\chapter{Quantum Gravity (IV)}
\begin{exercise}[Orthogonality relations]
Recall the definition of the $6j$ symbols:
$$
\xy
(-8,11)*{};(-4,4)**\dir{-} ?(.5)+(-2,-1)*{\scriptstyle\mathbf{a}};
(0,11)*{};(-4,4)**\dir{-} ?(.5)+(2,-1)*{\scriptstyle\mathbf{b}};
(8,11)*{};(0,-3)**\dir{-} ?(.5)+(2,-1)*{\scriptstyle\mathbf{c}};
(0,-3)*{};(0,-12)**\dir{-} ?(.5)+(2,0)*{\scriptstyle\mathbf{d}};
(-4,4)*{};(0,-3)**\dir{-} ?(.5)+(-2,-1)*{\scriptstyle\mathbf{j}};
\endxy
=\sum_i\left\{\matrix{a&b&i\cr
c&d&j\cr}\right\}
\xy
(-8,11)*{};(0,-3)**\dir{-} ?(.5)+(2,1)*{\scriptstyle\mathbf{a}};
(0,11)*{};(4,4)**\dir{-} ?(.5)+(2,1)*{\scriptstyle\mathbf{b}};
(8,11)*{};(4,4)**\dir{-} ?(.5)+(2,-1)*{\scriptstyle\mathbf{c}};
(0,-3)*{};(0,-12)**\dir{-} ?(.5)+(2,0)*{\scriptstyle\mathbf{d}};
(4,4)*{};(0,-3)**\dir{-} ?(.5)+(2,-1)*{\scriptstyle\mathbf{i}};
\endxy
$$
Express
$$
\xy
(-8,11)*{};(0,-3)**\dir{-} ?(.5)+(2,1)*{\scriptstyle\mathbf{a}};
(0,11)*{};(4,4)**\dir{-} ?(.5)+(2,1)*{\scriptstyle\mathbf{b}};
(8,11)*{};(4,4)**\dir{-} ?(.5)+(2,-1)*{\scriptstyle\mathbf{c}};
(0,-3)*{};(0,-12)**\dir{-} ?(.5)+(2,0)*{\scriptstyle\mathbf{d}};
(4,4)*{};(0,-3)**\dir{-} ?(.5)+(2,-1)*{\scriptstyle\mathbf{i}};
\endxy
=\sum_i\left\{\matrix{?&?&?\cr
?&?&?\cr}\right\}
\xy
(-8,11)*{};(-4,4)**\dir{-} ?(.5)+(-2,-1)*{\scriptstyle\mathbf{a}};
(0,11)*{};(-4,4)**\dir{-} ?(.5)+(2,-1)*{\scriptstyle\mathbf{b}};
(8,11)*{};(0,-3)**\dir{-} ?(.5)+(2,-1)*{\scriptstyle\mathbf{c}};
(0,-3)*{};(0,-12)**\dir{-} ?(.5)+(2,0)*{\scriptstyle\mathbf{d}};
(-4,4)*{};(0,-3)**\dir{-} ?(.5)+(-2,-1)*{\scriptstyle\mathbf{k}};
\endxy
$$
and derive a quadratic orthogonality relation that the $6j$ symbols
satisfy. [Hint: rotate the definition of the $6j$ symbols so that it
goes ``backwards''.]
\end{exercise}
\section{Topological Quantum Field Theories (IV)}
Recall the definitions of (U)TQFTs:
\begin{definition}
A {\bf topological quantum field theory\/} is a symmetric monoidal
functor $Z\colon n\Cob\rightarrow \Vect$.
\end{definition}
\begin{definition}
A {\bf unitary topological quantum field theory\/} is a symmetric,
monoidal $*$-functor $Z\colon n\Cob\rightarrow \Hilb$.
\end{definition}
Having analysed the categories of (un)oriented one-dimensional
cobordisms, we can characterise all one-dimensional (U)TQFTs.
\subsection{One-dimensional Topological Quantum Field Theories (II)}
In one dimension we have the categories $1\Cob$ and ${\rm Un}1\Cob$,
and TQFTs will map them into $\Vect$ or $\Hilb$.
