INEQUALITIES OF SMMETRIC FUNCTIONS JONATHAN D. LIMA Abstract. We prove several symmetric function inequalities and conjecture a partially proved comprehensive theorem. We also introduce the condition of -Positivity and prove several known inequalities in a new, stronger sense.. Introduction to Symmetric Functions [?] Definition.. A symmetric function in n variables is a function, f, such that f(x,..., x n ) = f(x π,..., x πn ), where π S n, is a permutation of n objects. Definition.2. A homogeneous symmetric polynomial of degree d, is a symmetric polynomial, such that, for each term, the sum of the exponents is equal to d. The homogeneous symmetric polynomials of degree d form a real vector space of dimension part(d), the number of non-increasing sequences of positive integers whose sum is d, under the expected operations of adding and scaling. These sequences are called partitions of d. We call this vector space Λ d. Because the dimension of this vector space is not immediately obvious, we first enumerate some of the common bases used in the study of symmetric functions. We begin by defining the monomial symmetric polynomials, which helps clarify the dimension of Λ d. For simplicity let us assume that the number of variables n is greater than d, and we allow our partitions λ of d, written λ d, to contain trailing zeroes in order to be of length n. For any partition λ = (λ,..., λ n ) we define the monomial symmetric function: () m λ (x,..., x n ) = σ x λ σ... x λn σ n where the sum is over distinct permutations σ of the n variables. For example, if x R 3, m (x) = x x 2 + x x 3 + x 2 x 3. The permutation σ Date: May 3, 200. A special thanks to Curtis Greene for showing me such beautiful math.
2 J. LIMA which permutes the first and second coordinates is not counted because it results in the same term as the identity permutation. The elementary symmetric polynomials also form a basis for Λ d. Before defining them we must introduce notation which will be used throughout the paper. In a subscript, a b will simply denote the sequence (a,..., a), of length b. We first define the elementary symmetric polynomial for a partition with only one nonzero part and define the function for arbitrary partitions multiplicatively. For r N, define (2) e r = m r (3) e λ = e λ e λ2... e λn For example if x R n, then e n (x) = x... x n. Thirdly, we have the power sum symmetric functions, which will be defined in a very similar fashion to the elementary basis. (4) p r = m r (5) p λ = p λ p λ2... p λn The complete homogeneous symmetric polynomials, often referred to as simply complete, are defined in, yet again, a similar manner. (6) h r = µ r m µ (7) h λ = h λ h λ2... h λn The fifth and final basis for Λ d, called the Schur basis, is significantly less straightforward than the other four. Because this basis is not the focus of this paper, its definition will be brief. Before introducing Schur functions we must introduce several combinatorial objects, called tableaux. Let λ be a partition of d, with length n. Definition.3. A standard oung tableau, abbreviated ST, of shape λ d is a two dimensional array of positive integers. It has n rows, the i th row having length λ i. The entries of the tableau are increasing from top to bottom in every column and from left to right in every row. A semi-standard oung tableau, SST, is defined similarly to an ST, but with only weak inequality required in rows.
INEQUALITIES OF SMMETRIC FUNCTIONS 3 ST 3 4 6 2 5 8 7 0 9 SST 2 2 4 5 3 3 5 4 5 7 6 The type of an SST is the vector containing the frequencies of the labels, often called parts. Let us call the SST above T. Then we have type(t ) = (, 2, 2, 2, 3,, ). So to parse our example completely, T is a SST of shape (5, 3, 3, ), type (, 2, 2, 2, 3,, ), with parts less than or equal to 7. Now that we have this terminology completed we can finally define our fifth basis. Let λ be a partition of d, and x R n. (8) s λ (x,..., x n ) = T x type(t ) In this sum T ranges over all SST of shape λ with largest part n. The resulting polynomial will have, for each monomial term, a coefficient which counts the number of λ-ssts of that type. The functions, s λ, are not obviously symmetric, but a proof of this fact can be found in [?]