Hope to add bits of mathematics or mathematical news I find interesting.
Friday, October 23, 2015
Monday, October 19, 2015
Wednesday, October 7, 2015
Wednesday, September 9, 2015
Friday, February 20, 2015
Springer Verlag is one fucked-up company: the Alzheimer years. (Please repost)
In an earlier post I described my experience with Springer's new editing practices. Briefly, they're cheap SOBs. ( SOB:= sonovabitch) The saga continues.
Two days ago I received this nice letter from them, thanking me for my refereeing services. Part of the thank-you was an electronic discount token, good for two weeks that I can use to purchase one book of my choice at 50% discount.
The problem is that their shopping website is either not working, or I may be blacklisted. For several months I could not purchase anything there. That did not bother me too much because I could always go to the more reliable Amazon site. I cannot do this with my stupid token: I have to use it on their website which is reliably not working. My attempts to contact the customer service were fruitless (as I write this).
Here is a company that forgot how books are edited, cannot sell them on electronic platforms, and cannot handle customer inquires.
I was pissed off, but then I became worried: maybe one of world's largest media company is run by people suffering from early onset Alzheimer. This seems the most plausible explanation for a behavior displaying forgetfulness of the most basic business practices. Or maybe their just bumbling idiots in charge of the Titanic.
Two days ago I received this nice letter from them, thanking me for my refereeing services. Part of the thank-you was an electronic discount token, good for two weeks that I can use to purchase one book of my choice at 50% discount.
The problem is that their shopping website is either not working, or I may be blacklisted. For several months I could not purchase anything there. That did not bother me too much because I could always go to the more reliable Amazon site. I cannot do this with my stupid token: I have to use it on their website which is reliably not working. My attempts to contact the customer service were fruitless (as I write this).
Here is a company that forgot how books are edited, cannot sell them on electronic platforms, and cannot handle customer inquires.
I was pissed off, but then I became worried: maybe one of world's largest media company is run by people suffering from early onset Alzheimer. This seems the most plausible explanation for a behavior displaying forgetfulness of the most basic business practices. Or maybe their just bumbling idiots in charge of the Titanic.
Wednesday, November 26, 2014
Thursday, November 13, 2014
Wednesday, November 5, 2014
Friday, October 31, 2014
A new method of constructing connections on vector bundles
This post was suggested by a question on the MathOverflow site. After I answered part of it I noticed that it is related to a recent work of mine of a probabilistic nature. What follows involves no probability. $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\pa}{\partial}$ As far as the terminology concerning connections, I'll stick to the terminology in Section 3.3. of my book.
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
$$T(x,y)= S_x^{-1}S_{x,y}: E_y\to E_x. $$
Fix an open coordinate patch $\newcommand{\eO}{{\mathscr{O}}}$ $\eO\subset M$ with coordinates $(x^i)_{1\leq i\leq m}$. Assume $\eO$ is sufficiently small so $E$ trivializes over $\eO$. Suppose that $\newcommand{\be}{\boldsymbol{e}}$ $\underline{\be}(x)=(\be_\alpha(x))_{1\leq \alpha\leq \nu}$ is a local frame of $E$ over $\eO$. We denote by $\newcommand{\ur}{\underline{\mathbb{R}}}$ $\ur_\eO$ the trivial vector bundle $\bR^\nu\times \eO\to\eO$.
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
$$\Gamma(\ube,x)=\sum_{i=1}^m \Gamma_i(\ube, x) dx^i=-d_x T(x,y)\bigl|_{y=x}\in \Endo\bigl(\;\bR^\nu\;\bigr)\otimes \Omega^1(\eO),$$
where $d_x$ denotes the differential (exterior derivative) with respect to the $x$-variables. $\newcommand{\bsf}{\boldsymbol{f}}$ $\newcommand{ubf}{{\underline{\boldsymbol{f}}}}$ If $\ubf$ is another local frame of $E_\eO$, then there exists a smooth map $g:\eO\to\Aut(\bR^\nu)$ such that
$$\Phi_x(\ubf) =\Phi_x(\ube)\circ g(x),\;\;\forall x\in\eO. $$
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
$$\Gamma_i(\ube,x)= -\pa_{x^i} S_\ube(x,x)^{-1}\bigl|_{y=x} S_\ube(x,x)-S_\ube(x,x)^{-1} \pa_{x^i}S_\ube(x,y)\bigl|_{y=x} $$
\begin{equation}
= 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}
\end{equation}
We have
$$\pa_{x^i}\Gamma_j(\ube,x)=\pa_{x^i}\Bigl(\; S_\ube(x,x)^{-1} \pa_{y^j}S_\ube(x,y)\bigl|_{y=x}\;\Bigr) $$
$$ = -S_\ube(x,x)^{-1}\Bigl(\pa_{x^i}S_\ube(x,x)\;\Bigr) S_\ube(x,x)^{-1} \pa_{y^j}S_\ube(x,y)\bigl|_{x=y}+ S_\ube(x,x)^{-1}\pa_{x^i}\Bigl( \; \pa_{y^j}S_\ube(x,y)\bigl|_{x=y}\;\Bigr) $$
\begin{equation}=- 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}
\end{equation}
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 origin 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
\begin{equation}
\Gamma(\ubf, 0)=0.
\label{2}
\end{equation}
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
\begin{equation}
\pa_{x^i}\Gamma_j(\ubf,0)=S_\ubf(0,0)^{-1} \pa^2_{x^iy^j}S_\ubf(0,0),
\label{3}
\end{equation}
and thus
\begin{equation}
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}
\end{equation}
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
\begin{equation}
\nabla^B \hat{B}=0.
\label{const}
\end{equation}
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
$$T(x,y)= S_x^{-1}S_{x,y}: E_y\to E_x. $$
Fix an open coordinate patch $\newcommand{\eO}{{\mathscr{O}}}$ $\eO\subset M$ with coordinates $(x^i)_{1\leq i\leq m}$. Assume $\eO$ is sufficiently small so $E$ trivializes over $\eO$. Suppose that $\newcommand{\be}{\boldsymbol{e}}$ $\underline{\be}(x)=(\be_\alpha(x))_{1\leq \alpha\leq \nu}$ is a local frame of $E$ over $\eO$. We denote by $\newcommand{\ur}{\underline{\mathbb{R}}}$ $\ur_\eO$ the trivial vector bundle $\bR^\nu\times \eO\to\eO$.
