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\begin{center}
{\large Classical Mechanics, Lecture 8 \\}
{\small February 4, 2008 \\}
{\small lecture by John Baez \\}
{\small notes by Alex Hoffnung}
\end{center}
\section{Poisson Manifolds}
We have described Poisson brackets for functions on $\R^{2n}$ - the phase space for a system whose configuration space is $\R^n$. Now let's generalize this to systems whose configuration space is any manifold, $M$. Here we will see that the phase space is the ``cotangent bundle" $T^*M$ and this is a ``Poisson manifold" - a manifold such that the commutative algebra of smooth real-valued functions on it, $C^\infty(T^*M)$ is equipped with Poisson bracket $\lbrace\cdot,\cdot\rbrace$ making it into a Poisson algebra - the Poisson algebra of ``observables" for our system.\\\\
{\bf Example}: a particle on a sphere $S^2$.\\
(picture of a sphere $ M = S^2 $ with point $ \q\in M$)\\
\noindent The position and momentum of this particle give a point in $T^*S^2: \q\in S^2, \p\in T_\q^*M$, so $(\q,\p)\in T^*S^2$.\\
Recall that a manifold $M$ is a topological space such that every point $\q\in M$ has a ``neighborhood that looks like $\R^n$." In other words, there is an open set $U\subset M$ with $\q\in U$ and a bijection
\[\phi\maps U\to\R^n.\]
Indeed we have a collection of these ($U_i,\phi_i\maps U_i\to\R^n$) and they are {\bf compatible}:\\
(picture of charts overlapping)\\
that is, $\phi_j\circ\phi_i^{-1}$ is smooth (infinitely differentiable) where defined. A collection of this sort is an {\bf atlas}, and the functions $\phi_i\maps U_i\to\R^n$ are called {\bf charts}. We will usually use a {\bf maximal} atlas, i.e. one containing all charts that are compatible with all charts in the atlas. So - a {\bf manifold} is a topological space with a maximal atlas.\\\\
If the manifold $M$ is the configuration space of some physical system, the a point $\q\in M$ describes the position of the system and a ``tangent vector" $v\in T_\q M$ describes its velocity, where $T_\q M$ is the {\bf tangent space} of $M$ at $\q$:\\
(picture of tangent space to $S^2$ at $\q$)\\
which can be defined in various ways:
\begin{enumerate}
\item A {\bf tangent vector} $v$ at the point $\q$ is an equivalence class of (smooth) curves
\[\gamma\maps\R\to M\]
such that $\gamma(0) = \q$,
where $\gamma_1\sim\gamma_2$ if and only if for every smooth function $f\in C^\infty(M)$ (smooth real-valued functions on $M$) we have
\[\frac{d}{dt}f(\gamma_1(t))|_{t=0} = \frac{d}{dt}f(\gamma_2(t))|_{t=0}.\]
You can show the set of such equivalence classes is an $n$-dimensional vector space, the {\bf tangent space} $T_\q M$. Given $v = [\gamma]\in T_\q M$, we can define the {\bf derivative} $vf\in\R$ for any $f\in C^\infty(M)$ by
\[vf = \frac{d}{dt}f(\gamma(t))|_{t=0}\]
\item A tangent vector $v$ at $\q\in M$ is a {\bf derivation}
\[v\maps C^\infty(M)\to\R\]
i.e., a map that is:
\begin{itemize}
\item $v(\alpha f+\beta g) = \alpha v(f) + \beta v(g)$ - linearity
\item $v(fg) = v(f)g(\q) + f(\q)v(g)$ - product rule
\end{itemize}
These clearly form a vector space, the {\bf tangent space} $T_\q M$.
\end{enumerate}
The set of all position-velocity pairs is a manifold, the {\bf tangent bundle} of $M$:
\[TM = \lbrace(\q,v): \q\in M, v\in T_\q M\rbrace\]
Naively, we might define momentum by $\p = mv$, in which case it would be a tangent vector. It is better to think of it is a ``cotangent vector". Every vector space $V$ has a {\bf dual} $V^*$:
\[V^* = \lbrace l\maps V\to\R : l ~linear\rbrace\]
The {\bf cotangent space} of $M$ at $\q\in M$ is:
\[T_\q^*M = (T_\q M)^*\]
and the {\bf cotangent bundle} of $M$ is:
\[T^*M = \lbrace (\q,\p):\q\in M, \p\in T_\q^*M\rbrace\]
So $T^*M$ will be the system's ``phase space" - space of position-momentum pairs.\\\\
What good are cotangent vectors, though?\\
The ``gradient" or ``differential" of a function $f\in C^\infty(M)$ at $\q\in M$ is a cotangent vector, $(df)_\q\in T_\q^*M$:
\[(df)_\q(v) = v(f), v\in T_\q M\]
(or in low-brow notation: $(\nabla f)(\q)\cdot v = v f$).\\
The potential energy for our system (e.g. a particle on $M$) is some fucntion of its position: $V\in C^\infty(M)$. We have seen already that ``$\nabla V = -F$" - but this really means
\[(dV)_\q = -F(\q)\]
where $F(\q)$, the {\bf force} at $\q\in M$, is a cotangent vector: $F(\q)\in T_\q^*M$.\\
A tangent vector looks like a little arrow:\\
(picture of a tangent vector)\\
A cotangent vector looks like a ``stack of hyperplanes" - its level surfaces:\\
(picture of level surfaces)\\
Together they give a number $l(v)\in\R$:\\
(picture of tangent vector crossing level curves)\\\\
In physics, the velocity $v\in T_pM$ is a tangent vector and the force $F\in T_\q^* M$ is a cotangent vector, so $F(v)\in\R$. We have seen this before in the formula:
\[\int_{t_1}^{t_2}F\cdot\dot{\q}(t)dt = work\]
Now we would say, given a particle's path $\q\maps\R\to M$, that work from $t_1$ to $t_2$ is equal to
\[\int_{t_1}^{t_2}F(\q(t))(\dot{\q}(t))dt\]
Since force is a cotangent vector, so is momentum, since:
\[\frac{d\p}{dt} = F\]
So: a position-momentum pair ($\q,\p$) is really a point in $T^* M$:
\[\q\in M, \p\in T_\q ^* M.\]
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