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In analogy with the equator of a 2-sphere (parametrized by 2 angles) being a 1-sphere (parametrized by one of them), js the equator of a 3-sphere (3 angles) a 2-sphere?

TheQuantumMan
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  • Yep.... it is... – David G. Stork Feb 04 '19 at 04:40
  • This should be easy to calculate, using the equation $x^2+y^2+z^2+w^2=1$ for a $3$-sphere sitting in Euclidean $4$-space. – Lubin Feb 04 '19 at 04:45
  • @Lubin I was not sure whether or not the equator is always considered to be a 1D subspace of an n-sphere or an (n-1)-dimensional subspace of it – TheQuantumMan Feb 04 '19 at 04:47
  • Well, I guess that is a matter of definitions, but I would suppose that the equator is something that divides the $3$-sphere into two equal parts: one on one side of it, the other part on the other side. That wouldn’t happen with a $1$-dimensional subspace. I suppose it’s a case where you pays your money and takes your choice. – Lubin Feb 04 '19 at 05:03

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We can define $S^2\subseteq \mathbb{R}^3$ as the vanishing set of $f(x,y,z)=x^2+y^2+z^2-1$. By equator you likely mean the intersection of $S^2$ with $P=\{(x,y,z)\in \mathbb{R}^3: z=0\}$ (for instance). In this case, the equator is the set of points satisfying $x^2+y^2+z^2=1$ and $z=0$, i.e. $x^2+y^2=1$. So, the equator in this case is a copy of $S^1$, the circle.

In the case of $S^3\subseteq \mathbb{R}^4$, we write $S^3$ as the vanishing set of $F(w,x,y,z)=w^2+x^2+y^2+z^2-1$, and repeat the same reasoning. We probably want to define the equator as the intersection of $S^3$ with $P'=\{(w,x,y,z): z=0\}$, which is the set of points cut out by $w^2+x^2+y^2=1$, namely a copy of $S^2$.

Alekos Robotis
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