\chapter{Schur's Lemma and Related Results}

\begin{theorem}[Schur's Lemma] 
If $T$ is an open spin network with two legs, then it is proportional
to the identity.
\end{theorem}

We have to show that
$$
\xy
 (0,0)*+{T}*\cir{}="T";
 (0,10)**\dir{-} ?(.5)+(2,0)*{\scriptstyle\mathbf{j}};
 "T";(0,-10)**\dir{-} ?(.5)+(2,0)*{\scriptstyle\mathbf{k}};
\endxy
=\alpha_{jk}\quad
\xy
 (0,10)*{};
 (0,-10)**\dir{-} ?(.5)+(2,0)*{\scriptstyle\mathbf{j}};
\endxy
\qquad{\rm with}\quad\alpha_{jk}=0\quad{\rm if}\quad j\neq k
$$
where $\xy(0,0)*+{T}*\cir{};\endxy$ represents a spin network. 

The proof is remarkably simple. We reexpress all vertices as triangles
and expand all symmetrisers---except those at the top and bottom---as
sums over permutations. The spin network falls apart into a sum of
diagrams consisting of ``spaghetti'' and ``spaghetti-ohs'':
$$
\sum_{l,m}
\xy
 (-5,10)*{};(1,10)**\crv{(-2,0)} ?(.5)+(0,-2)*{\scriptstyle l};
 (-3,10)*{};(-1,10)**\crv{(-2,4)};
 (-5,-10)*{};(1,-10)**\crv{(-2,0)} ?(.5)+(0,2)*{\scriptstyle m};
 (-3,-10)*{};(-1,-10)**\crv{(-2,-4)};
 (3,10)*{};(3,-10)**\dir{-} ?(.5)+(-2,0)*{\scriptstyle n};
 (5,10)*{};(5,-10)**\dir{-};
 (12,0)*\cir<14pt>{}*\cir<8pt>{};
 (19,0)*{\scriptstyle p};
 (-6,9)*{};(6,9)**\dir{-} ?(.5)+(0,3)*{\scriptstyle\\j};
 (-6,-9)*{};(6,-9)**\dir{-} ?(.5)+(0,-3)*{\scriptstyle\\k};
\endxy
$$
but
$$
\xy
 (-1,4)*{};
 (1,4)**\crv{(-1,-6)&(1,-6)};
 (-2,3)*{};(2,3)**\dir{-};
\endxy
=0
$$
implies that $T=0$ except possibly if $j=k$.

We will call this proof technique ``rewiring'' or ``rerouting''.

\begin{corollary}
If $T$ is an open spin network,
$$
\xy
 (0,0)*+{T}*\cir{}="T";
 (0,10)**\dir{-} ?(.5)+(2,0)*{\scriptstyle\mathbf{j}};
 "T";(0,-10)**\dir{-} ?(.5)+(2,0)*{\scriptstyle\mathbf{j}};
\endxy
=
{\xy
  (0,0)*+{T}*\cir{}="T";
  "T";"T";**\crv{(10,10)&(10,-10)} ?(.5)+(2,0)*{\scriptstyle\mathbf{j}};
 \endxy
 \over
 \xy
  (-2,-2)*\cir<8pt>{};
  (3,-2)*{\scriptstyle\mathbf{j}};
 \endxy}
\quad
\xy
 (0,10)*{};
 (0,-10)**\dir{-} ?(.5)+(2,0)*{\scriptstyle\mathbf{j}};
\endxy
$$
\end{corollary}

The formula follows by taking the trace of both sides of Schur's Lemma:
$$
\xy
 (0,0)*+{T}*\cir{}="T";
 "T";"T";**\crv{(10,10)&(10,-10)} ?(.5)+(2,0)*{\scriptstyle\mathbf{j}};
\endxy
=\alpha_j
\xy
 (0,0)*\cir<8pt>{};
 (3,0)*{\scriptstyle\mathbf{j}};
\endxy
$$
Now, we observe that a closed spin network is just a complex number,
so we can divide by it if it is nonzero! Since
$\xy(0,0)*\cir<4pt>{};(3,0)*{\scriptstyle\mathbf{j}};\endxy$ is nonzero
for nonnegative $j$, we can solve for $\alpha_j$ and the result
follows.

