Exercise 2.2.14

We define the weight of σ 1 a 1 σ n a n to be a 1 + 2 a 2 + + n a n .

(a)
Prove that σ 1 a 1 σ n a n is homogeneous and that its weight is the same as its total degree when considered as a polynomial in x 1 , , x n .
(b)
Let f = F ( x 1 , , x n ] be symmetric and homogeneous of total degree d . Show that f is a linear combination of products σ 1 a 1 σ n a n of weight d .

Answers

Proof.

(a)
By Ex. 2.2.13, each σ k being homogeneous of degree k , the product σ 1 a 1 σ n a n is homogeneous. As deg ( σ k ) = k , deg ( σ 1 a 1 σ n a n ) = a 1 + 2 a 2 + + n a n is equal to the weight of σ 1 a 1 σ n a n .
(b)
Since f is symmetric, f is a linear combination of products σ 1 a 1 σ n a n . These products being homogeneous of degree a 1 + 2 a 2 + + n a n , and f being homogeneous, by Ex 2.2.13(b), each term of this sum has degree d .

Conclusion : f is a linear combination of products σ 1 a 1 σ n a n of weight d .

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2022-07-19 00:00
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