Certain subsets of Euclidean space admit a suitable theory of calculus. The class of subsets we consider are the so-called sub-manifolds. Linear subspaces are a particular example, but not all (in fact most) sub-manifolds are not linear subspaces.
In particular, given a point \(p\) in a sub-manifold and a vector \(V\), the path \(p + tV\) is generally not in the sub-manifold. This means we need to develop a new notion of directional derivative. The definition we give for sub-manifolds is such that we can indeed develop such a notion. Indeed, from a certain perspective this is precisely what the definition aims to achieve.
A (smooth) local parametrisation is a \(C^{\infty}\) map \(\varphi : U \to \mathbb{R}^n\) where \(U \subseteq \mathbb{R}^k\) is an open set, with \(k \leq n\) and such that
Let \(M \subseteq \mathbb{R}^n\). We say \(M\) is an (embedded) sub-manifold if \(M\) may be covered by local parametrisations. That is, there are local parametrisations \(\{\varphi_{\alpha} : U_{\alpha} \to \mathbb{R}^n\}_{\alpha \in A}\) such that \[ M = \bigcup_{\alpha} \varphi_{\alpha}(U_{\alpha}) \] and each \(U_{\alpha} \subseteq \mathbb{R}^k\) is an open set for one fixed \(k\).
The range, \(M = \varphi(U)\) of a local parametrisation is a sub-manifold, but there are many examples of sub-manifolds that require more than one local parametrisation to cover them. For example, a theorem from topology states that the unit sphere, \(\mathbb{S}^n = \{x_1^2 + \cdots + x_{n+1}^2 = 1\} \subseteq \mathbb{R}^{n+1}\) is not homeomorphic to any open subset of Euclidean space. Allowing coverings by local parametrisations provides a rich array of examples that are often interesting in their own right and arise in applications (both within and without mathematics).
Let \(M\) be an embedded submanifold. The dimension of \(M\), \(\dim M = k\) where all of the local parametrisations are defined on an open subset \(U_{\alpha} \subseteq \mathbb{R}^k\).
Let \(M = \{x_1^2 + x_2^2 + x_3^2 = 1\} \subseteq \mathbb{R}^3\) be the unit sphere.
We will cover \(M\) with \(6\) local parametrisations, \(\varphi_N, \varphi_S, \varphi_E, \varphi_W, \varphi_F, \varphi_B\) each defined on the open set \(D = \{u_1^2 + u_2^2 < 1\} \subseteq \mathbb{R}^2\).
First, \[ \varphi_N (u_1, u_2) = \left(u_1, u_2, \sqrt{1-u_1^2-u_2^2}\right). \] Note that since \(u_1^2 + u_2^2 < 1\) the term under the square root is strictly positive. The positive square root function, \(\sqrt{z}\) is smooth for \(z > 0\) hence \(\sqrt{1 - u_1^2 - u_2^2}\) is smooth.
The range of \(\varphi_N\) is the "northern" hemisphere, \[ H_N = \left\{(x_1, x_2, x_3) \in \mathbb{R}^3 : x_1^2 + x_2^2 < 1, x_3 = \sqrt{1 - x_1^2 - x_2^2}\right\}. \]
On this set, inverse function is \[ \varphi_N^{-1} (x_1, x_2, x_3) = (x_1, x_2) \] which is continuous, hence \(\varphi_N\) is a homeomorphism.
For the differential, at \((u_1, u_2) \in D\),
\begin{equation*} d\varphi_N = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ - \frac{u_1}{\sqrt{1-u_1^2-u_2^2}} & - \frac{u_2}{\sqrt{1-u_1^2-u_2^2}} \end{pmatrix} \end{equation*}Already without even consider the third row we see that the columns are linearly independent hence \(d\varphi_N\) is injective.
