Preimage is $2$-dimensional torus?

Question

Let $H$ be a $3$-dimensional sphere in $\mathbb{R}^4$ centered at the origin an with radius $\sqrt{2h}$, with $h >0$. Denote by $S^2(\frac{h}{2})$ the 2-sphere of radius $\frac{h}{2}$ in $\mathbb{R}^3$ centered at the origin. Consider the map $$\pi: H(h) \to S^2(\frac{h}{2}):(p_1,p_2,q_1,q_2) \mapsto (I,J,K)$$ where $(I, J, K)=(\frac{1}{2}(p_1p_2+q_1q_2), \frac{1}{2}(q_2p_1-p_2q_1),\frac{1}{2}(p_1^2-p_2^2)+\frac{1}{2}(q_1^2-q_2^2))$. How does one show that the preimage $\pi^{-1}(S^2(\frac{h}{2}) \cap \lbrace K=k \rbrace)$ is a $2$-dimensional torus if $k \in (-\frac{h}{2},\frac{h}{2})$.

Answer 1

Note that we have $p_1^2-p_2^2+q_1^2-q_2^2 = 2k$ and $p_1^2+p_2^2+q_1^2+q_2^2 = 2h$. Let $x=(p_1,q_1)$ and $y=(p_2,q_2)$. Then you have $\|x\|^2+\|y\|^2=2h$ and $\|x\|^2-\|y\|^2 = 2k$. We infer that $\|x\|^2 = h+k$ and $\|y\|^2=h-k$. There's your $S^1\times S^1$. (I'm not sure why we have $|k|<h/2$ rather than $|k|<h$, but I haven't checked other details.)