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       \begin{center}
      {\large Classical Mechanics, Lecture 10 \\}  
      {\small February 12, 2008 \\}  
      {\small lecture by John Baez \\}  
      {\small notes by Alex Hoffnung}  
      \end{center}

\section{How Observables Generate Symmetries}
Hamilton's equations are first-order differential equations.
In the language of differential geometry, they are all about a
certain vector field and the `flow' it `generates':
\[\textrm{picture of manifold }X\textrm{ with vector field and integral curve}\]
A vector field $v$ on a manifold $X$, we say a smooth function or {\bf curve}
\[\gamma\maps\R\to X\]
is the {\bf integral curve} of $v$ through $x\in X$ if:
\begin{enumerate}
\item $\gamma (0) = x$
\item $\frac{d}{dt}\gamma (t) = v(\gamma (t)),~~\forall t\in\R$
\end{enumerate}
We say a vector field $v$ is {\bf integrable} if $\forall x\in X$ there exists an integral curve of $v$ through $x$.\\\\
{\bf Example} - $X = (0,1)$ and the vector field: $\frac{\partial}{\partial x}$.  If we try to get the integral curve through $x\in (0,1)$ we get
\[\gamma(t) = x + t\]
but this is not in $(0,1)$ for $t$ large!  So, this vector field is
not integrable.\\\\
{\bf Example} - $X = \R$.  This is secretly the same, but anyway: let
\[v = x^2\frac{\partial}{\partial x}\]
Here our integral curve would satisfy:
\[\frac{d}{dt}\gamma(t) = \gamma(t)^2\]
\begin{eqnarray*}
\frac{dy}{dt} &=& y^2\\
\int\frac{dy}{y^2} &=& \int dt\\
-\frac{1}{y} &=& t + C\\
y &=& -\frac{1}{t + C}\\
\end{eqnarray*}
i.e.,
\[\gamma(t) = -\frac{1}{t + C}\]
The problem is that this solution is not defined for all $t$ --- 
it blows up at $t = -C$.  So, this vector field is also not integrable.\\\\
Suppose $v$ is an integrable vector field on a manifold $X$.  Then:
\begin{thm}
for every $x\in X$ the integral curve of $v$ through $x$ is unique.
\end{thm}
This let's us define a function:
\[\phi\maps\R\times X\to X\]
by
\[(t,x)\mapsto\phi(t,x) = \phi_t(x)\]
such that $\phi_t(x)$ is the integral curve of $v$ through $x$.
\begin{thm}
$\phi\maps\R\times X\to X$ is smooth.
\end{thm}
Note also:
\[\phi_0(x) = x\]
and
\[\phi_s(\phi_t(x)) = \phi_{s+t}(x)\]
Mathematicians summarize these equations by saying ``$\phi$ is an action of the group $(\R,+,0)$ on $X$"; note they imply:
\[\phi_{-t}(x) = (\phi_t)^{-1}(x)\]
since
\[\phi_t\circ\phi_{-t} = \phi_0 = 1_X\] 
So: for any $t\in\R$,
\[\phi_t\maps X\to X\]
is smooth (by Theorem) with a smooth inverse, $\phi_{-t}$.  A smooth map $f\maps X\to Y$ with smooth inverse is called a {\bf diffeomorphism}.
\begin{defn}
If $\phi\maps\R\times X\to X$ is a smooth map such that
\begin{enumerate}
\item $\phi_0(x) = x$
\item $\phi_s(\phi_t(x)) = \phi_{s+t}(x)$
\end{enumerate}
we call $\phi$ a {\bf flow}.
\end{defn}
We've seen that any integrable vector field $v$ gives a flow $\phi$: we call $\phi$ the flow {\bf generated} by $v$.  Conversely, any flow $\phi$ is generated by a unique (integrable) vector field $v$:
\[v(x) = \frac{d}{dt}\phi_t(x)|_{t=0},~~~x\in X\]
Now suppose $X$ is a Poisson manifold.  If $H\in C^\infty(X)$ is any observable, thought of as the Hamiltonian, we get a vector field
\[\{H,\cdot\}\maps C^\infty(X)\to C^\infty(X)\]
also called $v_H$, the {\bf Hamiltonian vector field generated by $H$}.  If $v_H$ is integrable, it generates a flow
\[\phi\maps\R\times X\to X\]
called {\bf time evolution} or the {\bf flow generated by $H$}.  If our system is in the state $x\in X$ initially, then at time $t$ it will be at $\phi_t(x)\in X$.\\\\
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