Equivalence of generalized and ordinary Haantjes integrability for commuting operators

Let MM be a differentiable manifold and let A,B:TMTM\boldsymbol{A},\boldsymbol{B}:TM\rightarrow TM be two (1,1)(1,1) tensor fields. Assume that HA(X,Y)=HB(X,Y)=0\mathcal{H}_{\boldsymbol{A}}(X,Y)=\mathcal{H}_{\boldsymbol{B}}(X,Y)=0 and [A,B]=0[\boldsymbol{A},\boldsymbol{B}]=0. Equivalence conjecture.

HA,B(X,Y)=0HA+B(X,Y)=0.\mathcal{H}_{\boldsymbol{A},\boldsymbol{B}}(X,Y)=0 \Longleftrightarrow \mathcal{H}_{\boldsymbol{A+B}}(X,Y)=0.

This conjecture concerns the relationship between the generalized Haantjes binary tensor of two commuting Haantjes operators and the ordinary Haantjes tensor of their sum; the source says that the result is conjectural, without providing a resolution.

Sources & referencesView supporting material

Primary source

Piergiulio Tempesta and Giorgio Tondo, “Higher Haantjes Brackets and Integrability”, arXiv:1809.05908 (2021).

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