MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

My question is coming from the method Reid and Chris suggested in solving the problem here. Help on any point is greatly appreciated!

Question 1. For a real manifold $M$, consider $C^{\infty}(M,\mathbb{R})$. For a point $p\in M$, consider the ideal $I_p=\{ f : f(p)=0 \}$. Is $I_p^n$ equal to the set of smooth maps $f$ which have $n-1$st order contact (ref. Golubitsky, Guillemin pg. 37) with the $0$ function?

Question 2. Consider $C(\mathbb{R},\mathbb{R})$ and the ideal $I_0= \{ f : f(0)=0 \}$. Is it the case that $I_0^2 = \{ f : f(0)=f'(0)=0\}$? Is it the case that $I_0^n=\{ f : f(0)=f'(0)=...=f^{(n-1)}(0)=0\}$?

To question 2, the one inclusion is immediate. However the inclusion $ \{ f : f(0)=f'(0)=0\} \subseteq I_0^2 $ doesn't seem obvious to me right now. It seems like one could perhaps find a function which doesn't agree with it's Taylor series that satisfies the derivative condition, but not be in $I^2_0$.

I hope this isn't too elementary for MO. Thanks for your help!

share|cite|improve this question
Is $g(x)$ necessarily smooth? L'Hôpital's rule doesn't seem to be enough to guarantee that all higher derivatives are continuous at $x=0$. – Glen M Wilson Apr 16 '10 at 21:22
Thanks Leonid! This along with Robin's answer are a big help. – Glen M Wilson Apr 16 '10 at 23:37
up vote 4 down vote accepted

This is all fairly well-known stuff. In Q1, the ideal $I_p^n$ is the set of functions for which the derivatives up to order $n-1$ at $p$ vanish in a local coordinate system centred at $p$.

I'll just give a rough outline of how to reduce to the case where $M=\mathbb{R}^k$. Take a coordinate neighbourhood $U$ of $p$ where $p$ corresponds to $0$. It's clear that any function in $I_p^n$ has Taylor expansion in $U$ with no terms of total degree $< n$. Conversely suppose that $f$ is such a function. Using the result for $M=\mathbb{R}^k$ we can express $f$ restricted to $U$ as a finite sum of terms $g_1\cdots g_n$ where each $g_i$ vanishes at $0$. Multiply each $g_i$ by a smooth function $h$ vanishing on a neighbourhood of $0$ and equalling $1$ on a smaller neighbourhood. We can regard $h g_i$ as an element of $I_p\subseteq C^\infty(M)$. Then $f$ minus a linear combination of terms of the form $(hg_1)\cdots (hg_n)$ vanishes in a neighbourhood of $0$ and to complete the proof one has to show that all such functions are also in $I_p^n$ (which is straightforward).

share|cite|improve this answer
I think you mean: "Multiply each $g_i$ by a smooth function $h$ equalling 1 on a neighborhood of 0 and vanishing elsewhere. Then $hg_i\in I_p$, so that $f$ munius a linear combination of terms $(hg_1)\cdots (hg_n)$ vanishes in a neighborhood of $0$. – Glen M Wilson Apr 18 '10 at 21:57

I thought I would add a little more elaboration to the answers given above. There is a general result in the book “Stable Mappings and Their Singularities” by Golubitsky and Guillemin in §II.6 that provides an answer to my Q2. Restricted to my special case, one first observes that if $f$ is smooth and $f(0)=f'(0)=0$ that $g(x):=\frac{f(x)}{x}=\int_{0}^{1}f'(tx)dt$. The right hand side is easily seen to be smooth as we can differentiate under the integral sign. Also $g(0)=f'(0)=0$. Induction then provides the general result (for $\mathbb{R}$). The case when $f:\mathbb{R}^n\rightarrow \mathbb{R}$ is handled analogously.

As for the general statement, it looks like Robin has that taken care of! Thanks everyone!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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