Area excess of a spherical geodesic triangle

Use the chart $X(\theta ,\phi )=(R\sin \theta \cos \phi ,R\sin \theta \sin \phi ,R\cos \theta )$ with $0<\theta <\pi$ and $0<\phi <2\pi$, positively oriented. The induced round metric is $g=R^{2}d{\theta }^2+R^2{\sin }^2\theta d{\phi }^2$, so $({\mathrm{\partial }}_{\theta },{\mathrm{\partial }}_{\phi })$ has $g_{\theta \theta }=R^2$, $g_{\phi \phi }=R^2{\sin }^2\theta$ and $g_{\theta \phi }=0$; the fields $e_1=(1\mathrm{/}R){\mathrm{\partial }}_{\theta }$ and $e_2=(1\mathrm{/}(R\sin \theta )){\mathrm{\partial }}_{\phi }$ therefore form a smooth positive orthonormal frame on the chart.

The triangle $T$ is a compact regular oriented disk region with ordinary corners whose boundary is the cyclic concatenation of the three geodesic sides, so the geodesic-polygon formula [F6, F8] applies with the outward-normal-first boundary orientation: ${\int }_{T}KdA+{\alpha }_1+{\alpha }_2+{\alpha }_3=2\pi$ for the signed exterior angles.

Since all $x^i$-independent coefficients satisfy ${\mathrm{\partial }}_{\phi }{g}_{\theta \theta }={\mathrm{\partial }}_{\phi }{g}_{\phi \phi }=0$, and ${\mathrm{\partial }}_{\theta }{g}_{\phi \phi }=2R^2\sin \theta \cos \theta$, ${\mathrm{\partial }}_{\theta }{g}_{\theta \theta }=0$, formula [F2] gives ${\mathrm{\Gamma }}^{\theta }{}_{\phi \phi }=-R^2\sin \theta \cos \theta \cdot R^{-2}=-\sin \theta \cos \theta$ and ${\mathrm{\Gamma }}^{\phi }{}_{\theta \phi }={\mathrm{\Gamma }}^{\phi }{}_{\phi \theta }=\frac{1}{2}(R^2{\sin }^2\theta {)}^{-1}\cdot 2R^2\sin \theta \cos \theta =\cot \theta$, with all remaining symbols zero.

Consequently ${\mathrm{\nabla }}_{{\mathrm{\partial }}_{\phi }}{\mathrm{\partial }}_{\theta }=\cot \theta {\mathrm{\partial }}_{\phi }$, while ${\mathrm{\nabla }}_{{\mathrm{\partial }}_{\theta }}{\mathrm{\partial }}_{\theta }$ has vanishing $\theta$- and $\phi$-components; with $e_1=(1\mathrm{/}R){\mathrm{\partial }}_{\theta }$ and $e_2=(1\mathrm{/}(R\sin \theta )){\mathrm{\partial }}_{\phi }$ this gives ${\mathrm{\nabla }}_{e_1}{e}_1=0$ and ${\mathrm{\nabla }}_{e_2}{e}_1=(\cot \theta \mathrm{/}(R^2\sin \theta )){\mathrm{\partial }}_{\phi }=(\cot \theta \mathrm{/}R)e_2$. Hence the connection form satisfies $\omega (e_1)=0$ and $\omega (e_2)=\cot \theta \mathrm{/}R$, so $\omega =\cos \theta d\phi$.

Therefore $d\omega =-\sin \theta d\theta \wedge d\phi$, while the dual coframe $e^1=Rd\theta$, $e^2=R\sin \theta d\phi$ has $e^1\wedge e^2=R^2\sin \theta d\theta \wedge d\phi$, which is $dA$ by [F3]. So $d\omega =-(1\mathrm{/}{R}^2)dA$ on the chart.

The structure equation [F1] applied to the frame of step 1.1 gives $d\omega =-KdA$, so step 4.1 yields $K=1\mathrm{/}{R}^2$ on the chart; since the rotation axis of the construction is arbitrary and the sphere is covered by such charts, $K\equiv 1\mathrm{/}{R}^2$ on $S_R^2$ by smoothness.

Inserting $K\equiv 1\mathrm{/}{R}^2$ into step 1.2 gives $A\mathrm{/}{R}^2+{\alpha }_1+{\alpha }_2+{\alpha }_3=2\pi$, and each exterior angle is ${\alpha }_j=\pi -{\beta }_j$ by [F7] for the interior angles ${\beta }_j\in (0,2\pi )$; hence ${\beta }_1+{\beta }_2+{\beta }_3=3\pi -({\alpha }_1+{\alpha }_2+{\alpha }_3)=\pi +A\mathrm{/}{R}^2$.

The octant $O=S_R^2\cap \{x_1\ge 0,x_2\ge 0,x_3\ge 0\}$ is a simple geodesic triangle: its boundary consists of the three great-circle arcs joining the scaled basis vectors $R{e}_1,R{e}_2,R{e}_3$, which are geodesics by [F4], and each vertex has ordinary non-antipodal tangents. Its area is $A={\int }_0^{\pi \mathrm{/}2}\text{\hspace{0.17em}⁣}{\int }_0^{\pi \mathrm{/}2}{R}^2\sin \theta d\theta d\phi =R^2\cdot \frac{\pi }{2}=\pi R^2\mathrm{/}2$ in the chart of step 1.1, and its three interior angles are right angles: at each vertex the two inward boundary directions are two distinct standard basis vectors, which are orthonormal in the ambient inner product and hence, by [F5], orthonormal for the induced metric. So $3\pi \mathrm{/}2=\pi +A\mathrm{/}{R}^2$ holds for $O$, exhibiting a genuine spherical geodesic triangle whose angle sum exceeds $\pi$.