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Given a polynomial $p(x_1,x_2,\ldots,x_d)$ in $d$ variables, with maximum degree $k$, what is the maximum number of components of $\mathbb{R}^d$ minus $p(\ldots)=0$? In other words, into how many pieces can an implicit polynomial equation partition $\mathbb{R}^d$?

For example, the following three equations partition $\mathbb{R}^2$ or $\mathbb{R}^3$ into $3$, $4$, and $2$ pieces respectively (I think!):

$$x^3 y^2+x^3 -3 x^2 y -y^2 +4 x y+x=0$$

$$x^6 y^8+x^3+4 x y-y=0$$

$$x^4+3 \left(x^2+y^4+z\right)- \left(x^2+y^2+z^2\right)^2+y^3+z^5 + 2 xy=3$$

Of course the answer is $k+1$ in $\mathbb{R}^1$. I suspect this is well known for $\mathbb{R}^d$; if so, I would appreciate a pointer. Thanks!

Update. Greg Martin's idea (from the comments), using the 5th Chebyshev polynomial of the first kind:

As Aaron Meyerowitz points out, here the degree $k=10$, and the plane is partitioned into $28$ pieces. But using Pietro Majer's line-arrangement idea leads to (now corrected:) $58$ 56$pieces for a degree$10\$ polynomial.

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Given a polynomial $p(x_1,x_2,\ldots,x_d)$ in $d$ variables, with maximum degree $k$, what is the maximum number of components of $\mathbb{R}^d$ minus $p(\ldots)=0$? In other words, into how many pieces can an implicit polynomial equation partition $\mathbb{R}^d$?

For example, the following three equations partition $\mathbb{R}^2$ or $\mathbb{R}^3$ into $3$, $4$, and $2$ pieces respectively (I think!):

$$x^3 y^2+x^3 -3 x^2 y -y^2 +4 x y+x=0$$

$$x^6 y^8+x^3+4 x y-y=0$$

$$x^4+3 \left(x^2+y^4+z\right)- \left(x^2+y^2+z^2\right)^2+y^3+z^5 + 2 xy=3$$

Of course the answer is $k+1$ in $\mathbb{R}^1$. I suspect this is well known for $\mathbb{R}^d$; if so, I would appreciate a pointer. Thanks!

Update. Greg Martin's idea (from the comments), using the 5th Chebyshev polynomial of the first kind:

As Aaron Meyerowitz points out, here the degree $k=10$, and the plane is partitioned into $28$ pieces. But using Pietro Majer's line-arrangement idea leads to $58$ pieces for a degree $10$ polynomial.

2 added 253 characters in body

Given a polynomial $p(x_1,x_2,\ldots,x_d)$ in $d$ variables, with maximum degree $k$, what is the maximum number of components of $\mathbb{R}^d$ minus $p(\ldots)=0$? In other words, into how many pieces can an implicit polynomial equation partition $\mathbb{R}^d$?

For example, the following three equations partition $\mathbb{R}^2$ or $\mathbb{R}^3$ into $3$, $4$, and $2$ pieces respectively (I think!):

$$x^3 y^2+x^3 -3 x^2 y -y^2 +4 x y+x=0$$

$$x^6 y^8+x^3+4 x y-y=0$$

$$x^4+3 \left(x^2+y^4+z\right)- \left(x^2+y^2+z^2\right)^2+y^3+z^5 + 2 xy=3$$

Of course the answer is $k+1$ in $\mathbb{R}^1$. I suspect this is well known for $\mathbb{R}^d$; if so, I would appreciate a pointer. Thanks!

Update. Greg Martin's idea (from the comments), using the 5th Chebyshev polynomial of the first kind:

1

Partitions of $\mathbb{R}^d$ by implicit polynomial equations

Given a polynomial $p(x_1,x_2,\ldots,x_d)$ in $d$ variables, with maximum degree $k$, what is the maximum number of components of $\mathbb{R}^d$ minus $p(\ldots)=0$? In other words, into how many pieces can an implicit polynomial equation partition $\mathbb{R}^d$?

For example, the following three equations partition $\mathbb{R}^2$ or $\mathbb{R}^3$ into $3$, $4$, and $2$ pieces respectively (I think!):

$$x^3 y^2+x^3 -3 x^2 y -y^2 +4 x y+x=0$$

$$x^6 y^8+x^3+4 x y-y=0$$

$$x^4+3 \left(x^2+y^4+z\right)- \left(x^2+y^2+z^2\right)^2+y^3+z^5 + 2 xy=3$$

Of course the answer is $k+1$ in $\mathbb{R}^1$. I suspect this is well known for $\mathbb{R}^d$; if so, I would appreciate a pointer. Thanks!
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