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Let $f\in\mathbb{R}[\mathbf{z}]$ be an homogeneous $n$-variable polynomial of degree $r$ with real coefficients. Then the number of hyperbolicity cones of $f$ is at most $$\mathcal{H}(n,r)=\begin{cases} 2^r & \text{for }r\leq n\ ,\\ 2\sum^{n-1}_{k=0}\binom{r-1}{k} & \text{for }r>n\ .\end{cases}$$ The maximum is attained if and only if $f$ is a product of linear polynomials.

Notice that the saturation of the bound occurs precisely as conjectured for the number of connected components of the complement of an arrangement of $r$ hyperplanes of $\mathbb{R}^n$ in general position (see Boris Bukh's comment below). A connection between both numbers is reached through the reasoning discussed in Aaron's comment to his answer: let $$f(n,r)=\begin{cases}2^r &\text{for }r\leq n\ ,\\ \sum^{n}_{k=0}\binom{r}{k} &\text{for }r>n\ .\end{cases}$$ be the latter number, so that we clearly have $$\mathcal{H}(n,r)=\begin{cases} f(n,r) & \text{for }r\leq n\ ,\\ 2f(n-1,r-1) &\text{for }r>n\ .\end{cases}$$ Let us consider from now on the case $r>n$ - adding an $(r+1)$-th hyperplane $E$ in general position to the arrangement causes the intersection of the arrangement with $E$ to be an arrangement of $r$ hyperplanes in $\mathbb{R}^{n-1}$ in general position, whose complement has $f(n-1,r)$ connected components, each cutting a connected component of the previous arrangement in half and thus adding $$f(n-1,r)=f(n,r+1)-f(n,r)$$ connected components. This formula entails $$\begin{split} f(n,r)-f(n-1,r-1) &=f(n,r)-f(n-1,r)+f(n-1,r)-f(n-1,r-1) \\ &=f(n-1,r-1)+\binom{r}{n}-\binom{r-1}{n-1}\end{split}$$ and hence $$\begin{split} f(n,r) &=\mathcal{H}(n,r)+\binom{r}{n}-\binom{r-1}{n-1} \\ &=\mathcal{H}(n,r)+\binom{r-1}{n-1}\frac{r-n}{n}\ .\end{split}$$ Particularly, this entails that not all connected components of the complement of the zero level set of a product of $r$ linear polynomials in $\mathbb{R}^n$ are hyperbolicity cones if $r>n$, contrary to what happens for $r\leq n$. This was already to be expected since, as I discussed above, hyperbolicity cones always come in pairs and $f(r,n)$ can clearly fail to be even for $r>n$. This also entails that connected components of the complement of the zero set of an (homogeneous) hyperbolic polynomial may be convex cones without being hyperbolicity cones.

Of course, the bound $\mathcal{H}(n,r)$ is far tighter than the Milnor-Thom bound, but too tight for being the desired answer.

Let $P\in\mathbb{R}[\mathbf{z}]$ be an homogeneous $n$-variable polynomial of degree $r$ with real coefficients. Then the number of hyperbolicity cones of $P$ is at most $$\mathcal{H}(n,r)=\begin{cases} 2^r & \text{for }r\leq n\ ,\\ 2\sum^{n-1}_{k=0}\binom{r-1}{k} & \text{for }r>n\ .\end{cases}$$ The maximum is attained if and only if $P$ is a product of linear polynomials.

To get a feeling for the saturation of the above bound, we point out that $\mathcal{H}(n,r)$ is exactly the number of connected components of the complement of a central arrangement of hyperplanes in general position (here "central" means that all hyperplanes in the arrangement contain the origin) - the centrality comes from the fact that we are requiring here that $P$ is homogeneous. Notice that the saturation of the bound occurs precisely as conjectured in Boris Bukh's comment below (in the homogeneous case).

Of course, the conjecture of whether the above bound also applies to the total number of connected components of the complement of the zero set of an homogeneous polynomial $P$ remains. If so, the bound will also be sharp.

Notice that the saturation of the bound occurs precisely as conjectured for the number of connected components of the complement of an arrangement of $r$ hyperplanes of $\mathbb{R}^n$ in general position (see Boris Bukh's comment below). A connection between both numbers is reached through the reasoning discussed in Aaron's comment to his answer: if $r>n$, let $$f(n,r)=\sum^{n}_{k=0}\binom{r}{k}$$ be the latter number, so that $$\mathcal{H}(n,r)=2f(n-1,r-1)$$ in this case. Then adding an $(r+1)$-th hyperplane $E$ in general position to the arrangement causes the intersection of the arrangement with $E$ to be an arrangement of $r$ hyperplanes in $\mathbb{R}^{n-1}$ in general position, whose complement has $f(n-1,r)$ connected components, each cutting a connected component of the previous arrangement in half and thus adding $$f(n-1,r)=f(n,r+1)-f(n,r)$$ connected components. This formula entails $$\begin{split} f(n,r)-f(n-1,r-1) &=f(n,r)-f(n-1,r)+f(n-1,r)-f(n-1,r-1) \\ &=f(n-1,r-1)+\binom{r}{n}-\binom{r-1}{n-1}\end{split}$$ and hence $$\begin{split} f(n,r) &=\mathcal{H}(n,r)+\binom{r}{n}-\binom{r-1}{n-1} \\ &=\mathcal{H}(n,r)+\binom{r-1}{n-1}\frac{r-n}{n}\ .\end{split}$$ Particularly, this entails that not all connected components of the complement of the zero level set of a product of $r$ linear polynomials in $\mathbb{R}^n$ are hyperbolicity cones if $r>n$, contrary to what happens for $r\leq n$. This was already to be expected since, as I discussed above, hyperbolicity cones always come in pairs and $f(r,n)$ can clearly fail to be even for $r>n$. This also entails that connected components of the complement of the zero set of an (homogeneous) hyperbolic polynomial may be convex cones without being hyperbolicity cones.

