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Context. I'm looking for a "natural" definition of subdifferentials on riemannian manifolds.

Given a function $F:\mathbb R^m \to \mathbb R$, its Fréchet-subdifferential at a point $w \in \mathbb R^m$, denoted $\partial F(w)$, is defined by to be the following (possible empty) subset of $\mathbb R^m$ $$ \partial F(w) \mathrel{:=} \left\{w^\star \in \mathbb R^m \mid \liminf_{w' \to w}\frac{f(w') - f(w) - \langle w^\star,w'-w\rangle}{\|w'-w\|} \ge 0\right\}. $$

Note that if $F$ is convex, then this definition reduces to the familiar definition $\partial F(w) = \{w^\star \mid f(w') - f(w) \ge \langle w^\star,w'-w\rangle \;\forall w' \in \mathbb R^m\}$.

Extension to riemannian manifolds

Now, let $M$ be a riemannian submanifold of $\mathbb R^m$ and let $f:M \to \mathbb R$ be the restriction of $F$ on $M$.

Question. What is a natural way to define $\partial f$, so that it is "compatible" with the differential structure on $M$?

My attempt

$\partial f(w) \mathrel{:=} \{P_{T_wM}(w^\star) \mid w^\star \in \partial F(w)\}$,

where $P_{T_wM}:\mathbb R^m \to T_wM$ is the projection operator from $\mathbb R^m$ to the tangent space $T_wM$ of $M$ at $w$. This attempt is motivated by the fact that the ordinary gradient of $f$ (in case $f$ is differentiable), namely $\nabla f(w) \mathrel{:=} P_{T_wM}(\nabla F(w))$.

However

  • Is my proposal natural enough ?
  • Has it already been proposed / studied elsewhere ?

Any help / references would be greatly appreciated.

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    $\begingroup$ That works if the manifold is isometrically embedded in Euclidean space. If not you might use Definition 2.1. of researchgate.net/publication/… $\endgroup$
    – user35593
    Commented Nov 17, 2020 at 8:08
  • $\begingroup$ Thanks for the suggestion. Definnition 2.1 is referring to differentiability. I guess you where referring to something like $$ \partial f(w) := \left\{w^\star \in T_wM \mid \liminf_{x,y \to w,\; x \ne y}\frac{F(x) - F(y) -\langle w^\star, \exp^{-1}_w(y)-\exp^{-1}_w(x)\rangle}{\|\exp_w^{-1}(y) - \exp_w^{-1}(x)\|} \ge 0\right\} $$ $\endgroup$
    – dohmatob
    Commented Nov 17, 2020 at 8:22
  • $\begingroup$ BTW, I don't understand you remark about "isometrically embedded". Do you mean embedded in euclidean space of any dimension (this is always possible if $M$ is closed, thanks to the Nash embedding theorem) or embedded in $R^n$ (i.e a constraint on the embedding dimension) ? $\endgroup$
    – dohmatob
    Commented Nov 17, 2020 at 8:36
  • $\begingroup$ Yes its always possible but it might be difficult to find a concrete embedding for a specific manifold. Also it might be favorable to have a method that is independent of the embedding. Therefore the exponential map. $\endgroup$
    – user35593
    Commented Nov 17, 2020 at 8:41
  • $\begingroup$ On second thought, if $M$ is closed and sufficiently smooth, then it can be embedded into some euclidean space of dimension $m=O(n^3)$. Then one can lift $F$ from $\mathbb R^n$ to $\mathbb R^m$ in an obvious way. Then the tentative definition in my comment (based on inverse retractions) should reduce to the attempt in my question above, modulo an $\epsilon$ which we later limit to $0^+$. What do you think ? $\endgroup$
    – dohmatob
    Commented Nov 17, 2020 at 8:42

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