# 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 $$C^infty(M)$$?

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

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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