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A Problem about representations of group and particles

  1. Jul 16, 2015 #1
    I'm reading a paper these days,
    upload_2015-7-17_1-57-44.png
    How can I get 2.28?
    It seems for a D-dim SO(2) gauge field, we have spin2, spin1, as well as spin0 particles?
     
  2. jcsd
  3. Jul 17, 2015 #2

    fzero

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    For reference, the paper in question is http://arxiv.org/abs/1412.5606. Eq. (2.28) is a computation done for a graviton, not a ##SO(2)## gauge field.

    To understand the computation, you need to understand the discussion starting prior to equation (2.2). The authors want to compute a particular entanglement entropy using the partition function for free-field theories on Euclidean spacetimes of the form ##M= \mathbb{C}/\mathbb{Z}_N\times \mathbb{R}^{D-2}##. The coordinates are ##(x_0,x_1,x_2,\ldots, x_{D-1})##, where ##(x_0,x_1)## refer to ##\mathbb{C}## and ##(x_2,\ldots, x_{D-1})## to ##\mathbb{R}^{D-2}##. The orbifold group ##\mathbb{Z}_N## is a subgroup of the ##SO(2)\subset SO(D)## rotations that act only on ##(x_0,x_1)##.

    The ##D##-dimensional ##U(1)## gauge field is in the spin 1 representation of ##SO(D)##. In the paper, the components of the gauge field ##A_a## are numbered from ##1## to ##D## rather than from ##0## to ##D-1##. The ##SO(2)## subgroup acts on the ##A_1## and ##A_2## components, but not on ##A_3,\ldots A_D##. Hence, we determine that ##A_1## has ##SO(2)## spin ##s_1=1##, ##A_2## has ##s_2=-1##, while ##A_3,\ldots A_D## have ##s_a =0##.

    The graviton has components ##g_{ab}## (we'll use the Latin indices to match the index chosen on ##s_a##) and is in the spin 2 representation of ##SO(D)##. The ##SO(2)## subgroup acts as follows:
    $$ \begin{split}
    & g_{11}~~~ \text{with} ~~~ s=2,~~~1~\text{component}, \\
    & g_{22}~~~ \text{with} ~~~ s=-2,~~~1~\text{component}, \\
    & g_{1a} = g_{a1}, a\neq 1,2~~~ \text{with} ~~~ s=1,~~~D-2~\text{components}, \\
    & g_{2a} = g_{a2}, a\neq 1,2~~~ \text{with} ~~~ s=-1,~~~D-2~\text{components}, \\
    & g_{ab} , a,b\neq 1,2~~~ \text{with} ~~~ s=0,~~~\frac{D(D-3)}{2}~\text{components}.
    \end{split}$$
    Note, there is a typo in the last line of (2.28), so I should do the counting explicitly. Since the graviton is a symmetric, traceless tensor, the number of components of ##g_{ab}## with ##a,b\neq 1,2## is
    $$ \frac{(D-2)^2 - (D-2)}{2} + (D-2) -1 = \frac{D(D-3)}{2}.$$
    The calculation is as follows: the first term counts the number of independent off-diagonal elements, the second is the diagonal elements and the last removes the trace. You should also verify that the above counting gives the RHS of (2.29) correctly.
     
  4. Jul 17, 2015 #3
    fzero, Thank you. I got the idea. But I don't think the typo is there since they never introduce a traceless constraint.

    The other thing is why we can do that. Here by that I mean we divide the SO(D) into 3 parts.
    Can I say g11,g22 part is a representation of SO(2)
    g1a, g2a part is a vector representation of SO(D-2)
    gab part is a representation of SO(D-2) totally broken?

    Can I break the SO(D) group as I wish? For example, I'd like to have two diagonal blocks of SO(2)?
     
  5. Jul 17, 2015 #4

    fzero

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    The symmetric non-traceless representations of ##SO(D)## are reducible into the trace (singlet) and traceless irreducible representations. The elementary particle states always correspond to irreducible representations.

    Besides, the extra term is ##4/2=2## which can't be explained away by the trace.

    If you wanted to study some other theory, like the spacetime ##\mathbb{C}/{\mathbb{Z}_{N_1}}\times \mathbb{C}/{\mathbb{Z}_{N_2}} \times \mathbb{R}^{D-4}##, then you would have to consider ##SO(2)_1\times SO(2)_2## representations.
     
  6. Jul 21, 2015 #5
    Got it, thank you.
     
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