\begin{theorem}[Wigner-Eckart]
If $T$ is an open spin network with three legs, it is proportional to
the corresponding trivalent vertex.
$$
\xy
 (0,0)*+{T}*\cir{}="T";
 (-7,4)**\dir{-} ?(.67)+(-1,-2)*{\scriptstyle\mathbf{i}};
 "T";(7,4)**\dir{-} ?(.67)+(-1,2)*{\scriptstyle\mathbf{j}};
 "T";(0,-9)**\dir{-} ?(.67)+(2,0)*{\scriptstyle\mathbf{k}};
\endxy
=
{
\xy
 (-5,0)*+{T}*\cir{}="T";
 (5,0)*\dir{*}="o";**\dir{-} ?(.5)+(0,2)*{\scriptstyle\mathbf{j}};
 "T";"o";**\crv{(0,10)} ?(.5)+(0,2)*{\scriptstyle\mathbf{i}};
 "T";"o";**\crv{(0,-10)} ?(.5)+(0,2)*{\scriptstyle\mathbf{k}};
\endxy
\over
\xy
 (-5,0)*\dir{*}="*";
 (5,0)*\dir{*}="o";**\dir{-} ?(.5)+(0,2)*{\scriptstyle\mathbf{j}};
 "*";"o";**\crv{(0,10)} ?(.5)+(0,2)*{\scriptstyle\mathbf{i}};
 "*";"o";**\crv{(0,-10)} ?(.5)+(0,2)*{\scriptstyle\mathbf{k}};
\endxy
}~
\xy
 (0,0)*\dir{*}="Y";
 (-7,4)**\dir{-} ?(.5)+(-1,-2)*{\scriptstyle\mathbf{i}};
 "Y";(7,4)**\dir{-} ?(.5)+(-1,2)*{\scriptstyle\mathbf{j}};
 "Y";(0,-9)**\dir{-} ?(.5)+(2,0)*{\scriptstyle\mathbf{k}};
\endxy
$$
\end{theorem}

We have to prove two things, that $T$ is proportional to the vertex
and that the proportionality factor is as stated. Assuming that
proportionality holds, the proportionality factor follows by joining
the three free ends to a single vertex. Proportionality to the vertex
can be obtained almost trivially by rewiring (in essentially the same
way as the uniqueness of the vertex itself).

We expand each leg and each of the $\mathbf{j}$-lines inside $T$ into $2j$
spin-$1/2$ lines with a symmetriser across them. Then, the spin
network breaks up into
$$
\sum_{m_1,m_2,m_3}
\xy
 (-5,-8)*{};(5,-8)**\dir{-} ?(.5)+(1,-3)*{\scriptstyle\mathbf{k}};
 (-1,-9)*{};(1,-9)**\crv{(0,-2)};
 (-2,-9)*{};(2,-9)**\crv{(0,0)} ?(.5)+(0,2)*{\scriptstyle n_3};
 (-3,-9)*{};(-8.5,1)**\crv{(0,0)};
 (-4,-9)*{};(-9,0)**\crv{(-2,-1)} ?(.5)+(-2,-1)*{\scriptstyle m_2};
 (-4,8)*{};(-9,-1)**\dir{-} ?(.5)+(-2,2)*{\scriptstyle\mathbf{i}};
 (-7.5,3)*{};(-6.5,5)**\crv{(-2,1)};
 (-8,2)*{};(-6,6)**\crv{(0,0)} ?(.5)+(2,-1)*{\scriptstyle n_1};
 (-5.5,7)*{};(5.5,7)**\crv{(0,0)};
 (-5,8)*{};(5,8)**\crv{(0,2)} ?(.5)+(0,2)*{\scriptstyle m_3};
 (4,8)*{};(9,-1)**\dir{-} ?(.5)+(2,2)*{\scriptstyle\mathbf{j}};
 (7.5,3)*{};(6.5,5)**\crv{(2,1)};
 (8,2)*{};(6,6)**\crv{(0,0)} ?(.5)+(-2,-1)*{\scriptstyle n_2};
 (3,-9)*{};(8.5,1)**\crv{(0,0)};
 (4,-9)*{};(9,0)**\crv{(2,-1)} ?(.5)+(2,-1)*{\scriptstyle m_1};
 (15,0)*\cir<12pt>{}*\cir<8pt>{};
 (22,0)*{\scriptstyle p};
\endxy
$$
but, again,
$\xy(-1,2)*{};(1,2)**\crv{(-1,-4)&(1,-4)};(-2,1)*{};(2,1)**\dir{-};\endxy=0$
implies that all the terms vanish except those proportional to the
trivalent vertex.

The spin network
$$
\xy
 (-7,0)*\dir{*}="*";
 (7,0)*\dir{*}="o";**\dir{-} ?(.5)+(0,2)*{\scriptstyle\mathbf{j_2}};
 "*";"o";**\crv{(0,14)} ?(.5)+(0,2)*{\scriptstyle\mathbf{j_1}};
 "*";"o";**\crv{(0,-14)} ?(.5)+(0,2)*{\scriptstyle\mathbf{j_3}};
\endxy
=
\xy
 (1,3.5)*{\scriptstyle\\k_3};
 (0,1)*{};(0,1)**\crv{(-10,0)&(0,12)&(10,0)};
 (1,-3.5)*{\scriptstyle\\k_1};
 (0,-1)*{};(0,-1)**\crv{(-10,0)&(0,-12)&(10,0)};
 (0,0)*\cir<24pt>{};(11,0)*{\scriptstyle\mathbf{k_2}};
 (0,2)*{};(0,-2)**\dir{-};
 (0,5)*{};(0,9)**\dir{-};
 (0,-5)*{};(0,-9)**\dir{-};
\endxy
%={\sum(-1)^{a+b+c+g+h+i}(a+1)!\cdots(j+1)!\over\bigl(2(k_1+k_2)\bigr)!\bigl(2(k_1+k_3)\bigr)!\bigl(2(k_2+k_3)\bigr)!}
$$
is called ``$\theta$-net'' or, in physics, ``$3j$-coefficient''. 