Assume $Z\colon 1\Cob\rightarrow\Hilb$ is given. Then, we have
\begin{itemize}
\item $0$-dimensional oriented manifolds map to Hilbert spaces:
$$
Z(\bullet)=H\qquad{\rm and}\quad Z(\circ)=H'.
$$
We will be able to find a relation between $H$ and $H'$ when we
consider morphisms.
\item The only morphisms we have are the identity morphisms:
$$
\xy
(-9,-5)*{\bullet};(-9,5)*{\bullet};**\dir{-} ?(.6)*\dir{>};
\endxy
\buildrel Z\over\longrightarrow
1_H
\qquad
\xy
(-15,5)*{\circ};(-15,-5)*{\circ};**\dir{-} ?(.6)*\dir{>};
\endxy
\buildrel Z\over\longrightarrow
1_{H'}
$$
\item The union of manifolds maps to the tensor product:
$Z(\amalg)=\otimes$, so
$$
Z(n\bullet+m\circ)=H^{\otimes n}\otimes{H'}^{\otimes m}
$$
and the tensor product of identity morphisms is the identity
morphism. We also have $Z(\emptyset)=\K$, where $\K=\R$ or $\C$ is
the base field of our vector spaces.
\item Duality is given by $\bullet^*=\circ$, so
$H'=Z(\circ)=Z(\bullet^*)=Z(\bullet)^*=H^*$. We also have
$$
\xy
(-3,1)*{\circ};(3,1)*{\bullet};**\crv{(-0,-4)} ?(.6)*\dir{>};
\endxy
\buildrel Z\over\longrightarrow
\matrix{e_H\colon &H^*\otimes H&\rightarrow&\K\cr
&\alpha\otimes v &\mapsto &\alpha(v)\cr}
\qquad{\rm and}\quad
\xy
(9,-1)*{\bullet};(15,-1)*{\circ};**\crv{(12,4)} ?(.6)*\dir{>};
\endxy
\buildrel Z\over\longrightarrow
\matrix{i_H\colon &\K&\rightarrow&H\otimes H^*\cr
& 1&\mapsto &e_i\otimes e^i\cr}
$$
where $(e_i)$ is any basis of $H$ and $(e^i)$ is the associated
dual basis of $H^*$. Observe that
$$
\xy
(21,0)*\cir<8pt>{};
\endxy
\buildrel Z\over\longrightarrow
e^i(e_i)=\dim H=\dim H^*.
$$
\item So far the whole construction works for $\C\Vect$, but if we want
to preserve adjoints we must map into $\C\Hilb$. Recall that
adjoints are
$$
\xy
(-3,1)*{\circ};(3,1)*{\bullet};**\crv{(-0,-4)} ?(.6)*\dir{>};
\endxy
\buildrel \dagger\over\rightarrow
\xy
(15,-1)*{\bullet};(9,-1)*{\circ};**\crv{(12,4)} ?(.6)*\dir{>};
\endxy\qquad{\rm and}\quad
\xy
(9,-1)*{\bullet};(15,-1)*{\circ};**\crv{(12,4)} ?(.6)*\dir{>};
\endxy
\buildrel \dagger\over\rightarrow
\xy
(3,1)*{\circ};(-3,1)*{\bullet};**\crv{(-0,-4)} ?(.6)*\dir{>};
\endxy,
$$
so
$$
e_H^\dagger=i_{H^*}\qquad{\rm and}\quad i_H^\dagger=e_{H^*}.
$$
\end{itemize}
In conclusion, $1\Cob$ admits topological quantum field theories into
$\C\Vect$ or $\R\Vect$ which are determined by the single datum
$Z(\bullet)$. Unitary TQFTs incorporate the adjoint operation and map
into $\C\Hilb$ or $\R\Hilb$. The image of a TQFT can be given a
Hilbert space structure by adding the image of the $1\Cob$ adjoint,
i.e., the relations $e_H^\dagger=i_{H^*}$ and $i_H^\dagger=e_{H^*}$.
