When is a real-analytic variety a union of non-singular subvarieties? I have asked this before on MSE, but received no answer yet.
Say I have a set in $\mathbb{R}^n$ defined to be the zero set of an analytic function $F:\mathbb{R}^n\to\mathbb{R}^k$, $k<n$. Everywhere, where $DF$ is not singular, an analytic local parametrization is given simply by the implicit function theorem. But what about the points where $DF$ is singular?
So my question is, are there some standard tools/theorems which tell the zero set to still be an immersed manifold which just crosses/intersects itself in such a point?
More concretely, say I have an analytic function $F:\mathbb{R}^2\to\mathbb{R}$ such that $F(p)=0$ and $\nabla F(p)=0$ at some $p$. Then I can prove that, if the Hessian of $F$ in $p$ is indefinite, then there exists a neighborhood of $p$ and two smooth curves in this neigborhood intersecting only in $p$ such that $F$ vanishes on these curves. These curves have differently directed derivatives, so they indeed cross. However, while they may of course be analytically parametrized away from $p$, in $p$ itself I can only show smoothness, but no analyticity. Moreover, I cannot prove that there are only these two curves. The latter would follow from analyticity since all derivatives of multiple solution for one of the branches must coincide in all derivatives at $p$, but however, the Taylor coefficients are only implicitly given by the Taylor coefficients of $F$ and I cannot resolve it. So, since I can show for my particular $F$ that whenever $\nabla F=0$ then the Hessian is indefinite, my goal would be to state that the solution set of $F=0$ is just the union of the ranges of some analytic curves.
(1.) So, on the one hand, I'd be interested in some "standard theorem" or some "standard toolbox" to study the singular points of such $F$, which ideally tells me that I can analytically continue my analytic parametrization into the critical point.
(2.) Also I cannot figure out how this may be generalized to higher dimensions. I suspect that the higher dimensional analog of an indefinite Hessian should include an invertible Hessian to exclude $0$-eigenvalues (which in two dimensions is already granted). Probably an answer on (1.) would make an answer on (2.) redundant.
Poorly, a google research did not result in anything useful. The only text I found going in the right direction is this, approximating the solution curves by a Puiseux series expansion. However, I'm not too familiar with algebraic geometry, but I have the impression that this approach heavily relies on that there is no "second limit" if $F$ is only a polynomial.
Any help will be appreciated. Thanks!
 A: The general problem mentioned at the beginning of the question is extremely difficult, and, without more hypotheses, there is not that much that can be said.
The OP might be interested in this answer of mine, which addresses the question in the two variable case. Basically, if one has a real-analytic function $F(x,y)$ that satisfies $F(0,0)=0$ and the lowest order homogeneous term in the power series expansion at $(0,0)$ is of degree $d$ with $k\le d$ distinct linear real factors, then, near $(0,0)$, the locus $F=0$ is the union of $k$ smooth real-analytic curves that are pairwise transverse.
Here is how one can prove this:  One can assume, by a linear change of variables, that
$$
F(x,y) = x\,f_{d-1}(x,y) + f_{d+1}(x,y) + \cdots + f_m(x,y) + \cdots,
$$
where $f_i(x,y)$ is homogeneous of degree $i$ and $f_{d-1}(0,y) = y^{d-1}$ (i.e., $x$ is not a factor of $f_{d-1}(x,y)$).  Now make the substitution $x = yz$, yielding
$$
0 = F(yz,y) = yz\,f_{d-1}(yz,y) + f_{d+1}(yz,y) + \cdots + f_m(yz,y) + \cdots,
$$
which factors as
$$
F(yz,y) = y^d\bigl(z\,f^*_{d-1}(y,z) + y\,f^*_{d+1}(y,z) + \cdots + y^{m-d}\,f^*_m(y,z) + \cdots\bigr)
$$
where, $f^*_{m}(y,z) = f_{m}(yz,y)/y^m$ are polynomials in $(y,z)$.  Morover, by construction
$$
f^*_{d-1}(z,y) = 1 + a_1\,z + a_2\,z^2 + \cdots +a_{d-1} z^{d-1}
$$
for some constants $a_1,\ldots, a_{d-1}$.
