 #1
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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:
View attachment 8649
View attachment 8650
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:
View attachment 8651
View attachment 8652
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:
View attachment 8653
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:
View attachment 8655
View attachment 8656
View attachment 8657
Hope that helps ...
Peter
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:
View attachment 8649
View attachment 8650
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:
View attachment 8651
View attachment 8652
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:
View attachment 8653
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:
View attachment 8655
View attachment 8656
View attachment 8657
Hope that helps ...
Peter
Attachments

Dundas  1  Example 2.2.6 ... PART 1 ... .png11.1 KB · Views: 75

Dundas  2  Example 2.2.6 ... PART 2 .png11.6 KB · Views: 70

Dundas  1  Example 2.1.5 ... PART 1 ... .png8.1 KB · Views: 87

Dundas  2  Example 2.1.5 ... PART 2 ... .png17.4 KB · Views: 78

Dundas  1  Start of Section 2.1 ... Topological Manifolds ... PART 1 .png35.8 KB · Views: 80

Dundas  2  Start of Section 2.1 ... Topological Manifolds ... PART 2 .png58 KB · Views: 75

Dundas  1  Start of Section 2.2 ... Smooth Structures ... PART 1 ... .png17.5 KB · Views: 82

Dundas  2  Start of Section 2.2 ... Smooth Structures ... PART 2 ... .png35.9 KB · Views: 77

Dundas  3  Start of Section 2.2 ... Smooth Structures ... PART 3 .png32.8 KB · Views: 68

Dundas  Special  Topological Manifolds  2.png57.9 KB · Views: 68
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