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Calculation of the Pontryagin index

  1. Oct 5, 2012 #1
    When I read A.Zee's QFT in a Nutshell, he asked me to do the calculation of his exercise IV.4.6: (A.Zee called these the Pontryagin Index)

    Let g(x) be the element of a group G. The 1-form v = gdg† is known as the Cartan-Maurer form. Then tr v^N is trivially closed on an N-dimensional manifold since it is already an N-form. Consider Q = SN ∫tr vN with SN the N-dimensional sphere. Discuss the topological meaning of Q. These con- siderations will become important later when we discuss topology in field theory in chapter V.7. [Hint: Study the case N = 3 and G = SU(2).]

    I found that the result for G=SU(2) and N=3 is -24π^2, is it correct? I also calculated for the case N=1 and G=U(1), and i found the result is -2πi. What's on earth is the topological meaning of these results?

    Thanks.
     
    Last edited: Oct 5, 2012
  2. jcsd
  3. Oct 5, 2012 #2

    Bill_K

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    Isn't vN the volume element? Sounds like you're calculating the group volume.
     
  4. Oct 5, 2012 #3

    tom.stoer

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    Just to clarify the notation: you have to calculate

    [tex]v_i = g^\dagger \partial_{i}\,g [/tex]

    [tex]Q[g] = \frac{1}{c}\int_{S^N}d^N\Omega\,\epsilon^{i_1 i_2 \ldots i_N}\,\text{tr}(v_{i_1}\,v_{i_2}\ldots v_{i_N})[/tex]

    for various Lie groups G with a Lie group valued function

    [tex]g \in G [/tex]

    I guess the main problem you have is to find the correct constant c which obviously depends on the group G.
     
  5. Oct 5, 2012 #4

    tom.stoer

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    OK, let's continue with U(1) on S1. Let's write

    [tex]g(x) = e^{i\nu(x)}[/tex]

    with a function nu(x) respecting the S1 periodicity, i.e. with

    [tex]\nu(x+2\pi) = \nu(x)+n\;\Rightarrow\;g(x+2\pi) = g(x)+n[/tex]

    Calculating v(x) we get

    [tex]v_x = g^\dagger\,\partial_x\,g = i\,\partial_x\,\nu[/tex]

    For the integral we find

    [tex]-i\int_0^{2\pi}dx \, v_x = \int_0^{2\pi}dx \, \partial_x\,\nu = \nu(2\pi) - \nu(0) = n[/tex]

    That means that the function nu(x) runs from 0 to n when x runs from 0 to 2π; so g runs around the U(1) circle n-times when x runs around the S1 circle once. That means that

    [tex]Q[g] = n [/tex]

    is nothing else but the winding number of the map

    [tex]g: S^1 \to U(1)[/tex]
     
    Last edited: Oct 5, 2012
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