The Saxl conjecture for fourth powers via the semigroup property

Abstract

The tensor square conjecture states that for $n \ge 10$, there is an irreducible representation V of the symmetric group $S_n$ such that $V \otimes V$ contains every irreducible representation of $S_n$. Our main result is that for large enough n, there exists an irreducible representation V such that $V^{\otimes 4}$ contains every irreducible representation. We also show that tensor squares of certain irreducible representations contain $(1-o(1))$-fraction of irreducible representations with respect to two natural probability distributions. Our main tool is the semigroup property, which allows us to break partitions down into smaller ones.

Appendices

Appendix 1: Technical lemma on $\beta$-sum flexibility

1.1 Overview of the proof

We conclude by proving the key technical result used in the proof of Theorem 1.6. This result is restated below:

Theorem 4.2

For any $\beta >0$, let $P(n,\beta )$ denote the probability that a (Plancherel) random partition of n is $\beta$-sum flexible. Then we have

$$\begin{aligned} \lim _{n\rightarrow \infty } P(n,\beta ) = 1 \end{aligned}$$

for all $\beta$.

We will estimate the typical value for the maximum height among the smallest $n^{\alpha }$ columns, for $\frac{1}{6} < \alpha \le \frac{1}{2}$. These bounds will enable us to conclude Theorem 4.2. We first recall from, e.g. [4] that, for a random partition of n, the number $\lambda _1-\lambda _2$ of columns of size 1 is $\Theta (n^{1/6})$ with high probability. In fact, $\frac{\lambda _1-\lambda _2}{n^{1/6}}$ converges weakly to a non-trivial limiting density.

This result tells us that for any $\varepsilon$ there exists $\delta$ such that, for large n, we will have at least $\delta n^{1/6}$ parts of size 1 with probability at least $1-\varepsilon$. This means that the condition in Theorem 4.2 is safe for all columns of size at most $\delta n^{1/6}$. If the sum of all these columns is of size $\Theta ({n^{\alpha }})$, the condition in Theorem 4.2 is now safe for all columns with at most $c n^{\alpha }$ size, for some c. Our plan is to “bootstrap” in this manner up to $\Theta ({n^{1/2}})$. If we can achieve the condition up to this point, we will have proved (4), because for pieces of size $\Theta ({n^{1/2}})$ the limit shape easily implies Theorem 4.2.

We actually estimate a closely related quantity, for which explicit formulae exist. For a partition $\lambda$, we follow [4] in denoting by $D(\lambda )$ the set $\{\lambda _i - i\}$. We will also work with poissonized Plancherel measure $M^{\theta }$ instead of ordinary Plancherel measure $M_n$. At the end, we will depoissonize to recover information about the measures $M_n$. The use of Poissonized Plancherel measure is that useful exact formulae describe the behaviour of a random partition.

Definition 38

For $\theta \in \mathbb {R}^+$, the Poissonized Plancherel measure $M^{\theta }$ is a probability distribution over all partitions $\lambda$, with

$$\begin{aligned} M^{\theta }(\lambda )=e^{-\theta }\theta ^{|\lambda |}\left( \frac{\dim (\lambda )}{|\lambda |!}\right) ^2. \end{aligned}$$

Conceptually, we pick a $M^{\theta }$ partition by first taking n to be Poisson with mean $\theta$, and then picking a Plancherel random $\lambda \vdash n$.

The formulas to follow will allow us to estimate the mean and variance $\mu _{\theta , w}, \sigma _{\theta ,w}^2$ of $T(\lambda ,w)=|D(\lambda ) \cap \mathbb {Z}_{\ge w}|$, where $\lambda$ is taken from poissonized Plancherel measure with mean $\theta$. Because the set $D(\lambda )$ corresponds to the vertical border edges of the Young diagram, this approximately gives the height of the wth smallest column. By setting $w=2\sqrt{n}-Kn^{\alpha }$, knowledge of these quantities will allow us to understand roughly the value of the maximum-size column among the first $\Theta (n^{\alpha })$.

We now list the formulas and bounds we will use for these computations.

1.2 Formulas for poissonized Plancherel measure

Definition 39

The Bessel function of the first kind $J_{\nu }$ is defined as

$$\begin{aligned} J_{\nu }(x)=\sum _{m=0}^{\infty } \frac{(-1)^m (\frac{x}{2})^{2m+\nu }}{(m!)\varGamma ({m+\nu +1})}. \end{aligned}$$

Definition 40

The Airy function Ai is defined as

$$\begin{aligned} Ai(x)=\frac{1}{\pi } \int _{0}^{\infty } \cos \left( \frac{u^3}{3}+xu\right) du. \end{aligned}$$

Definition 41

For a partition $\lambda =(\lambda _1,\lambda _2,\ldots$, define $D(\lambda )$ to be the set $\{\lambda _i-i\}$.

Definition 42

Define the function $\mathbf J (x,y;\theta )$ as

$$\begin{aligned} \mathbf J (x,y;\theta )=\sum _{s=1}^{\infty } J_{x+s}(2\sqrt{\theta })J_{y+s}(2\sqrt{\theta }). \end{aligned}$$

Lemma 8.1

[4, Theorem 2, Proposition 2.9] Let $\lambda$ be chosen according to $M^{\theta }$. Then for any finite $X=\{x_1,x_2\ldots x_s\} \subseteq \mathbb {Z}$, the probability that $D(\lambda ) \supseteq X$ is

$$\begin{aligned} M^{\theta }\left( \{\lambda | D(\lambda ) \supseteq X\}\right) =\det \left[ \mathbf J (x_i,x_j,\theta )\right] _{1\le i,j \le s}. \end{aligned}$$

Lemma 8.2

[11] $J_{\nu }(x) \le x^{-1/3}$ and $J_{\nu }(x) \le \nu ^{-1/3}$.

