Let $M$ be a smooth manifold on which a Lie group $G$ acts properly and freely. Suppose there exists a set $D_M \subset M$ which is a smooth submanifold and which contains one and only one representative from each orbit of $M$. I am trying to see whether from a given chart on $M$ which is adapted to the group action, we can construct a chart on $D_M$.

Let $n = \dim M - \dim G$ and $k= \dim G$. Around each point in $M$ we can find an adapted chart of the form $(\tilde U, \tilde \phi = (x^1, \ldots, x^k, y^1 ,\ldots y^n))$ such that for any orbit $qG$ we either have $qG \cap \tilde U = \emptyset$ or: $$ qG \cap \tilde U = \{ y^i = c^i_{qG} \} $$ for $c^i_{qG}$ some constants depending on the orbit $qG$. This is shown in Theorem 21.10 in Lee's "Introduction to smooth manifolds".

Of course, the constants $c^i_{qG}$ are different for different orbits, because the orbits are either the same or disjoint.

Now let $p \in D_M$ and consider a chart $\tilde U$ as above around $p$. Let $\textrm{pr}_2: \mathbb R^k \times \mathbb R^n \rightarrow \mathbb R^n$ be the projection.

My question is: is it true that $\phi:=\textrm{pr}_2 \circ \tilde \phi: \tilde U \cap D_M \rightarrow \mathbb R^n$ is a chart in $D_M$ around $p$, eventually restricting $\tilde U$ to a smaller open subset? If so, how could we show it?

Since $D_M$ only contains points from different orbits, we have that $\phi: \tilde U \cap D_M \rightarrow \mathbb R^n$ is injective. It is also smooth, since $\textrm{pr}_2$ is smooth and the restriction of the smooth map $\tilde \phi$ to the submanifold $D_M \cap \tilde U$ is smooth.

But beyond this, I don't know how to show that $\phi$ is a diffeomorphism on its image. For example, I can't even show that $\phi$ is an open map.

  • $\begingroup$ @AntonioJPan I set $m = \dim M - \dim G$ for convenience, although it seems it just causes confusion since we're used to $m = \dim M$. I'll just set $n = \dim M - \dim G$. $\endgroup$ Commented Oct 17, 2022 at 12:05
  • $\begingroup$ Don't worry. It was my fault, sorry $\endgroup$ Commented Oct 17, 2022 at 12:07
  • $\begingroup$ Can you point exactly to what, in the post itself as opposed to the title, you mean by "fundamental domain"? It's an odd choice of terminology... for example, is $\mathbb R^2 \times \{0\}$ a fundamental domain for $\mathbb R$ acting on $\mathbb R^3$ by the formula $x \cdot (a,b,c) = (a,b,c+x)$? $\endgroup$
    – Lee Mosher
    Commented Oct 19, 2022 at 11:56
  • $\begingroup$ In the title I just used it to mean "a set containing a unique representative for each orbit of the action". I'll edit the title to be more precise. $\endgroup$ Commented Oct 19, 2022 at 12:02
  • $\begingroup$ Your question reminds me Olver and Fels "moving frames and coframes". Your submanifold $D_M$ plays the role of their "cross section". See, for example, researchgate.net/publication/… $\endgroup$ Commented Oct 19, 2022 at 14:48

2 Answers 2


I think this is a not difficult but very technical proof. Here is an outline of what I think you can do:

  • First, your map $\tilde{\phi}$ is a diffeomorphism from $\tilde{U}$ to $\tilde{\phi}(\tilde{U})$, obviously.

  • Second, the submanifolds are preserved under diffeomorphism. So since $D_M$ is a submanifold, $N:=\tilde{\phi}(D_M)$ is a submanifold of $\tilde{\phi}(\tilde{U})\subset R^{n+k}$. From now on, we will work, then, in $\tilde{\phi}(\tilde{U})$. enter image description here

  • Take a parametrization $\sigma: R^n \to N$ in a neighbourhood of $p$. If the map $f=pr_2 \circ \sigma$ is a diffeomorphism (even in a srunken domain) then you are done, because then $\sigma \circ f^{-1}$ gives you the desired chart (a kind of "inverse" of the projection). enter image description here

  • To show that $f=pr_2 \circ \sigma$ is a diffeomorphism, you have to assume an additional transversality hypothesis, I think. Let's see. If $\sigma(a)=\phi(p)\in\phi(D_M)$ then to show is a diffeomorphism you need to prove that $$ d(f)_a=d(pr_2 \circ \sigma)_a=d(pr_2)_{\phi(p)} \cdot d(\sigma)_a $$ is non-singular. But $d(pr_2)_{\phi(p)}=(0_{k\times n} | I_n)$, and $$ d(\sigma)_a=\begin{pmatrix} \dfrac{\partial (x_1,\ldots,x_k)}{\partial (t_1,\ldots,t_n)}\\ \dfrac{\partial (y_1,\ldots,y_n)}{\partial (t_1,\ldots,t_n)}\\ \end{pmatrix} $$ where $t_1,\ldots,t_n$ are the parameters for $\sigma$, so you have to require that $$ det(\dfrac{\partial (y_1,\ldots,y_n)}{\partial (t_1,\ldots,t_n)})\neq 0 $$ So assuming this last condition you are done. Anyway, my own picture tell me that it should be possible to prove your statement even without this requirement, but I don't get the way.

