An “elementary” problem

Posted on September 10, 2013 by Fabrice Baudoin

A few days ago, I was told the following problem:

Let E a finite set of points in the plane with the property that any line containing two points of E contains at least a third one. Show that E is included in a line.

I find this problem particularly interesting because the problem a priori reflects an affine property of the plane, however the only solution I know is very tricky and uses an additional structure of the plane…

Can you find the solution of this problem ?

Posted in Mathematicians | 5 Comments

Meet Cedric Villani

Posted on August 28, 2013 by Fabrice Baudoin

Cedric Villani is an extremely well-known mathematician in France, who was awarded the Fields medal in 2009. The following video shows well his personality and explains a little his conception about the work of mathematicians.

As he points out, mathematics have a very long tradition of excellence in France and some mathematicians, like Villani, are highly respected individuals who give interviews to mainstream medias, write books for the general public and give their opinions about all kind of things. I find that this sharply contrasts the situation in the US, where I think most of the people are not even aware that mathematician is a job or that there exists a research in mathematics. Valorizing mathematics and mathematicians in the mainstream media certainly encourages a lot of vocations in the youth and the US system could benefit from this. There are many other reasons, but I think, it partly explains the striking low proportion of mathematicians, who work in the US and were born, raised and educated here.

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Lecture 12. The heat semigroup on a compact Riemannian manifold

Posted on August 28, 2013 by Fabrice Baudoin

In this section we shall consider a smooth and complete Riemannian manifold (\mathbb{M},g) with dimension n. The Riemannian measure will be denoted by \mu.

The Laplace-Beltrami of \mathbb{M} will be denoted by L. Since \mathbb{M} is assumed to be complete, as we have seen in the previous section, the operator L is essentially self-adjoint on the space \mathcal{C}_c(\mathbb{M},\mathbb{M}. More precisely, there exists an increasing sequence h_n\in C_c(\mathbb{M},\mathbb{R}) such that h_n\nearrow 1 on \mathbb{M}, and ||\Gamma(h_n,h_n)||_{\infty} \to 0, as n\to \infty.

The Friedrichs extension of L, which is therefore the unique self-adjoint extension of L in \mathbf{L}^2_\mu (\mathbb{M},\mathbb{R}) will still be denoted by L and the domain of this extension is denoted by \mathcal{D}(L).

We shall not repeat the whole theory of diffusion semigroups on manifolds, since many of the results that were proved before are easily extended to manifolds. In particular, we may prove:

  • By using the spectral theorem for L in the Hilbert space \mathbf{L}^2_\mu (\mathbb{M},\mathbb{R}), we may construct a strongly continuous contraction semigroup (\mathbf{P}_t)_{t \ge 0} in \mathbf{L}^2_\mu (\mathbb{M},\mathbb{R}) whose infinitesimal generator is L;
  • By using the ellipticity of L, we may prove that (\mathbf{P}_t)_{t \ge 0} admits a heat kernel, that is: There is a smooth function p(t,x,y), t \in (0,+\infty), x,y \in \mathbb{M}, such that for every f \in \mathbf{L}_{\mu}^2 (\mathbb{M},\mathbb{R}) and x \in \mathbb{M} , \mathbf{P}_t f (x)=\int_{\mathbb{M}} p(t,x,y) f(y) d\mu (y). Moreover, the heat kernel satisfies the two following conditions:
    • (Symmetry) p(t,x,y)=p(t,y,x);
    • (Chapman-Kolmogorov relation) p(t+s,x,y)=\int_{\mathbb{M}} p(t,x,z)p(s,z,y)d\mu(z).
  • The semigroup (\mathbf{P}_t)_{t \ge 0} is a sub-Markov semigroup: If 0\le f \le 1 is a function in \mathbf{L}_{\mu}^2 (\mathbb{M},\mathbb{R}), then 0 \le \mathbf{P}_t f \le 1.
  • By using the Riesz-Thorin interpolation theorem, (\mathbf{P}_t)_{t \ge 0} defines a contraction semigroup on \mathbf{L}_{\mu}^p (\mathbb{M},\mathbb{R}), 1 \le p \le \infty.

Let us now assume that (\mathbb{M},g) is a compact Riemannian manifold. In that case, it obviously complete. As usual, we denote by (\mathbf{P}_t)_{t \ge 0} the heat semigroup and by p(t,x,y) the corresponding heat kernel. As a preliminary result, we have the following Liouville’s type theorem.

Lemma: Let f \in \mathcal{D}(L) such that Lf=0, then f is a constant function.

Proof: From the ellipticity of L, we first deduce that f is smooth. Then, since \mathbb{M} is compact, the following equality holds -\int_{\mathbb{M}} f Lf d\mu=\int_{\mathbb{M}} \Gamma(f,f) d\mu. Therefore \Gamma(f,f)=0, which implies that f is a constant function \square

In the compact case, the heat semigroup satisfies the so-called stochastic completeness (or Markov) property.

Proposition: For t \ge 0, \mathbf{P}_t 1=1.

Proof: Since the constant function 1 is in \mathbf{L}^2_\mu (\mathbb{M},\mathbb{R}), by compactness of \mathbb{M}, we may apply uniqueness of solutions of the heat equation in \mathbf{L}^2_\mu (\mathbb{M},\mathbb{R}) \square

It turns out that the compactness of \mathbb{M} implies the compactness of the semigroup.

Proposition: For t > 0 the operator \mathbf{P}_t is a compact operator on the Hilbert space \mathbf{L}^2_\mu (\mathbb{M},\mathbb{R}). It is moreover trace class and \mathbf{Tr} (\mathbf{P}_t)=\int_\mathbb{M} p(t,x,x)\mu(dx).

Proof: We shall provide two proofs of the fact that \mathbf{P}_t is a compact operator. You may observe that the first proof does not rely on the existence result of the heat kernel. The first proof stems from the local regularity theory of elliptic operators. Indeed, for t > 0, the operator \mathbf{P}_t: \mathbf{L}^2_\mu (\mathbb{M},\mathbb{R}) \rightarrow \mathcal{H}^0_1(\mathbb{M}) is bounded. Moreover, from Rellich’s theorem, the map \iota: \mathcal{H}^{0}_1 (\mathbb{M}) \rightarrow \mathcal{H}^{0}_0 (\mathbb{M})=\mathbf{L}^2_\mu (\mathbb{M},\mathbb{R}), is compact. Therefore by composition,
\mathbf{P}_t: \mathbf{L}^2_\mu (\mathbb{M},\mathbb{R}) \rightarrow \mathbf{L}^2_\mu (\mathbb{M},\mathbb{R}) is a compact operator.

The second proof is simpler and more direct. Indeed, from the existence of the heat kernel \mathbf{P}_t f (x)=\int_{\mathbb{M}} p(t,x,y) f(y) d\mu (y). But from the compactness of \mathbb{M}, \int_{\mathbb{M}}\int_{\mathbb{M}} p(t,x,y)^2 d \mu(x) d\mu(y) < +\infty. Therefore, the operator \mathbf{P}_t: \mathbf{L}^2_\mu (\mathbb{M},\mathbb{R}) \rightarrow \mathbf{L}^2_\mu (\mathbb{M},\mathbb{R})
is a Hilbert-Schmidt operator. It is thus in particular a compact operator.

