\chapter{Quantum Gravity}
The orthogonality relation gets its name from the following inner
product on the space of intertwining operators:
$$
\left\langle
\xy
(0,0)*+{T} *\cir{}="T";
(-5,5)**\dir{-};
(0,4)*{\ldots};
"T";(5,5)**\dir{-};
"T";(0,-5)**\dir{-};
\endxy
,
\xy
(0,0)*+{S} *\cir{}="T";
(-5,5)**\dir{-};
(0,4)*{\ldots};
"T";(5,5)**\dir{-};
"T";(0,-5)**\dir{-};
\endxy
\right\rangle
=
\xy
(0,5)*+{T^*} *\cir{}="T";
(0,-5)*+{S} *\cir{}="S";
"T";"S";**\crv{(-5,0)};
(0,0)*{\cdots};
"T";"S";**\crv{(5,0)};
"T";"S";**\crv{(5,-10)&(10,0)&(5,10)};
\endxy
$$
With respect to this inner product, the
$$
\xy
(-8,11)*{};(-4,4)**\dir{-} ?(.5)+(-2,-1)*{\scriptstyle\\a};
(0,11)*{};(-4,4)**\dir{-} ?(.5)+(2,-1)*{\scriptstyle\\b};
(8,11)*{};(0,-3)**\dir{-} ?(.5)+(2,-1)*{\scriptstyle\\c};
(0,-3)*{};(0,-12)**\dir{-} ?(.5)+(2,0)*{\scriptstyle\\d};
(-4,4)*{};(0,-3)**\dir{-} ?(.5)+(-2,-1)*{\scriptstyle\\j};
\endxy
$$
form an orthogonal basis of intertwiners. The orthogonality relation
just says that the $6j$ symbols are an orthogonal matrix.
\begin{exercise}
Show that the orthogonality relation for the $6j$ symbols corresponds
to this move on $3d$ triangulations:
$$
\xy
(-5,0)*\dir{*};
(0,10)**\dir{-} ?(1) *\dir{*};
(5,0)**\dir{-} ?(1) *\dir{*};
(0,-10)**\dir{-} ?(1) *\dir{*};
(-5,0)**\dir{-};(5,0)**\dir{-};
\endxy
\Leftrightarrow
\xy
(0,10)*\dir{*};
(5,0)**\dir2{-} ?(1) *\dir{*};
(0,-10)**\dir2{-} ?(1) *\dir{*};
(-5,0)**\dir2{-} ?(1) *\dir{*};
(5,0)**\crv{(0,-5)};(-5,0)**\dir{.};(0,10)**\dir2{-};(0,-10)**\dir2{-};
\endxy
$$
This is the $3d$ analogue of the following move in two dimensions:
$$
\xy
(0,10)*\dir{*};
(0,0)**\dir{-} ?(1) *\dir{*};
(0,-10)**\dir{-} ?(1) *\dir{*};
\endxy
\Leftrightarrow
\xy
(0,10)*\dir{*};
(5,0)**\dir{-} ?(1) *\dir{*};
(0,-10)**\dir{-} ?(1) *\dir{*};
(-5,0)**\dir{-} ?(1) *\dir{*};
(0,10)**\dir{-};(0,-10)**\dir2{-};
\endxy
$$
[Hint: translate the move into an equation between spin networks using
these rules:
\begin{enumerate}
\item one tet-net per tetrahedron
$$
\xy
(0,0)*\dir{*}="Y";
(-7,4)*\dir{*}="a";
**\dir{-} ?(.5)+(-1,-1)*{\scriptstyle\\i};
(0,-9)*\dir{*}="b";
"Y"**\dir{-} ?(.5)+(2,0)*{\scriptstyle\\q};
(7,4)*\dir{*}="c";
"Y"**\dir{-} ?(.5)+(-1,2)*{\scriptstyle\\m};
"a";"b"**\crv{(-7,-4)} ?(.5)+(-2,-1)*{\scriptstyle\\p};
"c";"b"**\crv{(7,-4)} ?(.5)+(2,-1)*{\scriptstyle\\k};
"c";"a"**\crv{(0,9)} ?(.5)+(0,2)*{\scriptstyle\\j};
\endxy
$$
\item the reciprocal of a theta net for each triangle
$$
{1\over
\xy
(-5,0)*\dir{*}="*";
(5,0)*\dir{*}="o";**\dir{-} ?(.5)+(0,-2)*{\scriptstyle\\j};
"*";"o";**\crv{(0,10)} ?(.5)+(0,-2)*{\scriptstyle\\a};
"*";"o";**\crv{(0,-10)} ?(.5)+(0,-2)*{\scriptstyle\\b};
\endxy}
$$
\item one loop per edge.
$$
\xy
(-2,0)*\cir<8pt>{};
(3,0)*{\scriptstyle\\j};
\endxy
$$
\end{enumerate}
Using the relation between tet nets and $6j$ symbols, show that this
move is equivalent to the orthogonality identity!]
