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The mapping $\phi$ is given by: $[0, 2\pi]^2 \to \mathbb{R}^4$, $\phi(\alpha, \beta) = (x(\alpha, \beta), y(\alpha, \beta), z(\alpha, \beta), u(\alpha, \beta))$, where:

$$x(\alpha, \beta) = (3 + cos \alpha) cos \beta$$ $$y(\alpha, \beta) = (3 + cos \alpha) sin \beta$$ $$z(\alpha, \beta) = sin \alpha cos\frac{\beta}{2} $$ $$u(\alpha, \beta) = sin \alpha sin\frac{\beta}{2}$$

Write a martix of such transforation.

Let $K$ be the image of $(0, 2 \pi)^2$ in this mapping.


Show that $K$ is a two-dimensional manifold and that $\phi$ is a parametrization, in particular that it is $1-1$ and that it has a maximum-order differential at every point


To show that we need 3 elements: local injectivity, continuity of $\phi$ and its inverse, differentiability of $\phi$ and its inverse, which is checked below.

Local injectivity - we want to show that for each point $(\alpha,\beta)$ in $[0,2 \pi]^2$, there exists a neighborhood around it such that $\phi$ maps each point in that neighborhood uniquely to a point in $\mathbb{R}^4$. That is the case if and only if the determinant of the matrix of transformation is nonzero.

$$ \begin{pmatrix} (3 + cos \alpha) cos \beta\\ (3 + cos \alpha) sin \beta\\ sin \alpha cos\frac{\beta}{2}\\ sin \alpha sin\frac{\beta}{2}\\ \end{pmatrix}^T \cdot \begin{pmatrix} (3 + cos \alpha) cos \beta\\ (3 + cos \alpha) sin \beta\\ sin \alpha cos\frac{\beta}{2}\\ sin \alpha sin\frac{\beta}{2}\\ \end{pmatrix} = $$ $$ = \begin{pmatrix} (3 + cos \alpha) cos \beta & (3 + cos \alpha) sin \beta & sin α cos\frac{\beta}{2} & sin \alpha sin\frac{\beta}{2} \end{pmatrix} \cdot \begin{pmatrix} (3 + cos \alpha) cos \beta\\ (3 + cos \alpha) sin \beta\\ sin (\alpha) cos \left( \frac{\beta}{2} \right)\\ sin (\alpha) sin \left( \frac{\beta}{2} \right)\\ \end{pmatrix} = $$ $$ =(3cos(\beta) + cos(\alpha)cos(\beta))^2 + (3sin(\beta) + cos(\alpha)sin(\beta))^2 + \left( sin (\alpha) cos \left( \frac{\beta}{2} \right) \right)^2 + \left( sin (\alpha) sin \left(\frac{\beta}{2} \right) \right)^2 = $$ $$ =9cos^2(\beta) + 6cos(\alpha)cos^2(\beta) + cos^2(\alpha)cos^2(\beta) + 9sin^2(\beta) + 6cos(\alpha)sin^2(\beta) + cos^2(\alpha)sin^2(\beta) + sin^2 (\alpha) cos^2 \left( \frac{\beta}{2} \right) + sin^2 (\alpha) sin^2 \left( \frac{\beta}{2} \right) = $$ $$ = 6cos(\alpha) + 10 $$ From that we get that: $$ Det(6cos(\alpha) + 10) = 6cos(\alpha) + 10 > 0 $$

So we have local injectivity.

Continuity of $\phi$ and its Inverse - is implied by differentiability of $\phi$ and its inverse, which is checked below.

Differentiability of $\phi$ and its Inverse - to demonstrate that $\phi$ is differentiable we need to check if each of its partial derivatives is nonzero and continous. Then to check that it has a differentiable inverse, we need to calculate the Jacobian matrix and show that its determinant is nonzero and continuous.

