Plane Partitions

PLANE PARTITIONS

AMY BECKER, LILLY BETKE-BRUNSWICK, ANNA ZINK

Abstract. Throughout our study of the enumeration of plane partitions we make use of bijective proofs to ﬁnd generating functions. In particular, we consider bounded plane partitions, symmetric plane partitions and weak reverse plane partitions. Using the combinatorial interpretations of Schur functions in relation to semistandard Young tableaux, we rely on the properties of symmetric functions. In our paper we will walk through some of these bijections and present the corresponding generating functions.

1. Introduction In order to understand and interpret plane partitions, it is important to have some background knowledge of various concepts. We will be reviewing the ba- sic ideas behind partitions, semistandard Young tableaux (SSYT), and symmetric functions– speciﬁcally Schur functions– to develop a strong base for the work we will be doing with plane partitions. We begin with the partition, the more com- monly seen 2-D version of a plane partition.

Deﬁnition 1.1. For any nonnegative integer n, a partition λ of n is a nonin- ∞ P∞ creasing sequence {λj}j=1 of nonnegative integers such that j=1 λj = n. If λ is a partition of n then we often write |λ| = n.

We represent these partitions visually with what we call Ferrers diagrams. There are various styles of Ferrers diagrams, but we will be using the English version, which is left-aligned and stacked from top to bottom.

λ1 λ2 λ3 Figure 1: Ferrers diagrams for a few partitions of n=5: λ1 = 4, 1, λ2 = 3, 2, and λ3 = 2, 2, 1.

Now, given a partition λ, we can fill it in with entries of numbers or even variables. A semistandard Young tableau is one type of filling of the Ferrers diagram. Definition 1.2. For any partition λ, a semistandard Young tableau (or SSYT) of shape λ is a filling of the Ferrers diagram of λ so that columns are strictly increasing from top to bottom and rows are weakly increasing from left to right. 1 2 AMY BECKER, LILLY BETKE-BRUNSWICK, ANNA ZINK

1 2 2 3 1 2 3 4 2 4 3 4 4 4

Figure 2: Two ﬁllings of the Ferrers diagram of λ = 4, 2, 1 with entries {1, 2, 3, 4}.

Given the Ferrers diagram of partition λ, we can see from our above example that there are various ways to fill it to create semistandard Young tableaux given a set of entries. We will let SSYT(λ) denote the set of all such fillings of λ with entries {1, 2, ..., n}. Within SSYT(λ), we can think of each filling or SSYT as corresponding to a term in a polynomial. In order to create the term, take a SSYT and let each entry i represent the variable xi. Then, multiplying these variables together we let the product be the resulting term.

1 2 x1 x2 2 3 x2 x3 3 x 2 2 =⇒ 3 =⇒ x1x2x3 Figure 3: How to convert a SSYT into a term in a polynomial.

The resulting polynomial is the Schur polynomial, sλ(x1, x2, ..., xn), where λ is the partition we are ﬁlling in with entries from {x1, x2, ..., xn}. Schur polynomials are in fact symmetric functions. Recall:

Deﬁnition 1.3. A function f(x1, x2, ..., xn) is symmetric whenever it does not change under any permutation of its variables.

The Schur function sλ(x1, x2, ..., xn) is a generating function for SSYT(λ). If we look at a term in the Schur function, the coefficient represents the number of ways to fill the partition λ to create a SSYT with the variables in the term. For example if a term in s2,1(x1, x2, ..., xn) is 2x1x2x3 then there are 2 ways to fill in the partition λ = 2, 1 with x1, x2, and x3.

We have seen that one way to fill in a Ferrers diagram is with entries that are weakly increasing across rows and strictly increasing down colums. If instead we fill the Ferrers diagram so that the rows are weakly decreasing and the columns strictly decreasing this is referred to as a reverse semistandard Young tableau. A third way to fill in the Ferrers diagram is with weakly decreasing rows and columns. Such a filling is a plane partition. We can think of the numbers in the filling as repre- senting the heights for stacks of blocks placed on each cell of the diagram. PLANE PARTITIONS 3

4 4 3 1 4 3 2 1 3 1 2 1 ⇒

Figure 4: Filling of a Ferrers diagram associated with a plane partition, π

More formally:

Deﬁnition 1.4. A plane partition is an array π = (πij)i,j≥1 of nonnegative integers such that π has ﬁnitely many nonzero entries and is weakly decreasing in rows and columns.

Note that although a plane partition is an inﬁnite array with ﬁnitely many nonzero entries, when writing it we don’t include the zero parts– they are implied.

