Lorentzian Polynomials · Notes for directed study
Finish presenting Chapter 2.1 and proof of Theorem 2.23 (beginning)
This week's event is hosted by Marcus Chen and Ivy Liu.
General content of Marcus' part
Today I've already presented all of Chapter 2.1. The last theorem we finished is Theorem 2.13: the closure of strictly Lorentzian polynomials in \((H_n^d)\) contains Lorentzian polynomials. It means \(\overline{\mathring{L}_n^d} \supseteq L_n^d\).
Our first short term goal is Thm 2.25: The closure of strictly Lorentzian polynomials in \((H_n^d)\) is Lorentzian polynomials.General content of Ivy's part
We break the proof of \(L_n^d\) being closed into two main pieces:
- Proving that polynomials in \(L_n^d\) are \(c\)-Rayleigh (Prop 2.19).
- Proving that being homogeneous and \(c\)-Rayleigh implies having an \(M\)-convex support (Theorem 2.23).
We successfully prove the first part. Here is a quick breakdown of how we got there:
- By looking at the \(d=2\) case, we found that if the Hessian matrix has exactly one positive eigenvalue, we directly get the \(1\)-Rayleigh property.
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The same logic generalizes perfectly to arbitrary dimensions to yield the \(2(1-1/d)\)-Rayleigh property.
Thus, the problem reduces to proving that the Hessian of any polynomial in \(L_n^d\) has exactly one positive eigenvalue (Thm 2.16(2)).
The homotopy defined in Thm 2.13 implies that \(\mathring{L}_n^d\) is connected, and thus we only needed to show two things:
- The Hessian is nonsingular everywhere in \(\mathring{L}_n^d\) (Thm 2.16(1)), and thus the eigenvalue cannot be 0.
- There exists at least one polynomial in \(\mathring{L}_n^d\) whose Hessian has exactly one positive eigenvalue. In fact, we showed that non-zero stable polynomials with non-negative coefficients have Hessians with exactly one positive eigenvalue (Prop 2.14).