5

I hope to find a second-order conic (SOC) relaxation for $z = xy$, but it seems very hard.

Rodrigo de Azevedo
  • 23,223
Shuai Lu
  • 53

1 Answers1

6

The 3x3 moment matrix is $\begin{pmatrix} 1 & x & y\\x & x^2 & xy\\y & xy & y^2\end{pmatrix}$ which is psd and rank-1. Introduce relaxation variables and your semidefinite relaxation is $\begin{pmatrix} 1 & x & y\\x & X_{20} & X_{11}\\y & X_{11} & X_{02}\end{pmatrix}\succeq 0$ (with $z$ thus being $X_{11}$). For this to be psd, all minors have to be psd, so a necessary condition is (i.e. relaxation)

$$\begin{pmatrix} 1 & x\\x & X_{20}\end{pmatrix}\succeq 0,\begin{pmatrix} 1 & y\\y & X_{02}\end{pmatrix}\succeq 0, \begin{pmatrix} X_{20} & X_{11}\\X_{11} & X_{02}\end{pmatrix}\succeq 0$$

A 2x2 semidefinite constraint $\begin{pmatrix} a &b\\b & c\end{pmatrix}\succeq 0$ can be represented with an SOCP by looking at the minors and after an SOCP representation of the quadratic constraint you have $\left|\left|\begin{matrix}2b\\a-c\end{matrix}\right|\right|\leq a + c, a\geq 0, c\geq 0$

Johan Löfberg
  • 9,737
  • 1
  • 16
  • 15
  • 2
    Nice answer. One thing to note: this relaxation is looser than the SDP relaxation, since it doesn't enforce the constraint that the determinant of the $3 \times 3$ matrix is non-negative. In general we can't enforce $3 \times 3$ SDP constraints using SOCP, as proved by Fawazi here. However, since there is a diagonal entry of "1" it might (I haven't checked) be possible to do a congruence transform from the moment matrix to one with arrow structure, and then enforce the SDP constraint (see reference 7 in the article I linked). – Ryan Cory-Wright Jul 06 '19 at 16:43
  • 2
    Good point, will make that more clear – Johan Löfberg Jul 06 '19 at 19:13
  • Thank you both very much. Your answers inspire me a lot. – Shuai Lu Jul 08 '19 at 02:02