\chapter{Diagrammatic Methods for Linear Algebra (III)}
We are now ready to answer the question why dualisation is represented
by a $180^\circ$ rotation rather than a reflection:
\begin{exercise}
Given $T\colon V\rightarrow W$, show that
$$
\xy
(0,0)*++{T}*\cir{}="t"; (-10,10)**\crv{(-5,-15)}
?(.5)*\dir{>}+(1,-3)*{\scriptstyle W}; "t";(10,-10)**\crv{(5,15)}
?(.5)*\dir{<}+(1,1)*{\scriptstyle V};
\endxy
=\quad
\xy
(0,0)*+{T^*}*\cir{}="t";
(-0,10)**\dir{-} ?(.6)*\dir{>}+(2,0)*{\scriptstyle W};
"t";(0,-10)**\dir{-} ?(.5)*\dir{<}+(2,0)*{\scriptstyle V};
\endxy
$$
\end{exercise}
We note in passing that with our notation it becomes obvious that the
dual of a tensor product is the tensor product of the duals in the
opposite order: $(V\otimes W)^*=W^*\otimes V^*$.
$$
\left(
\xy
(0,5)*++{T}*\cir{}="t";
(-6,-5)**\dir{-} ?(.67)*\dir{>}+(-2,0)*{\scriptstyle V};
"t";(6,-5)**\dir{-} ?(.67)*\dir{>}+(2,0)*{\scriptstyle W};
\endxy
\right)^*
=\quad
\xy
(0,-5)*+{T^*}*\cir{}="t";
(6,5)**\dir{-} ?(.67)*\dir{>}+(2,-1)*{\scriptstyle V};
"t";(-6,5)**\dir{-} ?(.67)*\dir{>}+(-2,-1)*{\scriptstyle W};
\endxy
$$
\section{Braidings}
There is an isomorphism
$$
\matrix{B_{V,W}\colon&V\otimes W&\rightarrow&W\otimes V\cr
&v\otimes w&\mapsto &w\otimes v\cr}
$$
called ``braiding'' which we draw as
$$
\xy
\vtwist[9]>>>{V}|>|{W}
\endxy
$$
The braiding has an inverse
$$
\matrix{B_{V,W}^{-1}\colon&W\otimes V&\rightarrow&V\otimes W&=\cr
&w\otimes v&\mapsto &v\otimes w&\cr}
\xy 0;
+(0,7),\vcross[9]<<<{V}|>|{W}
\endxy
$$
Note that, diagrammatically, $B^{-1}_{V,W}\neq B_{W,V}$ because
$$
\matrix{
\xy
(-4,8)*{};(-4,-8)**\dir{-} ?(.5)*\dir{>}+(2,0)*{\scriptstyle V};
(4,8)*{};(4,-8)**\dir{-} ?(.5)*\dir{>}+(2,0)*{\scriptstyle W};
\endxy
&=&
\xy /r2pc/:+(0,1),
{\vtwist|>|{W}\vcross><>{V}};
\endxy
&\neq&
\xy /r2pc/:+(0,1),
{\vtwist>>>{V}\vtwist<><{W}};
\endxy
\cr
\id_{V\otimes W}
&=&
B^{-1}_{V,W}B_{V,W}
&\neq&
B_{W,V}B_{V,W}\cr}
$$
The first equality is called ``the second Reidemeister move'' in knot
theory. This is an example of how to become famous by being the first
mathematician to state a trivality.