When discussing plane partitions, there are a few important properties that will come up. Deﬁnition 1.5. The size of a plane partition, |π|, is the sum of the entries. X |π| = πij. i,j≥1 Informally, we can think of |π| as the total number of blocks in the three- dimensional interpretation of the plane partition. For example, the size of the plane partition in Figure 4 is |π| = 29.

Deﬁnition 1.6. The shape of a plane partition, sh(π), is the partition whose Ferrers diagram is ﬁlled.

Deﬁnition 1.7. The max of a plane partition, max(π), is the largest part. We can think of max(π) as the height of the tallest stack. This is always the entry of the northwestern-most cell of the diagram.

The SSYT and plane partitions represent two different fillings of the Ferrers diagrams. Since sλ is the generating function for SSYT(λ), it would be convenient to relate SSYT to plane partitions in order to make use of Schur functions in our analysis of plane partititions. We will create a correspondence by once more thinking of SSYT as being filled with variables whose subscripts obey the column- strict and row-weak relations. We then let each variable represent a number. In order for this correspondence to result in a plane partition, we let subscript size be inversely related to numeric entry. That is, the variable with smallest subscript equals the largest number, and the variable with the largest subscript equals the smallest number, so that the order is preserved– in reverse.

In our example below we set xi = 10−i. Notice that the resulting plane partition will be column-strict since the SSYT is column-strict. 4 AMY BECKER, LILLY BETKE-BRUNSWICK, ANNA ZINK

x1 x1 x2 x3 x5 9 9 8 7 5 x3 x4 x4 x6 7 6 6 4 x4 x6 6 4 x5 x8 5 2 x7 ⇒ 3

2. Plane Partitions Bounded in a Box Now that we have defined plane partitions, it is natural to attempt to enumerate them. Initially, it is convenient to restrict our consideration to a bounded size. We define B(r, c, t) to be the set of all plane partitions with at most r rows and c columns and largest part (height) at most t. We can think of this as restricting our 3D plane partition to an r × c × t box. The following lemma provides a generating function for plane partitions confined to these bounds.

Lemma 2.1. r+1 − c X |π| ( 2 ) t+r 3 2 q = q shcr i(q , ..., q , q , q). π⊆B(r,c,t)

(Note: hcri is the partition (c, c, c, ..., c) of cr. The Ferrers diagram of hcri is an r × c rectangular grid.)

Proof. We establish the following bijection between bounded, column-strict plane partitions with a rectangular shape and ordinary bounded plane partitions. Let µ be a column-strict plane partition of shape hcri with max(µ) ≤ t + r where t, c and r are positive integers. Since each column of µ contains a decreasing list of positive integers, 1 ≤ µr,j < µr−1,j < ... < µ1,j ≤ t + r. So for all i, j, µi,j > r − i.

Let π be the plane partition deﬁned by πi,j = µi,j − (r + 1 − i). When πi,j = 0 we remove that cell from our Ferrers diagram. Consider the following example.

− =

µ ∈ B(r, c, t + r) π ∈ B(r, c, t)

9 9 8 7 5 4 4 4 4 4 5 5 4 3 1 7 6 6 4 4 3 3 3 3 3 4 3 3 1 1 6 4 3 2 2 2 2 2 2 2 4 2 1 5 3 1 1 1 1 1 1 1 1 4 2 PLANE PARTITIONS 5

Note that the resulting π is indeed a plane partition. Because we have not changed the relative values across any given row, πi,j − πi,j+1 = µi,j − µi,j+1 ≥ 0, so π too has weakly decreasing rows. Additionally, πi,j −πi+1,j = µi,j −µi+1,j −1 ≥ 0 since µ is column-strict. As such, π is weakly decreasing down columns.

Because we have only altered the heights of stacks, sh(π) ⊆ sh(µ). Furthermore, since µi,j ≤ t + r + 1 − i, then πi,j ≤ t for all i, j. Therefore π is in B(r, c, t)

The inverse of this correspondence is immediate. We can start with any ordinary 0 0 0 0 plane parition, π , in B(r, c, t) and construct µ such that µi,j = πi,j + (r + 1 − i) for all i and j with 1 ≤ i ≤ r and 1 ≤ j ≤ c. Now r c r c X X X X |π| = πi,j = (µi,j − (r + 1 − i)) i=1 j=1 i=1 j=1 = |µ| − rc − (r − 1)c − ... − 2c − c = |µ| − c(1 + 2 + ... + r) r + 1 = |µ| − c . 2

Therefore, by substitution we achieve the following:

X |π| X |µ|−c r+1 q = q ( 2 ) π⊆B(r,c,t) µ⊆B(r,c,t+r) sh(µ)=hcr i

r+1 −c ( 2 ) X = q q|µ| µ⊆B(r,c,t+r) sh(µ)=hcr i

r+1 −c ( 2 ) t+r 3 2 = q shcr i(q , ..., q , q , q).