. The focus of this paper will not be properties of the vector space Λ d but rather symmetric function inequalities. For this reason we will not prove that these functions form a basis for this space, but rather we will develop the idea of a symmetric function inequality. In Section 2, we will introduce several known inequalities and state the current conjecture of our research. In section 3, we will state a condition stronger than the ordinary inequality we ve defined. Lastly, in Sections 4 and 5, we will provide proofs of stronger versions of the inequalities stated in Section 2. While these inequalities, including their stronger analogues, were already proved, I have original bijective proofs to present. 2. Classical and Modern Inequalities Definition 2.. A symmetric function f is positive if f(x) 0 for all real x such that x i 0 for all i. We also say that f g, if f g is positive. One familiar example of a symmetric function inequality is the Arithmetic Geometric Mean Inequality, abbreviated the AGM inequality. Take x R n, such that x i 0 for all i. The AGM inequality, as in [?], states that
4 J. LIMA x + + x n (9) n n x... x n. Clearly these are both symmetric functions and the hypothesis on x is exactly the same as in our definition above. There are many proofs of the AGM inequality, but because it is a special case of a lesserknown inequality, called Maclaurin s inequality, we present one which may generalize. In order to introduce Maclaurin s inequality we first introduce some new notation which persists through the paper. Definition 2.2. [?]Let f be a homogeneous symmetric polynomial in n variables of degree d, with non-negative coefficients. We now define the term-weighted polynomial F, and finally the degree-weighted function F, corresponding to f. The latter is not a polynomial but this is not a problem as we will see in later sections. (0) F (x) = f(x) f( n ) () F(x) = d F (x) It is simple to see that the left hand side of equation (9) is just M (x), i.e. the degree weighted function corresponding to m (x), and the right hand side is M n(x). From this observation we proceed to the statement of the theorem. Theorem 2.3. (Maclaurin s Inequality) Let a, b be positive natural numbers with a b. Then M a M b. We do not prove this theorem in this section; however, we prove a stronger version of the inequality later in the paper. Before moving on to the more recent theory which encompasses Maclaurin s inequality we will first consider one more classical family of inequalities, attributed to Robert Franklin Muirhead. In order to do this we must first discuss a partial order on the set of partitions of d. Definition 2.4. Let λ and µ be partitions of d. We say that λ is majorized by µ, written λ µ if, for all i d we have, (2) i λ i i µ i. Theorem 2.5. (Muirhead s Inequality [?]) Let λ and µ be partitions of d with λ µ. Then, M λ M µ.
INEQUALITIES OF SMMETRIC FUNCTIONS 5 Once again we do not prove Muirhead s Inequality in this section, but we wait until we have developed more technical machinery, and then prove a stronger version. For now, we would like to find some common theory out of which both of these families of inequalities would fall. We first introduce a partial order on the infinite set of all partitions. Definition 2.6. [?]Let λ and µ be partitions of l and m, respectively. We say λ is weight-majorized by µ, written λ µ, if for all i, we have (3) i λ i l i µ i m. Secondly we say that λ is double-majorized by µ, written λ µ, if (4) λ µ, and λ T µ T, where λ T is the transpose partition reached by reflecting the diagram of λ. For example, the transpose of (3,2,2) is just (3,3,). Example 2.7. Let λ = (,, ) and µ = (). Then, λ µ Example 2.8. Let λ = (3, 2) and µ = (4, ). Then, λ µ Example 2.9. Let λ = (2, ) and µ = (3, ). Then, λ µ Example 2.0. Let λ = (4, 2) and µ = (5,, ). Then, λ and µ are not comparable. Before we state our main conjecture, we ought to form some sort of intuition about the double-majorization ordering. It is proved in [?] that, given two partitions, λ and µ, of the same number, λ µ if and only if µ T λ T. So, in the special case of two partitions of the same number, double-majorization is simply majorization. Additionally, it is pretty simple to see that if a b, then b a. We would like a comprehensive theorem, from which Muirhead s and Maclaurin s inequalities would result. So given these new definitions we can state our main conjecture, stated first in [?]. Conjecture 2.. Let λ and µ be partitions. Then M λ M µ if and only if λ µ. This conjecture is partially proved (completely in one direction and partially in the other) in [?] and is still not proved in its entirety. However, special cases of this inequality have been proved using a new technique which yields a stronger condition than inequality itself.
