# Quotient map is a submersion

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 $$\{(x,x)\}$$.

• 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 $\mathbb R$? May 26, 2021 at 13:14
• Isn't the quotient $\mathbb{R} / G$ rather the half-line $[0, \infty)$? (My definition of smooth manifolds is without boundary.) May 26, 2021 at 13:30
• Oops, you are both right of course. I edit.
– abx
May 26, 2021 at 13:51
• @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. 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 $$V/G$$.