# Hall-Littlewood polynomials and elementary symmetric functions-- Chapter III (2.8) in Macdonald's "Symmetric Functions and Hall Polynomials"

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 $$P_\lambda(x_1,...,x_n;t)=\frac{1}{v_\lambda(t)}\sum_{w\in S_n} w\left(x_1^{\lambda_1}...x_n^{\lambda_n}\prod_{i or equivalently $$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 $$r$$

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 $$f=P_{(1^r)}(t)$$
• Sorry for the late reply. I have put this question aside for a while and when I back to it today, I feel confused why I ask this. It seems $P_{(1^r)}(t)=e_r$ is always true (by the second formula in my question) no matter $n$ is less than $r$ or not. Maybe I just misunderstood the question then. Thank you all the same. Commented Jul 25, 2021 at 9:38