Is convex set still convex when endowed with a different Riemannian metric? Say $V \in \mathbb{R}^n$ is a closed convex set in the Euclidean space. And therefore I am free to define the projection of $x$ to $V$ as the point in $V$ that minimizes the distance to $x$. Now suppose  I endow $\mathbb{R}^n$ with a different riemannian metric $g$, and embed $(V, g)$ isometrically to $\mathbb{R}^N$ by nash embedding. Is the new $V$ still convex with respect to $\mathbb{R}^N$? And if so, let $d(\cdot, \cdot)$ be the distance induced by the trace metric, is the projection still $arg min \; d(x, V)$?
 A: Convexity is not preserved under a change of Riemannian metric. To see this consider the following example. Let $V \subset \mathbb{R}^2$ be the unit disk, clearly closed and convex. Now replace the flat metric on $V$ by a very steep hill (in a smooth way). In other words, distances within $V$ become much larger in the new metric. This embeds into $\mathbb{R}^3$ but we don't need that.
In the old metric the distance minimizer between two suitably chosen points outside $V$ will be a straight line passing through $V$. In the new metric, this line will not be a distance minimizer, the minimizer will pass around $V$.
A: The other answer is great. Here is another one which might maybe look like an overkill solution.
From the Riemann uniformization theorem, all simply connected open subset of $\mathbb{C}$ are biholomorphic. Consider $\Delta = \left\{ z \in \mathbb{C} \mid |z|<1\right\}$ the unit disk and $U = \left\{z=x+iy \mid y < x^2\right\}$. These two subsets are open and simply connected and hence there exists a biholomorphism $\varphi : \Delta \to U$.
Consider the euclidean metric $g$ on $\mathbb{C}$. Then $\Delta$ is convex with respect to $g$ and $U$ isn't. It follows that $\left(\Delta,g\right)$ is convex, but $\left(\Delta,\varphi^* g\right)$ isn't.
