\chapter{Diagrammatic Methods for Linear Algebra (II)}
The basic building blocks in the theory we are developing are linear
mappings between tensor products of vector spaces. For example:
$$
\xy
(0,0)*++{T}*\cir{}="t";
(-8,13)**\dir{-} ?(.5)*\dir{<}+(1,2)*{\scriptstyle V_1};
"t";(8,13)**\dir{-} ?(.5)*\dir{<}+(2,-1)*{\scriptstyle V_2};
"t";(-13,-8)**\dir{-} ?(.75)*\dir{>}+(-1,2)*{\scriptstyle V_3};
"t";(0,-15)**\dir{-} ?(.75)*\dir{>}+(-2,0)*{\scriptstyle V_4};
"t";(13,-8)**\dir{-} ?(.75)*\dir{>}+(-1,-2)*{\scriptstyle V_5};
\endxy
$$
\section{Degenerate Cases}
What does this diagram stand for?
$$
\xy
(0,-5)*++{f}*\cir{}="f";
(0,5)**\dir{-} ?(.5)*\dir{<}+(2,0)*{\scriptstyle V};
\endxy
$$
The key to answering this question is to give a meaning to ``the
tensor product of no spaces''. The only possibility is that it is the
base field, $\C$ in this case, and that makes sense because the base
field is the identity for the operation ``tensor product of vector
spaces''! That is, we have the following canonical linear isomorphism:
$$
\matrix{\C\otimes V &\rightarrow &V\cr
1\otimes v &\mapsto &v\cr}
$$
Therefore, we may write
$$
\xy
(0,0)*++{f}*\cir{}="f";
(0,10)**\dir{-} ?(.5)*\dir{<}+(2,0)*{\scriptstyle V};
\endxy
\in V^*
$$
Now, what is this?
$$
\xy
(0,0)*++{f}*\cir{}="f";
(0,-10)**\dir{-} ?(.75)*\dir{>}+(-2,0)*{\scriptstyle V};
\endxy
$$
It is a linear mapping $f\colon\C\rightarrow V$, which can be
characterised by giving $f(1)=v\in V$, so we conclude that
$$
\xy
(0,0)*++{f}*\cir{}="f";
(0,-10)**\dir{-} ?(.75)*\dir{>}+(-2,0)*{\scriptstyle V};
\endxy
\in V
$$
Despite the fact that our formalism involves only vector spaces and
linear maps, we now have a way to talk about individual elements of
vector spaces.
Finally, we should have no problem interpreting this:
$$
\xy
(0,0)*++{f}*\cir{}="f" ;
\endxy
$$
This is a linear mapping from $\C$ to $\C$, clearly a complex number!
This last example will prove especially useful in the sequel, since it
means that a diagram with no arrows sticking into or out of it
represents a complex number.
\section{Duality (II)}
Given a vector space $V$ we can construct its dual $V^*$, and it comes
along with two bilinear maps: the unit and the counit.
\subsection{The Counit}
This is a map from $V^*\otimes V$ to $\C$ defined in the obvious way:
$$
\matrix{e_V\colon &V^*\otimes V&\rightarrow &\C \cr
&f\otimes v &\mapsto &f(v)\cr}
$$
This mapping is called ``evaluation'' or ``dual pairing'' and we draw
it as
$$
\xy
(-6,4)*{};(6,4)*{};
**\crv{(0,-12)} ?(.20)*\dir{>}+(2,-1)*{\scriptstyle V}
?(.80)*\dir{>}+(-2,-1)*{\scriptstyle V};
\endxy
=
\xy
(0,-4)*+{e_V}*\cir{}="e" ;
(-6,4)**\dir{-} ?(.75)*\dir{>}+(-2,-1)*{\scriptstyle V};
"e";(6,4)**\dir{-} ?(.5)*\dir{<}+(2,-1)*{\scriptstyle V};
\endxy
$$
We do not draw a blob in it because the dual pairing is canonical,
just like we did not draw a blob for the identity map on $V$. Another
name for this operation is ``cup'', for obvious reasons. To understand
the name ``counit'' we'll have to wait until the ``unit'' has been
introduced.
