ThR37
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 Jun17 comment How does one denote the set of all positive real numbers? @YuvalFilmus Do not forget that this is just an english convention. In France for example, we usually say that 0 is both positive and negative. I have often seen $\mathbb{R}^+$ for all positive/null numbers and $\mathbb{R}^{\ast +}$ for all strictly positive numbers. Mar31 asked Quadratic minimization with real output Sep17 comment Smooth mapping between Manifolds Thanks for your answer. Using $V$ and $V'$ identification seems indeed to make sense and allows me to explain better why I can transport the structure ! Sep17 accepted Smooth mapping between Manifolds Sep17 comment Smooth mapping between Manifolds Ok thanks. Yes both spaces are diffeomorphic wrt the usual smooth structure but I wasn't 100% sure that it was enough to pull back the Riemanian metric (I am mainly into computer science and I lack of formalism when it comes to deal with Riemanian geometry) Sep17 comment Smooth mapping between Manifolds $f^{-1}$ is still linear. I have changed bijection into isomorphism to reflect this fact so $g_\mathcal{N}$ is bilinear. I am not sure if I can use $f$ to move the tangent spaces (point 2.). If it is the case, I should be able to pull back the vectors using $f^{-1}$. Sep17 comment Smooth mapping between Manifolds @AnthonyCarapetis I have edited the post following your first comment. Sorry for the lack of clarity. Sep17 revised Smooth mapping between Manifolds added 114 characters in body Sep17 revised Smooth mapping between Manifolds added 58 characters in body Sep17 asked Smooth mapping between Manifolds Aug7 awarded Editor Aug7 awarded Scholar Aug7 comment Geodesic complete subset of a connected manifold @DanielRust I have updated the question to reflect this case. Aug7 revised Geodesic complete subset of a connected manifold Updating question Aug7 accepted Geodesic complete subset of a connected manifold Aug6 awarded Student Aug6 comment Geodesic complete subset of a connected manifold @DanielRust Yes of course ! I must be tired and missing the most obvious thing. The set of geodesics works in this case. I guess we can find other examples by this idea of getting stuck to a dimension (like any plane $P$ in $\mathbf{R}^3$). This question was really silly actually :) Aug6 comment Geodesic complete subset of a connected manifold @hardmath For me an open convex subset $O$ of the plane $P$ embedded with a simple Euclidean metric is geodesic convex but not complete since for $x,y\in O$, $\gamma(t)=t*x+(1-t)*y$ will escape from $O$. It could be the case with some specific metric but could become unconnected in this case ? Aug6 asked Geodesic complete subset of a connected manifold May29 awarded Supporter