In the case of both spherical and hyperbolic geometry, the area of a triangle is proportional to the angular defect -- the difference of the sum of its angles and $\pi$.

In Euclidean (flat) geometry, one might argue that this is still true: $\frac{\pi-\pi}{A}$ is still a constant, it just so happens that that constant is $0$.

Is this true in the general case? Is some similar statement (proportional to a power of its area, proportional to the area times the curvature of the space) generally true? I ran across the Gauss-Bonnet theorem while doing preliminary Google checking, but I'm not enough of a topologist to work out its relevance -- I'm sure it's applicable but it's a tool I don't know how to use.

  • $\begingroup$ "the area of a triangle is proportional to its area." Huh? I guess "the answer to this question is the answer to this question." $\endgroup$ Commented Sep 23, 2020 at 6:01
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    $\begingroup$ What do you mean by "general case"? $\endgroup$ Commented Sep 23, 2020 at 7:01
  • $\begingroup$ @ChristianBlatter I don't really understand my question well enough to properly answer that! However, I'd start by asking "is it true for spaces of arbitrary constant curvature". $\endgroup$ Commented Sep 23, 2020 at 7:54

1 Answer 1


Basically Gauss-Bonnet theorem tells you that given some distorted triangle, if you integrate the "curvature of the plane the triangle lies in(hyperbolic, spherical or any)" on the triangle, plus the total "turning" you make along the rim of the triangle gives you the $2\pi$ times Euler characteristic of triangle, which is $2\pi$ (Euler characteristic of polygon is defined by v-e+f, in this case 3-3+1=1. Euler characteristic of polygon or polyhedra is easy to follow so I'm not gonna explain it too much. see here https://en.wikipedia.org/wiki/Euler_characteristic#Polyhedra)

Now in the case of constant curvature plane with triangle with straight edges, the first integral term becomes simply $KA$, curvature times the area of the triangle. Since you do not make any turning along the edges since the edges are straight, only turning you make along the rim of the triangle is the exterior angles at each vertices. Now the formula mentioned above becomes :

$KA+\epsilon_1+\epsilon_2+\epsilon_3 = 2\pi$

where $\epsilon_i$ denotes exterior angles at three vertices of the triangle.

Since interior angle $\theta_i$ at each vertex is the $\pi - \epsilon_i$ of that vertex, we can write down above formula as this :

$KA=\theta_1+\theta_2+\theta_3 - \pi$

The right hand side is the defect of the triangle, and this tells you that the area of a triangle is proportional to the angular defect and the proportional constant is the curvature of the plane you're dealing with. So yes! in general your initial claim is correct and it is a direct consequence of the famous Gauss-Bonnet theorem.

P.s If you want to study more about this theorem, I suggest you to look up differential geometry text book, not topology book.

  • $\begingroup$ For the derivation of $KA$ term from the integral term, think about integral of constant function on some interval. If $f=c$(constant), $\int_a^b f = c|b-a|$, which is constant value of the function times the length(1-dimentional volume) of the domain of integration. $\endgroup$
    – Kiyoon Eum
    Commented Feb 25, 2021 at 13:02

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