# 0708-1300/Class notes for Tuesday, September 25

Announcements go here

## Dror's Notes

• Class photo is on Thursday, show up and be at your best! More seriously -
• The class photo is of course not mandatory, and if you are afraid of google learning about you, you should not be in it.
• If you want to be in the photo but can't make it on Thursday, I'll take a picture of you some other time and add it as an inset to the main picture.
• I just got the following email message, which some of you may find interesting:
NSERC - CMS Math in Moscow Scholarships

The Natural Sciences and Engineering Research Council (NSERC) and the
Canadian Mathematical Society (CMS) support scholarships at \$9,000
each. Canadian students registered in a mathematics or computer
science program are eligible.

The scholarships are to attend a semester at the small elite Moscow
Independent University.

Math in Moscow program
www.mccme.ru/mathinmoscow/
Application details
www.cms.math.ca/bulletins/Moscow_web/

For additional information please see your department or call the CMS
at 613-562-5702.

Deadline September 30, 2007 to attend the Winter 2008 semester.


## 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.

### First hour

Recall from last class we were proving the equivalence of the two definitions for a smooth manifold. The only nontrivial point that remained to be proved was that if we started with the definition of a manifold in the sense of functional structures and produced charts ${\displaystyle \varphi ,\psi }$ that these charts would satisfy the property of a manifold, defined in the atlas sense, that ${\displaystyle \psi \circ \varphi ^{-1}}$ is smooth where defined.

Proof

${\displaystyle \psi \circ \varphi ^{-1}:\mathbb {R} ^{n}\rightarrow \mathbb {R} ^{n}}$ is smooth ${\displaystyle \Leftrightarrow }$ ${\displaystyle (\psi \circ \varphi ^{-1})_{i}:\mathbb {R} ^{n}\rightarrow \mathbb {R} }$ is smooth ${\displaystyle \forall }$ i ${\displaystyle \Leftrightarrow }$ ${\displaystyle \pi _{i}\circ \psi \circ \varphi ^{-1}}$ is smooth where ${\displaystyle \pi _{i}}$ is the ${\displaystyle i^{th}}$ coordinate projection map.

Now, since ${\displaystyle \pi _{i}}$ is always smooth, ${\displaystyle \pi _{i}\in F_{\mathbb {R} ^{n}}(U_{\psi }^{'})}$

But then we have ${\displaystyle \pi _{i}\circ \psi \in F_{M}(U_{\psi })}$ and so, by a property of functional structures, ${\displaystyle \pi _{i}\circ \psi |_{U_{\varphi }\bigcap U_{\psi }}\in F_{M}(U_{\varphi }\bigcap U_{\psi })}$ and hence ${\displaystyle \pi _{i}\circ \psi \circ \varphi ^{-1}\in F_{\mathbb {R} ^{n}}}$ where it is defined and thus is smooth. QED

Definition 1 (induced structure) Suppose ${\displaystyle \pi :X\rightarrow Y}$ and suppose Y is equipped with a functional structure ${\displaystyle F_{Y}}$ then the "induced functional structure" on X is

${\displaystyle F_{X}(U)=\{f:U\rightarrow \mathbb {R} \ |\ \exists g\in F_{Y}(V)\ such\ that\ V\supset \pi (U)\ and\ f=g\circ \pi \}}$

Claim: this does in fact define a functional structure on X

Definition 2 This is the reverse definition of that given directly above. Let ${\displaystyle \pi :X\rightarrow Y}$ and let X be equipped with a functional structure ${\displaystyle F_{X}}$. Then we get a functional structure on Y by ${\displaystyle F_{Y}(V)=\{g:V\rightarrow \mathbb {R} \ |\ g\circ \pi \in F_{X}(\pi ^{-1}(V)\}}$ Claim: this does in fact define a functional structure on X

Example 1 Let ${\displaystyle S^{2}=\mathbb {R} ^{3}-\{0\}/}$~ where the equivalence relation ~ is given by x~${\displaystyle \alpha }$x for ${\displaystyle \alpha }$>0 We thus get a canonical projection map ${\displaystyle \pi :\mathbb {R} ^{3}-\{0\}\rightarrow S^{2}}$ and hence, there is an induced functional structure on ${\displaystyle S^{2}}$. Claim: 1) This induced functional structure makes ${\displaystyle S^{2}}$ into a manifold 2) This resulting manifold is the same manifold as from the atlas definition given previously

Example 2 Consider the torus thought of as ${\displaystyle T^{2}=\mathbb {R} ^{2}/\mathbb {Z} ^{2}}$, i.e., the real plane with the equivalence relation that (x,y)~(x+n,y+m) for (x,y) in ${\displaystyle \mathbb {R} ^{2}}$and (n,m) in ${\displaystyle \mathbb {Z} ^{2}}$

