The Saxl Conjecture for Fourth Powers via the Semigroup Property Sammy Luo and Mark Sellke JMM 2016

Finite Group Representations Representation: map from group to 𝐺 𝐿 𝑛 ℂ Combine via direct sum, tensor product Can always decompose into irreducibles For 𝑆 𝑛 : irreducible rep ⇔ Young diagram ⇔ Partition 𝜆⊢𝑛 denotes partition of 𝑛 ⇔ 5,3,2

Examples of Representations Trivial Representation Alternating Representation Conjugation: tensor with alternating rep

Kronecker Coefficients 𝑔 𝜇𝜈 𝜆 = 1 𝑛! 𝜎∈ 𝑆 𝑛 𝜒 𝜆 (𝜎) 𝜒 𝜇 (𝜎) 𝜒 𝜈 (𝜎) 𝑔 𝜇𝜈 𝜆 is symmetric in 𝜆,𝜇,𝜈 𝑔 𝜇𝜈 𝜆 = 𝑔 𝜇 𝑡 𝜈 𝑡 𝜆 = 𝑔 𝜇 𝜈 𝑡 𝜆 𝑡 = 𝑔 𝜇 𝑡 𝜈 𝜆 𝑡 No known comb. interpretation, #P-hard Notation: 𝑐( 𝜆,𝜇,𝜈) means 𝑔 𝜇𝜈 𝜆 >0 Definition. The Kronecker Coefficient 𝑔 𝜇𝜈 𝜆 is the multiplicity of 𝜆 in the decomposition of 𝜇⊗𝜈. formula sucks

The Tensor Square Conjecture Staircase case - Saxl conjecture Convention: 𝑛= 𝑚+1 2 , so 𝜚 𝑚 ⊢𝑛 Conjecture. For every 𝑛 except 2, 4, 9 there exists a partition 𝜆⊢𝑛 such that 𝑐(𝜆,𝜆,𝜇) for all 𝜇⊢𝑛. 𝜚 5 = We’re going to focus on the saxl conjecture (but some of our results generalize to other things)

Some Prior Results 𝜆 is symmetric ⇔ 𝜆⊗𝜆 contains alt. rep. Hooks in 𝜚 𝑚 ⊗2 (Pak, Panova, Vallejo) 2 row partitions in 𝜚 𝑚 ⊗2 (Pak, Panova, Vallejo)

The Semigroup Property Horizontal and vertical sums + 𝐻 =

The Semigroup Property + 𝑉 =

The Semigroup Property Theorem (Christandl, Harrow, Mitchison). If 𝑐 𝜆 1 , 𝜆 2 , 𝜆 3 and 𝑐 𝜇 1 , 𝜇 2 , 𝜇 3 then also 𝑐 𝜆 1 + 𝐻 𝜇 1 , 𝜆 2 + 𝐻 𝜇 2 , 𝜆 3 + 𝐻 𝜇 3 . Corollary. If 𝑐 𝜆 1 , 𝜆 2 , 𝜆 3 and 𝑐 𝜇 1 , 𝜇 2 , 𝜇 3 then also 𝑐 𝜆 1 + 𝑉 𝜇 1 , 𝜆 2 + 𝑉 𝜇 2 , 𝜆 3 + 𝐻 𝜇 3 .

Dominance Order Definition. 𝜆=( 𝜆 1 ,⋯, 𝜆 𝑗 )⊢𝑛 dominates 𝜇=( 𝜇 1 ,⋯, 𝜇 𝑙 )⊢𝑛 if for all 𝑘≥1, 𝑖=1 𝑘 𝜆 𝑖 ≥ 𝑖=1 𝑘 𝜇 𝑖 . Theorem (Ikenmeyer). 𝜚 𝑚 ⊗ 𝜚 𝑚 contains all partitions 𝜆 which are dominance-comparable to 𝜚 𝑚 . Dominance is like how stretched / convexified you are

Probabilistic Approach Are most partitions contained in the tensor square? Uniform distribution: just pick a partition! Plancherel distribution: Pick a uniform pair of Young tableaux of the same shape 𝑃(𝜆)= dim 𝜆 2 𝑛! Pick a uniform permutation, use RSK

Uniform Limit Shape

Plancherel Limit Shape

+ 𝐻 = + 𝑉 + 𝐻 = = 𝜚 𝑘 + 𝐻 𝜚 𝑘+1 + 𝑉 𝜚 𝑘 + 𝐻 𝜚 𝑘 = 𝜚 2𝑘+1 𝜚 𝑘 + 𝐻 𝜚 𝑘+1 + 𝑉 𝜚 𝑘 + 𝐻 𝜚 𝑘 = 𝜚 2𝑘+1 𝜚 𝑘 + 𝐻 𝜚 𝑘−1 + 𝑉 𝜚 𝑘 + 𝐻 𝜚 𝑘 = 𝜚 2𝑘