Because we can easily convert between (reverse) semi-standard Young tableaux t+r 3 2 and column-strict plane partitions of the same shape, shcr i(q , ..., q , q , q) is the generating function for column-strict plane partitions of shape hcri with entries in n t+r 3 2 {1, 2, ..., t+r}. Note that the coeﬃcient of q in shcr i(q , ..., q , q , q) corresponds both to the number of such plane partitions of size n, and to the number of such (reverse) semistandard Young tableaux whose entries sum to n.

The above lemma gives us our desired generating function as a variation of a more familiar generating function (the Schur polynomial). We wish, ultimately, to develop a form for our generating function that is independent of the Schur functions. In order to do this we ﬁrst must deﬁne a few terms that are used in the proof. 6 AMY BECKER, LILLY BETKE-BRUNSWICK, ANNA ZINK

Deﬁnition 2.2. For any cell u in the ith row and jth column of the diagram of the partition, λ, the hook length of u, h(u), is the number of cells below and to the right of u including u itself.

0 h(u) = λi + λj − i − j + 1 0 (where λi is the length of row i and λj is the length of column j).

Deﬁnition 2.3. For any partition λ, ∀u ∈ λ if u = λi,j, then the content, c(u) is c(u) = j − i.

Deﬁnition 2.4. For any partition λ, let X b(λ) = (i − 1)λi.

Deﬁnition 2.5. For any positive integer n, let [n] = 1 − qn.

Lemma 2.6. For any partition λ, Y [n + c(u)] s (qn−1, ..., q2, q, 1) = qb(λ) . λ [h(u)] u∈λ

The details of this proof are beyond the scope of this paper, but we invite the reader to peruse Stanley’s proof [1, Thm. 7.21.2].

n−1 2 Note that sλ(q , ..., q , q, 1) counts semistandard Young tableaux with entries from {0, 1, 2, ..., n − 1}. In order to count those tableaux with entries from {1, 2, ..., n} we must multiply each term in the Schur function by q|λ| to account for adding one to each entry in the tableau. Therefore

n 3 2 |λ| n−1 2 sλ(q , ..., q , q , q) = q sλ(q , ..., q , q, 1).

Theorem 2.7. For all positive integers r, c, and t with r ≤ c we have X [t + 1][t + 2]2...[t + r]r[t + r + 1]r...[t + c]r[t + c + 1]r−1...[t + c + r − 1] q|π| = . [1][2]2...[r]r[r + 1]r...[c]r[c + 1]r−1...[c + r − 1] π⊆B(r, c, t) PLANE PARTITIONS 7

Proof. Combining Lemma 2.1 and Lemma 2.6 we have

r+1 − c X |π| ( 2 ) t+r 3 2 q = q shcr i(q , ..., q , q , q) π⊆B(r, c, t)

r+1 − c ( 2 ) cr t+r−1 2 = q q shcr i(q , ..., q , q, 1) r − c (2) t+r−1 2 = q shcr i(q , ..., q , q, 1) r − c (2) r Y [t + r + c(u)] = q qb(hc i) . [h(u)] u∈hcr i

X r Observe that b(hcri) = (i − 1)c = 0 + c + 2c + ... + (r − 1)c = c. 2 1≤i≤r

Therefore, X Y [t + r + c(u)] q|π| = . [h(u)] π⊆B(r,c,t) u∈hcr i

Isolating the numerators of the product, r c Y Y Y [t + r + c(u)] = [t + r + j − i]. u∈hcr i i=1 j=1

The [t + r + c(u)] for the cells of hcri are shown in the following grid:

[t+r] [t+r+1] [t+r+2] ... [t+c+r-2] [t+c+r-1] [t+r-1] [t+r] [t+r+1] ... [t+c+r-3] [t+c+r-2] [t+r-2] [t+r-1] [t+r] ... [t+c+r-4] [t+c+r-3] [t+r-3] [t+r-2] [t+r-1] ... [t+c+r-5] [t+c+r-4] ...... [t+3] [t+4] [t+5] ... [t+c+1] [t+c+2] [t+2] [t+3] [t+4] ... [t+c] [t+c+1] [t+1] [t+2] [t+3] ... [t+c-1] [t + c]

We note that [t + r + c(u)] is constant along diagonals going from bottom right to upper left. From this (recalling r ≤ c) we achieve our desired product by beginning in the lower left corner (i = r, j = 1) and reading oﬀ diagonals.