We do not have an analogue of Schur's Lemma for spin networks with
four legs, but we can come really close. 

First, we have
$$
\xy
 (0,9)*{}="Y";(-4,11)**\dir{-} ?(.5)+(-1,-2)*{\scriptstyle\mathbf{a}};
 "Y";(4,11)**\dir{-} ?(.5)+(-1,2)*{\scriptstyle\mathbf{b}};
 "Y";(0,2)**\dir{-}="A" ?(.5)+(2,0)*{\scriptstyle\mathbf{i}};
 "A";(0,-2)**\crv{(-4,0)}="Y2" ?(.5)+(-2,0)*{\scriptstyle\mathbf{a}};
 "A";"Y2"**\crv{(4,0)} ?(.5)+(2,0)*{\scriptstyle\mathbf{b}};
 "Y2";(0,-9)**\dir{-}="A2" ?(.5)+(2,0)*{\scriptstyle\mathbf{j}};
 "A2";(-4,-11)**\dir{-} ?(.5)+(-1,2)*{\scriptstyle\mathbf{a}};
 "A2";(4,-11)**\dir{-} ?(.5)+(-1,-2)*{\scriptstyle\mathbf{b}};
\endxy
=
\delta_{ij}
{\xy
  (-5,0)*\dir{*}="*";
  (5,0)*\dir{*}="o";**\dir{-} ?(.5)+(0,2)*{\scriptstyle\mathbf{i}};
  "*";"o";**\crv{(0,10)} ?(.5)+(0,2)*{\scriptstyle\mathbf{a}};
  "*";"o";**\crv{(0,-10)} ?(.5)+(0,2)*{\scriptstyle\mathbf{b}};
 \endxy
\over
 \xy
  (-2,-2)*\cir<8pt>{};
  (3,-2)*{\scriptstyle\mathbf{i}};
 \endxy}
\xy
 (0,5)*{}="Y";(0,-5)**\dir{-}="A" ?(.5)+(2,0)*{\scriptstyle\mathbf{i}};
 "Y";(-7,9)**\dir{-} ?(.5)+(0,2)*{\scriptstyle\mathbf{a}};
 "Y";(7,9)**\dir{-} ?(.5)+(0,2)*{\scriptstyle\mathbf{b}};
 "A";(-7,-9)**\dir{-} ?(.5)+(0,2)*{\scriptstyle\mathbf{a}};
 "A";(7,-9)**\dir{-} ?(.5)+(0,2)*{\scriptstyle\mathbf{b}};
\endxy
$$
(where we apply Schur's lemma to the intermediate ``bubble'') which
implies that
$$
{\xy
  (-2,-2)*\cir<8pt>{};
  (3,-2)*{\scriptstyle\\i};
 \endxy
\over
 \xy
  (-5,0)*\dir{*}="*";
  (5,0)*\dir{*}="o";**\dir{-} ?(.5)+(0,-2)*{\scriptstyle\mathbf{i}};
  "*";"o";**\crv{(0,10)} ?(.5)+(0,-2)*{\scriptstyle\mathbf{a}};
  "*";"o";**\crv{(0,-10)} ?(.5)+(0,-2)*{\scriptstyle\mathbf{b}};
 \endxy}
~
\xy
 (0,4)*{}="Y";(0,-4)**\dir{-}="A" ?(.5)+(2,0)*{\scriptstyle\mathbf{i}};
 "Y";(-4,6)**\dir{-} ?(.5)+(-1,-2)*{\scriptstyle\mathbf{a}};
 "Y";(4,6)**\dir{-} ?(.5)+(-1,2)*{\scriptstyle\mathbf{b}};
 "A";(-4,-6)**\dir{-} ?(.5)+(-1,2)*{\scriptstyle\mathbf{a}};
 "A";(4,-6)**\dir{-} ?(.5)+(-1,-2)*{\scriptstyle\mathbf{b}};
\endxy
$$
is a projector, and the product of different projectors vanishes. Now,

