# Conjectural identities for Young symmetrizers and Young-Jucys-Murphy elements

The following questions I have found in my own notes from about 3 years ago. Unfortunately, I lost much of the context; I believe I made these conjectures reading Okounkov-Vershik, arXiv:0503040v3, but I don't remember any details, including whether I was lacking a proof or I had a proof and was looking for a better one. What is sure that I am not able to prove them now.

Preliminaries and notation:

In the following, "Young tableau" will always mean "Young tableau of straight shape".

Let $n$ be a nonnegative integer. An $n$-permutational tableau will mean a Young tableau with $n$ cells, each of them being filled with an integer from the set $\left\lbrace 1,2,...,n\right\rbrace$, in such a way that every integer from the set $\left\lbrace 1,2,...,n\right\rbrace$ appears exactly once. Such a tableau doesn't, a priori, have to be standard.

When $T$ is an $n$-permutational tableau, we can define two subgroups $R_T$ and $C_T$ of the symmetric group $S_n$ as follows:

• We let $R_T$ be the group of all $\sigma \in S_n$ such that for every $i \in \left\lbrace 1,2,...,n\right\rbrace$, the integers $i$ and $\sigma\left(i\right)$ lie in the same row of $T$.

• We let $C_T$ be the group of all $\sigma \in S_n$ such that for every $i \in \left\lbrace 1,2,...,n\right\rbrace$, the integers $i$ and $\sigma\left(i\right)$ lie in the same column of $T$.

Now, we define three elements $a_T$, $b_T$ and $c_T$ of $\mathbb Q\left[S_n\right]$ by

$a_T = \dfrac{1}{\left|R_T\right|} \sum\limits_{r\in R_T} r$, $b_T = \dfrac{1}{\left|C_T\right|} \sum\limits_{c\in C_T} \left(-1\right)^c c$, and $c_T = a_T b_T$.

(Today I learned that not everybody is using the convention that $c_T = a_T b_T$, and some (particularly in England) prefer to set $c_T = b_T a_T$ instead. Apparently, there is even a disagreement about how to multiply permutations. I will stick to defining $c_T$ as $a_T b_T$, and defining products of permutations by $\left(\sigma \pi\right)\left(j\right) = \sigma\left(\pi\left(j\right)\right)$, since these are the notations I've been using for years. The questions at hand are confusing enough without nonstandard notations adding to the mess.)

It is known that for every $n$-permutational tableau $T$, the elements $a_T$ and $b_T$ of $\mathbb Q\left[S_n\right]$ are idempotents, while $c_T$ is a quasi-idempotent (that is, $c_T^2 = \lambda c_T$ for some rational $\lambda$). The element $c_T$ (or, occasionally, a scalar multiple of $c_T$ which actually is idempotent) is called the Young symmetrizer corresponding to the tableau $T$, and is sometimes denoted by $e_T$. Its main significance is that the left $\mathbb Q\left[S_n\right]$-module $\mathbb Q\left[S_n\right] c_T$ is (isomorphic to) the irreducible representation of $\mathbb Q\left[S_n\right]$ corresponding to the shape of $T$.

Now, let us talk about standard tableaux. If $T$ is a standard $n$-permutational tableau, then for every $i \in \left\lbrace 0,1,...,n\right\rbrace$, we can define the $i$-restriction $T\mid_{\leq i}$ to be tableau obtained by only keeping the cells of $T$ which are filled with integers $\leq i$, while removing all the other cells (along with the integers in them). This $T\mid_{\leq i}$ is a standard $i$-permutational tableau. Clearly, $T\mid_{\leq n} = T$.

One last notation. If $\lambda$ is a partition of $n$, the initial $\lambda$-tableau $T_{\lambda}$ will mean the tableau obtained by writing the integers $1$, $2$, ..., $n$ into the cells of the Young diagram of $\lambda$ in the usual order in which books are written in the Western world (i. e., filling the first row from left to right, then the second row from left to right, and so on). Formally, this can be defined as the tableau whose $i$-th row is $\left(k_{i-1}+1, k_{i-1}+2, ..., k_i\right)$ for every $i$, where $\lambda=\left(\lambda_1,\lambda_2,\lambda_3,...\right)$ and $k_i = \lambda_1+\lambda_2+...+\lambda_i$. Of course, this initial $\lambda$-tableau $T_{\lambda}$ is a standard $n$-permutational tableau.

If you have heard about Young symmetrizers, but you know them as being indexed by partitions rather than by tableaux, you are most likely used to only considering the $c_{T_{\lambda}}$'s. (There is not much lost by this restriction except for clarity, since all other $c_T$'s have the form $\omega c_{T_{\lambda}} \omega^{-1}$ for some $\omega\in S_n$ and some partition $\lambda$ of $n$. Indeed, any $n$-permutational tableau $S$ and any $\pi\in S_n$ satisfy $c_{\pi S} = \pi c_S \pi^{-1}$, where $\pi S$ means the tableau obtained by applying $\pi$ to all entries of $S$.)

