4.3 Description of the map ϕ
5.1.2 The strategy and the proof of Theorem 5.1.1
In this subsection, we will frequently use Definitions 4.2.2, and 5.1.6. The reader is urged to recall these definitions.
Letnbe a positive integer. LetU be a finite monomial inN(v)of cardinalityn. Then U can be regarded as apositive multiset onN2 of degreen . Now, we know from the note of §3.2.7 thatBRSK(U) =ι(BRSK(ι(U))) (where ι has stated in Definition 3.1.2). Let (P(n), Q(n))denoteBRSK(ι(U)). It then follows from the definition of ιthat for proving Theorem 5.1.1, it suffices to prove the following:
Theorem 5.1.7. With notation as in the statement of Theorem 5.1.1, the following two statements hold true for each r∈ {0,1,· · ·, m}:
(i) the column numbers of the elements of Swr consist of the (r+ 1)-th row entries of P(n).
(ii) The row numbers of the elements of Swr consist of the(r+ 1)-th row entries of Q(n).
Observe now (with the notation used in Theorem 5.1.1) that for eachr∈ {0,1,· · ·, m}, we have π(U(r)) = (wr, U(r+1)), and Swr is the distinguished monomial in N(v) corre- sponding to wr. It now follows from the description of the mapπ (on any finite monomial in N(v), as given in §4.2) that it suffices to prove the following:
Theorem 5.1.8. With notation as in the statement of Theorem 5.1.1, the following two statements hold true for each r∈ {0,1,· · ·, m}:
(i) the least column numbers of all blocks of U(r) consist of the (r+ 1)-th row entries of P(n).
(ii) The largest row numbers of all blocks of U(r) consist of the (r+ 1)-th row entries of Q(n).
We will prove Theorem 5.1.8 by induction on the cardinality n of U.
The induction hypothesis: Theorem 5.1.8 is true for all finite monomials U inN(v) having cardinality ≤n−1.
If n = 1, the theorem is obvious. So we assume that n > 1. Let ι be the involution map which was described in Definition 3.1.2. LetU be a monomial in N(v)of cardinality n. Arrange ι(U) in lexicographic order (which is stated in Definition 3.2.7), say, ι(U) = {(a1, b1),(a2, b2),· · ·,(an, bn)}. Let F = {(b1, a1),(b2, a2),· · ·,(bn−1, an−1)}. Then bn ≤ bn−1 and if bn =bn−1, we have an−1 ≥an (by the definition of lexicographic order). The element (bn, an) enters into F to make it U. Let BRSK(ι(F)) = (P(n−1), Q(n−1)). Then we have:
(P(n), Q(n)) =BRSK(ι(U)) = (P(n−1), Q(n−1))←bn an.
To explain the strategy and the proof further, first we need a definition (see Definition 5.1.9 below) and a Lemma (see Lemma 5.1.11 below).
Definition 5.1.9. Let{(R1, C1),(R2, C2),· · ·,(Rp, Cp)}be a blockBof some finite mono- mial S of N(v). Let bn ≤ R1 be such that bn ∈ [N]\v and bn > C1. Let an ∈v be such thatan ≤C1. Then we say that {(bn, an),(R1, C1),· · ·,(Rp, Cp)}is aleft concatenation of B by (bn, an).
Example 5.1.10 below illustrates the above definition of left concatenation of a block.
Example 5.1.10. Let d = 9, N = 17, and v = (1,2,4,6,7,11,12,13,15). Let S be the finite monomial in N(v) given by
S={(8,2),(8,2),(8,6),(8,6),(8,7),(9,2),(10,1),(14,13),(16,12),(16,13),(17,11)}.
