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Suppose a planar curve $C$ is defined by parametric equations in $t$: $x(t)$ and $y(t)$. Are there conditions on these two functions that guarantee that $C$ does not self-intersect?

For example, the Maclaurin trisectrix can be defined by $$x(t) = \frac{t^2-3}{t^2+1}, \;\;\;\;\;y(t)=\frac{t(t^2-3)}{t^2+1}$$ and it self-intersects:

                  MathWorld
So, rational functions do not suffice to imply non-self-intersection. Pointers would be appreciated—Thanks!

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    $\begingroup$ In fact, high degree rational functions (at least over complex numbers) will generally lead to curves with self-intersection. This is because a generic high-degree rational curve in $\mathbb P^2$ has many nodes. Thus to avoid the intersections in the affine patch, you would need to have all of the singularities at the line at infinity. $\endgroup$
    – Lev Borisov
    Commented Jun 8, 2014 at 0:22
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    $\begingroup$ The injectivity of both functions x(t) and y(t) provides a sufficient condition to ensure that the curve does not self-intersect. $\endgroup$
    – José Hdz. Stgo.
    Commented Jun 8, 2014 at 0:26
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    $\begingroup$ Thanks, @J.H.S., for the exponent correction in the trisectrix example. $\endgroup$
    – Joseph O'Rourke
    Commented Jun 8, 2014 at 0:27
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    $\begingroup$ @IanAgol: In order for the Gauss map criterion to work, you have to assume, in addition, that the curve has no cusps. Otherwise, you could have a curve with two cusps and one self-intersection that has Gauss map contained in an arbitrarily small neighborhood of a point. A suitable example can be made by setting $\bigl(x'(t),y'(t)\bigr) = \bigl((1-t^2), t(1-t^2)^2/h(t^2)\bigr)$ where $h$ is a positive function of $t^2$ that grows sufficiently fast. $\endgroup$
    – Robert Bryant
    Commented Jun 8, 2014 at 3:46
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    $\begingroup$ Ah, yes, Robert, I meant a curve in which the derivative is non-vanishing (I assumed this was implicit in defining the Gauss map - I was thinking of the Gauss map defined using the unit tangent vector, rather than the unit normal vector). Another way of phrasing it is that the tangent vector always has non-zero inner product with a particular vector, which means that it has 1-1 projection to a line parallel to that vector. So this is really just a variation on JHS's condition. $\endgroup$
    – Ian Agol
    Commented Jun 8, 2014 at 4:53

1 Answer 1

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By its very nature, this question cannot expect a definitive answer but here are some suggestions.

  1. For a curve with parametrisation of the form $(\int^t(u)du,f(t))$ for a function $f$ of one variable it is the case if $f$ is injective or, better, if and only if $\int^s f(u)du\neq \int^t f(u)du$ whenever $f(s)=f(t)$.

  2. Many important curves have parametrisations of this form (circle, cycloid, catenary, ....).

  3. In a certain sense "every" curve has such a parametrisation. More precisely, consider the curve with parametrisation $(x(s),y(s))$. Since self-intersection is preserved under diffeomorphisns, we can suppose that the curve lies in the upper half plane. Under the new parameter $t$ where the latter is, as a function of $s$, the primitive of $\frac{x'(s)}{y(s)}$, the parametrisation will have the above form.

  4. Of course, this will not work universally since this $t$ will not always be a reparametrisation but it will be in many concrete situations, e.g., if $x$ is strictly monotone.

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  • $\begingroup$ Just a small remark: the condition in point 1 is actually independent of the fact that $F$ is a primitive of $f$. $\endgroup$
    – Marco Golla
    Commented Jun 8, 2014 at 9:43
  • $\begingroup$ True, but the point of my remark was that this condition depends only on ONE function $f$, rather than on the pair of functions $x$ and $y$. I should probably have written $$\int^s f(u)dt \neq \int^t f(u)du.$$ $\endgroup$
    – janowski
    Commented Jun 8, 2014 at 13:10
  • $\begingroup$ have edited my answer to avoid the impression that it was tautological $\endgroup$
    – janowski
    Commented Jun 8, 2014 at 15:06

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