AN S3-SYMMETRIC LITTLEWOOD-RICHARDSON RULE
HUGH THOMAS AND ALEXANDER YONG
1. INTRODUCTION
Fix Young shapes λ, µ, ν ⊆ Λ := ℓ × k. Viewing the Littlewood-Richardson coeffi-
cients Cλ,µ,ν as intersection numbers of three Schubert varieties in general position in the
Grassmannian of ℓ-dimensional planes in Cℓ+k, one has the obvious S3-symmetries:
(1) Cλ,µ,ν = Cµ,ν,λ = Cν,λ,µ = Cµ,λ,ν = Cν,µ,λ = Cλ,ν,µ.
There is interest in the combinatorics of these symmetries; previously studied Littlewood-
Richardson rules for Cλ,µ,ν manifest at most three of the six, see, e.g., [BeZe91, KnTaWo01,
VaPa05, HeKa06] and the references therein. We construct a carton rule for Cλ,µ,ν that
transparently and uniformly explains all symmetries (1).
Figure 1 depicts a carton, i.e., a three-dimensional box with a grid drawn rectilinearly
on the six faces of its surface. Along the “∅, Tλ, Tµ” face, the grid has (|µ| + 1) × (|λ| + 1)
vertices (including ones on the bounding edges of the face); the remainder of the grid is
similarly determined. Define a carton filling to be an assignment of a Young diagram to
each vertex of the grid so the shapes increase one box at a time while moving away from
∅, and so that, for any subgrid
α − β
γ − δ
the Fomin growth assumptions hold:
(F1) if α is the unique shape containing γ and contained in β, then δ = α;
(F2) otherwise there is a unique such shape other than α, and this shape is δ.
Notice that these conditions are symmetric in α and δ.
Initially, assign the shapes ∅ and Λ to opposite corners, as in Figure 1. A standard Young
tableau T ∈ SYT(σ/π) of shape σ/π is equivalent to a shape chain in Young’s lattice, e.g.,
T = 1 5
↔ − − − − −
by starting with the unfilled inner shape, and adding one box at a time to indicate the
box containing 1, then 2, etc. Fix a choice of standard tableaux Tλ, Tµ and Tν of respective
shapes λ, µ and ν. Initialize the indicated edges with the shape chains for these tableaux.
Let CARTONSλ,µ,ν be all carton fillings with the above initial data.
Main Theorem. The Littlewood-Richardson coefficient Cλ,µ,ν equals #CARTONSλ,µ,ν.
This rule manifests bijections between CARTONSλ,µ,ν and CARTONSδ,ǫ,ζ for permutations
(δ, ǫ, ζ) of (λ, µ, ν). We remark that it can be stated as counting lattice points of a 0, 1-
polytope, and readily extends to the setting of [ThYo06, ThYo07].
Date: April 4, 2007.
http://arxiv.org/abs/0704.0817v1
FIGURE 1. The carton rule counts Cλ,µ,ν by assigning Young diagrams to
the vertices of the six faces
The Theorem is proved in Section 2, starting from the jeu de taquin formulation of the
Littlewood-Richardson rule, and using Fomin’s growth diagram ideas. In Figure 2 we
give an example of the Main Theorem. An extended example is given in Section 3.
ν = 1
FIGURE 2. The “front” three faces of a carton filling for C(2),(2,1),(1) = 1. The
choices of tableaux Tλ, Tµ and Tν are as shown.
2. TABLEAU FACTS AND PROOF OF THE MAIN THEOREM
2.1. Tableau sliding. There is a partial order ≺ on the rectangle Λ where x ≺ y if x is
weakly northwest of y. Given T ∈ SYT(ν/λ) consider x ∈ λ, maximal in ≺ subject to
being less than some box of ν/λ. Associate another standard tableau jdtx(T), called the
jeu de taquin slide of T into x: Let y be the box of ν/λ with the smallest label, among those
covering x. Move the label of y to x, leaving y vacant. Look for boxes of ν/λ covering y
and repeat, moving into y the smallest label among those boxes covering it. Then jdtx(T)
results when no further slides are possible. The rectification of T is the iteration of jeu de
taquin slides until terminating at a straight shape standard tableau rectification(T).
