# Alternative Geometries

In our world, the distance between two points (in 2d) is defined as $\sqrt{(\Delta x)^2 + (\Delta y)^2}$. Suppose that in an alternative geometry, it was defined as $\sqrt[p]{|\Delta x|^p + |\Delta y|^p}$. How would the shapes of geodesics change? How would angles be defined? Is there a way of visualizing such a geometry as a smooth surface? Is there an intuitive way of thinking about it? Could someone provide a reference to the subject?

• This question is not appropriate for this site, which is for research math; but see en.wikipedia.org/wiki/Lp_space#The_p-norm_in_finite_dimensions. Nov 26, 2015 at 6:31
• Equip the Euclidean plane with the p-norm. When $p$ is even, the geodesics are still the straight lines as can be seen from solving the Euler-Lagrange ODE for the length functional. I don't know enough to solve the functional for $p$ any other number since then the presence of the absolute value introduces irregular points.
– M.G.
Nov 26, 2015 at 9:36
• Also, as far as the general case is concerned, elementary facts about p-norms like their decreasing property do not seem to help establishing a suitable inequality. Note that, heuristically speaking, equivalence of norms (on finite-dim. spaces) alone is not expected to help here (e.g. in Riemannian geometry, different inner products give you different structure).
– M.G.
Nov 26, 2015 at 9:44

Using such a metric locally turns the Euclidean space into a Finsler manifold instead of a Riemannian manifold. If you are not familiar with differential geometry, you might want to ignore this. You should also check the link Noah Schweber gave in a comment: Wikipedia tells about $\ell^p$-norms in $\mathbb R^n$.
One striking difference is that there would be preferred directions, the coordinate axes. The usual metric is rotationally invariant and we can set coordinates quite freely. For the norm you propose this is no longer the case unless $p=2$. Due to this lack of symmetry balls will no longer be round, for example, and a rotated ball is typically not a ball.