\begin{theorem}[resolution of the identity]
$$
\xy
 (-4,6)*{};(-4,-6)**\dir{-} ?(.5)+(-2,0)*{\scriptstyle\mathbf{a}};
 (4,6)*{};(4,-6)**\dir{-} ?(.5)+(2,0)*{\scriptstyle\mathbf{b}};
\endxy
=
\sum_{i=\left|a-b\right|}^{a+b}
{\xy
  (-2,-2)*\cir<8pt>{};
  (3,-2)*{\scriptstyle\mathbf{i}};
 \endxy
\over
 \xy
  (-5,0)*\dir{*}="*";
  (5,0)*\dir{*}="o";**\dir{-} ?(.5)+(0,-2)*{\scriptstyle\mathbf{i}};
  "*";"o";**\crv{(0,10)} ?(.5)+(0,-2)*{\scriptstyle\mathbf{a}};
  "*";"o";**\crv{(0,-10)} ?(.5)+(0,-2)*{\scriptstyle\mathbf{b}};
 \endxy}
~
\xy
 (0,4)*{}="Y";(0,-4)**\dir{-}="A" ?(.5)+(2,0)*{\scriptstyle\mathbf{i}};
 "Y";(-4,6)**\dir{-} ?(.5)+(-1,-2)*{\scriptstyle\mathbf{a}};
 "Y";(4,6)**\dir{-} ?(.5)+(-1,2)*{\scriptstyle\\b};
 "A";(-4,-6)**\dir{-} ?(.5)+(-1,2)*{\scriptstyle\mathbf{a}};
 "A";(4,-6)**\dir{-} ?(.5)+(-1,-2)*{\scriptstyle\mathbf{b}};
\endxy
$$
where the sum is extended to the values of $i$ satisfying the
compatibility conditions with $a$ and $b$.
\end{theorem}

The right-hand side is a projector onto a subspace of
$\mathbf{a}\otimes\mathbf{b}$. All we need to do is prove that the
dimension of its image is the dimension of
$\mathbf{a}\otimes\mathbf{b}$. Calling the left-hand side $P$, we have
$$
\xy
  (0,0)*+{P}*\cir{}="T";
  "T";"T";**\crv{(10,10)&(10,-10)} ?(.5)+(2,0)*{\scriptstyle\mathbf{b}};
  "T";"T";**\crv{(-10,10)&(-10,-10)} ?(.5)+(-2,0)*{\scriptstyle\mathbf{a}};
\endxy
=
\sum_{i=\left|a-b\right|}^{a+b}
\xy
 (-2,0)*\cir<8pt>{};
 (3,0)*{\scriptstyle\\i};
\endxy
=
\sum_{i=\left|a-b\right|}^{a+b}
(-1)^{2i}(2i+1)
=
(-1)^{2(a+b)}(2a+1)(2b+1)=
\xy /r1.5pc/:
 (0,.5),\hcap-|{\\a}\huncross\hcap|{\mathbf{b}};
\endxy
$$

\begin{corollary}[tetravalent Schur's lemma]
If $T$ is a spin network with four legs,
$$
\xy
 (0,0)*+{T}*\cir{}="T";(-4,-6)**\dir{-} ?(.75)+(-2,1)*{\scriptstyle\mathbf{c}};
 "T";(-4,6)**\dir{-} ?(.75)+(-2,-1)*{\scriptstyle\mathbf{a}};
 "T";(4,6)**\dir{-} ?(.75)+(2,-1)*{\scriptstyle\mathbf{b}};
 "T";(4,-6)**\dir{-} ?(.75)+(2,1)*{\scriptstyle\mathbf{d}};
\endxy
=
\sum_i
{\xy
  (0,0)*+{T}*\cir{}="T";
  "T";"T";**\crv{(5,10)&(10,0)} ?(.17)+(-1,1)*{\scriptstyle\mathbf{d}} 
                                ?(.67)+(1,-1)*{\scriptstyle\mathbf{c}}
                                ?(.33)*\dir{*}="p";
  "T";"T";**\crv{(-5,10)&(-10,0)} ?(.17)+(1,1)*{\scriptstyle\mathbf{b}}
                                ?(.67)+(-1,-1)*{\scriptstyle\mathbf{a}}
                                ?(.33)*\dir{*};
  "p";**\crv{(0,10)} ?(.5)+(0,2)*{\scriptstyle\mathbf{i}};
 \endxy
~
 \xy
  (-2,0)*\cir<8pt>{};
  (3,0)*{\scriptstyle\mathbf{i}};
 \endxy
\over 
 \xy
  (-5,0)*\dir{*}="*";
  (5,0)*\dir{*}="o";**\dir{-} ?(.5)+(0,-2)*{\scriptstyle\mathbf{i}};
  "*";"o";**\crv{(0,10)} ?(.5)+(0,-2)*{\scriptstyle\mathbf{a}};
  "*";"o";**\crv{(0,-10)} ?(.5)+(0,-2)*{\scriptstyle\mathbf{b}};
 \endxy
~
 \xy
  (-5,0)*\dir{*}="*";
  (5,0)*\dir{*}="o";**\dir{-} ?(.5)+(0,-2)*{\scriptstyle\mathbf{i}};
  "*";"o";**\crv{(0,10)} ?(.5)+(0,-2)*{\scriptstyle\mathbf{c}};
  "*";"o";**\crv{(0,-10)} ?(.5)+(0,-2)*{\scriptstyle\mathbf{d}};
 \endxy}
~
\xy
 (0,4)*{}="Y";(0,-4)**\dir{-}="A" ?(.5)+(2,0)*{\scriptstyle\mathbf{i}};
 "Y";(-4,6)**\dir{-} ?(.5)+(-1,-2)*{\scriptstyle\mathbf{a}};
 "Y";(4,6)**\dir{-} ?(.5)+(-1,2)*{\scriptstyle\mathbf{b}};
 "A";(-4,-6)**\dir{-} ?(.5)+(-1,2)*{\scriptstyle\mathbf{c}};
 "A";(4,-6)**\dir{-} ?(.5)+(-1,-2)*{\scriptstyle\mathbf{d}};
\endxy
$$
\end{corollary}

