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%TBD numbering
\title{Ezra Getzler: Topics in QFT}
\author{Gabriel C. Drummond-Cole}

\begin{document}
\maketitle
\section{September 20,2011}

Welcome.  Today will be review for many of you.  Maybe it will be brand new for some of you.  I'll talk about the Hamiltonian formalism in terms of symplectic geometry, in the language of differential geometry.  If you're happy with differential geometry this will definitely help.  

We start with the idea of a symplectic manifold.  What we're doing here is developing the Hamiltonian approach to mechanics.  We'll spend most of our time using a different point of view, the Lagrangian point of view, which is more variational.  I wanted to start with Hamiltonian because it gives a clearer link between the classical and the quantum theories.

A symplectic manifold is a manifold $M$ with a $2$-form $\omega\in\Omega^2(M)$, which is:\begin{itemize}\item[i.] nondegenerate: there is a map $TM\to T^*M$ taking $X$ to $\iota(X)\omega$, this is an isomorphism of vector bundles.  This implies that $M$ is even dimensional.
\item[ii.]$d\omega=0$, $\omega$ is closed.\end{itemize}
I won't dwell too long on these manifolds, you could give a whole course on them.  The main interest for me is to understand a particular class of vector fields on them, called Hamiltonian vector fields.  So, in fact, what we have is a linear map from the smooth functions on $M$ to the vector fields on $M$, $C^\infty(M)\to Vec(M)$, and this the first term in the ``Poisson'' complex.  So $f\mapsto H_f$, with the formula as follows: $\iota(H_f)\omega=-df$.  I'll give examples in a moment but let me describe properties and I guess references.  I like Abraham and Marsden for this subject, or the graduate texts in mathematics by Arnold, one of the classics.

The key point about these vector fields, if the manifold is compact or the vector field generates a global flow, this vector field will preserve the symplectic form.
\begin{lemma}
The Lie derivative $\Ll(H_f)\omega=0$.  
\end{lemma}
\begin{proof}
I'll use Cartan's formula, which says that 
\[\Ll(H_f)=d\iota(H_f)+\iota(H_f)d\]
The second term kills $\omega$ which is closed, and so this is $d\iota(H_f)\omega=-d^2f=0$.
\end{proof}
A symplectic manifold automatically has a canonical volume form on it, called the Liouville measure, which gives a well-defined notion of how big a region is.  One way to write it is to exponentiate the symplectic form, $e^\omega=\sum\frac{1}{k!}\omega^{\wedge k}$, and this isn't a volume form, so we can extract the top degree form $\frac{1}{\frac{n}{2}!}\omega^{\wedge{(\frac{n}{2})}}$.  It's so much more elegant to write $e^\omega$.  When we integrate functions against a volume form. $\int_M f\mu=\int_M fe^\omega$ because, well, the form $e^\omega$ is inhomogeneous, and we define integration to throw away lower degree forms.  We change our perspective and define these integrals to be zero.

I did this integral without mentioning orientation, and it happens that symplectic manifolds are canonically oriented.  This is a fact that I could stop and prove, but I'd like to say that the symmetry group of a symplectic form is connected, so the determinant is always positive.

This has to be proved, it's a concrete fact of this structure.

We have a corollary, which is that:
\begin{cor} the flow associated to a Hamiltonian vector field preserves the Liouville measure.
\end{cor}
The proof, at this point, is embarrassingly simple.
\begin{proof}
We need to show that the Lie derivative of this top form is zero with respect to a Hamiltonian vector field.  $\frac{d}{dt}\Phi_t^*(e^\omega)=-\Phi_t^*(\Ll(H_f)e^\omega)$ where $\Phi_t$ is the flow generated by $H_f$.

I can check that $\Ll(H_f)e^\omega$ is $e^\omega\wedge \Ll(H_f)\omega$, which is zero by our lemma.
\end{proof}
This is all, we're covering the same material that is in any textbook on Hamiltonian mechanics.  Now the main thing we're interested in, oh, first let me do the Poisson bracket.  I had a map $C^\infty(M)\to Vec(M)$.  Now, vector fields form a Lie algebra, and we'll put a Lie structure on the smooth functions so as to make this a Lie algebra homomorphism.  This is called the Poisson bracket.  If $f$ and $g$ are two functions on $M$, then $\{f,g\}$ is the flow of $f$ applied to $g$, the Lie derivative along $H_f$ of $g$, $\Ll(H_f)g$.  A priori this has nothing to do with Lie algebras.  This is linear over the real numbers.  It satisfies the Poisson relation 
\[\{f,gh\}=\{f,g\}h+g\{f,h\}\]
Let's use the fact that our vector field is Hamiltonian and see how that allows us to analyze this Poisson bracket.  I want to prove it's a Lie bracket.  This is by no means obvious at this point.  It'll be obvious in ten seconds.  So, $\{f,g\}=\iota(H_f)dg$, but $dg$ can be rewritten and this is $-\iota(H_f)\iota(H_g)\omega$.  It's a basic fact of nature that if you exchange the arguments here the sign changes, this is $\iota(H_g)\iota(H_f)\omega=-\{g,f\}$.  So just as you have the Poisson relation in the second variable, you also have it in the first variable.

This leaves us with the Jacobi rule to establish.  This is pretty easy too if you know your calculus of differential forms.  I'll expand these in slightly different ways:
\[
\{f,\{g,h\}\}-\{g,\{f,h\}\}
=\Ll(H_f)\iota(H_g)dh-\iota(H_g)d\Ll(H_f)h
\]
Now, $d$ commutes with Lie derivative, so this is
\[
[\Ll(H_f),\iota(H_g)] dh
\]
and now it happens, and this is a standard calculation in, so, this is
\[\iota([H_f,H_g])dh\]
using the useful formula $[\Ll X,\iota Y]=\iota([X,Y])$.  
I need to check that $H_{\{f,g\}}=[H_f,H_g]$, and so, remember, we're contracting $\omega$ with these vector fields to see what we get.  So on the right we get $-\iota([H_f,H_g])\omega=-\Ll(H_f)\iota(H_g)\omega+\iota(H_g)\Ll(H_f)\omega$, the second term of which cancels.  The left hand side is $d\Ll(H_f)g$, which is $\Ll(H_g)dg$, which is equal to the first term.

There may be a more direct proof.

Now I want to observe something interesting about Poisson brackets.  We had $C^\infty(M)\to Vec(M)$, and the kernel is nontrivial, the kernel is the same as that of the de Rham differential.  That is locally constant functions, so if I assume that my manifold is connected, the constants $\rR$, and in general $H^0(M,\rR)$.  If I look at locally Hamiltonian vector fields, the ones where $\Ll(X)\omega=0$, if you contract with $\omega$ you get a closed one-form, those are isomorphic to $\Omega^1_{\text{closed}}(M)$, whereas the image are exact one-forms.  So there's a nontrivial cokernel, which is $H^1(M,\rR)$.  I want to observe that if you have a Poisson bracket wihch is compactly supported, so we can integrate it, and the integral will always be zero.  So if $\{f,g\}\in C_c^\infty(M)$, then $\int_M\{f,g\}e^\omega=0$.  To check this, we see that it's a consequence of the definition: $\int_M \Ll(H_f)g\wedge e^\omega$ which is an integration by parts.  Let me actually assume that $f$ or $g$ is compactly supported, we can use Stokes' theorem, this is $-\int_M g\Ll(H_f)e^\omega$.  It might not be enough if $\{f,g\}$ is compactly supported.

The interest of symplectic manifolds is when you have a Lie group acting in a way that preserves the symplectic form, such as $\rR$ with addition, acting as time evolution.

Let's start with the infinitessimal case.  We have a Lie algebra homomorphism from a Lie algebra $\gGg$ to $Vec(M,\omega)$, vector fields that preserve the symplectic form.  I didn't introduce a notation for it, $\xi\mapsto X_\xi$.  So $\Ll(X_\xi)\omega=0$.  Then under some hypotheses, my goal is to lift this homomorphism to a homomorphism $\gGg\to C^\infty(M)$.  If you take the composition of these two maps, that may not be zero.  So we'll assume instead that our Lie algebra maps to Hamiltonian vector fields, not just locally Hamiltonian vector fields.  A necessary condition is that $X_\xi$ is Hamiltonian for all $\xi\in \gGg$.  If I take a pair of vector fields that are locally Hamiltonian, then their bracket is automatically Hamiltonian.  Saying that Lie derivatives of closed one-forms are exact.  If $\gGg$ is semisimple, so $[\gGg,\gGg]=\gGg$,  then this obstruction vanishes.  You could also take the universal cover of $M$.  Having obtained that this obstruction vanishes, we can turn to lifting.

If the manifold is connected, we may get a nontrivial central extension by the real numbers.  In the first case we should introduce all lifts.  If it's semisimple, all lifts are trivial, and we can arrange to have a best lift.  So, to restate, potentially, when we try to lift the homomorphism $\gGg\to Vec(M,\omega)$ to $(C^\infty (M),\{\,,\,\})$ we get a central extension of $\gGg$ by $H^0(M,\rR)$.  Two options are if all extensions are trivial for $\gGg$, or if all the components of $M$ are compact, in which case I choose the one where, well, I get a $2$-cocycle on $\gGg$ where $c(\eta,\xi)=\{\mu_\eta,\mu_\xi\}-\mu_{[\eta,\xi]}\in H^0(M,\rR)$. 
\begin{enumerate}\item
If $\gGg$ is semisimple, then $c\in Z^2(\gGg)$ is a coboundary, $c(\eta,\xi)=\gamma([\eta,\xi]),$ where $\gamma$ is a linear map $\gGg\to H^0(M,\rR)$, and then adding (or subtracting) $\gamma$ to $\mu$ gives a homomorphism of LIe algebras.  This is known as the Marsden-Weinstein moment or momentum map.  This terminology squabble is momentous in mathematical physics.  I actually haven't looked into the history of this, so, it might be older.
\item If each component of $M$ is compact we may select the translate of $\mu_\xi$ such that $\int_{M_i} \mu_\xi e^\omega=0$ for each component $M_i$.  
\end{enumerate}
Then it's automatic that we get a homomorphism of Lie algebras.  In other situations you may well get a nontrivial central extension, such as non-semisimple ones, like affine, Kac-Moody Lie algebras, gauge transformations for principal bundles on the circle.  In this case it's not semisimple, it's not acting on a compact manifold (maybe not on a symplectic manifold at all, we'll talk about it later), so that's very interesting.

Finally, I get to some examples.  The first example I feel obliged to mention is $M=T^*X$, the cotangent bundle of a manifold $X$. This example, we'll come back to it from the perspective of Lagrangian mechanics later.  This is usually studied in terms of a coordinate system, let me remind you of what those coordinates are.  The key point is, there's a canonical one-form on $M$, in terms of which everything is defined.  It has the following property: A one-form will define a section of the cotangent bundle.  I can pull it back by the section, and this is characterized by the property that pulling it back gives the same one-form.  If $\theta\in\Omega^1(X)$, that is, then $\theta$ defines a section of $T^*X\to X$, then $\theta^*(\alpha)=\theta$.  In local coordinates, I can choose coordinates on $X$, $q^1,\ldots, q^m$, and then a differential form $\theta$ is $\sum \theta_i dq^i$, and now you see that the one-form $\alpha$ will be $p_i dq^i$, where $p_i$ are the values of $\theta_i$.  This gives us the remaining coordinates.

