## Wednesday, November 26, 2014

### Killing the Hydra

You might like this nice post on an old game with modern ramifications.

## Thursday, November 13, 2014

### Alexandre Grothendieck has passed

http://www.math.columbia.edu/~woit/wordpress/?p=7335

## Friday, October 31, 2014

### A new method of constructing connections on vector bundles


Suppose that $M$ is a smooth manifold of dimension $E\to M$ is a real, smooth vector bundle of rank $\nu$ over $M$.      We define  a pairing on $E$ to be  a section of the bundle $E^*\boxtimes E^*\to M\times M$, where  $E^*\boxtimes E^*$ is the vector bundle $\pi_1^* E^*\otimes \pi_2^*E^*$, $\pi_i(x_1,x_2)=x_i$,  $\forall (x_1,x_2)\in M\times M$, $i=1,2$.

For $x,y\in M$ we can view $B_{x,y}\in E_x^*\times E^*_y$ as a bilinear map

$$B_{x,y}: E_x\times E_y\to \bR.$$

This  induces a  linear map
$$S_{x,y}= S(B)_{x,y}: E_y\to E^*_x.$$
We say that the pairing is nondegenerate if for any $x\in M$ the bilinear map $B(x,x): E_x\times E_x\to \bR$ is nondegenerate. In particular, this induces an isomorphism
$$S_x=S_{x,x}: E_x\to E^*_x.$$

We obtain tunneling operators

$$T(x,y)= S_x^{-1}S_{x,y}: E_y\to E_x.$$


The local frame  $\underline{\be}$ $\newcommand{\ube}{{\underline{\boldsymbol{e}}}}$ defines a bundle isomorphism $\Phi(\underline{\be}):\ur_\eO\to E_\eO$.  In the local frame $\ube$ the tunelling are represented by$\DeclareMathOperator{\Endo}{End}$ $\DeclareMathOperator{\Aut}{Aut}$ a tunneling map

$$T_\ube:\eO\times \eO\to\Endo(\bR^\nu),\;\;T_\ube(x,y)= \Phi_x(\ube)^{-1}T(x,y)\Phi_y(\ube) .$$

Note that $T_\ube(x,x) = \mathbf{1}_{\bR^\nu}$.  For $i=1,\dotsc, m$ define

$$\Gamma_i(\ube):\eO\to \Endo(\bR^\nu),\;\;\Gamma_i(\ube,x)=-\pa_{x^i}T_\ube(x,y)\bigl|_{y=x}.$$

We set

$$\Gamma(\ube,x)=\sum_{i=1}^m \Gamma_i(\ube, x) dx^i=-d_x T(x,y)\bigl|_{y=x}\in \Endo\bigl(\;\bR^\nu\;\bigr)\otimes \Omega^1(\eO),$$
$$\Phi_x(\ubf) =\Phi_x(\ube)\circ g(x),\;\;\forall x\in\eO.$$
Then
$$T_\ubf(x,y)=g(x)^{-1} T_\ube(x,y) g(y),$$
$$\Gamma(\ubf,x) = -d_x\bigl(\; g(x)^{-1}\;\bigr)\bigl|_{y=x} T_\ube(x,x)g(yx+g^{-1}(x) \Gamma(\ube,x) g(x)=g(x)^{-1}dg(x)+g^{-1}(x) \Gamma(\ube,x) g(x).$$

This proves that the correspondence $\ube\mapsto \Gamma(\ube)$ defines a connection on $E$. We will denote it by $\nabla^B$ and we will refer to it as the connection  associated to the nondegenerate pairing $B$.

