\chapter{Diagrammatic Methods for Linear Algebra (V)}
So far we have studied the category of finite-dimensional complex
vector spaces with linear maps. Diagrammatically, the operations in
the category give rise to structures which have natural diagrammatic
representations of various dimensions:
\begin{itemize}
\item the basic operation in any {\bf category\/} is the {\bf
composition\/} of morphisms, which we can draw as a $1$-dimensional
diagram with nothing to the side:
$$
\xy
(0,7)*++{f}*\cir{}="f";
(0,17)**\dir{-} ?(.5)*\dir{<}+(2,0)*{\scriptstyle U};
"f";
(0,-7)*++{g}*\cir{}="g";
**\dir{-} ?(.5)*\dir{<}+(2,0)*{\scriptstyle V};
"g";
(0,-17)**\dir{-} ?(.75)*\dir{>}+(2,0)*{\scriptstyle W};
\endxy
$$
Since linear maps can be composed we say that vector spaces with
linear maps as morphisms form a category, which we call $\rm Vect$.
\item\textbf{products\/} in a category are represented by $2$-dimensional
diagrams like
$$
\xy
(-5,0)*++{f}*\cir{}="f"; (-5,10)**\dir{-}
?(.5)*\dir{<}+(2,0)*{\scriptstyle V_1}; (-5,-10)*{}; "f"**\dir{-}
?(.25)*\dir{<}+(3,0)*{\scriptstyle W_1}; (5,0)*++{g}*\cir{}="g";
(5,10)**\dir{-} ?(.5)*\dir{<}+(2,0)*{\scriptstyle V_2}; (5,-10)*{};
"g"**\dir{-} ?(.25)*\dir{<}+(3,0)*{\scriptstyle W_2};
\endxy
$$
A category with a product is called {\bf monoidal\/}, and since
tensoring of vector spaces is a product, we say that $\rm Vect$ is a
monoidal category.
\begin{itemize}
\item If we have a monoidal category in which every object has morphisms
to the tensor identity object, we say the category {\bf has duals\/}
and we can define ``cup'' and ``cap'' operators:
$$
\xy
(-6,4)*{};
(6,4)**\crv{(0,-12)} ?(.25)*\dir{<}+(-1,-2)*{\scriptstyle V}
?(.75)*\dir{<}+(1,-2)*{\scriptstyle V}
\endxy
\quad=\quad
\xy
(-6,4)*{};
(6,4)**\crv{(0,-12)} ?(.25)*\dir{>}+(-1,-2)*{\scriptstyle V^*}
?(.75)*\dir{<}+(1,-2)*{\scriptstyle V}
\endxy
\qquad{\rm and}\qquad
\xy
(-6,-4)*{};
(6,-4)**\crv{(0,12)} ?(.25)*\dir{<}+(-1,2)*{\scriptstyle V}
?(.75)*\dir{<}+(1,2)*{\scriptstyle V}
\endxy
\quad=\quad
\xy
(-6,-4)*{};
(6,-4)**\crv{(0,12)} ?(.25)*\dir{<}+(-1,2)*{\scriptstyle V}
?(.75)*\dir{>}+(2,2)*{\scriptstyle V^*}
\endxy
$$
The category $\rm Vect$ is a monoidal category with duals.
\end{itemize}
\item The product in a monoidal category depends on the order of the
factors, but if there may be a {\bf braiding\/}, that is, an
isomorphism between $V\otimes W$ and $W\otimes V$
$$
\xy
\vtwist[9]|>>>|{W}>{V};
\endxy
$$
which is drawn as the $2$-dimensional projection of a $3$~dimensional
structure. In this case the category is called {\bf braided\/}. Since
tensor products of vector spaces come equipped with a braiding, we say
that $\rm Vect$ is a braided monoidal category with duals.
