MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $E_7$.

share|cite|improve this question
up vote 4 down vote accepted

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 $\alpha_2$.

share|cite|improve this answer

This was worked out explicitly by A. A. Katanova in her paper Explicit form of certain multivector invariants in Advances in Soviet Mathematics 8 (1992), pp. 87-93. According to her calculations, the degrees of the generators are 2, 6, 8, 10, 12, 14, and 18, and these freely generate the ring of invariants. I suggest that you consult this paper for details, which are too long to write out here.

NB: It is possible that this is also done in the earlier paper by L. V. Antonyan, Classification of $4$-vectors of $8$-dimensional space, Trudy Sem. Vektor. Tenzor. Anal. 20 (1981), pp. 144-161, but I don't have access to this paper and, in any case, I don't read Russian. It could be that this paper just deals with the normal forms and doesn't discuss the ring of invariants.

share|cite|improve this answer
Yes I am studying this paper, but how he computed such numbers 2, 6, 8, 10, 12, 14? – baba ab dad Jan 22 '13 at 14:37
infact he left this question without detail and for the most of the papers only copy /pasted this fact – baba ab dad Jan 22 '13 at 14:40
I don't remember all the details myself, but I remember that the author (a woman, by the way, so 'she') got the degrees by transforming the problem to a problem about invariants on Lie algebras and then using a general result of Vinberg (her advisor) that computes the degrees using a combinatorial method based on invariants of the Weyl group. You should check her paper for the reduction to the Lie algebra case and see whether or not that does the trick. I don't know any easier way to do this; it's a nontrivial problem. (BTW, you should have mentioned that you knew her paper in your question.) – Robert Bryant Jan 22 '13 at 14:58
Thanks for your comment and corrections, but I am looking for a clear solution for this question, – baba ab dad Jan 22 '13 at 21:39
The 7 numbers 2,6,8,10,12,14,18 are the degrees of polynomial invariants of the Weyl reflection group of $E_7$. These generate (freely) the full ring of polynomial invariants of the group in the reflection rep. These can be calculated many ways; look at invariant theory of reflection or Coxeter groups. See also – Y Macdisi Jan 23 '13 at 6:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.