Chopping up the Uniform Shape

Idea for Probabilistic Results If 𝑐( 𝜚 𝑘 , 𝜚 𝑘 , 𝜆 1 ), 𝑐( 𝜚 𝑘 , 𝜚 𝑘 , 𝜆 2 ), 𝑐( 𝜚 𝑘 , 𝜚 𝑘 , 𝜆 3 ), 𝑐( 𝜚 𝑘+1 , 𝜚 𝑘+1 , 𝜆 4 ), and 𝜆= 𝜆 1 + 𝐻 𝜆 2 + 𝐻 𝜆 3 + 𝐻 𝜆 4 then 𝑐 𝜚 𝑘 + 𝐻 𝜚 𝑘 , 𝜚 𝑘 + 𝐻 𝜚 𝑘 , 𝜆 1 + 𝐻 𝜆 2 , 𝑐 𝜚 𝑘 + 𝐻 𝜚 𝑘+1 , 𝜚 𝑘 + 𝐻 𝜚 𝑘+1 , 𝜆 3 + 𝐻 𝜆 4 Vertical summing gives 𝑐( 𝜚 2𝑘+1 , 𝜚 2𝑘+1 ,𝜆)

Probabilistic Results Theorem (L., S.). With respect to both uniform and Plancherel measure, almost all partitions of 𝑛 appear in the tensor square 𝜚 𝑚 ⊗ 𝜚 𝑚 .

Deterministic Approach Is every partition close to some partition contained in 𝜚 𝑚 ⊗ 𝜚 𝑚 ? Blockwise distance Δ: # of squares to move Subadditive: Δ 𝜆 1 , 𝜆 2 +Δ 𝜇 1 , 𝜇 2 ≥Δ 𝜆 1 + 𝐻 𝜇 1 , 𝜆 2 + 𝐻 𝜇 2

The Standard Representation 𝜆⊗ 𝜏 𝑛 contains all representations within distance 1 of 𝜆 Standard representations let us smooth things out Is every partition contained in 𝜚 𝑚 ⊗ 𝜚 𝑚 ⊗ 𝜏 𝑛 ⊗𝑓 𝑚 (for some small 𝑓)? Definition. The standard representation is the 𝑛-dimensional representation 𝜏 𝑛 = 𝑛 ⊕ 𝑛−1,1 .

Staircase Identity Generalized + 𝐻 + 𝐻 + 𝐻 = + 𝑉 + 𝑉 + 𝐻 = + 𝑉 (4x4 analog, combine 3x3 first) Unlike in prob version, results are not immediately comparable to the corresponding staircases, and need to be resolved by breaking down further (via recursion). This is why it’s very important that staircases are breaking down into only more staircases. =

Deterministic Approach Split into pieces comparable to staircases Let 𝑀(𝑚) be the number of 𝜏 𝑛 ’s we need for 𝜚 𝑚 ⊗2 to contain everything. The recurrence: 𝑀 𝑚 ≤𝑀 3𝑚 4 +𝑀 𝑚 8 +𝑂 𝑚 . Theorem (L., S.). 𝑀 𝑚 =𝑂 𝑚 .

Results Lemma (L., S.). For any 𝑐, for large 𝑛= 𝑚 𝑚+1 2 , if 𝜆⊢𝑛 is contained in 𝜏 𝑛 ⊗ 𝑐𝑚 , then 𝜆 is contained in 𝜚 𝑚 ⊗2 . Theorem (L., S.). For sufficiently large 𝑚, 𝜚 𝑚 ⊗4 contains all partitions of 𝑛. Theorem (L., S.). For sufficiently large 𝑛, 𝜆 ⊗4 contains all partitions of 𝑛 for some “irregular staircase” 𝜆⊢𝑛.

Future Directions Reducing 4 to 2: no smoothing allowed Rectangles are the “hardest” case Are these methods strong enough?

Questions?

Ad-hoc solution showing a 6x6 square satisfies the Saxl conjecture Ad-hoc solution showing a 6x6 square satisfies the Saxl conjecture. All other cases up to this size follow from the dominance result and semigroup property.

Rectangles seem to be the hardest case for the Saxl conjecture. This picture shows that tensor cubes of staircases contain rectangles.