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The complex algebra graded by Z-2

  1. May 29, 2015 #1
    I'm trying to understand something in my notes here....

    So if we call the real part of the complex algebra 'even' and the imaginary part 'odd' then this graded algebra is communitive but NOT graded commutative. so ab = ba for all a and b in C.

    If we call the whole complex algebra 'even' and only zero (also the only element in the intersection) to be odd then it would be graded commutative.

    so ab = (-1)^(|b|*|a|)*ba

    but if the whole of C is even, won't the parity of |b| and |a| always be zero and therefore the multiplication would just be normal commutative?

    P.S. The whole idea of grading is still uneasy with me (obviously..)
     
  2. jcsd
  3. May 30, 2015 #2

    fzero

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    The complex numbers can be viewed as a ##\mathbb{Z}/2\mathbb{Z}## graded ring ##R_0\oplus R_1## with ##R_0=\mathbb{R}##, ##R_1=i\mathbb{R}##. This is a commutative ring.

    They can also be thought of as a ##\mathbb{Z}/2\mathbb{Z}## graded algebra over the reals with the same ##A_0\oplus A_1## structure. In the algebra picture, ##A_1=i\mathbb{R}## is naturally an ##\mathbb{R}##-module, so again we see that the multiplication has to be commutative.

    I don't think that your proposition to make zero odd helps too much. In that case the structure is ##A_0=\mathbb{C}## and ##A_1=\{0\}## is the trivial ##\mathbb{C}##-module. This isn't the graded structure that the example was supposed to represent.

    Maybe you want to get comfortable with the idea of graded rings before adding structure to get graded modules and algebras. Graded-commutativity is a step beyond the direct sum structure that characterizes a graded ring.
     
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