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\title{Notes from the Operator Algebras and CFT Workshop}
\author{Scott Carnahan}
\begin{abstract}
Notes August 16-21 2010. Other people's notes appear at:
\texttt{http://math.mit.edu/\~{}eep/CFTworkshop}
\end{abstract}
\maketitle
\tableofcontents
\section{Monday, 8/16/10}
\subsection{9:00am Andre Henriques, Overview and introduction}
Main question: What is Conformal Field Therory?
There are many possible answers to that question that live in very different areas of mathematics. One of these answers has to do with operator algebras. Other key words are "vertex algebras" and the formalism of Graeme Segal. There is an axiomatic setup that uses the formalism of operator algebras, more specifically von Neumann algebras, and these objects are called conformal nets.
A conformal field theory is some vague notion that some will call a conformal net, and others will define differently. There are parallel constructions, but no known functors.
Most prominent examples come from loop groups of compact Lie groups (
``loop group nets''), and Dirac free fermions. (Loop groups are only conjectured to satisfy Segal's CFT axioms.)
Start with $G$ a compact Lie group, (simply connected). $LG = Map_{C^\infty} (S^1,G)$. $\widetilde{LG}$ is a central extension of $LG$ by $S^1$. The set of such extensions is isomorphic to $\mathbb{Z}$, so this depends on a choice of element $\ell \in \mathbb{Z}$, called the level.
One has a representation of $\widetilde{LG}$ on a separable Hilbert space $H_0$. If $\ell < 0$, then such a representation does not exist (rather, negative level reps are not unitarizable).
Given $I \subset S^1$, consider $L_I G = \{ \gamma: S^1 \to G \mid \gamma(z) = e \, \forall z \not\in I \}$. Pull back the central extension to a central extension $\widetilde{L_I G}$. This group acts on $H_0$. We can then consider the von Neumann algebra $A(I)$ generated by all operators in $\widetilde{L_I G}$. (this includes taking the weak closure)
The assignment $I \mapsto A(I)$ is a conformal net. A conformal net is a Hilbert space, together with a collection of algebras of operators acting on the Hilbert space - satisfying some axioms.
$SL_2(\mathbb{R})$ also acts on $H_0$. There is an $SL_2(\mathbb{R})$-invariant vector $\Omega \in H_0$, called the vacuum. $SL_2(\mathbb{R})$ appears as the group of algebraic automorphisms of $S^1$ as a real projective line. It is a subgroup of all diffeomorphisms.
In summary, for every compact Lie group $G$ and every integer $\ell$, we have a conformal net. This is true in all setups of CFT (conjecturally in Segal's setup).
Wasserman doesn't talk about conformal nets in general. He works with $SU(N)$. Fortunately, most of his constructions generalize to conformal nets, i.e., there isn't much that is specific to $SU(N)$. When he describes representations of the loop group, we can replace it with representations of a conformal net.
What is a representation of a conformal net? Recall that for each interval $I$, we have an algebra $A(I)$, and for each inclusion $I \subset J$, we have an algebra map $A(I) \to A(J)$. A representation is a Hilbert space $H_\lambda$ together with a system of maps $A(I) \to B(H_\lambda)$, such that $A(I) \to A(J) \to B(H_\lambda)$ agrees with $A(I) \to B(H_\lambda)$. It is important to exclude $S^1$ from the possible intervals. For loop groups, the algebra of all operators is just $B(H)$, and it might not act on $H_\lambda$.
The local loop groups sit inside the algebras $A(I)$, so we get representations of all of these groups, and whatever they generate. It turns out the groups $\widetilde{L_I G}$ together generate a representation of $\widetilde{LG}$.
We get a 1-1 correspondence between representations of the loop group net and projective representations of $LG$ that are positive energy. This restricts correspondences between net reps of a specific level and projective reps of a given central charge.
What is Wasserman's big innovative idea? He defines a tensor product operation on these loop group representations using techniques from von Neumann algebras. Suppose we have a bimodule ${}_A H_B$ and another bimodule ${}_B K_C$. If these were just algebras, we could just do a tensor product of $H$ with $K$. In von Neumann algebras, there is an operation that behaves formally like tensor product, called Connes fusion. It is written ${}_A H \underset{B}{\boxtimes} K_C$.
Warning: It is not true that given $h \in H$ and $k \in K$, one obtains an element of $H \boxtimes K$.
More nets: Subintervals yield subalgebras, and disjoint intervals yields commuting algebras. Given left and right half-circles, we get two left actions by the algebras corresponding to intervals. We want a right action. Fortunately, if $I \cong J$ is orientation-reversing, we get $A(I) \cong A(J)^{op}$. We use reflection in the vertical line as our orientation-reversing isomorphism. We get $A(I)-A(I)$-bimodules, and given two of them, we can use Connes fusion to get a third bimodule.
We have a concrete question: $H_\lambda \boxtimes H_\mu \cong \bigoplus_\nu N_{\lambda\mu}^\nu H_\nu$, so what are the integers $N_{\lambda\mu}^\nu$?
The answer is quite pretty: $(Rep(LG_\ell),\boxtimes)$ is a quotient of $(Rep(G),\otimes)$. This is all very explicit, and we can compute the numbers explicitly. This is true in general for groups $G$ , done for vertex algebras. It is not known for general groups in conformal net language. It is difficult to do - not just because people are lazy. Conjecturally, we get the semisimplified representation category of a quantum group at a root of unity. The representation category is known to be modular, but the identification has not been done.
\subsection{10:15am Hiro, Representations of $SU(N)$, Pieri rule, fusion rules of $LSU(N)$}
Say we have two irreducible representations of a group $G$ ($=SU(N)$). Representations of a group $G$ are a semisimple category, so we can ask how to their tensor product decomposes into irreducible representations.
$V_f \otimes V_g = \bigoplus_h N^H_{fg} V_h$. This formula is in general kind of complicated, but we can say what it is for special $f$ and $g$.
We'd like to consider reps of a central extension of the loop group. We can label them by $f$ and $g$ as well. Consider $H_f \boxtimes H_g = \bigotimes_{h'} N^{h'}_{fg} sgn(\sigma) H_{h'}$. Here, $\sigma$ is an element of the affine Weyl group $\Lambda_0 \rtimes S_N$.
Crash course in the representation theory of groups, in particular, $SU(N)$:
Two principles in representation theory:
\begin{enumerate}
\item Complex representations of $G$ are in correspondence with complex representations of $\mathfrak{g}$.
\item Irreducible representations of $\mathfrak{g}$ correspond to complex representations of $\mathfrak{g} \otimes \mathbb{C}$.
\end{enumerate}
We will work with the representations of $SU(3)$. Hopefully, at the end of this, you will understand highest weight vectors and signatures.
\textbf{Example 1:} Take the Lie algebra $\mathfrak{su}(3)$. Its complexification is $\mathfrak{sl}_3(\mathfrak{C})$. Consider the action of this Lie algebra on itself, called the adjoint representation. $X(v) = [X,v] = Xv-vX$. This is a representation because of the Jacobi identity. There is a Cartan subalgebra $\mathfrak{h} \subset \mathfrak{sl}_3$, given by diagonal matrices. We can decompose $\mathfrak{sl}_3$ into eigenspaces under the action of $\mathfrak{h}$. The zero eigenspace is $\mathfrak{h}$ itself, of dimension 2. Other eigenspaces are spanned by $E_{ij}$ for $i \neq j$, that is one at the $(i,j)$ entry and zero elsewhere. These together with $\mathfrak{h}$, span $\mathfrak{sl}_3$. If $X = diag(\alpha_1, \alpha_2, \alpha_3)$, then $X(E_{ij}) = (\alpha_i-\alpha_j)E_{ij}$. We get 6 one-dimensional eigenspaces, and we can place them in a symmetrical arrangement in the plane $\mathfrak{h}^\vee$. This plane is called the space of weights, and we have a weight decomposition $\mathfrak{g} = \bigoplus_{\alpha \in \mathfrak{h}^\vee} \mathfrak{g}_\alpha$.
Off-diagonal matrices will shift weights around.
\textbf{Claim:} If $\beta$ is the weight corresponding to $E_{ij}$, then $v_\beta$ takes $\mathfrak{g}_\alpha$ to $\mathfrak{g}_{\alpha + \beta}$.
\begin{proof} Take $X \in \mathfrak{h}$.
\[ \begin{aligned}
X(v_\beta(v_\alpha)) &= v_\beta(X(v_\alpha)) + [X,v_\beta](v_\alpha) \\
&= v_\beta(\alpha(X)v_\alpha + \beta(X)v_\beta(v_\alpha) \\
&= (\alpha(X) + \beta(X))v_\beta(v_\alpha)
\end{aligned} \]
\end{proof}
Raising operators are those that take things up and to the right (by convention). They kill things that are up and to the right, because we run out of nonzero weight spaces. $E_{ij}$ for $i_k f} V_g$. Here, $g >_k f$ means $g$ is given by adding $k$ boxes to $f$, but no two in the same row. For example, $V_\square \otimes V_\square = V_(2,0,\dots,0) \oplus V_{1,1,0,\dots,0}$. When $N=2$, the latter summand is trivial.
Jump to section 34:
Note: when looking at representations of $\widetilde{LSU(N)}$ at level $\ell$, signatures need to be permissible, i.e., $f_1-f_N \leq \ell$. (This barrier is the far wall of the alcove.)
\textbf{Theorem:} (Verlinde formula)
If $V_f \otimes V_g = \bigoplus_h N^h_{fg} V_h$, then $H_f \boxtimes H_g = \bigoplus_h N_{fg}^h sign(\sigma_h) H_{h'}$.
Action of affine Weyl group: $\Lambda_0 = \{ (N+\ell)(m_i) \mid m_i \in \mathbb{Z}^n, \sum m_i = 0 \}$. $W_{aff} = \Lambda_0 \rtimes S_N$, where the action of the symmetric group permutes coordinates.
\subsection{11:30am Min Ro, Hilbert spaces, polar decomposition, spectral theorem, von Neumann algebras}
Let $H$ be a Hilbert space. $S \subset B(H)$ a set of bounded linear operators.
The commutant of $S$ is $S' = \{ a \in B(H) \mid ab=ba \, \forall b\in S \}$.
The adjoint of an operator $a$ is the unique operator $a^*$ that satisfies $(a\xi,\eta) = (\xi, a^* \eta)$ for all $xi,\eta \in H$. $a$ is self-adjoint if $a^* = a$. $a$ is normal if $a^*a = aa^*$. $a$ is unitary if $aa^* = a^*a = 1$. $a$ is a projection if $a^* = a = a^2$.
There is a rough dictionary, where self-adjoint operators correspond to real-valued functions, and projections correspond to characteristic functions.
Convergence: $a_\lambda \to a$ in strong operator topology if and only if we have pointwise convergence in norm: $\lim_\lambda \Vert a_\lambda \xi \Vert = \Vert a \xi \Vert$. $a_\lambda \to a$ in the weak operator topology if we have pointwise convergence in inner products: $\lim_\lambda |(a_\lambda \xi,eta) | = |(a\xi,\eta)|$ and $|(a\xi,\eta)| \leq \Vert a \xi \vert \cdot \Vert \eta \Vert$.
\textbf{Theorem} (DCT) If $M$ is a unital $*$-subalgebra of $B(H)$, then TFAE
\begin{enumerate}
\item $M'' = M$
\item $M$ is weak operator closed.
\item $M$ is strong operator closed.
\end{enumerate}
\textbf{Definition} A von Neumann algebra $M \subset B(H)$ is a unital $*$-subalgebra that satisfies $M'' = M$.
The center of $M$ is $Z(M) = M \cap M'$.
$M$ is a factor if $Z(M) = \mathbb{C} \cdot 1$.
Examples: $B(H)$ is a factor, and if $H$ is finite dimensional, this is just $M_n(\mathbb{C})$. We can also take direct limits of matrix algebras, e.g., for $F_n$ the Fibonacci sequence, take a limit of inclusions $(M_{F_{n-1}} \oplus M_{F_n} \to M_{F_n} \oplus M_{F_{n+1}}$ by $(a,b) \mapsto (b,\binom{a0}{0b})$.
If $S \subset B(H)$ is a self-adjoint subset, then $S'$ is a von Neumann algebra, and $S''$ is the smallest von Neumann algebra containing $S$. In particular, for any $a \in B(H)$, we can construct $W^*(a) = \{a,a^* \}''$.
Let $X$ be a $\sigma$-finite measure space $(X = \bigcup_{i=1}^\infty E_i)$, then $L^\infty(X)$ includes into $B(L^2(X))$, and this is a von Neumann algebra. Conversely, for any commutative von Neumann algebra $A$, there is a measure space $X$ such that $A \cong L^\infty(X)$.
Let $X$ be compact, $T^2$. $B_b(X)$ is the space of complex Borel bounded functions. $Sp(a) = \{ \lambda \in \mathbb{C} \mid \lambda \cdot 1 -a \text{ not invertible} \}$. This is a nonempty compact set.
\textbf{Theorem:} (Borel functional calculus)
Let $a \in B(H)$ be normal. There is a $*$-homomorphism $B_b(sp(a)) \to W^*(a)$ taking $f \mapsto f(a)$. If $(f_n) \subset B_b(sp(a))_{sa}$, and $f_n \to f$, then $\lim f_n(a) = f(a)$.
\textbf{Definition:} Let $X$ be compact Hausdorff. A spectral measure relative to $(X,H)$ is a map $E$ from the Borel sets of $X$ to the projections of $H$ such that
\begin{enumerate}
\item $E(\emptyset) = 0$, $E(X) = 1$
\item $E(\bigcup_{n=1}^\infty S_i)$ converges to $\sum_{i=1}^\infty E(S_i)$ in the strong operator topology, if $S_i$ are pairwise disjoint.
\item $E(S_1 \cap S_2) = E(S_1) E(S_2)$.
\end{enumerate}
\textbf{Definition:} $\int_X f(\lambda) d(E,\lambda)$ can be defined.
\textbf{Theorem:} (Spectral theorem) Let $a \in B(H)$ be normal. There exists a unique spectral measure relative to $(sp(a),H)$ such that $a = \int_{sp(a)} \lambda d(E,\lambda)$.
\textbf{Theorem:} (Polar decomposition) Let $a \in B(H)$ with $a^*a$ positive (i.e., its spectrum is contained in the nonnegative reals). Then there is a square root $(a^*a)^{1/2}$. There is a unique $u \in B(H)$ such that $u$ is a partial isometry (i.e., $u^*u$ and $uu^*$ are projections), $a=u(a^*a)^{1/2}$, and $\ker(u) = \ker(a^*a)^{1/2}$.
If $a$ is densely defined and closed (i.e., the graph is closed in $H \times H$), then this also works. (NB: there is also a notion of closable operator, whose graph is not closed, but has the property that the closure of the graph is the graph of something else.)
\textbf{Definition:} Let $S \subset B(H)$ be a self-adjoint subset. We consider two types of representation of $S$. ($K$ is some other Hilbert space.)
\begin{enumerate}
\item $\pi: S \to U(K)$ a group.
\item $\pi: S \to B(K)$ a $*$-homomorphism.
\end{enumerate}
We can associate a von Neumann algebra by $S'$ or $S''$. We call either a factor representation if $S'$ is a factor. This is an analogue of a piece of an isotypical decomposition - canonical, not necessarily irreducible.
If $M = S'$, then the projections in $M$ correspond one-to-one to subrepresentations of $(\pi,K)$.