Now, for any linear operator $f\colon
U\rightarrow V$ we have $B_{V,W}(f\otimes 1_W)=(1_W\otimes
f)B_{U,W}$. This is drawn
$$
\xy
(0,4)*++{f}*\cir{}="f";(0,13)**\dir{-} ?(.5)*\dir{<}+(2,0)*{\scriptstyle U};
\vtwist~{"f"}{(10,13)}{(0,-13)}{(10,-13)}|>|{W}>>>{V};
\endxy
=
\xy
(10,-4)*++{f}*\cir{}="f";(10,-13)**\dir{-} ?(.75)*\dir{>}+(2,0)*{\scriptstyle V};
\vtwist~{(0,13)}{(10,13)}{(0,-13)}{"f"}|><>|{W}<{U};
\endxy
$$
Replacing $f$ by $B_{U,V}\colon U\otimes V\to V\otimes U$, we obtain
the identity known as ``the third Reidemeister move'', which relates
two different ways to go from $U\otimes V\otimes W$ to $W\otimes
V\otimes U$ by repeated braiding:
$$
\xy /r1pc/:
\vtwist~{(-2,3)}{(0,3)}{(-2,0)}{(0,1)}|>>>|{V}>{U};
\vtwist~{(0,1)}{(2,3)}{(0,-1)}{(2,-3)}|>|{W};
\vtwist~{(-2,0)}{(0,-1)}{(-2,-3)}{(0,-3)};
\endxy
=
\xy /r1pc/:
\vtwist~{(0,3)}{(2,3)}{(0,1)}{(2,0)}|>>>|{W}>{V};
\vtwist~{(-2,3)}{(0,1)}{(-2,-3)}{(0,-1)}>>>{U};
\vtwist~{(0,-1)}{(2,0)}{(0,-3)}{(2,-3)};
\endxy
$$
Finally, the first Reidemeister move applies to just one space, and we
introduce it last because it involves the unit and counit:
$$
\xy /r1pc/:
\hover~{(1,-3)}{(-1,-1)}{(1,3)}{(-1,1)}<<|>>><{V^*}|{V}>{V^{**}};
\xcapv~{(-1,-1)}{(-2,-1)}{(-1,1)}{(-2,1)}<>><<{V^*}>{V};
\endxy
$$
If $V$ is reflexive (as is the case for the finite-dimensional spaces
we are considering) this represents the canonical isomorphism between
$V$ and $V^{**}$.
\begin{exercise}
Prove the first Reidemeister move:
$$
\xy /r1pc/:
\hunder~{(1,2)}{(-1,1)}{(1,-2)}{(-1,-1)}|<;
\xcapv~{(-1,1)}{(-2,1)}{(-1,-1)}{(-2,-1)};
\endxy
=
\xy
(0,10)*{};(0,-10)**\dir{-} ?(.5)*\dir{>}
\endxy
$$
\end{exercise}
The non-trivial thing that Reidemeister did was to prove the following
theorem:
\begin{theorem}
Given two two-dimensional projections of the same knot, one can be
obtained from the other by a composition of one-parameter
diffeomorphisms of the plane and the three Reidemeister moves.
\end{theorem}
As an aside, we have defined the cap and cup operations
$$
\xy
(-6,-4)*{};(6,-4)**\crv{(0,12)} ?(.25)*\dir{<} ?(.75)*\dir{<};
\endxy
\qquad{\rm and}\qquad
\xy
(-6,4)*{};(6,4)**\crv{(0,-12)} ?(.25)*\dir{<} ?(.75)*\dir{<};
\endxy
$$
but there are two other operations which can be
drawn as a cap and a cup. They can be defined from the usual ones by
duality
$$
\xy
(6,-4)*{};(-6,-4)**\crv{(0,12)} ?(.25)*\dir{<} ?(.75)*\dir{<};
\endxy
=\biggl(\,\xy
(-6,-4)*{};(6,-4)**\crv{(0,12)} ?(.25)*\dir{<} ?(.75)*\dir{<};
\endxy\,\biggr)^*
\qquad{\rm and}\qquad
\xy
(6,4)*{};(-6,4)**\crv{(0,-12)} ?(.25)*\dir{<} ?(.75)*\dir{<};
\endxy
=\biggl(\,\xy
(-6,4)*{};(6,4)**\crv{(0,-12)} ?(.25)*\dir{<} ?(.75)*\dir{<};
\endxy\,\biggr)^*
$$
or by braiding
$$
\xy
(6,-4)*{};(-6,-4)**\crv{(0,12)} ?(.25)*\dir{<} ?(.75)*\dir{<};
\endxy
=
\xy
+(0,3)\vtwist[9]\vcap[9] |>;
\endxy
\qquad{\rm and}\qquad
\xy
(6,4)*{};(-6,4)**\crv{(0,-12)} ?(.25)*\dir{<} ?(.75)*\dir{<};
\endxy
=
\xy
+(0,6)\vtwist[9];
+(0,-3)\vcap[-9] |<;
\endxy
$$
\begin{exercise}
Check that the two definitions coincide.
\end{exercise}
In linear algebra, the ``four-dimensional'' fact that
$B_{V,W}^{-1}=B_{W,V}$, drawn as
$$
\xy
+(0,5)\vcross[9] <<|> <{V}|{W}
\endxy
=
\xy
+(0,5)\vtwist[9] |>>> |{V}>{W}
\endxy
$$
is true because both map $w\otimes v$ to $v\otimes w$. This means the
category of vector spaces is a symmetric monoidal category, and we say
symmetry is ``four-dimensional'' because only in more than three
dimensions is the above diagram true.
