# understanding the torus described as a product of two circles

I would like to ask two (related) things about the torus:

The torus can be described as the cartesian product $S^1 \times S^1$ of two circles in $\mathbb{R}^3$. Then one can talk about meridional circles and longitudinal circles. I was wondering - is there a common way to associate the two factors with meridional and longitudinal circles ? That is, is for example the first factor in $S^1 \times S^1$ always used as the longitudinal one ?

Also, I am wondering if I identify two tori along the circle $S^1 \times \{x_0 \}$ - would such an identification be the same as the one obtained by an identification of $\{x_0 \} \times S^1$ ? This question is related to the first one in that it asks whether there is any kind of natural way to order the product, I guess.

Many thanks for your help!

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That depends: is the first factor in $\mathbb{R} \times \mathbb{R}$ the $x$-coordinate (horizontal) or the $y$-coordinate? Depends who you talk to. (And does $y$ point up or down?) As $S^1$ is a quotient of $\mathbb{R}$, these questions are very similar to yours ;). Of course, there's a homeomorphism that takes $(x,y)$ to $(y,x)$, so we can always swap coordinates if we need to. – Thomas Belulovich Apr 19 '12 at 14:42
@ThomasBelulovich you're absolutely right I was just wondering whether there is a common way to say the first factor in $S^1 \times S^1$ stands for the longitudinal cirlce, say. Like for example in the case of $\mathbb{R} \times \mathbb{R}$ where it is understood that $x$ stands for the first factor if not specified otherwise, and $y$ for the second factor. Of course these are only conventions, but it is important to be aware of these, if there are any. – harlekin Apr 19 '12 at 15:43
I don't know of any such convention, neither in the differential geometry, differential topology or Riemann surface contexts. If you need this distinction in a certain context I'm rather sure you may feel free to assign this to you liking. – user20266 Apr 19 '12 at 17:37
@Thomas: ok, thanks! – harlekin Apr 19 '12 at 22:49