Transition functions in manifold

Question

Following is given as the definition of a manifold in a book which I am reading.

Let $\{U_{\alpha}:\alpha\in I\}$ be an open covering of a Hausdorff topological space $X$ and $\phi_{\alpha}$ be homeomorphisms from $U_{\alpha}$ onto open subsets of $\mathbb{R}^m$ with $m$ a non negative integer. Suppose there exists a $k\in \mathbb{N}\cup\{\infty\}$ such that whenever $U_{\alpha}\cap U_{\beta}\neq \emptyset$, we have $$\phi_{\alpha}\circ \phi_{\beta}^{-1}:\phi_{\beta}(U_{\alpha}\cap U_{\beta})\to \phi_{\alpha}(U_{\alpha}\cap U_{\beta}) \text{ is } C^k$$ Then $(X,\{(U_{\alpha},\phi_{\alpha}):\alpha\in I\})$ is called a $C^k$ manifold.

I have somehow developed the intuition for the homeomorphisms $\phi_{\alpha}$ from $U_{\alpha}$ to an open subset of $\mathbb{R}^{m}$ from the wikipedia article. But I am unable to buid intuition for this bizarre condition $$\phi_{\alpha}\circ \phi_{\beta}^{-1}:\phi_{\beta}(U_{\alpha}\cap U_{\beta})\to \phi_{\alpha}(U_{\alpha}\cap U_{\beta}) \text{ is } C^k$$

Can someone help me to understand this?

Answer 1

There is a very visually appealing picture that goes along with this, but not being able to draw it on here I shall just describe it to you.

For each point $x\in X$ we know we can find some open set $U\subseteq X$ containing $x$ such that there exists an embedding $\phi:U\hookrightarrow\mathbb{R}^m$ which is homeomorphic to its image $U'$--an open subset of $\mathbb{R}^m$. Think visually that $U'$ sits above $X$ casting a shadow down onto $U$. Let's further cement this intuition by writing $\phi$ in the form

$$\begin{array}aU'& &\\ & _\phi\searrow & \\ & & U\subseteq X\end{array}$$

As you mentioned there is definitive intuition as to what this means. This means that $\phi$ allows us to give a sort of coordinate system for $U$ allowing us to geometrically think about it looking like $U'$.

The problem with the above is that this "$U$", or more precisely, what it represents, is not unique. We may equally well find some other open set $V\subseteq X$ containing $x$ and a homeomorphism $\psi:V\to V'\subseteq\mathbb{R}^m$ with $V'$ open such that we get a very similar picture

$$\begin{array} & & V'\\ & \swarrow^\psi & \\ X\supseteq V & & \end{array}$$

Once again, we can think of $V$, being just the shadow of $V'$, as allowing us to think locally around $x$ as just a normal subset of $\mathbb{R}^m$.

We still have not figured out exactly why we need this strange condition you mention. Well, the idea is simple. Imagine that we are doing math near $x$, and to make our lives simpler we would like to define coordinates locally around $x$ so that we can just pretend we are doing math in $\mathbb{R}^m$. We realize then that we are at a crossroads--which coordinate definition do we pick? Namely, we have a diagram of the form

$$\begin{array}\;\phi^{-1}(U\cap V) & & & \psi^{-1}(U\cap V)\\ & _\phi \searrow & \swarrow\psi &\\\ & \;\;\;\;\;\;U\cap V & & \end{array}$$

