Suppose that, instead of using the usual Euclidean norm on R^2, we use an arbitrary p-norm. Then the distance between two points x and y is given by \sqrt[p]{|x_1-y_1|^p + |x_2-y_2|^p}.

What are the geodesics in R^2 with the p-norm? (or in R^n, though I suspect that answering the question in R^2 answers the question in R^n).


5 Responses to “Geodesics”

  1. konradswanepoel Says:

    First note that the distance between two points (x_1,x_2) and (y_1,y_2) should be defined with absolute values: (|x_1-y_1|^p+|x_2-y_2|^p)^{1/p}.

    Then it follows that the geodesics are only straight lines when p>1. When p=1 there are infinitely many shortest between two points, unless they both lie on the same horizontal or vertical line.

    In fact,
    (1) for any norm, straight lines are always geodesics, and
    (2) given any strictly convex norm (the triangle inequality is strict: \|x+y\| < \|x\| + \|y\| whenever x and y are linearly independent) there are no other geodesics.

    Fact (1) is not difficult to prove (use the triangle inequality and the definition of arc length as a limit of approximating polygons, no integrals needed), and fact (2) is easy.

    Note that a norm is strictly convex if and only if there is no line segment on the boundary of the unit ball.

    For more on this, you can look at my paper with Horst Martini:

    The geometry of Minkowski spaces — a survey. Part I, Expositiones Mathematicae 19 (2001) 97-142.

  2. Randall J. Elzinga Says:

    Thanks for the response.

    Is it obvious that the only geodesics are straight lines when p>1?
    I don’t see it at the moment.

  3. konradswanepoel Says:

    Whatever your definition of the length r of a curve (polygonal approximation or integral) from a to b, the triangle inequality implies that

    (1) \|a-b\| \leq r,

    meaning that the straight line segment is a shortest path from a to b. From this and the strict triangle inequality (valid when the norm is strictly convex, see my previous comment), it follows that the only shortest curve between a and b is the straight line segment joining them (of length \|a-b\|).

    Proof: Suppose there is a curve of length r < \|a-b\| joining a and b. It cannot be the straight line segment, so there exists a point c on it that is not on the segment between a and b. Then a-c and c-b are linearly independent, so you can apply the strict triangle inequality:

    (2) \|a-b\| < \|a-c\| + \|c-b\|.

    Denote the length of the curve from a to c by r_a and the length of the curve from c to b by r_b. Then by (1),

    (3) \|a-c\| + \|c-b\| \leq r_a+r_b =r.

    Combining (2) and (3) contradicts r < \|a-b\|. QED

    It remains to show that the \ell_p norm is strictly convex when 1 < p < \infty. This follows if you analyze equality in the Minkowski inequality and the Hölder inequality on which it is based.

  4. konradswanepoel Says:

    Hi Randall

    Sorry, I made a mistake. In the proof, don’t suppose there is a shorter curve. Instead assume that there is a shortest curve (also of length r) that is not the straight line segment. Then there is a point c etc. Finally, combining (2) and (3) already contradicts (1) and QED.

    Hope this is not too confusing.

  5. Randall J. Elzinga Says:

    Hi Konrad,

    Makes sense.
    Thanks again.

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