So, this is a symplectic manifold, if I take the differential as my symplectic form, $\omega=d\alpha$.  The formalism will be the same basic procedure.  Indeed, you can see that with the coordinates I've chosen, $\sum dp_i\wedge dq^i$ is closed and nondegenerate.  It takes $\frac{\partial}{\partial q^i}\mapsto -dp_i$ and $\frac{\partial}{\partial p_i}\mapsto dq^i$. Then the Hamiltonian map takes
$p_i$ to $\frac{\partial}{\partial q^i}$ and $q_i$ to $\frac{\partial}{\partial p_i}$.  If you have a Riemannian manifold, if $X$ is Riemannian, then $TX$ and $T^*X$ are naturally isomorphic, and $M$ is naturally isomorphic to $TX$.  Now we can consider the function $f(q,p)=\frac{1}{2}|p|^2$, so $\frac{1}{2}g^{ij}(q)p_ip_j$, the Hamiltonian flow is a rescaling.  You can take the geodesic emanating from a point in the base you get a flow along the geodesic, and then if you take that over to $M$ you get a Hamiltonian flow.

This leads to something surprising, that the geodesic flow preserves the Liouville measure on the cotangent bundle.  In the world of Riemannian geometry, this is not obvious.  

So, I have lots more examples, but let me abbreviate a little.  We'll touch on the other examples but they're not the main point.  So the second example is a quasiprojective manifold, an algebraic subvariety of projective space or at least a Zariski open of such, or any open subset in the sense of ordinary topology.  I don't want to go into great detail right now.  Projective space has a canonical symplectic form, and on a complex submanifold it remains symplectic, something about the details.

The third example is a special case of both the others, a symplectic vector space.  It's always writeable as the cotangent space of a vector space, and on the other hand, once you've observed this, you can put an almost complex structure on it, it becomes a complex vector space and sits inside $\cC\pP^n$.  Here I'm thinking just in terms of the cotangent bundle because that's probably the wrong symplectic structure.

We don't understand quantization terribly well except in the vector space case.  So all of our rigorous theory is useless for calculations later.  There will be a lack of rigor in certain parts of this course, I should warn you.  I'll try to let you know when it's happening.

I want to talk now about Prequantization, due to Kostant and Souriau.  This is clarifying, not needed, but something that explains more clearly what we're doing.  There's a ready source of closed two-forms in geometry, the curvatures of complex line bundles.  You can put a connection on any complex line bundle, and its curvature will be a closed two-form.  So we want to work backwards and from a symplectic manifold find a Hermitian line bundle $L$ (structure group $U(1)$) with connection whose curvature is $2\pi i\omega$.  In some sense this is the first step in quantization.  Note that $-\frac{1}{2\pi i}$ times the curvature is an element of $H^2(M,\rR)$, this is the first Chern class of the line bundle.  It's proven that this element is the image of $c_1(L)$ in $H^2(M,\zZ)$.  We'll take a cover of the manifold, and the cover is chosen so that the transitions have logarithms on each open subset, which leads to the construction of a $2$-cocycle with integral coefficients, and its image in de Rham cohomology is what we said.  Once you've proved de Rham's theorem, very little extra work gets you this.  We'll see calculations close to that.  This is a necossary condition for our symplectic form to be, to have a Hermitian line bundle of this sort.  It has to have integral periods.  This will be a sufficient condition as well.  It's not a difficult theorem, done using the same techniques as de Rham's theorem.  

Where are we going?  Suppose I have a Lie algebra acting in a Hamiltonian action on our manifold?  We'd like to lift it to the line bundle, and we'll need the moment map, and adding that to the covariant derivative, this will give us the action.  That is, $D_{X_\xi}+\mu_\xi$ (maybe with a constant) will have Lie structure.  The pieces do not: $[D_{X_\eta},D_{X_\xi}]=D_{X_{[\eta,\xi]}}+curv(\eta,\xi)$.  So if we can lift to the line bundle, then we're ready to quantize.  We'll be getting at the kernel of a Dirac operator twisted by this line bundle.  

You can start thinking about this theorem for next time.  We'll start with a cocycle, take a partition of unity, and produce explicitly a line bundle with connection whose curvature will represent the desired cycle, and the difference will be an exact two-form whose integral can be added.  You can try doing that in the two days before Thursday.

There will be no class the next Thursday, or on the thirteenth of October.
\section{September 22}
What does it mean to be a de Rham cohomology class?  I'll write it in terms of Cech cohomology.  If you have a manifold, there is one cover that works for all cohomology classes.  That is called a good cover, and there are many.  Start with a collection of open subsets $U_\alpha$ of $M$ that is locally finite, and in fact the standard constructions of an open cover will produce one.  Triangulate the manifold and take the stars of all the vertices.

As usual, we have the intersections of the open sets $U_{\alpha_0}\cap\cdots U_{\alpha_n}$, and the cover is good if all these are diffeomorphic to a star-shaped, or even convex subset of $\rR^n$, or empty.  For example, the cover I told you about has this property.  Small geodesic balls, a locally finite subcover, that would work.  This is all standard stuff.

I'm going to assume I have a Cech two-cocycle, $C_{\alpha\beta\gamma}\in\zZ$, and vanishes if $U_\alpha\beta\gamma$ is empty.  To be a cocycle is the following equation:
\[c_{\beta\gamma\delta}-c_{\alpha\gamma\delta}+c_{\alpha\beta\delta}-c_{\alpha\beta\gamma}=0\]
this will be used over and over.

Let me pick a partition of unity $\{\phi_\alpha\in C^\infty_c(U_\alpha)|\alpha\in A\}$ with $\sum \phi_\alpha=1$ so $\sum d\phi_\alpha=0$.  Then
\[
\sum c_{\alpha\beta\gamma}\phi_\alpha d\phi_\beta d\phi_\gamma\in \Omega^2_{\text{closed}}(M)\]
is a representative of the image of the Cech cocycle $c$.

Our hypothesis from last time is that our symplectic form minus one of this type is an exact two-form.  Let me give a name to that exact two-form.  There is an integral cocycle such that $\omega=\sum c_{\alpha\beta\gamma}\phi_\alpha d\phi_\beta d\phi_\gamma+d\theta$, and here $\theta$ is a real one-form on $M$.  Now I'll construct a line bundle whose curvature is $2\pi i$ times this first form, and we'll add on $\theta$ at the end.  

For the line bundle, we need transition functions in the circle group.  The logarithms won't be uniquely defined but by convexity they'll at least be defined.  So let me write $g_{\alpha\beta}\in C^\infty(U_{\alpha\beta},U(1))$ then \[\log g_{\alpha\beta}=2\pi i\sum_\gamma c_{\alpha\beta\gamma}\phi_\gamma\]

So if I multiply by $\phi_\delta$, then I get \[2\pi i \sum c_{\beta\gamma\delta}\phi_\delta - 2\pi i\sum c_{\alpha\beta\delta}\phi_\delta+2\pi i\sum c_{\alpha\beta\delta} \phi_\delta-2\pi\sum c_{\alpha\beta\gamma}\phi_\delta\] and I can rewrite to get 
\[\log g_{\beta\gamma}-\log g_{\alpha \gamma}+\log g_{\alpha\beta}=2\pi i c_{\alpha\beta\gamma}\in \zZ(1)=2\pi i\zZ\]
So the connection, just following the proof of de Rham's theorem, is
\[A_\alpha=2\pi i \sum c_{\alpha \beta\gamma}\phi_\beta d\phi_\gamma\in \Omega^1(U_\alpha,i\rR)\]
So $A_\alpha-A_\beta=d\log g_{\alpha\beta}$
To see this, 
I multiply this time by $\phi_\gamma d\phi_\delta$ and get
\[2\pi i \sum c_{\beta\gamma\delta}\phi_\gamma d\phi_\delta - 2\pi i\sum
c_{\alpha\beta\delta}\phi_\gamma d\phi_\delta+2\pi i\sum c_{\alpha\beta\delta} \phi_\gamma d\phi_\delta-2\pi\sum c_{\alpha\beta\gamma}\phi_\gamma d\phi_\delta\].
The last term vanishes because it has a sum over $d\phi_\delta$ with no other delta in the sum.  Then these formulas give exactly the desired equation.

Then the the curvature of this connection is $d A_\alpha=\sum c_{\alpha\beta\gamma}d\phi_\beta d\phi_\delta$.  We also get $dA_\alpha=dA_\beta$ on $U_\alpha\beta$, and so these are equal to the restriction to $U_\alpha$ of the global form $\sum \phi_\alpha dA_\alpha$.  The conclusion is that $g_\alpha\beta=exp(2\pi i \sum c_\alpha\beta\gamma \phi_\gamma)$ and $\tilde{A}_\alpha=2\pi i\sum c_{\alpha\beta\delta}\phi_\beta d\phi_\delta + 2\pi i \theta|_{U_\alpha}$, this has curvature $2\pi i\omega$.  If I had two of these, well, you can invert these, you can add these, and so if I have two of these, their difference is a flat line bundle with structure group $U(1)$.  There are many solutions, then, in general, if the manifold is not simply connected.

Now we get some magic.  We can ignore where this comes from.  The big question is, how do we get a group acting on this line bundle.  I've already answered that if the group is a simply connected Lie group.  [missed a little bit]
\subsection{Prequantization of Hamiltonian group actions}
To recall from last time, we had a Lie algebra $\gGg$ and we assumed we had a Hamiltonian $\gGg\to C^\infty(M)$, to the Poisson algebra of $M$, and I denoted that by $\xi\mapsto \mu_\xi$ and then this had the property that you get an associated vector field $X_\xi$ on $M$, the Hamiltonian vector field associated to $\mu_xi$, called $H_{\mu_\xi}$.  Just to recall, the contraction $\iota(X_\xi)\omega=-d\mu_\xi$.  We have a differential operator acting on sections of our line bundle, covariant differentiation.  Now we make a sign error.

$\xi\mapsto D_{X_\xi}$, a first order differential operator on smooth sections of our prequantization line bundle, $\Gamma(M,L)$.  If you recall the definition of curvature, you'll know that these operators are almost a homomorphism.
\[ [D_{X_\xi},D_{X_\eta}]=D_{[X_\xi,X_\eta]}+2\pi i\iota(X_\xi)\iota(X_\eta)\omega\] but the last term is $2\pi i \iota(X_\xi)(-d\mu_\eta)$ which is $-2\pi i\{mu_\xi,\mu_\eta\}=-2\pi i\mu_{[\xi,\eta]}$.  The second to last term is $D_{X_{[\xi,\eta]}}$.  Now I make an ansatz, $L_\xi=D_\xi+2\pi i\mu_\xi$, then $[L_\xi,L_\eta]=[D_{X_\xi},D_{X_\eta}]+2\pi i\Ll(X_\xi)\mu_\eta -2\pi i \Ll(X_\eta)\mu_\xi$ and this comes out to $L_{[\xi,\eta]}$.  I think Hamiltonian mechanics is really about a Hermitian line bundle with a group action.  

By exponentiation, the simply connected Lie group associated to this action acts on our line bundle.  I just realized there's an important brick in the wall I forgot to check, that this preserves the connection as well.  This $G$-action preserves the inner product on $L$, I hope that's pretty evident, since $D_\xi$ preserves it and the other piece is pure imaginary.  It also preserves the connection.  I neglected to check that.  Can I have, well.  I think that's sort of obvious, but I'm going to give a bullshit proof.  The action preserves the curvature, and it takes our line bundle to another with the same curvature.  So these differ by a flat line bundle, which are a discrete set---no.  I think that the answer is that this isn't necessarily the case.  If it's not true, it's not true for some circle action on the torus.  I don't see why not.  This doesn't happen if our group is semisimple, because the error is a homomorphism from the Lie group to the Jacobean, and, well, somebody should check. 

Let's assume that we've gotten rid of yet another obstruction.  Then, well, that's kind of a problem.

