# Triangle inequality and the pseudo-metric of general relativity

The pseudo-metric, such as those used in general relativity, can be expressed as the following sum:

$$ds^2=\sum_{\nu}\sum_{mu}g_{\nu\mu}dX_\nu dX_{\mu}$$

The elements $$g_{\nu\mu}$$ can be organized as a matrix

$$g=\pmatrix{g_{00} & g_{01} & g_{02} & g_{03} \\ g_{10} & g_{11} & g_{12} & g_{13} \\ g_{20} & g_{21} & g_{22} & g_{23} \\ g_{30} & g_{31} & g_{32} & g_{33} }$$

The only restriction I am aware of is that all elements of $$g$$ must be functions of two vectors of the tangent space of the manifold to the real numbers : $$g_{\nu\mu}: V \times V \to \mathbb{R}$$. Otherwise, the value $$\mathbb{R}$$ can be $$0$$, negative or positive. This does not seem true to me for the following reasons:

Consider that $$ds^2$$ is a pseudo-metric, formally defined for all $$x,y,z$$ elements of the manifold $$M$$,

$$s(x,x)=0\\ s(x,y)=s(y,x)\\ s(x,y)\leq s(x,y) + s(y,z)$$

The last requirement, the triangle inequality seems to me, intuitively, to imply additional restrictions. It seems to me that the cross-elements cannot exceed some value. Explicitely, from the triangle inequality, does it follow that the coefficient of cross terms cannot exceed inequalities of this type:

$$g_{01}+g_{10} \leq 2g_{00}g_{11} \\ g_{02}+g_{20} \leq 2g_{00}g_{22} \\ g_{03}+g_{30} \leq 2g_{00}g_{33} \\ g_{21}+g_{12} \leq 2g_{11}g_{22}\\ \vdots$$

If these inequalities are violated, the triangle inequality is also violated.

## 2 Answers

Yes. If you take the pseudo-Riemannian manifold of general relativity and attempt to add extra structure to it in order to make it a pseudometric space, you will have nontrivial axioms to ensure this.

This is another way of saying that those are not the same notion; the notion of being a pseudometric space is not encapsulated in that of being a pseudo-Riemannian manifold.

To take a very simple example, in 1+1D Minkowski space your desire for $$g_{10} + g_{01} \le 2g_{00}g_{11}$$ would be violated by the metric tensor as it has $$g_{10} = g_{01} = 0$$ while $$g_{00}=1$$ and $$g_{11} = -1$$, and clearly $$-2$$ is not greater than or equal to zero.

To get a little bit more physical so that there is no question, the desire for $$s(x, z)\le s(x,y) + s(y, z)$$ fails in 1+1 Minkowski because I can realize either a timelike or spacelike separation $$s(x, z)$$ with light rays. For a spacelike separation, I can fire two light rays from spacelike separated points to the point halfway between them. For a timelike separation, I can fire a light ray out from the origin for half of the time difference, take that point $$(t/2, t/2)$$ in the space, and fire another light ray back to $$x=0$$ where it will arrive at time $$t$$. So whether you choose spacelike or timelike separations to be positive, one of those is positive but I can create it with null rays.

• Ignoring my inequalities (which might have errors), surely the shortest distance, in space-time, between me and you cannot be longer than the shortest distance between me and bob + bob and you? If it is, I can simply go from me to bob then from bob to you and this will become the shortest distance. – Alexandre H. Tremblay Mar 26 '19 at 22:46
• So on the $(+---)$ metric you are always free to cut two light years of distance off of any path by first teleporting to the nearest star in an instant, then teleporting back. I tell this to you because you appear to be thinking like a mathematician. If you want to think like a physicist you may need to consider only paths that stay locally within the light cone: that restriction makes things more interesting but also means that $s(x,y)$ is undefined for non-timelike-separated points. – CR Drost Mar 27 '19 at 6:17

The only restriction I am aware of is that all elements of $$g$$ must be functions of two vectors of the tangent space of the manifold to the real numbers : $$g_{\nu\mu}: V \times V \to \mathbb{R}$$.

No, the components are functions of what point P in spacetime we're talking about. The metric, considered as a linear transformation for fixed P, is a map that takes two vectors $$u$$ and $$v$$ to an output in $$\mathbb{R}$$. The components do not depend on $$u$$ and $$v$$. In general, the output of the metric depends on all three things: P, $$u$$, and $$v$$.

Also, re "The only restriction I am aware of...," there is an additional restriction, which is that the metric has to be symmetric (which is equivalent to saying that the metric's output must be invariant when you swap $$u$$ and $$v$$).

Consider that $$ds^2$$ is a pseudo-metric, formally defined for all $$x,y,z$$ elements of the manifold $$M$$,

$$s(x,x)=0\\ s(x,y)=s(y,x)\\ s(x,y)\leq s(x,y) + s(y,z)$$

The metric we talk about in GR is a function that takes as its inputs two tangent vectors ($$u$$ and $$v$$ in my notation), not two points in the manifold ($$x$$ and $$y$$ in your notation). This is different from the way we define a metric in metric geometry, where it tells you the distance between points. The property $$s(x,x)=0$$ doesn't even make sense as applied to the kind of metric we use in GR, because it takes vectors as inputs, not points.

In order to talk about triangle inequalities, you need to first fix a tangent space, which means essentially that you're talking about special relativity, not general relativity. There are triangle inequalities in SR, but they don't have the same form as the triangle inequalities in Euclidean geometry. If you want to see a treatment of triangle inequalities and Cauchy-Schwarz in SR, I have one written up in section 1.5 of my SR book, which is free online.

From a comment:

surely the shortest distance, in space-time, between me and you cannot be longer than the shortest distance between me and bob + bob and you? If it is, I can simply go from me to bob then from bob to you and this will become the shortest distance.

SR doesn't use a metric space, so we don't have notions such as shortest distance. For timelike paths from event A to event B, we have a longest distance, which is the distance along an inertial geodesic. This is precisely the twin paradox.

• Isn't the shortest interval/distance from event A to B, the path taken by light? – Alexandre H. Tremblay Mar 26 '19 at 23:30
• @alexandre-h-tremblay - the photon would follow the null geodesic. I prefer to refer to time like geodesics the ones with the longest proper time - and use proper length for space like curves. – Cinaed Simson Mar 27 '19 at 8:16
• @AlexandreH.Tremblay: Isn't the shortest interval/distance from event A to B, the path taken by light? For most choices of the events A and B, there is no lightlike geodesic connecting them. Keep in mind that when we calculate $\int ds$, we're finding a "distance" between events, not between points in space. – user4552 Mar 27 '19 at 19:14