The remaining local parametrisations are similar. Here's the complete list:
\begin{align*} \varphi_N (u_1, u_2) &= \left(u_1, u_2, \sqrt{1-u_1^2 - u_2^2}\right) \\ \varphi_S (u_1, u_2) &= \left(u_1, u_2, -\sqrt{1-u_1^2 - u_2^2}\right) \\ \varphi_E (u_1, u_2) &= \left(u_1, \sqrt{1-u_1^2 - u_2^2}, u_2\right) \\ \varphi_W (u_1, u_2) &= \left(u_1, -\sqrt{1-u_1^2 - u_2^2}, u_2\right) \\ \varphi_F (u_1, u_2) &= \left(\sqrt{1-u_1^2 - u_2^2}, u_1, u_2\right) \\ \varphi_B (u_1, u_2) &= \left(-\sqrt{1-u_1^2 - u_2^2}, u_1, u_2\right) \end{align*}Verify that \(\mathbb{S}^2\) is covered by the six parametrisations above.
It should be clear that the proof \(\varphi_N\) is a local parametrisation carries over to the other five parametrisations. However, it can be constructive to try write out a proof without simply copying the proof five times. One approach is to determine maps such that
\begin{align*} \varphi_N &= \tau_{NN} \circ \varphi_N\\ \varphi_S &= \tau_{SN} \circ \varphi_N \\ \varphi_E &= \tau_{EN} \circ \varphi_N \\ \varphi_W &= \tau_{WN} \circ \varphi_N \\ \varphi_F &= \tau_{FN} \circ \varphi_N \\ \varphi_B &= \tau_{BN} \circ \varphi_N \\ \end{align*}For example \(\tau_{NN} = \operatorname{Id}\), \(\tau_{SN} (x_1, x_2, x_3) = (x_1, x_2, -x_3)\), and \(\tau_{EN} (x_1, x_2, x_3) = (x_1, x_3, x_2)\). Write out the remaining cases and verify that they are all homeomorphisms \(\mathbb{R}^3 \to \mathbb{R}^3\) with invertible differentials \(d\tau_{\ast N}\) where \(\ast\) is a placeholder for \(N, S, E, \dots\).
Conclude that each \(d\varphi_{\ast}\) is a homeomorphism with injective differential.
Bonus points to write each \(\tau_{\ast N}\) as a rotation.
For higher dimensional spheres, covering by hemispheres is analogous but a little more cumbersome to write out. However, there is a way to cover a sphere of any dimension with only two local parametrisations.
Stereographic coordinates
Let \(\mathbb{S}^n = \{x_1^2 + \cdots + x_{n+1}^2 = 1\}\) be the unit sphere in \(\mathbb{R}^{n+1}\). Let \(N = (0, \dots, 0, 1)\) and \(S = -N = (0, \dots, 0, -1)\). Colloquially \(N\) and \(S\) are referred to as the north and south poles respectively.
The stereographic projection of a point \(P \in \mathbb{S}^n \backslash \{N\}\) onto \(\mathbb{R}^n\) is defined as the intersection of the line \(\overline{NP}\) with the \(\{x_{n+1} = 0\}\) hyperplane.
Let \(u = \pi_N(p)\) be this point. We can derive a formula for \(\pi_N\) as follows: parametrise the line by \[ t \mapsto N + t (P - N) = (1-t)N + tP. \] Writing \(P = (x_1, \dots, x_{n+1})\) we get \[ t \mapsto (0, \dots, 0, 1-t) + (tx_1, \dots, t x_{n+1}). \] The point of intersection occurs at the parameter value \(t_0\) where the last coordinate is zero: \[ 1 - t_0 + t_0 x_{n+1} = 0. \] That is \[ t_0 = \frac{1}{1-x_{n+1}}. \] Thus the intersection point is \[ (1-t_0) N + t_0 P = \left(\frac{x_1}{1-x_{n+1}}, \dots, \frac{x_n}{1-x_{n+1}}, 0\right). \]
In other words, we define, \[ \pi_N(p) = \pi_N(x_1, \dots, x_{n+1}) = \left(\frac{x_1}{1-x_{n+1}}, \dots, \frac{x_n}{1-x_{n+1}}\right), \] or \[ u_i = \frac{x_i}{1-x_{n+1}}. \] Notice this is well defined provided \(x_{n+1} \neq 1\). That is, provided \(p \neq N\). Moreover, \(\pi_N : \mathbb{S}^n \backslash \{N\} \to \mathbb{R}^n\) is continuous.