This time the proof follows by applying the resolution of the identity
to the top and bottom pairs of legs.

An important special case is 
$$
\xy
 (4,0)*{}="Y";(-4,0)**\dir{-}="A" ?(.5)+(0,2)*{\scriptstyle\mathbf{i}};
 "Y";(6,-4)**\dir{-} ?(.5)+(-2,-1)*{\scriptstyle\mathbf{d}};
 "Y";(6,4)**\dir{-} ?(.5)+(2,-1)*{\scriptstyle\mathbf{b}};
 "A";(-6,-4)**\dir{-} ?(.5)+(2,-1)*{\scriptstyle\mathbf{c}};
 "A";(-6,4)**\dir{-} ?(.5)+(-2,-1)*{\scriptstyle\mathbf{a}};
\endxy
=
\sum_j
\left\{
\matrix{a&b&i\cr
        c&d&j\cr}\right\}
\xy
 (0,4)*{}="Y";(0,-4)**\dir{-}="A" ?(.5)+(2,0)*{\scriptstyle\mathbf{j}};
 "Y";(-4,6)**\dir{-} ?(.5)+(-1,-2)*{\scriptstyle\mathbf{a}};
 "Y";(4,6)**\dir{-} ?(.5)+(-1,2)*{\scriptstyle\mathbf{b}};
 "A";(-4,-6)**\dir{-} ?(.5)+(-1,2)*{\scriptstyle\mathbf{c}};
 "A";(4,-6)**\dir{-} ?(.5)+(-1,-2)*{\scriptstyle\mathbf{d}};
\endxy
$$
where $\left\{a\atop c\right.\left.b\atop d\right.\left.i\atop
j\right\}$ are what spectroscopists call ``$6j$-symbols'' or ``Racah
$6j$ coefficients''. We leave it as an exercise to prove that
$$
\left\{
\matrix{a&b&i\cr
        c&d&j\cr}\right\}
\colon=
{\xy
  (0,0)*\dir{*}="Y";
  (-7,4)*\dir{*}="a";
  **\dir{-} ?(.5)+(-1,-1)*{\scriptstyle\mathbf{a}};
  (0,-9)*\dir{*}="b";
  "Y"**\dir{-} ?(.5)+(2,0)*{\scriptstyle\mathbf{j}};
  (7,4)*\dir{*}="c";
  "Y"**\dir{-} ?(.5)+(-1,2)*{\scriptstyle\mathbf{b}};
  "a";"b"**\crv{(-7,-4)} ?(.5)+(-2,-1)*{\scriptstyle\mathbf{c}};
  "c";"b"**\crv{(7,-4)} ?(.5)+(2,-1)*{\scriptstyle\mathbf{d}};
  "c";"a"**\crv{(0,9)} ?(.5)+(0,2)*{\scriptstyle\mathbf{i}};
 \endxy
\quad
 \xy
  (-2,0)*\cir<8pt>{};
  (3,0)*{\scriptstyle\mathbf{j}};
 \endxy
\over 
 \xy
  (-5,0)*\dir{*}="*";
  (5,0)*\dir{*}="o";**\dir{-} ?(.5)+(0,-2)*{\scriptstyle\mathbf{j}};
  "*";"o";**\crv{(0,10)} ?(.5)+(0,-2)*{\scriptstyle\mathbf{a}};
  "*";"o";**\crv{(0,-10)} ?(.5)+(0,-2)*{\scriptstyle\mathbf{b}};
 \endxy
\quad
 \xy
  (-5,0)*\dir{*}="*";
  (5,0)*\dir{*}="o";**\dir{-} ?(.5)+(0,-2)*{\scriptstyle\mathbf{j}};
  "*";"o";**\crv{(0,10)} ?(.5)+(0,-2)*{\scriptstyle\mathbf{c}};
  "*";"o";**\crv{(0,-10)} ?(.5)+(0,-2)*{\scriptstyle\mathbf{d}};
 \endxy}
$$
As is usual with these kinds of quantities, there are several
definitions in the literature which differ by signs or multiplicative
factors. In fact, if it wasn't for historical reasons and consistency
with the physics literature, it would certainly make sense to call the
tetrahedron diagram ``$6j$-symbol''.

