Is it possible to show that ${\gamma^5}^\dagger = \gamma^5$, where $$ \gamma^5 := i\gamma^0 \gamma^1 \gamma^2 \gamma^3,$$ using only the anticommutation relations between the $\gamma$ matrices, $$ \left\{\gamma^\mu,\,\gamma^\nu\right\}=2\eta^{\mu\nu}\,\mathbb{1},$$ and without using any specific representation of this algebra and a unitary invariance argument, as is usually done?
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As the comments explained, you need to know a few properties of the $\gamma$ matrices. First of all, from $$ \{\gamma_\mu, \gamma_\nu\} = 2 \eta_{\mu \nu} \mathbf{1}_4$$ you can infer that (depending on the metric but not on the representation of the dirac algebra!) in (+---) metric $\gamma_0$ is hermitian (hint: look at the $\mu = 0, \nu = 0$ component of the above equation), while the $\gamma_i$ ($i = 1, 2, 3$) are anti-hermitian. (In -+++ metric, this would be interchanged). And this should allow you to solve the problem.
The hermicity properties can be condensed into $$ \gamma_\mu^\dagger = \gamma_0 \gamma_\mu \gamma_0$$ which just reproduces the above if you take the commutation properties into account.
Neuneck
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Try using the definition of $\gamma^5$ and just apply the conjugation. Remember that conjugation flips the order of the matrices, which means you want to change their order before applying the conjugation.
Then realize that the anticommutation relations give you a way of interchanging two gamma matrices, giving only a minus sign (provided the indices are different).
Lastly, note that, independent of your convention, you have an uneven number of anti-hermitian matrices in this expression (while the rest is hermitian).
Wouter
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No, this can't be shown. Here's a counterexample: take $$γ^0=Γ^0, \quad γ^1=Γ^1, \quad γ^2=Γ^2, \quad γ^3 = Γ^3 \cosh α + Γ^5 \sinh α$$
where the $Γ^μ$ are standard generators and $α\ne 0$. These matrices satisfy the anticommutation relations, but $$γ^5 = Γ^5 \cosh α + Γ^3 \sinh α \ne Γ^5 \cosh α - Γ^3 \sinh α = (γ^5)^\dagger.$$
benrg
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