\item a braided monoidal category is called {\bf symmetric\/} if
$$
\xy
+(0,5),\vtwist[9]|>>>|{W}>{V};
\endxy
\quad =\quad
\xy
+(0,5),\vcross[9]<<|>|{V}<{W};
\endxy
$$
This diagram represents a Reidemeister move valid in more than three
dimensions, so we can say that the diagram is $4$-dimensional. Since
this is the case for vector spaces, the category $\rm Vect$ is a
symmetric monoidal category with duals.
\end{itemize}
It is generally true for any category that at the highest dimension we
have a property (i.e. an identity that the structures of lower
dimension have to satisfy) rather than a new structure. In the case
of $\rm Vect$ we do not have a new structure in four dimensions but a
property satisfied by the three-dimensional braiding.
As promised at the beginning, one starts to see geometry arising from
just algebraic diagrams. The fact that the category on which quantum
mechanics is based is four-dimensional like space-time (at least
macroscopically) may be just a coincidence, but we may also take it as
an indication that quantum gravity will turn out to be
four-dimensional.
We have also seen that some vector spaces (orthogonal and symplectic)
have a canonical isomorphism from $V$ to $V^*$
$$
\xy
(-6,4)*{};
(6,4)**\crv{(0,-12)}
?(.5)*\dir{*}+(0,-2)*{\scriptstyle g}
?(.1)*\dir{>}+(-2,0)*{\scriptstyle V}
?(.9)*\dir{<}+(2,0)*{\scriptstyle V};
\endxy
\Rightarrow
\xy
(0,10)*{};
(0,-10)**\dir{-}
?(.5)*\dir{*}+(-2,0)*{\sharp}
?(.25)*\dir{>}+(2,0)*{\scriptstyle V}
?(.75)*\dir{<}+(2,0)*{\scriptstyle V};
\endxy
\quad{\rm or}\quad
\xy
(-6,4)*{};
(6,4)**\crv{(0,-12)}
?(.5)*\dir{o}+(0,-2)*{\scriptstyle \omega}
?(.1)*\dir{>}+(-2,0)*{\scriptstyle V}
?(.9)*\dir{<}+(2,0)*{\scriptstyle V};
\endxy
\Rightarrow
\xy
(0,10)*{};
(0,-10)**\dir{-}
?(.5)*\dir{o}+(-2,0)*{\sharp}
?(.25)*\dir{>}+(2,0)*{\scriptstyle V}
?(.75)*\dir{<}+(2,0)*{\scriptstyle V};
\endxy
$$
This blurs the distinction between $V$ and $V^*$. In diagrammatic
terms, it allows us to eschew the arrows on the lines representing
the vector space $V$.
A metric has a symmetry property
$$
\xy
(-6,4)*{};(6,4)*{};
**\crv{(0,-12)} ?(.5)*\dir{*}
?(.10)*\dir{>}*{\strut^V~}
?(.90)*\dir{<}*{\strut_V~}
\endxy
=
\xy
+(0,6)\vtwist[9] |>>>|{V}>{V};
+(0,-3)\vcap[-9] |{*\dir{*}};
\endxy
$$
which is the first Reidemeister move for two-dimensional projections
of knots in three dimensions. A symplectic structure, on the other
hand, gives rise to diagrams violating the first Reidemeister move:
$$
\xy
(-6,4)*{};(6,4)*{};
**\crv{(0,-12)} ?(.5)*\dir{o}
?(.10)*\dir{>}*{\strut^V~}
?(.90)*\dir{<}*{\strut_V~}
\endxy
=-
\xy
+(0,6)\vtwist[9] |>>>|{V}>{V};
+(0,-3)\vcap[-9] |{*\dir{o}};
\endxy
$$
We can resolve this discrepancy by representing symplectic vector
spaces by ribbons rather than by strings; technically, we replace
knots by framed knots. We have \footnote{Obtain the diagrams!}
$$
(\hbox{diagram})
$$
for ribbons imbedded in any dimension for which the diagram makes
sense.