6 J. LIMA Before moving on, we end this section by stating two analogous results for the elementary and power-sum symmetric bases, proved in [?] Theorem 2.2. Let λ and µ be partitions. Then E λ E µ if and only if λ µ. Theorem 2.3. Let λ and µ be partitions. Then P λ P µ if and only if λ µ. 3. -Positivity and -positive results Definition 3.. Let f be a symmetric polynomial in variables x... x n. Make the substitutions: (5) x i = y i + y i+ + + y n. We say that f is -positive, if the resulting polynomial in y... y n has only positive coefficients. We also write f g if g f is -positive, and will refer to such statements as -inequalities. Because we are dealing with symmetric functions it is always safe to assume that x n x n x. So if we assume x n 0 then we have positivity for each of the y i. It follows that every -positive function is indeed a positive function. It is not the case, however, that every positive symmetric function is -positive. In fact, at a glance of the definition it may come as a surprise that there exist symmetric functions that are -positive, while not being obviously positive to begin with. In this section and the sections that follow, we will provide several -inequalities. One very important note, before we state and prove -inequalities, is that the terminology of -positivity requires that we deal exclusively with polynomials. So in this case, and in any case involving degreeweighted means, we will simply exponentiate to eliminate radicals. We will still refer to these results as -inequalities. As an illustration of these ideas, we will begin with a proof of the -positive analog of the AGM inequality. Theorem 3.2. (AGM -Inequality) (6) x + + x n n x x n n (x + + x n ) n n n (x x n )
INEQUALITIES OF SMMETRIC FUNCTIONS 7 Proof. We must first make the -substitution, yielding: (7) (y + 2y 2 + + ny n ) n n n (y + + y n ) (y n ). To show that the condition is satisfied we must prove this last inequality term by term. In order to keep track of indices we introduce a bit of notation. We will let i N n represent the n-multiset of indices for the monomial whose coefficient we are calculating. For example, if i = (,, 2, 2), then y i = y 2 y 2 2. We will always index i in non-decreasing order. The coefficient of y i is denoted c(i), with a subscript indicating which function. We will begin with the left hand side of equation (7) and call these coefficients c L (i). The coefficients of the left hand side we can calculate by simply counting the number of ways of arriving at our particular monomial, which involves taking all permutations of n indices, and adjusting for multiplicity. Using the multinomial theorem, we arrive at the formula: (8) c L (i) = n! p(i) n i j, where p(i) denotes the product of the multiplicities factorial of the multiset i. For example, c L (,, 3, 4, 4, 4) = 6!( 3 4 4 4). 2!!3! The right hand side of equation (7) is a bit more complex. In calculating the number of ways of arriving at our desired monomial when expanding this product it helps if we try to imagine putting pieces on a jagged chess board. The board has n rows and the j th row has j spaces. Each row corresponds to one of the factors on the right hand side, and each column refers to a y variable. For example the top row corresponds to the factor x n, and has only one space corresponding to y n, and the bottom row corresponds to x, and has n spaces representing y n through y. Placing an in the j th row and k th column represents pulling y k out of x j. j= x 5 x 4 x 3 x 2 x y 5 y 4 y 3 y 2 y
8 J. LIMA We begin by trying to place the smallest index i, i.e. marking a box in the column representing y i. There are i rows that have this column and therefore i choices for placement. When we go to place i 2, there are i 2 rows with the proper column, but one of them already has the previous index, so we have only i 2 choices. The number of choices for placing i j is independent of the previous choices because i j is larger than the previous indices. If we iterate this process and correct for multiplicity again by dividing by p(i), we have: (9) c R (i) = nn p(i) n (i j j + ). j= Before we proceed we should note that in some cases we will not be able to place all the indices. For example, if i = (,..., ), after marking the only box in the column for y i we will have no more positions. Equation (9) is still correct, and this simply means that there exists some j such that i j j + = 0. In these cases c R (i) = 0 and we need not compare them to c L (i) which are clearly non-negative. So for the remainder of the proof we will assume i j > j. Finally, after eliminating the multiplicity factors, and moving things around a bit we have one relatively simple inequality to prove, (20) n j= n i j n j= n j + i j j +. In order to complete the proof of Theorem 3.2, we will introduce a lemma which is as basic as it is essential to every proof in this paper. For the remainder of this paper, any result following from an application of this lemma will be referred to as trivial. The proof of this lemma is left to the child of the reader. Lemma 3.3. Let a, b, c R, with 0 a b c. Then, (2) c b c a b a. To conclude the proof of Theorem 3.2, we look at equation (20) factor by factor and apply the lemma, which applies because 0 j for all j. (22) n i j n (j ) i j (j ).
