What is the volume of the $3$-dimensional elliptope?

My question

Compute the following double integral analytically

$$\int_{-1}^1 \int_{-1}^1 2 \sqrt{x^2 y^2 - x^2 - y^2 + 1} \,\, \mathrm{d} x \mathrm{d} y$$

Background

The $3$-dimensional elliptope is the spectrahedron defined as follows

$$\mathcal E_3 := \Bigg\{ (x_{12}, x_{13}, x_{23}) \in \mathbb R^3 : \begin{bmatrix} 1 & x_{12} & x_{13}\\ x_{12} & 1 & x_{23}\\ x_{13} & x_{23} & 1\end{bmatrix} \succeq 0 \Bigg\}$$

Using Sylvester's criterion for positive semidefiniteness (i.e., all $2^3-1 = 7$ principal minors are nonnegative), we obtain $1 \geq 0$ (three times), the three quadratic inequalities

$$1 - x_{12}^2 \geq 0 \qquad \qquad \qquad 1 - x_{13}^2 \geq 0 \qquad \qquad \qquad 1 - x_{23}^2 \geq 0$$

and the cubic inequality.

$$\det \begin{bmatrix} 1 & x_{12} & x_{13}\\ x_{12} & 1 & x_{23}\\ x_{13} & x_{23} & 1\end{bmatrix} = 1 + 2 x_{12} x_{13} x_{23} - x_{12}^2 - x_{13}^2 - x_{23}^2 \geq 0$$

Thus, $\mathcal E_3$ is contained in the cube $[-1,1]^3$. Borrowing the pretty figure in Eisenberg-Nagy & Laurent & Varvitsiotis, here is an illustration of $\mathcal E_3$ What is the volume of $\mathcal E_3$?

Motivation

Why is $\mathcal E_3$ interesting? Why bother? Because $\mathcal E_3$ gives us the set of $3 \times 3$ correlation matrices.

My work

For convenience,

$$x := x_{12} \qquad\qquad\qquad y := x_{13} \qquad\qquad\qquad z := x_{23}$$

I started with sheer brute force. Using Haskell, I discretized the cube $[-1,1]^3$ and counted the number of points inside the elliptope. I got an estimate of the volume of $\approx 4.92$.

I then focused on the cubic surface of the elliptope

$$\det \begin{bmatrix} 1 & x & y\\ x & 1 & z\\ y & z & 1\end{bmatrix} = 1 + 2 x y z - x^2 - y^2 - z^2 = 0$$

which I rewrote as follows

$$z^2 - (2 x y) z + (x^2 + y^2 - 1) = 0$$

Using the quadratic formula, I obtained

$$z = x y \pm \sqrt{x^2 y^2 - x^2 - y^2 + 1}$$

Integrating using Wolfram Alpha,

$$\int_{-1}^1 \int_{-1}^1 2 \sqrt{x^2 y^2 - x^2 - y^2 + 1} \,\, \mathrm{d} x \mathrm{d} y = \cdots \color{gray}{\text{(magic happens)}} \cdots = \color{blue}{\frac{\pi^2}{2} \approx 4.9348}$$

I still would like to compute the double integral analytically. I converted to cylindrical coordinates, but did not get anywhere.

Other people's work

This is the same value Johnson & Nævdal obtained in the 1990s: Thus, the volume is

$$\left(\frac{\pi}{4}\right)^2 2^3 = \frac{\pi^2}{2}$$

However, I do not understand their work. I do not know what Schur parameters are.

Here's the script:

-- discretization step
delta = 2**(-9)

-- discretize the cube [-1,1] x [-1,1] x [-1,1]
grid1D = [-1,-1+delta..1]
grid3D = [ (x,y,z) | x <- grid1D, y <- grid1D, z <- grid1D ]

-- find points inside the 3D elliptope
points = filter (\(x,y,z)->1+2*x*y*z-x**2-y**2-z**2>=0) grid3D

-- find percentage of points inside the elliptope
p = (fromIntegral (length points)) / (1 + (2 / delta))**3


*Main> delta
1.953125e-3
*Main> p
0.6149861105903861
*Main> p*(2**3)
4.919888884723089


Hence, approximately $61\%$ of the grid's points are inside the elliptope, which gives us a volume of approximately $4.92$.

A new Buffon's needle

A symmetric $3 \times 3$ matrix with

• $1$'s on the main diagonal

• realizations of the random variable whose PDF is uniform over $[-1,1]$ on the entries off the main diagonal

is positive semidefinite (and, thus, a correlation matrix) with probability $\left(\frac{\pi}{4}\right)^2$. Estimating the probability, we estimate $\pi$. Using the estimate given by the Haskell script:

*Main> 4 * sqrt 0.6149861105903861
3.1368420058151125


References

• This is an astoundingly good question! Nice! – clathratus Apr 16 at 1:08

The integral can be separated:

$$I = 2\int_{-1}^1 \sqrt{1-x^2} dx \cdot \int_{-1}^1 \sqrt{1-y^2} dy = 2\left(\int_{-1}^1 \sqrt{1-t^2} dt\right)^2$$

This integral is straight-forward using the substitution $$t=\sin\theta$$:

$$\int_{-1}^1 \sqrt{1-t^2} dt = \int_{-\pi/2}^{\pi/2} \sqrt{1-\sin^2\theta} \cos\theta d\theta = \int_{-\pi/2}^{\pi/2} |\cos\theta|\cos\theta d\theta$$

$$=\int_{-\pi/2}^{\pi/2} \cos^2\theta d\theta = \dfrac{1}{2}\int_{-\pi/2}^{\pi/2} (1+\cos2\theta) d\theta = \dfrac{1}{2}\left(\theta + \dfrac{1}{2}\sin 2\theta\right)\Big|_{-\pi/2}^{\pi/2} = \dfrac{\pi}{2}$$

Therefore

$$I = 2\left(\int_{-1}^1 \sqrt{1-t^2} dt\right)^2 = 2\left(\dfrac{\pi}{2}\right)^2 = \dfrac{\pi^2}{2}$$

• Thanks. Shame on me for not realising that $(1-x^2) (1-y^2) = 1 - x^2 - y^2 + x^2 y^2$. I never expected the double integral to be this easy. – Rodrigo de Azevedo Oct 16 '16 at 19:19
• easy to get lost in the bigger picture, sometimes fresh eyes can help – David Peterson Oct 16 '16 at 19:21

Then integrand factors as $\sqrt{(1-x^2)(1-y^2)}=\sqrt{(1-x^2)}\sqrt{(1-y^2)}$ and every factor can be integrated separately. But you recognize the integral for the area of a half circle of radius $1$, hence

$$I=2\left(\frac\pi2\right)^2.$$