The moral of this story is that an $n{\rm TQFT}$ associates a complex
number to each closed $n$-dimensional manifold. This is called the
partition function of the manifold. If the manifold is self-adjoint,
this will be a real number. In the case of $1{\rm TQFTs}$, a closed
manifold is a disjoint union of circles, which are self-adjoint and
evaluate to the dimension of $H=Z(\bullet)$, which is a real number.
The case of unoriented cobordisms is simpler, and it can be obtained
from the previous one by adding relations:
\begin{itemize}
\item There is only one basis Hilbert space $H=Z(\bullet)$, which is
{\it self-dual\/}: $\bullet^*=\bullet$. This means that $H$ is
isomorphic to $H^*$, which requires $H$ to be a real Hilbert
space, as in the complex case $H$ is anti-isomorphic to $H^*$. Note
that in this case it cannot be merely a vector space.
\item The evaluation map is a bilinear inner product:
$$
\xy
(-3,1)*{\bullet};(3,1)*{\bullet};**\crv{(-0,-4)};
\endxy
\buildrel Z\over\longrightarrow
e_H=\colon H\otimes H\rightarrow\R
$$
and,
\item as we observed last time, duals and adjoint coincide, so $\R\Vect$
and $\R\Hilb$ coincide, where $\R\Hilb$ has the inner product
given by the evaluation.
\end{itemize}
Therefore, we have $Z\colon {\rm Un}1\Cob\rightarrow\R\Hilb=\R\Vect$.
\subsection{Associative algebras}
In any dimension there is a God-given closed manifold, and that is the
$n$-sphere (the boundary of an $(n+1)$-disk). The $0$-sphere is
$$
S^0\simeq
\xy
(0,0)*{\bullet};(5,0)*{\circ}
\endxy
\buildrel Z\over\longrightarrow
H\otimes H^*\simeq\End(H)\colon =\Hom(H,H).
$$
That is, $S^0$ maps to the associative algebra of endomorphisms of
$H=Z(\bullet)$. Multiplication is given by
$$
e_\bullet=\xy
(-3,1)*{\bullet};(3,1)*{\bullet};**\crv{(-0,-4)};
\endxy\qquad{\rm via}\quad
\xy
(-9,5)*{\bullet};(-3,-5)*{\bullet};**\crv{(-3,0)&(-9,0)} ?(.5) *\dir{<};
(3,5)*{\bullet};(-3,5)*{\circ};**\crv{0,0} ?(.5) *\dir{<};
(9,5)*{\circ};(3,-5)*{\circ};**\crv{(3,0)&(9,0)} ?(.5) *\dir{>};
\endxy
$$
and the unit is
$$
i_\bullet=\xy
(9,-1)*{\bullet};(15,-1)*{\circ};**\crv{(12,4)};
\endxy
$$
This defines an associative unitary algebra. We already proved this in
a different context when we were developing the diagrammatic methods
for linear algebra.
\subsection{Two-dimensional Topological Quantum Field Theories (I)}
Two-dimensional topological quantum field theories map (un)oriented
$2$-dimensional cobordisms to Vector or Hilbert spaces. Now, the
objects in $2\Cob$ are one-dimensional compact manifolds, that is,
disjoint unions of circles. By analogy with the $0$-spheres of the
previous paragraph, we would expect these circles to be elements of an
algebra. It turns out that all $2{\rm TQFTs}$ actually map not only
into Hilbert spaces, but into Hilbert algebras.
Let us now study oriented two-dimensional cobordisms and the
possibilities for $2{\rm TQFTs}$. First, a note about orientation. As
we know, an oriented $2$-cobordism will be an oriented surface with
boundary, which will be divided into an ``input'' and an ``output''
part. We orient the input in such a way that a positively oriented
basis for the input, followed by an outward vector, is a positively
oriented basis for the $2$-cobordism. The output is oriented
similarly, but with an inward vector completing the basis. In
pictures:
$$
\xy
(0,3)*{};(2,3)**\dir{-} ?(1) *\dir{>};
(0,3)*{};(0,5)**\dir{-} ?(1) *\dir{>};
(0,5)*{};(0,3)*{};{\xycircle(3,2){-}};
(0,0)*{};(2,0)**\dir{-} ?(1) *\dir{>};
(0,0)*{};(0,2)**\dir{-} ?(1) *\dir{>};
(-3,5)*{};(-3,-5)**\dir{-};
(3,5)*{};(3,-5)**\dir{-};
(0,-3)*{};(2,-3)**\dir{-} ?(1) *\dir{>};
(0,-3)*{};(0,-1)**\dir{-} ?(1) *\dir{>};
(0,-5)*{};(0,-3)*{};{\xycircle(3,2){-}};
\endxy
$$
has both circles positively oriented.
