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

[Jamz]

Hello,

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

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

Many thanks!

 

[andrewkirk]

how to map a region of the space $\mathbb{C}^2$ spanned by two complex conjugate variables to the real plane $\mathbb{R}^2$ .

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 $(a+bi,a-bi)$?

 

[Jamz]

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 $(a+bi,a-bi)$

 

[fresh_42]

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.

 

[andrewkirk]

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 $\mathbb R$.

 

[WWGD]

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.