Co-ordinate charts

Manifolds are sets on which we can define co-ordinates in such a way that we can do calculus. In general we don't expect to be able to define co-ordinates on all of a manifold. First we define:

Definition 2.1 (Co-ordinate charts.)   A co-ordinate chart on a set $M$ is a pair $(U, \psi)$ where $U \subset M$, $\psi \colon U \to \mathbb{R}^n$ is a bijection and $\psi(U) \subset \mathbb{R}^n$ is open.

If $(U, \psi)$ is a co-ordinate chart we call $U$ the domain of the co-ordinate chart and $\psi$ the co-ordinates. Notice that we do not say that $U$ is open in $M$ because $M$ is not a topological space yet; it is just a set.

Example 2.1   Let $1_{R^n} \colon \mathbb{R}^n \to \mathbb{R}^n$ be the identity map. That is $1_{R^n}(x) = (x^1, \dots, x^n)$. Then $( \mathbb{R}^n, 1_{R^n})$ is a co-ordinate chart on $\mathbb{R}^n$. We usually call these the standard, usual or natural co-ordinates.

Example 2.2   Let $U$ be any open subset of $\mathbb{R}^n$ and

$$\iota \colon U \to \mathbb{R}^n$$

the inclusion map defined by $\iota(x) = x$. Then clearly $\iota(U ) = U$ which is open so that $(U, \iota)$ is a co-ordinate chart on $\mathbb{R}^n$.

Example 2.3   Let $V$ be a finite dimensional vector space. Choose a basis $v^1, \dots, v^n$ for $V$ and define $\psi\colon V \to \mathbb{R}^n$ by

$$u = \sum_{i=1}^n \psi^i(u) v^i.$$

Then $\psi$ is a bijection, in fact a linear isomorphism. Indeed every linear isomorphism arises in this way as. If $\phi \colon V \to \mathbb{R}^n$ is a linear isomorphism we can take $w^i = \phi^{-1}(e^i)$ where $e^i$ is the vector with a $1$ in the $i$th place and zeros everywhere else. We leave it as an exercise to show that for every $u \in V$

$$u = \sum_{i=1}^n \phi^i(u) w^i.$$

Example 2.4   Let

$$U = \mathbb{R}^2 - \{(x, y) \colon y < 0\}$$

and define polar co-ordinates

$$(r, \theta) \colon U \to (0, \infty) \times (-\pi, \pi) \subset \mathbb{R}^2$$

as follows. We define $r(x, y) = \sqrt{x^2 + y^2}$ we define $\theta$ by the requirement that $x = r(x, y) \cos(\theta(x, y))$ and $y = r(x,y)\sin(\theta(x,y))$ and $\theta(x, 0) = 0$. Clearly $(r, \theta$ is a bijection on the given domain and range.

Example 2.5   Let $S^2$ be the set of all points in $\mathbb{R}^3$ of length one. Let

$$U_0 = S^2 - \{(0,0,1)\} \subset \mathbb{R}^2.$$

We can define co-ordinates on $U$ by stereographic projection from the point $(0,0,1)$ onto the $X$-$Y$ plane. That is if $p = (x, y, z) \in U_0$ it has co-ordinates $\psi(p) = (\psi_0^1(p), \psi^2_0(p))$ defined uniquely by the requirement that the line through $(0,0,1)$ and $p$ intersects the $X$-$Y$ plane at $(\psi_0^1(p), \psi^2_0(p), 0)$. So we must have

$$(x-0, y-0, z-1) = (x^1-0, x^2-0, 0-1)$$

and hence

$$\psi_0^1(x, y, z) = \frac{x}{1-z}$$

and

$$\psi_0^2(x, y, z) = \frac{y}{1-z}.$$

In general a manifold will have lots of co-ordinates. We don't expect a manifold to come with a given set of co-ordinates anymore that we expect an abstract vector space to come with a given basis. However not all co-ordinate charts will do. We want them to be able to fit together in some compatible way. The motivation for our definition comes from the desire to define differentiable functions on a manifold. Indeed we can regard co-ordinates as a device to decide which, of the many functions on $M$, are going to be differentiable. Let $(U, \psi)$ be a co-ordinate chart and let $f \colon U \to \mathbb{R}$ be a function. Then as $U$ is just a set it makes no sense to ask that $f$ be differentiable. However we can ask that $f$ be differentiable with respect to the co-ordinates. That is we consider