This means that vector spaces and linear maps are nicely suited to
(the generally rather boring) four-dimensional knot theory. To have an
interesting knot-theoretical application of our formalism we would
have to modify the identity above; in other words, we would have to
modify the braiding so that it is not symmetrical. This was realised
in the 1980's, when representations of quantum groups were used to
obtain invariants of knots.
Now, consider an endomorphism of $V$
$$
\xy
(0,0)*++{f}*\cir{}="f";
(0,10)**\dir{-} ?(.5)*\dir{<};
"f";(0,-10)**\dir{-} ?(.75)*\dir{>};
\endxy
$$
and stick a unit (cap) at the top and a ``pseudocup'' at the bottom:
$$
\xy
(0,0)*++{f}*\cir{}="f";
"f"**\crv{(0,15)&(15,0)&(0,-15)} ?(.25)*\dir{<} ?(.75)*\dir{<};
\endxy
$$
This, having no free lines, is obviously a complex number that we can
obtain from $T$ in a canonical way. The only likely candidate for this
is the trace of $T$, which we now check is true.
$$
\xy
(0,0)*++{f}*\cir{}="f";
"f"**\crv{(0,15)&(15,0)&(0,-15)} ?(.25)*\dir{<} ?(.75)*\dir{<};
(-5,9)*{};
(10,9)**{-};
(-5,3)*{};
(10,3)**{-};
(-5,-3)*{};
(10,-3)**{-};
(-5,-9)*{};
(10,-9)**{-};
(20,9)*{\C};
(20,3)*{V\otimes V^*};
(20,-3)*{V\otimes V^*};
(20,-9)*{\C};
(35,9)*{1};
(35,3)*{e_i\otimes e^i};
(35,-3)*{f(e_i)\otimes e^i};
(35,-9)*{e^i\bigl(f(e_i)\bigr)};
\endxy
$$
And to end, a puzzle. What is this?
$$
\xy
\hcap[-9]\hcap[9] |>|{V}
\endxy
$$
If we interpret this as the trace of the identity operator, we
conclude that it is nothing other than the dimension of $V$!
\chapter{Physics from Lagrangians (I)}
\section{Lagrangians in Particle Mechanics}
To illustrate the Lagrangian formulation of mechanics we will use the
simplest case of all, that of a particle moving in one dimension. The
trajectory of the particle will be a function $q\colon\R\rightarrow\R$
where the domain represents time and the codomain represents
one-dimensional space. A Lagrangian is then a function $L(q,\dot q;
t)$. In most applications the Lagrangian can be split into two parts,
$L=T(\dot q)-U(q;t)$, where $T$ is called kinetic energy and $U$
potential energy. For one-dimensional particle motion $L(q,\dot q;
t)={m\over 2}\dot q^2-V(q)$ for some $V\colon\R\rightarrow\R$.
Given a Lagrangian, the action for the trajectory is the functional
$$
S[q]=\int dt\,L(q,\dot q;t).
$$
To ensure that the action is finite we uaually restrict time to a
closed interval, so $q\colon[t_0,t_1]\rightarrow\R$ and
$$
S[q]=\int_{t_0}^{t_1} dt\,L(q,\dot q;t).
$$
Now, the physical trajectory going from an initial position $q_1$ at
$t=t_1$ to the final position $q_2$ at $t=t_2$ is somewhat
misteriously determined by the condition that it be a stationary point
of the action $S$ given the endpoints.
To formalise this, we consider ``variations of the trajectory'' of the
form $\tilde q=q+\delta q$, where $\delta
q\colon[t_1.t_2]\rightarrow\R$ is the ``variational field'' and is
required to vanish at the endpoints. We let
$q_\epsilon=q_0+\epsilon\delta q$. The condition that $q_0$ be a stationary
point of $S$ is that
$$
\delta S[q_0,\delta q]\colon=\left.d\over d\epsilon\right|_{\epsilon=0}S[q_\epsilon]=0
$$
for all variational fields $\delta q$.
