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Chapter 15: Problem 15

Is the map \(\alpha: \mathbb{R} \rightarrow \mathbb{R}^{2}\) given by \(x=\sin t, y=\sin 2 t\) (1) an immersion, (ii) an cmbedded submanifold?

Short Answer

Expert verified
The map \( \alpha: \mathbb{R} \rightarrow \mathbb{R}^{2}\) given by \(x=\sin t,\)\(y=\sin 2 t\) is neither an immersion nor an embedded submanifold.

Step by step solution

01

Calculate the Jacobian matrix

The derivative of \( \alpha(t) \) is given by the Jacobian matrix of the component functions. So,we must take the derivatives of \(x=\sin t\) and \(y=\sin 2 t\), obtaining \(\begin{bmatrix} x'\ y' \end{bmatrix} = \begin{bmatrix} \cos t \ 2\cos 2t \end{bmatrix}\).
02

Determine if the map is an immersion

For the map to be an immersion, the derivative, or Jacobian, must be injective. In this case, the Jacobian would be injective if the determinant of Jacobian matrix is not zero everywhere. However, it should be noted that the determinant of a matrix and whether a mapping is injective are related for square matrices only. And since in this case the Jacobian is a 1x2 matrix, it can't be injective, hence the map \( \alpha(t) \) is not an immersion.
03

Check if the map is an embedded submanifold

For a map to be an embedded submanifold, it must first be an immersion, which as we determined is not the case. Therefore, the map \( \alpha(t) \) is not an embedded submanifold.

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