$$f \circ \psi^{-1} \colon \psi(U) \to \mathbb{R}.$$

Now $f\circ \phi^{-1}$ is a function defined on an open subset of $\mathbb{R}^n$, namely $\psi(U)$ and we know what it means for such a function to be differentiable. Consider now what happens when we change co-ordinates to some other co-ordinate chart say $(V, \chi)$ for convenience assuming that $V = U$. Then it is possible that $f \circ \psi^{-1}$ is differentiable but $f \circ \chi^{-1}$ is not. To compare them we write

$$f \circ \psi^{-1} = f\circ \chi^{-1} \circ (\chi\circ \psi^{-1})$$

where

$$\chi\circ \psi^{-1}\colon \psi(U) \to \chi(V)$$

is a bijection between open subsets of $\mathbb{R}^n$. Then a sufficient condition for $f \circ \psi^{-1}$ to be differentiable if $f \circ \chi^{-1}$ is is that $\chi\circ \psi^{-1}$ is differentiable. As we want this to work both ways we also require that $\psi\circ \chi^{-1}$ be differentiable. In other words we require that $\chi\circ \psi^{-1}$ is a diffeomorphism. If we want this to be true for any $f$ then we have already seen in Lemma 1.1 that this becomes a necessary condition.

In practice we may not be able to find charts $(U, \psi)$ and $(V, \chi)$ with $U = V$ so in the definition we need to allow for this.

Definition 2.2 (Compatibility of charts)   A pair of charts $(U, \psi)$ and $(V, \chi)$ are called compatible if the sets $\psi(U \cap V)$ and $\chi(U \cap V)$ are open and the map

$$\chi\circ \psi^{-1}_{\vert\psi(U\cap V)} \colon \psi(U \cap V) \to \chi(U \cap V)$$

is a diffeomorphism.

Note that we need to restrict the map $\psi^{-1}$ to the set $\psi(U \cap V)$ so that it can be composed with $\chi$.

Example 2.6   If $U \subset \mathbb{R}^2$ is the set in example 2.4 on which polar co-ordinates are defined then it has two co-ordinate charts defined on it $(U, (r, \theta))$, and $(U, \iota)$. The polar co-ordinates and the inclusion. Notice that $U \cap U = U$ so that $\iota(U \cap U)$ and $(r, \theta)(U\cap U)$ are open by assumption.

If we calculate the composition

$$\iota \circ (r, \theta)^{-1} \colon (0, \infty) \times (-\pi, \pi) \to U$$

we obtain

$$\iota \circ (r, \theta)^{-1} (s, \phi) = (r\sin(\phi), s\cos(\phi))$$

which is a diffeomorphism. Hence $(U, (r, \theta))$ and $(U, \iota)$ are compatible.

Example 2.7   Let $V$ be a vector space and $v^1, \dots, v^n$ and $w^1, \dots, w^n$ bases defining co-ordinates $\psi$ and $\phi$ by

$$v = \sum_{i=1}^n \psi^i(v) v^i = \sum_{i=1}^n \phi^i(v) w^i.$$

Notice that both $\phi$ and $\psi$ are onto so that $\psi(V \cap V ) = \mathbb{R}^n$ is certainly open in $\mathbb{R}^n$ and likewise for $\phi$. If we define a matrix $X^i_j$ by

$$v^i =\sum_{j=1}^n X^i_j w^j$$

for all $i$ then

$$\sum_{i,j=1}^n \psi^i(v) X^i_j w^j = \sum_{j=1}^n \phi^j(v) w^j$$

so that

$$\phi^j (v) = \sum_{i=1}^n X^i_j \psi^i(v).$$

Another way of calculating this result is to observe that

$$\phi\colon V \to \mathbb{R}^n$$

and

$$\psi \colon V \to \mathbb{R}^n$$

are linear isomorphisms so that

$$\phi\circ\psi^{-1} \colon \mathbb{R}^n \to \mathbb{R}^n$$

is the linear isomorphism with matrix $X^i_j$. Being linear $\phi\circ\psi^{-1}$ is certainly smooth so that $(V, \phi)$ and $(V, \psi)$ are compatible.