Lemma 8.3

[4, Lemma 4.4] For $x \in \mathbb {R}$ we have

$$\begin{aligned} \left| n^{1/6} J_{2n^{1/2}+xn^{1/6}}(2n^{1/2})-Ai(x)\right| = O\left( n^{-1/6}\right) , n \rightarrow \infty \end{aligned}$$

where additionally the implicit constant in $O(n^{-1/6})$ is uniform for x in any compact set.

Lemma 8.4

[4, Proposition 4.3] For fixed $a \in \mathbb {R}$,

$$\begin{aligned} \lim _{n\rightarrow \infty } \left( \sum _{k=2\sqrt{n}+an^{\frac{1}{6}}}^{\infty } J(k,k;n)\right) = \int _0^{\infty } t\left( Ai(a+t)\right) ^2 \mathrm{d}t. \end{aligned}$$

Specializing to $a=0$ gives

Lemma 8.5

$$\begin{aligned} \lim _{n\rightarrow \infty } \left( \sum _{k=2\sqrt{n}}^{\infty } J(k,k;n)\right) = \int _0^{\infty } t(Ai(t))^2 \mathrm{d}t. \end{aligned}$$

Lemma 8.6

[4, Lemma 4.5] There exist $C_1, C_2, C_3, \varepsilon > 0$ such that for sufficiently large n, for any $A>0$, $s > 0$, we have

$$\begin{aligned}&\left| J_{r+Ar^{1/3}+s}(r)\right| \le C_1 r^{-1/3} \exp \left( -C_2(A^{\frac{3}{2}}+sA^{\frac{1}{2}}r^{-1/3})\right) , s \le \varepsilon \\&\left| J_{r+Ar^{1/3}+s}(r)\right| \le \exp \left( -C_3(r+s)\right) ,s \ge \varepsilon \end{aligned}$$

for all $r\gg 0$.

Setting $r=2n^{\frac{1}{2}}$ and $A=2^{-\frac{1}{3}}$ in Lemma 8.6, squaring, and adjusting the values $C_i$ as needed, we have the following for all s, for some fixed values $C_i, \varepsilon$, and large enough n.

Lemma 8.7

There exist $C_1, C_2, C_3, \varepsilon > 0$ such that for sufficiently large n, for any $s > 0$, we have

$$\begin{aligned}&\left( J_{2n^{1/2}+n^{1/6}+s}(2n^{1/2})\right) ^2 \le C_1 n^{-1/3} \exp \left( -C_2(1+sn^{-1/6})\right) , \quad s \le \varepsilon n^{1/2}\\&\left( J_{2n^{1/2}+n^{1/6}+s}(2n^{1/2})\right) ^2 \le \exp \left( -C_3(2n^{1/2}+s)\right) , \quad s \ge \varepsilon n^{1/2}. \end{aligned}$$

1.3 Proof of Theorem 4.2

We proceed as described in Sect. 8.1. First, direct application of Lemma 8.1 in the case $|X|=1$, combined with linearity of expectation, yields that

$$\begin{aligned} \mu _{\theta , w} = \sum _{y=w}^{\infty } J(y,y,\theta )=\sum _{y=w}^{\infty } \sum _{s=1}^{\infty } (J_{y+s}(2\sqrt{\theta }))^2 = \sum _{r=1}^{\infty } r (J_{w+r}(2\sqrt{\theta }))^2, \end{aligned}$$

where the last equality is a simple sum rearrangement. Now let $\theta =n, w=2n^{1/2}-Kn^{\alpha }, \frac{1}{6} < \alpha \le \frac{1}{2}$, as alluded to before. We will now establish the bounds on this expectation.

Lemma 8.8

$\mu _{n,(2n^{1/2}-Kn^{\alpha })}=\varOmega \left( n^{\alpha -\frac{1}{6}}\right)$.

Proof

Taking n sufficiently large, we have that

$$\begin{aligned} \mu _{n,(2n^{1/2}-Kn^{\alpha })}= & {} \sum _{r=1}^{\infty } r (J_{2n^{1/2}-Kn^{\alpha }+r}(2\sqrt{n}))^2\\\ge & {} \sum _{r=Kn^{\alpha }-n^{1/6}}^{Kn^{\alpha }+n^{1/6}} r (J_{2n^{1/2}-Kn^{\alpha }+r}(2\sqrt{n}))^2 \\\ge & {} \sum _{j=-n^{1/6}}^{n^{1/6}} \left( \frac{Kn^{\alpha }}{2}\right) (J_{2n^{1/2}+j}(2\sqrt{n}))^2\\= & {} \left( \frac{Kn^{\alpha }}{2}\right) \sum _{j=-n^{1/6}}^{n^{1/6}} (J_{2n^{1/2}+j}(2\sqrt{n}))^2. \end{aligned}$$

Because the Airy function has no zeros in the interval $[-1,1]$ and is continuous, the uniformity in Lemma 8.3 implies that each squared term in this last sum is of size $\Theta (n^{-1/3})$. Since we have $\Theta (n^{1/6})$ terms in the sum, we conclude that for large n,

$$\begin{aligned} \mu _{n,(2n^{1/2}-Kn^{\alpha })}\ge \left( \frac{Kn^{\alpha }}{2}\right) \sum _{j=-n^{1/6}}^{n^{1/6}} (J_{2n^{1/2}+j}(2\sqrt{n}))^2 =\Theta (n^{\alpha -\frac{1}{6}}). \end{aligned}$$

