Boy's surface: an immersion of the real projective plane in three-space

Put $d(w)=D_6(w)$. Its roots satisfy $w^3=u_{\pm }=(-\sqrt{5}\pm 3)\mathrm{/}2$. Exactly the three cube roots of $u_{+}$ lie in the open unit disk, and all are simple since $d^{\mathrm{\prime }}(w)=3w^2(2w^3+\sqrt{5})=9w^2$ there. Put $A=w\mathrm{/}d$, $B=w^5\mathrm{/}d$ and $C=(1+w^6)\mathrm{/}d$. If $g_1=g_2=0$ then $\mathrm{Im}(A-B)=0$ and $\mathrm{Re}(A+B)=0$, hence $B=-\bar{A}$. For $w\mathrm{\ne }0$ their absolute values force $\mathrm{\mid }w\mathrm{\mid }=1$. On that circle, writing $w=e^{i\theta }$ gives $C=2\cos 3\theta \mathrm{/}(\sqrt{5}+2i\sin 3\theta )$, so $\mathrm{\mid }\mathrm{Im}C\mathrm{\mid }=\mathrm{\mid }4\cos 3\theta \sin 3\theta \mathrm{\mid }\mathrm{/}(5+4{\sin }^{2}3\theta )\le 2\mathrm{/}5<1\mathrm{/}2$. Thus $g_3=\mathrm{Im}C-1\mathrm{/}2\mathrm{\ne }0$. At $w=0$ it is $-1\mathrm{/}2$. Consequently $g$ never vanishes at a finite non-pole.

Write $g=\mathrm{Re}h$, where $h=(\frac{3}{2}i(A-B),-\frac{3}{2}(A+B),-iC-\frac{1}{2})$ is holomorphic off the poles. Differentiation gives $h^{\mathrm{\prime }}=d^{-2}(\frac{3}{2}i{N}_1,-\frac{3}{2}{N}_2,-i{N}_3)$, with $N_1=w^{10}-2\sqrt{5}{w}^7-5w^6+5w^4-2\sqrt{5}{w}^3-1$, $N_2=-w^{10}+2\sqrt{5}{w}^7-5w^6-5w^4-2\sqrt{5}{w}^3-1$, and $N_3=3w^2(\sqrt{5}{w}^6-4w^3-\sqrt{5})$. The identities $N_2-N_1=-2w^4(w^3-\sqrt{5}{)}^2$ and $N_2+N_1=-2(\sqrt{5}{w}^3+1{)}^2$ give $\frac{9}{4}(N_2^2-N_1^2)=N_3^2$, hence $h^{\mathrm{\prime }}\cdot h^{\mathrm{\prime }}=0$ for the complex bilinear dot product. The first two numerators never vanish simultaneously: at $w=0$ both equal $-1$, and otherwise their sum and difference would require both $w^3=\sqrt{5}$ and $w^3=-1\mathrm{/}\sqrt{5}$. Thus $h^{\mathrm{\prime }}\mathrm{\ne }0$. Its real and imaginary parts are orthogonal and have the same positive norm, so the real derivatives $g_x=\mathrm{Re}{h}^{\mathrm{\prime }}$ and $g_y=-\mathrm{Im}{h}^{\mathrm{\prime }}$ are independent. Inversion $J(v)=v\mathrm{/}\mathrm{\mid }v{\mathrm{\mid }}^2$ has derivative $\mathrm{\mid }v{\mathrm{\mid }}^{-2}(I-2v{v}^{\ast }\mathrm{/}\mathrm{\mid }v{\mathrm{\mid }}^2)$, an invertible scaled reflection. By step 1.1, $P=J\circ g$ therefore has injective differential away from the poles.