Any two-dimensional cobordism can be constructed from the following
elements:
\begin{itemize}
\item Two units
$$
\xy
(0,0)*{};(12,0)*{};{\xycircle(6,4){-}};
(-6,0)*{};(6,0)**\crv{(0,20)};
(0,4)*\dir{>};
\endxy
\buildrel Z\over\longrightarrow
\xymatrix{\C\ar[d]^\iota\\
H\\}
\qquad{\rm and}\quad
\xy
(0,0)*{};(12,0)*{};{\xycircle(6,4){-}};
(-6,0)*{};(6,0)**\crv{(0,20)};
(0,4)*\dir{<};
\endxy
\buildrel Z\over\longrightarrow
\xymatrix{\C\ar[d]^\varepsilon\\
H^*\\}
$$
\item two multiplications
$$
\buildrel Z\over\longrightarrow
\xymatrix{H\otimes H\ar[d]^m\\
H\\}
$$
\end{itemize}
...etc...
\chapter{Quantum Mechanics from a Category Theoretic Viewpoint (IV)}
\section{Fields on $(1+1)$-dimensional spacetime (II)}
Last time we saw that (1+1) vacuum Maxwell equations on a cylinder
were secretly the same thing as the point particle on a line. (This is
a great thing because we know how to quantise the particle.) Now let's
play around with the theory. Some things we can do are changing the
gauge group or deforming the manifold.
We saw that, on $\R\times S^1(l)$ (a cylinder with perimeter $l$), the
Maxwell connection takes the form $A=(a+elt)dx$, where $e$ is a
constant. The connection has curvature $F=dA=el d\rm vol$, and
${*F}=e$. Apart from $e$ there is another gauge-invariant quantity,
which is the integral of $A$ around a loop going around the cylinder
once. We call this constant
$$
a=\oint_{t=t_0}A.
$$
We saw that, in temporal gauge, $a(t)=a+elt$, which comes from the
equations
$$
\dot a=el\qquad{\rm and}\quad\dot e=0.
$$
This motivates the identification with a point particle of position
$a$, momentum $e$ and mass $1/l$,
with equations
$$
\dot q=p\qquad{\rm and}\quad\dot p=0.
$$
The classical solution is determined by a pair $(a,e)$ or $(q,p)$
(initial conditions). This is also good because the Hamiltonians match
up, as we saw last time. The quantised maxwell equations are therefore
$$
i{\partial\over\partial t}\psi(a)=-{l\over 2}{\partial\over\partial
a}\psi(a).
$$
Now let's consider Maxwell's theory with gauge group $U(1)$ as opposed
to $\R^+$. If the connection is a $U(1)$ connection, on each point of
$M=\R\times S^1$ we have a fiber (not canonically) isomorphic to the
gauge group, and the connection tells us how to parallel transport
objects around a loop.
The only context in which we encountered this parallel transport was
in computing
$$
a=\oint_{t=t_0}A\in\frak u(1).
$$
If the group is $U(1)$, $a$ is only determined up to $2\pi i$, because
we can take $f(x)=e^{2\pi inx}$, and $A\rightarrow A'=A-2\pi i n$. In
fact, the holonomy of the connection around the loop is technically a
group element representing the effect of parallel transport on the
various representations of the group, and this is written as
$$
e^{\oint_{t=t_0}A}\in U(1).
$$
Let's just call this $a$. Then, the role of $\dot a$ is played by
$a^{-1}\dot a$.
A solution of Maxwell's $U(1)$ theory up to gauge transformations is,
then, $(a,e)\in(\R/2\pi\Z)\times\R$. This is a free point particle on
a circle or radius $1$ with the same Hamiltonian as the free particle,
but Schr\"odinger's equation will be able to tell the difference.