$\square$

Lemma 8.9

$\mu _{n,(2n^{1/2}-Kn^{\alpha })}=O\left( n^{2\alpha -\frac{1}{3}}\right)$. Further, given fixed $\alpha$, the implicit constant in $O(n^{2\alpha -\frac{1}{3}})$ is $O(K^2)$

Proof

Using Lemma 8.5, we see that

$$\begin{aligned} \mu _{n,(2n^{1/2}-Kn^{\alpha })}= & {} \sum _{y=2n^{1/2}-Kn^{\alpha }}^{\infty } J(y,y,n)\\= & {} \sum _{y=2n^{1/2}-Kn^{\alpha }}^{2n^{1/2}} J(y,y,n)+\sum _{y=2n^{1/2}}^{\infty } J(y,y,n) \\= & {} \sum _{y=2n^{1/2}-Kn^{\alpha }}^{2n^{1/2}} J(y,y,n)+O(1). \end{aligned}$$

Expanding out the definition of $\mathbf J (y,y,n)$ we have that

$$\begin{aligned} \sum _{y=(2n^{1/2}-Kn)^{\alpha }}^{2n^{1/2}} \mathbf J (y,y,n)= & {} \sum _{y=(2n^{1/2}-Kn^{\alpha })}^{2n^{1/2}} \sum _{s=1}^{\infty } (J_{y+s+1}(2n^{1/2}))^2 =\\&\quad \left( \sum _{\ell =1}^{Kn^{\alpha }} \ell (J_{\ell +s+1}(2n^{1/2}))^2 \right) + Kn^{\alpha } \sum _{m=2n^{1/2}}^{\infty } (J_m(2n^{1/2}))^2, \end{aligned}$$

where the last equality follows from another simple regrouping of terms. By Lemma 8.2, for any value of $\ell$ we have $J_{\ell +s+1}(2n^{1/2}) \le n^{-1/6}$. Thus, we have that the first sum is of appropriate size:

$$\begin{aligned} \left( \sum _{\ell =1}^{Kn^{\alpha }} \ell (J_{\ell +s+1}(2n^{1/2}))^2 \right) \le \left( \sum _{\ell =1}^{Kn^{\alpha }} \ell \right) n^{-1/3} \le K^2n^{2\alpha -\frac{1}{3}}. \end{aligned}$$

To establish Lemma 8.9, it remains to show that the latter term

$$\begin{aligned} Kn^{\alpha } \sum _{m=2n^{1/2}}^{\infty } (J_m(2n^{1/2}))^2 \end{aligned}$$

is of appropriate size. First, note that from Lemma 8.2 again,

$$\begin{aligned} Kn^{\alpha }\sum _{m=2n^{1/2}}^{2n^{1/2}+n^{1/6}} (J_m(2n^{1/2}))^2 \le Kn^{\alpha }n^{1/6} (n^{-1/3}) = Kn^{\alpha -\frac{1}{6}} = O(n^{2\alpha -\frac{1}{3}}), \end{aligned}$$

where the last equality follows from the assumption $\alpha > \frac{1}{6}$. We are left with upper-bounding the sum

$$\begin{aligned} Kn^{\alpha }\sum _{m=2n^{1/2}+n^{1/6}}^{\infty } (J_m(2n^{1/2}))^2. \end{aligned}$$

Now, using Lemma 8.7, take suitable constants $C_1, C_2, C_3, \varepsilon$, and let n be large enough, so that we have for all s

$$\begin{aligned}&(J_{2n^{1/2}+n^{1/6}+s}(2n^{1/2}))^2 \le C_1 n^{-1/3} \exp (-C_2(1+sn^{-1/6})), \quad s \le \varepsilon n^{1/2}\\&(J_{2n^{1/2}+n^{1/6}+s}(2n^{1/2}))^2 \le \exp (-C_3(2n^{1/2}+s)), \quad s \ge \varepsilon n^{1/2}. \end{aligned}$$

We claim that the above sum is also $O(n^{\alpha -\frac{1}{6}})$, or equivalently that

$$\begin{aligned} \sum _{m=2n^{1/2}+n^{1/6}}^{\infty } (J_m(2n^{1/2}))^2= & {} \sum _{m=2n^{1/2}+n^{1/6}}^{(2+\varepsilon )n^{1/2}+n^{1/6}} (J_m(2n^{1/2}))^2\\&+\sum _{m=(2+\varepsilon )n^{1/2}+n^{1/6}}^{\infty } (J_m(2n^{1/2}))^2 = O(n^{-1/6}). \end{aligned}$$

For the first of these 2 sums, we have

$$\begin{aligned}&\sum _{m=2n^{1/2}+n^{1/6}}^{(2+\varepsilon )n^{1/2}+n^{1/6}} (J_m(2n^{1/2}))^2= \sum _{s=0}^{\varepsilon n^{1/2}} (J_{2n^{1/2}+n^{1/6}+s}(2n^{1/2}))^2 \\&\qquad \le \sum _{s=0}^{\varepsilon n^{1/2}} C_1 n^{-1/3} \exp (-C_2(1+sn^{-1/6})) \\&\qquad \le C_1 n^{-1/3}\exp (-C_2) \sum _{s=0}^{\infty } (\exp (-C_2(n^{-1/6})))^s = \frac{C_1 n^{-1/3}\exp (-C_2)}{1-\exp (-C_2n^{-1/6})}\\&\qquad =\Theta \left( \frac{C_1n^{-1/3}\exp (-C_2)}{C_2n^{-1/6}}\right) = \Theta \left( \frac{C_1\exp (-C_2)}{C_2} n^{-1/6}\right) =O(n^{-1/6}). \end{aligned}$$

