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Let $h \in \mathbb{R}[x, y]$ be an irreducible cubic polynomial. Consider the affine variety$$\{(x, y) \in \mathbb{R}^2: h(x, y) = 0\}.$$What are the qualitatively different possibilities for what this affine variety looks like?

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    $\begingroup$ Could you be more precise about what you mean by "qualitatively different"? What about classification into orbits under the action of the affine group of $\mathbb{R}^2$, projecteuclid.org/euclid.rmjm/1250126126 ? $\endgroup$ – j.c. Sep 28 '15 at 23:20
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Let $\overline{h}$ be the homogenization of $h$. If all points of $\mathcal{V}_{\mathbb{R}}(\overline{h})=\left\{(x:y:z)\in \mathbb{P}^2(\mathbb{R}) :\overline{h}(x,y,z)=0\right\}$ are non-singular points of the projective variety $\mathcal{V}(\overline{h})$, then Harnack's theorem states that $V$ decomposes into at most two connected components. Each connected component $C$ is Nash-diffeomorphic to a circle and is called an oval. Two things can happen: Either the component $C$ hits the plane at infinity or it does not. Suppose that $C$ does not hit the plane at infinity. In this case we can assume that $C$ lies in the real affine set $\mathbb{R}^2$. In fact, $C$ can be viewed as an connected component of $\mathcal{V}_{\mathbb{R}}(h)=\left\{(x,y)\in \mathbb{R}^2:h(x,y)=0\right\}$. Note that although $h$ is irreducible, $V:=\mathcal{V}_{\mathbb{R}}(h)$ may still decompose into several connected components. Hence, if a connected component $C$ does not hit the plane at infinity, then $V$ might 'look' like a circle. Furthermore, by Bezout's theorem there can only be one 'circle' in the affine case. If $C$ hits the plane at infinity, then the cut with $\mathbb{A}^2(\mathbb{R})$ will not be an oval of course, but it will be the 'affine part' of the oval.

If $\mathcal{V}(h)$ has real singular points, then the above arguments cannot be applied. But at least, the singular points can be classified (see wikipedia)

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