Relationship between $p$-normed plane and Euclidean $3$-space

Question

This question arised from the comments on this one.

Let $(\Bbb R^2,d_p)$ be the Cartesian plane equipped with the $p$-distance, i.e., the distance given by

$$d_p(x,y):=(|x_1-y_1|^p+|x_2-y_2|^p)^{1/p},\qquad x,y\in\Bbb R^2,$$

where $p\ge2$, and let $(\Bbb R^3,d')$ be the Cartesian space equipped with the usual Euclidean distance,

$$d'(x,y):=(|x_1-y_1|^2+|x_2-y_2|^2+|x_3-y_3|^2)^{1/2},\qquad x,y\in\Bbb R^3.$$

1. If $p\ne2$, is it possible to define a smooth, isometrical embedding $e_p:(\Bbb R^2,d_p)\to(\Bbb R^3,d')$ ? Perhaps an immersion?

2. In either case, if the answer is yes, are there any constraints on the sectional curvature of the resulting submanifold $e(\Bbb R^2)\subset\Bbb R^3$ with respect to the Riemannian metric induced from the usual flat metric on $\Bbb R^3$ ?

UPDATE

Following the help from the comments below, I'll try to formulate the question in another way:

If $p>2$, is there a differentiable immersion $i:\Bbb R^2\to\Bbb R^3$ such that the pullback of the Riemannian metric induced on $i(\Bbb R^2)$ defines the $p$-distance, as described above, on $\Bbb R^2$ ?

Or even simpler:

If $p>2$, can the $p$-distance, as described above, be induced from a Riemannian metric on $\Bbb R^2$ ?

Answer 1

The answer to your question in both versions is negative.

1. There is no immersion $i: R^2\to R^3$ such that the pull-back of the flat Riemannian metric on $R^3$ yields the flat Finsler metric on $R^2$ given by a $p$-norm, $p\ne 2$.

To understand this, you have to think through definitions. The pull-back of a Riemannian metric, in general, is again a Riemannian metric. This is simply because a pull-back at each point is essentially the restriction of an inner product on $R^n$ to a linear subspace in $R^n$. But such a restriction is obviously again an inner product, hence, cannot be given by a norm which does not come from an inner product, such as a p-norm for $p\ne 2$.

2. Consider $R^m$ equipped with a p-norm ($p\ne 2$) and $R^n$ equipped with an inner product. Both define distance functions $d_p$ on $R^m, d_2$ on $R^n$. Then there is no isometric embedding $f: (R^m,d_p)\to (R^n,d_2)$, i.e. a map such that $$ d_2(f(x), f(y))=d_p(x,y), \forall x, y\in R^m. $$ Proof. Suppose that such a map exists. Then it has to send geodesics to geodesics (shortest paths). Unless $p=1$ or $p=\infty$, any two points in $(R^m, d_p)$ are connected by a unique (up to a reparameterization) geodesic. If the case $p=1, \infty$ uniqueness fails, which is why there is no distance-preserving map $f: (R^m,d_p)\to (R^n,d_2)$ in this case. Now, assume that $p\ne 1, \infty$. Then it is not hard to see that straight line segments are geodesics in $(R^m,d_p)$. Therefore, under a distance-preserving map $f$ they would go to straight line segments. Consider now a parallelogram $ABCD$ in $ (R^m,d_p)$, in which we draw the two diagonals: Its image under $f$ would have to be a parallelogram again since the diagonals would map to intersecting line segments, forcing the image to be contained in a 2-dimensional affine plane in $R^n$. Now, I use the parallelogram law in $(R^n,d_2)$: $$ 2 ( d^2_2(f(A), f(B)) + d^2_2(f(B), f(C)))= d^2_2(f(A), f(C)) + d^2_2(f(B), f(D)). $$ Since $f$ is an isometric embedding, this forces the parallelogram law in $(R^m,d_p)$: $$ 2 ( d^2_p(A, B) + d^2_p(B, C))= d^2_p(A, C) + d^2_p(B, D). $$ But then this means that $d_p$ comes from an inner product on $R^m$, implying that $p=2$.