\end{exercise}
This move together with the $2-3$ move are insufficient to go from any
triangulation to any other, because the number of vertices of the
triangulation is preserved. The $1-4$ move allows us to change the
number of vertices, but it turns out that, when it is translated into
spin networks it diverges because we get to sum over all
representations of $SL(2,\C)$. This problem is solved by introducing
quantum groups, which have only finitely many irreducible
representations.
\section{Topological Quantum Field Theories (V)}
\subsection{Two-Dimensional Topological Quantum Field Theories}
Suppose we have
$$
\matrix{Z\colon&2\Cob&\rightarrow&\Vect\cr
&\xy
(0,0)*\xycircle(3,2){};
\endxy&\mapsto &{\cal A}\cr
&{\xy
(-7,7)*\xycircle(3,2){};
(7,7)*\xycircle(3,2){};
(0,-7)*\xycircle(3,2){};
(-10,7)*{};(-3,-7)**\crv{(-11,3)&(-2,-3)};
(10,7)*{};(3,-7)**\crv{(11,3)&(2,-3)};
(-4,7)*{};(4,7)**\crv{(0,-1)};
\endxy}&\mapsto& m\colon {\cal A}\otimes {\cal A}\rightarrow {\cal A}\cr
&{\rm cap}&\mapsto&\iota\colon\C\rightarrow {\cal A}\cr
&{\rm cup}&\mapsto&\tau\colon {\cal A}\rightarrow\C\cr}
$$
This is an Abelian associative algebra with unit $\iota$, and the
additional structure $\tau$ gives rise to an inner product
$g=\tau\circ m$:
$$
(\hbox{diagram: U})
\buildrel Z\over\mapsto
g\colon {\cal A}\otimes {\cal A}\rightarrow\C.
$$
As usual, $g$ determines a mapping $\sharp\colon {\cal A}\rightarrow
{\cal A}^*$, and $g$ is non-degenerate if, and only if, $\sharp$ is an
isomorphism. The map $\sharp$ is obtained from $g$ by composition with
the unit endomorphism, which we represent as a bundt:
$$
\displaylines{
({\rm diagram})\cr
{\cal A}\simeq {\cal A}\otimes\C
\buildrel {1_{\cal A}\otimes i_{\cal A}}\over\longrightarrow
{\cal A}\otimes {\cal A}\otimes {\cal A}^*
\overbrace{\buildrel {m\otimes 1_{{\cal A}^*}}\over\longrightarrow
{\cal A}\otimes {\cal A}^*
\buildrel {\tau\otimes
1_{{\cal A}^*}}\over\longrightarrow}^{g\otimes 1_{\cal A}^*}
\C\otimes {\cal A}^*\simeq {\cal A}^*\cr}
$$
But now we observe that the cobordism corresponding to $\sharp$ is
invertible, and its inverse (corresponding to $\flat$) is
$$
\hbox{(diagram and proof)}
$$
Therefore, $g$ is a non-degenerate metric. An algebra with a
non-degenerate inner product is called a Frobenius algebra.
\begin{theorem}
If $Z$ is a $2dTQFT$, then $Z(\xy(0,0,)*\xycircle(3,2){}\endxy)$ is a
{\bf commutative Frobenius algebra\/} with the operations described
above. Conversely, give any commutative Frobenius algebra ${\cal A}$
there is a unique $2dTQFT$ such that
$Z(\xy(0,0,)*\xycircle(3,2){}\endxy)={\cal A}$.
\end{theorem}
Uniqueness is proven by showing that the multiplication on ${\cal
A}^*$ and the comultiplications on ${\cal A}$ can be obtained from the
product on ${\cal A}$ by means of $\sharp$ and $\flat$.
Existence is harder, because one has to show that if $M$ and $M'$ are
diffeomorphic cobordisms obtained by composing elementary cobordisms
in different ways, then $Z(M)=Z(M')$. This gives rise to all sorts of
identities that a Frobenius algebra must satisfy, for example,
$$
\hbox{flat torus}\simeq\hbox{vertical torus}\qquad\Leftrightarrow\quad
\xymatrix{{\cal A}\ar[d]^\delta&\C\ar[l]^\iota\ar[d]^{i_{\cal A}}\\
{\cal A}\otimes{\cal A}\ar[d]^m&{\cal A}\otimes {\cal A}^*
\ar[d]^{e_{\cal A}}\\\
{\cal A}\ar[r]^\tau&\C\\}
$$
The primary tool for carrying out this proof is ``serf theory''.
As an example, consider the matrix algebra ${\cal A}=M_n(\C)$. The
most general bilinear form is of the form $g(A,B)=\tr(\tau AB)$, where
$\tau$ is some matrix. We require that $g(A,B)=g(B,A)$, so
$\tr(\tau\bigl[A,B]\bigr)=0$ for all $A,B$. Let $\tau=\tau^j_i E^i_j$,
where $E^i_j$ are the elementary matrices of ${\cal A}$, and consider
$A=E^i_j$ and $B=E^k_l$. Then, $g(A,B)=\tr(\tau^h_m E^m_h E^i_j
E^k_l)=\tau^i_l\delta^k_j$, so
$\tau^i_l\delta^k_j=\tau^k_j\delta^i_l$. This implies that
$\tau^i_j=\alpha\delta^i_j$ for some $\alpha$, so we have
$\tau_\alpha(\cdot)=\alpha\tr(\cdot)$.
\begin{theorem}
Any Frobenius algebra is isomorphic to a direct sum of algebras of
this sort.