\textbf{Proposition:} If $(\pi,H)$ is a factor representation of $S$, and $\pi_1, K_1)$, $(\pi_2,K_2)$ are two subrepresentations, then
\begin{enumerate}
\item there is a unique $*$-isomorphism $\Theta: \pi_1(S)'' \to \pi_2(S)''$ such that $\Theta(\pi_1(x)) = \pi_2(x)$ for all $x\in S$.
\item For $X = Hom_S(K_1,K_2)$, then $\overline{XK_1} = K_2$.
\item $\Theta(a)T = Ta$ for all $a \in \pi_1(S)'', T \in X$.
\item If $X_0 \subset X$ such that $\overline{X_0K_1} = K_2$, then $\Theta(a)$ is the unique $b \in \pi_2(S)''$ such that $bT = Ta$ for all $T \in X_0$.
\end{enumerate}
\subsection{2:45pm Ryan Grady, Segal's quantization criterion, actions of $LSU(N)$ and of $Diff(S^1)$ on the fermionic Fock space}
We'll construct a representation of $LU(N)$ by considering sections of the vector bundle $S^1 \times \mathbb{C}^N \to S^1$, and studying Clifford algebras and Fock representations.
Let $(H, \langle, \rangle )$ be a Hilbert space, and let $Cliff(H)$ be the corresponding Clifford algebra. It is the unital $*$-algebra generated by $c(f)$ for all $f \in H$, subject to the relations $c(f) c(g) + c(g)c(f) = 0$, $c(f)c(g)^* + c(g)^*c(f) = \langle f, g \rangle$. There is an explicit realization as a quotient of the tensor algebra: $Cliff(H) = T(H)/(f \otimes f - \langle f,f \rangle)$.
$Cliff(H)$ acts on $\bigwedge H$ by the ``wave representation''. $\pi(c(f)) := f\wedge$. There is a distinguished vector $\Omega = 1 \in \bigwedge^0 H \cong \mathbb{C}$, called ``vacuum''. It is cyclic.
$a(f) = c(f)^*$, the annihilation operator. $a(f)(w_0\wedge \dots \wedge w_n) := \sum_{j=0}^n (-1)^j \langle f, w_j \rangle (w_0 \wedge \dots \wedge \widehat{w_j} \wedge \dots \wedge w_n)$.
\textbf{Fact:} $a(f)$ and $c(f)$ are adjoint with respect to the inner product $\langle w_0 \wedge \dots \wedge w_n , \eta_0 \wedge \dots \wedge eta_n \rangle = \det( \langle w_i, \eta_j \rangle )$.
\textbf{Proposition:} $\bigwedge H$ is irreducible as a $Cliff(H)$ representation.
\begin{proof}
Let $T$ be an endomorphism of $\bigwedge H$ that commutes with all $a(f)$s. Then $T\Omega = \lambda \Omega$ for some scalar $\lambda$, since $Omega$ spans the intersection of all kernels of annihilation operators. If $T$ also commutes with all creation operators, then $T = \lambda I$.
\end{proof}
\textbf{Unitary structures}
Let $(V, (,))$ be a real Hilbert space. A unitary structure is an element $J \in O(V)$ such that $J^2 = -I$. $V_J$ is then a complex Hilbert space, with Hermitian inner product $\langle v, w \rangle := (v,w) + i(v,Jw)$.
If $V$ already has a complex structure, we can define a projection operator $P_J = \frac12 (I - iJ) \in End(V_J)$. A choice of $P$ is equivalent to a choice of new complex structure. We will use this to define a representation much in the spirit of the wave representation.
\textbf{Definition:} The Fermionic Fock space is $\mathcal{F}_P = \bigwedge(PH) \hat{\otimes} \bigwedge(P^\perp H)^*$.
This is an irreducible representation of $Cliff(V_J)$ via $\pi_P(c(f)) = c(Pf) \otimes 1 + 1 \otimes c((P^\perp f)^*)^*$.
\textbf{Theorem:} (I. Segal-Shale equivalence criterion)
The Fock representation $\pi_P$ and $\pi_Q$ are unitarily equivalent if and only if $P-Q$ is Hilbert-Schmidt, i.e., for $\{ e_i \}$ a basis for $H$, $\sum_i \Vert (P-Q)e_i \Vert^2 < \infty$.
For $u \in U(H)$, we get an automorphism of $Cliff(H)$ by $c(f) \mapsto c(uf)$. $u$ is said to be implemented in $\pi_P$ (or $\mathcal{F}_P$) if $\pi_P(c(uf)) = U \pi_P(c(F))U^*$ for some unitary $U$ (in $U(\mathcal{F}_P)$, unique up to a scalar).
\textbf{Proposition:} $u \in U(H)$ is implemented if and only if $[u,P]$ is Hilbert-Schmidt.
\textbf{Definition:} The restricted unitary group $U_{res}(H) = \{ u \in U(H) \mid u \text{ is implemented in } \mathcal{F}_P \}$.
We have a representation $U_{res}(H) \to PU(\mathcal{F}_P)$, called the basic representation.
If $u \in U(H)$ and $[u,P] = 0$, then $u$ is implemented in $\mathcal{F}_P$, and is ``canonically quantized''.
Now we construct a representation of $LU(N)$. Let $H = L^2(S^1) \otimes \mathbb{C}^N$, and let $P: H \to H_{\geq 0}$. The target is the Hardy space of boundary values of functions that are holomorphic on the unit disc. For $f \in C^\infty(S^1, End(\mathbf{C}^n)$, we define an operator $m(f)$ on $H$ via multiplication.
\textbf{Fact:} $\Vert [P,m(f)] \Vert_2 \leq \Vert f' \Vert_2$. In particular, $f \in LU(N)$ is implemented in $\mathcal{F}_P$, so we get a projective representation of $LU(N)$ on $\mathcal{F}_P$, called the fundamental representation.
Let $SU_\pm(1,1) = \left\{ \begin{pmatrix} \alpha & \beta \\ \overline{\beta} & \overline{\alpha} \end{pmatrix} \mid |\alpha|^2 - |\beta|^2 = \pm 1 \right\}$. This is a semidirect product of $SU(1,1)$ with an orientation reversing transformation on the circle. Fractional linear transformations $z \mapsto \frac{\alpha z + \beta}{\overline{\beta}x + \overline{\alpha}}$ yield a unitary representation $H = L^2(S^1) \otimes \mathbb{C}^N$ via $(V_g \cdot f)(z) = \frac{f(g^{-1}(z))}{\alpha - \overline{\beta}z}$. Note that if $|z| < 1$ and $|\alpha| > |\beta|$, then $(\alpha - \overline{\beta}z)^{-1}$ is holomorphic. This implies $V_g$ commutes with $P$, and the action is canonically quantized.
For $F = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$, $(V_F \cdot f)(z) = \frac{f(z^{-1})}{z}$, so $V_F P V_F = I - P$.
There is an action of $U(1)$ on $H$, via multiplication by the constant function $z$. This is canonically quantized, so we let $U_Z$ be the action on $\mathcal{F}_P$. The $U(1)$ action is called the ``charge operator''. The inclusion of $PH$ into the Fock space has charge 1, and the inclusion of $P^\perp H$ has charge $-1$. Higher wedge powers have suitable charges.
\textbf{Proposition:} If $\pi$ is the representation of $LSU(N)$ on fermionic Fock space $\mathcal{F}_P$, and $U_z$ is the $U(1)$ action, then $\pi(g) U_z \pi(g)^* = U_z$.
\subsection{4:00pm Owen Gwilliam, The central extension of $LG$, positive energy representations, Lie algebra cocycles}
My goal today is to tell you what a positive energy representation is, and how to get a handle on one, and tomorrow, James will talk about positive energy reps of $LSU(N)$.
\textbf{Outline:}
\begin{enumerate}
\item Define and motivate positive energy representations
\item projective representations and central extensions
\item reducing questions to the loop algebra
\end{enumerate}
Let $G$ be a compact connected Lie group, and let $T$ be the circle as a group.
\textbf{Definition:} A positive energy representation of $LG$ is a topological vector space $E$ with
\begin{enumerate}
\item a projective representation of $LG$, i.e., for $\gamma \in LG$, get $U_\gamma \in GL(E)$, such that $U_\gamma U_{\gamma'} = c_{\gamma \gamma'} U_{\gamma \gamma'}$ for $c_{\gamma \gamma'} \in \mathbb{C}^\times$.
\item an intertwining action of $T$, i.e., an operator $R_\theta$ on $E$ for each $\theta \in T$, such that $R\theta U_\gamma R_\theta^{-1} = U_{R_\theta \gamma}$.
\item under the weight decomposition of $E$ by $T$, $E = \bigoplus_{n \in \mathbb{Z}} E(n)$ then $\dim E(n) < \infty$ and $E(n) = 0$ for $n < 0$.
\end{enumerate}
Concrete examples: $G = SU(N)$, $V = \mathbb{C}^N$, $H = L^2(S^1,V)$, and $P: H \to H$ is projection onto nonnegative Fourier modes. Then $\mathcal{F}_P = \bigwedge(PH) \hat{\otimes} \bigwedge(P^\perp H)^*$, the fermionic Fock space, is positive energy and level 1. The $\ell$th tensor power is level $\ell$.
\textbf{Theorem:} (Pressley-Segal 9.3.1) A positive energy representation of $GL$ is
\begin{enumerate}
\item completely reducible into irreducible positive energy representations
\item unitary
\item extend to holomorphic representations of $LG_\mathbb{C}$.
\item admit a projective intertwining action of $Diff_+(S^1)$.
\end{enumerate}
The first three are directly analogous to the theory of compact Lie groups.
Segal says you can think of positive energy representations as boundary conditions of holomorphic representations of the semigroup $\mathbb{C}^times_{\leq 1} \times \widetilde{LG_\mathbb{C}}$. Something about CFT on a cylinder.
More about projective representations:
\textbf{Definition:}
A projective unitary representation of a group $G$ on a Hilbert space $V$ is a continuous homomorphism $G \to PU(V) = U(V)/T$ (with strong operator topology on $U(V)$ - most of the other topologies on $B(V)$ yield the same topology on $U(V)$, but the norm topology doesn't admit enough representations).
We have $U(V)$ a central extension of $PU(V)$, and a map from $G$ to $PU(V)$, so we can pull back the central extension to get a group $\tilde{G}$ and an honest unitary representation. A central extension of $LG$ yields a circle bundle, and it has a first Chern class $c_1 \in H^2(LG,\mathbb{Z})$.
\textbf{Exercise:} Calculate $H^2(LSU(N),\mathbb{Z})$.
An alternative idea is to consider a central extension of the Lie algebra $L\mathfrak{g}$ by $\mathbb{R}$. Elements are pairs $(X,a)$ for $X \in L\mathfrak{g}$ and $a \in \mathbb{R}$. The bracket has the form $[(X,a),(Y,b)] = ([X,Y], \omega(X,Y))$ for $\omega: L\mathfrak{g} \times \mathfrak{g} \to \mathbb{R}$. The combination of skew-symmetry and Jacobi identity implies $\omega([X,Y],Z) + \omega([Y,Z],Z) + \omega([Z,X],Y) = 0$.
\textbf{Proposition:} (Pressley-Segal 4.2.4) For $\mathfrak{g}$ semisimple, every continuous $G$-invariant 2-cocycle $\omega$ for $L\mathfrak{g}$ has the form $\omega(X,Y) = \frac{1}{2\pi} \int_0^{2\pi} \langle X(\theta), Y'(\theta) \rangle d\theta$ with $\rangle,\langle$ a $G$-invariant symmetric bilinear form on $\mathfrak{g}$.
Note that $H^3(\mathfrak{g})$ is the space of $\mathfrak{g}$-invariant symmetric bilinear forms on $\mathfrak{g}$. Given $\langle,\rangle$, we get the cocycle that takes $X,Y,Z$ to $\langle X, [Y,Z] \rangle$. Given a 2-cocycle on $L\mathfrak{g}$, we get a 2-form on $LG$.
\textbf{Theorem:} (Pressley-Segal 4.4.1) If $G$ is simply connected, then
\begin{enumerate}
\item A 2-cocycle $\omega$ on $L\mathfrak{g}$ deines a group extension if and only if $[\omega/2\pi] \in H^2(LG,\mathbb{Z})$.
\item If we have a group extension, then it is unique (up to isom?)
\item Such an $\omega$ is an integral multiple of $\omega_{basic}$ (when $G$ is simple).
\end{enumerate}
$\omega_{basic}$ corresponds to the inner product on $\mathfrak{g}$ such that $\langle \theta,\theta \rangle = 2$, where $\theta$ is the highest weight.
\textbf{Example:} $\mathfrak{g} = \mathfrak{su}(N)$. $\langle X,Y \rangle = -tr(XY)$. For $N=2$, $\theta = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}$, so $-tr(\theta^2) = 2$.
\textbf{Definition:} The level of a central extension $\widetilde{LG}$ of $LG$ is the integer $\ell$ such that $\omega_{\widetilde{LG}} = \ell \omega_{basic}$.
Return to positive energy:
\textbf{Definition:} A positive energy representation of $LG$ is an honest representation of $\widetilde{LG} \rtimes T$, satisfying a positive energy condition. The $T$ on the right is the ``rotation circle'' (or ``energy circle'') $T_{rot}$, and the central circle is labelled $T_{CE}$.
We have a Cartan subgroup $T_{rot} \times T_G \times T_{CE} \subset \widetilde{LG} \rtimes T_{rot}$, where $T_G$ is a maximal torus for $G$. This gives us a decomposition $E = \bigoplus_{n,\lambda,\ell} E(n, \lambda, \ell)$
Try passing to the Lie algebra. $L^{poly}\mathfrak{g} = \{ \sum_{finite} X_n \}$, and $L^{poly}\mathfrak{g} \otimes \mathbb{C} \cong \mathfrak{g}_\mathbb{C}[z,z^{-1}]$. We want a correspondence between positive energy representations of $LG$ and positive energy representations of $L^{poly}\mathfrak{g}$. Energy is given by $\mathbb{R} \to PU(E)$ via $t \mapsto \pi(exp(tx))$. Abstractly, for $E$ a positive energy representation of $LG$, $L^{poly}\mathfrak{g} \to LG \to PU(E)$, where the first arrow is exponsntiation, and the second is the representation. For each $x \in L\mathfrak{g}$, we get a 1-parameter subgroup, and by Stone's theorem, we get $X \in End(E)$ such that $e^{tX} = \pi(e^{tx})$. This timples we get a projective representation $\rho: L^{poly}\mathfrak{g} \to End(E)$.
Let $d$ denote an infinitesimal generator of $T_{rot}$ on $E$. $d|_{E(n)} = n$ and $R_\theta = e^{i \theta d}$.
\textbf{Theorem:} (Wasserman) Let $E$ be a level $\ell$ representation of $LG$. Then
\begin{enumerate}
\item $E^{fin} := \bigoplus^{alg}_{n \geq 0} E(n) \subset E$ is preserved by $\rho$.
\item We can choose lifts such that $[d,\rho(X)] = i\rho(x')$ on $E(n)$.
\item $[\rho(x),\rho(y)] = \rho([x,y]) + i\ell \omega_{basic}(x,y)$.
