Projective-plane curvature via a hemisphere

For $i\in \{0,1,2\}$ put $H_i:=\{x\in S_R^2:x_i>0\}$ and $U_i:=\pi (H_i)$. Every antipodal pair meets ${\bigcup }_i{H}_i$, because some coordinate of $x$ is nonzero and exactly one of $x,-x$ has its $i$-th coordinate positive whenever $x_i\mathrm{\ne }0$; hence the $U_i$ cover ${\mathbb{R}\mathbb{P}}^2$. The restriction $\pi {\mathrm{\mid }}_{H_i}$ is injective: if $x,y\in H_i$ and $\pi (x)=\pi (y)$, then $y=\pm x$, and $x_i>0$, $y_i>0$ force $y=x$. Since $\pi$ is open by [F2], each $\pi {\mathrm{\mid }}_{H_i}$ is a homeomorphism onto $U_i$.

Define $p_i:{\mathbb{R}}^2\to H_i$ by $p_i(a,b):=R(1+a^2+b^2{)}^{-1\mathrm{/}2}{\tilde{x}}_i(a,b)$, where ${\tilde{x}}_i(a,b)\in {\mathbb{R}}^3$ has entries $a,b$ in the two slots other than $i$ and the entry $1$ in slot $i$. Then $p_i$ is a smooth bijection with smooth inverse $x\mapsto (x_j\mathrm{/}{x}_i{)}_{j\mathrm{\ne }i}$ on $H_i$, hence a diffeomorphism onto $H_i$. Therefore ${\mathrm{\Psi }}_i:=p_i^{-1}\circ (\pi {\mathrm{\mid }}_{H_i}{)}^{-1}:U_i\to {\mathbb{R}}^2$ is a chart with ${\mathrm{\Psi }}_i^{-1}=\pi \circ p_i$. On $U_i\cap U_j$, put $x=p_i(u)$ and $\epsilon (u)=\mathrm{sign}(x_j)$; then $x_j\mathrm{\ne }0$, the unique representative of $\pi (x)$ in $H_j$ is $\epsilon (u)x$, and the transition is ${\mathrm{\Psi }}_j\circ {\mathrm{\Psi }}_i^{-1}(u)=p_j^{-1}(\epsilon (u)p_i(u))=(x_k\mathrm{/}{x}_j{)}_{k\mathrm{\ne }j}$. The ratios are smooth wherever $x_j\mathrm{\ne }0$; equivalently, $\epsilon$ is constant on each overlap component. Thus the family $\{(U_i,{\mathrm{\Psi }}_i)\}$ is a smooth atlas. Moreover ${\mathrm{\Psi }}_i\circ \pi \circ p_i={\mathrm{i}\mathrm{d}}_{{\mathbb{R}}^2}$, so $\pi$ is smooth and has invertible differential on every $H_i$. For an $x$ outside their union, $-x$ belongs to some $H_i$ and $\pi \circ A=\pi$ with $A$ a diffeomorphism, so the same conclusion holds at $x$: the quotient map is a local diffeomorphism everywhere.

Define a bilinear form $g$ on ${\mathbb{R}\mathbb{P}}^2$ by $g_q(v,w):=g_R((d{\pi }_x{)}^{-1}v,(d{\pi }_x{)}^{-1}w)$, where $x\in S_R^2$ is any point with $\pi (x)=q$ and $d{\pi }_x$ is the isomorphism of step 2.1. This is well defined: the other preimage of $q$ is $y=A(x)=-x$, and $d{\pi }_y\circ d{A}_x=d{\pi }_x$ with $d{A}_x=-I$, so replacing $x$ by $y$ changes both arguments by the linear map $-I$, which preserves the ambient inner product and hence $g_R$ by [F1]. Symmetry and positive definiteness are inherited from $g_R$. In the chart ${\mathrm{\Psi }}_i$ the coefficient functions are $g_{ab}(u)=g_R(d(p_i{)}_u{\mathrm{\partial }}_a,d(p_i{)}_u{\mathrm{\partial }}_b)=(p_i^{\ast }{g}_R{)}_{ab}(u)$, because $d{\mathrm{\Psi }}_i^{-1}=d(\pi \circ p_i)=d\pi \circ d{p}_i$; these are smooth in $u$. So $g$ is a Riemannian metric by [F4], and ${\pi }^{\ast }g=g_R$, i.e. $\pi$ is a local isometry by [F3].