At each interior pole $w_j$, write $u=w-w_j$ and $h=c\mathrm{/}u+h_0(u)$. The residue vector $c$ is nonzero because the second numerator $w_j(1+w_j^4)$ is nonzero ($0<\mathrm{\mid }{w}_j\mathrm{\mid }<1$). The leading term in $h^{\mathrm{\prime }}\cdot h^{\mathrm{\prime }}=0$ gives $c\cdot c=0$. Thus $\mathrm{Re}c$ and $\mathrm{Im}c$ are independent, orthogonal and have common squared norm $\lambda >0$. Set $N(u)=\mathrm{Re}(c\bar{u})$ and $H(u)=\mathrm{Re}{h}_0(u)$; then $\mathrm{\mid }N(u){\mathrm{\mid }}^2=\lambda \mathrm{\mid }u{\mathrm{\mid }}^2$ and $g=N(u)\mathrm{/}\mathrm{\mid }u{\mathrm{\mid }}^2+H(u)$. It follows that $P=(N(u)+\mathrm{\mid }u{\mathrm{\mid }}^{2}H(u))\mathrm{/}(\lambda +2⟨N(u),H(u)⟩+\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{\mid }H(u){\mathrm{\mid }}^2)$ extends real-analytically to $u=0$, with value $0$ and differential $N\mathrm{/}\lambda$, which is injective. This verifies all three ends, including the two nonreal poles.

For $t=-1\mathrm{/}\bar{w}$, direct substitution gives $d(t)=-\overline{d(w)}\mathrm{/}{\bar{w}}^6$, $(t-t^5)\mathrm{/}d(t)=-\overline{(w-w^5)\mathrm{/}d(w)}$, $(t+t^5)\mathrm{/}d(t)=\overline{(w+w^5)\mathrm{/}d(w)}$, and $(1+t^6)\mathrm{/}d(t)=-\overline{(1+w^6)\mathrm{/}d(w)}$. Taking the indicated real and imaginary parts proves $g(t)=g(w)$ and $P(t)=P(w)$; on $\mathrm{\mid }w\mathrm{\mid }=1$ this is $P(-w)=P(w)$. Near the boundary these identities hold on a two-sided annulus, not merely on the circle. At infinity use the coordinate $t=-1\mathrm{/}\bar{w}$ near $0$; the same identity gives a smooth immersive extension there. Thus the extended map on the Riemann sphere is invariant under its free antipodal involution and descends through the local quotient charts to a smooth immersion $\beta :{\mathbb{R}\mathbb{P}}^2\to {\mathbb{R}}^3$. The disk model in [F2] is a fundamental domain for this involution.

The three interior poles are distinct points of the projective-plane domain: their antipodes lie outside the disk. All have image $0$ by step 3.1, so $\beta$ has a triple point and is not injective. This is the Bryant–Kusner Boy surface; Karcher's source, PDF p.2, describes the three antipodal pairs of planar ends and their common image after inversion. The formulas above verify its immersedness directly.

For any immersion $f:{\mathbb{R}\mathbb{P}}^2\to {\mathbb{R}}^3$, a global nonzero normal field would orient each tangent plane by the sign of $\det (d{f}_x{v}_1,d{f}_x{v}_2,n(x))$, continuously and consistently. This contradicts [F2], so its normal line is nontrivial by [F4]. For the characteristic-class description assume AC as in [F5]. The ambient volume form gives $\det (T{\mathbb{R}\mathbb{P}}^2)\otimes {\nu }_f\cong {\epsilon }^1$, so [F5] yields $w_1({\nu }_f)=w_1(T{\mathbb{R}\mathbb{P}}^2)\mathrm{\ne }0$. To identify this with the tautological class, represent a point by a unit $x\in S^2$ and tangent vectors by $v,w\in x^{\perp }$; the map $v\wedge w\mapsto \det (x,v,w)x$ identifies the determinant tangent line with the tautological line. It is unchanged under $(x,v,w)\mapsto (-x,-v,-w)$ and is a fibrewise isomorphism. Thus their $w_1$ classes agree by [F5], giving precisely the tautological degree-one class.