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On the unit sphere, use $\zeta =(x+iy)/(1-z)$ away from the north pole and $\eta =(x-iy)/(1+z)$ away from the south pole. Their overlap relation $\eta =1/\zeta$ is a holomorphic map, giving a holomorphic atlas for the Riemann sphere. The opposite sign of $iy$ in the second manifold chart ensures a holomorphic rather than antiholomorphic transition.

There is a small domain issue in the printed notation. If ${\mathbb{R}}^3/{\mathbb{R}}^2$ means a quotient vector space by the horizontal plane, the displayed formula does not descend to that quotient vector space: $(0,0,0)$ and $(1,0,0)$ represent the same class but give different values. The intended construction is stereographic projection, restricted to the unit sphere with the north pole removed. Interpreting the slash as removal of the plane $z=1$ also supplies a suitable ambient domain. The holomorphic stereographic atlas of the sphere is obtained as follows.

Write $N=(0,0,1)$ and $S=(0,0,-1)$. On $U_N=S^2\setminus \{N\}$ use $\zeta =(x+iy)/(1-z)$. Its inverse, with $\zeta =u+iv$, isThese formulas give a smooth manifold chart from $U_N$ onto $\mathbb{C}$. On $U_S=S^2\setminus \{S\}$ choose the second manifold chartIts inverse is $x=2\mathrm{Re}\eta /(1+|\eta {|}^2)$, $y=-2\mathrm{Im}\eta /(1+|\eta {|}^2)$ and $z=(1-|\eta {|}^2)/(1+|\eta {|}^2)$, so this is also a smooth manifold chart onto $\mathbb{C}$. The conjugation in this second stereographic projection is essential. On the overlap, $x^2+y^2=1-z^2$, soBoth directions of this transition are holomorphic maps on ${\mathbb{C}}^{\times }$, with nonzero derivative. The two manifold charts cover the sphere, hence define a holomorphic atlas, giving precisely the Riemann sphere. If one instead used $x+iy$ in both manifold charts, the transition would be $1/\bar{\zeta }$ and would not be holomorphic.

Orient the sphere by this holomorphic atlas. The given volume form is smooth at infinity: replacing $\zeta$ by $1/\eta$ in its exterior product gives the same expressionMoreover, $id\zeta \wedge d\bar{\zeta }=2du\wedge dv$, so the normalization of this volume form isThus the printed volume form has half the area of the standard round unit sphere; replacing its integral by $4\pi$ would introduce an erroneous factor of two.

For $k\ge 1$, the holomorphic map $f(\zeta )={\zeta }^k$ extends over infinity, since the target reciprocal coordinate is ${\eta }^k$ when the source reciprocal coordinate is $\eta$. Its pullback of a differential form isUsing $s=r^{2k}$ givesThe degree of a map between oriented manifolds is thereforeFor $k=0$, the formula on the finite manifold chart is the constant $1$. Its unique continuous extension is also $1$ at infinity, rather than an undefined expression ${\infty }^0$. Its pullback of a differential form is zero and its degree of a map between oriented manifolds is zero.

For the preimage calculation when $k\ge 1$, choose a regular value $w\in {\mathbb{C}}^{\times }$. There are exactly $k$ distinct roots of ${\zeta }^k=w$. At each root the real Jacobian determinant is $|k{\zeta }^{k-1}{|}^2>0$, so every local contribution to the degree of a map between oriented manifolds is $+1$. Their sum is $k$, agreeing with the integral. The exceptional values $0$ and infinity are avoided because they are branch values when $k>1$. For the constant map, any $w\ne 1$ is a regular value with no preimages, giving the same answer zero. This establishes the degree of a power map of the Riemann sphere for every allowed $k$.

A holomorphic atlas on a real smooth manifold of dimension $2n$ consists of compatible smooth charts ${\phi }_a:U_a\to V_a\subseteq {\mathbb{C}}^n$ whose transition maps ${\phi }_b{\phi }_a^{-1}$ are biholomorphic on their domains. An almost complex structure is a smooth real vector bundle endomorphism $J:TX\to TX$ with $J^2=-\mathrm{id}$. In each chart defineThe differential of a holomorphic map is a complex-linear map, so the transition differentials commute with multiplication by $i$. Consequently the local definitions agree on overlaps. They are smooth and square to $-\mathrm{id}$, yielding the natural almost complex structure induced by a complex atlas. This construction uses the atlas, rather than an arbitrary choice of coordinates on the underlying real smooth manifold.

Extend $J$ complex-linearly to the complexification of a real vector bundle $T_{\mathbb{C}}X=TX{\otimes }_{\mathbb{R}}\mathbb{C}$. Since $J^2=-\mathrm{id}$, its $+i$ and $-i$ eigenbundles have smooth projectionsEach has complex rank $n$: complex conjugation interchanges them, and together they have rank $2n$. These are the type decomposition of the complexified tangent bundle. For an arbitrary almost complex structure, they are smooth complex vector bundles; a holomorphic vector bundle structure requires integrability.

When $J$ comes from a holomorphic atlas, write a holomorphic coordinate as $z_j=x_j+i{y}_j$. Its Wirtinger derivatives areThe induced almost complex structure has $J{\partial }_{x_j}={\partial }_{y_j}$ and $J{\partial }_{y_j}=-{\partial }_{x_j}$. Hence the displayed vector fields are respectively $+i$ and $-i$ eigenvectors. They are linearly independent, and each collection has $n$ elements. Thus they give local frames for $T^{1,0}X$ and $T^{0,1}X$ respectively, with $T^{1,0}X$ the holomorphic tangent bundle. The repeated $T^{0,1}X$ in the first sentence of the printed item must be read as the two complementary eigenbundles.

With the normalization used here, $N=2N_J$, where $N_J$ is the standard Nijenhuis tensor. First check that it is a tensor: the Lie bracket identity $[fU,V]=f[U,V]-V(f)U$ givesSkew symmetry gives the same $C^{\infty }$-linearity in the second input. In coordinates from a holomorphic atlas, the real coordinate vector fields ${\partial }_{x_j},{\partial }_{y_j}$ commute, and $J$ sends each to another such field with a constant sign. Every Lie bracket in $N_J$ therefore vanishes on these coordinate fields. Tensoriality then gives $\boxed{N=0}$ on all smooth vector fields. The factor two in this paper does not change this vanishing conclusion.