Let us compute its  curvature $R^B$.  Using the local frame $\ube$ we can write
$$R^B= \sum_{1\leq i<j\leq m} R_{ij}(\ube, x)dx^i\wedge dx^j\in \Endo(\bR^\nu)\otimes\Omega^2(\eO),$$
where
$$R_{ij}(\ube,y)=\pa_{x^i}\Gamma_j(\ube,x)-\pa_{x^j}\Gamma_i(\ube,x)+[\Gamma_i(\ube,x),\Gamma_j(\ube,x)].$$

Using the local frame $\ube$ we represent $S_{x,y}$ as a $\nu\times \nu$-matrix

$$S_\ube(x,y)=\bigl(\; s_{\alpha\beta}(x,y)\,\bigr)_{1\leq\alpha,\beta\leq \nu},\;\;s_{\alpha\beta}(x,y)=B_{x,y}\bigl(\,\be_\alpha(x),\be_\beta(y)\;\bigr).$$
Then $T_\ube(x,y)=S_\ube(x,x)^{-1} S_\ube(x,y)$, and
$$\Gamma_i(\ube,x)= -\pa_{x^i} S_\ube(x,x)^{-1}\bigl|_{y=x} S_\ube(x,x)-S_\ube(x,x)^{-1} \pa_{x^i}S_\ube(x,y)\bigl|_{y=x}$$

= S_\ube(x,x)^{-1}\pa_{x^i} S_\ube(x,x)\bigl|_{x=y}-S_\ube(x,x)^{-1}  \pa_{x^i}S_\ube(x,y)\bigl|_{y=x} =S_\ube(x,x)^{-1}  \pa_{y^i}S_\ube(x,y)\bigl|_{y=x}.\label{gamma}

We have
$$\pa_{x^i}\Gamma_j(\ube,x)=\pa_{x^i}\Bigl(\; S_\ube(x,x)^{-1} \pa_{y^j}S_\ube(x,y)\bigl|_{y=x}\;\Bigr)$$

$$= -S_\ube(x,x)^{-1}\Bigl(\pa_{x^i}S_\ube(x,x)\;\Bigr) S_\ube(x,x)^{-1} \pa_{y^j}S_\ube(x,y)\bigl|_{x=y}+ S_\ube(x,x)^{-1}\pa_{x^i}\Bigl( \; \pa_{y^j}S_\ube(x,y)\bigl|_{x=y}\;\Bigr)$$

=- S_\ube(y,y)^{-1}\Bigl(\pa_{x^i}S_\ube(x,y)+\pa_{y^i}S_\ube(x,y)\;\Bigr)_{x=y}S_\ube(y,y)^{-1}   \pa_{y^j}S_\ube(x,y)\bigl|_{x=y} +S_\ube(y,y)^{-1} \pa^2_{x^iy^j}S_\ube(x,y)\bigr|_{x=y}. \label{1}

We can simplify the computations a bit if we choose the frame  $\ube$ judiciously.  Fix a distinguished point in $\eO$ and assume it is the orgin in the coordinates $(x^i)$. Note that if $x$ is sufficiently close to $0$, then $T(x,0)$ is an isomorphism $E_0\to E_x$. We set

$$\bsf_\alpha(x): = T(x,0)\be_\alpha(0).$$

More explicitly

$$\bsf_\alpha(x)=\sum_{\gamma,\lambda} s^{\gamma\lambda}(x)s_{\lambda \alpha}(x,0)\be_\gamma(x),$$

where $(s^{\gamma\lambda}(x))$ is the inverse of the matrix $(s_{\alpha\beta}(x) )$. $\newcommand{\one}{\mathbf{1}}$

In this frame we have  $T_\ubf(x,0)=\one$ and we deduce that

\Gamma(\ubf, 0)=0.
\label{2}

On the other hand,
$\Gamma_i(\ubf,0)=S_\ubf(0,0)^{-1}\pa_{y^i}S_\ubf(0,y)\bigr|_{y=0}.$
We deduce that for this special frame we have
$\pa_{y^i}S_\ubf(0,y)\bigr|_{y=0}=0.$
Using this in (\ref{1}) we deduce

\pa_{x^i}\Gamma_j(\ubf,0)=S_\ubf(0,0)^{-1} \pa^2_{x^iy^j}S_\ubf(0,0),
\label{3}

and thus

R_{ij}(\ubf,0)=S_\ubf(0,0)^{-1}\Bigl(\pa^2_{x^iy^j}S_\ubf(0,0)-\pa^2_{x^jy^i}S_\ubf(0,0)\Bigr).
\label{curv}

Remark.    We say that the pairing $B$ is symmetric  if  for any $x,y\in M$ and any $u\in Y_x$, $v\in E_y$ we have
$B_{x,y}(u,v)=B_{y,x}(v,u).$
Observe that the symmetry  condition is equivalent to requiring that  the tunneling $T_{x,y}: E_y\to E_x^*$ is  self-adjoint, i.e., the adjoint $T_{x,y}^*: (E_x^*)^*\to E_y^*$ coincides with $T_{y,x}$.

