Germs of smooth functions

Jordan Bell
April 4, 2016

1 Sheafs

Let M=m. For an open set U in M, write (U)=C(U), which is a commutative ring with unity 1M(x)=1. For open sets VU in M, define rU,V:(U)(V) by rU,Vf=f|V, which is a homomorphism of rings. is a presheaf, a contravariant functor from the category of open sets in M to the category of commutative unital rings. For to be a sheaf means the following:

  1. 1.

    If Ui, iI, is an open cover of an open set U and if f,g(U) satisfy rU,Uif=rU,Uig for all iI, then f=g.

  2. 2.

    If Ui, iI, is an open cover of an open set U and for each iI there is some fi(Ui) such that for all i,jI, rUi,UiUjfi=rUj,UiUjfj, then there is some f(U) such that rU,Uif=fi for each iI.

For the first condition, let pU. As Ui is an open cover of U, there is some i for which pUi. As f|Ui=g|Ui, f(p)=g(p). Therefore f=g. For the second condition, let pU. If pUi and pUj, then fi(p)=fj(p). This shows that it makes sense to define f:U by f(p)=fi(p), for any i such that pUi. Then f|Ui=fi, which implies that f(U): for each pU, there is some open neighborhood Ui of p on which f is smooth. Therefore is a sheaf.

2 Stalks and germs

For pM, let 𝒰p be the set of open neighborhoods of p. For U,V𝒰p, say UV when VU. For UVW and f(U),


For f(U) and g(V), say fpg if there is some W𝒰p, WU, WV, such that rU,Wf=rV,Wg. Let


and let p be the direct limit of the direct system (U), rU,V of commutative unital rings:


We call p the stalk of F at p. An element of p is called a germ of F at p. In other words, for fp, let [f]p be the set of those gp such that fpg, equivalently, f|UfUg=g|UfUg. A germ of at p is such an equivalence class [f]p, and


3 Maximal ideals

For pM, and f,gp with fpg, f(p)=g(p). Thus it makes sense to define evp:p by evp[f]p=f(p). Now, for [f]p,[g]pp,


evp[1M]p=1. This means that evp:p is a homomorphism of unital rings. It is straightforward that evp is surjective. Write 𝔪p=kerevp. By the first isomorphism theorem, there is an isomorphism of unital rings p/𝔪p. Therefore 𝔪p is a maximal ideal in p. Now, if [f]pp𝔪p then evp[f]p0, hence f(p)0. Then there is some U𝒰p such that f(x)0 for xU, and (1/f)(x)=1f(x) belongs to (U). Then [1/f]pp and [f]p[1/f]p=[f1/f]p=[1M]p, which shows that if [f]pp𝔪p then [f]p has an inverse [1/f]p in p. This means 𝔪p is the set of noninvertible elements of p, which means that p is a local ring.

For 1im define the coordinate function xi:M by xi(p)=pi, which belongs to (M). Because ev0xi=0, [xi]0𝔪0. We prove Hadamard’s lemma, that the ring 𝔪0 is generated by the germs of the coordinate functions at 0.11 1 Liviu Nicolaescu, An Invitation to Morse Theory, second ed., p. 14, Lemma 1.13.

Lemma 1 (Hadamard’s lemma).

The ideal m0 is generated by the set {[xi]0:1im}.


Let [f]0𝔪0 with f(Br) for some r>0. For yBr, using the fundamental theorem of calculus and using the chain rule,


and ui(Br). This means that [f]0=i=1m[xi]0[ui]0, which shows that [f]0 belongs to the ideal generated by the set {[xi]0:1im}. ∎

For a multi-index α0m, write




and say αβ if αiβi for each i. We shall use the fact that

Lemma 2.

For fR0, if (αf)(0)=0 for all |α|<k, then [f]0m0k.


For k=1, if (αf)(0)=0 for α=(0,,0) then ev0f=f(0)=0, hence [f]0𝔪0. Suppose the claim is true for some k1, and suppose that f0 and that (αf)(0)=0 for all |α|<k+1. A fortiori, (αf)(0)=0 for all |α|<k and then by the induction hypothesis we get [f]0𝔪0k. Now, Lemma 1 tells us that the ideal 𝔪0 is generated by the set {[xi]0:1im}, and then the product ideal 𝔪0k is generated by the set

{[xi1]0[xik]0:1i1,,ikm} ={[xi1xik]0:1i1,,ikm}

for xα=(x1)α1(xm)αm. As [f]0𝔪k, there are [uα]00, |α|=k, such that


For |α|=k, on some set in 𝒰0, using the Leibniz rule,


And for γβ, (γxβ)(0)=0, so


But (αf)(0)=0, so uα(0)=0, which means that uα𝔪0. And


so [uα]0[xα]0𝔪0k+1, showing that [f]0𝔪0k+1. This completes the proof by induction. ∎

4 Hessians

For an open set U in m and ϕ(U), ϕ:U(m,), and ϕ:Um satisfies


xU is a critical point of ϕ if ϕ(x)=0, equivalently ϕ(x)=0. Define Hessϕ:U(m,m) by


This satisfies22 2


A critical point x of ϕ is called nondegenerate if Hessϕ(x) is invertible in (m,m).

For ϕp, let Jϕ be the ideal in the ring p generated by the set


We call Jϕ the Jacobian ideal of ϕ at p. If p is a critical point of ϕ, then (iϕ)(p)=0 for each i, hence [iϕ]p𝔪p for each i.

If 0 is a nondegenerate critical point of ϕ, we prove that 𝔪0Jϕ.33 3 Liviu Nicolaescu, An Invitation to Morse Theory, second ed., p. 15, Lemma 1.15.

Theorem 3.

Let U be an open set in Rm containing 0 and let ϕF(U). If 0 is a nondegenerate critical point of ϕ, then Jϕ=m0.


Let f=ϕ, which is a smooth function Um. Because 0 is a nondegenerate critical point of ϕ, f(0) is invertible in (m,m) and hence by the inverse function theorem,44 4 Serge Lang, Real and Functional Analysis, third ed., p. 361, chapter XIV, Theorem 1.2. f is a local C isomorphism at x: there is some open set V, xV and VU, such that W=f(V) is open in m, and there is a smooth function g:WV such that gf=idV and fg=idW. ∎