\end{theorem}
\begin{theorem}[Wedderburn]
Any simple Frobenius algebra is isomorphic to a matrix algebra.
\end{theorem}
\begin{corollary}
Any commutative Frobenius algebra is of the form $\oplus(\C,\alpha)$,
with component-wise addition and multiplication; alternatively, any
commutative Frobenius algebra is an algebra of diagonal matrices with
an invertible diagonal matrix as its $\alpha$.
\end{corollary}
In higher dimensions, this ``hard knuckles'' approach of enumerating
all manifolds gets pretty hard pretty quick, so in higher dimensions
it is necessary to switch to ``lattice TQFTs''. This is a reflection
of the fact that only in two dimensions is the problem of classifying
all compact manifolds completely solved. In higher dimensions, one
resorts to various combinatorial-topological techniques like
triangulations, cellulations, CW-complexes, handle-body theory,
Heeg{\aa}rd decompositions, and so on, to obtain algorithms that allow
one to determine whether two manifolds can be homeomorphic. In a
similar spirit, we will specify TQFTs by giving an algorithm for
extracting an algebraic structure from a manifold specified
combinatorially. The trick will be to prove that the result does not
depend on the combinatorial presentation, and to this effect we will
have to take theorems like that of Pachner, which give conditions for
two combinatorial presentations to be equivalent, and translate the
theorems into relations that the algebraic objects must satisfy. This
is what we have been doing so far in our exercises about $6j$ symbols
and Pachner moves.
\section{Lattice Field Theory}
Until we get more sophisticated, we will just think of a {\bf lattice
field theory\/} ad a recipe to get an $nTQFT$ from a combinatorial
presentation (usually a triangulation) of an $n$-dimensional
cobordism. Triangulation independence has to be checked using the
Pachner moves. There is much more to lattice field theory than this,
but we have to start somewhere, and this seems like a good starting
point.
Let's illustrate the principle in the $2d$ case, where we already know
the answer, by computing $Z(M)\colon\C\rightarrow\C$ for a closed
$2$-manifold. Let's follow the procedure step by step:
\begin{enumerate}
\item Triangulate $M$:
$$
(\hbox{diagram: sphere})\Rightarrow
\xy
(0,6)*\dir{*};
(0,-6)**\dir{-} ?(1) *\dir{*};
(-6,0)**\dir{-} ?(.6) *\dir{>>} ?(1) *\dir{*};
(0,6)**\dir{-} ?(.5) *\dir{>} ?(1) *{};
(6,0)**\dir{-} ?(.5) *\dir{<} ?(1) *\dir{*};
(0,-6)**\dir{-} ?(.4) *\dir{<<} ?(1) *{};
\endxy
\qquad{\rm or}\quad
(\hbox{diagram: torus})\Rightarrow
\xy
(0,6)*\dir{*};
(0,-6)**\dir{-} ?(1) *\dir{*};
(-6,0)**\dir{-} ?(.5) *\dir{>} ?(1) *\dir{*};
(0,6)**\dir{-} ?(.6) *\dir{>>} ?(1) *{};
(6,0)**\dir{-} ?(.5) *\dir{<} ?(1) *\dir{*};
(0,-6)**\dir{-} ?(.4) *\dir{<<} ?(1) *{};
\endxy
$$
\item There are two things one can do at this stage, and they determine
how the next steps are phrased, but it will be clear that they are
actually the same thing:
\begin{itemize}
\item ``Explode the triangulation'' (following Fukuma and Hosano).
$$
\xy
(0,6)*\dir{*};
(0,-6)**\dir{-} ?(1) *\dir{*};
(-6,0)**\dir{-} ?(.6) *\dir{>>} ?(1) *\dir{*};
(0,6)**\dir{-} ?(.5) *\dir{>} ?(1) *{};
(6,0)**\dir{-} ?(.5) *\dir{<} ?(1) *\dir{*};
(0,-6)**\dir{-} ?(.4) *\dir{<<} ?(1) *{};
\endxy
\Rightarrow
\xy
(0,6)*\dir{*};
(0,-6)**\dir2{-} ?(1) *\dir{*};
(-6,0)**\dir{--} ?(1) *\dir{*};
(0,6)**\dir{.};
(6,0)**\dir{.} ?(1) *\dir{*};
(0,-6)**\dir{--};
\endxy
\qquad{\rm or}\quad
\xy
(0,6)*\dir{*};
(0,-6)**\dir{-} ?(1) *\dir{*};
(-6,0)**\dir{-} ?(.5) *\dir{>} ?(1) *\dir{*};
(0,6)**\dir{-} ?(.6) *\dir{>>} ?(1) *{};
(6,0)**\dir{-} ?(.5) *\dir{<} ?(1) *\dir{*};
(0,-6)**\dir{-} ?(.4) *\dir{<<} ?(1) *{};
\endxy
\Rightarrow
\xy
(0,6)*\dir{*};
(0,-6)**\dir2{-} ?(1) *\dir{*};
(-6,0)**\dir{--} ?(1) *\dir{*};
(0,6)**\dir{.};
(6,0)**\dir{--}?(1) *\dir{*};
(0,-6)**\dir{.};
\endxy
$$
\item Take the ``dual cellulation'' of the triangulation.
