# Implicit Möbius strip

Reading the Wolfram MathWorld™ article on the Möbius strip I came across an equation that was news to me.

After the familiar parametric form $\displaystyle x=(1+s\cos(t/2))\cos t,\quad y=(1+s\cos(t/2))\sin t, \quad z = s\sin(t/2) \qquad (1)$

it is said that “In this parametrization, the Möbius strip is therefore a cubic surface with equation…” $\displaystyle y(x^2+y^2+z^2-1)-2z(x^2+y^2+x) =0 \qquad\qquad\qquad(2)$

That’s a neat cubic equation for sure. But… wouldn’t the sign of the left hand side of (2) (call it $F$) distinguish the “positive” and “negative” sides of the surface, contrary to its non-orientability?

I went to SageMathCloud™ to check and got this plot of (2):

That’s… not exactly the Möbius strip as I know it. But it’s true that (1) implies (2): the cubic surface contains the Möbius strip. Here is the plot of (1) superimposed on the plot of (2).

The self-intersection arises where the “positive side” $F > 0$ suddenly transitions to negative $F< 0$.

## 2 thoughts on “Implicit Möbius strip”

1. David Speyer says:

Fun example! I enjoyed playing with it. I think that, if you let s range over all real numbers, you do sweep out the whole surface. The line of self intersection occurs because the lines for t and t+Pi lie in a common plane, and hence meet at some point. I get this this point is at (x,y,z) = (1, -tan(t), – tan(t)). So, as t varies, we get self intersection along the line x=1, y=z. If you restrict the range of s, then you get two line segments instead of two lines, and thus you don’t see that anything funny is happening along that self intersection curve.

Can you see any reason why the final formula for the self intersection came out so simple? It seems surprising to me.

1. David Speyer says:

(-1, – tan(t), – tan(t)), and hence x=-1, y=z. I got the sign wrong above.