Bourbaki

Exercise 15, § 6

Algebra, IV, p. 101

Let TT be a finite set with kk elements. We denote by DkD_k the set of elements (a1,…,ak)(a_1, \ldots, a_k) of Nk\mathbf{N}^k such that a1+…+ak=ka_1 + \ldots + a_k = k and a1≥a2≥…≥aka_1 \geq a_2 \geq \ldots \geq a_k. With every equivalence relation R in T we associate an element λ(R)\lambda(R) of DkD_k as follows: if U1,...,UlU_1, ..., U_l are the equivalence classes under R, numbered so that Card(Ui)≥...≥Card(Ul)\mathrm{Card}(U_i) \geq ... \geq \mathrm{Card}(U_l), put λ(R)=(α1,...,αk)\lambda(R) = (\alpha_1, ..., \alpha_k) with αi=Card(Ui)\alpha_i = \mathrm{Card}(U_i) if 1≤i≤l1 \leq i \leq l and αi=0\alpha_i = 0 if 1<i≤k1 < i \leq k. Let XnX_n (n∈Nn \in \mathbf{N}) be indeterminates; for each R∈H(T)R \in \mathscr{H}(T) put r(R)=∑f∏i∈IXf(i)r(R) = \sum_{f} \prod_{i \in I} X_{f(i)}, where the summation is extended to the set of mappings f:T→Nf : T \to \mathbf{N} such that R is the equivalence relation associated with ff. Put s(R)=∑S≠Rr(S)s(R) = \sum_{S \neq R} r(S).

a) Prove that r(R)=∏i=1kri.M(λ(R))r(R) = \prod_{i=1}^{k} r_i . M(\lambda(R)) if rir_i of the equivalence classes under R arc of cardinal i for 1≤i≤k1 \leq i \leq k. (See formula (11) of IV, p. 65 for the definition of M(α)M(\alpha).)

b) Prove the relation S(R)=∏i=1kpiriS(R) = \prod_{i=1}^{k} p_i^{r_i} where the rir_i are as in a) and pm=∑n∈NXnmp_m = \sum_{n \in \mathbf{N}} X_n^m for every integer m≥1m \geq 1.

c) Show that r(R)=∑S≠Rμ(R,S)s(S)r(R) = \sum_{S \neq R} \mu(R, S) s(S), where μ(R,S)\mu(R, S) is defined as in Exercise 14.

d) Hence deduce an expression of the sks_k as a polynomial in p1,...,pkp_1, ..., p_k with rational coefficients, where sk=∑i1<...<ikXi1...Xiks_k = \sum_{i_1 < ... < i_k} X_{i_1} ... X_{i_k}.

e) For a∈Dka \in D_k put S′(α)=∏i=1ksαiS'(\alpha) = \prod_{i=1}^{k} s_{\alpha_i}. Show that S′(α)=∑β∈Dkcαβ′.M(β)S'(\alpha) = \sum_{\beta \in D_k} c'_{\alpha \beta} . M(\beta), where the integer cαβ′c'_{\alpha \beta} is defined as follows: suppose that rir_i (resp. sis_i) of the components of aa (resp. β\beta) are equal to i (for 1≤1≤k1 \leq 1 \leq k). Then cαβ′c'_{\alpha \beta} is the product Nr,!...rk!s1!...sk!/k!Nr,!... r_k!s_1!... s_k!/k!, where N is the number of pairs of equivalence relations R, S on T such that λ(R)=a,λ(S)=β\lambda(R) = a, \lambda(S) = \beta and such that the intersection of an equivalence class under R with an equivalence class under S has at most one element. Deduce the symmetry relation cαβ′=cβα′c'_{\alpha \beta} = c'_{\beta \alpha}.

f) Show that there is an involution a↦α~a \mapsto \tilde{\alpha} in DkD_k, characterized by the fact that the relations i≤a,i \leq a, and j≤α~ij \leq \tilde{\alpha}_i are equivalent, for i, j in the range from 1 to k. Show that S′(α)=S(α~)S'(\alpha) = S(\tilde{\alpha}) and cαβ′=cα~β~c'_{\alpha \beta} = c_{\tilde{\alpha} \tilde{\beta}}. Hence obtain the relation cαβ=cα~β~c_{\alpha \beta} = c_{\tilde{\alpha} \tilde{\beta}} (with the notation of IV, p. 66).