We can impose a specific “inner” U ∈ SYT(λ) that encodes the order in which the jeu
de taquin slides are done. For example, if U = 1 2 3
T = 1 5
→ 1 5
→ 1 5
→ 1 5
→ 1 3 5
= rectification(T). Placing these chains one atop another (starting with T ’s and ending
with rectification(T)’s) gives a Fomin growth diagram, which in this case is given in
Table 1. Growth diagrams satisfy (F1) and (F2); see Fomin’s [St99, Appendix 2] for more.
(3, 1) (4, 1) (4, 2) (4, 3) (4, 3, 1) (5, 3, 1) (5, 3, 2)
(3) (4) (4, 1) (4, 2) (4, 2, 1) (5, 2, 1) (5, 2, 2)
(2) (3) (3, 1) (3, 2) (3, 2, 1) (4, 2, 1) (4, 2, 2)
(1) (2) (2, 1) (2, 2) (2, 2, 1) (3, 2, 1) (3, 2, 2)
∅ (1) (1, 1) (2, 1) (2, 1, 1) (3, 1, 1) (3, 2, 1)
TABLE 1. A growth diagram
Given a top row U and left column T , define infusion1(U, T) to be the bottom row of the
growth diagram, and infusion2(U, T) to be the right column. We write infusion(U, T)
for the ordered pair (infusion1(U, T), infusion2(U, T)). Growth diagrams are transpose
symmetric, because (F1) and (F2) are. So one has the infusion involution:
infusion(infusion(U, T)) = (U, T).
If U is a straight shape, infusion1(U, T) = rectification(T), while infusion2(U, T) en-
codes the order in which squares were vacated in the jeu de taquin process.
Also, given T ∈ SYT(ν/λ), consider x ∈ Λ\ν minimal subject to being larger than some
element of ν/λ. The reverse jeu de taquin slide revjdtx(T) of T into x is defined similarly
to a jeu de taquin slide, except we move into x the largest of the labels among boxes in ν/λ
covered by x. We define reverse rectification revrectification(T) similarly. The first
fundamental theorem of jeu de taquin asserts (rev)rectification is well-defined.
Fix Tµ ∈ SYT(µ). The number of T ∈ SYT(ν/λ) such that rectification(T) = Tµ
equals Cλ,µ,ν∨ = C
λ,µ where ν
∨ is the straight shape obtained as the 180-degree rotation
of Λ\ν. This is Schützenberger’s jeu de taquin formulation of the Littlewood-Richardson
rule, see [St99, Appendix 2].
By a slide we mean either kind of jeu de taquin slide. Consider two equivalence rela-
tions on a pair of tableaux T and U. Tableaux are jeu de taquin equivalent if one can be
obtained from the other by a sequence of slides. They are dual equivalent if any such se-
quence results in tableaux of the same shape [Ha92]. Facts: Tableaux of the same straight
shape are dual equivalent. A common application of slides to dual equivalent tableaux
produces dual equivalent tableaux. A pair of tableaux that are both jeu de taquin and
dual equivalent must be equal.
Recall Schützenberger’s evacuation map. For T ∈ SYT(λ), let T̂ be obtained by erasing
the entry 1 (in the northwest corner c) of T and subtracting 1 from the remaining entries.
Let ∆(T) = jdtc(T̂). The evacuation evac(T) ∈ SYT(λ) is defined by the shape chain
∅ = shape(∆|λ|(T)) − shape(∆|λ|−1(T)) − . . .− shape(∆1(T)) − T.
This map is an involution: evac(evac(T)) = T .
2.2. Proof of the rule. Given T ∈SYT(α) for a straight shape α, define T̃ ∈SYT(rotate(α))
where rotate(α) = Λ \ α∨ by computing evac(T) ∈ SYT(α), replacing entry i with
|α|− i+ 1 throughout and rotating the resulting tableau 180-degrees and placing it at the
bottom right corner of Λ.
We need the following well-known fact. The proof we give extends straightforwardly
to the setup of [ThYo06, ThYo07], allowing a “cominuscule” version of the Main Theorem.
Lemma 2.1. Suppose α, β and γ are shapes where
Tβ ∈ SYT(β), Tγ∨/α ∈ SYT(γ
∨/α) and rectification(Tγ∨/α) = Tβ.
Then revrectification(Tγ∨/α) = revrectification(Tβ) = T̃β.
Proof. It suffices to show that revrectification(Tβ) = T̃β. We induct on |β| = n.
