ldots,v_n\}\to \ell_p$ be an embedding
such that for every $x,y\in K_{n,n}$, $d(x,y)\le
\|f(x)-f(y)\|_p\le Ld(x,y)$. Then,
$$
\sum_{i=1}^n\sum_{j=1}^n(\|f(u_i)-f(u_j)\|_p^p+\|f(v_i)-f(v_j)\|_p^p)\ge
2n(n-1)2^p,
$$
and
$$
\sum_{i=1}^n\sum_{j=1}^n\|f(u_i)-f(v_j)\|_p^p\le n^2L^p.
$$
For $1\le p\le 2$ Corollary \ref{corr:roundness} gives:
$$
2n(n-1)^p2^p\le 2n^2L^p\Longrightarrow L\ge
2\left(\frac{n-1}{n}\right)^{1/p}.
$$
For $2\le p<\infty$ we get that:
$$
2n(n-1)2^p\le 2^{p-1}n^2L^p\Longrightarrow L\ge
2^{2/p}\left(\frac{n-1}{n}\right)^{1/p}.
$$
This proves the required lower bounds on $c_p(K_{n,n})$.

To prove the upper bound assume first that $p=2$ and denote by
$\{e_i\}_{i=1}^{\infty}$ the standard unit vectors in $\ell_2$.
Define $f:K_{n,n}\to \ell_2^{2n}$ by:

\begin{align*}
f(u_i) &=\sqrt{2}\Bigl(e_i-\frac{1}{n}\sum_{j=1}^ne_j\Bigr),\\
f(v_i)
&=\sqrt{2}\Bigl(e_{n+i}-\frac{1}{n}\sum_{j=1}^ne_{n+j}\Bigr).
\end{align*}
Then for $i\neq j$,
$\|f(u_i)-f(u_j)\|_2=\|f(v_i)-f(v_j)\|_2=2=d(u_i,u_j)=d(v_i,v_j)$.
On the other hand: {\setlength\arraycolsep{1pt}
\begin{eqnarray*}
\|f(u_i)-f(v_j)\|_2&=&\sqrt{\|f(u_i)\|_2^2+\|f(v_j)\|_2^2}\\
&=&\sqrt{4\left(1-\frac{1}{n}\right)^2+4(n-1)\cdot\frac{1}{n^2}}=2\sqrt{\frac{n-1}{n}}.
\end{eqnarray*}}
This finishes the calculation of $c_2(K_{n,n})$. For $1\le p<2$,
since for every $\epsilon>0$ and for every $k$, $\ell_p$ contains
a $(1+\epsilon)$ distorted copy of $\ell_2^k$, we get the estimate
$c_p(K_{n,n})\le 2\sqrt{\frac{n-1}{n}}$.

The case $2<p< \infty$ requires a different embedding. We begin by
describing an embedding with distortion $2^{2/p}$ and then explain
how to modify it so as to reduce the distortion by a factor of
$\left(1-\frac{1}{2n}\right)^{1/p}$. Let $z_1,\ldots,z_n$ be a
collection of $n$ mutually orthogonal $\pm 1$ vectors of dimension
$m =O(n)$. (For example the first $n$ rows in an $m \times m$
Hadamard matrix). In our first embedding we define $f(u_i)$ as the
$(2m)$-dimensional vector $(z_i,0)$, namely, $z_i$ concatenated
with $m$ zeros. Likewise, $f(v_i)= (0,z_i)$ for all $i$. Now
$\|f(u_i) - f(u_j)\|_p = 2 \left(\frac{m}{2}\right)^{1/p}$ and
$\|f(u_i) - f(v_j)\|_p = (2m)^{1/p}$, and so $f$ has distortion
$2^{2/p}$. To get the $\left(1-\frac{1}{2n}\right)^{1/p}$
improvement, note that for some $m\le 4n$ it is possible to select
the $z_i$ so that the $m$-th coordinate in all of them is $+1$.
Modify the previous construction to an embedding into $2m-1$
dimensions as follows: Now $g(u_i)$ is $z_i$ concatenated with
$m-1$ zeros, whereas $g(v_i)$ has zeros in the first $m-1$
coordinates, $1$ in the $m$-th and this is followed by the first
$m-1$ coordinates of the vector $z_i$. The easy details are
omitted.
\end{proof}

