0708-1300/Class notes for Thursday, October 4

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Movie Time

With the word "immersion" in our minds, we watch the movie "Outside In". Also see the movie's home, a talk I once gave, and the movie itself, on google video.

Class Notes

Definition

Let $\theta :M^{m}\rightarrow N^{n}\!$ be a smooth map between manifolds. If for each $p\in M\!$ the differential $d\theta _{p}:T_{p}M\rightarrow T_{\theta (p)}N\!$ is surjective, $\theta \!$ is called a submersion.

Theorem

If $\theta :M^{m}\rightarrow N^{n}\!$ is a smooth map between manifolds and for some $p\in M\!$ the differential $d\theta _{p}:T_{p}M\rightarrow T_{\theta (p)}N\!$ is surjective then there exist charts $\phi :U\rightarrow U'\subset \mathbb {R} ^{m}\!$ and $\psi :V\rightarrow V'\subset \mathbb {R} ^{n}\!$ on $M\!$ and $N\!$ respectively such that

1. $\phi (p)=0\!$
2. $\psi \left(\theta (p)\right)=0\!$
3. The diagram

   

   commutes, where $\pi :\mathbb {R} ^{m}=\mathbb {R} ^{n}\times \mathbb {R} ^{m-n}\rightarrow \mathbb {R} ^{n}\!$ is the canonical projection.

Proof

Since translations are diffeomorphisms of $\mathbb {R} ^{k}\!$ for every $k\in \mathbb {N} \!$, it is trivial to find charts $\phi _{0}:U_{0}\rightarrow U_{0}'\subset \mathbb {R} ^{m}\!$ and $\psi :V\rightarrow V'\subset \mathbb {R} ^{m}\!$ such that $\phi (p)=0\in \mathbb {R} ^{m}\!$ and $\psi (\theta (p))=0\in \mathbb {R} ^{n}\!$. Furthermore, since $V\!$ is open, $U_{0}\!$ is open and $\theta \!$ is continuous, $U_{0}\cap \theta ^{-1}\left(V\right)\subset M\!$ is open so that we may assume $U_{0}\subset \theta ^{-1}\left(V\right)\!$ without loss of generality.

Let $\theta _{0}=\psi \circ \theta \circ \phi _{0}^{-1}:U_{0}'\rightarrow V'\!$ be the local representative of $\theta \!$ and let $D_{0}:\mathbb {R} ^{m}\rightarrow \mathbb {R} ^{n}\!$ be the local representative of $d\theta _{p}\!$. Since $d\theta _{p}\!$ is onto, we may apply a change of basis $T:\mathbb {R} ^{m}\rightarrow \mathbb {R} ^{m}$ such that $D_{0}=T^{-1}\circ \pi \circ T$.

Let $\phi _{1}=T\circ \phi _{0}\!$. Then $\phi _{1}\!$ is a chart because $T\!$ is a diffeomorphism. Let $\theta _{1}=\psi \circ \theta \circ \phi _{1}^{-1}:U_{0}'\rightarrow V'\!$ be the corresponding local representative, define $\zeta :U_{0}'\rightarrow \mathbb {R} ^{m}\!$ by $\zeta (x,y)=(\theta _{1}(x,y),y)\!$, and let $D_{1}:\mathbb {R} ^{m}\rightarrow \mathbb {R} ^{n}\!$ be the differential of $\theta _{1}\!$ at $0\!$. Then, by construction of $T\!$ we have that $D_{1}=\pi \!$ and hence $d\zeta _{0}=\mathrm {id} _{\mathbb {R} ^{m}}\!$, which is invertible. Hence, the inverse function gives the existence of non-empty open set $U'\subset \zeta (U_{1}')\!$ such that $\zeta |_{\zeta ^{-1}(U')}\!$ is a diffeomorphism. Put $U=\phi _{1}^{-1}(\zeta ^{-1}(U'))\!$ and $\phi =\zeta \circ \phi _{1}|_{U}\!$. Then $\phi \!$ is a chart.

It remains to check that $\pi \circ \phi =\psi \circ \theta |_{U}\!$, but this is clear: if $(x,y)=\phi _{1}(q)\!$ for some $q\in U\!$, then $\pi \circ \phi (q)=\pi \circ \zeta (x,y)=\theta _{1}(x,y)=\psi (\theta (q))\!$ by definition of $\theta _{1}\!$. $\Box \!$