In the chart ${\mathrm{\Psi }}_i$, use polar coordinates $(a,b)=\tan \varphi (\cos \alpha ,\sin \alpha )$ locally on overlapping angular patches, with $0<\varphi <\pi \mathrm{/}2$ and each $\alpha$ in an open interval shorter than $2\pi$. These patches cover $U_i\setminus \{P_i\}$; no single angular interval is a coordinate chart for the whole punctured plane. On each patch $p_i(\varphi ,\alpha )$ is the spherical parametrization of the open hemisphere, so by [F1] the pulled-back round metric is $p_i^{\ast }{g}_R=R^2(d{\varphi }^2+{\sin }^2\varphi d{\alpha }^2)$. The centre $\varphi =0$ is the single point $P_i:=\pi (R{e}_i)$, and the limiting equator $\varphi =\pi \mathrm{/}2$ corresponds to $L_i:=\{[x]:x_i=0\}$, so $U_i={\mathbb{R}\mathbb{P}}^2\setminus L_i$.

Let $L:=\{[x]\in {\mathbb{R}\mathbb{P}}^2:x_2=0\}$. Then $L\subset U_0\cup U_1$, and the coordinate images are explicit, namely ${\mathrm{\Psi }}_0(L\cap U_0)=\{(a,0):a\in \mathbb{R}\}$ and ${\mathrm{\Psi }}_1(L\cap U_1)=\{(a,0):a\in \mathbb{R}\}$, because ${\mathrm{\Psi }}_0([x])=(x_1\mathrm{/}{x}_0,x_2\mathrm{/}{x}_0)$ and ${\mathrm{\Psi }}_1([x])=(x_0\mathrm{/}{x}_1,x_2\mathrm{/}{x}_1)$ vanish in their second entry precisely on $L$. Each of these is a line in ${\mathbb{R}}^2$, a Lebesgue-null Borel set.

In the chart ${\mathrm{\Psi }}_i$ the metric is $p_i^{\ast }{g}_R$ by step 3.1. Differentiating $p_i$ with $s=1+a^2+b^2$ gives $g_{aa}=R^2(1+b^2)\mathrm{/}{s}^2$, $g_{ab}=-R^{2}ab\mathrm{/}{s}^2$ and $g_{bb}=R^2(1+a^2)\mathrm{/}{s}^2$, so $\det G=R^4{s}^{-3}$ and the density coefficient is ${\rho }_i:=\sqrt{\det G}=R^2(1+a^2+b^2{)}^{-3\mathrm{/}2}\le R^2$ on the whole chart. The charts ${\mathrm{\Psi }}_i$ differ only by permuting the three ambient coordinates, and a coordinate permutation preserves the Euclidean inner product and commutes with $A$, so $p_i^{\ast }{g}_R$ is the same function of $(a,b)$ for every $i$.

On $U_i\setminus \{P_i\}$ the fields $e_1:=R^{-1}{\mathrm{\partial }}_{\varphi }$ and $e_2:=(R\sin \varphi {)}^{-1}{\mathrm{\partial }}_{\alpha }$ are a positive orthonormal frame for $R^2(d{\varphi }^2+{\sin }^2\varphi d{\alpha }^2)$ with dual coframe $e^1=Rd\varphi$, $e^2=R\sin \varphi d\alpha$, and the positively oriented area form is $dA=e^1\wedge e^2=R^2\sin \varphi d\varphi \wedge d\alpha$. Since only $g_{\alpha \alpha }=R^2{\sin }^2\varphi$ depends on the coordinates, the Christoffel formula gives ${\mathrm{\Gamma }}^{\varphi }{}_{\alpha \alpha }=-\sin \varphi \cos \varphi$, ${\mathrm{\Gamma }}^{\alpha }{}_{\varphi \alpha }={\mathrm{\Gamma }}^{\alpha }{}_{\alpha \varphi }=\cot \varphi$ and all other symbols zero, hence ${\mathrm{\nabla }}_{e_1}{e}_1=0$ and ${\mathrm{\nabla }}_{e_2}{e}_1=(\cos \varphi \mathrm{/}R)e_2$. Therefore $\omega (e_1)=0$, $\omega (e_2)=\cos \varphi \mathrm{/}R$ and $\omega =\cos \varphi d\alpha$.

Exterior differentiation in step 4.2 gives $d\omega =-\sin \varphi d\varphi \wedge d\alpha$, while $dA=R^2\sin \varphi d\varphi \wedge d\alpha$; comparing with $d\omega =-KdA$ from [F7] yields $K=1\mathrm{/}{R}^2$ on $U_i\setminus \{P_i\}$, and this holds for each $i$ because the metric expression of step 3.2 is the same in every chart. The polar frame of step 4.2 is undefined at $P_i$. To cover that point, apply smooth Gram–Schmidt to the coordinate basis $({\mathrm{\partial }}_a,{\mathrm{\partial }}_b)$ of the smooth metric from step 3.1 on all of $U_i$. Its positive orthonormal frame has a smooth connection form and a nowhere-zero smooth area form, so [F7] makes $K=-(d{\omega }_{\mathrm{c}\mathrm{a}\mathrm{r}\mathrm{t}})\mathrm{/}d{A}_{\mathrm{c}\mathrm{a}\mathrm{r}\mathrm{t}}$ smooth throughout $U_i$. By continuity, the equality $K=1\mathrm{/}{R}^2$ on the punctured chart extends to $P_i$, and the charts cover ${\mathbb{R}\mathbb{P}}^2$.

