PLANE PARTITIONS AMY BECKER, LILLY BETKE-BRUNSWICK, ANNA ZINK

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1 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.. 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 basic 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 commonly seen 2-D version of a plane partition. Definition.. For any nonnegative integer n, a partition λ of n is a nonincreasing sequence {λ j } j of nonnegative integers such that j λ j 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. λ λ 2 λ 3 Figure : Ferrers diagrams for a few partitions of n5: λ 4,, λ 2 3, 2, and λ 3 2, 2,. 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.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.

2 2 AMY BECKER, LILLY BETKE-BRUNSWICK, ANNA ZINK Figure 2: Two fillings of the Ferrers diagram of λ 4, 2, with entries {, 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 {, 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 x i. Then, multiplying these variables together we let the product be the resulting term x x 2 x 2 x 3 x 3 x x 2 2x 2 3 Figure 3: How to convert a SSYT into a term in a polynomial. The resulting polynomial is the Schur polynomial, s λ (x, x 2,..., x n ), where λ is the partition we are filling in with entries from {x, x 2,..., x n }. Schur polynomials are in fact symmetric functions. Recall: Definition.3. A function f(x, x 2,..., x n ) is symmetric whenever it does not change under any permutation of its variables. The Schur function s λ (x, x 2,..., x n ) 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 s 2, (x, x 2,..., x n ) is 2x x 2 x 3 then there are 2 ways to fill in the partition λ 2, with x, x 2, and x 3. 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 representing the heights for stacks of blocks placed on each cell of the diagram.

3 PLANE PARTITIONS Figure 4: Filling of a Ferrers diagram associated with a plane partition, π More formally: Definition.4. A plane partition is an array π (π ij ) i,j 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.5. The size of a plane partition, π, is the sum of the entries. π π ij. i,j Informally, we can think of π as the total number of blocks in the threedimensional interpretation of the plane partition. For example, the size of the plane partition in Figure 4 is π 29. Definition.6. The shape of a plane partition, sh(π), is the partition whose Ferrers diagram is filled. Definition.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 s λ 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 columnstrict 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 x i 0 i. Notice that the resulting plane partition will be column-strict since the SSYT is column-strict.

4 4 AMY BECKER, LILLY BETKE-BRUNSWICK, ANNA ZINK x x x 2 x 3 x 5 x 3 x 4 x 4 x 6 x 4 x 6 x 5 x 8 x 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.. π B(r,c,t) q ( r+ 2 )c s c r (q t+r,..., q 3, q 2, q). (Note: c r is the partition (c, c, c,..., c) of cr. The Ferrers diagram of c r 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 c r 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, µ r,j < µ r,j <... < µ,j t + r. So for all i, j, µ i,j > r i. Let π be the plane partition defined by π i,j µ i,j (r + i). When π i,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)

5 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, π i,j π i,j+ µ i,j µ i,j+ 0, so π too has weakly decreasing rows. Additionally, π i,j π i+,j µ i,j µ i+,j 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 µ i,j t + r + i, then π i,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 plane parition, π, in B(r, c, t) and construct µ such that µ i,j π i,j + (r + i) for all i and j with i r and j c. Now r c r c π π i,j (µ i,j (r + i)) i j i j µ rc (r )c... 2c c µ c( r) ( ) r + µ c. 2 Therefore, by substitution we achieve the following: π B(r,c,t) µ B(r,c,t+r) sh(µ) c r q c ( r+ 2 ) q µ c(r+ 2 ) µ B(r,c,t+r) sh(µ) c r q µ q c ( r+ 2 ) s cr (q t+r,..., q 3, q 2, q). Because we can easily convert between (reverse) semi-standard Young tableaux and column-strict plane partitions of the same shape, s cr (q t+r,..., q 3, q 2, q) is the generating function for column-strict plane partitions of shape c r with entries in {, 2,..., t + r}. Note that the coefficient of q n in s c r (q t+r,..., q 3, q 2, 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 6 AMY BECKER, LILLY BETKE-BRUNSWICK, ANNA ZINK Definition 2.2. For any cell u in the i th row and j th 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. h(u) λ i + λ j i j + (where λ i is the length of row i and λ j is the length of column j). Definition 2.3. For any partition λ, u λ if u λ i,j, then the content, c(u) is c(u) j i. Definition 2.4. For any partition λ, let b(λ) (i )λ i. Definition 2.5. For any positive integer n, let [n] q n. Lemma 2.6. For any partition λ, s λ (q n,..., q 2, q, ) q b(λ) u λ [n + c(u)]. [h(u)] The details of this proof are beyond the scope of this paper, but we invite the reader to peruse Stanley s proof [, Thm ]. Note that s λ (q n,..., q 2, q, ) counts semistandard Young tableaux with entries from {0,, 2,..., n }. In order to count those tableaux with entries from {, 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 s λ (q n,..., q 3, q 2, q) q λ s λ (q n,..., q 2, q, ). Theorem 2.7. For all positive integers r, c, and t with r c we have π B(r, c, t) [t + ][t + 2]2...[t + r] r [t + r + ] r...[t + c] r [t + c + ] r...[t + c + r ] [][2] 2...[r] r [r + ] r...[c] r [c + ] r....[c + r ]

