Fundamental theorem of invariant theory for $SO(m,n)$. Let $O(n)$ and $SO(n)$ denote the orthogonal and special orthogonal groups respectively on $\mathbb{R}^n$. In classical invariant theory, the first fundamental theorem of invariant theory states the following:
Theorem: The ring of invariant polynomials for $O(n)$ is generated by the metric. Likewise, the ring of invariant polynomials for $SO(n)$ is generated by the metric and the determinant.
I am interested in the case where we have a metric of indefinite signature. In particular, I would like to know whether the above theorem holds with $O(n)$ and $SO(n)$ replaced with $O(n,m)$ and $SO(n,m)$. 
I have only been able to find proofs of the positive-definite case, and it seems to be folk-lore that the indefinite case holds as well, but I would like to see a proof or a reference to a proof. Perhaps its a simple corollary of the positive definite case, but I am not knowledgeable enough to see it. Any help would be appreciated.
 A: Consider $SO(n)$ and $SO(n-m,m)$. It will suffice to show that the ring of invariants for $SO(n-m,m)$ is generated by the metric and the determinant if we can show each graded part of the invariant ring has the same dimension as the corresponding graded part of that for $SO(n)$. 
This can be accomplished by noticing that $SO(n)$ and $SO(n-m,m)$ have natural corresponding complex Lie groups, which we can call $SO(n,\mathbb{C})$ and $SO(n-m,m,\mathbb{C})$ which act on $V \otimes \mathbb{C} = \mathbb{C}^n$. Let $A_G$ denote the graded $k$-algebra of invariants for the real or complex lie group $G$ on its natural acting space, where $k = \mathbb{R},\mathbb{C}$ depending on $G$. It is clear that $\mathbb{C} \otimes A_{SO(n)} \subset A_{SO(n,\mathbb{C})}$, and in fact one can prove that equality holds. Similarly, $\mathbb{C}\otimes A_{SO(n-m,m)} = A_{SO(n-m,m,\mathbb{C})}$. However, the groups $SO(n-m,m,\mathbb{C})$ and $SO(n,\mathbb{C})$ are isomorphic, as are their corresponding representations. This follows as, for instance, the quadratic form $x_1^2+\cdots+x_{n-m}^2 - x_{n-m+1}^2 - \cdots x_n^2 = x_1^2 +\cdots + x_{n-m}^2 + (ix_{n-m+1})^2 + \cdots +(ix_{n})^2$, where $i=\sqrt{-1}$. Hence $A_{SO(n-m,m,\mathbb{C})} \cong A_{SO(n,\mathbb{C})}$. Hence $\mathbb{C}\otimes A_{SO(n-m,m)} \cong \mathbb{C}\otimes A_{SO(n)}$, and the dimension of the graded parts of $A_{SO(n-m,m)}$ and $A_{SO(n)}$ are equal, as required.
The same complexification argument works to prove the result for the orthogonal groups $O(n)$ and $O(n-m,m)$.