Matrix of partial derivatives (Jacobian): $$ \begin{pmatrix} -sin(\alpha) cos(\beta) & (3 + cos(\alpha)) sin(\beta) \\ -sin(\alpha) sin(\beta) & (3 + cos(\alpha)) -cos(\beta) \\ cos(\alpha) cos\left(\frac{\beta}{2}\right) & -\frac{1}{2} sin(\alpha) sin\left(\frac{\beta}{2}\right)\\ cos(\alpha) sin\left(\frac{\beta}{2}\right) & \frac{1}{2}sin(\alpha) cos\left(\frac{\beta}{2}\right)\\ \end{pmatrix} $$ As we can see, those are nonzero and are continous. As for the Jacobian: $$ \begin{pmatrix} -sin(\alpha) cos(\beta) & (3 + cos(\alpha)) sin(\beta)\\ -sin(\alpha) sin(\beta) & (3 + cos(\alpha)) (-cos(\beta))\\ cos(\alpha) cos\left(\frac{\beta}{2}\right) & -\frac{1}{2} sin(\alpha) sin\left(\frac{\beta}{2}\right)\\ cos(\alpha) sin\left(\frac{\beta}{2}\right) & \frac{1}{2}sin(\alpha) cos\left(\frac{\beta}{2}\right)\\ \end{pmatrix}^T \cdot \begin{pmatrix} -sin(\alpha) cos(\beta) & (3 + cos(\alpha)) sin(\beta)\\ -sin(\alpha) sin(\beta) & (3 + cos(\alpha)) (-cos(\beta))\\ cos(\alpha) cos\left(\frac{\beta}{2}\right) & -\frac{1}{2} sin(\alpha) sin\left(\frac{\beta}{2}\right)\\ cos(\alpha) sin\left(\frac{\beta}{2}\right) & \frac{1}{2}sin(\alpha) cos\left(\frac{\beta}{2}\right)\\ \end{pmatrix} = \begin{pmatrix} -sin(\alpha) cos(\beta) & -sin(\alpha) sin(\beta) & cos(\alpha) cos\left(\frac{\beta}{2}\right) & cos(\alpha) sin\left(\frac{\beta}{2}\right) \\ (3 + cos(\alpha)) sin(\beta) & (3 + cos(\alpha)) (-cos(\beta)) & -\frac{1}{2} sin(\alpha) sin\left(\frac{\beta}{2}\right) & \frac{1}{2}sin(\alpha) cos\left(\frac{\beta}{2}\right) \end{pmatrix} \cdot \begin{pmatrix} -sin(\alpha) cos(\beta) & (3 + cos(\alpha)) sin(\beta)\\ -sin(\alpha) sin(\beta) & (3 + cos(\alpha)) (-cos(\beta))\\ cos(\alpha) cos\left(\frac{\beta}{2}\right) & -\frac{1}{2} sin(\alpha) sin\left(\frac{\beta}{2}\right)\\ cos(\alpha) sin\left(\frac{\beta}{2}\right) & \frac{1}{2}sin(\alpha) cos\left(\frac{\beta}{2}\right)\\ \end{pmatrix} = $$

$$ = \begin{pmatrix} sin^2(\alpha)cos^2(\beta) + sin^2(\alpha)sin^2(\beta) + cos^2(\alpha) cos^2\left(\frac{\beta}{2}\right) + cos^2(\alpha) sin^2\left(\frac{\beta}{2}\right) & (-sin(\alpha)cos(\beta))(3 + cos(\alpha)) sin(\beta) + sin(\alpha)sin(\beta)(3+cos(\alpha)(cos(\beta))) - \frac{1}{2}sin(\alpha)cos(\alpha) sin\left(\frac{\beta}{2}\right)cos\left(\frac{\beta}{2}\right) + \frac{1}{2}sin(\alpha)cos(\alpha) sin\left(\frac{\beta}{2}\right)cos\left(\frac{\beta}{2}\right)\\ (3+cos(\alpha))sin(\beta)(-sin(\alpha)cos(\beta)) + (3+cos(\alpha))(−cos(\beta))(−sin(\alpha)sin(\beta)) + (−\frac{1}{2}sin(\alpha)sin\left(\frac{\beta}{2}\right))(cos(\alpha)cos\left(\frac{\beta}{2}\right)) + (\frac{1}{2}sin(\alpha)cos\left(\frac{\beta}{2}\right))(cos(α)sin\left(\frac{\beta}{2}\right)) & (3sin(\beta) + cos(\alpha)sin(\beta))^2 + (3cos(\beta) + cos(\alpha)cos(\beta))^2 + \frac{1}{4}sin^2(\alpha)sin^2\left(\frac{\beta}{2}\right) + \frac{1}{4}sin^2(\alpha)cos^2\left(\frac{\beta}{2}\right)\\ \end{pmatrix} $$

$$ = \begin{pmatrix} 1 & 0 \\ 0 & 10 + 6cos(\alpha)\\ \end{pmatrix} $$

$$|Det(J)| = 10 + 6cos(\alpha)$$

Determinant of that matrix is nonzero and continuous so we get differentiability of $\phi$ and its Inverse.

Is that correct? Is there any easier way to show that $K$ is a two-dimensional manifold and that $\phi$ is a parametrization (it is $1-1$ and that it has a maximum-order differential at every point)?

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  • $\begingroup$ Maybe a more insightful check of the injectivity would be to realize that in the $x,y$ plane, this is a rotation of the vector $((3 + \cos \alpha), 0)$. Then you are left with only two distinct values of $\alpha$ that might be mapped to the same point (those with the same $\cos$ value, but then you can check that those will have different values assigned in the $z,u$ plane (which is again a rotation, of the vector $(\sin \alpha, 0)$, but with half the frequency) $\endgroup$
    – a_student
    Commented May 6 at 10:08
  • $\begingroup$ Did you forget the $\frac{1}{2}$ factor, in the last calculation of the Jacobian determinant ? $\endgroup$
    – PTony
    Commented May 12 at 21:42

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I would check that $\phi : (0,2\pi)^2 \to K$ is injective and that $\frac{\partial{\phi}}{\partial{\alpha}}$ and $\frac{\partial{\phi}}{\partial{\beta}}$ are linearly independent for all $\alpha,\beta \in (0,2\pi)$. Moreover, $\phi$ is a smooth function so that, once the first two points are proved, the implicit function theorem ensures that for each $p \in (0,2\pi)^2$ there exists an open neighborhood $U$ of $p$ such that $\phi^{-1}|_{\phi(U)} : \phi(U) \to (0,2\pi)^2$ is smooth.

