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lets assume we have a real vectorspace $V$ and functions $f_1, \dots, f_k \colon V \to \mathbb{R}$ which are real-analytic (for instance, let them be polynomial). Furthermore we have an embedded real-analytic submanifold $M \subset V$ and a point $x \in M$, such that $$F := (f_1|_M, \dots, f_k|_M) \colon M \to \mathbb{R}^k$$ is a submersion in $x$, i.e. $$d_xF \colon T_xM \to T_{F(x)} \mathbb{R}^k$$ is surjective. Do we know something about the set of points $M_0 = \{ y \in M \ | \ F \text{ is a submersion in } y\}$?

Since $M$ is an embedded submanifold, $M_0$ is open in $M$. But is it dense?

Edit: Thanks to the comments of Willie Wong and Holonomia my question is answered, if I know, that $f_j|_M \colon M \to \mathbb{R}$ are real analytic. So if $y \in M$ is a point, we find a real analytic "submanifold" chart $(U,\phi)$, with $U$ open in $V$, s.th. $f \circ \phi^{-1} \colon \phi(U) \to \mathbb{R}$ is real analytic and such that $\phi(U\cap M) = \phi(U) \cap (\mathbb{R}^k \times \{0\})$. With the subspace-topology in $\mathbb{R}^n$, we can now conclude, that at each point $z \in \phi(U \cap M)$ it is locally (in $\phi(U\cap M)$) given as a convergent power series. Therefore $f|_M \colon M \to \mathbb{R}$ is real analytic.

Is this right? Do we find such "submanifold" charts?

Improve this question
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    $\begingroup$ Do you require $M$ to be real analytic? If not there is no reason for $M_0$ to be dense. For example, take $k = 1$. Let $f = f_1 = x_2$. Let $V = \mathbb{R}^2$. Let $\phi$ be a bump function on $\mathbb{R}$ such that $\phi(0) = 1$ and $\phi'(0) = 0$ and $\phi''(0) < 0$. Then take $M$ to be the graph of $\phi'$. $\endgroup$
    – Willie Wong
    Jun 15, 2016 at 3:28
  • $\begingroup$ Well, I have a compact Liegroup acting on my vectorspace with some linear action. Are these submanifolds then real anyltic? $\endgroup$
    – Feanoris
    Jun 15, 2016 at 9:25
  • $\begingroup$ Since your embedded submanifold $M$ is an orbit of a compact Lie group acting linearly on $V$ then $M$ is real analytic. Indeed, the manifold structure of Lie groups is known to be real analytic and the action is real analytic being the restriction of the action of $GL(V)$ on $V$, due that you say that the action is linear. $\endgroup$
    – Holonomia
    Jun 15, 2016 at 10:11
  • $\begingroup$ @Holonomia: And the restriction of $f_j \colon V \to \mathbb{R}$ to $M$ is still a real analytic function, since $M$ is a real analytic embedded submanifold? $\endgroup$
    – Feanoris
    Jun 15, 2016 at 16:00
  • $\begingroup$ Yes, this so. Here is the explanation: take a point of $p$ of $M$. Let $x_1,\cdots,x_n$ cartesian coordinates of $V$ w.r.t. some basis. The fact that $M$ is real analytic means that there are local coordinates of $M$ around $p$, say $t_1,\cdots,t_k$ such that the composition with the coordinates $x_j$ are real analytic functions i.e. $x_j(t_1,\cdots,t_k)$ are real analytic functions of $t_1,\cdots,t_k$. The restriction of the $f_j$ are real analytic since are composition of real analytic is real analytic i.e. $f_j(x_i(t_1,\cdots,t_k))$ $\endgroup$
    – Holonomia
    Jun 15, 2016 at 16:10

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