Proof of Theorem 10.3, Part 1 (VERSION of JULY 31, 2014, 11:05am (my time) )

By DZ

Theorem 10.3 Let λ ⊢ n and τ ⊢ n + 1.

1. Assume first that τ ≠ (μ, 1), namely τ = (τ_{1}, …, τ_{k}) where τ_{k} ≥ 2, then

∑_{ν∈λ^{+}} χ^{ν}(τ) = 0.

2. Let λ, μ ⊢ n then

∑_{ν∈λ^{+}} χ^{ν}(μ, 1) = (N_{1}(μ) + 1) · χ^{λ}(μ),

where N_{1}(μ) is the number of ones in μ.

Remark: A better way to state the theorem is:

∑_{ν∈λ^{+}} χ^{ν}(τ) = N_{1}(τ) · χ^{λ}(μ) ,

where μ = (τ_{1}, …, τ_{k−1}, τ_{k} − 1) (and remove the last 0 if τ_{k} = 1). The proof is by induction on the number ones in τ, and part 1 is the base case.

Important Definition Let F (x_{1}, …, x_{n}) be any Laurent polynomial. Then the constant term of F, denoted by CT_{x_{1},...,x_{n}}F, or simply CT F when there is no ambiguity, is the coefficient of x_{1}^{0} ⋯ x_{n}^{0} in F.

Examples:

CT_{x_{1},x_{2}}(3 + x_{1} + x_{2}^{−1}) = 3 ,
CT_{x_{1}}(3 + x_{1} + x_{2}^{−1}) = 3 + x_{2}^{−1} .

Proof of Part 1: One way to define the characters χ^{λ}(μ) is via the formula mentioned on the very last line of p. 114 of Maconad’s “Symmetric Functions and Hall Polynomials” (2nd ed.). Let’s rephrase it in terms of constant terms.

Important Fact: Let

Δ(x_{1}, …, x_{n}) = ∏_{1≤i<j≤n}(x_{i} − x_{j}) ,

p_{r}(x_{1}, …, x_{n}) = ∑_{i=1}^{n} x_{i}^{r} ,

Let n be a positive, integer and λ and τ partitions of n. Write τ = (τ_{1}, …, τ_{k}) and if λ has m < n parts, define λ_{m+1} = … = λ_{n} = 0. We have:

χ^{λ}(τ) = CT_{x_{1},...,x_{n}} \frac{Δ(x_{1}, …, x_{n}) ∏_{i=1}^{k} p_{τ_{i}}(x_{1}, …, x_{n})}{∏_{i=1}^{n} x_{i}^{λ_{i}+n−i}} .

Now let’s prove Theorem 10.3, part 1. If |τ| = n, then |λ| = n − 1, of course.

Let’s consider a typical member of λ^{+}. It is obtained by changing one of the parts, λ_{i} to λ_{i} + 1. In order for this to be legal, we need λ_{i−1} > λ_{i}. Doing this gives almost the same expression, but with the constant-termand multiplied by x_{i}^{−1} (the denominator gets multiplied by x_{i}). If it is not legal, then λ_{i−1} = λ_{i}, and the denominator (after multiplying it by x_{i}) has the degrees of x_{i−1} and x_{i} being the same. Of course, transposing x_{i−1} and x_{i} does not change the constant term, but it makes the consant-termand the negative of the original (since Δ(x_{1}, …, x_{n}) is anti-symmetric), so the constant term equals to its negative and hence is 0 in this (illegal) case.

Hence, we have to prove that, if τ_{k} ≥ 2, then

CT_{x_{1},...,x_{n}} \frac{Δ(x_{1}, …, x_{n}) (∏_{i=1}^{k} p_{τ_{i}}(x_{1}, …, x_{n})) · ( ∑_{i=1}^{n} x_{i}^{−1})}{∏_{i=1}^{n} x_{i}^{λ_{i}+n−i}} = 0 .

But since ∑_{i=1}^{n} x_{i}^{−1} = e_{n−1}(x_{1}, …, x_{n}) · ∏_{i=1}^{n} x_{i}^{−1}, the left side equals

CT_{x_{1},...,x_{n}} \frac{Δ(x_{1}, …, x_{n}) (∏_{i=1}^{k} p_{τ_{i}}(x_{1}, …, x_{n})) · e_{n−1}(x_{1}, …, x_{n})}{∏_{i=1}^{n} x_{i}^{λ_{i}+n−i+1}} ,

and we have to prove that it equals 0. But since λ is a partition of n − 1, λ_{n} = 0, and we claim the much stronger result

CT_{x_{n}} \frac{Δ(x_{1}, …, x_{n}) (∏_{i=1}^{k} p_{τ_{i}}(x_{1}, …, x_{n})) · e_{n−1}(x_{1}, …, x_{n})}{∏_{i=1}^{n} x_{i}^{λ_{i}+n−i+1}} = 0 .

The crucial fact is that the parts of τ are all ≥ 2. Let Junk_{i}(x_{1}, …, x_{n−1}), (i = 0, 1, 2) denote various polynomials that only depend on x_{1}, …, x_{n−1}. We have

∏_{i=1}^{k} p_{τ_{i}}(x_{1}, …, x_{n}) = Junk_{0} + O(x_{n}^{2}) ,

and we also have:

Δ(x_{1}, …, x_{n}) = Junk_{1}·( ∏_{i=1}^{n−1}(x_{i} − x_{n}) ) = Junk_{1}·(( ∏_{i=1}^{n−1} x_{i} ) · x_{n}^{0} − e_{n−2}(x_{1}, …, x_{n−1})x_{n} + O(x_{n}^{2})) .

Also

e_{n−1}(x_{1}, …, x_{n}) = ( ∏_{i=1}^{n−1} x_{i} ) · x_{n}^{0} + e_{n−2}(x_{1}, …, x_{n−1}) · x_{n}^{1} .

Now

CT_{x_{n}} [\frac{Δ(x_{1}, …, x_{n}) (∏_{i=1}^{k} p_{τ_{i}}(x_{1}, …, x_{n})) · e_{n−1}(x_{1}, …, x_{n})}{∏_{i=1}^{n} x_{i}^{λ_{i}+n−i+1}}]

= Junk(x_{1}, …, x_{n−1}) · CT_{x_{n}}[(Junk_{0} + O(x_{n}^{2}))·

((( ∏_{i=1}^{n−1} x_{i} ) · x_{n}^{0} − e_{n−2}(x_{1}, …, x_{n−1})x_{n}^{1} + O(x_{n}^{2})) · (( ∏_{i=1}^{n−1} x_{i} ) · x_{n}^{0} + e_{n−2}(x_{1}, …, x_{n−1})x_{n}^{1}) · \frac{1}{x_{n}} )]

and, thanks to (a − b)(a + b) = a^{2} − b^{2}, this equals:

Junk(x_{1}, …, x_{n−1}) · CT_{x_{n}}[(Junk_{0} + O(x_{n}^{2}))·

((( ∏_{i=1}^{n−1} x_{i} )^{2} · x_{n}^{0} − (e_{n−2}(x_{1}, …, x_{n−1}))^{2}x_{n}^{2} + O(x_{n}^{2})) · \frac{1}{x_{n}} )] .

and this is 0. .