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\documentclass{article} 
\usepackage{amsfonts,amssymb}
\usepackage[all]{xy}

	\textwidth 6in   
	\textheight 8.5in	\evensidemargin .25in   
	\oddsidemargin .25in   
	\topmargin .25in   
	\headsep 0in   
	\headheight 0in   
	\footskip .5in   
%If you want single spaced copy, delete the next two lines.   
%	\parskip 1.75\parskip plus 3pt minus 1pt   
%	\renewcommand{\baselinestretch}{1.5}   
	\pagestyle{empty}   
	\pagenumbering{none}   
	\begin{document}   
 
	\begin{center}   
	{\bf Nondegenerate Pairings \\}   
	{\small John\ C.\ Baez, November 9, 2004\\}   
	\vspace{0.2cm}   
        \end{center}

\noindent
{\it
First let's straighten out something that was explained rather
poorly in class.... 

\vskip 1em
\noindent
Let $\Vect$ be the category of finite-dimensional complex vector
spaces.  If $V \in \Vect$, we all know that a linear map 
\[       g \maps V \tensor V \to \C \]
yields a linear map
\[      \sharp \maps V \to V^\ast \]
as follows:
\[       \sharp (v)(w) = g(v \tensor w)  \]
for all $v,w \in V$.   But let's see how to do this using
string diagrams!  For this, we might as well work quite
generally.  So, assume 
$C$ is any monoidal category and $(x,x^\ast,i_x,e_x)$ is
any adjunction in $C$.
}

\vskip 1em
\noindent
1.  Given a morphism 
\[               g \maps x \tensor x \to 1 \]
in $C$, called a {\bf pairing}, construct a morphism 
\[   \sharp \maps x \to x^\ast \]
using $g$ and the adjunction $(x,x^\ast,i_x,e_x)$.   Make sure
that in the special case $C = \Vect$, this morphism 
$\sharp$ reduces to the linear operator given by the previous formula.

\vskip 1em
\noindent
2.  If the string diagram for $g$ looks like this:
\[
\xy
 (6,0)*{};
 (-6,0)*{}
   **\crv{(6,-12) & (-6,-12)} ?(.1)*\dir{>} ?(.92)*\dir{<};
 (1,-11)*{\scriptstyle g};
\endxy 
\]
what does the string diagram for $\sharp$ look like?  
{\it (Hint: if you have trouble drawing pictures in LaTeX,
you can just copy and modify some pictures from the LaTeX file of this or 
the previous homework assignment, available on my website.  Just
remember to include the line {\rm \texttt{$\backslash$usepackage[all]{xy}}}.)}

\vskip 1em
\noindent
{\it 
In this general situation, we say the pairing $g$
is {\bf nondegenerate} if $\sharp$ is an isomorphism.
We then denote the inverse of $\sharp$ by
\[         \flat \maps x^\ast \to x  \]
}

\noindent
3.  If $g \maps x \tensor x \to 1$ is nondegenerate,
use the morphism $\flat \maps x^\ast \to x$ and whatever
else we have at hand to construct a morphism
\[     \bar{g} \maps 1 \to x \tensor x  \]
such that $(x,x,\bar{g},g)$ is an adjunction.  In other
words, find $\bar{g}$ such that these identities
hold:  
\[
\xy0;/r.28pc/: (-6,-8)*{}="1E"; (-6,0)*{}="1"; (0,0)*{}="2";
(6,0)*{}="3"; (6,8)*{}="3B";
 "2";"1" **\crv{(0,10)& (-6,10)}
     ?(.02)*\dir{<}  ?(1)*\dir{>};
 "3";"2" **\crv{(6,-10)& (0,-10)} ;
 "1";"1E" **\dir{-};
 "3B";"3" **\dir{-}?(1)*\dir{>} ;
(-2.5,9)*{\scriptstyle \bar{g}}; (3.5,-9)*{\scriptstyle g};
\endxy
\qquad = \xy0;/r.28pc/: (-6,8)*{}; (0,8)*{}; (0,-8)*{}; **\dir{-}
?(.5)*\dir{<}; (6,-8)*{}?(.5)*\dir{<}+(3,1)*{\scriptstyle x};
\endxy
\qquad  
=
\qquad \xy0;/r.28pc/: (-6,8)*{}="1E"; (-6,0)*{}="1";
(0,0)*{}="2"; (6,0)*{}="3"; (6,-8)*{}="3B";
 "2";"1" **\crv{(0,-10)& (-6,-10)}
     ?(.02)*\dir{>}  ?(1)*\dir{<};
 "3";"2" **\crv{(6,10)& (0,10)}
       ;
 "1";"1E" **\dir{-};
 "3B";"3" **\dir{-}?(1)*\dir{<};
(-2.5,-9)*{\scriptstyle g}; (3.5,9)*{\scriptstyle \bar{g}};
\endxy
\]

