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?
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2$\begingroup$ 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. $\endgroup$– Noah SchweberNov 26, 2015 at 6:31
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$\begingroup$ 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. $\endgroup$– M.G.Nov 26, 2015 at 9:36
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$\begingroup$ 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). $\endgroup$– M.G.Nov 26, 2015 at 9:44
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Here are some thoughts on the subject.
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.
One way to rephrase is that if rotations are defined to be linear isometries, the only rotations are obtained by permuting and flipping the coordinate axes. Rotations only come in steps of 90 degrees and there are no small rotations. (The rotation group is finite and generated by 90 degree rotations between pairs of coordinate directions.)
answered Nov 26, 2015 at 9:27
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1$\begingroup$ IMHO, the wiki link is rather useless for addressing the question. I don't see how one can derive something reasonable from elementary facts about p-norms. The "orientation" of the p-norm inequalities is weirdly inconclusive when you write down the details :-) $\endgroup$– M.G.Nov 26, 2015 at 10:01