It is now easy to show that the function
$$
G(y,z) = z\,f^*_{d-1}(y, z) + y\,f^*_{d+1}(y,z) + \cdots + y^{m-d}\,f^*_m(y,z) + \cdots
$$
is real-analytic near $(y,z) =(0,0)$. [Use the estimates on the coefficients of $F$ provided by its convergence near $(x,y)=(0,0)$.] By construction, the power series expansion of $G$ is of the form
$$
G(y,z) = z + y\,f^*_{d+1}(0,0) + g_2(y,z) + g_3(y,z) + \cdots,
$$
where $g_k(y,z)$ is homogeneous of degree $k$ in $y$ and $z$.
By the implicit function theorem, $G(y,z)=0$ defines $z$ as an analytic function of $y$ in a neighborhood of $0$, say $z = \phi(y)$.  Thus, the equation $x = y\,\phi(y)$ defines a nonsingular analytic curve in the zero locus of $F$.  Consequently, $L(x,y) = \bigl(x-y\,\phi(y)\bigr)$ is a (prime) factor of $F(x,y)$ in the ring $\mathcal{R}$ of real-analytic functions on a neighborhood of $(0,0)$.
Now, using the assumption that the lowest order homogeneous term in $F$ is of degree $d$ with $k\le d$ distinct linear real factors and the fact that $\mathcal{R}$ is a UFD, one can write
$$
F(x,y) = L_1(x,y)\cdots L_k(x,y)\,H(x,y)
$$
where each $L_i$ has lowest order nonzero term of degree $1$ and $H$ is analytic near the origin and has lowest order nonzero term $h_{d-k}(x,y)$ of degree $d{-}k$.  Since $h_{d-k}(x,y)$ has no real linear factors, it is a product of quadratic factors that are irreducible over the reals.  Thus, each of these factors is definite, so that
$$
H(x,y) = h_{d-k}(x,y) + h_{d-k+1}(x,y) + \cdots
$$
where $h_{d-k}(x,y)\not=0$ when $(x,y)\not=0$.  It then follows (by an order-of-vanishing estimate) that $H$ is nonzero on a punctured neighborhood of the origin.
Consequently, the zero locus of $F$ near $(0,0)$ is the union of the $k$ pairwise transverse smooth real-analytic curves defined by $L_i(x,y) = 0$.
The case in which $f_{d}(x,y)$ has repeated real factors is more subtle, since the real prime factors of $F$ can be irreducible and of higher degree (e.g., $F(x,y) = x^d - y^{d+1}$) and the answer mentioned at the beginning gives some information about this case.
A: I write this as an answer since it is too long for a comment.
First of all, thank you very much, this is actually a really nice and instructive proof. I had a lot of stuff to do so I was just able to work through it now. However, I have a few followup questions:
At first, you suggested an alternative title "When is a real-analytic variety a union of smooth subvarieties?". Is it true, that if the locus $F=0$, $F$ analytic, is smooth, that then one already has an analytic parametrization?
At second: As I said, I think your proof is really instructive, even giving more or less explicitly an analytic parametrization. Basically, you zoom into the critical point by $x\mapsto yz$ and show that the presence of the respective factor $x$ in the lowest non-vanishing order yields some analytic solution curve through the origin, and then rescaling back to the original coordinated you obtain yields a solution curve for which, just as expected, $x(y)\in\mathcal{O}(y^2)$ so that the locus $x=0$ is already the linear approximation. Moreover, by this particular rescaling you remove every other curve of zeros, which is not linearly approximated by $x=0$, away from the critical point, and then, as expected, there remains only one solution curve allowing to apply the implicit function theorem.
However, I cannot figure out which exact step goes wrong for more variables. Say we have
$$F:\mathbb{R}^n\to\mathbb{R}$$
analytic and the lowest order homogeneous part, say of order $k$, again factors into $d\le k$ distinct linear polynomials, say exactly one factor is $x_1$. I would expect that there is a similarly suitable rescaling of that $x_1$ coordinate, or is there? Is there maybe a counterexample which should immediately come to one's mind?
At third, what are these "more hypotheses" you speak of in the beginning?
At last, you define $\mathcal{R}$ to be the ring of power series which converge with some positive radius, which is a UFD. This question might not be too related to the original one, but the subring of $\mathcal{R}$ for a fixed convergence radius fails to be a UFD, right? Just that I get the step in your proof of using the UFD property right in my head.
Thank you very much in advance.
