Complexes C and C^n as Vector Spaces.

[WWGD]

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).

 

[morphism]

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.

 

[WWGD]

Excellent, morphism, very helpful, thanks. I'll try to prove your why? after dinner; I may need a followup if you don't mind.

 

[morphism]

No problem!

 

[blue_raver22]

Hi there,

1) Yes, you are correct. The new basis you provided is a valid way of turning the complex vector space V over R into a vector space of R^(2*n) over R. This can be done for any n, not just n=2. Essentially, we are taking the real and imaginary parts of each complex number in the original basis and treating them as separate components in the new basis.

2) The concept of orientability/orientation in a complex vector space is slightly different from that of a real vector space. In a complex vector space, we can define the orientation as the direction in which the complex numbers are mapped under a linear transformation. So if we have two bases B1 and B2, and the matrix M takes B1 to B2, then the orientation is determined by the direction in which the complex numbers in B1 are mapped to B2. This direction can be determined by the argument (angle) of the complex numbers in B1. If the argument increases in the counterclockwise direction, then the orientation is positive, and if it decreases, then the orientation is negative.

3) When we say a complex vector space is positively-oriented, we mean that the orientation of the complex numbers in the basis is positive, as determined by the argument. This can also be thought of as having a positive determinant for the matrix M that takes one basis to another.

I hope this helps clarify things for you. Let me know if you have any other questions or if anything is unclear.