\end{enumerate}
Concretely, $H = L^2(S^1,V)$, $V = \mathbb{C}^N$. $Cliff(H)$ acts on $\mathcal{F}_P$, the fermionic Fock space. $L^{poly}V = \{$ polynomials in $e^{in\theta} \} \otimes V \to Cliff(H)$ by $\pi$. $e^{in\theta} \otimes v$ is taken to $c(e^{in\theta}v)$. $L^{poly}\mathfrak{su}(N) \to Cliff(H)$. The complexification $\mathfrak{sl}_N$ of $\mathfrak{su}(N)$ is generated by $E_{ij}$. We just write a formula $E_{ij}(n) = \sum_{m >0} e_i(n-m) e_j(-m)^* - \sum_{m \geq 0} e_j(m)^* e_i(m+n)$.
\textbf{Theorem:} (Wasserman)
\begin{enumerate}
\item $[X(m), a(f)] = a(Xe^{im\theta} f)$ for $f \in L^{poly}V$.
\item $[d,X(m)] = -mX(m)$.
\item $[X(n),Y(m)] = [X,Y](n+m) + m \langle X,Y \rangle \delta_{n+m,0}$.
\end{enumerate}
\section{Tuesday, 8/17/10}
\subsection{9:00am Andre, General physics-y explanation of where it all comes from}
We start with a Minkowski space we will call Mink, and a Hilbert space of everything $H_0$. In quantum field theory, we have a bunch of fields. These are often defined as operator-valued distributions on Mink. Such a field will take a smooth function and produce an unbounded operator on $H_0$, i.e., a linear map defined on a dense subspace of $H_0$. They must be subject to some relation with respect to ``composition'': note that distributions do not always multiply well, e.g., $\delta_0$ doesn't have a square.
What is Minkowski space? It is $\mathbb{R}^4$. If you try to write examples of quantum field theory on $\mathbb{R}^4$, you find that it is hard. People have been unable to write interesting examples. It is easier to write examples in two dimensions.
There is an action of the Poincar\'e group, made of translations and ``rotations'', on Minkowski space, and it acts on $H_0$. If you want to do conformal field theory, you need to add additional symmetries.
The group of conformal maps of $\mathbb{R}^2$ has a description as $Diff(\mathbb{R}) \times Diff(\mathbb{R})$. Each coordinate corresponds to a lightlike (i.e., norm zero) direction. If I have a quantum field theory on the reals, and another quantum field theory on the reals, then I can take their tensor product, and get a conformal field theory on Minkowski space. These are the chiral conformal field theories. The non-chiral conformal field theories are those that do not split as a tensor product.
The last step is to say that $\mathbb{R}$ can be replaced by $S^1$, its compactification. Given a QFT on $S^1$, we can restrict it to a QFT on $\mathbb{R}$. This is why we study quantum field theories with actions of $Diff(S^1)$.
\textbf{Warning:} There are anomalies: $H_0$ may only admit a projective action of $Diff(S^1)$.
\textbf{Example:} Dirac field. We write $\psi(z)$, subject to the relations:
\begin{enumerate}
\item $[\psi(z),\psi(w)]_+ = 0$.
\item $[\psi^*(z), \psi^*(w)]_+ = 0$.
\item $[\psi(z), \psi^*(w)]_+ = \delta(z-w)$.
\end{enumerate}
$H = L^2(S^1)$, $\mathcal{F}_P = \bigwedge(PH) \otimes \bigwedge(PH^\perp)^*$, and $H_0$ is the Fock space. Here, $PH$ is the Hardy space, spanned by $\{ z^n\}_{n \geq 0}$. We write $\psi: C^\infty(S^1) \to B(H_0)$, taking $f \mapsto \psi(f) = a(f)$ (also written $c(f)$ yesterday). It is creation on the $PH$ side and annihilation on the $PH^\perp$ side. We have the relation $[a(f),a^*(g)]_+ = \langle f,g \rangle$.
Formally, $\psi(z) = \psi(\delta_z)$. We can write $a(f) = \int f(z) \psi(z)$. This is called a smeared field: we take a distributional field $\psi$, and smear it with the smooth function $f$. Let's check the commutation relation:
\[ \begin{aligned}
[a(f), a^*(g)]_+ &= [\int f(z) \psi(z), \int g(w)^* \psi(w)^*]_+ \\
&= \int \int f(z) \overline{g(w)} [\psi(z), \psi^*(w)]_+ \\
&= \int f(z) \overline{g(z)} \\
&= \langle f,g \rangle
\end{aligned} \]
For the loop groups, take $\mathfrak{g}$, and let $\{ X_\alpha \}$ be a basis, with structure constants $[ X_\alpha, X_\beta] = \sum_\gamma c^\gamma_{\alpha \beta} X_\gamma$. Then
\[ [J_\alpha(z), J_\beta(w)] = \sum_\gamma c^\gamma_{\alpha \beta} J_\gamma(w) \delta(z-w) + \ell \langle X_\alpha, X_\beta \rangle \frac{\partial}{\partial w} \delta(z-w). \]
Smeared fields here have the form $J_\alpha(f) = \int f(z) J_\alpha(z)$. $f(z) X_\alpha \in L\mathfrak{g} = C^\infty(S^1,\mathfrak{g})$.
If $G$ is not simply connected, then $LG$ is not connected, and you seem to get a different CFT.
\subsection{10:15am James Tener, Classification of $LG$-reps, weight polytopes for the characters of $LG$-reps}
The goal of this talk is to establish a classification of irreducible positive energy representations. This will mirror the classification of irreducible representations of $SU(N)$ that we saw yesterday. Note: today, $N=3$.
Recall from yesterday, $G = SU(3)$. We accepted that we could classify representations of the complexified Lie algebra. Here's the standard picture of the weight space. Irreducible representations are classified by highest weight, and the highest weight of the adjoint representation was $L_1 - L_3$. We thought of the highest weight as a signature in $\mathbb{Z}^3$, and the adjoint representation had signature $(1,0,-1)$.
\textbf{Theorem:} The possible highest weights are $(f_1, \dots, f_N)$, where $f_1 \geq f_2 \geq \dots \geq f_N \geq 0$. By adding constant vectors (i.e., tensoring with determinant = trivial) we can set $f_N = 0$. The lattice points lie in the positive Weyl chamber. [shaded region in picture]
For $LSU(3)$, we ask:
\begin{enumerate}
\item what is the analogue of signature?
\item what are the possible values?
\end{enumerate}
\textbf{Setup:} $(\pi, H)$ an irreducible positive energy representation of $LG$ at level $\ell$. $\pi$ is a projective representation of $LG \rtimes T_{rot}$ (honest on $T_{rot}$), satisfying the properties that $H = \bigoplus_{n \geq 0} H(n)$ with $\dim H(n) < \infty$, $T_{rot}$ acts on $H(n)$ by $z\cdot \xi = z^n \xi$, or $r_\theta(\xi) = e^{in\theta}\xi$.
\textbf{Theorem:}
\begin{enumerate}
\item $H(n)$ is an irreducible $SU(N)$-module, where $SU(N) \subset LSU(N)$ is the subgroup of constant loops.
\item The signature of $H(0)$ satisfies $f_1 - f_N \leq \ell$.
\item If $f$ is such a signature, there is an irreducible positive energy representation of $LG$ with $H(0) = V_f$.
\item This representation is unique up to isomorphism.
\end{enumerate}
Owen told us yesterday about representations of the Lie algebra, and I want to describe where it appears in the proof of the classification.
\textbf{Question:} Why is $H(0)$ invariant under $SU(N)$?
Recall that there exists a representation $\rho$ of $L^{poly}\mathfrak{g} \rtimes \mathbb{R}$ such that:
\begin{enumerate}
\item $\pi(e^x) = e^{\rho(x)}$
\item if $X(n) = \rho(e^{-int}X)$ for $X \in \mathfrak{g}$, then $[X(n), Y(m)] = [X,Y](n+m) + m\ell \rangle X, Y \langle$
\item $\rho(\mathbb{R}) = \mathbb{R}D$, where $D$ acts on $H(n)$ by multiplication by $n$.
\item $[X(n),D] = -nX(n)$.
\end{enumerate}
Note that if $\xi \in H(j)$, then $DX(n)\xi = X(n)D\xi - nX(n)\xi = (j-n)X(n)\xi$. In particular, $H(j)$ is an $\mathfrak{su}(N)$-module. There is a weight diagram, looks like a parabola.
I'll describe condition 3 - we just write down the modules, and argue that they are everything we want. In Wasserman's paper, the irreducible positive energy representations are defined as subrepresentations of the $\ell$th tensor power of the fundamental representation $\mathcal{F}_P$. This means $H(0) \subset \mathcal{F}_P(0)$, and we should look for $SU(N)$-modules in $\mathcal{F}_P^{\otimes \ell}(0)$. We want to find an irreducible representation of $SU(N)$ with signature $(f_1,\dots, f_{N-1},0)$ with $f_1 \leq \ell$.
$\mathcal{F}_P = \bigwedge H_P$, with $H_P = L^2(S^1,V)$, with $V = \mathbb{C}^3$. In $H_P$, we have $V$ as constant loops, and in $\mathcal{F}_P^{\otimes \ell}(0)$, we have $\bigwedge V)^{\otimes \ell}$. Recall that $e_1^{\otimes (f_1-f_2)} \otimes (e_1 \wedge e_2)^{\otimes(f_2-f_3)} \otimes \dots \otimes (e_1 \wedge \dots \wedge e_{N-1})^{\otimes (f_{N_1} - f_N)} \in \lambda^1 V^{\otimes (f_1-f_2)} \otimes \dots \otimes (\lambda^{N_1}V)^{\otimes (f_{N-1} - f_N)}$ generate an irreducible $SU(N)$ representation of signature $f$.
$(\bigwedge V)^{\otimes \ell} = (\lambda^1 V \oplus \dots \oplus \lambda^N V) \otimes \dots \otimes (\lambda^1 V \oplus \dots \oplus \lambda^N V)$. We can generate modules as begore, if $f_1 \leq \ell$.
If $H$ and $H'$ are two irreducible positive energy representations of $LG$ at level $\ell$ such that $H(0) \cong H'(0)$ as $SU(N)$-modules, then there is a $U: H \to H'$ intertwining the projective actions of $LG \rtimes T_{rot}$.
\subsection{11:30am Michael Hartglass, Tomita-Takesaki theory and the KMS condition}
A. Wasserman defines an operator $\Delta^{it}$, but he doesn't define $\Delta$. I feel like that's cheating. The standard reference for this subject is chapter 12 and 13 in Vaughan Jones's lecture notes, available on his web page.
Notation: $\phi$ is a faithful normal state on a von Neumann algebra $M$. By state, I mean $\phi: M \to \mathbb{C}$ is a continuous linear functional satisfying $\phi(1) = 1 = \Vert \phi \Vert$ and $\phi(x^* x) \geq 0$. Faithful means $\phi(x^*x) = 0$ iff $x=0$. Normal means ultra-weakly continuous, where $x_\lambda \to x$ ultraweakly if $\sum_{i=1}^\infty \langle x_\lambda \xi_i, \eta_i \rangle \to \sum_i \langle x \xi_i, \eta_i \rangle$ for all pairs $\xi_i, \eta_i$ satisfying $\sum_i \Vert \xi_i \Vert^2 < \infty$ and $\sum_i \Vert \eta_i \Vert^2 < \infty$. This has something to do with $M$ having a predual.
We put an inner product on $M$ by $\langle x, y \rangle = \phi(y^*x)$. this gives a new norm $\Vert \cdot \Vert_2$ on $M$. Completing $M$ with respect to this norm yields the Hilbert space $L^2(M)$. Set $\Omega$ to be the image of 1 in $L^2(M)$. $M$ acts on $L^2(M)$ by $x(y\Omega) = xy\Omega$, and extend by continuity. $\Omega$ is cyclic and separating for $M$ acting on $L^2(M)$.
Look at $S_0: M\Omega \to M\Omega$ by $S_0(x\Omega) = x^*\Omega$, conjugate-linear. $S_0$ does not have to be bounded, but parts of its polar decomposition may be of interest to us. We can also define $F_0: M'\Omega \to M'\Omega$ by the same formula $F_0(x'\Omega) = (x')^*\Omega$. By $M'$, we are referring to the commutant with respect to the action on $L^2(M)$.
\textbf{Notation:} If $A$ and $B$ are unbounded, we write $A \subset B$ if the domain of $A$ is contained in the domain of $B$ and $B|_{dom A} = A$.
If $A$ is unbounded, the domain of $A^*$ is $\{ \eta \mid \langle A\xi, \eta \rangle$ is a bounded function of $\xi \}$. We have $\langle \xi, A^*\eta \rangle = \langle A \xi, \eta \rangle$.
\textbf{Fact:} $S_0 \subset F_0^*$ and $F_0 \subset S_0^*$.
\begin{proof}
Let $a \in M$, $a' \in M'$. Then $\langle S_0(a\Omega, a'\Omega \rangle = \langle a^*\Omega, a'\Omega \rangle = \langle (a')^*\Omega, a\Omega \rangle$. This is clearly a bounded function of $a\Omega$. this implies the domain of $S_0$ is a subset of the domain of $F_0^*$, and $F_0^*$ restricted to the domain of $S_0$ is $S_0$. The other direction is similar.
\end{proof}
Since $\Omega$ is cyclic and separating for $M$ (hence $M'$), $S_0$ and $F_0$ are densely defined. Since they are contained in each other's adjoints, they are closable. (note that the adjoint of any operator is closed)
If $x_n \to x$ and $S_0(x_n)$ converges, then it converges to $S_0(x)$.
\textbf{Theorem:} Let $S$ and $F$ be the closures of $S_0$ and $F_0$, respectively. Then $S = F^*$ and $F = S^*$.
We won't prove this, but it will be very useful in computations.
$S$ and $F$ are closed and densely defined. This implies we can give a polar decomposition of $S$. $S = J\Delta^{1/2}$, where $J$ is a conjugate-linear isometry, and $\Delta^{1/2}$ is unbounded, positive, and linear.
\textbf{Claim:} $J \Delta^{1/2} J = \Delta^{-1/2}$.
Why is this? $S = S^{-1}$, so we use $S^{-1} = (J\Delta^{1/2})^{-1} = \Delta^{-1/2}J$. $J$ is its own inverse.
We want to show that $JMJ = M'$. This is one of the main results of Tomita-Takesaki theory.
\textbf{Lemma:} With $\phi$ as before, and $\psi \in M_*$ in the predual, such that $|\psi(y^*x) |^2 \leq \phi(x^*x)\phi(y^*y)$ (analogue of Cauchy-Schwarz), and given $\lambda>0$, then there exists $a$ with $\Vert a \Vert < 1/2$ such that $\psi(x) = \lambda \phi(ax) + \lambda^{-1}\phi(xa)$.
\textbf{Lemma:} Let $\lambda$ be as above. Given $a' \in M'$, there exists $a \in M$ such that $a\Omega$ is in the domain of $F$, and $a'\Omega = (\lambda S + \lambda^{-1}F)a\Omega$,
\begin{proof}
$\phi(x) = \langle x\Omega, \Omega \rangle$. Using the previous lemma, we can write
\[ \langle x\Omega, a'\Omega \rangle = \lambda \rangle ax\Omega, \Omega \rangle + \lambda^{-1} \rangle xa\Omega, \Omega \rangle, \]
assuming $\Vert a \Vert < 1$.
Checking everything is in the right domain, this is equal to $\lambda \rangle x \Omega, Sa\Omega \rangle + \lambda^{-1} \langle a\Omega, S(x\Omega) \rangle$. The last expression is $\lambda^{-1} \langle x\Omega, f(a\Omega) \rangle$.