Let $(\alpha_1,\beta_1)$ and $(\alpha_2,\beta_2)$ such that $\phi(\alpha_1,\beta_1)=\phi(\alpha_2,\beta_2)$.

From the first and second equation and the inequality $3+\cos(\alpha)>0$, we get $3+\cos(\alpha_1)=\sqrt{x^2+y^2}=3+\cos(\alpha_2)$ so that $\alpha_1= \alpha_2$ or $\alpha_1= 2 \pi - \alpha_2$. Again using the first two equation you get $\beta_1=\beta_2$ since $\sin(\beta_1)=\sin(\beta_2)$ and $\cos(\beta_1)=\cos(\beta_2)$.

I use the third equation to check whether $\alpha_1= 2 \pi - \alpha_2$ and $\beta_1=\beta_2$ gives the same point in $K$. You get that $\sin(\alpha_1)\cos(\beta_1/2)$ is the opposite of $\sin(\alpha_2)\cos(\beta_2/2)$ so that they do not have the same image through $\phi$. The only remaining possibility is $(\alpha_1,\beta_1)=(\alpha_2,\beta_2)$ which means that $\phi$ is injective.

For the linear independence of $\frac{\partial{\phi}}{\partial{\alpha}}$ and $\frac{\partial{\phi}}{\partial{\beta}}$ you can check the determinant of the 2-minors of the Jacobien of $\phi$. You can also check that the Gram matrix (ordinary scalar product) is non-singular (as you have done) since in this case $\frac{\partial{\phi}}{\partial{\alpha}}$ and $\frac{\partial{\phi}}{\partial{\beta}}$ are independent. You should find $$ \begin{pmatrix} \langle \frac{\partial{\phi}}{\partial{\alpha}} , \frac{\partial{\phi}}{\partial{\alpha}}\rangle & \langle \frac{\partial{\phi}}{\partial{\alpha}} , \frac{\partial{\phi}}{\partial{\beta}} \rangle \\ \langle \frac{\partial{\phi}}{\partial{\beta}} , \frac{\partial{\phi}}{\partial{\alpha}} \rangle & \langle \frac{\partial{\phi}}{\partial{\beta}} , \frac{\partial{\phi}}{\partial{\beta}}\rangle\\ \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & (3+\cos(\alpha))^2+\frac14 \sin^2(\alpha)\\ \end{pmatrix} $$

Since $(3+\cos(\alpha))^2+\frac14 \sin^2(\alpha) \geq (2)^2 - \frac14 > 0$ we have the result. The form of the Gram matrix says that $\frac{\partial{\phi}}{\partial{\alpha}}$ and $\frac{\partial{\phi}}{\partial{\beta}}$ are orthogonal and $\frac{\partial{\phi}}{\partial{\alpha}}$ has norm 1.

This shows that $K$ is a smooth manifold (locally diffeomorphic to $\Bbb R^2$) without self intersections, but not necessarily an (embedded) submanifold of $\Bbb R^4$.

In this particular case, $\phi \left([0,2\pi]^2 \right)$ happens to be an (embedded) submanifold of $\Bbb R^4$ since $\phi$ is a known parametrization of the klein bottle wikipedia Klein bottle. You can also look at embedding klein bottles or embedding klein bottle

Roughly speaking, $\phi$ identifies opposite sides of the square, but is injective in its interior and two points can have the same image through $\phi$ only if $\beta_1 = 2 k \pi + \beta_2$ (as seen before). Let $M:=\phi ([0;2 \pi]^2)$ with $\phi : [0;2 \pi]^2 \to M$ (obviously) surjective on $M \subseteq \Bbb R^4$. Since $[0;2 \pi]^2$ is compact, $\phi$ continuous and onto $M$ (Hausdorff), $\phi$ is a quotient map.

$U=(0;2 \pi)^2$ is saturated (because no points on the sides of $[0;2 \pi]^2$ can have the same image as a point in $(0;2 \pi)^2$) and open, so that $\phi|_U$ is still a quotient map. An injective quotient map is an homeomorphism so $(\phi|_U)^{-1}$ is a global chart of $K$.

For $M$ you can use 9 (or 4) squares, obtained translating horizontally or vertically $(0;2 \pi)^2$ by $\pi$, that cover all the sides of $[0;2 \pi]^2$ (of $(0;2 \pi]^2$), with the same function $\phi$ (transition functions are smooth).

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