The theory is quantised by taking the Hilbert space ${\cal
H}=L^2(S^1)$ and the Hamiltonian $H=-{l\over 2}{d^2\over
da^2}$. Schr\"odinger's equation is
$$
\Bigl(i{\partial\over\partial t}+{l\over 2}{\partial^2\over \partial
a^2}\Bigr)\psi(t,a)=0 \qquad{\rm with}\quad \psi(t,a)=\psi(t,a+2\pi).
$$
So the exponentials $\psi(t,x)=\exp\bigl(-i(Et-kx\bigr)$ are solutions if
$$
E={l\over 2}k^2\qquad{\rm with}\quad k\in\Z.
$$
Note how this differs from the solution in the case of a particle in a
line, with a continuous spectrum and generalised eigenvectors.
We therefore have two theories that no local experiments can
distinguish, but quantum mechanics or global experiments can. Of
course, the energy $E$ in this case is the energy of the whole
universe, so that is not a local observable and the fact that quantum
gravity can tell the difference does not contradict the fact that
$\R^+$ and $U(1)$ theory are locally indistinguishable.
We now want to relax the three fixed structures we have: the topology,
the metric and the foliation into space-like slices. First notice that
classically, $2D$ vacuum Maxwell theory has a lot of symmetry that is
not entirely obvious: the whole theory is invariant under all
area-preserving diffeomorphisms of space-time.
For a proof, all we need to show is that the action itself is
invariant under area-preserving diffeomorphisms.
$$
S={1\over 2}\int F\wedge{*F}.
$$
Under a diffeomorphism $\phi$, $A$ is pulled back to $\phi^*(A)$ and
so is the curvature: $F\rightarrow\phi^*(F)$. So we need to calculate
$$
S={1\over 2}\int_M F\wedge{*F}={1\over 2}\int_M
\phi^*(F')\wedge{*\phi^*(F')}
$$
but notice that any $2$-form is equal to a function times the volume
form, so we have $F'=E'd{\rm Vol}'$ and, since $\phi$ is
area-preserving, $\phi^*(d{\rm Vol}')=d{\rm Vol}$. Therefore,
\begin{eqnarray*}
S &=& {1\over 2}\int_M
\phi^*(F')\wedge{*\phi^*(F')}={1\over 2}\int_M
\phi^*(E'd{\rm Vol}')\wedge{*\phi^*(E'd{\rm Vol}')}={1\over 2}\int_M
\phi^*(E')^2d{\rm Vol}=\\
&=& {1\over 2}\int_M
\phi^*({E'}^2)d{\rm Vol}={1\over 2}\int_M
\phi^*({E'}^2d{\rm Vol}'),\\
\end{eqnarray*}
since $*d{\rm Vol}=1$.
We now use
$$
\int_M\phi^*\omega=\int_{\phi^{-1}(M)}\omega,
$$
and we obtain
$$
S={1\over 2}\int_{\phi(M)}{E'}^2d{\rm Vol}'={1\over 2}\int_MF'\wedge{*F'}.
$$
Now, there is a theorem of Moser asserting that, if there are two
non-intersecting circles, there is an area-preserving diffeomorphism
that takes them to two ``nice'' ($t=\rm const$) circles. So trying to
understand time evolution between any two closed curves is not harder
than on a very simple cylinder.
Therefore the time-evolution operator $U\colon L^2(\R)\rightarrow
L^2(\R)$ corresponding to time evolution from now to then, is just a
function of the area of the spacetime between ``now'' and ``then''!
$$
U=e^{-iAH},
$$
where $A$ is the area of the spacetime enclosed by the two
curves. So, in $(1+1)$-Yang-Mills theory, time {\it is\/} area!
What we will do next is do $t\rightarrow it$, which will turn our
theory into statistical mechanics or stochastic processes... Then we
will be able to study more general spacetimes, like the ones we were
talking about on track $1$:
$$
(\hbox{diagram: trousers})
$$
This will give us something not unlike the topological quantum field
theories of Track~$1$.