# 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. 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. 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.

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|, 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|. A fortiori, $(\partial^{\alpha}f)(0)=0$ for all $|\alpha| 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 satisfies2

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

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 $f\circ g=\mathrm{id}_{W}$. ∎