Germs of smooth functions

Jordan Bell

April 4, 2016

1 Sheafs

Let $M=\mathbb{R}^{m}$ . For an open set $U$ in $M$ , write $\mathcal{F}(U)=C^{\infty}(U)$ , which is a commutative ring with unity $1_{M}(x)=1$ . For open sets $V\subset U$ in $M$ , define $r_{U,V}:\mathcal{F}(U)\to\mathcal{F}(V)$ by $r_{U,V}f=f|_{V}$ , which is a homomorphism of rings. $\mathcal{F}$ is a presheaf, a contravariant functor from the category of open sets in $M$ to the category of commutative unital rings. For $\mathcal{F}$ to be a sheaf means the following:

1.

If $U_{i}$ , $i\in I$ , is an open cover of an open set $U$ and if $f,g\in\mathcal{F}(U)$ satisfy $r_{U,U_{i}}f=r_{U,U_{i}}g$ for all $i\in I$ , then $f=g$ .
2.

If $U_{i}$ , $i\in I$ , is an open cover of an open set $U$ and for each $i\in I$ there is some $f_{i}\in\mathcal{F}(U_{i})$ such that for all $i,j\in I$ , $r_{U_{i},U_{i}\cap U_{j}}f_{i}=r_{U_{j},U_{i}\cap U_{j}}f_{j}$ , then there is some $f\in\mathcal{F}(U)$ such that $r_{U,U_{i}}f=f_{i}$ for each $i\in I$ .

For the first condition, let $p\in U$ . As $U_{i}$ is an open cover of $U$ , there is some $i$ for which $p\in U_{i}$ . As $f|_{U_{i}}=g|_{U_{i}}$ , $f(p)=g(p)$ . Therefore $f=g$ . For the second condition, let $p\in U$ . If $p\in U_{i}$ and $p\in U_{j}$ , then $f_{i}(p)=f_{j}(p)$ . This shows that it makes sense to define $f:U\to\mathbb{R}$ by $f(p)=f_{i}(p)$ , for any $i$ such that $p\in U_{i}$ . Then $f|_{U_{i}}=f_{i}$ , which implies that $f\in\mathcal{F}(U)$ : for each $p\in U$ , there is some open neighborhood $U_{i}$ of $p$ on which $f$ is smooth. Therefore $\mathcal{F}$ is a sheaf.

2 Stalks and germs

For $p\in M$ , let $\mathcal{U}_{p}$ be the set of open neighborhoods of $p$ . For $U,V\in\mathcal{U}_{p}$ , say $U\leq V$ when $V\subset U$ . For $U\leq V\leq W$ and $f\in\mathcal{F}(U)$ ,

(r_{V,W}\circ r_{U,V})(f)=r_{V,W}f|_{V}=f_{W}=r_{U,W}f.

For $f\in\mathcal{F}(U)$ and $g\in\mathcal{F}(V)$ , say $f\sim_{p}g$ if there is some $W\in\mathcal{U}_{p}$ , $W\geq U$ , $W\geq V$ , such that $r_{U,W}f=r_{V,W}g$ . Let

\mathcal{R}_{p}=\bigsqcup_{U\in\mathcal{U}_{p}}\mathcal{F}(U),

and let $\mathcal{F}_{p}$ be the direct limit of the direct system $\mathcal{F}(U)$ , $r_{U,V}$ of commutative unital rings:

\mathcal{F}_{p}=\mathcal{R}_{p}/\sim_{p}.

We call $\mathcal{F}_{p}$ the stalk of $\mathcal{F}$ at $p$ . An element of $\mathcal{F}_{p}$ is called a germ of $\mathcal{F}$ at $p$ . In other words, for $f\in\mathcal{R}_{p}$ , let $[f]_{p}$ be the set of those $g\in\mathcal{R}_{p}$ such that $f\sim_{p}g$ , equivalently, $f|_{U_{f}\cap U_{g}}=g|_{U_{f}\cap U_{g}}$ . A germ of $\mathcal{F}$ at $p$ is such an equivalence class $[f]_{p}$ , and

\mathcal{F}_{p}=\left\{[f]_{p}:f\in\mathcal{R}_{p}\right\}.

