\chapter{A bit of manifolds} Last chapter, we stated Stokes' theorem for cells. It turns out there is a much larger class of spaces, the so-called \emph{smooth manifolds}, for which this makes sense. Unfortunately, the definition of a smooth manifold is \emph{complete garbage}, and so by the time I am done defining differential forms and orientations, I will be too lazy to actually define what the integral on it is, and just wave my hands and state Stokes' theorem. \section{Topological manifolds} \prototype{$S^2$: ``the Earth looks flat''.} Long ago, people thought the Earth was flat, i.e.\ homeomorphic to a plane, and in particular they thought that $\pi_2(\text{Earth}) = 0$. But in fact, as most of us know, the Earth is actually a sphere, which is not contractible and in particular $\pi_2(\text{Earth}) \cong \ZZ$. This observation underlies the definition of a manifold: \begin{moral} An $n$-manifold is a space which locally looks like $\RR^n$. \end{moral} Actually there are two ways to think about a topological manifold $M$: \begin{itemize} \ii ``Locally'': at every point $p \in M$, some open neighborhood of $p$ looks like an open set of $\RR^n$. For example, to someone standing on the surface of the Earth, the Earth looks much like $\RR^2$. \ii ``Globally'': there exists an open cover of $M$ by open sets $\{U_i\}_i$ (possibly infinite) such that each $U_i$ is homeomorphic to some open subset of $\RR^n$. For example, from outer space, the Earth can be covered by two hemispherical pancakes. \end{itemize} \begin{ques} Check that these are equivalent. \end{ques} While the first one is the best motivation for examples, the second one is easier to use formally. \begin{definition} A \vocab{topological $n$-manifold} $M$ is a Hausdorff space with an open cover $\{U_i\}$ of sets homeomorphic to subsets of $\RR^n$, say by homeomorphisms \[ \phi_i : U_i \taking\cong E_i \subseteq \RR^n \] where each $E_i$ is an open subset of $\RR^n$. Each $\phi_i : U_i \to E_i$ is called a \vocab{chart}, and together they form a so-called \vocab{atlas}. \end{definition} \begin{remark} Here ``$E$'' stands for ``Euclidean''. I think this notation is not standard; usually people just write $\phi_i(U_i)$ instead. \end{remark} \begin{remark} This definition is nice because it doesn't depend on embeddings: a manifold is an \emph{intrinsic} space $M$, rather than a subset of $\RR^N$ for some $N$. Analogy: an abstract group $G$ is an intrinsic object rather than a subgroup of $S_n$. \end{remark} \begin{example}[An atlas on $S^1$] Here is a picture of an atlas for $S^1$, with two open sets. \begin{center} \begin{asy} size(8cm); draw(unitcircle, black+2); label("$S^1$", dir(45), dir(45)); real R = 0.1; draw(arc(origin,1-R,-100,100), red); label("$U_2$", (1-R)*dir(0), dir(180), red); draw(arc(origin,1+R,80,280), blue); label("$U_1$", (1+R)*dir(180), dir(180), blue); dotfactor *= 2; pair A = opendot( (-3, -2), blue ); pair B = opendot( (-1, -2), blue ); label("$E_1$", midpoint(A--B), dir(-90), blue); draw(A--B, blue, Margins); draw( (-1.25, -0.2)--(-2,-2), blue, EndArrow, Margins ); label("$\phi_1$", (-1.675, -1.1), dir(180), blue); pair C = opendot( (1, -2), red ); pair D = opendot( (3, -2), red ); label("$E_2$", midpoint(C--D), dir(-90), red); draw(C--D, red, Margins); draw( (1.25, -0.2)--(2,-2), red, EndArrow, Margins ); label("$\phi_2$", (1.672, -1.1), dir(0), red); \end{asy} \end{center} \end{example} \begin{ques} Where do you think the words ``chart'' and ``atlas'' come from? \end{ques} \begin{example} [Some examples of topological manifolds] \listhack \begin{enumerate}[(a)] \ii As discussed at length, the sphere $S^2$ is a $2$-manifold: every point in the sphere has a small open neighborhood that looks like $D^2$. One can cover the Earth with just two hemispheres, and each hemisphere is homeomorphic to a disk. \ii The circle $S^1$ is a $1$-manifold; every point has an open neighborhood that looks like an open interval. \ii The torus, Klein bottle, $\RP^2$ are all $2$-manifolds. \ii $\RR^n$ is trivially a manifold, as are its open sets. \end{enumerate} All these spaces are compact except $\RR^n$. A non-example of a manifold is $D^n$, because it has a \emph{boundary}; points on the boundary do not have open neighborhoods that look Euclidean. \end{example} \section{Smooth manifolds} \prototype{All the topological manifolds.