Since \mathbf{P}_t=\mathbf{P}_{t/2} \mathbf{P}_{t/2}, \mathbf{P}_t is a product of two Hilbert-Schmidt operators. It is therefore a class trace operator and \mathbf{Tr} ( \mathbf{P}_t)=\int_{\mathbb{M}}\int_{\mathbb{M}} p(t/2,x,y)p(t/2,y,x) d \mu(x) d\mu(y). We conclude then by applying the Chapman-Kolmogorov relation \square

In this compact framework, we have the following theorem

Theorem: There exists a complete orthonormal basis (\phi_n)_{n \in \mathbb{N}} of \mathbf{L}^2_\mu (\mathbb{M},\mathbb{R}), consisting of eigenfunctions of -L, with \phi_n having an eigenvalue \lambda_n with finite multiplicity satisfying 0=\lambda_0 < \lambda_1\le \lambda_2 \le \cdots \nearrow +\infty. Moreover, for t > 0, x,y \in \mathbb{M}, p(t,x,y)=\sum_{n=0}^{+\infty} e^{-\lambda_n t} \phi_n (x) \phi_n (y), with convergence absolute and uniform for each t > 0.

Proof: Let t > 0. From the Hilbert-Schmidt theorem for the non negative self adjoint compact operator \mathbf{P}_t, there exists a complete orthonormal basis (\phi_n(t) )_{n \in \mathbb{N}} of \mathbf{L}^2_\mu (\mathbb{M},\mathbb{R}) and a non increasing sequence \alpha_n(t) \ge 0, \alpha_n(t) \searrow 0 such that \mathbf{P}_t \phi_n(t)=\alpha_n(t) \phi_n (t). The semigroup property \mathbf{P}_{t+s}=\mathbf{P}_t \mathbf{P}_s implies first that for k \in \mathbb{N}, k \ge 1, \phi_n(k)=\phi_n(1), \alpha_n(k)=\alpha_n (1)^k. The same result is then seen to hold for k \in \mathbb{Q}, k > 0 and finally for k \in \mathbb{R}, due to the strong continuity of the semigroup. Since the map t \to \| P_t \|_2 is decreasing, we deduce that \alpha_n (1) \le 1. Thus, there is a \lambda_n \ge 0 such that \alpha_n(1)=e^{-\lambda_n}. As a conclusion, there exists a complete orthonormal basis (\phi_n)_{n \in \mathbb{N}} of $\mathbf{L}^2_\mu (\mathbb{M},\mathbb{R})$, and a sequence \lambda_n satisfying 0 \le \lambda_0 \le \lambda_1\le \lambda_2 \le \cdots \nearrow +\infty such that \mathbf{P}_t \phi_n =e^{-\lambda_n t} \phi_n. Since \mathbf{P}_t 1=1, we actually have \lambda_0=0. Also, if f \in \mathbf{L}^2_\mu (\mathbb{M},\mathbb{R}) is such that \mathbf{P}_t f=f, it is straightforward that f \in \mathcal{D}(L) and that Lf=0, so that thanks to Liouville theorem, f is a constant function. Therefore \lambda_1 > 0.

Since \mathbf{P}_t \phi_n =e^{-\lambda_n t} \phi_n, by differentiating as t \to 0 in \mathbf{L}^2_\mu (\mathbb{M},\mathbb{R}), we obtain furthermore that \phi_n \in \mathcal{D}(L) and that L\phi_n=-\lambda_n \phi_n. The family (x,y)\to \phi_n(x) \phi_m(y) forms an orthonormal basis of \mathbf{L}^2_{\mu \otimes \mu} (\mathbb{M}\times \mathbb{M},\mathbb{R}). We therefore have a decomposition in \mathbf{L}^2_{\mu \otimes \mu} (\mathbb{M}\times \mathbb{M},\mathbb{R}), p(t,x,y)=\sum_{m,n \in \mathbb{M}} c_{mn} \phi_m(x) \phi_n(y). Since p(t,\cdot,\cdot) is the kernel of \mathbf{P}_t, it is then straightforward that for m \neq n, c_{mn}=0 and that c_{nn}=e^{-\lambda_n t}. Therefore in \mathbf{L}^2_\mu (\mathbb{M},\mathbb{R}), p(t,x,y)=\sum_{n=0}^{+\infty} e^{-\lambda_n t} \phi_n (x) \phi_n (y). The continuity of p, together with the positivity of \mathbf{P}_t imply, via Mercer’s theorem that actually, the above series is absolutely and uniformly convergent for t > 0 \square

As we stressed it in the statement of the theorem, in the decomposition p(t,x,y)=\sum_{n=0}^{+\infty} e^{-\lambda_n t} \phi_n (x) \phi_n (y), the eigenvalue \lambda_n is repeated according to its multiplicity. It is often useful to rewrite this decomposition under the form p(t,x,y)=\sum_{n=0}^{+\infty} e^{-\alpha_n t} \sum_{k=1}^{d_n} \phi^n_k (x) \phi^n_k (y), where the eigenvalue \alpha_n is not repeated, that is 0=\alpha_0 < \alpha_1 < \alpha_2 < \cdots In this decomposition, d_n is the dimension of the eigenspace \mathcal{V}_n corresponding to the eigenvalue \alpha_n and (\phi^n_k)_{1 \le k \le d_n } is an orthonormal basis of \mathcal{V}_n. If we denote, \mathcal{K}_n(x,y)= \sum_{k=1}^{d_n} \phi^n_k (x) \phi^n_k (y), then \mathcal{K}_n is called the reproducing kernel of the eigenspace \mathcal{V}_n. It satisfies the following properties whose proofs are let to the reader:

Proposition:

  • \mathcal{K}_n does not depend on the choice of the basis (\phi^n_k)_{1 \le k \le d_n } ;
  • If f \in \mathcal{V}_n, then \int_{\mathbb{M}} \mathcal{K}_n (x,y) f(y) d\mu(y) =f(x).

From the very definition of the reproducing kernels, we have p(t,x,y)=\sum_{n=0}^{+\infty} e^{-\alpha_n t} \mathcal{K}_n(x,y).

The compactness of \mathbb{M} also implies the convergence to equilibrium for the semigroup with an exponential rate.

Proposition: Let f \in \mathbf{L}^2_\mu (\mathbb{M},\mathbb{R}), then uniformly on \mathbb{M}, when t \to +\infty, \mathbf{P}_t f \rightarrow \frac{1}{\mu(\mathbb{M})} \int_\mathbb{M} f d\mu.