I'm going to quantize using Dirac operators.  There will be a reference on blackboard.  I want to make a differential operator, and define the kernel as the quantization.  This will require some auxilliary data.  Under what conditions do you get a projectively flat \wh from this technique.  We'll see some examples.  The basic structure I'll put on my symplectic manifold is a compatible almost complex structure.  So $M$ has an $Sp(n)$-structure.  The tangent bundle is associated to an $Sp(n)$ principle bundle $P\times_{Sp(n)} TM$.  There's a standard basis we'll talk about.  We'll consider a reduction of this group to the unitary group $U(n)$.  The unitary group $U(m)$ is the intersection of $Sp(n,\rR)$ with a correctly configured orthogonal group.  So in the first place we have $\cC^m$ with, there are linear transformations preserving the imaginary part for $Sp(n,\rR)$, and $O(n)$ preserve the real part.  To preserve the whole form is to preserve both the imaginary and real part.  One can prove that the unitary group is the maximal compact subgroup of $Sp(n)$, so \wh corresponds to sections $\Gamma(M,P/U(m))$, which is a fibel bundle with fiber $Sp(n,\rR)/U(m)$, which is a contractible manifold.  The example you're familiar with, $Sp(2,\rR)\cong SL(2,\rR)$, and $SL(2,\rR)/U_1$ is the upper half plane.

Since I haven't said anything about the structure of symplectic vector spaces, I thought I would take a detour to describe why two symplectic vector spaces are isomorphic.  

Let's start with the notion of an isotropic subspace of a symplectic vector space.  $W\subset V$ is isotropic if all the vectors are orthogonal to each other.  So if $v,w$ are in the subspace, then $\langle v,w\rangle=0$.  The main construction you can do if you have one of these is to take the orthogonal complement.  So $W^\perp$ is the set of $v\in V$ so that for all $w\in W$, $\langle v,w\rangle=0$.  Let's make some observations.  First, $W\subset W^\perp$, that's the isotropic hypothesis, then $W^\perp/W$ is a symplectic vector space.  Finally, using nondegeneracy, the sum of the dimensions of $W$ and $W^\perp$ is the dimension of $V$.  What this immediately shows is that an isotropic subspace can't have a dimension greater than half the dimension of $V$.  If that is attained, we say that $W$ is a Lagrangian subspace.  These are maximal isotropic subspaces, and you can see that $W^\perp=W$ in those cases, and $V/W$ is naturally isomorphic to $W^*$.  In fact, we can choose an isomorphism between $V$ and $T^*W$.  The other thing, I haven't checked it, but we see from these observations that every isotropic subspace is contained in a Lagrangian.  In particular, if we choose a basis of $W$ and a dual basis of $W^*$, which we lift to $V$, we get a so-called Darboux basis.  Explicitly if $W$ has a basis $\{e_i\}$, with dual basis $\{\bar{f}^i\}$, we can lift the latter to vectors $f^i\in V$, and we get a Darboux basis $\{e_i,f^i\}$ so that $\langle e_i,f_j\rangle=\delta_i^j$.  The principal bundle I was talking about is the set of all Darboux bases.  This is a manifold, a torsor, acted on simply transitively by the symplectic group. Thus the principal bundle $P$ is the set of Darboux bases of the tangent spaces.

Right. Now, given, just a couple of words about what kinds of transformations give rise to a compatible almost-complex structure (reduction to $U(m)$).  An almost complex structure $J:TM\to TM$ has $J^2=-I$.  You can ask whether they're complex or integrable.  The condition I want is that $(v,w)=\langle Jv,w\rangle$ is an inner product.  There will be a condition to make this positive definite, $\langle Jv,v\rangle>0$ if $v\ne 0$, and also symmetric, so that $\langle Jv,w\rangle=-\langle v,Jw\rangle$.

We can consider the action induced by $J$ on the differential forms, a graded derivation of the differential forms.  To define this it suffices to define it on $1$-forms (and functions).  On functions it will vanish.  $J^*$ is the adjoint of $J$ on $1$-forms.  On a symplectic manifold, this is beside the point, forget it, we could just as easily have described this in terms of $J^*$.  What I mean by derivation is just that $J^*(\alpha\wedge \beta)=J^*\alpha \wedge \beta + \alpha\wedge J^*\beta$.  This is an interesting operator.

What are the eigenvalues of this operator?  $J^*$ splits the cotangent bundle tensored with $\cC$ into two subspaces, the $\pm i$ eigenspaces.  Let me call them $\wedge^{1,0}$ and $\wedge^{0,1}$, respectively.  Because we have a derivation, the maximum eigenvalue you could get would be $ki$ if these are all in the $+i$ eigenspaces, the minimum is $-ki$, and these can differ by multiples of $2i$.  The $-ki$ eigenspace is the one we're interested in, the $0,k$-forms $\Omega^{0,k}$.  The spinors are $\sum_k \Omega^{0,k}$.  i'll take spinors with values in our prequantization line bundle next time and make an operator.  We'll take the differential.  Let me start with a $0,0$ form, then its de Rham differential is in the sum of $1,0$ and $0,1$ forms.  If I take $d$ of a $0,1$ form it will be in $1,1$ and $0,2$.  I'l pull out the projection $\Omega^{0,k}\to\Omega^{0,k+1}$, and that will be half of my Dirac operator, the other half being its formal adjoint.  You can take the graded commutator of two derivations.  The manifold is complex if $d^c=[J^*,d]$ squares to zero.  Then there's a major theorem that if this condition holds, there's a local coordinate system made from holomorphic functions, where holomorphic functions are those whose image under $d$ is entirely in $1,0$ forms.

If my prequantization connection is holomorphic, then the manifold turns out to be a projective variety (if the manifold is compact).
\section{September 27}
Last time I was befuddled, I needed to check that it was automatic that, we constructed a line bundle with connection the symplectic form and a Hamiltonian action, it's automatic that the connection is invariant.
\begin{prop}
The connection on a prequantization line bundle is invariant under the lift we constructed of a Hamiltonian group action.
\end{prop}
If you have such an action, then we've lifted the action of the Lie algebra to the line bundle, so a priori we have a central extension acting.

The proof is immediate, and similar to but simpler than what we were doing.  For $\xi\in \gGg$, I introduced $L_\xi=D_{X_\xi}+2\pi i \mu_\xi$.  I need to show that the commutator relations are correct with respect to the connection, that $[L_\xi, D_Y]=D_{[X_\xi,Y]}$.  This is the infinitessimal form of equivariance, and if we integrate this we'll get equivariance under the action of group elements close to the identity.

To prove this, we substitute the formulas and plug away:

\[[D_{X_\xi}+2\pi i \mu_3, D_Y]=D_{[X_\xi,Y]}-2\pi i \iota(X_\xi)\iota(Y)\omega-2\pi i \Ll(Y)\mu_\xi\]
and these two last terms cancel, since $\Ll(Y)\mu_\xi=\iota(Y)d\mu_\xi=-\iota(Y)\iota(X_\xi)\omega$.

Now at the next stage it will become very much a matter of, I remember reading once that, this was a statement related to something we'll be studying, second quantization.  First quantization is an art, and second quantization is a functor.  We'll see very much how it's an art.  We'll see that there are many ways to quantize.  Let me talk more about these Dirac operators.

Let me review what we were doing.  We had this almost complex structure $J$ on our symplectic manifold $X$.  This almost complex structure makes $X$ into a Riemannian manifold.  It also, remember, before the tangent bundle has a symplectic structure, it actually reduces the structure group to the unitary group, from $Sp(n)$ to $U(n)$.  It's a special case of what is known as a $Spin_c$ structure, the exact situation in which we can construct a Dirac operator.  Don't worry too much about the group $Spin_c$.  What we need are two vector bundles on our manifold.  The first is the spinor bundle, 
\[S=\bigoplus_{q=0}^n \wedge^{0,q} T^*X\]
where $\wedge^{0,q} T^*X\subset \wedge^q T^*_\cC X$, well, $J^*$ has eigenvalues from $2qi$ down to $-2qi$, and this is the eigenvalue we're interested in.  We could also say that this is the $q$th exterior power of $\wedge^{0,1}T^*X$.  The complexification splits into two eigenspaces, and this is the power of the one with value $-i$.  The spinor bundle has rank $2^m$, and breaks into two pieces $S_+$ and $S_-$, where $S_+$ is the sum of the even exterior powers and $S_-$ is the sum of the odd exterior powers.  Both of these have rank $2^{m-1}$.

This is all a bit confusing.  Let me discuss an example where $X$ is the complex line $\cC$.  There we have coordinates $x$ and $y$, and we write $z=x+iy$.  I should have mentioned, $S$ has induced on it an inner product, it's a complex vector bundle, and these subspaces for different $q$ are orthogonal, and the inner product is induced by the Riemannian structure on $X$.  In our example, $\wedge^{0,0}=X\times\cC$, $\wedge^{0,1}=X\times \cC d\bar{z}$, where $d\bar{z}=dx-idy$ and $|d\bar{z}|=\sqrt{2}$.

So we can write the $\bar{\del}$ operator as a matrix with respect to the orthonormal frame given by $\{1,\frac{1}{\sqrt{2}}d\bar{z}\}$.  Then our operator $\bar{\del}$ is $\displaystyle\left[\begin{array}{cc}0&0\\\sqrt{2}\frac{\del}{{\del\bar{z}}}&0\end{array}\right]$.
Here we have $\frac{\del}{\del\bar{z}}=\frac{1}{2}\left(\frac{\del}{\del x}+i\frac{\del}{\del y}\right)$.  The formal adjoint has $(\bar{\del}^*\alpha,\beta)=(\alpha,\bar{\del}\beta)$ where $\alpha$ and $\beta$ are smooth sections of $S$ with compact support.  Note that both sides will vanish unless $\alpha\in \Omega^{0,q}(X)$ and $\beta\in\Omega^{0,q-1}(X)$.  So $\bar{\del}^*:\Omega^{0,q}(X)\to \Omega^{0,q-1}(X)$, and working this out explicitly is a little long, but focusing on the leading term, the so-called symbol, we evaluate
\[\int_X (\alpha_x,(\bar{\del}\beta)_x)dVd(x)\]
and we have to integrate by parts.  In the case of the complex line, 
\[\bar{\del}^*=\left[\begin{array}{cc}0&-\sqrt{2}\frac{\del}{{\del z}}\\0&0\end{array}\right]\]
Now we see that $\bar{\del}+\bar{\del}^*$, our Dirac operator, which I'll denote $\mathbb{D}$, squares to $-2 \frac{\del^2}{\del\bar{z}\del{z}}\otimes\id=-\frac{1}{2}\left(\frac{\del^2}{\del x^2}+\frac{\del^2}{\del y^2}\right)$ and so we multiply $\mathbb{D}$ by $\sqrt{2}$ and then end up without the factor.  

We're not there yet because we haven't used our prequantization line bundle, but before I get there I want to talk some about Dirac operators in general.  In general we're going to be interested in a manifold with what I'll call a Clifford module on it.  So let $X$ be a Riemannian manifold with a Clifford module $V$.  This is a complex vector bundle with connection and a Clifford action.  By this, I mean there's a linear action by the vector fields (or one-forms depending on your preference).  To package it all together, we have $T_\cC X\otimes_\cC V\to V$, which action is denoted by $c(X)v$, or equivalently, we may think of this as an action of the cotangent bundle, since $TX$ and $T^*X$ are isomorphic by the Riemannian metric.  This should satisfy two conditions.  The first is the Clifford relation: $c(X)c(Y)+c(Y)c(X)=-2(X,Y)$, where $(X,Y)$ is the Riemannian inner product, for these real vector fields.  This is equivalent by polarization to $c(X)^2=-|X|^2\id$.  The second condition is that Clifford multiplication is skew adjoint: $c(X)^*=-c(X)$.  Also, the connection should be compatible: $[\nabla_X^V,c(Y)]=c(\nabla_X^{LC}Y)$ with respect to the Levi-Civita connection.  We've constructed such a guy, the spinors that I described are such an example.  You need to cook up the Clifford action.  The connection is easy. This is a subbundle of the exterior bundle of the cotangent bundle, so you can extend the Levi-Civita connection, uh-oh, that doesn't work.  There will be a connection.  I hope the one I just told you will be the one I'm about to say.  

