Chapter 15: Problem 16

Show that the map $\alpha: \hat{R}^{2} \rightarrow \mathbb{R}^{3}$ defined by $$ u=x^{2}+y^{2}, \quad v=2 x y, \quad w=x^{2}-y^{2} $$ is an immersion Is it an embedded submanifold?

Short Answer

Expert verified

The map $\alpha$ is an immersion but not an embedded submanifold.

Step by step solution

Compute the Jacobian matrix

Begin by computing the Jacobian matrix of the map $\alpha$, which is the matrix of the first order partial derivatives of $\alpha$. The Jacobian matrix is thus given by \[J_{\alpha} =\begin{bmatrix}\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\\frac{\partial w}{\partial x} & \frac{\partial w}{\partial y}\end{bmatrix} = \begin{bmatrix}2x & 2y \2y & 2x \2x & -2y\end{bmatrix}, \]provided that $x$ and $y$ are not both zero, since in that case the derivative is not defined.

Check the rank of the Jacobian matrix

We then need to check that the Jacobian matrix $J_{\alpha}$ has full rank for all $(x,y) \in \mathbb{R}^2$. We can achieve this by computing the determinant of the Jacobian matrix $J_{\alpha}$, given by\[\text{det}(J_{\alpha}) = 2x \cdot 2x - 2y \cdot 2y = 4x^{2} - 4y^{2} = 4(x^{2} - y^{2}).\]We get that $\text{det}(J_{\alpha}) \neq 0$, hence the Jacobian matrix has full rank $2$ everywhere except at the origin, and $\alpha$ is an immersion.

Check if the map is an embedded submanifold

To check if the map $\alpha$ represents an embedded submanifold, we need to ensure that the image of the map is a topological submanifold and the differential of the map is a homeomorphism. The latter is guaranteed by the fact that $\alpha$ is an immersion. However, the former condition is, in general, harder to verify. In our case, the map $\alpha$ is not a homeomorphism on its image, due to the zero Jacobian at the origin, thus it does not represent an embedded submanifold.