Alternative proof of Girard's theorem

Question

I am looking for an alternative proof of Girard's theorem. The standard proof, which is almost trivial, relies too much on visualizing spherical triangles on the sphere. Is there a more algebraic proof, or even a proof by straightforward integration of the area of the triangle?

Answer 1

An alternative proof can be obtained using the properties of surface curvatures. The Gauss-Bonnet theorem states that, given a domain $D$ on a compact two-dimensional Riemannian manifold $M$ (e.g., a region of a surface in the three-dimensional space), the integral of the Gaussian curvature over $D$ and that of the geodesic curvature over the domain boundary $\partial D$ satisfy the relation

$$\int_{D} K dA+\int_{\partial D} k_g ds=2 \chi \pi$$

where $K$ is the Gaussian curvature, $dA$ represents the area element of the domain $D$, $k_g$ is the geodesic curvature, $ds$ is the line element of the boundary $\partial D$, and $\chi$ is the Euler characteristic of the domain.

To explain this formula: the meaning of the first integral in the LHS is directly given by the standard definition of Gauss curvature as the determinant of the differential of the Gauss map. In practice, integrating the curvature $K$ over a domain $D$ оп а surface $M$, we get (modulo sign) the area of the image of $D$ by the Gauss mapping to the unit sphere.

The meaning of the second integral in the LHS can be better explained if we consider that any domain $D$ with boundary $\partial D$ оп a surface can be approximated by а geodesic n-gon. As $n \rightarrow \infty$, the sum of the extrinsic angles of the n-gon tends to approach the integral of $\frac{d \theta }{ds}$, i.e. the rate by which the tangent vector field along $\partial D$ rotates with respect to the parallel vector field along the same trajectory. The function $\frac{d \theta}{ds}$ is called geodesic curvature, and is usually denoted by $k_g$. Clearly, if we consider a geodesic segment, then $\frac{d \theta }{ds}=0$.

The Euler characteristic in the RHS is the classical measure most often used for the surfaces of polyhedra, according to the formula $\chi=V-E+F$, where $V$, $E$, and $F$ are the numbers of vertices, edges, and faces in the polyhedron, respectively. The Euler characteristic of a spherical triangle is $1$.

Taking into account all these considerations, for а domain $D$ on the unit sphere, we can directly write

$$\displaystyle A(D)=\int_{ D} K dA$$ $$\int_{ D} K dA + \int_{\partial D} k_g ds=2 \pi$$

and then

$$\displaystyle A(D) =2 \pi- \int_{\partial D} k_g ds$$

where $A(D)$ is the area of the domain $D$.

Now if we apply the last equation to a spherical triangle whose angles are $\alpha_1$, $\alpha_2$, and $\alpha_3$, we have to consider that for geodesics the only contribution to $\displaystyle \int_{\partial D} k_g ds$ is given by $\displaystyle \sum_{i=1}^3 (\pi-\alpha_i)$, i.e. by the sum of the three extrinsic angles at the vertices of the triangle. So we get

$$\displaystyle A(D) =2 \pi- \sum_{i=1}^3 (\pi-\alpha_i)$$

and then

$$\displaystyle A(D) =\alpha_1+\alpha_2+\alpha_3-\pi$$

which generalized to non-unitary spheres gives the same result of the Girard's theorem

$$\displaystyle A =R^2(\alpha_1+\alpha_2+\alpha_3-\pi)$$

Answer 2

It is also possible to get another alternative proof that does not require the Gauss-Bonnet theorem, as requested. Let us call $A, B,C$ the vertices of a spherical triangle on the surface of a unit sphere with center $O$, and let us consider the tetrahedron $OABC$. Let $\phi_A$ be the dihedral angle between the planes identified by the tetrahedral faces $OAB$ and $OAC$, and thus corresponding to the edge $OA$. Similarly, let us define $\phi_B$ and $\phi_C$ the dihedral angles corresponding to the edges $OB$ and $OC$, respectively. A property of tetrahedra is that summing $\pi$ to the amplitude of the trihedral solid angle corresponding to a vertex (in steradians) we obtain the sum of the three dihedral angles originating from that vertex (this property is also illustrated by the known relation $\displaystyle \sum_{i=1}^4 \theta_i+4 \pi=2 \cdot \sum_{i=1}^6 \phi_i$, where $\theta_i$ represents the four trihedral angles of the tetrahedron and $\phi_i$ indicates the six dihedral angles). Applying the rule to the tetrahedron $OABC$, we get that the trihedral angle $\theta$ in the center $O$ satisfies

$$\theta+\pi=\phi_A+\phi_B+\phi_C$$

where $\theta$ is measured in steradians and the three dihedral angles are measured in radians.

