q-Binomial Coefficients and the q-Binomial Theorem Yan Sun and Pete McNeely Math 333 December 1, 2012

1

1

The q-Binomial Numbers

Let n, e be non-negative integers. We begin our study of q-binomial numbers by recalling that the number of ways travel from (0,0) to (n, e) using only north steps (0,1) and east steps (1,0) is

ton+e n+e equal to e = n . This is because there are n + e total steps to take, and we are allowed to decide where to place the n north steps (leaving the remaining steps to be e east steps). We call these paths lattice paths
and it is easy to see for n = e = 2 that the number of lattice paths from 2+2 (0,0) to (2, 2) is 2 = 6.

Figure 1: The six lattice paths from (0,0) to (2,2) Now suppose we were to place unit squares above and to the left of these lattice paths. What results is all the different ways to fit Ferrers diagrams with largest part at most e and with at most n parts into an n × e box. In other words, we generate all of the integer partitions with largest part at most e and with at most n parts. Figure 2 illustrates the n = e = 2 case.

∅

1

1+1

2

2+1

2+2

Figure 2: The six Ferrers diagrams with n = e = 2 This leads us to our definition of a q-binomial number.   Definition 1.1. For non-negative integers n,m, the q-binomial number, n+m m , is the generating function (in the variable q) for the number of integer partitions with at most m parts, largest part at most n. More formally,   X n+m = p(t| ≤ m parts, each ≤ n)q t (1) m t≥0

Notice that when q = 1, we get the following corollary. Corollary 1.2. For all n, m ≥ 0 we have X t≥0

  n+m p(t| ≤ m parts, each ≤ n) = . m

(2)

Many other interesting concepts arise when studying the q-binomial numbers. We list a few: Theorem 1.3. For all n, m ≥ 0 we have     n+m n+m = m n

2

(3)

  Proof. Take an integer partition generated by n+m and represent it with a Ferrers diagram. This m is a Ferrers diagram with at most m parts, largest part at most n. Take the diagram’s conjugation.   The result is a Ferrers with at most n parts, largest part at most m, which is generated by n+m n . Inversely, take the diagram’s conjugation and conjugate it again. We get the original Ferrers diagram again. This notion of symmetry between q-binomial numbers illustrates identities to those

similar found when working with binomial coefficients. For instance, we know that n0 = nn . In fact, this identity transfers to the q-analog of the binomial coefficients, which leads us to our next corollary. Corollary 1.4. For all n ≥ 0 we have

hni 0

=

hni

(4)

n

Our first proof of Corollary 1.4. Let m = 0. The result immediately follows using Theorem 1.3. Our second proof of Corollary 1.4. hni X hni X = p(t| ≤ 0 parts, each ≤ n)q t = q 0 = 1 = q 0 = p(t| ≤ n parts, each ≤ 0)q t = . n 0 t≥0

t≥0

Seeing as though there are similar identities between the binomial coefficients, and the qbinomial numbers, we ask ourselves if there is a similar Pascal’s identity. This leads us to our next theorem. Theorem 1.5. For all n, m ≥ 0 we have       n+m n+m−1 m n+m−1 = +q (5) m−1 m m   either has exactly m parts, or less than m parts. Proof. An integer partition generated by n+m m The partitions with less than m parts are Ferrers diagrams that can fit into a n × (m − 1) box; h i n+m−1 therefore, we get the term m−1 . The partitions with exactly m parts look like Ferrers diagrams whose leftmost column has m m parts, and whose remaining diagram fits into  n+m−1  a (n − 1) × m box; therefore, we get the term q for the leftmost column, and the term . The result follows. m n

m

n

7→

empty →

Figure 3: Partitions with < m parts.

3

m−1

n

n−1

7→

m

+

m

non-empty → Figure 4: Partitions with exactly m parts.