\end{proof}
\textbf{Lemma:} Let $\lambda, a, a'$ be as before. If $\xi, \eta \in dom(F) \cap dom(S)$, then $\lambda \langle SaS\xi, \eta \rangle + \lambda^{-1} \langle FaF\xi, \eta \rangle = \langle a'\xi, \eta \rangle$.
We will use the following complex analysis fact:
\[ \int_{-\infty}^\infty \frac{f(it+1/2) + f(it-1/2)}{2 \cosh(\pi t)} dt = f(0) \]
provided $f$ is bounded, and holomorphic on the vertical strip in question.
\textbf{Proposition:} Let $\lambda, a, a'$ be as before. Then
\[ \begin{aligned}
a &= \int_{-infty}^\infty \lambda^{2it} \frac{\Delta^{it} JaJ\Delta^{-it}}{2 \cosh (\pi t) } dt \\
&= \lambda^{2it}\left( \lambda \langle (\Delta^{it} a \Delta^{-it})\Delta^{-1/2}\xi, \Delta^{1/2}\eta \rangle + \lambda^{-1} \langle \Delta^{it} a \Delta^{-it}\Delta^{1/2}\xi, \Delta^{-1/2} \eta \rangle \right)
\end{aligned} \]
\textbf{Theorem:}
$JMJ = M'$ and $\Delta^{it}M\Delta^{-it} = M$ for all $t$.
For $u \in M'$, the Fourier transforms of $u^* \Delta^{it} Ja'J\Delta^{-it} u$ and $\Delta^{it}Ja'J\Delta^{it}$ are equal, so the operators are equal. By symmetry, blah.
\textbf{Example 1:} $M = M_2(\mathbb{C})$ and $\phi(x) = tr(ax)$, where $a = \begin{pmatrix} \mu_1 & 0 \\ 0 & \mu_2 \end{pmatrix}$, $\mu_i < 0$, and $mu_1 + \mu_2 = 1$. In this example, $J(e_{ij}) = \sqrt{\frac{\mu_j}{\mu_i}}e_{ji}$, and $\Delta(e_{ij}) = \frac{\mu_i}{\mu_j}e_{ij}$. Everything is bounded.
\textbf{Example 2:} $M = \bigotimes_{i=1}^\infty M_2(\mathbb{C}$. Consider a state $\phi = \bigotimes_{i=1}^\infty \phi_i$, with $\phi_i(x) = tr(xa_i)$, where $a_i = \begin{pmatrix} \mu_{1i} & 0 \\ 0 & \mu_{2i} \end{pmatrix}$. Let $\mu_{1i} \to 0$ and $\mu_{2i} \to 1$ as $i \to \infty$. We get to say something about unboundedness (?).
In the $M_2$ example, we can write down $F(z) = \langle \Delta^{it} y \Omega, x\Omega \rangle$ where $z\in \mathbb{C}, x,y \in M_2(\mathbb{C})$, $F(t) = \phi(\Delta^{it}xDelta^{-it}y) =: \phi(\sigma_t^\phi(x)y)$, $F(t+i) = \phi(y\sigma_t^\phi(x))$.
\textbf{Theorem:} (KMS condition)
Define $F(z) = \langle \Delta^{-iz} x\Omega, y\Omega \rangle$ for $x,y \in M$. Then $F(t) = \phi(\sigma_t^\phi(x)y)$ and $F(t+i) = \phi(y\sigma_t^\phi(x))$. Furthermore, if $\alpha_t$ is a 1-parameter automorphism group that is strongly continuous and satisfies $\phi \circ \alpha_t = \phi$ and there exists holomorphic $G$ satisfying $G(t+i) = \phi(y\alpha_t(x))$ and $G(t) = \phi(\sigma_t^\phi(x)y)$, then $\alpha_t = \sigma_t^\phi$ for all $t$.
Idea: use stone's theorem to show that the one-parameter subgroup comes from exponentiating a self-adjoint operator. Subtract the two holomorphic functions to get something that vanishes on the real line. Then get equality.
\textbf{Corollary:} The following are equivalent for $a \in M$.
\begin{enumerate}
\item $\phi(ax) = \phi(xa)$ for all $x \in M$.
\item $\sigma_t^\phi(a) = a$ for all $t \in \mathbb{R}$.
\end{enumerate}
($1 \Rightarrow 2$): If $X \in M$, then
\[ \begin{aligned}
\langle x^*\Omega, a\Omega \rangle &= \langle \Omega, xa \Omega \rangle \\
&= \langle \Omega, ax \Omega \rangle \\
&= \langle a^*\Omega, x\Omega \rangle \\
&= \langle S(a\Omega), x\Omega \rangle \
\end{aligned} \]
Therefore, $a \Omega$ lies in the domainof $S^*$ and $S^*(a\Omega) = a^*\Omega$. This means, since $S^* = \Delta^{1/2}J$, $a^*\Omega$ has to be fixed by $Delta$. Hence, $a\Omega$ is fixed by $\Delta$, and $a\Omega$ is fixed by $\Delta^{it}$.
($2 \Rightarrow 1$): $\phi(\sigma_t^\phi(x)a) = \phi(\sigma_t^\phi(xa)) = \phi(xa)$. This implies $f(a)$ is constant along the real axis, so $f(z)$ is constant in the strip $0 \leq Im(z) \leq 1$. Plugging in $t = 0 + i$, we get $\phi(xa) = \phi(ax)$.
\subsection{2:00pm Dmitri Pavlov, Tomita Takesaki theory for fermions}
My talk today consists of 2 parts. First, I'll offer an alternative perspective on Tomita-Takesaki theory. Second, I'll give some examples of how Tomita-Takesaki theory appears in field-theoretic constructions: fermions and Segal's CFT.
There is a unified picture of the objects in the previous talk, given by noncommutative $L^p$ spaces. We will use the convention $L_p = L^{1/p}$. For example, $L_0 = L^{\infty}$, bounded measurable functions, $L_1 = L^1$, complex-valued measures on a measurable space, and $L_{1/2} = L^2$ a Hilbert space.
In the commutative setting, if we fix a measure $\mu$, then $L_p(\mu) = \{ f \mid \int |f|^{1/Re(p)} < \infty \}$. This only depends on the real part of $p$. If we change the measure, we take a map $L_p(\mu) \to L_p(\nu)$ by $f \mapsto f(D\nu:D\mu)^p$, where $(D\nu:D\mu)$ is the Radon-Nikodym derivative. This uses the imaginary part. For real part zero, we take all bounded functions. Measure-independent $L_p$ spaces are canonically isomorphic to the spaces with respect to a given measure via $f \in L_p(\mu) \mapsto f\mu^p \in L_p$.
\textbf{Definition/Theorem:} Let $M$ be a $W^*$-algebra. Then $L_*(M)$ is a $\mathbb{C}_{Re \geq 0}$-graded complex unital $*$-algebra. There are subspaces $L_p(M)$ for complex $p$ with nonnegative real part. There is a multiplication operation $L_p(M) \times L_q(M) \to L_{p+q}(M)$. There is a $*$-operation $L_p(M) \to L_{\bar{p}}(M)$. We can identify $L_0(M) = M$.
$M$ has a predual $M_*$, so $(M_*)^* = M$. We have a $M$-$M$-bimodule isomorphism $L_1(M) \cong M^*$. The bimodule structure on $M_*$ is given by $(mf)(x) = f(xm)$, $(fm)(x) = f(mx)$, where $f \in M_*$, $m \in M$. There is a trace operation $tr: L_1 \to \mathbb{C}$, satisfying $tr(xy - yx) = 0$ for $x \in L_p$, $y \in L_q$ for $p+q=1$.
For $p \in \mathbb{R}$, we say that $z \in L_p$ is positive (i.e., $z \in L_p^+$) if there is $y$ such that $y^*y = z$. If $p \in \mathbb{C}_{Re \geq 0}$ and $q \in \mathbb{R}_{\geq 0}$, then there is an operation $L_q^+(M) \to L_{qp}(M)$ given by $z \mapsto z^p$. If $p \in \mathbb{R}$, then this is a bijection $L_q^+ \to L_{pq}^+$.
We can complete $L_q^+$ to a space $\widehat{L_q^+}$. Elements of $\widehat{L_1^+}$ are called weights, and the completion passes from all positive bounded measures to positive unbounded measures.
The most general definition of modular automorphism group is:
\textbf{Definition:} $M$ a $W^*$-algebra, $\phi \in \widehat{L_1^+}(M)$. Let $t \in I := \{ x \in \mathbb{C} \mid Re(x) = 0 \}$. We define $\sigma_t^\phi(x) = \phi^t x \phi^{-t}$, a modular automorphism of $L_0(M) = M$.
$\sigma_s^\phi(xy) = \phi^s xy \phi^{-s} = \phi^s x\phi^{-s} \phi^s y \phi^{-s} = \sigma_s^\phi(x) \sigma_s^\phi(y)$. Similarly, $\sigma_s^\phi(\sigma_t^\phi(x)) = \sigma_{s+t}^\phi(x)$.
When $p=0$ we have $\sigma_t^\phi$, and when $p=1/2$, we get $(\Delta_\phi)^t$. Note that $t$ is imaginary in this talk, so this automorphism matches that in the previous talk.
\textbf{Definition:} $\phi, \psi \in L_1^+$, and $t \in I$, then $(D\phi:D\psi)_t := \phi^t\psi^{-t} \in L_0$.
We need the $t$ parameter to capture some extra information that pops up in the noncommutative setting (?). It is easier to work with than unbounded operators (?)
\textbf{Theorem:} (KMS condition - after Kubo-Martin-Schwinger)
For any $M$, there exists a bijection between $\phi \in \widehat{L_1^+}(M)$ and one-parameter groups of elements in $L_t: t \in I \mapsto u(t) \in L_t(M)$, satisfying $u(s+t) = u(s)u(t)$ and $u(s)^* = u(-s)$. In this correspondence, $u(t) = \phi^t$.
We now pass to the part of the talk concerning the example.
We'll explain how the modular automorphism relates to Clifford algebras. We usually see Clifford algebras in type I or II, but these will be type III (?).
Our setup: Let $H$ be a complex Hilbert space, $K \subset H$ a closed real subspace, such that $K \cap iK = 0$ and $K+iK$ is dense in $H$. The algebraic Clifford algebra $Cl^{alg}(K)$ acts on $\bigwedge H$. $Cl(K)$ is the $W^*$-algebra generated by $Cl^{alg}(K)$. There is also an action of $Cl(K^\perp)$, so $\bigwedge H$ is a $Cl(K)$-$Cl(K^\perp)$-bimodule. $\Omega$ is cyclic and separating. This implies $\bigwedge H \cong L_{1/2}(Cl(K))$. $\Omega$ is positive, so it is equal to $\phi^{1/2}\phi$ for some $\phi$.
\textbf{Example:} $H = L_{1/2}(M)$, $\phi \in \widehat{L_1^+}$, $K = M_{sa}\phi^{1/2} \cap L_{1/2}(M)$. $\sigma_t^\phi \in Aut(L_{1/2}(M) = Aut(H)$, $* \in \widetilde{Aut}(L_{1/2}(M)) = \widetilde{Aut}(H)$.
\textbf{Theorem:} (Jones, Wasserman)
\begin{enumerate}
\item $\bigwedge H$ is an invertible bimodule in the category of bimodules, i.e., $Cl(K) \to End(\bigwedge H_{Cl(K^\perp))})$ is an isomorphism, and $\bigwedge H \cong L_{1/2}(Cl(K),\phi) \cong L_{1/2}(Cl(K))$.
\item $*$ acts antilinearly on $L_{1/2}(Cl(K)) = \bigwedge H$. For $\psi \in \bigwedge H$, $\psi = a \wedge b \wedge \dots \wedge z$, $\psi^* = z^* \wedge \dots \wedge b^* \wedge a^*$.
\item $\sigma_t^\phi$ acts on $\bigwedge H$ componentwise
\end{enumerate}
Explicit geometric example. Segal wants a functor from conformal bordisms to hilbert spaces and bounded maps. Extending to points, we want von Neumann algebras, bimodules, and intertwiners. In the extended situation, we want to glue boundary circles along subsegments. Boundaries are assigned spaces $L_{1/2}(Y_i,S)$. Have a Dirac operator $D^+: C^\infty(\Sigma, S^+) \to C^\infty\Sigma, S^+)$. Let $W = ker D^+$. $L: W \to L_{1/2}(\partial \Sigma, S)$. Get a type III factor. Nice construction due to Jones, Wasserman. $\sigma^t$ acts by ``dilations'' on $P^1(\mathbb{R})$.
\subsection{3:15pm Corbett Redden, Conformal nets}
Outline:
\begin{enumerate}
\item More M\"obius group
\item Conforma Nets: definition and example
\item Properties
\item Representations
\end{enumerate}
The M\"obius group is the group of conformal automorphisms of $D \subset \mathbb{C}$. It is isomorphic to $PSU(1,1)$ and $PSL_2(\mathbb{R})$. There is an $S^1$ picture and an $\mathbb{R}$-picture, related by stereographic projection. The groups are generated by translations on $\mathbb{R}$, dilations on $\mathbb{R}$, and rotations on $S^1$. In fact, Iwasawa decomposition gives us a manifold decomposition $SU(1,1) \cong D \times T \times R$.
\textbf{Corollary:} $SU(1,1)$ acts transitively on the set of (connected open non-dense) intervals.
Stabilizer of a point is conjugate to $D \times T$, and the stabilizer of a pair of points is conjugate to $D$.
\textbf{Definition:} A (vacuum) conformal net (written CN or VCN) is a rule that assigns to each open connected non-dense interval $I$ a von Neumann algebra $A(I)$ acting on a fixed Hilbert space $H$, satisfying:
\begin{enumerate}
\item (Isotony) $I \subset J \Rightarrow A(I) \subset A(J)$.
\item (Locality) $I \subset J' \Rightarrow A(I) \subset A(J)'$, where $J' = int(S^1 \setminus J)$.
\item (M\"obius coherence) There is a representation $\pi: PSU(1,1) \to U(H)$ (only guaranteed to be projective away from vacuum) such that $\pi(g)A(I)\pi(g)^* = A(gI)$.
\item (Positive energy - for VCN) $R \subset PSU(1,1)$ should be positive energy.
\item (Vacuum - for VCN) There exists a unique (up to scalar) vacuum vector $\Omega \in H$, that is invariant under the M\"obius action, and cyclic under the algebra generated by all $A(I)$.
\end{enumerate}
\textbf{Example:} Let $\pi: LG_\ell \to U(H)$ be an irreducible positive energy representation. We set $A(I) = \pi(\widetilde{L_I(G)})''$.
\textbf{Theorem:} $A$ is a conformal net. If $\pi$ is the vacuum representation, then $A$ is a vacuum conformal net.
\begin{proof}
\begin{enumerate}
\item If $I \subset J$, then $\widetilde{L_I(G)} \subset \widetilde{L_J(G)}$, so $\pi(\widetilde{L_I(G)})'' \subset \pi(\widetilde{L_J(G)})''$.
\item If $I \cap J = \emptyset$, then $[\widetilde{L_I(G)},\widetilde{L_J(G)}] = 1$, so $A(I)$ commutes with $A(J)$.
\item $PSU(1,1)$ acts conformally on $D \supset S^1$, so the representation is canonically implemented.
\item This is an assumption.
\item Also taken care of by the definition.
\end{enumerate}
\end{proof}
\textbf{Theorem:} (Reeh-Schlieder) If $A$ is a vacuum conformal net, then $\overline{A(I)\Omega} = H$.