Let U be the tableau obtained by removing the box labeled 1 from Tβ. So U is of skew
shape β/(1). Let V = rectification(U), a tableau of shape κ ⊂ β. By induction,
revrectification(V) = Ṽ . Now revrectification(Tβ) and revrectification(V) agree
except that the former has a label 1 in a box that does not appear in the latter. In [ThYo06,
Proposition 4.6], we showed that the reverse rectification of a tableau of shape β is nec-
essarily of shape rotate(β). Thus, the location of 1 in revrectification(Tβ) must be the
box κ∨/β∨. This is the 180-degree rotation of the box β/κ in Λ, the latter being the position
of n in evac(Tβ). Thus, 1 is located in the desired position in revrectification(Tβ). The
rest of the statement follows from the fact that the rest of revrectification(Tβ) agrees
with Ṽ , and the entries of evac(Tβ) other than n agree with evac(V). �
Corollary 2.2. Fix a carton filling. The uninitialized corners of the “∅, Tλ, Tµ” and “∅, Tν, Tλ”
faces are ν∨ and µ∨ respectively. The remaining “sixth” corner (the unique one not visible in
Figure 1) is λ∨. Thus, we can speak of the edges λ∨−Λ, µ∨−Λ and ν∨−Λ. These represent the
shape chains of T̃λ, T̃µ and T̃ν respectively.
Proof. Think of the union of the faces “∅, Tµ, Tλ” with the adjacent face involving Λ and the
“sixth corner” as a single growth diagram. This diagram computes revrectification(Tλ)
and records the result along the edge involving Λ and diagonally opposite to the Tλ edge.
But the lemma asserts this result is T̃λ and hence the “sixth corner vertex” is assigned λ
The other conclusions are proved similarly. �
Corollary 2.2 affords us the convenience of referring to a face by its corner vertices.
Note any carton filling gives a growth diagram on the face ∅ − µ − ν∨ − λ for which
the edge λ− ν∨ is a standard tableau of shape ν∨/λ rectifying to Tµ. By the jeu de taquin
Littlewood-Richardson rule, fillings of this face count Cλ,µ,ν. Hence it suffices to show that
any such growth diagram for this face extends uniquely to a filling of the entire carton.
If such an extension exists, it is unique: λ− ν∨ and Λ− ν∨ determine, by (F1), (F2) and
the Corollary, the face λ−µ∨−Λ−ν∨. Similarly, these two faces determine the remaining
faces (in order):
∅− ν− µ∨− λ =⇒ ν− λ∨−Λ− µ∨ =⇒ µ− λ∨−Λ− ν∨ =⇒ ∅− ν− λ∨ − µ.
We now show that with a filling of ∅−µ−ν∨−λ (together with the extra edges given in
the Corollary) one can extend it to fill the carton, using the conclusions of Corollary 2.2.
∅− µ− ν∨− λ: This is given by labeling λ − ν∨ with Tν∨/λ ∈ SYT(ν
∨/λ) that witnesses
Cλ,µ,ν = C
λ,µ, i.e., it rectifies to Tµ. By the infusion involution, the edge µ−ν
∨ representing
Tν∨/µ := infusion2(Tλ, Tν∨/λ) ∈ SYT(ν
∨/µ) witnesses Cν
λ− µ∨−Λ− ν∨: Build the growth diagram using Tν∨/λ and T̃ν from the edges λ−ν
∨ and
ν∨ − Λ respectively. The only boundary condition we need to check is that
infusion2(Tν∨/λ, T̃ν) = T̃µ, which is true by the Lemma. The newly determined edge
λ − µ∨ is infusion1(Tν∨/λ, T̃ν). This is a tableau Tµ∨/λ ∈ SYT(µ
∨/λ) which rectifies to Tν,
i.e., one that witnesses C
∅− ν− µ∨− λ: Using the determined edges λ−µ∨ and ∅−λ we obtain a growth diagram
with edge ∅ − ν representing infusion1(Tλ, Tµ∨/λ), which equals Tν, as desired. By the
infusion involution, ν− µ∨ represents a tableau Tµ∨/ν ∈ SYT(µ
∨/ν) that witnesses C
ν− λ∨ −Λ− µ∨: This time, we grow the face using the edges ν−µ∨ and µ∨−Λ. The edge
λ∨ − Λ is thus infusion2(Tµ∨/ν, T̃µ) which indeed equals T̃λ, by the Lemma. The newly
determined edge ν−λ∨ is Tλ∨/ν := infusion1(Tµ∨/ν, T̃µ) ∈ SYT(λ
∨/ν) that witnesses Cλ
µ− λ∨−Λ− ν∨: Growing the face using µ−ν∨ and ν∨−Λ we find that the edge λ∨−Λ
equals infusion2(Tν∨/µ, T̃ν), which is the already determined T̃λ. The newly determined
edge µ− λ∨ is Tλ∨/µ := infusion1(Tν∨/µ, T̃ν) ∈ SYT(λ
∨/µ) that witnesses Cλ
∅− ν− λ∨− µ: We grow this final face using ∅ − ν and ν − λ∨, but need to make two
consistency checks. First the edge ∅− µ is infusion1(Tν, Tλ∨/ν) which clearly is Tµ.
