# Limit of Riemannian metrics on the disk.

I'm working through Burago, Burago and Ivanov's book A Course in Metric Geometry and I'm trying to solve the following excercise:

If we denote by $D^2$ the standard unit ball in $\mathbb{R}^2$, then the metric of $\mathbb{R}^2_1$ (that is, $\mathbb{R}^2$ with the norm $\| (x,y) \| = |x| + |y|$) restricted to $D^2$ can be obtained as a uniform limit of Riemannian metrics.

My idea is that if we consider the $p$-norms on $\mathbb{R}^2$ then these converge to the norm mentioned above and since these are equivalent to the $2$-norm, then they come from inner products and hence they are Riemannian. I'm not entirely sure if this argument is correct. Also, I don't really understand the hint provided:

Hint: If we take grids of side $\lambda \rightarrow 0$ in $\mathbb{R}^2$ with the induced metric, then these grids converge to $\mathbb{R}^2_1$. The hint is to "fill the cells".

• The only $p$-norm that comes from an inner product is $p=2$ - you can prove this by showing the failure of the parallelogram law for other $p$. The equivalence of norms is pretty weak geometrically - it does not preserve the property of arising from an inner product. – Anthony Carapetis Nov 6 '13 at 10:11

The $\ell_1$ norm is colloqually known is the taxicab distance: the shortest driving distance in a city laid out in a square grid. The Riemannian metric is obtained by infimizing integrals of $g_{ij}$ over connecting curves. We need some $g_{ij}$ which will force minimizing curves to travel along the streets, and not over the buildings. The intuition suggests how to go about: make it costly to crawl over buildings, i.e., make $g_{ij}$ large there.
Following the hint, let $\lambda>0$ be a small number and draw the square grid $G$ of side length $\lambda$ (by $G$ I mean the $1$-skeleton of the grid). The metric will be $g_{ij}= \rho(x,y)\delta_{ij}$, i.e., a conformal deformation of the Euclidean metric. There is a smooth function $\rho$ such that $\rho(x,y )=1$ when $(x,y)\in G$ and $\rho(x,y)\ge 3$ when $\operatorname{dist}((x,y), G)\ge \lambda^2$.
Geodesics with respect to this metric will not go into the square "buildings" where $\rho\ge 3$, because it's easier to go around the building. (The path around a square is at most twice as long as the straight line across). This makes the metric $g$ approximate the desired taxicab distance. Much more needs to be done to make this rigorous, of course.