# 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 $\theta:X\rightarrow Y$ 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 $\theta_*$ will denote the push forward, and $\theta^*$ will denote the pullback.

1) points pushforward $x\mapsto\theta_*(x) := \theta(x)$

2) Paths $\gamma:R\rightarrow X$, ie a bunch of points, pushforward, $\gamma\rightarrow \theta_*(\gamma):=\theta\circ\gamma$

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

4) A measures $\mu$ pushforward via $\mu\rightarrow (\theta_*\mu)(B) :=\mu(\theta^*B)$

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

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

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

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 $D\rightarrow (\theta_*D)(f):= D(\theta^*f)$

CHECK: This definition satisfies linearity and Liebnitz property.

Theorem 1

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

Proof:

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

$\theta_*D_{\gamma}f = D_{\theta_*\gamma}f$

for functions $f:Y\rightarrow R$

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

Q.E.D

Functorality

let $\theta:X\rightarrow Y, \lambda:Y\rightarrow Z$

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

$\lambda_*(\theta_*s) = (\lambda\circ\theta)_*s$

and

$\theta^*(\lambda^*u) = (\lambda\circ\theta)^*u$

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

Let us consider $\theta_*$ on $T_pM$ given a $\theta:M\rightarrow N$

We can arrange for charts $\varphi$ on a subset of M into $R^m$ (with coordinates denoted $(x_1,...,x_m)$)and $\psi$ on a subset of N into $R^n$ (with coordinates denoted $(y_1,...,y_n)$)such that $\varphi(p) = 0$ and $\psi(\theta(p)=0$

Define $\theta^o = \psi\circ\theta\circ\varphi^{-1} = (\theta_1(x_1,...,x_m),...,\theta_n(x_1,...,x_m))$

Now, for a $D\in T_pM$ we can write $D=\sum_i a_{i=1}^m\frac{\partial}{\partial x_i}$

So, $(\theta_*D)(f) = \sum_{i=1}^m a_i\frac{\partial}{\partial x_i}(f\circ\varphi) = \sum_{i=1}^m a_i \sum_{j=1}^n\frac{\partial f}{\partial y_j}\frac{\partial\theta_j}{\partial x_i}=$ $=\begin{bmatrix} \frac{\partial f}{\partial y_1} & ... & \frac{\partial f}{\partial y_n}\\ \end{bmatrix} \begin{bmatrix} \frac{\partial \theta_1}{\partial x_1} & ... & \frac{\partial \theta_1}{\partial x_m}\\ ...& & ...\\ \frac{\partial \theta_n}{\partial x_1} & ... & \frac{\partial \theta_n}{\partial x_m}\\ \end{bmatrix} \begin{bmatrix} a_1\\ ...\\ a_m\\ \end{bmatrix}$

Now, we want to write $\theta_*D = \sum b_j\frac{\partial}{\partial y_j}$

and so, $b_k = (\theta_*D)y_k =\begin{bmatrix} 0&...,i,...&0\\ \end{bmatrix} \begin{bmatrix} \frac{\partial \theta_1}{\partial x_1} & ... & \frac{\partial \theta_1}{\partial x_m}\\ ...& & ...\\ \frac{\partial \theta_n}{\partial x_1} & ... & \frac{\partial \theta_n}{\partial x_m}\\ \end{bmatrix} \begin{bmatrix} a_1\\ ...\\ a_m\\ \end{bmatrix}$

where the i is at the kth location.

So, $\theta_* = d\theta_p$, i.e., $\theta_*$ is the differential of $\theta$ at p

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

### Second Hour

Defintion 1

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

Example 1

Consider the canonical immersion, for m<n given by $\iota:(x_1,...,x_m)\mapsto (x_1,...,x_m,0,...,0)$ with n-m zeros.

Example 2

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

Example 3

Consider the map from R to $R^2$ that looks like a check mark. While this map itself is 1:1, $\theta_*$ 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 $\lambda(x) = e^{-1/x^2}$ for x>0 and 0 otherwise.

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

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 $R^3$ is an immersion

Now, consider the 1:1 linear map $T:V\rightarrow W$ where V,W are vector spaces that takes $(v_1,...,v_m)\mapsto (Tv_1,...,Tv_m) = (w_1,..,w_m,w_{m+1},...,w_n)$

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 $\iota:R^m\rightarrow R^n$.

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

$\begin{matrix} U&\rightarrow^{\phi}&U'\subset R^n\\ \downarrow_{\theta} &&\downarrow_{\iota} \\ V& \rightarrow^{\psi}& V'\subset R^n\\ \end{matrix}$

that is, $\iota\circ\varphi = \psi\circ\theta$

Definition 2

M is a submanifold of N if there exists a mapping $\theta:M\rightarrow N$ such that $\theta$ 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 $R^3$ is 1:1

Definition 3

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

Example 7

Consider the map from $R\rightarrow R^2$ 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 $f:M\rightarrow R$ and $\forall p\in M$ there exists a neighborhood V of $\theta(p)$ and a smooth $g:N\rightarrow R$ such that $g|_{\theta(M)\bigcap U} = f|_U$

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 $R^m\subset R^n$. We then consider some function defined on M, and hence on the subset in 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 $\theta:M\rightarrow N$ with M,N manifolds with atlases containing $(\varphi_0,U_)0)$ and $(\psi_0, V_0)$ respectively. We also expect that for $p\in U_0, \varphi_0(p) = \psi_0(\theta(p)) = 0$. 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.

$\begin{matrix} M\supset U_0 & \rightarrow^{\varphi_0} & U_1\subset R^m & \rightarrow^{Id} & U_2 = U_1 \\ \downarrow_{\theta} & &\downarrow_{\theta_1} & &\downarrow_{\iota}\\ N\supset V_0 & \rightarrow^{\psi_0} & V_1\subset R^n & \leftarrow^{\xi} & V_2\\ \end{matrix}$

It is very important to note that the $\varphi_0$ and $\psi_0$ 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, $\theta_1 = \psi_0\circ\theta\circ\varphi^{-1}$. So, $\theta_1(0) = 0$ and $(d\theta_1)_0 = i$. Note the $\theta_1$, being merely the normal composition with the appropriate charts, does not fundamentally add anything. What makes this theorem work is the function $\xi$

We consider the map $\xi:V_2\rightarrow V_1$ given $(x,y)\mapsto \theta_1(x) + (0,y)$. We note that $\xi$ corresponds with the idea of "lifting" a flattened image back to its original height.

Claims:

1) $\xi$ is invertible near zero. Indeed, computing $d\xi_0 = I$ which is invertible as a matrix and hence $\xi$ is invertible as a function near zero.

2) Take an $x\in U_2$. There are two routes to get to $V_1$ 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 $\iota$ and the charts are the compositions going between $U_0$ to $U_2$ and $V_0$ to $V_2$

Q.E.D