\input robmacro
\documentstyle{article}
%\texwidth
%\oddsidemargin=1-(8.5-\texwidth)
\begin{document}

{\small
\ni{Lectures by Robert Geroch}
\ni{{\TeX}ed and supplemented by R. Salgado}
}

\ni{\bf REPRESENTATIONS}
\bigskip

We now turn to the subject of representations.

\smallskip
\ni\underline{UNITARY REPRESENTATIONS OF GROUPS}
\smallskip

Let $G$ be a group.  By a {\bf (unitary) representation}
\marginpar{\bf (unitary)\\representation} of $G$, we mean:
     \begin{itemize}
          \item a Hilbert space $H$
          \item a map that assigns, to each element $g$ of $G$,
               a unitary-operator, $U_g$, on $H$, satisfying the following
               condition:
               \tab For any $g,g'\in G,\ U_gU_{g'}=U_{gg'}$.
     \end{itemize}
In other words, a representation ``represents'' `the group' as
`operators on a Hilbert space', in such a way that `the group product'
corresponds to `the composition of operators'.  It follows immediately
that, for $U_g$ a representation,
$$U_e=I \mbox{ (since $U_eU_e=U_{ee}=U_e$)}$$
$$U_{g^{-1}}=(U_g)^{-1} \mbox{ (since $U_gU_{g^{-1}}=U_{gg^{-1}}=U_e=I$).}$$

\ni EXAMPLES

For $G$ any group, and $H$ any Hilbert space, the choice $U_g=I$ for all
$g$ is a representation.

Let $G$ be the additive group of reals.  Then $H=L^2(\Re{})$, and,
for $a\in G$ and $\phi\in H$, $(U_a\phi)(x)=\phi(x-a)$ is a representation
of $G$.  Also, $H$ the space of states of any quantum system, and,
for $t\in G$ $U_t$ the unitary time-evolution operator, is a
representation of $G$.

Let $G$ be the rotation group.  Then $H=L^2(\Re{3})$, and, for $R\in G$
a rotation and $\phi\in H$, $(U_R\phi)(x)=\phi(Rx)$ (where ``$Rx$''
is the result of applying the rotation $R$ to the position $x$ of \Re{3})
is a representation of $G$.

Let $G$ be the permutation group $\pi_3$.  Then
$H=L^2(\Re{})\otimes L^2(\Re{})\otimes L^2(\Re{})$, with, for $\pi \in G$
and $\phi\in H$, $(U_\pi \phi)(x,y,z)=\phi(\pi(x,y,z))$ (\ie, ``permute
the variables $x$, $y$, $z$ as prescribed by the permutation $\pi$'')
is a representation of $G$.

\pagebreak
\ni\underline{HERMITIAN REPRESENTATIONS OF LIE-ALGEBRAS}
\smallskip

Let $L$ be a lie-algebra.  By a {\bf (hermitian) representation}
\marginpar{\bf (hermitian)\\representation} of $L$, we mean:
     \begin{itemize}
          \item a Hilbert space $H$
          \item a map that assigns, to each element $l$ of $L$,
               a hermitian-operator, $A_l$, on $H$, satisfying the following
               two conditions:
               \begin{enumerate}
                    \item (linearity) For any $b\in\Re{}$ and $l,l'\in L$,
                           $A_{bl+l'}=bA_l+A_{l'}$.
                    \item (bracket) For any $l,l' \in L$,
                           $A_l A_{l'}-A_{l'} A_l = \frac{\hbar}{i} A_{[l,l']}$
               \end{enumerate}
     \end{itemize}
In other words, a representation ``represents'' `the lie-algebra'
as `hermitian-operators on a Hilbert space', in such a way that linearity
is preserved, and `the bracket' goes over to `the commutator of Hermitian
operators'.  The case of unbound operators is allowed (indeed, is quite common),
but then one must be careful to choose a domain such that the operator
products in condition (2) make sense.

\ni EXAMPLES

For $L$ any lie-algebra, and $H$ any hilbert space, the choice $A_l=0$
for all $l$ is a representation.

