The polynomial inverse-factorization alternative for invertible noncommutative polynomial matrices

At least 5 years old · documented by

Let K⟨x‾⟩\mathbb K\langle\underline{x}\rangle denote the free noncommutative polynomial algebra, let r=(pij)∈K⟨x‾⟩L×L\mathfrak r=(p_{ij})\in\mathbb K\langle\underline{x}\rangle^{L\times L} have noncommutative polynomial entries, and let a‾=(a1,…,ad)∈Ad\underline{\mathfrak a}=(\mathfrak a_1,\ldots,\mathfrak a_d)\in\mathcal A^d. Assume that

r(a‾)=(pij(a‾))i,j∈AL×L\mathfrak r(\underline{\mathfrak a})=(p_{ij}(\underline{\mathfrak a}))_{i,j}\in\mathcal A^{L\times L}

is invertible. Polynomial inverse-factorization conjecture. There exist either noncommutative rational expressions S11,…,SLLS_{11},\ldots,S_{LL} satisfying

a‾∈⋂i,jdom⁡A(Sij),\underline{\mathfrak a}\in\bigcap_{i,j}\operatorname{dom}_{\mathcal A}(S_{ij}),

or noncommutative polynomials S11,…,SLLS_{11},\ldots,S_{LL}, such that (Sij(a‾))i,j(S_{ij}(\underline{\mathfrak a}))_{i,j} is invertible and there exist 1≤i,j≤L1\leq i,j\leq L for which

∑k=1Lpik(a‾)Skj(a‾)\sum_{k=1}^L p_{ik}(\underline{\mathfrak a})S_{kj}(\underline{\mathfrak a})

is invertible in A\mathcal A. The assertion is presented as an alternative formulation needed to prove the preceding matrix-inverse conjecture.

References

Primary source

Motke Porat and Victor Vinnikov, “Realizations of non-commutative rational functions around a matrix centre, II: The lost-abbey conditions”, arXiv:2009.08527 (2022).

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.