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?


2 Answers 2


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)\} $.

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    $\begingroup$ 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, 2021 at 13:14
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    $\begingroup$ 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, 2021 at 13:30
  • $\begingroup$ Oops, you are both right of course. I edit. $\endgroup$
    – abx
    May 26, 2021 at 13:51
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    $\begingroup$ @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, 2021 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 $V/G$.


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