It remains to check that
(2) infusion2(Tν, Tλ∨/ν) = Tλ∨/µ.
(Notice that infusion2(Tν, Tλ∨/ν) is a filling of λ
∨/µ that rectifies to Tν, just as Tλ∨/µ does.
However it is not clear a priori that they are the same.)
To prove (2), we need a definition. For tableaux A and B of respective (skew) shapes α
and γ/α let A ⋆ B be their concatenation as a (nonstandard) tableau. If C is a tableau of
shape Λ/γ and α is a straight shape, A ⋆ B ⋆ C is a layered tableau of shape Λ.
Example 2.3. Let α = (2, 1) and γ = (4, 2, 1). We have
A = 1 2
, B = 2 3
, C =
1 4 5
, and A ⋆ B ⋆ C = 1 2 2 3
3 1 2 3
4 1 4 5
We have used boldface and underlining to distinguish the entries from A,B and C.
Let I1 and I2 be operators on layered tableaux defined by
I1 : A ⋆ B ⋆ C 7→ infusion1(A,B) ⋆ infusion2(A,B) ⋆ C
I2 : A ⋆ B ⋆ C 7→ A ⋆ infusion1(B,C) ⋆ infusion2(B,C).
The infusion involution says I21 and I
2 are the identity operator. In fact, the following
crucial “braid identity” holds, showing I1 and I2 generate a representation of S3:
Proposition 2.4. I1 ◦ I2 ◦ I1 = I2 ◦ I1 ◦ I2.
In view of the Proposition, (2) follows by setting α = λ, γ = ν∨, A = Tλ, B = Tλ∨/ν and
C = T̃ν, since the assertion merely says the “middle” tableau in
(3) I1 ◦ I2 ◦ I1 ◦ I2(Tλ ⋆ Tν∨/λ ⋆ T̃ν) and I2 ◦ I1(Tλ ⋆ Tν∨/λ ⋆ T̃ν)
are the same, whereas we even have equality of the two layered tableaux.
Proof of Proposition 2.4: By the Lemma it follows that
(4) C̃ ⋆ B ⋆ Ã := I1 ◦ I2 ◦ I1(A ⋆ B ⋆ C), and
(5) C̃ ⋆ B̂ ⋆ Ã := I2 ◦ I1 ◦ I2(A ⋆ B ⋆ C),
where the shapes of C̃, B and Ã are respectively the 180 degree rotations of the shapes of
A, B and C, and we know the fillings of C̃ and Ã in terms of evacuation. We thus also
know the shapes of B and B̂ are the same. It remains to show B equals B̂.
We first study (4). Let B ′ = infusion1(A,B), and A
′ = infusion2(A,B). Now
(6) I1 ◦ I2(B
⋆A ′ ⋆ C) = C̃ ⋆ B ⋆ Ã,
and the skew tableau B ⋆ Ã is obtained by treating B ′ ⋆ A ′ as a single standard tableau by
valuing an entry i of A ′ as |B ′| + i, evacuating, rotating the result 180 degrees and finally
sending the entry i from A ′ (respectively B ′) to |A ′| − i+ 1 (respectively |B ′| − i+ 1).
Example 2.5. Continuing the previous example,
B ′ ⋆A ′ = 1 2 3 2
and evac(B ′ ⋆A ′) = A ′′ ⋆ B ′′ = 1 3 3 4
7→ B ⋆ Ã = 3
1 2 1 3
Let us focus on evac(B ′⋆A ′) = A ′′⋆B ′′ and momentarily ignore the subsequent rotation
and complementation of entries. Fomin has shown in [St99, Appendix 2] that a fruitful
way to think of evac(X) is with triangular growth diagrams obtained by placing the shape
chain of X, then ∆(X), ∆2(X), . . . on top of one another (slanted left to right); see Figure 3.
