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

In Nagata's "Lectures on the 14th problem of Hilbert" I found a reference to Weyl's "Classical Groups". Nagata writes that Weyl gives a positive answer to the original problem

If $G\subseteq\operatorname{GL}_n(\mathbb{C})$ is acting on $\mathbb{C}[x_1,\dots,x_n]$ via $x\mapsto A\cdot x$, is $\mathbb{C}[x_1,\dots,x_n]^G$ a finitely generated $k$-algebra?

in the special case that $G$ is a semisimple connected Lie group.

Now I have no familiarity whatsoever with Lie groups, and I would just like to reference this result. But I can't seem to find it in Weyl's book in this form. I guess it is formulated in a different way and the above is a more modern formulation. It would be nice if anyone familiar with Weyl's book or Lie groups could help me with this and maybe tell me which theorem this is. Also, what "known" groups fall under the notion "connected semisimple Lie group"?

Thank you in advance and sorry for the vague formulation (again)!

share|cite|improve this question

migrated from Jul 4 '13 at 12:19

This question came from our site for people studying math at any level and professionals in related fields.

marked as duplicate by Willie Wong, Dan Petersen, Andrey Rekalo, Ben Webster Jul 4 '13 at 14:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

the complex simple groups you may know about are $SL_n({\mathbb C})$, $Sp_{2g}({\mathbb C})$, and $SO_n({\mathbb C})$. Apart from these (and finite covers and factors by finite central subgroups), there are only finitely many complex groups (the exceptional groups). Semi-simple groups are roughly, products of complex simple groups.

In Hermann Weyl's group, there is reference to the "unitarian trick". This means, in more "modern" language, that a complex semi-simple group $G$ contains a Zariski dense compact subgroup $K$ and hence $G$ invariants are the same as $K$ invariants.

The finite generation of the ring of $K$ invariants is proved using the Hilbert basis theorem (and the fact that $K$ invariants are obtained by integrating over $K$ (remember $K$ is compact). For details, you may see Hermann Weyl's book, where he does this for $SL_n({\mathbb C})$ (and $K=SU(n)$).

share|cite|improve this answer

Claudio Procesi, Lie groups, An Approach Through Invariants and Representations, p. 555, theorem 1. The compact simple Lie groups are explained in detail in Wikipedia

share|cite|improve this answer

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