# 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 $$arg min \; d(x, V)$$?

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 $$V$$.
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.