# Sum of Young symmetrisers of a given shape

Preliminaries and notation:

Let $n\in \mathbb{Z}_{>0}$ and $\lambda=(\lambda_1,\lambda_2,\dots,\lambda_s)\vdash n$ be a partition. Given a Young diagram of shape $\lambda$, we can associate it with a Young tableau with $1,2,\dots,n$ filled in the Young diagram. A Young tableau is called standard if both the row and column entries in the Young diagram are strictly increasing.

Now let ${\rm Tab}(\lambda)$ be the set of all Young tableaux of shape $\lambda$, and let ${\rm Std}(\lambda)$ be the set of all standard Young tableaux of shape $\lambda$.

Given $\mathfrak{t}\in {\rm Tab}(\lambda)$, define the Young Symmetriser $$c_{\lambda}(\mathfrak{t}):=x_{\lambda}(\mathfrak{t})y_{\lambda}(\mathfrak{t}),$$ where $x_{\lambda}(\mathfrak{t})=\sum_{\sigma\in R(\mathfrak{t})}\sigma, y_{\lambda}(\mathfrak{t})=\sum_{\sigma\in C(\mathfrak{t})}\epsilon(\sigma)\sigma$ with $R(\mathfrak{t})$ and $C(\mathfrak{t})$ being respectively the set of row stablisers and column stablises of $\mathfrak{t}$.

Let $h_\lambda={\prod_{i,j} h_{ij}^{\lambda}}$, where $h_{ij}^{\lambda}=\lambda_{i}+\lambda^{T}_{j}-i-j+1$ is the hook length of the $(i,j)$-box in the Young diagram of shape $\lambda$. $\lambda^{T}$ is the conjugate partition of $\lambda$.

Question

Does the following identity hold? $$\sum_{\mathfrak t\in {\rm Tab}(\lambda)} c_{\lambda}(\mathfrak{t})=h_{\lambda} \sum_{\mathfrak t\in {\rm Std}(\lambda)} c_{\lambda}(\mathfrak{t}).$$

One can verify this directly for all partitions $\lambda\vdash n$ when $n=2,3$. But I have no idea for the general proof.

• Let $L$ be the left hand side of your identity. For every $\mathfrak{s} \in \operatorname{Tab}\left(\lambda\right)$, we have $L = \sum_{\sigma \in S_n} \sigma c_\lambda \left(\mathfrak{s}\right) \sigma^{-1}$. If we sum this fact over all $\mathfrak{s} \in \operatorname{Std}\left(\lambda\right)$, we obtain $\left|\operatorname{Std}\left(\lambda\right)\right| L = \sum_{\sigma \in S_n} \sigma \sum_{\mathfrak{s} \in \operatorname{Std}\left(\lambda\right)} c_\lambda\left(\mathfrak{s}\right) \sigma^{-1}$. But ... Jun 17, 2016 at 18:42
• ... every $\sigma \in S_n$ satisfies $\sigma \sum_{\mathfrak{s} \in \operatorname{Std}\left(\lambda\right)} c_\lambda\left(\mathfrak{s}\right) \sigma^{-1} = \sum_{\mathfrak{s} \in \operatorname{Std}\left(\lambda\right)} c_\lambda\left(\mathfrak{s}\right)$ (since $\sum_{\mathfrak{s} \in \operatorname{Std}\left(\lambda\right)} c_\lambda\left(\mathfrak{s}\right)$ is a projector onto the $\lambda$-isotypic component, maybe up to scalar factor, and thus lies in the center of the group algebra and therefore commutes with $\sigma$). Hence, the equation I got ... Jun 17, 2016 at 18:43
• ... becomes $\left|\operatorname{Std}\left(\lambda\right)\right| L = n! \sum_{\mathfrak{s} \in \operatorname{Std}\left(\lambda\right)} c_\lambda\left(\mathfrak{s}\right)$. Dividing this equation by $\left|\operatorname{Std}\left(\lambda\right)\right| = \dfrac{n!}{h_\lambda}$, we obtain $L = h_\lambda \sum_{\mathfrak{s} \in \operatorname{Std}\left(\lambda\right)} c_\lambda\left(\mathfrak{s}\right)$, which is precisely your claim. Jun 17, 2016 at 18:44
• My above "proof" is false. See the answer below. Jun 17, 2016 at 22:21

Oops! Despite my comments, this is actually false. (I was wrong in claiming that $\sum_{\mathfrak{s} \in \operatorname{Std}\left(\lambda\right)} c_\lambda\left(\mathfrak{s}\right)$ is a projector onto the $\lambda$-isotypic component.)

