I'm confused about the proof of Chapter III (2.8), page 209 in Macdonald's book, see
proof of (2.8).
Here is the background. Let $\Lambda_n$ be the ring of symmetric polynomials in $r$ variables, i.e. $\Lambda_n=\mathbb{Z}[x_1,…,x_n]^{S_n}$, being the fixed point of the symmetric group $S_n$. Let $\lambda$ be a partition of length $l \leq n$ (we set $\lambda_i =0$ if $i \geq l$). In the picture, $P_\lambda(x,t)$ is Hall-Littlewood polynomial (which is an element of the inverse limit of $\Lambda_n[t]$, with $n^{th}$ coordinate $P_\lambda(x_1,…,x_n;t)$, and the projection of the inverse limit being letting extra variables to be zero). Where
$$P_\lambda(x_1,…,x_n;t)=\frac{1}{v_\lambda(t)}\sum_{w\in S_n} w\left(x_1^{\lambda_1}…x_n^{\lambda_n}\prod_{i<j}\frac{x_i-tx_j}{x_i-x_j}\right)$$
or equivalently
$$P_\lambda(x_1,…,x_n;t)=\sum_{w\in S_n/S_n^\lambda} w\left(x_1^{\lambda_1}…x_r^{\lambda_n}\prod_{\lambda_i>\lambda_j}\frac{x_i-tx_j}{x_i-x_j}\right)$$
And $e_r$ is the $r^{th}$ elementary symmetric function. And (2.5), which is used in the proof, asserts that the inverse limit described above is well-defined, i.e. $$P_\lambda(x_1,…,x_n,0;t)=P_\lambda(x_1,…,x_n;t)$$
My question is as follows. I understand (2.5) but
(1) I can't see why $P_{(1^r)}$ is uniquely determined by its image in $\Lambda_r[t]$
(2) Even if $P_{(1^r)}$ is uniquely determined by its image in $\Lambda_r[t]$, I don't see why this implies $P_{(1^r)}=e_r$ in the case of variables strictly more than $r$
Best Answer
Let $f\in \Lambda^r[t]$ be such that $f(x_1,...,x_r;t)=e_r(x_1,...,x_r)$. Write $f$ as $$ f=\sum_{|\mu|=r} a_\mu(t) e_\mu $$ then $$e_r(x_1,...,x_r)=f(x_1,...,x_r;t)=\sum a_\mu(t) e_\mu(x_1,...,x_r)$$, but since $e_\mu(x_1,...,x_r)$ are linearly independent for $|\mu|=r$, we have $a_\mu(t)=0$ for all $\mu \neq (1^r)$ and $a_{(1^r)}(t)=1$. So, $f=e_r$.
Now apply this to $f=P_{(1^r)}(t)$