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
$$\Gamma(\ube,x)=\sum_{i=1}^m \Gamma_i(\ube, x) dx^i=-d_x T(x,y)\bigl|_{y=x}\in \Endo\bigl(\;\bR^\nu\;\bigr)\otimes \Omega^1(\eO),$$
where $d_x$ denotes the differential (exterior derivative) with respect to the $x$-variables. $\newcommand{\bsf}{\boldsymbol{f}}$ $\newcommand{ubf}{{\underline{\boldsymbol{f}}}}$ If $\ubf$ is another local frame of $E_\eO$, then there exists a smooth map $g:\eO\to\Aut(\bR^\nu)$ such that
$$\Phi_x(\ubf) =\Phi_x(\ube)\circ g(x),\;\;\forall x\in\eO. $$
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
$$\Gamma_i(\ube,x)= -\pa_{x^i} S_\ube(x,x)^{-1}\bigl|_{y=x} S_\ube(x,x)-S_\ube(x,x)^{-1} \pa_{x^i}S_\ube(x,y)\bigl|_{y=x} $$
\begin{equation}
= 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}
\end{equation}
We have
$$\pa_{x^i}\Gamma_j(\ube,x)=\pa_{x^i}\Bigl(\; S_\ube(x,x)^{-1} \pa_{y^j}S_\ube(x,y)\bigl|_{y=x}\;\Bigr) $$
$$ = -S_\ube(x,x)^{-1}\Bigl(\pa_{x^i}S_\ube(x,x)\;\Bigr) S_\ube(x,x)^{-1} \pa_{y^j}S_\ube(x,y)\bigl|_{x=y}+ S_\ube(x,x)^{-1}\pa_{x^i}\Bigl( \; \pa_{y^j}S_\ube(x,y)\bigl|_{x=y}\;\Bigr) $$
\begin{equation}=- 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}
\end{equation}
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 origin 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
\begin{equation}
\Gamma(\ubf, 0)=0.
\label{2}
\end{equation}
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
\begin{equation}
\pa_{x^i}\Gamma_j(\ubf,0)=S_\ubf(0,0)^{-1} \pa^2_{x^iy^j}S_\ubf(0,0),
\label{3}
\end{equation}
and thus
\begin{equation}
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}
\end{equation}
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
\begin{equation}
\nabla^B \hat{B}=0.
\label{const}
\end{equation}
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
The Unreasonable Effectiveness of Mathematics in the Natural Sciences
Monday, August 25, 2014
Friday, August 15, 2014
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
Quillen Notebooks | Clay Mathematics Institute
Sunday, April 13, 2014
The unspoken stresses of a research career
http://www.theguardian.com/higher-education-network/blog/2014/apr/05/academics-anonymous-research-stressful-job-depression
Friday, April 4, 2014
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
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. $\newcommand{\bR}{\mathbb{R}}$ $\DeclareMathOperator{\area}{Area}$
Suppose that $C$ is a compact convex region in the plane with nonempty interior. Assume that the origin is in the interior of $C$ and $\area(C)=1$. Choose $n$ points $P_1,\dotsc, P_n\in C$ randomly, independently and uniformly distributed. (In technical terms, we choose $n$ independent $\bR^2$-valued random variables with probability distribution $I_Cdxdy$, where $I_C$ is the characteristic function of $C$.) We denote by $\Delta_n=\Delta_n(P_1,\dotsc, P_n)$ the convex hull of this collection of points. This is a convex polygon and we denote by $V_n=V_n(P_1,\dotsc, P_n)$ its number of vertices, by $L_n=L_n(P_1,\dotsc, P_n)$ its perimeter and by $A_n=A_n(P_1,\dotsc, P_n)$ its area. These are random variables and we denote by$\newcommand{\bE}{\mathbb{E}}$ $\bE(V_n)$, $\bE(L_n)$ and respectively $\bE(A_n)$ their expectations. Reny and Sulanke asked to describe the behavior of these expectations as $n\to\infty$. $\newcommand{\bP}{\mathbb{P}}$
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
\begin{equation}
V_n=\sum_{i<j} I_{E_{ij}}.
\end{equation}
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
\begin{equation}
\bE(V_n)=\binom{n}{2} p_n.
\end{equation}
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
\[
\bP(P_1,P_2)= a(P_1,P_2)^{n-2}+\bigl(\;1-a(P_1,P_2)\;\bigr)^{n-2},
\]
\begin{equation}
\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.
\end{equation}
Since $a(P,Q)^{n-2}\leq \frac{1}{2^{n-2}} $, we deduce that
\begin{equation}
\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}
\end{equation}
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.
\begin{equation}
\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.
\end{equation}
Observing that for any $\ell>0$ we have
\[
\int_{[0,\ell]\times[0,\ell]}|x-y|dxdy=\frac{ \ell^3}{3}
\]
we deduce
\begin{equation}
\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.
\end{equation}
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.
Fix $\theta\in [0,2\pi]$ and set $\newcommand{\ii}{\boldsymbol{i}}$ $\newcommand{\jj}{\boldsymbol{j}}$ $\bn(\theta)=\cos \theta \ii +\sin\theta\jj\in\bR^2$. Since the curvature of $\pa C$ is strictly positive there exists a unique point $P(\theta)\in \pa C$ such that the unit outer normal to $\pa C$ at $P(\theta)$ is $\bn(\theta)$.
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.
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.
\begin{equation}
\si_0<\inf_{\theta\in [0,2\pi]} p_0(\theta) .
\end{equation}
\begin{equation}
a(p,\theta)=A(p,\theta),\;\;\forall p\in [\si_0,p_0(\theta)].
\end{equation}
\begin{equation}
c<a(p,\theta)\leq\frac{1}{2},\;\;\forall p\in [0,s_0].
\label{low}
\end{equation}
\begin{equation}
\frac{d\ell}{dp}<0\;\;\mbox{on $[\si_0,p_0]$}.
\end{equation}
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
\begin{equation}
I_n(\theta)\sim\frac{n^2}{6} J_n(\theta),\;\;n\to\infty.
\end{equation}
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
\begin{equation}
\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}
\end{equation}
We deduce
\begin{equation}
J_n(\theta)=2\int_0^{t_0} \bigl( 1-a(t)\bigr)^{n-2} \ell(t)^3tdt,\;\;t_0=\sqrt{\tau_0}.
\end{equation}
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
\begin{equation}
\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}
\end{equation}
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.
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. $\newcommand{\bR}{\mathbb{R}}$ $\DeclareMathOperator{\area}{Area}$
Suppose that $C$ is a compact convex region in the plane with nonempty interior. Assume that the origin is in the interior of $C$ and $\area(C)=1$. Choose $n$ points $P_1,\dotsc, P_n\in C$ randomly, independently and uniformly distributed. (In technical terms, we choose $n$ independent $\bR^2$-valued random variables with probability distribution $I_Cdxdy$, where $I_C$ is the characteristic function of $C$.) We denote by $\Delta_n=\Delta_n(P_1,\dotsc, P_n)$ the convex hull of this collection of points. This is a convex polygon and we denote by $V_n=V_n(P_1,\dotsc, P_n)$ its number of vertices, by $L_n=L_n(P_1,\dotsc, P_n)$ its perimeter and by $A_n=A_n(P_1,\dotsc, P_n)$ its area. These are random variables and we denote by$\newcommand{\bE}{\mathbb{E}}$ $\bE(V_n)$, $\bE(L_n)$ and respectively $\bE(A_n)$ their expectations. Reny and Sulanke asked to describe the behavior of these expectations as $n\to\infty$. $\newcommand{\bP}{\mathbb{P}}$
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
\begin{equation}
V_n=\sum_{i<j} I_{E_{ij}}.
\end{equation}
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
\begin{equation}
\bE(V_n)=\binom{n}{2} p_n.
\end{equation}
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
\[
\bP(P_1,P_2)= a(P_1,P_2)^{n-2}+\bigl(\;1-a(P_1,P_2)\;\bigr)^{n-2},
\]
\begin{equation}
\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.
\end{equation}
Since $a(P,Q)^{n-2}\leq \frac{1}{2^{n-2}} $, we deduce that
\begin{equation}
\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}
\end{equation}
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.
\begin{equation}
\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.