Example 2.8   If we consider again the example of $S^2$ we had defined a co-ordinate chart $(U_0, \psi_0)$ taking

$$U_0 = S^2 - \{(0,0,1)\}$$

and

$$\psi_0(x, y, z) = (\frac{x}{1-z}, \frac{y}{1-z}).$$

If we stereographically project from the point $(0,0,-1)$ then we get co-ordinates

$$\psi_1(x, y, z) = (\frac{x}{1+z}, \frac{x}{1+z}).$$

defined on

$$U_1 = S^2 - \{(0,0,-1)\}$$

We want to check that these are compatible. Note first that both $\psi_0(U_0 \cap U_1)$ and $\psi_1(U_0 \cap U_1)$ are equal to $\mathbb{R}^2 - \{(0,0,0)\}$ which is open in $\mathbb{R}^2$. Then an easy calculation shows that

$$\psi_0\circ \psi_1^{-1} (x^1 , x^2) = \bigl(\frac{x^1}{(x^1)^2 + (x^2)^2}, \frac{x^2}{ (x^1)^2 + (x^2)^2}\bigr)$$

which is a smooth map on $\mathbb{R}^2 - \{(0,0,0)\}$. Similarly for $\psi_1\circ \psi_0^{-1}$.

To make $M$ into a manifold we need to be able to cover it with compatible co-ordinate charts.

Definition 2.3 (Atlas)   An atlas for a set $M$ is a collection $\{(U_\alpha, \psi_\alpha)\mid \alpha \in I\}$ of co-ordinate charts such that:

(i) for any $\alpha$ and $\beta$ in $I$, $(U_\alpha, \psi_\alpha)$ and $(U_\beta, \psi_\beta)$ are compatible and;

(ii) $M = \cup_{\alpha\in I}U_\alpha$.

Then we have

Definition 2.4 (Manifold)   A manifold is a set $M$ with an atlas $\cal A$. We call the choice of an atlas $\mathcal{A}$ for a set $M$ a choice of differentiable structure for $M$.

Example 2.9   If there is a co-ordinate chart with domain all of $M$ then this, by itself defines an atlas and makes $M$ a manifold. For example $( \mathbb{R}^n,$   id$)$ makes $\mathbb{R}^n$ a manifold and if $U$ is open in $\mathbb{R}^n$ then $(U, \iota)$ makes $U$ a manifold.

Example 2.10   If $V$ is a vector space then any linear isomorphism from $V$ to $\mathbb{R}^n$ makes $V$ a manifold. The vector space $V$ has other atlases such as the atlas of all linear isomorphisms

$$\{(V, \phi)\mid \phi\colon V \to\mathbb{R}^n$$    a linear isomorphism$$\}.$$

Example 2.11   The charts $(U_0, \psi_0)$ and $(U_1, \psi_1)$ are compatible and have domains that cover $S^2$ so they make it into a manifold. It is not difficult to show that we cannot make $S^2$ into a manifold with only one chart $(S^2, \chi)$ if we require that $\chi$ is continuous. Indeed if $\chi$ is continuous then because $S^2$ is compact we must have $\chi(S^2) \subset \mathbb{R}^n$ compact and hence closed but $\chi(S^2)$ is open so this is not possible unless $\chi(S^2) = \mathbb{R}^n$ but then it is not compact.

Example 2.12   Consider the set $\mathbb{R}P_n$ of all lines through the origin in $\mathbb{R}^{n+1}$. We shall show that this is a manifold. This manifold is called real projective space of dimension $n$. If $x = (x^0, \dots, x^n)$ is non-zero vector in $\mathbb{R}^{n+1}$ we denote by $[x] = [x^0, \dots, x^n]$ the line through it. The numbers $x = (x^0, \dots, x^n)$ are often called the homogeneous co-ordinates of the line $[x]$. It is important to note that they are not uniquely determined by knowing the line. Indeed we have that $[x] = [y]$ if and only if there is a non-zero real number $\lambda$ such that $x = \lambda y$. The numbers $x = (x^0, \dots, x^n)$ are often called the homogeneous co-ordinates of the line $[x]$. Define a subset $U_i \subset \mathbb{R}P_n$ by

$$U_i = \{ [x] \mid x^i \neq 0 \}$$

for each $i = 0, \dots, n$ and notice that these subsets cover all of $\mathbb{R}P_n$. Define maps

$$\begin{array}{ccccc} \psi_i &\colon & U_i & \to & \mathbb{R}... ...}, \frac{x^{i+1}}{x^i}, \dots, \frac{x^n}{x^i}). \end{array}$$

Notice that we need to check that these maps are well defined but that follows from the fact that $[x] = [y]$ only if $x$ is a scalar multiple of $[y]$. It also straightforward to check that the $\psi_i$ are bijections onto $\mathbb{R}^n$ and hence define co-ordinates. Lastly it is straightforward to check that these co-ordinate charts are all compatible and hence make $\mathbb{R}P_n$ into a manifold.

