On the diameter of left-invariant sub-Riemannian structures on a compact Lie group

Question

Let $G$ be a compact connected Lie group with Lie algebra $\mathfrak g$ of dimension $m$. We fix an inner product $\langle\cdot,\cdot\rangle$ on $\mathfrak g$.
We may assume (in case is necessary) that $G$ is semisimple and/or $\langle\cdot,\cdot\rangle$ is $\textrm{Ad}(G)$-invariant.

Given any subspace $\mathcal H$ of $\mathfrak g$, $(\mathcal H,\langle\cdot,\cdot\rangle|_{\mathcal H})$ induces a left-invariant sub-Riemannian structure on $G$. The corresponding sub-Riemannian manifold $(G,\mathcal H,\langle\cdot,\cdot\rangle|_{\mathcal H})$ has associated a distance $d_{\mathcal H}$ on $G$ known as the Carnot–Carathéodory metric. For $a,b\in G$, $$d_{\mathcal H}(a,b)=\inf_{\gamma} \int_0^1 \langle \gamma'(t),\gamma'(t)\rangle^{1/2}\, dt,$$ where the infimum is taken over all smooth curves $\gamma:[0,1]\to G$ such that $\gamma(0)=a$, $\gamma(1)=b$, and $\gamma'(t)\in \mathcal H$ for all $t$. Chow's theorem ensures that $d_{\mathcal H}(a,b)<\infty$ for all $a,b\in G$ if $\mathcal H$ is bracket-generating (i.e. the only subalgebra of $\mathfrak g$ containing $\mathcal H$ is $\mathfrak g$). I am interested in the diameter of the metric space $(G,d_{\mathcal H})$, i.e. $$\textrm{diam}(G,d_{\mathcal H})=\max_{a,b\in G}d_{\mathcal H}(a,b).$$

Let $\textrm{Gr}_{\mathfrak g}(k)$ denote the set of $k$-dimensional subspaces of $\mathfrak g$. This is a symmetric space called Grassmannian. We will only use the topology on it induced by the distance metric associated with the Riemannian symmetric metric.

Is the map $\Phi: \textrm{Gr}_{\mathfrak g}(k)\to \mathbb R_{\geq 0}\cup\{\infty\}$ given by $\Phi(\mathcal H)= \textrm{diam}(G,d_{\mathcal H})$ continuous?

A reference or a proof would be very appreciated.

Answer 1

From semicontinuity of length in $(G,\langle\ ,\ \rangle)$, we get that diameter is lower semicontinuous. It remains to show that it is upper semicontinuous.

Suppose $\mathcal H_n\to \mathcal H_\infty$ as $n\to\infty$. We can assume that $H_\infty$ is bracket-generating, otherwise $(G,\mathcal H_\infty,\langle\ ,\ \rangle)$ has infinite diameter --- so, there is nothing to prove.

Suppose two points $x$ and $y$ are connected by a path $\gamma$ of length $\ell$. We can assume that $\gamma_\infty$ is Lipschitz, in particular it is differentiable almost everywhere and it completely defined by its derivative $\gamma_\infty'\colon t\to H_\infty$. Denote by $\pi_n$ the orthogonal projection $\mathfrak{g}\to \mathcal H_n$. Consider the curve $\gamma_n$ that starts at $x$ and $\gamma'_n(t)=\pi_n\circ\gamma_\infty'(t)$ for any $t$. Observe that $$\mathrm{lenght}\,\gamma_n\leqslant \mathrm{lenght}\,\gamma_\infty$$ for any $n$. Further note that $\gamma_n\to \gamma_\infty$ as $n\to\infty$; in particular $y_n=\gamma_n(1)\to y$ as $n\to\infty$.

Since $H_\infty$ is bracket-generating, so is every $H_n$ for every large $n$. Moreover, for any $\varepsilon>0$ we can choose a neighborhood $N$ of $y$ such that for any large $n$, the distance in $(G,\mathcal H_n,\langle\ ,\ \rangle)$ from $y$ to any $z\in N$ is smaller than $\varepsilon$. It follows that given $\varepsilon>0$, the distance from $x$ to $y$ in $(G,\mathcal H_n,\langle\ ,\ \rangle)$ exceeds the distance in $(G,\mathcal H_\infty,\langle\ ,\ \rangle)$ is at most $\ell+\varepsilon$. Hence the upper semicontinuity follows.