$$
\xy
(0,6)*\dir{*};
(0,-6)**\dir{-} ?(1) *\dir{*};
(-6,0)**\dir{-} ?(.6) *\dir{>>} ?(1) *\dir{*};
(0,6)**\dir{-} ?(.5) *\dir{>} ?(1) *{};
(6,0)**\dir{-} ?(.5) *\dir{<} ?(1) *\dir{*};
(0,-6)**\dir{-} ?(.4) *\dir{<<} ?(1) *{};
\endxy
\Rightarrow
\xy
(0,6)*\dir{o};
(0,-6)**\dir{.} ?(1) *\dir{o};
(-6,0)**\dir{.} ?(1) *\dir{o};
(0,6)**\dir{.};
(6,0)**\dir{.}?(1) *\dir{o};
(0,-6)**\dir{.};
(2,0)*\dir{*};
(4,2)*{};**\dir{-};
(1,-5)*{};**\dir{-};
(-2,0)**\dir{-} ?(1) *\dir{*};
(-4,2)*{};**\dir{-};
(-1,-5)*{};**\dir{-};
\endxy
\qquad{\rm or}\quad
\xy
(0,6)*\dir{*};
(0,-6)**\dir{-} ?(1) *\dir{*};
(-6,0)**\dir{-} ?(.5) *\dir{>} ?(1) *\dir{*};
(0,6)**\dir{-} ?(.6) *\dir{>>} ?(1) *{};
(6,0)**\dir{-} ?(.5) *\dir{<} ?(1) *\dir{*};
(0,-6)**\dir{-} ?(.4) *\dir{<<} ?(1) *{};
\endxy
\Rightarrow
\xy
(0,6)*\dir{o};
(0,-6)**\dir{.} ?(1) *\dir{o};
(-6,0)**\dir{.} ?(1) *\dir{o};
(0,6)**\dir{.};
(6,0)**\dir{.}?(1) *\dir{o};
(0,-6)**\dir{.};
(2,0)*\dir{*};
(4,2)*{};**\dir{-};
(5,-1)*{};**\dir{-};
(-2,0)**\dir{-} ?(1) *\dir{*};
(-1,5)*{};**\dir{-};
(-2,-4)*{};**\dir{-};
\endxy
$$
(Here we are trying to use the distance from the intersection
of primal and dual edges to the primal vertices as a way to
indicate how the edges of the primal triangulation are to be
identified. We see that for the triangulation of the sphere
there are three dual faces, three dual edges and two dual
vertices; while for the triangulation of the torus there is
one dual face, three dual edges and two dual vertices. Note
that the dual vertices are trivalent.)
\end{itemize}
\item Choose a vector space $V$, and elements $c\in V\otimes V\otimes V$
and $g\in V^*\otimes V^*$. If you are a physicist, or if you are
using abstract index notation, you are allowed to write $c^{ijk}$
and $g_{ij}$.
\item Now we come to defining the linear operators associated to each
triangulation. Both approaches, the primal and the dual,
illuminate each other.
\begin{enumerate}
\item Label the three edges around each triangle with
$i_\Delta,j_\Delta,k_\Delta$, where $\Delta$ is a label
indicating to which triangle the edge belongs. This is why we
exploded the triangulation by duplicating the edges. Then, for
each triangle $\Delta$ write $c^{i_\Delta j_\Delta k_\Delta}$,
if $(i,j,k)$ is the order in which you encounter the labels as
you go around the positively oriented boundary of the
triangle, and for each exploded edge shared by triangles
$\Delta$ and $\Delta'$ write $g_{i_\Delta l_{\Delta'}}$:
$$
\xy
(0,6)*\dir{*};
(0,-6)**\dir2{-} ?(.5) +(-2,0)*{\scriptstyle i}
?(.5) +(2,0)*{\scriptstyle i'}
?(1) *\dir{*};
(-6,0)**\dir{-} ?(.5) +(-1,-1)*{\scriptstyle j}
?(1) *\dir{*};
(0,6)**\dir{-} ?(.5) +(-1,1)*{\scriptstyle k}
?(1) *{};
(6,0)**\dir{-} ?(.5) +(1,1)*{\scriptstyle k'}
?(1) *\dir{*};
(0,-6)**\dir{-} ?(.5) +(1,-1)*{\scriptstyle j'}
?(1) *{};
\endxy
\Rightarrow c^{ijk}c^{i'k'j'}g_{ii'}g_{jj'}g_{kk'}
\qquad{\rm or}\quad
\xy
(0,6)*\dir{*};
(0,-6)**\dir2{-} ?(.5) +(-2,0)*{\scriptstyle i}
?(.5) +(2,0)*{\scriptstyle i'}
?(1) *\dir{*};
(-6,0)**\dir{-} ?(.5) +(-1,-1)*{\scriptstyle j}
?(1) *\dir{*};
(0,6)**\dir{-} ?(.5) +(-1,1)*{\scriptstyle k}
?(1) *{};
(6,0)**\dir{-} ?(.5) +(1,1)*{\scriptstyle j'}
?(1) *\dir{*};
(0,-6)**\dir{-} ?(.5) +(1,-1)*{\scriptstyle k'}
?(1) *{};
\endxy
\Rightarrow c^{ijk}c^{i'j'k'}g_{ii'}g_{jj'}g_{kk'}
$$
We now observe that there is an ambiguity: apart from the
cyclic order of the edges around a triangle, how does the
triangulation know the difference between $(i,j,k)$ and
$(j,k,i)$? We conclude that we must impose the condition
$c^{ijk}=c^{jki}=c^{kij}$. Similarly, because there is no
reason to enumerate the triangles in one order or another, we
have to impose the condition $g_{ij}=g_{ji}$.
