Press "Enter" to skip to content

Libres pensées d'un mathématicien ordinaire Posts

Lutte des classes et domination sociale

Cohen brothers movie "A serious man"

Voici deux thèmes à propos de certaines attitudes paradoxales :

  • Salaires des universitaires. Le pouvoir d’achat des universitaires n’a cessé de se dégrader depuis plusieurs dizaines d’années. Malgré une revalorisation récente, les salaires des maîtres de conférences (et des chargés de recherches) débutants restent particulièrement bas étant donné leur haut degré de formation et la difficulté du concours auquel ils ont réussi. Le moindre diplômé d’un bon M2 ou d’une grande école gagnera autant ou plus, immédiatement ou assez rapidement. Cela commence à avoir un impact sur les recrutements en région parisienne et pour les étrangers les plus brillants. La situation a peu de raison de s’améliorer dans les années qui viennent, car bon nombre d’universitaires pensent qu’ils sont trop payés, qu’il font déjà partie des mieux payés dans la population, que leur salaire est indécent étant donné leur liberté, voire que les salaires devraient être tous les mêmes dans une société idéale. Tout cela accompagne lentement mais sûrement une tendance générale de dévalorisation de plus en plus marquée du métier d’universitaire dans la société, surtout dans les classes populaires. Mais après tout, peu importe, car voilà déjà des années que les classes populaires ne produisent plus vraiment d’universitaires. À l’opposé, dans certaines disciplines, les universitaires arrondissent leurs fins de mois en monnayant leur expertise dans le secteur privé.
  • Évaluation pédagogique. En mathématiques comme dans d’autres disciplines, les activités de recherche des enseignants-chercheurs sont évaluées et cela est bon pour la recherche. Les activités d’enseignement sont également évaluées dans bon nombre d’établissements prestigieux, comme par exemple à l’École Polytechnique, et cela est bon pour la formation des élites. En revanche, à l’université, les activités d’enseignement sont peu ou pas évaluées. Mais après tout, peu importe, car les étudiants des universités ne font pas vraiment partie des élites (sauf peut être dans certaines disciplines comme le droit et la médecine). Tout se passe comme si une dévalorisation de l’université était intégrée dans les esprits et les attitudes, à l’insu des acteurs eux-mêmes. Il est bien naturel de relier tout cela à un phénomène de domination sociale et de reproduction des dominants. En mathématiques comme dans d’autres disciplines, l’essentiel des universitaires ne viennent pas de l’université et ne souhaitent pas y mettre leur enfants. D’autre part, un système de promotion juste nécessiterait une évaluation des enseignants et des chercheurs, et non pas seulement des enseignements et des recherches. Nous le faisons déjà pour la recherche, mais nous avons grand peine à le faire pour l’enseignement.

 

Leave a Comment

Carré du champ

Charles-Augustin de Coulomb

This tiny post is about the physical origin of a term used by mathematicians in analysis, probability, and geometry. If ${(X_t)}_{t\geq0}$ is a continuous time Markov process with state space $E$ and infinitesimal generator $L$ then its carré du champ operator is defined for any $f:E\to\mathbb{R}$ by

$$\Gamma f=\frac{1}{2}(L(f^2)-2fLf).$$ It appears naturally in the analysis of the conditional laws of the Markov process. Namely, denoting ${(P_t)}_{t\geq0}={(\mathrm{e}^{tL})}_{t\geq0}$ the Markov semigroup, we get, informally, for any $f$ and $t\geq0$,
$$\mathrm{Var}_{P_t}(f)=P_t(f^2)-P_t(f)^2=(f^2+tL(f^2)+\cdots)-(f+tL+\cdots)^2=2t\Gamma(f)+o(t)$$ (and the higher order terms lead to the more general notion of $\Gamma_n$). In the same spirit,
$$\mathrm{Var}_{P_t}(f)=P_t(f^2)-P_t(f)^2=\int_0^t\!\partial_sP_s(P_{t-s}(f)^2)\,\mathrm{d}s=2\int_0^t\! P_s(\Gamma(P_{t-s}(f)))\,\mathrm{d}s.$$ In the basic example of the standard Brownian Motion on $E=\mathbb{R}^d$, we have $L=\frac{1}{2}\Delta$, $P_t(f)(x)=\mathbb{E}(X_t\mid X_0=x)=\mathbb{E}(f(x+\sqrt{t}G))$, $G\sim\mathcal{N}(0,I_d)$, and we get

