Ronny Bergmann

You are again a little imprecise here. So if you have `p1` as a point on the manifold and `Xa` is a tangent vector at `p1` then sure `exp)M,p1,Xa)` is...

> > So if you have p1 as a point on the manifold and Xa is a tangent vector at p1 then sure exp)M,p1,Xa) is a point on the manifold...

The answer is already directly above (1) turn your Lie algebra elements (tangent vectors at the identity) into points on the manifold and apply the group action then. That was...

Great, looking forward to the update. I would prefer though, if it was self-contained and that it does not need links to comments/issues.

> * T(n) x SO(n) a direct product manifold (aka "more Riemannian") and I think Manifolds.jl in general is coming from this angle vs. This is `N = ProductManifold(TranslationGroup(3), SpecialOrthogonal(3))`...

I came along that one some time ago, but haven't taken much attention yet. Just a short question on the exponential barycenter – can you link to more details? Concerning...

Ah, Section 3.3 is our implementation of the mean, and we also do that without the square, which is our median.

Ok, I will check the documentation somewhen today and add that explanation and references to the general median and mean docs.

If I understand the landmark space correctly it should be what we do with `PowerManifold(M,n)` where our `n::Int` is the number of landmarks? We also have a first version of...

Thanks for the links, I will surely have a look