In this case  it is not easy to prove that the bilinear form $\hat{B}\in C^\infty(E^*\otimes E^*)$, $\hat{B}_x=B_{x,x}$, is  covariant constant

\nabla^B \hat{B}=0.
\label{const}

## Wednesday, October 15, 2014

### The Unreasonable Effectiveness of Mathematics in the Natural Sciences

This a good read for any person interested in Math.

The Unreasonable Effectiveness of Mathematics in the Natural Sciences

## Friday, August 15, 2014

### ICM2014 — Bhargava laudatio | Gowers's Weblog

ICM2014 — Bhargava laudatio | Gowers's Weblog

## Saturday, August 9, 2014

### A nice piece on the Fields medal

http://www.nytimes.com/2014/08/10/opinion/sunday/how-math-got-its-nobel-.html?hp&action=click&pgtype=Homepage&module=c-column-top-span-region®ion=c-column-top-span-region&WT.nav=c-column-top-span-region&_r=1

## Friday, May 16, 2014

### Quillen Notebooks | Clay Mathematics Institute

The Clay institute is making Quillen's notebooks available to the public!

Quillen Notebooks | Clay Mathematics Institute

## Thursday, March 13, 2014

### How to Fix Issues with MathJax and Blogger Preview

If you used MathJax on blogger you may have noticed that the preview  does not render the LaTex    output.  At the link below  you can find a simple way to fix it.

Clueless Fundatma: Issues with MathJax and Blogger Preview

## Wednesday, March 12, 2014

### Random convex polygons I.

It's been a long time since I last posted something here;    busy, not  having  something relevant to say, you name it.    lately I've been  excited by  all things probabilistic. Somehow I find this area fresh. The  fact that I am novice   may contribute to  this excitement.

A few months ago, in the coffee room of our department,  I stumbled   on an older Bulletin A.M.S and, since I did not have anything pressing to do, I opened it  and saw a nice survey by a  Hungarian mathematician   called  Imre Barany.   I was intrigued by the  title of this beautiful survey: Random points and lattice points in convex bodies.     I found many marvelous  questions there, questions that   never came close to my mind.

One of the problems mentioned  in that survey  was a problem posed and  solved by Renyi and Sulanke sometime in the 60s.    Unfortunately, their results were written  in German which for me  is synonym with  Verboten.     I tried to find  an English   source for this paper, and all my Google searches  were fruitless.     Still,  I was very curious how they did it.  So, armed with Google Translate, patience and curiosity I proceeded to read the first of their 3 papers.  What follows is  an exposition of a small part of the first paper.  For more details   you can look in the first paper of Renyi-Sulanke, that is, if you  know enough German to read a math paper.  First the problem. $\newcommand{\bR}{\mathbb{R}}$ $\DeclareMathOperator{\area}{Area}$


Surprisingly, the answer depends in a  dramatic fashion on the regularity  of the boundary $\newcommand{\pa}{\partial}$ $\pa C$ of $C$.   Here I only want to investigate the behavior of $\bE(V_n)$, the expected number of vertices of $\Delta_n$   when $\pa C$ is smooth and has positive curvature at every point.   The final answer is the asymptotic relation (\ref{RSv}).

For two points $P,Q\in C$ we denote by $L(P,Q)$ the line determined by them.  Denote by $\bP(P,Q)$ the probability that $(n-2)$  points chosen randomly  from $C$ lie on the side of $L(P,Q)$.  For $i\neq j$ we denote by $E_{ij}$  the event that all the points $P_k$, $k\neq i,j$ lie on the same side of  of $L(P_i,P_j)$.

