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

\ni{\bf MANIFOLDS}
\bigskip

A {\bf manifold} consists of a set $X$, a non-negative integer $n$,
and a collection ${\cal F}$ of real-valued functions on the set $X$,
satisfying the following three conditions:
\begin{enumerate}
     \item Let $f_1,f_2,\ldots,f_s$ be any $s$ functions in the collection
           ${\cal F}$, and $g$ any $C^\oo$, real-valued function of $s$
           variables.  Then the function $g(f_1,f_2,\ldots,f_s)$ is
           also in the collection ${\cal F}$.
     \item Let $f$ be any real-valued function on $X$ having the
           following property:
           \begin{quotation}
                  \noindent Given any point $x\in X$, there is a neighborhood
                  $U$ of $x$ and a function $g$ in the collection ${\cal F}$
                  with $f=g$ in the neighborhood~$U$.
           \end{quotation}
           Then $f$ itself is in the collection ${\cal F}$.
     \item
Let $x$ be any point of $X$. Then there is a neighborhood $U$ of $x$
           and $n$ functions, $f_1,f_2,\ldots,f_n$, in the collection ${\cal F}$
           such that
           \begin{enumerate}
               \item the mapping $(f_1,f_2,\dots,f_n)$ from $U$ to \Re{n}
                     is one-to-one from $U$ onto an open subset of \Re{n}
               \item any function $f$ in the collection ${\cal F}$ can be
                     written in the form $f=g(f_1,f_2,\ldots,f_n)$ in the
                     neighborhood $U$, for some $C^\oo$ function $g$
                     of $n$ variables.
          \end{enumerate}
\end{enumerate}

This definition is, to say the very least, forbidding!  But, fortunately,
this is one of those cases in which the basic idea is much simpler than
the formal definition.  Manifolds just take a little getting used to.
As a first step in this process, let us briefly discuss these three
conditions in turn.

\smallskip
\underline{Condition 1.}  Here, $g(f_1,f_2,\ldots,f_s)$ means that
function on $X$ whose value at the point $x\in X$ is the number
$g(f_1(x),f_2(x),\ldots,f_s(x))$.  In other words, given the point $x$,
you first evaluate the $s$ functions $f_1,f_2,\ldots,f_s$ at this point,
to obtain $s$ numbers.  Then, you evaluate $g$ (a function of $s$ variables)
on these $s$ numbers.  The condition demands that whenever you do this
(for the $f$'s in ${\cal F}$ and $g$ an ordinary $C^\oo$ function of
$s$ variables), the resulting function is again back in the collection
${\cal F}$.

In short, this condition demands that
\begin{center}
{\em smooth functions of `functions in the collection $\cal F$'\\
are again in the collection $\cal F$}.
\end{center}

\pagebreak
As an example,
\begin{itemize}
     \item let $g$ be the function of two variables given by
     $g(a,b)=a+b$ ---certainly a $C^\oo$ function.  Then $g(f_1,f_2)=f_1+f_2$,
     \ie, is the function on $X$ that results from adding the functions
     $f_1$ and $f_2$ on $X$.
     In this example, then, condition 1 is demanding that `the sum
     of two functions-in-the-collection-${\cal F}$ is again
     again in the collection-$\cal F$.
     \item Similarly, from the choice $g(a,b)=ab$, we learn that the
     `the product of two functions-in-$\cal F$ must be again in $\cal F$;
     \item from the choice $g(a)=6$ (the function of one variable $a$ with
     constant value ``$6$''), we learn that `the function-on-$X$ with constant
     value ``$6$'' must be in the collection $\cal F$'
     \item from the choice $g(a)=\exp(\sin a)$, we learn that `for any function
     $f$ in the collection $\cal F$, the function $\exp(\sin f)$ must also be
     in $\cal F$'
     \item from $g(a)=|a|$, we learn nothing from condition 1, for this function
     of one variable is not $C^\oo$.
\end{itemize}
Clearly, there are lots of $C^\oo$-functions $g$ around, and
lots of functions $f_1,f_2,\ldots,f_s$ in $\cal F$, and so
condition 1 demands that lots and lots of functions on $X$ in fact
be in the collection-$\cal F$.