Lurking in all this is the associative property of the tensor
product,
$\mathbf{i}\otimes(\mathbf{j}\otimes\mathbf{k})\simeq(\mathbf{i}\otimes\mathbf{j})\otimes\mathbf{k}$. The linear
operator which effects the isomorphism is the ``associator'', drawn as
$$
\xy
 (-4,12)*{};(-4,-12)**\dir{-} ?(.5)+(2,0)*{\scriptstyle\mathbf{i}};
 (0,12)*{};(0,-12)**\dir{-} ?(.5)+(2,0)*{\scriptstyle\mathbf{j}};
 (4,12)*{};(4,-12)**\dir{-} ?(.5)+(2,0)*{\scriptstyle\mathbf{k}};
\endxy
=
\sum_{m,p,q}C_{mpq}
\xy
 (0,12)*{};
 (4,10)*\dir{*}="a";**\dir{-} ?(.5)+(0,2)*{\scriptstyle\mathbf{j}};
 "a";(0,5)*\dir{*}="b";**\dir{-} ?(.5)+(-2,1)*{\scriptstyle\mathbf{m}};
 "b";(0,-4)*\dir{*}="c";**\dir{-} ?(.5)+(2,0)*{\scriptstyle\mathbf{p}};
 "c";(-4,-9)*\dir{*}="d";**\dir{-} ?(.5)+(-2,1)*{\scriptstyle\mathbf{q}};
 "d";(0,-11)**\dir{-} ?(.5)+(0,-2)*{\scriptstyle\mathbf{j}};
 "a";(8,12)**\dir{-} ?(.5)+(0,2)*{\scriptstyle\mathbf{k}};
 "b";(-8,12)**\dir{-} ?(.5)+(-1,-1)*{\scriptstyle\mathbf{i}};
 "c";(8,-11)**\dir{-} ?(.5)+(-1,-1)*{\scriptstyle\mathbf{k}};
 "d";(-8,-11)**\dir{-} ?(.5)+(0,-2)*{\scriptstyle\mathbf{i}};
\endxy
$$
where the $C_{mpq}$ stand for a coefficient too tedious to calculate
here and to cumbersome to print, although as an exercise, the reader
is encouraged to evaluate it as a function of bubbles, thetas and
tetrahedrons. In fact, everything we will do next quarter on $3D$
quantum gravity will be based on the associator!

The amazing thing that happens is that each term on the
right-hand-side of the definition of the associator can be closed up
to give a tetrahedron diagram
$$
\xy
 (0,0)*\dir{*}="Y";
 (-7,4)*\dir{*}="a";
 **\dir{-} ?(.5)+(-1,-1)*{\scriptstyle\mathbf{i}};
 (0,-9)*\dir{*}="b";
 "Y"**\dir{-} ?(.5)+(2,0)*{\scriptstyle\mathbf{q}};
 (7,4)*\dir{*}="c";
 "Y"**\dir{-} ?(.5)+(-1,2)*{\scriptstyle\mathbf{m}};
 "a";"b"**\crv{(-7,-4)} ?(.5)+(-2,-1)*{\scriptstyle\mathbf{p}};
 "c";"b"**\crv{(7,-4)} ?(.5)+(2,-1)*{\scriptstyle\mathbf{k}};
 "c";"a"**\crv{(0,9)} ?(.5)+(0,2)*{\scriptstyle\mathbf{j}};
\endxy
$$
Then it is possible to triangulate a three-dimensional manifold, label
all the edges by half-integer numbers and multiply together the
complex numbers associated to all tetrahedra. This will turn out to be
an invariant of three-manifolds. This last fact follows
if we can show that the resulting number is independent of the
triangulation. 

