Solutions for the problems of problem set 3
1.
4
1
4 8
1 2
Partial Young-tableauxs (P1 , Q1 ) and (P2 , Q2 ).
3 8
4
1 8
3
4
1 2
3
1 2
3
4
Partial Young-tableauxs (P3 , Q3 ) and (P4 , Q4 ).
1 7
3 8
4
1 2
3 5
4
1 7 11
3 8
4
1 2 6
3 5
4
Partial Young-tableauxs (P5 , Q5 ) and (P6 , Q6 ).
1 5 11
3 7
4 8
1 2 6
3 5
4 7
1 5 6
3 7 11
4 8
1 2 6
3 5 8
4 7
Partial Young-tableauxs (P7 , Q7 ) and (P8 , Q8 ).
1 5 6 9
3 7 11
4 8
1 2 6 9
3 5 11
4 7
8
1 2 6 9
3 5 8
4 7
1 2 6 9
3 5 8
4 7
10
Partial Young-tableauxs (P9 , Q9 ) and (P10 , Q10 ).
1 5 6 9 12
3 5 11
4 7
8
1 2 6 9 11
3 5 8
4 7
10
1 2 6 9 10 1 2 6 9 11
3 5 11 12
3 5 8 12
4 7
4 7
8
10
Partial Young-tableauxs (P11 , Q11 ) and (P12 , Q12 ).
1
2
2. We prove the statement by induction on n. For n = 1 the statement is
trivial. Given a permutation π ∈ Sn+1 , then deleting n + 1 from it we get
a permutation π 0 ∈ Sn . Then I(π) = I(π 0 ) + (n + 1 − π(n + 1)). Clearly,
if we have a π 0 ∈ Sn we can insert n + 1 into n + 1 places thus obtaining a
permutation π with I(π) = I(π 0 ) + n + 1 − k where k = π(n + 1) runs between
0 and n. Then
n+1
X
X
X
X
0
I(π)
I(π 0 )
q
=
q
q n+1−k = (1 + q + · · · + q n )
q I(π )
π 0 ∈Sn
π∈Sn+1
k=1
π 0 ∈Sn
By induction this is equal to
(1 + q)(1 + q + q 2 ) . . . (1 + q + q 2 + · · · + q n−1 )(1 + q + · · · + q n ) = (n + 1)q !.
3. We prove the statement by induction on n. The claim is trivial for n = 1.
We will use that
n+1
n
n
n+1−k
=
+q
.
k
k q
k−1 q
q
Hence
!
n+1
X
k
n
+
1
n
n
q( )
q (2)
tk =
+ q n+1−k
tk =
k
k
k
−
1
q
q
q
k=0
k=0
n+1
X
k
2
n+1
X
k−1
k
n
n k
n
(
)
(
)
2
2
tk−1 =
t +q t
q
=
q
k
−
1
k
q
q
k=1
k=0
n
n
X k n
X k n
tk =
tk + q n t
q (2)
=
q (2)
k q
k q
k=0
k=0
n
X
= (1 + t)(1 + qt) . . . (1 + q n−1 t) + q n t(1 + t)(1 + qt) . . . (1 + q n−1 t) =
= (1 + t)(1 + qt) . . . (1 + q n−1 t)(1 + q n t).
n
n
k
k−1
We used that n+1
=
=
0
and
+
n
+
1
−
k
=
+ n.
−1 q
2
2
q
4. (a) Assume that given a partition of n with exactly t pieces of 1’s, then
after erasing the 1’s one by one, we get a partition of n − 1, n − 2, . . . , n − t.
And of course, given a partition of some m we can add n − m 1’s to get a
partition of n. Hence sn (1) = p(n − 1) + p(n − 2) + · · · + p(0).
(b) The analogous argument shows that sn (2) = p(n − 2) + p(n − 4) + · · · +
p(n−2bn/2c). And in general, sn (t) = p(n−t)+p(n−2t)+· · ·+p(n−tbn/tc).
3
(c) Note that
np(n) =
n
X
isn (i),
i=1
since both sides counts the sum of numbers in all partitions of n. Then
np(n) =
n
X
i (p(n − i) + p(n − 2i) + · · · + p(n − ibn/ic)) .
i=1
Now
n
X
i (p(n − i) + p(n − 2i) + · · · + p(n − ibn/ic)) =
i=1
n
X
σ(k)p(n − k)
k=1
since a term p(n − k) appears at every divisor of k. Hence
np(n) =
n
X
σ(k)p(n − k).
k=1
5. Let λ = (λ1 , . . . , λm ) be a partition of n, so λ1 ≥ . . . λm and
Let
k = max{i | λi ≥ i}.
Pm
i=1
λi = n.
