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Let $S^3 = \{(z_1, z_2) \in \mathbb{C}^2 : |z_1|^2 + |z_2|^2 = 1\}$ and $S^1 = \{z \in \mathbb{C} : |z| = 1\}$. Let $S^1$ act on $S^3$ by $z \cdot (z_1, z_2) = (zz_1, \overline{z}z_2)$. Let $X = S^3/S^1$. Consider $\tilde{f}, \tilde{g}, \tilde{h} : S^3 \rightarrow \mathbb{R}$ given by $$\tilde{f}(z_1, z_2) = z_1z_2 + \overline{z_1z_2}$$ $$\tilde{g}(z_1, z_2) = i(z_1z_2 - \overline{z_1z_2})$$ $$\tilde{h}(z_1, z_2) = |z_1|^2 - |z_2|^2$$

Show that $\tilde{f}, \tilde{g}, \tilde{h}$ induce functions $f, g, h : X \rightarrow \mathbb{R}$. If I am not mistaken, the orbit of a point $(z_1, z_2)$ consists of all points $(z_1^*, z^*_2)$ such that $|z_1^*| = |z_1|$ and $|z_2^*| = |z_2|$. Let $[(z_1, z_2)]$ denote the equivalence class of $(z_1, z_2)$. Would it be correct to say that $\tilde{f}: S^3 \rightarrow \mathbb{R}$ induces the function $f: X \rightarrow \mathbb{R}$ given by $f([(z_1, z_2)]) = 2\operatorname{Re} (z_1z_2)$, or have I completely misunderstood what I am supposed to do? (Note that the last part comes from letting $z_1 = x_1 + iy_1$ and $z_2 = x_2 + iy_2$ and then expanding $z_1z_2 + \overline{z_1z_2}$ and finding that this is just $2(x_1x_2 - y_1y_2) = 2\Re(z_1z_2))$

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    Your description of the orbit of $(z_1,z_2)$, as consisting of all points $(z_1^,z_2^)$ with $|z_1^|=|z_1|$ and $|z_2^|=|z_2|$, seems wrong. For example, it would put $(1,-1)$ into the orbit of $(1,1)$, whereas in fact the orbit of $(1,1)$ consists only of the points $(z,\overline z)$. – Andreas Blass Jun 03 '13 at 13:45

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Hint: Since $X$ is a quotient space and you want to define a function on $X$ induced by $\tilde f$ you should show that $\tilde f$ coincides on every point of the orbit of $(z_1,z_2)$ and so you can choose any point of the orbit to define $ f$ on it.

More precisely, you have to show that $\tilde f(z_1,z_2)=\tilde f(z\cdot (z_1,z_2))$. So you have a well defined function $f$ on the equivalence classes.

Sigur
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