\end{equation}
Observing that for any $\ell>0$ we have
\[
\int_{[0,\ell]\times[0,\ell]}|x-y|dxdy=\frac{ \ell^3}{3}
\]
we deduce
\begin{equation}
\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.
\end{equation}
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.
Fix $\theta\in [0,2\pi]$ and set $\newcommand{\ii}{\boldsymbol{i}}$ $\newcommand{\jj}{\boldsymbol{j}}$ $\bn(\theta)=\cos \theta \ii +\sin\theta\jj\in\bR^2$. Since the curvature of $\pa C$ is strictly positive there exists a unique point $P(\theta)\in \pa C$ such that the unit outer normal to $\pa C$ at $P(\theta)$ is $\bn(\theta)$.
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 $\bn(\theta)$. |
\begin{equation}
\si_0<\inf_{\theta\in [0,2\pi]} p_0(\theta) .
\end{equation}
\begin{equation}
a(p,\theta)=A(p,\theta),\;\;\forall p\in [\si_0,p_0(\theta)].
\end{equation}
\begin{equation}
c<a(p,\theta)\leq\frac{1}{2},\;\;\forall p\in [0,s_0].
\label{low}
\end{equation}
\begin{equation}
\frac{d\ell}{dp}<0\;\;\mbox{on $[\si_0,p_0]$}.
\end{equation}
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
\begin{equation}
I_n(\theta)\sim\frac{n^2}{6} J_n(\theta),\;\;n\to\infty.
\end{equation}
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
\begin{equation}
\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}
\end{equation}
We deduce
\begin{equation}
J_n(\theta)=2\int_0^{t_0} \bigl( 1-a(t)\bigr)^{n-2} \ell(t)^3tdt,\;\;t_0=\sqrt{\tau_0}.
\end{equation}
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
\begin{equation}
\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}
\end{equation}
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.
Sunday, January 26, 2014
Wednesday, January 8, 2014
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/
Tuesday, September 3, 2013
Academy Fight Song |Thomas Frank | The Baffler
This is an interesting take of what's going on now in higher education.
Academy Fight Song | Thomas Frank | The Baffler
Academy Fight Song | Thomas Frank | The Baffler
Thursday, August 22, 2013
Friday, August 9, 2013
The Fulton-MacPherson compactification of a configuration space
$\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bZ}{\mathbb{Z}}$ $\DeclareMathOperator{\Bl}{\boldsymbol{Bl}}$
Suppose that $M$ is a real analytic manifold of dimension $m$. Fix a finite set $L$ of labels. For any subset $S\subset L$ we define the following objects.
The space of configurations $M(L)$ is an open subset of $M^L$. We want to construct a certain completion $M[L]$ of $M(L)$ as a manifold with corners. This completion is known as the Fulton-MacPherson compactification of $M$. The completion $M[L]$ is compact when $M$ is compact. We follow closely the approach of Axelrod and Singer, Chern-Simons perturbation theory.II, J. Diff. Geom., 39(1994), 173-213.
We begin with some simple observations. Observe that if $S\subset S'$, then we have a natural projection $\pi_S: M^{S'}\to M^S$ which associates to a map $S'\to S$ its restriction to $S$
$$ M^{S'}\ni x_{S'}\mapsto x_{S}\in M^S. $$
$\newcommand{\hra}{\hookrightarrow}$ We set
$$ \Delta_S^L=\pi_S^{-1}(\Delta_S)\subset M^L.$$
More explicitly, $\Delta_S^L$ consists of the maps $L\to M$ which are constant on $S$. Observe that
$$ M^L\setminus M(L)=\bigcup_{|S|\geq 2} \Delta_S^L. $$
For $x\in M$ and $S\subset M$ we denote by $x^S$ the constant map $S\to\lbrace x\rbrace$ viewed as an element in $M^S$. We can identify $x^S$ with a point in $x\in M$ so
Suppose that $M$ is a real analytic manifold of dimension $m$. Fix a finite set $L$ of labels. For any subset $S\subset L$ we define the following objects.
- The manifold $M^S$ consisting of maps $S\to M$. We will indicate a point in $M^S$ as a collection $x_S:=(x_s)_{s\in S}$, $x_s\in M$, $\forall s\in S$.
- The configuration space $M(S)\subset $ consisting of injective maps $S\to M$.
- The thin diagonal $\Delta_S\subset M^S$ consisting of the constant maps $S\to M$.
The space of configurations $M(L)$ is an open subset of $M^L$. We want to construct a certain completion $M[L]$ of $M(L)$ as a manifold with corners. This completion is known as the Fulton-MacPherson compactification of $M$. The completion $M[L]$ is compact when $M$ is compact. We follow closely the approach of Axelrod and Singer, Chern-Simons perturbation theory.II, J. Diff. Geom., 39(1994), 173-213.
We begin with some simple observations. Observe that if $S\subset S'$, then we have a natural projection $\pi_S: M^{S'}\to M^S$ which associates to a map $S'\to S$ its restriction to $S$
$$ M^{S'}\ni x_{S'}\mapsto x_{S}\in M^S. $$
$\newcommand{\hra}{\hookrightarrow}$ We set
$$ \Delta_S^L=\pi_S^{-1}(\Delta_S)\subset M^L.$$
More explicitly, $\Delta_S^L$ consists of the maps $L\to M$ which are constant on $S$. Observe that
$$ M^L\setminus M(L)=\bigcup_{|S|\geq 2} \Delta_S^L. $$
For $x\in M$ and $S\subset M$ we denote by $x^S$ the constant map $S\to\lbrace x\rbrace$ viewed as an element in $M^S$. We can identify $x^S$ with a point in $x\in M$ so
$$ T_{x^S}M^S\cong (T_xM)^S. $$
The thin diagonal $\Delta_S$ is a submanifold in $M^S$ of codimension $m(|S|-1)$. We denote by $\newcommand{\eN}{\mathscr{N}}$ $\eN_S$ the normal bundle of the embedding $\Delta_S\hra M^S$. The fiber $\eN_S(x)$ of the normal bundle $\eN_S=T_{x^S}M^S/T_{x^S}\Delta_S$ at a point $x^S$ is the quotient of $(T_xM)^S$ modulo the equivalence relation $\newcommand{\Llra}{\Longleftrightarrow}$
$$ (u_S)_{s\in S}\in (T_xM)^S\sim (v_S)_{s\in S}\in (T_xM)^S\;\stackrel{def}{\Llra}\; (u_{s_0}-u_{s_1})=v_{s_0}-v_{s_1},\;\;\forall s_0,s_1\in S. $$
We identify $\eN_S(x)$ with the subspace of $Z_S(x)\subset (T_xM)^L$ consisting of vectors $v_L=(v_\ell)_{\ell\in L}$, $v_\ell\in T_xM$ such that
$$v_\ell=0,\;\;\forall \ell\in L\setminus S,\;\; \sum_{s\in S} v_s=0. \tag{1}\label{1}$$
We have a natural $\newcommand{\bsP}{\boldsymbol{P}}$ projector
$$ \bsP_S: (T_x M)^L\to Z_S(x) $$
defined as follows. For a vector $\vec{v}=(v_\ell)_{\ell\in L} \in (T_xM)^L$, we denote by $\newcommand{\bb}{\boldsymbol{b}}$ $\bb_S(\vec{v})\in T_xM$ the barycenter of its $S$-component
$$\bb_S(\vec{v}):=\frac{1}{|S|}\sum_{s\in S} v_s\in T_x M, $$
and we set
$$\bsP_S(\vec{v}) =(\bar{v}_s)_{s\in S},\;\;\bar{v}_s:=v_s-\bb_S(\vec{v}),\;\;\forall s\in S,\;\;v_\ell=0. $$
Note that for $u_S\in (T_xM)^S$ we have $u_S\sim \bsP_S u_S$.
