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Consider two real polynomials in three variables, defined on the 3-sphere, $S^3$. Is there some Bezout-type theorem, relating the intersection of two closed surfaces defined by these polynomials and their degrees?

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  • $\begingroup$ What's a tree variable? $\endgroup$ Jan 21, 2015 at 0:46
  • $\begingroup$ @DavidHandelman, I mean "three" of course) $\endgroup$
    – Gauss
    Jan 21, 2015 at 0:49
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    $\begingroup$ What do you call the 3-sphere? The unit sphere in the 3-dimensional real space is usually called 2-sphere. $\endgroup$
    – YCor
    Jan 21, 2015 at 7:03

1 Answer 1

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You can take the embedding

$$\begin{array}{cccc} S^3&\to& X\\ (x,y,z)&\to & [1:x:y:z]\end{array}$$ where $X\subset \mathbb{P}^3$ is the surface given by $w^2=x^2+y^2+z^2$. The complement of the image is the intersection of $X$ with the hyperplane $w=0$, which is a conic, having no real point.

Using Bezout on $\mathbb{P}^3$, the intersection of two curves on $S^3$ given by polynomials of degree $d_1$ and $d_2$ is at most $2d_1d_2$. This number is in fact equal to the point of intersection, viewed on the complex points of $X$, up to multiplicity. Hence, as you consider the real points of this intersection, you could have less points but always an even number.

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  • $\begingroup$ In fact I was wondering in the following example: a sphere and an elipsoid in ordinary real 3-space "genericaly" intersects in four closed curves as can be easily seen. So does the fact that in a generic case the number of curves of intersection is less then 4=2*2 is a consequence of some real Bezout-type theorem? $\endgroup$
    – Gauss
    Jan 21, 2015 at 14:05

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