Recall that 
\[\sqrt{2}(\bar{\del}+\bar{\del}^*)=\left[\begin{array}{cc}0&-\frac{\del}{\del x}+i\frac{\del}{\del y}\\\frac{\del}{\del x}+i\frac{\del}{\del y}\end{array}\right]\]
so this is 
\[\twoby{0}{-1}{1}{0}\otimes\delby{x}+\twoby{0}{i}{i}{0}\otimes\delby{y}.\]These matrices are 
$\sqrt{2}\epsilon(d\bar{z})-\sqrt{2}\iota\left(\delby{\bar{z}}\right)$ and 
$\sqrt{2}\epsilon(d\bar{z})+\sqrt{2}i\iota\left(\delby{\bar{z}}\right)$, and this gives a Clifford action.  We have $c(dx)=\twoby{0}{-1}{1}{0}$, $c(dy)=\twoby{0}{i}{i}{0}$, $J\delby{x}=\delby{y}$,$J\delby{y}=-\delby{x}$, $J^*dy=dx$, and $J^*dx=-dy$.  In a general situation, these are local formulas, and the space we're working on looks like a direct sum of planes that look like this.  If you have $E\oplus F$ where these are symplectic vector spaces with almost complex structures, then the associated spinor bundles are in fact tensored together, and the Clifford multiplications by elements of $E$ and $F$ in $S_{E\oplus F}=S_E\otimes S_F$ each acts only on its own space.  

I'm just about to use a $\zZ_2$ grading.  I can't give the formulas correctly until I address the grading.  So $S=S_+\oplus S_-$, and we're assuming that $c(\alpha)$ maps $S_\pm\to S_\mp$.  The point is that, when you tensor two bundles together you get the correct grading, so $S_{E\oplus F,+}=S_{E,+}\otimes S_{F,+}\oplus S_{E,-}\otimes S_{F,-}$ whereas $S_{E\oplus F,-}=S_{E,+}\otimes S_{F,-}\oplus S_{E,-}\otimes S_{F,+}$, and now if $v\in S_{E,\pm}$, $w\in S_F$< $x\in E$ and $y\in F$ then $c(x)(v\otimes w)=c(x)v\otimes w$ and $c(y)(v\otimes w)=\pm v\otimes c(y)w$.

Okay.  The abstract definition of our Dirac operator is that $\mathbb{D}_V=\sum c(dx^i)\nabla^V_{\frac{\partial}{\partial x^i}}$, or you could write this as, for $e_a$ a frame of the tangent bundle $\sum g^{ab} c(e_b)\nabla^V_{e_a}$.  The first uses Clifford multiplication of one-forms, the second Clifford multiplication on the other side, that's why you get the metric.  You can check that $\dD_V$ is formally self-adjoint, and that its symbol is the same as that of the Laplacian.  What I mean by that is that $\Dd_V^2=g^{ij}(x)\frac{\del^2}{\del x^i\del x^j}$ plus lower order terms.

Notice that this operator takes sections of $S_\pm$ to $S_\mp$, so its square preserves $\Gamma(X,S_\pm)\subset \Gamma(X,S)$.

[Longish discussion of the need for the almost complex structure]

Our next step is to twist the spinors with our line bundle.  We're interested in the Clifford module $S\otimes L$.  In other words, $\bigoplus \wedge^{0,q} T^*M\otimes_\cC L$.  What has changed, other than the global structure, is that the connection is combined.  In local coordinates we can choose a frame of the line bundle $L$, a single section.  Let $s$ be a nowhere vanishing section of $L$ (only defined locally, obviously).  When I constructed $L$, I gave an explicit formula for this.  My original line bundle was delivered with these frames.  The covariant derivative of $S$ is a one-form $2\pi i \alpha$, tensored with $s$, where $\alpha$ is a real one-form.  Now our condition is that $d\alpha=\omega$.  Now we get in the local frame an explicit formula for our dirac operator, it's the original plus Clifford multiplication by this guy.  So, the twisted Dirac operator that we're interested in is $\dD^{S\otimes L}=\dD+2\pi i c(\alpha)$.

Notice by the way that Clifford multiplication by a real form is skew-adjoint, and multiplying it by $i$ makes it self-adjoint.  So
\[\dD^{S\otimes L}(v\otimes s)=\dD^S(v)\otimes s+2\pi i c(\alpha)v\otimes s\] where $v\in\Gamma(X,S)$ and $s$ is our frame of $L$.

If the manifold is not compact, the details of the connection will be absolutely essential in defining quantization. 

Roughly speaking, there are two types of group actions we'd like to understand.  There are those corresponding to space-time symmetries.  In quantum mechanics, this refers to the function on $X$ called the Hamiltonian.  This is basically the energy function.  In classical mechanics, it's the sum of a kinetic and a potential energy.  There, $X=T^*M$ and $H=\frac{1}{2}|\xi|^2+V(x)$ where the first term is kinetic and the second is potential.  Emmy Noether realized that understanding the time evolution involves the same mathematics as understanding internal symmetries.  We have internal symmetries corresponding to a Hamiltonian action of a Lie group $G$.  What I mean is that $H$ is invariant under the symmetry.  Then the Poisson bracket of the momenta, they commute with $H$, so are invariant under the Hamiltonian flow of $H$.  Not only is $\Ll(X_\xi) H =0$ but $\Ll(H_H)\mu_\xi=0$, the momenta are invariant under the time evolution of the system.  

There are somehow two very different situations, and quantizing them are different problems.  In the first case we think the group on the left is $\rR$ and on the right we have a compact group.  The problem of understanding the flow is different than understanding the way a group acts.  With our symmetry group, let's assume our almost complex structure is invariant under our symmetry group.  We'll lose this kind of assumption in quantum field theory.  We won't be able to construct a Hilbert space invariant with respect to the gauge group action.  You need to avoid the hypothesis that we have an almost complex structure invariant under the group in such a situation. But ideally, we would like this, which is the same as saying that $G$ acts by isometries.

\section{October 4}
So, last time, you may have been wondering, how does a vector field act on spinors?  The answer is, it acts by a Lie derivatives, but this is only when the vector field is an infinitesimal isometry, that is, a Killing field.  So we may want to focus on vector fields that are simultaneously Hamiltonian and Killing, which is pretty rare.  Such a vector field induces an isometry, preserving the real part of the metric and also the symplectic, imaginary, part, so it's Hermitian, it commutes with everything.  Here is a formula from my book, from Bismut, for the action, it should have the same symbol as covariant differentiation.

So $\Ll(H)X$ is going to be the covariant derivative $D_H X + \mu^{TM}(H)X$ where $\mu^{TM}(H)$ is a skew-symmetric endomorphism of our tangent bundle $TX$ (oops! $X$ is both a vector field and a symplectic manifold).  Our Levi-Civita connection is torsion free, so $[H,X]=D_H X - D_X H$, so we can rewrite $\mu^{TM}(H)=-D_X H$.  This is a skew-symmetric endomorphism because $H$ is a Killing field, that's pretty much the equation for generating an isometry. That is, this is a section of $\mathfrak{so}(TX)$ precisely whene $H$ is Killing.  You can promote this to anything associated to the tangent bundle.  The spinors is not quite, it depends on choices, but we can transport the formula over locally.  When the dust settles, what we see is, well, I'll just present the answer, and you can extract this from scattered pages in my book.  Transporting to the spinors, this formula becomes 
\[\Ll^S(H)=-\frac{1}{2}[\mathbb{D},c(H)].\]
This is how a Killing vector field works, this hasn't used Hamiltonian yet.  Now we have our ansatz for how to quantize this.  My ansatz was to twist by our connection on our line bundle plus $2\pi i$ times the Hamiltonian.  Twisting by $L$ and adding $2\pi i f$ where $f$ is the Hamiltonian, we get $-\frac{1}{2}[\mathbb{D}^{S\otimes L},c(H)]+2\pi i f$.  These are well-motivated if the Hamiltonian is Killing.  The bracket is two terms: $\mathbb{D} c(H) + c(H)\mathbb{D}$.  The latter term vanishes on the kernel of the Dirac operator.  We could project back into the kernel, but if we adopt the general ansatz in general, then we're looking just at $2\pi i f$.  This is a bullshitty way of motivating the idea of quantization as this kind of operator, Toeplitz operators.  If $P$ is projection from $L^2$ sections of the spinor bundle to the kernel of the Dirac operator, which may not be defined (but is by elliptic regularity if $X$ is compact), well, we propose to quantize the Hamiltonian $f$ by $P\circ \left(-\frac{1}{2}[\mathbb{D},c(H)]+2\pi i f \right)\circ P$ which is $P(2\pi i f) P= 2\pi i T_f$ where the T stands for Toeplitz.  These  don't satisfy the right bracket, but that's the idea of quantization.  

Having said that, I want to give a quick introduction to the idea of classical limit and asymptotics, a crude discussion of that, one idea, and then the rest of the lecture will be calculation.  What I want to point out is that the classical limit is taking our prequantization line bundle and raising it to the $n$th power, and let $n$ go to $\infty$.  Maybe I should call this quasi-classical, I'm hazy on the difference.  

The curvature was $2\pi i \omega$.  Locally, taking a frame of $L$ we get a frame of $L^n$, and the connection form is multiplied by $n$, so we get curvature $2\pi i n\omega$.  We've taken our symplectic manifold and replaced it by one where the symplectic form is multiplied by $n$.  Then the Poisson brackets get divided by $n$.  That is, $\{f,g\}_n=\frac{1}{n}\{f,g\}$. The smaller the Poisson brackets, the closer we are to the classical regime.  I think that this means Planck's constant was misnamed, we should have one over Planck's constant, which is the Chern class of the line bundle.  

You may be concerned by the integrality here.  If the Chern class of the line bundle is zero, we may be able to raise it to fractional powers.  For the rest of the talk, our line bundle will be trivial with a nontrivial connection, and our base space will be $\rR^2$.  I hope Steve likes this.

Maybe I should say that if you attend faithfully all of the events going on this year, microlocal analysis, you may learn tools for calculating asymptotical expansions of things like $T_f\circ T_g$ as $n$ goes to $\infty$.  This is almost the universal problem. 

We'll do canonical quantization today.  Our symplectic manifold is the first one anyone would ever think of, $\cC$ or $\rR^2$, with coordinates $z=q+ip$ and $\bar{z}$, so often physicists will think of $q$ as position and $p$ as momentum, but we'll leave that behind.  This will be symmetric with respect to these, so we won't distinguish them.  Roughly nothing will change if you act by an element of $SL(2,\rR)$.  Our form is $\omega=dq\wedge dp$.  We will start finding a connection on a line bundle with this as its curvature.  So $L$ will always be the trivial line bundle over $\cC$, as there are no other choices.  So we could, well, we have the metric and the connection $1$-form.  We can always take the global frame, which is just the section $1$.  For the metric, we can specify $|1|^2$, so if it's $1$, then the connection $1$-form could be $2\pi i pdq$.  We could instead try $2\pi i p dz$, which is $2\pi i pdq- 2\pi p dp$.  That won't be compatible with the metric where $|1|^2=1$, but we can see what metric it will be compatible with by looking at $d |s|^2=2\Re (Ds,s)$.  So then $|1|^2=e^{-2\pi p^2}$, a Gaussian in the $p$ direction.  You can think of things on $T^*\rR$ that are $L^2$ in the $q$ direction but only with respect to this metric in the $p$ direction, and we could take real powers of this line bundle. 

