# 0708-1300/Class notes for Tuesday, October 2

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## English Spelling

Many interesting rules about 0708-1300/English Spelling

## 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.

2) A questionnaire was passed out in class

3) Homework one is due on thursday

### First Hour

Today's Theme: Locally a function looks like its differential

Pushforward/Pullback

Let ${\displaystyle \theta :X\rightarrow Y}$ be a smooth map.

We consider various objects, defined with respect to X or Y, and see in which direction it makes sense to consider corresponding objects on the other space. In general ${\displaystyle \theta _{*}}$ will denote the push forward, and ${\displaystyle \theta ^{*}}$ will denote the pullback.

1) points pushforward ${\displaystyle x\mapsto \theta _{*}(x):=\theta (x)}$

2) Paths ${\displaystyle \gamma :\mathbb {R} \rightarrow X}$, ie a bunch of points, pushforward, ${\displaystyle \gamma \rightarrow \theta _{*}(\gamma ):=\theta \circ \gamma }$

3) Sets ${\displaystyle B\subset Y}$ pullback via ${\displaystyle B\rightarrow \theta ^{*}(B):=\theta ^{-1}(B)}$ Note that if one tried to pushforward sets A in X, the set operations compliment and intersection would not commute appropriately with the map ${\displaystyle \theta }$

4) A measures ${\displaystyle \mu }$ pushforward via ${\displaystyle \mu \rightarrow (\theta _{*}\mu )(B):=\mu (\theta ^{*}B)}$

5)In some sense, we consider functions, "dual" to points and thus should go in the opposite direction of points, namely ${\displaystyle \theta ^{*}f=f\circ \theta }$

6) Tangent vectors, defined in the sense of equivalence classes of paths, [${\displaystyle \gamma }$] pushforward as we would expect since each path pushes forward. ${\displaystyle [\gamma ]\rightarrow \theta _{*}[\gamma ]:=[\theta _{*}\gamma ]=[\theta \circ \gamma ]}$

CHECK: This definition is well defined, that is, independent of the representative choice of ${\displaystyle \gamma }$

7) We can consider operators on functions to be in a sense dual to the functions and hence should go in the opposite direction. Hence, tangent vectors, defined in the sense of derivations, pushforward via ${\displaystyle D\rightarrow (\theta _{*}D)(f):=D(\theta ^{*}f)=D(f\circ \theta )}$

CHECK: This definition satisfies linearity and Liebnitz property.

Theorem 1

The two definitions for the pushforward of a tangent vector coincide.

Proof:

Given a ${\displaystyle \gamma }$ we can construct ${\displaystyle \theta _{*}\gamma }$ as above. However from both ${\displaystyle \gamma }$ and ${\displaystyle \theta _{*}\gamma }$ we can also construct ${\displaystyle D_{\gamma }f}$ and ${\displaystyle D_{\theta _{*}\gamma }f}$ because we have previously shown our two definitions for the tangent vector are equivalent. We can then pushforward ${\displaystyle D_{\gamma }f}$ to get ${\displaystyle \theta _{*}D_{\gamma }f}$. The theorem is reduced to the claim that:

${\displaystyle \theta _{*}D_{\gamma }f=D_{\theta _{*}\gamma }f}$

for functions ${\displaystyle f:Y\rightarrow \mathbb {R} }$

Now, ${\displaystyle D_{\theta _{*}\gamma }f={\frac {d}{dt}}f\circ (\theta _{*}\gamma )|_{t=0}={\frac {d}{dt}}f\circ (\theta \circ \gamma )|_{t=0}={\frac {d}{dt}}(f\circ \theta )\circ \gamma |_{t=0}=D_{\gamma }(f\circ \theta )=\theta _{*}D_{\gamma }f}$

Q.E.D

Functorality

let ${\displaystyle \theta :X\rightarrow Y,\lambda :Y\rightarrow Z}$

Consider some "object" s defined with respect to X and some "object u" defined with respect to Z. Something has the property of functorality if

${\displaystyle \lambda _{*}(\theta _{*}s)=(\lambda \circ \theta )_{*}s}$

and

${\displaystyle \theta ^{*}(\lambda ^{*}u)=(\lambda \circ \theta )^{*}u}$

Claim: All the classes we considered previously have the functorality property; in particular, the pushforward of tangent vectors does.