In the Young–diagram of λ, you can see k as follows: it is the size of the largest
square you can put into the Young–diagram of λ. (This square is called Durfee–
square.) Then one can decompose the Young–diagram into three parts: (1) a
square of size k ×k, (2) a partition with at most k parts: λ02 = (λ1 −k, . . . , λk −
k), (3) a partition whose parts have size at most k: λ03 = (λk+1 , . . . , λm ). Hence
! ∞
!
∞
∞
∞
X
X
X
X
2
xk
p≤k (n2 )xn2
p(n)xn =
p≤k (n3 )xn3 =
n=0
n2 =0
k=0
=
∞
X
k=0
n3 =0
2
xk
.
((1 − x)(1 − x2 ) . . . (1 − xk ))2
On the other hand,
∞
X
n=0
n
p(n)x =
∞
Y
1
.
n
1
−
x
n=1
4
6. First solution: inductive proof. The statement is trivial for n = 0, 1.
Assume that the statement holds true till n. Then for n + 1 we have
!
n X
n
(x + y)n+1 = (x + y)n (x + y) =
xk y n−k (x + y) =
k
q
k=0
n
n
X
X
n
n
=
q n−k xk+1 y n−k +
xk y n+1−k =
k
k
q
q
k=0
k=0
!
n+1
n+1 X
X
n
n
n+1
n−k+1
k n+1−k
=
q
+
x y
=
xk y n+1−k .
k
−
1
k
k
q
q
q
k=0
k=0
In the last step we used the recursion formula for q-binomial coefficients.
Second solution: combinatorial proof. Let us consider a term w in the expansion of (x + y)n which contains k pieces of x’s and (n − k) pieces of y’s, the set
of these words will be denoted by Sk . Then w = q A(w) xk y n−k , where A(w) can
be computed as follows: for each y we count the number of x’s after y, and we
add these numbers together. If we reverse w and replace each x by a letter r
("right") and each y be a letter u ("up"). Then this new word will describe
a walk in the (n − k) × k from the bottom-left corner to the up-right corner.
Then A(w) counts the number of boxes above the walk, because every time
we step upward the number of boxes in the corresponding row is exactly the
number of right steps before this up step (so originally these were the number
of x’s after this y). Hence
n X
n X
X
X
q A(w) xk y n−k =
w=
(x + y)n =
k=0 w∈Sk
k=0 w∈Sk
=
n
X


X
q

k=0
since we learned that
|λ| 
k n−k
x y
k=0
λ<(n−k)×k
n
k q
=
P
7. We know that
λ<(n−k)×k
Q
λ
f = n!
=
n X
n
k
xk y n−k
q
q |λ| .
i<j ((λi − i) − (λj −
Qm
i=1 (λi + m − i)!
j))
.
To prove that f λ = n! det(A), we will use the following fact: let V = V (x1 , . . . xn )
be the m × m matrix such that Vij = xj−1
. This
i
Q is the so-called Vandermonde
determinant, and it is known that det V = i<j (xj − xi ). A simple observation is the following: if P0 , P1 , . . . , Pm−1 are monic polynomials such that
5
deg Pi = i − 1, then for theQ
matrix S = S(x1 , . . . xm ) for which Sij = Pj (xi ) we
still have det S = det V = i<j (xj − xi ). It is true for the following reason: if
P
Pi (x) = xi−1 + k<i−1 ak xk , then one can obtain S from V by simple column
operations: just add ak times the k + 1.th column to the i.the column.
If we multiply the i.th row of A with (λi + m − i)! then we get that
1
det B,
i=1 (λi + m − i)!
det(A) = Qm
where Bij = Pj (xi ), where xi = λi + m − i and P1 (x) = x(x − 1) . . . (x − m + 2),
P2 (x) = x(x − 1) . . . (x − m + 3), . . . , Pm (x) = 1. Hence the polynomials Pj
are monic polynomials such that deg Pj = m − j. Thus we can apply our
previous observation. Note that the order of columns is reversed compared to
the Vandermonde-matrix, so
Y
Y
det B = (−1)m(m−1)/2 (xj − xi ) =
(xi − xj ) =
i<j
=
i<j
Y
Y
((λi + m − i) − (λj + m − j)) =
((λi − i) − (λj − j)).
i<j
i<j
Hence
Q
n! det A = n!
i<j ((λi − i) − (λj −
Qm
i=1 (λi + m − i)!
j))
= f λ.
8. First solution: spectral graph theory. Recall that when we determined
the eigenvalues of the bipartite graph Gn induced by two neighboring levels,
Yn−1 ∪ Yn , of the Young-lattice then we developed a method finding the eigenvalues and eigenvectors of Gn+1 from Gn : if (xn , xn−1 ) is an eigenvector of
Gn belonging to λ, then ( √λ12 +1 Un xn , xn ) is an eigenvector of Gn+1 belong√
ing to λ2 + 1. In particulat, this means that the eigenvector corresponding
1
to the largest eigenvalue of Gn+1 is (U n e∅ , √n+1
U n+1 e∅ ) corresponding to the
√
eigenvalue n + 1. We have seen at the lecture that
X
U n e∅ =
f λ eλ
λ`n
and
U n+1 e∅ =
X
µ`n+1
f µ eµ .
