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In Lie theory, one often asks about alternating forms on $\mathbb{R}^n$ which are invariant under some particular subgroup $G\subseteq GL_n(\mathbb{R})$, and there is always some algebra of invariant forms associated to $G$. For example, $SO(n)$ preserves the algebra generated by $1\in \Lambda^0(\mathbb{R}^n)$ and $*1\in \Lambda^n(\mathbb{R}^n)$ (and nothing else).

However, taking the algebra and calculating the group which leaves it invariant may yield a strictly larger group than the original one. Of course, another way of looking at it is that sometimes a subgroup of $G$ will still have the same algebra of invariant forms. I've been told, although I have no concrete examples, that these subgroups may not even be nested.

So, I'm wondering if there is some nice algebraic conditions that govern this correspondence. It would be nice to have some characterization along the lines of the basic facts in algebraic geometry that for any subset $T\subseteq A$, $Z(T)=Z((T))$, that for any ideal $a\subseteq A$, $I(Z(a))=\sqrt{a}$, etc.

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Not an answer, but you should check out Cvitanovic's book, available online at nbi.dk/GroupTheory --- he classifies simple Lie algebras over C in terms of the forms they preserve. –  Theo Johnson-Freyd Nov 18 '09 at 16:40

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I don't have a full answer yet. Some notation: let's write $\mathrm{Inv}(G)$ for the collection (algebra) of all invariant tensors for $G \subseteq GL_n$, and $\mathrm{Grp}(I)$ for the group of matrices that leave invariant some collection $I$ of tensors.

Then a necessary condition for $G = \mathrm{Grp}(\mathrm{Inv}(G))$ is for $G$ to be Zariski-closed in $GL_n$. (Recall that $GL_n$ is a codimension-one Zariski-closed subset of affine $(n^2 +1)$-space, where the first $n^2$ coordinates are the matrix coefficients, and the last one is the inverse determinant; $GL$ is cut out by the condition that the actual determinant times the last coefficient is unity.) Indeed, $\mathrm{Grp}(I)$ is Zariski-closed for any $I$, because it is the intersection of $\mathrm{Grp}(i)$ over all $i\in I$, and fixing a tensor is a Zariski-closed condition, because $GL_n \to \mathrm{End}(V)$ is polynomial for $V$ and tensor product of the $n$-dimensional representation and its dual.

So this rules out things like the irrational line in the torus (diagonal two-by-two matrices with eigenvalues $\exp(x)$ and $\exp(\pi x)$ as $x$ ranges over the field, and $\pi$ is your favorite irrational number).

I think that a sufficient condition is for $G$ to be compact. This is because if you know all the tensor invariants, then you know the full subcategory of representations that are tensor-generated by the defining representations, and in fact all the subrepresentations of these, and if $G$ is compact then this category is equivalent to the full category of representations and knows the group by Tannakian arguments. But this is much too strong --- $SL_n(\mathbb{R})$ is not compact, for example.

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Thanks! You lose me a little in the last paragraph though, is there an easy way of rephrasing what you're saying without (much) reference to category theory? –  Aaron Mazel-Gee Nov 24 '09 at 7:06
    
All I'm saying in the last paragraph is that for compact (real) groups, if you know all their finite-dimensional representations then you know the group, and that the invariant tensors give enough data. To make this precise requires categories and functors. –  Theo Johnson-Freyd Nov 24 '09 at 17:28

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