\vskip 1em
\noindent
{\it There, now that's straightened out....}

\vfill\eject

\noindent
4.  Suppose $V \in \Vect$, and let $A = \End(V)$ be the 
vector space of linear transformations of $V$, made into an
algebra in the usual way.  Define the pairing 
\[               g \maps A \tensor A \to \C \]
by 
\[               g(a \tensor b) = \tr(L_a L_b) \]
where $L_a$ is left multiplication by $a \in A$:
\[   
\begin{array}{rcl}
   L_a \maps A &\to& A  \\
        b &\to& ab   
\end{array}
\]
Show that $g$ is nondegenerate.

\vskip 1em
\noindent
{\it Any algebra has a pairing of the above form; if the 
pairing is nondegenerate the algebra is {\bf semisimple}.
This is either a definition or a theorem depending on
your taste: if we define a semisimple algebra to be a direct
sum of algebras with no nontrivial two-sided ideals, it's
a theorem.   The algebras $\End(V)$ are usually called 
{\bf matrix algebras}, since $\End(\C^n)$ is the algebra of
$n \times n$ complex matrices.  So, we're seeing that
matrix algebras are semisimple.  A harder theorem is that
any semisimple algebra is a direct sum of matrix algebras!

\vskip 1em 
\noindent
By the way, if we think of $A = \End(V)$ as $V \tensor V^\ast$,
we can express matrix multiplication in terms of string diagrams
as follows:
\[
\xy
 (-6,2)*{};(6,2)*{};
 **\crv{(0,-14)} ?(.75)*\dir{>} ?(.25)*\dir{>};
 (-12,2)*{};(-3,-14)**\dir{-} ?(.5)*\dir{>};
 (0,-8)*{\scriptstyle e_V};
 (12,2)*{};(3,-14)**\dir{-} ?(.5)*\dir{<};
\endxy
\quad
\xymatrix{A\otimes A\ar[d]^m\\
          A\\}
\]
and then there's an easy diagram proof that matrix multiplication
is associative:
\[
\xy 
 (-3,-12)*{};(-21,12)**\dir{-} ?(.5)*\dir{<};
 (-15,12)*{};(-3,12)*{};
 **\crv{(-6,0)} ?(.25)*\dir{>} ?(.75)*\dir{>};
 (3,12)*{};(15,12)*{};
 **\crv{(-9,-16)} ?(.25)*\dir{>} ?(.75)*\dir{>};
 (21,12)*{};(3,-12)**\dir{-} ?(.5)*\dir{<};
\endxy
=
\xy 
 (-3,-12)*{};(-21,12)**\dir{-} ?(.5)*\dir{<};
 (-15,12)*{};(-3,12)*{};
 **\crv{(6,-16)} ?(.25)*\dir{>} ?(.75)*\dir{>};
 (3,12)*{};(15,12)*{};
 **\crv{(9,0)} ?(.25)*\dir{>} ?(.75)*\dir{>};
 (21,12)*{};(3,-12)**\dir{-} ?(.5)*\dir{<};
\endxy
\]
In physics we sometimes use $V$ as the Hilbert space of states
for a particle and $V^\ast$ as the space of states for its
corresponding antiparticle.  If this particle happens to be a
quark, $V \tensor V^\ast$ is the Hilbert space for a quark-antiquark
pair, otherwise known as a {\bf meson}.  Then this:
\[
\xy
 (-6,2)*{};(6,2)*{};
 **\crv{(0,-14)} ?(.75)*\dir{>} ?(.25)*\dir{>};
 (-12,2)*{};(-3,-14)**\dir{-} ?(.5)*\dir{>};
 (12,2)*{};(3,-14)**\dir{-} ?(.5)*\dir{<};
\endxy
\]
is the Feynman diagram for an interaction where
two mesons collide and become one.  This process indeed occurs
in nature... perhaps we should call it `meson multiplication'!
}




\end{document}


