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 $\lambda_2(\mathcal{S}_1)=\pi$.


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.


$\def\CC{\mathbb{C}}\def\RR{\mathbb{R}}$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 $\prod_{j=1}^n \tfrac{\pi}{j} = \tfrac{\pi^n}{n!}$.

  • $\begingroup$ 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

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