Map from space spanned by 2 complex conjugate vars to R^2

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Hello,

I would like your help understanding how to map a region of the space [itex] \mathbb{C}^2 [/itex] spanned by two complex conjugate variables to the real plane [itex] \mathbb{R}^2 [/itex] .

Specifically, let us think that we have two complex conugate variables [itex] z [/itex] and [itex] \bar{ z} [/itex] and we define a triangle in the space represented schematically by having [itex] z [/itex] in the abscissa and [itex] \bar{z} [/itex] in the ordinate. I know this [itex] \mathbb{C}^2 [/itex] space shold be isomorphic to [itex] \mathbb{R}^4[/itex] , but considering the constraint that the variables are conjugate, I am hopping one can map such region to a representation in [itex] \mathbb{R}^2 [/itex] .

Many thanks!
 

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  • #2
andrewkirk
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how to map a region of the space [itex] \mathbb{C}^2 [/itex] spanned by two complex conjugate variables to the real plane [itex] \mathbb{R}^2 [/itex] .
Crucial information is missing from the question. When you talk about 'spanning' it sounds like you want to consider ##\mathbb C^2## as a vector space. If so, what is the related scalar field - ##\mathbb R## or ##\mathbb C##? Each one leads to a different answer.

Also, what do you mean by 'spanned by two complex conjugate variables'? An element of ##\mathbb C^2## is an ordered pair of complex numbers ##(z_1,z_2)##. Are you referring to an ordered pair of the form ##(a+bi,a-bi)##?
 
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Crucial information is missing from the question. When you talk about 'spanning' it sounds like you want to consider ##\mathbb C^2## as a vector space. If so, what is the related scalar field - ##\mathbb R## or ##\mathbb C##? Each one leads to a different answer.

Also, what do you mean by 'spanned by two complex conjugate variables'? An element of ##\mathbb C^2## is an ordered pair of complex numbers ##(z_1,z_2)##. Are you referring to an ordered pair of the form ##(a+bi,a-bi)##?
You are right, sorry for the misuse. I thought that seeing it as vector space with basis ##\{ \partial_{z},\partial_{\bar{z}} \}## would mean the same. But let us forget about that. What I am saying is precisely what you wrote:

I am referring to an ordered pair of the form ##(a+bi,a-bi)##
 
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You are right, sorry for the misuse. I thought that seeing it as vector space with basis ##\{ \partial_{z},\partial_{\bar{z}} \}## would mean the same. But let us forget about that. What I am saying is precisely what you wrote:

I am referring to an ordered pair of the form ##(a+bi,a-bi)##
You haven't answered the question about the scalar field.
If so, what is the related scalar field - ##\mathbb R## or ##\mathbb C##? Each one leads to a different answer.
 
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andrewkirk
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A ##n##-dimensional vector space over field ##F## is isomorphic to the vector space ##F^n##.

##\mathbb C^2## can be considered as a two-dimensional vector space over scalar field ##\mathbb C## and a four-dimensional vector space over scalar field ##\mathbb R##. In the latter case it is isomorphic to ##\mathbb R^4## over scalar field ##\mathbb R##.

The question is about the vector space generated by element ##(a+bi,a-bi)##, which is a single element of whatever vector space we are considering. So it generates a one-dimensional subspace.

Using the theorem of the first paragraph, over scalar field ##\mathbb C## that one-dimensional subspace will be isomorphic to ##\mathbb C^1=\mathbb C##. And over ##\mathbb R## it will be isomorphic to ##\mathbb R^1=\mathbb R##.

Either way it will not be isomorphic to ##\mathbb R^2##. However in the first case (ie over ##\mathbb C##), there will be a natural, intuitive bijection between the subspace and ##\mathbb R^2##, in the same way that there is a natural, intuitive bijection between ##\mathbb C## and ##\mathbb R##.
 
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  • #6
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To summarize : span is ## \{ c1(a+bi)+c2(a-bi)\} ## and it will depend on whether ##c_1, c_2 ## are Complex or Real scalars.
 
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