INEQUALITIES OF SMMETRIC FUNCTIONS 9 While this example shows the power of -positivity as a condition, it also demonstrates the complexity of computing these coefficients. Because we do not want to go through these steps every time we prove -positivity, it would behoove us to develop a standard method of computing -coefficients for certain families of symmetric polynomials. We will describe such a formula for our term-weighted monomial symmetric polynomials, written M λ. Let λ be a partition of d with l parts, and let n N. Theorem 3.4. (Computation of -Coefficients) (23) M λ (x,..., x n ) xi y i + +y n = i Φ λ (i, n)y i where the sum is over all d-multisets i of the indices... n, y i = y i y id, and Φ is defined as follows. Let σ be a multiset permutation of i. Define σ,..., σ l as the sequence of multiset permutations obtained by breaking σ into l consecutive blocks of sizes λ,..., λ l. Define b = (b,..., b l ), where b k = Min(σ k ), and b = (b,..., b l ) to be the result of sorting b k in increasing order. Now, define φ λ (σ, n) to be (24) φ λ (σ, n) = b (b 2 )... (b l l + ) n(n )... (n l + ). And finally, we compute the coefficient, (25) Φ λ (i, n) = σ φ λ (σ, n) where the sum is over all distinct permutations σ of the multiset i. Proof. We begin the proof by rewriting the formula for term-weighted monomial symmetric polynomials. For λ = (λ,..., λ l ), (26) M λ (x,..., x n ) = x λ j (n) x λ 2 j 2... x λ l j l l The above sum is over all distinct sequences j,..., j l, and (n) l is the falling factorial, defined as n!/(n l)!. Now, let i be a d-multiset of indices, and let σ be a permutation of this multiset, with σ defined as in the statement of the theorem. We must first derive an expression for the number of tuples, (x j,..., x jl ), whose expansion contains the desired monomial (y σ y σd ), in this
0 J. LIMA order. So for example y σ must appear as a summand in the expression for x j. Below we demonstrate this condition given the partition λ = (3, 2, ). x j x j x j x j2 x j2 x j3 y σ y σ2 y σ3 y σ4 y σ5 y σ6 σ σ 2 σ 3 Recall from our definition of the -substitution that y k is a summand in the expression for x j, if and only if j k. Therefore the condition above is equivalent to x jk Min(σ k ). Now we define b k = Min(σ k ), for k l. We need to identify the l-tuples, (j,..., j l ), such that j k b k, for k l. We do this in a similar manner as in the proof of the AGM -inequality. Consider a chessboard with l rows, with length b k, k l. If we take any configuration of l non-attacking rooks on this board, rooks in different columns and rows, this will correspond to exactly one choice of tuples. The non-attacking condition guarantees that we do not pull two terms from the same factor, and also guarantees that the tuple has distinct entries. Clearly, a non-attacking rook configuration is invariant under reordering of the rows, so there is a simple equation for counting nonattacking rook configurations, which is easily proved using the logic of the AGM proof. The number of choices is (b )(b 2 )... (b l l + ), where we define b = (b,..., b l ) by sorting b i in increasing order. After including the falling factorial coefficient in equation (26), we have the formula (27) φ λ (σ, n) = (b )(b 2 )... (b l l + ) n(n )... (n l + ). So now we can rewrite M λ as a function of y, (28) M λ = φ λ (σ, n)y i, i σ where σ ranges over all permutations of the multiset i. Now we can clearly remove the y-monomial from the interior sum and bring the falling factorial inside to obtain,
INEQUALITIES OF SMMETRIC FUNCTIONS (29) M λ = y i φ λ (σ, n) = i σ i This completes the proof. Φ λ (i, n)y i Example 3.5. Computing the coefficient, Φ (2,) ((2, 2, 3), 4) We are given: λ = (2, ), i = (2, 2, 3), and n = 4. There are three permutations of this multiset and we must first calculate the coefficient of each. σ: (2, 2, 3) (2, 3, 2) (3, 2, 2) σ: 2 2 3 2 3 2 3 2 2 b: (2, 3) (2, 2) (2, 2) b: (2, 3) (2, 2) (2, 2) φ: 2 2 4 3 The above figure shows all but the last step of the computation which is simply adding the pieces to arrive at Φ λ (i, n) = 2 3. Example 3.6. Computing Coefficients for f = M λ M µ It will become clear in later sections that we must extend the computation of -coefficients to products of term-weight monomial symmetric polynomials. This is not a problem, however, and we will demonstrate without proof the calculation of such a coefficient. We will begin with a particular multiset permutation, σ, and calculate only the corresponding partial, φ, coefficient. To obtain the full coefficient, Φ, we would sum individual terms like these. In this example we will also introduce some self-explanatory notation. 2 4 3 σ: (7, 4, 9, 2, 0, 3, 4, 0) λ: (3, 2) µ: (2, ) n: 5 σ λ : 7 4 9 2 0 σ µ : 3 4 0 b λ :(4, 2) b µ :(4, 0) b λ :(2, 4) b µ :(4, 0) φ λ : 2 3 5 4 φ µ : 4 9 5 4 2 4 3
