After seeing this picture of the figure eight knot:
Why isn't the figure eight knot considered a $(2,3)$-torus knot?
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3$\begingroup$ Isn't the $(2,3)$ torus knot the trefoil knot? The trefoil isn't equivalent to the figure-8. I think you have to study the definition of torus knot carefully. $\endgroup$– Gerry MyersonCommented Mar 26, 2013 at 12:44
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1$\begingroup$ You could cite a theorem of Thurston classifying knots into hyperbolic, torus, and satellite. The fig-8 is hyperbolic, hence not torus. (I don't know its proof, though.) This seems to be a decent survey paper: arxiv.org/pdf/math/0309466.pdf $\endgroup$– NealCommented Mar 26, 2013 at 12:58
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$\begingroup$ Sorry I mean T(3,2) $\endgroup$– mbidwdCommented Mar 26, 2013 at 13:26
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$\begingroup$ The (p,q)-torus knot is equivalent to the (q,p)-torus knot, so T(3,2) is the same as T(2,3) and is still the trefoil. $\endgroup$– wilsonwCommented Jun 15, 2013 at 6:25
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2 Answers
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The $(3,2)$ torus knot is also known as the "trefoil" knot. It admits a tricoloring, while the figure eight knot (that you drew) does not. Here is a reference: http://en.wikipedia.org/wiki/Tricolorability
To answer your question directly: The diagram you drew certainly looks like it lies on the standard two-torus in $\mathbb{R}^3$. But stare at one of four arcs that connect an outer crossing with an inner crossing. The crossing pattern at the ends of the arc prevent it from lying in the two-torus. At one of the ends, it will have to go into (or out of, respectively) the two-torus, to avoid crashing into another arc.
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The knot drawn is an alternating knot that cannot be embedded in the surface of a torus. It needs at least a double torus to be drawn crossing free on that surface.
