Is there any geometry where the triangle inquality fails? - MathOverflow

Is there any geometry where the triangle inquality fails?

2010-04-30T14:37:24Z

We know from elementary school that the triangle inequality holds in Euclidean geometry. Some where in High School or in Univ., we come across non-Euclidean geometries (hyperbolic and Riemannian) and Absolute geometry where in both the inequality holds.

I am curious whether the triangle inequality is made to hold in any geometry( from the beginning) or is a consequence of some axioms. Presumably, the denial of the inequality would create havoc in that conceivable geometry.

Thanks.

Answer by Tom Smith for Is there any geometry where the triangle inquality fails?

2010-04-30T14:56:04Z

What do you mean by "denial"? There is the hoary old story of the researcher who wrote an entire thesis on the properties of antimetric spaces, ie spaces where $d(a,b) \ge d(a,c) + d(b,c)$ for all $a,b$ and $c$... without realising that any such space must consist of at most a single point!

Answer by gowers for Is there any geometry where the triangle inquality fails?

2010-04-30T15:36:05Z

There are people who seriously study quasi-normed spaces. The most natural examples are $\ell_p$ spaces for p strictly between 0 and 1 (the "norm" given by the usual formula and the distance given by the norm of the difference). Although these spaces do not satisfy the triangle inequality, you get an inequality of the form $\|x+y\|\leq C(\|x\|+\|y\|)$.

Answer by Joseph Malkevitch for Is there any geometry where the triangle inquality fails?

2010-04-30T15:39:30Z

In the Euclidean plane when equality holds in the triangle inequality the points lie along a line of the geometry (degenerate triangle). However, for the triangle inequality in the Taxicab Plane (distance given by sums of absolute values of the differences in the coordinates) points which do not lie along a line of the geometry can have the sum of two sides of a triangle with equal length to the third.

Answer by S. Carnahan for Is there any geometry where the triangle inquality fails?

2010-04-30T17:40:08Z

Any geometry modelled on non-positive definite spaces will yield an example. As Sergei Ivanov mentioned in the comments, an $n$-dimensional Lorentz geometry is modelled on the indefinite space $\mathbb{R}^{1,n-1}$, and there are tangent vectors of negative norm, hence paths of negative length squared. These appear a lot in special and general relativity, where one direction is time, and the rest are space.

Another example is the root space of an affine Kac-Moody Lie algebra, which has a singular metric (i.e., there is a line that is perpendicular to everything). It embeds as the codimension 1 subspace of a Lorentz space that is perpendicular to a lightlike (i.e., nonzero norm 0) vector. Its geometry comes into play when considering the affine Weyl group, which acts on this space by reflections. For reasons of sanity, one typically studies the action by using the Lorentz embedding to extend it to a group of hyperbolic reflections that fix a boundary point, and considering the induced action on a horocycle (which has Euclidean geometry).

Answer by marc for Is there any geometry where the triangle inquality fails?

2010-04-30T19:56:29Z

In Information Geometry, the Kullback-Liebler divergence is commonly used in the manner of a metric, but it does not satisfy the triangle inequality. It localizes to the Fisher Information Metric on the probability simplex, the interior of which is diffeomorphic to the geometry of the sphere with the inherited metric from Euclidean space under the map $x \mapsto 2\sqrt{x}.$ This geometry is used in evolutionary game theory to study natural selection.

There are many other information divergences that are not symmetric and do not satisfy the triangle inequality. You can form a divergence that does satisfy the triangle inequality.

Answer by Eric O. Korman for Is there any geometry where the triangle inquality fails?

2010-05-01T12:30:41Z

I just wanted to add that in the Lorentzian case, the triangle inequality gets reversed for a certain class of vectors (those with positive time component). Since the norm of a vector corresponds to the elapsed time measured by a clock moving along the vector, a physical consequence of this is the twin's paradox: by flying out in a spaceship from earth and then returning, you age less than if you had just stayed still.

Answer by Peter Luthy for Is there any geometry where the triangle inquality fails?

2010-05-01T15:38:54Z

The weak $L^p$ spaces, often denoted $L^{p,\infty}$, provide another class of commonly occurring function spaces for which the triangle inequality fails to hold. The most common of these is weak $L^1$ which is the set of all functions so that $\sup_{\alpha>0}\alpha\textrm{ }\mu(x:|f(x)|>\alpha)$ is finite. This quantity is called the weak-$L^1$ "norm" of $f$, though it is not a true norm because it does not satisfy the triangle inequality. The term "weak" is appropriate as every $L^1$ function is automatically in weak $L^1$ by Chebyshev's inequality; the function $1/x$ is in weak $L^1$, however.