We denote by $\Bl(S,M)$ the radial blowup of $M^S$ along $\Delta_S$. This is manifold with boundary whose interior is naturally identified with $M_*^S:=M^S\setminus \Delta_S$. $\newcommand{\bsS}{\boldsymbol{S}}$ Its boundary is $\bsS(\eN)$, the "unit" sphere bundle bundle of $\eN$. Equivalently we identify the fiber of $\bsS(\eN)$ at $x^S$ with the quotient
$$ \bsS(\eN(x^S))=\bigl(\; Z_S(x)\setminus 0\; \bigr)/\propto, $$
where
$$ u_S\propto v_S \Llra \exists c>0:\;\; v_S=c u_S. $$
Following Fulton and MacPherson we will refer to the elements in $Z_S(x)$ as $S$-screens at $x$. Up to a a positive rescaling, an $S$-screen at $x^S\in \Delta_S$ is a collection of points $u_S\in (T_xM)^S\setminus 0^S$ with barycenter at the origin. If we fix a metric $g$ on $M$, then the fiber $\bsS\bigl(\;\eN(x^S)\;\bigr)$ that can be identified with the collection $v_L\in (T_xM)^L$ satisfying (\ref{1}) and
$$ \max_{s\in S}|v_s|=1. \tag{3}\label{3} $$
There is a natural smooth surjection (blow-down map)
$$\beta_S:\Bl(S,M)\to M^S$$
whose restriction to the interior of $\Bl(S,M)$ $\DeclareMathOperator{\int}{\boldsymbol{int}}$ induces a diffeomorphism to $M^S_*$ We denote by $\beta_S^{-1}$ the inverse
$$\beta_S^{-1}: M_*^S\to \int \Bl(S,M). $$
For $x_S\in M_*^S$ we set $\newcommand{\ve}{{\varepsilon}}$
$$\hat{x}_S:=\beta_S^{-1}(x_S). $$
If
$$[0,\ve)\ni t\mapsto x_S(t) \in M^S,\;\; x_S(0)=x_0^S\in\Delta_S $$
is a real analytic path such that $x_S(t)\in M_*^S$ for $t>0$, then the limit $\lim_{t\searrow 0}\hat{x}_S(t)$ can be described as follows.
We have a natural map $\newcommand{\eX}{\mathscr{X}}$
$$\gamma: M(L)\to \eX(M,L):=M^L\times\prod_{|S|\geq 2} \Bl(S,M),\;\; M(L)\ni x_L\mapsto \gamma(x_L):=\Bigl(\; x_L;\;\;(\hat{x}_S)_{|S|\geq 2}\;\Bigr)\in \eX(M,L). $$
The Fulton-MacPherson compactification of $M(V)$ is the closure of $\gamma\bigl(\;M(L)\;\bigr)$ in $\eX(M,L)$.
We want to give a more explicit description of this closure. Observe first that $\eX(M,L)$ and thus any point in the closure of $\gamma(M[L])$ can be approached from within $\gamma(M[L])$ along a real analytic path. Suppose that $(0,\ve)\ni t \mapsto x_L(t)$ is a real analytic path such that $\gamma\bigl(\; x_L(t)\;\bigr)$ approaches a point $\gamma^0\in \eX(M,L)$. The limit point is a collection $( x^*_L, (y(S))_{|S|\geq 2})\in \eX(M,L)$.
To the point $x^*_L\in M^L$ we associate an equivalence relation on $L$
$$\ell_0\sim_0 \ell_1 \Llra x^*_{\ell_0}=x^*_{\ell_1}. $$
Denote by $\newcommand{\eC}{\mathscr{C}}$ $\eC_0\subset 2^S$ the collection of equivalence classes of $\sim_0$ of cardinality $\geq 2$.
The subsets $S$ of $L$ of cardinality $\geq 2$ are of two types.
The thin diagonal $\Delta_S$ is a submanifold in $M^S$ of codimension $m(|S|-1)$. We denote by $\newcommand{\eN}{\mathscr{N}}$ $\eN_S$ the normal bundle of the embedding $\Delta_S\hra M^S$. The fiber $\eN_S(x)$ of the normal bundle $\eN_S=T_{x^S}M^S/T_{x^S}\Delta_S$ at a point $x^S$ is the quotient of $(T_xM)^S$ modulo the equivalence relation $\newcommand{\Llra}{\Longleftrightarrow}$
$$ (u_S)_{s\in S}\in (T_xM)^S\sim (v_S)_{s\in S}\in (T_xM)^S\;\stackrel{def}{\Llra}\; (u_{s_0}-u_{s_1})=v_{s_0}-v_{s_1},\;\;\forall s_0,s_1\in S. $$
We identify $\eN_S(x)$ with the subspace of $Z_S(x)\subset (T_xM)^L$ consisting of vectors $v_L=(v_\ell)_{\ell\in L}$, $v_\ell\in T_xM$ such that
$$v_\ell=0,\;\;\forall \ell\in L\setminus S,\;\; \sum_{s\in S} v_s=0. \tag{1}\label{1}$$
We have a natural $\newcommand{\bsP}{\boldsymbol{P}}$ projector
$$ \bsP_S: (T_x M)^L\to Z_S(x) $$
defined as follows. For a vector $\vec{v}=(v_\ell)_{\ell\in L} \in (T_xM)^L$, we denote by $\newcommand{\bb}{\boldsymbol{b}}$ $\bb_S(\vec{v})\in T_xM$ the barycenter of its $S$-component
$$\bb_S(\vec{v}):=\frac{1}{|S|}\sum_{s\in S} v_s\in T_x M, $$
and we set
$$\bsP_S(\vec{v}) =(\bar{v}_s)_{s\in S},\;\;\bar{v}_s:=v_s-\bb_S(\vec{v}),\;\;\forall s\in S,\;\;v_\ell=0. $$
Note that for $u_S\in (T_xM)^S$ we have $u_S\sim \bsP_S u_S$.
