Sections of a polar action are totally geodesic This question is a repost of the following: https://math.stackexchange.com/questions/4195805/sections-of-a-polar-action-are-totally-geodesic. I've decided it to post it here because it didn't seem to get much traction there.
Suppose $G\curvearrowright M$ is an isometric action of a Lie group on a complete Riemannian manifold $M$, and assume it is polar. This means that the action is proper and there exists a closed (hence complete) embedded submanifold $\Sigma\subseteq M$ (called a section) which meets all orbits orthogonally. It is well known that $\Sigma$ is totally geodesic, but I have not found a convincing proof of that fact.
There is a special case where the last statement is easy to prove: the second fundamental form vanishes at regular (and exceptional) points. Indeed, given any $p\in \Sigma$ such that $G\cdot p$ has maximal dimension, $v\in T_{p}(\Sigma)=\nu_{p}(G\cdot p)$ and $\xi\in T_{p}(G\cdot p)$, we can find an element $X$ in the Lie algebra $\mathfrak{g}$ of $G$ such that
$$\xi=X^{*}(p), \quad X^{*}(q)=\dfrac{d}{dt}\bigg|_{t=0}\operatorname{Exp}(tX)\cdot q.$$
The vector field $X^{*}$ is Killing, so that its covariant derivative is antisymmetric. Let $\mathbb{II}$ be the second fundamental form of $\Sigma$. Then $\mathbb{II}(v,v)$ is tangent to $G\cdot p$ and $\langle \mathbb{II}(v,v),\xi \rangle=-\langle v,\nabla_{v}X^{*} \rangle=0$, so $\mathbb{II}(v,v)=0$. Polarizing, we get $\mathbb{II}=0$.
The usual argument for proving that sections are totally geodesic at all points revolves around the fact that regular points are dense in $\Sigma$. My problem is that all proofs that I found seem to lack crucial details for it. Here are some examples:

*

*In "Lie Groups and Geometric Aspects of Isometric Actions", by Alexandrino and Bettiol, it is stated in Exercise 4.9 that the density follows from Kleiner's Lemma (cf Lemma 3.70), but I can't get the connection between the lemma and this fact.

*In "Critical Point Theory and Submanifold Geometry", by Palais and Terng, the authors state that density follows from the theory of Riemannian submersions, without giving further details.

*In "Polar Manifolds and Actions", by Grove and Ziller, the authors state that singular points are isolated along any geodesic, because of the Slice Theorem. This is because if $\gamma$ is any geodesic of $\Sigma$, and $t_{0}$ lies in the closure of $\{ t\in \mathbb{R}\colon \gamma(t)\in M_{R} \}$, then $\gamma(t_{0}-\varepsilon)$ and $\gamma(t_{0}+\varepsilon)$ have the same isotropy for sufficiently small $\varepsilon>0$ (again, because of the Slice Theorem), but I don't understand why this is the case.

*Another method of proof is proposed in the book "Submanifolds and Holonomy", by Berndt, Console and Olmos. I have an (almost full) solution, which needs to prove the following crucial fact: if $p\in \Sigma$ and there is an open subset $\Omega\subseteq \Sigma$ such that all orbits of the points in $\Omega$ have the same (nonprincipal) type, then $T_{p}(\Sigma)$ is pointwise fixed by the slice representation. If somebody could give a proof of this last fact, I would also accept it as an answer.

Could somebody elaborate on any of the methods of proof proposed above (preferably the first or the third), or give a reference to a detailed proof of the fact that regular points are dense?
Thank you in advance!
 A: I will try to answer your question posed in (4). From now on, $G$ is a Lie group acting properly and isometrically on a connected Riemannian manifold $M$. First, we need a little preparation.
Fact. Let $Q \subseteq M$ be an orbit of $G$. There exists $\varepsilon > 0$ such that the exponential map is defined on $N^\varepsilon Q$ and sends it diffeomorphically onto a neighborhood of $Q$, which we denote $U^\varepsilon Q$ (here $NQ$ is the normal bundle of $Q$ and $N^\varepsilon Q$ denotes the subset of vectors in it of length $< \varepsilon$). Given any such $\varepsilon$ and $p \in Q$, we write $S^\varepsilon_p$ for $\exp(N^\varepsilon_p Q)$. It then follows that $S^\varepsilon_p$ is a slice at $p$.
