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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.

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    $\begingroup$ $G$ is $\mathbb{Z}/2$, $X$ is $\mathbb{R}^3$ with $G$ acting by negation, $Y$ is $\mathbb{R}$ with $G$ acting trivially, $f(x,y,z) = x^2+y^2-z^2$. $\endgroup$
    – David E Speyer
    Commented Jan 27, 2012 at 1:29
  • $\begingroup$ 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. $\endgroup$
    – Brett Parker
    Commented Jan 27, 2012 at 3:47

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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.

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  • $\begingroup$ 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 math.kyoto-u.ac.jp/~fukaya/foZ.pdf 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: people.maths.ox.ac.uk/~joyce/dmanifolds.html $\endgroup$
    – Brett Parker
    Commented Jan 30, 2012 at 0:13
  • $\begingroup$ 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 arxiv.org/pdf/1105.5124.pdf $\endgroup$
    – Brett Parker
    Commented Jan 31, 2012 at 2:47
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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.

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