Proof: It is obvious from the previous proposition and from spectral theory that in \mathbf{L}^2_\mu (\mathbb{M},\mathbb{R}), \mathbf{P}_t f converges to a constant function that we denote \mathbf{P}_\infty f . The convergence is also uniform, because for s,t,T > 0,
\| \mathbf{P}_{t+T} f -\mathbf{P}_{s+T} f \|_\infty
= \sup_{x \in \mathbb{M}} \left| \mathbf{P}_{T} ( \mathbf{P}_{t} f -\mathbf{P}_{s} f) (x) \right|
= \sup_{x \in \mathbb{M}} \left|\int_{\mathbb{M}} p(T,x,y) ( \mathbf{P}_{t} f -\mathbf{P}_{s} f) (y) d\mu(y) \right|
\le \left( \sup_{x \in \mathbb{M}} \sqrt{ \int_{\mathbb{M}} p(T,x,y)^2 d\mu(y)}\right) \| \mathbf{P}_{t} f -\mathbf{P}_{s} f \|_2.
Moreover, for every t \ge 0, \int_{\mathbb{M}} \mathbf{P}_t f d\mu =\int_{\mathbb{M}} f d\mu. Therefore \int_{\mathbb{M}} \mathbf{P}_\infty f d\mu =\int_{\mathbb{M}} f d\mu. Since \mathbf{P}_\infty f is constant, we finally deduce the expected result \mathbf{P}_\infty f=\frac{1}{\mu(\mathbb{M})} \int_\mathbb{M} f d\mu \square

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Lecture 11. The Laplace-Beltrami operator on a Riemannian manifold

Posted on August 28, 2013 by Fabrice Baudoin

In this lecture we extend the previous results in the framework of smooth manifolds. The main idea to extend those results is that, similar computations may be performed in local coordinates charts and then we use a partition of unity.

Lemma: Let \mathbb{M} be a paracompact manifold. Let (U_i)_{i \in I} be a locally finite covering of \mathbb{M} such that each \bar{U}_i is compact. Then, there exists a system (\phi_i)_{i \in I} of smooth functions on \mathbb{M} such that:

  • Each \phi_i has a compact support contained in U_i,
  • \phi_i \ge 0, \sum_{i \in I} \phi_i =1.

 

We recall that on a topological space \mathbb{M}, a covering (U_i)_{i \in I} is said to be locally finite if each m\in \mathbb{M} has a neighborhood that intersects only finitely many of the sets U_i‘s. The space \mathbb{M} is said to be paracompact if for each covering (U_i)_{i \in I} of \mathbb{M}, there is a locally finite covering of \mathbb{M} which is a refinement of (U_i)_{i \in I}.

From now on, in this lecture \mathbb{M} will be a smooth manifold with dimension n.

Definition: A Riemannian structure g on \mathbb{M} is a smooth, symmetric and positive (0,2) tensor on \mathbb{M}.

In other words, a Riemannian structure g induces for each x \in \mathbb{M} an inner product g_x on the tangent space \mathbf{T}_x \mathbb{M} and the dependence x \rightarrow g_x is required to be smooth.

Unlike the case of \mathbb{R}^n, in general we may not define a Riemannian structure on a manifold by using global frames. For instance on the two-dimensional sphere \mathbb{S}^2, it is impossible to find smooth vector fields (V_1,V_2) such that for every x \in \mathbb{S}^2, (V_1(x),V_2(x)) is a basis of \mathbf{T}_x \mathbb{S}^2. However, of course, we may always deal with local orthonormal frames: That is, if \mathbb{M} is a smooth Riemannian manifold (i.e. a smooth manifold endowed with a Riemannian structure), for every x in \mathbb{M}, we can find an open set x \in U \subset \mathbb{M} and smooth vector fields (V_1,\cdots,V_n) on U such that for every y \in U, (V_1(y),\cdots, V_n(y)) is an orthonormal basis of the tangent space \mathbf{T}_y \mathbb{M} for the inner product g_y.

From now on we consider a smooth Riemannian manifold (\mathbb{M},g). It is possible to find a locally finite covering of \mathbb{M} by local coordinate charts (U_i,\phi_i)_{i \in I} and smooth vector fields (V_1^i,\cdots,V_n^i) on U_i such that for every x \in U_i, (V^i_1(x),\cdots, V^i_n(x)) is an orthonormal basis of the tangent space \mathbf{T}_x \mathbb{M} for the inner product g_x. Let (\psi_i)_{i \in I} be a partition of unity subordinated to this covering.

Our first goal is to define the canonical Riemannian measure on \mathbb{M}. The vector fields (V_1^i,\cdots,V_n^i) induce smooth vector fields (\tilde{V}_1^i,\cdots,{V}_n^i) on \phi_i(U_i). Without loss of generality, we may assume that on \phi_i(U_i), \mathbf{det}(\tilde{V}_1^i,\cdots,\tilde{V}_n^i) >0. Consider on \phi_i(U_i) the Borel measure with density d\mu_i = \frac{1}{\mathbf{det} (\tilde{V}_1^i,\cdots,\tilde{V}_n^i)} d x, where dx is the Lebesgue measure on \mathbb{R}^n. If f:\mathbb{M} \rightarrow \mathbb{R} is a non negative Borel function with a compact support included in U_i, it is natural to define \mu(f)= \int_{\phi_i(U_i)} f \circ \phi_i^{-1} d\mu_i. We observe that if the support of f is included in U_i \cap U_j, \int_{\phi_i(U_i)} f \circ \phi_i^{-1} d\mu_i= \int_{\phi_j(U_j)} f \circ \phi_j^{-1} d\mu_j, so that \mu is well defined. Now, for a general non negative Borel function f:\mathbb{M} \rightarrow \mathbb{R}, we define \mu (f)= \sum_{i \in I} \mu(\psi_i f), where (\psi_i)_{i \in I} is the partition of unity subordinated to the covering (U_i)_{i \in I}. This defines a Borel measure \mu on \mathbb{M} which is called the Riemannian measure.

The same idea allows to construct the Laplace-Beltrami operator on \mathbb{M}. If f:\mathbb{M} \rightarrow \mathbb{R} is a smooth function on U_i, we define L^i f=\sum_{k=1}^n (V_k^i)^2 f+V_0^i f, where V_0^i is the smooth vector field on the open set U_i constructed as in the linear case. Let us now observe that, on U_i \cap U_j, we have L^i f=L^jf. This leads to the following definition of the Laplace-Beltrami operator on \mathbb{M}: If f: \mathbb{M} \rightarrow \mathbb{R} is a smooth function, Lf= \sum_{i \in I} L^i(\psi_i f), where (\psi_i)_{i \in I} is the partition of unity subordinated to the covering (U_i)_{i \in I}.

Exercise: Show that the Laplace-Beltrami operator L is symmetric with respect to the Riemannian measure \mu.

