MHB Isometries .... Garling, Example 11.5.2 .... ....

  • Thread starter Thread starter Math Amateur
  • Start date Start date
  • Tags Tags
    Example
Math Amateur
Gold Member
MHB
Messages
3,920
Reaction score
48
I am reading D. J. H. Garling's book: "A Course in Mathematical Analysis: Volume II: Metric and Topological Spaces, Functions of a Vector Variable" ... ...

I am focused on Chapter 11: Metric Spaces and Normed Spaces ... ...

I need some help in order to understand Example 11.5.2 on a linear isometry ... ... I wish to prove the mapping given is a linear isometry ... but I am not sure I understand the context of the example/problem ...

The start of Section 11.5 defining isometries plus example 11.5.2 ... ... reads as follows:
View attachment 8978In Example 11.5.2 we are given $$f: \mathbb{R}^2 \to \mathbb{C}$$ ... where $$f(x,y) = x + iy$$ ...

I wish to show that $$f$$ is a linear isometry ... but how do I proceed ...

Basically I am unsure how to go about considering $$\mathbb{C}$$ as a real vector space ... how do we go about this ... ?

Is it just a matter of considering the scalars as real numbers?Help will be appreciated ...

Peter
 

Attachments

  • Garling - Example 11.5.2 & Start of Secton 11.5 on Isometries  .png
    Garling - Example 11.5.2 & Start of Secton 11.5 on Isometries .png
    26.2 KB · Views: 123
Physics news on Phys.org
I think the main trick here is that if you have a vector $z=x+iy\in\mathbb{C},$ you write its norm as $\|z\|=\sqrt{\bar{z}z},$ which is always a non-negative real number. You carry this over into the induced metric as well. Intuitively, the $\mathbb{R}^2$ vector $(x,y)$ corresponds to real part $x$ and imaginary part $y$ of a complex number.

Does that answer your question?
 


Hello Peter,

I haven't personally read that book, but I can try to help you with understanding Example 11.5.2. In this example, we are given a mapping f from the real vector space \mathbb{R}^2 to the complex vector space \mathbb{C}. The mapping is defined as f(x,y) = x + iy, where x and y are real numbers.

To show that f is a linear isometry, we need to prove two things: first, that f is a linear map (preserves vector addition and scalar multiplication), and second, that f preserves distances between vectors (is an isometry).

To prove linearity, you can use the properties of complex numbers to show that f satisfies the definition of a linear map. For example, f(u+v) = (u+v) + i(u+v) = (u + iy) + (v + iy) = f(u) + f(v), showing that f preserves vector addition. Similarly, you can show that f(ca) = c(a + iy) = cf(a), where c is a scalar and a is a vector, proving that f preserves scalar multiplication.

To show that f is an isometry, you can use the definition of distance in a complex vector space, which is given by ||u|| = \sqrt{u \cdot \bar{u}}, where u is a complex vector and \bar{u} is its complex conjugate. You can then show that ||f(u)|| = \sqrt{(x+iy) \cdot (x-iy)} = \sqrt{x^2 + y^2} = ||u||, proving that f preserves distances between vectors.

As for considering \mathbb{C} as a real vector space, yes, you can think of the scalars as real numbers. This is because a complex number can be written as a sum of a real and imaginary part, and the real part can be thought of as the scalar in the vector space.

I hope this helps. Let me know if you have any further questions. Good luck with your studies!
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
Back
Top