The first crucial observation, one that I missed because I am still not thinking as a probabilist,  is that

V_n=\sum_{i<j} I_{E_{ij}}.

In particular,
$\bE(V_n)=\sum_{i<j}\bE( I_{E_{ij}})= \sum_{i<j}\bP(E_{ij}).$
Since the points $\{P_1,\dotsc, P_n\}$ are independent and identically distributed  we deduce that
$\bP(E_{ij})=\bP(E_{i'j'}),\;\;\forall i<j,\;\;i'<j'.$
We denote by $p_n$ the  common probability of the events $E_{ij}$. Hence

\bE(V_n)=\binom{n}{2} p_n.

To compute $p_n =\bP(E_{12})$ we observe that
$\bP(E_{12}) =\int_C\int_C \bP(P_1,P_2) dP_1dP_2.$
The line $L(P_1,P_2)$ divides   the region $C$ into two regions $R_1, R_2$ with areas $a_1(P_1,P_2)$ and $a_2$.   The probability  that   $n-2$  random independent points from  $C$   lie in $R_1$ is  $a_1(P_1,P_2)^{n-2}$    and  the probability  $n-2$  random independent points from $C$  lie in $R_2$ is $a_2(P_1,P_2)^{n-2}$.  We set
$a(P_1,P_2):=\min\bigl\{ a_1(P_1,P_2),\;a_2(P_1,P_2)\}$
and we deduce that
$\bP(P_1,P_2)= a(P_1,P_2)^{n-2}+\bigl(\;1-a(P_1,P_2)\;\bigr)^{n-2},$

\bE(V_n)=\binom{n}{2}\int_C\int_C\Bigl\{ \; a(P,Q)^{n-2}+\bigl(\;1-a(P,Q)\;\bigr)^{n-2}\;\bigr\} dPdQ.

Since $a(P,Q)^{n-2}\leq \frac{1}{2^{n-2}}$,  we deduce that

\bE(V_n)\sim \binom{n}{2} \int_C\int_C \bigl(1-a(P,Q)\bigr)^{n-2} dPdQ\;\;\mbox{as $n\to\infty$}.
\label{4}

To proceed further we need to use a formula from integral geometry.     Renyi and Sulanke refer to another  German source, a book of integral geometry by  Blaschke.  Fortunately, there is a  very good English substitute to Blaschke's book  that contains a myriad of   exotic  formulas. I am referring of course to  Luis  Santalo's   classical monograph   Integral Geometry and Geometric Probability.  (Santalo was Blaschke's student.)

In  Chapter 4, section 1,  Santalo  investigates the   density of pairs of points, more precisely  the  measure $dPdQ$ used in (\ref{4}). More precisely,  he discusses a clever choice of coordinates  that is particularly useful in integral geometry.

The  line $L(P,Q)$ has a normal $\newcommand{\bn}{\boldsymbol{n}}$ $\bn=\bn(p,q)$
$\bn =(\cos \theta,\sin \theta), \;\;\theta\in [0,2\pi],$
and it is described by a linear equation.
$x\cos \theta+y\sin \theta = p,\;\;p\geq 0.$
Once we  fix a linear isometry $T: L(P,Q)\to \bR$, we can identify  $P,Q$ with  two points  $t_1,t_2\in \bR$. Note that $|dt_1dt_2|$ and $|t_1-t_2|$ are independent of the choice of $T$.    Santalo  op. cit.  shows that
$|dPdQ|=|t_1-t_2| |dp d\theta dt_1dt_2|.$
Now observe that the line  $L(P,Q)$ is determined only by the two parameters $p,\theta$ so we will denote it by $L(p,\theta)$.     Similarly, $a(P,Q)$ depends only on $p$ and $\theta$ and we will denote it by $a(p,\theta)$. We set
$p_0(\theta)=\max\{ s;\;\;s\geq 0, s\bn(\theta)\in C\,\bigr\}.$
We denote  by $S(p,\theta)$ the segment on $L(p,\theta)$ cut-out by  $C$ and by $\ell(p,\theta)$ its length.