Conventions (cf. page IV, 66).

A symbol such as pα⋅qβ…p^\alpha \cdot q^\beta \ldots, where p,q,α,β,…p, q, \alpha, \beta, \ldots are integers ≥1\geq 1, denotes an element of DkD_k where k=pα+qβ+…k = p\alpha + q\beta + \ldots having α\alpha components equal to pp, β\beta components equal to qq, etc. Thus 3⋅123 \cdot 1^2 denotes the element of D5D_5 equal to (3,1,1,0,0)(3, 1, 1, 0, 0), while 141^4 denotes the element (1,1,1,1)(1, 1, 1, 1) of DrD_r.

On the right of the row of index α\alpha of the matrix C the expression of S(α)S(\alpha) as monomial in s1,…,sks_1, \ldots, s_k is given. For example, for k=3k = 3 and α=2.1\alpha = 2 . 1 we have

S(2.1)=s1s2=M(2.1)+3M(13) S(2 . 1) = s_1 s_2 = M(2 . 1) + 3M(1^3)

We have used a corresponding convention for the columns of the matrix D. For example, for k=4k = 4 and α=22\alpha = 2^2 we have

M(22)=2s4−2s1s3+s22. M(2^2) = 2s_4 - 2s_1 s_3 + s_2^2 .

We recall that M(22)M(2^2) is the sum of all monomials of the form Xi2Xj2X_i^2 X_j^2 for 1≤i<j≤n1 \leq i < j \leq n in the ring Z[X1,…,Xn]\mathbf{Z}[X_1, \ldots, X_n].

An empty box indicates a zero element of the matrices C and D.

k = 2 2 1^2 2 1 2 S_1^2 1^2 1 S_2

2 1^2 2 1 -2 1^2 1 S_1^2 S_2

k = 3 3 2.1 1^3 3 1 3 6 S_1^3 2.1 1 3 S_1 S_2 1^3 1 S_3

3 2.1 1^3 3 1 -3 3 2.1 1 -3 1^3 1 S_1^3 S_1 S_2 S_3

k = 4 4 3.1 2^2 2.1^2 1^4 4 1 4 6 12 24 S_1^4 3.1 1 2 5 12 S_1^2 S_2 2^2 1 2 6 S_2^2 2.1^2 1 4 S_1 S_3 1^4 1 S_4

Matrix C

4 3.1 2^2 2.1^2 1^4 4 1 -4 2 4 -4 3.1 1 -2 -1 4 2^2 1 -2 2 2.1^2 1 -4 1^4 1

S_1^4 S_1^2 S_2 S_2^2 S_1 S_3 S_4 Matrix D

TABLE

k = 5 5 4.1 3.2 3.1^2 2^2.1 2.1^3 1^5 5 1 5 10 20 30 60 120 S_1^5 4.1 1 3 7 12 27 60 S_1^3 S_2 3.2 1 2 5 12 30 S_1 S_2^2 3.1^2 1 2 7 20 S_1^2 S_3 2^2.1 1 3 10 S_2 S_3 2.1^3 1 5 S_1 S_4 1^5 1 S_5

Matrix C

5 4.1 3.2 3.1^2 2^2.1 2.1^3 1^5 5 1 -5 5 5 -5 -5 5 4.1 1 -3 -1 5 1 -5 3.2 1 -2 -1 5 -5 3.1^2 1 -2 -1 5 2^2.1 1 -3 5 2.1^3 1 -5 1^5 1

S_1^5 S_1^3 S_2 S_1 S_2^2 S_1^2 S_3 S_2 S_3 S_1 S_4 S_5 Matrix D

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.