Y [t + r + c(u)] = [t+1][t+2]2...[t+r]r[t+r+1]r...[t+c]r[t+c+1]r−1...[t+c+r−1]. u∈hcr i 8 AMY BECKER, LILLY BETKE-BRUNSWICK, ANNA ZINK

Similarly, isolating the denominators of the product, r c Y Y Y [h(u)] = [r + c − i − j + 1]. u∈hcr i i=1 j=1

The [h(u)] for the cells of hcri are therefore the following:

[c+r -1] [c+r -2] [c+r -3] ... [r+1] [r] [c+r-2] [c+r -3] [c+r -4] ... [r] [r-1] [c+r -3] [c+r -4] [c+r -5] ... [r-1] [r-2] [c+r -4] [c+r -5] [c+r -6] ... [r-2] [r-3] ...... [c+2] [c+1] [c] ... [4] [3] [c+1] [c] [c-1] ... [3] [2] [c] [c-1] [c-2] ... [2] [1]

We note that hook-lengths are constant along diagonals going from bottom left to upper right. From this (recalling r ≤ c) we achieve our desired product by beginning in the lower right corner (i = r, j = c) and reading oﬀ diagonals.

Y [h(u)] = [1][2]2 ··· [r]r[r + 1]r ··· [c]r[c + 1]r−1 ··· [c + r − 1]. u∈hcr i

Combining the numerator and denominator products,

X Y [t + r + c(u)] q|π| = . [h(u)] π⊆B(r,c,t) u∈hcr i

[t + 1][t + 2]2 ··· [t + r]r[t + r + 1]r ··· [t + c]r[t + c + 1]r−1 ··· [t + c + r − 1] = . [1][2]2 ··· [r]r[r + 1]r ··· [c]r[c + 1]r−1 ··· [c + r − 1]

We can ﬁnd a more concise product form for our generating function as follows.

Corollary 2.8.

r c t X Y Y Y [i + j + k − 1] q|π| = . [i + j + k − 2] π⊆B(r,c,t) i=1 j=1 k=1 PLANE PARTITIONS 9

Proof. From the previous proof we have:

r c X Y [t + r + c(u)] Y Y [t + r + j − i] q|π| = = [h(u)] [r + c − i − j + 1] π⊆B(r,c,t) u∈hcr i i=1 j=1 r c Y Y [t + r + j − i] = . [r + c − i − (c + 1 − j) + 1] i=1 j=1 This last step is justified by the commutativity of multiplication which allows us to iterate through the columns in any order and achieve the same product. Specifically our choice of iteration order in the numerator is independent of that of the denominator. We have thus effectively traversed the columns forwards for the numerator and backwards for the denominator. Cancelation yields:

r c X Y Y [t + r + j − i] q|π| = [r + j − i] π⊆B(r,c,t) i=1 j=1

r c Y Y [t + r + j − i] [t − 1 + r + j − i][t − 2 + r + j − i] ··· [1 + r + j − i] = · [r + j − i] [t − 1 + r + j − i][t − 2 + r + j − i] ··· [1 + r + j − i] i=1 j=1

r c Y Y [t + r + j − i][t − 1 + r + j − i][t − 2 + r + j − i] ··· [1 + r + j − i] = [t − 1 + r + j − i][t − 2 + r + j − i] ··· [1 + r + j − i][r + j − i] i=1 j=1

r c t r c t Y Y Y [k + r + j − i] Y Y Y [k + j + (r + 1 − i) − 1] = = . [k + r + j − i − 1] [k + j + (r + 1 − i) − 2)] i=1 j=1 k=1 i=1 j=1 k=1

And by equivalently iterating through the rows in the reverse order we arrive at the following representation of our generating function:

r c t X Y Y Y [i + j + k − 1] q|π| = . [i + j + k − 2] π⊆B(r,c,t) i=1 j=1 k=1

From this representation, we readily see the symmetry of our bounding dimensions, since the commutativity of multiplication allows us to permute the axes of our plane partition. X X X q|π| = q|π| = q|π| π⊆B(r,c,t) π⊆B(r,t,c) π⊆B(c,r,t) X X X = q|π| = q|π| = q|π|. π⊆B(c,t,r) π⊆B(t,r,c) π⊆B(t,c,r)

Furthermore, from the above product form, we can immediately find generating functions for unbounded plane partitions. By letting each of r, c, and t tend to infinity, we find the generating function for all plane partitions. 10 AMY BECKER, LILLY BETKE-BRUNSWICK, ANNA ZINK

Corollary 2.9. Let P be the set of all plane partitions. Then we have

X Y 1 q|π| = . (1 − qi)i π∈P i≥1 Proof. ∞ ∞ ∞ X Y Y Y [i + j + k − 1] q|π| = [i + j + k − 2] π∈P i=1 j=1 k=1

∞ ∞ Y Y [i + j][i + j + 1][i + j + 2]... = [i + j − 1][i + j][i + j + 1][i + j + 2]... i=1 j=1 ∞ ∞ Y Y 1 = [i + j − 1] i=1 j=1

∞ Y 1 = [i][i + 1][i + 2][i + 3]... i=1 Y 1 Y 1 = = . [i]i (1 − qi)i i≥1 i≥1

The previous equation may look familiar to a combinatorialist due to its simi- larity to the generating function for all ordinary partitions. In fact, we can derive this latter function by setting t = 1 and letting r and c tend to inﬁnity, since a partition is essentially a plane partition with constant height one.

