Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 15 / 4

A subset $Z$ of an $m$-dimensional smooth manifold $M$ is an $r$-dimensional embedded submanifold if, for each $p\in Z$, some coordinate chart $x:U\to V\subset {\mathbb{R}}^m$ containing $p$ satisfiesThe induced topology on $Z$ is the subspace topology, and the first $r$ coordinates restricted to $Z$ give its smooth atlas. To see their smooth transition maps agree, restrict the ambient smooth transitions to the coordinate planes; their inverses restrict in the same way. The inclusion then has injective differential of a smooth map and is a homeomorphism onto its image. Conversely this is the local coordinate description of a smooth smooth embedding. An embedded submanifold need not be a closed subset of $M$.

Put $m=\dim M$ and $n=\dim N$. The hypothesis says that $q$ is a regular value of $F$. Since its preimage is nonempty and the differential of a smooth map is surjective there, $m\ge n$.

We use the following finite-dimensional inverse function theorem: if a smooth map $H:O\to {\mathbb{R}}^m$, with $O\subset {\mathbb{R}}^m$ open, has invertible derivative at $a$, there are open neighborhoods of $a$ and $H(a)$ on which $H$ is a diffeomorphism, with smooth inverse.

Fix $p\in F^{-1}(q)$. Choose coordinate charts centered at $p$ and $q$, and shrink the source neighborhood so that its image under $F$ lies in the target chart. Its coordinate expression is $f=(f^1,\dots ,f^n)$ with $f(0)=0$ and derivative of rank $n$. Choose $n$ domain coordinates giving an invertible $n\times n$ minor, and relabel them as $x^1,\dots ,x^n$. DefineIts derivative at zero has block formso it is invertible. The inverse function theorem makes $u=H(x)$ a new coordinate chart near $p$. In this chart $F=q$ is exactly $u^1=\cdots =u^n=0$, because the target chart sends $q$ to zero. These are slice charts for an embedded submanifold; reordering coordinates if desired puts the free coordinates first. Giving the level set the subspace topology, these restricted charts are smoothly compatible, as in part (a). This proves the regular level set theorem rather than merely invoking it.

The level set is an embedded submanifold of dimension $\boxed{m-n}$. The same coordinates show $T_{p}Z=\ker d{F}_p$: tangent vectors to the slice have zero first $n$ components, which are exactly the components of $d{F}_p$. The empty-dimensional case $m=n$ gives a discrete zero-dimensional embedded submanifold.

Let $m=\dim M$, $n=\dim N$ and $k={\mathrm{codim}}_{N}S$. Fix $p\in f^{-1}(S)$ and choose the allowed slice chart for an embedded submanifold on a neighborhood $U$ of $f(p)$, so that $S\cap U$ is given by $x^1=\cdots =x^k=0$. On $f^{-1}(U)$ define the smooth mapThe derivative $L=d(x^1,\dots ,x^k)$ at a point of $S\cap U$ is surjective and has kernel $TS$, by the slice description. The printed sum-of-spaces condition givesIndeed any target tangent vector is a sum of an image vector and a tangent vector to $S$, and $L$ kills the latter. By the chain rule, $d{G}_r=L\circ d{f}_r$ is therefore surjective. This condition is precisely transversality of a map to a submanifold; the original sum need not be direct.

We have $G^{-1}(0)=f^{-1}(S)\cap f^{-1}(U)$. Part (b), applied to $G$, gives slice charts for an embedded submanifold of codimension $k$ around every point of this zero set. Since the construction is available around every $p\in f^{-1}(S)$, the charts give its global structure as an embedded submanifold with the subspace topology. Its dimension is $\boxed{m-k=m-n+\dim S}$. Moreover,This follows either from the slice coordinates or from $T_p{G}^{-1}(0)=\ker d{G}_p$ and $\ker L=T_{f(p)}S$. No injectivity of $df$, closedness of $S$, or direct-sum hypothesis is required.