7 PLANE PARTITIONS 7 Proof. Combining Lemma 2. and Lemma 2.6 we have π B(r, c, t) q ( r+ 2 )c s c r (q t+r,..., q 3, q 2, q) q ( r+ 2 )c q cr s c r (q t+r,..., q 2, q, ) q ( r 2)c s c r (q t+r,..., q 2, q, ) q ( r 2)c q b( c r ) [t + r + c(u)]. [h(u)] u c r Observe that b( c r ) i r (i )c 0 + c + 2c (r )c ( ) r c. 2 Therefore, π B(r,c,t) u c r [t + r + c(u)]. [h(u)] Isolating the numerators of the product, c [t + r + c(u)] [t + r + j i]. u c r i j The [t + r + c(u)] for the cells of c r are shown in the following grid: [t+r] [t+r+] [t+r+2]... [t+c+r-2] [t+c+r-] [t+r-] [t+r] [t+r+]... [t+c+r-3] [t+c+r-2] [t+r-2] [t+r-] [t+r]... [t+c+r-4] [t+c+r-3] [t+r-3] [t+r-2] [t+r-]... [t+c+r-5] [t+c+r-4] [t+3] [t+4] [t+5]... [t+c+] [t+c+2] [t+2] [t+3] [t+4]... [t+c] [t+c+] [t+] [t+2] [t+3]... [t+c-] [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 ) and reading off diagonals. u c r [t + r + c(u)] [t+][t+2] 2...[t+r] r [t+r+] r...[t+c] r [t+c+] r...[t+c+r ].

8 8 AMY BECKER, LILLY BETKE-BRUNSWICK, ANNA ZINK Similarly, isolating the denominators of the product, c [h(u)] [r + c i j + ]. u c r i j The [h(u)] for the cells of c r are therefore the following: [c+r -] [c+r -2] [c+r -3]... [r+] [r] [c+r-2] [c+r -3] [c+r -4]... [r] [r-] [c+r -3] [c+r -4] [c+r -5]... [r-] [r-2] [c+r -4] [c+r -5] [c+r -6]... [r-2] [r-3] [c+2] [c+] [c]... [4] [3] [c+] [c] [c-]... [3] [2] [c] [c-] [c-2]... [2] [] 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. [h(u)] [][2] 2 [r] r [r + ] r [c] r [c + ] r [c + r ]. u c r Combining the numerator and denominator products, π B(r,c,t) u c r [t + r + c(u)]. [h(u)] [t + ][t + 2]2 [t + r] r [t + r + ] r [t + c] r [t + c + ] r [t + c + r ] [][2] 2 [r] r [r + ] r [c] r [c + ] r. [c + r ] We can find a more concise product form for our generating function as follows. Corollary 2.8. π B(r,c,t) c t i j k [i + j + k ] [i + j + k 2].

9 PLANE PARTITIONS 9 Proof. From the previous proof we have: π B(r,c,t) u c r c i j [t + r + c(u)] [h(u)] c i j [t + r + j i] [r + c i (c + j) + ]. [t + r + j i] [r + c i j + ] 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: c i j i j c [t + r + j i] [r + j i] π B(r,c,t) c i j [t + r + j i] [r + j i] [t + r + j i][t 2 + r + j i] [ + r + j i] [t + r + j i][t 2 + r + j i] [ + r + j i] [t + r + j i][t + r + j i][t 2 + r + j i] [ + r + j i] [t + r + j i][t 2 + r + j i] [ + r + j i][r + j i] c t i j k [k + r + j i] [k + r + j i ] c t i j k [k + j + (r + i) ] [k + j + (r + i) 2)]. And by equivalently iterating through the rows in the reverse order we arrive at the following representation of our generating function: π B(r,c,t) c t i j k [i + j + k ] [i + j + k 2]. 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. π B(r,c,t) π B(c,t,r) π B(r,t,c) π B(t,r,c) π B(c,r,t) 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. q π q π.

