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Let $L=\mathfrak{sl}(2,\mathbb{F})$, $H$ a borel subalgebra, $\Delta=\{\alpha\}$ a base of the corresponding root system and $W$ the Weyl group.
Let $\lambda=\frac{1}{2}\alpha$ be the fundamental dominant weight.
I read, that $\lambda^{2}$ generates the algebra of $W$-invariant polynomial functions $\mathfrak{P}(H)^{W}$, but I don´t get, why this should be true.
I though of the following:
I already know that algebra of polynomial functions $\mathfrak{P}(H)$ is generated by the $\lambda^{k}$ $k\in\mathbb{Z}^{+}$ and that $W$ is spaned by the reflexion $\sigma_{\alpha}$, so it is enought to check, what happenes, when $\sigma_{\alpha}$ acts on the $\lambda^{k}$. Since $\sigma_{\alpha}$ sends $\alpha$ to $-\alpha$ it sends $\lambda$ to $-\lambda$, so this is not an element of $\mathfrak{P}(H)^{W}$.
But I do not know how to continue for $k\geq2$.
Thank you for helping me.

Idun E.
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    The Weyl group acts on the algebra of polynomial functions by algebra automorphisms. So because $(-\lambda)^2=\lambda^2$ it follows that $\lambda^2$ is invariant under $\sigma_\alpha$. – Jyrki Lahtonen Jun 12 '15 at 19:12
  • Ok I get that, so no $\lambda^{k}$ for k uneven is in $\mathfrak{P}(H)^{W}$, correct? But what about those for k even? – Idun E. Jun 12 '15 at 19:15
  • $$\sigma_\alpha(\lambda^{2t})=\sigma_\alpha(\lambda)^{2t}=(-\lambda)^{2t}=\cdots$$ – Jyrki Lahtonen Jun 12 '15 at 19:16
  • This means in general, that $\sigma_{\alpha}$ sends $\lambda^{k}$ to $(-1)^{k}\lambda^{k}$. So only those $\lambda^{k}$ for k even lie in $\mathfrak{P}(H)^{W}$. Doesn´t that mean, that $\mathfrak{P}(H)^{W}$ is generated by $\lambda^{k}$, k even and not only $\lambda^{2}$? – Idun E. Jun 12 '15 at 19:25
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    I would say that it is spanned by $\lambda^k$, $k$ even, (here spanned = generated as a vector space) and generated by $\lambda^2$ (here generated = generated as a subalgebra). – Jyrki Lahtonen Jun 12 '15 at 19:28

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