\item Label each vertex of the dual triangulation with a $c$ and
each edge of the dual triangulation with a $g$. Clearly $c$
must be cyclic-symmetric and $g$ must be symmetric. Note also
that any cyclic-symmetric $c\in V\otimes V\otimes V$ is the
average of a completely symmetric element and a completely
antisymmetric element.
$$
\xy
(0,6)*\dir{o};
(0,-6)**\dir{.} ?(1) *\dir{o};
(-6,0)**\dir{.} ?(1) *\dir{o};
(0,6)**\dir{.};
(6,0)**\dir{.}?(1) *\dir{o};
(0,-6)**\dir{.};
(2,0)*\dir{*};
(4,2)*{};**\dir{-};
(1,-5)*{};**\dir{-};
(-2,0)**\dir{-} ?(1) *\dir{*};
(-4,2)*{};**\dir{-};
(-1,-5)*{};**\dir{-};
\endxy
\Leftrightarrow
\xy
(2,0)*\dir{*}="A" +(2,0)*{\scriptstyle c};
(-2,0)*\dir{*}="B" +(-2,0)*{\scriptstyle c};
"A";"B";**\dir{-} ?(.5) +(0,2)*{\scriptstyle g};
"B";"A";**\crv{(4,-3)&(0,-9)&(-4,-3)} ?(.5) +(0,2)*{\scriptstyle g};
"A";"B";**\crv{(-4,3)&(0,9)&(4,3)} ?(.5) +(0,2)*{\scriptstyle g};
\endxy
\qquad{\rm or}\quad
\xy
(0,6)*\dir{o};
(0,-6)**\dir{.} ?(1) *\dir{o};
(-6,0)**\dir{.} ?(1) *\dir{o};
(0,6)**\dir{.};
(6,0)**\dir{.}?(1) *\dir{o};
(0,-6)**\dir{.};
(2,0)*\dir{*};
(4,2)*{};**\dir{-};
(5,-1)*{};**\dir{-};
(-2,0)**\dir{-} ?(1) *\dir{*};
(-1,5)*{};**\dir{-};
(-2,-4)*{};**\dir{-};
\endxy
\Leftrightarrow
\xy
(2,0)*\dir{*}="A" +(-1,-1)*{\scriptstyle c};
(-2,0)*\dir{*}="B" +(-2,0)*{\scriptstyle c};
"A";"B";**\dir{-} ?(.5) +(0,2)*{\scriptstyle g};
"B";"A";**\crv{(4,3)&(10,-6)&(-6,-6)} ?(.5) +(1,-1)*{\scriptstyle g};
"A";"B";**\crv{(-6,6)&(10,6)&(4,-3)} ?(.5) +(1,1)*{\scriptstyle g};
\endxy
$$
Translating the diagrams into linear algebra, the same
expressions as above are obtained. Notice the similarity with
spin networks, and also the differences. The $\SL(2,\C)$ spin
networks have labels on the edges, which is the same as to say
that we can assign different vector spaces $V$ to each edge of
the triangulation. This means that there is some extra
structure associated to spin network models, over and above
the topological structure that TQFTs are able to see. The key
here is provided by the compatibility conditions for
intertwining operators. Recall that the compatibility was
somewhat mysteriously related to the triangle
inequality. Notice that the $\\j$ label edges of the dual
cellulation, which correspond to sides of triangular faces of
the primal triangulation, and that the compatibility
conditions just say that the three $\\j$ on the sides of a
triangle must be valid lengths for the sides of a flat
triangle. In the topological setting there is no notion of
length, so we have no labels.
\end{enumerate}
\end{enumerate}
Now we come to the final question that we need to ask to validate the
model, and that is, is this computation triangulation-independent, and
if so, how do we prove it?
