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

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The discussion centers on understanding Example 11.5.2 from D. J. H. Garling's book, which involves proving that the mapping f: ℝ² → ℂ defined by f(x,y) = x + iy is a linear isometry. To demonstrate this, one must show that f is a linear map by verifying it preserves vector addition and scalar multiplication. Additionally, it is necessary to prove that f maintains distances between vectors, using the norm defined in the complex vector space. The conversation also clarifies that when considering ℂ as a real vector space, scalars can be treated as real numbers, highlighting the relationship between the real and imaginary components of complex numbers. Overall, the thread provides guidance on approaching the proof of linear isometry for the given mapping.
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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
 

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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!
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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