Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $V$ and $W$ be two complex vector spaces with an action of a finite group $G$. The $G$-equivariant polynomial maps from $V$ to $W$ are finitely generated as a module over the ring of $G$-invariant polynomials on $V$. In other words, there exist $G$-equivariant polynomial maps $p_1,\dotsc,p_n$ so that any $G$-equivariant polynomial map may be written as $$q_1p_1+\dotsb +q_np_n$$ where the $q_i$ are $G$-invariant polynomials on $V$.

Is it true that the $G$-equivariant holomorphic maps $V\longrightarrow W$ are finitely generated as a module over the ring of $G$-invariant holomorphic functions on $V$, and that the generators may be taken as $G$-equivariant polynomial maps? In other words, may we also write any $G$-equivariant holomorphic map as $$f_1p_1+\dotsb +f_np_n$$ where the $f_i$ are now $G$-invariant holomorphic functions?

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

I believe the answer is yes for $C^\infty$ maps and actions of compact (not necessarily finite) Lie groups. I think it is due to Poénaru and can be found in his book Singularités $C^\infty$ en présence de symétrie Lecture Notes in Mathematics, Vol. 510.

See also Lemma 6.6.1 in Dynamics and symmetry by Michael Field. (ICP Advanced Texts in Mathematics, 3. Imperial College Press, London, 2007. xiv+478 pp. ISBN: 978-1-86094-828-2)

(edit to reply to Brett's comment): Poénaru's theorem is not holomorphic. However, I believe it should not be hard to mimic its proof to extract the holomorphic version. I should note that I am not much of an expert on this area of mathematics and I know it more or less as a collection of black boxes. My impression, however, is that in going from polynomial versions the results (which is classical invariant theory) to $C^\infty$ version the main difficulty is in dealing with smooth invariant functions that vanish to infinite order. Going from polynomials to power series is not hard. And holomorphic maps from $V$ to $W$ are power series, aren't they?

Note also that in your example there is a big difference between complex $\mathbb Z/3$ invariant polynomials on $\mathbb C$ and real invariant polynomials on $\mathbb C$: $\mathbb C [\mathbb C]^{\mathbb Z/3}$ is generated by $z^3$ while $\mathbb R[\mathbb C]^{\mathbb Z/3}$ is generated by $Re(z^3), Im (z^3)$ and $|z|^2$.

share|improve this answer
    
Thanks Eugene - I assume you mean that $C^\infty$ equivariant maps are finitely generated as a module over $C^\infty$ equivariant functions. On the other hand, I am pretty sure that in general the generators can't be complex polynomial maps. The counter example is as follows: $V$ and $W$ are $\mathbb C$ with an action of $\mathbb Z_3$ by multiplication by $e^{2\pi i/3}$ and $e^{4\pi i/3}$ respectively. $z\mapsto \bar z$ is not equal to a sum of smooth functions times equivariant complex polynomial maps, because those are generated by the map $z\mapsto z^2$. –  Brett Parker Aug 3 '12 at 0:23
    
Brett, I will edit my answer rather than writing a comment, since the math won't fit in the comment –  Eugene Lerman Aug 3 '12 at 14:15
    
I emailed Mike Field, who pointed me to the paper `Lifting smooth isotopies of orbit spaces', Publ. I.H.E.S. (51) (1980), 37-135 by GW Schwarz. Proposition 6.8 from there gives that the G-equivariant holomorphic maps are generated by the G-equivariant polynomial maps. –  Brett Parker Aug 6 '12 at 4:01
add comment

Your Answer

 
discard

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.