10 0 AMY BECKER, LILLY BETKE-BRUNSWICK, ANNA ZINK Corollary 2.9. Let P be the set of all plane partitions. Then we have Proof. π P ( q i ) i. π P i i j k i j i j [i + j + k ] [i + j + k 2] [i + j][i + j + ][i + j + 2]... [i + j ][i + j][i + j + ][i + j + 2]... [i + j ] [i][i + ][i + 2][i + 3]... i [i] i ( q i ) i. i i The previous equation may look familiar to a combinatorialist due to its similarity to the generating function for all ordinary partitions. In fact, we can derive this latter function by setting t and letting r and c tend to infinity, since a partition is essentially a plane partition with constant height one. Corollary 2.0. We have q λ q i, i where the sum on the left is over all partitions. λ Proof. q λ λ i j k i j i i [i + j + k ] [i + j + k 2] [i + j] [i + j ] [i + ][i + 2][i + 3]... [i][i + ][i + 2][ + 3]... [i] q i. i

11 PLANE PARTITIONS 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.. A plane partition π is symmetric if π i,j π j,i for all i, j. In a symmetric plane partition, the entry in the i th row and j th column is equal to the entry in the j th row and i th column. Symmetric plane partitions have a plane of symmetry along the diagonal of the entries in the partition 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. s λ (q 2r, q 2r 3,..., q 3, q). π S λ t r 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 2 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 j th row and j th 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

13 PLANE PARTITIONS 3 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 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 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 4 AMY BECKER, LILLY BETKE-BRUNSWICK, ANNA ZINK 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. That is, the coefficient of q n in π S q π is the number of symmetric plane partitions of size n. The coefficient of q n in λ t r s λ(q 2r, q 2r 3,..., q 3, q) is the number of reverse SSYT with odd parts whose shape is a subset t r and whose parts sum to n. By our bijection we know that these counts are identical, giving π S λ t r s λ (q 2r, q 2r 3,..., q 3, q). 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 [] [2]: Lemma 3.3. s λ (x,..., x n ) λ n i x i i<j n x i x j. It is beyond the scope of this paper to provide a proof of this equation.

15 PLANE PARTITIONS 5 Theorem 3.4. π S i q 2i i<j r q 2(i+j ). This formula can be derived computationally but does not have a simple bijective proof. However, we can examine this product to see what the factors intuitively construct. The first factor generates the main diagonal for an r r Ferrers diagram where the entries in the diagonal are numbers that are isible by increasing odd numbers. The second factor in the product generates the remainder of the Ferrers diagram, where the entries above the diagonal are isible by numbers increasing up to r. This factor creates a symmetry across the diagonal of the Ferrers diagram. by by 2 by 3 by 3... by r by 4... by r+ by 5... by r by 2r- by a a 2... a 4 a by 3 a 3... a 5 a 2 a 3 by 5... a a 4 a 5 a 6... by 2r- 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: i q 2i i<j r q 2(i+j ). 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 c r base. This was convenient because we associate 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 nonrectangular 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 6 AMY BECKER, LILLY BETKE-BRUNSWICK, ANNA ZINK π a c e f b d π (a + 3) (b + 3) (c + 2) (d + 2) (e + ) (f) π π Figure 9: Example of why the correspondence used in the B(r,c,t) proof fails: π 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. π a b c d e f g h i j π (a + 4) (b + 2) (c + ) (d + 3) (e + ) (f) (g + 2) (h) (i + ) (j) π π Figure 0: Example of how another potential bijection fails: π 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.. For a partition λ, q π u λ [h(u)]. π Here π is a weak reverse plane partition of shape λ and h(u) represents the hook length of u, a cell in λ.

