RSK via local transformations [expository notes]

17 Pages·2017·0.316 MB·English

by  Sam Hopkins

RSK VIA LOCAL TRANSFORMATIONS
SAM HOPKINS
1. Introduction
The Robinson–Schensted–Knuth correspondence is a map from n×n N-
matrices (i.e., matrices with nonnegative integer entries) to pairs of semi-
standard Young tableaux of the same shape. For basic definitions of and
backgroundontableauxandtheRSKalgorithm,see[18,Ch. 7]. Abstractly,
we may regard RSK as a bijection
(cid:26) (cid:27)
{n×n N-matrices} −R−S−K→ (P,Q): P,Q are SSYT with sh(P) = sh(Q)
and whose entries are in [n] := {1,...,n}
RSK
A (cid:55)−−−→ (P,Q)
with the property that type(P) = col(A) and type(Q) = row(A). Here for
a SSYT T with entries in [n], the type of T is defined to be
type(T) := (type(T) ,...,type(T) ),
1 n
where
type(T) := the number of entries of T equal to i,
i
and for a n×n N-matrix B = (b ) the column and row sums of B are
ij
n n
(cid:88) (cid:88)
col(B) := (col(B) ,...,col(B) ) := ( b ,..., b )
1 n i1 in
i=1 i=1
n n
(cid:88) (cid:88)
row(B) := (row(B) ,...,row(B) ) := ( b ,..., b ).
1 n 1i ni
i=1 i=1
RSK is normally defined in terms of a somewhat complicated procedure
called row-insertion. This definition unfortunately masks the remarkable
symmetry property that if A (cid:55)−R−S−K→ (P,Q) then At (cid:55)−R−S−K→ (Q,P). Here
we present an alternate description of RSK via local transformations that
makes the symmetry obvious. Our approach is thus similar in spirit to the
growth diagrams of Fomin [6]. In order to motivate this description, we first
examinetheGelfand-TsetlinpatternsassociatedtotheYoungtableauxthat
result from RSK, and show how RSK can be thought of as a map between
matrices.
Date: July, 2014.
1
2 SAMHOPKINS
Definition1. AGelfand-Tsetlinpattern isatriangulararrayofnonnegative
integers
g g g ··· g
1,1 1,2 1,3 1,n
g g ··· g
2,2 2,3 2,n
g ··· g
3,3 3,n
. .
. .
. .
g
n,n
such that g ≥ g ≥ g for all 1 ≤ i ≤ j ≤ n−1. Gelfand-Tsetlin
i,j i+1,j+1 i,j+1
patterns with n rows are in bijection with semi-standard Young tableaux
with entries among [n] in a natural way, as explained below. (There seems
to be no consensus in the literature about how Gelfand-Tsetlin patterns are
indexed so we are merely choosing one convention here.)
RSK
Suppose A (cid:55)−−−→ (P,Q). We associate an n-row Gelfand-Tsetlin pattern
to P, denoted GT(P), as follows. Define the series P1,...,Pn of SSYT by
• P1 := P;
• Pi+1 is obtained from Pi by removing all entries of n+1−i.
Then GT(P) is just the Gelfand-Tsetlin pattern whose i-th row is sh(Pi).
WedefineGT(Q)similarly. Inotherwords,foraSSYTT withentriesin[n],
we define GT(T) to be the Gelfand-Tsetlin pattern with n rows defined by
g := number of entries ≤ n+1−i in row j −i+1 of T.
i,j
It is easy to check that the necessary inequalities on the entries hold. Note
also that because sh(P) = sh(Q), the first rows of GT(P) and GT(Q) are
the same. So it is possible to glue GT(P) and GT(Q) along their first
row to form a single n×n N-matrix that is weakly increasing in rows and
columns. Formally, suppose GT(P) = (g ) and GT(Q) = (h ); then the
ij ij
matrix A(cid:98)= ((cid:98)aij) is given by
(cid:26)
g if i ≥ j;
a := i−j+1,n+1−j
(cid:98)ij h otherwise.
j−i+1,n+1−i
This series of combinatorial objects is best understood through an example.
Example 2. Suppose
 
1 0 2
A = 0 2 0.
