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Let $V$ be a finite dimensional euclidean space and let $G\subset O(V)$ be a finite (or compact) group, let $\mathbb{R}[V]$ be the algebra of polynomial functions on $V$. If $E\subset \mathcal{C}^\infty(V)$ is a sub algebra, we denote by $E^G$ the sub-algebra of $G$-invariant function of $E$.

A theorem of Schwarz says:

Theorem : Let $(P_1,\dotsc,P_k)$ be a system of generators of $\mathbb{R}[V]^G$ as an algebra over $\mathbb{R}$, then $\mathcal{C}^\infty(V)^G$ is "smoothly generated" by $(P_1,\dotsc,P_k)$ in the sense that the map:

$$\begin{align}\mathcal{C}^\infty(\mathbb{R}^k)&\to \mathcal{C}^\infty(V)^G\\\\ g&\mapsto g(P_1,\dots,P_k) \end{align}$$ is surjective (and continuous in the right topologies).

There is also another proof by Bierstone in case $G$ is finite (although the proof only works at the level of germs of smooth functions if I understand correctly).

My question relates to the extension of this kind of results to subalgebras of smooth functions.

Question

Let $E\subset \mathcal{C}^\infty(V)$ be such that $(E\cap \mathbb{R}[V])^G$ is generated as an algebra over $\mathbb{R}$ by $(Q_1,\dots,Q_k)$, under what conditions on $E$ and $G$ is $E^G$ "smoothly generated" by $(Q_1,\dotsc,Q_k)$ ?

I'm currently trying to read through the proofs of Schwarz and Bierstone but have yet to understand if and how they can be adapted to this setting. The case of a finite group $G$ is already very interesting for what I have in mind. If this helps, one can also assume that $E$ is actually an ideal.

An obvious restriction should be that $E$ and $E^G$ contain enough polynomials, and probably density of $\mathbb{R}[V]\cap E$ in $E$ at least for the $\mathcal{C}_\text{loc}^0$ topology should be required.

Additional remark :

I'm realising that there is a perhaps more general question lurking around which has nothing to do with group actions: let $E$ be a subalgebra of $\mathcal{C}^\infty(V)$ such that $E\cap\mathbb{R}[V]$ is finitely generated, under what condition on $E$ do the generators of $E\cap\mathbb{R}[V]$ smoothly generate $E$ ? Schwarz theorem shows this is true for $E=\mathcal{C}^\infty(V)^G$ where $G$ is a compact Lie group acting on $V$ (there are actually counterexamples when $G$ is non compact).

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  • $\begingroup$ I'm a little confused. If $f\in E^G$ it is both smooth and invariant, so by Schwartz/Bierstone it must be of the form $f = g(P_1,\ldots,P_k)$, which is already what you are after. What am I missing? $\endgroup$
    – Igor Khavkine
    Commented Jun 4 at 8:14
  • $\begingroup$ My notation may be a bit confusing, I changed them, the point is that I want to express a smooth function in E^G in terms of polynomials in E^G. $\endgroup$
    – Thomas Richard
    Commented Jun 4 at 14:19
  • $\begingroup$ @ThomasRichard could you please provide at least some of these counterexamples when $G$ is non compact? Do they include, for instance, some pseudo-orthogonal group? $\endgroup$
    – Pedro Lauridsen Ribeiro
    Commented Jun 4 at 17:12
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    $\begingroup$ I added a link to the paper on the non compact counter example. $\endgroup$
    – Thomas Richard
    Commented Jun 4 at 17:20
  • $\begingroup$ @PedroLauridsenRibeiro Specifically about pseudo-orthogonal groups, see the discussion at this older question. The upshot is that some distinct real orbits, which can be connected through a single complex orbit, cannot be separated by invariant polynomials. $\endgroup$
    – Igor Khavkine
    Commented Jun 4 at 17:45

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