How Can I solve this problem on Hopf Map?

Question

I want to solve this:

Let $\pi : S^3\to S^2$ be the Hopf map. Show that for all $a\in (-1,1)$, the pre-image $\pi ^{-1} (Z_a)\subset S_3$ is the common solution set of the equations $|z|^2 + |w|^2 = 1$, and $|z|^2 - |w|^2 = a$. Use an explicit parametrization of the solution set to conclude that $\pi ^{-1}(Z_a)$ is a 2- torus (i.e. it is $S^1\times S^1$).

Note: Take $Z_a = \{(x_0,x_1,x_2)\in S^2|x_2=a\}$

Attempt

The Hopf map is given by:

$$\pi : S^3\to S^2,\quad \pi (z,w) = \frac{1}{|z|^2 + |w|^2} (2\Re(w,\overline{z}),2\Im(w\overline{z}), |z|^2 - |w|^2)$$

Now, we have: $|z|^2 + |w|^2 = 1$ and $|z|^2 - |w|^2 = a$. By adding these equations together, we will get $|z|^2 = \frac{1+a}{2}$. If we let $z = x + iy$, then we will get $$x^2 + y^2 = \frac{1+a}{2}\quad\color{blue}\blacktriangle$$ By subtracting the second equation from the first, and doing the substitution $ w = u + iv$, we will get $$u^2 + v^2 = \frac{1-a}{2}\quad \color{red}\ast$$ By adding $\color{blue}\blacktriangle$ and $\color{red}\ast$ together, we will have:

$$x^2 + y^2 + u^2 + v^2 = 1$$

Questions:

This is how far I could get. So:

(1) how does this answer the question, "for all $a\in (-1,1)$, show that the pre-image $\pi ^{-1} (Z_a)$ is the common solution set of the equations $|z|^2 + |w|^2 = 1$, and $|z|^2 - |w|^2 = a$"?

(2) how can I use an explicit parametrization of the solution set to conclude that $\pi ^{-1}(Z_a)$ is a$2$-torus (i.e. it is $S^1\times S^1$)?

Answer 1

Haven't you answered both questions already?

For (1), you even wrote the conditions out yourself.

For (2), $|z|^2=\frac{1+a}2$ and $|w|^2=\frac{1-a}2$ define two circles and hence the solution set is the product as a 2-torus. (If it is not clear to you yet, $S^3\subset {\mathbb C}^2$ and $z$ and $w$ are the coordinates on them.)

Let me add a bit more than what the question asks for, since it is interesting. We can ask for the the inverse image $\pi^{-1}(p_\theta)$ of a point $$ p_\theta = (\sqrt{1-a^2}\cos\theta, \sqrt{1-a^2}\sin\theta, a)\in Z_a\subset S^2, $$ where $\theta\in [0, 2\pi]$. Then the condition would be that for $(z, w)\in \pi^{-1}(Z_a)$ with $$ z = \sqrt{\frac{1+a}2}e^{i\phi},\quad w=\sqrt{\frac{1-a}2}e^{i\psi}, $$ we should have $$ \phi - \psi = \theta. $$

Therefore the inverse image of a point is a circle on a torus $T^2=S^1\times S^1$ with coordinates $(\phi, \psi)$ defined by that their difference is a constant.

These are parallel lines with $45^\circ$ angles if you represent your torus by a square. On the torus, they are circles linking with each other. Here is a picture.

Answer 2

The Hopf map is given by

$$\pi : S^3\to S^2,\quad \pi (z,w) = (2\Re(w\overline{z}),2\Im(w\overline{z}), |z|^2 - |w|^2 ) .$$

Thus for all $a \in [-1,1]$ $$\pi ^{-1}(Z_a) = \{(z,w) \in S^3 \mid|z|^2 - |w|^2 = a\} = \{(z,w) \in \mathbb C^2 \mid|z|^2 + |w|^2 = 1, \ z|^2 - |w|^2 = a\} .$$ This shows that $\pi ^{-1} (Z_a)$ is the common solution set of the equations $|z|^2 + |w|^2 = 1$ and $|z|^2 - |w|^2 = a$, where $z, w \in \mathbb C$.

With $z = x + iy$ and $w = u + iv$ you correctly get the equations $$x^2 + y^2 = \frac{1+a}{2} , \tag{1}$$ $$u^2 + v^2 = \frac{1-a}{2} . \tag{2} $$ For $a = 1$ we get $u = v = 0$ and $x^2 + y^2 = 1$, thus $$\pi ^{-1}(Z_1) = S^1 \times \{0\}.$$

For $a = -1$ we get $x= y = 0$ and $u^2 + v^2 = 1$, thus $$\pi ^{-1}(Z_{-1}) = \{0\} \times S^1.$$

For $a \in (-1,1)$ both $r_1 = \frac{1+a}{2}$ and $r_2 = \frac{1-a}{2}$ lie in $(0,1)$, thus $(1)$ and $(2)$ desrcibe circles $S^1_{r_1}$ and $S^1_{r_2}$ with center $0$ and radius $r_1$ and $r_2$, respectively. Hence $$\pi ^{-1}(Z_a) = S^1_{r_1} \times S^1_{r_2} .$$ This means that $\pi ^{-1}(Z_a)$ is a $2$-torus. It degenerates to a circle for $a = \pm 1$.