We need to now deal with a technical problem raised by the definition of atlas. We often want to work with co-ordinate charts that are not in the atlas $\mathcal{A}$ used to define the differentiable structure. For example if $M = \mathbb{R}^2$ we might take $\mathcal{A}= \{ 1_{ \mathbb{R}^n} \}$. Then in a particular problem we might want to work with polar co-ordinates. But are they somehow compatible with the differentiable structure already imposed by $\mathcal{A}$ ? The definition of what compatibility is in this sense is easy. We could say that another co-ordinate chart is compatible with the given atlas if when we add it to the atlas we still have an atlas. In other words it is compatible with all the charts already in the atlas. We will take a different, but equivalent, approach via the notion of a maximal atlas containing $\mathcal{A}$ to explain these notions. We define;

Definition 2.5 (Maximal atlas.)   An atlas $\bar \mathcal{A}$ for a set $M$ is a maximal atlas for an atlas $\mathcal{A}$ if $\mathcal{A}\subset {\bar \mathcal{A}}$ and for any other atlas $\cal B$ with $\mathcal{A}\subset \cal B$ we have $\cal B \subset \bar \mathcal{A}$.

We then have

Proposition 2.1   For any atlas $\mathcal{A}$ on a set $M$ there is a unique maximal atlas $\bar \mathcal{A}$ containing $\mathcal{A}$. The maximal atlas consists of every chart compatible with all the charts in $\mathcal{A}$.

Proof. Define the set $\bar \mathcal{A}$ to the set of all charts which are compatible with every chart in $\mathcal{A}$. Then clearly if $\cal B$ is another atlas for $M$ with $\mathcal{A}\subset \cal B$ then we must have $\cal B \subset \bar \mathcal{A}$. What is not immediate is that $\bar \mathcal{A}$ is an atlas. The problem is that we do not know that the charts in $\bar A$ are compatible with each other. So let $(U, \psi)$ and $(V, \chi)$ be charts in $\bar \mathcal{A}$. We need to show that $(U, \psi)$ is compatible with $(V, \chi)$. Recall from the definition that this is true if the sets $\psi(U \cap V)$ and $\chi(U \cap V$ are open and

$$\chi\circ\psi^{-1}_{\vert\psi(U\cap V)}\colon \psi(U \cap V) \to \chi(U \cap V$$

is a diffeomorphism. Notice that to prove this it suffices to show that for every $x$ in $U \cap V$ we can find a $W$ with $x \in W \subset U \cap V$ such that $\psi(W)$ and $\chi(W)$ are open and such that

$$\chi\circ\psi^{-1}_{\vert\psi(W)}\colon \psi(W) \to \chi(W)$$

is a diffeomorphism.

To find $W$ choose a co-ordinate chart $(Z, \phi)$ in $\mathcal{A}$ with $x \in Z$. This is possible as the domains of the charts in an atlas cover $M$. Then let $W= U \cap V \cap Z$. Now $(U, \psi)$ is compatible with $(Z, \phi)$ so that $\phi(U \cap Z)$ is open. Similarly $\phi(V \cap Z)$ is open so that

$$\phi(W) = \phi(U \cap Z) \cap \phi(V \cap Z)$$

is open. Using compatibility again we that

$$\psi \circ \phi^{-1}_{\vert\phi(U \cap Z)} \colon \phi(U \cap Z) \to \psi(U \cap Z)$$

is a diffeomorphism and hence a homeomorphism so that

$$\psi (W ) = \psi \circ \phi^{-1}(\phi(W))$$

is open as required. A similar argument shows that $\chi(W)$ is open. Then the chain rule shows that

$$\chi\circ\psi^{-1}_{\vert\psi(W)} = (\chi\circ\phi^{-1}_{\vert\phi(W)}) \circ (\phi\circ \psi^{-1}_{\vert\psi(W)})$$

is a diffeomorphism. $\qedsymbol$

Finally we have

Definition 2.6   If $M$ is a manifold with atlas $\mathcal{A}$ we define a co-ordinate chart on the manifold $M$ to be a co-ordinate chart on the set $M$ which is in the maximal atlas $\bar \mathcal{A}$.

It should be noted that having defined an atlas we tend not to refer to it very much. We usually say $(U, \psi)$ is a co-ordinate chart on a manifold $M$ rather than $(U, \psi)$ is a member of the atlas $\mathcal{A}$ for a manifold $(M , \mathcal{A})$. The situation is similar to that for a topological space $X$ with topology $\mathcal{T}$. We rarely refer to the topology $\mathcal{T}$ by name. We say $U$ is an open subset of $X$ rather than $U \in \mathcal{T}$.