\paragraph{Remark:} The upper bounds in Proposition~\ref{prop:knn} were not used in the proof of
Theorem~\ref{thm:<2}. Apart from their intrinsic interest, these
upper estimates show that the above technique cannot prove an
upper bound of $O(\log n)$ on $R_2({2-\epsilon},n)$ which is
independent of $\epsilon$. In fact, this can never be achieved
using $\{0,1,2\}$ metric spaces due to the following proposition.

\begin{proposition}
Let $X$ be an $n$-point $\{0,1,2\}$ metric space. Then $c_2(X)\le
2\sqrt{\frac{n-1}{n}}$.
\end{proposition}
\begin{proof}
We think of $X$ as a metric on $\{1,\ldots,n\}$ and denote
$d(i,j)=d_{ij}$. Define an $n\times n$ matrix $A=(a_{ij})$ as
follows:
 \[ a_{ij}=\begin{cases} 2 & \text{if}\ i=j \\
                                           0 & \text{if}\  d_{ij}=2\\
                                           \frac{2}{n} & \text{if}\ d_{ij}=1
                            \end{cases} .\]

We claim that $A$ is positive semidefinite. Indeed, for any $z\in
\R^n$ {\setlength\arraycolsep{1pt}
\begin{eqnarray*}
\langle Az,z\rangle&=&\sum_{i=1}^n\sum_{j=1}^n a_{ij}z_iz_j\\
&\ge& \sum_{i=1}^n 2z_i^2-\sum_{i \neq j}\frac{2}{n}|z_i|\cdot|z_j|\\
&\ge&  \sum_{i=1}^n 2z_i^2-\sum_{i=1}^n\sum_{j=1}^n\frac{2}{n}|z_i|\cdot|z_j|\\
&=&2\|z\|_2^2-\frac{2}{n}\|z\|_1^2\ge
2\|z\|_2^2-\frac{2}{n}n\|z\|_2^2=0.
\end{eqnarray*}}
In particular it follows that $A$ has a square root, denoted
$A^{1/2}$. Let $e_1,\ldots,e_n$ be the standard unit vectors in
$\R^n$. Define $f:X\to \R^n$ by $f(i)=A^{1/2}e_i$. Now,
$$
\|f(i)-f(j)\|_2^2=\langle Ae_i,e_i\rangle+\langle
Ae_j,e_j\rangle-2\langle Ae_i,e_j\rangle=a_{ii}+a_{jj}-2a_{ij},
$$
so that if $d_{ij}=1$ then $\|f(i)-f(j)\|_2=\sqrt{4-\frac{4}{n}}$
and if $d_{ij}=2$ then $\|f(i)-f(j)\|_2=2$. It follows that
$$
\mathrm{dist}(f)=2\sqrt{\frac{n-1}{n}}.
$$
\end{proof}

\section{The Isometric Ramsey Problem}\label{section:isometric}

In this section we prove that for $n\ge 3$, $1<p<\infty$,
$R_p(n)=R_p(1,n)=3$. In
fact, we show that this is true for any uniformly convex normed space. We
begin by sketching an argument that is specific to $\ell_2$:

\begin{proposition}
$R_2(n) = 3$ for $n\ge 3$.
\end{proposition}
\begin{proof}
That $R_2(n)\geq 3$ follows since any metric space on 3
points embeds isometrically in $\ell^2_2$. To show that $R_2(n) \le
3$, we construct a metric space on $n>3$
points, no $4$-point subspace of which
embeds isometrically in $\ell_2$. Fix an integer $n>3$ and let
$\{a_i\}_{i=0}^n$ be an increasing sequence such that $a_0=0$, $a_1=1$ and for $1\le
i<n$, $a_{i+1}\ge 2(n+1)a_i$. Fix some $0<\epsilon<1/(2a_n)$. It is
easily verified that
$d(i,j)=|i-j|-\epsilon a_{|i-j|}$ is a metric on $\{1,2,\ldots,n\}$. We
show that for
$\epsilon$ small enough no four points in $(\{1,\ldots,n\},d)$ embed
isometrically in $\ell_2$. Fix four integers $1\le i_1<i_2<i_3<i_4\le n$ and
set $j=i_2-i_1$, $k=i_3-i_2$, $l=i_4-i_3$. Suppose that for every $\epsilon>0$
there exists an isometric embedding $f:(\{i_1,i_2,i_3,i_4\},d) \rightarrow
\ell_2^3$. Without loss of generality we may assume that
$f(i_1)=(\alpha,\beta,\gamma)$, $f(i_2)=(0,0,0)$, $f(i_3)=(k- \epsilon
a_k,0,0)$ and $f(i_4)=(p,q,0)$. Then: {\setlength\arraycolsep{1pt}
\begin{eqnarray*}
2\alpha(k-\epsilon a_k)&=&2\langle
f(i_1),f(i_3)\rangle\\
&=&\|f(i_1)-f(i_2)\|_2^2+\|f(i_3)-f(i_2)\|_2^2-\|f(i_3)-f(i_1)\|_2^2\\
&=& (j-\epsilon a_j)^2+(k-\epsilon a_k)^2-(j+k-\epsilon
a_{j+k})^2.
\end{eqnarray*}}
Hence,
$$
\alpha\leq
-j+\frac{\epsilon}{k}[(k+j)a_{k+j}-ja_j-ka_k-ja_k]+O(\epsilon^2).
$$
Similarly:
$$
p \geq (k+l)+\frac{\epsilon}{k}[(k+l)a_k-(k+l)a_{k+l}-ka_k+la_l]+O(\epsilon^2).
$$
Now:
{\setlength\arraycolsep{1pt}
\begin{eqnarray*}
j+k+l&-&\epsilon a_{j+k+l}=
\|f(i_4)-f(i_1)\|_2\\
&\ge&
p-\alpha\\
&\geq&j+k+l+\frac{\epsilon}{k}[(k+l)a_k-(k+l)a_{k+l}+la_l-(k+j)a_{k+j}+ja_j+ja_k]+O(\epsilon^2).
\end{eqnarray*}}
Letting $\epsilon$ tend to zero we deduce that:
{\setlength\arraycolsep{1pt}
\begin{eqnarray*}
a_{j+k+l}\le
\left(1+\frac{j}{k}\right)a_{k+j}+\left(1+\frac{l}{k}\right)a_{k+l}-\frac{l}{k}a_l-\frac{j}{k}a_j-
\frac{j+k+l}{k}a_k<2(n+1)a_{j+k+l-1},
\end{eqnarray*}}
which is a contradiction.
\end{proof}

The argument above is quite specific to $\ell_2$, and so we now consider any
uniformly convex normed space. The modulus of
uniform convexity of a normed space $X$ is defined by:

$$
\delta_X(\epsilon)=\inf\left\{1-\frac{\|a+b\|}{2};\ \|a\|,\|b\|\le 1
\quad \mathrm{and}\quad \|a-b\|\ge \epsilon\right\}.
$$
$X$ is said to be uniformly convex if $\delta_X(\epsilon)>0$ for
every $0<\epsilon\le 2$. The $L_p$ spaces $1<p<\infty$, are known
to be uniformly convex. For a uniformly convex space $X$,
$\delta_X$ is known to be continuous and strictly increasing on
$(0,2]$.

