Littlewood–Richardson coefficients and products
Count Schur-product multiplicities using skew tableaux with a fully specified ballot reading convention.
On this page
The question for today
What you’ll understand
- Apply the top-to-bottom right-to-left Littlewood–Richardson reading rule.
- Compute a coefficient greater than one by explicit ballot fillings.
- Use Pieri as a multiplicity-free special case and distinguish it from general products.
Multiplying Schur functions produces nonnegative integer coefficients with several interpretations: counts of restricted skew tableaux, multiplicities in induced symmetric-group representations, and tensor multiplicities for polynomial general linear representations in a sufficient number of variables. The Littlewood–Richardson rule explains the counting interpretation. Because different books use different reading conventions, we fix English diagrams, weakly increasing rows, strictly increasing columns, and a reading word obtained by scanning each row right to left, starting with the top row.
Skew boxes carry a content and a ballot condition
If the diagram of λ is contained in ν, the skew shape ν/λ consists of boxes in ν that are outside λ. Fill only those boxes, comparing entries in the usual skew tableau row and column positions; deleted boxes receive no entries and impose no artificial inequalities. Content μ means that exactly μi entries equal i. Necessarily the number of skew boxes equals the size of μ. These elementary support conditions eliminate many candidate coefficients before any word is tested.
Read the filled skew diagram in the declared order. The word is lattice, or ballot, if every initial segment contains at least as many 1s as 2s, at least as many 2s as 3s, and so on. Checking only the whole word is insufficient: its total content is already a partition. The first symbol must therefore be 1. The LR theorem states that the coefficient cλμν counts exactly the semistandard skew fillings satisfying this prefix condition. We state the theorem and verify small cases rather than derive its general proof.
Pieri exposes an accessible boundary case
When μ is a single row of length r, every skew entry is 1. Column strictness then allows at most one added box in each column, while the word is automatically ballot. Thus multiplication by hr adds a horizontal r-strip, each allowable shape with coefficient one. Horizontal refers to the no-two-in-one-column condition; the added boxes need not lie in a single row or form a connected region. For er, the corresponding theorem adds a vertical strip, with at most one new box in each row.
General LR products are not multiplicity-free. Distinct ballot fillings of the same skew shape and content can yield coefficients two or larger. Commutativity of symmetric functions gives cλμν=cμλν even though the chosen tableau description treats λ as removed boxes and μ as content. This equality is not visually automatic from the definition. A bijection or algebraic argument must explain it; exchanging shape and content in a single filling without changing anything else is not a valid proof.
Stable products and representation multiplicities
The product identity lives in the stable ring over Z. After specialization to N variables, terms whose shapes have more than N rows vanish. Therefore one cannot infer that their stable coefficient was zero by observing a small-variable polynomial product. In complex representation theory the surviving coefficients describe polynomial GLN tensor products when both input shapes fit in N rows; the stable rule and the finite-dimensional interpretation remain consistent through this vanishing mechanism.
A useful verification workflow combines three independent constraints. First check sizes and containment. Then enumerate row- and column-admissible fillings for the required content. Finally test every word prefix, not just final totals. All-ones specialization provides a separate dimension check on a full product, but it does not isolate individual coefficients. This distinction becomes important for positivity conjectures: an integer can be nonnegative for representation-theoretic reasons even when no efficient positive counting formula has yet been found.
Work it through
A coefficient with two different witnesses
Compute c₂₁,₂₁³²¹. The skew boxes of (3,2,1)/(2,1) are at coordinates (row,column)=(1,3),(2,2),(3,1).
The skew shape has three boxes and content (2,1) requires two 1s and one 2. No two skew boxes share a row or column, so all three distinct assignments are semistandard.
The reading word lists the boxes in the stated coordinate order. The possible words are 112,121,211. The last word fails immediately because its first prefix has one 2 and no 1.
For 112, prefix counts are (1,0),(2,0),(2,1); for 121 they are (1,0),(1,1),(2,1). Every required inequality holds in both words, leaving two witnesses.
The coefficient c₂₁,₂₁³²¹ equals 2. The multiplicity comes from two ballot fillings, not from counting identical entries as separately labeled.
The living notebook
Explore the model
Interactive lab · probability
Prefix constraints filter a content class
Pause, reset, then set p=0.5 and n=3. Predict the one-success bar 3/8, counting the three words with two 1s and one 2. Check the words 112,121,211 manually against the prefix rule; exactly two survive, so their unconditional binary-word probability is 2/8. The bar groups by final content and cannot enforce ballot prefixes or skew column conditions.
- Mean · np
- 5
- Variance · np(1−p)
- 2.5
- Total probability
- 1
These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA final-content generating function needs extra structure to count Littlewood–Richardson witnesses.
Pause & reason
Check your understanding
Your turn
Try it yourself
Use Pieri to compute s₂₁s₁. List every allowed shape and explain why there are no multiplicities.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Add one box while keeping a partition; any single added box is a horizontal one-strip.
Compare with the solution
Starting from rows of lengths (2,1), a box may extend the first row to (3,1), extend the second to (2,2), or create a third row to (2,1,1). These are all addable corners. Every skew shape has one box, uniquely filled by 1, so each coefficient is one. Thus s₂₁s₁=s₃₁+s₂₂+s₂₁₁. In two variables the final term vanishes, but it remains present in the stable identity.
Keep the idea
Take it with you
- LR coefficients count skew semistandard fillings satisfying every ballot prefix.
- Pieri strips are multiplicity-free special cases of general Schur products.
- Small alphabets can hide positive stable coefficients by making their shapes vanish.
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