Now we'll come to the one I'm interested in.  Let's assume that the prequantization line bundle is holomorphic.  What is a holomorphic line bundle?  I'll do it really quickly.  We have our connection, $D$, and that will split into $D^{1,0}+D^{0,1}$.  It's holomorphic if $(D^{0,1})^2=0$.  I assume that the underlying manifold is complex so $\bar{\del}^2=0$.  In that case, one can locally construct a holomorphic frame, in the kernel of $D^{0,1}$.  The curvature is in (this is the same), $\Omega^{2,0}+\Omega^{1,1}$.  Now  let's assume that it's also Hermitian, the line bundle, and see what compatibility of a holomorphic connection with a Hermitian line bundle means.  That means that the curvature is skew-adjoint.  Adjointness interchanges $\Omega^{0,2}$ with $\Omega^{2,0}$, so the curvature will actually be in $\Omega^{1,1}$.  Actually, it's very important to know that there is a unique holomorphic connection compatible with a Hermitian line bundle on a holomorphic manifold.  The explicit formula for the connection $1$-form for a Hermitian holomorphic line bundle with respect to a holomorphic frame $s\in \Gamma(X,L)$ (which only exists locally) is as follows:
\[
\alpha=\frac{(Ds,s)}{|s|^2}\]
This is just equal to the $(1,0)$ component of $d\log |s|^2$, using the compatibility of the connection with the Hermitian inner product.  

So back to our story of $\cC$.  I'm interested in the connection form $-\pi \bar{z}dz$.  Taking $d$ of this gives $-\pi d\bar{z}\wedge dz$, which is $-\pi(dq-idp)\wedge (dq+idp)$.  Then the metric, $|1|=e^{-\pi |z|^2}$, the Bagram realization.  I want to write down the eigenspaces of the Dirac operator, which will be tedious for those of you who have taken a course in quantum mechanics.  We have a bit of futzing around.  We have this notion of canonical commutation relations.  You usually see these in week two of a good course on quantum mechanics. So $[a,a^*]=1$, the pair is not always adjoints (they will be in our case).  This is just to be sure of the situation, this means $aa^*-a^*a$.  There is the Fermionic version, named after Enrico Fermi, in my opinion the greatest intellectual ever to toil in Chicago, and I pray they will rename Balboa avenue after him.  So now we have $b,b^*$, which interchange the even and odd parts: $H_\pm\to H_\mp$.  Now it's an anticommutation relation: $[b,b^*]=1$ so $bb^*+b^*b=1$.  Let $A^*$ be a graded algebra, and then if $x\in A^i$ and $y\in A^j$, then $[x,y]$ is defined to be $xy - (-1)^{ij} yx$.  It's okay if the grading is modulo $2$ or any even integer.  Then $[x,yz]=[x,y]z+(-1)^{ij}y[x,z]$.  That sign is because we had to move $y$ past $x$.  There's a principle that you need this sign to move things past each other like this.  We'll have a pair of operators of each type, because our space is two dimensional.  

Another thing to say is that $a$ and $b$ are called annihilation operators, and $a^*$ and $b^*$ are called creation operators.  We'll see this in action.  I'll give you the explicit formulas now.  There are some fudge factors so that I get exactly $1$ and not another number.  I chose
\begin{multline*}
a=\frac{1}{\sqrt{\pi}} \delby{z},\quad \bar{a}=\frac{1}{\sqrt{\pi}}\delby{\bar{z}}
a^*=\frac{1}{\sqrt{\pi}}\left(-\delby{\bar{z}}+\pi z\right),\bar{a}^*=\frac{1}{\sqrt{\pi}}\left(-\delby{z}+\pi \bar{z}\right)
\end{multline*}
So why \wh, well, it's integration by parts:
\[f\bar{g}e^{-\pi |z|^2}\omega\]
By the way, as I wrote it, we have a probability measure.  Integration by parts is very simple in this situation, the adjoint is the negative.  So 
\[\int \frac{\del f}{\del z}\bar{g}e^{-\pi |z|^2}\omega
= -\int f\frac{\del \bar{g}}{\del z} e^{-\pi |z|^2}\omega+\int f\bar{g}(\pi \bar{z})e^{-\pi |z|^2}\omega\]
and then $\frac{\del\bar{g}}{\del z}=\left(\frac{\del g}{\del\bar{z}}\right)^{-}$ and $\bar{z}\bar{g}=(zg)^{-}$.  I'll construct the Dirac operator, calculate its square, and get a harmonic operator [some fast side comments]

I don't have $b$ and $b^*$ but I do have $\bar{b}$ and $\bar{b}^*$.  We have twice as many bosons as fermions.  Half of the bosons get canceled against the fermions.  We've got these generators of an associative algebra, and there are twice as many of the even kind as the odd kind, and we'll see a quasiisomorphism of algebras.  So $\bar{b}$ maps $\Omega^{0,1}(\rR^2,L)\to \Omega^{0,0}(\rR^2,L)$, so its adjoint will go back in the opposite direction.  We think of this as creation because it raises the number of indices. So 
\[\bar{b}=\sqrt{2}\iota\left(\delby{\bar{z}}\right);\quad \bar{b}^*=\frac{1}{\sqrt{2}}\epsilon(d\bar{z})
\]
where $\epsilon$ is multiplication
So $\bar{b}(d\bar{z})=\sqrt{2}$ and $\bar{b}(1)=0$ and then $\bar{b}^*(1)=\frac{1}{\sqrt{2}}d\bar{z}$ and $\bar{b}^*(d\bar{z})=0$.  We get the commutation $[b,b^*]=1$, as desired.

So the $\bar{\del}$ operator is $\sqrt{2\pi}\bar{b}^*\bar{a}$, and we can immediately see that $\bar{\del}^2=0$.  Let me show you how to do this with commutator calculus.  This square is $\frac{1}{2}[\bar{\del},\bar{\del}]$.  This is, well, $\pi[\bar{b}^*\bar{a},\bar{b}^*\bar{a}]=0$.  This kind of calculation works in infinite dimensions too.  To take adjoints, we should commute the operators, but still, we should be careful.  So 
\[\bar{\del}^*=\sqrt{2\pi}\bar{a}^*\bar{b}\]
and then $[\bar{\del},\bar{\del}^*]$ is proportional to $\mathbb{D}^2$.  I forget and don't need it because I'm going to project to the kernel.  So this is $2\pi [\bar{b}^*\bar{a},\bar{a}^*,\bar{b}]$.  Maybe I'd better take the slow route; this is:
\[2\pi [\bar{b}^*\bar{a},\bar{a}^*]\bar{b}+2\pi a^*[b^*\bar{a},\bar{b}]\]
I'm going to use the other Poisson relation, which you can figure out in the same way.  There will be two terms in each case
\[2\pi \bar{b}^*[\bar{a},\bar{a}^*]\bar{b}+2\pi[\bar{b}^*,\bar{a}^*]\bar{a}\bar{b}+2\pi\bar{a}^*[\bar{b}^*,\bar{b}]\bar{a}+2\pi\bar{a}^*\bar{b}^*[\bar{a},\bar{b}]\]
So the second and fourth are zero and then we can simplify the brackets and get \[
[\bar{\del},\bar{\del}^*]=2\pi (\bar{a}^*\bar{a}+\bar{b}^*\bar{b})\]
and we can calculate the eigenvalues.  This is similar to the harmonic operator.

[Some bigger picture description that I missed]

The physicists often talk about the vacuum using the term $\Omega$.  I won't say anything about the relation to the standard meaning of vacuum.  In this case, it's the constant function $1$.  This is indeed in the kernel of the Dirac operator.  So now, if you apply powers of $a^*$, we get holomorphic polynomials, and that's a dense subspace, orthogonal subspace (orthogonal in the Hilbert space which is the kernel of the Dirac operator).  The monomials $(a^*)^n\Omega=\pi^{-\frac{n}{2}}z^n$ form an orthogonal basis of the kernel of $\mathbb{D}$.  As long as there's a mass gap, the lowest nonzero eigenvalue of the Dirac operator is a discrete distance, we'll be fine.  I don't know how, next week's speaker has done a lot of work on extending this model to a general sympletic manifold.

We've got this, and if you want to understand the spectral theory of the Dirac operator, we have that \[[\mathbb{D}^2,\bar{a}]=-2\pi\bar{a},\ [\mathbb{D}^2,\bar{a}^*]=-2\pi \bar{a},\ [\mathbb{D}^2,\bar{a}^*]=2\pi \bar{a}^*, [\mathbb{D}^2,\bar{b}]=-2\pi \bar{b},\ [\mathbb{D}^2,\bar{b}^*]=2\pi \bar{b}^*,\] where $\mathbb{D}=\bar{\del}+\bar{\del}^*$.

A basis of the Hilbert space is given by $(\bar{b}^*)^k(\bar{a}^*)^\ell(a^*)^m\Omega$.  We move around the kernel with the first and then change the eigenvalue with $k+\ell$, the value is $2\pi (k+\ell)$.

I actually have finished my lecture.  I could try to say something more.  The one thing I want to say, I did this for $\rR^2$, and all this works the same for $\rR^{2m}$, you'll have $a_1$ through $a_m$, and $\bar{a}_1$ through $\bar{a}_m$, and the commutation is that $[a_i,a^*_j]=\delta_{ij}$.  Maybe next time I'll show how the Lie algebra of the symplectic group acts on the Hilbert space.  I can write down the projection, but maybe, we'll see.
\section{October 6}

I thought of a really good homework, a good project, since I left it open, is to write the explicit formula for the projection.  $\{z_n|n\ge 0\}$ form an orthogonal basis, and rempmber the measure is $\int f\bar{g}e^{-\pi|z|^2}\omega$.  You can use Graham-Schmidt, and write $P(z,\bar{z},\omega,\bar{\omega})$ or $P(q,p,q',p')$, we want a kernel so that if I insert something in the span of the $z^n$ I get it back, but if I put anything else I get $0$.  So $\int P(q,p;q'p')f(q,p)e^{-\pi(q'^2+p'^2)}dq'dp'$, and the full kernel is the $P$ times the exponential.  The first exercise is to calculate this kernel.  The second exercise, I talked about Toeplitz operators, you should calculate $T_{\bar{z}}$ or if that's too easy $T_{\bar{z}^2}$.  You want to see what you get when you apply the kernel.  These are explicit and not too complicated operators.  If you want $T_{z^2\bar{z}}$ that's more complicated.  This is explained by Berezin (sp?) in a book on quantization.

Til now I've talked about the Hamiltonian formalism, but now I want to turn to the Lagrangian formalism, because that's the most effective way to build a bridge between the classical and the quantum theories.  I'll be making various standard observations, which will help anyone who wants to read the definitive account, a chapter by Deligne and Freed (I'll put this on the blackboard page).  They give far more detail but go fast and assume sophistication.

There's a famous line: ``physics is where the action is.''  The action will be a function of the ``fields'' in the theory: $S(\Phi)$, so $\Phi$ represents all the different fields.  The one to bear in mind most of the time is Minkowski space $\rR^{1,d-1}$, we'll usually use the following conventions, that the Lorentzian metric $\eta_{\mu\nu}$ is a diagonal matrix with entries $(1,-1,\ldots, -1)$, so the coordinates go $x^0,\ldots, x^{d-1}$.  That's our spacetime.  if $d=1$ then spacetime is the real line, and quantum field theory becomes the study of paths in a space, and we're back at the study of loop spaces and path spaces, so, quantum mechanics.  We want to move our formalism away from quantum mechanics.  We also have a target manifold, and the space of fields will be differential maps from our Minkowski space to our manifold.  So we have $M$ and the a space of fields $Map(\rR^{1,d-1},M)$, and I want to be vague about my maps.  Maybe I could say compact support, but that's not terribly physical.  I want to write local expressions, so I won't worry yet about any of that.  We'll choose coordinates on $M$ of the form $X^a$.  We can't generally do this globally.  You'd better make sure that your action is covariant, well-defined whatever the coordinate system is.  A typical example of an action $\Phi$ is the one appropriate for harmonic (after Wick rotation) maps.  We'll have a Riemannian metric $g_{ab}(X)$, and finally our action is:
\[
S(\Phi)=-\frac{1}{2}\int_{\rR^{1,d-1}}\eta^{\mu\nu}g_{ab}(X(x))\delby{\mu}X^a(x)\delby{\nu} X^b(x)dx\]
We'll discern standard features.  It depends in a smooth fashion on the values of the field at a point, and in a polynomial fashion on its derivatives.  If you, there's a further generalization in which Minkowski space is replaced by a manifold with a Minkowskian metric, and being polynomial in iterated partial derivatives will be equivalent to being polynomial in covariant derivatives.  