Assume that $X$ is a uniformly convex normed space and $a,b\in X \setminus\{0\}$. Then:
{\setlength\arraycolsep{1pt}
\begin{eqnarray*}
\left\|\frac{a}{\|a\|}+\frac{b}{\|b\|}\right\| &=&
\left\|\left(\frac{1}{\|a\|}+\frac{1}{\|b\|}\right)(a+b)-\frac{a}{\|b\|}-\frac{b}{\|a\|}
\right\|\\
&\ge& \left(\frac{1}{\|a\|}+\frac{1}{\|b\|}\right)\|a+b\|-\frac{\|a\|}{\|b\|}-\frac{\|b\|}
{\|a\|}\\
&=&2-\left(\frac{1}{\|a\|}+\frac{1}{\|b\|}\right)(\|a\|+\|b\|-\|a+b\|).
\end{eqnarray*}}
Now,
{\setlength\arraycolsep{1pt}
\begin{eqnarray*}
\delta_X\left(\left\|\frac{a}{\|a\|}-\frac{b}{\|b\|}\right\|\right)
\le 1-\frac{1}{2}\cdot
\left\|\frac{a}{\|a\|}+\frac{b}{\|b\|}\right\|
\le\frac{1}{2}\cdot\left(\frac{1}{\|a\|}+\frac{1}{\|b\|}\right)(\|a\|+\|b\|-\|a+b\|).
\end{eqnarray*}}
Hence
$$
\left\|\frac{a}{\|a\|}-\frac{b}{\|b\|}\right\|\le\delta_X^{-1}\left(
\frac{1}{2}\cdot\left(\frac{1}{\|a\|}+\frac{1}{\|b\|}\right)(\|a\|+\|b\|-\|a+b\|)\right).
$$
Take $x,y,z\in X$ and apply this inequality for $a=x-y$, $b=y-z$. It follows that:
\begin{eqnarray}\label{eq:use}
\nonumber\left\|y-\left(\frac{\|y-z\|}{\|x-y\|+\|y-z\|}\right.\right.\!\!\!\!\!\!\!&\cdot&\!\!\!\!\!\!\!\left.\left.
x+\frac{\|x-y\|}{\|x-y\|+\|y-z\|}
\cdot z\right)\right\| \\
&\le& \frac{\|x-y\|\cdot\|y-z\|}{\|x-y\|+\|y-z\|}\cdot
\delta_X^{-1}
\left(\frac{\|x-y\|+\|y-z\|-\|x-z\|}{\min\{\|x-y\|,\|y-z\|\}}\right).
\end{eqnarray}
This inequality is the way uniform convexity is going to be applied in the sequel. Indeed, we
have the following ``metric'' consequence of it:
\begin{lemma}\label{convexity}
Let $X$ be a uniformly convex normed space and $x_1,x_2,x_3,x_4\in X$ be distinct. Then:
{\setlength\arraycolsep{1pt}
\begin{eqnarray*}
\frac{\|x_1-x_2\|+\|x_2-x_3\|-\|x_1-x_3\|}{2\|x_2-x_3\|}
&\le&\delta_X^{-1}
\left(\frac{\|x_1-x_3\|+\|x_3-x_4\|-\|x_1-x_4\|}{\min\{\|x_1-x_3\|,\|x_3-x_4\|\}}\right)
+\\
&\phantom{\le}&+ \delta_X^{-1}
\left(\frac{\|x_2-x_3\|+\|x_3-x_4\|-\|x_2-x_4\|}{\min\{\|x_2-x_3\|,\|x_3-x_4\|\}}\right).
\end{eqnarray*}}
\end{lemma}
Lemma~\ref{convexity} is a quantitative version of the fact that
in a uniformly convex space, if $\|x_1-x_4\|$ is approximately
$\|x_1-x_3\|+ \|x_3-x_4\|$ and $\|x_2-x_4\|$ is approximately
$\|x_2-x_3\|+ \|x_3-x_4\|$ then $\|x_1-x_3\|$ is approximately
$\|x_1-x_2\|+ \|x_2-x_3\|$. This fact is geometrically evident
since the first assumption implies that $x_3$ is almost on the
line segment connecting $x_1$ and $x_4$ and $x_2$ is almost on the
line segment connecting $x_1$ and $x_3$. It follows that $x_2$ is
almost on the line segment connecting $x_1$ and $x_3$, as
required. Since we are dealing with bi-Lipschitz embeddings, we
must formulate this phenomenon without referring to``line
segments".

