Polynomial lower-bound conjecture for least singular values of random matrix polynomials

From papers

Let P(U1,,Uk,U1,,Uk)P(U_1,\ldots,U_k,U_1^*,\ldots,U_k^*) be a noncommutative nonzero *-polynomial with complex coefficients, let λ0\underline{\lambda}\neq 0 be a signature, and let U1,,UkU_1,\ldots,U_k be independently Haar-distributed elements of U(n)U(n). Write πλ\pi_{\underline{\lambda}} for the corresponding representation and let σmin\sigma_{\min} denote the least singular value. Polynomial lower-bound conjecture. There is a constant c=c(P,λ)>0c=c(P,\underline{\lambda})>0, depending only on PP and λ\underline{\lambda}, such that

P[(U1,,Uk)U(n)k: σmin(P(πλ(U1),,πλ(Uk)))>1nc]n1.\mathbb{P}\left[(U_1,\ldots,U_k)\in U(n)^k:\ \sigma_{\min}\left(P(\pi_{\underline{\lambda}}(U_1),\ldots,\pi_{\underline{\lambda}}(U_k))\right)>\frac{1}{n^c}\right]\xrightarrow{n\to\infty}1.

The conjecture asks for a uniform polynomial lower bound on the least singular value of a noncommutative random matrix polynomial. The surrounding discussion explains that such control is needed for estimates involving singular bundles and that even simpler random-matrix questions can be difficult; the source gives no resolution status.

Progress summary

Open

No public discussion or published progress was found, so the conjecture appears open.

No public discussion or published progress concerning this conjecture was found.

Current status (as of August 2026): No result settling the conjecture is recorded, and the conjecture remains open.

Sources & referencesView supporting material

Primary source

Masoud Zargar, “Random flat bundles and equidistribution”, arXiv:2210.09547 (2023).

Solutions 1

Counterexample

The conjecture is false for a nonzero polynomial that is not a unitary relation and for a nontrivial fixed balanced signature.

Take k=1k=1 and

P(X)=XI.P(X)=X-I.

For every n2n\ge2, take the dominant highest weight

λn=(1,0,,0,1),\lambda_n=(1,0,\ldots,0,-1),

obtained by the usual zero-extension of the fixed nonzero balanced signature (1,1)(1,-1). The corresponding irreducible representation of U(n)U(n) is its adjoint action on

sln(C)={HMn(C):trH=0},πλn(U)H=UHU1.\mathfrak{sl}_n(\mathbb C) =\{H\in M_n(\mathbb C):\operatorname{tr}H=0\}, \qquad \pi_{\lambda_n}(U)H=UHU^{-1}.

Given any UU(n)U\in U(n), choose a unitary diagonalization U=VDVU=VDV^* and let

H=Vdiag(1,1,0,,0)V.H=V\operatorname{diag}(1,-1,0,\ldots,0)V^*.

Then H0H\ne0, trH=0\operatorname{tr}H=0, and UH=HUUH=HU. Consequently

P(πλn(U))H=(πλn(U)I)H=0.P(\pi_{\lambda_n}(U))H =(\pi_{\lambda_n}(U)-I)H=0.

In fact, the traceless diagonal matrices in the same eigenbasis provide a kernel of dimension at least n1n-1. Therefore

σmin ⁣(P(πλn(U)))=0for every UU(n), n2.\sigma_{\min}\!\left(P(\pi_{\lambda_n}(U))\right)=0 \qquad\text{for every }U\in U(n),\ n\ge2.

For every c>0c>0, it follows that

P[σmin ⁣(P(πλn(U)))>nc]=0,\mathbb P\left[ \sigma_{\min}\!\left(P(\pi_{\lambda_n}(U))\right)>n^{-c} \right]=0,

whereas the conjecture requires this probability to converge to 11.

More generally, the source already observes that for every nontrivial balanced signature, πλ(U)\pi_\lambda(U) has eigenvalue 11 for every UU. Thus P(X)=XIP(X)=X-I disproves the assertion throughout the balanced-signature regime.

0 endorsements
Shivam Patel ·