Let $A\subseteq U_i$ be Borel. A chart partition of ${\mathbb{R}\mathbb{P}}^2$ may be refined so that $U_i$ is a union of its parts; applying [F5] to these parts, and using that the density coefficient is the same smooth function ${\rho }_i$ in the coordinates ${\mathrm{\Psi }}_i$, gives ${\mu }_g(A)={\int }_{{\mathrm{\Psi }}_i(A)}{\rho }_{i}d{\lambda }_2$, and by the bound ${\rho }_i\le R^2$ of step 4.1 together with monotonicity of the integral, ${\mu }_g(A)\le R^2{\lambda }_2({\mathrm{\Psi }}_i(A))$. Consequently ${\mu }_g(L\cap U_0)={\mu }_g(L\cap U_1)=0$ by step 3.3 and [F8], and since $L=(L\cap U_0)\cup (L\cap U_1)$ with ${\mu }_g$ a measure, ${\mu }_g(L)\le {\mu }_g(L\cap U_0)+{\mu }_g(L\cap U_1)=0$, so ${\mu }_g(L)=0$. The same bound with $A=\{q\}$ shows that every point $q$ of ${\mathbb{R}\mathbb{P}}^2$ is ${\mu }_g$-null, since some chart $U_i$ contains $q$ and the image of a point under ${\mathrm{\Psi }}_i$ is a Lebesgue-null singleton.

The two sets $U_2=\pi (H_2)$ and $L={\mathbb{R}\mathbb{P}}^2\setminus U_2$ are disjoint and exhaust ${\mathbb{R}\mathbb{P}}^2$, so additivity of the measure and step 5.2 give ${\mu }_g({\mathbb{R}\mathbb{P}}^2)={\mu }_g(U_2)+{\mu }_g(L)={\mu }_g(U_2)$.

By steps 5.2 and 4.1, ${\mu }_g(U_2)={\int }_{{\mathrm{\Psi }}_2(U_2)}{\rho }_{2}d{\lambda }_2={\int }_{{\mathbb{R}}^2}{R}^2(1+a^2+b^2{)}^{-3\mathrm{/}2}dadb$. The integrand is radial, so the polar-coordinate formula [F6] evaluates it as ${\mu }_g(U_2)=R^2\sigma (S^1){\int }_0^{\mathrm{\infty }}r(1+r^2{)}^{-3\mathrm{/}2}dr=R^2\cdot 2\pi \cdot [-(1+r^2{)}^{-1\mathrm{/}2}{]}_0^{\mathrm{\infty }}=2\pi R^2$, the antiderivative being checked by differentiation and its limit at infinity being $0$. With step 6.1 this gives ${\mathrm{area}}_g({\mathbb{R}\mathbb{P}}^2)={\mu }_g({\mathbb{R}\mathbb{P}}^2)=2\pi R^2$.

Since $\{P_0,P_1,P_2\}$ is ${\mu }_g$-null by step 5.2 and $K=1\mathrm{/}{R}^2$ off that set by step 5.1, the integrands $K$ and the constant $1\mathrm{/}{R}^2$ agree ${\mu }_g$-almost everywhere; by [F8] their integrals against ${\mu }_g$ coincide, and the constant is integrable because ${\mu }_g$ is finite on the compact surface ${\mathbb{R}\mathbb{P}}^2$ with total mass $2\pi R^2$ from step 7.1. Hence ${\int }_{{\mathbb{R}\mathbb{P}}^2}K{\mu }_g=R^{-2}{\mu }_g({\mathbb{R}\mathbb{P}}^2)=2\pi$.

The surface ${\mathbb{R}\mathbb{P}}^2$ is closed and compact, and $\pi$ is a two-to-one local isometry from the connected sphere, so ${\mathbb{R}\mathbb{P}}^2$ is nonorientable by [F9]; [F10] therefore applies to $(M,g)=({\mathbb{R}\mathbb{P}}^2,g)$ and gives ${\int }_{{\mathbb{R}\mathbb{P}}^2}K{\mu }_g=2\pi \chi ({\mathbb{R}\mathbb{P}}^2)$. Comparing with step 8.1 yields $\chi ({\mathbb{R}\mathbb{P}}^2)=1$, and the total curvature is $2\pi$. The full-choice assumption entered only through [F10] and the polar formula of [F6].