Complexes C and C^n as Vector Spaces.

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Complex vector spaces can be transformed into real vector spaces through a process known as "realification," which involves creating a new basis that includes both the original complex vectors and their imaginary counterparts. For example, a complex basis {v1, v2} translates to a real basis in R^4 as {(a,b,0,0), (-b,a,0,0), (0,0,c,d), (0,0,-d,c)}. The orientation of a complex vector space is determined by considering its realification, where the change of basis matrix has a positive determinant, indicating a positively-oriented structure. This means that every complex vector space has a canonical orientation when viewed as a real vector space. The discussion highlights the importance of understanding the relationship between complex and real vector spaces in terms of orientation and basis transformation.
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Hi, Everyone:

Just curious about two things:

1) if we are given the complexes as a vector space V over R , so that

z1,..,zn are a basis, I heard there is a "natural way" of turning this

into a vector space of R^{2*n} over R; IIRC , is this how it is done:

{ z1,iz1,z2iz2,...,zn,izn}is a new basis.

Does this mean that , for, say, n=2 , if v1=a+ib and v2=c+id are

a basis, then

{ (a,b,0,0), (-b,a,0,0), (0,0,c,d), (0,0,-d,c)}

Is the associated basis for R^4:=R^{2*2} over R (or over any other

field)?

2) How do we define orientability/orientation of C^n over R (over F):

In the case of a vector space of R over F , we say that two bases

B1 and B2 have the same orientation if the matrix M taking

(the rows/columns of ) B1 to B2 has positive determinant. BUT**

if the basis vectors are complex, M is a complex matrix, and the

determinant may not be real.

3)Given 2: what do we mean when we say every complex vector space

is positively-oriented?

Thanks (sorry if post is too long).
 
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1) Correct. (Careful with the "other field" comment though. This only applies to R and C. The process is called "realification".)

2) To discuss the orientability of a complex vector space (with a given ordered basis), you think of it as a real vector space (i.e. you consider its realification). I'll elaborate on this below.

3) Using 2), it's an easy exercise in linear algebra that the realification of a complex vector space has a canonical orientation.

Here's what this means. Suppose you start with a complex vector space V with a basis z_1, ..., z_n. Then V can be viewed as a real vector space V_R with basis z_1, ..., z_n, iz_1, ..., iz_n. Suppose now that you choose another basis w_1, ..., w_n for V, and let T be the change of basis matrix z_i -> w_i. Then T will look like A+iB for some matrices A, B.

Now note that the change of basis matrix from the basis z_i, iz_i to w_i, iw_i will be
\begin{pmatrix}A & -B \\ B & A\end{pmatrix}.
(Why?) You can show that the determinant of this matrix is |det T|^2 > 0.

So, any given complex basis z_i has a (canonically) associated real basis z_i, iz_i that is positively oriented.
 
Excellent, morphism, very helpful, thanks. I'll try to prove your why? after dinner; I may need a followup if you don't mind.
 
No problem!
 
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