Notice that the saturation of the bound occurs precisely as conjectured for the number of connected components of the complement of an arrangement of $r$ hyperplanes of $\mathbb{R}^n$ in general position (see Boris Bukh's comment below). A connection between both numbers is reached through the reasoning discussed in Aaron's comment to his answer: let $$f(n,r)=\begin{cases}2^r &\text{for }r\leq n\ ,\\ \sum^{n}_{k=0}\binom{r}{k} &\text{for }r>n\ .\end{cases}$$ be the latter number, so that we clearly have $$\mathcal{H}(n,r)=\begin{cases} f(n,r) & \text{for }r\leq n\ ,\\ 2f(n-1,r-1) &\text{for }r>n\ .\end{cases}$$ Let us consider from now on the case $r>n$ - adding an $(r+1)$-th hyperplane $E$ in general position to the arrangement causes the intersection of the arrangement with $E$ to be an arrangement of $r$ hyperplanes in $\mathbb{R}^{n-1}$ in general position, whose complement has $f(n-1,r)$ connected components, each cutting a connected component of the previous arrangement in half and thus adding $$f(n-1,r)=f(n,r+1)-f(n,r)$$ connected components. This formula entails $$\begin{split} f(n,r)-f(n-1,r-1) &=f(n,r)-f(n-1,r)+f(n-1,r)-f(n-1,r-1) \\ &=f(n-1,r-1)+\binom{r}{n}-\binom{r-1}{n-1}\end{split}$$ and hence $$\begin{split} f(n,r) &=\mathcal{H}(n,r)+\binom{r}{n}-\binom{r-1}{n-1} \\ &=\mathcal{H}(n,r)+\binom{r-1}{n-1}\frac{r-n}{n}\ .\end{split}$$ Particularly, this entails that not all connected components of the complement of the zero level set of a product of $r$ linear polynomials in $\mathbb{R}^n$ are hyperbolicity cones if $r>n$, contrary to what happens for $r\leq n$. This was already to be expected since, as I discussed above, hyperbolicity cones always come in pairs and $f(r,n)$ can clearly fail to be even for $r>n$. This also entails that connected components of the complement of the zero set of an (homogeneous) hyperbolic polynomial may be convex cones without being hyperbolicity cones.

Of course, the bound $\mathcal{H}(n,r)$ is far tighter than the Milnor-Thom bound, but too tight for being the desired answer.

Update: This is not exactly an answer to my question (hence it does not appear as a separate answer) but rather to the original question that motivated it - to wit, a sharp upper bound on the number of hyperbolicity cones of an (homogeneous) hyperbolic polynomial. Since the result is connected to Aaron Meyerowitz's bounty-awarded answer, I feel it is worthwhile to bring it up here.

Thorsten Jörgens and Thorsten Theobald have just posted (March 15th, 2017) the arXiv preprint Hyperbolicity Cones and Imaginary Projections, arXiv:1703.04988 [math.AG].

The main result of the paper (Theorem 1.1) states (I have retained only the part of the statement that interests us here and added a short explanation on the notation):

Let $f\in\mathbb{R}[\mathbf{z}]$ be an homogeneous $n$-variable polynomial of degree $r$ with real coefficients. Then the number of hyperbolicity cones of $f$ is at most $$\mathcal{H}(n,r)=\begin{cases} 2^r & \text{for }r\leq n\ ,\\ 2\sum^{n-1}_{k=0}\binom{r-1}{k} & \text{for }r>n\ .\end{cases}$$ The maximum is attained if and only if $f$ is a product of linear polynomials.

Notice that the saturation of the bound occurs precisely as conjectured for the number of connected components of the complement of an arrangement of $r$ hyperplanes of $\mathbb{R}^n$ in general position (see Boris Bukh's comment below). A connection between both numbers is reached through the reasoning discussed in Aaron's comment to his answer: if $r>n$, let $$f(n,r)=\sum^{n}_{k=0}\binom{r}{k}$$ be the latter number, so that $$\mathcal{H}(n,r)=2f(n-1,r-1)$$ in this case. Then adding an $(r+1)$-th hyperplane $E$ in general position to the arrangement causes the intersection of the arrangement with $E$ to be an arrangement of $r$ hyperplanes in $\mathbb{R}^{n-1}$ in general position, whose complement has $f(n-1,r)$ connected components, each cutting a connected component of the previous arrangement in half and thus adding $$f(n-1,r)=f(n,r+1)-f(n,r)$$ connected components. This formula entails $$\begin{split} f(n,r)-f(n-1,r-1) &=f(n,r)-f(n-1,r)+f(n-1,r)-f(n-1,r-1) \\ &=f(n-1,r-1)+\binom{r}{n}-\binom{r-1}{n-1}\end{split}$$ and hence $$\begin{split} f(n,r) &=\mathcal{H}(n,r)+\binom{r}{n}-\binom{r-1}{n-1} \\ &=\mathcal{H}(n,r)+\binom{r-1}{n-1}\frac{r-n}{n}\ .\end{split}$$ Particularly, this entails that not all connected components of the complement of the zero level set of a product of $r$ linear polynomials in $\mathbb{R}^n$ are hyperbolicity cones if $r>n$, contrary to what happens for $r\leq n$. This was already to be expected since, as I discussed above, hyperbolicity cones always come in pairs and $f(r,n)$ can clearly fail to be even for $r>n$. This also entails that connected components of the complement of the zero set of an (homogeneous) hyperbolic polynomial may be convex cones without being hyperbolicity cones.