Corollary 2.10. We have X Y 1 q|λ| = , 1 − qi λ i≥1 where the sum on the left is over all partitions.

Proof. ∞ ∞ 1 X Y Y Y [i + j + k − 1] q|λ| = [i + j + k − 2] λ i=1 j=1 k=1 ∞ ∞ Y Y [i + j] = [i + j − 1] i=1 j=1 ∞ Y [i + 1][i + 2][i + 3]... = [i][i + 1][i + 2][1 + 3]... i=1 ∞ Y 1 Y 1 = = . [i] 1 − qi i=1 i≥1 PLANE PARTITIONS 11

We can use a similar process to ﬁnd generating functions for plane partitions with only some bounded dimensions. For example, we can acquire the generating function for all plane partitions with at most r rows. In addition to bounding our plane partitions in each dimension, we can impose restrictions on the type of plane partitions we consider.

3. Symmetric Plane Partitions Just as we did with bounded plane partitions, we can use Schur functions to ﬁnd the generating function for symmetric plane partitions. We do this by creating a bijection between symmetric plane partitions and SSYT with odd entries. Before we explain this bijective proof, we deﬁne symmetric plane partitions.

Deﬁnition 3.1. A plane partition π is symmetric if πi,j = πj,i for all i, j.

In a symmetric plane partition, the entry in the ith row and jth column is equal to the entry in the jth row and ith column. Symmetric plane partitions have a plane of symmetry along the diagonal of the entries in the partition.

4 4 3 2 2 4 4 3 2 1 3 3 2 2 1 2 2 2 1 2 1 1 Figure 5: A symmetric plane partition in three-dimensional and two-dimensional form with the plane of symmetryshown .

In this discussion of symmetric plane partitions, the plane partitions are bounded by an r × r × t box. Since symmetric plane partitions have the same number of rows and columns, they are contained within B(r, r, t).

Letting S denote the set of symmetric plane partitions that are subsets of B(r, r, t), we arrive at the following generating function:

Theorem 3.2. X |π| X 2r−1 2r−3 3 q = sλ(q , q , ..., q , q). π∈S λ⊆htr i

Proof. We follow an example symmetric plane partition of size |π| through the bijection to create a reverse SSYT with odd parts and size |π|, and then discuss how the bijection works in reverse. We use the plane partition above, π, for the example.

We begin by decomposing the partition into levels. A level is made up of all 12 AMY BECKER, LILLY BETKE-BRUNSWICK, ANNA ZINK the cubes at the same height in the partition. We decompose π into four levels, since the maximum height of π is four.

−→

Figure 6: A symmetric plane partition decomposed into levels

We then decompose each level into angles. Each angle is composed of the cubes from the jth row and jth column of the level. The angles are the outside edges of each level, and we can see how they are constructed by peeling oﬀ each outer L-shaped edge. Note that levels of a symmetric plane partition are self conjugate partitions. We can see that the process of constructing angles from these levels is a bijection from self conjugate partitions to partitions with distinct, odd parts.

−→

Figure 7: Levels decomposed into angles

Now that we can see how levels and angles are constructed from an example symmetric plane partition, we can make some general statements about levels and PLANE PARTITIONS 13 angles. To begin, the number of cubes in each angle is odd. Since angles are constructed to be symmetric, each angle has the same number of rows and columns, but the cube at the corner is counted twice. Next, the number of cubes in each successive angle of a level is strictly decreasing. This is because the angles are nested within one another. If they were not strictly decreasing, the original symmetric plane partition would not be a partition. There are at most r angles per level. The largest possible level is the r × r rectangle. This level would have r angles. There are at most 2r −1 cubes per angle, because the largest angle possible is the outer angle of a level that extends to r rows and r columns. There are 2r − 1 cubes in this angle. Finally, there are at most t levels, since the symmetric plane partition had a height at most t.

We now construct a new plane partition where each of the levels forms a column and each angle is a part in the column. To do this, we take the cubes in each angle and stack them on top of each other, then put the stacked angles together to form a column. We do this for each level in the original symmetric plane partition. Then we combine these columns to form a new plane partition. In our example, we see how this process results in the formation of the new plane partition.

