Generic rank equality for odd-order Hankel tensors

Let m5m\geq 5 be odd, and let HHm(Cn)\mathcal{H}\in\operatorname{H}^m(\mathbb{C}^n) be a generic Hankel tensor. The generic Hankel rank conjecture. The answer to whether

rank(H)=rankS(H)=brank(H)=brankS(H)=rankV(H)\operatorname{rank}(\mathcal{H})=\operatorname{rank}_S(\mathcal{H})=\operatorname{brank}(\mathcal{H})=\operatorname{brank}_S(\mathcal{H})=\operatorname{rank}_V(\mathcal{H})

is yes. The equality is known for generic Hankel tensors of even order or order three, but the rank relations for generic odd-order Hankel tensors with m5m\geq 5 remain unresolved; the corresponding assertion for all Hankel tensors is false.

Sources & referencesView supporting material

Primary source

Jiawang Nie and Ke Ye, “Hankel tensor decompositions and ranks”, arXiv:1706.03631 (2019).

Additional references

2 papers in this index state this conjecture (2008–2017). The statement above is taken from the most recent of them; the others are arXiv:0805.3777.

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