A slice orthogonal to each orbit Assume that a compact (connected) Lie group $G$ acts on a manifold $M$. We choose a $G$-invariant Riemannian metric on $M$ and a point $p \in M$. Then using the exponential map at $p$, we can obtain a slice $U \subset M$ at $p$ so that $T_pU \perp T_p(G\cdot p)$. Namely the slice is perpendicular to the orbit through $p$.
Questions:


*

*Is $T_qU$ is perpendicular to the orbit $G \cdot q$ for each $q \in U$?

*If not, is it possible to choose a slice so that it is perpendicular to orbits through the slice?


Notes: 


*

*We may choose another $G$-invariant Riemannian metric on $M$ if necessary.

*We may assume that $G$-action is locally free so that Killing vector fields form a vector subbundle of the tangent bundle $TM$.

 A: An extension of my comments above:
Considering a fixed metric the answer to both questions is NO:
Consider the Hopf Action of $\mathsf S^1 = \{e^{i\theta}| \theta \in \mathbb R\}$ on $\mathbb S^3$, that is $\mathsf S^1$ acts on $\mathbb S^3 \subset \mathbb C^2$ via left multiplication and is thus free.
To the first question: A slice obtained via the normal exponential map of the orbit $\mathsf S^1 \ast p$ is isometric to (a part of) a totally geodesic $\mathbb S^2$ contained in $\mathbb S^3$. Further all orbits of the Hopf action are great circles. Since two great circles in a sphere which are perpendicular to a codimension 1 sphere intersect it follows that $\mathsf S^1 \ast p$ is the only orbit perpendicular to the slice.
To the second question: A slice which is perpendicular to all orbits intersecting it is necessarily totally geodesic (compare the notion of a 'polar' action, that is one admitting a submanifold which is perpendicular to all orbits. Such a section is known to be totally geodesic. See for example 
http://www.math.upenn.edu/~wziller/papers/Polar.pdf
). Hence such a slice must again be a part of a 2-sphere and the same argument applies.
As you note one may further ask whether for a given action of a Lie Group $G$ on $M$ there exists an invariant metric such that at all points there exists a slice which is perpendicular to orbits. Such an action can be thought of as being a local notion of a polar action. Propably, compare proposition 1.3 of the reference above, it follows that $M/G$ is an orbifold as it is the case for a polar action. So we obtain a restriction for the existence of such a metric (note that the property of $M/G$ of being an orbifold is independent of the choosen invariant metric).
Unfortunately, this does not help for the above example (or in general free actions), since $\mathbb S^3 / \mathbb S^1 \cong \mathbb S^2$ is clearly an orbifold. 
