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Let $G$ be a Lie group and $H$ a Lie subgroup of $G$.

Let $M$ be a smooth manifold.

Let $\theta$ be a left smooth action of $G$ on $M$.

Let $S=\{p\in M| G_p=H\}$, where $G_p$ is the isotropy group of $p$.

Is $S$ a smooth submanifold of $M$?

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    $\begingroup$ No: you need compactness assumptions. Take an R action with a nasty fixed point set, and H=G=R $\endgroup$
    – Thomas Rot
    Commented Oct 6, 2019 at 11:28

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More details in Thomas Rot's comment. Every closed subset of any manifold is the set of zeroes of a vector field, even one which is complete (for example, bounded in a complete Riemannian metric). So the vector field generates a group action, whose fixed point set is that closed set. So if we take $G$ to be the real line, and $H=G$, we get any closed set arising as $S$, for some vector field. To be more explicit, if $f(x)$ is bounded and vanishes to all orders at the origin, and nowhere else, then $\sin(1/x)f(x)\partial_x$ can be our vector field on the real line, with fixed points $x$ at $1/x \in 2\pi \mathbb{Z}$.

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