0708-1300/Class notes for Tuesday, October 2
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English Spelling
0708-1300/English_Spelling 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.
General class comments
1) The class photo is up, please add yourself
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 [math]\displaystyle{ \theta:X\rightarrow Y }[/math] 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 [math]\displaystyle{ \theta_* }[/math] will denote the push forward, and [math]\displaystyle{ \theta^* }[/math] will denote the pullback.
1) points pushforward [math]\displaystyle{ x\mapsto\theta_*(x) := \theta(x) }[/math]
2) Paths [math]\displaystyle{ \gamma:R\rightarrow X }[/math], ie a bunch of points, pushforward, [math]\displaystyle{ \gamma\rightarrow \theta_*(\gamma):=\theta\circ\gamma }[/math]
3) Sets [math]\displaystyle{ B\subset Y }[/math] pullback via [math]\displaystyle{ B\rightarrow \theta^*(B):=\theta^{-1}(B) }[/math] Note that if one tried to pushforward sets A in X, the set operations compliment and intersection would not commute appropriately with the map [math]\displaystyle{ \theta }[/math]
4) A measures [math]\displaystyle{ \mu }[/math] pushforward via [math]\displaystyle{ \mu\rightarrow (\theta_*\mu)(B) :=\mu(\theta^*B) }[/math]
5)In some sense, we consider functions, "dual" to points and thus should go in the opposite direction of points, namely [math]\displaystyle{ \theta^*f = f\circ\theta }[/math]
6) Tangent vectors, defined in the sense of equivalence classes of paths, [[math]\displaystyle{ \gamma }[/math]] pushforward as we would expect since each path pushes forward. [math]\displaystyle{ [\gamma]\rightarrow \theta_*[\gamma]:=[\theta_*\gamma] = [\theta\circ\gamma] }[/math]
CHECK: This definition is well defined, that is, independent of the representative choice of [math]\displaystyle{ \gamma }[/math]
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 [math]\displaystyle{ D\rightarrow (\theta_*D)(f):= D(\theta^*f) }[/math]
CHECK: This definition satisfies linearity and Liebnitz property.
Theorem 1
The two definitions for the pushforward of a tangent vector coincide.
Proof:
Given a [math]\displaystyle{ \gamma }[/math] we can construct [math]\displaystyle{ \theta_{*}\gamma }[/math] as above. However from both [math]\displaystyle{ \gamma }[/math] and [math]\displaystyle{ \theta_*\gamma }[/math] we can also construct [math]\displaystyle{ D_{\gamma}f }[/math] and [math]\displaystyle{ D_{\theta_*\gamma}f }[/math] because we have previously shown our two definitions for the tangent vector are equivalent. We can then pushforward [math]\displaystyle{ D_{\gamma}f }[/math] to get [math]\displaystyle{ \theta_*D_{\gamma}f }[/math]. The theorem is reduced to the claim that:
[math]\displaystyle{ \theta_*D_{\gamma}f = D_{\theta_*\gamma}f }[/math]
for functions [math]\displaystyle{ f:Y\rightarrow R }[/math]
Now, [math]\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\gamma |_{t=0} = D_{\gamma}(f\circ\theta) =\theta_*D_{\gamma}f }[/math]
Q.E.D
Functorality
let [math]\displaystyle{ \theta:X\rightarrow Y, \lambda:Y\rightarrow Z }[/math]
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
[math]\displaystyle{ \lambda_*(\theta_*s) = (\lambda\circ\theta)_*s }[/math]
and
[math]\displaystyle{ \theta^*(\lambda^*u) = (\lambda\circ\theta)^*u }[/math]
Claim: All the classes we considered previously have the functorality property; in particular, the pushforward of tangent vectors does.
Let us consider [math]\displaystyle{ \theta_* }[/math] on [math]\displaystyle{ T_pM }[/math] given a [math]\displaystyle{ \theta:M\rightarrow N }[/math]
We can arrange for charts [math]\displaystyle{ \varphi }[/math] on a subset of M into [math]\displaystyle{ R^m }[/math] (with coordinates denoted [math]\displaystyle{ (x_1,...,x_m) }[/math])and [math]\displaystyle{ \psi }[/math] on a subset of N into [math]\displaystyle{ R^n }[/math] (with coordinates denoted [math]\displaystyle{ (y_1,...,y_n) }[/math])such that [math]\displaystyle{ \varphi(p) = 0 }[/math] and [math]\displaystyle{ \psi(\theta(p)=0 }[/math]
Define [math]\displaystyle{ \theta^o = \psi\circ\theta\circ\varphi^{-1} = (\theta_1(x_1,...,x_m),...,\theta_n(x_1,...,x_m)) }[/math]
Now, for a [math]\displaystyle{ D\in T_pM }[/math] we can write [math]\displaystyle{ D=\sum_i a_{i=1}^m\frac{\partial}{\partial x_i} }[/math]
So, [math]\displaystyle{ (\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}= }[/math] [math]\displaystyle{ =\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} }[/math]
Now, we want to write [math]\displaystyle{ \theta_*D = \sum b_j\frac{\partial}{\partial y_j} }[/math]
and so, [math]\displaystyle{ 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} }[/math]
where the i is at the kth location.
So, [math]\displaystyle{ \theta_* = d\theta_p }[/math], i.e., [math]\displaystyle{ \theta_* }[/math] is the differential of [math]\displaystyle{ \theta }[/math] at p
We can check the functorality, [math]\displaystyle{ (\lambda\circ\theta)_* = \lambda_*\circ\theta_* }[/math], then [math]\displaystyle{ d(\lambda\circ\theta) = d\lambda\circ d\theta }[/math]
This is just the chain rule.
