7
$\begingroup$

Denote $\pmb{a}=(a_1,\dots,a_d)\in\mathbb{R}^d$ and consider the set $$\mathcal{E}_d=\{\pmb{a}\in\mathbb{R}^d: \text{each root $\xi$ of $x^d+a_dx^{d-1}+\cdots+a_2x+a_1=0$ lies in $\vert\xi\vert<1$}\}.$$ In the reference shown below, Fam proved that the $d$-dimensional Lebesgue measure satisfies $$\lambda_d(\mathcal{E}_d)=2^d\prod_{k=1}^{\lfloor\frac{d}2\rfloor}\left(1+\frac1{2k}\right)^{2k-d}.$$ I'd like to propose a complex version here. Denote $\pmb{c}=(c_1,\dots,c_d)\in\mathbb{C}^d$ and consider the set $$\mathcal{S}_d=\{\pmb{c}\in\mathbb{C}^d: \text{each root $\xi$ of $x^d+c_dx^{d-1}+\cdots+c_2x+c_1=0$ lies in $\vert\xi\vert<1$}\}.$$ Now, let's ask:

QUESTION. What is the $2d$-dimensional Lebesgue measure $$\lambda_{2d}(\mathcal{S}_d)?$$

For contrast, $\lambda_1(\mathcal{E}_1)=2$ while $\lambda_2(\mathcal{S}_1)=\pi$.

Reference.

A. T. Fam,The volume of the coefficient space stability domain of monic polynomials, Proc. IEEE Int. Symp.Circuits and Systems, 2 (1989), pp. 1780–1783.

$\endgroup$
5
$\begingroup$

$\def\CC{\mathbb{C}}\def\RR{\mathbb{R}}$The answer is $\tfrac{\pi^n}{n!}$. It is certainly a surprise to have the answer come out so simple!

Let $\phi : \CC^n \to \CC^n$ be the map which takes $(z_1, z_2, \ldots, z_n)$ to the elementary symmetric functions $(e_1, e_2, \ldots, e_n)$ where $e_k = \sum_{1 \leq i_1 < i_2 < \cdots < i_k \leq n} z_{i_1} z_{i_2} \cdots z_{i_n}$. Let $D$ be the unit disc in $\CC$. So you want to compute the volume of $\phi(D^n)$, also known as $\int_{\phi(D^n)} \mathrm{Vol}$.

The map $D^n \to \phi(D^n)$ is $n!$ to $1$ and, since $\phi$ is complex analytic, $\phi$ is orientation preserving. So $$\int_{\phi(D^n)} \mathrm{Vol} = \frac{1}{n!} \int_{D^n} \phi^{\ast}(\mathrm{Vol}) = \frac{1}{n!} \int_{D^n} \det J^{\RR}_{\phi}$$ where $J^{\RR}_{\phi}$ is the Jacobian of $\phi$ considered as a smooth map $\RR^{2n} \to \RR^{2n}$. I will write $J^{\CC}_{\phi}$ when I instead want the $n \times n$ matrix of complex numbers coming from thinking of $\phi$ as a complex analytic map $\CC^n \to \CC^n$.

The relation between these two notions is this: $\det J^{\RR}_{\phi} = |\det J^{\CC}_{\phi}|^2$. (This is just linear algebra -- if $L/K$ is a degree $d$ field extension, $f: L^n \to L^n$ is a linear map and $g: K^{dn} \to K^{dn}$ is the linear map gotten by identifying $L$ with $K^d$, then $\det g = N_{L/K}(\det f)$.) We have the well known identity $$\det J^{\CC}_{\phi}(z_1, \ldots, z_n) = \prod_{i<j} (z_i - z_j) = \sum_{w \in S_n} (-1)^w z_1^{w(1)-1} z_2^{w(2)-1} \cdots z_n^{w(n)-1}.$$ Here is the first reference I found.

So we need to compute $$\frac{1}{n!} \int_{D^n} \sum_{u \in S_n} (-1)^u z_1^{u(1)-1} z_2^{u(2)-1} \cdots z_n^{u(n)-1} \overline{\sum_{v \in S_n} (-1)^v z_1^{v(1)-1} z_2^{v(2)-1} \cdots z_n^{v(n)-1}}.$$

We can distribute the product to get a sum of $(n!)^2$ terms of the form $\int_{D^n} \prod z_j^{a_j} \overline{z_j}^{b_j}$. If $a_j \neq b_j$, the integral on $z_j$ is zero, so we reduce to $$\frac{1}{n!} \sum_{u \in S_n} \int_{D^n} (z_1 \overline{z_1})^{u(1)-1} (z_2 \overline{z_2})^{u(2)-1} \cdots (z_n \overline{z_n})^{u(n)-1}.$$ All $n!$ terms have the same integral, so we are reduced to the one integral $$\int_{D^n} \prod (z_j \overline{z_j})^{j-1} = \prod_j \int_D (z \overline{z})^{j-1}.$$ Switching to polar coordinates, $$\int_D (z \overline{z})^{j-1} = 2 \pi \int_{r=0}^1 r^{2j-1} dr = \frac{\pi}{ j}.$$ So the final answer is $\prod_{j=1}^n \tfrac{\pi}{j} = \tfrac{\pi^n}{n!}$.

$\endgroup$
4
  • $\begingroup$ This is cool. Two things came to mind while reading your response: (1) somewhere it seems you had $det(A)=\pmb{Pfaff}(A)^2$; (2) the integral you wrote $\int \det$ when restricted to the hyperspheres may be translated as the "degree of a map" - I wonder if this has some relation here. Note: $\pmb{Pfaff}$ stands for Pfaffian of a matrix. $\endgroup$ – T. Amdeberhan Feb 20 '20 at 19:03
  • $\begingroup$ (1) I don't think I have any Pfaffians. The key fact is that, if $A : \mathbb{C}^n \to \mathbb{C}^n$ is a $\mathbb{C}$-linear map and $A^{\mathbb{R}}$ is the same map considered as an $\mathbb{R}$-linear map $\mathbb{R}^{2n} \to \mathbb{R}^{2n}$ then $\det A^{\RR} = |\det A|^2$. A determinant equaling a square is reminiscient of a Pfaffian, but I don't see one here. (2) Well, I use that $\phi$ has degree $n!$. I'm not sure what else to say. $\endgroup$ – David E Speyer Feb 20 '20 at 19:45
  • $\begingroup$ Cool. $\pi^n/n!$ is also the volume of $D^n$ in $\mathbb{C}^n / S_n$ just computed naively. $\endgroup$ – Nate Feb 20 '20 at 19:59
  • $\begingroup$ @Nate I noticed that but couldn't think what to say about it! $\endgroup$ – David E Speyer Feb 20 '20 at 20:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.