# Equators and meridians on a discrete torus

Consider the 4 × 4 grid graph:

Now torify it, i.e. connect its opposing vertices:

How can one tell the difference between a “meridian” and an “equator”?

The difference seems clear when looking at a “continuous” torus embedded in $\mathbb{R}^3$:

meridians

vs. equators

In the discrete case there seems to be no difference at all.

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The only reason the difference seems clear in the continuous case is that you made a choice of which pair of opposing sides of a square to connect first. –  Gerry Myerson Dec 3 '12 at 23:22
But why can you make this choice in the continuous case - yielding a difference - but not in the discrete case? This is my question. –  Hans Stricker Dec 3 '12 at 23:25
The embedding into $R^3$ preserves distances in one direction but not in the perpendicular direction. If you consider the continuous torus as a subset of $R^4$, the asymmetry between parallels and meridians disappears and the structure is completely symmetric. For example, an equation for a torus as a subset of $R^4$ is $x^2 + y^2 = R^2, z^2 + t^2 = r^2$, where $R$ and $r$ are the two radii. Similarly, the topological structure of the torus is just $S^1\times S^1$, where one $S^1$ is the bundle of parallels and one is the bundle of meridians, but there's no way to decide which is which. –  MJD Dec 3 '12 at 23:27
@MJD: Why do you call "equators" "parallels" opposed to "meridians"? Do you refer to some literature? (I'd really like to know, because meridians seem to be as parallel as equators.) –  Hans Stricker Dec 3 '12 at 23:35
In the continuous case, you are gluing things together. In the discrete, you are just drawing lines to indicate which things are adjacent, rather than identifying things that were previously distinct. If you really torify a grid, if you actually glue pairs of vertices, you'll fold it up into a torus with a grid on it, and you'll have the same meridian/equator distinction as continuously. –  Gerry Myerson Dec 4 '12 at 0:27