1 1 0
Then
(cid:32) (cid:33)
1 1 2 2 1 1 1 3
RSK
A (cid:55)−−−→ P = ,Q = ,
2 3 2 2
3 3
RSK VIA LOCAL TRANSFORMATIONS 3
and
4 2 1
GT(P) = 4 1
2
4 2 1
GT(Q) = 3 2
3
so
 
1 2 3
A(cid:98)= 1 2 3.
2 4 4
Indeed, A(cid:98) is weakly increasing along rows and columns. (cid:3)
We have built up a series of maps between combinatorial objects
A (cid:55)→ (P,Q) (cid:55)→ (GT(P),GT(Q)) (cid:55)→ A(cid:98).
By ignoring the intermediary steps, we can really think of RSK as a map
(cid:26) (cid:27)
{n×n N-matrices} −R−S−K→ n×n N-matrices that are weakly
increasing in rows and columns
RSK
A (cid:55)−−−→ A(cid:98).
Because each intermediary step was bijective, it is clear that this map is a
bijection. There are a few more properties of this map worth noting. First
of all, the diagonal sums of A(cid:98)record information about the row and column
sums of A.
Definition 3. For an n×n N-matrix B = (b ) and some i,j ∈ [n], define
ij
the diagonal sum at entry (i,j), denoted diag (i,j), to be
B
min(i,j)−1
(cid:88)
diag (i,j) := b .
B i−k,j−k
k=0
RSK
Using the above notation, when A (cid:55)−−−→ A(cid:98), we have for any i ∈ [n]
i
(cid:88)
diag (n,i) = col(A) ;
A(cid:98) k
k=1
i
(cid:88)
diag (i,n) = row(A) .
A(cid:98) k
k=1
To see why this is the case, observe that diag (n,i) is just the sum of the
A(cid:98)
entries in row (n + 1) − i of GT(P). But the sum of this row of GT(P)
is the number of 1’s, 2’s, ..., and i’s in P. But this is just (cid:80)i type(P) ,
k=1 k
and because type(P) = col(A), the diagonal sum is as claimed. Similarly
4 SAMHOPKINS
for Q and row sums. The final key property of our map is that if A (cid:55)→ A(cid:98)
then At (cid:55)→ A(cid:98)t, which follows from the above-mentioned symmetry property
of the map to Young tableaux.
To summarize, RSK is secretly a map A (cid:55)−R−S−K→ A(cid:98) from n×n N-matrices
to n×n N-matrices that are weakly increasing in rows and columns such
that
(a) the map is bijective;
(b) diag (n,i) = (cid:80)i col(A) and diag (i,n) = (cid:80)i row(A) for i ∈ [n];
A(cid:98) k=0 k A(cid:98) k=0 k
(c) At (cid:55)−R−S−K→ A(cid:98)t.
Our aim is to give an explicit construction of such a map where these essen-
tial properties are immediate from the construction.
Remark 4. These notes are not intended for publication. They record
some ideas that were explained by Alex Postnikov during the MIT combina-
torics preseminar. The ideas presented here are closely related to a certain
birational lifting of RSK originally studied by Kirillov [10] which takes ma-
trices with entries in R to other matrices of this form (or more generally,
>0
to certain three-dimensional arrays with entries in R ). Birational RSK
>0
has recently seen significant interest; see, e.g., [13], [2], [4], [1], [15], [14].
The work of Danilov and Koshevoy [2] [3] in particular is closely related
to the presentation of RSK given here because they explicitly observe the
connection between RSK and the octahedron recurrence. Although every-
thing written here naturally generalizes to the birational setting we choose
to work in the combinatorial max-plus semiring setting because we are just
trying to give an alternative description of classical RSK. In particular, our
aim is to show how many classical combinatorial results (for instance, the
“hook-length formula”) follow easily from this treatment of RSK, and at a
greaterlevelofgeneralityaswell. ThankstoDarijGrinbergforproofreading
and helpful comments.
Remark 5. In May of 2017, Igor Pak pointed out his paper [16] to me, in
which he develops essentially the same piecewise-linear “toggle” description
of RSK that we will explain in the next section. The toggles can be seen
specifically in §4 of [16]. That paper also cites an earlier paper of Berenstein
and Kirillov [9], who independently discovered a similar toggle map.
