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Problem: Consider the set $E = \{(x_1,x_2,x_3,x_4) \in \mathbb{R}^4: \sum_i x_i^2 = 1, x_1x_3 = x_2x_4\}$ (imagine a $2\times2$ real matrix with norm 1 and determinant 0). Show that the set $E$ is a 2-torus.

Attempt at solution: The problem seems to indicate that a 2-torus is embedded in $S^3$, which intuitively makes sense, however I am finding it difficult to prove that this set is actually a torus. I tried parametrizing the set in two coordinates:

It seems that the points in $E$ can be parametrized in terms of two variables $\theta, \phi \in \mathbb{R}$ by $x_1 = cos\theta \cdot cos\phi$, $x_2 = cos\theta \cdot sin\phi$, $x_3 = sin\theta \cdot cos\phi$, and $x_4 = sin\theta \cdot sin\phi$. However, I am not seeing how this parametrization describes a 2-torus though... any help is appreciated!

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  • $\begingroup$ In what possible way does it "intuitively makes sense" that the torus is embedded in the sphere?! $\endgroup$ – Mariano Suárez-Álvarez Feb 18 '18 at 4:52
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The Guide for the Perplexed::::: ::: ::: https://en.wikipedia.org/wiki/Clifford_torus#Formal_definition

This is, in effect, about (orthogonally) diagonalizing a quadratic form. Introduce $$ p = \frac{x_1 + x_3}{\sqrt 2} \; , $$ $$ q = \frac{x_1 - x_3}{\sqrt 2} \; , $$ $$ r = \frac{-x_2 + x_4}{\sqrt 2} \; , $$ $$ s = \frac{x_2 + x_4}{\sqrt 2} \; . $$

You still have $$ p^2 + q^2 + r^2 + s^2 = 1 $$ but now $$ p^2 - q^2 = -r^2 + s^2 \; ,$$ or $$ p^2 + r^2 = q^2 + s^2 = \frac{1}{2} \; \; .$$

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  • $\begingroup$ I don't think the new variables satisfy $pr=qs$ or $ps=qr$, which was part of the restrictions in set $E$. $\endgroup$ – Longti Feb 18 '18 at 3:43
  • $\begingroup$ @Longti no, they don't. Can you see how to parametrize the set of $(p,q,r,s)$ so as to make a torus? You can then solve for $x_1, x_2,x_3,x_4.$ $\endgroup$ – Will Jagy Feb 18 '18 at 3:45
  • $\begingroup$ I believe having trouble with how to parametrize a 2-torus in $\mathbb{R}^4$ is part of my confusion with this problem... could you elaborate a bit more please? Thank you. $\endgroup$ – Longti Feb 18 '18 at 3:47
  • $\begingroup$ @Longti see en.wikipedia.org/wiki/Clifford_torus#Formal_definition $\endgroup$ – Will Jagy Feb 18 '18 at 3:48
  • $\begingroup$ Ok thanks. I will proceed and see what the results are. $\endgroup$ – Longti Feb 18 '18 at 3:50
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À two-by-two singular matrix of norm 1 has columns two vectors in the plane which are not simultaneously zero but which are linearly dependent. Each such matrix thus determines a line in the plane (which is a point in the real projective line, which is a circle) and the extended scalar which is the ratio of the first Vector by the second one. If we view extended scalars as elements of the real projective line in the obvious way and then as elements of $S^1$, we see that each such matrix determines a point in a torus.

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