Prove that Hopf maps on $S^3, S^2$ and $\mathbb{C} \mathbb{P}^1$ are smooth submersions

Endow $S^3$ with its standard smooth structures from the stereographic projections, i.e. $$\chi_1:S^3 \setminus \{(0,0,0,1) \} \to \mathbb{R}^3, \chi(x,y,z,t) = \frac{1}{1-t}(x,y,z) $$ and $$\chi_2: S^3 \setminus \{(0,0,0,-1) \} \to \mathbb{R}^3, \chi_2(x,y,z,t) = \frac{1}{1+t}(x,y,z).$$

Do the same for $S^2$, call the charts $\eta_1, \eta_2$ (in the same order as the ones for $S^3$). Now, consider the complex projective space $\mathbb{C} \mathbb{P}^1$ endowed with the smooth structure given by the charts $\{(U_0, \varphi_0), (U_1, \varphi_1)\},$ $$U_0 = \{[z_0:z_1] \ \mid \ z_0 \neq 0 \} \text{ and } U_1 = \{[z_0: z_1] \ \mid \ z_1 \neq 0 \},$$ where $[x:y]$ represents the complex line through the point $(x,y)$ and $0$ in $\mathbb{C}^2$, and $$\varphi_0([z_0: z_1]) = \frac{z_1}{z_0} \text{ and } \varphi_1([z_0: z_1]) = \frac{z_0}{z_1}. $$

Now define the Hopf maps: $$h: S^3 \to S^2, h(x,y,z,t) = (x^2+y^2-z^2-t^2, 2(yz-xt), 2(xz+yt)) $$ and $$H:S^3 \to \mathbb{C} \mathbb{P}^1, H(x,y,z,t) = [(x+iy):(z+it)]. $$

Is there an easier way to prove that $h$ is a well-defined smooth submersion and that $H$ is a smooth submersion (with respect to the respective smooth structures)? By this I mean we can of course compute $\eta_i \circ h \circ \chi_j^{-1}, \forall i,j \in \{1,2 \}$ and $\varphi_k \circ H \circ \chi_l^{-1}, \forall k,l \in \{1,2 \}$ and prove that they are smooth (in the classical real-analysis sense) and that they are submersions, but this is a very tedious process.

So, are there easier ways to do this? Will this maybe be easier if we switch the smooth structure of the unit spheres from stereographic projections to, say, normal projections (though we increase the number of combinations of charts to verify from 4 to 48), since they induce the same smooth structure?

Solution 1:

All you need to know is that $S^{n-1}$ is a smooth submanifold of $\mathbb R^n$ and thereby receives a canonical smooth structure. It is well-known that this smooth structure is also generated by the smooth atlas given by your two stereographic projections $\chi^{n-1}_N$ and $\chi^{n-1}_S$ from north pole $N$ and south pole $S$. There are also other smooth atlases which are simpler, but have more charts. As an example take the $2n$ charts living on $U_i^\epsilon = \{ (x_1,\dots,x_n) \in S^{n-1} \mid (-1)^\epsilon x_i > 0 \}$, with $i = 1,\dots,n$ and $\epsilon =\pm 1$, and projecting $U_i^\epsilon$ onto the standard open $(n-1)$-disk by omitting the $i$-th coordinate.

Let us observe that the "stereographic atlas" $\mathcal S_{n-1}$ can be modified by choosing any linear automorphism $L : \mathbb R^{n-1} \to \mathbb R^{n-1}$ and defining $\mathcal S^L_{n-1} =\{L \circ \chi^{n-1}_N, \chi^{n-1}_S \}$. This atlas clearly gives the same smooth structure.

Now write $\mathbb R^4 = \mathbb C^2$ and $\mathbb R^3 = \mathbb C \times \mathbb R$. Then $S^3 \subset \mathbb C^2$ and $S^2 \subset \mathbb C \times \mathbb R$. Let us redefine the Hopf map $h : S^3 \to S^2$ by $$h(z_0,z_1) = (2\bar z_0 z_1,\lvert z_0 \rvert^2 - \lvert z_1 \rvert^2 ). \tag{1}$$ Note that this differs from your definition by the coordinate permutation $\pi : S^2 \to S^2, \pi(x_1,x_2,x_3) = (x_3,x_2,x_1)$. The latter is a diffeomorphism (it is the restriction of the corresponding coordinate permutation on $\mathbb R^3$ to the submanifold $S^2$). Clearly our new $h$ is a smooth submersion iff the original $h$ is one. The purpose of this redefinition was to obtain the above elegant formula for $h(z_0,z_1)$.

Now define

$$\rho : \mathbb{CP}^1 \to S^2, \rho([z_0:z_1]) = \frac{1}{\lvert z_0 \rvert^2 + \lvert z_1 \rvert^2}(2\bar z_0 z_1,\lvert z_0 \rvert^2 - \lvert z_1 \rvert^2 ) . $$ This is a well-defined continuous map. Note that $\rho([1:0]) = N$. We shall show that it is a bijection, hence a homeomorphism because $\mathbb{CP}^1$ is compact and $S^2$ Hausdorff. Define $$\sigma : S^2 \to \mathbb{CP}^1,\sigma(z,t) = \begin{cases} [\bar z:1-t] & (z,t) \ne N \\ [1:0] & (z,t) = N \end{cases}$$ Note that for $(z,t) \in S^2 \subset \mathbb R \times \mathbb R$ we have $t = 1$ iff $(z,t) = (0,1) = N$ = north pole of $S^2$. Also observe that each element of $\mathbb{CP}^1 $ has a representative of the form either $[z_0:1]$ or $[1:0]$. Using this it is easy to verify that $\sigma \circ \rho = id$ and $\rho \circ \sigma = id$.

On $S^2$ let us consider the smooth atlas $\mathcal S'_2 = \{L \circ \chi^2_N, \chi^2_S \}$, where $L$ is complex conjugation. It is easy to check that $$L \circ \chi^2_N \circ \rho = \varphi_1 \quad, \quad \chi^2_S \circ \rho = \varphi_0 .$$ This shows that $\rho$ is a diffeomorphism. By construction of $\rho$ we have $\rho \circ H = h$. It therefore suffices to show that one of $H$ and $h$ is a smooth submersion.

For $h$ have a look at Proving that the Hopf map $S^3 \to S^2$ is a submersion.

For $H$ it is easier. Let $S^3_k = \{(z_0,z_1) \in S^3 \mid z_k \ne 0 \}$, $k= 0,1$. These are open subsets of $S^3$ which cover $S^3$. Consider the maps $$H_k : S^3_k \stackrel{H}{\to} U_k \stackrel{\varphi_k}{\to} \mathbb C$$ which are given by $H_0(z_0,z_1) = \dfrac{z_1}{z_0}$ and $H_1(z_0,z_1) = \dfrac{z_0}{z_1}$. It suffices to show that these are submersions (smoothness is obvious). We shall do that for $H_0$, the other case is similar.

Given a point $(z_0,z_1) \in S^3_0$, define $$\phi : \mathbb C \to \mathbb C^2, \phi(w) = \frac{1}{\lVert (z_0, z_0 w) \rVert}(z_0,z_0 w) .$$ This is a smooth map with image contained in $S^3_0$. Since $S^3_0$ is a submanifold of $\mathbb C^2$, the map $$\psi : \mathbb C \stackrel{\phi}{\to} S^3_0$$ is smooth. We have $H_0 \circ \psi = id$ and therefore the differential of $H_0$ must be surjective in all points of $\psi(\mathbb C)$. But clearly $(z_0,z_1) = \psi(\dfrac{z_1}{z_0}) \in \psi(\mathbb C)$.