Chapter 15: Problem 5

Let \(\mathbb{R}^{\prime}\) be the marifold consisting of \(\mathbb{R}\) with differentiable structure generated by the chart \(\left(R ; y=x^{3}\right)\). Show that the identity map id \(\mathbb{R}: \mathbb{R}^{\prime} \rightarrow \mathbb{R}\) is a differentiable homeomorphism, which is not a diffeomorphism.

Short Answer

Expert verified

The identity map id :\(\mathbb{R}^{\prime}\rightarrow \mathbb{R}\) is a differentiable homeomorphism, because both the identity map and its inverse are continuous and the identity map is differentiable. However, it is not a diffeomorphism because the inverse function, or the transition map from \(\mathbb{R}\) to \(\mathbb{R}^{\prime}\), is not differentiable at 0.

Step by step solution

Check if id :\(\mathbb{R}^{\prime}\rightarrow \mathbb{R}\) is continuous

The identity map is always continuous, because it is defined as id:\(\mathbb{R}^{\prime}\rightarrow \mathbb{R}\), \(x\rightarrow x\). For the continuity of a real-valued function id\(x\), limit of id\(x_{n}\) as \(n\) approaches infinity is equal to id\(x\), for every sequence \(x_{n}\) converging to \(x\). Therefore id :\(\mathbb{R}^{\prime}\rightarrow \mathbb{R}\) is continuous.

Check if the inverse map id :\(\mathbb{R}\rightarrow \mathbb{R}^{\prime}\) is continuous

The inverse map is also the identity, because it is defined as id:\(\mathbb{R}\rightarrow \(\mathbb{R}^{\prime}\), \(x\rightarrow x\). And using the similar reasoning as in Step 1, id :\(\mathbb{R}\rightarrow \mathbb{R}^{\prime}\) is continuous.

Check the differentiability of id :\(\mathbb{R}^{\prime}\rightarrow \mathbb{R}\)

Since \(\mathbb{R}^{\prime}\) has the differentiable structure generated by the chart \(\left(R ; y=x^{3}\right)\), we need to check differentiability in this chart. The derivative of the identity function on \(\mathbb{R}\) in the chart is always 1. Hence, the map id :\(\mathbb{R}^{\prime}\rightarrow \mathbb{R}\) is differentiable.

Check if id :\(\mathbb{R}\rightarrow \mathbb{R}^{\prime}\) is differentiable

To be a diffeomorphism, the inverse function also needs to be differentiable. Transition map from \(\mathbb{R}\) to \(\mathbb{R}^{\prime}\) is \(x^{3}\), and its derivative at 0 does not exist. Hence, id :\(\mathbb{R}\rightarrow \mathbb{R}^{\prime}\) is not differentiable.