Okay.  While I'm at it, there's an easy modification of this that one wants to study, this is analogous to where we have the Schrodinger equation, which has the Laplacian and a potential, and here we're free to add a potential function $V(X)$, if here $V$ is a smooth function on $M$.  This is thought of as a potential energy or interaction term.  Actually, this is two of the terms that occur in the action.  Later in the course, there will be a different type, topological terms in the action.  They won't be given by local expressions.  They're analogous to holonomy of a connection.  You can write down the holonomy around a loop in a space with connection, this is like the logarithm of the holonomy.  These will be multivalued, and for quantum theory that's not a problem.

Putting topological terms aside, we are interested in actions obtained by integrating a density, a $d$-form on Minkowski space $\rR^{1,d-1}$ which has a local expression $C^\infty$ in the value of the field and polynomial in its partial derivatives.  In fact, all the actions we'll write are going to be first order, will only depend, all the actions I've ever heard of involve at worst second partials, such as the Hilbert action, which uses curvature.  You can, however, and this is an important discovery of Elie Cartan, reexpress general relativity in a way that only involves first derivatives.

If $d=1$, this is the energy of a path in our Riemannian manifold. 

This action is the integral of a density, which motivates the introduction of the $d$-form on the product of Minkowski space and the space of fields.  Somehow the words Lagrangian and action are mixed up a bit.  So $\Ll\in \Omega^d(\rR^{1,d-1}\times Map(\rR^{1,d-1},M))$.  This looks silly, but this point of view, which goes back to Emmy Noether, is quite useful.  Then $S=\int_{\rR^{1,d-1}}\Ll$.  This makes it easy to explain Wick rotation.  This is an analytic continuation in which we consider what happens when the time coordinate becomes imaginary.  I'd better introduce $x^0=i\xi^0$ and $x^u=\xi^u$ otherwise.  This replaces $\eta$ by the negative of the Kronecker delta, and the Lebesgue measure with $i$ times it.    $\Ll=\frac{i}{2}\delta^{\mu\nu}g_{ab}(X(\xi))\delby{\mu} X^a(\xi)\delby{\nu} X^b(\xi)d\xi+i V(X(\xi))d\xi$
This has made it easier, taken us from our hyperbolic metric into something we understand better.  So, there's, I think I'll put a $-$ sign in a couple of places.

This is become, in the 80s and afterward, the most important idea in quantum field theory.  When differential geometers talk about quantum field theory, they do so after this rotation, and then they can study elliptic things and not hyperbolic ones.  The Einstein \wh are firmly in the hyperbolic camp.  A supersymmetric field theory, after Wick rotation, is not generally field theory.  Spinors are completely different depending on the metric on spacetime.  Whether spinors are real or only Dirac is interrupted by Wick rotation. 

Another thing to point out, we don't know how to apply this in \wh.  Quantum gravity will be thorny, and I'd like to talk about it but I don't think so.  

Okay.  Physicists call the Minkowksi version the nonlinear $\sigma$-model, and one can study the special case when the manifold $M$ is $\rR^N$, the linear $\sigma$-model.  The action is exactly what I wrote down, but the metric is constant, $g^{ab}=\delta^{ab}$.  [What if $X$ is a bundle over spacetime?]  We can trivialize the bundle.  In this generalization, you replace $\delby{\mu}$ by $D_\mu$, you have covariant derivatives, that's called coupling the linear model to a magnetic field.

I'm actually only going to talk about free theories, where the aciton is quadratic in the fields.  If the action is quadratic, in a moment, I haven't said this, but we'll exponentiate the action and try to integrate over the field space.  If it's quadratic we can do it and will get Gaussian integrals.  If we're nonlinear asking this doesn't make sense, it's not free.

An example is the linear $\sigma$-model with quadratic (or zero) $V$.  A quadratic $V$ is called a mass term.  The things that are natural to mathematicians are also natural to physics.  Why we're using mass here is related to classical mechanics.

Now comes the punchline.  Hang on, first I should talk about the Euler-Lagrange equation.  Because of the hypothese I made about the nature of the action.  Being a critical point of the action is the same as solving a certain nonlinear partial differential equation.  Apart from the free case, where it's easy to calculate, the next easiest is when $d=1$, where you have the energy of a path, and then the thing we find is a geodesic.  Doing this from stationary position is not entirely trivial and requires knowledge of differential geometry.  The free case in the geodesics, you get straight lines.  You find you're describing Newton, particles which don't experience force continue at the same speed.  We more generally have geodesics at $d=1$.

So how do mathematicians and physicists find the Euler-Lagrange equations?  We put in a parameter $\epsilon$, take a derivative, and then take integrals by parts.  Let me do this calculation, even throwing in a potential.  Let me write $X(x)+\epsilon Y(x)$, and we'll insert this into the action and take the derivative with respect to $\epsilon$ at $\epsilon=0$.  We see that this is $S(X)+\epsilon\int -\eta^{\mu\nu}\delta_{ab}\delby{\mu} X^b(x)\delby{\nu} Y^a(x)dx+\delby{a} V(X(x)Y^a(x))gx+O(\epsilon^3)$.

Now I do my integration by parts ignoring boundary conditions, which makes for the deepest suspicions.  We'll get the famous Klein-Gordon equation:
\[
\frac{d}{d\epsilon}S(X+\epsilon Y)=\int \eta^{\mu\nu}\delta_{ab}\del_\mu\del_\nu X^b(x)+\del_a V(X(x)))Y^2(x)dx\]

We're now on page one of any textbook on quantum field theory:
\[\square X+\del_a V(X)=0\] where $\square$ is the ``D'Alambertian.''  If we have a quadratic potential, then Euler Lagrange equation is linear.  I'm not here to study Euler Lagrange equations.  I don't know the first thing about quasi-linear \wh equations.  We'll be saying more about these as we go on, but now I can make the main point about the bridge between this and classical field theory.  The study of solutions to the Euler Lagrange equations and their symmetries, well, this is Feynmann's idea, oh, I should say that {\em classical} field theory is the study of solutions to the Euler Lagrange equations and its symmetries; {\em quantum} field theory is the study of the measure $e^{2\pi i}S(\Phi)/h$, where $\hbar=\frac{h}{2\pi}$.  Then you put $D\Phi$ for the measure, indicating that physicists don't know what volume on the field space is.  Despite many efforts, this expression doesn't make sense.  It's a beautiful way to find out what quantum theory is on about, but not good for studying it.

After Wick rotation, $S$ is replaced, and we get an entirely different measure which has, in some cases, been constructed.  The space of fields is quite different, but something like this.  In certain cases,
$e^{-2\pi S^{\text{Euclidean}}(\Phi)/h}D\Phi$.  This works in $d=1$, $d=2$, and massive theories, $d=3$ (in the linear $\sigma$-model and some others).  I'll leave open what that means.  No one can analyze $d=2$ with vanishing potential.  If you have the target the $2$-sphere for instance.  Feynmann-Kac (Wiener,It$\hat{\text{o}}$) did it in dimension one.

Once I've performed the rotation, you can see why there's a relationship at all, and you have to cross your fingers that this is similar to reeal quantum field theory.  You'll get the critical points with the lowest value of the action as $h$ goes to $0$.  That's the strong interactin limit of the theory.  The stationary phase evaluation is what you would call this before Wick rotation.

[Some side discussion]

Incidentally, I can now return to my mention of topological terms.  These will lead us to considerations like those with line bundles.  What makes sense is the whole expression.  You don't want to to think of a topological term as the exponential of something divided by Planck's constant.  You want to think of it as maps from the field space to $U(1)$.  It's not so straightforward to apply a Wick rotation to a topological term; how do you rotate something which isn't local? 

I should mention the most famous example, if $d=2$ and $M$ is a compact Lie group, in that case, if we have an inner product, a biinvariant metric on $G$, or an invariant inner product on the Lie algebra, in modern terms what we learn is that provided a condition is satisfied by this metric, it has to be quantized in a certain, very precise, sense.  There is a discrete set of values that this metric can take, and you can make something like a line bundle on $G$, but with curvature a closed $3$-form.  Then there's an analogue wher ewe can pull this back to $\rR^{1,1}$, like holonomy, for these things, called gerbes (on a surface).  We essentially take the non-linear $\sigma$-model and multiply it by a topological term.  This is called the $WZW$-model (sometimes with an N).  This uses affine (Kac-Moody) Lie algebras.  There's a quantization in the labelling of representations, just as in the case of Lie algebras of compact Lie groups, which corresponds to the question as to whether you get a map to $U(1)$ or a cover of $U(1)$.  The representations of affine Lie algebras exactly matches up to \wh.

That's the introduction to quantum field theory.  [Historical notes]

Now we'll think about the following thing: I had to perform an integration by parts, now I'll try to pay attention to that.  One example, if my space time is this picture, and here's my initial and final times, I can't handle the situation where time goes to $\pm \infty$, I've got these propagating waves and my action will diverge, but now my spacetime ha a boundary, I've done something wrong, done an integration by parts without using Stokes' theorem.  This is what was done by Emmy Noether.  I know this from a paper of Greg Zuckerman.  This was only taken on board by the mathematicians I know in the mid-80s.  This is done in the chapter of Deligne and Freed.  They copy it from Zuckerman's paper.  Witten wrote a paper on this in '84 as if this was unknown.  Maybe I should have worked this out before I presented it.  

Let's keep track of the integration by parts.  I want to think of differential forms now on that whole space of fields, we'll work on $\Omega^*(\rR^{1,d-1}\times Map (\rR^{1,d-1},M=\rR^N))$.  Now exterior differentiation breaks up into two pieces.  We could call them horizontal and vertical, but I'll use $D$ for the total differential, $d$ for the de Rham differential in the spatial directions, and $\delta$ for the variational derivative, the de Rham differential in the mapping space.  If we take Banach maps or \wh maps, then the mapping space will be a Banach manifold.  We were calculating the differential and we threw away a term, and I'll go back and write down the piece I threw away, and we'll see the Hamiltonian formalism.  

So $D\Ll$, well, it's a $d$-form in the horizontal direction, this is, I should have used notation $\Omega^{i,j}\subset \Omega^{i+j}$ which is an $i$-form in the horizontal direction and $j$ in the vertical.  Since $\Ll$ is already $d,0$, the $d$ term will take it to zero so
\[D\Ll=\delta\Ll=\-\eta^{\mu\nu}\delta_{ab}\delby{\mu} X^a(x)\delby{\nu} X^b(x) dx\]
I'll continue this on Tuesday

\section{October 11}
[Came in late because Ezra was starting early]


\[\Ll=\frac{-1}{2}\eta^{\mu\nu}\del_\mu\phi\del_\nu\phi dx\in \Omega^{d,0}\]
By solving the equation $\delta\Ll+d\gamma=(\ )\delta\phi\wedge dx$, then $\delta\gamma=\omega\in\Omega^{d-1,2}$  restricting it to the space of solutions to the Klein-Gordon equation, then integrating we get a closed two-form on the space of solutions.
\[\gamma=-\eta^{\mu\nu}\del_\mu\phi\delta\phi\wedge\iota(\del_\nu) dx\in \Omega^{d-1,1}\]
\[\omega=-\eta^{\mu\nu}\delta(\del_\mu\phi)\wedge\delta\phi \iota(\del_\nu)dx\]
$d=1, x^0=t$, $-\delta\phi\wedge \delta\phi$
\\
$d=2, x^0=t,x^1=x$,  $-\delta\phi_t\wedge \delta\phi dx-\delta\phi_x\wedge\delta\phi dt$

Consider the set of solutions to the Euler Lagrange equations and suppose that the boundary of $X$ is empty, then automatically, the integral of $\omega$ over $X$ is a closed two-form on the solutions to the equation (on shell).  This is just classical field theory.  This only involves solutions of these equations.  We've seen how to get the equations, and how to get a symplectic form once we have them.  

