Is this a submanifold?

Question

Let $(M,g)$ be a compact Riemannian manifold with an isometric action $\rho : G \to \mathrm{Iso}(M)$ by a compact Lie group $G$. There is a natural extension of $\rho$ to $TM$ given by: $$\psi : G \times TM \to TM$$ $$\psi(g,p,X) := (\rho(g)p,d\rho(g)X).$$

I would like to know if the set: $$TM \supset \tilde S := \{(p,X) \in TM : G_X = G_p\}$$ is a submanifold of $TM$, and further, if $\pi : TM \to M$ is the natural projection, then $\pi(\tilde S)$ is a submanifold of $M$.

I tried the following approach:

For each $g\in G$ consider $\eta_g(p,X) \equiv \eta(g,p,X) := d^2_{TM}\left((p,X),\psi(g,p,X)\right).$ Then, one has: $$\eta_g^{-1}(0) = \{(p,X) :(\rho(g)p,d\rho(g)X) = (p,X)\}.$$ So, $$\tilde S = \bigcup_{g\in G}\eta_g^{-1}(0). $$

But according to my calculation $0$ is not a regular value of $\eta_g$.

I appreciate any help.

EDIT

Thanks to all the answers and comments. I shall change a little the candidate to manifold to another that will be more helpful to me.

I would like to know if one denotes by $\cal H_p$ the orthogonal complement to $T_pG\cdot p$ on the $g$-metric, then the set $$S := \{p \in M : \exists X \in \mathcal H_p : G_X = G_p\}$$ is a submanifold of $M$. In fact, this is what I was trying to prove at first.

Answer 1

Your definition implies that $$ \tilde S = \bigcup_{p\in M} T_pM^{G_p}. $$ In particular, $\pi(\tilde S) = M$, and $\tilde S$ will be a submanifold of $TM$ iff the dimension of $T_pM^{G_p}$ is the same for all $p\in M$.

$G$ being connected won't necessarily make this happen: Another counterexample is $S^1$ acting by rotation on $S^2$, fixing the north and south poles, $n$ and $s$. $\tilde S$ is then $T(S^2\setminus\{n,s\})\cup \{n,s\}$, and the points $n$ and $s$ do not have neighborhoods homeomorphic to open balls.

EDIT

To address the modified question: If $p\in S$, then the orbit $Gp$ has a neighborhood homeomorphic to $G\times_{G_p} \mathcal H_p$ and we're assuming that $\mathcal H_p^{G_p} \neq 0$. If $q\in \mathcal H_p$ is any point, then $\mathcal H_p^{G_p}$ will be fixed by $G_q$ and still be orthogonal to $Gq$, hence $q$ and every point in $Gq$ will be in $S$. This shows that a neighborhood of $Gp$ is contained in $S$. Thus $S$ is an open submanifold of $M$.

Answer 2

For the obvious reflection action of $\mathbb{Z}/2\mathbb{Z}$ on $S^1$ this set $\tilde S $ has two singular points at the zero vectors tangent at points $p=(1,0)$ and $q=(-1,0)$. Then $\tilde S$ is $T(S^1\setminus \{p,q\}) \cup \{p_0,q_0\}$ where $p_0, q_0$ are zero vectors at $p,q$. This is a connected set and after removing $p_0,q_0$ we obtain a disconnected set. So obviously it is not a manifold because every connected manifold of dimension at least $2$, remains connected after removing a finite set.