For the second, we have

$$\begin{aligned}&\sum _{m=(2+\varepsilon )n^{1/2}+n^{1/6}}^{\infty } (J_m(2n^{1/2}))^2=\sum _{s=\varepsilon n^{1/2}}^{\infty } (J_{2n^{1/2}+n^{1/6}+s}(2n^{1/2}))^2 \\&\quad \le \sum _{s=\varepsilon n^{1/2}}^{\infty } \exp (-C_3(2n^{1/2}+s)) \\&\quad \le \sum _{s=0 }^{\infty }\exp (-C_3(2n^{1/2}+s)) = \frac{\exp (-2C_3n^{1/2})}{1-exp(-C_3)} = \exp (\Theta (-n^{1/2}))=O(n^{-1/6}). \end{aligned}$$

Note that only the first part of our bounding affects the eventual constant in the $O(n^{2\alpha -\frac{1}{3}})$ of the lemma statement, because the other terms are $O(n^{\alpha -\frac{1}{6}})=o(n^{2\alpha -\frac{1}{3}})$. So, combining these separate bounds, we have established the lemma.$\square$

Now we verify that $T(n,2n^{1/2}-Kn^{\alpha })$ is concentrated around its mean.

Lemma 8.10

If $\lambda$ is distributed according to poissonized $M^n$, then $T=T(\lambda ,2n^{1/2}-Kn^{\alpha })$ is concentrated near its mean $\mu =\mu _{n,(2n^{1/2}-Kn^{\alpha })}$, in the sense that for all $\varepsilon > 0$,

$$\begin{aligned} \lim _{n \rightarrow \infty } \mathbb {P}[|T-\mu |>\varepsilon \mu |] =0. \end{aligned}$$

Proof

To show the lemma, we estimate the variance $\sigma ^2$ of $T(\lambda ,2n^{1/2}-Kn^{\alpha })$. Perhaps surprisingly, this is very easy. We will show that

$$\begin{aligned} \sigma ^2=\sigma _{n,(2n^{1/2}-Kn^{\alpha })}^2 \le \mu = \mu _{n,(2n^{1/2}-Kn^{\alpha })}, \end{aligned}$$

which suffices by the Chebyshev inequality, since $\mu$ grows to infinity with n by Lemma 8.8. To show this, we note the following general fact (Lemma 8.11): if $X=\sum _{j=0}^{\infty }x_j$ is a finite-expectation sum of Bernoulli (0 or 1) random variables with pairwise non-positive covariances, then $\sigma _X^2 \le \mu _X$.

In our case, this lemma easily implies the result: when $\lambda$ is distributed according to poissonized $M^n$ we have

$$\begin{aligned} T(\lambda ,2n^{1/2}-Kn^{\alpha })=\sum _{w=2n^{1/2}-Kn^{\alpha }}^{\infty } I(w \in D(\lambda )), \end{aligned}$$

where the indicator variables I are 0 or 1 according to the truth value of their argument. We need only check that ${{\mathrm{Cov}}}(I(x \in D(\lambda )),I(y \in D(\lambda ))) \le 0$ for $x \ne y$. We have

$$\begin{aligned} {{\mathrm{Cov}}}(I(x \in D(\lambda )),I(y \in D(\lambda )))= \mathbb {P}[\{x,y\} \subseteq D(\lambda )]-\mathbb {P}[x \in D(\lambda )]\mathbb {P}[y \in D(\lambda )]. \end{aligned}$$

Using Lemma 8.1, this is

$$\begin{aligned} \det {\left( \begin{matrix} J(x,x;n) &{}J(x,y;n) \\ J(y,x;n)&{}J(y,y;n)\end{matrix}\right) } - J(x,x;n)J(y,y;n)= & {} -J(x,y;n)J(y,x;n)\\ {}= & {} -J(x,y;n)^2\le 0, \end{aligned}$$

since J(x, y; n) is symmetric in x and y. Therefore, the covariances are all negative, so it only remains to verify Lemma 8.11 $\square$

Lemma 8.11

$X=\sum _{j=0}^{\infty }x_j$ is a finite-expectation sum of Bernoulli variables with pairwise non-positive covariances, then $\sigma _X^2 \le \mu _X$

Proof

To prove this, we simply compute $\sigma _X^2=\mathbb {E}[X^2]-\mathbb {E}[X]^2$. The condition on covariances is equivalent to $\mathbb {E}[x_ix_j] \le \mathbb {E}[x_i]\mathbb {E}[x_j]$ for all $i \ne j$. Because $\mathbb {E}[x]=\sum _{i=0}^{\infty } \mathbb {E}[x_i]$ is finite, its square is finite, so we have

$$\begin{aligned} (\mathbb {E}[x])^2=\left( \sum _{i=0}^{\infty } (\mathbb {E}[x_i])^2\right) + \left( \sum _{i>j \ge 0}^{} \mathbb {E}[x_i]\mathbb {E}[x_j]\right) < \infty . \end{aligned}$$

Because $\mathbb {E}[x_ix_j] \le \mathbb {E}[x_i]\mathbb {E}[x_j]$, we find that

$$\begin{aligned} \mathbb {E}[x]^2+\mu _X= & {} \mathbb {E}[x]^2+\mathbb {E}[x]=\left( \sum _{i=0}^{\infty } (\mathbb {E}[x_i])^2+\mathbb {E}[x_i]\right) + \left( \sum _{i>j \ge 0}^{} \mathbb {E}[x_i]\mathbb {E}[x_j]\right) \\&\ge \left( \sum _{i=0}^{\infty } (\mathbb {E}[x_i^2])\right) + \left( \sum _{i>j \ge 0}^{} \mathbb {E}[x_ix_j]\right) =\mathbb {E}[x^2]. \end{aligned}$$

