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 \begin{document}

 \begin{center}
 \textbf {Groups as Categories}\\
 {\small Questions by: John C. Baez, 2004 September 30}\\
 {\small Answers by: Toby Bartels\footnote
{I reserve no legal rights whatsoever to any of my creative work;
see \texttt {http://toby.bartels.name/copyright/}.}, 2004 October 7}\\
 \vspace{0.3cm}
 \end{center}

\begin {commentary}
Here's a little homework just to make sure you understand the
concepts of \emph {category}, \emph {functor}
and \emph {natural transformation},
as defined in the handout `Some Definitions Everyone Should Know'.

Recall that a set with an associative binary product and an element
serving as the unit for this product is called a \strong {monoid}:
examples include \((\N,+,0)\) and \((\N,\cdot,1)\).  A monoid where
every element has a two-sided inverse is called a \strong {group}.

A category \(C\) with only one object (say \(\ast\)) is the same thing as
a monoid, since all \(C\) has is a set of morphisms \(f \maps \ast \to \ast\)
that can be composed associatively, together with
a morphism \(1_\ast \maps \ast \to \ast\) serving as the unit for composition.

Similarly, a category with only one object and all morphisms invertible
is the same as a group!

So, among other things, category theory is a massive generalization
of group theory. This means that whenever you encounter a definition
in category theory, you should figure out what it amounts to in the
case of groups.
\end {commentary}

\begin {instruction}
In what follows, you can either do problems 1--5 or problem 6.
I greatly prefer answers in LaTeX.
\end {instruction}

\begin {question}
Suppose that
\(G\) and \(H\) are groups, and regard them as one-object categories
with all morphisms invertible.
Figure out what a functor \(F \maps G \to H\) amounts to.
What are such functors usually called?
\end {question}

\begin {answer}
\( F \) consists of a function \( F \maps G \to H \),
such that \( F ( { 1 _ G } ) = 1 _ H \)
and for any pair of elements \( f , g \) of \( G \),
\( F ( { f g } ) = F ( f ) F ( g ) \).
This is usually called a \strong {group homomorphism} from \( G \) to~\( H \).
\end {answer}

\begin {question}
Suppose \(G\) and \(H\) are groups regarded as categories, and
let \(F,F' \maps G \to H\) be a pair of functors. Figure out what a
natural transformation \(\alpha \maps F \To F'\) amounts to.
\end {question}

\begin {answer}
\( \alpha \) consists of an element \( \alpha \) of \( H \),
such that for any element \( f \) of \( G \),
\( F ( f ) \alpha = \alpha F ' ( f ) \).
\end {answer}

\begin {question}
Suppose \(G\) is a group regarded as a category and let
\(1_G \maps G \to G\) be the identity functor. Figure out what
a natural transformation \(\alpha \maps 1_G \To 1_G\) amounts to.
What is the set of all such natural transformations usually called?
\end {question}

\begin {answer}
\( \alpha \) consists of an element \( \alpha \) of \( G \),
such that for any element \( f \) of \( G \), \( f \alpha = \alpha f \).
The set of such \( \alpha \) is usually called
the \strong {centre} (or \strong {center}) of~\( G \).
\end {answer}

\begin {question}
Let \(\Vect\) be the category of vector spaces over your favorite
field, where the morphisms are linear transformations. Suppose
\(G\) is a group regarded as a category. Figure out what a functor
\(F \maps G \to \Vect\) amounts to. What is such a functor usually
called?
\end {question}

\begin {answer}
\( F \) consists of an object \( F \) of \( \Vect \)
and a function \( F \maps G \to \hom ( F , F ) \),
such that \( F ( { 1 _ G } ) = 1 _ F \)
and for any pair of elements \( f , g \) of \( G \),
\( F ( { f g } ) = F ( f ) F ( g ) \).
This is usually called a \strong {linear representation} of \( G \).
\end {answer}

\begin {question}
Suppose \(G\) is a group regarded as a category and let
\(F,F' \maps G \to \Vect\) be functors. Figure out what a natural
transformation \(\alpha \maps F \To F'\) amounts to. What is such
a natural transformation usually called?
\end {question}

\begin {answer}
\( \alpha \) consists of a linear operator \( \alpha \maps F \to F ' \),
such that for any element \( f \) of \( G \),
this diagram commutes:
\[ \xymatrix { F \ar [d] _ \alpha \ar [r] ^ { F ( f ) } & F \ar [d] ^ \alpha \\
F ' \ar [r] _ { F ' ( f ) } & F ' } \]
\( \alpha \) is usually called
an \strong {intertwining operator} from \( F \) to~\( F ' \).
\end {answer}

\begin {question}
Suppose \(G\) is a Lie group, regarded as a one-object
category where the morphisms form a manifold. Let \(\Aut(G)\)
be the category whose objects are smooth invertible functors
\(F \maps G \to G\) and whose morphisms are smooth invertible
natural transformations \(\alpha \maps F \to F'\).
The objects of \(\Aut(G)\) form a Lie group. Any object
\(F\) in \(\Aut(G)\) gives a subset \([F]\) consisting
all objects that are isomorphic to it. What do these subsets
look like for \(G = \SO(3)\)? How about for \(G = \SU(2)\)?
\end {question}

\begin {answer}
First, let me state some facts about automorphisms of Lie groups.
To begin with, every element of \( G \)
defines an automorphism of \( G \) by conjugation.
This defines a group homomorphism \( G \to \Ob ( \Aut ( G ) ) \);
let \( \Inn ( G ) \) be the image of this homomorphism.
Then \( \Inn ( G ) \) is normal; let \( \Out ( G ) \) be the quotient group.
If \( G \) is a connected compact real form of a simple Lie group,
then \( \Out ( G ) \) is isomorphic to
the symmetry group of the Dynkin diagram of \( G \).
Also, \( \Inn ( G ) \) is isomorphic to \( G / Z \),
where \( Z \) is the centre of \( G \).

Now, both \( \SO ( 3 ) \) and \( \SU ( 2 ) \)
are connected compact real forms of the Lie group \( A _ 1 \):
\[ \b \]
The symmetry group of this Dynkin diagram is trivial,
so \( \Ob ( \Aut ( G ) ) = \Inn ( G ) \) in both cases.
Since \( Z ( \SO ( 3 ) ) \) is trivial,
I have \( \Ob ( \Aut ( G ) ) = \Inn ( G ) \iso \SO ( 3 ) \) in both cases.

So, suppose \( F \) and \( F ' \) are elements of \( G \),
and let \( \alpha \) be \( F ( F ' ) ^ { - 1 } \).
Then \[ F ( f ) \alpha = F f F ^ { - 1 } F ( F ' ) ^ { - 1 }
= F f ( F ' ) ^ { - 1 }
= F ( F ' ) ^ { - 1 } F ' f ( F ' ) ^ { - 1 } = \alpha F ' ( f ) \].
Thus, \( F \iso F ' \) in \( \Aut ( G ) \).
Therefore, there is a unique isomorphism class \( [ F ] \),
which is all of \( \Ob ( \Aut ( G ) ) \iso \SO ( 3 ) \).
\end {answer}

\end{document}
