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Let $M,N$ be smooth manifolds, with a smooth $G$-action on them, by some Lie group $G$. Suppose also that $M$ has a finite number of orbits under $G$'s-action.

Let $f:M \to N$ be a smooth, equivariant, injective immersion.

Is $f(M)$ a weakly embedded submanifold of $N$?

Weakly embedded here means that for every manifold $Q$ and for every smooth map $h:Q \to N$, with $h(Q)\subset f(M)$,the associated map $h:Q\to f(M)$ is also smooth. In other words, it's always valid to restrict the range.

It is known that it suffices to prove that $h:Q\to f(M)$ is continuous. Note that in general $f(M)$ is only an immersed submanifold. In particular, it can have more open sets than those that come from the subspace topology.

Weakly embedded is a notion which is between "immersed" and "embedded". It is also known that every Lie subgroup is weakly embedded.

A famous example for a weakly embedded submanifold, which is not embedded is the dense curve on the torus. (In that case, there is also a Lie group action in the background, by $\mathbb{R}$).

Asaf Shachar
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  • At least when $M$ is connected, must any $G$ action on $M$ with finitely many orbits be transitive? This is certainly true if $G$ is compact. Also, is there any easy example of an immersed unmanifold which is not weakly embedded? – Jason DeVito - on hiatus Jun 12 '18 at 12:07
  • @JasonDeVito I think the picture is supposed to be of $\Bbb C^\times$ acting on $\Bbb{CP}^1$, which has three orbits. –  Jun 12 '18 at 12:10
  • @JasonDeVito No, the action does not need to be transitive. Take $M$ to be the space of all real $d \times d$ matrices of rank greater than $r$. This space is connected, since you can fall in the rank via a continuous path. This space has a natural $GL(\mathbb{R}^d) \times GL(\mathbb{R}^d)$ action, via left and right multiplication. It is known that every matrix of constant rank $r$ is equivalent to a diagonal matrix whose first $r$ diagonal elements are $1$'s and all the rest are zero. Hence, all matrices of rank $k$ form a single orbit... – Asaf Shachar Jun 12 '18 at 12:16
  • @JasonDeVito and of course you can have more than one, depending on $r$. – Asaf Shachar Jun 12 '18 at 12:16
  • @JasonDeVito An example of an immeresed submanifold which is not weakly embedded is the figure eight curve: http://mathworld.wolfram.com/FigureEight.html. (John Lee discusses this in his book on smooth manifolds, see example 5.28 there, and also theorem 5.29 and the discussion after that). – Asaf Shachar Jun 12 '18 at 12:22
  • @Mike and Asaf: Thanks! I am very happy to see all your counterexamples involve noncompact group actions - I thought I was losing my mind there for a second. I loaned my copy of Lee's book to a student a long time ago and no longer remember who has it :(. (I should probably pick up the second edition anyway). What does is mean for a map into a non-smooth manifold (like $f(M)$) to be smooth? (Sorry to be a bother, especially because I don't think my questioning will lead me to provide anything helpful on your question!) – Jason DeVito - on hiatus Jun 12 '18 at 14:18
  • @Jason Because $f$ is injective, at the set level it makes sense to lift a smooth map with image in $f(M)$ to a map to $M$ itself. The question is whether this lift is even continuous - if it is, it is smooth. In the figure eight case, identifying the figure 8 as the image of $\Bbb R$, you will see a problem trying to lift a circle parameterizing the figure 8 to a continuous map to R. –  Jun 12 '18 at 14:28
  • And you have company: I seem to lose my mind at least twice a day on MSE lately. :) –  Jun 12 '18 at 14:29
  • @Mike: I think I understand the question now! (Not that I can answer it). To be clear, $f$ is merely locally injective, so you have lots of choices in the lift - and the question is if any one of these can be made continuous or smooth or whatever. – Jason DeVito - on hiatus Jun 12 '18 at 14:34
  • @Jason You do want $f$ to be literally injective; there shouldn't be a choice in the lift. For the figure 8, the way we identify it as the injective image of R is to 'trace it out' so that near t = 0, you are tracing out the line y = x, and as t goes to infinity, you slowly slowly approach 0 along the line y = -x. –  Jun 12 '18 at 14:41
  • @MIke: Thanks for clarifying! I was imagining $f$ as wrapping $\mathbb{R}$ around the figure 8 many times. I now see that Asaf explicitly asked that $f$ be injective. Asaf: sorry to hijack your question! I'm done now. – Jason DeVito - on hiatus Jun 12 '18 at 14:46
  • One can find $f$ which satisfies everything except for being an immersion and an immersion $g : M \to N$ such that $g(M) = f(M)$ is a figure 8. This example suggests to me that your question is answerable in the affirmative, and I would try to prove it as follows. $M$ having finitely many orbits would imply that it has open orbits whose reunion is dense. Equivariance would force $f$ to be non-immersive on non-open orbits, so there's none. $f(M)$ is connected iff it's only 1 orbit, so it suffices to know that orbits are weakly embedded, claimed here – Jordan Payette Jun 13 '18 at 13:48
  • @JordanPayette Thanks. However, I don't understand something: Why should $f$ be non-immersive on non-open ortbis? (I don't see how this follows from equivariance). – Asaf Shachar Jun 13 '18 at 14:18
  • You are right, that is wrong and stupid: the identity map $M \to M$ is equivariant and a diffeomorphism. I had a bad intuition on how $df$ would map the 'transverse' directions to the reunion of lower-dimensional orbits: I thought it needs to vanish on those directions, but that is wrong. Truly sorry. – Jordan Payette Jun 13 '18 at 16:15