Figure 5.1 shows the monomial Sand its block decomposition. The dark circles denote the elements of the monomial S, and the numbers written near these dark circles denote the
5.1. Relation between the maps BRSK and eπ for a monomial in N(v) 39
multiplicities with which these elements occur in the monomialS. The dark line segments (in case of blocks consisting of more than one distinct elements) and the dark circles (in case of blocks consisting of a single element possibly appearing more than once) are the blocks ofS.
p p p p p p p
p p p p p p p
p p p p p p p
p p p p
p p p p
p p
p p
•
•
• • •
•
• •
• 17
16 14 10 9 8 5
3 1 2
4
6 7
11 12 13 15 1 1
2 2 1
1 1 1 1
Figure 5.1: The monomial Sand its block decomposition
Consider the blockB of Sgiven byB={(8,6),(8,6),(8,7)}. Thenp= 3, (R1, C1) = (R2, C2) = (8,6), and (R3, C3) = (8,7). Let (bn, an) = (8,4). Then bn ≤R1, bn ∈[N]\v, and bn > C1. Also, an ∈ v and an ≤ C1. Observe that {(8,4),(8,6),(8,6),(8,7)} is a left concatenation of the block B by (bn, an) = (8,4) (see Figure 5.2 for an illustration).
The dark circles in Figure 5.2 denote the elements of the monomial S∪ {(8,4)}, and the numbers written near these dark circles denote the multiplicities with which these elements occur in the monomialS∪ {(8,4)}. The dark line segments (in case of blocks consisting of more than one distinct elements) and the dark circles (in case of blocks consisting of a single element possibly appearing more than once) are the blocks ofS∪ {(8,4)}.
p p p p p p p
p p p p p p p
p p p p p p p
p p p p
p p p p
p p
p p
•
•
• • •
•
• •
•
•
17 16 14 10 9 8 5
3 1 2
4
6 7
11 12 13 15 1
1 1
2 2 1
1 1 1 1
Figure 5.2: Left concatenation of a block
Lemma 5.1.11. Let U = {(b1, a1),(b2, a2),· · ·,(bn, an)} be a finite monomial in N(v) such that ι(U) is in lexicographic order (which is stated in Definition 3.2.7). Let F = {(b1, a1),(b2, a2),· · ·,(bn−1, an−1)}. The element(bn, an)enters intoF to make itU. Then either the singleton set {(bn, an)}is itself a block in U or the element (bn, an) left concate- nates a block of F.
Proof. Suppose the singleton set {(bn, an)} is not a block in U and the element (bn, an) does not left concatenate any block of F as well.
Then there exists a blockB ofF such that in the process of forming the monomial U from the monomialF, the element(bn, an)gets added to the blockBof F, but not at the leftmost end. Observe now thatbn is the least possible row number among all elements of U. All these facts put together imply that there exists an element (bn, p)∈B such that p < an. This in turn implies that there exist(s) element(s) in F having row number bn. So, we must have bn−1 =bn. But then since (bn, p)∈ F, it follows from the definition of lexicographic order that an≤p. This contradicts the fact that p < an.
Recall that our goal is to prove Theorem 5.1.8 above. Let F and U be as in the statement of Lemma 5.1.11 above. Then we can easily see (from the statement of Lemma 5.1.11) that we can divide the proof of Theorem 5.1.8 into 2 cases:
Case I: when the singleton set {(bn, an)} is itself a block in U. Case II: when the element (bn, an) left concatenates a block of F.
Case I
In this case, we can easily see that all the blocks ofU other than the block{(bn, an)} are same as all the blocks of F. In fact, the block {(bn, an)} is the topmost block of U of some depth (say, k). This is simply because bn is the least possible row number among all elements of U.
LetU(0) =U and π(U(r)) = (wr, U(r+1))for allr ∈ {0,1,· · ·, m} (where the integerm is as given in the statement of Theorem 5.1.1). Similarly, it also holds true for F. Then it is easy to see that for each r ∈ {0,1,· · ·, m}, all the blocks of U(r+1) are going to be the same as all the blocks of F(r+1).
Clearly, in this case,(bn, an)becomes an element ofSw0. Hence we must have that an is an entry in the (0 + 1)-th row of P(n) and bn is an entry in the (0 + 1)-th row of Q(n). All other entries of(P(n), Q(n))remain the same as in (P(n−1), Q(n−1))(in the same rows).
So, to prove Theorem 5.1.8, all that we need to show is the following:
• during the process P(n−1) ←bn an of bounded insertion, the elementan gets placed in the first row of P(n−1)<bn and an bumps nothing.