quantum mechanics - Can one show that ${gamma^5}^dagger = gamma^5$ directly from the anticommutation relations?

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?

Answer

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.