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I don't see why an equivariant $G$-map $f: M \rightarrow M$, where $M$ is a compact homogeneous space, is necessarily a difeomorphism. Any idea?

Jotabeta
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2 Answers2

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I just wanted to complement freakish's (perfectly fine) answer with details about smoothness. To that end, recall that any homogeneous space $M$ is equivariantly diffeomorphic to a coset space $M\cong G/H$, where the $G$ action on $G/H$ is just left multiplication. I'll use the notation $N_G(H)$ for the normalizer of $H$ in $G$.

Lemma: Suppose $f:G/H\rightarrow G/H$ is equivariant. Then there is a $g\in N_G(H)$ for which $f(kH) = kgH$ for all $kH\in G/H$.

Proof: Choose $g\in G$ with $f(eH) = gH$ (with $e\in G$ the identity). Then, for any coset $kH$, we have $f(kH) = f(keH) = kf(eH) = kgH$.

To see $g\in N_G(H)$, note that for any $h\in H$, $f(kH) = f(khH)$, which implies that $kg = khg h'$ for some $h'\in H$. Canceling $k$ and rearranging gives $g^{-1} h^{-1} g \in H$, which then easily implies $g\in N_G(H)$. $\square$

Now, if $\pi:G\rightarrow G/H$ is the obvious projection and $R_g$ is right multiplication by $g$, it follows that $f\circ \pi = \pi \circ R_g$. The map $\pi\circ R_g$ is obviously smooth, so $f\circ \pi $. Because $\pi$ is a submersion, smoothness of $f\circ \pi$ implies smoothness of $f$. (See this MSE question for a proof of this last statement.)

Finally, to see $f$ is a diffeomorphism, simply note that if $f(kH) = kgH$, then $f^{-1}(kH) = kg^{-1} H$, so it's smooth for the same reason $f$ is.

Jason DeVito - on hiatus
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  • Great answer. Just to fill one detail in case someone wonders where compactness matters: otherwise $M$ need not be equivariantly diffeomorphic to $G/H$. – freakish Oct 13 '21 at 06:00
  • @freakish: I typically never think about non-compact Lie groups. Do you have an example in mind where compactness is important? – Jason DeVito - on hiatus Oct 13 '21 at 17:43
  • Consider discrete $\mathbb{Z}$ acting on $S^1\subseteq\mathbb{C}$ via $(n,z)\mapsto e^{2\pi i\theta n}\cdot z$ where $\theta$ is irrational, i.e. $\mathbb{Z}$ acts on $S^1$ via irrational rotation. Note that the orbit of say $1$ is an infinite countable dense subset of $S^1$, and so not discrete, unlike any quotient of $\mathbb{Z}$. So these aren't even homeomorphic. However the orbit is not a manifold. Now that I think about it I'm not sure if there is a counterexample when dealing with homogenous manifolds. – freakish Oct 14 '21 at 07:17
  • Yeah, when dealing with homogenous manifolds this seems to hold for any Lie group: https://math.stackexchange.com/questions/4187728/exact-requirements-for-stabilizer-orbit-theorem-for-lie-groups – freakish Oct 14 '21 at 07:27
  • @freakish: Thanks for the clarification! – Jason DeVito - on hiatus Oct 14 '21 at 11:27
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Partial answer:

So $f(gx)=gf(x)$ for any $g\in G$, $x\in M$. Since $M$ is homogenous this means that $f$ is uniquely determined by a single value on a single element $x_0\in M$, say $y_0=f(x_0)$. Therefore we can construct the inverse $h:M\to M$ by $h(y_0)=x_0$ which uniquely extends to entire $M$ via $h(gy_0):=gh(y_0)$.

And so $f$ is a continuous bijection. Which means it is a homeomorphism since $M$ is compact.

Now you also ask about diffeomorphism so I assume that $M$ is additionally a smooth manifold. Typically smooth homeomorphisms need not be diffeomorphisms. The $x\mapsto x^3$ map is a smooth map that is not a diffeomorphism (the derivative at $0$ is $0$). I think we can turn that example into compact case by considering $[-1,1]$ interval and glueing ends into $S^1$. But this example does not include the action of $G$ on $M$, so I might actually be wrong here.

freakish
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