1
$\begingroup$

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".

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Anthony Carapetis Nov 6 '13 at 10:11
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.