Harish Chandra isomorphism:Invariant polynomial functions I am trying to read the complete proof of Harish Chandra Isomorphism theorem from the book of Humphreys.
Notations:
$L$ is a finite dimensional semisimple Lie algebra with Cartan subalgebra $H$. $G$ is the inner automorphism group of $L$ i.e. the subgroup of $\operatorname{Aut}(L)$ generated by $\exp(ad~x)$ with $x\in L$ and $x$ is nilpotent.
$\mathcal{W}$ is the Weyl group. $\mathcal{B}(L)$ and $\mathcal{B}(H)$ are the symmetric algebras of $L^*$ and $H^*$ respectively. Then $G$ and $\mathcal{W}$ have natural actions on these algebras. $\mathcal{B}(L)^G$ denotes the elements of $\mathcal{B}(L)$ fixed by every element of $G$ by the action and similarly for the other one.
$\Lambda$ is the subspace of $H^*$ consisting of integral weights.
To show $\theta:\mathcal{B}(L)^G\to \mathcal{B}(H)^{\mathcal{W}}$ is a surjective map there it is assumed (rather referred to an exercise) that pure powers $\lambda^k$ for $\lambda \in \Lambda$ span $\mathcal{B}(H)$. Humphreys called the process as the process of polarization. Since $\Lambda $ is closed under addition it suffices to show that any monomial in $\mathcal{B}(H)$ is a $\mathbb{C}$ linear combination of such powers of linear functions with variables from $\Lambda$.
I could do it for $2$ only.
Like $(ab)=\frac{(a+b)^2}{2}-\frac{a^2}{2}-\frac{b^2}{2}$.
I think by induction we should proceed but have no idea how to..
Thank you for your help in advance.
 A: Note that $\Lambda$ is an additive subgroup of $H^*$, containing all roots. Thus $ \Lambda $ spans $H^*$. Let $\{\lambda_1, \dots, \lambda_l \}$ be a basis for $H^*$. Then  $F[\lambda_1, \dots, \lambda_l]$ = $B(H^*)$, where $F$ is the base field of characteristic zero.
Denote $P^l_d$ by the subspace of $F[\lambda_1, \dots, \lambda_l]$ consisting of the  homogeneous elements of degree $d$. We can prove the following claim by the induction on $l$.
Claim. $P^l_d$ is spanned by $\{\eta^d : \eta \in P^l_1 \}$
proof. The case $l=1$ is trivial. Suppose $l \geq 2$ and let $d \geq 1$ be given. Fix $\lambda \in P^{l-1}_1$. It suffices to show that $\lambda^k \lambda_l^m \in \mbox{span}\{\eta^d  : \eta \in P^l_1\}$ for every nonnegative integers $k, m$ such that $k+m=d$. Consider a $F$-vector subspace $W$ generated by $\mathfrak{B}=\{\lambda^d,~\lambda^{d-1}\lambda_l, ~\dots~, ~ \lambda \lambda^{d-1}_l, ~\lambda^d_l\} $. By construction, $\dim W= d+1$. Now consider $d+1$ vectors $\lambda^d,(\lambda + \lambda_l)^d, \dots, (\lambda + d\lambda_l)^d$ of $W$. With respect to the basis $\mathfrak B$, the $(d+1) \times (d+1)$ matrix having $(\lambda+i\lambda_l)^d$ as $(i+1)$-th column can be written by $$ \begin{bmatrix} 1 & 1 & 1& \cdots  & 1 \\ \binom{d}{1}0 & \binom{d}{1}1 & \binom{d}{1}2& \cdots &\binom{d}{1}d \\ \binom{d}{2}0^2 & \binom{d}{2}1^2 &  \binom{d}{2}2^2&\cdots & \binom{d}{1}d^2 \\ \vdots & \vdots  & \ddots & \vdots \\ \binom{d}{d}0^d & \binom{d}{d}1^d & \binom{d}{d}2^d &\cdots & \binom{d}{d}d^d \end{bmatrix} $$
whose determinant is $\prod_{i=1}^d \binom{d}{i} \prod_{0 \leq i < j \leq d}(j-i) \neq 0$. (See the determinant of Vandermonde matrix.) In other words, $(\lambda + i\lambda_l)^d$'s are linearly independent, so forms a basis for $W$. Now $$\mathfrak B \subset W = \mbox{span} \{(\lambda + i\lambda_l)^d : 0 \leq i \leq d\} \subset \mbox{span}\{\eta^d : \eta \in P^l_1 \}$$
as desired. $\blacksquare$
Alternatively, we can prove the result by the Jacobi's bialternant formula; see my another answer. This approach does not use induction.