Our local parametrisation is the inverse of this map, \(\varphi_N = \pi_N^{-1}\). It is given by \[ \varphi_N(u) = \frac{1}{\|p\|^2 + 1}\left(2u_1, \dots, 2u_n, \|p\|^2 - 1\right). \] The domain of \(\varphi_N\) is \(\mathbb{R}^n\) which is indeed an open subset of \(\mathbb{R}^n\)!
The function \(\varphi_N\) is continuous, and we have already established that it's inverse \(\pi_N\) is also continuous. Thus \(\varphi_N\) is a homeomorphism.
One can directly compute \(d\varphi_N\) to verify that it is indeed injective, but this is rather tedious. Instead, we can use the chain rule. Observe that the formula for \(\pi_N\) defines a smooth function from the open set \(\mathbb{R}^{n+1} \backslash \{x_{n+1} = 1\}\) to \(\mathbb{R}^n\). For a point \(u \in \mathbb{R}^n\) we have that \[ \pi_N \circ \varphi_N (u) = u = \operatorname{Id} (u). \] By the chain rule, \[ d\pi_N \circ d\varphi_N = \operatorname{Id}. \] This implies that \(d\varphi_N\) is injective (and \(d\pi_N\) is surjective though we don't need that here).
Thus \(\varphi_N : \mathbb{R}^n \to \mathbb{S}^n \backslash \{N\}\) is a local parametrisation.
Since \(N \notin \mathbb{S}^n \backslash \{N\}\), \(\varphi_N\) alone does not cover all of \(\mathbb{S}^n\). But to remedy the situation we need only repeat the construction replacing \(N\) by \(S = (0, \dots, 0, -1)\) obtaining a local parametrisation \(\varphi_S : \mathbb{R}^n \to \mathbb{S}^n \backslash \{S\}\). Together \(\varphi_N, \varphi_S\) cover all of \(\mathbb{S}^n\) hence \(\mathbb{S}^n\) is an sub-manifold.
Verify that \(\varphi_N\) is indeed the inverse of \(\pi_N\). Also, try to see if you can derive the formula yourself!
Compute \(d\varphi_N\).
Let \(T : V \to W\) and \(S : W \to V\) be linear maps such that \[ T \circ S = \operatorname{Id}_W. \] Show that \(S\) is injective and \(T\) is surjective.
Determine local parametrisations for \(\mathbb{S}^1 \subseteq \mathbb{R}^2\).
Let \(f : U \to \mathbb{R}\) be a \(C^{\infty}\) function with \(U \subseteq \mathbb{R}^n\) an open set. The graph of \(f\) is the set \[ \operatorname{Graph} f = \{(p, f(p)) \in \mathbb{R}^{n+1} : p \in U\}. \]
For any \(C^{\infty}\) function \(f : U \to \mathbb{R}\), the graph of \(f\) is an embedded sub-manifold.
We only need one local parametrisation to cover \(\operatorname{Graph} f\), namely \[ \varphi(p) = (p, f(p)), \quad p \in U. \]
Since \(f\) is \(C^{\infty}\), \(\varphi\) is also \(C^{\infty}\). The inverse is the restriction to \(\operatorname{Graph} f\) of the orthogonal projection \(\pi\) onto the first \(n\) variables: \[ \varphi^{-1} (x_1, \dots, x_{n+1})= \pi|_{\operatorname{Graph} f} (x_1, \dots, x_{n+1}) = (x_1, \dots, x_n). \] Since \(\pi\) is continuous, \(\varphi^{-1}\) is continuous and \(\varphi\) is a homeomorphism onto it's image.