The answer is provided by Pachner's theorem. This theorem is valid in
any dimension, and we have been exploring the three-dimensional case
in the exercises. The two-dimensional case was probably proved by
Alexander in the 1920's, and is as follows:
\begin{theorem}[Alexander, Pachner]
Two triangulations of the same two-dimensional compact manifold can be
obtained from each other by a succession of moves of the form:
$$
\displaylines{
\xy
(-4,-2)*{};(0,5)**\dir{-};(4,-2)**\dir{-};(-4,-2)**\dir{-};
\endxy
\Leftrightarrow
\xy
(0,0)*{};(-4,-2)**\dir2{-};(0,5)**\dir{-};(4,-2)**\dir{-};(-4,-2)**\dir{-};
(0,0)*{};(0,5)**\dir2{-};
(0,0)*{};(4,-2)**\dir2{-};
\endxy
\qquad\hbox{($1-3$ move)}\cr
\xy
(-4,0)*{};(4,0)**\dir2{-};(0,4)**\dir{-};
(-4,0)**\dir{-};(0,-4)**\dir{-};(4,0)**\dir{-};
\endxy
\Leftrightarrow
\xy
(0,-4)*{};(0,4)**\dir2{-};(4,0)**\dir{-};
(0,-4)**\dir{-};(-4,0)**\dir{-};(0,4)**\dir{-};
\endxy
\qquad\hbox{($2-2$ move)}\cr}
$$
\end{theorem}
There are two things to notice in the statement of this theorem. The
first is that both moves can be obtained from a tetrahedron by
colouring the faces of the tetrahedron with two colours, say, black
and white. Then, the triangulations on either side of
$\Leftrightarrow$ in each move are simply the black and white
triangulations. Similarly, the $(1-4)$ and $(2-3)$ moves in three
dimensions are the two ways of bi-colouring the five tetrahedral faces
of a four-simplex.
\begin{exercise}
State Pachner's theorem in one dimension and use it to prove that the
only one-dimensional triangulable manifolds are the circle and the
segment. [Hint: a two-dimensional simplex is a triangle.]
\end{exercise}
The second thing is that, taking duals of the Alexander-Pachner moves,
we have
$$
\displaylines{
\xy
(0,0)*{};
(-7,4)*{};**\dir{-};
(7,4)*{};**\dir{-};
(0,-9)*{};**\dir{-};
\endxy
\Leftrightarrow
\xy
(-7,4)*{};(-4,2)**\dir{-};(4,2)**\dir{-};(7,4)**\dir{-};
(-4,2)*{};(0,-5)**\dir{-};(4,2)**\dir{-};
(0,-5)*{};(0,-9)**\dir{-};
\endxy
\qquad\hbox{(Star-triangle relation)}\cr
\xy
(-8,7)*{};(-4,0)**\dir{-};(5,0)**\dir{-};(9,-7)**\dir{-};
(-8,-7)*{};(-4,0)**\dir{-};
(9,7)*{};(5,0)**\dir{-};
\endxy
\Leftrightarrow
\xy
(7,-8)*{};(0,-4)**\dir{-};(0,5)**\dir{-};(-7,9)**\dir{-};
(-7,-8)*{};(0,-4)**\dir{-};
(7,9)*{};(0,5)**\dir{-};
\endxy
\qquad\hbox{(Crossing symmetry)}\cr}
$$
These diagrams crop up virtually everywhere in statistical physics and
field theory.
\begin{theorem}
Given the $(2-2)$ move, the $(1-3)$ move is equivalent to
$$
\xy
(0,5)*\dir{*};(0,-5)*\dir{*};**\dir{-};
\endxy
\Leftrightarrow
\xy
(0,0)*\dir{*};(0,-5)*\dir{*};**\dir{-};
(0,5)**\dir{-} ?(1) *\dir{*};
(0,-5)**\crv{(10,0)};
(0,5)**\crv{(-10,0)};
\endxy
\qquad\hbox{(Bubble move)}
$$
\end{theorem}
The proof is simple, but it rests on the assumption that the bubble
move must be interpreted as being inside some triangulation, that is,
the initial edge must be a side of some triangle in the
triangulation. First, $(1-3)$ implies $(\rm bubble)$:
$$
\xy
(-4,-2)*{};(0,5)**\dir{-};(4,-2)**\dir{-};(-4,-2)**\dir{-};
\endxy
\Leftrightarrow
\xy
(0,0)*{};(-4,-2)**\dir{-};(0,5)**\dir{-};(4,-2)**\dir{-};(-4,-2)**\dir{-};
(0,0)*{};(0,5)**\dir{-};
(0,0)*{};(4,-2)**\dir{-};
\endxy
\Leftrightarrow
\xy
(-4,-2)*{};
(0,5)**\dir{-};(4,-2)**\dir{-};(-4,-2)**\crv{(0,5)};
(0,0)**\dir{-};(4,-2)**\dir{-};(-4,-2)**\dir{-};
\endxy
$$
And now, $(\rm bubble)$ implies $(1-3)$:
$$
\xy
(-4,-2)*{};(0,5)**\dir{-};(4,-2)**\dir{-};(-4,-2)**\dir{-};
\endxy
\Leftrightarrow
\xy
(0,-2)*\dir{*};(4,-2)*{};**\dir{-};
(-4,-2)**\dir{-};(0,5)**\dir{-};(4,-2)**\dir{-};
(-4,-2)**\crv{(0,3)};(4,-2)**\crv{(0,-5)};
\endxy
\Leftrightarrow
\xy
