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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???

  • 4
    [Here's what an equator is](http://en.wikipedia.org/wiki/Equator), [here's what a 3-sphere is](http://en.wikipedia.org/wiki/3-sphere), please adjust your question so that it makes some sense.2012-11-04

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