\begin{proof}[Proof of Lemma~\ref{convexity}]
Define:
$$
\lambda=\frac{\|x_3-x_4\|}{\|x_1-x_3\|+\|x_3-x_4\|}\quad \mathrm{and}\quad
\mu=\frac{\|x_3-x_4\|}{\|x_2-x_3\|+\|x_3-x_4\|}.
$$
An application of \eqref{eq:use} twice gives:
$$
\|x_3-(\lambda x_1+(1-\lambda) x_4)\|\le \frac{\|x_1-\|x_3\|\cdot\|x_3-x_4\|}
{\|x_1-x_3\|+\|x_3-x_4\|}
\cdot \delta_X^{-1}
\left(\frac{\|x_1-x_3\|+\|x_3-x_4\|-\|x_1-x_4\|}{\min\{\|x_1-x_3\|,\|x_3-x_4\|\}}\right),
$$
and
$$
\|x_3-(\mu x_2+(1-\mu) x_4)\|\le
\frac{\|x-2-x_3\|\cdot\|x_3-x_4\|}{\|x_2-x_3\|+\|x_3-x_4\|} \cdot
\delta_X^{-1}
\left(\frac{\|x_2-x_3\|+\|x_3-x_4\|-\|x_2-x_4\|}{\min\{\|x_2-x_3\|,\|x_3-x_4\|\}}\right).
$$
By symmetry, we may assume without loss of generality that $\lambda\le \mu$. Now,
{\setlength\arraycolsep{1pt}
\begin{eqnarray*}
\left\|x_2-\frac{\lambda(1-\mu)}{\mu(1-\lambda)}x_1-
\frac{\mu-\lambda}{\mu(1-\lambda)}x_3\right\|
&=&\frac{1}{\mu}\left\|\mu
x_2+(1-\mu)x_4-x_3+\frac{1-\mu}{1-\lambda}(x_3-\lambda x_1-
(1-\lambda)x_4)\right\|\\
&\le&
\frac{1}{\mu}\|x_3-\mu x_2-(1-\mu)x_4\|+\frac{1-\mu}{\mu(1-\lambda)}\cdot\|x_3-\lambda x_1-
(1-\lambda)x_4\|\\
&\le&
\frac{\|x_2-x_3\|+\|x_3-x_4\|}{\|x_3-x_4\|}\cdot\frac{\|x_2-x_3\|\cdot\|x_3-x_4\|}
{\|x_2-x_3\|+\|x_3-x_4\|}\cdot \\
&\phantom{\le}&\cdot\delta_X^{-1}
\left(\frac{\|x_2-x_3\|+\|x_3-x_4\|-\|x_2-x_4\|}{\min\{\|x_2-x_3\|,\|x_3-x_4\|\}}\right)+\\
&\phantom{\le}&+\frac{\|x_2-x_3\|}{\|x_3-x_4\|}\frac{\|x_1-x_3\|+\|x_3-x_4\|}{\|x_1-x_3\|}
\frac{\|x_1-\|x_3\|\cdot\|x_3-x_4\|}
{\|x_1-x_3\|+\|x_3-x_4\|}
\cdot \\
&\phantom{\le}&\cdot\delta_X^{-1}
\left(\frac{\|x_1-x_3\|+\|x_3-x_4\|-\|x_1-x_4\|}{\min\{\|x_1-x_3\|,\|x_3-x_4\|\}}\right)\\
&=&
\|x_2-x_3\|\delta_X^{-1}
\left(\frac{\|x_1-x_3\|+\|x_3-x_4\|-\|x_1-x_4\|}{\min\{\|x_1-x_3\|,\|x_3-x_4\|\}}\right)+ \\
&\phantom{\le}&+\|x_2-x_3\|\delta_X^{-1}
\left(\frac{\|x_2-x_3\|+\|x_3-x_4\|-\|x_2-x_4\|}{\min\{\|x_2-x_3\|,\|x_3-x_4\|\}}\right)
.
\end{eqnarray*}}
Additionally,
{\setlength\arraycolsep{1pt}
\begin{eqnarray*}
\|x_2-x_1\|  &\le&
\left\|x_2-\frac{\lambda(1-\mu)}{\mu(1-\lambda)}x_1-\frac{\mu-\lambda}
{\mu(1-\lambda)}x_3\right\|+
\left\|x_1-\frac{\lambda(1-\mu)}{\mu(1-\lambda)}x_1-\frac{\mu-\lambda}{\mu(1-\lambda)}x_3
\right\|\\
 &=& \left\|x_2-\frac{\lambda(1-\mu)}{\mu(1-\lambda)}x_1-\frac{\mu-\lambda}
{\mu(1-\lambda)}x_3\right\| +\frac{\mu-\lambda}
{\mu(1-\lambda)}
\|x_1-x_3\|,
\end{eqnarray*}}
and
{\setlength\arraycolsep{1pt}
\begin{eqnarray*}
\|x_2-x_3\|  &\le&
\left\|x_2-\frac{\lambda(1-\mu)}{\mu(1-\lambda)}x_1-\frac{\mu-\lambda}
{\mu(1-\lambda)}x_3\right\|+
\left\|x_3-\frac{\lambda(1-\mu)}{\mu(1-\lambda)}x_1-\frac{\mu-\lambda}
{\mu(1-\lambda)}x_3\right\|\\
 &=&\left\|x_2-\frac{\lambda(1-\mu)}{\mu(1-\lambda)}x_1-\frac{\mu-\lambda}
{\mu(1-\lambda)}x_3\right\|+\frac{\lambda(1-\mu)}
{\mu(1-\lambda)}
\|x_1-x_3\|.
\end{eqnarray*}}
Summing up these estimates gives the required result.
\end{proof}

We can now prove the main result of this section:

\begin{theorem}\label{th:convex}
Let $X$ be a uniformly convex normed space with $dim(X)\ge 2$. Then for every
$n\ge 3$, $R_{X}(n)=3$. Moreover, for every $\delta:(0,2]\to(0,\infty)$ which
is continuous, increasing and $\delta\le \delta_{\ell_2}$, let $UC_{\delta}$
be the class of all normed spaces $X$ with $\delta_X\ge \delta$. Then for
each $n\ge 3$ there is a constant $\epsilon_n(\delta)>0$ such that $R_{
UC_{\delta}}({1+\epsilon_n(\delta)},n)=3$.
\end{theorem}

The proof of Theorem~\ref{th:convex} proceeds by constructing a
space in which each quadruple violates the conclusion of
Lemma~\ref{convexity}. The construction is done iteratively, by
adding one point at a time.

\begin{proof}[Proof of Theorem~\ref{th:convex}] That $R_{X}(n)\ge 3$ follows since
any $3$ point metric embeds isometrically into any $2$ dimensional normed space,
by a standard continuity argument.

