# Gamma 5 matrix

1. Jun 24, 2014

### ChrisVer

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:
$\Gamma^{2}_{chir} =1$
Do I have to take:
$\Gamma_{chir} \Gamma_{chir}$

or
$\Gamma_{chir} \Gamma_{chir}^{\dagger}$
?

2. Jun 25, 2014

### ofirg

$\gamma_{5}$ is an hermitain matrix.

so $\gamma_{5}=\gamma_{5}^{\dagger}$

so both options are the same

3. Jun 25, 2014

### ChrisVer

is that true in general dimensions? For example D= 5 dims?

4. Jun 25, 2014

### samalkhaiat

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

And in D = 5, $\gamma_{ ( 5 ) } \equiv \gamma^{ 4 }$

Last edited: Jun 25, 2014
5. Jun 25, 2014

### ChrisVer

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

$\Gamma = i^{a} \gamma_{0} \gamma_{1} ... \gamma_{D-1}$

So far I've proven that $\Gamma$ anticommutes with $\gamma^{M}$ 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 $\Gamma \propto 1$
And I'm trying to find the constraint on $a$ so that I can fulfill the requirement that $\Gamma^{2}=1$
However I am not sure if by the square they mean $\Gamma \Gamma$ or $\Gamma \Gamma^{\dagger}$...

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

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

However if $\Gamma^{2}= \Gamma \Gamma^{\dagger}$
and by supposing that:
$\gamma_{0}^{\dagger}= \gamma_{0}$
$\gamma_{i}^{\dagger}= \gamma_{0} \gamma_{i} \gamma_{0}$

I have:

$\Gamma \Gamma^{\dagger}= (i)^{2a} (-1)^{a} \gamma_{0} \gamma_{1} ... \gamma_{D-1}\gamma_{D-1}^{\dagger} ... \gamma_{1}^{\dagger} \gamma_{0}^{\dagger}$

now inserting the above assumption:

$\Gamma \Gamma^{\dagger}= (i)^{2a} (-1)^{a} \gamma_{0} \gamma_{1} ... \gamma_{D-1} \gamma_{0} \gamma_{D-1} \gamma_{0} ... \gamma_{0} \gamma_{1} \gamma_{0} \gamma_{0}$

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

$\Gamma \Gamma^{\dagger}= (-1)^{2a} \gamma_{0} \gamma_{1} ... \gamma_{D-1} \gamma_{0} \gamma_{D-1}... \gamma_{1}$

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

$\Gamma \Gamma^{\dagger}= (-1)^{2a+1}$

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

*eg
$D=1, ~~00 =-1$
$D=3, ~~ 012 012= +1$
$D=5, ~~ 0123401234= -1$
etc
So for general D I have $(-1)^{\frac{D+1}{2}}$

**eg
$D=3, ~~ 012 2'1'0' = 012 0 200100= 012021 * (-1)^{2}$
$D=5, ~~ 01234 4'3'2'1'0'=012340400300200100= 0123404321 * (-1)^{4}$
$D=7, ~~ 0123456 6'5'4'3'2'1'0'= 01234560600500400300200100 = 01234560654321 * (-1)^{6}$

Last edited: Jun 25, 2014