First note that the distance between two points and should be defined with absolute values: .
Then it follows that the geodesics are only straight lines when . When 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: whenever and 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.
Whatever your definition of the length of a curve (polygonal approximation or integral) from to , the triangle inequality implies that
(1) ,
meaning that the straight line segment is a shortest path from to . 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 and is the straight line segment joining them (of length ).
Proof: Suppose there is a curve of length joining and . It cannot be the straight line segment, so there exists a point on it that is not on the segment between and . Then and are linearly independent, so you can apply the strict triangle inequality:
(2) .
Denote the length of the curve from to by and the length of the curve from to by . Then by (1),
(3) .
Combining (2) and (3) contradicts . QED
It remains to show that the norm is strictly convex when . This follows if you analyze equality in the Minkowski inequality and the Hölder inequality on which it is based.
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 ) that is not the straight line segment. Then there is a point etc. Finally, combining (2) and (3) already contradicts (1) and QED.
![\sqrt[p]{|x_1-y_1|^p + |x_2-y_2|^p}](http://s0.wp.com/latex.php?latex=%5Csqrt%5Bp%5D%7B%7Cx_1-y_1%7C%5Ep+%2B+%7Cx_2-y_2%7C%5Ep%7D&bg=ffffff&fg=333333&s=0 "\sqrt[p]{|x_1-y_1|^p + |x_2-y_2|^p}")
2007 August 20 at 3:42 pm |
First note that the distance between two points
and 
should be defined with absolute values: 
.
Then it follows that the geodesics are only straight lines when
. When 
there are infinitely many shortest between two points, unless they both lie on the same horizontal or vertical line.
In fact,
whenever 
and 
are linearly independent) there are no other geodesics.
(1) for any norm, straight lines are always geodesics, and
(2) given any strictly convex norm (the triangle inequality is strict:
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.
2007 August 22 at 12:36 am |
Thanks for the response.
Is it obvious that the only geodesics are straight lines when
?
I don’t see it at the moment.
2007 August 22 at 12:53 pm |
Whatever your definition of the length
of a curve (polygonal approximation or integral) from 
to 
, the triangle inequality implies that
(1)
,
meaning that the straight line segment is a shortest path from
to 
. 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 
and 
is the straight line segment joining them (of length 
).
Proof: Suppose there is a curve of length
joining 
and 
. It cannot be the straight line segment, so there exists a point 
on it that is not on the segment between 
and 
. Then 
and 
are linearly independent, so you can apply the strict triangle inequality:
(2)
.
Denote the length of the curve from
to 
by 
and the length of the curve from 
to 
by 
. Then by (1),
(3)
.
Combining (2) and (3) contradicts
. QED
It remains to show that the
norm is strictly convex when 
. This follows if you analyze equality in the Minkowski inequality and the Hölder inequality on which it is based.
2007 August 22 at 1:53 pm |
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
) that is not the straight line segment. Then there is a point 
etc. Finally, combining (2) and (3) already contradicts (1) and QED.
Hope this is not too confusing.
2007 September 1 at 2:53 pm |
Hi Konrad,
Makes sense.
Thanks again.