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Nov 20, 2020 at 11:52 history edited Nicolas Boumal CC BY-SA 4.0
Added link to relevant new book on the topic of the OP's question.
Jun 13, 2017 at 14:41 comment added Nicolas Boumal It is difficult to say if second-order retractions could help reduce the total number of required iterations. However, it is known that using second-order retractions can be useful to guarantee convergence to second-order critical points (as shown by my colleagues and myself in arxiv.org/abs/1605.08101). Also, computing second-order retractions is often inexpensive: Malick and Absil showed that, for Riemannian submanifolds of R^n, orthogonal projection to the manifold is a second-order retraction (see also example above for the sphere): epubs.siam.org/doi/abs/10.1137/100802529.
Jun 5, 2017 at 18:28 comment added Mark L. Stone How difficult is it to compute 2nd order retractions, at least for some cases? Even if it takes much longer to compute than 1st order retraction, might it possibly save enough top level algorithm iterations as to speed up the solution, and maybe even make the solution process more robust? Even though 1st order retraction is enough to produce quadratic convergence of a (trust region, say) Newton method, might 2nd order retraction improve robustness and performance, especially for the most of the algorithm solution path when quadratic convergence has not yet kicked in?
Mar 25, 2017 at 6:04 vote accept Markus Sprecher
Nov 1, 2016 at 13:23 review First posts
Nov 1, 2016 at 13:25
Nov 1, 2016 at 13:21 history answered Nicolas Boumal CC BY-SA 3.0