3 Maximal ideals

For $p\in M$ , and $f,g\in\mathcal{R}_{p}$ with $f\sim_{p}g$ , $f(p)=g(p)$ . Thus it makes sense to define $\mathrm{ev}_{p}:\mathcal{F}_{p}\to\mathbb{R}$ by $\mathrm{ev}_{p}[f]_{p}=f(p)$ . Now, for $[f]_{p},[g]_{p}\in\mathcal{F}_{p}$ ,

\mathrm{ev}_{p}([f]_{p}+[g]_{p})=\mathrm{ev}_{p}([f+g]_{p})=(f+g)(p)=f(p)+g(p)% =\mathrm{ev}_{p}[f]_{p}+\mathrm{ev}_{p}[g]_{p},

\mathrm{ev}_{p}([f]_{p}[g]_{p})=\mathrm{ev}_{p}([fg]_{p})=(fg)(p)=f(p)g(p)=% \mathrm{ev}_{p}[f]_{p}\cdot\mathrm{ev}_{p}[g]_{p},

$\mathrm{ev}_{p}[1_{M}]_{p}=1$ . This means that $\mathrm{ev}_{p}:\mathcal{F}_{p}\to\mathbb{R}$ is a homomorphism of unital rings. It is straightforward that $\mathrm{ev}_{p}$ is surjective. Write $\mathfrak{m}_{p}=\ker\mathrm{ev}_{p}$ . By the first isomorphism theorem, there is an isomorphism of unital rings $\mathcal{F}_{p}/\mathfrak{m}_{p}\to\mathbb{R}$ . Therefore $\mathfrak{m}_{p}$ is a maximal ideal in $\mathcal{F}_{p}$ . Now, if $[f]_{p}\in\mathcal{F}_{p}\setminus\mathfrak{m}_{p}$ then $\mathrm{ev}_{p}[f]_{p}\neq 0$ , hence $f(p)\neq 0$ . Then there is some $U\in\mathcal{U}_{p}$ such that $f(x)\neq 0$ for $x\in U$ , and $(1/f)(x)=\frac{1}{f(x)}$ belongs to $\mathcal{F}(U)$ . Then $[1/f]_{p}\in\mathcal{F}_{p}$ and $[f]_{p}\cdot[1/f]_{p}=[f\cdot 1/f]_{p}=[1_{M}]_{p}$ , which shows that if $[f]_{p}\in\mathcal{F}_{p}\setminus\mathfrak{m}_{p}$ then $[f]_{p}$ has an inverse $[1/f]_{p}$ in $\mathcal{F}_{p}$ . This means $\mathfrak{m}_{p}$ is the set of noninvertible elements of $\mathcal{F}_{p}$ , which means that $\mathcal{F}_{p}$ is a local ring.

For $1\leq i\leq m$ define the coordinate function $x^{i}:M\to\mathbb{R}$ by $x^{i}(p)=p_{i}$ , which belongs to $\mathcal{F}(M)$ . Because $\mathrm{ev}_{0}x^{i}=0$ , $[x^{i}]_{0}\in\mathfrak{m}_{0}$ . We prove Hadamard’s lemma, that the ring $\mathfrak{m}_{0}$ 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 $\mathfrak{m}_{0}$ is generated by the set $\{[x^{i}]_{0}:1\leq i\leq m\}$ .

Proof.