This is a special case of the tubular neighborhood theorem, when our submanifold is homogeneous, meaning that for any $p, q \in Q$ there exists an isometry $\varphi \in I(M)$ with $\varphi(Q) = Q$ that sends $p$ to $q$. Generally, unless $Q$ is compact, we need to allow our tubular neighborhood to have varying thickness, but if $Q$ is homogeneous, we can make the neighborhood uniform. This is the content of exercise 2.11.2 in the book "Submanifolds and Holonomy" that you mentioned in your question (see also definition 6.2 in Michor's notes). I can sketch a proof here if needed.
Let's now agree that whenever we say the word "slice", we mean a slice of the form $S^\varepsilon_p$ as above. Here is a simple but vital
Corollary. Let $p \in M$ be any, and let $S^\varepsilon_p$ be a slice at $p$. Then $G_q \subseteq G_p$ for any $q \in S^\varepsilon_p$.
Proof. Take any $\varphi \in G_q$. Since isometries commute with the exponential map, $\varphi(S^\varepsilon_p) = S^\varepsilon_{\varphi(p)}$. These two slices have a point in common, namely $\varphi(q) = q$, so they coincide, i.e. $\varphi$ preserves $S^\varepsilon_p$. But $G \cdot p \cap S^\varepsilon_p = \{p\}$, so $\varphi$ preserves $p$. $\square$
Now we assume that $M$ is complete and the action of $G$ is polar. Let $\Sigma$ be a section. Let me remark here that some authors only require a section to be an immersed complete submanifold, not necessarily closed or embedded (for example, C. Gorodski does so in this paper). As far as I understand, such a pathology can occur, for example, for certain polar actions on symmetric spaces of compact type. In any case, everything works even for such a general definition because any immersed submanifold is locally (in its own topology!) an embedded one. So let us assume there is an open subset $\Omega \subset \Sigma$ consisting of nonregular points of the same type. WLOG, we may assume $\Omega$ is an embedded submanifold of $M$. Pick $p \in \Omega$ and let $S^\varepsilon_p$ be a slice at $p$. Shrinking $\Omega$ if needed, we may assume $\Omega \subseteq U^\varepsilon Q$, where $Q = G \cdot p$. A posteriori, $\Omega$ simply lies in $S^\varepsilon_p$ because it intersects $Q$ orthogonally and is totally geodesic, but we don't know that at this point, so we need to somehow "project" $\Omega$ to $S^\varepsilon_p$. Consider the map $\pi \colon G \times S^\varepsilon_p \twoheadrightarrow U^\varepsilon Q, (g, q) \mapsto gq$. This is a smooth surjective submersion and in fact it is just quotienting by the free proper right action
$$
G_p \curvearrowright G \times S^\varepsilon_p, \quad h \cdot (g,q) = (gh,h^{-1}q).
$$
In other words, $\pi$ is a principal $G_p$-bundle (in fact, it is one of the chief properties of slices that $\pi$ realizes $U^\varepsilon Q$ as $G \times_{G_p} S^\varepsilon_p = (G \times S^\varepsilon_p)/G_p$, which is the fiber bundle over $G/G_p \cong Q$ with a fiber $S^\varepsilon_p$ associated to the principal $G_p$-bundle $G \twoheadrightarrow G/G_p \cong Q$). Our principal bundle $\pi$ can be restricted to the submanifold $\Omega$ of $U^\varepsilon Q$:
$$
\widehat{\pi} \colon \; \pi^{-1}(\Omega) \twoheadrightarrow \Omega.
