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Suppose $f:\mathbb{R}^d\to\mathbb{R}$ is a polynomial of degree $\ell$. Then, the number of connected components of its zero set $\{a\in\mathbb{R}^d : f(a) = 0\}$ is bounded by roughly $\ell^{d}$. I've seen this result attributed to Warren, Milnor-Thom, and stated to be a consequence of Bezout's theorem (a concrete reference would be much appreciated, especially one that derives this using Bezout's theorem).

Onto my main question: suppose $f$ is known to only have $k$ terms. Is there a better bound on the number of connected components of the zero set of $f$ that depends on $k$? I suspect that there might be a reference that answers this, but I don't know much about this area and would appreciate any suggestions on where to look. Thanks in advance!

tc1729
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For $d=1$, you can use Descartes' rule of signs.

The number of positive roots is upper bounded by $k$ (since there are at most $k$ sign changes).

The number of negative roots is the number of positive roots of $f(-x)$ which is at most $k$ for the same reason.

Thus, we get a total bound of $2k+2$ (there are $2k+1$ real roots in total (including possibly $0$)).

mathworker21
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  • I imagine this approach works incrementally: each time you add a dimension (a new variable $x_{l}$) you can look at "slices" where all the variables except $x_{l}$ are constant to "reduce" to the $d = 1$ case and argue by continuity that the number of connected components is multiplied by at most $(2k+2)$ in each new dimension, and that's how you get the $\mathcal{O}(l^d)$ bound in the general case. – Rivers McForge Apr 27 '21 at 18:24
  • @RiversMcForge do you have a rigorous proof or was that just a thought? – mathworker21 Apr 28 '21 at 01:48
  • I'm pretty sure the idea in my comment could be turned into a rigorous proof, but I didn't bother working out the details, hence why I wrote a comment and not an answer, lol. – Rivers McForge Apr 28 '21 at 20:32