Let $[f]_{0}\in\mathfrak{m}_{0}$ with $f\in\mathcal{F}(B_{r})$ for some $r>0$ . For $y\in B_{r}$ , using the fundamental theorem of calculus and using the chain rule,

f(y)=f(y)-f(0)=\int_{0}^{1}\frac{d}{ds}f(sy)ds=\int_{0}^{1}\sum_{i=1}^{m}x^{i}% (y)(\partial_{i}f)(sy)ds=\sum_{i=1}^{m}x^{i}(y)u_{i}(y),

and $u_{i}\in\mathcal{F}(B_{r})$ . This means that $[f]_{0}=\sum_{i=1}^{m}[x^{i}]_{0}[u_{i}]_{0}$ , which shows that $[f]_{0}$ belongs to the ideal generated by the set $\{[x^{i}]_{0}:1\leq i\leq m\}$ . ∎

For a multi-index $\alpha\in\mathbb{Z}_{\geq 0}^{m}$ , write

|\alpha|=\sum_{i=1}^{m}\alpha_{i},\qquad\alpha!=\alpha_{1}!\cdots\alpha_{m}!

and

\partial^{\alpha}=\partial_{1}^{\alpha_{1}}\cdots\partial_{m}^{\alpha_{m}},% \qquad x^{\alpha}=(x^{1})^{\alpha_{1}}\cdots(x^{m})^{\alpha_{m}},

and say $\alpha\leq\beta$ if $\alpha_{i}\leq\beta_{i}$ for each $i$ . We shall use the fact that

\partial^{\alpha}x^{\beta}=\begin{cases}\frac{\beta!}{(\beta-\alpha)!}x^{\beta% -\alpha}&\alpha\leq\beta\\ 0&\textrm{otherwise}.\end{cases}

Lemma 2.

For $f\in\mathcal{R}_{0}$ , if $(\partial^{\alpha}f)(0)=0$ for all $|\alpha|<k$ , then $[f]_{0}\in\mathfrak{m}_{0}^{k}$ .

Proof.

For $k=1$ , if $(\partial^{\alpha}f)(0)=0$ for $\alpha=(0,\ldots,0)$ then $\mathrm{ev}_{0}f=f(0)=0$ , hence $[f]_{0}\in\mathfrak{m}_{0}$ . Suppose the claim is true for some $k\geq 1$ , and suppose that $f\in\mathcal{R}_{0}$ and that $(\partial^{\alpha}f)(0)=0$ for all $|\alpha|<k+1$ . A fortiori, $(\partial^{\alpha}f)(0)=0$ for all $|\alpha|<k$ and then by the induction hypothesis we get $[f]_{0}\in\mathfrak{m}_{0}^{k}$ . Now, Lemma 1 tells us that the ideal $\mathfrak{m}_{0}$ is generated by the set $\{[x^{i}]_{0}:1\leq i\leq m\}$ , and then the product ideal $\mathfrak{m}_{0}^{k}$ is generated by the set

	$\displaystyle\{[x^{i_{1}}]_{0}\cdots[x^{i_{k}}]_{0}:1\leq i_{1},\ldots,i_{k}% \leq m\}$	$\displaystyle=\{[x^{i_{1}}\cdots x^{i_{k}}]_{0}:1\leq i_{1},\ldots,i_{k}\leq m\}$
		$\displaystyle=\{[x^{\alpha}]_{0}:\|\alpha\|=k\},$

for $x^{\alpha}=(x^{1})^{\alpha_{1}}\cdots(x^{m})^{\alpha_{m}}$ . As $[f]_{0}\in\mathfrak{m}^{k}$ , there are $[u_{\alpha}]_{0}\in\mathcal{F}_{0}$ , $|\alpha|=k$ , such that

[f]_{0}=\sum_{|\alpha|=k}[u_{\alpha}]_{0}[x^{\alpha}]_{0}.

For $|\alpha|=k$ , on some set in $\mathcal{U}_{0}$ , using the Leibniz rule,

\partial^{\alpha}f=\sum_{|\beta|=k}\partial^{\alpha}(u_{\beta}x^{\beta})=\sum_% {|\beta|=k}\sum_{\gamma\leq\alpha}\binom{\alpha}{\gamma}(\partial^{\alpha-% \gamma}u_{\beta})(\partial^{\gamma}x^{\beta}).

