Partitions, permutations and posets Péter Csikvári

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1 Partitions, permutations and posets Péter Csivári In this note I only collect those things which are not discussed in R Stanley s Algebraic Combinatorics boo Partitions For the definition of (number) partition, Ferrers diagram and Young tableaux, conugate partition see Chapter of R Stanley s Algebraic Combinatorics boo This section is based (more or less) on the treatment of the boo A Course in Combinatorics by Van Lint and Wilson (Chapter 5) Proposition Let p (n) be the number of partitions of n into at most parts Let p (n) be the number of partitions of n with at most parts Then p (n) p (n) Proof There is a natural biection between the two sets: associate the conugate partition to a partition Next let us understand the generating function of the seuence (p (n)) Theorem We have where p (0) Proof Note that p (n)x n x i, x ( + x i + x i + x 3 i + ) If we expand this product, the coefficient of x n will come from the products of the form x m x m x m, where m,, m 0 and m + +m n Note that this naturally correspond to the partition in which we have m s, m s,,m s and vice versa each partition naturally corresponds to such a term in the expansion The same idea helps us to understand the generating function of all partitions Theorem 3 We have p(n)x n x i,

2 where p(0) Proof As before x ( + x i + x i + x 3 i + ) i It might be scary to consider an infinite product, but observe that if you want to compute the coefficient of x n then you always have to choose the term from the terms + x i + x i + x 3 i + when i n + Let us introduce the notation [x n ]f(x) for a n if f(x) n a nx n Then n [x n ] (+x i +x i +x 3 i + ) [x n ] (+x i +x i +x 3 i + ) p n (n) p(n) by the previous theorem and the fact that the largest part in a partition of n is at most n Hence p(n)x n x i One can thin to generating functions a n x n in two different ways: (i) they are algebraic obects which form a ring, you can manipulate them algebraically, but you cannot plug any number (different form 0) into them, (ii) they are analytic functions with some convergence radius The function n!x n is a good example for the difference between (i) and (ii) Since the convergence radius is 0 for this function, you will hardly be able to do anything with it analytically, but this is a completely eligible algebraic expression, a "prominent" element of a ring Theorem 4 Let p o (n) be the number of partitions of n into odd parts Let p u (n) be the number of partitions of n into uneual parts Then (a) p o (n)x n x i (b) p u (n)x n ( + x i )

3 (c) p o (n) p u (n) Proof The proof of part (a) and (b) goes as before We only concentrate to part (c) Note that hence ( + x i ) + x i xi x i, x i x i x i since the terms x will cancel from the denominator and the enumerator Hence p o (n) p u (n) Second proof for part (c) We will give a biection between the set of partitions of n into odd parts and the set of partitions of n into uneual parts The ey ingredient of this biection will be the observation that any number can be uniuely written into the form (t + ), where, t 0 So let (λ,, λ m ) be a partition of n such that λ > > λ m Let λ i i (t i + ) and replace λ i by i pieces of t i + Then clearly we obtained a partition of n into odd parts Now we show that we can decode the original partition Let s count the number of parts t i + in a partition of n into odd parts Assume that there r i pieces of t i + Then r i can be uniuely written in base, ie, there are uniue s > s > > s such that r i s + + s Now replace the r i pieces of t i + with elements s n (t i + ), where n Hence we gave a biection between the set of partitions of n into odd parts and the set of partitions of n into uneual parts and so p o (n) p u (n) An example for this proof is the following Consider the partition , then 8 3, 3, 4, 3 3 and Hence the corresponding partition into odd parts will contain pieces of s + 3 pieces of 3 s And if you get the partition of 3 s and 3 pieces of 3 s then we now that we have to decompose 3 into -powers which can be uniuely done as , and similarly 3 + so we get bac the original partition Now let us consider the generating function as an analytic function Theorem 5 For n > we have π p(n) < e π 3 n (n ) 3

4 4 Remar Hardy and Ramanuan proved that p(n) 4 3n eπ 3 n, so, where f(n) 4 eπ 3 n As we can see our upper bound 3n agree with this function in the main term (and we won t need to wor very hard for this bound) lim n p(n) f(n) Proof (Van Lint) Recall that P (t) t p()t We will see that P (t) is convergent if t < Actually, we will choose t to be 0 < t < later The idea is the following, we will give an upper bound for P (t) and we will choose a t such that p(n)t n dominates the terms in P (t) Note that Then log P (t) log log log P (t) Now let 0 < t <, then and so Then log P (t) ( t ) log t t log( t ) t t t t t + t + t + + t > t, t t < t ( t)t t t t t < t t t t t t π t t Now we give a lower bound to P (t) Note that p(n) is a monoton increasing seuence (why?), so P (t) p()t p()t p(n) t p(n) tn t n n