Fix some $\delta:(0,2]\to (0,\infty)$ which is continuous, increasing and
$\delta\le \delta_{\ell_2}$. We shall construct inductively a sequence
$\{M_n\}_{n=3}^{\infty}$ of
metric spaces and numbers $\{\eta_n\}_{n=3}^{\infty}$ such that:

\medskip
\noindent{\bf a)} For every $n\ge 3$, $\eta_n>0$. Each $M_n$ is a metric on $\{1,\ldots,n\}$, and we denote $d^n_{ij}=
d_{M_n}(i,j)$.

\medskip
\noindent{\bf b)} For every $1\le i<j<k \le n$,
$$
d_{i,j}^n+d_{jk}^n-d_{i,k}^n-\eta_n\ge
2d_{j,k}^n\left[\delta^{-1}\left(\frac{d_{i,k}^n+d_{k,n}^n-
d_{i,n}^n}{\min\{d_{i,k}^n,d_{k,n}^n\}}\right)+\delta^{-1}\left(\frac{d_{j,k}^n+d_{k,n}^n-
d_{j,n}^n}{\min\{d_{j,k}^n,d_{k,n}^n\}}\right)\right].
$$
Lemma~\ref{convexity} immediately implies that there is
a constant $\epsilon_n(\delta)>0$ such that for every $1\le i<j<k<l\le n$ and
for every normed space $X$ with $\delta_X\ge \delta$:
$$
c_X(\{i,j,k,l\},d_{M_n})\ge 1+\epsilon_n(\delta),
$$
as required.

$M_3$ is the equilateral metric on $\{1,2,3\}$, in which case
$\eta_3=1$. We construct $M_{n+1}=(\{1,\ldots,n+1\},d^{n+1})$
as an extension of $M_n$, by setting

$$
d^{n+1}_{n,n+1}=1-s/2 \quad \text{and} \quad \forall 1\le i< n, \ d^{n+1}_{i,n+1}=d^n_{in}+
1-s.
$$
This is indeed a definition of a metric as long as
$0< s \leq \min \{1,2\min_{1\leq i<n}d^n_{i,n}\}$ (this
fact follows from a simple case analysis).

We are left to check condition ${\bf b)}$.
Fix $1\le i<j<k\le n$. If $k\neq n$ then:
{\setlength\arraycolsep{1pt}
\begin{eqnarray*}
d_{i,j}^{n+1}+d_{j,k}^{n+1}-d_{i,k}^{n+1}-
\eta_n&=&d_{i,j}^n+d_{j,k}^n-d_{i,k}^n-\eta_n\\
&\ge&
2d_{j,k}^n\left[\delta^{-1}\left(\frac{d_{i,k}^n+d_{k,n}^n-
d_{i,n}^n}{\min\{d_{i,k}^n,d_{k,n}^n\}}\right)+\delta^{-1}\left(\frac{d_{j,k}^n+d_{k,n}^n-
d_{j,n}^n}{\min\{d_{j,k}^n,d_{k,n}^n\}}\right)\right]\\
&\ge&
2d_{j,k}^n\left[\delta^{-1}\left(\frac{d_{i,k}^n+(d_{k,n}^n+1-s)-
(d_{i,n}^n+1-s)}{\min\{d_{i,k}^n,d_{k,n}^n+1-s\}}\right)\right.\\
&\phantom{\le}&\left.+\delta^{-1}
\left(\frac{d_{j,k}^n+(d_{k,n}^n+1-s)-
(d_{j,n}^n+1-s)}{\min\{d_{j,k}^n,d_{k,n}^n+1-s\}}\right)\right]\\
&=&
2d_{j,k}^{n+1}\left[\delta^{-1}\left(\frac{d_{i,k}^{n+1}+d_{k,n+1}^{n+1}-
d_{i,n+1}^{n+1}}{\min\{d_{i,k}^{n+1},d_{k,n+1}^{n+1}\}}\right)+
\delta^{-1}\left(\frac{d_{j,k}^{n+1}+d_{k,n+1}^{n+1}-
d_{j,n+1}^{n+1}}{\min\{d_{j,k}^{n+1},d_{k,n+1}^{n+1}\}}\right)\right].
\end{eqnarray*}}