Diffusion operators on manifolds are intrinsically defined as follows:

Definition: Let \mathcal{C}^{\infty} (\mathbb{M}, \mathbb{R}) be the set of smooth functions \mathbb{M} \rightarrow \mathbb{R} and \mathcal{C} (\mathbb{M}, \mathbb{R}) be the set of continuous functions \mathbb{M} \rightarrow \mathbb{R}. A diffusion operator L is an operator
L: \mathcal{C}^{\infty} (\mathbb{M}, \mathbb{R}) \rightarrow \mathcal{C} (\mathbb{M}, \mathbb{R}) such that:

  • L is linear;
  • L is a local operator; That is, if f,g \in \mathcal{C}^{\infty} (\mathbb{M}, \mathbb{R}) coincide on a neighnorhood of x, then Lf(x)=Lg(x);
  • If f \in \mathcal{C}^{\infty} (\mathbb{M}, \mathbb{R}) has a local minimum at x, Lf (x) \ge 0.

 

And it is easily seen, that the Laplace-Beltrami is a diffusion operator. It is moreover elliptic in the sense that if (\phi,U) is a local coordinate chart, then the operator L read in this chart is an elliptic operator on \phi(U).

As usual, we associate to L the differential bilinear form \Gamma(f,g) =\frac{1}{2}(L(fg)-fLg-gLf). The bilinear form \Gamma is related to the notion of Riemannian gradient.

Definition: Let f:\mathbb{M} \rightarrow \mathbb{R} be a smooth function. There is a unique smooth vector field on \mathbb{M} which is denoted by \nabla f and that is called the Riemannian gradient that satisfies for every x \in \mathbb{M} and u \in \mathbf{T}_x \mathbb{M}, df_x(u)=g_x( \nabla f (x), u), where df is the differential of f.

If U is an open set of \mathbb{M}, and V_1,\cdots, V_n are smooth vector fields on U such that for x \in U, (V_1(x),\cdots, V_n(x)) is an orthonormal frame of \mathbf{T}_x \mathbb{M}, it is readily checked that \nabla f (x) =\sum_{i=1}^n V_if(x) V_i (x), \quad x \in U.

The bilinear form \Gamma is related to the Riemannian gradient by the following formula:

Lemma: Let f,h :\mathbb{M} \rightarrow \mathbb{R} be smooth functions. We have \Gamma (f,h)=g(\nabla f, \nabla h).

Proof: Let x \in \mathbb{M}. Let U be an open neighborhood of x and V_1,\cdots, V_n smooth vector fields on U such that for y \in U, (V_1(y),\cdots, V_n(y)) is an orthonormal frame of \mathbf{T}_y \mathbb{M}. On U, we have \Gamma(f,h) =\frac{1}{2}(L(fh)-fLh-hLf)=\sum_{i=1}^n V_i f V_i f=g(\nabla f, \nabla h), so that in particular \Gamma (f,h)(x)=g(\nabla f, \nabla h)(x). Since x is arbitrary, the proof is complete \square

The last result we wish to extend to manifolds is the relation between completeness and essential self-adjointness of the Laplace-Beltrami operator.

The Riemannian distance on a Riemannian manifold is defined exactly as in \mathbb{R}^n.

Given an absolutely continuous curve \gamma: [0,1] \rightarrow \mathbb{M}, we define its Riemannian length by L_g (\gamma)=\int_0^1 \sqrt{g_{\gamma(s)} (\gamma'(s),\gamma'(s) )} ds. If x,y \in \mathbb{M}, let us denote by \mathcal{C}(x,y) the set of absolutely continuous curves \gamma: [0,1] \rightarrow \mathbb{M} such that \gamma(0)=x, \gamma(1)=y. The Riemannian distance between x and y is then defined by d(x,y)=\inf_{\gamma \in \mathcal{C}(x,y)} L_g(\gamma). The Hopf-Rinow theorem also holds on manifolds:

Theorem:(Hopf-Rinow theorem on manifolds) The metric space (\mathbb{M},d) is complete if and only the compact sets are the closed and bounded sets.

We then have the following expected theorem:

Theorem: If the metric space (\mathbb{M},d) is complete, then the Laplace-Beltrami operator L is essentially self-adjoint on the space of smooth and compactly supported functions f:\mathbb{M} \rightarrow \mathbb{R}.

Posted in Curvature dimension inequalities | 2 Comments

How to write mathematics badly

Posted on August 17, 2013 by Fabrice Baudoin

It is not always clear how to write a mathematics paper but it is much more clear how not to write it !

This funny video by Jean-Pierre Serre shows some of the things one should avoid.

 

 

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Creativity in mathematics

Posted on July 7, 2013 by Fabrice Baudoin

Richard Feynman was obviously an extraordinary physicist and a charismatic lecturer. In the following video he discusses the differences and the interplay between mathematics and physics:

His opening statement is: “Mathematicians only are dealing with the structure of the reasoning.” Actually, as many other mathematicians, I believe that dealing with the structure of the reasoning is a very secondary (and boring) aspect of my work. From a personal perspective, I think that, ideally, research in mathematics is more about creativity: establishing connections between ideas and concepts based on a criterium of elegance and beauty.

The psychology of the invention in mathematics has been adressed by one of the greatest French mathematician, Henri Poincaré, in the following famous lecture before the Société de Psychologie in Paris: Mathematical creation.

This lecture inspired the beautiful essay by Jacques Hadamard: The psychology of invention in the mathematical field, that I recommend to anyone interested by the research in mathematics. A very interesting discussion is also given by Paul Halmos in the following paper: Mathematics as a creative art.

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S.S. Chern

Posted on July 7, 2013 by Fabrice Baudoin

The following video contains interesting testimonies about the differential geometer Shiing-Shen Chern.

Source

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Lecture 10. The distance associated to Laplace-Beltrami operators

Posted on July 3, 2013 by Fabrice Baudoin

In order to apply the diffusion semigroup theory developed in the first lectures and construct without ambiguity the semigroup associated to the Laplace-Beltrami operator L, we need to know if L is essentially self-adjoint. Interestingly, this property of essential self-adjointness is clodely related to a metric property of the underlying Riemannian structure: The completeness of the associated distance.

Given an absolutely continuous curve \gamma: [0,T] \rightarrow \mathbb{R}^n, we define its Riemannian length by L_g (\gamma)=\int_0^T \sqrt{g_{\gamma(s)} (\gamma'(s),\gamma'(s) )} ds. If x,y \in \mathbb{R}^n, let us denote by \mathcal{C}(x,y) the set of absolutely continuous curves \gamma: [0,1] \rightarrow \mathbb{R}^n such that \gamma(0)=x, \gamma(1)=y. The Riemannian distance between x and y is defined by d(x,y)=\inf_{\gamma \in \mathcal{C}(x,y)} L_g(\gamma). By using reparametrization, we may define the Riemannian distance in a equivalent way by using the notion of sub-unit curve. Let \gamma: [0,T] \to \mathbb{R}^n be an absolutely continuous curve. Since the vector fields V_1,\cdots,V_n‘s form a basis of \mathbb{R}^n at each point, we may find continuous functions \alpha_i:[0,T] \to \mathbb{R}^n such that \gamma'(t)=\sum_{i=1}^n \alpha_i(t) V_i (\gamma(t)). The curve \gamma is then said to be sub-unit if for almost every t \in [0,T], \sum_{i=1}^n \alpha_i(t)^2 \le 1. By using reparametrization, it is easily seen that for x,y \in \mathbb{R}^n, d(x,y)=\inf \left\{ T \text{ such that there exists a sub-unit curve } \gamma:[0,T] \to \mathbb{R}^n, \gamma(0)=x, \gamma(1)=y \right\}.

