# Manifolds – How to Show an Image of Transformation is a Manifold

manifoldssolution-verification

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)?

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).