0708-1300/Class notes for Thursday, October 4

0708-1300/Navigation Panel   Add your name / see who's in! # Week of... Links Fall Semester 1 Sep 10 About, Tue, Thu 2 Sep 17 Tue, HW1, Thu 3 Sep 24 Tue, Photo, Thu 4 Oct 1 Questionnaire, Tue, HW2, Thu 5 Oct 8 Thanksgiving, Tue, Thu 6 Oct 15 Tue, HW3, Thu 7 Oct 22 Tue, Thu 8 Oct 29 Tue, HW4, Thu, Hilbert sphere 9 Nov 5 Tue,Thu, TE1 10 Nov 12 Tue, Thu 11 Nov 19 Tue, Thu, HW5 12 Nov 26 Tue, Thu 13 Dec 3 Tue, Thu, HW6 Spring Semester 14 Jan 7 Tue, Thu, HW7 15 Jan 14 Tue, Thu 16 Jan 21 Tue, Thu, HW8 17 Jan 28 Tue, Thu 18 Feb 4 Tue 19 Feb 11 TE2, Tue, HW9, Thu, Feb 17: last chance to drop class R Feb 18 Reading week 20 Feb 25 Tue, Thu, HW10 21 Mar 3 Tue, Thu 22 Mar 10 Tue, Thu, HW11 23 Mar 17 Tue, Thu 24 Mar 24 Tue, HW12, Thu 25 Mar 31 Referendum,Tue, Thu 26 Apr 7 Tue, Thu R Apr 14 Office hours R Apr 21 Office hours F Apr 28 Office hours, Final (Fri, May 2) Register of Good Deeds Errata to Bredon's Book

Announcements go here

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

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

During the previous class, we discussed immersions---smooth maps whose differentials are injective. This class deals with the dual notion of submersions, defined as follows:

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.

Remarks

We had previously seen that immersions induce "nice" coordinate charts---ones where the immersion looks like the canonical inclusion $\iota :\mathbb {R} ^{m}\rightarrow \mathbb {R} ^{n}$ (where $n\geq m\!$). The proof of this theorem made use of the Inverse Function Theorem on a function defined on a chart of $N\!$. In the case of submersions, there is a similar theorem. Submersions locally look like the canonical projection $\pi :\mathbb {R} ^{m}=\mathbb {R} ^{n}\times \mathbb {R} ^{m-n}\rightarrow \mathbb {R} ^{n}\!$, and the proof of this fact makes use of the Inverse Function Theorem for a function define on a chart of $M\!$ (duality!). However, before we can prove this theorem, we will need the following lemma.

Lemma

Let $V\!$ and $W\!$ be finite-dimensional vector spaces over $\mathbb {R} \!$ and let $T:V\rightarrow W\!$ be a surjective linear map. Then there exist bases $v=\{v_{1},\ldots ,v_{m}\}\!$ for $V\!$ and $w=\{w_{1},\ldots ,w_{n}\}\!$ for $W\!$ such that the matrix representative of $T\!$ with respect to $v\!$ and $w\!$ is that of the canonical projection $\pi :\mathbb {R} ^{m}\rightarrow \mathbb {R} ^{n}\!$.

Proof

Let $w=\{w_{1},\ldots ,w_{n}\}\!$ be any basis for $W\!$ and choose $\{v_{1},\ldots ,v_{n}\}\subset V\!$ such that $T(v_{i})=w_{i}\!$ for each $i\in \{1,\ldots ,n\}\!$ (this may be done since $T\!$ is surjective). We claim that the set $v'=\{v_{1},\ldots ,v_{n}\}\subset V\!$ is linearly independent. Suppose it were not. Then there would exist $\{c_{1},\ldots ,c_{n}\}\subset \mathbb {R} \!$ with $\sum _{i=1}^{n}c_{i}v_{i}=0\!$ and $c_{i}\neq 0\!$ for some $i\in \{1,\ldots ,n\}\!$. But then $0=T\left(\sum _{i=1}^{n}c_{i}v_{i}\right)=\sum _{i=1}^{n}c_{i}T\left(v_{i}\right)=\sum _{i=1}^{n}c_{i}w_{i}\!$ by linearity of $T\!$, contradicting the assumption that $w\!$ is a basis.

Note that $\mathrm {dim} (\mathrm {ker} (T))=m-n\!$. Hence we may find a basis $v''=\{v_{n+1},\ldots ,v_{m}\}\subset V\!$ for $\mathrm {ker} (T)\!$. Since $\mathrm {span} (v')\cap \mathrm {ker} (T)=\{0\in V\}\!$, the set $v=v'\cup v''$ must be linearly independent and hence form a basis for $V\!$. We then have $w_{i}=\sum _{i=j}^{m}\delta _{ij}T(v_{j})\!$, so that the matrix representative of $T\!$ is $\left(I_{n\times n}|0_{n\times (m-n)}\right)\!$, which is the matrix representative of $\pi \!$. $\Box \!$

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, $V\!$ is open, $U_{0}\!$ is open and $\theta \!$ is continuous, so $U_{0}\cap \theta ^{-1}\left(V\right)\subset M\!$ is open and contains $p\!$. Hence, 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 the previous lemma to obtain a change of coordinates $T:\mathbb {R} ^{m}\rightarrow \mathbb {R} ^{m}$ such that $D_{0}=\pi \circ T$.

Let $\phi _{1}=T\circ \phi _{0}\!:U_{0}\rightarrow U_{1}'\subset \mathbb {R} ^{m}$. Then $\phi _{1}\!$ is a chart because Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T\!} is a diffeomorphism. Let $\theta _{1}=\psi \circ \theta \circ \phi _{1}^{-1}:U_{1}'\rightarrow V'\!$ be the corresponding local representative, define $\zeta :U_{1}'\rightarrow \mathbb {R} ^{m}\!$ 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 Theorem 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 $\phi _{1}(q)=(x,y)\!$ 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 $\theta _{1}\!$. Hence, $\pi \circ \phi =\psi \circ \theta |_{U}\!$ and the proof is complete.$\Box \!$