The differential is
\begin{equation*} d\varphi = \begin{pmatrix} 1 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & 1 \\ \partial_1 f & \cdots & \partial_n f \end{pmatrix} \end{equation*}which is injective since the columns are linearly independent.
Another way to write it is
\begin{align*} d\varphi_p (X) &= \partial_t (p + t X, f(p + t X)) \\ &= (X, D_X f (p)) \\ &= (X, 0) + D_X f (p) e_{n+1}. \end{align*}Here \((X, z) = (X^1, \dots, X^n, z)\) for \(z = D_X f\) in the second line and for \(z = 0\) in the last line.
Succinctly \[ d\varphi (X) = \iota (X) + df(X) e_{n+1} \] where \(\iota : \mathbb{R}^n \to \mathbb{R}^{n+1}\) is the inclusion \[ \iota(X^1, \dots, X^n) = (X^1, \dots, X^n, 0). \]
We can easily verify injectivity. Let \(\pi : \mathbb{R}^{n+1} \to \mathbb{R}^n\) be the orthogonal projection \[ \pi(X^1, \dots, X^{n+1}) = (X^1, \dots, X^n). \] Then \(\pi \circ \iota = \operatorname{Id}_n\) and \(\pi(df(X) e_{n+1}) = 0\). Thus \(\pi \circ d\varphi = \operatorname{Id}_n\) hence \(d\varphi(X) = d\varphi(Y)\) implies that \[ X = \pi \circ d\varphi (X) = \pi \circ d\varphi(Y) = Y. \]
The paraboloid of revolution is the set \[ M = \{(x_1, \dots, x_n, x_1^2 + \dots + x_n^2) : (x_1, \dots, x_n) \in \mathbb{R}^n\} \]
This is the graph of the smooth function \[ f(x_1, \dots, x_n) = x_1^2 + \dots + x_n^2 \] hence is an embedded submanifold.
A Hemisphere .
Let \(D \subseteq \mathbb{R}^n\) be the open unit disc, \[ D = \{p \in \mathbb{R}^n : \|p\| < 1\} \] and let \(f : D \to \mathbb{R}\) be the smooth function, \[ f(p) = \sqrt{1 - \|p\|^2} \]
Then since \(\|p\| < 1\), \(f\) is \(C^{\infty}\) and hence the open hemisphere \[ M = \{(p, \sqrt{1-\|p\|^2}) : p \in D\} \] is an embedded submanifold.
Let \(P = \{(x_1, \dots, x_n, 0) : (x_1, \dots, x_n) \in \mathbb{R}^n\}\). This is the hyperplane through the origin perpendicular to the \(x_{n+1}\)-axis. For short we write \(P = \{x_{n+1} = 0\}\). \(P\) is the graph of the constant function \(f(x_1, \dots, x_n) = 0\) and is hence a submanifold.
Let \(P\) be a hyperplane through the point \(p\) with normal vector \(N\). Show that \(P\) is an embedded submanifold.
Let \(F : U \to \mathbb{R}^k\) be a smooth function with \(U \subseteq \mathbb{R}^n\) an open set. Show that the graph of \(F\), \[ \operatorname{Graph} F = \{(p, F(p)) \in \mathbb{R}^{n+k} : p \in U\} \] is an embedded sub-manifold.
Let \(M \subseteq \mathbb{R}^n\) be a union of graphs of the same dimension. Show that \(M\) is an embedded sub-manifold.