Let us consider ${\displaystyle \theta _{*}}$ on ${\displaystyle T_{p}M}$ given a ${\displaystyle \theta :M\rightarrow N}$

We can arrange for charts ${\displaystyle \varphi }$ on a subset of M into ${\displaystyle \mathbb {R} ^{m}}$ (with coordinates denoted ${\displaystyle (x_{1},\dots ,x_{m})}$)and ${\displaystyle \psi }$ on a subset of N into ${\displaystyle \mathbb {R} ^{n}}$ (with coordinates denoted ${\displaystyle (y_{1},\dots ,y_{n})}$)such that ${\displaystyle \varphi (p)=0}$ and ${\displaystyle \psi (\theta (p))=0}$

Define ${\displaystyle \theta ^{o}=\psi \circ \theta \circ \varphi ^{-1}=(\theta _{1}(x_{1},\dots ,x_{m}),\dots ,\theta _{n}(x_{1},\dots ,x_{m}))}$

Now, for a ${\displaystyle D\in T_{p}M}$ we can write ${\displaystyle D=\sum _{i=1}^{m}a_{i}{\frac {\partial }{\partial x_{i}}}}$

So,

 ${\displaystyle (\theta _{*}D)(f)\!}$ ${\displaystyle =D(\theta ^{*}f)\!}$ ${\displaystyle =\sum _{i=1}^{m}a_{i}{\frac {\partial }{\partial x_{i}}}(f\circ \theta )}$ ${\displaystyle =\sum _{i=1}^{m}a_{i}\sum _{j=1}^{n}{\frac {\partial f}{\partial y_{j}}}{\frac {\partial \theta _{j}}{\partial x_{i}}}}$ ${\displaystyle ={\begin{bmatrix}{\frac {\partial f}{\partial y_{1}}}&\cdots &{\frac {\partial f}{\partial y_{n}}}\\\end{bmatrix}}{\begin{bmatrix}{\frac {\partial \theta _{1}}{\partial x_{1}}}&\cdots &{\frac {\partial \theta _{1}}{\partial x_{m}}}\\\vdots &&\vdots \\{\frac {\partial \theta _{n}}{\partial x_{1}}}&\cdots &{\frac {\partial \theta _{n}}{\partial x_{m}}}\\\end{bmatrix}}{\begin{bmatrix}a_{1}\\\vdots \\a_{m}\\\end{bmatrix}}}$

Now, we want to write ${\displaystyle \theta _{*}D=\sum b_{j}{\frac {\partial }{\partial y_{j}}}}$

and so, ${\displaystyle b_{k}=(\theta _{*}D)y_{k}={\begin{bmatrix}0&\cdots &1&\cdots &0\\\end{bmatrix}}{\begin{bmatrix}{\frac {\partial \theta _{1}}{\partial x_{1}}}&\cdots &{\frac {\partial \theta _{1}}{\partial x_{m}}}\\\vdots &&\vdots \\{\frac {\partial \theta _{n}}{\partial x_{1}}}&\cdots &{\frac {\partial \theta _{n}}{\partial x_{m}}}\\\end{bmatrix}}{\begin{bmatrix}a_{1}\\\vdots \\a_{m}\\\end{bmatrix}}}$

where the 1 is at the kth location. In other words, ${\displaystyle \theta _{*}D=\sum _{j=1}^{n}\sum _{i=1}^{m}a_{i}{\frac {\partial \theta _{j}}{\partial x_{i}}}{\frac {\partial }{\partial y_{j}}}}$

So, ${\displaystyle \theta _{*}=d\theta _{p}}$, i.e., ${\displaystyle \theta _{*}}$ is the differential of ${\displaystyle \theta }$ at p

We can check the functorality, ${\displaystyle (\lambda \circ \theta )_{*}=\lambda _{*}\circ \theta _{*}}$, then ${\displaystyle d(\lambda \circ \theta )=d\lambda \circ d\theta }$ This is just the chain rule.