The cup operation was introduced into physics by Feynman, in
connection with his idea that antiparticles could be interpreted as
particles moving backwards in time. In this language, the cup diagram
is interpreted as the annihilation of a particle/anti-particle
pair.
$$
\xy
(-6,8)*{};
(6,8)**\crv{(0,-8)}
?(.25)*\dir{<}
?(.25)+(-2,-1)*{\scriptstyle \bar p}
?(.75)*\dir{<}
?(.75)+(2,-1)*{\scriptstyle p};
\endxy
$$
The reader may be wondering what ever happened to the photons
that are produced by such an annihilation event. The answer is that in
quantum field theory particles are described by representations of a
group including the Poincar\'e group and hence time-translation, But
time-translation symmetry implies energy conservation, so a
particle/anti-particle pair cannot annihilate into nothing.
$$
\xy
(-6,8)*{};
(6,8)**\crv{(0,-8)}
?(.25)*\dir{<}
?(.25)+(-2,-1)*{\scriptstyle \bar p}
?(.75)*\dir{<}
?(.75)+(2,-1)*{\scriptstyle p};
(0,0)*{};
(0,-8)**\dir{~} ?(.5)+(2,0)*{\scriptstyle \gamma};
\endxy
$$
With the tools at our disposal we still cannot talk about
time-translation invariance and so we are allowed to draw diagrams
such as the cup with nothing coming out at the bottom.
\subsection{The Unit}
The unit is a map $i_V\colon\C\rightarrow V\otimes V^*$. Now,
$V\otimes V^*\equiv\left\{{\rm linear~}T\colon V\rightarrow
V\right\}=\End(V)$, since to any $v\otimes f\in V\otimes V^*$ we can
assign a unique endomorphism
$$
\begin{array}{rccc}
T_{v\otimes f}\colon&V&\rightarrow&V\\
&w&\mapsto&vf(w)\\
\end{array}
$$
Now, the simplest endomorphism is the identity, and so we define
$$
\matrix{i_V\colon &\C &\rightarrow &V\otimes V^*\cr
&1 &\mapsto &\id_V \cr}
$$
The unit is drawn as
$$
\xy
(0,4)*+{i_V}*\cir{}="e" ;
(6,-4)**\dir{-} ?(.5)*\dir{<}+(3,-1)*{\scriptstyle V};
"e";(-6,-4)**\dir{-} ?(.75)*\dir{>}+(-2,1)*{\scriptstyle V};
\endxy
=
\xy
(-6,-4)*{};(6,-4)*{};
**\crv{(0,12)} ?(.20)*\dir{>}+(2,1)*{\scriptstyle V}
?(.80)*\dir{>}+(-2,1)*{\scriptstyle V};
\endxy
$$
and is called ``cup''. Note that this is not the dual of the counit,
as they are obtained from one another by reflection and not by
rotation.
Now that we have the unit and counit we can combine them in different
ways. For example, we have
$$
\xy
(-10,-10)*{};(10,10)*{};
**\crv{(5,-30)&(-5,30)} ?(.15)*\dir{>} ?(.5)*\dir{>} ?(.9)*\dir{>};
\endxy
\mathrel{\mathop{\Rightarrow}^?}\quad
\xy
(0,10)*{};(0,-10)**\dir{-} ?(.5)*\dir{>};
\endxy
$$
which we would like to equate to the identity on $V$ by
``straightening out'' the diagram. Let us prove that we can actually
do that:
$$
\xy
(-12,-24)*{};(12,24)*{};
**\crv{(6,-70)&(-6,70)} ?(.15)*\dir{>} ?(.5)*\dir{>} ?(.9)*\dir{>};
(-15,20)*{} ;
(15,20)**{-} ;
(-15,10)*{} ;
(15,10)**{-} ;
(-15,0)*{} ;
(15,0)**{-} ;
(-15,-10)*{} ;
(15,-10)**{-} ;
(-15,-20)*{} ;
(15,-20)**{-} ;
\endxy
\quad
\xy
(0,15)*+{j\colon v\mapsto 1\otimes v}="1"*\frm<5pt>{-};
(0,5)*+{i_V\otimes\id_V}="2"*\frm<6pt>{-};
