# Irreducible hyperbolic polynomials of any degree over the reals

Is it true that there are irreducible hyperbolic polynomials $p(x,y,z) \in \mathbb{R}[x,y,z]$, $p$ homogeneous of any degree? Are there even concrete examples for such polynomials?

I know that there are irreducible polynomials of any degree and that there are hyperbolic polynomials of any degree in this setting, but I do not see whether also polynomials which have both properties exist of any degree.

Edit: By hyperbolic I mean the following:

$p$ is hyperbolic with respect to $\textbf{e} \in \mathbb{R}^3$ if $p(\textbf{e}) > 0$ and for all vectors $\textbf{x} \in \mathbb{R}^3$ the univariate polynomial $t \mapsto p(\textbf{x} - t\textbf{e})$ has only real roots.

• Could you define "hyperbolic polynomial"? – paf Aug 15 '16 at 16:34

Is it true that there are irreducible hyperbolic polynomials $p(x,y,z) \in \mathbb{R}[x,y,z]$, $p$ homogeneous of any degree? Are there even concrete examples for such polynomials?

Yes, there are. The set of reducible polynomials in the set $V$ of real homogeneous polynomials of degree $d$ in $n$ variables (in your case $n = 3$) is "thin" in the sense that it does not contain any subset which is open with respect to the Euclidean topology (this is a basic fact from algebraic geometry). On the other hand, the set of all hyperbolic polynomials in $V$ has nonempty interior — see this paper of Nuij here.

Nuij, Wim. A note on hyperbolic polynomials. Math. Scand. 23 1968 69–72 (1969).

Thus, there must be irreducible hyperbolic polynomials. Let me just add briefly a few ways to construct explicit examples.

1. Take a directional derivative of a reducible hyperbolic polynomial. For example, one could take the reducible hyperbolic polynomial given by a product of $d + 1$ linear forms. A directional derivative of this polynomial (with respect to a point not on any of the hyperplanes defined by these linear forms. A directional derivative of this polynomial (with respect to a point not on any of the hyperplanes defined by these linear forms) should be irreducible.

One really nice example of this is the elementary symmetric functions, which can be obtained by taking directional derivatives of the product $x_1x_2\ldots x_n$ with respect to the all ones vector $(1, 1, \ldots, 1)$.

2. Randomly pick $n - 1$ real symmetric matrices of size $d \times d$, say $A_2, \ldots, A_n$, and let $A_1$ be the $d\times d$ identity matrix. The polynomial given by the determinant of the matrix$$x_1A_1 + x_2A_2 + \ldots + x_nA_n$$will have degree $d$ and be hyperbolic with respect to the coordinate vector $(1, 0, \ldots, 0)$. By the argument I described above, for "almost all" choices of matrices $A_2, \ldots, A_n$, this polynomial will also be irreducible.

Do you have any reference for the set of reducible polynomials being thin?

It follows from a count of dimensions. This may be an overly complicated explanation, but the dimension of the vector space of polynomials of derggee at most $d$ in $n$ variables is the binomial coefficient$$\binom{n+d}{n}.$$For $n > 2$ and any $1 \le k \le d$, this is more than the dimension of the set of product of polynomials of degree $k$ and $d - k$. The dimension of products should be$$\binom{n + k}{n} + \binom{n + d - k}{n} - 1.$$

Thinking about your answer again, I realize that I do not really understand the last part of it, since I fail to see what we count the dimensions of. I see that we define a map$$\phi: \mathbb{R}^{\le k}[x_1, \ldots, x_n] \times \mathbb{R}^{\le d - k}[y_1, \ldots, y_n] \mapsto \mathbb{R}^{\le d}[x_1, \ldots, x_n], \quad \phi(p, r) = p \cdot r.$$On the left hand side, we have vector spaces, which I can count the dimension of as you explained, but the reducible polynomials do not form a subspace of the polynomials of degree smaller equal $d$, so why does this dimension counting help?

That is true. The dimension I am referring to is not the vector space dimension, but rather the dimension of an algebraic variety in the sense of algebraic geometry.

For the definition of that dimension, I'll refer you to textbooks on algebraic geometry. However, it has the following two properties that I used for my argument.

1. If $X$ is some set of dimension $d$ and $\phi$ a polynomial map, then $\phi(X)$ has dimension at most $d$.
2. If the dimension of a set $X$ in $\mathbb{R}^n$ (or $\mathbb{C}^n$) is less than $n$, then $X$ is contained in a hypersurface. In other words, there is some nonzero polynomial that vanishes identically on $X$. In particular, this means that $X$ is "thin". (For instance, it has measure $0$.)

Furthermore, we are looking here at polynomials of degree lesser equal $d$. However, we only know by Nuij that the hyperbolic polynomials have nonempty interior in the homogeneous polynomials of degree $d$, so we do not know if they also have nonempty interior $\mathbb{R}^{\le d}[x_1, \ldots, x_n]$, do we? I think I am somehow missing your main idea.

You are right. One should work inside the space of homogeneous polynomials of degree $d$. But the dimension argument will still work in the same way as before.

1. If $X$ is some set of dimension $d$ and $\phi$ a polynomial map, then $\phi(X)$ has dimension at most $d$.

In fact, there is one more thing going on here: If a polynomial is a product of the form of degree $k$ and $d - k$, then this decomposition is not unique, only up to a scalar. So (if you don't know about projective spaces, then) the appropriate version of the statement would be "If $X$ is some set of dimension $d$ and $\phi$ a polynomial map, then $\phi(X)$ has dimension at most $d - e$, where $e$ is the dimension of the general fiber.

Example: Let $V_i$ be the space of forms of degree $i$ in three variables. Then $\dim V_1 = 3$ and $\dim V_2 = 6$. So the the map $V_1 \times V_1 \to V_2$ that is given by multiplication of two linear forms goes from something of dimension $6$ to something of dimension $6$ with general fiber having dimension $1$. Thus the image has dimension $5$ and thus a "thin" set in the sense used before.

• Thank you, this looks convincing. Do you have any reference for the set of reducible polynomials being thin? (Despite this being a basic fact, I was so far unable to find it, but maybe I just did not know how to look) – Andreas B Aug 16 '16 at 13:15
• I see, thank you very much – Andreas B Aug 18 '16 at 9:29
• Thinking about your answer again, I realize that I do not really understand the last part of it, since I fail to see what we count the dimensions of. I see that we define a map $\phi: \mathbb{R}^{\leq k}[x_1,\ldots, x_n] \times \mathbb{R}^{\leq d-k}[y_1,\ldots, y_{n}] \mapsto \mathbb{R}^{\leq d}[x_1,\ldots, x_n], \phi(p, r) = p \cdot r$. On the left hand side, we have vector spaces, which I can count the dimension of as you explained, but the reducible polynomials do not form a subspace of the polynomials of degree smaller equal $d$, so why does this dimension counting help? – Andreas B Aug 25 '16 at 9:41
• Furthermore, we are looking here at polynomials of degree lesser equal $d$. However, we only know by Nuij that the hyperbolic polynomials have non-empty interior in the homogeneous polynomials of degree $d$, so we do not know if they also have non-empty interior $\mathbb{R}^{\leq d}[x_1, \ldots, x_n]$, do we? I think I am somehow missing your main idea. – Andreas B Aug 25 '16 at 9:44
• Thanks again, I think I got it now – Andreas B Aug 31 '16 at 13:43