2 J. LIMA Finding the coefficient for this term is simply taking the product of these two coefficients as we would in multiplying any two polynomials. So finally, we have φ λµ (σ, n) = 2 3 4 9 = 6. 5 4 5 4 225 Now that we have an algorithm for computing -coefficients we will use this algorithm to prove the -positivity of both Muirhead s and Maclaurin s inequalities. 4. Bijective proof of Maclaurin s Inequality In this section we will prove Maclaurin s inequality and several generalizations. The proof will involve constructing a bijection between multiset permutations. We will proceed by stating the inequality and generalizations, constructing the bijection, and finally proving the theorems. We should note that Maclaurin s inequality has already been proved. It has also been proved as a -inequality in [?], but without a bijection. This paper introduces a bijective proof of this inequality, and a generalization of the result found at the end of the section. Theorem 4.. Maclaurin s Inequality (-version). Let a, b N, with a b. Then, (30) M b M a. The AGM -inequality is just the special case when a = and b = n, the number of variables. Because the definition of -inequality only makes sense for polynomials, what we are actually saying in this case is that: (3) (M b) a (M a) b. This is now simply a -inequality between polynomials of degree ab. Upon noticing that as partitions b a majorizes a b we will generalize a bit. A non--positive version of the following theorem is presented and proved in [?]. Theorem 4.2. Let λ and µ be partitions of d such that λ µ. Then, (32) i M µ i i M λ i The condition λ µ, as proved in [?], is equivalent to the condition that µ can be obtained from λ by repeatedly subtracting one from the i th part, and adding it to some part j i. We will use this equivalence
INEQUALITIES OF SMMETRIC FUNCTIONS 3 to simplify Theorem 4.2 into a statement involving products of only two polynomials. Lemma 4.3. Let a, b N, with 0 a b. Then, (33) M b+m a M bm a. Our proof consists of constructing a bijection between multiset permutations that map terms on the right to terms on the left with a smaller coefficient. Let us recall several steps involved in the calculation of -coefficients. For a given multiset of indices, i, we consider a permutation σ. We then place σ into chunks according to the parts of the partitions defining the function, forming σ. Because we always do computations with a fixed underlying multiset, we will make some simplifications. Rather than recording the actual multiset permutation of the indices and constructing the bijection we will order the indices from smallest to largest and record this order. Because the indices in the multiset need not be distinct we will be counting each permutation p(i) times, but all this will do is multiply the two coefficients we are trying to compare by the positive constant p(i). So, just as in the AGM proof, this factor will not matter. Also, because n is fixed throughout these calculations we will drop n from our notation. In this case we are dealing exclusively with partitions whose parts are all ones. So placing them into chunks is straightforward. Minimizing each chunk is simple as well because each chunk has only one element. For example let a = 3, b = 6, and σ = (2, 5,, 6, 9, 4, 8, 3, 7). 2 5 6 9 4 8 3 7 The diagram above could be used to represent σ, σ, or b. We first construct a correspondence between tableau fillings and sequences of length a + b, called rank sequences, whose i th term is a positive integer less than or equal to i. Given a tableau filling, T, we will call its corresponding rank sequence r(t ). We construct this sequence iteratively. In order to find the last term of the sequence we label the positions of the tableau from left to right then from bottom to top in increasing order, assuming that the rows of the tableau are sorted by length. We then record the position of the largest element of our filled tableau and then remove this piece from the tableau entirely. At this point if the rows are no longer ordered by length we rearrange them and repeat. An example of such a calculation is shown below.
4 J. LIMA Example 4.4. Let b = 3, a = 2, and σ = (, 4, 5, 3, 2) T = 4 5 3 2 Filling Ranks 4 5 3 2 3 4 5 2 5 4 3 2 3 4 2 4 Rearranged 3 2 2 3 2 2 2 2 The resulting sequence r(t ) is (,2,2,4,5). If the tableau filling T is obtained by filling a tableau of shape λ with the permutation σ we define φ(t ) = φ λ (σ, n). One can see that the process described above can be inverted, that is, given a rank sequence and a shape, one can fill in the shape by placing the largest index first and repeating. We can also see that there are n! rank sequences, and likewise, n! tableau fillings of each shape. We would now like to define a bijection f between tableau fillings of shape ( b, a ) and tableau fillings of shape ( b+, a ). We do so using the rank sequences, (34) f(t ) = T if r(t ) = r(t ) where T is a tableau filling of shape ( b+, a ). So we would simply like to show that φ(t ) φ(f(t )) for all tableau fillings T. In order to do that we prove certain properties of rank sequences, namely that the canonical order on rank sequences, pointwise domination, agrees with the ordering we get from φ.