It remains to check ${\bf b)}$
for the quadruple $\{i,j,n,n+1\}$. Condition ${\bf b)}$ for $M_n$ implies that:
$$
d_{ij}^{n+1}+ d_{jn}^{n+1}-d_{in}^{n+1}\ge \eta_n.
$$
On the other hand,
\begin{multline*}
2d_{j,n}^{n+1}\left[\delta^{-1}\left(\frac{d_{i,n}^{n+1}+d_{n,n+1}^{n+1}-
d_{i,n+1}^{n+1}}{\min\{d_{i,n}^{n+1},d_{n,n+1}^{n+1}\}}\right)+\delta^{-1}
\left(\frac{d_{j,n}^{n+1}+d_{n,n+1}^{n+1}-
d_{j,n+1}^{n+1}}{\min\{d_{j,n}^{n+1},d_{n,n+1}^{n+1}\}}\right)\right]=\\
=2d_{j,n}^n\left[\delta^{-1}\left(\frac{s/2}{\min\{d_{i,n}^n,1-s/2\}}\right)+\delta^{-1}
\left(\frac{s/2}{\min\{d_{j,n}^n,1-s/2\}}\right)\right],
\end{multline*}
so that condition ${\bf b)}$ will hold
when $s$ is small enough such that the quantity
above is at most $\eta_n/2$ and with $\eta_{n+1}=\eta_n/2$.
\end{proof}

\begin{corollary}\label{cor:isometric-lp}
For all $1<p<\infty$, $R_p(n) = 3$ for $n\ge 3$.
\end{corollary}
\bigskip

We end this section with a simple lower bound for the isometric
Ramsey problem for graphs. We do not know the
asymptotically tight bound in this setting.

\begin{proposition}
Let $G$ be an unweighted graph of order $n$. Then there is a set
of $\Omega\left(\sqrt{\frac{\log n}{\log \log n}}\right)$ vertices
in $G$ whose metric embeds isometrically into $\ell_2$.
\end{proposition}
\begin{proof}
Let $\Delta$ be the diameter of $G$. The
shortest path between two diameterically far vertices is isometrically embeddable in
$\ell_2$. On the other hand, the Bourgain,Figiel,
Milman theorem \cite{bfm}
yields that for every $0<\epsilon<1$ a subset $N\subset V$ which is
$(1+\epsilon)$ embeddable in Hilbert space and
$|N|=\Omega\left(\frac{\epsilon}{\log(2/\epsilon)}\log n \right)$.
When $\epsilon=\frac{1}{2\Delta}$, such an embedding is an isometry.
Hence we can
always extract a subset of $V$ which is isometrically embeddable
in $\ell_2$ with cardinality
$$
\Omega\left(\max\left\{\Delta,\frac{\log n}{\Delta\log \Delta}\right\}\right)=\Omega
\left(\sqrt{\frac{\log n}{\log \log n}}\right),
$$
as claimed.
\end{proof}

\bigskip
\noindent{\bf Acknowledgments:} The authors would like to express
their gratitude to Guy Kindler for some helpful discussions.

\bibliographystyle{plain}


TITLE: Resolvent Operator in QM
QUESTION [5 upvotes]: In a lecture, the tutor mentioned that

"when the discrete energy spectrum becomes continuous and the poles of the resolvent shrink into a continuous line. Therefore it becomes a branch cut".

This is not clear to me. I understand that the poles of the resolvent are the energy eigenvalues, but how come the singularity become a branch point in the continuous case? It would be great if someone could make this clear to me.

REPLY [3 votes]: As nicely explained by Ron Maimon in this answer, one can think of a branch cut as a continuous line of poles each with an infinitesimal residue. For example, quoting him, $$\int_a^b \frac{1}{z-u} \mathrm du= \log \left( \frac{z-a}{z-b} \right)$$
with the latter indeed having a branch cut between $z =a $ and $z=b$.
For example if $b= - a$, the multi-valued function on the right-hand side looks like

(image courtesy of MIT OpenCourseWare)


TITLE: How to solve recurrence relations by the generalized hypergeometric series
QUESTION [5 upvotes]: I am reading methods of solving recurrence relation on Wikipedia. There is one method:

Many linear homogeneous recurrence
  relations may be solved by means of
  the generalized hypergeometric series.
  Special cases of these lead to
  recurrence relations for the
  orthogonal polynomials, and many
  special functions. For example, the
  solution to $$J_{n+1}=\frac{2n}{z}J_n-J_{n-1}$$
  is given by $$J_n=J_n(z), \,$$ the Bessel function.

