# Difference between revisions of "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/

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 $\varphi,\psi$ that these charts would satisfy the property of a manifold, defined in the atlas sense, that $\psi\circ\varphi^{-1}$ is smooth where defined.

Proof

$\psi\circ\varphi^{-1}$:Rn$\rightarrow$Rn is smooth $\leftrightarrow$ $(\psi\circ\varphi^{-1})_i$:Rn$\rightarrow$R is smooth $\forall$ i $\leftrightarrow$ $\pi_i\circ\psi\circ\varphi^{-1}$ is smooth where $\pi_i$ is the $i^{th}$ coordinate projection map.

Now, since $\pi_i$ is always smooth, $\pi_i\in F_{R^{n}}(U^{'}_{\psi})$

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

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

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

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

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

Example 1 Let $S^{2} = R^{3}-\{0\}/$~ where the equivalence relation ~ is given by x~$\alpha$x for $\alpha$>0 We thus get a canonical projection map $\pi:R^{3}-\{0\}\rightarrow S^{2}$ and hence, there is an induced functional structure on $S^{2}$. Claim: 1) This induced functional structure makes $S^{2}$ 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 $T^{2} = R^{2}/Z^{2}$, i.e., the real plane with the equivalence relation that (x,y)~(x+n,y+m) for (x,y) in $R^{2}$and (n,m) in $Z^{2}$

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

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

Proof of Claim

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

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

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

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

Definition 3 Product Manifolds

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

Claim: This does in fact yield a manifold

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

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

Definition 2 Smooth Function, Functional Structure Sense

Let $M^{m}$ and $N^{n}$ be manifolds. We say that a function $f:M^{m}\rightarrow N^{n}$ is smooth if $\forall h:V\rightarrow R$ such that h is smooth, $h\circ f$ 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(x-> y) (these would be functions for sets, continuous functions for topological spaces, smooth functions for manifolds, etc.) along with

1)Composition maps $\circ$:mor(x->y) X mor(y->z) --> mor (x->z)

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

such that several axioms are satisfied, including

a) Associativity of the composition map

b) $1_{x}$ 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 f:M-> N and g:N->L are smooth maps between manifolds then gof:M->L is smooth

2) $1_{M}$ 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 $\{\gamma:R\rightarrow M\ |\ s.t.\ \gamma\ smooth,\ \gamma(0) = p\}$

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

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 R.

such that,

1) D(af +bg) = aDf + bDg for a,b constants in 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, $T_{p}M$ = the set of all tangent vectors at a point p

Claim 3

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

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

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

Definition 7 (atlas sense)

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

Definition 8 (functional structure sense)

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

Proof of Theorem 1

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

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

It is clear that this map is linear is f and satisfies the leibnitz rule.

Claim: $\Phi$ is a bijection.

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

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

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

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

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

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

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

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

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

$\Phi$ is trivially 1:1.