Facts:

In the following, whenever $m\leq n$, we view $\mathbb Q \left[S_m\right]$ as a subring of $\mathbb Q \left[S_n\right]$ in the standard way.

Scroll down to "Conjectures" if you don't care for the little that has been shown.

Lemma 1. Let $n$ be a nonnegative integer. Let $S$ and $T$ be two Young tableaux of size $n$ such that the shape of $S$ is greater than the shape of $T$ in lexicographic order (this makes sense because shapes of Young tableaux are partitions). Then, $a_S \mathbb Q\left[S_n\right] b_T = 0$ and $b_T \mathbb Q\left[S_n\right] a_S = 0$.

Proof sketch. Lemma 4.41 in Etingof et al, arXiv:0901.0827v5 shows that if two partitions $\lambda$ and $\mu$ of $n$ satisfy $\lambda > \mu$ in lexicographic order, then $a_{T_{\lambda}} \mathbb Q\left[S_n\right] b_{T_{\mu}} = 0$. From this it is easy to deduce that $a_S \mathbb Q\left[S_n\right] b_T = 0$ (recalling that every $c_T$ has the form $\omega c_{T_{\lambda}}$ for some $\omega\in S_n$, where $\lambda$ is the shape of $T$), or alternatively one can notice that $a_S \mathbb Q\left[S_n\right] b_T = 0$ follows by the same argument as Lemma 4.41.

Applying the antipode of the Hopf algebra $\mathbb Q\left[S_n\right]$ to the equality $a_S \mathbb Q\left[S_n\right] b_T = 0$, we obtain $b_T \mathbb Q\left[S_n\right] a_S = 0$, whence Lemma 1 follows.

Lemma 2. Let $n$ be a nonnegative integer. Let $S$ and $T$ be two Young tableaux of size $n$ having different shapes. Then, $c_S \mathbb Q\left[S_n\right] c_T = 0$.

Proof sketch. Either the shape of $S$ is greater than the shape of $T$ in lexicographic order, or the shape of $S$ is smaller than the shape of $T$ in lexicographic order. In the first case, Lemma 2 follows from $a_S \mathbb Q\left[S_n\right] b_T = 0$, which is a direct application of Lemma 1. In the second case, Lemma 2 follows from $b_S \mathbb Q\left[S_n\right] a_T = 0$, which in turn follows from Lemma 1 with $S$ and $T$ switched. In either case, Lemma 2 is thus proven.

Proposition 3. Let $n$ be a nonnegative integer. Let $\lambda$ be a partition of $n$, and let $S$ be a standard $n$-permutational tableau distinct from $T_{\lambda}$. Then, $c_S c_{T_{\lambda}} = 0$.

Proof sketch.

• In the case when the shape of $S$ is distinct from $\lambda$, the product $c_S c_{T_{\lambda}}$ is $0$ because Lemma 2 yields $c_S \mathbb Q\left[S_n\right] c_{T_{\lambda}} = 0$.

• In the case when the shape of $S$ is $\lambda$, one can show the stronger claim that $b_S a_{T_{\lambda}} = 0$. This follows from proving that there exist two distinct integers which are in the same row of $T_{\lambda}$ and in the same column of $S$ (akin to the proof of Lemma 4.41 in Etingof et al, arXiv:0901.0827v5). Alternatively, this case follows from Lemma 3.1.20 in James/Kerber, "The Representation Theory of the Symmetric Group", 1981.

Proposition 4. Let $n$ be a nonnegative integer. Let $\lambda$ be a partition of $n$, and let $S$ be a standard $n$-permutational tableau distinct from $T_{\lambda}$. Then, $c_{S\mid_{\leq n}} c_{S\mid_{\leq n-1}} ... c_{S\mid_{\leq 1}} \cdot c_{T_{\lambda}} = 0$.

Proof sketch. Again, the case when the shape of $S$ is distinct from $\lambda$ can be easily dealt with: in this case, Lemma 2 yields $c_S Q\left[S_n\right] c_{T_{\lambda}}$, but since $S = S\mid_{\leq n}$, this becomes $c_{S\mid_{\leq n}} Q\left[S_n\right] c_{T_{\lambda}}$. So we are done in this case.

What remains is the case when the shape of $S$ is $\lambda$. Let $w_S$ denote the word obtained by reading the entries of the tableau $S$ row by row, from top to bottom, where the English convention is used in describing Young tableaux (so the top row is the longest). Since $S \neq T_{\lambda}$, we have $w_S \neq 1 2 ... n$. Thus, there exists a $k\in\left\lbrace 1,2,...,n\right\rbrace$ such that the $k$-th letter of $w_S$ is $\neq k$. Pick the smallest such $k$. Of course, the first $k-1$ letters of $w_S$ are $1 2 ... \left(k-1\right)$ then. This yields easily that the shape of $T_{\lambda}\mid_{\leq k}$ is greater than the shape of $S_{\leq k}$ in the lexicographic ordering. Hence, Lemma 1 yields $a_{T_{\lambda}\mid_{\leq k}} \mathbb Q\left[S_n\right] b_{S\mid_{\leq k}} = 0$ and $b_{S\mid_{\leq k}} \mathbb Q\left[S_k\right] a_{T_{\lambda}\mid_{\leq k}} = 0$.