−→ −→

−→ −→ 14 AMY BECKER, LILLY BETKE-BRUNSWICK, ANNA ZINK

9 9 5 3 7 5 3 1 5 3 + + + = = 1

Figure 8: Constructing a new plane partition using the cubes in each angle to form an entry in a column.

In the resulting plane partition each entry is odd since the number of cubes in each angle is odd. The columns are strictly decreasing since the number of cubes in each angle is strictly decreasing. There are at most t columns and r rows, since the largest possible number of levels is t and angles is r. Therefore the resulting plane partition is a reverse SSYT, bounded by an r × t rectangle with odd parts. Since we used all the cubes of the symmetric plane partition to create this SSYT, we see that the resulting partition is the same size as the original partition.

To describe the inverse of this bijection, we begin with a SSYT bounded by an r × t rectangle with odd parts. Since the entries in each column are odd and strictly decreasing, we can create angles and nest the angles to form a symmetric level. Each column creates one level and we can stack the levels on top of each other to form a symmetric plane partition within B(r,r,t). Since we have found a bijection between symmetric plane partitions of size |π| and a reverse SSYT with odd parts where the sum of the entries is |π|, we know that the generating function for symmetric plane partitions is equal to the generating function for reverse SSYT with odd parts.

n P |π| That is, the coeﬃcient of q in π∈S q is the number of symmetric plane n P 2r−1 2r−3 3 partitions of size n. The coeﬃcient of q in λ⊆htr i sλ(q , q , ..., q , q) is the number of reverse SSYT with odd parts whose shape is a subset htri and whose parts sum to n. By our bijection we know that these counts are identical, giving

X |π| X 2r−1 2r−3 3 q = sλ(q , q , ..., q , q). π∈S λ⊆htr i To construct a product form of the generating function for symmetric plane partitions, we use the equation derived above in addition to the following result [1] [2]: Lemma 3.3. n X Y 1 Y 1 sλ(x1, ..., xn) = . 1 − xi 1 − xixj λ i=1 1≤i

Theorem 3.4. r X Y 1 Y 1 q|π| = . 1 − q2i−1 1 − q2(i+j−1) π∈S i=1 1≤i

div div div div div by 1 by 2 by 3 ... by r by 1 a1 a2 ... a4 div div div div by 3 by 4 ... by r+1 a1 by 3 a3 ... a5 div div div by 5 ... by r+2 a2 a3 by 5 ... a6 ...... div div by 2r-1 a4 a5 a6 ... by 2r-1

Proof. : Beginning with our original generating function and using Lemma 3.3, we substitute q for x in the equation. Since we want the generating function for reverse SSYT with odd entries and at most r rows, we change the bounds on the product. We then have that the generating function for reverse SSYT with odd parts and thus for symmetric plane partitions:

r Y 1 Y 1 . 1 − q2i−1 1 − q2(i+j−1) i=1 1≤i

4. Reverse Plane Partitions and the Hillman-Grassl Correspondence We have studied generating functions for plane partitions confined within B(r, c, t). But can we find generating functions for other plane partition cofinements? In our previous proof we created a correspondence between bounded plane partitions and column strict partitions with a hcri base. This was convenient because we asso- ciate column strict plane partitions with SSYT which have Schur functions as their generating function. Naturally, we might wonder if this type of correspondence exists for a non- rectangular base-shape. That is, can we develop a similar bijection between column strict plane partitions of shape λ and ordinary plane partitions whose shape is a subset of λ to find a Schur-based generating function for these plane partitions? It turns out there is not a similar correspondence. For example, if we consider the correspondence we used in the last proof where we peeled off blocks row by row, we will see that this does not work. 16 AMY BECKER, LILLY BETKE-BRUNSWICK, ANNA ZINK

a b (a + 3) (b + 3) c d (c + 2) (d + 2) π = ⇔ π’ = e (e + 1) f (f)

6 2 3 -1 4 1 2 -1 π’ = ⇔ π = 2 1 1 1 Figure 9: Example of why the correspondence used in the B(r,c,t) proof fails: π0 maps to a ﬁlling with negative integers.

We can consider alternative bijections, such as the following, which accounts for column-strictness independently in each column. But we’ll see that this correspondence also fails.

a b c (a + 4) (b + 2) (c + 1) d e f (d + 3) (e + 1) (f) π = g h ⇔ π’ = (g + 2) (h) i (i + 1) j (j)

5 5 2 1 3 1 4 3 1 1 2 1 π’ = 3 2 ⇔ π = 1 2 2 1 1 1

Figure 10: Example of how another potential bijection fails: π0 maps to a ﬁlling without increasing rows.

