How can we find the degrees of the invariants for the action of $SL(V)$ on $\wedge^4V$ , $dimV=8$ by the model in the Lie algebra $E\_7$.



degrees of the invariants for the action of $SL(V)$ on $\wedge^4V$ 
You don't really need Vinberg theory to find the degrees of the invariants here. This representation of $SL\_8=SL(V)$ can be constructed by considering a maximal rank involution of an algebraic group of type $E\_7$, which has fixed points isomorphic to $SL(V)/\{\pm I\}$ and such that the $(-1)$ eigenspace in the Lie algebra is exactly $\Lambda^4(V)$.


Well-known results about representations associated to symmetric spaces, as found for example in Kostant-Rallis' 1971 paper in the American Journal of Mathematics, then tell you that the invariants are given by restricting the invariants for the adjoint representation of the full $E\_7$ group, and that these restrictions are non-zero. So the degrees are exactly the degrees of the invariants for the adjoint representation of $E\_7$.


If you want to know how to see $SL\_8/\{\pm I\}$ inside the group of type $E\_7$, just take the subgroup whose Lie algebra is spanned by a Cartan subalgebra and all root spaces which have an even coefficient of $\alpha\_2$ (in Bourbaki numbering); the representation $\Lambda^4(V)$ is just the span of all the root-spaces which have an odd coefficient of $\alpha\_2$.