} Let $M$ be a topological $n$-manifold with atlas $\{U_i \taking{\phi_i} E_i\}$. \begin{definition} For any $i$, $j$ such that $U_i \cap U_j \neq \varnothing$, the \vocab{transition map} $\phi_{ij}$ is the composed map \[ \phi_{ij} : E_i \cap \phi_i\im(U_i \cap U_j) \taking{\phi_i\inv} U_i \cap U_j \taking{\phi_j} E_j \cap \phi_j\im(U_i \cap U_j). \] \end{definition} Sorry for the dense notation, let me explain. The intersection with the image $\phi_i\im(U_i \cap U_j)$ and the image $\phi_j\im(U_i \cap U_j)$ is a notational annoyance to make the map well-defined and a homeomorphism. The transition map is just the natural way to go from $E_i \to E_j$, restricted to overlaps. Picture below, where the intersections are just the green portions of each $E_1$ and $E_2$: \begin{center} \begin{asy} size(8cm); draw(unitcircle, black); draw(arc(origin, 1, 80, 100), heavygreen+2); draw(arc(origin, 1, -100, -80), heavygreen+2); label("$S^1$", dir(45), dir(45)); real R = 0.1; draw(arc(origin,1-R,-100,100), red); label("$U_2$", (1-R)*dir(0), dir(180), red); draw(arc(origin,1+R,80,280), blue); label("$U_1$", (1+R)*dir(180), dir(180), blue); dotfactor *= 2; pair A = opendot( (-3, -2), blue ); pair B = opendot( (-1, -2), blue ); label("$E_1$", midpoint(A--B), dir(-90), blue); draw(A--B, blue, Margins); draw( (-1.25, -0.2)--(-2,-2), blue, EndArrow, Margins ); label("$\phi_1$", (-1.675, -1.1), dir(180), blue); pair C = opendot( (1, -2), red ); pair D = opendot( (3, -2), red ); label("$E_2$", midpoint(C--D), dir(-90), red); draw(C--D, red, Margins); draw( (1.25, -0.2)--(2,-2), red, EndArrow, Margins ); label("$\phi_2$", (1.672, -1.1), dir(0), red); draw(A--(0.7*A+0.3*B), heavygreen+2, Margins); draw(B--(0.7*B+0.3*A), heavygreen+2, Margins); draw(C--(0.7*C+0.3*D), heavygreen+2, Margins); draw(D--(0.7*D+0.3*C), heavygreen+2, Margins); draw(B--C, heavygreen, EndArrow, Margin(4,4)); label("$\phi_{12}$", B--C, dir(90), heavygreen); \end{asy} \end{center} We want to add enough structure so that we can use differential forms. \begin{definition} We say $M$ is a \vocab{smooth manifold} if all its transition maps are smooth. \end{definition} This definition makes sense, because we know what it means for a map between two open sets of $\RR^n$ to be differentiable. With smooth manifolds we can try to port over definitions that we built for $\RR^n$ onto our manifolds. So in general, all definitions involving smooth manifolds will reduce to something on each of the coordinate charts, with a compatibility condition. AS an example, here is the definition of a ``smooth map'': \begin{definition} \begin{enumerate}[(a)] \ii Let $M$ be a smooth manifold. A continuous function $f \colon M \to \RR$ is called \vocab{smooth} if the composition \[ E_i \taking{\phi_i\inv} U_i \injto M \taking f \RR \] is smooth as a function $E_i \to \RR$. \ii Let $M$ and $N$ be smooth with atlases $\{ U_i^M \taking{\phi_i} E_i^M \}_i$ and $\{ U_j^N \taking{\phi_j} E_j^N \}_j$, A map $f \colon M \to N$ is \vocab{smooth} if for every $i$ and $j$, the composed map \[ E_i \taking{\phi_i\inv} U_i \injto M \taking f N \surjto U_j \taking{\phi_j} E_j \] is smooth, as a function $E_i \to E_j$. \end{enumerate} \end{definition} \section{Regular value theorem} \prototype{$x^2+y^2=1$ is a circle!} Despite all that I've written about general manifolds, it would be sort of mean if I left you here because I have not really told you how to actually construct manifolds in practice, even though we know the circle $x^2+y^2=1$ is a great example of a one-dimensional manifold embedded in $\RR^2$. \begin{theorem} [Regular value theorem] Let $V$ be an $n$-dimensional real normed vector space, let $U \subseteq V$ be open and let $f_1, \dots, f_m \colon U \to \RR$ be smooth functions. Let $M$ be the set of points $p \in U$ such that $f_1(p) = \dots = f_m(p) = 0$. Assume $M$ is nonempty and that the map \[ V \to \RR^m \quad\text{by}\quad v \mapsto \left( (Df_1)_p(v), \dots, (Df_m)_p(v) \right) \] has rank $m$, for every point $p \in M$. Then $M$ is a manifold of dimension $n-m$. \end{theorem} For a proof, see \cite[Theorem 6.3]{ref:manifolds}. One very common special case is to take $m = 1$ above. \begin{corollary} [Level hypersurfaces] Let $V$ be a finite-dimensional real normed vector space, let $U \subseteq V$ be open and let $f \colon U \to \RR$ be smooth. Let $M$ be the set of points $p \in U$ such that $f(p) = 0$. If $M \ne \varnothing$ and $(Df)_p$ is not the zero map for any $p \in M$, then $M$ is a manifold of dimension $n-1$. \end{corollary} \begin{example} [The circle $x^2+y^2-c=0$] Let $f(x,y) = x^2+y^2 - c$, $f \colon \RR^2 \to \RR$, where $c$ is a positive real number. Note that \[ Df = 2x \cdot dx + 2y \cdot dy \] which in particular is nonzero as long as $(x,y) \ne (0,0)$, i.e.\ as long as $c \ne 0$. Thus: \begin{itemize} \ii When $c > 0$, the resulting curve --- a circle with radius $\sqrt c$ --- is a one-dimensional manifold, as we knew. \ii When $c = 0$, the result fails. Indeed, $M$ is a single point, which is actually a zero-dimensional manifold! \end{itemize} \end{example} We won't give further examples since I'm only mentioning this in passing in order to increase your capacity to write real concrete examples. (But \cite[Chapter 6.2]{ref:manifolds} has some more examples, beautifully illustrated.) \section{Differential forms on manifolds} We already know what a differential form is on an open set $U \subseteq \RR^n$. So, we naturally try to port over the definition of differentiable form on each subset, plus a compatibility condition. Let $M$ be a smooth manifold with atlas $\{ U_i \taking{\phi_i} E_i \}_i$. \begin{definition} A \vocab{differential $k$-form} $\alpha$ on a smooth manifold $M$ is a collection $\{\alpha_i\}_i$ of differential $k$-forms on each $E_i$, such that for any $j$ and $i$ we have that \[ \alpha_j = \phi_{ij}^\ast(\alpha_i). \] \end{definition} In English: we specify a differential form on each chart, which is compatible under pullbacks of the transition maps. \section{Orientations} \prototype{Left versus right, clockwise vs.\ counterclockwise.} This still isn't enough to integrate on manifolds. We need one more definition: that of an orientation. The main issue is the observation from standard calculus that \[ \int_a^b f(x) \; dx = - \int_b^a f(x) \; dx. \] Consider then a space $M$ which is homeomorphic to an interval. If we have a $1$-form $\alpha$, how do we integrate it over $M$? Since $M$ is just a topological space (rather than a subset of $\RR$), there is no default ``left'' or ``right'' that we can pick. As another example, if $M = S^1$ is a circle, there is no default ``clockwise'' or ``counterclockwise'' unless we decide to embed $M$ into $\RR^2$. To work around this we have to actually have to make additional assumptions about our manifold. \begin{definition} A smooth $n$-manifold is \vocab{orientable} if there exists a differential $n$-form $\omega$ on $M$ such that for every $p \in M$, \[ \omega_p \neq 0. \] \end{definition} Recall here that $\omega_p$ is an element of $\Lambda^n(V^\vee)$. In that case we say $\omega$ is a \vocab{volume form} of $M$. How do we picture this definition? If we recall that an differential form is supposed to take tangent vectors of $M$ and return real numbers. To this end, we can think of each point $p \in M$ as having a \vocab{tangent plane} $T_p(M)$ which is $n$-dimensional. Now since the volume form $\omega$ is $n$-dimensional, it takes an entire basis of the $T_p(M)$ and gives a real number. So a manifold is orientable if there exists a consistent choice of sign for the basis of tangent vectors at every point of the manifold. For ``embedded manifolds'', this just amounts to being able to pick a nonzero field of normal vectors to each point $p \in M$. For example, $S^1$ is orientable in this way. \begin{center} \begin{asy} size(5cm); draw(unitcircle, blue+1); label("$S^1$", dir(100), dir(100), blue); void arrow(real theta) { pair P = dir(theta); dot(P); pair delta = 0.5*P; draw( P--(P+delta), EndArrow ); } arrow(0); arrow(50); arrow(140); arrow(210); arrow(300); \end{asy} \end{center} Similarly, one can orient a sphere $S^2$ by having a field of vectors pointing away (or towards) the center. This is all non-rigorous, because I haven't defined the tangent plane $T_p(M)$; since $M$ is in general an intrinsic object one has to be quite roundabout to define $T_p(M)$ (although I do so in an optional section later). In any event, the point is that guesses about the orientability of spaces are likely to be correct. \begin{example} [Orientable surfaces] \listhack \begin{enumerate}[(a)] \ii Spheres $S^n$, planes, and the torus $S^1 \times S^1$ are orientable. \ii The M\"obius strip and Klein bottle are \emph{not} orientable: they are ``one-sided''. \ii $\CP^n$ is orientable for any $n$. \ii $\RP^n$ is orientable only for odd $n$. \end{enumerate} \end{example} \section{Stokes' theorem for manifolds} Stokes' theorem in the general case is based on the idea of a \vocab{manifold with boundary} $M$, which I won't define, other than to say its boundary $\partial M$ is an $n-1$ dimensional manifold, and that it is oriented if $M$ is oriented. An example is $M = D^2$, which has boundary $\partial M = S^1$. Next, \begin{definition} The \vocab{support} of a differential form $\alpha$ on $M$ is the closure of the set \[ \left\{ p \in M \mid \alpha_p \neq 0 \right\}. \] If this support is compact as a topological space, we say $\alpha$ is \vocab{compactly supported}. \end{definition} \begin{remark} For example, volume forms are supported on all of $M$. \end{remark} Now, one can define integration on oriented manifolds, but I won't define this because the definition is truly awful. Then Stokes' theorem says \begin{theorem} [Stokes' theorem for manifolds] Let $M$ be a smooth oriented $n$-manifold with boundary and let $\alpha$ be a compactly supported $n-1$-form. Then \[ \int_M d\alpha = \int_{\partial M} \alpha. \] \end{theorem} All the omitted details are developed in full in \cite{ref:manifolds}. \section{(Optional) The tangent and contangent space} \prototype{Draw a line tangent to a circle, or a plane tangent to a sphere.} Let $M$ be a smooth manifold and $p \in M$ a point. I omitted the definition of $T_p(M)$ earlier, but want to actually define it now. As I said, geometrically we know what this \emph{should} look like for our usual examples. For example, if $M = S^1$ is a circle embedded in $\RR^2$, then the tangent vector at a point $p$ should just look like a vector running off tangent to the circle. Similarly, given a sphere $M = S^2$, the tangent space at a point $p$ along the sphere would look like plane tangent to $M$ at $p$. \begin{center} \begin{asy} size(5cm); draw(unitcircle); label("$S^1$", dir(140), dir(140)); pair p = dir(0); draw( (1,-1.4)--(1,1.4), mediumblue, Arrows); label("$T_p(M)$", (1, 1.4), dir(-45), mediumblue); draw(p--(1,0.7), red, EndArrow); label("$\vec v \in T_p(M)$", (1,0.7), dir(-15), red); dot("$p$", p, p, blue); \end{asy} \end{center} However, one of the points of all this manifold stuff is that we really want to see the manifold as an \emph{intrinsic object}, in its own right, rather than as embedded in $\RR^n$.\footnote{This can be thought of as analogous to the way that we think of a group as an abstract object in its own right, even though Cayley's Theorem tells us that any group is a subgroup of the permutation group. Note this wasn't always the case! During the 19th century, a group was literally defined as a subset of $\text{GL}(n)$ or of $S_n$. In fact Sylow developed his theorems without the word ``group'' Only much later did the abstract definition of a group was given, an abstract set $G$ which was independent of any \emph{embedding} into $S_n$, and an object in its own right.} So, we would like our notion of a tangent vector to not refer to an ambient space, but only to intrinsic properties of the manifold $M$ in question. \subsection{Tangent space} To motivate this construction, let us start with an embedded case for which we know the answer already: a sphere. Suppose $f \colon S^2 \to \RR$ is a function on a sphere, and take a point $p$. Near the point $p$, $f$ looks like a function on some open neighborhood of the origin. Thus we can think of taking a \emph{directional derivative} along a vector $\vec v$ in the imagined tangent plane (i.e.