$\square$

We now know that the bounds in lemmas 8.8, 8.9 apply to almost all partitions $\lambda$, not only “average” partitions: for some constants $B_1(K,\alpha ), B_2(K,\alpha )$ where $B_2=O(K^2)$ for fixed $\alpha$, if $\lambda$ is distributed according to $M^n$ we have

$$\begin{aligned} B_1(K,\alpha )n^{\alpha -\frac{1}{6}} \le T(\lambda ,2n^{1/2}-Kn^{\alpha }) \le B_2(K,\alpha )n^{2\alpha -\frac{1}{3}} \end{aligned}$$

with probability $1-o(1)$ as $n \rightarrow \infty$. We now depoissonize the result. The key observation, as described in [8], is that simple Plancherel measures $M_n$ may be viewed as samplings at time n of a growth process on partitions, in which we begin with an empty partition and add additional blocks randomly at each step. It is clear that adding a block to $\lambda$ cannot decrease the value of $T(\lambda ,w)$. So because we can embed these Plancherel measures into a growth process, the expectations are monotone: for $\ell <n$ we have

$$\begin{aligned} \mathbb {P}\left[ \lambda ^{(\ell )}< B_1(K,\alpha )n^{\alpha -\frac{1}{6}}\right] \ge \mathbb {P}\left[ \lambda ^{(n)} < B_1(K,\alpha )n^{\alpha -\frac{1}{6}}\right] \end{aligned}$$

and for $\ell > n$ we have

$$\begin{aligned} \mathbb {P}\left[ \lambda ^{(\ell )}> B_2(K,\alpha )n^{2\alpha -\frac{1}{3}}\right] \ge \mathbb {P}\left[ \lambda ^{(n)} > B_2(K,\alpha )n^{2\alpha -\frac{1}{3}}\right] . \end{aligned}$$

We claim that, for n large, the probability of a Poisson variable with mean n to be less than n converges to $\frac{1}{2}$. Indeed, this is an immediate result of the Central Limit Theorem for independent, identically distributed random variables applied to the Poisson random variable with mean 1. Therefore,

$$\begin{aligned}&\limsup _{n \rightarrow \infty } M_n\left( \left\{ \lambda |T(\lambda ,2n^{1/2}-Kn^{\alpha })< B_1(K,\alpha )n^{\alpha -\frac{1}{6}}\right\} \right) \\&\quad \le 2 \limsup _{n \rightarrow \infty } M^n\left( \left\{ \lambda |T(\lambda ,2n^{1/2}-Kn^{\alpha }) < B_1(K,\alpha )n^{\alpha -\frac{1}{6}}\right\} \right) =0, \end{aligned}$$

and similarly for the upper bound; a low probability of deviation for poissonized Plancherel measure directly implies the same bound (up to a factor of 2) for standard Plancherel measure. So, we have

$$\begin{aligned} B_1(K,\alpha )n^{\alpha -\frac{1}{6}} \le T(\lambda ,2n^{1/2}-Kn^{\alpha }) \le B_2(K,\alpha )n^{2\alpha -\frac{1}{3}} \end{aligned}$$

with probability $1-o(1)$ for $\lambda$ distributed according to $M_n$ measure.

Now we establish the connection between $T(\lambda ,2n^{1/2}-Kn^{\alpha })$ and the column sizes. It will be valuable to visualize a Young diagram (in English coordinates) as a block-walk beginning from the far-right of the top border line, with coordinates (n, 0). In this model, every down-move corresponds to a value $(\lambda _i-i)$ and every left-move corresponds to the end of a column. With this in mind, $T(\lambda ,2n^{1/2}-Kn^{\alpha })$ is essentially the height of the largest column so far after $n-2n^{1/2}+Kn^{\alpha }$ moves from the starting point. The idea is that our upper bound lets us move far out without fearing that our largest column is too big. Our lower bound tells us that many of the columns we have formed are large, implying that the sum of their sizes is large. The only caveat is that if we are moving vertically, T measures the current vertical distance moved, but may not exactly measure a column height; we may be in the middle of a column. However, we will circumvent this issue by using multiple values of K simultaneously, and showing that in between there must be many horizontal moves.

Pick an $\varepsilon > 0$. We will give a bootstrapping argument that gives result (4) for parameter $\beta$ with probability $1-6\varepsilon$. First, because the number of size-1 columns $\lambda _1-\lambda _2$ satisfies $\frac{\lambda _1-\lambda _2}{n^{1/6}}$ converges in distribution to a limiting density, there exists $\delta _1$ such that (for large n) with probability at least $1-\varepsilon$, there are at least $\frac{\delta _1}{\beta }n^{1/6}$ parts of size 1. Using lemmas 8.8 and 8.9, since $\frac{1}{4}-\frac{1}{6}=\frac{1}{12}$ and $\frac{2}{4}-\frac{1}{3}=\frac{1}{6}$, we may pick $\delta _2,$ and then $\delta _3$, such that for large n, with probability at least $1-\varepsilon$,

$$\begin{aligned} T(\lambda ,2n^{1/2}-\delta _2n^{1/4}) \ge {\delta _3}n^{\frac{1}{12}} \end{aligned}$$

and

$$\begin{aligned} T(\lambda ,2n^{1/2}-2\delta _2n^{1/4}) \le \delta _1n^{\frac{1}{6}} \end{aligned}$$

