If $f ∈ C^∞(M)$ has vanishing first-order Taylor polynomial at $p$, is it a finite sum of $gh$ for $g, h ∈ C^∞(M)$ that vanish at $p$?

Mathematics Asked by Fred Akalin on January 7, 2022

This is 11-4(a) in Lee’s "Introduction to Smooth Manifolds":

Let $M$ be a smooth manifold with or without boundary and $p$ be a point of $M$. Let $mathcal{I}_p$ denote the subspace of $C^infty(M)$ consisting of smooth functions that vanish at $p$, and let $mathcal{I}_p^2$ be the subspace of $mathcal{I}_p$ spanned by functions of the form $fg$ for some $f, g in mathcal{I}_p$.

(a) Show that $f in mathcal{I}_p^2$ if and only if in any smooth local coordinates, its first-order Taylor polynomial at $p$ is zero.

The $Rightarrow$ direction is easy enough, but the $Leftarrow$ direction has a complication in that Taylor’s theorem only tells you that $f$ is a finite sum of products of pairs of functions in $C^infty(U)$ for some open $U$ around $p$. How can I extend those functions so that $f$ is a finite sum of products of pairs of functions in $C^infty(M)$?

f(x) = sum_{i,j} c_{ij}(x) (x^i – p^i) (x^j – p^j)

for some smooth functions $c_{ij} colon U to mathbb{R}$. I tried using a smooth bump function to have $c_{ij}(x) (x^i – p^i)$ and $x^j – p^j$ go to $0$ outside of $U$ except for $x^0 – p^0$ which goes to $1$, and then try to extend $c_{00}(x)(x^0 – p^0)$ to be $f$ outside of $U$, but ran into difficulties getting things to equal when the bump function is between $0$ and $1$. It feel like this is the wrong track.

On the other hand, the usual formulation of this problem has $mathcal{I}_p$ be an ideal of $mathcal{O}_p$, the ring of germs of functions at $p$, in which case there’s no need to extend the functions globally. I’m wondering if the $Leftarrow$ direction is even true as stated.

Am I missing something?

(This question was asked before, but the answers only work in a single chart.)

One Answer

You only need a single chart.

By standard construction, there is a smooth function $varphicolon Mto[0,1]$ which is supported on a coordinate neighbourhood of $p$, $varphi(x)=1$ for all $x$ near $p$, namely, doing the construction on a coordinate neighbourhood of $p$ and extend by 0 outside that neighbourhood. You can write $$ f=varphi f+(1-varphi)f. $$ The $(1-varphi)f$ is already in $mathcal{I}_p^2$ since $1-varphi$ and $f$ both vanish at $p$ by supposition. The function $varphi f$ is identical to $f$ near $p$, so you reduce to the case of $mathbb{R}^n$.

Answered by user10354138 on January 7, 2022

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