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Here is a counter-example with $G = (\mathbb{R}, +)$, $M = \mathbb{R}$ and $N = \mathbb{R}^2$. I shall specify the actions later on.

Consider on $N$ (equipped with cartesian coordinates $(u,v)$) the smooth function $h(u,v) = v(u^4 + v^4 + u^2 - v^2)$. Here is a figure of some of its level sets. We notice in particular that there is a figure 8 inside the level $[h=0]$; we shall denote this subset $8$. To be safe, add to $h$ a smooth nonnegative exhausting function which vanishes identically on a ball which contains $8$ (so as to leave unchanged the local picture), but which grows sufficiently fast to ensure that the resulting function (call it also $h$) only has compact level sets.

Consider now the (smooth) vector field $X$ on $N$ which is the symplectic gradient of $h$, that is the (usual) gradient vector field everywhere rotated (clockwise, say) by 90°. The field $X$ is everywhere tangent to the levels sets of $h$. The flow of $X$ turns out to be complete (since the levels sets are compact and $X$ smooth) and smooth; it determines a (smooth) $G$-action $\alpha_N : G \times N \to N : (g, (u,v)) \mapsto \Phi^{g}_{X}(u,v)$.

Let $f : M \to N$ be any smooth injective immersion whose image is $8$; it is not a weak embedding. We can pullback $X$ to $M$ along $f$ to get a (smooth and complete) vector field $Y$ which turns out to vanish in only finitely many points. Consider the $G$-action $\alpha_M : G \times M \to M : (g, x) \mapsto \Phi^g_Y(x)$.

By construction, $\alpha_M$ has only finitely many orbits and $f$ is $G$-equivariant since $X$ and $Y$ are $f$-related.

Jordan Payette
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  • Very nice! (and some more characters – Jason DeVito - on hiatus Jun 13 '18 at 17:18
  • Thanks! Your answer is indeed very nice. I have two questions: (1) How do you know $X$ has only a finite number of points where it vanishes? (Do you "see" it from the picture $8$? I guess that every time the direction of the $8$ curve "switches" there should be a zero point of its tangent vector field, but I am not sure). (2) The reason the pullback of $X$ has the same properties of $X$ (completeness and finite number of zero points), is that we can consider it as a pullback via a diffeomorphism, that is we think of $f:M \to f(M) =8$, right? – Asaf Shachar Jun 15 '18 at 05:29
  • @AsafShachar Thanks for your appreciation. (1) Because of the exhausting function, it's not so clear that $X$ vanishes only at finitely many points, but a direct computation shows that the (symplectic) gradient of the polynomial I used vanishes in only finitely many points, which is sufficient to guarantee that $Y$ has finitely many orbits (which is what matters). (2) Exactly. Remark: The overall multiplicative $v$ in the polynomial I used is an artefact of a previous attempt and may be forgotten. In that case the gradient vanishes only at the origin on $8$, hence $M$ has 3 orbits. – Jordan Payette Jun 15 '18 at 12:51
  • @AsafShachar Again about (1): By the implicit function theorem, a level set is locally a manifold around a point where the gradient does not vanish. In particular here, the origin was necessarily a critical point because of the self-intersection of $8$, but there might be other critical points. That's why a computation is in order (there is most probably some conceptual algebraic geometric arguments to justify why there are only finitely many zeroes for the gradient, but whatever...). The $X$ in the actual answer vanishes also everywhere along $v=0$, but this contributes 1 point on $8$. – Jordan Payette Jun 15 '18 at 13:01