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Let $\Sigma$ be a closed Riemannian surface. I read a paper which said:

We let $\Sigma_k$ denote the family of formal sums $$ \Sigma_k=\sum_{i=1}^k t_i \delta_{x_i} ; \quad t_i \geq 0, \quad \sum_{i=1}^k t_i=1 ; \quad x_i \in \Sigma, $$ endowed with the weak topology of distributions. This is known in literature as the formal set of barycenters of $\Sigma$ (of order $k$ ).

I have no idea about this paragraph, what is this $\delta_{x_i}$?

And what is endowed with the weak topology of distributions?

At first I thought it was a Dirac function but it seems that $\Sigma_k$ is a set not a function. Could you give me some clear definitions? I checked the definition on wiki but there is no introduction on this.

Elio Li
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1 Answers1

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The whole discussion has nothing to do with the fact that $\Sigma$ is a Riemann surface: it applies to any manifold. For $x\in \Sigma$, $\delta_x$ is the Dirac mass at $x$, that is the distribution $\delta_x \in \mathcal{D}(\Sigma) = \mathcal{C}^{\infty}(\Sigma)^*$ (the dual of the space of smooth functions), acting on smooth functions $f \in \mathcal{C}^{\infty}(M)$ by $$ \langle \delta_x,f\rangle = f(x). $$ Note that $\delta_x$ is not a function, it is a distribution. This way, we have a correspondence $x \leftrightarrow \delta_x$. Through this correspondence, a barycenter "$\sum t_i x_i$", which makes no sense in $\Sigma$, can be thought of as $\sum t_i \delta_{x_i}$. If $$ \Sigma_k = \left\{ \sum_{i=1}^k t_i x_i \mid x_i \in \Sigma, t_i \in [0,1], \sum_{i=1}^n t_i = 1\right\} \subset \mathcal{D}(\Sigma) $$ is the set of all of those formal barycenters of $k$ points, we can endow it with a topology called the weak topology, which is the smallest topology such that for any $T \in \Sigma_k$, the map $$ \langle T,\cdot \rangle \colon f \in \mathcal{C}^{\infty}(\Sigma) \longmapsto \langle T,f\rangle \in \Bbb R $$ is continuous.

Didier
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  • Thanks for your answer! Let $M_1= \delta_{x_i}$, $x_i \in M$, I read about that $M_1$ is a smooth manifold in $C^1(M)^$. For $|f|_{C^1(M)} \leq 1$, the distance is defined by $ dist(\sigma, \hat{\sigma})=\sup _f|(\sigma, f)-(\hat{\sigma}, f)| .$ Then how to prove the Dirac distribution on a manifold $M$ is a smooth manifold in $C^1(M)^$, and what is its dimension? Is it dependent on the dimension of the manifold $M$? Can you give me some hints or lectures on it? I have not learned many knowledges on differential geometry, I met this when reading the min-max scheme in PDE on manifold. – Elio Li Oct 26 '23 at 07:42
  • @YuxuanLi Hi. You should ask this in another question, providing the references where you have read such things so that specialists of the field (which I am not!) can answer you – Didier Oct 26 '23 at 07:50
  • Thanks for your suggestion! I will post another question. – Elio Li Oct 26 '23 at 09:16