Here to say \(M\) is a union of graphs of the same dimension is to say that there are open sets \(U_{\alpha} \subseteq \mathbb{R}^k\) for \(k < n\), and smooth functions \(F_{\alpha} : U_{\alpha} \to \mathbb{R}^{n-k}\) such that \[ M = \bigcup_{\alpha} \operatorname{Graph} F_{\alpha}. \] Here the phrase "same dimension" means that there is one fixed \(k\) such that each function \(F_{\alpha}\) is defined on an open subset of \(\mathbb{R}^k\). Note carefully that the definition of embedded sub-manifold requires that the local parametrisations are all defined on open subsets of Euclidean space of a fixed dimension.
Let \(f : (a, b) \to \mathbb{R}\) be a \(C^{\infty}\), positive function. The surface of revolution \(M \subseteq \mathbb{R}^3\) of \(f\) around the \(e_3\)-axis is the set \[ M = \{(x, y, z) : x^2 + y^2 = f(z), a < z < b\}. \]
Let \(f : (a, b) \to \mathbb{R}\) be a \(C^{\infty}\), positive function. The surface of revolution \(M \subseteq \mathbb{R}^3\) of \(f\) around the \(e_3\)-axis is an embedded sub-manifold.
For each fixed \(r \in (a, b)\), we have that \(M \cap \{z = r\}\) is a circle of radius \(f(r)\) in the \(\{z=r\}\) plane. We can parametrise this circle by \[ \theta \mapsto (f(r) \cos\theta, f(r) \sin\theta, r) \] with \(\theta \in [0, 2\pi)\). The main issue to deal with here is that \([0, 2\pi)\) is not an open interval. To remedy this we can restrict to \(\theta \in (0, 2\pi)\) but now we have the problem that the point \((f(r), 0, r)\) corresponding to \(\theta = 0\) is not in the image. To fix this last we problem we simply introduce another parametrisation using the same formula but with \(\theta \in (\pi, 3\pi)\).
In summary, we have two parametrisations of \(M \cap \{z = r\}\):
\begin{align*} c_1(\theta) &= (f(r) \cos\theta, f(r) \sin\theta, r), \quad 0 < \theta < 2 \pi \\ c_2(\theta) &= (f(r) \cos\theta, f(r) \sin\theta, r), \quad \pi < \theta < 3 \pi \end{align*}Then finally to parametrise \(M\) we just allow \(r\) to vary:
\begin{align*} \varphi_1(r, \theta) &= (f(r) \cos\theta, f(r) \sin\theta, r), \quad (r, \theta) \in (a, b) \times (0, 2 \pi) \\ \varphi_2(r, \theta) &= (f(r) \cos\theta, f(r) \sin\theta, r), \quad (r, \theta) \in (a, b) \times (\pi, 3 \pi) \end{align*}Now we verify these are local parametrisations:
Each parametrisation is injective: for example, if \(\varphi_1(r, \theta) = \varphi_1(s, \phi)\), then equating third coordinates, \(s = r\) and hence also \(f(r) = f(s)\). Then \(\cos \theta = \cos \phi\) and \(\sin \theta = \sin \phi\) which in turn implies \(\theta = \phi + 2\pi m\) for some \(m \in \mathbb{Z}\). But \(\theta, \phi \in (0, 2\pi)\) hence \(\theta = \phi\). The case of \(\varphi_2\) is the same except that \(\theta,\phi \in (\pi, 3\pi)\) and the conclusion is the same.
Thus for \(i = 1, 2\), \(\varphi_i\) is a bijection onto it's image and hence the inverse function \(\varphi_i^{-1}\) is defined on the image. To show \(\varphi_i^{-1}\) is continuous is a little awkward. It can be done by using the function atan2 which provides a continuous inverse to \(\theta \mapsto (\cos \theta, \sin \theta)\). The link provides some details. Alternatively, one can use the inverse function theorem which we will do in another section.