$$
Clearly, $\widehat{\pi}$ is a principal $G_p$-bundle in its own right. In particular, it is a smooth surjective submersion, which we'll use in a second. Now it's time to use the corollary above. I claim that if $(g,q), (g',q') \in \pi^{-1}(\Omega)$ lie over the same point in $\Omega$, then $q = q'$. Indeed, since $q \in S^\varepsilon_p$, we have $G_q \subseteq G_p$ by the corollary above. On the other hand, $q$ has the same orbit type as $gq = \pi(g,q) \in \Omega$, which, by our assumption, coincides with the orbit type of $p$. By definition, it means that $G_q$ and $G_p$ are conjugate, so we deduce that $G_q = G_p$. But $(g',q')$ and $(g,q)$ differ by some element $h$ from $G_p = G_q$: $(g',q') = (gh, h^{-1}q) = (gh, q)$, so the assertion follows. Now, consider the composite map
$$
\pi^{-1}(\Omega) \hookrightarrow G \times S^\varepsilon_p \twoheadrightarrow S^\varepsilon_p.
$$
We've just proved that it is constant on the fibers of $\widehat{\pi}$, hence, by the characteristic property of smooth surjective submersions, it factors through $\widehat{\pi}$ to a smooth map $f \colon \Omega \to S^\varepsilon_p$. One readily checks that $f$ is an injective immersion. This map $f$ is our desired "projection" of $\Omega$ to $S^\varepsilon_p$. Write $\Omega_1 \subseteq S^\varepsilon_p$ for the image of $f$. As we have already shown in the assertion above, the action of $G_p$ on $S^\varepsilon_p$ fixes $\Omega_1$ pointwise, hence the slice representation $G_p \curvearrowright N_pQ$ fixes $T_p\Omega_1$ pointwise as well. Actually, since
$$
\dim \Omega_1 = \dim \Sigma = \mathrm{cohom}(G \curvearrowright M),
$$
this suffices to derive a contradiction, for the subspace of invariants of the slice representation at a nonregular point cannot be that large, but let me formally finish the proof by showing that $T_p\Omega_1 = T_p\Sigma$. Denote $\mathrm{Lie}(G_p) = \mathfrak{k} \subseteq \mathfrak{g} = \mathrm{Lie}(G)$, and let $\sigma \colon \mathfrak{g} \twoheadrightarrow T_pQ$ stand for the differential of the orbit map $G \twoheadrightarrow Q, \; g \mapsto gp,$ at $e$ (it sends $X \in \mathfrak{g}$ to the value of the corresponding Killing vector field at $p$). Clearly, $\ker \sigma = \mathfrak{k}$. We make two simple observations. First,
$$
T_{(e,p)} \pi^{-1}(\Omega) = \mathfrak{k} \oplus T_p \Omega_1 \subseteq \mathfrak{g} \oplus N_pQ = T_{(e,p)} (G \times S^\varepsilon_p).
$$
Second, the differential of $\pi$ at $(e,p)$ is given simply by
$$
\sigma \oplus \mathrm{Id}_{N_pQ} \colon \; \mathfrak{g} \oplus N_pQ \to T_pQ \oplus N_pQ = T_pM.
$$
Therefore, this differential sends $T_{(e,p)} \pi^{-1}(\Omega)$ onto $T_p\Omega_1$. But at the same time,
$$
d\pi(T_{(e,p)} \pi^{-1}(\Omega)) = d\widehat{\pi}(T_{(e,p)} \pi^{-1}(\Omega)) = T_p \Omega = T_p \Sigma,
$$
which completes the proof. $\square$
Remark. What you're saying in (3) is true for geodesics in $\Sigma$, i.e. geodesics normal to the orbits, but it is false for general geodesics. Consider the action of $\mathrm{SO}(2)$ on $\mathbb{R}^3$ by rotations around the $z$-axis. It is a polar action, and its singular orbits are nothing but points on the $z$-axis. So the geodesic running along the $z$-axis consists entirely of singular points.
A: This isn't an answer to the question, but more of a completion of Ivan Solonenko's answer, giving more details and insight to his proof.
Firstly, I'm giving the proof of a claim I made in the original question:
Lemma 1: If $\Sigma$ is a section whose regular points are not dense, then there is a nonempty open subset $\Omega\subseteq \Sigma$ such that all points of $\Omega$ have the same orbit type, and it is nonprincipal.