The above theorem leads us to the following corollary. Corollary 1.6. For all n, m ≥ 0 we have       n+m−1 n+m n n+m−1 + =q m−1 m m

(6)

Proof. This result follows from the conjugation of Ferrers diagrams and a similar argument found in Theorem 1.5. A recurrence relation tells us a lot of information about these q-binomial numbers, but it would be nice to have an explicit formula for nk . We now have the tools that allow us find such a formula. Theorem 1.7. For n, k ≥ 1 we have hni k

=

(1 − q n )(1 − q n−1 )(1 − q n−2 ) · · · (1 − q n−k+1 ) (1 − q k )(1 − q k−1 )(1 − q k−2 ) · · · (1 − q)

(7)

Proof. We prove our theorem by induction on k. Base Case: k = 1 : hni n − 1 + 1 X = = p(t| ≤ 1 part, each ≤ n − 1)q t = 1 1 t≥0

(1 − q n ) . (1 − q) i   h   k n−1 . And by induction, we know Inductive Step: k ≥ 2 : By Theorem 1.5, nk = n−1 k−1 + q k (1 + q + q 2 + q 3 + · · · + q n−1 ) =

4

h

n−1 k−1

i

=

hni k

(1−q n−1 )(1−q n−2 )···(1−q n−k+1 ) (1−q k−1 )(1−q k−2 )···(1−q)

and

 n−1  k

=

(1−q n−1 )(1−q n−2 )···(1−q n−k ) . (1−q k )(1−q k−1 )···(1−q)

   n−1 k n−1 +q k k−1 n−1 )(1 − q n−2 ) · · · (1 − q n−k ) n−1 n−2 (1 − q )(1 − q ) · · · (1 − q n−k+1 ) k (1 − q + q (1 − q k−1 )(1 − q k−2 ) · · · (1 − q) (1 − q k )(1 − q k−1 ) · · · (1 − q) n−1 n−2 n−k+1 k n−1 )(1 − q n−2 ) · · · (1 − q n−k ) (1 − q )(1 − q ) · · · (1 − q )(1 − q ) k (1 − q + q (1 − q k )(1 − q k−1 )(1 − q k−2 ) · · · (1 − q) (1 − q k )(1 − q k−1 ) · · · (1 − q) (1 − q n−1 )(1 − q n−2 )(1 − q n−3 ) · · · (1 − q n−k+1 )(1 − q k + q k (1 − q n−k )) (1 − q k )(1 − q k−1 ) · · · (1 − q) (1 − q n−1 )(1 − q n−2 )(1 − q n−3 ) · · · (1 − q n−k+1 )(1 − q k + q k − q n−k+k ) (1 − q k )(1 − q k−1 ) · · · (1 − q) (1 − q n−1 )(1 − q n−2 )(1 − q n−3 ) · · · (1 − q n−k+1 )(1 − q n ) (1 − q k )(1 − q k−1 ) · · · (1 − q) (1 − q n )(1 − q n−1 )(1 − q n−2 )(1 − q n−3 ) · · · (1 − q n−k+1 ) . (1 − q k )(1 − q k−1 ) · · · (1 − q)

 = = = = = = =

 

Notice how similar nk is to nk . Whereas nk = n (1−qn )(1−qn−1 )(1−qn−2 )(1−qn−3 )···(1−qn−k+1 ) . k = (1−q k )(1−q k−1 )···(1−q)

2

Therefore,

n(n−1)(n−2)···(n−k+1) k(k−1)(k−2)···(2)(1) ,

The q-Binomial Theorem

When studying the binomial coefficients, we proved a powerful theorem called the Binomial Theorem. It is powerful because it allows us to easily find many more binomial coefficient identities. Here we are going to find the q-analog of the Binomial Theorem, aptly named the q-Binomial Theorem. Theorem 2.1. (The q-Binomial Theorem) For all n ≥ 1 we have n Y

j

(1 + xq ) =

j=1

n X k=0

q

k(k+1)/2

  n k x . k

(8)

We prove Theorem 2.1 in two ways. The first is a proof by induction using the recurrence relation for the q-binomial numbers (Theorem 1.5). The second is a combinatorial proof using generating functions. First Proof of Theorem 2.1. The result is easy to check for n = 1 and n = 2, so suppose n ≥ 3.

5

We argue by induction on n. So, assume n Y

j

Qn−1

j=1 (1

n

n−1 Y

n

n−1 X

(1 + xq ) = (1 + xq )

j=1

+ xq j ) =

Pn−1 k=0

n−1 k

xk .

(1 + xq j )

j=1

= (1 + xq )

q

k(k+1)/2

k=0

=

n−1 X

q

k(k+1)/2



k=0



 n−1 k x k

   n−1 n − 1 k X k(k+1)/2 n n − 1 k+1 x + x . q q k k k=0

Reindex the second sum by substituting k for k + 1, then notice use the reccurrence relation or Pascal’s identity to get n Y

q k(k+1)/2

n X

(k−1)k 2

=

k(k+1) 2

− k, and finally

   n n − 1 k X k(k+1)/2 n−k n − 1 k (1 + xq ) = q x + q q x k k−1 j=1 k=0 k=1     n X n−1 n−k n − 1 k(k+1)/2 xk +q = q k−1 k k=0   n X k(k+1)/2 n xk = q k j

k(k+1)/2



k=0

Second Proof of Theorem 2.1. Consider the generating function for partitions with distinct parts each ≤ n. Recall the exponent of q contributes to the number of boxes, and the exponent of x contributes to the number of parts in the partition. For example, xk q j indicates the partitions of j with exactly k parts. For each j with 1 ≤ j ≤ n, a part of size j can be either in the partition or not in the partition, therefore the generating function for it is 1 + xq j . By the multiplication principle for the ordinary generating functions, the generating function for the partitions with distinct parts each ≤ n is Qn j j=1 (1 + xq ). On the other hand, let ak (q) be the generating function for partitions with exactly k distinct parts each ≤ n. Notice k ≤ n, otherwise by pigeonhole principle the parts cannot Pnbe distinct. Therefore, the generating function for the partitions with distinct parts each ≤ n is k=0 ak (q)xk . The generating functions for the same series must be the same, so we have n Y

(1 + xq j ) =

j=1

n X

ak (q)xk .

k=0

Now we only need to find ak (q). Given a partition with k distinct parts each ≤ n, we remove 1 from the smallest part, remove 2 from the second smallest part, and in general remove i from the i-th smallest part. Then we will get an arbitrary partition with at most k parts each ≤ n − k: In this way, each partition with k distinct parts each ≤ n can be constructed uniquely by first setting up the triangular set of boxes in the left, then choosing a partition with at most k parts each ≤ n − k. The left part has 1 + 2 + · · · + k = k(k+1) boxes, so the generating function for this 2 k(k+1)/2 part is q . By definition, the generating function for partitions with at most k parts each

6

n

n−k

k

7→

k

+

k

Figure 5: The process of removing i from the i-th smallest part. ≤ n − k is have

n−k+k k

=

n

Now it follows that

k

. By the multiplication principle for ordinary generating functions, we   n . ak (q) = q k(k+1)/2 k n Y

j

(1 + xq ) =

j=1

n X

q

k(k+1)/2

k=0

  n k x . k

With the q-Binomial Theorem, we can conveniently derive many amazing results, and the proof for them is extremely easy. Of course each of the following theorems has a combinatorial proof as well, but here we only present the algebraic proof to emphasize the convenience of using the q-Binomial Theorem. Corollary 2.2. (The Binomial Theorem) For all n ≥ 1 we have n   X n k x . (1 + x) = k n

(9)

k=0

Proof. Set q = 1 in (8).
Theorem gave us the necessary tools to discover the remarkable identity that PnThen
Binomial 2 2n = k=0 k n . Not surprisingly, we consider the possibility of a q-analog of such an identity. Proposition 2.3. For all n ≥ 0 we have n X k=0

