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Let $(M,g,m_0)$ be a pointed-Hadamard manifold with Riemmanian distance function $d_g$, $(X,\Sigma,\mu)$ be a finite measure space. We use $L^2(\mu;M,m_0)$ to denote the metric space consisting of all equivalence classes of measurable functions $g:X\rightarrow M$ with finite the distance $D(m_0,g)$ from the constant function $m_0$ where $D$ is the metric: $$ D(g,h):=\int_{m \in M} d_g^2(g(m),h(m))d\mu(m) = \int_{m \in M} \|\operatorname{exp}_{g(m)}^{-1}(h(m))\|^2d\mu(m). $$

Fix some non-constant $f \in L^2(\mu;M,m_0)$.

Let $c\in M$ and let $C\subseteq L^2(\mu;M,m_0)$ be proper and non-empty. Under what joint conditions on $C$ and $c$ can we guarantee that $$ \operatorname{argmin}_{g \in C} D(f,g) = \operatorname{argmin}_{g \in C} \int_{m \in M} \|\exp_{c}^{-1}(f(m))-\operatorname{exp}_{c}^{-1}(g(m))\|^2d\mu(m) . $$

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  • $\begingroup$ What do you mean by $\exp_c$? Also, is $g$ supposed to be $h$? $\endgroup$
    – Moishe Kohan
    Commented Jan 4, 2021 at 17:56
  • $\begingroup$ @MoisheKohan I mean the Riemannian exponential at $c$''s inverse function (which is globally defined) $\endgroup$
    – Catologist_who_flies_on_Monday
    Commented Jan 4, 2021 at 18:44
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    $\begingroup$ unless I'm overlooking something, probably this equality happens only in the "most boring" of settings. E.g., if the manifold under consideration is the manifold of psd matrices under its usual affine invariant metric, then the two sides will be equal in general only for subsets of simultaneously commuting matrices.... $\endgroup$
    – Suvrit
    Commented Jan 4, 2021 at 19:49
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    $\begingroup$ @Topology_Catologist unclear: $c$ is s not in $M$. $\endgroup$
    – Moishe Kohan
    Commented Jan 4, 2021 at 22:35
  • $\begingroup$ @MoisheKohan Thanks for noticing the misprint. Indeed $c\in M$ is correct. $\endgroup$
    – Catologist_who_flies_on_Monday
    Commented Jan 5, 2021 at 8:31

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