The variation of the action is
\begin{eqnarray*}
\delta S[q_0,\delta q]\colon&=&\left.d\over
d\epsilon\right|_{\epsilon=0}\int_{t_0}^{t_1}dt\,\left[{m\over 2}\dot
q_\epsilon^2-V(q_\epsilon)\right]=\int_{t_0}^{t_1}dt\left.d\over
d\epsilon\right|_{\epsilon=0}\left[{m\over 2}\dot
q_\epsilon^2-V(q_\epsilon)\right]=\\
&=&\int_{t_0}^{t_1}dt\,\Bigl[m\dot q_0\delta\dot q-\left.dV\over
dq\right|_{q_0}\delta q\Bigr]=\underbrace{\left.mq_0\delta
q\right|_{t_0}^{t_1}}_{=0}-\int_{t_0}^{t_1}dt\,\Bigl[m\ddot
q_0+\left.dV\over dq\right|_{q_0}\Bigr]\delta q.\\
\end{eqnarray*}
For this to be zero for all $\delta q$ we must have
$$
m\ddot q_0+V'(q_0)=0,
$$
which is Newton's law with force $F=-V'$.
\section{Lagrangians in Field Theory (I)}
Let us now apply these ideas to the field theory Lagrangians we
introduced previously.
\subsection{The First Chern Theory}
Let $M$ be a $2D$ orientable manifold. From the lagrangian ${\cal
L}=\tr F$, define the action
$$
S=\int_M\tr F.
$$
Then we do $A\rightarrow \tilde A=A+\epsilon\delta A$, where $\delta
A$, being the difference between two connections, is an $Ad(P)$-valued
$1$-form. Now, from $F=dA+{1\over 2}[A,A]$ we get
$$
\delta S[A,\delta A]=\int_M\tr(\delta F)=\int_M \tr\Bigl\{d(\delta A)+{1\over
2}\bigl([\delta A,A]+[A,\delta A]\bigr)\Bigr\}.
$$
Now, $\tr$ and $d$ commute because the first acts on the
$Ad(P)$-valued part and the second on the $2$-form part. Also,
$[A,\delta A]=[\delta A,A]$ because $[A,\delta A]=[A_idx^i,\delta
A_jdx^j]=[A_i,\delta A_j]dx^i\wedge dx^j$, which is
symmetric. Therefore
$$
\delta S[A,\delta A]=\int_M \bigl\{d\tr(\delta
A)+\underbrace{\tr[\delta A,A]}_{=0}\bigr\}=\int_{\partial M}\tr(\delta
A).
$$
We have used $\tr[A,\delta A]=0$, which follows from the cyclic
property of the trace. For matrices, we can prove the cyclic property
of the trace as follows:
$$
\xy
(0,5)*+{Y}*\cir{}="f";
(0,-5)*+{X}*\cir{}="g";
**\crv{(0,-15)&(15,0)&(0,15)} ?(.25)*\dir{>} ?(.75)*\dir{>};
"f";"g"**\dir{-} ?(.5)*\dir{>};
\endxy
=
\xy
(-5,0)*+{Y}*\cir{}="f";
(5,0)*+{X^*}*\cir{}="g";
**\crv{(5,10)&(-5,10)} ?(.5)*\dir{>};
"f";"g"**\crv{(-5,-10)&(5,-10)} ?(.5)*\dir{>};
\endxy
=
\xy
(0,5)*+{X}*\cir{}="f";
(0,-5)*+{Y}*\cir{}="g";
**\crv{(0,-15)&(15,0)&(0,15)} ?(.25)*\dir{>} ?(.75)*\dir{>};
"f";"g"**\dir{-} ?(.5)*\dir{>};
\endxy
$$
so $\tr(XY)=\tr(YX)$ and therefore $\tr[X,Y]=0$. One can see from the
diagram that ``cyclic'' is a fitting name for this property.
In conclusion, for the first Chern theory the stationary action
condition is
$$
0=\delta S[A,\delta A]=\int_{\partial M}\tr(\delta A),
$$
which does not depend on $A$ at all! This means that either all
conections are admissible solutions or none is. One way to ensure that
there are stationary solutions is to impose the condition that
$\left.\delta A\right|_{\partial M}=0$, which is analogous to the
variational field $\dot q$ vanishing at the endpoints of the
trajectory. This condition holds trivially if $M$ does not have a
boundary.
We anticipate once more that $2D$ general relativity is an instance of
the first Chern theory, so that the Einstein equations in 2D are
vacuous. 2DGR is therefore rather boring, at least until one quantizes
it.