# 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 $n$-permutational tableaux 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 $n$-permutational tableaux 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 $c_S \mathbb{Q}\left[S_n\right] c_{T_{\lambda}} = 0$, but since $S = S\mid_{\leq n}$, this becomes $c_{S\mid_{\leq n}} \mathbb{Q}\left[S_n\right] c_{T_{\lambda}} = 0$. 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 $S\mid_{\leq k}$ in the lexicographic ordering. Hence, Lemma 1 yields $a_{T_{\lambda}\mid_{\leq k}} \mathbb{Q} \left[S_k\right] b_{S\mid_{\leq k}} = 0$ 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, $a_{T_{\lambda}\mid_{\leq k}}$ divides $a_{T_{\lambda}}$ in the group algebra $\mathbb{Q}\left[S_n\right]$ (since $R_{T_{\lambda}\mid_{\leq k}}$ 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!

EDIT: Conjecture 5 follows from Theorem 1.2 in Claudiu Raicu, Products of Young symmetrizers and ideals in the generic tensor algebra, 1301.7511v2 (after some straightforward transformations like transposing tableaux). I would still enjoy a simpler proof (Raicu's one is about 10 pages long).

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 $y_j$ ?)

EDIT: Conjecture 6 can be strengthened to the statement that $c_{T\mid_{\leq n}} c_{T\mid_{\leq n-1}} ... c_{T\mid_{\leq 1}}$ is a scalar multiple of the seminormal idempotent $e\left(T\right)$ (where I am using the notations of Adriano Garsia, Young seminormal representation, Murphy elements and content evaluations). Thanks to Andrew Mathas for suggesting this statement! It can be proven by induction over $n$ using Conjecture 5 (actually, using the statement that $c_{T\mid_{\leq k}} c_T$ is a scalar multiple of $c_T$ for every $k\leq n$; but this statement isn't hard to derive from Conjecture 5).

So, technically, all conjectures above are settled; but any ideas for short proofs are welcomed.

[Obsolete remark: It seems possible that §3.2 of James/Kerber (which is about Young's seminormal form) has something to tell about this, but my knowledge of that part of $S_n$ theory is negligible.]

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.

I think that these conjectures are essentially equivalent to results in the literature, but the translation is not straightforward. I know corresponding results for the Hecke algebras of the symmetric groups. When you work in this generality you have to change some of the statements a little in order to get them to work. Primarily this is because your column stabliser $C_T$ will not usually be a standard parabolic subgroup so there is no corresponding natural subalgebra of the Hecke algebra to work with. To get around this you take the corresponding standard parabolic, for which the Hecke algebra does have a subalgebra, and then in the Hecke algebra you conjugate by $T_d$ where $d$ is a minimal length coset representative which makes the two parabolics conjugate. Ultimately, this will prove the results that you want, but it gives an additional layer of complications, which add to the notational confusion.

I think that the easiest way to approach these questions is using seminormal forms. As you probably know, Okounkov and Vershik [1] have given a really nice approach to this subject.

The main fact we need is that over $\Bbb{Q}$ the Specht module $S^\lambda$ has a seminormal basis $\{v_T\}$ indexed by the standard $\lambda$-tableaux with the property that $$y_k v_T = c_k(T) v_T.\qquad\qquad(1)$$ Here $c_k(T)$ is the content of $k$ the tableau $T$. (This conflicts slightly with your notation, but it shouldn't cause any problems.) That is, $c_k(T)$ is equal to the column index of $k$ in $T$ minus the row index of $k$ in $T$. (If I have misread your conventions above then you will need to replace $c_k(T)$ with $-c_k(T)$.)

Each seminormal basis element $v_T$ is a simultaneous eigenvector for $y_1,\dots,y_n$. So $v_T$ is uniquely determined up to multiplication by a non-zero scalar. The exact choice of seminormal basis $\{v_T\}$ doesn't matter in what follows. Now, for each standard tableau $S$ define $$F_S = \prod_{k=1}^n\prod_{c\ne c_k(S)}\frac{y_k-c}{c_k(S)-c},$$ where in the second product $c\in\{c_m(V)\}$ where $1\le m\le n$ and $V$ ranges over all of the standard tableaux of size $n$ (so any shape). Acting on our seminormal basis using (1) shows that $$F_S v_T = \delta_{ST}v_T.$$ To see this you only need to observe that if $S$ and $T$ are two standard tableaux, not necessarily of the same shape, then $S=T$ if and only if $c_m(S)=C_m(T)$ for $1\le m\le n$. In turn, this is easily proved by induction on $n$. Letting $\lambda$ range over all partitions of $n$, and using (1) again, you can now prove that $\{F_S\mid S\text{ standard}\}$ is a complete set of pairwise orthogonal primitive idempotents in $\Bbb{Q}S_n$.