INEQUALITIES OF SMMETRIC FUNCTIONS 5 Theorem 4.5. Given a shape λ = ( b, a ), with a b, and two rank sequences α and β, let T α and T β be the tableau fillings of shape λ that correspond to these rank sequences. If α i β i, for all i with i a + b, then φ(t α ) φ(t β ). Proof. Of course, it suffices to prove this result for two rank sequences, α and β, which agree for all terms except one. Then we could simply form a descending sequences of inequalities from one to the other. For example, if we let α = (,, 2, ) and β = (, 2, 3, 3), we could show: (35) φ(t (,2,3,3) ) φ(t (,,3,3) ) φ(t (,,2,3) ) φ(t (,,2,) ). Also, given two rank sequences which agree after a certain term we can essentially ignore all the terms beyond that point. This is because when we evaluate φ on the respective tableau fillings the terms corresponding to the remainder of the rank sequences will be the same on both sides. For example, let b = 5, a = 2, α = 32244, and β = 32544. Then for T α and T β we have, T α = 7 6 3 5 2 4 T β = 4 7 6 3 2 5 When we apply φ to both sides we can clearly remove the term, (i 7 4)(i 6 3) (36) (n 4)(n 3) from both sides and proceed only comparing the shortened rank sequences and the smaller shape. For these two reasons it suffices to consider only rank sequences which agree on all terms except the last. The weak inequality in Theorem 4.5 is obviously equality if either α a+b β a+b a or a + α a+b β a+b, because the largest index will then be sorted to the same position and if we were to ignore this position the remaining shapes will be identical. Therefore after sorting the rows the two fillings will agree. Therefore we focus on the specific circumstance in which α and β agree at every term except the (a+b) th term, α a+b a, and a+ β a+b. The proof will be divided into three cases based on α a+b. We will rely heavily on the inductive assumption that the result holds for all rank sequences and shapes of length and size less than a + b. In the first case we assume that α a+b = β a+b a. This means that the second largest entry is in the bottom row of both T α and T β. In the diagram below we outline a sequence of inequalities, with and
6 J. LIMA representing the largest and second largest indices, respectively, and the blank spaces will be filled according to identical rank sequences of length (a + b 2). The diagrams represent the position of these indices after sorting the rows of each tableau, so the positions of these indices represent their algebraic contribution to the calculation of the φ function. T β T α The first inequality holds because once the φ function is applied, the contribution of and is larger on the left then on the right, as a result of Lemma 3.3. (37) (i a+b b + )(i a+b a + ) (n b + )(n a + ) (i a+b a + )(i a+b b + ) (n b + )(n a + ) For the remainder of the proof, this type of inequality is referred to as a trivial swap. The second inequality holds because of our inductive assumption. If we remove the contribution of from both sides, i.e. divide by (i a+b a + )/(n a + ), then what remains is simply two tableaux of the same shape, filled in such away that their rank sequences differ only in the last term. In the second case we assume that α a+b = β a+b = a. This condition translates to the second largest index being placed on the bottom row of T β and the top row of T α. Following the exact same assumptions we have only one simple equality to show, and it results from a trivial swap. T β T α The third and final case assumes α a+b = β a+b a +. This results in the second largest index being placed on the top row of both T α and T β. In the diagram below the first inequality follows from our inductive assumption after removing the contribution made by, and the second inequality is a trivial swap.