\bE(V_n)\sim \binom{n}{2}\int_0^{2\pi}\int_0^{p_0(\theta)}\bigl(1-a(p,\theta)\;\bigr)^{n-2}\left(\int_{S(s,\theta)\times S(s,\theta)} |t_1-t_2|dt_1dt_2\right) dp d\theta.

Observing that for any $\ell>0$ we have
$\int_{[0,\ell]\times[0,\ell]}|x-y|dxdy=\frac{ \ell^3}{3}$
we deduce

\bE(V_n)\sim \frac{1}{3}\binom{n}{2}\int_0^{2\pi}\int_0^{p_0(\theta)}\bigl(1-a(p,\theta)\;\bigr)^{n-2}\ell(s,\theta)^3dp d\theta.

Now comes the  analytical part.   For each $\theta\in [0,2\pi]$ we set
$I_n(\theta):=\frac{1}{3}\binom{n}{2}\int_0^{p_0(\theta)}\bigl(1-a(p,\theta)\;\bigr)^{n-2}\ell(s,\theta)^3dp ,$
so that
$\bE(V_n)\sim \int_0^{2\pi} I_n(\theta) d\theta.$
Renyi and Sulanke  find the asymptotics of  $I_n(\theta)$ as $n\to \infty$ by a  disguised version of the  old reliable Laplace method.


For simplicity  we set $a(p):=a(p,\theta)$.  For  $p\in [0,p_0(\theta)]$ we denote by $A(p)=A(p,\theta)$  the  area of the cap of $C$  determined by the  line $L(p,\theta)$ and the tangent line to $\pa C$ at $P(\theta)$  $x\cos\theta+y\sin\theta=p_0(\theta)$. In  Figure 1 below,  this cap is the yellow region between the green and the red line.

 Figure 1: Slicing a convex region in directions perpendicular to $\bn(\theta)$. The circle in red is the osculating circle to the boundary at the unique  point $P(\theta)$ where the unit outer normal is $\bn(\theta)$.

Observe that $A(p)=a(p)$ as long as $A(p)\leq \frac{1}{2}$.  It could happen that $A(p)> \frac{1}{2}\area(C)=\frac{1}{2}$ for some $p \in [0,p_0(\theta)]$. In any case, the uniform convexity  of $\pa C$ shows that  we can find  $\newcommand{\si}{\sigma}$  $\si_0> 0$  and  $0<c<\frac{1}{2}$ with the following properties.

\si_0<\inf_{\theta\in [0,2\pi]} p_0(\theta) .

a(p,\theta)=A(p,\theta),\;\;\forall  p\in [\si_0,p_0(\theta)].

c<a(p,\theta)\leq\frac{1}{2},\;\;\forall  p\in [0,s_0].
\label{low}

\frac{d\ell}{dp}<0\;\;\mbox{on $[\si_0,p_0]$}.

We have
$I_n(\theta)\sim\frac{n^2}{6}\int_0^{p_0(\theta)}\bigl(1-a(p)\,\bigr)^{n-2} \ell(p) dp$
$=\underbrace{\frac{n^2}{6}\int_0^{\si_0}\bigl(1-a(p)\,\bigr)^{n-2} \ell(p) dp}_{=:I_n^0(\theta)}+\frac{n^2}{6}\underbrace{\int_{\si_0}^{p_0(\theta)}\bigl(1-a(p)\,\bigr)^{n-2} \ell(s) dp}_{=J_n(\theta)}.$
The condition (\ref{low})  implies that as $n\to\infty$ we have $I_n^0(\theta)=o(1)$, uniformly in $\theta$.  Thus

I_n(\theta)\sim\frac{n^2}{6} J_n(\theta),\;\;n\to\infty.

We will use   Laplace's method  to   estimate $J_n(\theta)$.      We introduce a new variable $\tau=\tau(p)=p_0(\theta)-p$, $s\in [\si_0, p_0(\theta)]$ so that $\tau\in [0,\tau_0(\theta)]$, $\tau_0(\theta)=p_0(\theta)-\si_0$.   Geometrically, $\tau$ denotes the distance to the tangent line $L(p_0(\theta),\theta)$.

