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I am trying to understand how is it that we can define a Torus as a cartesian product of two circles, that is: $T^2 = S^1 \times S^1 \subset R^2 \times R^2 = R^4$
*First, the torus is well defined in $R^3$ by the following formula : $((x^2+y^2)^{1/2}-R)+z^2 = r^2$, so how is it that the torus, is somehow simpler to understand as a cartesian product in $R^4$
*Second, could i get some explanation as why the product with two circles is true?

daruma
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rannoudanames
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    The torus (not taurus) is topologically equivalent (homeomorphic) to a (topological) product of two circles. Have you studied homeomorphisms and products of topological spaces yet? – Mark S. Jan 09 '17 at 21:10
  • No for products of topological spaces, i have not studied that yet unfortunatly, any good references for me to read up on? – rannoudanames Jan 09 '17 at 21:16
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    Munkres is standard, but maybe Topology without Tears is better. See http://math.stackexchange.com/questions/7520/best-book-for-topology – Mark S. Jan 09 '17 at 21:20
  • Assuming, i knew about the above, how can we justify the definition of the tOrus? – rannoudanames Jan 09 '17 at 21:26
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    @rannoudanames: Do you see a circle rotating around another circle? Can you coordinatize the points by angles along each of those circles? – Ted Shifrin Jan 09 '17 at 23:39

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A point on a circle $x^2+y^2=r^2$ is defined by an angle $\theta $ via $x=r\cos\theta $ and $y=r\sin\theta $. And it wraps around as $\theta$ goes beyond $2\pi $. A point with angle just under $2\pi $ is close to a point with angle $0$.

A point on the embedded torus $\left (\sqrt {x^2+y^2}-R\right)^2+z^2=r^2 $ is defined by two angles, $\theta $ and $\phi $, via $x=(R+r\cos\theta)\cos\phi $, $y=(R+r\cos\theta)\sin\phi $, $z=r\sin\theta $. And you go around the torus (in two independent directions) as an angles goes beyond $2\pi $.

By the shape of a torus, two points on the torus are close in 3d space when the corresonding pairs of angles $(\theta,\phi) $ for those points are close (an angle just under $2\pi$ being close to $0$). That's why a torus is topologically (rough shape wise) like a product of circles.

Mark S.
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