INEQUALITIES OF SMMETRIC FUNCTIONS 7 T β T α This leaves us only with the task of justifying the base cases. The two base cases that need to be considered are a = and a = b. It is quite easily seen, however, that in the first case we have trivial inequality because placing the largest index in the lone box of the second row would minimize the function. Any trivial swap to the row above would increase the evaluation of the φ function on the tableau. In the latter case one can see that if a = b given two rank sequences which differ only in the last term, their corresponding fillings would be the same after simply switching the rows. Therefore this completes the proof of the theorem. Now that we have proved Theorem 4.5, we can use this property to prove Lemma 4.3. Given a b and a certain tableau filling T of ( b, a ), we find f(t ), a tableau filling of ( b+, a ) by constructing the filling of this shape corresponding to the same rank sequence. Now, we add an extra box to the top row of T and the bottom row of f(t ), and place in this box a new greater index a + b +, (we may assume for instance that i a+b i a+b+ = n). Now we are left with two fillings, which we call T and f(t ), of the same shape. Their rank sequences agree at every term except the last, at which the rank sequence of T will be greater than that of f(t ), because in the former the largest element is in the longer row and in the latter it is in the shorter. We therefore know that φ(t ) φ(f(t ) ). The contribution of n in each of these calculations will be equal to, (n b + )/(n b + ) in the former and (n a + )/(n a + ) in the latter. So the resulting inequality is, (38) φ(t ) = φ(t ) φ(f(t ) ) = φ(f(t )). Therefore, the bijection f has precisely the property we desired and Lemma 4.3 is proved, from which Theorems 4.2 and 4. immediately follow. Before moving on we should first take note of an interesting phenomenon. If we think of the tableau fillings given above as the indices remaining after the minimization of larger cells, we could form a bijection between fillings of shapes, λ, containing all cells of any constant
8 J. LIMA size. Moving these cells together we can use precisely the bijection f defined. For example, if we let that constant equal 2, let b = 4, and a = 2, we have: 8 4 5 2 3 0 6 8 4 5 2 3 0 6 7 2 9 7 2 9 min 4 2 3 6 9 f min 4 2 3 6 If we let k N, and define in this way a bijection f from fillings of (k b, k a ) to fillings of (k b+, k a ), we can go back and check that every step of in the proof of Theorem 4.5 and Lemma 4.3 holds for this new more general statement. We therefore end this section with a statement of several general implications, analogues of the theorems proved above. As far as we know, these results are new, not only in the -positive sense, but also as ordinary inequalities. Theorem 4.6. Let a, b, k N with a b. Also let λ µ be partitions of some positive integer d. The following inequalities, as symmetric function inequalities in any finite number of variables. 9 (39) M k b M k a (40) i M k µ i i M k λ i (4) M k b+m k a M k bm k a 5. Bijective proof of Muirhead s Inequality In this section we prove Muirhead s inequality using a bijection of rank sequences once again. We begin by stating Muirhead s inequality. Theorem 5.. Muirhead s Inequality (-version). partitions of d with λ µ. Then, M λ M µ. Let λ and µ be As we have stated before, because λ and µ are partitions of the same number, the final statement is equivalent to M λ M µ. The setup for this proof is precisely the same as for Maclaurin s inequality. In order to prove Theorem 5., we must show that for all multisets i and positive integers n, Φ λ (i, n) Φ µ (i, n). Therefore, we assume a fixed
INEQUALITIES OF SMMETRIC FUNCTIONS 9 i and n, and apply Theorem 3.4 to determine these values. We use tableaux to represent different permutations of i. We have replaced the elements of our multiset with indices representing their order from smallest to largest. We will now have n! permutations of these indices, despite the fact that the elements of our multiset need not be distinct. This is not a problem, however, because we are proving that p(i)φ λ (i, n) p(i)φ µ (i, n). Just as in the proof of Maclaurin s inequality, we begin by constructing a map from tableau fillings to rank sequences. The process is iterative in the same way, but because we have a new type of shape, the numbering does not work the same way. Whereas, in the previous proof, we were only concerned with two-rowed tableaux, with cells uniformly equal to, in this case we are concerned with single-rowed tableaux, with arbitrary cell size, or essentially, partitions. In order to construct rank sequences from tableau fillings, we now number the boxes from largest part to smallest part, and record the position of the largest index. After recording this position, we remove this index from our tableau filling and repeat. If ever the cells of the tableau filling are not in weakly decreasing order, we rearrange them. The example below shows the constructon of a rank sequence from a tableau. Example 5.2. Let λ = (3, 2), and σ = (2, 4, 5,, 3) T = 5 4 2 3 Filling 5 4 2 3 Ranks 3 2 5 3 4 4 2 3 2 4 3 2 3 Rearranged 2 2 3 2 2 2 2