We denote by $\Bl(S,M)$ the radial blowup of $M^S$ along $\Delta_S$. This is manifold with boundary whose interior is naturally identified with $M_*^S:=M^S\setminus \Delta_S$. $\newcommand{\bsS}{\boldsymbol{S}}$ Its boundary is $\bsS(\eN)$, the "unit" sphere bundle bundle of $\eN$. Equivalently we identify the fiber of $\bsS(\eN)$ at $x^S$ with the quotient
$$ \bsS(\eN(x^S))=\bigl(\; Z_S(x)\setminus 0\; \bigr)/\propto, $$
where
$$ u_S\propto v_S \Llra \exists c>0:\;\; v_S=c u_S. $$
Following Fulton and MacPherson we will refer to the elements in $Z_S(x)$ as $S$-screens at $x$. Up to a a positive rescaling, an $S$-screen at $x^S\in \Delta_S$ is a collection of points $u_S\in (T_xM)^S\setminus 0^S$ with barycenter at the origin. If we fix a metric $g$ on $M$, then the fiber $\bsS\bigl(\;\eN(x^S)\;\bigr)$ that can be identified with the collection $v_L\in (T_xM)^L$ satisfying (\ref{1}) and
$$ \max_{s\in S}|v_s|=1. \tag{3}\label{3} $$
There is a natural smooth surjection (blow-down map)
$$\beta_S:\Bl(S,M)\to M^S$$
whose restriction to the interior of $\Bl(S,M)$ $\DeclareMathOperator{\int}{\boldsymbol{int}}$ induces a diffeomorphism to $M^S_*$ We denote by $\beta_S^{-1}$ the inverse
$$\beta_S^{-1}: M_*^S\to \int \Bl(S,M). $$
For $x_S\in M_*^S$ we set $\newcommand{\ve}{{\varepsilon}}$
$$\hat{x}_S:=\beta_S^{-1}(x_S). $$
If
$$[0,\ve)\ni t\mapsto x_S(t) \in M^S,\;\; x_S(0)=x_0^S\in\Delta_S $$
is a real analytic path such that $x_S(t)\in M_*^S$ for $t>0$, then the limit $\lim_{t\searrow 0}\hat{x}_S(t)$ can be described as follows.
- Fix local (real analytic) coordinate near $x_0$ so that the points $x_{s}(t)$ can be identified with points in a neighborhood of $0\in\bR^m$.
- For $t>0$ denote by $\bb(t)$ the barycenter of the collection $(x_s(t))_{s\in S}\subset\bR^m$, $$\bb(t)=\frac{1}{|S|}\sum_{s\in S} x_s(t). $$
- For $t>0$ and $s\in S$ define $\bar{x}_s(t) =x_{s}(t)-\bb(t)$, $m(t)=\max_s|\bar{x}_s(t)|$.
Then $\lim_{t\searrow 0}\hat{x}_S(t)$ can be identified with the vector
$$\lim_{t\searrow 0}\frac{1}{m(t)} \bigl(\;\bar{x}_s(t)_{s_\in S}\;\bigr)\in \eN(x_0^S). $$
We have a natural map $\newcommand{\eX}{\mathscr{X}}$
$$\gamma: M(L)\to \eX(M,L):=M^L\times\prod_{|S|\geq 2} \Bl(S,M),\;\; M(L)\ni x_L\mapsto \gamma(x_L):=\Bigl(\; x_L;\;\;(\hat{x}_S)_{|S|\geq 2}\;\Bigr)\in \eX(M,L). $$
The Fulton-MacPherson compactification of $M(V)$ is the closure of $\gamma\bigl(\;M(L)\;\bigr)$ in $\eX(M,L)$.
We want to give a more explicit description of this closure. Observe first that $\eX(M,L)$ and thus any point in the closure of $\gamma(M[L])$ can be approached from within $\gamma(M[L])$ along a real analytic path. Suppose that $(0,\ve)\ni t \mapsto x_L(t)$ is a real analytic path such that $\gamma\bigl(\; x_L(t)\;\bigr)$ approaches a point $\gamma^0\in \eX(M,L)$. The limit point is a collection $( x^*_L, (y(S))_{|S|\geq 2})\in \eX(M,L)$.
To the point $x^*_L\in M^L$ we associate an equivalence relation on $L$
$$\ell_0\sim_0 \ell_1 \Llra x^*_{\ell_0}=x^*_{\ell_1}. $$
Denote by $\newcommand{\eC}{\mathscr{C}}$ $\eC_0\subset 2^S$ the collection of equivalence classes of $\sim_0$ of cardinality $\geq 2$.
The subsets $S$ of $L$ of cardinality $\geq 2$ are of two types.
- The set $S$ is not contained in any of the equivalence classes in $\eC_0$, i.e., $\exists s_0,s_1\in S$ such that $x^*_{s_0}\neq x^*_{s_1}$. We will refer to such subsets as separating subsets.Then $y_S=\hat{x}_S^*$.
- The subset $S$ is contained in an equivalence class $C\in \eC_0$. In other words there exists $x^*(C)\in M$ such that $x^*_{s}=x^*(C)\in M$, $\forall s \in C$. We will refer to such a subset as non-separating. Then $y(S)$ is an $S$-screen at $x^*(C)$, $y(S)=\bigl(\;y(S)_s\;\bigr)_{s\in S}$.
Fix an equivalence class $C\in\eC_0$. Here is how one computes $y_S$ for $S$ non-separating, $S\subset C$. The point $x^*(C)^S\in\Delta^S$ is approached along the real analytic path
$$(0,\ve)\ni t\mapsto x_S(t)\in M(S). $$
Choose real analytic local coordinates at $x^*(C)$ so a neighborhood of this point in $M$ is identified with a neighborhood of $0$ in $\bR^m$. We have Taylor expansions
$$ x_s(t) = v_s(1) t+v_2(2)t^2+\cdots ,\;\; s\in S. $$
For $k\geq 1$ and $s\in S$ we denote by $[x_s(t)]_k$ the $k$-th jet of $x_s(t)$ at $0$
$$[x_s(t)]_k:=\sum_{j=1}^k v_s(j) t^j. $$
For each $k\geq 1$ we have an equivalence relation $\sim_k$ on $S$ given by
$$s \sim_k s'\Llra [x_s(t)]_k=[x_{s'}(t)]_k. $$.
We denote by $\sim_0$ the trivial equivalence relation on $C$ with a single equivalence class $C$. Let $k=k_C(S)$ be the smallest $k$ such that $\sim_k$ is a nontrivial equivalence relation on $S$. The integer $k_C(S)$ is called the separation order of $S$. Then the $S$-screen $y(S)$ is described as the projection
$$y(S)\propto \bsP_S v_S(k),\;\; v_S(k)=\bigl(\; v_s(k)\;\bigr)_{s\in S}. $$
Remark 1. Suppose $S\subset S'\subset C\in \eC_0$ and $|S|\geq 2$. Then $k_C(S) \geq k_C(S')$. Moreover
$$ k_C(S)=k_C(S') \Llra \bsP_Sy_{S'} \neq 0 \Llra y(S)\propto \bsP_S y(S'). $$
The condition $\bsP_S y(S)\neq 0$ signifies that there exist $s_0,s_1\in S'$ such that
$$ y(S')_{s_0}\neq y(S')_{s_1}. $$
Note that if $S_0,S_1\subset C$, $|S_0|,|S_1|\geq 2$ then
$$ k_C(S_0\cup S_1)\leq \min\bigl\lbrace\; k_C(S_0),k_C(S_1)\;\bigr\rbrace. $$
Recall that we have a sequence of equivalence relations $\sim_k$ on $C\in \eC_0$. They are finer and finer $\sim_k\prec \sim_{k+1}$, i.e.