\smallskip
\underline{Condition 2.}  Here, we begin with some fixed real-valued function
$f$ on the set $X$.  We do not assume that this function $f$ is in the
collection $\cal F$.  But we do assume, instead, that $f$ ``locally
coincides with functions in the collection $\cal F$''.  In more detail,
we assume that every point has a neighborhood within which the function $f$
is identical with some function $g$ that \underline{is} in the collection
$\cal F$.  Of course, different neighborhoods may require different $g$'s.
It is only required that you give me the point $x$, and I will find the
neighborhood $U$ of $x$ and the $g$ in the collection $\cal F$ all such
that $f=g$ in $U$.  Condition 2 demands that, under these circumstances,
the function $f$ was actually in the collection $\cal F$ all the time.
Condition 2, in short, demands that
\begin{center}
{\em functions on $X$ that look locally like
     `functions-in-the-collection-$\cal F$' \\
     are themselves in the collection $\cal F$}.
\end{center}

In order to complete the statement of condition 2, we still
have to specify what these ``neighborhoods'' are.  The object $X$, after
all, comes to us only as a set, and as such is not automatically equipped
with neighborhoods.  The strategy is to define the neighborhoods in $X$
using the only other object at our disposal: the collection $\cal F$
of real-valued functions on $X$.

\underline{Definition.} Fix a point $x$ of $X$.  A {\bf neighborhood}
\marginpar{\bf neighborhood} of $x$ consists of all points $y$ of $X$
such that $|f_1(y)-f_1(x)|<\epsilon$, $|f_2(y)-f_2(x)|<\epsilon$,
$\ldots$, and $|f_s(y)-f_s(x)|<\epsilon$, where $f_1, f_2, \ldots, f_s$
are any $s$ functions in the collection $\cal F$, and $\epsilon$ is
any positive number.

The idea is that a neighborhood of point $x$ consists of those points
$y$ at which each of the functions $f_1, f_2, \ldots, f_s$ takes a
value within $\epsilon$ of the value taken at $x$.  In particular, then,
$x$ itself is in every neighborhood of $x$.  Roughly speaking, points of $X$
are ``close together'' if ``the functions in the collection $\cal F$ assume
close values on them''.  Different choices of ``$s$'',
of ``$f_1,f_2,\ldots,f_s$'', and of ``$\epsilon$'' yield, of course,
different neighborhoods.  For, say, ``$s=2$ and $\epsilon=100,00$'',
one would normally obtain a very large neighborhood of $x$, while
``$s=138$ and $\epsilon=10^{-5}$'' would yield a ``smaller neighborhood''.
(In more formal language, we are imposing a topology on the set $X$, namely
the coarsest one in which all the functions in the collection $\cal F$
are continuous.)


\smallskip
\underline{Condition 3.}  The first two conditions require that every function
on $X$ constructed in a certain way (condition 1), and every function
having a certain property (condition 2) must be in the collection $\cal F$.
The third condition is somewhat more complicated than these.  It requires,
roughly speaking, that ``$X$ look locally like Euclidean $n$-space, \Re{n},
with the functions in $\cal F$ looking locally like $C^\oo$-functions
of $n$ variables''.  The functions $f_1, f_2,\ldots, f_n$ play the role
of coordinates on the neighborhood $U$.  Thus, $(f_1, f_2,\ldots, f_n)$ is
the following mapping from $U$ to \Re{n}: it assigns to each point $y$ of $U$
the $n$-tuple of real-numbers given by $(f_1(y), f_2(y),\ldots, f_n(y))$.
In other words, this mapping assigns to each point of $U$, the ``coordinates''
of that point.  Now, the condition under i) is that this mapping be
one-to-one and onto from $U$ to some open subset in \Re{n}.  In other words,
we must have a one-to-one-correspondence between $U$ and this open subset.
It is just this correspondence that ``makes $X$ locally (\ie, in $U$) look
like \Re{n} (more specifically, like some open set in \Re{n})''.

