Equators in the $4$-dimensional sphere.

Question

Are there infinitely many disjoint equators (centrally symmetric circles) on the surface of the $4$-dimensional sphere? There are at least two of them, namely $[0,0,x,\sqrt{1-x^2}]$ and $[x,\sqrt{1-x^2},0,0] $ , $ x\in [-1,1]$. (Note that we may have to adjust the sign of the squareroot, but that just confuses the reader.)

I tried to take the mean of their coordinates, and rescale it to get points on he sphere, but i do not know how to show that some parametrized curve is a circle in $4$-dimensions.

score 8 · Accepted Answer · answered Dec 18 '13 at 15:12

8

Yes -- in fact the 3-sphere can be completely covered by disjoint great circles, the Hopf fibration.

(Terminology nitpick: The subset $\{ (x,y,z,w)\in\mathbb R^4\mid x^2+y^2+z^2+w^2=1\}$ is usually known as the three-sphere, counting the dimension of the surface itself rather than the space it happens to embed into).

answered Dec 18 '13 at 15:12

hmakholm left over Monica

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Is it clear that we have infinitely many circles of the type asked by the OP in the Hopf fibration ? – Jeremy Daniel Dec 18 '13 at 15:23
@Jeremy Yes: http://en.wikipedia.org/wiki/Hopf_fibration I learned about Hopf fibration a few years ago. I am a little bit ashamed that i did not remember. Thank you for the answer! – Daniel Soltész Dec 18 '13 at 15:26

score 1 · Answer 2 · answered Dec 18 '13 at 15:18

Firstly, you should speak of the $3$-dimensional sphere, viewed in the $4$-dimensional euclidian space.

Secondly, the term equator in this situation should certainly refer to the intersection in $\mathbb{R}^4$ of the sphere $S^3$ and a hyperplane, that is a $2$-dimensional object.

For your question, consider the vectors $(\cos(\phi)\cos(\theta), \sin(\phi)\cos(\theta), \cos(\phi)\sin(\theta), \sin(\phi)\sin(\theta))$. Then, when you fix $\phi$, you get a circle with the coordinate $\theta$, and if I'm not mistaken the circles corresponding to $\phi_1$ and $\phi_2$ are disjoint is $\phi_1 \neq \phi_2$.

score 0 · Answer 3 · answered Dec 18 '13 at 15:10

0

Yes. Take any orthonormal basis of $\mathbb{R}^4,$ call it $v_1, v_2, v_3, v_4$ Then the circles $\cos t v_1 + \sin t v_2$ will be disjoint from $\cos s v_3 + \sin s v_4.$

answered Dec 18 '13 at 15:10

Igor Rivin

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I think that what is wanted is an infinite family of disjoint cirlces and if I understand correctly your answer, this is not what you give. – Jeremy Daniel Dec 18 '13 at 15:12
@JeremyDaniel You understand _in_correctly. Notice I said any orthonormal basis. – Igor Rivin Dec 18 '13 at 15:12
What you give is an infinite family of two disjoint circles. I think that the OP wants an infinite family of disjoint circles. – Jeremy Daniel Dec 18 '13 at 15:19
@JeremyDaniel Ah, I see. Then you are right. – Igor Rivin Dec 18 '13 at 15:21

Equators in the $4$-dimensional sphere.

3 Answers3