$$ s\sim_{k+1}s'\Rightarrow s\sim_k s'. $$
Observe that
$$ S\subset C,\;\;|S|\geq 2,\;\; k_C(S)>k \Llra \mbox{$S$ is contained in an equivalence class of $\sim_k$.} \tag{4}\label{4} $$
Equivalently
$$ S\subset C,\;\;|S|\geq 2,\;\; k_C(S)\leq k \Llra \mbox{exist distinct equivalence classes $S_0,S_1$ of $\sim_k$ such that}\;\; S\cap S_0, S\cap S_1\neq \emptyset \tag{4'}\label{4'} $$
Let $N_C$ denote the smallest $N$ such that all the equivalence classes of $\sim_N$ consists of single points., i.e.,
$$ s \sim_N s'\Llra s=s'. $$
Consider $\newcommand{\eS}{\mathscr{S}}$ the collection $\eS_C$ of all the equivalence classes of cardinality $\geq 2$ of the relations $\sim_k$, $k\geq 0$ on $C$ . This is a nested family of subsets of $C$ i.e., if $S_0, S_1\in\eS_C$, then
$$ S_0\cap S_1 \neq \emptyset \Llra S_0\subset S_1 \;\;\mbox{or}\;\;S_1\subset S_1. $$
Moreover $C\in \eS_C$. Observe that if $S_0,S_1\in \eS_C$ and $S_0\subsetneq S_1$, then $k_C(S_0)> k_C(S_1)$. Using Remark 1 we deduce
$$ S_0,S_1\in \eS_C,\;\;S_0\subsetneq S_1 \Rightarrow \bsP_{S_0}y(S_1)=0. \tag{5}\label{5} $$
Suppose now that $S\subset C$ and $|S|\geq 2$. We set
$$\hat{S}=\bigcap_{S\subset S' \in\eS_C} S'. $$
In other words, $\hat{S}$ is the smallest subset in $\eS_C$ containing $S$.
Lemma 2. We have $k_C(S)= k_C(\hat{S})$.
Proof. Observe first we have $k_C(S)\leq k_C(\hat{S})$. Set $k_0 :=k_C(S)$.
If $k_C(\hat{S})> k_0$, then (\ref{4}) implies $\hat{S}$ is contained in an equivalence class of $\sim_{k_0}$. On the other hand $k_C(S)=k_0$ (\ref{4'}) implies $S_0$ intersects nontrivially two equivalence classes of $\sim_{k_0}$. This contradicts the condition $S\subset \hat{S}$. qed
Using Remark 1 we deduce
$$ S\subset C,\;\;|S|\geq 2 \Rightarrow y(S)\propto \bsP_S y(\hat{S}). \tag{6}\label{6} $$
The conditions (\ref{5}), (\ref{6}) describe some compatibility conditions satisfied by the screens $y(S)$, $S\subset L$ non-separating.
We can now form the family of subsets of $L$
$$\eS=\bigcup_{C\in\eC_0} \eS_C. $$
This also a nested family of subsets of cardinality $\geq 2$. A subset $S\subset L$ of cardinality $\geq 2$ is called $\eS$-separating if it is not contained in any of the sets of $\eS$. Otherwise it is called nonseparating. For any separating set $S$ we denote by $\hat{S}$ the smallest subset of $\eS$ containg $S$. The limit point
$$c:= \Bigl(\;x^*(L), \bigl(\;y(S)\;\bigr)_{S\subset L,\;|S|\geq 2}\;\Bigr)\in\eX(M,L) $$ satifies the following conditions.
$$ y(S)\in \beta^{-1}_S\bigl(\;M^S_*\;\bigr),\;\; \mbox{if $S$ is separating}. \tag{$C_1$} \label{C1} $$
$$ y(S) \;\;\mbox{is an $S$-screen if $S$ is non-separating}. \tag{$C_2$} \label{C2} $$
$$ S_0,S_1\in \eS,\;\;S_0\subset S_1\Rightarrow \bsP_{S_0}y(S_1)=0. \tag{$C_3$}\label{C3} $$
$$ S\;\;\mbox{nonseparating} \Rightarrow y(S)\propto \bsP_S y(\hat{S}). \tag{$C_4$}\label{C4} $$
Comments. (a) Let us recall that (\ref{C3}) signifies that the components $y(S_1)_s$ $s\in S_0$ are identical.
(b) Let me say a few words about the interpretation of the nested family $\eS$. A set $S$ corresponds to a collection of distinct points in $(x_s)_{s\in S}$ in $M$ that is clustering ner a point $x^*$. A subset $S'$ corresponds to a subcollection of the above collection that is clustering at a faster rate.
Running the above arguments in revers one can show that a collection
$$ \Bigl(\;x^*(L), \bigl(\;y(S)\;\bigr)_{S\subset L,\;|S|\geq 2}\;\Bigr)\in\eX(M,L) $$
belongs to the closure of $\gamma\bigl(\;M(L)\;\bigr)$ in $\eX(M, L)$ if and only if there exists a nested collection $\eS$ of subsets of $L$ of cardinality $\geq 2$ such that satisfying the compatibility conditions (\ref{C1}-\ref{C4}) are satisfied. The set $\eS$ is called the type of the limit point. For a nested family $\eS$ of subsets of $L$ of cardinality $\geq 2$ we denote Define $M^(\eS)$ the collection of points of type $\eS$.
The stratum $M(\eS)$ has codimension $|\eS|$. This can be seen after a tedious computation that takes into account a (\ref{C1}-\ref{C4}) . To explain introduce a notation. Given $S, S'\in \eS$ we say that $S$ precedes $S'$ and we write this $S\lessdot S'$, if $S$ is maximal amomgst the subsets of $\eS$ contained but not equal to $S'$. Denote by $\eS_{\max}$ the collection of maximal sets in $\eS$. (The collection $\eS_{\max}$ coincides with the collection $\eC_0$ in the above discussion.) The, if we recall that $\dim M=m$ and $|L|=n$ we deduce
$$\dim M(\eS)^* =m\left(\; n-\sum_{S\in\eS_{\max}}(|S|-1)\;\right) +\sum_{S\in \eS}\left[\;\;m\left(\;(|S|-1)-\sum_{S'\lessdot S}\bigl(\;|S'|-1\;\bigr)\right)-1\;\right]$$
To understand this formula let us consider a point
$$ c =\Bigl(\;x^*(L), \bigl(\;y(S)\;\bigr)_{S\subset L,\;|S|\geq 2}\;\Bigr)\in M(\eS)^*. $$
The coordinates of $x^*(L)$ are described by $nm$ parameters Each $S\in \eS_{\max}$ introduces the constraints
$$x^*(L)_{s_1}=x^*(L)_{s_2},\forall s_1,s_2\in S. $$
If $S=\lbrace s_1,\dotsc,s_N\rbrace$ we see that the above constraints are consequences of the linearly independent ones
$$ x^*(L)_{s_1}-x^*(L)_{s_2}= \cdots =x^*(L)_{s_{N-1}}-x^*(L)_{s_N}=0. $$
These cut down the number of parameters required to describe $x^*(L)$ by $m(N-1)=m(|S|-1)$.
