# Locating the quadrant containing a point on an n-sphere

Suppose I have a point $x \in \mathbb{R}^n$ on an n-sphere. Suppose I divide the n-sphere into 4 sections (I think this makes sense in $n$ dimensions), how do I know which section $x$ lies on?

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No, it doesn't make sense in n dimensions unless n=2. What are the 4 quadrants in 3 dimensions? You can talk of the 8 octants instead, and the 2^n orthants in n dimensions (it's an $(n-1)$-sphere, BTW), and if that's what you want, then the answer to your question is simply to look at the signs of the coordinates in x. – ShreevatsaR Aug 5 '10 at 20:13
@Shreevatsa does it make sense to say "if I split a sphere into 8 equal hemispheres" ? – Jonathan Fischoff Aug 5 '10 at 20:16
@Shreevatsa: In 3 dimensions, you have 2 axis aligned planes dividing the sphere into 4 quadrants (one plane divides it into two hemispheres and the other divides each hemisphere). – Jacob Aug 5 '10 at 20:17
@Shreevatsa: Also, is (n-1) in LaTeX? If so, how did you format it? I couldn't find any formatting tips regarding LaTeX! – Jacob Aug 5 '10 at 20:19
@Jacob: Actually if you wait for 4 seconds the equations will appear. – kennytm Aug 5 '10 at 20:24

This is just a rephrasing of ShreevastasR's answer; no credit to me. It does make sense to divide an $n$-sphere into quadrants, as you explain in $\mathbb{R}^3$: partition by two coordinate planes. But then deciding which quadrant is, as ShreevastasR says, simply looking at the signs of the coordinates of $x$. If $x_1$ and $x_2$ are both positive, you are in the first, $++$, quadrant; if $x_1$ is negative and $x_2$ positive, you are in the second, $-+$, quadrant. And so on. If instead you partition the sphere into $2^n$ orthants, then you consider all the signs of the coordinates of $x$.
My understanding from you question is that you have $x$ in your hands, presumably in coordinate form. So you look at the first coordinate, the second, the third, etc. I guess I don't understand your question! – Joseph O'Rourke Aug 5 '10 at 21:20