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Let $M$ be a (usual, finite dimensional) smooth manifold, and $C^\infty(M)$ the algebra of (real valued) smooth functions on $M$. For each point $a\in M$ and for each subalgebra $A$ in $C^\infty(M)$ (I mean $A$ contains constants and is closed under summing and multiplication) let us denote by $I_a[A]$ the ideal in $A$ consisting of functions which vanish in $a$: $$ f\in I_a[A]\quad \Longleftrightarrow\quad f\in A\quad \&\quad f(a)=0. $$ Let us consider also the ideal $I_a^2[A]$ in $A$ defined by the following rule: $$ f\in I_a^2[A]\quad \Longleftrightarrow\quad \exists g_1,...g_n,h_1,...,h_n\in I_a[A]:\quad f=\sum_{i=1}^ng_i\cdot h_i. $$

From the Hadamard lemma it follows that in the simplest situation, when $A=C^\infty(M)$ the following criterion holds:

Theorem. If $A=C^\infty(M)$, then $f\in I_a^2[A]$ if and only if $f\in I_a[A]$ and $f$ "has zero derivatives in all directions" in $a$ (i.e. $t(f)=0$ for each tangent vector $t\in T_a(M)$).

A question: is it possible that the same true for any subalgebra $A\subseteq C^\infty(M)$?

One can simplify a little bit this question to the following

Hypothesis 1: For each (unital) subalgebra $A\subseteq C^\infty(M)$ $$ f\in I_a^2[A]\quad \Longleftarrow\quad \Big( f\in I_a[A]\quad\&\quad \forall t\in T_a(M) \quad t(f)=0 \Big). $$

Or to the following

Hypothesis 2: For each (unital) subalgebra $A\subseteq C^\infty(M)$ $$ I_a^2[A]\supseteq A\cap I_a^2[C^\infty(M)]. $$

I am not sure if this is important, but my algebra $A$ differs the points of $M$ $$ a\ne b\in M \quad \Longrightarrow\quad \exists f\in A\quad f(a)\ne f(b), $$ and the tangent vectors in each point $$ 0\ne t\in T_a(M) \quad \Longrightarrow\quad \exists f\in A\quad 0\ne t(f). $$

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If teo functions vanish to order 1 at some point, then their product vanishes to order 2 at that point. Hence both Hypotheses are true. – Matthias Ludewig Jul 8 '13 at 15:28
Excuse me, what is teo function? – Sergei Akbarov Jul 8 '13 at 15:57
@Kofi, I don't see how your comment addresses the question. The question is whether elements of A that vanish to order 2 at a point can be factorized as products of elements of $I_a[A]$ – Yemon Choi Jul 8 '13 at 18:46
Yes, I misunderstood the question, I am afraid. – Matthias Ludewig Jul 8 '13 at 20:24
up vote 2 down vote accepted

Edit: modified to satisfy the last condition.

Consider the subalgebra of $C^\infty(\mathbb{R})$ generated by two elements $x^2,xe^x$. This algebra separates points and "sees" any nonzero tangent vector. The ideal $I_0[A]$ has the functions $x^{2k+l}e^{lx}$ where $k,l\in\mathbb{N}$ and not both $k$ and $l$ are zero as a vector space basis and $I^2_0[A]$ is hence spanned by products of such elements. Using this one verifies that the function $x^2$ is not in $I^2_0[A]$. On the other hand it is contained in the rhs of both of the hypothesis.

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But in this case, both hypotheses are true, if I am not mistaken! – Matthias Ludewig Jul 8 '13 at 15:26
Yes, thank you! – Sergei Akbarov Jul 8 '13 at 19:30
I had a mistake in the last one, but the two functions $x^2,xe^x$ should work (hope I got it right this time) – Michael Bächtold Jul 8 '13 at 19:32
A mistake? What was wrong? I don't see a mistake... – Sergei Akbarov Jul 8 '13 at 19:34
With $x^2,e^x$ you can write $x^2=x^2(1-e^x)+x^2e^x\in I^2$. – Michael Bächtold Jul 8 '13 at 19:36

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