Thus the number of parameters need to describe $x^*(L)$ is
$m\left(\; n-\sum_{S\in\eS_{\max}}(|S|-1)\;\right)$
From (\ref{C3}) and (\ref{C4}) we deduce that the collection
$$ \bigl(\;y(S)\;\bigr)_{S\subset L,\;|S|\geq 2}$$
is uniquely determined by the subcollection
$$ \bigl(\;y(S)\;\bigr)_{S\in\eS} . $$
The screen $y(S)$ belongs to the unit sphere $\bsS(\eN(x_S))$ which has dimension
$$\dim M^S-\dim\delta_S-1= m(|S|-1)-1. $$
Thus we need $m(|S|-1)$ parameters to describe the screen $y(S)$. However, the condition (\ref{C3}) shows that any $S'\lessdot S$ induces $m(|S'|-1)$ linearly independent constraints on these parameters so that $y(S)$ has a total of
$$m\left(\;(|S|-1)-\sum_{S'\lessdot S}\bigl(\;|S'|-1\;\bigr)\right)-1 $$
degrees of freedom.
We want to describe a neighborhood of $M(\eS)$ in $M[L]$. We will achieve this via an explicit map
$$\Psi : M(\eS)\times \bR_{\geq 0}^{\eS} \to M[L] $$
defined as follows. Denote by $\vec{t}=(t_S)_{s\in\eS}$ the coordinates on $\bR^{\eS}_{\geq 0}$. For $S\in \eS$ we set
$$T_S=\prod_{\eS\ni S'\supseteq S} t_{S'}. $$
If
$$c = (x(c), (y(S,c))_{S\in\eS})\in M(\eS),$$ then
$$\Psi(c, \vec{t})= \bigl( x_\ell (c,\vec{t})\;\bigr)_{\ell \in L}, $$
where
$$ x_\ell(c,\vec{t})= x(c)_\ell +\sum_{\ell\in S\in \eS} T_S y(S,c)_\ell. $$
In the above formula $y(S)$ is assumed to be a vector of norm $1$ in $Z_S(x_\ell)$.
Let us convince ourselves that for fixed $c_0\in M(\eS)$ there exists a small neighborhood $U$ of $c_0$ in $M(\eS)$ and a neighborhood $V$ of $0\in\bR^{\eS}_{\geq 0}$ such that $\Psi$ maps $U\times V_{>0}$ into $M(L)$. Here $V_{>0}-V\cap \bR^{\eS}_{>0}$.
Thus we have to show that if $i,j\in L$, $i\neq j$, then for $c$ close to $c_0$ and $\vec{t}$ close to $0$.
$$ x_i(c,\vec{t})\neq x_j(c,\vec{t}) $$
Note that a set $S\in\eS$ that contains $i$ is either contained in $S_0$ or contains $S_0$. A similar fact is true for $j$. Observe that if $S\supset\neq S_0$ then $y(S,c)_i=y(S,c)j$. Thus
$$ x(c,\vec{t})_i-x(c,\vec{t})_j =\sum_{S\subsetneq S_0} T_S\bigl(\; y(S)_i-y(S)_j\;\bigr)+ T_{S_0}(y(S_0)_i-y(S_0)_j $$
$$ =T_{S_0}\left(\sum_{S\subsetneq S_0} \tau _S\bigl(\; y(S)_i-y(S)_j\;\bigr)+ (y(S_0)_i-y(S_0)_j\;\right), $$
where
$$\tau_S=\prod_{S\subset S'\subset\neq S_0} t_{S'}. $$
The conclusion follows by observing that $y(S)_i\neq y(S)_j$.
We denote by $M[\eS]$ the closure of $M(\eS)$ in $M[L]$. Observe that
$$ M(\eS')\subsetneq M[\eS] \Llra \eS'\supsetneq \eS. $$
We can now form the family of subsets of $L$
$$\eS=\bigcup_{C\in\eC_0} \eS_C. $$
This also a nested family of subsets of cardinality $\geq 2$. A subset $S\subset L$ of cardinality $\geq 2$ is called $\eS$-separating if it is not contained in any of the sets of $\eS$. Otherwise it is called nonseparating. For any separating set $S$ we denote by $\hat{S}$ the smallest subset of $\eS$ containg $S$. The limit point
$$c:= \Bigl(\;x^*(L), \bigl(\;y(S)\;\bigr)_{S\subset L,\;|S|\geq 2}\;\Bigr)\in\eX(M,L) $$ satifies the following conditions.
$$ y(S)\in \beta^{-1}_S\bigl(\;M^S_*\;\bigr),\;\; \mbox{if $S$ is separating}. \tag{$C_1$} \label{C1} $$
$$ y(S) \;\;\mbox{is an $S$-screen if $S$ is non-separating}. \tag{$C_2$} \label{C2} $$
$$ S_0,S_1\in \eS,\;\;S_0\subset S_1\Rightarrow \bsP_{S_0}y(S_1)=0. \tag{$C_3$}\label{C3} $$
$$ S\;\;\mbox{nonseparating} \Rightarrow y(S)\propto \bsP_S y(\hat{S}). \tag{$C_4$}\label{C4} $$
Comments. (a) Let us recall that (\ref{C3}) signifies that the components $y(S_1)_s$ $s\in S_0$ are identical.
(b) Let me say a few words about the interpretation of the nested family $\eS$. A set $S$ corresponds to a collection of distinct points in $(x_s)_{s\in S}$ in $M$ that is clustering ner a point $x^*$. A subset $S'$ corresponds to a subcollection of the above collection that is clustering at a faster rate.
Running the above arguments in revers one can show that a collection
$$ \Bigl(\;x^*(L), \bigl(\;y(S)\;\bigr)_{S\subset L,\;|S|\geq 2}\;\Bigr)\in\eX(M,L) $$
belongs to the closure of $\gamma\bigl(\;M(L)\;\bigr)$ in $\eX(M, L)$ if and only if there exists a nested collection $\eS$ of subsets of $L$ of cardinality $\geq 2$ such that satisfying the compatibility conditions (\ref{C1}-\ref{C4}) are satisfied. The set $\eS$ is called the type of the limit point. For a nested family $\eS$ of subsets of $L$ of cardinality $\geq 2$ we denote Define $M^(\eS)$ the collection of points of type $\eS$.
The stratum $M(\eS)$ has codimension $|\eS|$. This can be seen after a tedious computation that takes into account a (\ref{C1}-\ref{C4}) . To explain introduce a notation. Given $S, S'\in \eS$ we say that $S$ precedes $S'$ and we write this $S\lessdot S'$, if $S$ is maximal amomgst the subsets of $\eS$ contained but not equal to $S'$. Denote by $\eS_{\max}$ the collection of maximal sets in $\eS$. (The collection $\eS_{\max}$ coincides with the collection $\eC_0$ in the above discussion.) The, if we recall that $\dim M=m$ and $|L|=n$ we deduce
$$\dim M(\eS)^* =m\left(\; n-\sum_{S\in\eS_{\max}}(|S|-1)\;\right) +\sum_{S\in \eS}\left[\;\;m\left(\;(|S|-1)-\sum_{S'\lessdot S}\bigl(\;|S'|-1\;\bigr)\right)-1\;\right]$$
To understand this formula let us consider a point
$$ c =\Bigl(\;x^*(L), \bigl(\;y(S)\;\bigr)_{S\subset L,\;|S|\geq 2}\;\Bigr)\in M(\eS)^*. $$
The coordinates of $x^*(L)$ are described by $nm$ parameters Each $S\in \eS_{\max}$ introduces the constraints
$$x^*(L)_{s_1}=x^*(L)_{s_2},\forall s_1,s_2\in S. $$
If $S=\lbrace s_1,\dotsc,s_N\rbrace$ we see that the above constraints are consequences of the linearly independent ones
$$ x^*(L)_{s_1}-x^*(L)_{s_2}= \cdots =x^*(L)_{s_{N-1}}-x^*(L)_{s_N}=0. $$
These cut down the number of parameters required to describe $x^*(L)$ by $m(N-1)=m(|S|-1)$.