Exercise: Let \theta_1, \cdots , \theta_n \in \mathbb{R}. Show that for every x \in \mathbb{R}^n, d \left(x, e^{\sum_{i=1}^n \theta_i V_i } x \right) \le \sum_{i=1}^n \theta_i^2.

An important fact is that d hence defined is indeed a distance that induces the usual topology of \mathbb{R}^n.

Definition: The function d defined above is a distance that induces the usual topology of \mathbb{R}^n.

Proof: Since any curve can be parametrized backwards and forwards, we have d(x,y)=d(y,x). The triangle inequality is easily proved by using juxtaposition of curves. Plainly d(x,x)=0, so it remains to prove that if x \neq y then d(x,y) > 0. Let x,y \in \mathbb{R}^n such that x \neq y. Let us denote R=\| x-y \|. The closed Euclidean ball \bar{B}_e (x,R) is compact, therefore there exist two constants \alpha, \beta > 0 such that for every z \in \bar{B}_e (x,R) and u \in \mathbb{R}^n, \alpha^2 \| u \|^2 \le g_z(u,u) \le \beta^2 \| u \|^2. Let now \gamma:[0,1]\to \mathbb{R}^n be an absolutely continuous curve such that \gamma(0)=x, \gamma(1)=y. Let \tau =\inf \{ t, \| \gamma(t) \|=R \}.
We have
L_g(\gamma)
=\int_0^1 \sqrt{g_{\gamma(s)} (\gamma'(s),\gamma'(s) )} ds
\ge \int_0^\tau \sqrt{g_{\gamma(s)} (\gamma'(s),\gamma'(s) )} ds
\ge \alpha \int_0^\tau \| \gamma'(s) \| ds
\ge \alpha \int_0^\tau \| \gamma'(s) \| ds
\ge \alpha \| \gamma(\tau) - \gamma (0) \|
\ge \alpha \| x -y \|.
As a consequence, we deduce that d(x,y) \ge \alpha^2 \| x -y \| > 0. Therefore d is indeed a distance. Moreover, it is shown as above that for every z \in \mathbb{R}^n, R> 0, there are constants C_1,C_2> 0 such that for every x,y \in \bar{B}_e (z,R), C_1 \| x-y \| \le d(x,y) \le C_2 \| x-y \|. This implies that d induces the usual topology of \mathbb{R}^n \square

As shown in the following proposition, the distance d is intrinsically associated to the Laplace-Beltrami operator.

Proposition: For x,y \in \mathbb{R}^n, we have d(x,y) =\sup \{ | f(x)-f(y) |, f \in \mathcal{C}_c^\infty(\mathbb{R}^n,\mathbb{R}), \| \Gamma(f,f) \|_\infty \le 1 \}.

Proof: Let x,y \in \mathbb{R}^n. We denote \delta (x,y)=\sup \{ | f(x)-f(y) |, f \in \mathcal{C}_c^\infty(\mathbb{R}^n,\mathbb{R}), \| \Gamma(f,f) \|_\infty \le 1 \}. Let \gamma: [0,T] \to \mathbb{R} be a sub-unit curve such that \gamma(0)=x, \gamma(T)=x. We can find \alpha_1,\cdots,\alpha_n:[0,T] \to \mathbb{R}^n such that \gamma'(t)=\sum_{i=1}^n V_i (\gamma(t)) \alpha_i (t), and \sum_{i=1}^n \alpha_i^2 \le 1. Let now f\in \mathcal{C}^\infty(\mathbb{R}^n,\mathbb{R}), \| \Gamma(f,f) \|_\infty \le 1. From the change of variable formula we have, f(\gamma(T))= f(\gamma(0)) + \sum_{i=1}^n \int_0^T V_i f(\gamma(s)) \alpha_i(s) ds. Therefore, from Cauchy-Schwarz inequality, \left| f(y)-f(x) \right| \le T. As a consequence \delta (x,y) \le d(x,y).

We now prove the converse inequality which is trickier. The idea is to use the function f(y)=d(x,y) that satisfies | f(x) -f(y)|=d(x,y) and "\Gamma(f,f) =1". However, giving a precise meaning to \Gamma(f,f) =1 is not so easy, because it turns out that f is not differentiable at x. It suggests to use an approximation of the identity to regularize f and avoid the discussion of this differentiability issue. More precisely, fix x_o ,y_o\in \mathbb{R}^n, and for N \ge 1, consider the function \Psi_N(y)= \eta \left( \int_{\mathbb{R}^n} \rho_N (t) d(x_o,y-t) dt \right), where \rho \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}), \rho \ge 0, \int_{\mathbb{R}^n} \rho=1, \rho_N(t)=N^n \rho(N t) and \eta \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}), \eta \ge 0, has the property that \| \eta ' \|_\infty \le 1 and \eta (u)=u for u \in [0,1+d(x_o,y_o)]. Since for any \theta \in \mathbb{R}^{n}, \| \theta \|=1, t \ge 0 | d(x_o,y)-d\left(x_o, e^{t \sum_{i=1}^n \theta_i V_i} y\right)| \le d\left(y, e^{t \sum_{i=1}^n \theta_i V_i} y\right) \le t, it is easy to see that \Gamma(\Psi_N,\Psi_N) \le 1+\frac{C}{N}, for some constant > 0. Hence \delta(x_o,y_o) \ge \lim \inf_{N \to +\infty} | \Psi_N(y_o) -\Psi_N(x_o) |=d(x_o,y_o) \square

The following theorem is known as the Hopf–Rinow theorem, it provides a necessary and sufficient condition for the completeness of the metric space (\mathbb{R}^n,d).

Proposition: The metric space (\mathbb{R}^n,d) is complete (i.e. Cauchy sequences are convergent) if and only the compact sets are the closed and bounded sets.

Proof: It is clear that if closed and bounded sets are compact then the metric space (\mathbb{R}^n,d) is complete; It comes from the fact that Cauchy sequence are convergent if and only if they have at least one cluster value. So, we need to prove that closed and bounded sets for the distance d are compact provided that (\mathbb{R}^n,d) is complete. To check this, it is enough to prove that closed balls are compact. Let x \in \mathbb{R}^n. Observe that if the closed ball \bar{B}(x, r) is compact for some r> 0, then \bar{B}(x, \rho) is closed for any \rho < r. Define R=\sup \{ r> 0, \bar{B}(x, r) \text{ is compact } \}. Since d induces the usual topology of \mathbb{R}^n, R>0. Let us assume that R < +\infty and let us show that it leads to a contradiction. We first show that \bar{B}(x, R) is compact. Since (\mathbb{R}^n,d) is assumed to be complete, it suffices to prove that \bar{B}(x, R) is totally bounded: That is, for every \varepsilon > 0 there is a finite set S_\varepsilon such that every point of \bar{B}(x, R) belongs to a \varepsilon-neighborhood of S_\varepsilon.