Let $L$ be the lie-algebra of all pairs, $(\vec v,f)$, consisting of
a smooth vector-field and a smooth scalar-field, on \Re{3}, with bracket
$$[(\vec v,f),(\vec v',f')]=
     ([\vec v,\vec v'], (\vec v\cdot\grad)f'-(\vec v'\cdot\grad)f).$$
Then $H=L^2(\Re{3})$, with $A_{(\vec v,f)}=P_{\vec v}+Q_f$, is a representation
of this lie-algebra.  [We may choose for the domain, \eg, smooth wavefunctions
of compact support.]
\tech{Special case:  Let $L$ be the lie-algebra that is a 3-dimensional
     vector space with bracket $[\vec u,\vec u']=\vec u \times \vec u'$.
     Then $H=L^2(\Re{3})$ with, for $\vec u \in L$,
     $A_{\vec u}=P_{\vec r\times\vec u}$ (where $\vec r\times\vec u$ is
     the vector field on \Re{3} given by the cross-product of the
     position, $\vec r$, with the fixed vector $\vec u$) is a
     representation.}


\pagebreak
\ni\underline{The connection between the representations}
\smallskip

Since lie-groups and lie-algebras are closely related, one might expect
that representations of lie-groups will be closely related to representations
of lie-algebras.  They are.

Fix a lie-group $G$, and $L$ be its lie-algebra.  Now fix also a
representation, $U_g$, of $G$ on some hilbert-space $H$.  We wish
to construct, from this representation of the group $G$, a representation
of its lie-algebra $L$.  The idea is quite simple.  Let $\gamma$ be
a smooth curve in $G$ through its identity $e$, so the tangent to $\gamma$
at $e$ defines an element, $l$, of $L$.  The idea is simply to set
$$A_l=\frac{\hbar}{i}\dd{a}\left.(U_{\gamma(a)})\right|_{a=0}.$$
That is, we evaluate $U_g$, of the $G$-representation, along the curve
$\gamma$, to obtain, for each number $a$, a unitary operator $U_{\gamma(a)}$.
We then take the derivative of this family with respect to $a$ at
$a=0$ (\ie, take
     $\lim_{\Delta a\rightarrow 0} \frac{1}{\Delta a}
          (U_{\gamma(\Delta a)}-U_{\gamma(0)})$ ).
Finally, we multiply by $\frac{\hbar}{i}$ to obtain a hermitian-operator
corresponding to the tangent, $l$, to the curve $\gamma$.

Unfortunately, a technical issue intervenes to make this construction---so
simple in principle---more complicated in practice.  The issue is that the
representation $U_g$ may have $U_g$ depending on $g$ in a way insufficiently
smooth that the derivative, $\dd{a}(U_{\gamma(a)})$, fails to exist.
Or, it may turn out that the derivative exists, but in some weaker sense.
This issue is closely related to the possible unboundedness of the $A_l$.
\tech{
   For example, let $G$ be the additive group of reals, $H$ the hilbert space
   $L^2(\Re{})$, and $(U_a\psi)(x)=\psi(x-a)$. Then the lie-algebra of $G$
   is the one-dimensional vector space $L$ with all brackets zero, and the
   corresponding representation of $L$ has
   $(A_b\psi)(x)=\frac{\hbar}{i} b \frac{d\psi}{dx}$.
   Here, $\dd{a} U_{\gamma(a)}$ is not defined on all wavefunctions $\psi$,
   but rather only on those in the domain of the unbounded operator $A_b$.
   }

In any case, we have a construction that suggests, given a representation
of a lie-group, how to get a representation of its lie-algebra.


\ni EXAMPLES

As an example, consider the case in which $\hat\gamma(a)=(\gamma(a))^{-1}$
for every $a$.  Then, the product curve, $\phi=\gamma\hat\gamma$, has
$\gamma(a)=e$ for every $a$, and so has tangent vector zero.  Applying
to this case the result above for the tangent of the product cruve, we
conclude: {\em the tangent vector to the curve $\gamma^{-1}(a)$ is
minus the tangent vector to $\gamma(a)$.}