6
1
Since (U n e∅ , √n+1
U n+1 e∅ ) corresponding to the eigenvalue
X
f µ = (n + 1)f λ .
√
n + 1 we have
µ>λ
µ`n+1
Second solution: counting walks. We learned that for a an operator w =
A1 A2 . . . Ak , where each Ai are U or D, the number α(w, λ) counting the
number of walks of type w starting at ∅ and ending at λ is equal to cw f λ ,
where
Y
cw =
(bi − ai ),
i∈Sw
where Sw is the set of indices i for which Ai = D, and bi − ai is the the level
where the walk is before we make the step Ai = D. In particular,
α(DU n+1 , λ) = (n + 1)f λ .
On the other hand,
α(DU n+1 , λ) =
X
α(U n+1 , µ) =
µ>λ
µ`n+1
X
f µ.
µ>λ
µ`n+1
Hence
X
f µ = (n + 1)f λ .
µ>λ
µ`n+1
Third solution: RSK algorithm. Let (F, k) be a pair where F is a standard
Young-tableaux and 1 ≤ k ≤ n + 1. Clearly, the number of such pairs is
exactly (n + 1)f λ . Let us increase each elements of F by one which are at least
k, and then let us insert k into F as we learned at the RSK-algorithm. Then
we get a standard Young-tableaux of shape µ
Pwhere µµ > λ. Of course, the
number of these standard Young-tableauxs is
µ>λ f . It is easy to see that
µ`n+1
this is a bijection: from the extra square we can carry out the RSK algorithm
backwards to learn which k was inserted.
9. We will show that
X
f (λ)f (λ) = (mn + 1)f m×n ,
λ
where f m×n is the number of standard Young-tableuaxs with shape of the m×n
rectangle. Indeed, the right hand side counts the standard Young-tableauxs
with shape of the m × n rectangle together with a number k between 0 and
7
nm; this gives us two standard Young-tableauxs, the standard Young-tableaux
with boxes containing 1, 2, . . . , k and the standard Young-tableaux with boxes
containing k + 1, k + 2, . . . , mn. The left hand side counts it as follows: fix
k, take a standard Young-tableaux with shape λ of k boxes, then choose a
standard Young-tableaux with shape λ and switch each number i to nm+1−i.
Putting together the two standard Young-tableauxs we get a standard Youngtableuax with shape of the m × n rectangle (why?).
10. First solution: combinatorial proof. Let q(n) be the number of partitions
whose Young–tableaux are self-conjugate. Then clearly p(n) ≡ q(n) (mod 2)
since every other partition can be paired with its conjugate.
The surprising fact is that q(n) = r(n) which can be seen as follows. Decompose the self-conjugate Young-tableauxs by central hooks: hooks belonging to
some square (i, i). The central hooks have odd lengths and they have different
sizes. This gives a partition of n into distinct odd numbers. And of course,
the process can be reversed: from a partition of n into distinct odd numbers
one can construct a self-conjugate partition.
Second solution: generating functions. We count everything mod 2. For instance,
∞
∞
∞
Y
X
Y
1
1
n
p(n)x =
=
n
1−x
1 + xn
n=1
n=0
n=1
since −1 = 1 in F2 . Note that
1
1−x
since every number can be uniquely written up as a sum of distinct 2-powers.
Similarly, for every (odd) m we have
∞
∞
Y
X
1
m2i
(1 + x ) =
xkm =
.
1 − xm
i=0
k=0
(1 + x)(1 + x2 )(1 + x4 )(1 + x8 ) · · · = 1 + x + x2 + · · · =
Hence
∞
Y
∞
Y Y
Y
Y
1
1
m
(1
−
x
)
=
(1 + xm ).
=
=
i
n
m2
1+x
1+x
n=1
m odd i=0
m odd
m odd
In the last step we again used that we count in F2 . Since
∞
Y
X
m
(1 + x ) =
r(n)xn
m odd
we are done.
n=0
8
11. Note that we only need to determine the eigenvalues of Gn,k for 2k ≤ n
since Gn,k is isomorphic to Gn,n−k−1 . For 2k ≤ n we will show by induction
that the eigenvalues of Gn,k are the following: let
v
u k
uX
aj = t (n − 2i),
i=j
n
then the eigenvalues are ±aj with multiplicity nj − j−1
for j = 0, 1, . . . , k,
n
n
n
and 0 with multiplicity k+1 − k . (Note that −1 = 0 which means that the
n
multiplicity of a0 = 1, and if n = 2k + 1 then k+1
− nk = 0, so 0 is actually
not an eigenvalue in this case.)