\begin{theorem}
Any two triangulations of the same three-dimensional manifold can be
obtained from one another by a sequence of ``Pachner moves'':
$$
\displaylines{
\xy
 (5,1)*\dir{*}="1";
 (1,5)*\dir{*}="2"**\dir{-};
 (-5,1)*\dir{*}="3"**\dir{-};
 (1,-5)*\dir{*}="4"**\dir{-};
 "1"**\dir{-};
 "3"**\dir{.};
 "2";"4"**\dir{-};
\endxy
\quad\Leftrightarrow\quad
\xy
 (0,0)*\dir{o}="0";
 (5,1)*\dir{*}="1"**\dir{.};
 (1,5)*\dir{*}="2"**\dir{-};
 (-5,1)*\dir{*}="3"**\dir{-};
 (1,-5)*\dir{*}="4"**\dir{-};
 "0"**\dir{.};
 "2"**\dir{.};
 "4"**\dir{-};
 "1"**\dir{-};
 "3"**\dir{.};
 "0"**\dir{.};
\endxy
\qquad\hbox{($1-4$ move)}\cr
\xy
 (0,9)*\dir{*}="b";(7,4)*\dir{*}="c"**\dir{-};
 (7,-4)*\dir{*}="d"**\dir2{-};
 (0,-9)*\dir{*}="e"**\dir{-};
 (-7,0)*\dir{*}="a"**\dir{-};
 "b"**\dir{-};
 "d"**\dir{-};
 "a"**\dir2{-};
 "c"**\dir2{.};
 "e"**\dir{.};
\endxy
\quad\Leftrightarrow\quad
\xy
 (0,9)*\dir{*}="b";(7,4)*\dir{*}="c"**\dir2{-};
 (7,-4)*\dir{*}="d"**\dir{-};
 (0,-9)*\dir{*}="e"**\dir2{-};
 (-7,0)*\dir{*}="a"**\dir2{-};
 "b"**\dir2{-};
 "d"**\dir2{-};
 "a"**\dir{-};
 "c"**\dir{.};
 "e"**\dir2{.};
 "b"**\dir3{.};
\endxy
\qquad\hbox{($2-3$ move)}\cr}
$$
\end{theorem}

All we need to do is evaluate the spin networks on both sides and show
that they are equal\footnote{In fact, the $1-4$ move leads to a
divergent sum, and this will be one of the major arguments for the
introduction of ``quantum (or cosmological)'' groups.}.

\chapter{Physics from Lagrangians (V)}

\section{Lagrangians in Field Theory (V)}

We've seen the following theories and equations:
\begin{itemize}
\item Chern Theories:
$$
S=\int_{M^{2n}}\tr(\wedge^n F)\longrightarrow\delta S=0\Leftrightarrow
\cases{0=0 & if $n=1$\cr
       (F^{\wedge(n-2)})\wedge d_AF=0 & if $n>1$,\cr}
$$
which hold identically by the Bianchi identity $d_AF=0$.
\item Electromagnetism:
$$
S=\int_M\tr(F\wedge *F)\longrightarrow \delta S=0\Leftrightarrow d{*F}=0.
$$
We will see that, in the non-abelian Yang--Mills case,
$$
S=\int_M\tr(F\wedge *F)\longrightarrow \delta S=0\Leftrightarrow d_A{*F}=0,
$$
which reduces to Maxwell's equations in the abelian case.
\item $EF$ theory:
$$
S=\int_M\tr(E\wedge F)\longrightarrow \delta S=0\Leftrightarrow\cases{F=0\cr
                                                                    d_A E=0}
$$
\end{itemize}

To finish up this story, we will now derive the equations of motion
for the rest of the Lagrangians we wrote at the beginning, and next
quarter we will quantise them.

\subsection{$EF$ Theory with Cosmological Constant}

We recall that, in dimensions less than 5, The $EF$ Lagrangian admits
additional terms not involving $F$ at all. We will ignore the $2D$
case for the moment, since it involves infinitely many coupling
constants.

In $4D$ we have
$$
S=\int_M\tr(E\wedge F+\lambda E\wedge E)
$$
The possible Chern term $F\wedge F$ does not affect the classical
equations of motion, so we ignore it. On quantisation, or on manifolds
with a boundary, this term does have an effect. The variation of the
action is
$$
\delta S=\int_M\tr(\delta E\wedge F)+\tr(E\wedge\delta F)+\lambda\tr\bigl[(\delta
E\wedge E)+(E\wedge\delta E)\bigr]=\int_M\tr(\delta E\wedge
F)+\tr(E\wedge d_A \delta A)+2\lambda\tr(\delta
E\wedge E),
$$
where $E$ and $\delta E$ are $2$-forms so their wedge product is
symmetric. We now integrate by parts:
\begin{eqnarray*}
\delta S&=&\int_M\tr(\delta E\wedge
F)+\tr(E\wedge d_A \delta A)+2\lambda\tr(\delta E\wedge
E)=\\
&=&\int_{\delta M}\tr(E\wedge\delta A)+\int_M\tr(\delta E\wedge
F)-\tr(d_A E\wedge F)+2\lambda\tr(\delta E\wedge E)
\\
\end{eqnarray*}
and the equations of motion turn out to be
$$
\delta S=0\Leftrightarrow\cases{d_A E=0\cr
                                F+2\lambda E=0\cr}
$$
If $\lambda\neq 0$, the second equation just says $E=-{1\over
2\lambda}F$, in which case the first equation is Bianchi's
identity. The equations of motion turn out to be vacuous again, and
this is just like Chern theory in that the space of solutions is the
same, namely the space of all connections. The equations of motion
also imply that $\lambda$ is a number, not a function on the manifold.