\textbf{Corollary:} $\Omega$ is cyclic, and separating for each $A(I)$.
\begin{proof}
$\Omega$ is separating for $A(I)$ if and only if $\Omega$ is cyclic for $A(I)'$. $\Omega$ is cyclic for $A(I') \subset A(I)'$.
\end{proof}
Now we can use Tomita-Takesaki: $S_I(A\Omega):= A^*\Omega$, and extend to $S_I$ unbounded on $H$. Get modular operators: $J_IA(I)J_I = A(I)'$, and $\Delta_I^{it} A(I) \Delta_I^{-it} = A(I)$.
\textbf{Theorem:}
\begin{itemize}
\item If $A$ is a vacuum conformal net, then all $A(I)$ are type $III_1$ factors.
\item If [inequality] (almost always satisfied), then $A(I)$ is hyperfinite.
\end{itemize}
For the first statement, need to take a look at the modular group, and show that its action is ergodic.
Let $j_{S_+} \in SU_-(1,1)$ be the flip map that reflects the upper half circle in the $x$-axis. Recall that $SU_\pm(1,1) = SU(1,1) \rtimes \{1,j_{S_+}\}$.
\textbf{Theorem:} (Geometric modular operations)
\begin{enumerate}
\item $\pi$ extends to $PSU_\pm(1,1) \to U_\pm(H)$, such that $J_{S_+} =: \pi(j_{S_+})$.
\item $\Delta_{S_+}^{it} = \pi(D_t)$.
\end{enumerate}
\begin{proof}
For first claim, check on components $D,R,T$ of Iwasawa decomposition. For the second claim, work with equivalent properties.
\end{proof}
\textbf{Theorem:} (Haag duality)
If $A$ is a vacuum conformal net, then $A(I') = A(I)'$.
\begin{proof}
Because of M\"obius covariance, it suffices to prove it for $S_+$. We have:
\[ A({S_+}') = \pi(j_{S_+})A(S_+)\pi(j_{S_+})^* = J_{S_+}A(S_+)J_{S_+} = A(S_+)'. \]
\end{proof}
\textbf{Definition:} A representation of a conformal net $A$ on $H_\pi$ is a collection of representations $\{ \pi_I\}_{I \subset S^1}$, where $\pi_I: A(I) \to B(H_\pi)$ satisfies the following conditions:
\begin{enumerate}
\item (Consistency) For $I \subset J$, $\pi_I = \pi_J \mid_{A(I)}$.
\item There exists a representation $\pi^m: PSU(1,1) \to PU(H_\pi)$, such that $\pi^m(g) \pi_i(-) \pi^m(g)^* = \pi_{gI}(\alpha_g-)$, where the $\alpha_g$ denotes conjugation using the M\"obius representation on $A$.
\item Rotations in $\pi^m$ are generated by a positive operator.
\end{enumerate}
Example: identity representation. $\pi(A(I)) = A(I)$.
Example: If $A_0$ is the vacuum conformal net from the level $\ell$ irreducible vacuum positive energy representation, an $\pi: \widetilde{LG_\ell} \to U(H_\pi)$ is an irreducible positive energy representation. Then we obtain a representation of $A_0$.
I'm over time, so just let me write the idea. We start with $\widetilde{L_I(G)}$, and we have a map $\pi_0(\widetilde{L_I(G)}) \to \pi(\widetilde{L_I(G)})$. We want to extend it a map $\pi_0(\widetilde{L_I(G)})'' \to \pi(\widetilde{L_I(G)})''$. Since $\pi_0$ and $\pi$ are subrepresentations of $\pi^{\otimes \ell}$, then the local equivalence property (from Min's talk yesterday) guarantees a map.
\subsection{4:30pm Christoph Solveen, The Bisognano-Wichman theorem and nets on $\mathbb{R}^4$}
The first example of a BW theorem was seen in the last talk ($\Delta_{S_+}^{it} = \pi(D_t)$, describing a geometric interpretation of the modular operators on $S^1$.
Outline: \begin{enumerate}
\item Minkowski spacetime
\item Axioms for QFT
\item BW-theorems
\item BW and physics
\end{enumerate}
Why do we model our universe after $\mathbb{R}^4$ with a signature $(1,3)$ metric? Our story starts in the 1830s with Faraday. He wanted some way to describe electricity and magnetism that involved non-instantaneous action. One could have particles passing between bodies mediating forces, but one could also think of fields of force. In modern language, a field is a section of a bundle over space and time.
Maxwell formulated equations for the behavior of the electromagnetic field. They were a set of hyperbolic PDEs, and they are invariant under the Lorentz group, not the Galilean group. Einstein proposed that we should stop following Newton's picture, and promote the two principles:
\begin{enumerate}
\item The laws of physics are invariant under the action of the Lorentz group.
\item The perturbations in the EM field propogate at a constant speed, regardless of reference frame.
\end{enumerate}
Minkowski proposed a picture that unified space and time, so they were not treated as separate entities. We cease to have a meaningful notion of simultaneity, but we have a coordinate-independent notion of past and future of any particular event. We have an axiom, that there exists a preferred class of coordinates, the ``inertial'' coordinates, such that the components of the metric are diagonal $(1,-1,-1,-1)$.
The Lorentz distance between $x, \tilde{x} \in \mathbb{R}^4$ is given by $(x-\tilde{x})^2 = \eta_{\mu\nu}(x^\mu-\tilde{x}^\mu)(x^\nu-\tilde{x}^\nu)$. The group of diffeomorphisms of $\mathbb{R}^4$ that leave the Lorentz distance invariant is called the Poincar\'e group, and can be written $\mathbb{R}^4 \rtimes O(1,3)$. The proper orthochronous Poincar\'e group is $\mathbb{R}^4 \rtimes SO^+(1,3)$.
The Lorentz group is generated by boosts and rotations of 3-space. The boosts are the discrepancy between the coordinate frames of different inertial observers. An example boost in the $x$-direction is:
\[ \Lambda(s) = \begin{pmatrix} \cosh(s) & -\sinh(s) & 0 & 0 \\ -\sinh(s) & \cosh(s) & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \]
where $s$ is a measure of rapidity.
Now, I will draw a very na\"ive picture of what quantum physics is, and why we are dealing with operator algebras.
What is physics. We have an evaluation $(A,\omega) \mapsto \omega(A) \in \mathbb{R}$. $\omega$ is an ensemble of physical configurations (i.e., a state), and $A$ is an observable. We get a real number as the expectation value of the measurement.
If we pass to an idealization, we assume we have state about which we have perfect knowledge about the configuration. These are called pure states - no dirt in our apparatus.
We assume that the set of observables is inside a $C^*$ algebra $\mathcal{A}$. States are positive linear functionals $\mathcal{A}$, i.e., points in $\mathcal{A}^{*1}_+$. This is a convex set, and pure states are extremal points.
We can ask how far our measurements deviate from the expectation, i.e., second moment, or standard deviation. $(\Delta_\omega A)^2 = \omega(A^2) - \omega(A)^2$.
In classical physics, a pure state implies the standard deviation vanishes for all $A \in mathcal{A}$. This is the case when $\mathcal{A}$ is commutative. In quantum physics, even the pure states can have strictly positive deviation. This happens when $\mathcal{A}$ is noncommutative.
\textbf{Haag-Kastler axioms:}
We consider a net $\mathcal{O} \mapsto R(\mathcal{O})$ of von Neumann algebras on $\mathbb{R}^4$, where $\mathcal{O}$ ranges over open bounded subregions of $\mathbb{R}^4$. We decree that they satisfy:
\begin{enumerate}
\item (Isotony) If $\mathcal{O}_1 \subset \mathcal{O}_2$, then $R(\mathcal{O}_1) \subset R(\mathcal{O}_2)$.
\item (Locality) If $\mathcal{O}_1 \subset \mathcal{O}'_2$, then $R(\mathcal{O}_1) \subset R(\mathcal{O}_2)'$, where $\mathcal{O}'$ denotes the causal complement, i.e., the set of points that are spacelike separated from all points in $\mathcal{O}$.
\item (Poincar\'e symmetry) There is an automorphic action of the autochronous Poincar\'e group $P_+^\uparrow$ on the net, so $\alpha_g(R(\mathcal{O})) = R(g\mathcal{O})$.
\end{enumerate}
A representation $\pi_0$ of a Haag-Kastler net is called a vacuum representation if:
\begin{enumerate}
\item[4] There is a strongly continuous representation $U$ of $P_+^\uparrow$ that implements
\[ \alpha_g: U(g) \pi_0(R(\mathcal{O}))U(g)^* = \pi(\alpha_g R\mathcal{O}). \]
\item[5] The joint spectrum of the generators of $U\mid_{\mathbb{R}^4}$ lies in the forward light cone.
\item[6] There exists a unique (up to phase) vacuum vector $\Omega$ that is invariant under $U\mid_{\mathbb{R}^4}$.
\end{enumerate}
There is another framework called the Wightman axioms. It is not clear how it is related to Haag-Kastler. You can get a set of von Neumann algebras by applying bounded functional calculi. The BW property was proved in the Wightman framework during the 1970s. In the Haag-Kaster framework, it has not been proved. However, if you add conformal symmetry, then you get BW.
Consider the wedge region $W = \{ x \in \mathbb{R}^4 \mid x^1 > |t| \}$. This is boost-invariant. The BW property is: $Jw = \Theta U(R_{23}(\pi))$, where $\Theta$ is the PCT operator. $\Delta^{it} = U(\Lambda(-2\pi t))$, where $\Lambda$ denotes boost.
Physical input: KMS states are thermal equilibria.
There is a relation between the modular group and the KMS condition ($\sigma_\tau = \Delta^{i\tau} \cdot \Delta^{-i\tau}$), which says that the modular parameter $\tau$ is related to physical time by a factor of $-1/\beta$.
It's related to the Unruh effect, where you get a thermal equilibrium in a uniformly accelerated frame.
\section{Wednesday, 8/18/10}
\subsection{9:00am Andre}
Gradings, and confusions related to gradings, groups acting on what, which actions are projective and which are honest. Basically, I'm going to list a bunch of facts.
Here is the Dirac Sea model of Fock space. $\mathcal{F}_p = \bigwedge(PH) \otimes \overline{\bigwedge(PH^\perp)}$. Write $H = L^2(S^1)$ Basis is indexed by $\mathbb{Z}$, the lattice of characters. The nonnegative part is covered by $PH$, the Hardy space, and the negative integers correspond to $PH^\perp$. [Draws a number line]
A basis vector of the Dirac sea is given by a subset of basis vectors of H that are all but finitely many basis vectors of $PH^\perp$ and only finitely many basis vectors of $PH$. Example: [darkens some dots in the line] $e_3 \wedge e_0 \wedge e_{-2} \wedge e_{-3} \wedge e_{-5} \wedge \dots$. This denotes the basis vector $(\overline{e_{-4} \wedge e_{-1}}) \otimes (e_0 \wedge e_3)$.
One slight disadvantage of this model is that it gives the impression that it depends on a choice of basis. In fact, all we need is the energy grading (i.e., the weight of the rotational $S^1$ action).
There is a grading called ``charge'' that is the number of dots (i.e., the lowest empty spot you get if you crunch everything down). In our example, the charge is zero. Charge is always an integer.
There is a grading called ``energy'': sum the positive particles, subtract the negative holes. In our case, we have 3 + 0 - (-1) - (-4).
There are clear reasons why this is the wrong way to compute energy. For example, the zero energy space is not spanned by the vacuum, since adding a particle in the zero spot doesn't change the enegy. The correct method is to label the basis vectors with elements of $\mathbb{Z} + 1/2$, so $PH$ is spanned by strictly positive elements, while $PH^\perp$ is spanned by strictly negative elements. In this case, our energy is $7/2 + 1/2 - (-1/2) - (-7/2) = 8$. The two energy computations yielded identical answers because we have charge zero. In general they will differ by half of the charge. Energy always lies in $\frac{1}{2}\mathbb{N}$. This comes from the action of the circle that lives in $SL_2(\mathbb{R})$, rather than $PSL_2(\mathbb{R})$.
Note that our Dirac sea has a one-dimensional space in each spot. When we work with $SU(N)$, we find that each graded piece of the Dirac sea has dimension $N$, namely a copy of the standard representation.
Here is a table of actions on the Fock space $\mathcal{F}_P$, for $G = SU(N)$:
\begin{tabular}{r|c|c}
& projective & honest \\ \hline
$LG$ & x & \\
$\widetilde{LG}$ & & x \\
$LG \rtimes T_{rot}$ & x & \\
$\widetilde{LG} \rtimes \widetilde{T_{rot}}$ & & x \\
$PSL_2(\mathbb{R})$ & x & \\
$SL_2(\mathbb{R})$ & & x
\end{tabular}
What is the $SL_2(\mathbb{R})$ action? For $g: S^1 \to S^1$, $(v_g \cdot f)(z) = (g'(g^{-1}z))^{-1/2}f(g^{-1}z)$. The factor in the front is to preserve the $L^2$ norm. In terms of matrices in $SU(1,1)$, we have:
\[ \left( \begin{pmatrix} \alpha & \beta \\ \overline{\beta} & \overline{\alpha} \end{pmatrix} \cdot f \right) (z) = \frac{ f\left( \frac{\overline{\alpha}z-\beta}{-\overline{\beta}z + \alpha} \right) }{-\overline{\beta}z+\alpha} . \]
\textbf{Fact:} The (projective) action of $LG$ preserves the charge grading, and the constant loops action of $G$ preserves the energy grading, also.
Suppose $H_\lambda$ is an irreducible positive energy representation of $LG$, and suppose it is not the vacuum representation. We can shift the energy grading on $H_\lambda$ around to get an honest action of $SL_2(\mathbb{R})$, and this corresponds to making the above mistake about energy grading. There is a correct energy grading, and you get an honest representation of some finite cover $\widetilde{SL_2(\mathbb{R})}$.
\subsection{9:45am Anatoly Preygel, The Knizhnik-Zamolodchikov equation}
So, I'll be talking about the KZ equation. Now, I have this thing where I don't think one should give talks about equations, so I'll be going under some contortions to avoid talking about this equation.
Why do we want to get a differential equation of some sort? How does it fit in with Wasserman's picture?
\textbf{Reminder:}
Let $\lambda$ be a signature. As we saw in Hiro's talk, this gives rise to an irreducible representation of $SU(N)$. If $\lambda$ is admissible for level $\ell$, then we get $H_\lambda$, an irreducible representation of $\widetilde{LG}_\ell \rtimes S^1_{rot,n}$, with $H_\lambda(0) = V_\lambda$.
\textbf{Primary fields:}
We have a vector space of intertwiners $Hom_G(V_\lambda \otimes W, V_\mu)$. Under favorable circumstances, we get from an intertwiner something called a primary field, which is an element of $Hom_{\widetilde{LG}_\ell \rtimes S^1_{rot}}(H_\lambda \otimes C^\infty(S^1,W), H_\mu)$. Think of an element $\phi$ as a $W^\vee \otimes Hom_{unbdd}(H_\lambda,H_\mu)$-valued distribution.
Fourier modes can be expressed heuristically as $\phi(n) = \int_{S^1} \phi(z) z^{-n} \frac{dz}{2\pi z} = \phi(z^{-n})$. How bad can this be? Arbitrarily many of them can be unbounded. Infinitely many can be nonzero. A not-so-bad property is that the Fourier modes behave well with energy: $\phi(n)(w)$ is a map from $H_\lambda(k)$ to $H_\mu(k-n)$.