Now, we need to prove that $c_{S\mid_{\leq n}} c_{S\mid_{\leq n-1}} ... c_{S\mid_{\leq 1}} \cdot c_{T_{\lambda}} = 0$. Of course, in order to do this, it is enough to show that $c_{S\mid_{\leq k}} c_{S\mid_{\leq k-1}} ... c_{S\mid_{\leq 1}} \cdot c_{T_{\lambda}} = 0$. Thus, it is enough to show that $c_{S\mid_{\leq k}} \mathbb Q\left[S_k\right] c_{T_{\lambda}} = 0$ (since all the factors in $c_{S\mid_{\leq k-1}} c_{S\mid_{\leq k-2}} ... c_{S\mid_{\leq 1}}$ lie in $\mathbb Q\left[S_k\right]$). Since $c_{S\mid_{\leq k}} = a_{S\mid_{\leq k}} b_{S\mid_{\leq k}}$ and $c_{T_{\lambda}} = a_{T_{\lambda}} b_{T_{\lambda}}$, this boils down to proving that $b_{S\mid_{\leq k}} \mathbb Q\left[S_k\right] a_{T_{\lambda}} = 0$.

But if $H$ is a subgroup of a group $G$, then the sum of all elements of $H$ divides the sum of all elements of $G$ (both from the left and from the right) in the group algebra $\mathbb Q\left[G\right]$. Hence, $a_{T_{\lambda}\mid_{\leq n}}$ divides $a_{T_{\lambda}}$ in the group algebra $\mathbb Q\left[S_n\right]$ (since $R_{T_{\lambda}\mid_{\leq n}}$ is a subgroup of $R_{T_{\lambda}}$). Thus, $b_{S\mid_{\leq k}} \mathbb Q\left[S_k\right] a_{T_{\lambda}} = 0$ follows immediately from $b_{S\mid_{\leq k}} \mathbb Q\left[S_k\right] a_{T_{\lambda}\mid_{\leq k}} = 0$. Proposition 4 is proven.

Conjectures:

Conjecture 5. Let $n$ be a nonnegative integer. Let $\lambda$ be a partition of $n$. Let $T = T_{\lambda}$. Then, $c_{T\mid_{\leq 1}} c_{T\mid_{\leq 2}} ... c_{T\mid_{\leq n}}$ is an integer multiple of $c_T$.

A stronger conjecture: $c_{T\mid_{\leq n-1}} c_T = \kappa_{T\mid_{\leq n-1}} c_T$, where $\kappa_S$ denotes the product of the hook lengths of the shape of a tableau $S$.

Unless I have done a mistake, in proving the latter (stronger) conjecture one can WLOG assume that $T$ is a two-column tableau, with both columns having the same length. I am not fully sure about it, and more vexingly, I am not able to prove it in this seemingly simple case!

Conjecture 6. Let $n$ be a nonnegative integer. Let $\lambda$ be a partition of $n$. Let $T = T_{\lambda}$. Then, $c_{T\mid_{\leq n}} c_{T\mid_{\leq n-1}} ... c_{T\mid_{\leq 1}}$ commutes with all the Young-Jucys-Murphy elements $y_1$, $y_2$, ..., $y_n$ in $\mathbb Q\left[S_n\right]$. Here, the Young-Jucys-Murphy elements $y_1$, $y_2$, ..., $y_n$ are defined by

$y_i = \left(1,i\right) + \left(2,i\right) + ... + \left(i-1,i\right)$ (a sum of $i-1$ transpositions)

for every $i \in \left\lbrace 1,2,...,n\right\rbrace$. As a consequence, $c_{T\mid_{\leq n}} c_{T\mid_{\leq n-1}} ... c_{T\mid_{\leq 1}}$ lies in the commutative subalgebra of $\mathbb Q\left[S_n\right]$ generated by $y_1$, $y_2$, ..., $y_n$. (This is a "consequence" because that commutative subalgebra is maximally commutative. But how to write it explicitly as a polynomial in the $y_j$ ?)

I made no real progress towards showing any of these. Sage verified both conjectures for $n\leq 6$. It seems possible that §3.2 of James/Kerber (which is about Young's seminormal form) has something to tell about this, but I my knowledge of that part of $S_n$ theory is negligible. Any help is welcome, particularly in the form of nice and readable proofs :)

Remark: It is not generally true that $c_S \mathbb Q\left[S_n\right] c_{T_{\lambda}} = 0$ in the context of Proposition 4.

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