So, let \varepsilon > 0 small enough. By definition of R, the ball \bar{B}(x, R-\varepsilon / 4) is compact; It is therefore totally bounded. We can find a finite set S=\{ y_1,\cdots,y_N\} such that every point of \bar{B}(x, R-\varepsilon / 4) lies in a \varepsilon / 2-neighborhood of S. Let now y \in \bar{B}(x, R). We claim that there exists y' \in \bar{B}(x, R-\varepsilon / 4) such that d(y,y') \le \varepsilon /2 . If y \in \bar{B}(x, R-\varepsilon / 4), there is nothing to prove, we may therefore assume that y \notin \bar{B}(x, R-\varepsilon / 4). Consider then a sub-unit curve \gamma: [0,R+\varepsilon / 4] such that \gamma(0)=x, \gamma(R+\varepsilon/4)=y. Let \tau =\inf \{t, \gamma(t) \notin \bar{B}(x, R-\varepsilon / 4) \}. We have \tau \ge R-\varepsilon / 4. On the other hand, d(\gamma(\tau), \gamma(R+\varepsilon/4)) \le R+\varepsilon/4 -\tau. As a consequence, d(\gamma(\tau),y) \le \varepsilon /2. In every case, there exists therefore y' \in \bar{B}(x, R-\varepsilon / 4) such that d(y,y') \le \varepsilon /2 . We may then pick y_k in S such that d(y_k,y') \le \varepsilon / 2. From the triangle inequality, we have d(y,y_k) \le \varepsilon. So, at the end, it turns out that every point of \bar{B}(x, R) lies in a \varepsilon-neighborhood of S. This shows that \bar{B}(x, R) is totally bounded and therefore compact because (\mathbb{R}^n,d) is assumed to be complete. Actually, the previous argument shows more, it shows that if every point of \bar{B}(x, R) lies in a \varepsilon /2-neighborhood of a finite S, then every point of \bar{B}(x, R+\varepsilon/4) will lie \varepsilon-neighborhood of S, so that the ball \bar{B}(x, R+\varepsilon/4) is also compact. This contradicts the fact the definition of R. Therefore every closed ball is a compact set, due to the arbitrariness of x \square

Checking that the metric space (\mathbb{R}^n,d) is complete is not always easy in concrete situations. From the Hopf-Rinow theorem, it suffices to prove that the closed balls are compact. The following proposition is therefore useful.

Proposition: Suppose that the vector fields V_1,\cdots,V_n‘s have globally Lipschitz coefficients . Then the closed ball \bar B(x,R) is compact for every x \in \mathbb{R}^n and R > 0. As a consequence the metric space (\mathbb{R}^n,d) is complete.
Proof: By the hypothesis on the V_j‘s there exists a constant M > 0 such
that \|V(x)\|=\left(\sum_{j=1}^n \|V_j(x)\|^2\right)^\frac{1}{2}\leq M(1+\|x\|) for any x\in\mathbb{R}^n. Fix x_o,y\in\mathbb{R}^n and let \gamma:[0,T]\to\mathbb{R}^n, be a sub-unit curve such that \gamma(0)=x_o, \gamma(T)=y. Letting y(t)=\|\gamma(t)\|^2 we obtain y'(t)=2\leq 2 \|\gamma(t)\|\|\gamma'(t)\| \leq 2\|\gamma(t)\|\|V(\gamma(t))\|. We infer that y'(t)\leq C\left(\sqrt{y(t)}+y(t) \right), for some C>0 depending only on M. Integrating the latter inequality one has \|\gamma(t)\| \leq (A+\| x_o \|) e^{\frac{C}{2}T},\quad t\in [0,T]. for some constant A >0. The previous estimate shows in particular, that B(x_o,R)\subset B_e(0, (A+\| x_o \|) e^{CR}). We conclude that \bar B(x_o,R) is Euclidean compact. Since the metric and the Euclidean topology coincide, it is also d-compact \square

Completeness of the metric space (\mathbb{R}^n,d) is related to the essential self-adjointness of the Laplace-Beltrami operator.

Theorem: If the metric space (\mathbb{R}^n,d) is complete, then the Laplace-Beltrami operator L is essentially self-adjoint.

Proof: We know that if there exists an increasing sequence h_n\in C_c(\mathbb{R}^n,\mathbb{R}) such that h_n\nearrow 1 on \mathbb{R}^n, and ||\Gamma(h_n,h_n)||_{\infty} \to 0, as n\to \infty, then the operator L is essentially self-adjoint. We are therefore reduced to prove the existence of such a sequence. Let us fix a base point x_0\in\mathbb{R}^n. We can find an exhaustion function \rho\in C^\infty(\mathbb{R}^n,\mathbb{R}) such that |\rho - d(x_0,\cdot)| \le L,\ \ \ \ \ \ |\Gamma(\rho,\rho)|\le L \ \ \text{on } \mathbb{R}^n. By the completeness of (\mathbb{R}^n,d) and the Hopf-Rinow theorem, the level sets \Omega_s = \{x\in \mathbb{M} \mid \rho(x) < s\} are relatively compact and, furthermore, \Omega_s \nearrow \mathbb{R}^n as s\to \infty. We now pick an increasing sequence of functions \phi_n\in C^\infty([0,\infty)) such that \phi_n\equiv 1 on [0,n], \phi_n \equiv 0 outside [0,2n], and |\phi_n'|\le \frac{2}{n}. If we set h_n(x) = \phi_n(\rho(x)), then we have h_n\in C_c(\mathbb{R}^n,\mathbb{R}), h_n\nearrow1 on \mathbb{R}^n as n\to \infty, and ||\Gamma(h_n,h_n)||_{\infty} \le \frac{2L}{n} \square

 

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Lecture 9. Laplace-Beltrami operators on Rn

Posted on July 1, 2013 by Fabrice Baudoin

In this lecture we define Riemannian structures and corresponding Laplace–Beltrami operators. We first study Riemannian structures on Rn to avoid technicalities in the presentation of the main ideas and then, in a later lecture, will extend our results to the manifold case.

We start with the following basic definition:

Definition: A Riemannian structure on \mathbb{R}^n is a smooth map g from \mathbb{R}^n to the set of symmetric positive matrices.

In other words, a Riemannian structure induces at each point x \in \mathbb{R}^n an inner product g_x, and the dependence x \rightarrow g_x is asked to be smooth.

A natural way to define Riemannian structures, is to start from a family of smooth vector fields V_1,\cdots, V_n such that for every x \in \mathbb{R}^n, (V_1(x),\cdots, V_n(x)) is a basis of \mathbb{R}^n. It is then easily seen that there is a unique Riemannian structure on \mathbb{R}^n that makes for x \in \mathbb{R}^n, (V_1(x),\cdots, V_n(x)) an orthonormal basis.