(0,-2)*{};(-4,-2)**\dir{-};(0,5)**\dir{-};(4,-2)**\dir{-};
(-4,-2)**\crv{(0,-5)};
(0,-2)*{};(0,5)**\dir{-};
(0,-2)*{};(4,-2)**\dir{-};
\endxy
$$
\begin{exercise}
Prove the following moves on two-dimensional triangulations:
$$
\displaylines{
\xy
(0,5)*{};(0,-5)**\dir{-};
\endxy
\Leftrightarrow
{\xy
(0,0)*{};(0,-5)**\dir{-};(0,5)**\crv{(5,0)};(0,-5)**\crv{(-5,0)};
(0,-5)**\crv{(-5,5)&(5,5)};
\endxy}
\qquad\hbox{(Cone move)}\cr
\xy
(0,0)*\dir{*};(0,5)*\dir{*};**\dir{-};(0,-5)*\dir{*};**\dir{-};
\endxy
\Leftrightarrow
\xy
(0,5)*{};(0,-5)**\dir{-};(5,0)**\dir{-};
(0,5)**\dir{-};(-5,0)**\dir{-};(0,-5)**\dir{-};
\endxy
\qquad\hbox{(Other Bubble move)}\cr}
$$
\end{exercise}
\chapter{Quantum Mechanics from a Category Theoretical Viewpoint}
We are going to continue sneaking up on TQFTs from the field theory
side.
In $2d$ vacuum electromagnetism, we have worked out the time evolution
for the classical theory, and we have seen that it is determined by
the area enclosed by the initial and final circles.
$$
(\hbox{diagram: cylinder with wiggly boundaries})
$$
Area and time are in fact so similar in this theory that we are going
to call the area $t$. Then we have seen that $(a,e)$ is changed to
$(a+et,e)$. Similarly, the quantum version is as follows.
$$
\xymatrix{\psi\in L^2(\R)\ar[d]^{e^{-itH}}\\
\psi\in L^2(\R)\\}
$$
where $H$ is the Hamiltonian
$$
(H\psi)(a)={-1\over 2}{d^2\over da^2}\psi(a)
$$
More generally, if $g$ is any Lorentzian metric on $[0,1]\times S^1$
such that the boundary circles are space-like, then there is an
area-preserving diffeomorphism taking it to the standard metric
$ds^2=-dt^2+dx^2$ on $[0,t]\times S^1$, where $t$ is the area of the
original manifold. Since vacuum electromagnetism is invariant under
area-preserving diffeomorphisms, the same time-evolution operator is
valid.
But we would like to work with more interesting two-dimensional
topologies like the ones we are considering in Track 1.
Now for more general spacetimes (i.e. cobordisms). The problem shows
up right away when one realises that there is no way to put a
Lorentzian metric on
$$
(\hbox{diagram: upside-down trousers})
$$
that makes all the boundary circles space-like. The reason for this is
that the light cones don't match up globally. If it were possible,
there would be a nowhere-vanishing time-like vector field, but by the
Poincar\'e-Hopf theorem any vector field of the trousers must vanish at
a point.
So we switch to studying Riemannian metrics of the form
$ds^2=dt^2+dx^2$.
For example, our cylinder admits the metric
$ds^2=dt^2+dx^2$. Formally, all we have done is replace $t$ by $-it$,
and this means that time evolution of the state in quantum theory is
given by
$$
\xymatrix{\psi\in L^2(\R)\ar[d]^{e^{-tH}}\\
\psi\in L^2(\R)\\}
$$
which is a solution of the heat equation:
$$
{\partial\over\partial t}\psi(a)={1\over 2}{d^2\over da^2}\psi(a).
$$
If ``spacetime'' is a Riemannian cylinder, we get
$$
\xymatrix{\psi\in L^2(\R)\ar[d]^{e^{-tH}}\\
\psi\in L^2(\R)\\}
$$
But now we can think of what happens if ``spacetime'' has a more
interesting topology. For example:
$$
(\hbox{diagram: upside-down trousers})\qquad
\xymatrix{\psi\in L^2(\R)\otimes L^2(R)\ar[d]^T\\
\psi\in L^2(\R)\\}
$$
What form can the time evolution operator take in this case?
$$
(\hbox{diagram: upside-down trousers with seams: two circles to a
circle})\qquad
\xymatrix{L^2(\R)\otimes L^2(\R)\ar[d]^{e^{-t_1H}\otimes e^{-t_2H}}\\
L^2(\R)\otimes L^2(\R)\ar[d]^m\\
L^2(\R)\ar[d]^{e^{-t_3H}}\\
L^2(\R)\\}
$$
We can assume that $T$ is of this form, but we have to impose the
consistency condition that it cannot depend on how the $t=t_1+t_2+t_3$
is divided.