20 J. LIMA The resulting sequence, r(t ), is (,2,2,2,3). Now, we would like to prove the analogous version of Theorem 4.5. We want inequality in our rank sequence to imply inequality for the φ function evaluated on tableau fillings. Theorem 5.3. Given a partition λ of l, and two rank sequences α and β, define T α and T β, as the tableau fillings of shape λ which correspond to these rank sequences. If α i β i, for all i with i l, then φ(t α ) φ(t β ). Proof. As in the previous proof we need only concern ourselves with rank sequences which differ in only the last term. We are also uninterested in comparing sequences in which the largest index is placed in the same part of λ, although they may be in different boxes in this part. So we will focus on the specific case in which T α has the largest index in λ i, and T β has the largest index in λ j, with i j, that is λ = (λ,..., λ i,..., λ j,... ), so λ i λ j. Also, we assume that λ i is strictly greater than λ j, otherwise, under the circumstances listed above, it is straight forward to check that φ(t α ) = φ(t β ) in this case. As with the last proof, this proof will be split into three cases based on α l. The proof also depends on our inductive hypothesis that Theorem 5.3 holds for all partitions µ of m, with m < l. In the first case we assume that α l = β l λ + + λ i. This translates into the second largest index being placed in a cell larger than the cells containing the largest index in both T α and T β. Once again, we diagram several inequalities using, l, to represent the largest index and to represent the second largest, l. We have the following sequence of inequalities, beginning with φ(t β ) and ending with φ(t α ). T β λ i λ j λ i λ j
INEQUALITIES OF SMMETRIC FUNCTIONS 2 = T α λ i λ j λ i λ j Now we explain the inequalities represented in the diagram above. While we can make no assumptions about λ j, we know that λ i > λ j. So λ i and any part to the left of λ i is greater than. The first inequality in the diagram above is what we refer to in this proof as a trivial swap. Because the part containing is greater than, disappears during the minimization. either disappears during minimization, or, if λ j =, makes a contribution during the evaluation of the function. Either way, switching the position of and decreases the evaluation of the φ function. The second inequality is simply our inductive assumption. We can remove because it is in a part greater than and therefore will not be expressed. So we are left with a shape of size l and two different placements of the now largest element. The last step is trivial equality because both parts are greater than, therefore both the and disappear in minimization, so their order does not matter. In the second case we assume λ + + λ i α l = β l λ + + λ j. This translates to the second largest index being placed among the parts between λ i and λ j, inclusively. T β λ i λ j λ i λ j
22 J. LIMA = λ i λ j The first inequality is a trivial swap. As for the second inequality, is now in a part which is greater than and can therefore be ignored, leaving us to our inductive hypothesis. The last step is trivial equality because both and are in parts greater than and will therefore both be ignored after minimization. The last case occurs when λ + +λ j α l = β l. This translates into the second largest index being placed into a part to the right of λ j. λ i λ j T β T α λ i λ j In this case we must consider two possibilities. If λ j = then we can first switch the contribution made by and, (i.e. if λ has d parts multiply φ(t β ) by (i l d+2)(i l d+)/(i l d+)(i l d+2) ). Then we can divide both sides of the equation by the contribution made by and we are reduced to an inequality given by the assumption with remaining the largest of the l indices. If λ j then we need only divide both sides by the contribution made by and employ the inductive assumption. Between these three cases we prove the entirety of the inductive step and the base case is simply the trivial case of the sole partition of. This completes the proof. Once again, in order to prove our desired result, we begin with T, a filling of the partition of l, λ. If we have a partition µ obtained from λ by subtracting from the i th part of λ and adding to the j th part with j < i. We define f(t ) by the filling of the partition µ, such that λ i λ j
INEQUALITIES OF SMMETRIC FUNCTIONS 23 r(t ) = r(f(t )). Now, we consider the partition of l +, λ +, defined by taking the largest parts from λ and µ. We define T, a filling of λ +, by filling it with T and placing the index l + in the empty box, again we assume i l+ = n. We define f(t ) by filling λ + with f(t ) and placing the index l + in the empty box. Following immediately from our hypothesis, and due to the fact that any contribution made by n will be equal to, we have: (42) φ(t ) = φ(t ) φ(f(t ) ) = φ(f(t )). Because we have proved the result for the coverings in the majorization order, it follows in general. This completes the proof of Muirhead s Inequality. These two -inequalities, being special cases of Conjecture 2.0, suggests that, not only is Conjecture 2.0 indeed a -inequality, but perhaps there is a straightforward bijective solution to this problem. Such a bijection would not only be useful in proving the main theory, but would also be interesting on its own merit, due to the nature of the φ function itself. References. A. Cuttler, C. Greene, M. Skandera, Inequalities for symmetric means, to appear in European Journal of Combinatorics. 2. R. P. Stanley, Enumerative Combinatorics, Vol., Cambridge University Press, Cambridge, 997. 3. R. P. Stanley, Enumerative Combinatorics, Vol. 2, Cambridge University Press, Cambridge, 999. 4. C. Greene, Jonathan D Lima, The Poset of 2-Rowed Standard Tableaux, in progress. 5. G. H. Hardy, J. E. Littlewood, G. Polya, Inequalities, Cambridge University Press, Cambridge 967. Department of Mathematics, Haverford College, Haverford PA 904 E-mail address: jlima@haverford.edu