(0,-5)*+{\id_V\otimes i_V}="3"*\frm<7pt>{-};
(0,-15)*+{\pi\colon v\otimes 1\mapsto v}="4"*\frm<8pt>{-};
"1";(0,25)**\dir{-} ?(.5)*\dir{<}+(-2,0)*{\scriptstyle V};
"2";"1";**\dir{-} ?(.5)*\dir{>}+(-4,0)*{\scriptstyle \C\otimes V};
"3";"2";**\dir{-} ?(.5)*\dir{>}+(-7,0)*{\scriptstyle V\otimes V^*\otimes V};
"4";"3";**\dir{-} ?(.5)*\dir{>}+(-4,0)*{\scriptstyle V\otimes\C};
"4";(0,-25)**\dir{-} ?(.5)*\dir{>}+(-2,0)*{\scriptstyle V};
\endxy
$$
We want to prove that $\pi(\id_V\otimes
e_V)(i_V\otimes\id_V)j=\id_V$. Pick $v\in V$. Then $j(v)=1\otimes
v\in\C\otimes V$. Now, $i_V(1)$ is the identity endomorphism on $V$,
which we can write as $e_i\otimes f^i$, where $f^i\in V^*$ is defined
by $f^i(e_j)=\delta^i_j$. Then $(i_V\otimes\id_V)(1\otimes
V)=e_i\otimes f^i\otimes v$. Now, $e_V(f^i\otimes v)=f^i(v)$, so
$(\id_V\otimes e_V)(e_i\otimes f^i\otimes v)=e_i\otimes v^i$, where
$v^i=f^i(v)\in\C$ are the components of $V$ with respect to the basis
$\{e_i\}$, as we now show:
$f^j(e_iv^i)=f^j(e_i)v^i=\delta^j_iv^i=v^j$. Finally, $\pi(e_i\otimes
v^i)=e_iv^i=v$ and we are done.
\begin{exercise}
Rigorously straighten out the following diagram:
$$
\xy
(-10,10)*{};(10,-10)*{};
**\crv{(5,30)&(-5,-30)} ?(.15)*\dir{>} ?(.5)*\dir{>} ?(.9)*\dir{>};
\endxy
\Rightarrow\quad
\xy
(0,10)*{}="v";(0,-10)**\dir{-} ?(.5)*\dir{<};
\endxy
$$
\end{exercise}
\section{Matrix Algebra}
An (associative) algebra is a (complex) vector space $A$ with a
bilinear ``multiplication'' $m\colon A\otimes A\rightarrow A$ and a
unit element $i\colon\C\rightarrow A$ satisfying
\begin{itemize}
\item the associative law:
$m\bigl(m(a,b),c\bigr)=m\bigl(a,m(b,c)\bigr)$, drawn as
$$
\xy
(-5,5)*+{m}*\cir{}="1";
(-10,15)**\dir{-} ?(.5)*\dir{<};
"1";(0,15)**\dir{-} ?(.5)*\dir{<};
(3,-5)*+{m}*\cir{}="2";
(10,15)**\dir{-} ?(.5)*\dir{<};
"1";"2";**\dir{-} ?(.5)*\dir{<};
"2";(0,-15)**\dir{-} ?(.75)*\dir{>};
\endxy
=
\xy
(5,5)*+{m}*\cir{}="1";
(10,15)**\dir{-} ?(.5)*\dir{<};
"1";(0,15)**\dir{-} ?(.5)*\dir{<};
(-3,-5)*+{m}*\cir{}="2";
(-10,15)**\dir{-} ?(.5)*\dir{<};
"1";"2";**\dir{-} ?(.5)*\dir{<};
"2";(0,-15)**\dir{-} ?(.75)*\dir{>};
\endxy
$$
and
\item the left and right unit laws, $m(a,i)=a=m(i,a)$, which we draw as
$$
\xy
(-5,10)*++{i}*\cir{}="i";
(0,0)*+{m}*\cir{}="m";**\dir{-} ?(.5)*\dir{<};
"m";(5,10)**\dir{-} ?(.5)*\dir{<};
"m";(0,-10)**\dir{-} ?(.75)*\dir{>};
\endxy
=
\xy
(0,10)*{};(0,-10)**\dir{-} ?(.5)*\dir{>};
\endxy
=
\xy
(5,10)*++{i}*\cir{}="i";
(0,0)*+{m}*\cir{}="m";**\dir{-} ?(.5)*\dir{<};
"m";(-5,10)**\dir{-} ?(.5)*\dir{<};
"m";(0,-10)**\dir{-} ?(.75)*\dir{>};
\endxy
$$
\end{itemize}
For the associative law to be true, we must extend the notion of
equivalence of diagrams to include the new operation ``sliding one
multiplication blob over another''. The unit laws can be interpreted
as ``unit and multiplication blobs cancel each other out''. Now, in
the case of matrix algebra we do not need to introduce these new
rules, but they follow from the ones we know already.
