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Gamma 5 matrix

  1. Jun 24, 2014 #1


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    I am sorry, this is a rather dump question- more like a notation question since the sources I've looked into don't specify it... But in case I want to calculate (in general dimensions) the:
    [itex] \Gamma^{2}_{chir} =1 [/itex]
    Do I have to take:
    [itex] \Gamma_{chir} \Gamma_{chir}[/itex]

    [itex] \Gamma_{chir} \Gamma_{chir}^{\dagger}[/itex]
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  3. Jun 25, 2014 #2
    [itex]\gamma_{5}[/itex] is an hermitain matrix.

    so [itex]\gamma_{5}=\gamma_{5}^{\dagger}[/itex]

    so both options are the same
  4. Jun 25, 2014 #3


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    is that true in general dimensions? For example D= 5 dims?
  5. Jun 25, 2014 #4


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    There is no [itex]\gamma_{ ( 5 ) }[/itex]" in odd dimensional spacetime. In fact, in odd dimensions, It is one of the fundamental Gammas. For examples: D=3,
    [tex] \gamma_{ ( 5 ) } \equiv \sigma_{ 3 } = - i \sigma_{ 1 } \sigma_{ 2 } , \ \ -i \sigma_{ 1 } \sigma_{ 2 } \sigma_{ 3 } = I[/tex]

    And in D = 5, [itex] \gamma_{ ( 5 ) } \equiv \gamma^{ 4 }[/itex]
    Last edited: Jun 25, 2014
  6. Jun 25, 2014 #5


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    In general D dimensions you can always define a gamma 5:

    [itex]\Gamma = i^{a} \gamma_{0} \gamma_{1} ... \gamma_{D-1} [/itex]

    So far I've proven that [itex]\Gamma[/itex] anticommutes with [itex]\gamma^{M}[/itex] if D=even and commutes if D=odd... In the case of D=odd, because it commutes with all the gamma matrices (I think because of Schur's Lemma) you have [itex] \Gamma \propto 1 [/itex]
    And I'm trying to find the constraint on [itex]a[/itex] so that I can fulfill the requirement that [itex]\Gamma^{2}=1 [/itex]
    However I am not sure if by the square they mean [itex]\Gamma \Gamma[/itex] or [itex] \Gamma \Gamma^{\dagger}[/itex]...

    In [itex]\Gamma \Gamma[/itex] case I have:
    [itex] \Gamma \Gamma= (i)^{2a} \gamma_{0} \gamma_{1} ... \gamma_{D-1} \gamma_{0} \gamma_{1} ... \gamma_{D-1} [/itex]
    By doing the commutations properly I get:*
    [itex] \Gamma \Gamma= (i)^{2a} (-1)^{\frac{D+1}{2}} 1 = (-1)^{a+\frac{D+1}{2}} [/itex]

    So if I want to get the identity matrix, I must ask for the exponent to be even.
    [itex]a+\frac{D+1}{2}= 2n [/itex]
    [itex] a= \frac{4n-D-1}{2} [/itex]
    In the simplest case [itex]n=0[/itex] so that [itex](-1)^0=+1[/itex] and we have:
    [itex] a= - \frac{D+1}{2}[/itex]

    However if [itex]\Gamma^{2}= \Gamma \Gamma^{\dagger}[/itex]
    and by supposing that:
    [itex] \gamma_{0}^{\dagger}= \gamma_{0}[/itex]
    [itex] \gamma_{i}^{\dagger}= \gamma_{0} \gamma_{i} \gamma_{0} [/itex]

    I have:

    [itex]\Gamma \Gamma^{\dagger}= (i)^{2a} (-1)^{a} \gamma_{0} \gamma_{1} ... \gamma_{D-1}\gamma_{D-1}^{\dagger} ... \gamma_{1}^{\dagger} \gamma_{0}^{\dagger} [/itex]

    now inserting the above assumption:

    [itex]\Gamma \Gamma^{\dagger}= (i)^{2a} (-1)^{a} \gamma_{0} \gamma_{1} ... \gamma_{D-1} \gamma_{0} \gamma_{D-1} \gamma_{0} ... \gamma_{0} \gamma_{1} \gamma_{0} \gamma_{0} [/itex]

    Now inside D points, you have D-1 regions (in this case it means D-1 [itex]\gamma_{0}^{2}[/itex])**. Since D is odd, D-1 is even and thus the result is just a +.

    [itex] \Gamma \Gamma^{\dagger}= (-1)^{2a} \gamma_{0} \gamma_{1} ... \gamma_{D-1} \gamma_{0} \gamma_{D-1}... \gamma_{1}[/itex]

    Since all the gammas commute with [itex]\Gamma[/itex] I can move the middle [itex]\gamma_{0}[/itex] to the 1st place, without a problem, where I'll get [itex]\gamma_{0}^{2}=-1[/itex] and the rest gammas will start cancelling each other one after the other without changing anything (+)(+)(+) etc....

    [itex] \Gamma \Gamma^{\dagger}= (-1)^{2a+1} [/itex]

    again asking for the power to be even:
    [itex] 2a+1 =2n[/itex]
    [itex] a= \frac{2n-1}{2}[/itex]
    Again in the simplest case [itex] a= \pm \frac{1}{2}[/itex] ( [itex]\pm[/itex] because I don't know if it's needed to be positive or negative, for - n=0, for + n=1)

    [itex]D=1, ~~00 =-1 [/itex]
    [itex]D=3, ~~ 012 012= +1 [/itex]
    [itex]D=5, ~~ 0123401234= -1 [/itex]
    So for general D I have [itex] (-1)^{\frac{D+1}{2}}[/itex]

    [itex]D=3, ~~ 012 2'1'0' = 012 0 200100= 012021 * (-1)^{2} [/itex]
    [itex]D=5, ~~ 01234 4'3'2'1'0'=012340400300200100= 0123404321 * (-1)^{4} [/itex]
    [itex]D=7, ~~ 0123456 6'5'4'3'2'1'0'= 01234560600500400300200100 = 01234560654321 * (-1)^{6}[/itex]
    Last edited: Jun 25, 2014
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