### Second Hour

Defintion 1

An immersion is a (smooth) map ${\displaystyle \theta :M^{m}\rightarrow N^{n}}$ such that ${\displaystyle \theta _{*}}$ of tangent vectors is 1:1. More precisely, ${\displaystyle d\theta _{p}:T_{p}M\rightarrow T_{\theta (p)}N}$ is 1:1 ${\displaystyle \forall p\in M}$

Example 1

Consider the canonical immersion, for m<n given by ${\displaystyle \iota :(x_{1},...,x_{m})\mapsto (x_{1},...,x_{m},0,...,0)}$ with n-m zeros.

Example 2

This is the map from ${\displaystyle \mathbb {R} }$ to ${\displaystyle \mathbb {R} ^{2}}$ that looks like a loop-de-loop on a roller coaster (but squashed into the plane of course!) The map ${\displaystyle \theta }$ itself is NOT 1:1 (consider the crossover point) however ${\displaystyle \theta _{*}}$ IS 1:1, hence an immersion.

Example 3

Consider the map from ${\displaystyle \mathbb {R} }$ to ${\displaystyle \mathbb {R} ^{2}}$ that looks like a check mark. While this map itself is 1:1, ${\displaystyle \theta _{*}}$ is NOT 1:1 (at the cusp in the check mark) and hence is not an immersion.

Example 4

Can there be objects, such as the graph of |x| that are NOT an immersion, but are constructed from a smooth function?

Consider the function ${\displaystyle \lambda (x)=e^{-1/x^{2}}}$ for x>0 and 0 otherwise.

Then the map ${\displaystyle x\mapsto {\begin{bmatrix}(\lambda (x),\lambda (x))&x>0\\(0,0)&x=0\\(-\lambda (-x),\lambda (-x))&x<0\\\end{bmatrix}}}$

is a smooth mapping with the graph of |x| as its image, but is NOT an immersion.

Example 5

The torus, as a subset of ${\displaystyle \mathbb {R} ^{3}}$ is an immersion

Now, consider the 1:1 linear map ${\displaystyle T:V\rightarrow W}$ where V,W are vector spaces that takes ${\displaystyle (v_{1},...,v_{m})\mapsto (Tv_{1},...,Tv_{m})=(w_{1},..,w_{m},w_{m+1},...,w_{n})}$

From linear algebra we know that we can choose a basis such that T is represented by a matrix with 1's along the first m diagonal locations and zeros elsewhere.

Theorem 2

Locally, every immersion looks like the inclusion ${\displaystyle \iota :\mathbb {R} ^{m}\rightarrow \mathbb {R} ^{n}}$.

More precisely, if ${\displaystyle \theta :M^{m}\rightarrow N^{n}}$ and ${\displaystyle d\theta _{p}}$ is 1:1 then there exist charts ${\displaystyle \varphi }$ acting on ${\displaystyle U\subset M}$ and ${\displaystyle \psi }$ acting on ${\displaystyle V\subset N}$ such that for ${\displaystyle p\in U,\varphi (p)=\psi (\theta (p))=0}$ such that the following diagram commutes:

${\displaystyle {\begin{matrix}U&\rightarrow ^{\varphi }&U'\subset \mathbb {R} ^{m}\\\downarrow _{\theta }&&\downarrow _{\iota }\\V&\rightarrow ^{\psi }&V'\subset \mathbb {R} ^{n}\\\end{matrix}}}$

that is, ${\displaystyle \iota \circ \varphi =\psi \circ \theta }$

Definition 2

M is a submanifold of N if there exists a mapping ${\displaystyle \theta :M\rightarrow N}$ such that ${\displaystyle \theta }$ is a 1:1 immersion.

Example 6

Our previous example of the graph of a "loop-de-loop", while an immersion, the function is not 1:1 and hence the graph is not a sub manifold.

Example 6

The torus is a submanifold as the natural immersion into ${\displaystyle \mathbb {R} ^{3}}$ is 1:1

Definition 3

The map ${\displaystyle \theta :M\rightarrow N}$ is an embedding if the subset topology on ${\displaystyle \theta (M)}$ coincides with the topology induced from the original topology of M.