These spaces arise when studying, for example, maximal operators, e.g. Hardy-Littlewood, and singular integral operators, e.g. the Hilbert transform. Both the Hardy-Littlewood maximal operator and the Hilbert transform are bounded from $L^p$ to $L^p$ for $p\in(1,\infty)$ but are not bounded $L^1$ to $L^1$. They are, however, bounded from $L^1$ to weak $L^1$. The weak Lebesgue spaces are very useful substitutes for the usual Lebesgue spaces because some interpolation theorems, e.g. the Marcinkiewicz interpolation theorem, allow one to interpolate between two weak estimates to produce strong estimates in between. So the fact that the Hardy-Littlewood maximal operator is bounded from $L^1$ to weak $L^1$ and $L^\infty$ to $L^\infty$ is enough to prove that it is bounded from $L^p$ to $L^p$ for $p\in(1,\infty)$.

Answer by Tom LaGatta for Is there any geometry where the triangle inquality fails?

2010-05-01T20:25:09Z

Last-passage percolation is a model of random geometry on the lattice which satisfies a superadditive inequality (a reverse triangle inequality). On the lattice $\mathbb Z^2$, define a passage time $t_b \ge 0$ for each bond $b$ of the lattice. These are typically chosen to be i.i.d. random variables, but they don't have to be.

Define the up-right ordering $\le$ on $\mathbb Z^2$ by $z \le z'$ if $z_1 \le z_1'$ and $z_2 \le z_2'$. Then the last-passage time between two points $z \le z'$ on $\mathbb Z^2$ is $$\tau(z,z') = \max_{\gamma} \sum_{b \in \gamma} t_b,$$ where the maximum is taken over all directed up-right paths between $z$ and $z'$. The last-passage time satisfies a reverse triangle inequality: $$\tau(z,z') \ge \tau(z,w) + \tau(w,z')$$ if $z \le w \le z'$.

Answer by David Sprehn for Is there any geometry where the triangle inquality fails?

2010-05-01T20:30:36Z

There are a number of good answers from differential geometry, analysis, etc. But I suspect the questioner had axiomatic geometries in mind. So here's another take from that perspective.

There are two statements of the triangle inequality in plane geometry. (1) If A,B,C are noncollinear points, then $AC\lt AB+BC$; (2) If A,B,C are any three points, then $AC\leq AB+BC$.

In any system which includes a Ruler Postulate, (1) is stronger than (2). In neutral geometry (which includes all of the axioms of Euclidean geometry except the parallel postulate), statement (1) can be proven pretty directly from the SAS congruence postulate. (Does anyone know if they're equivalent in the presence of the other neutral geometry axioms?) Statement (2), on the other hand, is the one needed to show that the plane is a metric space, and is strictly weaker, as shown by the "taxicab geometry" mentioned above, which satisfies all of the postulates for neutral geometry except for SAS, and has property (2) but not property (1).

So, to summarize, the triangle inequality is true in neutral geometry, so any model of it (including the Euclidean and hyperbolic planes, etc.) will satisfy the triangle inequality. But of course we can consider weaker axiom systems in which models do not satisfy it (like taxicab).

Answer by Gil Kalai for Is there any geometry where the triangle inquality fails?

2010-05-15T09:38:46Z

A famous example of a geometry which violates the triangle inequality is $\ell_2^2$, namely the distance between two points is defined as the square of their Euclidean distance. There is also much interesrt in metric spaces that do satisfy the triangle inequality which are subsets of $\ell_2^2$. Those are called "metric spaces of negative types". (Indeed, sumetimes the notation $\ell_2^2$ refer to a metric space which is a subset of $\ell_2^2$ as defined above.) There was a conjecture by Goemens and by Linial asserting that every metric space in $\ell_2^2$ can be embedded with constant distortion into $\ell_1$. This was disproved in a spectacular paper by Khot and Vishnoi.

Answer by Ivan Meir for Is there any geometry where the triangle inquality fails?

2012-05-19T09:44:28Z

The reverse triangle inequality|x+y|>|x|+|y| holds in the Minkowski Space of Special Relativity for two timelike vectors in the same direction. So geometries that deny the triangle inequality in some signficant way can be very important I guess.