which represents the space where the shadows created by $U'$ and $V'$ intersect and the parts of $U'$ and $V'$ which are creating these shadows. So, which do we pick? Doing mathematics we'd like for us to not really care which we pick--I mean, it's all just arbitrary coordinaization. But, what exactly should it mean "it doesn't matter". Of course, it doesn't mean that they are literally the same, but what it should mean is that any information made by choosing one coordinazation over the other is both true, and translatable to the case where we make the opposite coordinaization. Of course, the translation means that we have a map $\phi^{-1}(U\cap V)\to \psi^{-1}(U\cap V)$ that corresponds to the coordinization these spaces represent. The diagram gives us precisely how to make this map, namely by going down to $U\cap V$ via $\phi$ and then back up to $\psi^{-1}(U\cap V)$ by $\psi^{-1}$ and thus we get our map $\psi^{-1}\circ\phi:\phi^{-1}(U\cap V)\to\psi^{-1}(U\cap V)$ which we can roughly think about as the dictionary between the two coordinazations of $X$ locally around $x$. But, we don't want these to be any old maps. Indeed, if we are doing things like calculus, to transfer ideas through this dictionary it shouldn't just be a dictionary with set-words in it, but a dictionary with calculus-words in it. In other words, all the calculus statements we can make in $\phi^{-1}(U\cap V)$ should be able to be translated via this dictionary to those in $\psi^{-1}(U\cap V)$. Of course, a set-map isn't going to give us this--it could completely mess up notions of differentiability, etc. Thus, we want maps that won't change any words involving calculus, we want maps that respect it. Of course, these are $C^k$ maps (where the $k$ is to your taste in calculus, mine is $k=\infty$).

Answer 2

The function $\phi_\alpha$ imposes a coordinate structure like that of $\mathbb{R}^m$ on $U_\alpha$, and similarly for $\phi_\beta$ and $U_\beta$. The sets $U_\alpha$ and $U_\beta$ may overlap, and if they do, the maps $\phi_\alpha$ and $\phi_\beta$ impose different coordinate structures on $U_\alpha\cap U_\beta$, the set where they overlap. The composite function $\phi_\alpha\circ\phi_\beta^{-1}$ converts from the $\beta$ coordinate system on $U_\alpha\cap U_\beta$ to the $\alpha$ coordinate system on the overlap.

To require that this change-of-coordinate map be $C^k$ is merely to require that the two coordinate systems be related in a ‘nice’ way $-$ nicer and nicer as $k$ increases. For $k=0$ it just says that the change of coordinates is continuous: the two coordinate systems aren’t completely out of synch, to put it very informally. As $k$ increases, the change of coordinates can be carried out more and more smoothly.

Answer 3

Without that condition, what we have is just a topological manifold (though note that many people studying differentiable manifolds will add paracompactness to the definition).

Topological manifolds are nice and all, but we are very interested in getting "calculus" set up on our manifolds. To do so, we need a notion of what it means for a continuous map from a manifold $M$ to manifold $N$ to be "differentiable", or more generally, $C^k$. We already know what those terms mean for maps between Euclidean spaces, so we "transport" them to the manifold setting by declaring $f:M\to N$ to be $C^k$ when, for any $p\in M$, there exist charts $\phi_\alpha$ on $M$ and $\psi_i$ on $N$,$$\phi_\alpha:U_\alpha\to \phi(U_\alpha)\subseteq\mathbb{R}^{m}\quad\text{and}\quad \psi_i:V_i\to\psi_i(V_i)\subseteq\mathbb{R}^{n},$$ such that $p\in U_a$ and $f(U_\alpha)\subseteq V_i$, and such that the composition $$\phi_\alpha(U_\alpha)\xrightarrow{\;\;\phi_\alpha^{-1}\;\;}U_\alpha\xrightarrow{\;\;f\;\;} V_i\xrightarrow{\;\;\psi_i\;\;} \psi_i(V_i),$$ which is a map between open subsets of Euclidean spaces, is $C^k$. Seems reasonable enough, but in fact, this definition may not be consistent; what if, choosing $\phi_\alpha$ and $V_i$, the map $f$ appears to be $C^k$, and choosing different $\phi_\beta$ and $V_j$, the map $f$ appears not to be $C^k$?

The condition you refer to is precisely what is necessary to guarantee that we have a well-defined notion of $C^k$ function on a manifold, because it says that whenever we have two charts that cover some of the same points, the map that converts "looking in the first chart" to "looking in the second chart" is $C^k$ itself, and since compositions of $C^k$ functions are $C^k$, we will always come up with the same answer regardless of which chart we choose.