(for the second inequality, we use the fact that the constant factor in Lemma 8.9 is $O(K^2)$). Now, for large n, we have $\delta _2n^{1/4} -\delta _1n^{\frac{1}{6}} \ge \frac{\delta _2}{2}n^{1/4}$. This implies that $D(\lambda )$ cannot contain more than $\frac{\delta _2}{2}n^{1/4}$ of the interval of integers $[2n^{1/2}-2\delta _2n^{1/4}, 2n^{1/2}-\delta _2n^{1/4}]$ as this would violate the latter inequality above. In view of our block-walking model, this is equivalent to at least $\frac{\delta _2}{2}n^{1/4}$ horizontal move being made in between the steps $n-2n^{1/2}+\delta _2n^{1/4}, n-2n^{1/2}+2\delta _2n^{1/4}$ meaning that at least $\frac{\delta _2}{2}n^{1/4}$ columns in this range have height at least $\delta _3n^{\frac{1}{12}}$ (this lower bound on the height comes from the first inequality above). Because the largest column among these is smaller than $\delta _1n^{\frac{1}{6}}$, the condition in result (4) holds for all columns up to size $\delta _1n^{\frac{1}{6}}$. Therefore, by bootstrapping, it holds for all columns with size at most $\beta$ times the sum of their sizes, or at most

$$\begin{aligned} \beta \left( \frac{\delta _2}{2}n^{\frac{1}{4}}\right) \left( \delta _3n^{\frac{1}{12}}\right) =\frac{\beta \delta _2\delta _3}{2}n^{\frac{1}{3}}=\delta _4n^\frac{1}{3}. \end{aligned}$$

Now we repeat the argument once more. Again using Lemma 8.9, we pick $\delta _5$ such that with probability $1-\varepsilon$,

$$\begin{aligned} T\left( \lambda , 2n^{\frac{1}{2}}-2\delta _5n^{\frac{1}{3}}\right) \le \frac{\delta _5}{2}n^{\frac{1}{3}} \end{aligned}$$

and

$$\begin{aligned} \delta _5 \le \delta _4. \end{aligned}$$

The first is possible because the constant in Lemma 8.9 is $O(K^2)$, implying that it is o(K) for K small. Given $\delta _5$ we can then pick $\delta _6$ such that with probability $1-\varepsilon$,

$$\begin{aligned} T(\lambda ,2n^{1/2}-\delta _5n^{\frac{1}{3}})\ge \delta _6n^{1/6}. \end{aligned}$$

The first inequality above implies that of the $\delta _5n^{1/3}$ block moves from step $n-2n^{1/2}+\delta _5n^{1/3}$ to step $n-2n^{1/2}+2\delta _5n^{1/3}$, we may make at most $\frac{\delta _5}{2}n^{1/3}$ vertical moves. Hence, we make at least $\frac{\delta _5}{2}n^{1/3}$ horizontal moves in this span. By the definition of $\delta _6$, we make a column of size at least $\delta _6n^{\frac{1}{6}}$ each time, giving a total column size sum of $\frac{\delta _5\delta _6}{2}n^{1/2}$.

Now we are almost done. Our bootstrapping has shown Theorem 4.2 for columns of size at most $\frac{\beta \delta _5\delta _6}{2}=\delta _7n^{1/2}$ with probability at least $1-4\varepsilon$. Bootstrapping once more, we can now sum up all the column lengths of size at most $\delta _7n^{1/2}$. At scale $\Theta (n^{1/2})$, we can simply apply the limit shape theorem to conclude that for almost all partitions, the sum of such column lengths is $\Theta (n)$, a constant fraction of the total size, because we are collecting a positive-area amount of the limit shape. Because almost all partitions have all columns of size $O(\sqrt{n})$, we have bootstrapped to completion with error probability $6\varepsilon$ and so the result is proved.

Appendix 2: Generalization of dominance result

We extensively used Theorem 2.6 stating that if $\rho _m,\lambda$ are dominance comparable, then $c(\rho _m,\rho _m,\lambda )$. We present a generalization.

Theorem 9.1

For partitions $\mu ,$ $\nu \vdash n$, if $\mu$ has distinct row lengths and $\nu \succeq \mu$, then $c(\mu ,\mu ,\nu )$.

This is strictly more general than Theorem 2.6 because in the case when $\mu =\rho _m$ we have $c(\mu ,\mu ,\nu )$ for all $\nu \succeq \mu$. Since $\mu =\mu '$, conjugating gives $c(\mu ,\mu ,\nu ')$ as well, and so we have recovered both cases of Theorem 2.6.

Proof

The proof of this generalization requires only slight modification of the proof of the original result in [10]. We quote from there the following definition.$\square$

Definition 43

Let $d=|\lambda |=|\mu |=|\nu |$. A Young hypergraph H of type $(\lambda ,\mu ,\nu )$, is a hypergraph with d vertices such that

• 1. The are three layers of hyperedges $E_{\lambda },E_{\mu },E_{\nu }$.

• 2. Each of $E_{\lambda },E_{\mu },E_{\nu }$ contains each vertex in exactly 1 hyperedge.

• 3. There is a bijection between the vertices of H and the boxes of $\lambda$ such that 2 vertices lie in a common hyperedge of $E_{\lambda }$ iff the corresponding boxes in $\lambda$ lie in the same column. Analogously for $E_{\mu }$ and $\mu$ and for $E_{\nu }$ and $\nu$.

Given a Young hypergraph of type $(\lambda ,\mu ,\nu )$, we consider the ways to label its vertices with positive integers. In particular, we have the following definitions.