\textbf{Notation:}
\begin{itemize}
\item $\lambda_1, \dots, \lambda_2, \mu$ are tableaux
\item $W_1, W_2$ are $G$-representations
\item $\mathcal{U} = Hom_G(V_{\lambda_1} \otimes W_1 \otimes W_2, V_{\lambda_2})$.
\item We write primary fields using the notation $\phi^{\text{charge}}_{\text{target, source}}$.
\end{itemize}
If we have $\phi^1: V_{\lambda_1} \otimes W_1 \to V_\mu$ and $\phi^2: V_\mu \otimes W_2 \to V_{\lambda_2}$, then $\phi^2 \circ \phi^1 \in \mathcal{U}$. We can promote these intertwiners to primary fields. For a pair of test functions $(f,g) \in C^\infty(S^1,W_2) \times C^\infty (S^1,W_1)$, we have $\phi^2(f) \circ \phi^1(f)$. We find that:
\[ \mathcal{U} \cong \sum_{\mu'} Hom_G(V_{\mu'} \otimes W_1, V_{\lambda_2}) \otimes Hom_G(V_{\lambda_1} \otimes W_2, V_{\mu'}). \]
We can write a composition of intertwiners as
\[ \phi^2 \circ \phi^1 = \sum_{\mu'} \phi^{W_1}_{\lambda_2,\mu'} \otimes \phi^{W_2}_{\mu',\lambda_1} \]
This can be turned into a sum of tensors of smeared primary fields.
\textbf{Interlude:} If we write a test function $f$ as a power series $\sum f_n z^n$, we can expand a primary field $\phi$ in terms of its modes $\phi(f_n)$. We write $\underline{\phi}$ to be the induced action on the lowest energy space $H_\cdot(0)$, i.e., include $H_\cdot(0) \hookrightarrow H_\cdot$, then apply $\phi(f)$, then project back to $H_\cdot(0)$.
\textbf{4-point functions:}
We define the 4-point function $F_\mu = \underline{\phi^2(f) \circ \phi^1(g)}$. Formally, this is a function of $z/w$ with coefficients $f_n g_{-n}$ such that [some conditions]. With traditional notation, $\langle \phi^2(w_2 \cdot f) \circ \phi^2(w_1 \cdot g)v_{\lambda_1}, v_{\lambda_2} \rangle$. Here, $w_1 \in W_1$, $w_2 \in W_2$, $v_{\lambda_i} \in H_{\lambda_i}(0)$.
We can expand: $F_\mu(z) = \sum_{n \geq 0} \underline{\phi^2(n) \circ \phi^1(-n)} z^n$, and it is a holomorphic function on the unit disc.
\textbf{Lemma:} $\underline{\phi^2(f) \circ \phi^1(g)} \in \mathcal{U}$, and it is equal to $\int_{S^1 \setminus \{1\} } (\tilde{f} * g)(z) F_\mu(z) \frac{dz}{2\pi z}$ if $f$ and $g$ have disjoint support. Furthermore, $F_\mu$ extends continuously to $S^1 \setminus \{ 1 \} $.
We get the identity $(\tilde{f} * g)(z) = (\tilde{g} * f)(z^{-1})$. $F_\mu$ lives on the unit disk, and the analogous function from switching, $G_{\mu'}(z^{-1}$, lives on the exterior, and both extend continuously to unit circle away from 1. Relations between these functions yield relations for $\phi^2 \circ \phi^1$.
\textbf{Generalities on ODEs with regular singularities:}
We want a ``nice'' ODE satisfied by $F_\mu$ and $G_{\mu'}(z^{-1})$. The basic ODE has the form
\[ \frac{df}{dz} = \left( \frac{P}{z} + \frac{Q}{1-z} \right) f, \]
where $f$ is a $\mathcal{U}$-valued function, $P$ and $Q$ are endomorphisms of $\mathcal{U}$. This is linear, first order, and has regular singularities. For example, suppose $\mathcal{U} = \mathbb{C}$. We get
\[ \frac{df}{f} = \frac{dz}{z} \left( P + Q\frac{z}{1-z} \right) \]
The part in the parentheses is holomorphic near zero. Near zero we find that $f$ is $z^P$ near zero. The resulting function is holomorphic near zero, and unique up to a scalar.
Suppose $P$ has eigenvalues that do not differ by integers. Then the solutions near zero look like $f(z) = \sum_i a_i \tilde{f}_i(z) \xi_i z^{\lambda_i}$, where $\xi_i$ is the $i$th eigenvector in $\mathcal{U}$, and $\lambda_i$ is its eigenvalue.
\textbf{General Fact:} We have correspondences between:
\begin{enumerate}
\item nice differential equations on $\mathbb{CP}^1$ with regular singularities at 0,1, and $\infty$
\item Local systems on $\mathbb{CP}^1 \setminus \{0,1,\infty\}$ (with values in $\mathcal{U}$).
\item Representations $F_2 \cong \pi_1(\mathbb{CP}^1 \setminus \{0,1,\infty\}) \to GL(\mathcal{U})$.
\end{enumerate}
\textbf{Why do we get an ODE?}
Suppose we have a machine that takes $n+1$ points in $\mathbb{CP}^1$ with coordinates $D \to \mathbb{CP}^1$ at each point, and produces a vector space noncanonically isomorphic to $\mathcal{U}$, and suppose that changes of coordinates (preserving centers) yield isomorphisms [in a manner compatible with composition]. We get a vector bundle on the moduli space of points $\mathcal{M}_{0,n+1}$. If we specify that changes of coordinates that do not preserve the center yield isomorphisms, we get a notion of parallel transport on the moduli space, and a connection on the vector bundle.
When $n=3$, we get 4 points, and $\mathcal{M}_{0,4} \cong \mathbb{CP}^1 \setminus \{ 0,1,\infty \}$. The KZ equation that Wasserman describes is the connection we get from this procedure (but in an obscure way).
\textbf{KZ setup:}
$G = SU(N)$, $\{ X_i \}$ an orthonormal basis for the Lie algebra $\mathfrak{g}$. $\Omega$ (not the vacuum) is the Casimir operator $-\sum X_k \otimes X_k$. It is in the center of the universal enveloping algebra. If we have an irreducible representation $V_\lambda$, then the Casimir acts by a scalar $\Delta_\lambda$. We write $\Omega_{ij}$ for the action of the Casimir on the $i$ and $j$ slots of $\mathcal{U} = Hom(V_{\lambda_1} \otimes W_1 \otimes W_2 \otimes \dots \otimes W_{n-1}, V_{\lambda_2})$. In coordinates, this is $\Omega_{ij} = -\sum_k pi_i(X_k) \otimes \pi_j(X_k)$.
\textbf{Theorem:} $F_\mu$ satisfies the formula
\[ (N+\ell)\frac{dF_\mu}{dz} = \left( \frac{\Omega_{23} - (\Delta_\mu - \Delta_{\omega_1} - \Delta_{\omega_2})/2}{z} + \frac{\Omega_{12}}{z-1} \right) F_\mu \]
\textbf{Theorem:} $f_\mu = z^{\Delta_\mu - \Delta_{\omega_1} - \Delta_{\omega_2})/2(N+\ell)}F_\mu(z)$ satisfies
$ (N+\ell)\frac{df_\mu}{dz} = \left( \Omega_{23}{z} + \frac{\Omega_{12}}{z-1} \right) f_\mu$.
\subsection{11:00am Daniel Moseley, The hypergeometric function}
Recall from last time, we had a function $F_\mu$ on the lower half-sphere, and $G_{\mu'}$ on the upper half-sphere, and we wanted to relate $F_\mu$ with the $G_{\mu'}$ on the mutual boundary. They satisfied a nice ODE, called the KZ equation. In this talk, we will forget the interpretation of the equation in terms of four-point functions, and just study the equation by itself.
\begin{enumerate}
\item Hypergeometric function/equation.
\item Transport coefficients for the ``Basic ODE''.
\end{enumerate}
The (Gauss) Hypergeometric equation is the following 2nd-order ODE with 3 regular singular points $\{0,1,\infty \}$:
\[ z(1-z)f'' + [c-(1+a+b)z]f' - abf = 0 \]
Its solutions have the form
\[ {}_2F_1\left( \begin{array}{c} a,b \\ c \end{array} ; z\right) = \sum_{n \geq 0} \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!} \]
Here $(a)_n = (a)(a+1)\dots(a+n-1)$ is the Pochammer symbol for the rising factorial. We find that vector solutions $F = \binom{f(z)}{zf'(z)}$ satisfy $F'(z) = (A/z + B/(1-z))F(z)$, where $A = \begin{pmatrix} 0 & 1 \\ 0 & 1-c \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 0 \\ -ab & 1+a+b-c \end{pmatrix}$.
In general, if $A$ and $B$ are $n \times n$ matrices, then we call the vector equation an ``abstract hypergeometric equation'' (Hille).
I'd like to introduce some notation. Recall that $(a)_n = \Gamma(a+n)/\Gamma(a)$, where $\Gamma(x) = \int_0^\infty t^x e^{-t} \frac{dt}{t}$. If $x$ is a positive integer, then $\Gamma(x) = (x-1)!$.
Say $f: \mathbb{C} \to V = \mathbb{C}^N$ satisfies $f'(z) = \frac{Pf}{z} + \frac{Qf}{1-z}$, for $P,Q \in End(V)$. Suppose the eigenvalues of $P$ differ mod 1, and $Q$ is a nonzero multiple of a rank 1 idempotent (say, $Q^2 = \delta Q)$. Then the trace of $Q$ is $\delta$. Since it has rank one, there exists some $\phi \in V^*$ and $v \in V$ such that $Q(x) = \phi(x)v$, and $\phi(v) = \delta$. There is an additional niceness condition: $Q$ is in general position with respect to $P$ if $v = \sum \delta_i \xi_i$, where $\delta_i \neq 0$, where $\xi_i$ are a basis of eigenvectors of $P$. Let us choose eigenvectors so that $\phi(\xi) = 1$. Let $R = Q-P$, and suppose $R$ satisfies the same conditions as $P$ with respect to $Q$. Let $(\zeta_j, -\mu_j)$ be the normalized eigenvectors/values of $R$.
Write $f_i(z) = \sum_n \xi_{i,n} z^{\lambda_i+n}$, with $\xi_{i,0} = \xi_i$. This converges in the ``disc with slit'' $\{z \mid |z|<1, z \not\in [0,1) \}$. Now we write solutions near infinity: $g_j(z) = \sum_n \zeta_{j,n} z^{\mu_j-n}$, where $\zeta_{j,0} = \zeta_j$. Let's extend analytically and compare on the mutual boundary: $f_i(z) = \sum_{ij} c_{i,j} g_j(z)$. Our goal is to compute the transport coefficients $c_{i,j}$. The formula is:
\[ c_{i,j} = e^{i\pi(\lambda_i - \mu_j)} \frac{\prod_{k \neq i} \Gamma(\lambda_i - \lambda_k + 1) \prod_{\ell \neq j} \Gamma (\mu_j - \mu_\ell)}{\prod_{\ell \neq j} \Gamma(\lambda_i - \mu_\ell) \prod_{k \neq i} \Gamma(\mu_j - \lambda_k)} \]
\textbf{Fact 1:} The transport matrix depends only on the eigenvalues of $P+P-Q$. This dependence is holomorphic.
\textbf{Fact 2:} For $\sigma, \tau \in S_N$, $c_{i,j}(\lambda_1,\dots,\lambda_N,\mu_1,\dots,\mu_N) = c_{\sigma i,\tau j}(\lambda_{\sigma 1}, \dots, \lambda_{\sigma N},\mu_{\tau 1} ,\dots,\mu_{\tau N} )$.
Look at $\phi(f_i(z))$. $\phi(f_i(z)) = \sum_j c_{i,j} \phi(g_j(z))$. Writing $f_i(z) = \sum \xi_{i,n} z^{\lambda_i + n}$, we get:
\[ \begin{aligned} \sum_{n \geq 0} (n + \lambda_i) \xi_{i,n} z^n &= \sum_{n \geq 0} p \xi_{i,n} z^n + Q(1 + z + z^2 + \dots)\sum_{n \geq 0} \xi_{i,n}z^n \\
(n + \lambda_1 - P)\xi_{1,n} &= Q(\xi_{1,0} + \dots + \xi_{1,n-1}) \\
\alpha_{i,n} &= \phi(\xi_{1,0} + \dots + \xi_{1,n-1})
\alpha_{i,0} &= 1\\
\alpha_{i,n} &= \prod_{j = 1}^N \prod_{m=1}^n \frac{m+\lambda_i - \mu_j}{m + \lambda_i - \lambda_j} \\
\frac{\phi(f_1(z))}{z^{\lambda_1}(1-z)} = \sum_{n \geq 0} \alpha_{1,n} z^n &= \sum_{n \geq 0} z^n \prod_{j = 1}^N \prod_{m=1}^n \frac{m+\lambda_i - \mu_j}{m + \lambda_i - \lambda_j}
\end{aligned} \]
Restrict the $\lambda_i$s and $\mu_j$s to real numbers, and write them in order: $\lambda_i > \lambda_{i+1}$. $\lambda_1 + 1 > \mu_j > \lambda_j$ for all $j$. Recall the beta function: for $a,b > 0$, $\frac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)} = \int_0^1 t^{a-1}(1-t)^{b-1}dt$. Then
\[ \phi(f_1(z)) = (1-z)z^{\lambda_1} K \int\int \dots \int (1-zt_2\dots t_n)^{\mu_1-\lambda_1 - 1} \cdot \prod_{j \neq 1} t_j^{\lambda_1 - mu_j}(1-t_j)^{\mu_j - \lambda_j - 1} dt_j \]
Here $K = \prod_{j \neq 1} \frac{ \Gamma(\lambda_1 - \lambda_j + 1}{\Gamma(\lambda_1 - \mu_j + 1) \Gamma(\mu_j - \lambda_j)}$.
$\phi(g_j(z))$ acts like $|z|^{\mu_j} d^{\pi i\mu_j}$, where $g_j(z) = \sum \zeta_{j,n} z^{\mu_j-n}$. If $z \in \mathbb{R}$ is a large negative, then $\phi(f_i(z) \sim c_{1,1}|z|^{\mu_1}e^{\pi i\mu_j}$.
Then:
\[ c_{1,1} = e^{i\pi(\lambda_1 - \mu_1)} \prod_{j \neq 1} \frac{ \Gamma(\lambda_1 - \lambda_j + 1) \Gamma (\mu_1 - \mu_j)}{\Gamma(\lambda_1 - \mu_j) \Gamma(\mu_1 - \lambda_j)} \]
Get the rest by symmetry.
\section{Thursday, 8/19/10}
\subsection{9am, Andre}
One thing that I wanted to get on the board yesterday and I didn't get the time. It is a point about $SL_2(\mathbb{R})$ representations.
$SL_2(\mathbb{R})$ acts on the Fock space $(\mathcal{F}_P)^{\otimes \ell}$ (which can also be constructed by the exterior algebra construction on $L^2(S^1, \mathbb{C})^{\ell N}$ - part of rank-level duality). Every level $\ell$ representation of $LG$ lies in $(\mathcal{F}_P)^{\otimes \ell}$.