As in the previous example, the torus inherits a functional structure from the real plane we must again check that 1) We get a manifold 2) This is the same manifold as we had previously with the atlas definition

Example 3 Let ${\displaystyle CP^{n}}$ denote the n dimensional complex projective space, that is, ${\displaystyle CP^{n}=\mathbb {C} ^{n+1}-\{0\}/}$~ where ${\displaystyle [z_{0},...,z_{n}]}$ ~ ${\displaystyle [\alpha z_{0},...,\alpha z_{n}]}$ where ${\displaystyle \alpha \in \mathbb {C} }$

Again, this space inherits a functional structure from ${\displaystyle \mathbb {C} ^{n+1}}$ and we again need to claim that this yields a manifold.

Proof of Claim

We consider the subsets ${\displaystyle CP^{n}\supset U_{i}=\{[z_{0},...,z_{n}]\ |\ z_{i}\neq 0\}}$ for ${\displaystyle 0\leq i\leq n}$

Clearly ${\displaystyle \bigcup U_{i}=CP^{n}}$

Now, for each ${\displaystyle p\in U_{i}}$ there is a unique representative for its equivalence class of the form ${\displaystyle [z_{0},...,1,...,z_{n}]}$ where the 1 is at the ith location.

We thus can get a map from ${\displaystyle \varphi _{i}:U_{i}\rightarrow \mathbb {C} ^{n}=\mathbb {R} ^{2n}}$ by ${\displaystyle p\mapsto [z_{0}/z_{i},...,z_{i-1}/z_{i},z_{i+1}/z_{i},...,z_{n}/z_{i}]}$ Hence we have shown (loosely) that our functional structure is locally isormorphic to ${\displaystyle (\mathbb {R} ,C^{\infty })}$

Definition 3 Product Manifolds

Suppose ${\displaystyle M^{m}}$ and ${\displaystyle N^{n}}$ are manifolds. Then the product manifold, on the set MxN has an atlas given by ${\displaystyle \{\varphi \times \psi :U\times V\rightarrow U'\times V'\in \mathbb {R} ^{m}\times \mathbb {R} ^{n}\ |\varphi :U\rightarrow U'\subset \mathbb {R} ^{m}\ and\ \psi :V\rightarrow V'\subset \mathbb {R} ^{n}}$ are charts in resp. manifolds}

Claim: This does in fact yield a manifold

Example 4 It can be checked that ${\displaystyle T^{2}=S^{1}\times S^{1}}$ gives the torus a manifold structure, by the product manifold, that is indeed the same as the normal structure given previously.

### Second Hour

Aim: We consider two manifolds, M and N, and a function f between them. We aim for the analogous idea of the tangent in the reals, namely that every smooth ${\displaystyle f:M\rightarrow N}$ has a good linear approximation. Of course, we will need to define what is meant by such a smooth function, as well as what this linear approximation is.

Definition 1 Smooth Function, Atlas Sense

Let ${\displaystyle M^{m}}$ and ${\displaystyle N^{n}}$ be manifolds. We say that a function ${\displaystyle f:M^{m}\rightarrow N^{n}}$ is smooth if ${\displaystyle \psi \circ f\circ \varphi ^{-1}}$ is smooth, where it makes sense, ${\displaystyle \forall }$ charts ${\displaystyle \psi ,\varphi }$

Definition 2 Smooth Function, Functional Structure Sense

Let ${\displaystyle M^{m}}$ and ${\displaystyle N^{n}}$ be manifolds. We say that a function ${\displaystyle f:M^{m}\rightarrow N^{n}}$ is smooth if ${\displaystyle \forall h:V\rightarrow \mathbb {R} }$ such that h is smooth, ${\displaystyle h\circ f}$ is smooth on subsets of M where it is defined.

Claim 1

The two definitions are equivalent

Definition 3 Category (Loose Definition)

A category is a collection of "objects" (such as sets, topological spaces, manifolds, etc.) such that for any two objects, x and y, there exists a set of "morphisms" denoted mor(${\displaystyle x\rightarrow y}$) (these would be functions for sets, continuous functions for topological spaces, smooth functions for manifolds, etc.) along with

1)Composition maps ${\displaystyle \circ }$:mor(${\displaystyle x\rightarrow y}$) ${\displaystyle \times }$ mor(${\displaystyle y\rightarrow z}$) ${\displaystyle \rightarrow }$ mor (${\displaystyle x\rightarrow z}$)

2) For any x there exists an element ${\displaystyle 1_{x}\in }$mor(${\displaystyle x\rightarrow x}$)

such that several axioms are satisfied, including

a) Associativity of the composition map

b) ${\displaystyle 1_{x}}$ behaves like an identity should.