There are no description regarding how to use the method of "generalized hypergeometric series", nor can I find some on the article for generalized hypergeometric series or on Bessel function. I was wondering if someone here can explain somehow or gives some references about that? Thanks and regards!

REPLY [9 votes]: See the (on-line, downloadable) book

A = B, by Petkovsek, Wilf, and Zeilberger

It gives all sorts of links between hypergeometric series and recurrence relations.


TITLE: Proving oscillation of function on subset of domain
QUESTION [3 upvotes]: Let $ \emptyset\ne A\subset \mathbb{R}^n$ and let $f:A\to \mathbb{R}$ be a bounded function. There is the following definition:
For every non-empty subset $B$ of $A$, the oscillation of $f$ on $B$ is the number osc$(f)_B:=\sup(f)_B - \inf(f)_B$.
I want to prove that osc$(f)=\sup\{|(f(x))-(f(y))|:x,y \in B\}.$
I'm not quite sure how to go about a proof, but here's my attempt.
$\sup_B(f)-\inf_B(f)=\sup_B(f)+\sup_B(-f)=\sup\{f(x):x\in B\}+\sup\{-f(y):y\in B\}$
$=\sup\{f(x)+(-f(y)):x,y\in B\}$
$=\sup\{|f(x)-f(y)|:x,y\in B\}$ (the absolute value can be applied, since $\sup_B(f)\ge \inf_B(f)$).
Do you think this is correct / rigorous enough?

REPLY [2 votes]: According to "sequence", I give a restatement of his result, and then give a proof. Please check if my proof is right.
Lemma:
Let $f\colon X\to\mathbb{R}$ and $A\subset X$ with $A\neq \emptyset.$ If $f$ is bounded on $A,$ then
\begin{gather*}
 \sup\{f(x)\mid x\in A\}-\inf\{f(x)\mid x\in A\}=\sup\{\left|f(x)-f(y)\right| \mid x,y\in A\}.
\end{gather*}
Proof:
    Let $x\in A$ and $y\in A.$ Since $f$ is bounded on $A,$  we have $f(x)\leq \sup\{f(t)\mid t\in A\}$ and $f(y)\geq \inf\{f(t)\mid t\in A\}.$ Thus $-f(y)\leq -\inf\{f(t)\mid t\in A\}.$ It follows that $f(x)-f(y)\leq \sup\{f(t)\mid t\in A\}-\inf\{f(t)\mid t\in A\}.$ Similarly  (or just by swapping $x$ and $y$), we also have $f(y)-f(x)\leq \sup\{f(t)\mid t\in A\}-\inf\{f(t)\mid t\in A\}.$ Hence $|f(x)-f(y)|\leq \sup\{f(t)\mid t\in A\}-\inf\{f(t)\mid
t\in A\}.$ So we deduce that $\sup\{|f(x)-f(y)|\mid x,y\in A\}\leq \sup\{f(t)\mid t\in A\}-\inf\{f(t)\mid t\in A\}.$ Conversely,
for each $u$ and $v$ in $A,$ we have $f(u)=f(u)-f(v)+f(v)\leq |f(u)-f(v)|+f(v)\leq \sup\{|f(x)-f(y)|\mid x,y\in A\}+f(v).$ Because
$u$ is arbitrary, we have $\sup\{f(u)\mid u\in A\}\leq \sup\{\left|f(x)-f(y)\right|\mid x,y\in A\}+f(v),$ which implies that
$f(v)\geq \sup\{f(u)\mid u\in A\}-\sup\{|f(x)-f(y)|\mid x,y\in A\}$ and so $\inf\{f(v)\mid v\in A\}\geq \sup\{f(u)\mid u\in A\}-\sup\{|f(x)-f(y)|\mid x,y\in A\}.$ By rearrangement, we have $\sup\{f(u)\mid u\in A\}-\inf\{f(v)\mid v\in A\}\leq \sup\{|f(x)-f(y)|\mid x,y\in A\}.$ Therefore w have shown that $\sup\{f(u)\mid u\in A\}-\inf\{f(v)\mid v\in A\}= \sup\{|f(x)-f(y)|\mid x,y\in A\}.$