 2   n 2n q = . n k k2

(10)

Q Proof. Consider the coefficient polynomial of xn in the product  2n (1 + xq i ). By the q-Binomial i=1  Q 2n Theorem (Theorem 2.1) this coefficient polynomial is q n(n+1)/2 n . But we also have 2n i=1 (1 + Q Q xq i ) = nj=1 (1 + xq j ) nk=1 (1 + xq k+n ), and by applying Theorem 2.1 twice on the right we find   !     2n n n Y X X n n (1 + xq i ) =  q j(j+1)/2 xj  q k(k+1)/2 (xq n )k . j k i=1

j=0

k=0

7

A term in this product will be a multiple of xn and contribute to the coefficient of xn if and only if k + j = n, which happens if and only if j = n − k. Therefore the coefficient polynomial of xn in this product is n X

q (n−k)(n−k+1)/2



k=0

     n n n n k X n(n+1)/2 k2 n 2 q k(k+1)/2 (q ) = . q q n−k k k k=0

Cancelling q n(n+1)/2 on both sides, we have the equation we desired n X k=0

 2   n 2n q = . k n k2

Corollary 2.4. For all n ≥ 1 we have n Y

j

(1 + q ) =

j=1

n X

q

  n . k

(11)

  n (−1)k . k

(12)

k(k+1)/2

k=0

Proof. Set x = 1 in (8). Corollary 2.5. For all n ≥ 1 we have n Y

(1 − q j ) =

j=1

n X

q k(k+1)/2

k=0

Proof. Set x = −1 in (8).

3

The q-Binomial Series

The q-Binomial Series is very similar to the q-Binomial Theorem, since the combinatorial proof for the former is almost the same as for the latter. Theorem 3.1. (The q-Binomial Series) For all n ≥ 1 we have n Y j=1

  ∞ X 1 k n+k−1 = q xk . 1 − xq j k

(13)

k=0

For simplicity, we only prove Theorem 3.1 combinatorially, while the induction proof also works. Readers are encouraged to prove Theorem 3.1 by induction. Proof. Consider the generating function for partitions with each part ≤ n. For each j with 1 ≤ j ≤ 1 n, the generating function for it is 1 + xq j + x2 q 2j + · · · = 1−xq j . By the multiplication principle for the ordinary generating functions, the generating function for the partitions with each part ≤ n is Qn 1 . j=1 1−xq j

8

On the other hand, let bk (q) be the generating function for partitions with exactly k parts each P ≤ n. Then the generating function for the partitions with each part ≤ n is ∞ b (q)xk . The k k=0 generating functions for the same series must be the same, so we have n Y j=1

∞

X 1 = bk (q)xk . 1 − xq j k=0

Now we only need to find bk (q). Given a partition with k parts each ≤ n, we remove 1 from each part. Then we will get a partition with at most k parts each ≤ n − 1: n

n−1

7→

k

+

k

Figure 6: The process of removing 1 from each part. In this way, each partition with k parts each ≤ n can be constructed uniquely by first setting up a column of boxes in the left, then choosing a partition with at most k parts each ≤ n − 1. The left part has k boxes, so the generating function for this part is q k . By definition, the generating n+k−1 function for partitions with at most k parts each ≤ n − 1 is . By the multiplication principle k for ordinary generating functions, we have   k n+k−1 bk (q) = q . k Now it follows that

n Y j=1

  ∞ X 1 k n+k−1 = q xk . 1 − xq j k k=0

Similarly, with the q-Binomial Series, we can derive many amazing results immediately. Corollary 3.2. (The Binomial Series) For all n ≥ 1 we have  ∞  X 1 n+k−1 k = x . (1 − x)n k

(14)

k=0

Proof. Set q = 1 in (13). Corollary 3.3. For all n ≥ 1 we have n Y j=1

  ∞ X 1 k n+k−1 q . = 1 − qj k k=0

Proof. Set x = 1 in (13). 9

(15)

Corollary 3.4. For all n ≥ 1 we have n Y j=1

  ∞ X 1 k n+k−1 = q (−1)k . 1 + qj k

(16)

k=0

Proof. Set x = −1 in (13).