$$\Gamma(f)=\frac{\Delta(f^2)-2f\Delta f}{2}=\frac{(\partial_1f)^2+\cdots+(\partial_df)^2}{2}=\frac{1}{2}\left\Vert\nabla f\right\Vert^2.$$ The name carré du champ comes from the mathematical modeling of electrostatics. Namely if $\mu$ is a distribution of charges in $\mathbb{R}^d$ then the Coulomb potential at point $x\in\mathbb{R}^d$ is given by

$$U^\mu(x)=(\mu*g)(x)=\int g(x-y)\mu(\mathrm{d}y)$$ where $*$ is the convolution and $g$ is the Coulomb (or Newton) kernel given for $d\geq2$ by

$$g(x)=c_2\mathbf{1}_{d=2}\log\frac{1}{|x|}+c_3\mathbf{1}_{d\geq3}\frac{1}{{\left\Vert x\right\Vert}^{d-2}}.$$The Coulomb electric force generated by the distribution $\mu$ at point $x$ for a unit charge is $\nabla U^\mu(x)$. Also the quantity $\left\Vert\nabla U^\mu\right\Vert^2$ is the square of the (norm) of the (electric) field: carré du champ in French, see [1].

The electrostatic energy of self-interaction of this distribution of charges is given by

$$\mathcal{E}(\mu)=\int\!U^\mu(x)\mu(\mathrm{d}x).$$ Since $g$ is the fundamental solution of the Laplace equation we have $\Delta g=-c_d\delta_0$ in distributional sense and thus, by an integration by parts

$$\mathcal{E}(\mu)=-c_d\int\!U^\mu(x)\Delta U^\mu(x)\mathrm{d}x=c_d\int\!\left\Vert\nabla U^\mu(x)\right\Vert^2\mathrm{d}x.$$ The carré du champ plays a central role for the analysis and geometry of Markov diffusion operators in the Bakry-Émery theory, see [2]. The electrostatic Coulomb energy is a central concept in potential theory. The electrostatic Coulomb energy plays also a central role in the asymptotic analysis of Coulomb gases, notably those emerging from Random Matrix Theory, see for instance the modest contribution [3] and references therein. The funny thing is that most specialists of the analysis and geometry of Markov diffusion operators are not aware of the Coulomb physical (hi)story about their carré du champ!

Further reading.

Leave a Comment

Mathematics in Open Access

MathOA

Do you know MathOA? MathOA is a new project for mathematical publishing, inspired by the LingOA project in linguistics. MathOA aims to facilitate the transition of journals to fair open access. Take a look at the website: www.mathoa.org.

Leave a Comment

Back to basics – irreducible Markov kernels

Markov chain

Let \( {{(X_n)}_{n\geq0}} \) be a Markov chain on an at most countable state space \( {E} \), with transition kernel \( {\mathbf{P}} \). For any \( {x,y\in E} \), the number of passages in \( {y} \) before returning to \( {x} \) is given by

\[ \mu_x(y)=\mathbb{E}_x\left(\sum_{n=0}^{T_x-1}\mathbf{1}_{X_n=y}\right) \quad\mbox{where}\quad T_x=\inf\{n>0:X_n=x\}. \]

Note that \( {\mu_x(x)=1} \) and \( {\mu_x(E)=\mathbb{E}(T_x)} \).

I like very much the following classical theorem about irreducible kernels:

Irreducible kernels. Suppose that \( {\mathbf{P}} \) is irreducible. Then every invariant measure \( {\mu} \) has full support and satisfies \( {\mu(y)\geq\mu(x)\mu_x(y)} \) for any \( {x} \) and \( {y} \) in \( {E} \). Moreover there are three cases:

  1. \( {\mathbf{P}} \) is transient. Then \( {\mu(E)=\infty} \) for every invariant measure \( {\mu} \), \( {E} \) is infinite, and there is no invariant probability measure;
  2. \( {\mathbf{P}} \) is recurrent. Then \( {\mu_x} \) is invariant for every \( {x\in E} \); the invariant measures are all proportional; and if \( {\mu} \) is invariant then \( {\mu(y)=\mu(x)\mu_x(y)} \) for any \( {x,y\in E} \). In particular \( {\mu_x(y)\mu_y(x)=1} \). There are two sub-cases:
    1. \( {\mathbf{P}} \) is positive recurrent. Then there exists a unique invariant probability measure \( {\mu} \) given by \( {\mu(x)=1/\mathbb{E}_x(T_x)} \) for every \( {x\in E} \);
    2. \( {\mathbf{P}} \) is null recurrent. Then \( {\mu(E)=\infty} \) for every invariant measure \( {\mu} \), \( {E} \) is infinite, and there is no invariant probability measure.