Example 7

Consider the map from ${\displaystyle \mathbb {R} \rightarrow \mathbb {R} ^{2}}$ whose graph looks like the open interval whose two ends have been wrapped around until they just touch (would intersect at one point if they were closed) the points 1/3 and 2/3rds of the way along the interval respectively. The map is both 1:1 and an immersion. However, any neighborhood about the endpoints of the interval will ALSO include points near the 1/3rd and 2/3rd spots on the line, i.e., the topology is different and hence this is not an embedding.

Corollary 1 to Theorem 2

The functional structure on an embedded manifold induced by the functional structure on the containing manifold is equal to its original functional structure.

Indeed, for all smooth ${\displaystyle f:M\rightarrow \mathbb {R} }$ and ${\displaystyle \forall p\in M}$ there exists a neighborhood V of ${\displaystyle \theta (p)}$ and a smooth ${\displaystyle g:N\rightarrow \mathbb {R} }$ such that ${\displaystyle g|_{\theta (M)\bigcap U}=f|_{U}}$

Proof of Corollary 1

Loosely (and a sketch is most useful to see this!) we consider the embedded submanifold M in N and consider its image, under the appropriate charts, to a subset of ${\displaystyle \mathbb {R} ^{m}\subset \mathbb {R} ^{n}}$. We then consider some function defined on M, and hence on the subset in ${\displaystyle \mathbb {R} ^{n}}$ which we can extend canonically as a constant function in the "vertical" directions. Now simply pullback into N to get the extended member of the functional structure on N.

Proof of Theorem 2

We start with the normal situation of ${\displaystyle \theta :M\rightarrow N}$ with M,N manifolds with atlases containing ${\displaystyle (\varphi _{0},U_{0})}$ and ${\displaystyle (\psi _{0},V_{0})}$ respectively. We also expect that for ${\displaystyle p\in U_{0},\varphi _{0}(p)=\psi _{0}(\theta (p))=0}$. I will first draw the diagram and will subsequently justify the relevant parts. The proof reduces to showing a certain part of the diagram commutes appropriately.

${\displaystyle {\begin{matrix}M\supset U_{0}&\rightarrow ^{\varphi _{0}}&U_{1}\subset \mathbb {R} ^{m}&\rightarrow ^{Id}&U_{2}=U_{1}\\\downarrow _{\theta }&&\downarrow _{\theta _{1}}&&\downarrow _{\iota }\\N\supset V_{0}&\rightarrow ^{\psi _{0}}&V_{1}\subset \mathbb {R} ^{n}&\leftarrow ^{\xi }&V_{2}\\\end{matrix}}}$

It is very important to note that the ${\displaystyle \varphi _{0}}$ and ${\displaystyle \psi _{0}}$ are NOT the charts we are looking for , they are merely one of the ones that happen to act about the point p.

In the diagram above, ${\displaystyle \theta _{1}=\psi _{0}\circ \theta \circ \varphi ^{-1}}$. So, ${\displaystyle \theta _{1}(0)=0}$ and ${\displaystyle (d\theta _{1})_{0}=i}$. Note the ${\displaystyle \theta _{1}}$, being merely the normal composition with the appropriate charts, does not fundamentally add anything. What makes this theorem work is the function ${\displaystyle \xi }$

We consider the map ${\displaystyle \xi :V_{2}\rightarrow V_{1}}$ given ${\displaystyle (x,y)\mapsto \theta _{1}(x)+(0,y)}$. We note that ${\displaystyle \xi }$ corresponds with the idea of "lifting" a flattened image back to its original height.

Claims:

1) ${\displaystyle \xi }$ is invertible near zero. Indeed, computing ${\displaystyle d\xi _{0}=I}$ which is invertible as a matrix and hence ${\displaystyle \xi }$ is invertible as a function near zero.

2) Take an ${\displaystyle x\in U_{2}}$. There are two routes to get to ${\displaystyle V_{1}}$ and upon computing both ways yields the same result. Hence, the diagram commutes.

Hence, our immersion looks (locally) like the standard immersion between real spaces given by ${\displaystyle \iota }$ and the charts are the compositions going between ${\displaystyle U_{0}}$ to ${\displaystyle U_{2}}$ and ${\displaystyle V_{0}}$ to ${\displaystyle V_{2}}$

Q.E.D