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My question is about to what extent can we extend the implicit function theorem to Riemannian manifolds. In the Euclidean space, consider a bivariate function $F \colon \Theta \times \mathcal{X} \rightarrow \mathbb{R}$ where $\Theta$ and $\mathcal{X}$ are Euclidean spaces. For any $\theta \in \Theta$, assume that $F(\theta, \cdot)$ is differentiable and strongly convex on $\mathcal{X}$ such that there exists a unique vector $x^\star(\theta)$ satisfying

$$ x^{\star }(\theta) = \arg\min_{x\in\mathcal{X} } F(\theta, x). $$ We are interested in finding $\nabla_{\theta} x^\star (\theta)$, i.e., how the global minimizer is sensitive with respect to the change in $\theta$. What we do is to differentiate the first-order condition. In particular, $x^{\star }(\cdot)$ satisfies $$ \nabla_{x} F(\theta, x) \big\vert_{x = x^\star (\theta)} = 0 \qquad \forall \theta\in \Theta. $$ Taking derivatives with respect to $\theta$ yields $$ \nabla^2 _{x \theta} F(\theta, x) \big\vert_{x = x^\star (\theta)} + \Big[ \nabla^2 _{x x} F(\theta, x) \big\vert_{x = x^\star (\theta)} \Big] \nabla_{\theta} x^\star (\theta) = 0. $$ Solving this equation yileds $\nabla_{\theta} x^\star (\theta)$.

Now suppose $\Theta \subseteq \mathbb{R}^d$, $\mathcal{X}$ is a Riemannian manifold, and $F(\theta, \cdot)$ is geodesically strongly convex on $\mathcal{X}$, can we generalize such reasoning using the Riemannian gradient and Hessian?

Furthermore, suppose now $\mathcal{X}$ is a Wasserstein space, i.e., $\mathcal{X}$ contains all probability densities defined on $\mathbb{R}^p$ and is equipped with the second-order Wasserstein distance $W_2$. And $F(\theta, \cdot)$ is a strongly convex functional with respect to $W_2$, how to compute $\nabla_{\theta} x^\star (\theta)$?

Finally, when $\Theta$ itself is a Riemannian manifold, can we generalize the above idea to compute $\mathrm{grad} (x^\star(\theta))$ where $\mathrm{grad}$ stands for the Riemannian gradient.

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  • $\begingroup$ For any smooth manifold the implicit function theorem works in the same way as in the Euclidean case since we have local charts. The only difference is that instead of the Euclidean inner product you have to use the Riemannian metric, i.e. smooth mapping with values in symmetric positive definite matrices. $\endgroup$ Commented May 6, 2023 at 10:04
  • $\begingroup$ Thank you @VítTuček $\endgroup$
    – Steve
    Commented May 17, 2023 at 0:27

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