We will denote by $L(\tau)$ the line  $L(p,\theta)$. Thus, as $\tau$ increases the line $L(\tau)$ moves away from the boundary point $P(\theta)$ and towards the origin.  Similarly, we set $a(\tau)=a(p)=a(p,\theta)$ etc. Hence
$J_n(\theta)=\int_0^{\tau_0} \bigl(1-a(\tau))^{n-2}\ell(\tau)^3 d\tau.$
Observe first that for  along the interval $[0,\tau_0]$ we have
$\frac{d a}{d\tau}=\frac{dA}{d\tau}=\ell(\tau).$
The line  $L(\tau)=L(p,\theta)$  intersects the osculating  circle   to $\pa C$ at $P(\theta)$ along a chord of length $\bar{\ell}(\tau)$.     We set $t:=\sqrt{\tau}$. From the definition of the osculating circle we deduce
$\ell(t)=\bar{\ell}(t)(1+o(1)),\;\;\frac{d\ell}{dt}=\frac{d\bar{\ell}}{dt}(1+o(1))\;\;\mbox{as t\to 0}.$
If $r=r_\theta$ denotes the radius of the osculating circle so that $\frac{1}{r_\theta}$ is the curvature of $\pa C$ at $P(\theta)$, then
$\bar{\ell}(\tau)= 2\sqrt{r^2-(r-\tau)^2}=2\sqrt{2r\tau-\tau^2}= 2\sqrt{\tau}\sqrt{2r-\tau}=2t\sqrt{2r-t^2}.$
Hence
$\frac{d\bar{\ell}}{dt}|_{t=0}=2\sqrt{2r}$
We denote the $t$-derivative by an upper dot $\dot{}$. Note that

\dot{a}=\frac{d\tau}{dt}\frac{da}{d\tau}= 2t\ell(t),\;\;\dot{\ell}(t)=\dot{\ell}(0)+O(t)=2\sqrt{2r}+O(t).
\label{dota}

We deduce

J_n(\theta)=2\int_0^{t_0} \bigl( 1-a(t)\bigr)^{n-2} \ell(t)^3tdt,\;\;t_0=\sqrt{\tau_0}.