We will completely mimic the proof which we learned when we computed
the eigenvalues of the bipartite graphs induced by the Young-lattice. For
every set S, let us introduce a vector eS such that all these vectors are linearly
independent. Let RPk be the vectorspace induced by the vectors eS , where
|S| = k. Let us consider the operators Uk and Dk such that
X
Uk eS =
eT ,
S⊂T
|T |=k+1
and
Dk eS =
X
eT .
T ⊂S
|T |=k−1
We have seen at the lecture that
Dk+1 Uk − Uk−1 Dk = (n − 2k)I.
Note that if x = (xk , xk+1 ) is an eigenvector of the adjacency matrix of the
graph Gn,k corresponding to some eigenvalue λ, then it is equivalent with
saying that
Uk xk = λxk+1
and
Dk+1 xk+1 = λxk .
Dk xk . Then
This means that if we introduce the vector xk−1 = √ 2 1
λ −(n−2k)
p
Dk xk = λ2 − (n − 2k)xk−1 ,
and
1
1
Uk−1 xk = p
Uk−1 Dk xk = p
(Dk+1 Uk −(n−2k)I)xk =
λ2 − (n − 2k)
λ2 − (n − 2k)
p
1
=p
(λ2 − (n − 2k))xk = λ2 − (n − 2k)xk .
λ2 − (n − 2k)
9
This meanspthat (xk−1 , xk ) is an eigenvector of Gn,k−1 corresponding to the
eigenvalue λ2 − (n − 2k).
This construction works backwards: if (xk−1 , xk ) is an eigenvector of Gn,k−1
corresponding to the eigenvalue µ then (xk , √ 2 1
Uk xk ) is an eigenvector
µ +(n−2k)
Gn,k . Indeed, if xk+1
=√
1
Uk xk
µ2 +(n−2k)
Uk xk =
then
p
µ2 + (n − 2k)xk+1 ,
and
Dk+1 xk+1 = p
1
µ2
1
Dk+1 Uk xk = p
(Uk−1 Dk +(n−2k)I)xk =
2
+ (n − 2k)
µ + (n − 2k)
p
1
=p
(µ2 + (n − 2k))xk = µ2 + (n − 2k)xk .
µ2 + (n − 2k)
Note that this backward direction
means that if µ is an eigenvalue of Gn,k−1
p
2
with multiplicity mµ , then µ + (n − 2k) is an eigenvalue of Gn,k with multiplicity at least mµ . (As we will see, it is enough to consider only this backward
direction, but the other direction help us to see that actually there is a bijection
between the positive eigenvalues and eigenspaces of Gn,k and the non-negative
eigenvalues and eigenspaces of Gn,k−1 .)
Note that for any bipartite graph G with classes A and B for which |B| ≥
|A|, the rank of the adjacency matrix of G is at most 2|A|, consequently, the
multiplicity of 0 is at least |A| + |B| − 2|A| = |B| − |A|.
Now we are ready √to prove our claim. Gn,0 is a star on n + 1 vertices so
its eigenvalues
are ± n with multiplicity 1, and 0 with multiplicity n − 1 =
n
n
−
.
So
the
statement is true for k = 0. Assume that the statement holds
1
0
true for k − 1. Then the eigenvalues of Gn,k−1 are
v
u k−1
uX
±t (n − 2i)
i=j
n
n
with multiplicity j − j−1
for j = 0, 1, . . . , k − 1, and 0 with multiplicity
n
n
− k−1 . Then the eigenvalues of Gn,k are
k
v
u k
uX
±t (n − 2i)
i=j
10
n
with multiplicity at least nj − j−1
for j = 0, 1, . . . , k, and 0 with multiplicity
n
n
at least k+1 − k . But since
k X
n
n
n
n
n
n
−
+2
−
=
+
= |V (Gn,k )|,
k+1
k
j
j−1
k+1
k
j=0
we have found all eigenvalues and multiplicities:
v
u k
uX
±t (n − 2i)
i=j
with multiplicity exactly
n
exactly k+1
− nk .
n
j
−
n
j−1
for j = 0, 1, . . . , k, and 0 with multiplicity
12. Let m = α1 + · · · + αr , and let G = Sα1 × · · · × Sαr ≤ Sm embedded into Sm
such that Sα1 acts on {1, . . . , α1 }, Sα2 acts on {α1 +, . . . , α1 + α2 } etc. Then
the poset Sm /G is isomorphic to the poset of divisors of n (why?). Hence we
can apply all the theorems we learned: the poset is rank-symmetric, unimodal,
Sperner.
This problem has an elementary solution going as follows: if you have two
posets with one-one symmetric chain decompositions then you can create a
symmetric chain decomposition to the product poset. Note that all you have
to do is to create a symmetric chain decomposition of the products of two
chains and it is quite easy.