To connect with what you are saying let me set $a_\lambda=a_T$ and $b_\lambda=b_T$, where $T=T_\lambda$ is your initial tableau. Also let $T^\lambda$ be the final tableau, that is, the unique standard $\lambda$-tableau which has the numbers $1,2,\dots,n$ entered in order from top to bottom and then left to right down the columns of $\lambda$. (To add to the notational confusion, in my papers, I call these tableaux $t^\lambda$ and $t_\lambda$ respectively.)

The connection with your questions is that $a_\lambda b_\lambda$ is a scalar multiple of $F_{T^\lambda}$ and the multiple is exactly the product of the hook lengths. As far as I am aware, the first place that this appears in the literature is in Murphy's paper [3]. After translation, (I believe!) this follows from the displayed formula at the bottom of page 512. In Prop. 4.4 of my paper [2] you will find a statement more in keeping with the notation above, except that this paper considers the cyclotomic Hecke algebra case (so more notational complications).

Once you have this fact, then your conjecture 6 follows and conjecture 5 should follow from the definition of $F_s$. In the mathematical-physics literature, I think that your conjecture 5 is often referred to as a fusion system -- actually, this is not quite the same thing, but it is very similar.

Unfortunately, translating all of the notation back to your setting is slightly painful. Perhaps there are some (old?) papers in the representation theory of the symmetric groups that deal with seminormal forms which just give the result you are interested in, but I am not aware of any. I suspect that it might be easier to just re-prove these things in your setting using the seminormal idempotents $\{F_T\}$.

References

1. A. Okounkov and A. Vershik, A new approach to representation theory of symmetric groups, Selecta Math. (N.S.), 2 (1996), 581–605.
2. Mathas, Andrew Matrix units and generic degrees for the Ariki-Koike algebras. J. Algebra 281 (2004), no. 2, 695–730. arXiv:0108164
3. G.E. Murphy, On the representation theory of the symmetric groups and associated Hecke algebras, J. Algebra 152 (1992) 492–513.
• Sorry for the long silence, and welcome to MathOverflow, Andrew! I'm slowly getting friendly with the seminormal basis (though I'm reading Garsia's notes on it, and there he has a somewhat different formula for it in terms of YJMs than you do -- I assume your additional factors don't matter much). Unfortunately I have some troubles verifying the statement that $a_\lambda b_\lambda$ is a scalar multiple of $F_{T^\lambda}$. For $\lambda = (2,1)$, the element $a_\lambda b_\lambda$ has four terms, while all elements of the seminormal basis have six apart of two which have zero id-coefficient... – darij grinberg Jan 26 '14 at 11:42
• @darij Hi Darij, I will add some more details when I have time. Rereading my post it seems that I have only given background to answering your question rather than the actual answer:(. I've only just returned to Sydney, however, so it will be at least a few days until I will have time to write ore details. I haven't seen Garsia's notes, but the extra terms don't matter as they all act as the identity on the corresponding seminormal basis element. It's a good question as to what the "smallest" possible product is for the Young idempotents. – Andrew Jan 28 '14 at 0:14
• No problem -- I'm not in a hurry with this question; thanks for the education! – darij grinberg Jan 28 '14 at 0:21
• I think you meant that $c_{T\mid_{\leq n}} c_{T\mid_{\leq n-1}} ... c_{T\mid_{\leq 1}}$ is a scalar multiple of the seminormal idempotent $e\left(T\right)$ (I am using the notations of Garsia's notes on Young's Seminormal Representation -- math.ucsd.edu/~garsia/somepapers/Youngseminormal.pdf ). Actually, I see how to prove this, using induction and Theorem 1.1 of Claudiu Raicu's arXiv:1301.7511v2. OK, so no more conjectures, in the strict sense of this word, left. But my head is spinning from all this overkill... – darij grinberg Mar 25 '14 at 4:33