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Let $0<b<a,(u,v) \in \mathbb{R} \times \mathbb{R}$. Then the map $g(u,v):=((a+b\cos u)\cos v,(a+b\cos u)\sin v,b\sin u)$ defines a torus. I wonder for $g$ to be a surface does it really need $(u,v) \in \mathbb{R} \times \mathbb{R}$?

If let $(u,v) \in U \times V$ where $U,V$ are both open in $R$ , will $g$ still be a surface in the context of differential geometry? (I know it may not be a surface of revolution though)

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A continuously differentiable map $${\bf f}: \quad\Omega\to{\mathbb R}^3,\quad (u,v)\mapsto{\bf x}(u,v)=\bigl(x_1(u,v),x_2(u,v),x_3(u,v)\bigr)$$ defined on an open set $\Omega\subset{\mathbb R}^2$ is called an immersion, if for all $(u,v)\in\Omega$ the differential $d{\bf f}(u,v)$ has rank $2$, i.e., if ${\bf x}_u\times{\bf x}_v\ne{\bf 0}$ for all $(u,v)$.

When this "technical condition" is fulfilled then ${\bf f}$ maps sufficiently small pieces of $\Omega$ bijectively onto small pieces of a (large) smooth surface $S:={\bf f}(\Omega)$. But globally the map ${\bf f}$ need not be one-one: There might be self-intersections, or some parts of $S$ are covered several times.

Such is the case in your example: One computes $${\bf g}_u\times{\bf g}_v=(a+b\cos u)\bigl(-b\cos u\cos v, -b \cos\sin v,-b\sin u\bigr)\ ,$$ which is easily seen to be $\ne{\bf 0}$ for all $(u,v)\in{\mathbb R}^2$. Therefore this ${\bf g}$ is indeed an immersion. But the function ${\bf g}$ is doubly periodic; whence any two points $(u,v)$ differing by $(2k\pi,2\ell\pi)$ are mapped to the same point of $S$. Nevertheless it makes sense to leave the map ${\bf g}$ as it stands, because restricting $u$, $v$ to a period square introduces artificial seams.

It is another matter when you want to compute the area of $S$. In this case you have to make sure that $S$ is covered exactly once by the parametrization (up to said seams, which form a set of measure zero).

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By definition $df_u$ must be injective for all $(u,v) \in U \times V$ for $g$ to be a surface. It seems by making $U$ and $V$ to be arbitray (disjoint) union of open intervals will result $g$ to be a disjoint union of surfaces of revolution? –  user31899 Sep 2 '12 at 12:57
    
@user31899: The object $g$ is a map, not a surface. When the domain of $g$ consists of many pieces then the image set $S$ (commonly called a surface) may also consist of several disjoint pieces. –  Christian Blatter Sep 2 '12 at 13:27
    
Thanks! The result is a surface due to disjoint union of surfaces is still a surface. –  user31899 Sep 2 '12 at 22:18
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