## Question

Show that the two-dimensional torus

*T*^{2}is a manifold, by explicitly constructing an appropriate atlas. (Not a maximal one, obviously).## Answer

Fig 1: A Torus, with help from Wikipedia |

We can define a torus as points in

**R**^{3}that lie on the surface created by rotating a small circle of radius r round a larger circle of radius R. The larger circle is in the X-Y plane. Any point on the torus can be given by (*θ , κ*) as shown in Fig 1. There are no limits on*θ , κ*.
I realised that I
could map to infinite cylinders touching the inner and outer edges of the torus
and from them to annuli as in Exercise 2.01. No doubt our cunning author had
set this trap. Having been on that wild goose chase, I came to a
linear map from

*T*^{2}to the annuli and realised that we can also map from*T*^{2}to two open rectangles much as one can map from*S*^{1}to**R**^{1}. Fig 2 shows a section at some*θ*through our torus. We almost have the two charts*U*(Outer) and_{O}*U*(Inner) we just need to cut them and unroll them to make open rectangles. We cut the outer_{I}*U*at_{O}*θ*= 0 and the inner*U*at_{I}*θ*= -π. If we cut them both at the same*θ*it would lead to disaster.Fig 2: Section through torus and rectangular maps. |

This is where it gets complicated and we end up with a few maps, so;

Fig 4: Showing the transformations. T^{2} is exactly U ∩ _{I}U. _{O} |

We now want to check that the charts obey the strict
conditions, as given in the book before Fig 2.13 at the bottom of page 29, to
make an atlas from

*T*^{2}to**R**^{2}. I have adapted them for the exercise.
1. "The union of the charts

*U*,_{I}*U*is equal to_{O}*T*^{2}; that is,*U*,_{I}*U*cover_{O}*T*^{2}."
We have two charts

*U*,_{I}*U*; their union manifestly covers_{O}*T*^{2}by construction. ✓
We could check that there are no cracks in the union by
combining the limits on ϕ

*, ϕ*_{I}*. We have -π< θ<2π and -π<κ<2π. That covers*_{O}*T*^{2}. ✓
2. "The charts are smoothly sewn together. More
precisely, if two charts overlap, (ϕ

*U*∩_{I}*U*≠∅, then the map (ϕ_{O}_{I }∘ ϕ_{O}^{-1}) takes points in ϕ_{O}(*U*∩_{I}*U*) ⊂_{O}**R**^{2}*onto** an open set ϕ_{I}(*U*∩_{I}*U*) ⊂_{O}**R**^{2}, and all these maps must be*C*^{∞}. The reverse for_{O }∘ ϕ_{I}^{-1}) also applies.
* We remind ourselves that onto means each point of the
target has at least one point of the source mapped onto it.

The first part was easy as shown. The second part was fiddly and you can read the detail and wild goose chase in Ex 2.03 2D Torus is a manifold.pdf. It has helped me greatly to understand manifolds.

What on earth are your maps. This is not a proof that the torus is a manifold.

ReplyDelete