---
tag: Q490XBH1Q2
kind: statement
role: theorem
name: Littlewood–Richardson rule
msc: ["05E05", "05E10", 20C30]
uses: [43VFMJCDYJ, Q6TH9PH2R7, 096AHMB3P4, 0JV30XN73Z]
about: [43VFMJCDYJ]
sources:
  - url: https://en.wikipedia.org/wiki/Littlewood%E2%80%93Richardson_rule
same_as:
  wikidata: Q15304008
  wikipedia: Littlewood–Richardson rule
---

## Statement

Let $\lambda$ and $\mu$ be [[partitions|0JV30XN73Z]], and let $s_\lambda$ denote the [[Schur polynomial|43VFMJCDYJ]] (in any number of variables, or the Schur function).
Then
$$s_\lambda \, s_\mu = \sum_{\nu} c^{\nu}_{\lambda \mu} \, s_\nu,$$
where $c^{\nu}_{\lambda\mu}$ is the number of Littlewood–Richardson tableaux of shape $\nu / \lambda$ and content $\mu$.
Such a tableau is a filling $T$ of the skew shape $\nu / \lambda$ with these properties.
Here $\nu / \lambda$ is the set of cells of the [[Young diagram|096AHMB3P4]] of $\nu$ that are not in the diagram of $\lambda$.

1. The entries weakly increase along rows and strictly increase down columns.
2. For each $k$, exactly $\mu_k$ entries are equal to $k$.
3. Read the rows from right to left, from the top row down. In every initial segment of this word, for each $k \ge 1$, the entry $k$ occurs at least as often as $k+1$.

## Remarks

In particular $c^{\nu}_{\lambda\mu} = 0$ unless the diagram of $\lambda$ lies in the diagram of $\nu$ and $\abs{\nu} = \abs{\lambda} + \abs{\mu}$.
The coefficients are symmetric: $c^{\nu}_{\lambda\mu} = c^{\nu}_{\mu\lambda}$.
Modern proofs use the [[Robinson–Schensted–Knuth correspondence|8JW8F4VVED]] or Schützenberger's jeu de taquin, which turns each tableau of shape $\nu/\lambda$ into a straight-shape [[Young tableau|Q6TH9PH2R7]].

Example: take $\lambda = (1)$ and $\mu = (1, 1)$.
For $\nu = (2, 1)$ the two cells of $\nu / \lambda$ are in different rows and columns.
The reading word must start with $1$, so the only tableau has $1$ in the first row and $2$ in the second row.
For $\nu = (1, 1, 1)$ the two cells form a column filled with $1$ above $2$, and the reading word $1\,2$ is allowed.
For $\nu = (3)$ the cells form a row $1\,2$ whose reading word $2\,1$ fails clause 3.
Hence $s_{(1)} s_{(1,1)} = s_{(2,1)} + s_{(1,1,1)}$.

## Explanations

### L1 human

Schur polynomials are building blocks, and the product of two of them is again a sum of them.
The rule says how many times each building block appears: count certain fillings of a skew shape with numbers.
The reading condition makes sure that, at every point while reading, the $1$s are ahead of the $2$s, the $2$s ahead of the $3$s, and so on.

### L2 human

The numbers $c^\nu_{\lambda\mu}$ have several meanings at once.
They are the multiplicities in the tensor product of two irreducible polynomial representations of the general linear group.
They are the multiplicities of the irreducible representations of $S_{\abs\nu}$ in the representation induced from the irreducible representation of $S_{\abs\lambda} \times S_{\abs\mu}$ indexed by the pair $(\lambda, \mu)$.
They are also the structure constants of the cohomology ring of a Grassmannian in the basis of Schubert classes.
When $\mu = (k)$ is a single row, the rule reduces to Pieri's rule.
It says that $s_\lambda s_{(k)}$ is the sum of $s_\nu$ over the $\nu$ obtained by adding $k$ cells to $\lambda$, no two in the same column.