Calculating $\Gamma^2_{chir}$ in General Dimensions

[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}
?

 

[ofirg]

\gamma_{5} is an hermitain matrix.

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

so both options are the same

 

[ChrisVer]

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

 

[samalkhaiat]

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

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 }

 

[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}