Answer 4

Maybe we should boldly ask: Why are the $\phi_\alpha$ required to be homeomorphiss and not just set-theoretic bijections in the first place? The idea is that the $\phi_\alpha$ carry the local look and feel of $X$ to something we (believe to) understand: good old $\mathbb R^n$.

Via $\phi_\alpha$ we can transport information from $\mathbb R^n$ to $X$ (piecewise). For example we can declare that $f\colon X\to Y$ is continuous iff $f|_{U_\alpha}=f_\alpha\circ\phi_\alpha$ with continuous $f_\alpha$ for all $\alpha$. However, if the transiiton functions were not continuous, less $f$ than "expected" may turn out to be continuous this way (maybe only constant functions) because what is continuous when using one map may not be continuous when using another.

Similarly, we might declare that $f\colon X\to \mathbb R^m$ is affine iff $f|_{U_\alpha}=f_\alpha\circ\phi_\alpha$ with affine maps $f_\alpha\colon \mathbb R^n\to\mathbb R^m$ for all $\alpha$. Again, this produces a nontrivial amount of affine functions on $X$ only if the transition functions themselves are affine. That is: By ripping a few pages out of our atlas, we might increase the set of affine funcitons on $X$. We could keep ripping out pages until only such local maps remain for which the transitions are affine. It seems that such a smaller atlas would be the best way of giving $X$ a structure where it makes sense of a function $X\to\mathbb R^m$ being called affine.

Back from this made-up example, the same happens when we want to introduce the notion of a function on $X$ to be smooth. As long as transitions are not themselves smooth, a function that looks smooth via one map may not look smooth via another. Getting rid of any non-smooth transition in our atlas removes an obstacle to some functions being smooth. If we only have smooth transitions, then we obtaoin the maximla posible set of smooth functions on $X$ and at the same time allows us to check smoothness by checking any map around each point of $X$, not needing to check all local maps.

Answer 5

A smooth manifold by definition admits a set of charts whose transition maps are diffeomorphisms. The manifold itself may be scrunched up like wastepaper and so not smooth in its ambient space, but scrunched paper despite its sharp creases is still intrinsically smooth.

The transition maps are best understood by what they do to points, vectors and differential forms. Suppose a point on the manifold maps to $x$ and $y$ on two overlapping charts. Let $y = f(x)$. By definition, $f$ is smooth with full-rank Jacobean $J_f$.

Let $w$ be a differential form of full dimension on the manifold. It’s easy to show such smooth forms exist using partition of unity. Let $w$ translate to $a(x)dx$ and $b(y)dy$ in the two charts. Pull back $b(f(x))dy$ using change of variables formula to $b(f(x))|J_f(x)|dx$ where $|.|$ is the determinant. This should be equal to $a(x)dx$ cuz $w$ can only have one representative at $x$. So $a(x) = b(f(x))|J_f(x)|$. (Assume positive orientation so that the determinant |.| sign is not important.)

Next, using chain rule once again, push-forward a vector $v$ in local coordinates at $x$ using the Jacobian matrix $J_f$ to the vector $J_fv$ in local coordinates at $y$. In other words, $d/dx = J_fd/dy$. Notice how this is opposite to how forms get pulled back.

Everything snaps together when we realize differential forms are alternating tensor duals of vectors. Let $V = v_1, …, v_n$ be a matrix of vectors in local coordinates where $n$ is the dimension of the manifold. Then the alternating tensor $dx(V) = |V|$, the determinant. $dy(J_fV) = |J_fV| = |J_f||V|$. Thus at $y=f(x)$, $$a(x)dx(V) = b(y)dy(J_fV)$$

This is why you cannot integrate a function on a manifold using charts. It doesn’t obey change of variables. By contrast, a density, e.g., the square root of the determinant of the Riemann metric matrix, does, ie, $p(x) = |J_f|p(y)$, so you can integrate it (even though its value at a point depends on the chart). Eg, $d/dx = J_fd/dy$, so:

$$p^2(x) = |(d/dx,d/dx)| = |J_f|^2|(d/dy,d/dy)| = |J_f|^2p^2(y)$$