$SL_2(\mathbb{R})$ acts projectively on $H_\lambda$. The $\lambda$-isotypical component of $(\mathcal{F}_P)^{\otimes \ell}$ can be written as $H_\lambda \otimes V$, where $V$ a multiplicity space. $SL_2(\mathbb{R})$ acts projectively on $V$, but the tensor product has an honest action. This is an example of cancellation of anomalies.
Question from Hiro: How do we get a braided structure on something that is defined in terms of intervals on the circle?
We have exact sequences $\mathbb{Z} \to \widetilde{SL_2(\mathbb{R})} \to SL_2(\mathbb{R})$ and $\mathbb{Z} \to \mathbb{R} \to S^1$ that form pullback diagrams. $SL_2(\mathbb{R})$ and $S^1$ act projectively on $H_\lambda$, and we have honest actions of the corresponding universal covers.
Consider $H_\lambda$ in a disc, with intervals $I$ and $-I$ on opposite sides of the boundary. Rotation by $\pi$ induces an isomorphism between $\mathcal{A}(I)$ and $\mathcal{A}(-I)$. We define $H_\lambda^{(\pi)}$ to be $H_\lambda$ with action twisted by this isomorphism. The half-twist $\theta$ is an $LG$-equivariant map $H_\lambda \to H_\lambda^{(\pi)}$. $\beta: H_\lambda \boxtimes H_\nu \to H_\nu \boxtimes H_\lambda$ is given by $\theta_{H_\lambda \boxtimes H_\nu}^{-1}(\theta_{H_\lambda} \boxtimes \theta_{H_\nu})$.
Noah's comment: cactus condition is simply that the square of the half-twist is one, but there doesn't seem to be a similarly simple explanation of the twist in a braided category.
Question: what is a primary field in the Segal picture? A primary field is an (unbounded) $\widetilde{LG} \rtimes T_{rot}$-equivariant map $C^\infty(S^1, V_k) \otimes H_j \to H_i$. The first space is not positive energy, and the action is level zero (unsurprising - levels add for ordinary tensor products). A primary field encodes the same information as a map $H_k \boxtimes H_j \to H_i$, where $H_k$ is the corresponding positive energy representation, and $\boxtimes$ denotes Connes fusion. Pick a half-circle $I$ along which we do the fusion. A fusion map is determined by $Hom_{\widetilde{L_IG}}(H_0, H_k) \otimes H_j \to H_i$.
Given $f \in C^\infty(S^1,V_k)$, the primary field produces a map $H_j \to H_i$. If we already assume the $j=0, i=k$ case ofthe correspondence, then $f$ induces a map $H_0 \to H_k$. If $f$ is supported on $I'$, then the map $H_0 \to H_k$ is $\widetilde{L_IG}$-equivariant, and we get the fusion.
\subsection{10:15am Nick Rozenblyum, Sugawara's formula and the action of $Diff(S^1)$ on positive energy representation}
This talk will not have any operator algebras of conformal nets.
Let $G$ be a simple simply connected Lie group. $\widetilde{LG}$ is a central extension of the loop group. The category of positive energy representations is very nice. There is an action of $Diff(S^1)$ on any positive energy representation.
Following a pattern we've see before, rather than talk about the diffeomorphism group, we'll pass to the Lie algebra, which is made of smooth vector fields on $S^1$. If we ignore questions of convergence, elements have the form $\sum_n a_n e^{in\theta} \frac{d}{d\theta}$. We'll restrict to polynomial vector fields, i.e., those sums that have finitely many nonzero terms. It is spanned by $L_n = -i e^{in\theta}\frac{d}{d\theta}$, which is $t^{n+1}\frac{d}{dt}$ underthe transformation $t = e^{i\theta}$ [note: this is nonstandard - one typically sees $L_n = -t^{n+1}\frac{d}{dt}$]. These satisfy the commutation relation $[L_n,L_m] = (m-n)L_{m+n}$ [note: one usually sees $(n-m)$ here).
There is an algebraic geometry version of this. Let $\mathcal{O}$ denote the power series ring $\mathbb{C}[[t]]$, and view this as the ring of functions on the ``formal disk''. It is the algebraic geometer's version of the unit disk, but it is really small. The analogue of the boundary circle is the formal punctured disk, whose ring of functions is $\mathcal{K} = \mathbb{C}((t))$, the quotient field of $\mathcal{O}$. Vector fields on the formal punctured disk are described by $Der \mathcal{K} = \mathbb{C}((t))\frac{d}{dt}$.
There exists a universal one dimensional central extension of $Der \mathcal{K}$, called the Virasoro algebra. In other words, there is an exact sequence $0 \to \mathbb{C}\cdot K \to Vir \to Der\mathcal{K} \to 0$. $Vir$ is spanned by $\{L_n \}_{n \in \mathbb{Z}}$ together with a central vector $K$, and they satisfy the commutation relation:
\[ [L_m, L_n] = (m-n)L_{m+n} + \frac{m^3-m}{12} \delta_{-m,n}K \]
[Warning: this is nonstandard due to our choice of signs for $L_n$. In particular, the $K$ here is minus the usual $K$.]
The segal-Sugawara construction is a way to produce an action of Virasoro on a positive energy representation of $\widetilde{LG}$. Recall that positive energy representations have an energy decomposition $V = \bigoplus N(v)$, where $V(n) = 0$ for $n<0$. For $L_0 = \frac{d}{d\theta}$ a generator of the Lie algebra of $T_{rot}$, $[\frac{d}{d\theta}, X(m)] = -mX(m)$, so for $v \in V(n)$, $X(m)v \in V(n-m)$. There is an operator called the Casimir. We can write it down by choosing an invariant inner product $\langle, \rangle$ on $\mathfrak{g}$ such that $\langle \theta, \theta \rangle = 2$, when $\theta$ is the longest root (this condition fixes the inner product up to a sign). For $\{ X_j \}$ an orthonormal basis of $\mathfrak{g}$, $\Omega = \sum_j X_j^2$. This is an element of the tensor square of $\mathfrak{g}$, and is a central element in the second filtered piece of the universal enveloping algebra.
In the adjoint representation, $\Omega = 2h^\vee \cdot Id$. Here $h^\vee$ is the dual Coxeter number, and for $G=SU(N)$, the dual Coxeter number is $N$. In general, if $V$ is a highest weight representation of highest weight $\lambda$, then $\Omega = \langle \lambda, \lambda + 2\rho \rangle \cdot Id$, where $\rho$ is one half the sum of the positive roots.
Let's try to make a Casimir for a loop group. We encounter the sum $\sum_{j,n \in \mathbb{Z}} X_j(n) X_j(-n)$, which diverges. We can separate the divergent part, and manipulate some commutators to get
\[ \begin{aligned}
\sum_{j,n} : X_j(-n)X_j(n): + \sum_{j,n} [X_j(-n),X_j(n)] &= \sum_{j,n} : X_j(-n)X_j(n): + \sum_{n >0} \ell n \dim(\mathfrak{g}) \\
\text{where } : X(m)Y(n): &= \begin{cases} X(m)Y(n) & n \geq m \\ Y(n)X(m) & n < m \end{cases}
\end{aligned} \]
Let $\Delta_0 = \sum_{j,n \in \mathbb{Z}} : X_j(-n)X_j(n):$. For $V$ a positive energy representation of $LG$, $\Delta_0$ acts on $V$, but is not central. If we write $[Y, X_j = \sum_k \alpha_{jk}X_k$ and let $Z_n = \sum_{j,k} \alpha_{jk} :X_k(n)X_j(m-n):$, then
\[ [Y(m), \sum_j: X_j(-n)X_j(n): = \begin{cases} Z_{m+n}-Z_n & m \neq \pm n \\ Z_{m+n} - Z_n + mY(m) I & m=\pm n \end{cases} \]
where $I$ is a central element of $\widetilde{LG}$.
After suitable work, we find that
\[ \begin{aligned}
[Y(m),\Delta_0] &= m(2YI + \Omega Y)(m) \\
&= 2m(\ell + h^\vee)Y(m) \\
&= 2(\ell + h^\vee)\frac{d}{d\theta}Y(m)
\end{aligned} \]
We learn two things:\begin{enumerate}
\item We can get a central elements $\Delta = \Delta_0 + 2(\ell + h^\vee) \frac{d}{d\theta}$
\item $L_0 = \frac{d}{d\theta} = -\frac{1}{2(\ell + h^\vee)}\Delta_0$.
\end{enumerate}
This yields $L_0$, internally generated. We can apply similar sums for other terms:
\[ \Delta_m = \sum_{j,n \in \mathbb{Z}} :X_j(m-n) X_j(n): \]
Then $L_m = -\frac{1}{2(\ell+h^\vee)}\Delta_m$.
\textbf{Theorem:} These elements of the universal enveloping algebra satisfy the Virasoro relations with central charge $K = \frac{\ell \dim(\mathfrak{g})}{\ell + h^\vee}$.
\subsection{11:30am Arturo Prat-Waldron, Primary fields, boundedness of smeared primary fields}
We will study intertwiners. Let $H_\lambda$ and $H_\mu$ be irreducible positive energy representations of $\widetilde{LG \rtimes T_{rot}}$ at level $\ell$. Let $I$ be an interval in $S^1$. We consider the space $Hom^{bdd}_{\widetilde{L_IG}}(H_\lambda,H_\mu)$, and we will get explicit elements.
\textbf{Example:} $G = SU(N)$, $V = V_\square = \mathbb{C}^N$, the vector representation. $H = L^2(S^1,V)$, and $Cliff(H)$ acts on $\mathcal{F}_V$. We have a sequence of maps:
\[ LG \rtimes T_{rot} \to U_{res}(H) \to PU(\mathcal{F}_V) \]
$H_\lambda$ and $H_\mu$ lie in $\mathcal{F}_V^{\otimes \ell}$, and we have equivariant projections $P_\lambda$ and $P_\mu$. The Clifford construction takes $f \in L^2(S^1,V_\square)$ and produces a creation operator $a(f) \ in B(\mathcal{F}_V^{\otimes \ell})$.
For $g \in \widetilde{LG \rtimes T_{rot}}$, the representation $\pi$ on $\mathcal{F}_V$ satisfies $\pi(g) \phi(f) \pi(g)^* = \phi(gf)$. We define $\phi_{\lambda\mu}(f) = P_\mu a(f) P_\lambda^* \in Hom^{bdd}(H_\lambda, H_\mu)$. If $g \in \widetilde{L_I G}$ and $f$ is supported in $I'$, then $gf = f$. This means $\pi(g)\phi(f)\pi(g)^* = \phi(f)$, so $\phi(f) \in Hom_{\widetilde{L_I G}}(H_\lambda, H_\mu)$, and $\Vert \phi(f) \Vert \leq \Vert f Vert_{L^2}$.
Let $V$ be a $G$-module, and $H_\lambda$, $H_\mu$ irreducible primary fields. We write $V^{fin} = V[z,z^{-1}]$, and it has an action of $L^{pol}\mathfrak{g}_\mathbb{C} \rtimes i\mathbb{R}d$. A primary field of charge $V$ and level $\ell$ is a linear map
\[ \phi: V[z,z^{-1}] \otimes H_\lambda^{fin} \to H_\mu^{fin} \]
which is $L^{pol}\mathfrak{g}_\mathbb{C} \rtimes i\mathbb{R}d$-equivariant.
Decompose $H_\lambda^{fin} = \bigotimes_{n \geq 0}^{alg} H_\lambda(n)$. We have a mode decomposition of the primary field $\phi$ by $\phi(v,n) = \phi(v \otimes z^n): H_\lambda^{fin} \to H_\mu^{fin}$. For any $X \in \mathfrak{g}_\mathbb{C}$, we have two conditions:
\begin{enumerate}
\item $[X(m), \phi(v,n)] = \phi(Xv,m+n)$
\item $[d, \phi(v,n)] = -n\phi(v,n)$
\end{enumerate}
These together are equivalent to the map $\phi$ being $L^{pol}\mathfrak{g}_\mathbb{C} \rtimes i\mathbb{R}d$-equivariant.
The second condition implies $\phi(v,n)$ lowers energy by $n$. In particular, $\phi_0: V \otimes V_\lambda \to V_\mu$ is the initial term of $\phi$. The first condition implies $\phi_0$ is $\mathfrak{g}$-equivariant.
\textbf{Lemma:} The map
\[ Hom_{L^{pol}\mathfrak{g}_\mathbb{C} \rtimes i\mathbb{R}d}(V^{fin} \otimes H_\lambda^{fin}, H_\mu^{fin}) \to Hom_G(V \otimes V_\lambda, V_\mu) \]
is injective.
\begin{proof} $H^{fin}_\lambda$ is generated by $H_\lambda(0)$ as a $L^{pol}\mathfrak{g}_\mathbb{C}$-module.
\end{proof}
Let's write $Hom_G^\ell(V \otimes V_\lambda, V_\mu)$ for $Hom_{L^{pol}\mathfrak{g}_\mathbb{C} \rtimes i\mathbb{R}d}(V^{fin} \otimes H_\lambda^{fin}, H_\mu^{fin})$.
\textbf{Proposition:}
$Hom_G^\ell(V_\lambda \otimes V_\mu, V_\nu) = Hom_G(V_\lambda \otimes V_\mu, V_\nu)$ if $\lambda, \mu, \nu$ are admissible at level $\ell$, and at least one of them is minuscule [?].
We specialize to the vector representation of $G = SU(N)$, which is minuscule. I.e., we study $Hom_{SU(N)}(V_\square \otimes V_f, V_g)$. We know that $V_\square \otimes V_f = \bigoplus_{g \in f + \square} V_g$.
Suppose $W = V \otimes \mathbb{C}^\ell$. Then $\bigwedge W = (\bigwedge V)^{\otimes \ell}$. We have an operation $S: W \otimes (\bigwedge W) \to \bigwedge W$ by $w \otimes x \mapsto w \wedge x$. If $V = V_\square$, then for admissible $f$ and $g$, $V_f$ and $V_g$ are subrepresentations of $(\bigwedge V)^{\otimes \ell}$.
\textbf{Lemma:} Let $T \in Hom_{SU(N)}(V_\square \otimes V_f, V_g)$ be a nonzero intertwiner. We can find $SU(N)$-equivariant projections $P_\square: W \to V_\square$, $P_f: \bigwedge W \to V_f$ and $P_g: \bigwedge W \to V_g$, such that $T = P_g S(P_\square^* \otimes P_f^*)$.
\begin{proof} $f = (f_1 \geq f_2 \geq \dots \geq f_N)$, with $f_1 - f_N \leq \ell$. Let $(e_1, \dots, e_N)$ be a basis of $V_\square$. Then
\[ e_f = e_1^{\otimes (f_1-f_2)} \otimes (e_1 \wedge e_2)^{\otimes (f_2-f_3)} \otimes \dots \otimes (e_1 \wedge \dots \wedge e_{N-1})^{(f_{N-1} - f_N)} \otimes (e_1 \wedge \dots \wedge e_N)^{\ell-f_1+ f_N} \in (\bigwedge V)^{\otimes \ell} \]
$e_g$ is defined similarly. If $g_i = f_i$ for $i \neq k$ and $g_k = f_k+1$, then they differ by $\pm e_k$ in the $(f_1-f_k)$ copy of $\bigwedge W$.
