Let $G$ be a Lie group acting smoothly on a smooth manifold $M$. Suppose that the orbit space $M / G$ is a topological manifold, and is endowed with a smooth structure such that:

  1. the quotient map $\pi : M \to M / G$ is smooth, and
  2. the pullback map $\pi^* : C^\infty(M / G) \to C^\infty(M)^G$ is an isomorphism, where $C^\infty(M)^G$ is the space of $G$-invariant smooth functions on $M$.

Is $\pi$ necessarily a submersion?


Edit: This answer is wrong, I was misled by the complex case. The map $\pi$ is not surjective, its image is a semi-algebraic set. In fact it would seem that $M/G$ is never a manifold.

No. Take $M=\mathbb{R}^2$, $G=\mathbb{Z}/2$ acting by swapping the coordinates. The quotient is isomorphic to $\mathbb{R}^2$, with $\pi :\mathbb{R}^2\rightarrow \mathbb{R}^2$ given by $\pi (x,y)=(x+y,xy)$. By a general result of G. Schwarz, $\pi^*:C^{\infty}(\mathbb{R}^2)\rightarrow C^{\infty}(\mathbb{R}^2)^G$ is an isomorphism. But $\pi $ is not a submersion along the diagonal $\{(x,x)\} $.

  • 2
    $\begingroup$ I'm confused because your quotient map $\pi:\mathbb R\to\mathbb R:x\mapsto x^2$ isn't surjective. Did you perhaps mean $\mathbb C$ instead of $\mathbb R$? $\endgroup$ May 26 '21 at 13:14
  • 2
    $\begingroup$ Isn't the quotient $\mathbb{R} / G$ rather the half-line $[0, \infty)$? (My definition of smooth manifolds is without boundary.) $\endgroup$
    – Oscar
    May 26 '21 at 13:30
  • $\begingroup$ Oops, you are both right of course. I edit. $\endgroup$
    – abx
    May 26 '21 at 13:51
  • 2
    $\begingroup$ @abx: Actually, it is possible for $M/G$ to be a manifold. For example, if $M=\mathbb{C}$ and $G = \{\pm1\}$ acting by multiplication on $M$, then $M/G$ is homeomorphic to $\mathbb{C}$ and the map $\pi:M\to M/G$ given by $\pi(z) = z^2$ is smooth. However, this doesn't satisfy the OP's Condition 2. Another example is $M=\mathbb{C}^2$ and $G =\mathrm{U}(1)$, with $M/G$ homeomorphic to $\mathbb{R}^3$ via the smooth map $$\pi(z,w) = \bigl(\,|z|^2{-}|w|^2,\,2\mathrm{Re}(z\bar w),\,2\mathrm{Im}(z\bar w)\,\bigr),$$ (which also does not satisfy Condition 2). There are many other examples. $\endgroup$ May 29 '21 at 10:27

The answer is 'yes' when $G$ is compact. (More generally, if the $G$-stabilizer of $m\in M$ is a compact group, $K\subset G$, then $\pi'(m):T_mM\to T_[m](M/G)$ is surjective; see the remark below.) Here is an outline of the argument:

Note that all of the $G$-orbits in $M$ are closed in $M$. For $m\in M$, let $U\subset M/G$ be any open neighborhood of $\pi(m)\in M/G$, then $\pi^{-1}(U)\subset M$ is a $G$-invariant open neighborhood of the $G$-orbit of $M$, and we can replace $M$ by this open set without changing the problem (or any of the hypotheses). Thus, we can assume that $M/G\simeq \mathbb{R}^q$ for some $q>0$, and that $\pi = (x_1,\ldots, x_q)$ for smooth $G$-invariant functions $x_1,\ldots x_q$ on $M$ that vanish at $m\in M$.

Next, by the Slice Theorem, by shrinking to a smaller $G$-invariant neighborhood of $M$ if necessary and fixing a $G$-invariant metric on $M$, we can exponentiate the normal subspace $W\subset T_mM$ to the tangent space of the $G$-orbit of $m$. Then, letting $K\subset G$ be the closed subgroup that fixes $m$, we can reduce the question to the case of the compact group $K$ acting linearly on the $K$-representation $W$ and $\pi:W\to\mathbb{R}^q$.

Let $W = W_0\oplus V$ where $W_0$ is a trivial representation of $K$ and $V$ is a $K$-representation with no trivial summand. We can factor out the $W_0$ from the problem and reduce to the case of $K$ acting on $V$ without any trivial representation.

Now, I claim that this situation never satisfies Conditions 1 and 2, with $V/K$ being a topological manifold with a smooth structure such that $\pi=(x_1,\ldots,x_q):V\to V/K=\mathbb{R}^q$ is smooth.

The reason is as follows: The ring $R$ of $K$-invariant polynomials on $V$ is finitely generated (Hilbert) in degrees greater than or equal to $2$. Since $K$ is compact, there is at least one quadratic $K$-invariant polynomial, say, $Q$, that is positive definite. By Condition $2$, there is a smooth function $F:\mathbb{R}^q\to\mathbb{R}$ satisfying $F(0)=0$ but $F(x)>0$ for $x\not=0$ such that $$ Q = F(x_1,\ldots, x_q). $$ Let us write $F$ in the form $$ Q = F = c_1\,x_1+\cdots + c_q\,x_q + H(x_1,\ldots,x_q) $$ for constants $c_1,\ldots,c_q$ and where $H$ vanishes to order at least $2$ at $0\in\mathbb{R}^q$. Now, since there are no nonzero $K$-invariant linear functions on $V$, all of the $x_i$ vanish to order at least $2$ at $0\in V$, and hence $H(x_1,\ldots,x_q)$ must vanish to order at least $4$ at $0\in V$. Consequently, the $c_i$ cannot all be zero since $Q$ is a positive definite quadratic form on $V$. Thus, the smooth function $F:\mathbb{R}^q\to\mathbb{R}$ has a nonvanishing first derivative at $0\in\mathbb{R}^q$ while $0$ is a strict local minimum of $F$, which is impossible.

Thus, we must have $V=0$, i.e, $K$ must act trivially on $W$, and in this case, $M/K = W$, so Conditions 1 and 2 are fullfilled and, indeed, $\pi$ is a submersion.

Remark 1: The above argument basically relies on the Slice Theorem to reduce to the case of a smooth surjective map $\pi:T_m/T_m(G{\cdot} m)\to\mathbb{R}^q$ whose fibers are the orbits of the closed subgroup $K\subset G$ that fixes $m\in M$, and then uses the compactness of $K$ to show that Conditions $1$ and $2$ imply that the action of $K$ on the vector space $T_m/T_m(G{\cdot} m)$ must be trivial. Thus, the only place that compactness is needed is that $K$, the $G$-stabilizer of $m\in M$ should be compact.

Returning to the more general case, it's clear that, if the answer is going to be 'yes' for general $G$-actions, then the answer would at least have to be 'yes' for the simple case of $G$ represented faithfully on a vector space $V$ with the property that the space of orbits $V/G$ has the structure of a smooth manifold in such a way that the quotient map $\pi:V\to V/G$ is smooth and such that every $G$-invariant smooth function on $V$ is the $\pi$-pullback of a smooth function on $V/G$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.