If $\{X_t\}$ is a family of hypersurfaces, then $\int_{X_1}\omega-\int_{X_0}\omega=\delta\int_{\{X_t\}}\gamma$ on $\{EL=0\}$

[missed something trying to catch up on the formulas]

Once you have infinite dimensional symmetry groups, you won't have nondegeneracy, that's an interesting issue.

So we want to talk about symmetries, let's write some down.  By the way, to understand what I'm saying, it's quite useful to think about the $d=1$ case.  You can say ``at time $0$'' so you're just evaluating at that time on geodesics.  This is the usual symplectic form on the cotangent (in this case tangent) space of a Riemannian manifold.  It's harder to describe hypersurfaces in a higher dimensional space.

I need to talk a little about symmetries.  A free scalar field doesn't have many but it has translation invariance.  We'll see the basic idea that to a symmetry is associated a Hamiltonian, and those for translation invariance will \wh a vector, and we'll have the Hamiltonian component and the momentum component.  So $P_\mu=(\underbrace{P_0}_{\text{Hamiltonian}},\underbrace{P_1,\ldots, P_{d-1}}_{\text{momentum}})$ and we'll produce densities, this is the key thing.

So first, what is the symmetry?  We're using a crippled notation, $\Pi_\mu=\del_\mu\phi\delby{\phi}-\delby{x^\mu}$, this is a vector field on $\rR^{1,d-1}\times Map(\rR^{1,d-1},\rR)$.  One way to describe this is by saying what it does to the coordinates, and that will, of course, determine it.  So the Lie derivative $L(\del_\mu\phi\delby{\phi})\phi=\del_\mu\phi$ and $L(\del_\mu\phi\delby{\phi})x^\nu=0$.  On the other hand, $L(\del_\mu)\phi=\del_\mu\phi$ and $L(\del_\mu)x^\nu=\delta_\mu^\nu$, so that $L(\del_\mu)dx^\nu=0$.



For those who know about jet spaces, instead of fields you can work with jets, and if you start taking partial derivatives, then these commute with the Lie derivative. %$L(\Pi_\mu)\del_{\mu_1}\cdots\del_{\mu_\ell}\phi=\del_{\mu_1}\cdots\del_{\mu_\ell}L(\Pi_\mu)\phi=
%\del_{\mu_1}\cdots\del_{\mu_\ell}\del_\mu\phi$
%and the other thing we need to know is that Lie derivatives commute with total derivatives, and we also have Cartan's formula for the Lie derivative on differential forms:
%\[
%L(\Pi_\mu)\circ D=D\circ L(\Pi_\mu);\quad L(\Pi_\mu)=D\circ \iota(\Pi_\mu)+\iota(\Pi_\mu)\circ D\]

Wait, I've made a huge mistake.  Sorry.

[parts he took back removed]

Also, $\iota(\Pi)EL(\Ll)\sim 0$, this will vanish onshell if the last thing I do is a contraction.  

I've shown now that $L(\Phi_\mu)\Ll=0$.  Also $L(\Pi_\mu)\gamma=0$.  So $\iota(\Pi_\mu)\circ D(L+\gamma)+D\circ \iota(\Pi_\mu)(\Ll+\gamma)=0$.  This is promising.  So $D(\Ll+\gamma)$ is the Euler Lagrange form plus $\omega$.  The contraction, just like in the Hamiltonian framework, of $\Pi_\mu$ here, this has vertical and horizantal pieces, $\iota(\Pi_\mu)\omega\in \Omega^{d-2,2}+\Omega^{d-1,1}$, and onshell, it's just $D$ of the $d-1$ form $\iota(\Pi_\mu)(\Ll+\gamma)$.

So the contraction of our ``symplectic form'' with $\Pi_\lambda$ is $d+\delta$ of this form, the contraction with $\Ll$ is $0$, and $\gamma$ is a one-form, and we replace $\delta\phi$ with the partial derivative, and we get $(d+\delta)(-\eta^{\mu\nu}\del_\lambda\phi\del_\mu\phi\iota(\del_\nu)dx+\frac{1}{2}\eta^{\mu\nu}\del_\mu\phi\del_\nu\phi\iota(\del_\lambda)dx-\eta^{\mu\nu}\delta\phi\del_\mu\phi\iota(\del_\lambda\iota(\del_\nu)dx)$
The first two terms are $d-1,0$, forms and the last is a $d-2,1$ form.  The first two make up the Noether density.  The Hodge star is the current associated to these symmetries.  The $d-1,0$ component is also called the stress energy tensor.  At the moment it's a smooth map from the space of solutions to the Euler-Lagrange equations to sections over spacetime $\rR^{1,d-1}$ of $T^*\otimes \wedge^{d-1}T^*$, and when you rewrite this, it looks like $T^*\otimes T^*$, and you get a symmetric $2$-tensor, that's the discovery.  It sits in Einstein's equation, a remarkable coincidence, maybe I should be presenting it, that's the best theorem in classical field theory.  

We've said, if we minimally couple the Klein-Gordon equation to a Riemannian metric, and take the functional derivatives of the aciton, we essentially get this.  That's a theorem, it's not at all obvious.  It seems to me that it's not relevant to our considerations.  Having non-flat background metrics is an extra twist I'd rather not discuss.  Very simply, this is a replacement for the momentum.  When $\nu$ is zero, we get the Hamiltonian.  This is called a manifest symmetry because the Lie derivative of $\Ll+\gamma$ is zero (in Deligne and Freed).  This is not necessary to have a symmetry in field theory or even classical mechanics.  A non-manifest symmetry, the Lie derivative is exact, and that's a sufficient condition.  You'd have to find the integral and add it to get the momentum density.  Most of the symmetries we'll discuss are going to be manifest.  It's not possible to find realizations of supersymmetry in general that satisfy the manifest condition.

The condition before was $\iota(\Pi)\omega$ being exact.  If you have this, with a symmetry you'll get $\gGg\to \Omega^{d-1}(\rR^{1,d-1}\times\{EL=0\})$ so you might want to add that this action is $G$-equivariant.  These conditions are extremely difficult to enforce.  I'm trying to keep ahead of where I'm going, and discovering you get tremendous central extensions and $L_\infty$ algebras and so on, because the right hand side is part of a complex.  For now, let's just work out what happens, let's say we're in two dimensions, $t,x$, and look at $P_0$.  So we get \[
-\phi_t^2 dx -\phi_t \phi_x dt + \frac{1}{2}(\phi_t^2-\phi_x^2)
\]
I'm missing a factor of a half.  If we integrate over a spacelike slice, like $t=0$, we're just left with the integral over the slice of the velocity squared, up to a constant that I'm blanking out on.  All the other terms are just to make this covariant.

We can look at the Lie derivatives with respect to our vector fields $\Pi$, but they're all zero.  This is clearly, at the level of Poisson brackets, a representation.  If we were going to define it, it would probably be something like $\{P_\mu,P_\nu\}=\frac{1}{2}L(\Pi_\mu)P_\nu-\frac{1}{2}L(\Pi_\nu)P_\mu$.

[Is this a symmetric tensor?]
If I try to use the Hodge star, I get $P_\lambda=\eta^{\kappa\rho}P_{\kappa\lambda}\iota(\del\rho)dx$ and $P_{\kappa\lambda}=\frac{1}{2}\eta^{\mu\nu}\del_\mu\phi\del_\nu\phi-\eta\del_\kappa\phi\del_\lambda\phi$.  The first one is $\eta_{\kappa\lambda}$, and so both of these are symmetric.

The theorem, I don't know whose, is that this quantity is the variational derivative of the free field action with respect to the background metric.  We need to be coupled correctly to the background metric for that to be true.  Actually the action is easy to transfer, it's the integral with respect to the metric, times the length squared of the gradient of $\phi$.  So the coupling is very simple.  The first term is the variation in the measure and the second the variation in the Lagrangian density.  So if you hadn't met that, you wouldn't have learned it from what I just said.  

Moving on, I'm spending too much time on the free field, and I thought it would be nice to use the non-free field.  I want to exhibit all the features, almost all, of complications we can reach.  This is Chern Simons. This is a theory in $\rR^3$, and let me be vague about the metric.  There will be one field, it's a connection for a trivial $G$-bundle.  The Lie algebra of $G$ I'll call $\gGg$, and assume it has an invariant inner product.  Actually, what am I saying?  The example to bear in mind is $G$ is $GL(n)$ and $\gGg=M_n$.  A connection on a trivial bundle is the same as a one-form with values in $\gGg$, and we'll write those $A_\mu dx^\mu$.  As you know, the curvature is a two-form with values in the Lie algebra, given by the equation $F=dA+\frac{1}{2}[A,A]$.  We're looking for a Lagrangian density, whose Euler Lagrange equation is $F=0$.  We want onshell to be flat.  It turns out that $\Ll=\frac{1}{2}(A,dA)+\frac{1}{6}(A,[A,A])\in\Omega^{3,0}$.  This is a good place to stop.  See what partial derivatives you have to form to get $F=0$, and then we'll get the symplectic form on the space of flat connections, but it won't be symplectic, it will be zero in the gauge directions.  There are a couple of ways to get around this.  That is more like what people in field theory study.  We'll show, essentially, the curvature turns out to be the momentum for the gauge transformations.  We'll see all that next time.


\section{October 18}

Okay, so I said I was going to talk a little about Chern-Simons theory in the context of the formalism I was talking about.  I'll simplify things by assuming that our principal bundle is trivial.  We have a compact Lie group $G$, with a choice of invariant metric $(\ ,\ )$, and with Lie algebra $\gGg$, and I'll be thinking first of all of spacetime $\rR^{1,2}$, and later I'll change things around to make more contact with geometry.  In this simple setting our space of fields is $G$-connections on $\rR^{1,2}$, in other words $\Omega^1(\rR^{1,2},\gGg)$.  To consider more general bundles, you need to be able to write down.  We have our Chern-Simons form $L=\frac{1}{2}(A,dA)+\frac{1}{6}(A,[A,A])$, which is a form in $\Omega^{3,0}(\rR^{1,3}\times\text{connections})$.  In general, you need a flat $G$-bundle to write this.  We go through the procedure of calculating everything.  The variation in the Lagrangian density is $\delta L=(\delta A,F)-d\gamma$ with another term $\frac{1}{2}(A,d\delta A)$, which is $-d\frac{1}{2}(A,\delta A)+\frac{1}{2}(dA,\delta A)$.  So let me remind you, $F=dA+\frac{1}{2}[A,A]$ and $\gamma=\frac{1}{2}(A,\delta A)$.

So we get $\omega=\delta\gamma=\frac{1}{2}(\delta A,\delta A)\in \Omega^{2,2}$.  The Euler Lagrange equation is $F=0$, and let me replace non-compact spacetime with $\rR\times \Sigma$ where $\Sigma$ is a Liemann surface.  A connection on the product is a path in the space of connections on $\Sigma$ and a one-form in the $\rR$ direction times a gauge transformation.  So $A$ decomposes into $A_x(t)dx + A_y(t)dy+A_t(t)dt$.  The curvature being zero means that the connections $A_x$ and $A_y$ are flat at all time, it's enough to say it at zero, and if you take the derivative with respect to $t$ at zero, you get an infinitesimal gauge transformation.  So essentially you get a decomposition of the the space of solutions into a product: \[\{A\in \Omega^1(\Sigma,\gGg):F_\Sigma=0\}\times Map(\rR,\Omega^0(\Sigma,\gGg))\]
We get a standard closed two-form on the first space and the second part doesn't interact at all.  I want to relate this to symplectic reduction.  Please retain the formulas for $L$ and $\gamma$ in your mind.

