Smooth Structures .... Dundas, Example 2.2.6 .... ....

[Math Amateur]

I am reading Bjorn Ian Dundas' book: "A Short Course in Differential Topology" ...

I am focused on Chapter 2: Smooth Manifolds ... ...

I need help in order to fully understand Example 2.2.6 ... ... Example 2.2.6 reads as follows:
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My questions are as follows:
Question 1

In the above text from Bjorn Dundas we read the following:

" ... ... First we calculate the inverse of \(\displaystyle x^{ 0,0 }\) : Let \(\displaystyle p = ( p_1, \ ... \ ... \ p_n )\) be a point in the open disk \(\displaystyle E^n\), then \(\displaystyle ( x^{ 0,0 } )^{ -1 } (p) = ( \sqrt{ 1 - \mid p \mid^2 } , p_1, \ ... \ ... \ p_n )\) ... ... "
Can someone please show the details of how Dundas gets

\(\displaystyle ( x^{ 0,0 } )^{ -1 } (p) = ( \sqrt{ 1 - \mid p \mid^2 } , p_1, \ ... \ ... \ p_n )\) ... ... ?Question 2

In the above text from Bjorn Dundas we read the following:

" ... ... Finally we get that if \(\displaystyle p \in x^{ 0,0 } (U)\) then

\(\displaystyle x^{ 1,1 } ( x^{ 0,0 } )^{ -1 } (p) = ( \sqrt{ 1 - \mid p \mid^2 } , \widehat{p_1} , \ ... \ ... \ p_n )\) ... ... "Can someone please show/explain the details of how Dundas gets

\(\displaystyle x^{ 1,1 } ( x^{ 0,0 } )^{ -1 } (p) = ( \sqrt{ 1 - \mid p \mid^2 } , \widehat{p_1} , \ ... \ ... \ p_n )\) ... ... "Further, why/how are we sure that this is a smooth map ...?Peter

==========================================================================================The above post refers to Bjorn Dundas' Example 2.1.5 ... so I am providing access to this example as follows:
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It may be very helpful to MHB readers of the above post to have access to the start of Dundas' Section on topological manifolds ... so I am providing the same as follows:
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View attachment 8658It may also be very helpful to MHB readers of the above post to have access to the start of Dundas' Section on smooth structures ... so I am providing the same as follows:

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Hope that helps ...

Peter

 

[jvicens]

Hello Peter,

Thank you for reaching out for help with understanding Example 2.2.6 in Bjorn Dundas' book on Differential Topology. I will do my best to explain the details and provide further clarification on your questions.

First, let's review the definition of a smooth manifold. A smooth manifold is a topological space that is locally homeomorphic to Euclidean space. In other words, for each point on the manifold, there exists a neighborhood that can be mapped smoothly to a corresponding neighborhood in Euclidean space. Now, let's dive into the details of Example 2.2.6.

Question 1:

In order to find the inverse of x^{0,0}, we start by considering a point p=(p_1,...,p_n) in the open disk E^n. This point can also be written as p=(0,p_1,...,p_n) since the first coordinate is always 0 in the open disk. Now, we need to find the inverse of x^{0,0} at this point p.

Recall that x^{0,0} is defined as x^{0,0}(p_1,...,p_n)=(0,p_1,...,p_n). So the inverse of this map will take a point (0,p_1,...,p_n) and map it back to (p_1,...,p_n). In other words, it will remove the first coordinate of the point and give us the remaining coordinates.

Using this information, we can see that (x^{0,0})^{-1}(p)=(\sqrt{1-|p|^2},p_1,...,p_n). This is because the first coordinate of p is always 0 in the open disk, so we can simply replace it with \sqrt{1-|p|^2} to get the inverse.

Question 2:

Now, let's consider the map x^{1,1}. This map takes a point (p_1,...,p_n) and maps it to (\sqrt{1-|p|^2},\widehat{p_1},...,p_n). Here, \widehat{p_1} represents the unit vector in the direction of p_1.

To get the composition x^{1,1}(x^{0,0})^{-1}(p), we simply plug in the inverse of x^{0,0} that we found in Question