The Torus vs Its Embedding in Three-Space

July 3, 2016

As a visual person, this passage from Geometry, Topology and Physics by Nakahara Nakahara, Mikio. Geometry, Topology and Physics. CRC Press, 2003 blew my mind:

The Torus $T^2$ is a product manifold of two circles $T^2 = S^1 \times S^1$. If we denote the polar angle of each circle as $\theta_i \mod 2 \pi \quad (i = 1,2)$, the coordinates of $T^2$ are $(\theta_1, \theta_2)$. Since each $S^1$ is embedded in $\mathbb{R}^2$, $T^2$ may be embedded in $\mathbb{R}^4$. We often imagine $T^2$ as the surface of a doughnut in $\mathbb{R}^3$, in which case we inevitable have to introduce bending of the surface. This is an extrinsic feature brought by the ‘embedding’. When we say ‘a torus is a flat manifold’, we refer to the flat surface embedded in $\mathbb{R}^4$.

The bending of the torus is an extrinsic feature brought by the embedding! This makes a lot of sense: I couldn’t subconsciously associate the product of two circles with the torus in 3D, because they are in fact different: the 3D torus is an embedding of the cartesian product of two circles in 3D, not the object itself.

Another reminder that visualizing things, while helpful, can be misleading.