We have a metric space with the line element

$$ ds^2=(a^2)d\phi^2\ + (a^2 \sin^2{\phi})d\theta^2. $$

This line element can be used to find the distance, $L$, between any two points, $A$ and $B$, in the metric space with the path integral

$$ L=\int_{A}^{B}\sqrt{ds^2}=\int_{A}^{B}\sqrt{(a^2)d\phi^2+(a^2 \sin^2{\phi})d\theta^2}. $$

This 2D space can also be embedded as a spherical surface with radius $a$ into 3 dimensions and the distance between two points can be found in the same way.

The question arises from wanting to find angles in this metric space. I believe one way that an angle on a spherical surface is defined is by finding the standard (3D Euclidean) angle between vectors constructed as tangent to the sphere at the vertex of interest.

My question: Is there a way to think about defining an angle in a metric space (in particular, the metric space I specified) without first embedding it into a higher dimension, or is that analogy with embedding it into Euclidean space the exact notion by which angles are defined in a non-Euclidean space?

Can I define an angle in a metric space without using this notion of the angle between tangent vectors in the higher-dimensional space?

  • $\begingroup$ Your formula for $ds$ should be the square root in your integrand, not what you started with. But, to answer your question, a very, very deep theorem, called the Nash embedding theorem, says that any Riemannian metric on a manifold can be obtained by suitably embedding in some (very high-dimensional) Euclidean space. $\endgroup$ Nov 17, 2017 at 5:38
  • $\begingroup$ I've fixed the line element definition in the way I think you meant. I'm thinking I may have posed my question incorrectly, since I'm really interested in the difference between defining angles with the metric itself vs by embedding it. Indeed, after glancing over the Nash embedding theorem, it does seem very deep (and too deep for me and what I'm seeking). My focus is on defining angles and how that differs between the two methods of thinking of the space. $\endgroup$
    – xish
    Nov 17, 2017 at 6:03

1 Answer 1


Your metric defines angles and not just arclengths. No need to have your space embedded anywhere. You just need the pair of tangent vectors at some point, represented in the $(\phi,\theta)$ coordinate system. Remember that dot products of vectors allow you to compute angles, using $\cos\alpha = X\cdot Y/\|X\|\|Y\|$.


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