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Equation 3.5 of this paper by Gross and Klebanov makes the following interesting claim.

Take a group $U(N)$, with $N$ large, and consider the reducible representation $\mathcal{H}_{fund}^{\otimes k}$ where $\mathcal{H}_{fund}$ is the fundamental representation and $1 \ll k \ll N$.

It can be decomposed into irreps,

$$\mathcal{H}_{fund}^{\otimes k} = \bigoplus_{i} \alpha_i\mathcal{H}_i .$$

So far, very simple. Equation 3.5 says that

$$\text{dim } \mathcal{H}_i = \frac{N^k}{k!} \alpha_i + O(N^{k-1}),\tag{3.5}$$

where the same multiplicity $\alpha_i$ turn up in both equations!

How does one derive this?

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  • $\begingroup$ This might be well-known to combinatorics mavens; but you could see that k Young boxes combine to irreps whose dimensionality tends to $N^k/k!$ for each by the hook rule, to leading order in N and k, so what survives is the multiplicities of each resulting irrep α, as given, in the preservation of states. Explicitly trying k = 4 or 5 for huge N might give you insight... $\endgroup$ Commented Dec 29, 2021 at 21:55

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Here's the answer.

A Young diagram (YD) with $k \le N$ boxes is a partition of $k$, $[\lambda_1... \lambda_k | \lambda_1 \ge \lambda_2 ... \ge \lambda_k]$; these are in 1-1 correspondence with the irreps we find in $\mathcal{H}_{fund}^{\otimes k}$. We denote it by $[\lambda]$; it turns up in $\mathcal{H}_{fund}^{\otimes k}$ $\alpha_{[\lambda]}$ times.

Let's use this short note as a reference on how to calculate the dimension of the irrep corresponding to a Young diagram. At leading order in large $N$, the dimension of $[\lambda]$ is $$\text{dim} \mathcal{H}_{[\lambda]} = \frac{N^k}{hl([\lambda])},$$ where $hl([\lambda])$ is the hook length of the YD. Here, we have approximated the numerator by $N^k$ but left the denominator exact; this will be justified a posteriori. What we want to prove, then, is $$\alpha_{[\lambda]} = \frac{k!}{hl([\lambda])}.$$ Notice that this is trivially true for $k=1$. We prove it for generl $k$ by induction.

From a YD with $k$ boxes, we can make one with $k-1$ boxes by removing a "corner" box, which corresponds to reducing one of the $\lambda_i$ by $1$. Denote such a young diagram with $[\lambda]\setminus i$. (If $\lambda_i = 0$ then $[\lambda] \setminus i$ doesn't exist. We assume that these are removed in the discussion below.) $[\lambda] \setminus i$ appears in $\mathcal{H}_{fund}^{\otimes k-1}$ $\alpha_{[\lambda] \setminus i}$ times. Since $\mathcal{H}_{fund}^{\otimes k} = \mathcal{H}_{fund}^{\otimes k-1} \otimes \mathcal{H}_{fund}$, all the ways to arrive at the irrep $[\lambda]$ correspond to adding a box in the $i^{th}$ row to $[\lambda]\setminus i$, giving $$\alpha_{[\lambda]} = \sum_{i=1}^{k} \alpha_{[\lambda] \setminus i} \quad \Rightarrow \quad \sum_i \frac{\alpha_{[\lambda]\setminus i}}{\alpha_{[\lambda]}} = 1.$$

It turns out that this is a rather involved statement to prove, but it is one of the fundamental facts about the hook length, and the proof, via a "Hook walk" is sketched on wikipedia (Original paper by Greene, Nijenhuis, Wilf). $e_{\lambda}$ on wikipedia and $F$ in the paper is our $\alpha_{[\lambda]}$.

I thank Cosmas Zachos for their comment, which helped me put the pieces together (esp. for making me realise that large $k$ isn't necessary).

An interesting side-note: this formula relates the dimension of the irrep to the number of standard Young tableaux (sYTs), whereas states are indexed by semi-standard Young tableaux. The dimension can be written as ${N \choose k} (\# \text{sYTs with $k$ boxes})$, i.e. we pick $k$ numbers from $1... N$ and then use these as $k$ labels for making sYTs.

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