0708-1300/Class notes for Tuesday, September 25

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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 \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}:\mathbb{R}^n\rightarrow \mathbb{R}^n is smooth \Leftrightarrow (\psi\circ\varphi^{-1})_i:\mathbb{R}^n\rightarrow \mathbb{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_{\mathbb{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_{\mathbb{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 \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 \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 \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 S^{2} = \mathbb{R}^{3}-\{0\}/~ where the equivalence relation ~ is given by x~\alphax for \alpha>0 We thus get a canonical projection map \pi:\mathbb{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} = \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 \mathbb{R}^{2}and (n,m) in \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 CP^{n} denote the n dimensional complex projective space, that is, CP^{n} = \mathbb{C}^{n+1}-\{0\}/~ where [z_{0},...,z_{n}] ~ [\alpha z_{0},...,\alpha z_{n}] where \alpha\in \mathbb{C}

Again, this space inherits a functional structure from \mathbb{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 \mathbb{C}^{n} = \mathbb{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 isormorphic to (\mathbb{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 \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 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 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 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 \mathbb{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\rightarrow 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\rightarrow y) \times mor(y\rightarrow z) \rightarrow mor (x\rightarrow z)

2) For any x there exists an element 1_{x}\inmor(x \rightarrow 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\rightarrow N and g:N\rightarrow L are smooth maps between manifolds then g\circ f:M\rightarrow 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:\mathbb{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 \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 \mathbb{R}.

such that,

1) D(af +bg) = aDf + bDg for a,b constants in \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, 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 D_{\gamma} is linear and satisfies the leibnitz rule.

Claim: \Phi is a bijection.

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

Hadamard's Lemma

If f:\mathbb{R}^{m}\rightarrow \mathbb{R} is smooth near 0 then there exists smooth g_{i}:\mathbb{R}^{m}\rightarrow \mathbb{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:\mathbb{R}\rightarrow \mathbb{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.

Smoot

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