Skip to main content
added 421 characters in body
Source Link
L.F. Cavenaghi
  • 1.8k
  • 11
  • 17

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.

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.

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.

Became Hot Network Question
Source Link
L.F. Cavenaghi
  • 1.8k
  • 11
  • 17

Is this a submanifold?

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.