The differential is
\begin{equation*} d\varphi_i = \begin{pmatrix} f' \cos \theta & -f \sin \theta \\ f' \sin \theta & f \cos \theta \\ 1 & 0 \end{pmatrix} \end{equation*}The columns are linearly independent, since writing \(X_1, X_2\) for the columns if \(c^1 X_1 + c^2X_2 = 0\), then the third component becomes \(c^1 1 + c^2 0 = 0\) and hence \(c^1 = 0\). But then the first two components become
\begin{align*} -c^2 f \sin \theta &= 0 \\ c^2 f \cos \theta & = 0 \end{align*}If \(c^2 \neq 0\), since \(f > 0\), we must have \(\sin\theta = \cos \theta = 0\) but there is no such \(\theta\). Thus \(c^2 = 0\) and the columns are linearly independent.
Alternatively, rather than parametrising \(\mathbb{S}^1\) with \((\cos\theta, \sin\theta)\), \(0 < \theta < 2\pi\) and \((\cos \theta, \sin \theta)\), \(\pi < \theta < 3\pi\) we could parametrise \(\mathbb{S}^1\) by
\begin{align*} c_1(\theta) &= (\cos \theta, \sin \theta) \\ c_2(\theta) &= (\cos (\theta + \pi), \sin (\theta + \pi)) \\ \end{align*}with \(0 < \theta < 2\pi\) in both cases. There are infinitely many more ways to do it.
A cylinder
Let \(f (z) = 1\) for \(z \in (-1, 1)\). The surface of revolution of \(f\) around the \(z\) axis is the cylinder \[ M = \{(x, y, z) : x^2 + y^2 = 1\}. \] It may be parametrised by
\begin{align*} \varphi_1(r, \theta) &= (\cos \theta, \sin \theta, r) \\ \varphi_2(r, \theta) &= (\cos (\theta + \pi), \sin(\theta + \pi), r) \end{align*}for \(-1 < r < 1\) and \(0 < \theta < 2\pi\).
Sphere minus poles
Let \(f (r) = \sqrt{1-r^2}\) for \(r \in (-1, 1)\). The surface of revolution of \(f\) around the \(z\) axis is the cylinder \[ M = \{(x, y, z) : x^2 + y^2 = \sqrt{1-r^2}\}. \] It may be parametrised by
\begin{align*} \varphi_1(r, \theta) &= (\sqrt{1-r^2} \cos \theta, \sqrt{1-r^2} \sin \theta, r) \\ \varphi_2(r, \theta) &= (\sqrt{1-r^2} \cos (\theta + \pi), \sqrt{1-r^2} \sin(\theta + \pi), r) \end{align*}for \(-1 < r < 1\) and \(0 < \theta < 2\pi\).
A goblet like shape may be obtained by taking \(f(r) = 2 + \cos(r)\).
Let \(V \in \mathbb{R}^3\) be an arbitrary non-zero vector and \(p \in \mathbb{R}^3\) be an arbitrary point. Let \(f : (a, b) \to \mathbb{R}\) be a \(C^{\infty}\), positive function with \(a < 0 < b\). Define the surface of revolution about the axis \(V\) passing through the point \(p\) at \(r = 0\).
Let \(f : (a, b) \to \mathbb{R}\) be a smooth positive function. Show that the hypersurface of revolution , \[ M = \{(p, z) \in \mathbb{R}^n \times \mathbb{R} : \|p\| = f(z)\} \] is an embedded sub-manifold of dimension \(n\) in \(\mathbb{R}^{n+1}\).
Hint : generalise the parametrisations \((\cos \theta, \sin \theta)\) used for \(\mathbb{S}^1\) to parametrisations for \(\mathbb{S}^n\).
A torus is obtained by revolving one circle around another.