2. The RSK algorithm via toggles
Infact,ourmapwillworkinaslightlymoregeneralsettingthanmatrices.
Definition 6. An N-tableau of shape λ is an assignment of nonnegative
integers to the boxes of some partition λ. A (weak) reverse plane partition
of shape λ is an N-tableau of shape λ that is weakly increasing in rows and
columns. We will always use matrix coordinates for the boxes of partitions
(which, as is already evident, we write in English notation). This means
RSK VIA LOCAL TRANSFORMATIONS 5
that the upper-left box, which belongs to every partition except the empty
partition, is (1,1).
The map we are about to describe takes N-tableaux of shape λ to (weak)
RSK
reverseplanepartitionsofshapeλ. WewillwriteT (cid:55)−−−→ T(cid:98)withcalligraphic
scripttodifferentiatethisprocedurefromclassicalRSK.Thesereverseplane
partitions are termed weak because we allow 0 as the value of a box, but
we will drop this adjective from now on. Items (a) and (c) in the above list
of desiderata are clear enough in this context, but we need to explain what
exactly (b) should mean. We need a little terminology for this purpose
Definition 7. For any partition λ, a border box of λ is a box (i,j) such
that (i+1,j +1) is not a box. A corner box of λ is a box (i,j) such that
both(i+1,j)and(i,j+1)arenotboxes. Whenwespeakofborderorcorner
boxes of matrices, we mean when they are considered as square partitions.
Given an N-tableau T = (t ) of shape λ, we define the row and column
kl
sums exactly as above for matrices. For some box (i,j) of λ, we define the
diagonal sum at (i,j) as above for matrices. Finally, we define the rectangle
sum at (i,j) to be
i j
(cid:88)(cid:88)
rect (i,j) := t .
T kl
k=1 l=1
Note that if (i,j) is a border box (which is the case that will concern us
most often) then
j n−i
(cid:88) (cid:88)
rect (i,j) = col(T) − row(T) .
T k (n+1)−k
k=1 k=1
For any matrix A the sum of all the row sums equals the sum of all the
column sums. So we can think of (b) as saying that the diagonal sum of A(cid:98)
at some border box (i,j) of the matrix can be obtained by traveling through
border boxes from (n,0) to (i,j), where each time you step rightwards you
add the corresponding column sum, and each time you step upwards you
subtract the corresponding row sum. (We start at (n,0), which is not ac-
tually a box of A, because the first step rightward to (n,1) corresponds to
adding the first column sum.) That is, statement (b) is equivalent to
j n−i
(cid:88) (cid:88)
diag (i,j) = col(A) − row(A) = rect (i,j)
A(cid:98) k (n+1)−k A
k=1 k=1
for every i,j ∈ [n] with at least one of i or j equal to n. A good sanity
check is to verify that A(cid:98) has the claimed diagonal sums in Example 2.
The translation of (b) for N-tableaux T is thus: for any border box (i,j)
of λ := sh(T) we need diag (i,j) = rect (i,j).
T(cid:98) T
WearepreparedtogiveouraccountofRSK. Theideaistoworkbox-by-
box. Given an N-tableau T(cid:48) we will construct T(cid:98)(cid:48) with T(cid:48) (cid:55)−R−S−→K T(cid:98)(cid:48). If T(cid:48) = ∅
6 SAMHOPKINS
... ...
... β2α2
γ β α
2 1 1
γ ∗
1
Figure 1. The N-tableau T(cid:98), where position (i,j) has been
marked by a star.
then T(cid:98)(cid:48) := ∅. So say T(cid:48) is obtained from T by adding a single corner
RSK
box (i,j) whose entry is x. And say T (cid:55)−−−→ T(cid:98). Then from T(cid:98) we obtain T(cid:98)(cid:48)
as follows. Let βk be the entry of T(cid:98) at position (i−k,j −k), let γk be the
entry at position (i−k+1,j−k), and let α be the entry at (i−k,j−k+1)
k
for all k ≥ 1. If any of these boxes do not exist in T(cid:98) we regard these
values as 0; so eventually they are all 0 for the nonpositive positions. These
entries occupy three consecutive diagonals in T(cid:98), as Figure 1 illustrates. Our
procedure only modifies the values of the boxes that lie along the diagonal
wherethenewboxistobeinserted(whichwerepreviouslytheβ ). Observe
k
that because T(cid:98) is weakly increasing in rows and columns, we have for all k
the inequality
max(α ,γ ) ≤ β ≤ min(α ,γ ).