Having set up this correspondence in part i), then part ii) comes into play.
It requires that, via this correspondence, the functions in the collection
$\cal F$, when restricted to $U$, go over to precisely the $C^\oo$ functions
on \Re{n}.  In more detail, it requires that every function in $\cal F$,
when restricted to $U$, be merely some $C^\oo$ function of $n$ variables,
applied to the $n$ coordinate functions, $f_1, f_2,\ldots,f_n$.
Part ii), in short, is a restriction on the functions in the collection
$\cal F$ (for it requires that all functions in $\cal F$ have a
certain property).  To summarize, condition 3 requires that
\begin{center}
{\em the set $X$ can be made, locally, to look like \Re{n} {\rm (\,part i)\,)}\\
     in such a way that\\
     the `functions-in-$\cal F$' look locally like $C^\oo$-functions of
     $n$ variables {\rm (\,part ii)\,)}.}
\end{center}

\medskip
To specify a manifold, one must, of course, say what the set $X$ is,
what the non-negative integer $n$ is, and which functions on $X$ are
to be included in the collection $\cal F$.  Then, one must check the three
conditions.

\pagebreak
We give an example.

Let $X$ be the usual plane, \ie, the set of all pairs $(a,b)$ of real numbers.
Let $n=2$.  Now, a real-valued function on $X$ is (because $X$ is just
the set of all pairs of numbers) a real-valued function of the two real
varibales $a$ and $b$.  Let $\cal F$ consist of `all such functions $f(a,b)$
that are $C^\oo$'.  This, we claim, is a manifold.

Let us check the three conditions.

Condition 1 requires, in the present example, that every $C^\oo$ function
of $s$ variables, when applied to $s$
$C^\oo$-functions-of-two-variables-$a$-and-$b$, yield a function of $a$
and $b$ that is itself $C^\oo$.  But this is just true (as one learned in
calculus: `smooth functions of smooth functions are smooth').

Condition 2 requires that, given function $f(a,b)$ such that every point
of the plane has a neighborhood in which $f$ is a $C^\oo$-function of its
variables, then $f$ is itself a $C^\oo$-function of its variables.  But this
also is true (again, as one learned in calculus: `Smoothness is a local
property').

For condition 3, let $f_1$ and $f_2$ be the functions with $f_1(a,b)=a$ and
$f_2(a,b)=b$.  These are certainly in $\cal F$.  (That is, we choose
--not surprisingly-- our ``coordinates'' to be $a$ and $b$ themselves.)
Now the mapping $(f_1,f_2)$ just sends a point of our plane to the pair
of numbers $(a,b)$ that originally defined that point!  Part i) clearly
holds.  But so does part ii), for we originally defined the functions in
$\cal F$ to be precisely those that are $C^\oo$-functions of $(a,b)$.
This manifold is commonly called, need we add, `the plane'.

Had we chosen, instead, $\cal F$ to be only the functions $f(a,b)$ that
are polynomials in $a$ and $b$, then condition 1 would have failed
(for $C^\oo$-functions of polynomials are not, in general, polynomials),
while condition 2 would still have held (for functions that are locally
polynomials are globally polynomials).

Had we chosen, instead, $\cal F$ to be only the functions $f(a,b)$
that are $C^\oo$ and approach a constant at infinity, the condition 1
would still have held (for $C^\oo$ functions of such-functions are again
such-functions), while condition 2 would have failed (\eg, for the function
f(a,b)=a).

Had we chosen, instead, $\cal F$ to be all functions that are $C^3$, then
conditions 1 and 2 would have held.  Further, condition 3i) would also have
held (again, using $a$ and $b$ as coordinates).  But condition 3ii), would
have failed (for, \eg, the function $f(a,b)=|a|^{3.5}$, a member of the
collection $\cal F$ cannot be written in the required form).

Finally, had we chosen $X$ and $\cal F$ as originally given, but $n=5$
(say, instead of `$n=2$'), then conditions 1 and 2 would of course have
held (for they do not involve $n$), but condition 3i) would have failed.



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