Given a vector space $V$, we can get an algebra by considering
$\End(V)$ with the operation of composition of maps. Now, we have
$\End(V)=V\otimes V^*$, so the unit is a certain element of $V\otimes V^*$
$$
\xy
(-6,-10)*{};(6,-10)*{};
**\crv{(0,6)} ?(.25)*\dir{>}+(2,1)*{\scriptstyle V}
?(.75)*\dir{>}+(-2,1)*{\scriptstyle V};
\endxy
\quad
\xymatrix{\C\ar[d]^{i_V}\\
V\otimes V^*\rlap{=A}\\}
$$
and matrix multiplication is based on the counit:
$$
\xy
(-6,2)*{};(6,2)*{};
**\crv{(0,-14)} ?(.75)*\dir{>} ?(.25)*\dir{>};
(-12,2)*{};(-3,-14)**\dir{-} ?(.5)*\dir{>};
(12,2)*{};(3,-14)**\dir{-} ?(.5)*\dir{<};
\endxy
\quad
\xymatrix{A\otimes A\ar[d]^m\\
A\\}
$$
The associative property can be drawn as
$$
\xy
(-3,-12)*{};(-21,12)**\dir{-} ?(.5)*\dir{<};
(-15,12)*{};(-3,12)*{};
**\crv{(-6,0)} ?(.25)*\dir{>} ?(.75)*\dir{>};
(3,12)*{};(15,12)*{};
**\crv{(-9,-16)} ?(.25)*\dir{>} ?(.75)*\dir{>};
(21,12)*{};(3,-12)**\dir{-} ?(.5)*\dir{<};
\endxy
=
\xy
(-3,-12)*{};(-21,12)**\dir{-} ?(.5)*\dir{<};
(-15,12)*{};(-3,12)*{};
**\crv{(6,-16)} ?(.25)*\dir{>} ?(.75)*\dir{>};
(3,12)*{};(15,12)*{};
**\crv{(9,0)} ?(.25)*\dir{>} ?(.75)*\dir{>};
(21,12)*{};(3,-12)**\dir{-} ?(.5)*\dir{<};
\endxy
$$
where equality follows by ``shifting''. Similarly, the unit laws can
be drawn as
$$
\xy
(-3,-15)*{};(6,15)**\crv{(0,-5)&(-9,5)&(-6,15)&(3,5)} ?(.125)*\dir{<}
?(.5)*\dir{<}
?(.7)*\dir{<}
?(.875)*\dir{<};
(3,-15)*{};(12,15)**\dir{-} ?(.33)*\dir{>}
?(.75)*\dir{>};
\endxy
=\quad
\xy
(-3,-15)*{};(-3,15)**\dir{-} ?(.5)*\dir{<};
(3,-15)*{};(3,15)**\dir{-} ?(.5)*\dir{>};
\endxy
\quad =
\xy
(3,-15)*{};(-6,15)**\crv{(0,-5)&(9,5)&(6,15)&(-3,5)} ?(.125)*\dir{<}
?(.5)*\dir{<}
?(.7)*\dir{<}
?(.875)*\dir{<};
(-3,-15)*{};(-12,15)**\dir{-} ?(.33)*\dir{>}
?(.75)*\dir{>};
\endxy
$$
which is true because we can ``straighten out the bends''.