Here are my three favorite examples with infinite countable state space:

Random walk on \( {\mathbb{Z}} \). The transition kernel is given by \( {\mathbf{P}(x,x+1)=p} \) and \( {\mathbf{P}(x,x-1)=1-p} \) for every \( {x\in\mathbb{Z}} \), with \( {p\in(0,1)} \). The kernel is irreducible. A measure \( {\mu} \) on \( {\mathbb{Z}} \) is invariant for \( {\mathbf{P}} \) iff for every \( {x\in\mathbb{Z}} \),

\[ \mu(x)=\sum_{y\in\mathbb{Z}}\mu(y)\mathbf{P}(y,x)=p\mu(x-1)+(1-p)\mu(x+1). \]

This gives that all invariant measures take the form

\[ \mu(x)=a+b\left(\frac{p}{1-p}\right)^x,\quad x\in\mathbb{Z}, \]

where \( {a,b\geq0} \) are parameters such that \( {\mu(0)=a+b>0} \). We recover the counting measure if \( {b=0} \) and a sort of geometric measure when \( {a=0} \) which is reversible. If \( {p=1/2} \) then both terms are the same and the counting measure is reversible.

The chain never admits an invariant probability measure. If \( {p=1/2} \) then the chain is null recurrent, all invariant measures are proportional. If \( {p\neq 1/2} \) then the chain is transient, there is a two parameters family of invariant measures.

Reflected random walk on \( {\mathbb{N}} \). The transition kernel is \( {\mathbf{P}(x,x+1)=p} \) and \( {\mathbf{P}(x,x-1)=1-p} \) for any \( {x\in\{1,2,\ldots\}} \) and \( {\mathbf{P}(0,1)=1} \). Here again \( {p\in(0,1)} \). The kernel is irreducible. This time the reflection at zero kills the counting measure. All invariant measures are of geometric type, are reversible, and take the form

\[ \mu(x)=\mu(0)\left(\frac{p}{1-p}\right)^x,\quad x\in\mathbb{N}. \]

If \( {p<1/2} \) then the chain is positive recurrent and there is a unique invariant probability measure which is the geometric distribution with parameter \( {p/(1-p)} \). If \( {p=1/2} \) then the chain is null recurrent and the invariant measures are all proportional to the counting measure. If \( {p>1/2} \) then the chain is transient and all invariant measures are proportional to a geometric measure which cannot be normalized into a probability measure.

Growth and collapse walk on \( {\mathbb{N}} \). The transition kernel is given by \( {\mathbf{P}(x,x+1)=p_x} \) and \( {\mathbf{P}(x,0)=1-p_x} \), for every \( {x\in\mathbb{N}} \), where \( {p_0=1} \), \( {p_1,p_2,\ldots\in[0,1]} \). The kernel is irreducible iff

\[ \{x\in\mathbb{N}:p_x>0\}=\mathbb{N} \quad\text{and}\quad \mbox{Card}\{x\in\mathbb{N}:p_x<1\}=\infty. \]

The very special growth-collapse nature of \( {\mathbf{P}} \) gives

\[ \mathbb{P}_0(T_0>n)=\prod_{x=0}^n\mathbf{P}(x,x+1)=p_0p_1\cdots p_n. \]

Also, if the kernel is irreducible, then it is recurrent iff

\[ \lim_{x\rightarrow\infty}p_0p_1\cdots p_x=0. \]

If \( {\mu} \) is invariant then for any \( {x\in\mathbb{N}} \) with \( {x>0} \),

\[ \mu(x)=\sum_{y\in\mathbb{N}}\mu(y)\mathbf{P}(y,x)=\mu(x-1)p_{x-1}=\mu(0)\cdots=p_0p_1\cdots p_{x-1}. \]

It follows that if the kernel is irreducible, then it is positive recurrent iff

\[ \sum_{x=0}^\infty p_0p_1\cdots p_x<\infty, \]

and in this case there exists a unique invariant probability measure given by

\[ \mu(x)=\frac{1}{\mathbb{E}_x(T_x)},\quad x\in\mathbb{N}. \]

Note that if \( {p=\prod_{x\in\mathbb{N}}p_x\in(0,\infty)} \) and if \( {\mu} \) is invariant then \( {\mu(x)\geq p\mu(0)} \), and since \( {\mu(0)=\sum_{x\in\mathbb{N}}(1-p_x)\mu(x)} \), we get \( {\mu(0)\geq p\mu(0)\sum_{x\in\mathbb{N}}(1-p_x)} \), which forces either \( {\mu(0)=0} \) or \( {\mu(0)=\infty} \), and therefore there is no invariant measure in this case.

Further reading. Markov chains, by James Norris.

Leave a Comment