Conversely, given a Riemannian structure on \mathbb{R}^n, it is possible to find smooth vector fields V_1,\cdots, V_n on \mathbb{R}^n such that for x \in \mathbb{R}^n, (V_1(x),\cdots, V_n(x)) an orthonormal basis for this Riemannian structure.

In this course, we shall mainly deal with such a point of view on Riemannian structures and use as much as possible the language of vector fields. This point of view is not restrictive and will allow more easy extensions to the sub-Riemannian case in a later part of the course.

Let us consider a family of smooth vector fields V_1,\cdots, V_n such that for every x \in \mathbb{R}^n, (V_1(x),\cdots, V_n(x)) is a basis of \mathbb{R}^n. Without loss of generality we may assume that \mathbf{det} ( V_1(x),\cdots, V_n(x)) > 0. Our goal is to associate to this Riemannian structure a canonical diffusion operator.

As a first step, we associate with the vector fields V_1,\cdots, V_n a natural Borel measure \mu which is the measure with density d\mu =\frac{1}{ \mathbf{det} ( V_1(x),\cdots, V_n(x))} dx with respect to the Lebesgue measure. This is the so-called Riemannian measure. The diffusion operator we want to consider shall be symmetric with respect to this measure.

Remark: Let (U_1,\cdots,U_n) be another system of smooth vector fields on \mathbb{R}^n such that for every x \in \mathbb{R}^n, (U_1(x),\cdots,U_n(x)) is an orthonormal basis with respect to the inner product g_x. The systems of vector fields (U_1,\cdots,U_n) and (V_1,\cdots,V_n) are related one to each other through an orthogonal mapping. This implies that |\mathbf{det} ( V_1(x),\cdots, V_n(x))|=| \mathbf{det} ( U_1(x),\cdots, U_n(x))|. In other words, the Riemannian measure \mu only depends on the Riemannian structure g.

Due to the fact that for every x \in \mathbb{R}^n, (V_1(x),\cdots, V_n(x)) is a basis of \mathbb{R}^n, we may find smooth functions \omega^{k}_{ij}‘s on $\mathbb{R}^n$ such that [V_i,V_j]=\sum_{k=1}^n \omega_{ij}^k V_k. Those functions are called the structure constants of the Riemannian structure. Every relevant geometric quantities may be expressed in terms of these functions. Of course, these functions depend on the choice of the vector fields $V_i$’s and thus are not Riemannian invariants, but several combinations of them, like curvature quantities, are Riemannian invariants.

The following proposition expresses the formal adjoint with respect to the Lebesgue measure of the vector field V_i which is seen as an operator acting on the spave of smooth and compactly supported functions.

Proposition: If f,g\in \mathcal{C}_c(\mathbb{R}^n, \mathbb{R}) are smooth and compactly supported functions, then we have \int_{\mathbb{R}^n} (V_i f ) g d\mu =\int_{\mathbb{R}^n} f (V_i^* g) d\mu, where V_i^*=-V_i+ \sum_{k=1}^n \omega_{ik}^k.

Proof: Let us denote V_i=\sum_{j=1}^n v_i^j \frac{\partial}{\partial x_j}, and m(x)= \mathbf{det} ( V_1(x),\cdots, V_n(x)). We have \int_{\mathbb{R}^n} (V_i f ) g d\mu =\int_{\mathbb{R}^n}\sum_{j=1}^n v_i^j \frac{\partial f}{\partial x_j} g \frac{dx}{m} =-\sum_{j=1}^n\int_{\mathbb{R}^n} f \left( m \frac{\partial}{\partial x_j} \frac{1}{m} g v_i^j \right) d\mu. We now compute \sum_{j=1}^n m \frac{\partial}{\partial x_j} \frac{1}{m} v_i^j=-\sum_{j=1}^n v_i^j \frac{1}{m} \frac{\partial m}{\partial x_j} +\sum_{j=1}^n\frac{\partial v_i^j}{\partial x_j}. We then observe that \frac{\partial m}{\partial x_j} = \frac{\partial}{\partial x_j} \mathbf{det} ( V_1(x),\cdots, V_n(x)) =\sum_{k=1}^n \mathbf{det} \left( V_1(x),\cdots,\frac{\partial V_k}{\partial x_j}(x),\cdots, V_n(x)\right).
Thus, we obtain
-\sum_{j=1}^n v_i^j \frac{\partial m}{\partial x_j} +\sum_{j=1}^n m \frac{\partial v_i^j}{\partial x_j} = \sum_{k=1}^n \mathbf{det} \left( V_1(x),\cdots,-\sum_{j=1}^n v_i^j \frac{\partial V_k}{\partial x_j}(x),\cdots, V_n(x)\right)
+\sum_{k=1}^n\mathbf{det} \left( V_1(x),\cdots,\frac{\partial v_i^j}{\partial x_j} V_k(x) ,\cdots, V_n(x)\right)
=-\sum_{k=1}^n\mathbf{det} \left( V_1(x),\cdots,[V_i,V_k](x) ,\cdots, V_n(x)\right)
=-\sum_{k=1}^n\mathbf{det} \left( V_1(x),\cdots,\sum_{j=1}^n \omega_{ik}^j(x) V_j(x) ,\cdots, V_n(x)\right)
=-\sum_{k=1}^n \omega_{ik}^k (x)\mathbf{det} \left( V_1(x),\cdots,V_k(x) ,\cdots, V_n(x)\right)=-m\sum_{k=1}^n \omega_{ik}^k. \square

With this integration by parts formula in hands we are led to the following natural definition
Definition: The diffusion operator L =-\sum_{i=1}^n V_i^* V_i =\sum_{i=1}^n V_i^2 -\sum_{i,k}^n \omega_{ik}^k V_i is called the Laplace-Beltrami operator associated with the Riemannian structure g.

The following straightforward properties of L are let as an exercise to the reader:

  • L is an elliptic operator;
  • The Riemannian measure \mu is invariant for L;
  • The operator L is symmetric with respect to \mu.

Exercise: Let (U_1,\cdots,U_n) be another system of smooth vector fields on \mathbb{R}^n such that for every x \in \mathbb{R}^n, (U_1(x),\cdots,U_n(x)) is an orthonormal basis with respect to the inner product g_x. Show that \sum_{i=1}^n V_i^* V_i=\sum_{i=1}^n U_i^* U_i. In other words, the Laplace-Beltrami operator is a Riemannian invariant: It only depends on the Riemannian structure g.

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Lecture 8. The heat semigroup on the circle

Posted on June 28, 2013 by Fabrice Baudoin

We now turn to a new part in this course. The first few lectures were devoted to the study of diffusion operators and the construction of associated semigroups. The goal of this new part of the course will be to construct the heat semigroup on a Riemannian manifold. We shall see that given a Riemannian structure on a differentiable manifold, it is possible to canonically associate to it a diffusion operator which is called the Laplace-Beltrami operator of the Riemannian structure.