We begin with the data B ′ ⋆ A ′ along the left side of the diagram. This is given as a
concatenation of two shape chains, one from ∅−µ and then one from µ−γ where µ is the
shape of B ′. Applying (F1) and (F2) from Section 1, we obtain A ′′ ⋆ B ′′ = evac(B ′ ⋆ A ′),
∅ ∅ ∅ ∅ ∅ ∅ ∅ ∅
B ′ evac(B ′)
A ′′A
FIGURE 3. A triangular growth diagram to compute evac(B ′ ⋆A ′)
given along the righthand side as a pair of chains connected at α. It follows from the
construction of growth diagrams that the thick lines represent evac(B ′) and A = evac(A ′′)
(the latter being a consequence of the Lemma).
These thick lines together with the vertex γ define a rectangular growth diagram, so:
(7) infusion(A,B ′′) = (evac(B ′), A ′).
In particular, the shape of rectification(B ′′) = µ. On the other hand, by definition
(8) infusion(A,B) = (B ′, A ′),
and B rectifies to a tableau of shape µ also.
By (7) and (8) combined with the aforementioned results of [Ha92] we see B and B ′′
are dual equivalent, as they both are obtained by an application of the same sequence
of slides (encoded by A ′) to a pair of tableaux of the same straight shape. Moreover, by
(7) B ′′ is jeu de taquin equivalent to evac(B ′) = evac(rectification(B)). Carrying out
the analogous “reverse” analysis on I2 ◦ I1 ◦ I2(A ⋆ B ⋆ C) we let C
◦ = infusion1(B,C)
and B◦ = infusion2(B,C) and study I2 ◦ I1(A ⋆ C
⋆ B◦) = C̃ ⋆ B̂ ⋆ Ã. Parallel to (6) we
consider the “reverse evacuation” B◦◦ ⋆ C◦◦ of C◦ ⋆ B◦ using reverse jeu de taquin slides.
Then we similarly conclude B◦◦ is dual equivalent to B (and thus B ′′). Also B◦◦ is jeu de
taquin equivalent to the reverse evacuation of revrectification(B). Thus by the lemma,
B ′′ and B◦◦ are jeu de taquin equivalent. Hence B ′′ = B◦◦ and so B = B̂, as required.
Example 2.6. Revisiting Example 2.3, C◦⋆B◦ = 2 3
4 5 3
1 1 2 4
with reverse evacuation 3 4
1 1 2
2 3 4 5
and B◦◦ (boldface in the latter tableau) is also B ′′ from Example 2.5. Rotation and comple-
mentation gives B̂, which is B from Example 2.5, as desired. �
3. AN EXTENDED EXAMPLE OF THE MAIN THEOREM
Let Λ = 3 × 4 = , λ = (2, 1) = , µ = (3, 1) = , ν = (3, 2) = .
Therefore λ∨ = (4, 3, 2), µ∨ = (4, 3, 1) and ν∨ = (4, 2, 1). Also
Tλ = 1 2
, Tµ = 1 2 3
, Tν = 1 2 3
, T̃λ =
, T̃µ =
1 2 4
, and T̃ν =
1 4 5
Here Cλ,µ,ν = 1, and we now give the unique carton filling. We begin with Tν∨/λ =
. Then we have the following sides of the carton, described as growth diagrams
with the obvious identifications of boundaries (the partitions correspond to shapes placed
on the vertices of the carton and the diagrams have been oriented to be consistent with
Figure 1):
(2, 1) = λ (2, 1, 1) (3, 1, 1) (4, 1, 1) (4, 2, 1) = ν∨
(2) (2, 1) (3, 1) (4, 1) (4, 2)
(1) (1, 1) (2, 1) (3, 1) (3, 2)
∅ (1) (2) (3) (3, 1) = µ
TABLE 2. ∅− µ− ν∨− λ
(4, 3, 1) = µ∨ (4, 3, 2) (4, 3, 3) (4, 4, 3) (4, 4, 4) = Λ