4

An Application to the q, t-Catalan Numbers

This section serves to illustrate that the q-binomial numbers have applications in other topics in combinatorics. For instance, suppose we are given a bounce path with bounce k from (0,0) to (n, n). The q-binomial numbers allows us to generate how many Catalan paths from (0,0) to (n, n) satisfy this bounce. Theorem 4.1. Let a bounce path from (n, n) to (0,0) with bounce k be given. Realize that k = a0 + a1 + a2 + · · · + am where aj , (0 ≤ j ≤ m) is the x value of the point (aj , aj ) that is on both y = x and our bounce path (where the bounce path touches y=x). Also, a0 = 0, and am = n. Define A(q) to be the generating function for Catalan paths from (0,0) to (n, n) that satisfy the given bounce path with bounce k. ( 1 if m = 0 or 1 Qm−2 ai+2 −ai −1 (17) A(q) = if m ≥ 2 i=0 ai+1 −ai Proof. If m = 0, we have n = am = a0 = 0. Therefore, we have the empty Catalan path. Clearly, there is only one way to construct it. The generating function of this construction is q 0 = 1. If m = 1, we have n = am = a1 and a0 = 0. Therefore, our bounce path only touches y = x at the point (n, n) and the point (0, 0). There is only one way to construct a Catalan path that satisfies this bounce; n steps north, and n steps east. The generating function of this construction is q 0 = 1. If m ≥ 2 Suppose we are given a bounce path with bounce k like that depicted in Figure 7. am . . . am−1 a3 a2 a1 a0 Figure 7: The bounce path with bounce k A bounce path with bounce k tells us that we need to have a1 north (0,1) steps from (0,0) to (0, a1 ) and (am − am−1 ) east (1,0) steps from (am−1 , am ) to (am , am ). Additionally, in order to get bounce k, there needs to be a north step from (aj , aj+1 − 1) to (aj , aj+1 ) for each aj where 0 ≤ j ≤ m. We can think of these required north steps as rectangles with no length and east steps as rectangles with no width. Therefore, the generating function for these required north steps  in our Catalan path is 1+0 = 1 and the generating function for these required east steps in our 0 0+1 Catalan path is 1 = 1. 10

am . . . am−1 a3 a2 a1 a0 Figure 8: The required structure of Catalan paths that satisfy this bounce path We are given some freedom when filling in the remainder of the Catalan path. Specifically, we need to fill in the (ai+1 −ai )×(ai+2 −ai+1 −1) box for each i where 0 ≤ i ≤ m−2. Conveniently, the q-binomial number tells us exactly how to fill in these boxes! The ways of creating lattice  different ai+2 −a  i+2 −ai+1 −1 i −1 paths within an (ai+1 −ai )×(ai+2 −ai+1 −1) box is generated by ai+1 −aia+a = . −a a −a i+1 i i+1 i Finally, by the multiplication rule of ordinary generating functions, we take the product of all the 1’s that generate the necessary north and east steps and multiply that by the product over all possible values of i that generate the boxes we need to fill in. The result follows.

5

Conclusion

We have now explored the world of the q-analogs to the binomial coefficient, theorem, and series. It is quite fascinating that conclusions we find using binomial numbers have a way of generalizing with these analogs. The readers are encouraged to explore further q-analog identities of their own.

11

References [1] Andrews, George E., and Kimmo Eriksson. Integer Partitions. Cambridge, UK: Cambridge UP, 2004. N. pag. Sections 7.2, 7.3, 7.4. [2] Bressoud, David M. Proofs and Confirmations: The Story of the Alternating Sign Matrix Conjecture. Cambridge: Cambridge UP, 1999. Section 3.1.

12