As an appetizer, we first study the heat semigroup on the simplest (non Euclidean) Riemannian manifold: the circle \mathbb{S}^1= \left\{ e^{i\theta}, \theta \in \mathbb{R} \right\}. The Laplace operator on \mathbb{R}^n, \Delta=\sum_{i=1}^n \frac{\partial^2}{\partial x_i^2} is the canonical diffusion operator on \mathbb{R}^n. A natural question to be asked is: in the same way, is there a canonical diffusion operator on \mathbb{S}^1. A first step, of course, is to understand what is a diffusion operator on \mathbb{S}^1. We characterized diffusion operators as linear operators on the space of smooth functions that satisfy the maximum principle. Once a notion of smooth functions on \mathbb{S}^1 is understood, this maximum principle property can be taken as a definition. The circle \mathbb{S}^1 may be identified with the quotient space \mathbb{R} / 2\pi \mathbb{Z} . More precisely, it is easily shown that a smooth function, f :\mathbb{R} \rightarrow \mathbb{C} which is 2\pi periodic, i.e. f(\theta+2\pi)=f(\theta), can be written as f(\theta)=g\left( e^{i\theta} \right), for some function g: \mathbb{S}^1 \rightarrow \mathbb{C}. Conversely, any function g: \mathbb{S}^1 \rightarrow \mathbb{C} defines a 2\pi periodic function on \mathbb{R} by setting f(\theta)=g\left( e^{i\theta} \right). So, with this in mind, we simply say that g: \mathbb{S}^1 \rightarrow \mathbb{C} is a smooth function if f is. With this identification between the set of smooth 2\pi periodic functions on \mathbb{R} and the set of smooth functions on \mathbb{S}^1, it then immediate that the canonical diffusion operator \Delta_{\mathbb{S}^1} on \mathbb{S}^1 should write, \Delta_{\mathbb{S}^1} g ( e^{i\theta})=f''(\theta). The corresponding diffusion semigroup is also easily computed from the heat semigroup on \mathbb{R}. Indeed, a natural computation leads to
\left( e^{t \Delta_{\mathbb{S}^1} }g \right) ( e^{i\theta}) =\left( e^{t \frac{d^2}{d\theta^2} }f \right) ( \theta)
= \int_{-\infty}^{\infty} f(y+\theta) \frac{e^{-\frac{y^2}{4t} } }{\sqrt{4\pi t}} dy
=\sum_{k \in \mathbb{Z}} \int_{2k \pi }^{2k\pi +2\pi} f(y+\theta) \frac{e^{-\frac{y^2}{4t} } }{\sqrt{4\pi t}} dy
=\sum_{k \in \mathbb{Z}} \int_{0 }^{2\pi} f(y-2k\pi+\theta) \frac{e^{-\frac{(y-2k\pi)^2}{4t} } }{\sqrt{4\pi t}} dy
= \int_{0 }^{2\pi} f(y+\theta) p(t,y) dy,
where p(t,y)=\frac{1}{\sqrt{4\pi t}} \sum_{k \in \mathbb{Z}} e^{-\frac{(y-2k\pi)^2}{4t} }. This allows to define the heat semigroup on \mathbb{S}^1 as the family of operators defined by \mathbf{P}_t g (e^{i\theta} )= \int_{0 }^{2\pi} g(e^{i\nu }) p(t,\theta-\nu) d\nu. The natural domain of this operator is \mathbf{L}^2_\mu(\mathbb{S}^1,\mathbb{R}) where \mu is the measure on \mathbb{S}^1 which is defined through the property \int_{\mathbb{S}^1} g d\mu = \int_0^{2\pi} f(\theta) d\theta. The reader may then check the following properties for this semigroup of operators:

  • (Semigroup property) \mathbf{P}_{t+s} =\mathbf{P}_t \mathbf{P}_s;
  • (Strong continuity) The map t \rightarrow \mathbf{P}_t is continuous for the operator norm on \mathbf{L}^2_\mu(\mathbb{S}^1,\mathbb{R});
  • (Contraction property) \|\mathbf{P}_t g \|_{\mathbf{L}^2_\mu(\mathbb{S}^1,\mathbb{R}) } \le \| g \|_{\mathbf{L}^2_\mu(\mathbb{S}^1,\mathbb{R}) };
  • (Self-adjointness) For g_1,g_2 \in \mathbf{L}^2_\mu(\mathbb{S}^1,\mathbb{R}), \int_{\mathbb{S}^1} (\mathbf{P}_t g_1) g_2 d\mu=\int_{\mathbb{S}^1} g_1(\mathbf{P}_t g_2) d\mu
  • (Markov property) If g \in \mathbf{L}^2_\mu(\mathbb{S}^1,\mathbb{R}) is such that 0 \le g \le 1, then 0 \le \mathbf{P}_t g \le 1.

Exercise.

  • Prove the Poisson summation formula: If f:\mathbb{R} \rightarrow \mathbb{R} is a smooth and rapidly decreasing function, then \sum_{m \in \mathbb{Z}} f(m) e^{im \theta}=\sum_{k \in \mathbb{Z}} \hat{f} (\theta -2k\pi).
  • Deduce that the heat kernel on \mathbb{S}^1 may also be written p(t,y)=\frac{1}{2\pi}\sum_{m \in \mathbb{Z}} e^{-m^2 t} e^{im y}.

Exercise. From the previous exercise, the heat kernel on \mathbb{S}^1 is given by p(t,y) =\frac{1}{2\pi}\sum_{m \in \mathbb{Z}} e^{-m^2 t} e^{im y} =\frac{1}{\sqrt{4\pi t}} \sum_{k \in \mathbb{Z}} e^{-\frac{(y 2k\pi)^2}{4t} }.

  • By using the subordination identity e^{-\tau | \alpha | } =\frac{\tau}{2\sqrt{\pi}} \int_0^{+\infty} \frac{e^{-\frac{\tau^2}{4t}-t \alpha^2}}{t^{3/2}} dt, \quad \tau \neq 0, \alpha \in \mathbb{R}, show that for \tau > 0, \frac{1+e^{-2\pi \tau}}{1-e^{-2\pi \tau}} =\frac{1}{2\pi} \sum_{k \in \mathbb{Z}} \frac{2\tau}{\tau^2+n^2}
  • The Bernoulli numbers B_k are defined via the series expansion \frac{x}{e^x -1}=\sum_{k=0}^{+\infty} B_k \frac{x^k}{k!}. By using the previous identity show that for k \in \mathbb{N}, k \neq 0, \sum_{n=1}^{+\infty} \frac{1}{n^{2k}} =(-1)^{k-1} \frac{(2\pi)^{2k} B_{2k} }{2(2k)!}.

Exercise. Show that the heat kernel on the torus \mathbb{T}^n=\mathbb{R}^n / (2 \pi \mathbb{Z})^n is given by p(t,y) = \frac{1}{(4\pi t)^{n/2}} \sum_{k \in \mathbb{Z}^n} e^{-\frac{\|y+2k\pi\|^2}{4t} }=\frac{1}{(2\pi)^n} \sum_{l\in \mathbb{Z}^n} e^{i l \cdot y -\| l \|^2 t}.

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