## Littlewood-Richardson Coefficients

Recall that the Littlewood-Richardson coefficient satisfies:

By the previous article, for any SSYT of shape , is the number of skew SSYT of shape whose rectification is Since this number is independent of our choice of as long as its shape is we take the most natural one.

Definition. An SSYT is said to becanonicalif its -th row comprises of all For example, the following is a canonical SSYT of shape (5, 4, 3).

Note that an SSYT is canonical if and only if its shape is the same as its type. Thus is the number of skew SSYT of shape such that its rectification is canonical of shape

## Condition for Canonical SSYT

Definition. A word is areverse lattice wordif, for each , the substring has at least as many 1’s as it has 2’s, at least as many 2’s as it has 3’s, etc.A skew SSYT is called a

Littlewood-Richardson tableauif is a reverse lattice word.

For example, the following skew SSYT is a Littlewood-Richardson tableau since its word is (1,3,2,2,1,2,1,1).

Lemma 1. An SSYT is a Littlewood-Richardson tableau if and only if it is canonical.

**Proof**

Indeed, must end in 1, so the last entry of the first row is 1. Since each row is weakly increasing, the whole of first row is 1. Now the last entry of the second row is strictly greater than 1, but since is a reverse lattice word, that entry can only be 2. Thus the whole of second row comprises of 2’s. And so on. ♦

Lemma 2. If and are Knuth-equivalent words, then is a reverse lattice word if and only if is.

**Proof**

We may assume is obtained from via R1 or R2. For convenience, we say that a word is *OK* if it has at least as many as it has , for each

Consider R1; the other case is similar. We find a consecutive triplet in with then replace to form Let us show that, for any word the four words on the left are OK if and only if the four words on the right are OK.

We only need to check the third word in each case. Suppose the LHS is OK.

- If
*b*,*v*is not OK, since*v*is OK there must be a*d*<*b*such that*d*and*b*occur an equal number of times in*v*. - Since
*b*,*c*,*v*is OK, we must have*d*=*c*, and so any element between*c*and*b*(inclusive) occurs an equal number of times in*v*. - In particular,
*a*,*b*,*c*each occurs an equal number of times in*v*. - Now if
*a*<*b*, then*b*,*c*,*v*has more*b*‘s than*a*‘s, and if*a*=*b*, then*a*,*b*,*c*,*v*has more*a*‘s than*c*‘s, which contradicts the fact that the LHS is OK.

Now suppose the RHS is OK. If *c*,*v* is not OK, there is a *d* < *c* which occurs as many times in *v* as *c*. But now *c*,*b*,*v* is not OK, which gives a contradiction. ♦

For example, check that the rectification of the above Littlewood-Richardson tableau is:

**Exercise**

Write out the proof for R2.

Now, putting the above two lemmas together, we have:

Corollary. A skew SSYT is a Littlewood-Richardson tableau if and only if its rectification is canonical.

**Proof**

Indeed if is its rectification, then is Littlewood-Richardson if and only if is a reverse lattice word; by lemma 2, this holds if and only if is a reverse lattice word. By lemma 1, this holds if and only if is canonical. ♦

We thus obtain:

Littlewood-Richardson Rule. is the number of Littlewood-Richardson skew SSYT of shape and type

## Example 1: Expanding Skew Schur Polynomials

Consider the skew Schur polynomial where and To express as a linear sum of Schur polynomials, we only need to consider partitions of Thus:

We can also compute it directly by finding for various , which gives:

And the Schur polynomials give:

## Example 2: Expanding Product of Schur Polynomials

Let us express as a linear sum of Schur polynomials. For each partition of 6, we have:

Swapping the two partitions (3) and (2,1) gives us an identical expansion, as expected.