Let \(0 < r < R\). The inner radius is \(r\) and the outer radius is \(R\). Let \(C\) be the circle of radius \(r\) in the \(xz\) plane with centre \((R, 0, 0)\). That is, \[ C = \{(x, 0, z) \in \mathbb{R}^3 : (x-R)^2 + z^2 = r^2\}. \]
We can parametrise \(C\) by
\begin{align*} c_1 (\theta) &= \left(R + r\cos\theta, 0, r \sin\theta\right), \quad 0 < \theta < 2\pi \\ c_2 (\theta) &= \left(R + r\cos\theta, 0, r \sin\theta\right), \quad \pi < \theta < 3\pi \\ \end{align*}On way to discover this parametrisation is to let \(u = x - R\). Then \(u^2 + z^2 = r^2\) hence we may write \(u = r \cos \theta, z = r \sin \theta\). But this expression is not unique because of the periodicity of trig functions! If we restrict to the open set \(0 < \theta < 2\pi\), then we cover all points with \(u^2 + z^2 = r^2\) except for the point \((u, z) = (r, 0)\). If we restrict to the open set \(\pi < \theta < 3\pi\) then we cover all points with \(u^2 + z^2 = r^2\) except \((u, z) = (-r, 0)\). Thus together we cover all points with \(u^2 + z^2 = r^2\) and that leads to \(\psi_+\) and \(\psi_-\).
Now we rotate \(C\) about the \(z\)-axis so that the centre of \(C\) moves along the circle of radius \(R\) centred on the origin in the \(xy\) plane. To achieve this, observe that a rotation around the \(z\)-axis of angle \(\varphi\) corresponds to the transformation of \(\mathbb{R}^3\) that leaves the \(z\) coordinate unchanged and rotates the \(x, y\) coordinates by the angle \(\varphi\):
\begin{equation*} T_{\varphi} = \begin{pmatrix} \cos \varphi & -\sin\varphi& 0 \\ \sin\varphi & \cos\varphi & 0 \\ 0 & 0 & 1 \end{pmatrix} \end{equation*}The torus of inner radius \(r > 0\) and outer radius \(R > 0\) is the set \[ \mathbb{T}^2 = \{T_{\varphi} (C) : \varphi \in [0, 2\pi)\} \]
To parametrise \(\mathbb{T}^2\) we will have to deal with the fact that \(\varphi\) ranges over \([0, 2\pi)\) which is not open. For this we take \(\varphi\) in the range \(0 < \varphi < 2\pi\) and \(\pi < \varphi < 3\pi\). For this define,
\begin{align*} T_1(\varphi) &= T_{\varphi}, \quad 0 < \varphi < 2\pi \\ T_2(\varphi) &= T_{\varphi}, \quad \pi < \varphi < 3\pi. \end{align*}Putting everything together we obtain four parametrisations:
\begin{align*} \tau_{11} (\theta, \varphi) &= T_1(\varphi) (c_1(\theta)) \\ \tau_{12} (\theta, \varphi) &= T_1(\varphi) (c_2(\theta)) \\ \tau_{11} (\theta, \varphi) &= T_2(\varphi) (c_1(\theta)) \\ \tau_{11} (\theta, \varphi) &= T_2(\varphi) (c_2(\theta)) \end{align*}By \(T_1(\varphi) (c_1(\theta))\) for example, this means apply the transformation \(T_1(\varphi)\) to the vector \(c_1(\theta)\). Explicitly,
\begin{align*} T_1(\varphi) (c_1(\theta)) &= \begin{pmatrix} \cos \varphi & -\sin\varphi & 0 \\ \sin\varphi & \cos\varphi & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} R + r\cos\theta \\ 0 \\ r \sin\theta \end{pmatrix} \\ &= \begin{pmatrix} \cos \varphi (R + r \cos\theta) \\ \sin \varphi (R + r \cos\theta) \\ r \sin \theta \end{pmatrix} \end{align*}defined on the domain \[ U_{11} = \{0 < \theta < 2\pi, \quad 0 < \varphi < 2\pi\}. \]
The other domains \(U_{12}\), \(U_{21}\), \(U_{22}\) are defined similarly. Here we have the interesting situation that our four parametrisations are all defined by the same formula - the difference here is in the domains!