\chapter{Lagrangians for Field Theories (II)}
\section{Gauge-invariant Lagrangians for Gauge Theories (II)}
So far we have seen that, starting from just $A$ on a $2n$-dimensional
manifold, we can obtain the gauge-invariant Lagrangian $\tr(F^{\wedge
n})$. This seems to be the end of the story unless we endow the
manifold with additional structure.
\subsection{Yang--Mills Theory}
If $M$ is equipped with a metric (i.e. a symmetric, non-degenerate
$2$-form\footnote{Do we need to mention that the signature need not be
Euclidean?}), we can construct a Hodge $*$ operator which turns
$p$-forms into $(n-p)$-forms, so $*F$ is an $Ad(P)$-valued
$(n-2)$-form---locally a $\frak g$-valued $(n-2)$-form. Then, $F\wedge
*F=*F\wedge F$ is an $n$-form whose trace is gauge-invariant!
$$
\tr(F'\wedge *F')=\tr(gFg^{-1}\wedge *gFg^{-1})=\tr(gFg^{-1}\wedge
g*Fg^{-1})=\tr(gF\wedge *F g^{-1})=\tr(F\wedge *F),
$$
where $\wedge$ and $*$ act on the differential form part only and $g$
acts on the $Ad(P)$ part only, so $g$ can jump over the other
operators. This is the so-called Yang--Mills Lagrangian, and as far as
we know it describes all known forces except gravity. (Note: because
both $*$ and the measure change sign under a change of orientation,
the Yang--Mills action is independent of the orientation of the
manifold.) If we want to construct a theory of gravity we don't want
the metric to be fixed {\it a priori\/}, but the Yang--Mills
Lagrangian hints at a way to obtain new gauge-invariant lagrangians by
adding more fields.
\subsection{$EF$ Theory}
The properties of $*F$ that we needed in order to prove that the
Yang--Mills Lagrangian is gauge-invariant are that: 1) it is an
$(n-2)$-form so $\cal L$ can be integrated over an $n$-dimensional
manifold; and 2) it is $Ad(P)$ valued, so it has the proper behaviour
under gauge transformations. Accordingly, we will assume that the
manifold $M$ is equipped not only with a connection, but also with a
new field $E$ which is an $Ad(P)$-valued $(n-2)$-form. Now we can form
the lagrangian $\tr(E\wedge F)$, which gives rise to the ill-named
``$BF$~theory''. Originally, $E$ was ``wrongly'' named $B$ because it
was supposed to be analogous to the magnetic field, when in reality it
is closer to the electric field, as we will see later on. We will use
the name ``$EF$ theory'' throughout.
$EF$~theory leads to general relativity in $3D$ if we choose $G=\SO(3)$
(Riemannian gravity) or $G=\SO(2,1)$ (Lorentzian gravity). Now we
observe that, in $3D$, $E$ is a $1$-form, so $\tr(E\wedge E\wedge E)$
is also a valid Lagrangian. Let us now list all the $EF$ Lagrangians
in various dimensions:
$$
\centerline{
\vbox{
\halign{\hfil$#D$\quad&&\quad\hfil $#\relax$\hfil\cr
\noalign{\smallskip\hrule\smallskip}
2&\tr(F) &\tr(E\wedge F)&\tr(E\wedge E\wedge F)&\tr(E\wedge
E\wedge E\wedge F)&\ldots\cr
3& &\tr(E\wedge F)& &\tr(E\wedge
E\wedge E)\cr
4&\tr(F\wedge F)&\tr(E\wedge F)&\tr(E\wedge E)\cr
(2m-1)& &\tr(E\wedge F)\cr
(2m)&\tr(\wedge^m F)&\tr(E\wedge F)\cr
\noalign{\smallskip\hrule\smallskip
\hbox{$E$ is an $(n-2)$-form in $nD$; $m\ge 3$.}}
}}}
$$
In all these cases $\wedge$ is taken to mean ``product in $Ad(P)$''
and ``wedging of the differential forms''\footnote{This is a place
where the abstract index notation for internal indices resolves
ambiguities}. We observe that in $2D$ there is an infinite collection
of linearly independent $EF$ Lagrangians, since $E$ is, in this case,
a $0$-form (an $Ad(P)$-valued function) and can be wedged with $F$
arbitrarily many times to give a $2$-form. In $3D$ and $4D$ there are
terms depending only on $E$. These terms are called ``cosmological''
because $3D$ general relativity with cosmological constant $\lambda$
follows from the Lagrangian $\tr(E\wedge F)+\lambda\tr(E\wedge E\wedge
E)$. In $5D$ or higher dimension there cannot be a cosmological term.