And for $\gamma\neq\beta$ , $(\partial^{\gamma}x^{\beta})(0)=0$ , so

\partial^{\alpha}f\in u_{\alpha}\partial^{\alpha}x^{\alpha}+h,\qquad[h]_{0}\in% \mathfrak{m}_{0}.

But $(\partial^{\alpha}f)(0)=0$ , so $u_{\alpha}(0)=0$ , which means that $u_{\alpha}\in\mathfrak{m}_{0}$ . And

[x^{\alpha}]_{0}=[x^{1}]_{0}^{\alpha_{1}}\cdots[x^{m}]_{0}^{\alpha_{m}}\in% \mathfrak{m}_{0}^{|\alpha|}=\mathfrak{m}_{0}^{k},

so $[u_{\alpha}]_{0}[x^{\alpha}]_{0}\in\mathfrak{m}_{0}^{k+1}$ , showing that $[f]_{0}\in\mathfrak{m}_{0}^{k+1}$ . This completes the proof by induction. ∎

4 Hessians

For an open set $U$ in $\mathbb{R}^{m}$ and $\phi\in\mathcal{F}(U)$ , $\phi^{\prime}:U\to\mathscr{L}(\mathbb{R}^{m},\mathbb{R})$ , and $\nabla\phi:U\to\mathbb{R}^{m}$ satisfies

\left\langle\nabla\phi(x),v\right\rangle=\phi^{\prime}(x)(v),\qquad x\in U,% \quad v\in\mathbb{R}^{m}.

$x\in U$ is a critical point of $\phi$ if $\phi^{\prime}(x)=0$ , equivalently $\nabla\phi(x)=0$ . Define $\mathrm{Hess}\,\phi:U\to\mathscr{L}(\mathbb{R}^{m},\mathbb{R}^{m})$ by

\mathrm{Hess}\,\phi=(\nabla\phi)^{\prime}.

This satisfies²² 2 http://individual.utoronto.ca/jordanbell/notes/gradienthilbert.pdf

\phi^{\prime\prime}(x)(u)(v)=\left\langle v,\mathrm{Hess}\,\phi(x)(u)\right% \rangle,\qquad x\in U,\qquad u,v,\in\mathbb{R}^{m}.

A critical point $x$ of $\phi$ is called nondegenerate if $\mathrm{Hess}\,\phi(x)$ is invertible in $\mathscr{L}(\mathbb{R}^{m},\mathbb{R}^{m})$ .

For $\phi\in\mathcal{R}_{p}$ , let $J_{\phi}$ be the ideal in the ring $\mathcal{F}_{p}$ generated by the set

\{[\partial_{i}\phi]_{p}:1\leq i\leq m\}.

We call $J_{\phi}$ the Jacobian ideal of $\phi$ at $p$ . If $p$ is a critical point of $\phi$ , then $(\partial_{i}\phi)(p)=0$ for each $i$ , hence $[\partial_{i}\phi]_{p}\in\mathfrak{m}_{p}$ for each $i$ .

If $0$ is a nondegenerate critical point of $\phi$ , we prove that $\mathfrak{m}_{0}\subset J_{\phi}$ .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 $\mathbb{R}^{m}$ containing $0$ and let $\phi\in\mathcal{F}(U)$ . If $0$ is a nondegenerate critical point of $\phi$ , then $J_{\phi}=\mathfrak{m}_{0}$ .

Proof.

Let $f=\nabla\phi$ , which is a smooth function $U\to\mathbb{R}^{m}$ . Because $0$ is a nondegenerate critical point of $\phi$ , $f^{\prime}(0)$ is invertible in $\mathscr{L}(\mathbb{R}^{m},\mathbb{R}^{m})$ 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 $C^{\infty}$ isomorphism at $x$ : there is some open set $V$ , $x\in V$ and $V\subset U$ , such that $W=f(V)$ is open in $\mathbb{R}^{m}$ , and there is a smooth function $g:W\to V$ such that $g\circ f=\mathrm{id}_{V}$ and $f\circ g=\mathrm{id}_{W}$ . ∎