17 PLANE PARTITIONS 7 Recall the definition of h(u) from the previous section: Figure : Hook length for each cell in the partition λ 4, 2,. 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 π u λ f(u)h(u). In this bijection, we create a sequence of ordered pairs (π i, f i ) for i 0,,..., k in which we input our initial plane partition, π π 0, and after a series of intermediate steps, our output is f f k. 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 π 0 π and f 0 to be a Ferrers diagram of shape λ with all zero entries. Starting with i 0, we will construct a lattice path in π i to determine each subsequent π i+ given a set of rules: () Begin a path, L i, at x i, the southwestern-most nonzero entry of π i. (2) If the entry immediately north of x i 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, y i be the end of your path, L i. (5) Let π i+ be the result of subtracting from each entry along L i. (6) To get f i+, add to the entry, z i in f i that corresponds with the column of x i and the row of y i. (7) Repeat steps -6 until you reach π k, an empty partition. The resulting f k 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 to f i in a column to the left of the rightmost nonzero column of f i. By our rules, that would indicate that the most recent path, L i, started in a column farther to

18 8 AMY BECKER, LILLY BETKE-BRUNSWICK, ANNA ZINK the left than some previous path, L j, 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 L j 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 π i 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 L i+ and L i that start in the same column in which L i ends in a row above L i+. As such, we add to an entry in f i+ 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. L i L i+ must be nonempty. To see this, suppose L i L i+ is empty. We know that L i must start below L i+ because it starts with the Southwestern-most nonzero entry. So L i starts below L i+ but ends in a row above it. But since L i+ must end at the end of a row, L i must intersect L i+ in order to end in a higher row since it is only permited north and east steps. Since L i L i+ is nonempty and L i and L i+ don t end in the same row, we know that they must intersect and erge at least once. Let s look at their last point of ergence at cell v meaning that v L i, L i+ but no later cells are in both lattice paths. There are two cases: () L i takes a north step and L i+ takes an east step from v Let v be the cell above v and val(v)a, val(v )b in step i. Since L i takes a north step at v, ab. So during stepi+,val(v) val(v ) a b nonzero. These entries cannot be zero since v L i+. So by our rules L i+ must take a north step from v. (2) L i takes an east step and L i+ takes a north step from v Similarily to our argument for proving L i L i+ is nonempty, if L i+ 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 L i to end in a higher row than L i+, it will have to cross L i again.

19 PLANE PARTITIONS 9 Since neither case is possible, L i and L i+ cannot erge 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 f k f and find the topmost nonzero entry, call it z k, in the rightmost nonzero column. This will tell us the starting and ending points of the path L k since z k is in the column of the starting entry and row of the ending entry of L k. We can reconstruct L k by starting in y k, the easternmost cell in the row of z k 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 x k, the southernmost cell in the column of z k, we augment each entry of π k along this path by, and let π k be the result. We subtract from the value of z k to get f k. Continuing in this vein we recreate the coordinate pairs starting with i k until we reach i 0, as evidenced by f i 0 for all cells, and we have π 0 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 π u λ f(u)h(u). Proof. In our construction of f, for every i, we add to the entry u in f in the column of the starting entry and row of the ending entry of our path, L i. 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 to u, we are using f i+ to keep track of the total amount we have removed from π i. 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 π i f i

20 20 AMY BECKER, LILLY BETKE-BRUNSWICK, ANNA ZINK Now that we have created our bijection with the desired property that π f(u)h(u), let s revisit the original theorem: u λ Theorem 4.6. [h(u)]. π u λ Here λ n and π is a reverse plane partition with shape λ. Proof. u λ ( [h(u)] q h(u) ) ( q h(u2) ) ( q h(un) ) ( + q h(u) + q 2h(u) +...)(+q h(u2) +q 2h(u2) +...) (+q h(un) +q 2h(un) +...). This last step is justified by our ability to rewrite each of the factors as a geometric series. Making use of the bijective nature of Hillman-Grassl, we can imagine constructing a function f by choosing each f (u i ) from N for all i {, 2,..., n}. For each i we select the corresponding term from the i th factor of the above product. That is, we select q f(ui)h(ui) from the series. The product of all of these selections is q u λ f(u)h(u) Expanding the overall product, therefore, yields a sum with one term corresponding to each possible function f : λ N. [h(u)] u λ f q ( u λ f(u)h(u)).

21 PLANE PARTITIONS 2 Now applying the correspondence between weak reverse plane partitions, π, and functions, f, we substitute exponents using Lemma 4.4 ( π u λ f(u)h(u)), and achieve the desired equation. [h(u)] π u λ q 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 especially Eric Egge for all their support and help during this process. References [] R. Stanley. Enumerative combinatorics, Volume 2. Cambridge University Press, Cambridge, 999. [2] D. Bressoud. Proofs and confirmations: the story of the alternating sign matrix conjecture. Mathematical Association of America/Cambridge University Press, 999.