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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Sign up to join this communityLet $X$ 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?
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.
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.