Proof: By our assumption, there must be a nonempty open subset $\Xi\subseteq \Sigma$ such that no point of $\Xi$ is regular. Let $p\in \Xi$ and $O\subseteq M$ be an open subset such that $\Xi=O\cap \Sigma$. We choose $\varepsilon>0$ such that $S_{p}^{\varepsilon}$ is a geodesic slice at $p$, there is a finite amount of orbit types in $G\cdot S^{\varepsilon}_{p}$ and $\operatorname{exp}_{p}(B(0,\varepsilon))\subseteq O$. The existence of $\varepsilon$ is given by Theorem 6.16 in Michor's notes (already mentioned by Ivan). Choose $x\in S^{\varepsilon}_{p}\cap \Sigma$ such that its orbit type is maximal in $G\cdot S^{\varepsilon}_{p}$. Now, we choose a geodesic slice $S_{x}^{\delta}$ contained in the open subset $G\cdot S_{p}^{\varepsilon}$. As proved by Ivan, $G_{q}\subseteq G_{x}$ whenever $q\in S_{x}^{\delta}$, but since $S_{x}^{\delta}\subseteq G\cdot S_{p}^{\varepsilon}$, we know by maximality $G_{x}$ is conjugate to a subgroup of $G_{q}$, so $G_{x}=G_{q}$. To summarize, all points of the nonempty open set $G\cdot S_{p}^{\varepsilon}\cap G\cdot S_{x}^{\delta}$ have the same orbit type, and it is not maximum. Therefore, $\Omega=G\cdot S_{p}^{\varepsilon}\cap G\cdot S_{x}^{\delta} \cap \Sigma$ is the desired set (notice it is nonempty since it contains $x$). $\Box$
Next, here is a known result which Ivan referenced in order to derive the contradiction at the end:
Lemma 2: Let $p\in M$ be such that $G\cdot p$ is a nonprincipal orbit and $V\subseteq \nu_{p}(G\cdot p)$ be the subspace of all vectors which are fixed under the slice representation. Then $\dim V$ is strictly smaller than the cohomogeneity of the action $G\curvearrowright M$.
Proof: Let $x\in \nu_{p}(G\cdot p)$ be a regular point of the slice representation (which is not fixed since the slice representation is not trivial), and write it as $x=y+z$, where $y\in V$ and $z\in V^{\perp}$. The codimension of $G_{p}\cdot x$ in $\nu_{p}(G\cdot p)$ is equal to the cohomogeneity of the slice representation, which is also the cohomogeneity of the action. Notice that $G\cdot x = \{ y \} + G\cdot z $, so $G_{p}\cdot x$ and $G_{p}\cdot z$ have the same dimension. Also, since $G_{p}$ acts trivially on $V$, we deduce that $G_{p}\cdot V^{\perp}=V^{\perp}$. All in all, this means that $G_{p}\cdot z \subseteq V^{\perp}$, so $\dim V$ is less or equal to $\operatorname{codim}(G_{p}\cdot z)=\operatorname{cohom}(G\curvearrowright M)$. Since $G_{p}\cdot z$ is compact and $V^{\perp}$ is connected, either $G_{p}\cdot z=V^{\perp}$ or $\dim G_{p}\cdot z < \dim V^{\perp}$. If $G_{p}\cdot z=V^{\perp}$, then $V^{\perp}$ is compact, so it must be the zero subspace, contradicting the fact that $G\cdot p$ is a nonprincipal orbit. Therefore, $\dim V^{\perp}> \dim G_{p}\cdot z$, and by taking codimensions, $\dim V< \operatorname{cohom}(G\curvearrowright M)$. $\Box$
Finally, notice that the fact that $T_{p}(\Omega_{1})=T_{p}(\Sigma)$ implies that $f_{*p}$ is an injective map. This is enough for what we need, since even if we don't know if $f$ is globally an injective immersion, by shrinking $\Omega$ around $p$, we can guarantee it, so no more computations are needed.
If you see anything incorrect or incomplete (or if you wish for me to add more details), please let me know.