k+1 k+1 k k k
The trick is to “toggle” βk between these two extrema. That is, we set in T(cid:98)(cid:48)
the value of the box previously occupied by β to be
k
β(cid:48) := max(α ,γ )+min(α ,γ )−β ,
k k+1 k+1 k k k
for all 1 ≤ k < min(i,j). Finally note that the value of box (i,j) must be
greater than or equal to max(α ,γ ); so we set it equal to max(α ,γ )+x.
1 1 1 1
Example 8. Before we prove the correctness of this algorithm, we give a
few simple examples of how it behaves. First of all, for the simplest case of
a line RSK merely records partial sums:
RSK
a a a ... (cid:55)−−−→ a a +a a +a +a ...
1 2 3 1 1 2 1 2 3
Similarly for a hook:
a b b ... a a+b a+b +b ...
1 2 1 1 2
c a+c
1 RSK 1
(cid:55)−−−→
c a+c +c
2 1 2
. .
. .
. .
RSK VIA LOCAL TRANSFORMATIONS 7
Let us consider the first non-hook shape, a 2×2 square. We already know
that
a b RSK a a+b
(cid:55)−−−→
c a+c
So say we add a box in (2,2) with entry d. We then must “toggle” the value
of (1,1); its previous value was a and it is subject to the inequality
0 ≤ a ≤ min(a+b,a+c),
so we set it equal to min(a+b,a+c)−a = min(b,c). Finally, we set the
value of (2,2) to be max(a+b,a+c)+d = a+max(b,c)+d. Thus
a b RSK min(b,c) a+b
(cid:55)−−−→
c d a+c a+max(b,c)+d
It is easily verified that the diagonal sums are correct. (cid:3)
Proposition 9. The RSK map defined above is well-defined; that is, the
order in which we decompose an N-tableaux T does not matter.
Proof: Suppose sh(T) = λ and let P be the poset on the boxes of λ
λ
whereby (i,j) ≤ (k,l) iff i ≤ k and j ≤ l. So the upper-left box is minimal
in P . The RSK algorithm described above proceeds by adding boxes one
λ
at a time, starting from the empty tableau, according to any order that
is a linear extension of P . We need to show that all choices of linear
λ
extensions of P give the same result. As any linear extension of a finite
λ
poset can be obtained from any other via a series of adjacent transpositions
of incomparable elements, it suffices to show that if T(cid:48)(cid:48) is obtained from T(cid:48)
by adding entries in positions (i,j) and (i(cid:48),j(cid:48)), both of which are corner
boxes of T(cid:48)(cid:48), then the order in which we add these two entries does not
affect how RSK acts on T(cid:48)(cid:48). But note that because (i,j) and (i(cid:48),j(cid:48)) are
both corner boxes, the diagonals on which (i,j) and (i(cid:48),j(cid:48)) lie cannot be
adjacent. Therefore, the values of the diagonals adjacent to the diagonal
on which (i(cid:48),j(cid:48)) lies are not modified when we add some entry at (i,j)
and carry out the local transformations, and vice-versa. But these local
transformations only depend on adjacent diagonals. So indeed the order in
which we add the two entries does not matter. (cid:3)
Proposition 10. RSK is a bijection between N-tableaux of shape λ and
reverse plane partitions of shape λ.
Proof: We can explicitly give the inverse map T(cid:98)(cid:48) (cid:55)−R−−SK−−−→1 T(cid:48). The crucial
fact is that the “toggle” steps are reversible. If T(cid:98)(cid:48) is empty, then so is T(cid:48).
So consider any corner box (i,j) of T(cid:98)(cid:48). Let x(cid:98) be the value of T(cid:98)(cid:48) in (i,j);
then define x to be the x(cid:98) minus the maximum of the values of T(cid:98)(cid:48) in the
two boxes (i − 1,j) and (i,j − 1) (where these are 0 if there are no such
boxes). Let T(cid:98) be obtained from T(cid:98)(cid:48) by removing (i,j) and re-toggling along
its diagonal. Suppose T(cid:98) (cid:55)−R−−SK−−−→1 T. Then define T(cid:48) to be the tableau
8 SAMHOPKINS
obtained from T by adding the entry x to position (i,j). This procedure
is easily seen to locally reverse the RSK procedure, so it indeed gives the
inverse. (cid:3)
RSK
Proposition 11. If T (cid:55)−−−→ T(cid:98), then for any border box (a,b) of sh(T) we
have diag (a,b) = rect (a,b).
T(cid:98) T
Proof: Again we consider the local transformations. Suppose T(cid:48) (cid:55)−R−S−→K T(cid:98)(cid:48),
with T(cid:48) obtained from T by adding an entry of x in a corner box (i,j)
RSK
and T (cid:55)−−−→ T(cid:98). We may assume inductively that the result holds for T
and T(cid:98). First observe that the right-hand side of this equality only changes
for (i,j) because for any other border box (i(cid:48),j(cid:48)) of sh(T(cid:48)), either j(cid:48) ≥ j
and i(cid:48) < i and the contribution of x cancels between col(T) and −row(T) ,
j i
or j(cid:48) < j and i(cid:48) ≥ i and there is no contribution of x at all. But observe
further that the left-hand side of the equality also only changes for (i,j),
because this is the only diagonal whose values we have modified. Thus we
need only check that the equality holds at (i,j). Let α ,β ,γ be defined
k k k
as in Figure 1. Then
(cid:88)
diag (i,j) = x+ (max(α ,γ )+min(α ,γ )−β )
T(cid:99)(cid:48) k k k k k
k≥1
(cid:88)
= x+ (α +γ −β )
k k k
k≥1
= x+diag (i,j −1)+diag (i−1,j)−diag (i−1,j −1).
T(cid:98) T(cid:98) T(cid:98)
But observe that
x+diag (i−1,j)−diag (i−1,j −1) = x+col(T) = col(T(cid:48)) ,
T(cid:98) T(cid:98) j j
by our induction hypothesis, so
diag (i,j) = diag (i,j −1)+col(T(cid:48)) ,
T(cid:99)(cid:48) T(cid:98) j
and by induction indeed this diagonal sum is as claimed. (cid:3)
Proposition 12. If T (cid:55)−R−S−→K T(cid:98), then Tt (cid:55)−R−S−→K T(cid:98)t.
Proof: The local transformations evidently commute with transposition, so
the whole map does as well. (cid:3)
3. An application: hook-length formulas
We have succeeded in describing via simple local transformations a map
with the desired properties (a)–(c). In fact, one further nice property comes
along for free. Recall that for a partition λ, the hook of λ at (i,j), which we
denote H (i,j), is the set of all boxes weakly below or to the right of (i,j),
λ
and the hook-length is given by h (i,j) := |H (i,j)|. (Note that if (i,j) is
λ λ
not a box of λ, then H (i,j) = ∅ and h (i,j) = 0.) And recall that the
λ λ
weight of a reverse plane partition T, denoted |T|, is the sum of all of its
entries.
RSK VIA LOCAL TRANSFORMATIONS 9
RSK
Proposition 13. Suppose T (cid:55)−−−→ T(cid:98) with T = (tij) and sh(T) = λ. Then
(cid:88)
|T(cid:98)| = tijhsh(T)(i,j).
(i,j)∈λ
Proof: As always, we work inductively with the same setup involving the
playersT, T(cid:48), T(cid:98), T(cid:98)(cid:48), xand(i,j). WhenmovingfromT toT(cid:48), theright-hand
sideoftheequalityincreasesbyrow(T) +col(T) +x. Fortheleft-handside
i j
of the equality, let α ,β ,γ be defined as in Figure 1. Let β(cid:48) be defined
k k k k
as in the description of RSK above. Then the left-hand side of the equality
increases by x+max(α ,γ )+(cid:80) (β(cid:48) −β ), which is equal to
1 1 k≥1 k k
x+diag (i,j −1)+diag (i−1,j)−2·diag (i−1,j −1).
T(cid:98) T(cid:98) T(cid:98)
But by Proposition 11,
diag (i,j−1)+diag (i−1,j)−2·diag (i−1,j−1) = row(T) +col(T) .