Claim 2

The collection of smooth manifolds with smooth functions form a category. This (loosely) amounts to (easily) checking the two claims that

1) if ${\displaystyle f:M\rightarrow N}$ and ${\displaystyle g:N\rightarrow L}$ are smooth maps between manifolds then ${\displaystyle g\circ f:M\rightarrow L}$ is smooth

2) ${\displaystyle 1_{M}}$ is smooth

We now provide two alternate definition of the tangent vector, one from each definition of a manifold.

Definition 4 Tangent Vector, Atlas Sense

Consider the set of curves on the manifold, that is ${\displaystyle \{\gamma :\mathbb {R} \rightarrow M\ |\ s.t.\ \gamma \ smooth,\ \gamma (0)=p\}}$

We define an equivalence relation between paths by ${\displaystyle \gamma _{1}}$ ~ ${\displaystyle \gamma _{2}}$ if ${\displaystyle \varphi \circ \gamma _{1}}$ is tangent to ${\displaystyle \varphi \circ \gamma _{2}}$ as paths in ${\displaystyle \mathbb {R} ^{n}\ \forall \varphi }$.

Then, a tangent vector is an equivalence class of curves.

Definition 5 Tangent Vector, Functional Structure Sense

A tangent vector D at p, often called a directional derivative in this sense, is an operator that takes the smooth real valued functions near p into ${\displaystyle \mathbb {R} }$.

such that,

1) D(af +bg) = aDf + bDg for a,b constants in ${\displaystyle \mathbb {R} }$ and f,g such functions

2) D(fg) = (Df)g(p) + f(p)D(g) Leibniz' Rules

Theorem 1 These two definitions are equivalent.

Definition 6 Tangent Space

The tangent space at a point, ${\displaystyle T_{p}M}$ = the set of all tangent vectors at a point p

Claim 3

We will show later that in fact ${\displaystyle T_{p}M}$ is a vector space

Consider a map ${\displaystyle f:M\rightarrow N}$. We are interested in defining an associated map between the tangent spaces, namely, ${\displaystyle df_{p}:T_{p}M\rightarrow T_{f(p)}N}$

We get two different definitions from the two different definitions of a tangent vector

Definition 7 (atlas sense)

${\displaystyle (df_{p})([\gamma ]):=[f\circ \gamma ]}$

Definition 8 (functional structure sense)

${\displaystyle ((df_{p})(D))(h):=D(h\circ f)}$

Proof of Theorem 1

We consider a curve ${\displaystyle \gamma }$ and denote its equivalence class of curves by ${\displaystyle [\gamma ]}$

We consider the map ${\displaystyle \Phi :[\gamma ]\rightarrow D_{\gamma }}$ defined by ${\displaystyle D_{\gamma }(f)=(f\circ \gamma )'(0)}$

It is clear that ${\displaystyle D_{\gamma }}$ is linear and satisfies the leibnitz rule.

Claim: ${\displaystyle \Phi }$ is a bijection.

We will need to use the following lemma, which will not be proved now:

If ${\displaystyle f:\mathbb {R} ^{m}\rightarrow \mathbb {R} }$ is smooth near 0 then there exists smooth ${\displaystyle g_{i}:\mathbb {R} ^{m}\rightarrow \mathbb {R} }$ such that ${\displaystyle f(x)=f(0)+\Sigma _{i=1}^{n}x_{i}g_{i}(x)}$

Now, let D be a tangent vector. One can quickly see that D(const) = 0.

So, by Hadamard's lemma, ${\displaystyle Df=D(f(0)+\Sigma _{i=1}^{n}x_{i}g_{i}(x))=D(\Sigma _{i=1}^{n}x_{i}g_{i}(x))=\Sigma _{i=1}^{n}(Dx_{i})g_{i}(0)+\Sigma _{i=1}^{n}x_{i}(0)Dg_{i}}$ = ${\displaystyle \Sigma _{i=1}^{n}(Dx_{i})g_{i}(0)}$ as ${\displaystyle x_{i}(0)=0}$

and hence, Df is a linear comb of fixed quantities.

Now, let ${\displaystyle \gamma :\mathbb {R} \rightarrow \mathbb {R} ^{n}}$ be such that ${\displaystyle \gamma (0)=0}$

${\displaystyle D_{\gamma }f=\Sigma _{i=1}^{n}(D_{\gamma }x_{i})g_{i}(0)}$

but ${\displaystyle D_{\gamma }(x_{i})=(x_{i}\circ \gamma )'(0)=\gamma _{i}'(0)}$

We can now claim that ${\displaystyle \Phi }$ is onto because given a D take ${\displaystyle \gamma _{i}(t)=D(x_{i})t}$

${\displaystyle \Phi }$ is trivially 1:1.

## Smoot

In this class It was riced the question of what a "Smoot" is. Here it is 0708-1300/Smoot