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

Suppose that a finite group acts on two vector spaces $X$ and $Y$, and that $f:X\longrightarrow Y$ is an equivariant polynomial map which is homogeneous of degree $n$, and that there does not exist any nonzero equivariant polynomial map $X\longrightarrow Y$ with smaller degree.

What does $f^{-1}(0)$ look like?

Can you think of an example in which $f^{-1}(0)$ is not a finite union of linear subspaces?

For what groups must $f^{-1}(0)$ be a finite union of linear subspaces? In what situations must $f^{-1}(0)$ be nice? (For example, maybe for some groups, if the action on $Y$ is irreducible, $f^{-1}(0)$ a is finite union of submanifolds).

I am interested in the case where $X$ and $Y$ are real vector spaces, but if you know the answer for complex vector spaces, I would like to know that too.

The slightly more difficult question I am really interested in is the following: Given a generic equivariant smooth map $f:X\longrightarrow Y$, what does $f^{-1}(0)$ look like? If it is a (locally) finite union of submanifolds, then I would be surprised and happy.

share|cite|improve this question
$G$ is $\mathbb{Z}/2$, $X$ is $\mathbb{R}^3$ with $G$ acting by negation, $Y$ is $\mathbb{R}$ with $G$ acting trivially, $f(x,y,z) = x^2+y^2-z^2$. – David Speyer Jan 27 '12 at 1:29
This is not quite an example, because whenever Y has the trivial action, I would consider constant maps to be the only maps of minimal degree. – Brett Parker Jan 27 '12 at 3:47

The answer to the question in my title is no. An example is $\mathbb Z/5$ acting on $\mathbb C^3$ by multiplying each coordinate by $e^{2\pi i/5}$, and acting on $\mathbb C$ by multiplication by $e^{4\pi i/5}$. In this case there are no constant or linear equivariant maps $\mathbb C^3\longrightarrow \mathbb C$, but any homogeneous quadratic polynomial is equivariant.

share|cite|improve this answer
The question of `what does $f^{-1}(0)$ look like for a generic equivariant smooth map is studied a bit by Fukaya and Ono in Also, Joyce answers when a generic equivariant map is transverse to $0$ in the usual sense in Remark 11.59 of his book on derived orbifolds. Posted on the web here: – Brett Parker Jan 30 '12 at 0:13
Also, Fukaya, Oh, Ohta and Ono give a construction of a triangulation of $f^{-1}(0)$ for a type of piecewise smooth equivariant $f$ in – Brett Parker Jan 31 '12 at 2:47

You may want to have a look at the paper I wrote with M. Helmer and J.S.W. Lamb

On the zero set of G-equivariant maps

in the Mathematical Proceedings of the Cambridge Philosophical Society, 2009.

share|cite|improve this answer
Thanks Luciano, I'll take a look at it. – Brett Parker Apr 2 '13 at 3:53

Your Answer


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.