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This page claims that the proper lines for taxicab geometry should be the same lines as for Euclidean geometry (with the $L^2$ norm on $\mathbb{R}^2$ instead of the $L^1$ norm for taxicab geometry).

Lines in the Taxicab Plane look like lines in the Euclidean plane...

The same claim also appears to be implicit in the Wikipedia page for taxicab geometry, on this webpage, on this one, and also in the book by Millman and Parker, Geometry: A Metric Approach with Models. All of the sources claim as a result that taxicab satisfies all of the same axioms as Euclidean geometry except for the SAS postulate.

Question: Isn't this a very bad, artificial, and unnatural definition of lines for taxicab geometry?

$\mathbb{R}^2$ with the metric induced by the $L^1$ norm is a length space, according to Exercise 11 here, so the lengths of its geodesics do correspond to the distance between points -- so why shouldn't the "lines" in taxicab geometry be the geodesics with respect to the metric induced by the $L^1$ norm?

If we want to study taxicab geometry, and not Euclidean geometry, shouldn't we only study structures which arise naturally from the geometry's definition, rather than definitions of angle, line, and triangle which are unnaturally "imported" from Euclidean geometry unchanged?

Motivation: The answers to this question will allow me to answer my previous question.

In particular, because of the choice of lines for taxicab geometry, even though it is a length space, this is not a case where $d(A,B)+d(B,C) = d(A,C)$ implies that $A,B,C$ are collinear (although it would be if we chose the lines to be the geodesics of the length space). Thus, in order to define a notion of betweenness, we need to say that $B$ is between $A$ and $C$ if and only if (1) $A,B,C$ are collinear, and (2) $d(A,B) + d(B,C) = d(A,C)$. With a more natural definition of line, like that in Euclidean space, the first condition would be redundant, and the notion of betweenness more intimately and naturally related with our choice of metric.

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Geodesics in the Taxicab plane are not unique. A trivial example of this is given by the points $(0, 0)$ and $(1, 1)$ and the curves formed by going from $(0, 0)$ to $(0, 1)$ then $(1, 1)$ and the other as going from $(0, 0)$ to $(1, 0)$ then $(1, 1)$. The two segments forming the curves are easily seen to both have length 1, and the sum total of those lengths is then 2, so these curves both have length 2. But the distance between $(0, 0)$ and $(1, 1)$ is $|1 - 0| + |1 - 0| = 2$ so both are geodesics. In general, any "stair-step" which always moves parallel to either the $x$- or $y$-axes with no backtracking is a geodesic as can be seen by translating its component segments to match these simpler geodesics. Such a path is a continuous-space version of a Dyck path as described here:

http://mathworld.wolfram.com/DyckPath.html

Thus if we define a line as a geodesic it is subject to a great deal of ambiguity, in particular there are $\beth_1$ geodesics between any two points (it can't be more since geodesics must be continuous). In particular, there would be tremendous ambiguity in the "betweenness" relation as you define it -- I'd believe that in fact any point lying in the rectangle with points $A$ and $C$ would end up "between" in this definition.

Yup, it would: Let $A = (x_A, y_A)$, $C = (x_C, y-C)$, and $B = (x_B, y_B)$ with $x_A < x_B < x_C$ and $y_A < y_B < y_C$. Then

$$d(A, B) = |x_B - x_A| + |y_B - y_A|$$ $$d(B, C) = |x_C - x_B| + |y_C - y_B|$$

so

$$ \begin{align} d(A, B) + d(B, C) &= |x_B - x_A| + |y_B - y_A| + |x_C - x_B| + |y_C - y_B|\\ &= (x_B - x_A) + (y_B - y_A) + (x_C - x_B) + (y_C - y_B)\\ &= x_B - x_A + y_B - y_A + x_C - x_B + y_C - y_B\\ &= x_C - x_A + y_C - y_A\\ &= (x_C - x_A) + (y_C - y_A)\\ &= |x_C - x_A| + |y_C - y_A|\\ &= d(A, C) \end{align} $$

So yeah, betweenness this way works terribly. Unless perhaps you want to argue that Taxicab geometry should be a "geometry of solid rectangles" ... but then how is it any different from Euclidean geometry of rectangles I wonder?

(FWIW the midset of Taxicab geometry, that is, the "perpendicular bisector" is FREAKY!!! Scared and disturbed me when I first saw it! I can still feel that fweeky tingley little heebiejeebies feeling a little coming on thinking about it reading right now I'm also getting that with what I just proved above, there is something conceptually horrifying about the idea that my brain makes out that this is like a line that is like, eeeevshz, infinitely swollen or something...)

So we have to narrow things down. It turns out though that Euclidean lines are also a geodesic. The most general definition of arc length of a curve to a metric space $M$ given as $C: [0, 1] \rightarrow M$ is

$$L = \sup \sum_{i=1}^{N} d(C(t_{i+1}), C(t_i))$$

where the supremum is over all partitions $0 = t_1 < t_2 < \cdots < t_N = 1$ of $[0, 1]$. So let $C(t) = (t, kt)$ be a Euclidean line in Taxicab space that is not vertical (vertical can be dispensed with a change of coordinates). Then $d(C(t_{i+1}), C(t_i)) = |t_{i+1} - t_i| + k|t_{i+1} - t_i| = t_{i+1} - t_i + k(t_{i+1} - t_i) = (1 + k)(t_{i+1} - t_i)$ and

$$ \begin{align} L &= \sup \sum_{i=1}^{N} (1 + k) (t_{i+1} - t_i)\\ &= \sup (1 + k) \sum_{i=1}^{N} t_{i+1} - t_i\\ &= \sup (1 + k) (1)\\ &= \sup (1 + k)\\ &= 1 + k \end{align} $$

which is easily seen to be the distance between the start and end points. Note that we used telescoping sum in the above. This can be visualized as the "fallacious" method used to get "$\pi = 4$" in the following answer:

Is value of $\pi = 4$?

This method is correct for Taxicab geometry, and the image correctly shows that the ratio of the circumference of a Euclidean circle to its diameter in taxicab geometry is 4.

Thus, to get a sensible geometry, we pick the Euclidean line as the choice geodesic for the taxicab system.

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  • $\begingroup$ Thank you again so much for your answer! I am just putting this here for (my) future reference, to note that the non-uniqueness issue even holds for all p-norms with $p\not=2$, not just $p=1$, so the issue you raise is a concern even more generally than how I scoped my question: math.stackexchange.com/questions/641368/… $\endgroup$ – Chill2Macht Jul 5 '17 at 10:46

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