Definition 44

We call a labelling of the vertex set of a Young hypergraph of type $(\lambda ,\mu ,\nu )$ $\lambda$-permuting if for each hyperedge $e_{\lambda }$ of $\lambda$ with $|e_{\lambda }|=k$, the labels of the vertices of $e_{\lambda }$ are a permutation of $\{1,2,\ldots ,k\}$. We define $\mu$-permuting and $\nu$-permuting analogously.

Definition 45

We call a labelling of the vertex set of a Young hypergraph of type $(\lambda ,\mu ,\nu )$ $\lambda$-distinct if for each hyperedge $e_{\lambda }$ of $\lambda$ with $|e_{\lambda }|=k$, the labels of the vertices of $e_{\lambda }$ consist of distinct numbers. We define $\mu$-distinct and $\nu$-distinct analogously.

Definition 46

We call a labelling of the vertex set of a $(\lambda ,\mu ,\nu )$ Young hypergraph H perfect if it is $\lambda$-permuting, $\mu$-permuting, and $\nu$-distinct.

Then the finish of the proof of [10][2.1] (see Section 5) amounts to the following lemma.

Lemma 9.2

If there exists a Young hypergraph such that there is exactly 1 perfect labelling of its vertices then $c(\lambda ,\mu ,\nu )$ holds.

The proof in [10] uses as Young hypergraphs for $\rho _m$ the rows and columns of the Ferrers diagram of $\rho _m$, which works because $\rho _m$ is symmetric. We now adapt this to an arbitrary $\mu$ with distinct row lengths; for any $\nu \succeq \mu$ we show there exists a perfect Young hypergraph of type $(\mu ,\mu ,\nu )$. We use the Ferrers diagram for $\mu$ as the vertex set. For the first hypergraph $H_1$ corresponding to $\mu$, we simply take the columns of $\mu$ as the hyperedges. Note that in English coordinates this amounts to, for each hyperedge, greedily taking the left-most vertex in each available row. For our second hypergraph $H_2$ corresponding to $\mu$ we greedily take vertices from the right of each row instead. For example, when $\mu =\rho _m$ this amounts to taking the diagonals as hyperedges. It is easy to see that $H_2$ is also a hypergraph of type $\mu$. We claim that $H_1$ and $H_2$ already limit the possible vertex labellings to a unique one, namely the labelling which assigns to each vertex its row number.

Lemma 9.3

There is only one vertex labelling of the Ferrers diagram of $\mu$ which is $\mu$-permuting with respect to both $H_1$ and $H_2$. This labelling assigns to each vertex its row number.

Proof

We first make an easy observation. For any vertex v of our hypergraph, the hyperedge of $H_2$ containing v also contains a vertex in each higher row, all of which are strictly to the right of v. This is clear because the rows of $\mu$ have distinct lengths.

Now we inductively show that any such labelling must just be the row numbering. We induct on columns, starting from the furthest right. Clearly the rightmost column’s unique vertex is labelled 1, because it is contained in a size 1 hyperedge of $H_1$.

For each subsequent column, assume that all columns to the right are labelled by row numbers. We do a further induction within the column, starting from the bottom vertex. Call the column under consideration C, and say its vertices are $v_1$ in row 1, and so on to $v_k$ in row k. For some vertex $v_i$ in C, assume that $v_j$ is labelled j for all $j>i$; we show $v_i$ is labelled as i.

To do so, note that by our initial observation, $v_i$ must be labelled at least i; its hyperedge in $H_2$ contains labels from 1 to $(i-1)$ already. However, its hyperedge in $H_1$ is simply C, which already contains all labels greater than i. Therefore the only choice for $v_i$ is to be labelled i.

This completes the induction. Clearly the described labelling indeed satisfies the given conditions, so the lemma is proved.$\square$

Now the combination of the above lemmas implies that if we can find a perfect $(\mu ,\mu ,\nu )$ Young hypergraph for the above vertex labelling extending $H_1, H_2$ above, then we have $c(\mu ,\mu ,\nu )$. In fact, we can do so for precisely those $\nu$ which dominate $\mu$. It is clear that to find a $E_{\nu }$ hypergraph which yields a perfect Young hypergraph, it is equivalent to find a filling of $\nu$ with content $\mu$ such that each column has distinct entries. (By such a filling of $\nu$ with content $\mu$ we mean a labelling of the boxes of $\nu$ such that the number of k labels is the size $\mu _k$ of the kth row of $\mu$.) By [10][4.1], the existence of such a filling is equivalent to $\nu \succeq \mu$, so the proof of Theorem 9.1 is complete. $\square$

Appendix 3: The generalized semigroup property and tensor cubes

1.1 The semigroup property for many partitions

We first generalize $c(\cdot ,\cdot ,\cdot )$ to longer sequences of representations.

Definition 47

For k a positive integer let

$$\begin{aligned} c(\lambda _1,\lambda _2,\ldots ,\lambda _k) \end{aligned}$$

denote the assertion that $\lambda _1$ is a constituent of $\lambda _2\otimes \lambda _3\otimes \ldots \lambda _k.$

As in the $k=3$ case, c is symmetric because it simply asserts the positivity of

$$\begin{aligned} \frac{1}{n!}\sum _{\sigma \in S_n} {\chi ^{\lambda _1}(\sigma )}{\chi ^{\lambda _2}(\sigma )}\ldots {\chi ^{\lambda _k}(\sigma )}. \end{aligned}$$

We now show that the semigroup property still applies for longer sequences using induction.