Although we cannot ﬁnd a correspondence between column strict plane partitions of shape λ and regular partitions whose shape is contained in λ, we are able to ﬁnd a generating function for weak reverse column-strict plane partitions. A reverse plane partition is a plane partition that is weakly increasing across rows and down columns. If we allow 0 as a part, then we call it a weak reverse plane partition.

Theorem 4.1. For a partition λ,

X 1 q|π| = . Q [h(u)] π u∈λ Here π is a weak reverse plane partition of shape λ and h(u) represents the hook length of u, a cell in λ. PLANE PARTITIONS 17

Recall the deﬁnition of h(u) from the previous section:

6 4 2 1 3 1 ⇒ 1

Figure 11: Hook length for each cell in the partition λ = 4, 2, 1.

To ﬁnd this generating function we use the Hillman-Grassl Correspondence. We want to create a correspondence between weak reverse plane partitions π and functions f : λ → N (each f assigns a natural number to each cell in λ) such that X |π| = f(u)h(u). u∈λ

In this bijection, we create a sequence of ordered pairs (πi, fi) for i = 0, 1, ..., k in which we input our initial plane partition, π = π0, and after a series of intermediate steps, our output is f = fk. We demonstrate the Hillman Grassl Correspondence through an example.

The Hillman-Grassl Correspondence We will begin by letting π be a weak reverse plane partition of shape λ and setting π0 = π and f0 to be a Ferrers diagram of shape λ with all zero entries. Starting with i = 0, we will construct a lattice path in πi to determine each subsequent πi+1 given a set of rules:

(1) Begin a path, Li, at xi, the southwestern-most nonzero entry of πi.

(2) If the entry immediately north of xi is the same, take a north step.

(3) Otherwise, if the entry immediately east is nonzero, take an east step.

(4) Repeat steps 2 and 3 until you cannot move north or east. Then let your current cell, yi be the end of your path, Li.

(5) Let πi+1 be the result of subtracting 1 from each entry along Li.

(6) To get fi+1, add 1 to the entry, zi in fi that corresponds with the column of xi and the row of yi.

(7) Repeat steps 1-6 until you reach πk, an empty partition.

The resulting fk is the function f : λ → N that we are seeking.

In order to show that the Hillman-Grassl correspondence is a bijection, we note that we ﬁll in f from left to right over columns. Suppose to the contrary we add 1 to fi in a column to the left of the rightmost nonzero column of fi. By our rules, that would indicate that the most recent path, Li, started in a column farther to 18 AMY BECKER, LILLY BETKE-BRUNSWICK, ANNA ZINK the left than some previous path, Lj, which started in the rightmost nonzero column. But by our rules, each lattice path starts at the southwestern-most nonzero entry so this would mean Lj started at an entry that was not the soutwestern-most nonzero one, which is a contradiction.

Not only are we ﬁlling in f left to right, but in each column we ﬁll f from bottom to top. Before proving this, we establish the following lemma:

Lemma 4.2. Every lattice path ends at the end of a row.

Proof. Since πi is a weak reverse plane partition, the rows are weakly increasing which means that you can never have a nonzero entry followed by a zero entry thus you can always move east to the last entry in a row.

Lemma 4.3. We are filling in f bottom to top in each column. Proof. Suppose to the contrary that there exists a column that is not filled from bottom to top. Therefore, there must be two lattice paths Li+1 and Li that start in the same column in which Li ends in a row above Li+1. As such, we add 1 to an entry in fi+1 that already has a nonzero entry above it . If no such i exists, then f is indeed filled within each column from bottom to top.

Li ∩ Li+1 must be nonempty.

To see this, suppose Li ∩ Li+1 is empty. We know that Li must start below Li+1 because it starts with the Southwestern-most nonzero entry. So Li starts below Li+1 but ends in a row above it. But since Li+1 must end at the end of a row, Li must intersect Li+1 in order to end in a higher row since it is only permited north and east steps.

Since Li ∩ Li+1 is nonempty and Li and Li+1 don’t end in the same row, we know that they must intersect and diverge at least once. Let’s look at their last point of divergence at cell v meaning that v ∈ Li,Li+1 but no later cells are in both lattice paths. There are two cases:

(1) Li takes a north step and Li+1 takes an east step from v Let v0 be the cell above v and val(v)=a, val(v0)=b in step i. Since Li takes a north step at v, a=b. So during stepi+1,val(v)= val(v0) = a−1 = b−1 = nonzero. These entries cannot be zero since v ∈ Li+1. So by our rules Li+1 must take a north step from v. →←

(2) Li takes an east step and Li+1 takes a north step from v Similarily to our argument for proving Li ∩ Li+1 is nonempty, if Li+1 goes north from v then it must go east at some point to the end of a row. But this means that in order for Li to end in a higher row than Li+1, it will have to cross Li again. →← PLANE PARTITIONS 19

Since neither case is possible, Li and Li+1 cannot diverge which means they must be the same path which is a contradiction of our initial assumption that they start in the same column but end in a diﬀerent row. Our result follows from this contradiction.