\end{proof}
\textbf{Example:} $SU(3)$, $\ell=4$. $f = (4,2,1)$, $g = (4,3,1)$. Then
\[ \begin{aligned}
e_f &= e_1 \otimes e_1 \otimes (e_1 \wedge e_2) \otimes (e_1 \wedge e_2 \wedge e_3) \\
e_g &= e_1 \otimes (e_1 \wedge e_2) \otimes (e_1 \wedge e_2) \otimes (e_1 \wedge e_2 \wedge e_3)
\end{aligned} \]
$P_\square: W = V_\square \otimes \mathbb{C}^\ell \to V_\square$ is projection onto the $(f_1-f_k)$ copy of $V_\square$. $SU(N)e_f$ spans $V_f$, $SU(N) e_g$ spans $V_g$, both as subrepresentations of $\bigwedge W$, and $P_f$ and $P_g$ map back. The map $S(P_\square^* \otimes I): V_\square \otimes \bigwedge W \to \bigwedge W$ induces exterior multiplication by an element of $V_\square$ in the $f_1-f_k$ copy of $\bigwedge V \subset (\bigwedge V)^{\otimes \ell}$.
\textbf{Theorem:} Any $SU(N)$-intertwiner $\phi_0: V_\square \otimes H_\lambda(0) \to H_\mu(0)$ is the initial term of a (unique) vector primary field. All vector primary fields arise as ``compressions of fermions'' so they satisfy $\Vert \phi(f) \Vert \leq C \Vert F \Vert_2$ for all $f \in V[z,z^{-1}]$ and some constant $C$. The map extends continuously to $L^2(S^1, V_\square)$, and satisfies the global equivariance relation.
\begin{proof}
$\bigwedge W = \mathcal{F}_W(0)$, and $W \subset L^2(S^1,W)$ as constant functions. $V_\lambda, V_\mu \subset W$, and $H_\lambda, H_\mu \subset \mathcal{F}_W$, so we have projections $P_\lambda, P_\mu$. We find that
\[\phi_{\lambda\mu}(v,n) := P_\mu a(v \otimes z^n)P_\lambda^*. \]
\end{proof}
\subsection{2:45pm Yoh Tanimoto, Connes fusion}
We already talked about fusion of bimodules. Now I will talk about fusion on the level of von Neumann algberas, together with some discussion of endomorphisms.
We have to work with von Neumann algebras. We can't do it in full generality. Let $M$ be a type $III$ factor.
\textbf{Fact:} Any representation of $M$ on a separable Hilbert space (with a regularity condition) is implemented by a unitary operator (from the Hilbert space that comes with $M$ and the chosen one).
This is a particular property of Type $III$ factors. It does not always happen for other types.
\textbf{Definition:} An $M$-$M$-bimodule is a Hilbert space $X$ with commuting actions of $M$ and $M^{op}$.
If $M$ is represented in $L^2(M)$, then $M^{op} \cong M'$ by $x \mapsto Jx^*J$.
\textbf{Definition:} An endomorphism of $M$ ($*$-,unital) is a homomorphism of $M$ into $M$.
\textbf{Example:} $L^2(M)$ is a trivial bimodule (i.e., structure follows easily from the multiplicative structure in $M$). For $x,y \in M$, $\xi \in L^2(M)$, then $x\cdot \xi\cdot y = xJy^*J\xi$. If $\rho$ is an endomorphism of $M$, then $\rho(x) \cdot \xi \cdot y = \rho(x)Jy^*J\xi$. We call it $X_\rho$.
\textbf{Proposition:} Any bimodule $X$ is unitarily equivalent to some of $X_\rho$.
\begin{proof} As modules of $M^{op}$, $X$ and $L^2(M)$ are equivalent. We may assume that $X = L^2(M)$ as $M^{op}$ modules. The action of $M$ commutes with that of $M^{op}$. Since $(M^{op})' = M$, the image in $M$ is $M$.
\end{proof}
\textbf{Prop:} $X_{\rho_1} \cong X_{\rho_2}$ if and only if there exists $u \in U(M)$ such that $u\rho_1(x)u^* = \rho_2(x)$.
\begin{proof}
($\Rightarrow$) Anything in $M' = M^{op}$ commutes with unitaries in $M$.
($\Leftarrow$) Let $u$ implement the equivalence. $u$ must commute with $M^{op} = M'$.
\end{proof}
Here is a table of putting things together. We use the notation $P \overset{v}{\sim} I$ do say that $v$ implements the isomorphism between the projection $P$ and the identity $I$, which always exists in type $III$.
\begin{tabular}{c|c|c|c}
& direct sum & subobject & \\ \hline
bimodule & $X \oplus Y$ & invariant subspace & fusion \\ \hline
endomorphism & \begin{tabular}{c} $P_1 \perp P_2$, $P_1 + P_2 = I$ \\ $P_1 \overset{v_1}{\sim} I$, $P_2 \overset{v_2}{\sim} I$ \\ $v_1P_1v_1^* + v_2P_2v_2^*$ \end{tabular} & \begin{tabular}{c} $P \in M$, $[P,\rho(x)]=0$ \\ $P \overset{v}{\sim} I$, $vPv^*$ \end{tabular} & composition $\rho_2 \circ \rho_1$
\end{tabular}
Let $X$,$Y$ be bimodules. $\mathcal{X} = Hom(L^2(M)_M,X_M)$, and $\mathcal{Y} = Hom(_ML^2(M),_M Y)$. We consider $\mathcal{X} \otimes \mathcal{Y}$ with an inner product $\langle x_1 \otimes y_1, x_2 \otimes y_2 \rangle = \langle x_2^*x_1y_2^*y_1\Omega, \Omega \rangle$. Here the $x_2^*x_1 \in M$ and $y_2^*y_1 \in M^{op} = M'$. Recall that if $x \in Hom(L^2(M)_M, X_M)$ and $x^* \in Hom(X_M,L^2(M))$, then $x^*x \in Hom(L^2(M)_M,L^2(M)_M)$. Endomorphisms that commute with the right action of $M$ are in $M$.
We still have to check that the inner product is in fact an inner product.
\textbf{Lemma:} $\langle,\rangle$ on $\mathcal{X} \otimes \mathcal{Y}$ is positive definite.
\begin{proof} If $z = \sum_i y_i \otimes y_i$, then
\[ \langle z,z\rangle = \sum_{i,j=1}^n \langle x_i^*x_jy_i^*y_j\Omega, \Omega \rangle. \]
Let $x = x_i^*x_j \in M_n(M)$, so
\[ x = \begin{pmatrix} x_1^* \\ x_2^* \\ \vdots \\ x_n^* \end{pmatrix} \begin{pmatrix} x_1 & x_2 & \cdots & x_n \end{pmatrix} \]
Therefore, $x$ is positive in $M_n(M)$. Applying polar decomposition, $x = a^*a$ where $a \in M_n(M)$. Similarly we can write $y = b^*b$ for some $b \in M_n(M')$. Then
\[ \begin{aligned}
\langle z, z^* \rangle &= \sum_{i,j} \langle x_i^*x_j y_i^*y_j \Omega, \Omega \rangle \\
&= \sum_{i,j} \sum_{p,q} \langle a_{p,i}^*a_{p,j} b_{q,i}^* b_{q,j} \Omega, \Omega \rangle \\
&= \sum_{p,q} \sum_{i,j} \langle a_{p,j} b_{q,j} \Omega, a_{p,i} b_{q,i} \Omega \rangle \\
&= \sum_{p,q} \left\Vert \sum_j a_{p,j} b_{q,j} \Omega \right\Vert^2 \geq 0
\end{aligned}\]
\end{proof}
We define on $\mathcal{X} \otimes \mathcal{Y}$ actions of $M$, and $M^{op}$ by $a \cdot x \otimes y \cdot b = ax \otimes Jb^*Jy$ for $a, b \in M$, $x \otimes y \in \mathcal{X} \otimes \mathcal{Y}$. [here, $ax := x \circ a^*$]
\textbf{Proposition:} These actions are well-defined.
Call the completion of $\mathcal{X} \otimes \mathcal{Y}$ the fusion of $X$ and $Y$. It is written $X \boxtimes Y$.
\textbf{Theorem:} Let $\rho_1, \rho_2$ be endomorphisms of $M$. Then $X_{\rho_1} \boxtimes X_{\rho_2} \cong X_{\rho_1 \circ \rho_2}$.
\begin{proof}
The operator $V: x \otimes y \mapsto \rho_2(x) y \Omega = yx\Omega$ is unitary. We want the show that the action intertwines.
\[ \begin{aligned}
V \cdot a \cdot x \otimes y \cdot b &= V \cdot \rho_1(a)x \otimes Jb^*Jy \\
&= \rho_2(\rho_1(a)x)Jb^*Jy\Omega \\
&= \rho_1(\rho_1(a)\rho_2(x)Jb^*Jy \Omega \\
&= \rho_2\rho_1(a)Jb^*J \rho_2(x)y\Omega \\
&= \rho_2\rho_1(x) \cdot Jb^*J \cdot V x\otimes y
\end{aligned} \]
\end{proof}
\textbf{Corollary:}
If we have three endomorphisms, we get an associator:
\[ X_{\rho_1} \boxtimes (X_{\rho_2} \boxtimes X_{\rho_3}) \cong X_{\rho_3\rho_2 \rho_1} \cong (X_{\rho_1} \boxtimes X_{\rho_2}) \boxtimes X_{\rho_3} \]
\subsection{4:00pm Josiah Thornton, Quantum dimension}
Outline:
\begin{enumerate}
\item[0] Example
\item[1] Semisimple ribbon categories
\item[1'] Example(s)
\item[2] Define Quantum dimension
\item[3] Properties
\item[4] Computations with irreducible positive energy representations of $LSU(2)$.
\end{enumerate}
Let $V = \mathbb{C}^2$. Choose a basis $\{e_1, e_2 \}$, and a dual basis $\{ \epsilon_1, \epsilon_2 \}$ for $V^*$. We have an evaluation map $e: V \otimes V^* \to \mathbb{C}$ given by $\sum a_i \otimes \alpha_i \to \sum\alpha_i(a_i)$. We also have a coevaluation $i: \mathbb{C} \to V \otimes V^*$ given by $1 \mapsto e_1 \otimes \epsilon_1 + e_2 \otimes \epsilon_2$. Composing yields $\mathbb{C} \to V \otimes V^* \to \mathbb{C}$ given by $1 \mapsto e_1 \otimes \epsilon_1 + e_2 \otimes \epsilon_2 \mapsto \epsilon_1(e_1) + \epsilon_2(e_2) = 2$. We find that the dimension of $V$ is 2.
Let $\mathcal{C}$ be a complex linear abelian category which we will give some structure:
\begin{itemize}
\item $\otimes: \mathcal{C} \times \mathcal{C} \to \mathcal{C}$ a bilinear functor.
\item $\alpha_{U,V,W}: (U \otimes V) \otimes W \to U \otimes (V \otimes W)$ functorial isomorphisms
\item An object $I$ with $End(I) = \mathbb{C}$, and functorial isomorphisms $\lambda_V: I \otimes V \to V$ and $\rho_V: V \otimes I \to I$
\end{itemize}
These data describe a monoidal structure [once we have some compatibility, like a pentagon axiom for the associator].
Examples include finite dimensional complex vector spaces, or finite dimensional representations of $SU(N)$. The tensor product on vector spaces gives a monoidal structure. The tensor product on Hilbert spaces with group action satisfies $g(x \otimes y) = gx \otimes gy$.
We can have more structure, such as a braiding: $\sigma_{VW}: V \otimes W \overset{\sim}{\longrightarrow} W \otimes V$. Again this needs to satisfy some commutative diagrams. A monoidal category with well-behaved braiding is called a braided monoidal category.
\textbf{Examples:} We get a trivial braiding induced by the switch map $\tau: V \times W \to W \times V$ given by $(v,w) \mapsto (w,v)$. This is the usual braiding in the cases of complex vector spaces and representations of a group.
There is a notion of rigidity that pertains to duals. For right duals, we have an evaluation map $e_V: V^* \otimes V \to I$ and a coevaluation $i_V: I \to V \otimes V^*$. Right duals have to satisfy the conditions that $V \overset{i_V \otimes id_V}{\longrightarrow} V \otimes V^* \otimes V \overset{id_V \otimes e_V}{\longrightarrow} V$ is the identity map on $V$ and $V^* \to V^* \otimes V \otimes V^* \to V^*$ is the identity on $V^*$.
For the left dual ${}^*V$, we have $e_V': V \otimes {}^*V \to I$ and $i_V': I \to {}^*V \otimes V$. A monoidal category is rigid if it has left and right duals.
Examples: For complex vector spaces, $V^* = Hom(V,\mathbb{C})$. $e_V$ is given by $\sum \alpha_i \otimes a_i \mapsto \sum \alpha_i(a_i)$, and $i_V$ is given by $a \mapsto \sum e_i \otimes \epsilon_i$. For representations of $G$, $(g \cdot f)(v) = f(g^{-1}v)$.
Given a map $f: U \to V$, we get a dual map $f^*$ by $V^* \to V^* \otimes U \otimes U^* \to V^* \otimes V \otimes U^* \to U^*$.
\textbf{Definition:} A ribbon category is a rigid braided tensor category with a functorial isomorphism $\delta_V: V to V^{**}$, such that $\delta_{V \otimes W} = \delta_V \otimes \delta_W$, $\delta_I = id_I$, and $\delta_{V^*} = (\delta_V^*)^{-1}$.
By taking left duals, we get a natural identification between left and right duals.
Let $\mathcal{C}$ be a semisimple [all objects are finite sums of simple objects] ribbon category.
\textbf{Definition:} Let $V \in \mathcal{C}$ and $f \in End(V)$. Define $tr(f)$ to be $I \to V \otimes V^* \overset{f \otimes id}{\longrightarrow} V \otimes V^* \to V^{**} \otimes V^* \to I$.
Trace is multiplicative: $tr(f \otimes g) = tr(f) \cdot tr(g)$.
\textbf{Definition:} The quantum dimension of an object $V$ is defined to be $tr(id_V)$
We will assume that we have finitely many simple objects, and then the quantum dimension of an object is real, but not necessarily positive. When we work with loop groups, our categories will have positive quantum dimension.
\textbf{Calculations:}
For three objects $V_f, V_g, V_h$, we define $N^h_{f,g}:= \dim Hom(V_h, V_f \otimes V_g)$.
\textbf{Theorem:} (Wasserman) $H_f \boxtimes H_g = \bigoplus N^h_{f,g} sign(\sigma_n) H_{h'}$, where $h' = \sigma(h + \delta)-\delta$ is permissible.
\textbf{Corollary:} $H_\ell \boxtimes H_\ell = H_0$, where $\ell$ here is the largest permissible highest weight.
\textbf{Theorem:} For permissible signatures $f$, $H_\square \boxtimes H_f = \bigoplus_{g \in f + \square} H_g$.
\textbf{Corollary:} For $G = SU(2)$ and $1 \leq \lambda \leq \ell-1$, $H_\square \boxtimes H_\lambda = H_{\lambda-1} \oplus H_{\lambda + 1}$.
Now, we consider all irreducible positive energy representations of $LSU(2)$ at level $\ell$, written $H_0, H_1, \dots, H_\ell$. We define $d_i = \dim H_i$.
\textbf{Proposition:} For $0 leq i \leq \ell$, $d_i = d_{\ell-i}$. Also, $d_1 d_i = d_{i-1} + d_{i+1}$.
This can be used to determine all dimensions at all levels. For level 1, get 1 and 1. For level 2, get 1, $\sqrt{2}$, and 1. For level 3, get two simple objects of dimension $\frac{1+\sqrt{5}}{2}$.
\section{Friday, 8/20/10}
\subsection{9:00am Andre}
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