Lemma 10.1

If $c(\lambda _1, \ldots \lambda _k)$ and $c(\mu _1,\ldots \mu _k)$ then also $c(\lambda _1+_{_H}\mu _1, \ldots , \lambda _k +_{_H}\mu _k).$

Proof

First, the result is trivially true for $k\le 2$ since $c(\lambda _1)=1 \iff \lambda _1$ is trivial and $c(\lambda _1,\lambda _2) \iff \lambda _1=\lambda _2$. For $k\ge 3$ we induct from base case of 2.4. So take $k>3$ and assume the result for $k-1$.

Because $c(\lambda _1, \ldots \lambda _k)$ there exists a partition $\alpha$ such that $c(\lambda _1,\lambda _2,\ldots \lambda _{(k-2)},\alpha )$ and $c(\lambda _{(k-1)},\lambda _k,\alpha )$. Similarly there exists $\beta$ such that $c(\mu _1,\ldots , \mu _{(k-2)},\beta )$ and $c(\mu _{(k-1)},\mu _k,\beta )$. We now conclude using the inductive hypothesis that

$$\begin{aligned} c(\lambda _1+_{_H}\mu _1, \ldots , \lambda _{(k-2)} +_{_H}\mu _{(k-2)},\alpha +_{_H}\beta ) \end{aligned}$$

and

$$\begin{aligned} c(\lambda _{(k-1)} +_{_H}\mu _{(k-1)},\lambda _k +_{_H}\mu _k,\alpha +_{_H}\beta ). \end{aligned}$$

Together these imply $c(\lambda _1+_{_H}\mu _1, \ldots , \lambda _k +_{_H}\mu _k)$ as desired. $\square$

As in the $k=3$ case, we may vertically add any even number of partitions in applying the semigroup property: this is because conjugating an even number of the partitions does not change the truth of $c(\lambda _1,\ldots \lambda _k)$.

1.2 Rectangles appear in $\rho _m^{\otimes 3}$

As mentioned in the remarks, rectangles are difficult to control using the semigroup property because they can only be broken up into smaller rectangles. This suggests that rectangles should be the hardest case for the Saxl conjecture. In this section, we show that despite this, rectangles appear in the tensor cube of the staircase. In fact, the proof is a fairly simple induction.

Definition 48

The rectangle partition R(a, b) is the rectangular Young diagram $(a,a,\ldots a)$ with b rows.

Theorem 10.2

Any rectangular partition $\lambda =R(a,b)$ of size $\left( {\begin{array}{c}m+1\\ 2\end{array}}\right)$ is a constituent in the tensor cube $\rho _m^{\otimes 3}$. Equivalently, if $ab=\left( {\begin{array}{c}m+1\\ 2\end{array}}\right)$ then $c(\rho _m,\rho _m,\rho _m,R(a,b))$.

Proof

We induct on m. Assume WLOG that $a\ge b$. If $a\ge m$ then $b \le \frac{m+1}{2}$ and by Lemma 6.2, $\lambda$ and $\rho _m$ are dominance comparable. Thus, the result follows in this case.

Thus, we may assume that $\frac{m+1}{2}<a<m$. Then we have that $(2a-m)$ and $(2m-a-1)$ are positive. We prove some simple lemmas. $\square$

Lemma 10.3

$b+2a-2m-1\ge 0$ and $a(b+2a-m-1)=|\rho _{2a-m-1}|$.

Proof

$0\le \frac{(2a-m)(2a-m-1)}{2}=\frac{m^2+m}{2}+2a^2-2am-a=ab+2a^2-2am-a=a(b+2a-2m-1)$. Because $a\ge 0$, $b+2a-2m-1\ge 0$ follows.$\square$

Lemma 10.4

If $\mu =R(2a-m,2m-a-1)$ then

$$\begin{aligned} c(\mu ,\mu ,\mu ,\mu ). \end{aligned}$$

Proof

In fact this holds for any $\mu$ at all. We have

$$\begin{aligned} k(\mu ,\mu ,\mu ,\mu )=\langle \mu ^{\otimes 2},\mu ^{\otimes 2}\rangle >0 \end{aligned}$$

$\square$

Lemma 10.5

$$\begin{aligned} c(\rho _{(2m-2a)},\rho _{(2m-2a)},\rho _{(2m-2a)},R(m-a,2m-2a+1)). \end{aligned}$$

Proof

This follows from the inductive hypothesis because the total number of blocks in the rectangle and staircases are clearly equal. $\square$

Lemma 10.6

$$\begin{aligned} c\left( \rho _{(2a-m-1)}, \rho _{(2a-m-1)}, \rho _{(2a-m-1)}, R(a,b+2a-2m-1)\right) . \end{aligned}$$

Proof

This follows from the inductive hypothesis and Lemma 10.3.$\square$

Now note that

$$\begin{aligned}&(R(2a-m,2m-2a+1)+_{_H}R(m-a,2m-2a+1))+_{_V}R(a,b+2a-2m+1)\\&\qquad =R(a,2m-2a-1)+_{_V}R(a,b+2a-2m-1)=R(a,b) \end{aligned}$$

while

$$\begin{aligned} (R(2a-m,2m-2a+1)+_{_H}\rho _{(2m-a)})+_{_V}\rho _{(2a-m-1)}=\rho _m. \end{aligned}$$

This latter identity is a rewriting of the geometrically obvious

$$\begin{aligned} (R(x,y)+_{_H}\rho _y )+_{_V}\rho _x =\rho _{(x+y)}. \end{aligned}$$

Theorem 10.1 applied to these lemmas and identities gives the result; because 4 is even, it is permissible to vertically add all 4 partitions when using the semigroup property. Geometrically, we are simply combining shapes as below (Fig. 13).

Fig. 13

A geometric proof that rectangles are contained in $\rho _m^{\otimes 3}$