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Preimage is $2$-dimensional torus?

Discussion in 'Mathematics' started by J.Bosser, Oct 8, 2018.

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  1. J.Bosser

    J.Bosser Guest

    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 $k \in (-\frac{h}{2},\frac{h}{2})$.

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