Open Problems in Integrable Systems — Nijenhuis operators: singular points and global properties

Definitions and mathematical statement

Shared notation.

$\mathbb N=\{1,2,\ldots\}$, and $\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$ denote the integers, rationals, reals and complex numbers. Any other conventions are specified locally.

A smooth endomorphism field $L:TM\to TM$ is Nijenhuis if, for all smooth vector fields $X,Y$, its torsion \[N_L(X,Y)=L^2[X,Y]+[LX,LY]-L[LX,Y]-L[X,LY]\] vanishes identically.

An endomorphism $A$ of an $n$-dimensional real vector space is algebraically regular if its centralizer $\{B:AB=BA\}$ has dimension $n$, the minimum possible dimension. Equivalently, over $\mathbb C$ there is one Jordan block for each eigenvalue. Eigenvalues need not be real or distinct. A closed manifold is compact and has no boundary.

Statement recorded in the supplied source.

Describe precisely the closed smooth manifolds $M$ which admit a smooth Nijenhuis endomorphism field $L$ such that $L_x:T_xM\to T_xM$ is algebraically regular at every $x\in M$. [S2]

Short English statement

Which closed manifolds admit a Nijenhuis tensor that is algebraically regular everywhere, allowing both nontrivial Jordan blocks and complex eigenvalues?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, UnsolvedMath: A Curated Collection of Open Mathematics Problems, record 2100508 (AMR-020-0508). CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Alexey Bolsinov, Vladimir S. Matveev, Eva Miranda and Serge Tabachnikov, Open Problems, Questions, and Challenges in Finite-Dimensional Integrable Systems (2018), Problem 5.10 and its complete preceding definitions. Original question and full discussion. https://arxiv.org/abs/1804.03737.