(4, 2, 1) (4, 2, 2) (4, 3, 2) (4, 4, 2) (4, 4, 3)
(4, 2) (4, 2, 1) (4, 3, 1) (4, 4, 1) (4, 4, 2)
(3, 2) (3, 2, 1) (3, 3, 1) (4, 3, 1) (4, 3, 2)
(2, 2) (2, 2, 1) (3, 2, 1) (4, 2, 1) (4, 2, 2)
(2, 1) = λ (2, 1, 1) (3, 1, 1) (4, 1, 1) (4, 2, 1) = ν∨
TABLE 3. λ− µ∨−Λ− ν∨
(4, 3, 1) = µ∨ (4, 2, 1) (4, 2) (3, 2) (2, 2) (2, 1) = λ
(4, 2, 1) (4, 1, 1) (4, 1) (3, 1) (2, 1) (2)
(3, 2, 1) (3, 1, 1) (3, 1) (2, 1) (1, 1) (1)
(3, 2) = ν (3, 1) (3) (2) (1) ∅
TABLE 4. ∅− ν− µ∨− λ
(4, 3, 1) = µ∨ (4, 3, 2) (4, 3, 3) (4, 4, 3) (4, 4, 4) = Λ
(4, 2, 1) (4, 2, 2) (4, 3, 2) (4, 4, 2) (4, 4, 3)
(3, 2, 1) (3, 2, 2) (3, 3, 2) (4, 3, 2) (4, 4, 2)
(3, 2) = ν (3, 2, 1) (3, 3, 1) (4, 3, 1) (4, 3, 2) = λ∨
TABLE 5. ν− λ∨−Λ− µ∨
ACKNOWLEDGMENTS
HT was partially supported by an NSERC Discovery Grant and AY by NSF grant
0601010. This work was partially completed while HT was a visitor at the Centre de
Recherches Mathématiques and the University of Minnesota, and while AY was at the
Fields Institute and the University of Michigan, Ann Arbor. We thank the Banff Interna-
tional Research Station for a common visit during the “Schubert calculus and Schubert
(4, 4, 4) = Λ (4, 4, 3) (4, 4, 2) (4, 3, 2) (4, 2, 2) (4, 2, 1) = ν∨
(4, 4, 3) (4, 4, 2) (4, 4, 1) (4, 3, 1) (4, 2, 1) (4, 2)
(4, 3, 3) (4, 3, 2) (4, 3, 1) (3, 3, 1) (3, 2, 1) (3, 2)
(4, 3, 2) = λ∨ (4, 2, 2) (4, 2, 1) (3, 2, 1) (3, 1, 1) (3, 1) = µ
TABLE 6. λ∨−Λ− ν∨− µ
(3, 2) = ν (3, 2, 1) (3, 3, 1) (4, 3, 1) (4, 3, 2) = λ∨
(3, 1) (3, 1, 1) (3, 2, 1) (4, 2, 1) (4, 2, 2)
(3) (3, 1) (3, 2) (4, 2) (4, 2, 1)
(2) (2, 1) (2, 2) (3, 2) (3, 2, 1)
(1) (1, 1) (2, 1) (3, 1) (3, 1, 1)
∅ (1) (2) (3) (3, 1) = µ
TABLE 7. ∅− ν− λ∨− µ
geometry” workshop organized by Jim Carrell and Frank Sottile. We also thank Sergey
Fomin, Allen Knutson, Ezra Miller, Vic Reiner, Luis Serrano, David Speyer and Dennis
Stanton for helpful discussions.
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DEPARTMENT OF MATHEMATICS AND STATISTICS, UNIVERSITY OF NEW BRUNSWICK, FREDERICTON,
NEW BRUNSWICK, E3B 5A3, CANADA
E-mail address: hugh@math.unb.ca
DEPARTMENT OF MATHEMATICS, UNIVERSITY OF MINNESOTA, MINNEAPOLIS, MN 55455, USA
E-mail address: ayong@math.umn.edu
	1. Introduction
	2. Tableau facts and proof of the Main Theorem
	2.1. Tableau sliding
	2.2. Proof of the rule
	3. An extended example of the main theorem
	Acknowledgments
	References

  The classical Littlewood-Richardson coefficients C(lambda,mu,nu) carry a
natural $S_3$ symmetry via permutation of the indices. Our "carton rule" for
computing these numbers transparently and uniformly explains these six
symmetries; previously formulated Littlewood-Richardson rules manifest at most
three of the six.

Introduction
	2. Tableau facts and proof of the Main Theorem
	2.1. Tableau sliding
	2.2. Proof of the rule
	3. An extended example of the main theorem
	Acknowledgments
	References
