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Complexes C and C^n as Vector Spaces.

  1. Jan 25, 2012 #1

    WWGD

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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).
     
  2. jcsd
  3. Jan 25, 2012 #2

    morphism

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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
    [tex]\begin{pmatrix}A & -B \\ B & A\end{pmatrix}.[/tex]
    (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.
     
  4. Jan 25, 2012 #3

    WWGD

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    Excellent, morphism, very helpful, thanks. I'll try to prove your why? after dinner; I may need a followup if you don't mind.
     
  5. Jan 25, 2012 #4

    morphism

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    No problem!
     
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