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Let $X$ be a (singular, reducible) affine plane real algebraic curve of degree $d$.

How we can estimate maximal number of connected components of it's complement in $R^2$ in terms of degree?

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This is surely related to Hilbert's 16th problem, which must have a Wikipedia page, and Harnack's upper bound for the number of components. –  Mariano Suárez-Alvarez Jun 27 '11 at 18:49
    
Yes, of course it's related. But Harnack bound is available(in terms of degree),as far as i know, only for irreducible nonsingular curves. And the first part of Hilbert's 16-th problem is far more deep because it deals with full description of realizable topological configurations of plane curves. –  probably Jun 27 '11 at 18:56
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2 Answers

Surely, this number is at least $\frac{n^2+n+2}{2}$, because a generic collection of lines cut the plane in this amounts of pieces. This is a bit more than Harnak's bound and I would be surprised if it can be beaten.

ADDED.

One can prove a quadratic bound on the number of regions (quite possibly close to $\frac{n^2+n+2}{2}$) by the following argument:

We will assume for simplicity that $X$ is in $\mathbb RP^2$. Take a generic point $p$ on $\mathbb RP^2$ and consider a pencil of lines through $p$. Consider the projection of $X$ to $\mathbb RP^1$. Notice that the number of components of $\mathbb RP^2\setminus X$ is related to the number of critical points of this projection. (If $x\in X$ is a singular point of $X$ then the $x$ is also considered to be critical). Indeed if there where no critical points at all, the number of components of $\mathbb RP^2\setminus X$ would be at most $n$. So the number of components is proportional to the number of critical points of the projection. The last number is at most quadratic.

It should be possible to make the above sloppy argument precise with a precise bound.

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Can we beat this bound using Harnack itself? We can take Harnack bound(which holds for projective, not affine, curves) and assume that all components have nontrivial intersection with the line at infinity. Therefore we double the Harnack bound(and for sufficiently big $n$), In fact for $n>6$, we have beaten $\frac{n^2+n+2}{2}$. –  probably Jun 28 '11 at 6:04
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I don't think all components of a Harnack curve can have nontrivial intersection with the line at infinity for high degrees. Can you check it for d=4? –  quim Jun 28 '11 at 6:08
    
Yes, you are right, because of Besout's theorem. Only $n$ components can(theoretically) have nontrivial intersection and it is not sufficient to beat $\frac{n^2+n+2}{2}$. –  probably Jun 28 '11 at 6:18
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Dmitri -- re "I would be surprised if it can be beaten": Ilya Itenberg told me once that hyperplanes in general position don't always give the maximal number of connected components of the complement. I don't remember though in which dimension this happens. –  algori Sep 13 '11 at 21:11
    
Alogri, thanks! Ask Ilya next time that you meet him. I might ask Orevkov if meet him at Steklov next week. –  Dmitri Sep 14 '11 at 7:50
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up vote 1 down vote accepted

We can also prove that the maximal number of components for reducible nonsingular curve is $\frac{n^2+n+2}{2}$ by the following computation.

Let $m$ be a partition of $n$, where $m_i$ is the degree of $i$-th irreducible component of our curve. Denote $s=\{i|m_i=1\}$, $t=\{i|m_i>1\}$. Then, using Harnack inequality and Besout's theorem(for intersections with infinite line and for intersections between irreducible components) we can write that the number of connected components is lesser than $$ \max_{m\vdash n}(\sum_{i}(\frac{(m_i-1)(m_i-2)}{2}+1)+1+n-s+\sum_{i\neq j}m_im_j)=$$ $$=\frac{n^2}{2}-n+1+\max_{m\vdash n}(2t+s))\leq $$ $$\leq \frac{n^2}{2}-n+1+ \max_{0\leq 2x+y\leq n,0\leq x,0\leq y}(2x+y)=\frac{n^2+n+2}{2} $$

Therefore the only question we have is the following: why do Harnack bound (in terms of degree) holds for singular curves as well as for nonsingular?

The proof could be obtained using First Harnack Theorem.

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