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Recently, I have been caught in the question: What kind of algebraic curves in $\mathbb{R}^2$ can one get?

For instance, how do I know that there is no polynomial $P(x, y) \in \mathbb{R}[x, y]$ whose graph looks like the graph of the $y = \cos x$? My intuition is that the graph $y = \cos x$ has infinitely many turning points; which might differentiate it from graph of any polynomial. Obviously, one needs to define the meaning of "two curves that look like each other"; which is somewhat intuitive but might be very subjective. (Due to transcendence of $\cos x$, no polynomial can achieve the graph identical to that of $y = \cos x$.)

I conjectured that the compact component of any graph of polynomials in $\mathbb{R}[x, y]$ must have winding number 1 (more or less, circle-like). Unfortunately, this is easily seen to be false. Is there any known result along this direction?

I once found an equation of form $P^2 + Q^2 = 1$, along the line of $(x^2 - y + y^4)^2 + (y^2 - x - x^4)^2 = 1$, whose graph looks fairly close to a rectangle. (Of course, there can be no polynomial whose graph is a rectangle due to sharp corners. I do not know of easy formal argument for this though one can work it out by noting that such polynomial must basically be product of linear factors but then we have infinite lines in the graph.) It is suggestive that we can use higher and higher degree polynomial to get arbitrarily close to some (hence, arbitrary via linear change of coordinate) rectangle $R$. However, we want to exclude the case where one picks out point-by-point on the $R$ using polynomials $\prod_{a, b \in S} ((x-a)^2 + (y-b)^2)$ where $S \subset R$ finite. It is thus my wishful thinking to have a measure $\mu(\gamma, \delta)$ that measures similarity between two curves $\gamma, \delta$ on the plane.

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  • $\begingroup$ Relevant to the question in the second paragraph: two algebraic curves in $\mathbf R^2$ cannot meet in a countably infinite set of points, because of Bezout's theorem. This shows e.g. that a small deformation of $y=\cos x$cannot be algebraic. $\endgroup$ Apr 18, 2016 at 8:39
  • $\begingroup$ By "graph" you actually mean "zero set", right? $\endgroup$
    – Qfwfq
    May 18, 2016 at 12:25
  • $\begingroup$ Also, by Stone-Weierstrass type theorems, on any fixed compact I think you can approximate any continuous function by polynomials. By "approximate" here I mean in the $\mathcal{C}^0$ norm. Could $\mathcal{C}^0$-closeness of the defining functions be considered as a measure of how their zero sets are "similar"? $\endgroup$
    – Qfwfq
    May 18, 2016 at 12:30

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One way to measure the distance between two compact curves is Gromov-Hausdorff distance. This simply formalizes the idea that two curves are close to each other if each point on one curve is close to a point on the other curve. With this definition I would expect that any reasonable closed curve can be approximated arbitrarily well by algebraic curves. Of course the degree of the algebraic curve will get higher if you want a better approximation.

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