The bottom line is that as soon as there is a nonzero cosmological
constant the character of the theory changes drastically.

Let us now look at $3D$ $EF$ theory with cosmological constant. We
have
$$
S=\int_M\tr(E\wedge F+\lambda E\wedge E\wedge E).
$$
It is interesting that in lower dimension we can do more things with
the $E$. Each dimension has a distinct personality as far as the
available Lagrangians go, and this is related to how topology is very
different in different dimensions. Also, things become boring in
high-enough dimensions. We have seen reflections of these facts
already in track 1. 

The variation of the action is
$$
\delta S=\int_M\tr(\delta E\wedge F)+\tr(E\wedge\delta
F)+3\lambda\tr(\delta E\wedge E\wedge E)=
$$
using the cyclic property of the trace and the fact that cyclic
permutations of three $1$-forms do not involve a change of
sign. 
\begin{eqnarray*}
\delta S&=&\int_M\tr(\delta E\wedge F)+\tr(E\wedge\delta
F)+3\lambda\tr(\delta E\wedge E\wedge E)=\\
&=&\int_M\tr(\delta E\wedge
F)+\tr(E\wedge d_A \delta A)+3\lambda\tr(\delta E\wedge E\wedge
E)=\\
&=&-\int_{\delta M}\tr(E\wedge\delta A)+\int_M\tr(\delta E\wedge
F)+\tr(d_A E\wedge\delta A)+3\lambda\tr(\delta E\wedge E\wedge E),\\
\end{eqnarray*}
integrating by parts at the last step. The equations of motion are
$$
\delta S=0\Leftrightarrow\cases{d_A E=0\cr
                                F+3\lambda E\wedge E=0\cr}
$$
If we differentiate the second equation and use Bianchi's identity,
$$
0=3\lambda(d_A E\wedge E-E\wedge d_A E).
$$
Since $E$ is a $2$-form, its wedge products are symmetric, so this is
zero independently of whether $d_AE$ vanishes. Therefore, the two
equations turn out to be independent. 

We can now justify our previous statement that, if $G=\SU(2)$ or
$G=\SO(3)$, these are the equations of motion for Riemannian gravity
with cosmological constant. $E$ is a frame field, so it gives a metric
on the manifold!  Then $d_A E$ ends up meaning that $A$ is the
Levi-Civita connection of that metric. Lorentzian general relativity
can be obtained by making $G=\SO(2,1)$ or $\SL(2,\R)$

\subsection{$4D$ General Relativity}

To finish, let's derive the equations of motion for $4D$ general
relativity. We have a connection $A$ and an $\SO(4)$ frame field
$e$. The lagrangian is ${\cal L}=e\wedge e\wedge F$. With a
cosmological constant,
$$
S=\int_M\tr(e\wedge e\wedge F+\lambda e\wedge e\wedge e\wedge e)
$$
Here we are using the embedding of $\Lambda^2\R^4$ as antisymmetric
matrices, which are isomorphic to $\so(4)$. Therefore, we can pretend
that $e\wedge e$ is an $Ad(P)$-valued $2$-form. The action of a gauge
transformation on $e\wedge e$ is $e'\wedge e'=ge\wedge eg^{-1}$. This
is called the Palatini formulation of general relativity.

The variation of the action is
$$
\delta S=\int_M 2\tr(\delta e\wedge e\wedge F)+\tr(e\wedge
e\wedge\delta F)+4\lambda\tr(\delta e\wedge e\wedge e\wedge e)
$$
because $e\wedge f=f\wedge e$, as $\wedge$ is antisymmetric both on
$\R^4$ and on the $1$-form part. Substituting $\delta F=d_A\delta A$ and
integrating by parts,
$$
\delta S=\int_{\partial M}\tr(e\wedge e\wedge\delta A)+\int_M
2\tr(\delta e\wedge e\wedge F)-\tr\bigl(d_A(e\wedge
e)\wedge\delta A\bigr)+4\lambda\tr(\delta e\wedge e\wedge e\wedge e)
$$
so the equations of motion are
$$
\delta S=0\Leftrightarrow\cases{d_A(e\wedge e)=0\cr
                                e\wedge F+2\lambda e\wedge e\wedge e=0\cr}
$$
The first equation characterises $A$ as the Levi-Civita connection of
the metric $e\wedge e$, and the second equation is Einstein's equation
in the presence of a cosmological constant.

To finish, recall that the equations of motion of $4D$ $EF$ theory are
$$
\delta S=0\Leftrightarrow\cases{d_A E=0\cr
                                F+2\lambda E=0,\cr}
$$
which means that solutions of general relativity are solutions of $EF$
theory. This suggest the following way to generate solutions of $GR$:
We ``just'' have to find solutions to $EF$ theory in which $E=e\wedge
e$.