Notice that if we take all of the space of $A$, this is roughly speaking a symplectic vector space.  I want to put this in quotes, ``symplectic,'' we get a dense embedding into the dual, not an isomorphism.  By worrying about Sobolev spaces you might fix this issue.  This has an action of the gauge group restricted to actions on space, not spacetime.  The group is $Map(\Sigma, G)$, and the algebra is $g=\Omega^0(\Sigma,\gGg)$.  A Hamiltonian action corresponds to a function on a symplectic manifold, and that map may be taken to be linear in which we can express it as a map from the symplectic manifold to $g^*$, which has $\Omega^2(\Sigma,\gGg)$ as a dense subspace, since I can wedge together, and use the metric to produce a number.  In fact, there's a map from $\Omega^1(\Sigma,\gGg)$ to $\Omega^2(\Sigma,\gGg)$, which is the curvature.  The point of view is due to Atiyah and Bott, or maybe Donaldson.  What we're doing is mapping $A_\Sigma\to F_\Sigma$.  Actually, if you go through the calculation, the function $F_\Sigma$, I can pair with an element of the Lie algebra and get a function on my space.  If we have $\phi\in g$, then what is the Hamiltonian vector field associated, it's $H_\phi(A)$, which is given by $d\phi+[A_\Sigma,\phi]$, and the vector field is indeed associated to the differential of the moment map $\Omega^1\to \Omega^2$.  The contraction $\iota(H_\phi)\frac{1}{2}(\delta A,\delta A)$ is $(d_AF_\Sigma,\phi)$.  You can check that, it's pretty simple.  What's beautiful is that these have global meaning in terms of vector bundles.

Now there's a fundamental idea, reduction, where if you have a moment map like this, then you can look at the inverse image of $0\in F_\Sigma$, and the Hamiltonian action will reduce to an action on this level set by Noether's theorem, and we can quotient by this action, getting a new symplectic manifold.  The inverse image of the zero set has leaves, it's foliated by isotropic subspaces of the tangent space.  The quotient of that foliation is a symplectic manifold, the one we have in mind in Chern-Simons theory.

Having spoken about this Hamiltonian action, I'd like to discuss it in the Noether formalism.  This will be a bit confusing.  Remember my prescription, well, let's see what happens.

It's confusing because now I want to move to the gauge group of spacetime acting on fields on spacetime, instead of space and space.  If you recall, I said that we had, well, here's our vector field.  Let $\eta\in \Omega^0(\rR^{1,2},\gGg)$.  So we want to understand $H_\eta$.  We know that $\iota(H_\eta)\delta A=d\eta+[A,\eta]=D_A\eta$.  Let's show that this is Hamiltonian.  If you'll recall, let's see, on the one hand, we'll do the calculation by taking $L+\gamma$ and calculating its Lie derivative with respect to this vector field:
\[\Ll(H_\eta)(L+\gamma)=(d+\delta)\iota(H_\eta)(L+\gamma)+\iota(H_\eta)(d+\delta)(L+\gamma)\] restricted to the locus where $F=0$.  In the first case $L$ has no appropriate indices, and in the second case we get the form $\omega$.  We're left with
\[-(d+\delta)\frac{1}{2}(A,d\eta+[A,\eta])+(d\eta+[A,\eta],\delta A)\]
I'll go in mad circles if I try to go too far.  When $\delta$ comes in, it hits $A$ or the other $A$, and they give the same term, and that cancels a piece of the seconp part.  On the other hand, [other cancellation].  I get $\frac{1}{2}(d\eta,\delta A)$, and [some confusion in the calculation].  The end answer is that the monentum density is $(\eta,\delta A)$.  In the case of translations of spacetime, that was a $d-1$ form, that you could integrate over the spacelike slice and get the monementum.  Here you'll get zero, you have the wrong dimension.  The gauge transformations are acting trivially is what we get, then.  This is a spacetime $1$-form, so the actual momentum for gauge transformations on the Euler-Lagrange locus is zero.  This reflects that these directions are isotropic for the presymplectic form.

If you do the calculations more carefully without restricting to $\{F=0\}$ you get an additional term from the curvature, and if you integrate you get our original formula.

I told you in the first part that we should choose an almost complex structure to quantize.  There are extremely natural almost complex structures here related to algebraic geometry.  The point is, if you have a conformal structure on a surface, you have an almost complex structure on the surface, which is complex, and that's rotation by $\frac{\pi}{2}$.  This is related to the fact that the Hodge star on the middle dimension of an even dimensional manifold is a conformal invariant.  That is, $*$ is defined using a conformal structure, not a full Riemannian structure.  The tangent vectors to our presymplectic manifold are in fact one-forms, with values in the Lie algebra, so the tangent space $T_{A_\Sigma}\{F_\Sigma=0\}$ is the space $\{\alpha\in \Omega^1(\Sigma,\gGg)|d\alpha+[A,\alpha]=0\}$.  I forgot to mention that $*^2=-1$, so it's an almost complex structure on $1$-forms, and restricts to \wh on harmonic forms.  This is $Z^1(\Sigma,\gGg,D_A)$, which is, by the Hodge decomposition, $H^1(\Sigma,\gGg;D_A)\oplus B^1(\Sigma,\gGg;D_A)$.  This latter is the isotropic subpspace of infinitesimal gauge transformations.  $H^1$ is the space of forms $\alpha$ so that $D_A \alpha$ and $D_A *\alpha$ are zero.  This is a, this subbundle is almost complex, and on the quotient manifold, it defines a complex structure.  Inside our locus we should be restricting attention to $\{A_\Sigma\in \Omega^1(\Sigma,\gGg)|H^0(\Sigma,\gGg;D_A)=0\}$, the space of infinitesimal gauge transformations that don't act effectively.  This is a condition; then by Poincar\'e duality, $H^2$ will be dual.  The Euler characteristic of the complex is that of $\Sigma$ times the dimension of $\gGg$.  The complex dimension is half that.  We get a complex vector bundle of dimension $\dim(\gGg)(g-1)$ over this open subset.  As I said, the thoerem is that this vector bundle, which is constant along the leaves of our foliation, defines a complex manifold structure on the quotient.  On the leaf space of the isotropic foliation, we get an almost complex structure on the tangent bundle.  I guess I should announce the famous theorem, this is a quasiprojective variety.  Nature delivers a natural prequantization bundle.  This isn't exactly the same thing as quantizing.  There's some piece of this space of connections that's very bad.  Part of the idea of quantization is finding something to do when it's so singular.  The BRS formalism is another approach.  One task of theoretical physics is to see if these give the same answer.

Now to change topics, to get to the heart of the subject, which is functional integration.  This doesn't come down to quantizing symplectic manifolds but using a path integral formalism.  In fact, in the first stages, this is tautological, you construct things that have the correct answers.  Basically, more or less, this is restricted to free field theories.  There is expertise usable on understanding interacting theories, but it's difficult to think of path integrals in the same way as the free field case.  My immediate goal is to explain the path integral proof of the Atiyah-Singer index theorem in a heuristic way.  That will be a very simple case of path integration because we're doing quantum mechanics, real dimension one.  I should say, in quantum mechanics we can deal with field theories that aren't free, but we're restricted to the Euclidean formalism, after a Wick rotation.  We should imagine, we're working in imaginary time.  That's all a grave difficulty and moves us a distance away from quantum field theories of interest to physicists.  We have to bear in mind how distant these are from the kinds of things that physicists are interested in.  I thought we'd start with Wiener measure.  I'll be thinking in terms of, let's say that I have a time interval $[0,T]$ in the real line, and I'll look at maps $[0,T]\to \rR^n$, with imaginary time.  The idea is, roughly speaking, to pretend that this space of maps has Lebesgue measure, and multiply $d\phi$ by $e^{\frac{-1}{\hbar}S}$.  This is the Wick rotation of $e^{\frac{i}{\hbar}S(\phi)}d\phi$.  We're not interested in the overall normalization, and can at least intuitively convert it into a probability measure:
\[d\mu=\frac{e^{-\frac{1}{\hbar}S(\phi)}d\phi}{\int e^{-\frac{1}{\hbar}S(\phi)}d\phi}\]
What is $S(\phi)$?  It is $\int_0^T L(\phi)=\frac{1}{2}\int_0^T|d\phi|^2dt$, that is, energy.  As $\hbar$ goes to zero, this measure peaks around the classical solutions, geodesics.  The key idea is that as $\hbar$ moves away from zero, this formalism gives us the correct answer.  It's too difficult to check this in any real situation except maybe the one I'm about to say.  Steve is an expert on checxing whether there's a relation between these points of view on quantum mechanics.  Elton also works on this kind of integral.  We observe that $S(\phi)$ is quadratic.  The best way to study the exponential of a quadratic is to use a Lapalace transform.  Let's see.

This will give a Gaussian on the dual vector space.  If I call my mapping space $H=H^1$ and think of it as a Hilbert space of finite energy paths, then we have its dual $H'=H^{-1}$, elements of which are called sources.  From a mathematical point of view we'd think of them as generating functions.  We think of $\int e^{(\phi,J)}d\mu$, which can be integrated by completing the square.  This is 
\[\int e^{-\frac{1}{\hbar}\int_0^T \left(\frac{1}{2}(\dot{\phi})^2-\hbar\phi J\right)dt}d\phi\]
and the idea is to translate, we have the Green's kernel on the interval, given by 
$G(J)(t)=\int_0^t sJ(s)ds$
%$G(s,t)=|s-t|$.  The second derivative of this is the $\delta$ function, so this kernel \wh.  
%we have $AJ(t)=\int_0^t J(s)ds$

So I can write this as
\[\int e^{-\frac{1}{\hbar}\int_0^T \frac{1}{2}S\left( {\phi}-\hbar GJ\right)dt-\frac{\hbar}{2}(J,GJ)}d\phi
\]
At least intuitively, since we renormalized the measure, well, the first part cancels.
This is
\[e^{-\frac{\hbar}{2}(J,GJ)}\] or $e^{-\frac{\hbar}{2}H^{-1}}$.  In a finite dimensional case, these would generate a dense subalgebra of the continuous functions, and we'd see that there was only one measure that could satisfy these conditions.

I think I've written the wrong formula for $GJ$, it should have derivative $AJ$.

There's an extremely general theorem, that says that whenever you have a bilinear form, so basic that it's in Bourbaki, that this measure is a Radon measure on the space of distributions, the unique positive such measure on the space of distributional maps from the interval to $\rR^n$.

In fact, one could do much better, diagonalize the operator and work out what the measures look like in each coordinate and you can actually see that the measure of the Sobolev space $H^{\frac{1}{2}-\epsilon}$ is $1$ for each $\epsilon>0$, and $0$ otherwise.  This is the abstract formulation of the construction of Wiener measure.

This carries over to the free field in dimension $d$, and there, I'm obviously getting in a twist.  I've been working with a massless field.  Instead let's work with $\frac{1}{2}\int \left(|d\phi|^2+m^2|\phi|^2\right)dx$.  The analog of covariance, replace it with $\frac{1}{2}(\phi,(\Delta+m^2)^{-1}\phi)$.  You can define this if you like with Fourier transform.  This is the inner product on $H^{-1}$, and the story is similar except that the indices all change.  I don't know an analogue of the second result, but
$\mu(H^{1-\frac{d}{2}-\epsilon}(\rR^d,\rR^n))$ is $1$ if $\epsilon>0$ and $0$ otherwise.  As soon as the dimension of the spacetime is greater than one, we're no longer discussing functions.  The typical path isn't even a continuous function.  This is the fundamental problem with quantum field theory.  The typical paths don't have a well-defined value of the action.  Basically what's gone wrong are that the formulas we have are only true in infinite dimensions in very special cases.

Okay, so we'll pretend otherwise and see how far we can get.  Next time I'll construct a measure on the space of paths in a Riemannian manifold.  I won't give you details but some ideas of the properties.  Then I'll introduce fermions and formulate supersymmetric quantum mechanics.  This course is walking a fine line between rigor and physics.
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