T(cid:98) T(cid:98) T(cid:98) i j
So by induction the two sides of the equality agree. (cid:3)
One easily obtains as a corollary of this last proposition the generating
function for reverse plane partitions in terms of hook-lengths (see [8] or [18,
§7.22]):
Corollary 14. Let λ be a partition. Then
(cid:88) (cid:89) 1
q|T| = ,
(1−qhλ(i,j))
T (i,j)∈λ
where the sum is over all reverse plane partitions T of shape λ.
Proof: The coefficient of qn on the right-hand side is the number of ways to
choose t ∈ N such that n = (cid:80) t h (i,j). But by the previous propo-
ij (i,j)∈λ ij λ
sition, the RSK bijection takes such assignments of nonnegative integers to
the boxes of λ to reverse plane partitions T of shape λ whose entries sum
to n. The number of such reverse plane partitions is clearly what is counted
by the coefficient of qn on the left-hand side. (cid:3)
Actually a careful analysis of our procedure yields a more refined version
of this result. Fix some list x = (...,x ,x ,x ,...) of commuting indeter-
−1 0 1
minatesindexedbyZ; wethinkofthex beingnonnegativerealsthatassign
i
weights to the diagonals of our partition, where x corresponds to the main
0
diagonal, x the diagonal above the main diagonal, x the diagonal below
1 −1
it, and so on. We define the x-hook-length of λ at a box (i,j) to be
(cid:88)
h (i,j;x) := x .
λ j(cid:48)−i(cid:48)
(i(cid:48),j(cid:48))∈H (i,j)
λ
Similarly we define the x-weight of a reverse plane partition T = (t ) to be
ij
(cid:88)
|T| := t x .
x ij j−i
(i,j)∈λ
10 SAMHOPKINS
Observe that taking x = (...,1,1,1,...) recovers the normal hook-lengths
and weights. Recall that the content of a box (i,j) ∈ λ of a partition
is defined to be j − i, so these can be thought of as “content-weighted”
analogues of the classical notions. If f = (cid:80) c x where the c ∈ N and all
i i i i
but finitely many are zero, let us use the notation zf := (cid:81) zci, where the z
i i i
are some new set of commuting variables also indexed by Z. Exactly the
same proofs as above give us the following.
RSK
Proposition 15. Suppose T (cid:55)−−−→ T(cid:98) with T = (tij) and sh(T) = λ. Then
(cid:88)
|T(cid:98)|x = tijhsh(T)(i,j;x).
(i,j)∈λ
Corollary 16. Let λ be a partition. Then
(cid:88) (cid:89) 1
z|T|x = ,
(1−zhλ(i,j;x))
T (i,j)∈λ
where the sum is over all reverse plane partitions T of shape λ.
The classical generating function for reverse plane partitions in terms
of hook-lengths implies the so-called “hook-length formula” for standard
Young tableaux by taking the limit q → 1 and applying some results from
the theory of P-partitions (see [18, §7.22]). Our content-weighted version
similarly implies a content-weighted hook-length formula.
Let T be a standard Young tableau with sh(T) (cid:96) n. For any k ∈ [n], let
us use (i (k),j (k)) to denote the box whose entry is k. Then define
T T
n
(cid:89) 1
T := .
x (cid:80)k x
k=1 l=1 jT(n+1−l)−iT(n+1−l)
Example 17. If T is the SYT given by
1 3 4
T = 2 5
then we have
(cid:16) (cid:17)−1
T = x (x +x )(x +x +x )(x +x +x +x )(x +2x +x +x ) .
x 0 0 2 0 1 2 −1 0 1 2 −1 0 1 2
Theorem 18. Let λ (cid:96) n be a partition. Then
(cid:88) (cid:89) 1
T = ,
x
h (i,j;x)
λ
T (i,j)∈λ
where the sum is over all standard Young tableaux T of shape λ.
Proof: Let Pop be the order dual to the poset defined in the proof of
λ
Proposition 9; that is, Pop is the poset on the boxes of λ with (i,j) ≤ (k,l)
λ
iff i ≥ k and j ≥ l. So the upper-left box is maximal in Pop. We recall
λ
the main generating function for P-partitions [19, Theorem 3.15.5]: for any