Now fix two curves, $\gamma$ and $\hat\gamma$, and let $\phi$ be the
curve given by
$\phi(a)=\gamma(a)\hat\gamma(a)\gamma^{-1}(a)\hat\gamma^{-1}(a)$.
Then, for the tangent vector to this curve $\phi$, we have:
\begin{eqnarray*}
     \mbox{tang}(\phi)&=&\mbox{tang}(\gamma)+\mbox{tang}(\hat\gamma)
                     +\mbox{tang}(\gamma^{-1})+\mbox{tang}(\hat\gamma^{-1})\\
                      &=&\mbox{tang}(\gamma)+\mbox{tang}(\hat\gamma)
                     -\mbox{tang}(\gamma)-\mbox{tang}(\hat\gamma)\\
                      &=& 0,
\end{eqnarray*}
where we used the result for the tangent of the product of two curves
in the first step, and that for the tangent of the inverse of a curve
in the second.  So, $\phi$ has tangent zero.  We now \underline{define}
the bracket of the elements $\mbox{tang}(\gamma)$ and $\mbox{tang}(\hat\gamma)$
of $L$ to be the second derivative (``acceleration'') of the curve $phi$
at $a=0$:
$$[\mbox{tang}(\gamma), \mbox{tang}(\hat\gamma)]=
     \frac{1}{2}\left.\frac{d^2 y^i}{da^2}\right|_{a=0},$$
where $y^i(a)$ describes the curve $\phi$ near $a=0$.
To determine the properties of this bracket, let us write it out in terms
of coordinates.  Let $x^i(a)$ and $\hat x^i(a)$ describe the respective
curves $\gamma(a)$ and $\hat\gamma(a)$, we have
$$y^i(a)=w^i\left(x^j(a), w^k(\hat x^l(a),w^m(z^n(x^p(a)),z^q(x^r(a)))\right)$$
where $w$ is the $n$ functions-of-$2n$-variables describing the group
product, and $z$ the $n$ functions-of-$n$-variables describing the group
inverse.  Taking the second derivative at $a=0$, we obtain:
$$[\left.\frac{dx^j}{da}\right|_0,\left.\frac{d\hat x^j}{da}\right|_0]^i
     = (\frac{\p^2 w^i}{\p s^j \p t^k}-\frac{\p^2 w^i}{\p s^k \p t^j})
       \left.\frac{dx^j}{da}\right|_0 \left.\frac{d\hat x^k}{da}\right|_0
$$
where, on the right, we have labeled the $2n$ variables for $w$ thus:
$w^i(s^j,t^k)$.  It is immediate from this expression that the bracket is
linear, and antisymmetric.  To obtain the Jacobi relation, first note
that group associativity becomes, in terms of our coordinates,
$$w^i(s^j,w^k(t^l,u^m))=w^i(w^k(s^j,t^l),u^m)).$$
Taking $\frac{\p^3}{\p s^j \p t^l \p m}$ of this equation, and evaluating
at $s^j=0$, $t^l=0$, $u^m=0$, the Jacobi relation follows.

What we have shown, then, is that, given any Lie-group, the tangent-vectors
to smooth curves through the identity element acquire the structure of
a Lie-algebra.  [In fact, there is a sort of converse of this construction:
{\em Every finite-dimensional Lie-algebra arises, via this construction,
from some Lie-group}---indeed, a unique one under appropriate global
conditions on the Lie-group.]

There is also a sense in which one can---subject to the same techincal
issue---invert this construction.  Again, fix Lie-group $G$, with its
Lie-algebra $L$.  Fix also a representation, $A_l$, of $L$ on some
Hilbert space $H$.  A smooth curve $\gamma$ in $G$, with $\gamma(0)=e$,
is called a {\bf one-parameter subgroup} if, for all $a,b\in\Re{}$,
$\gamma(a)\gamma(b)=\gamma(a+b)$.  (For example, the rotations through
various angles about a fixed axis form a one-parameter subgroup of the
rotation group.)  Fix such a one-parameter subgroup, and let $l$ be
its tangent-vector.  We now set $U_{\gamma(a)}=\exp(\frac{i}{\hbar}a A_l)$.
That is, we define $U_g$, for group-elements $g$ on this curve, by the
exponential of the operator corresponding to the tangent of the curve.
(This will be recognized as the reverse of the construction of the
previous paragraph.)  The technical problem, of course, is that, since
the $A_l$ may be unbounded, we may have difficulty defining
``$\exp(\frac{i}{\hbar}a A_l)$''.  Of course, this formula only defines
$U_g$ for those group elements lying on one-parameter subgroups.
However, using ``$U_g U_{g'}=U_{gg'}$'', we can extend the definition
to all products of such elements.  For a connected group, this is in fact
the entire group.

This construction is frequently used to find the representations of Lie-groups.
One first passes to the Lie Algebra.  It is far simpler to work with, and its
representations are normally easier to find.  One then ``exponentiates''
those representations, as above, to find the representation of the
original Lie-group.





\end{document}
\end