\ some partial derivative). For a fixed $\vec v$ this partial derivative is a linear map \[ D_{\vec v} \colon C^\infty(M) \to \RR. \] It turns out this goes the other way: if you know what $D_{\vec v}$ does to every smooth function, then you can recover $v$. This is the trick we use in order to create the tangent space. Rather than trying to specify a vector $\vec v$ directly (which we can't do because we don't have an ambient space), \begin{moral} The vectors \emph{are} partial-derivative-like maps. \end{moral} More formally, we have the following. \begin{definition} A \vocab{derivation} $D$ at $p$ is a linear map $D \colon C^\infty(M) \to \RR$ (i.e.\ assigning a real number to every smooth $f$) satisfying the following Leibniz rule: for any $f$, $g$ we have the equality \[ D(fg) = f(p) \cdot D(g) + g(p) \cdot D(f) \in \RR. \] \end{definition} This is just a ``product rule''. Then the tangent space is easy to define: \begin{definition} A \vocab{tangent vector} is just a derivation at $p$, and the \vocab{tangent space} $T_p(M)$ is simply the set of all these tangent vectors. \end{definition} In this way we have constructed the tangent space. \subsection{The cotangent space} In fact, one can show that the product rule for $D$ is equivalent to the following three conditions: \begin{enumerate} \ii $D$ is linear, meaning $D(af+bg) = a D(f) + b D(g)$. \ii $D(1_M) = 0$, where $1_M$ is the constant function on $M$. \ii $D(fg) = 0$ whenever $f(p) = g(p) = 0$. Intuitively, this means that if a function $h = fg$ vanishes to second order at $p$, then its derivative along $D$ should be zero. \end{enumerate} This suggests a third equivalent definition: suppose we define \[ \km_p \defeq \left\{ f \in C^\infty M \mid f(p) = 0 \right\} \] to be the set of functions which vanish at $p$ (this is called the \emph{maximal ideal} at $p$). In that case, \[ \km_p^2 = \left\{ \sum_i f_i \cdot g_i \mid f_i(p) = g_i(p) = 0 \right\} \] is the set of functions vanishing to second order at $p$. Thus, a tangent vector is really just a linear map \[ \km_p / \km_p^2 \to \RR. \] In other words, the tangent space is actually the dual space of $\km_p / \km_p^2$; for this reason, the space $\km_p / \km_p^2$ is defined as the \vocab{cotangent space} (the dual of the tangent space). This definition is even more abstract than the one with derivations above, but has some nice properties: \begin{itemize} \ii it is coordinate-free, and \ii it's defined only in terms of the smooth functions $M \to \RR$, which will be really helpful later on in algebraic geometry when we have varieties or schemes and can repeat this definition. \end{itemize} \subsection{Sanity check} With all these equivalent definitions, the last thing I should do is check that this definition of tangent space actually gives a vector space of dimension $n$. To do this it suffices to show verify this for open subsets of $\RR^n$, which will imply the result for general manifolds $M$ (which are locally open subsets of $\RR^n$). Using some real analysis, one can prove the following result: \begin{theorem} Suppose $M \subset \RR^n$ is open and $0 \in M$. Then \[ \begin{aligned} \km_0 &= \{ \text{smooth functions } f : f(0) = 0 \} \\ \km_0^2 &= \{ \text{smooth functions } f : f(0) = 0, (\nabla f)_0 = 0 \}. \end{aligned} \] In other words $\km_0^2$ is the set of functions which vanish at $0$ and such that all first derivatives of $f$ vanish at zero. \end{theorem} Thus, it follows that there is an isomorphism \[ \km_0 / \km_0^2 \cong \RR^n \quad\text{by}\quad f \mapsto \left[ \frac{\partial f}{\partial x_1}(0), \dots, \frac{\partial f}{\partial x_n}(0) \right] \] and so the cotangent space, hence tangent space, indeed has dimension $n$. %\subsection{So what does this have to do with orientations?} %\todo{beats me} \section\problemhead \begin{problem} Show that a differential $0$-form on a smooth manifold $M$ is the same thing as a smooth function $M \to \RR$. \end{problem} \todo{some applications of regular value theorem here}