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!

  • 4
    $\begingroup$ By the way, your title "Analytically submersed manifold still immersed" does not really convey a good idea of what your question is about. For example, one more suggestive title might be, "When is a real-analytic variety a union of smooth subvarieties?", but others could be devised. $\endgroup$ – Robert Bryant Jun 14 at 14:27
  • $\begingroup$ Thanks for your suggestion, for sure my title is not the best. But if I search for analytic parametrizations, am I then asking for analytic subvarieties? These may again have branching points, right? As mentioned, I'm not too familiar with the language of algebraic geometry... However, I need to work through your answer below, thank you very much, but I'm a bit busy at the moment. But my first question, what exactly do you mean by $distinct$ linear real factors? $\endgroup$ – nicrot000 Jun 14 at 17:16
  • $\begingroup$ If I understand it correctly, an indefinite Hessian implies that I can assume the function to be of the shape $$(x,y)\mapsto x^2-y^2+\sum_{k=3}^\infty(k-\text{homogeneous terms}),$$ where the lowest order may be rewritten into $(x-y)(x+y)$, yielding the diagonals as linear approximation. Is this decomposition what you mean by linear factors? $\endgroup$ – nicrot000 Jun 14 at 17:20
  • 1
    $\begingroup$ Yes. What i mean by distinct linear factors is this: A polynomial $p(x,y)$ with real coefficients can be factored (over the reals) into linear and quadratic factors. For example, $x^3+y^3 = (x+y)(x^2-xy+y^2)$, so it has one linear factor and one quadratic factor. Meanwhile, $x^2y$ has three linear factors, but two of them are equal (i.e., not distinct). And $x^4-y^4 = (x-y)(x+y)(x^2+y^2)$ has two distinct linear factors and one quadratic factor. Indeed, your example, $x^2-y^2$ has two distinct linear factors. $\endgroup$ – Robert Bryant Jun 14 at 17:23
  • 1
    $\begingroup$ Yes. Actually, I realized that my previous comment left out a word: It should have read "A homogeneous polynomial $p(x,y)$ with real coefficients can be factored..." $\endgroup$ – Robert Bryant Jun 16 at 17:13

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 $$ G(y,z) = z\,f^*_{d-1}(yz) + y\,f^*_{d+1}(y,z) + \cdots + y^{m-d}\,f^*_m(y,z) + \cdots $$ 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.


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.

  • 2
    $\begingroup$ This is backwards: Your answer is actually more questions, and this comment is all answers: Yes, to your first question, see the MO answer I already referenced. Second, the argument I gave is a 'poor man's blow-up'. If you look up 'blow-up' in analytic geometry, you'll see how it generalizes to higher dimensions. Third, in higher codimension, i.e., $F:\mathbb{R}^m\to\mathbb{R}^k$ with $k>1$, you are asking about the structure of singular analytic varieties, and this is an enormous field. Finally, they are both UFDs, but the germs are the more natural place where these questions are posed. $\endgroup$ – Robert Bryant Jul 5 at 14:02
  • $\begingroup$ Okay, thank you very much again, that helped me alot. $\endgroup$ – nicrot000 Jul 5 at 19:20
  • $\begingroup$ This blow-up goes a bit over my head. However, after thinking about it for a while I'm not so sure if this is true anymore, rather I am sure that on cannot straight forwardedly generalize the proof. Even if you just multiply the defining equation of two intersecting, non-equal planes in $\mathbb{R}$, the analog blow-up along one oft the orthogonal complements does not completely get rid of the other plane, preventing an application of the implicit function theorem. Can you confirm this train of thought? $\endgroup$ – nicrot000 Jul 5 at 19:35
  • 2
    $\begingroup$ Well, I should have been more precise. I am talking about what people call the "Nash blow-up", which for a single equation $F=0$, means that you replace the ambient $n$-space space with the space of pairs $(p,H)$ where $H$ is a hyperplane through $p$, and then you look at the closure of the space of pairs $(p, T_pZ)$ where $p$ is a smooth point of the zero locus of $F$ and $T_pZ$ is its tangent space at $p$. In certain (well-defined and common) situations, this will resolve the singularity of $F=0$, but it's certainly a much more complicated story than the 1-dimensional case. $\endgroup$ – Robert Bryant Jul 5 at 19:53
  • $\begingroup$ Okay, thank you very much. $\endgroup$ – nicrot000 Jul 5 at 20:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.