Difference between revisions of "0708-1300/Class notes for Tuesday, October 2"

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==Class Notes==
 
==Class Notes==
 
<span style="color: red;">The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.</span>
 
<span style="color: red;">The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.</span>

Revision as of 12:42, 2 October 2007

Announcements go here

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 \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.