  • $\begingroup$ If you consider the domain of $\sigma$ to be $\mathbb R^n$ and $\sigma$ to be a parametrization, wouldn't the transversality condition you mention be satisfied automatically? But that's just my issue, namely why can we see $D_M$ as "living inside" the $\mathbb R^n$ part. I mean, $\phi$ injects each orbit into $\mathbb R^k \times \{ (c_1, \ldots, c_n) \}$ for some $c_i$, but how does it follow from this that I can see $D_M$ as injected into $\{ x_1, \ldots x_k \} \times \mathbb R^n$ for some fixed $(x_1, \ldots, x_k)$? $\endgroup$ Commented Oct 18, 2022 at 9:35
  • $\begingroup$ Also, a seemingly obvious thing which I glossed over but which I don't actually know how to prove is: if $D_M$ is, as assumed, a submanifold which contains one and only one representative of each orbit, why is it that $\dim D_M = n$? $\endgroup$ Commented Oct 18, 2022 at 9:37
  • $\begingroup$ I don't think the transversality is always satisfied. Think of a vertical tangent in my picture. Anyway, there were a mistake, I mixed $\tilde{\phi}$ with $\phi$. Sorry. Maybe it is now clear. $\endgroup$ Commented Oct 18, 2022 at 11:09
  • $\begingroup$ I have added one previous picture, maybe it helps you $\endgroup$ Commented Oct 18, 2022 at 14:33

Inspired by the answer of A.J. Pan-Collantes, I looked into how a transversality condition might appear and came up with the following line of reasoning. Basically, what I'm asking is not true, but we can construct an adapted chart for the action which also descends to a chart around $p$ on $D_M$. I'm leaving the argument here in case anybody needs it. It is basically just retracing Theorem 21.10 in Lee's "Introduction to smooth manifolds" but with the submanifold $D_M$ in mind.

Let's suppose additionally to my hypotheses that for any $p \in D_M$, $T_p D_M \oplus T_p (G p) = T_p M$.

Recall that in general if we have a distribution $D$ of rank $k$ we say that a smooth coordinate chart $(U, \phi)$ on $M$ is flat for $D$ if $\phi(U)$ is a cube in $\mathbb R^m$ and $\phi_* D$ is spanned by $\partial /\partial x^1, \ldots, \partial/\partial x^k$. Now the Frobenius theorem asserts that involutivity of a distribution is equivalent to integrability and to the existence of such flat charts (Theorem 19.12). A corollary of it is that if $S$ is an embedded submanifold of $M$ such that at some $p \in S$ we have $T_pS \oplus D_p = T_pM$ for $D$ an integrable/involutive distribution, then there exists a chart on $M$ around $p$ which is flat for $D$ and on which $S$ is described as $\phi(S \cap U) = \{ x^1 = \ldots = x^k = 0 \} \cap \phi(U)$.

Now consider the distribution $D:= \sqcup_{p \in M} T_p(pG)$ of tangent spaces to the orbits. It is smooth because it can be seen that a frame in it consists of fundamental vector fields. It is moreover an integrable distribution because by definition it consists of tangent spaces to orbits, and since the action is free and proper, by proposition Proposition 21.7 the orbits are all smooth submanifolds.

Recall that a foliation is collection of connected disjoint immersed $k$-dimensional submanifolds whose union is $M$ and such that around each point in $M$ there is a cubical chart such that each submanifold (in our case, orbit) intersects $U$ in at most a countable union of $k$-dimensional slices of the form $\{ y^1 = c^1, \ldots, y^s = c^s \}$.

Now, the orbits of $G$ are closed, so they are maximal connected submanifolds which are integral to the integrable distribution $D$ (see the proof of Theorem 21.10), and thus they are a foliation on $M$ by Theorem 19.21.

Suppose that for $p \in D_M$ we have $TpD_M \oplus D_p = T_pM$ i.e. $T_p D_M \oplus T_p(pG) = T_p M$. Then because the orbits constitute a foliation we find a cubical chart $(U, \phi=(x^1, \ldots, x^k, y^1, \ldots, y^n)$ around $p$ such that $\phi(U)$ intersects the image through $\phi$ of each orbit in at most a countable union of slices. An analysis of the proof of Theorem 19.21 shows that actually we can shrink any chart to arrive at such a chart $(U, \phi)$, so we can consider the chart given by the corollary to theorem Theorem 19.12 i.e. we can also suppose that $\phi(D_M \cap U) = \{ x^1 = \ldots x^k = 0 \} \cap \phi(U)$.

In summary, we have a chart $(U, \phi)$ such that $\phi(D_M \cap u) = \{ x^1 = \ldots x^k = 0 \}$ and $U$ intersects each orbit in at most a countable union of slices of the form $\{ y^1 = c^1, \ldots, y^s = c^s \}$. Now the exact argument used in the proof of Theorem 21.10 shows that we can use the fact that the action is free and proper to further reduce $U$ to a chart centered at $p$ which intersects each orbit in at most one slice of the form $\{ y^1 = c^1, \ldots, y^s = c^s \}$.

Finally we arrive where we wanted: we have a chart $(U, \phi)$ such that $\phi(U)$ is a cube centered at $\phi(p) = 0$ and for each $q \in U$ the orbit $qG$ is a slice of the form $\{y^i = c_{qG}^i \}$ and moreover $\phi(D_M \cap U)$ is a slice of the form $\{ x^1 = \ldots = x^k = 0 \}$ in the cube $\phi(U)$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .