The deletion game #
The global, purely combinatorial step of a border-rank lower bound of about
(1 + 2D/3) m for tight tensors with m = 2k + 1 slices Fin (2 * k + 1); for
D = (2p + 1) / (p + 1) this is (7/3 - 2 / (3 (p + 1))) m, so 19/9 m at p = 2. Border
substitution (Algebraic.Tensor3.Tight.exists_list_add_borderRank_le) gives an ordering of the
slices along which deleting the first q slices lowers the border rank by at least q. The
deletion game shows that for every ordering some stage q has q plus a lower bound LB on
the border rank of the surviving slices at least (1 + 2D/3) m - D E - (8 L + R m / L + 2).
The local input is the paired-cluster hypothesis DeletionGame.PairedClusterBound on LB
(see Complexitylib.Algebraic.LinearAlgebra.Tensor.DeletionGame.Defs), which this file takes
as a hypothesis: whenever a positive and a negative block of length L each keep R
survivors, LB is at least D (Φ_m(r, s) - E) for offsets r and -s in the two blocks.
- The width function.
Φ_n(r, s) = Φ_n(s, r)(DeletionGame.phi_comm),Φ_n(r, s) ≥ 2n/3(DeletionGame.two_mul_le_three_mul_phi), andΦ_nis2-Lipschitz in each argument (DeletionGame.phi_le_add). - Two-interval lemma (
DeletionGame.filter_phi_le_subset,DeletionGame.two_mul_card_filter_phi_le). Forn = 2k + 1ands ≥ 0, the positionsr ∈ [1, k]withΦ_n(r, s) ≤ Tlie in two integer intervals and number at most(3T - 2n + 3) / 2. - The count at the stopping time (
DeletionGame.le_sub_card_add_of_lost). If one side keeps a good block while the other loses its last good block at the next deletion, the survivors number at most about3T/2 - m + O(L + R m / L), whereD (T - E)is the paired-cluster bound; withT ≥ 2m/3andD ≥ 3/2this gives the bound. - The deletion game (
DeletionGame.PairedClusterBound.exists_stage). Stop at the last stage at which both signs have a good block. The first blocks are full before any deletion, so this needs1 ≤ R ≤ L ≤ k. - Combined with border substitution (
Tight.le_borderRank_of_pairedClusterBound,Tight.not_borderRankLE_of_pairedClusterBound). A tight tensor with nonzero slices, satisfying the paired-cluster hypothesis for the border ranks of its slice restrictions, has border rank at least(1 + 2D/3) m - D E - (8 L + R m / L + 2).
The width function #
The two-interval lemma #
Two-interval lemma. For n = 2k + 1, the positions r ∈ [1, k] with Φ_n(r, s) ≤ T
lie in the two intervals [n - T, ⌊T/2⌋ - s] and [⌈n - T/2⌉ - s, min (k, T - s)].
Two-interval lemma, counting form. For n = 2k + 1, s ≥ 0, and 2n ≤ 3T + 3, at
most (3T - 2n + 3) / 2 positions r ∈ [1, k] have Φ_n(r, s) ≤ T. (If some position
qualifies, then 2n ≤ 3T by two_mul_le_three_mul_phi.)
The deletion game #
The paired-cluster hypothesis passes to any larger function.
The count at the stopping time. Let f be injective with values in [-k, k]. Its
blocks sideBlock f L i form the winning side and the blocks sideBlock (-f) L j the losing
side. Suppose that after deleting at most one element of S, leaving S', no losing block is
good, that the winning block i₁ is good for S, and that every good winning block i for
S pairs with a fixed losing block j₀ as in the paired-cluster hypothesis, with value V.
Then (1 + 2D/3) n - D E - (8 L + R n / L + 2) ≤ (n - card S) + V for n = 2k + 1.
The deletion game. Assume the paired-cluster hypothesis for LB with D ≥ 3/2, and
1 ≤ R ≤ L ≤ k. Then for every ordering l of the m = 2k + 1 slices (l.Nodup and
l.toFinset = univ), some stage q ≤ m satisfies
(1 + 2D/3) m - D E - (8 L + R m / L + 2) ≤ q + LB S_q, where S_q is the set of slices
left after deleting the first q elements of l.
The deletion game with border substitution. Let T be tight with m = 2k + 1 nonzero
A-slices, and suppose the border ranks of its slice restrictions satisfy the paired-cluster
hypothesis with D ≥ 3/2, for 1 ≤ R ≤ L ≤ k. Then
borderRank T ≥ (1 + 2D/3) m - D E - (8 L + R m / L + 2).
The deletion game with border substitution, as a non-membership statement: under the
hypotheses of Tight.le_borderRank_of_pairedClusterBound, T does not have border rank at
most any r < (1 + 2D/3) m - D E - (8 L + R m / L + 2).