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.


2 Answers 2


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:


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 as being "between" in this definition, in the sense that "there would exist a line (geodesic) from $A$ to $C$ that passes through the given point".

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

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


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

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:

The staircase paradox, or why $\pi\ne4$

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.

  • $\begingroup$ 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/… $\endgroup$ Jul 5, 2017 at 10:46

If one considers general $L^p$ norms where the distance between two points, $\mathbf{x}$ and $\mathbf{y}$, is defined by $$ d(\mathbf{x},\mathbf{y}) = \left(\sum_{i}\left| x_i - y_i\right|^p\right)^{1/p}, $$ one can show that for all $p>1$, the geodesics are straight, i.e. Euclidean, lines.

The taxicab metric is the $L^p$ metric with $p=1$. Since all the $L^p$ metrics with $p>1$ have straight lines as geodesics, and, although there are no unique geodesics in taxicab geometry, straight lines do give the shortest distance in the taxicab metric, one could make the argument that straight lines are the natural geodesics for the taxicab geometry that follow from its metric. At least if taxicab geometry is seen as a limiting case of case of $L^p$ geometries with $p>1$.

In order to show that the geodesics for $L^p$ metrics are straight lines, we need to use the calculus of variations. Let $\mathbf{x}(t)=(x_1(t),..., x_n(t))$ be a parametrization of a curve in $\mathbb{R}^n$ . First we need to calculate the distance for two points infinitesimally close on the path $\mathbf{x}(t)$

$$d(\mathbf{x}(t+dt),\mathbf{x}(t)) = \left(\sum_{i}\left| x_i(t+dt) - x_i(t)\right|^p\right)^{1/p} = \left(\sum_{i}\left| x_i+\frac{dx}{dt} dt- x_i\right|^p\right)^{1/p} =\left(\sum_{i}\left| \frac{dx}{dt}\right|^p \right)^{1/p}dt.$$

Now let's define the function

$$f(\mathbf{x},\dot{\mathbf{x}}) \equiv \left(\sum_{i}\left|\dot{x}_i\right|^p \right)^{1/p}$$

where $\dot{x_i}=\frac{dx_i(t)}{dt}$. We now need to minimize the integral

$$\int f\left(\mathbf{x}(t),\dot{\mathbf{x}}(t)\right)dt = \int \left(\sum_{i}\left|\dot{x}_i\right|^p \right)^{1/p}dt.$$

This can be done via the Euler–Lagrange equations:

$$\frac{\partial f}{\partial x_i} - \frac{d}{dt}\left(\frac{\partial f}{\partial \dot{x_i}}\right)=0.$$

For our problem, the Euler–Lagrange equations give

$$0-\frac{d}{dt}\left(\dot{x}_i |\dot{x}_i|^{p-2}\left(\sum_{j}\left| \dot{x}_j\right|\right)^{1/p-1}\right)=0.$$

Integrating these equations with respect to $t$, and dividing pairs of them gives

$$ \frac{a_i}{a_j}=\frac{\dot{x}_i}{\dot{x}_j} \left|\frac{\dot{x}_i}{\dot{x}_j} \right|^{p-2}$$ where the $a_i$ are integration constants. There are two cases to check, depending on the sign of $\frac{\dot{x}_i}{\dot{x}_j}$, but it is pretty easy to see that in both cases the equations reduce to $$\dot{x}_j=\left(\frac{a_i}{a_j}\right)^\frac{1}{p-1}\dot{x}_i$$

(maybe up to a minus sign). These equations can be immediately integrated, and the results are lines. The only issue is when $p=1$, where the equations becomes singular.


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