# The $\mathbf{F}$-metric induces the weak topology on the set of bounded varifolds

Some preliminary definitions and notation:

(1) Given a vector space $\mathbb{V}$, we denote by $G_k(\mathbb{V})$ the $k$-grassmannian of $\mathbb{V}$, i.e. the set of all $k$-dimensional vector subspaces of $\mathbb{V}$;

(2) Given a differential (or riemannian) manifold $M$, we denote by $G_k(M)$ the $k$-grasmannian bundle on $M$, i.e. the bundle with fibers $G_k(T_pM)$;

(3) A $k$-dimensional varifold on $M$ is any Radon measure $V$ on $G_k(M)$;

(4) We denote by $\mathcal{V}_k(M)$ the set of all $k$-dimensional varifolds on $M$;

(5) The mass of a varifold $V\in \mathcal{V}_k(M)$ is defined by $\|V\|=\int_{G_k(M)}dV$.

One can define a topology - the weak topology - on $\mathcal{V}_k(M)$ as follows. Let $J:=\mathcal{C}_c(G_k(M))$ be the set of compactly supported continuous functions $f:G_k(M)\to \mathbb{R}$ and consider the family of functions $\{\varphi_f\,:\, f\in J\}$ where $$\begin{matrix} \varphi_f:&\mathcal{V}_k(M)&\to&\mathbb{R}\\ &V&\mapsto&\int_{G_k(M)}f\,dV. \end{matrix}$$ The weak topology on $\mathcal{V}_k(M)$ is the least topology on $\mathcal{V}_k(M)$ for which every $\varphi_f$ is continuous or, in other words, is the topology on $\mathcal{V}_k(M)$ generated by the subbasis $$\beta=\{\varphi_f^{-1}(\Omega)\,:\, f\in J,\,\Omega\subset_{op} \mathbb{R}\}.$$

One can ask if the weak topology on $\mathcal{V}_k(M)$ is metrizable. Well, define $\mathbf{F}:\mathcal{V}_k(M)\times \mathcal{V}_k(M)\to [0,+\infty]$ by $$\mathbf{F}(V,W)=\sup E(V,W),$$ with $$E(V,W)=\left\{\int_{G_k(M)}f\,dV-\int_{G_k(M)}f\,dW\,:\, f\in J,\,|f|\leq 1,\,\mathrm{Lip}(f)\leq 1\right\}.$$

One verifies that $\mathbf{F}$ has the three axioms of a metric, although it can be $+\infty$. But given any $c\geq 0$, $\mathbf{F}$ is always finite on $$\mathcal{V}_{k,c}(M):=\{V\in \mathcal{V}_k(M)\,:\, \|V\|\leq c\}.$$

Indeed, for any $f\in J$, $|f|\leq 1$, $\mathrm{Lip}(f)\leq 1$, $$\int_{G_k(M)}f\,dV-\int_{G_k(M)}f\,dW\leq \int_{G_k(M)}\,dV+\int_{G_k(M)}\,dW=\|V\|+\|W\|\leq 2c.$$

Therefore, $\mathbf{F}$ is a finite genuine metric on $\mathcal{V}_{k,c}(M)$.

I want to prove that:

($\ast$) The weak topology and the $\mathbf{F}$-induced topology coincide on $V_{k,c}(M)$.

This $\mathbf{F}$-metric appears in several papers in the internet (generally talking about minimal surfaces) but always with its definition and the claim ($\ast$) above, followed by some sort of "it is easy to prove" or "one easily verifies".

Is it really that easy? I guess I could prove that $\tau_w\subset \tau_{\mathbf{F}}$ but, given an open $\mathbf{F}$-ball $B_\epsilon(V)$, how do I find a weak-basic open $U$ with $V\in U\subset B_\epsilon(V)$? (such open must be a finite intersection of elements of $\beta$. Which $f_i$'s and real opens $I_i$'s to choose?)

• Just a curiosity: where did you learn varifolds from? – Eduardo Longa Apr 5 '18 at 1:09
• @EduardoLonga I am still a beginner on the subject. :) – Ders Apr 5 '18 at 9:49
• Are either of these obviously equivalent to the notion of weak (*?) convergence of measures? – Ryan Unger Apr 5 '18 at 18:47
• @0celo7 In this case, weak convergence of Radon measures. Are weak and weak* convergence equal for measures? I find both names but the definitions seem to be always the same... – Ders Apr 10 '18 at 18:01

I'm pretty sure this is false in general. Consider the degenerate case where $M = \mathbb{R}$ and $k=0$, then $\mathcal{V}_0(\mathbb{R})$ is the space of Radon measures on $\mathbb{R}$, which includes the dirac delta measures $\delta_x : f \mapsto f(x)$. Now note that for any $f$ with compact support:
$$\lim_{x \to \infty} f(x) = 0$$
hence $\lim_{x\to\infty} \delta_x = 0$ in the weak*-topology. And yet:
$$\lim_{x \to \infty} \mathbf{F}(\delta_x, 0) = 1$$
so the $\delta_x$ do not converge in the $\mathbf{F}$ topology (or equivalently $\mathbf{F}$ is not continuous in the weak*-topology). Restricting to a subspace of measures $V$ with $\|V\| \leq c$ does not help since $\|c \delta_x \| = c$.