Now we can note that the angle $\angle BAC$ of our spherical triangle $ABC$ is defined by the two planes passing through the center $O$ and containing the sides $AB$ and $AC$, so that it is equal to the dihedral angle between the tetrahedral faces $OAB$ and $OAC$. Thus, we have $\angle BAC=\phi_A$. Similarly, we get $\angle ABC=\phi_B$ and $\angle ACB=\phi_C$. We can then rewrite the equation above as

$$\theta=\angle BAC+\angle ABC+\angle ACB-\pi$$

where the RHS corresponds to the spherical excess of the Girard's formula. Reminding that the amplitude in steradians of a solid trihedral angle in the unit sphere is by definition given by the portion of the spherical surface subtended by the angle itself, we directly obtain that $\theta$ is equal to the area $A$ of the triangle $ABC$, and so

$$A=\angle BAC+\angle ABC+\angle ACB-\pi$$

Answer 3

I found this question because I am interested in a basic calculus solution. I didn't finish anything along Leo's advice, unfortunately. Here's as far as I got before I post a true solution with as close a calculus flavor as I can muster.

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As for describing a great circle $C$ analytically, we just need to obtain two orthogonal points on $C$, say $Q,R$,

$$C(t)=Q\cos t+R\sin t$$

If you only have a pole $P$ you can get these two points using rectangular coordinates. Note that $Q=(-P_1,P_2,0)/\sqrt{P_1^2+P_2^2}$ is orthogonal to $P$. Then, once you have $Q$, you get $R=Q\times P$. Or, if you have two points $R,Q\in C$ that are not orthogonal, you can find a pole by $$P=Q\times R / |Q\times R|=Q\times R / \sin(QR) = Q\times R/\sin c$$ so that $C(t)=Q\cos t+P\times Q\sin t$

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I will use rectangular coordinates for now and geocentric language where $(0,0,1)$ is the north pole. Start with a right triangle $\triangle ABC$, placing $A=(0,\sin a, \cos a), B=(\sin b, 0, \cos b), C=(0,0,1)$.

The main task is describing the interior of $\triangle ABC$, i.e. finding the intersection of side $c$ with great circles through the north pole (meridians). Side $c$ can be described in rectangular coordinates as $B\cos t+B\times(A\times B)\sin t/\sin c$ (with $0\le t\le c$)and a meridian $\theta=\theta_0$ is $(\sin\phi\cos\theta_0,\sin\phi\sin\theta_0,\cos\phi)$. Here are some useful relations

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some of Napier's rules for spherical right triangles

$\sin c= \sin a/\sin A = \sin b/\sin B$ (Law of Sines)

$\cos c=\cos a\cos b$ (Pythagorean Theorem)

$\cos A = \sin B\cos a$

$\cos B = \sin A\cos b$

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$A\times B=(\cos b\sin a,\sin b\cos a,-\sin a\sin b)$

$B\times (A\times B) = (-\cos a\cos b\sin b, \sin^2 b\sin a+\cos^2 b\sin a, \sin^2 b\cos a)=(-\cos c\sin b, \sin a, \sin^2 b\cos a)$

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Finally, the intersection of side $c$ with a meridian at some constant polar angle $\theta$ gives the boundary condition for a double integral (along meridians first). Again, point $B$ is $B=(\sin b,0,\cos b)$ and side $c$ is paramatrized by $B\cos t+B\times(A\times B)\sin t/\sin c$ for $0\le t\le c$.

$$\sin b\cos t-\cos b\cos A\sin t=\sin\phi\cos\theta$$

$$\sin A \sin t = \sin\phi\sin\theta$$

$$\cos b\cos t+\sin b\cos A\sin t=\cos\phi$$

Combining the first and third to cancel $\cos t$, I get

$$\cos A\sin t = \sin b\cos\phi-\cos b\cos\theta\sin\phi$$

Then, combine this with the second equation to eliminate $\sin t$,

$$\cos A\sin\phi\sin\theta=\sin A(\sin b\cos\phi-\cos b\cos\theta\sin\phi)$$ $$\dots = \sin a\sin B\cos\phi-\cos B\cos\theta\sin \phi$$

and reach

$$\tan\phi=\frac{\sin a\sin B}{\cos A\sin\theta+\cos B\cos\theta}$$

Call $\phi(\theta)$ this solution along side $c$ as a function of $\theta$.

The area element in spherical coordinates is $\sin\varphi d\varphi d\theta$, so

$$[ABC]=\int_0^{\pi/2}\int_0^{\phi(\theta)}\sin\varphi d\varphi d\theta$$

$$\dots =\int_0^{\pi/2} 1-\cos(\phi(\theta))d\theta$$

Changing the $\tan\phi$ to $\cos\phi$, and using $\cos A=\cos a\sin B$,

$$\cos(\phi(\theta))=\frac{\cos A\sin\theta+\cos B\cos\theta}{\left(\sin^2 B-\cos^2 A+(\cos A\sin\theta+\cos B\cos\theta)^2\right)^{1/2}}$$

The denominator nicely reduces to $$(1-(\cos A\cos\theta -\cos B\sin\theta)^2)^{1/2}$$

which can integrate after subbing $u=\cos A\cos\theta -\cos B\sin\theta$ then $u=\cos x$.

If you place $A$ at the north pole, you get an alternative form

$$[ABC]=A-\int_0^A\frac{\cos(B)\cos(\theta)d\theta}{(\sin^2(A)-\cos^2(B)\sin^2(\theta))^{1/2}}$$

Finally, any spherical triangles can be subdivided into two right triangles and the additivity is trivial.