This would be a natural thing to do if we want to use $\rm Vect$ to
represent spin-$1/2$ particles in quantum mechanics (e.g. electrons)
since rotating one such particle by $360^\circ$ results in multiplying
the state by $-1$. We have
$$
(\hbox{diagram: twist})
$$
In four dimensions this implies
$$
(\hbox{diagram: }-1)
$$
PROOF
To describe physical particles of spin $1/2$ we want to use symplectic
vector spaces. A vector space with a non-degenerate symplectic form
must be even dimensional, so the simplest nontrivial complex vector
space admitting a symplectic structure is $\C^2$. The symplectic
structure $\alpha\colon\C^2\otimes\C^2\rightarrow\C$ is determined by
the quantity $\alpha(e_i,e_j)=\alpha_{ij}=-\alpha_{ji}$, so
$\alpha=\alpha_{12}\bigl({0\atop-1}{1\atop 0}\bigr)$.
We want to find now an operator
$\beta\colon\C\rightarrow\C^2\otimes\C^2$ such that the the following
identities are satisfied:
\begin{enumerate}
\item antisymmetry
$$
(\hbox{diagram})
$$
\item
$$
(\hbox{bubble diagram }=-2)
$$
\item ``binor identity'', due to Penrose and named by him by
analogy with the term ``spinor''.
$$
(\hbox{diagram})
$$
\end{enumerate}
We know that any solution to $1)$ is of the form
$\alpha=A\bigl({0\atop-1}{1\atop 0}\bigr)$. This will be a symplectic
structure if and only if $A\neq 0$. Applying the binor identity to the
$\beta$ ``cap'', we obtain
$$
(\hbox{diagram})
$$
so $\beta$ is antisymmetric and it is of the form
$\beta=B\bigl({0\atop-1}{1\atop 0}\bigr)$. Finally, the bubble diagram
evaluates to
$$
-2=(\hbox{bubble})=2AB,
$$
so $AB=-1$ and $\alpha$ is non-degenerate and therefore a symplectic
structure. As a consistency check, we can see that the binor identity
is satisfied for any choice of $A\neq 0$.
To sum up, given a symplectic structure $\alpha$ on $\C^2$ we have a
cap $\beta=\alpha^T$ such that the above equations are satisfied, and
any solution to those equations is of this form.
\chapter{Physics from Lagrangians(III)}
\section{Lagrangians in Field Theory(III)}
So far all our equations of motion have turned out to be
vacuous, but this is not entirely bad news: we have seen that given a
principal $G$-bundle $P\rightarrow M$ with a connection $A$ on a
$2n$-dimensional manifold, we can define the gauge-invariant action
$$
S[A]=\int_M\tr(F^{\wedge n}).
$$
Then, $\delta S[A,\delta A]=0$ for all $A$ and $\delta A$, although we
have only proved it for $n\ge 2$.
\begin{exercise}
Derive the equations of motion for the $n$-th Chern theory and show
that they reduce to the Bianchi identity for all $n\ge 1$.
\end{exercise}
This is not entirely bad news, as it means that $S[A]$ depends only on
the bundle and not on the connection. The numbers $S[A]$ are
invariants called Chern numbers which can be used to classify
principal $G$-bundles.
For example, if $G=U(1)$ and $M$ is two-dimensional manifolds the
first Chern number classifies principal bundles in the sense that, if
$P\rightarrow M$ and $P'\rightarrow M$ have the same first Chern
number, then they are isomorphic. If $G=SU(2)$ and $M$ is
four-dimensional, the second Chern number, which is related to the
``$\theta$ angle'' of the standard model of particle physics,
classifies the principal $G$-bundles.
\subsection{Maxwell Theory}Before taking up nonabelian Yang--Mills
theory, we consider the abelian case (Maxwell's theory).
