# How many “equators” and “poles” 4-sphere has?

I mean 3-sphere (normal, like Earth) has 3 euators: namely equator, 0h meridian circle and 6h meridian circle.

So, "pole" is a point, where all coordinates equal zero, except one, which equals to sphere radius.

Can't factor out, how many such equators 4-sphere has?

On 3-sphere each equator intersects with 2 other equators in 4 poles. In each pole 2 equators intersect.

On 4-sphere there should be 3 equators intersecting in a pole. These 3 equators should also intersect at opposite pole.

So we have

$E_1=\{P_1, \bar{P_1},...\}$

$E_2=\{P_1, \bar{P_1},...\}$

$E_3=\{P_1, \bar{P_1},...\}$

where equator $E_i$ is represented with a set of poles it contains, while pole is denoted by $P_j$, having $\bar{P_j}$ as opposite pole.

There should be at least one more equator, which intersects with three previous:

$E_4=\{P_2, \bar{P_2}, P_3, \bar{P_3}, P_4, \bar{P_4},...\}$

poles $P_2...P_4$ should be on previous equators, so we have

$E_1=\{P_1, \bar{P_1}, P_2, \bar{P_2}, ...\}$

$E_2=\{P_1, \bar{P_1}, P_3, \bar{P_3}, ...\}$

$E_3=\{P_1, \bar{P_1}, P_4, \bar{P_4},...\}$

What we should have at ellipsis? Seems that it should be

$E_1=\{P_1, \bar{P_1}, P_2, \bar{P_2}, P_3, \bar{P_3}\}$

$E_2=\{P_1, \bar{P_1}, P_3, \bar{P_3}, P_4, \bar{P_4}\}$

$E_3=\{P_1, \bar{P_1}, P_4, \bar{P_4}, P_2, \bar{P_2}\}$

but I can't imagine, how 2 equators can intersect 4 times???

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Here's what an equator is, here's what a 3-sphere is, please adjust your question so that it makes some sense. – Raskolnikov Nov 4 '12 at 13:46

Shouldn't there only be four equators, one for each pair $p,p'$ of opposite poles? That is, if an equator is to be the set of points on the sphere which are equidistant between the two opposite poles, and the poles are taken to be points with all but one coordinate 0 and the other coordinate 1 or -1. – coffeemath Nov 4 '12 at 13:54