Classically, $m$ just corresponds to summing the holonomies:
$(a_1,e_1)\otimes(a_2,e_2)\mapsto(a_1+a_2,e+1,e+2)$. Now, there is a
natural way to take an operation on a space and obtain an operation on
functions on that space. From $f(a,a')$, we get
$f^*(\psi)(a,a')=\psi\bigl(f(a,a')\bigr)$. In other words, if the
classical configuration spaces are mapped as
$$
\xymatrix{\R\times\R\ar[d]^f\\
\R\\}
$$
the quantum version turns out to be the pull-back of $f$:
$$
\xymatrix{L^2(\R)\\
L^2(\R)\otimes L^2(\R)\ar[u]^{f^*}\\}
$$
and it represents not time evolution on upside-down trousers, but on
the regular ones.
So, in fact, we can hope that the following (where $f(a,a')=a+a'$) is
the right time evolution for a different spacetime:
$$
(\hbox{diagram: trousers with seams})\qquad
\xymatrix{L^2(\R)\ar[d]^{e^{-t_3H}}\\
L^2(\R)\ar[d]^{f^*}\\
L^2(\R)\otimes L^2(\R)\ar[d]^{e^{-t_1H}\otimes e^{-t_2H}}\\
L^2(\R^2)\\}
$$
But there is a problem: $f^*(\psi)$ is not in $L^2(\R^2)$, as we can
readily see!
$$
\left\|f^*(\psi)\right\|^2_{L^2(\R^2)}=\int
da_1\,da_2\,\psi(a_1+a_2)=\int d(a_1+a_2){d(a_1-a_2)\over
2}\psi(a_1+a_2)={1\over 2}\left\|\psi\right\|_{L^2(\R^2)}\int_\R dx=\infty.
$$
The reason, as is evident from the above calculation, is that the
measure of configuration space is infinite!
There are two ways out of this quandary:
\begin{itemize}
\item relax, stop worrying and generalise the heck out of quantum
mechanics so that it works on vector spaces rather than on Hilbert
spaces.
We can still explore the structure of the theory in this way, and
it may be instructive even if we cannot calculate finite
probabilities.
\item switch from $R^+$ to $U(1)$, so that the infinity does not arise
because everything is the same, except that the circle has finite
measure.
Let's do this in detail, because there are subtleties associated
with the need to take sums modulo~$2\pi i$. In fact, I am going to
use multiplicative notation so $z_1=e^{a_1}$ and $z_2=e^{a_2}$ are
complex numbers of norm $1$, $\psi$ is a function defined on
$S^1=\{z\in\C\colon |z|=1\}$ and $f(z_1,z_2)=z_1z_2$. Note that
the uniform unit measure on the circle is $dz\over 2\pi i z$. Now
we can calculate
$$
\left\|f^*(\psi)\right\|^2_{L^2(S^1\times S^1)}=\int_{S^1\times
S^1}{dz_1\over 2\pi i z_1}{dz_2\over 2\pi i z_2}\psi(z_1z_2)
$$
Let $u=z_1z_2$ and $v=z_1/z_2=z_1z_2^*$. Then,
$$
\cases{{du\over 2\pi i u}={dz_1\over 2\pi i z_1}+
{dz_2\over 2\pi i z_2}\cr
{dv\over 2\pi i v}={dz_1\over 2\pi i z_1}-
{dz_2\over 2\pi i z_2}\cr}
\qquad{\rm implies}\quad
\cases{{dz_1\over 2\pi i z_1}={1\over 2}\bigl({du\over 2\pi i u}+
{dv\over 2\pi i v}\bigr)\cr
{dz_2\over 2\pi i z_2}={1\over 2}\bigl({du\over 2\pi i u}-
{dv\over 2\pi i v}\bigr)\cr}
$$
so ${dz_1\over 2\pi i z_1}\wedge{dz_2\over 2\pi i z_2}={1\over
2}{dv\over 2\pi i v}\wedge{du\over 2\pi i u}$. Therefore,
$$
\left\|f^*(\psi)\right\|^2_{L^2(\S^1\times S^1)}={1\over
2}\int_{S^1\times S^1}{dv\over 2\pi i v}{du\over 2\pi i
u}\psi(u)={1\over
2}\left\|\psi\right\|^2_{L^2(S^1)}\int_{S^1}{dv\over 2\pi i
v}={1\over 2}\left\|\psi\right\|^2_{L^2(S^1)}.
$$
and if we want $m$ to be normalised we must let
$m(\psi)(z_1,z_2)={1\over\sqrt 2}\psi(z_1z_2)$ or, equivalently,
$m(\psi)(a_1,a_2)={1\over\sqrt 2}\psi\bigl((a_1+a_2)\mod 2\pi
i\bigr)$.
\end{itemize}
$$
(\hbox{diagram: trousers with seams})\qquad
\xymatrix{L^2(S^1)\ar[d]^{e^{-t_3H}}\\
L^2(S^1)\ar[d]^m\\
L^2(S^1\times S^1)\ar[d]^{e^{-t_1H}\otimes e^{-t_2H}}\\
L^2(S^1)\otimes L^2(S^1)\\}
$$
Moral: compactness is a good thing!
Next, we will have to check the consistency condition, and explore
other space-times.