Given f, we set fk = f and ﬁnd the topmost nonzero entry, call it zk−1, in the rightmost nonzero column. This will tell us the starting and ending points of the path Lk−1 since zk−1 is in the column of the starting entry and row of the ending entry of Lk−1. We can reconstruct Lk−1 by starting in yk−1, the easternmost cell in the row of zk−1 and reversing the rules listed above i.e. moving south if the immediately southern entry is equal (they can be zero) and west otherwise, etc. Once we retrace our steps back to xk−1, the southernmost cell in the column of zk−1, we augment each entry of πk along this path by 1, and let πk−1 be the result. We subtract 1 from the value of zk−1 to get fk−1. Continuing in this vein we recreate the coordinate pairs starting with i = k until we reach i = 0, as evidenced by fi = 0 for all cells, and we have π0 = π, our original partition.

Lemma 4.4. Let π be a weak reverse plane partition of shape λ and f be the corresponding function given by the Hillman-Grassl correspondence. Then X |π| = f(u)h(u). u∈λ

Proof. In our construction of f, for every i, we add 1 to the entry u in f in the column of the starting entry and row of the ending entry of our path, Li. But by observation, we can conclude that h(u) is in fact the length of this path. So for each i, when we subtract one for each entry in the lattice path and then add 1 to u, we are using fi+1 to keep track of the total amount we have removed from πi. So to ﬁnd the size of the original partition we know that each entry u in f tells us the number of times we have removed h(u) from our count. Then by summing f(u)h(u) over all cells in λ we get the size of π by our construction of f. That is, f counts the size of π in units of hook-lengths.

Example 4.5.

i πi fi

0 2 4 0 0 0 0 1 2 0 0 2 0

0 2 4 0 0 0 1 1 2 0 0 1 1 20 AMY BECKER, LILLY BETKE-BRUNSWICK, ANNA ZINK

0 1 3 1 0 0 2 0 1 0 0 0 1

0 0 2 1 1 0 3 0 0 0 0 0 1

0 0 1 1 1 1 4 0 0 0 0 0 1

0 0 0 1 1 2 5 0 0 0 0 0 1

Now that we have created our bijection with the desired property that |π| = P u∈λ f(u)h(u), let’s revisit the original theorem:

Theorem 4.6. X Y 1 q|π| = . [h(u)] π u∈λ Here |λ| = n and π is a reverse plane partition with shape λ.

Proof. Y 1 1 1 1 = ··· [h(u)] 1 − qh(u1) 1 − qh(u2) 1 − qh(un) u∈λ

= (1 + qh(u1) + q2h(u1) + ...)(1+qh(u2) +q2h(u2) +...) ··· (1+qh(un) +q2h(un) +...).

This last step is justiﬁed by our ability to rewrite each of the factors as a geo- metric series. Making use of the bijective nature of Hillman-Grassl, we can imagine constructing a function f1 by choosing each f1(ui) from N for all i ∈ {1, 2, ..., n}. For each i we select the corresponding term from the ith factor of the above product. That is, we select qf1(ui)h(ui) from the series. The product of all of these selections P f (u)h(u) is q u∈λ 1 Expanding the overall product, therefore, yields a sum with one term corresponding to each possible function f : λ → N.

Y 1 X (P f(u)h(u)) = q u∈λ . [h(u)] u∈λ f PLANE PARTITIONS 21

Now applying the correspondence between weak reverse plane partitions, π, and X functions, f, we substitute exponents using Lemma 4.4 (|π| = f(u)h(u)), and u∈λ achieve the desired equation.

Y 1 X = q|π|. [h(u)] u∈λ π 5. Future Work We are interested in investigating types of symmetric plane partitions other that those we discussed above. For example, plane partitions can have rotational symmetry, as well as combinations of reﬂexive and rotational symmetry. With further study, we would like to relate Schur functions to other symmetric plane partitions and explore new bounding conditions. Acknowledgements We thank the Carleton College Math Department, friends and familiy, and es- pecially Eric Egge for all their support and help during this process.

References [1] R. Stanley. Enumerative combinatorics, Volume 2. Cambridge University Press, Cambridge, 1999. [2] D. Bressoud. Proofs and conﬁrmations: the story of the alternating sign matrix conjecture. Mathematical Association of America/Cambridge University Press, 1999.