Note that
$\ddot{a}(t)=2\ell(t)+2t\dot{\ell}(t),\;\;\frac{d^3a}{dt^3}=4\dot{\ell}(t)+2t\ddot{\ell}(t),$
so that
$a(0)=\dot{a}(0)=\ddot{a}(0)=0, \;\;\frac{d^3 a}{dt^3}|_{t=0}= 4\dot{\ell}(0)=8\sqrt{2r}.$
We deduce
$a(t)=\frac{8\sqrt{2r}}{6}t^3+O(t^4)= \underbrace{\frac{4\sqrt{2r}}{3}}_{=:C(r)}\;\;t^3+O(t^4).$
We set
$\nu:=(n-2),\;\; w_\nu(t)= \bigl( 1-a(t)\bigr)^{\nu} \ell(t)^3t,$
$\frac{u}{\nu}:=C(r)t^3\iff t=\left(\frac{u}{C(r)\nu}\right)^{\frac{1}{3}}.$
Note that
$\ell(t)^3=\bigl( 2\sqrt{2r}t+O(t^2)\;\bigr)^3=(6C(r) t)^3 + O(t^4).$
Hence
$w_\nu(t)dt = \left( 1-\frac{u}{\nu} +O\left(\frac{u}{\nu} \right)^{4/3}\;\right)^\nu \left(\frac{((6C(r))^3}{C(r)\nu} u+ O\left(\frac{u}{\nu}\right)^{4/3}\;\right) \left(\frac{u}{C(r)\nu}\right)^{1/3}\left(\frac{1}{C(r)\nu}\right)^{1/3}\frac{1}{3}u^{-2/3} du$
$=\frac{(6C(r))^3}{3C(r)^{5/3}\nu^{5/3}} \left( 1-\frac{u}{\nu} +O\left( \frac{u}{\nu} \right)^{4/3}\;\right)^\nu u^{2/3} \left( 1+ O\left(\frac{u^{1/3}}{\nu^{1/3}}\right)\;\right) du$
$=\frac{6^3C(r))^{4/3}}{3\nu^{5/3}} \left( 1-\frac{u}{\nu} +O\left( \frac{u}{\nu} \right)^{4/3}\;\right)^\nu u^{2/3} \left( 1+ O\left(\frac{u^{1/3}}{\nu^{1/3}}\right)\;\right) du$
Hence
$J_n(\theta)=\frac{6^3C(r))^{4/3}}{3\nu^{5/3}}\underbrace{\int_0^{u_\nu} \left( 1-\frac{u}{\nu} +O\left( \frac{u}{\nu} \right)^{4/3}\;\right)^\nu u^{2/3} \left( 1+ O\left(\frac{u^{1/3}}{\nu^{1/3}}\right)\;\right) du}_{=:\hat{J}_\nu},\;\; u_\nu=\nu C(r)t_0^3.$
Now observe that
$\frac{6^3C(r)^{4/3}}{3}= \underbrace{\frac{1}{3}6^3 \left(4\sqrt{2}{3}\right)^{4/3}}_{=:Z_1} r^{2/3}.$
and
$\lim_{\nu\to\infty} \hat{J}_\nu=\int_0^\infty e^{-u} u^{2/3}=\Gamma(5/3).$
Thus
$J_n(\theta) \sim Z_1\Gamma(5/3)r^{2/3}\nu^{-5/3}\sim Z_1\Gamma(5/3)r^{2/3}n^{-5/3},$
$I_n(\theta) \sim \frac{n^2}{6}J_n(\theta)\sim \frac{Z_1\Gamma(5/3)r^{2/3}}{6} n^{1/3}.$
Now observe that the curvature  at the point $P(\theta)$ is $\kappa(\theta)=\frac{1}{r_\theta}$.  hence
$I_n(\theta) \sim \frac{n^2}{6}J_n(\theta)\sim \frac{Z_1\Gamma(5/3)\kappa(\theta)^{-2/3}}{6} n^{1/3}$
If we denote by $ds$ the arclength on $\pa C$, then, by definition
$\frac{d\theta}{ds}=\kappa(\theta)\iff d\theta=\kappa ds.$
Thus

\bE(V_n)\sim \int_0^{2\pi} I_n(\theta) d\theta \sim  \frac{Z_1\Gamma(5/3)}{6} n^{1/3}\int_{\pa C} \kappa^{1/3} ds.
\label{RSv}

This is one of Renyi-Sulanke's result.

Remark.  Before I close, let me mention that  the asymptotics of $\bE(V_n)$ for  $n$ large depends dramatically on  the regulariti of the boundary of $C$. For example, if $C$ itself is a convex polygon with $r$-vertices, then  Renyi-Sulanke show that
$\bE(V_n) \sim\frac{2r}{3}\log n.$
Compare this with the smooth case  when the convex hull is expected to have many more  vertices $\approx n^{1/3}$.

There are more to say about this  story. As the title indicates, I plan to return to it in a later post.

## Wednesday, January 8, 2014

### Great Lakes Geometry Conference

This will be a great event!

Great Lakes Geometry Conference

## Wednesday, January 1, 2014

### Why boycotting the assholes at Elsevier should be one of your New Year resolutions

We finally have a confirmation of the reason why Elsevier muzzles the libraries  concerning  the price they have to pay for an Elsevier subscription. Free markets work best when information flows freely. Apparently Elsevier believes that the more we know about their bussines, the more they would stink. Wouldn't it be awesome if some  economist submitted to  an Elsevier journal a research on the anti-market attitudes of  Elsevier, and then have this  rejected. That would be so meta. In any case,  below is Elsevier in its own words (hat tip to Tim Gowers)

http://svpow.com/2013/12/20/elseviers-david-tempest-explains-subscription-contract-confidentiality-clauses/