Gauss-Bonnet for a spherical cap

On overlapping polar patches away from the north pole, the parametrization $X(\theta ,\phi )=R(\sin \theta \cos \phi ,\sin \theta \sin \phi ,\cos \theta )$ has $X_{\theta }=R(\cos \theta \cos \phi ,\cos \theta \sin \phi ,-\sin \theta )$ and $X_{\phi }=R\sin \theta (-\sin \phi ,\cos \phi ,0)$, so by [F7] the induced metric is $g=R^{2}d{\theta }^2+R^2{\sin }^2\theta d{\phi }^2$ with $g_{\theta \theta }=R^2$, $g_{\phi \phi }=R^2{\sin }^2\theta$ and $g_{\theta \phi }=0$. Hence $e_1=(1\mathrm{/}R){\mathrm{\partial }}_{\theta }$ and $e_2=(1\mathrm{/}(R\sin \theta )){\mathrm{\partial }}_{\phi }$ form a smooth positive orthonormal frame on each such patch, with dual coframe $e^1=Rd\theta$, $e^2=R\sin \theta d\phi$. The polar frame is undefined at the pole, but the induced metric is smooth there in ordinary surface coordinates.

The cap carries the finite curvilinear triangulation by the three meridian arcs from the pole to the three boundary points at $\phi =0,2\pi \mathrm{/}3,4\pi \mathrm{/}3$ together with the three boundary arcs joining consecutive ones: the vertices are the pole and the three boundary points, so $V=4$; the edges are the three meridians and the three boundary arcs, so $E=6$; and the faces are the three closed lune triangles between consecutive meridians, each an embedded closed triangular disk meeting the others exactly along common full edges, so $F=3$. Therefore $\chi (D)=4-6+3=1$ by [F8].

Since only $g_{\phi \phi }=R^2{\sin }^2\theta$ depends on the coordinates, [F3] gives ${\mathrm{\Gamma }}^{\theta }{}_{\phi \phi }=-\sin \theta \cos \theta$ and ${\mathrm{\Gamma }}^{\phi }{}_{\theta \phi }={\mathrm{\Gamma }}^{\phi }{}_{\phi \theta }=\cot \theta$, all other symbols vanishing; in particular ${\mathrm{\nabla }}_{{\mathrm{\partial }}_{\phi }}{\mathrm{\partial }}_{\theta }=\cot \theta {\mathrm{\partial }}_{\phi }$ and ${\mathrm{\nabla }}_{{\mathrm{\partial }}_{\theta }}{\mathrm{\partial }}_{\theta }=0$. Consequently ${\mathrm{\nabla }}_{e_1}{e}_1=(1\mathrm{/}{R}^2){\mathrm{\nabla }}_{{\mathrm{\partial }}_{\theta }}{\mathrm{\partial }}_{\theta }=0$ and ${\mathrm{\nabla }}_{e_2}{e}_1=(1\mathrm{/}(R\sin \theta )){\mathrm{\nabla }}_{{\mathrm{\partial }}_{\phi }}((1\mathrm{/}R){\mathrm{\partial }}_{\theta })=(\cot \theta \mathrm{/}R)e_2$.

On the latitude circle $\theta ={\theta }_0$ the field $e_2$ restricts to a unit tangent field, and the curve $\gamma (\phi )=X({\theta }_0,\phi )$ has speed $R\sin {\theta }_0$; parametrized by arclength its unit tangent is $T=e_2$. The outward normal of the cap at $\theta ={\theta }_0$ points in the direction of increasing $\theta$, that is along $e_1$, and $(e_1,e_2)$ is positively oriented, so this parametrization is the positively oriented boundary and has length $2\pi R\sin {\theta }_0$.

By step 2.1, $\omega (e_1)=g({\mathrm{\nabla }}_{e_1}{e}_1,e_2)=0$ and $\omega (e_2)=g((\cot \theta \mathrm{/}R)e_2,e_2)=\cot \theta \mathrm{/}R$; since $\omega$ is a one-form on the chart with $\omega =\omega (e_1)e^1+\omega (e_2)e^2$, this gives $\omega =\cos \theta d\phi$.

From [F6] and step 3.1, ${\mathrm{\nabla }}_{T}T={\mathrm{\nabla }}_{e_2}{e}_2=-\omega (e_2)e_1=-(\cot {\theta }_0\mathrm{/}R)e_1$ along the circle, and $JT=J{e}_2=-e_1$; hence by [F5] the signed geodesic curvature of the positively oriented boundary is $k_g=g(-(\cot {\theta }_0\mathrm{/}R)e_1,-e_1)=\cot {\theta }_0\mathrm{/}R$.

Exterior differentiation of step 3.1 gives $d\omega =-\sin \theta d\theta \wedge d\phi$, while $dA=e^1\wedge e^2=R^2\sin \theta d\theta \wedge d\phi$ by [F4]; comparing with $d\omega =-KdA$ from [F2] yields $K=1\mathrm{/}{R}^2$ on the polar patches away from the pole. In an ordinary smooth chart about the pole, Gram–Schmidt gives a smooth positive orthonormal frame for the induced metric; its smooth connection form and nowhere-zero area form make $K=-(d\omega )\mathrm{/}dA$ continuous there by [F2]. Thus $K=1\mathrm{/}{R}^2$ also at the pole.

The pole has zero area because the smooth area density is bounded in an ordinary chart around it. Integrate first over $\epsilon \le \theta \le {\theta }_0$ using the polar patches and let $\epsilon \downarrow 0$; then ${\int }_{D}KdA=(1\mathrm{/}{R}^2){\int }_0^{2\pi }{\int }_0^{{\theta }_0}{R}^2\sin \theta d\theta d\phi =2\pi (1-\cos {\theta }_0)$.

By step 2.2 and step 4.1, ${\int }_{\mathrm{\partial }D}{k}_{g}ds=(\cot {\theta }_0\mathrm{/}R)\cdot 2\pi R\sin {\theta }_0=2\pi \cos {\theta }_0$.

Steps 5.1 and 5.2 give ${\int }_{D}KdA+{\int }_{\mathrm{\partial }D}{k}_{g}ds=2\pi (1-\cos {\theta }_0)+2\pi \cos {\theta }_0=2\pi$, which by [F1] equals $2\pi \chi (D)$; step 1.2 identifies $\chi (D)=1$. The boundary sign was fixed by the outward-normal-first convention of [F1] and [F5]: the positively oriented latitude circle carries $JT=-e_1$, the inward unit conormal. The full-choice assumption entered only through the parent corollary [F1].