Assume that $G=U(1)$ and the manifold $M$ is equipped with a
metric. Since $G$ is abelian, the adjoint representation coincides
with the trivial (scalar) representation, so $Ad(P)$-valued objects
are invariant. Therefore, $d_A=d$ on the adjoint representation; one
can also argue that the term $[A,A]$ vanishes because the Lie algebra
of $G=U(1)$ is $\frak g\cong i\R$. On the fundamental representation
we still have $d_A=d+A\wedge$, and the curvature $2$-form is $F=dA$
because $A$ is an ordinary $1$-form and $A\wedge A=0$. Also, in the
presence of a metric we have a Hodge $*$ operator mapping $p$-forms to
$(n-p)$-forms in such a way that, for any $p$-form $F$ and
$(n-p)$-form $G$,
$$
F\wedge G=\left<*F,G\right>{\rm vol},
$$
where $\rm vol$ is the metric-induced volume $n$-form and
$\left<~,~\right>$ acts on $p$-forms by
$$
\left<\alpha_1\wedge\cdots\wedge\alpha_p,\beta_1\wedge\cdots\wedge\beta_p\right>_p=\det\left<\alpha_i,\beta_j\right>_1.
$$
So, $*F$ is an $(n-2)$-form and we can form the action
$$
S[A]=\int_M\tr(F\wedge *F)=\int_M F\wedge *F,
$$
which describes a universe made purely of light. The variation of the
action is
\begin{eqnarray*}
\delta S[A,\delta A]&=&\delta\int_M F\wedge *F=\int_M\delta(F\wedge
*F)=\int_M(\delta F)\wedge *F+F\wedge\delta{*F}=\int_M(\delta F)\wedge
*F+F\wedge*\delta F\\
&=&\int_M(\delta F)\wedge *F+F\wedge*\delta F\\
\end{eqnarray*}
The Hodge operator and wedge product have the following symmetry
property\footnote{This solves the exercise posed in the lecture: show
that this does not depend on the signature of the metric}: if $F,G$
are $p$-forms,
$$
F\wedge *G=\left<*F,*G\right>{\rm vol}=\left<*G,*F\right>{\rm
vol}=G\wedge *F.
$$
Therefore,
\begin{eqnarray*}
\delta S[A,\delta A]&=&\int_M(\delta F)\wedge
*F+F\wedge *\delta F=2\int_M(\delta F)\wedge *F=2\int_M (d_A\delta
A)\wedge *F=2\int_M (d\delta A)\wedge *F=\\
&=&2\int_M d(\delta A\wedge
*F)+\delta A\wedge d{*F}=2\int_{\partial M}\delta A\wedge
*F+2\int_M\delta A\wedge d{*F}. \\
\end{eqnarray*}
As usual, we can force the first integral to vanish by requiring
$\delta A$ to vanish on $\partial M$, and
$$
0=\delta S[A,\delta A]=2\int_M\delta A\wedge d{*F}
$$
implies that the (vacuum) Maxwell's equations
$$
d{*F}=0
$$
hold.
To see what this has to do with Maxwell's equations, assume that
$M=\R\times S$, where $S$ is an $(n-1)$-dimensional manifold
(space). Then $F=dt\wedge E+B$, where $E$ (resp. $B$) is a $1$-form
(resp. $2$-form) on $S$. Now, if $G$ is a $p$-form on $S$,
$$
-\left{\rm vol}=\left<*_S G,*_S
G\right>_S dt\wedge{\rm vol}_S=dt\wedge G\wedge *_S G=(-1)^p G\wedge
dt\wedge *_S G
$$
Equating the first and last terms, we have
$$
*G=(-1)^{p+1}dt\wedge{*_S G}.
$$
Similarly, equating the middle terms one obtains
$$
*(dt\wedge G)=*_S G.
$$
Therefore, $*F=-dt\wedge *_SB+*_SE$, and $**F=-(dt\wedge
*_S*_SE+*_S*_SB)=-F$ since $*_S*_S=\id$. Then, $0=d{*F}=dt\wedge
d_S{*_SB}+dt\wedge \partial_t{*_SE}+d_S{*_SE}$ so
$$
*_Sd_S{*_SB}+\partial_tE=0\qquad{\rm and}\quad *_S d_S{*_S E}=0.
$$