Incidentally, the cosmological constant $\lambda$ of $3D$ general
relativity is closely related to the ``$q$ parameter'' appearing in
the theory of quantum groups, so maybe ``quantum'' is also an
inappropriate adjective and these groups should be called
``cosmological groups'' instead! I don't expect this terminology
to catch on, but we'll see it makes sense.
\subsection{$4D$ General Relativity}
$4D$ $EF$~theory is not equivalent to general relativity for any
choice of the gauge group, so for this purpose we need something more
complicated. The Lagrangian for $4D$ general relativity follows from
the following trick, which works for all the $\SO(p,q)$ groups\footnote{In
fact, with a slight modification it works for all symplectic and
unitary groups, too.}.
For simplicity, let's consider $\SO(n)$ and let $V=\R^n$ with its
usual inner product. The lie algebra $\so(V)$ consists of the linear
transformations of $V$ that are skew-adjoint with respect to the inner
product. This is a Lie subalgebra of $\End(V)=V\otimes V^*$---i.e., it
is a subspace that is closed under commutators (Lie brackets). Now,
the inner product on $V$ provides a canonical isomorphism between $V$
and $V^*$, so we can consider $\so(V)$ imbedded in $V\otimes V$ as
$\Lambda^2V$ (skew-symmetric $2$-tensors, or ``bivectors'').
The trick is to use this identification in the opposite
direction. Suppose we have a $V$-valued \hbox{$1$-form} $e$ on
spacetime. Then we can define $e \wedge e$ to be a $\Lambda^2
V$-valued 2-form in the following way: if $e=e_i dx^i$, where
$\{e_i\}$ are vectors in $V$, $e\wedge f=(e_i\wedge f_j)dx^i\wedge
dx^j$, which is symmetric in $e,f$ so that $e\wedge e=(e_i\wedge
e_j)dx^i\wedge dx^j$ does not necessarily vanish. By the above
identification, we can reinterpret this as an $\so(V)$-valued 2-form.
This is very much like the $E$ field in 4-dimensional $EF$ theory with
gauge group $\SO(n)$!
This allows us to write down an $EF$-like Lagrangian for $4D$
Riemannian general relativity in terms of two basic fields:
\begin{itemize}
\item an $\SO(4)$ connection $A$, and
\item the {\sl cotetrad} field $e$, which is locally an
$\R^4$-valued 1-form.
\end{itemize}
The Lagrangian looks like this: $\tr(e \wedge e \wedge F)$, where
$\tr$ is the ``trace'' on $Ad(P)$
%Observe that the groups $\SO(p,q)$, $\Sp(n)$ are
%invariance groups of bilinear forms $\langle~,~\rangle$ on a vector
%space $V$ ($U(n)$ is the invariance group of a sesquilinear
%form). Therefore they induce canonical isomorphisms from $V$ to $V^*$
%(antiisomorphisms in the case of $U(n)$), which allow us to transform
%the $V$-valued vierbein $1$-form into a $V^*$-valued $1$-form. Now,
%the invariance group $G$ acts on $V$ by left multiplication and, if we
%write $\langle w,v\rangle=w^*(v)$, we have
%$$
%(gw)^*(gv)=\langle gw,gv\rangle=\langle
%w,v\rangle=w^*(v)=w^*(g^{-1}gv)\Leftrightarrow (gw)^*=w^*g^{-1}.
%$$
%so it follows that $G$ acts on $V^*$ by right multiplication by the
%inverse. Therefore, $G$ acts on $V\otimes V^*$ by conjugation and, if
%$e$ and $f$ are cotetrad fields, $e\wedge f^*$ is an $Ad(P)$-valued
%$2$-form.