To get a counterexample for $n=5$, let me observe that the left hand side of your identity equals $\sum_{\sigma \in S_n} \sigma c_\lambda\left(\mathfrak{t}\right) \sigma^{-1}$ for any arbitrary tableau $\mathfrak{t} \in \operatorname{Tab}\lambda$; thus, it is invariant under conjugation by any $\sigma \in S_n$, and therefore is a central element of the group algebra $\mathbb{Q}\left[S_n\right]$. Hence, your conjecture implies that the sum $\sum_{\mathfrak{s} \in \operatorname{Std}\left(\lambda\right)} c_\lambda\left(\mathfrak{s}\right)$ is a central element of the group algebra $\mathbb{Q}\left[S_n\right]$. But not even the sum of these sums over all $\lambda \vdash 5$ is central:

#sagemath code
from sage.combinat.symmetric_group_algebra import a, b, e
QS5 = SymmetricGroupAlgebra(QQ, 5)
s = sum(e(T) for T in StandardTableaux(5))
g = QS5(Permutation([1,3,2,4,5]))
s*g - g*s


returns

[1, 2, 4, 5, 3] + [1, 2, 5, 3, 4] + [1, 3, 4, 5, 2] + [1, 3, 5, 2, 4] - [1, 4, 2, 5, 3] - [1, 4, 3, 5, 2] - [1, 5, 2, 3, 4] - [1, 5, 3, 2, 4] + [2, 1, 4, 5, 3] - [2, 1, 5, 3, 4] + 2*[2, 1, 5, 4, 3] + [2, 3, 4, 5, 1] + [2, 3, 5, 4, 1] - [2, 4, 1, 5, 3] - [2, 4, 3, 5, 1] + [2, 4, 5, 1, 3] - [2, 4, 5, 3, 1] + [2, 5, 1, 3, 4] - 2*[2, 5, 1, 4, 3] - [2, 5, 3, 4, 1] - [2, 5, 4, 1, 3] + [2, 5, 4, 3, 1] + [3, 1, 4, 5, 2] - [3, 1, 5, 2, 4] + 2*[3, 1, 5, 4, 2] + [3, 2, 4, 5, 1] + [3, 2, 5, 4, 1] - [3, 4, 1, 5, 2] - [3, 4, 2, 5, 1] + [3, 4, 5, 1, 2] - [3, 4, 5, 2, 1] + [3, 5, 1, 2, 4] - 2*[3, 5, 1, 4, 2] - [3, 5, 2, 4, 1] - [3, 5, 4, 1, 2] + [3, 5, 4, 2, 1] - [4, 1, 5, 2, 3] - [4, 1, 5, 3, 2] - 2*[4, 2, 5, 1, 3] - 2*[4, 3, 5, 1, 2] + [4, 5, 1, 2, 3] + [4, 5, 1, 3, 2] + 2*[4, 5, 2, 1, 3] + 2*[4, 5, 3, 1, 2] + [5, 1, 2, 3, 4] - [5, 1, 2, 4, 3] + [5, 1, 3, 2, 4] - [5, 1, 3, 4, 2] - [5, 1, 4, 2, 3] - [5, 1, 4, 3, 2] - [5, 2, 1, 3, 4] + [5, 2, 1, 4, 3] - 2*[5, 2, 4, 3, 1] - [5, 3, 1, 2, 4] + [5, 3, 1, 4, 2] - 2*[5, 3, 4, 2, 1] + [5, 4, 1, 2, 3] + [5, 4, 1, 3, 2] + 2*[5, 4, 2, 3, 1] + 2*[5, 4, 3, 2, 1]


EDIT: However, your conjecture is true for any $\lambda$ of size $\leq 4$, and also for any partition $\lambda$ of hook shape. The first of these two claims can be proven computationally. To prove the second claim, I will show that if $\lambda$ is a partition of hook shape, then my false proof (sketched in the comments) actually works. Why? Well, assume that $\lambda$ is of hook shape. Then, it is easy to see that if $\mathfrak{s}$ and $\mathfrak{s}'$ are two distinct standard tableaux of shape $\lambda$, then there exist two distinct numbers lying in the same row of $\mathfrak{s}$ (namely, the first row) and also lying in the same column of $\mathfrak{s}'$ (namely, the first column). Hence, $c_\lambda\left(\mathfrak{s}\right) c_\lambda\left(\mathfrak{s}'\right) = 0$. Thus, the sum $\sum_{\mathfrak{s} \in \operatorname{Std}\left(\lambda\right)} h_\lambda c_\lambda\left(\mathfrak{s}\right)$ is a sum of pairwise orthogonal idempotents. Each of these idempotents projects onto an irreducible $S_n$-submodule of the group ring $\mathbb{Q}\left[S_n\right]$ lying in the $\lambda$-isotypic component. These $S_n$-submodules are linearly disjoint (since the idempotents are orthogonal), and therefore cover the whole $\lambda$-isotypic component (by dimension reasons: the number of these idempotents is exactly the number of irreducible submodules in the $\lambda$-isotypic component of $\mathbb{Q}\left[S_n\right]$). Hence, the sum of these idempotents must be the idempotent projector onto the whole $\lambda$-isotypic component; hence, it is clearly a central element of $\mathbb{Q}\left[S_n\right]$. Now, the argument in the comments goes through.