Thus the number of parameters need to describe $x^*(L)$ is
$m\left(\; n-\sum_{S\in\eS_{\max}}(|S|-1)\;\right)$
From (\ref{C3}) and (\ref{C4}) we deduce that the collection
$$ \bigl(\;y(S)\;\bigr)_{S\subset L,\;|S|\geq 2}$$
is uniquely determined by the subcollection
$$ \bigl(\;y(S)\;\bigr)_{S\in\eS} . $$
The screen $y(S)$ belongs to the unit sphere $\bsS(\eN(x_S))$ which has dimension
$$\dim M^S-\dim\delta_S-1= m(|S|-1)-1. $$
Thus we need $m(|S|-1)$ parameters to describe the screen $y(S)$. However, the condition (\ref{C3}) shows that any $S'\lessdot S$ induces $m(|S'|-1)$ linearly independent constraints on these parameters so that $y(S)$ has a total of
$$m\left(\;(|S|-1)-\sum_{S'\lessdot S}\bigl(\;|S'|-1\;\bigr)\right)-1 $$
degrees of freedom.
We want to describe a neighborhood of $M(\eS)$ in $M[L]$. We will achieve this via an explicit map
$$\Psi : M(\eS)\times \bR_{\geq 0}^{\eS} \to M[L] $$
defined as follows. Denote by $\vec{t}=(t_S)_{s\in\eS}$ the coordinates on $\bR^{\eS}_{\geq 0}$. For $S\in \eS$ we set
$$T_S=\prod_{\eS\ni S'\supseteq S} t_{S'}. $$
If
$$c = (x(c), (y(S,c))_{S\in\eS})\in M(\eS),$$ then
$$\Psi(c, \vec{t})= \bigl( x_\ell (c,\vec{t})\;\bigr)_{\ell \in L}, $$
where
$$ x_\ell(c,\vec{t})= x(c)_\ell +\sum_{\ell\in S\in \eS} T_S y(S,c)_\ell. $$
In the above formula $y(S)$ is assumed to be a vector of norm $1$ in $Z_S(x_\ell)$.
Let us convince ourselves that for fixed $c_0\in M(\eS)$ there exists a small neighborhood $U$ of $c_0$ in $M(\eS)$ and a neighborhood $V$ of $0\in\bR^{\eS}_{\geq 0}$ such that $\Psi$ maps $U\times V_{>0}$ into $M(L)$. Here $V_{>0}-V\cap \bR^{\eS}_{>0}$.
Thus we have to show that if $i,j\in L$, $i\neq j$, then for $c$ close to $c_0$ and $\vec{t}$ close to $0$.
$$ x_i(c,\vec{t})\neq x_j(c,\vec{t}) $$
Note that a set $S\in\eS$ that contains $i$ is either contained in $S_0$ or contains $S_0$. A similar fact is true for $j$. Observe that if $S\supset\neq S_0$ then $y(S,c)_i=y(S,c)j$. Thus
$$ x(c,\vec{t})_i-x(c,\vec{t})_j =\sum_{S\subsetneq S_0} T_S\bigl(\; y(S)_i-y(S)_j\;\bigr)+ T_{S_0}(y(S_0)_i-y(S_0)_j $$
$$ =T_{S_0}\left(\sum_{S\subsetneq S_0} \tau _S\bigl(\; y(S)_i-y(S)_j\;\bigr)+ (y(S_0)_i-y(S_0)_j\;\right), $$
where
$$\tau_S=\prod_{S\subset S'\subset\neq S_0} t_{S'}. $$
The conclusion follows by observing that $y(S)_i\neq y(S)_j$.
We denote by $M[\eS]$ the closure of $M(\eS)$ in $M[L]$. Observe that
$$ M(\eS')\subsetneq M[\eS] \Llra \eS'\supsetneq \eS. $$
Monday, June 17, 2013
Wednesday, May 22, 2013
Timeless advise from Gian-Carlo Rota
Most of you may have read Rota's "Ten lessons I wish I had been taught". However, if you were not of drinking age when Rota regaled us with his wisdom, please check the link below.
alumni.media.mit.edu/~cahn/life/gian-carlo-rota-10-lessons.html#toc
alumni.media.mit.edu/~cahn/life/gian-carlo-rota-10-lessons.html#toc
Monday, May 20, 2013
Remarkable and unexpected progress in prime gap theory
This is a fascinating story about a mathematician "past his prime" taking the math world by surprise.
Yitang Zhang Proves 'Landmark' Theorem in Distribution of Prime Numbers | Simons Foundation
Thursday, May 16, 2013
Plagiarism in Romanian academia
This is a piece that I wrote one my non-mathematical blog and I thought it deserves some exposure to a mathematical audience.
EpsilonBee: Plagiarism in Romanian academia
EpsilonBee: Plagiarism in Romanian academia
Tuesday, May 14, 2013
The weak Goldbach conjecture is settled.
Quoting Terry Tao: "Busy day in analytic number theory; Harald Helfgott has complemented his previous paper http://arxiv.org/abs/1205.5252 (obtaining minor arc estimates for the odd Goldbach problem) with major arc estimates, thus finally obtaining an unconditional proof of the odd Goldbach conjecture that every odd number greater than five is the sum of three primes. "
Thursday, May 9, 2013
On the pitfalls of "Open Access" publishing
The e-mail below from our library says it all. (Emphasis is mine.)
"Dear Liviu Nicolaescu,
A request you have placed:
Journal of advanced research in statistics and probability
3 3 2011
Title: On the central limit theorem for $m$-dependent random variables with unbounded $m$
Author: Shang, Yilun
TN: 685774
has been cancelled by the interlibrary loan staff for the following reason:
We have exhausted all possible sources.
This is so frustraing! [sic] We haven't been able to find any library that has this journal, Per the journal's website it is supposed to be Open Access meaning we should be able to get any of the articles online at no charge but their "archive" returns an empty screen with no way ot getting to the older journal issues. We tried this on different days just to make sure it wasn't a one-day system problem. We have not been able to find an email address for the author so that we could ask him directly. We've found a number of other articles by this author on this subject but not this one. We're just out of options on this one."
"Dear Liviu Nicolaescu,
A request you have placed:
Journal of advanced research in statistics and probability
3 3 2011
Title: On the central limit theorem for $m$-dependent random variables with unbounded $m$
Author: Shang, Yilun
TN: 685774
has been cancelled by the interlibrary loan staff for the following reason:
We have exhausted all possible sources.
This is so frustraing! [sic] We haven't been able to find any library that has this journal, Per the journal's website it is supposed to be Open Access meaning we should be able to get any of the articles online at no charge but their "archive" returns an empty screen with no way ot getting to the older journal issues. We tried this on different days just to make sure it wasn't a one-day system problem. We have not been able to find an email address for the author so that we could ask him directly. We've found a number of other articles by this author on this subject but not this one. We're just out of options on this one."
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