5 Hence In other words, log p(n) + log log p(n) π tn t Now let u t t, then t +u Then log p(n) π < log P (t) < π t t t n log t + log( t) t u n log + u + log u + u π + (n ) log( + u) + log u u Note that log( + u) < u as + u < e u + u + u + Hence log p(n) < π Now let us choose u such a way that π optimal choice, then u + (n )u + log u u u π (n ) (n )u as it will be the (almost) Then we have log p(n) < (n )u + log u π 3 (n ) + log π (n ) In other words, p(n) < π (n ) e π 3 (n ) Euler s "pentagonal numbers" theorem Let us consider the partitions of n into uneual parts, and let p e (n) be the number of partitions of n into even number of uneual parts, and let p o (n) be the number of partitions of n into even number of uneual parts The following theorem is due to Euler Theorem We have p e (n) p o (n) { ( ) if n 3 ±, 0 otherwise 5

6 Proof We define two transformations on partitions with uneual parts They will be almost biection between partitions of n into even and odd number of uneual parts Let λ > > λ m be a partition of n into uneual parts The dots in the last row of the Ferrers diagram is called the base Its size is denoted by b, clearly b λ m Let s be the largest integer for which it is true that λ + λ + λ + The number s is the size of the slope of the partition: on the Ferrers diagram, the slope can be seen as follows: draw a 45 line in the direction NE-SW through the upper-right dot of the Ferrers diagram, then the dots on this line is the slope Its size is clearly s Now we give the two transformations: Transformation I: if b s then delete the base from the Ferrers diagram and add dots to the first b rows, this way we created a new slope This transformation results a new partition into uneual parts except in one case: if the original slope and base had a common dot and b s, then this transformation cannot be applied In this exceptional case n b + (b + ) + + (b ) 3b b, note that the number of parts in this case is b too Transformation II: if b > s then delete the slope from the Ferrers diagram and add a new base of size s to the Ferrers diagram This transformation results a new partition into uneual parts except in one case: if the original slope and base had a common dot and b s +, then this transformation cannot be applied In this exceptional case: n b + (b + ) + + (b 3) + (b ) 3(b ) +(b ), note that the number of parts in this case is b Figure Transformation II Both Transformation I and II change the parity of the number of parts and we can apply exactly one of them to a non-exceptional partition, and for the resulting partition we can only apply the other transformation which gives bac the original partition This shows that if n 3 ± then we get a biection between partitions of n into even and odd number of uneual parts, and if n 3 ± we will have

7 7 Figure Exceptional Ferrers-diagrams exactly one exceptional partition without pair and it has parts Hence we proved the theorem Note that we can easily give the generating function of p e (n) p o (n) as follows: (p e (n) p o (n))x n ( x i ) Indeed, if we expand the right hand side then a partition of n into uneual parts will contribute ( ) to the coefficient of n Combining this obseervation with Euler s theorem we get the following corollary Corollary We have ( x n ) + n ( ) ( ) x (3 )/ + x (3 +)/ The corollary of this corollary is a very fast way to compute the seuence (p(n)) Corollary 3 For n we have ( ) p(n) ( ) (p + n 3 Proof Recall that p(n)x n Now if we multiply it with ( x n ) + n ( )) + p n 3 + x i ( ) ( ) x (3 )/ + x (3 +)/, and compare the coefficient of n we get that for n we have ( ) ( )) p(n) + ( ) (p n 3 + p n which is euivalent with the statement of the corollary

8 8 3 Partitions fitting into a rectangle and -binomial numbers Let (n) n, and (n)! (n) (n ) () Finally, let ( ) n (n)! ()!(n )! Note that and so ( ) n It turns out that ( ) n following recursion formula Theorem 3 We have ( ) n (n) n, (n ) ( n + ) ( ) ( ) is actually a polynomial in This follows from the ( ) n ( ) n + n Proof See Lemma 5 in R Stanley s Algebraic Combinatorics Let p i (m, n) be the number of partitions of i with largest part at most n and at most m parts So this is the number of partitions of i such that the Young diagram of the partition fits into the m n rectangle Theorem 3 We have (n ) + m m mn i0 p i (m, n) i Proof See Theorem in R Stanley s Algebraic Combinatorics It is easy to see that (n ) + m m ( ) n + m mn m / from the definition It means that the coefficients are symmetric: p i (m, n) p mn i (m, n) Actually, this can be seen uite easily by deleting a Youngdiagram from an m n rectangle and then rotating the obtained diagram by 80, thus obtaining a Young diagram of mn i Theorem 33 Let be a prime power and F be the field with elements Then the number N (n, ) of -dimensional subspaces of the n-dimensional vectorspace F n is ( ) n

9 Proof Let us compute the number of elements of the following sets in two different ways: S {(v,, v ) v,, v F n are linearly independent vectors} We can choose v in n different ways, because we can choose any vector different from 0 Then we can choose v in n ( n ) different ways as we can choose everything except cv Having chosen v,, v t we can choose v t from F n v,, v t so we can choose v t in n t t ( n t+ ) ways Hence S ( )/ ( n )( n ) ( n + ) On the other hand, we can first choose the -dimensional subspace V induced by v,, v in N (n, ) ways and then inside V we can choose v,, v in ways Hence Hence we get that ( )/ ( )( ) ( ) S N (n, ) ( )/ ( )( ) ( ) N (n, ) ( ) n 9