I have the following question: Let $R \subset \Lambda^{\*}\mathbb{R}^{2n}$ be the sub-ring of forms which are preserved by $SU(n)$. How can one show that this subring is generated by $\Omega\_{0}$ and $\omega\_0$ where $\Omega\_{0}=dz^{1}\wedge ... \wedge dz^{n}$ and $\omega\_{0}=\frac{i}{2}\sum\_{i=1}^{n}dz^{i}\wedge d\overline{z}^{i}$?


monica



$SU(n)$-invariant subring of $\Lambda^{*}\mathbb{R}^{2n}$
Actually, as you have written it, the ring $R$ is generated by $\omega\_0$ and the real and imaginary parts of $\Omega\_0$, since $R$ consists of real-valued $1$-forms. You should probably consider, instead the ring $R^\mathbb{C} = \mathbb{C}\otimes R$ of complex-valued exterior forms on $\mathbb{R}^{2n}=\mathbb{C}^n$. Then the proof goes in two stages:


First, consider the forms that are invariant under the diagonal maximal torus in $\mathrm{SU}(n)$, which acts as $g\cdot dz^i = e^{\lambda\_i} dz^i$ and $g\cdot d\bar z^i = e^{-\lambda\_i} d\bar z^i$ where $\lambda\_i$ are purely imaginary and satisfy $\lambda\_1+\cdots+\lambda\_n=0$. It is easy to see that the only complex-valued forms that are invariant under this torus action are the ones generated by $\omega\_i = dz^i\wedge d\bar z^i$ and $\Omega\_0$ and $\overline{\Omega\_0}$. In particular, this shows that $R$ lies in the ring generated by the $(1,1)$-forms and $\Omega\_0$ and $\overline{\Omega\_0}$. Note that $\Omega\_0$ (and $\overline{\Omega\_0}$, too) wedged with any $(1,1)$-form must vanish, so it's clear that, outside of $\Omega\_0$ and $\overline{\Omega\_0}$, the ring $R$ must lie in the ring that is the sum of the $(p,p)$-forms. 


Now, each $(p,p)$-form $\Psi$ can be written uniquely in the form
$$
\Psi = \Psi\_p + \omega\_0\wedge\Psi\_{p-1} + {\omega\_0}^2\wedge\Psi\_{p-2} + \cdots + {\omega\_0}^p\wedge\Psi\_0
$$
where each $\Psi\_j$ is an $\omega\_0$-*primitive* $(j,j)$-form, and these primitive subspaces are $\mathrm{SU}(n)$-irreducible. (This follows from a root space representation.) In particular, if $\Psi$ is invariant under $\mathrm{SU}(n)$, then $\Psi = {\omega\_0}^p\wedge \Psi\_0$, so $\Psi$ is a complex multiple of ${\omega\_0}^p$.


