How to derive the form of the invariant spinor inner product? So we have gamma matrices that satisfy the spacetime algebra relations, $\{\gamma^\mu, \gamma^\nu\} = 2 \eta^{\mu\nu}$. We know that if we set $\sigma^{\mu\nu} = \frac{1}{4}[\gamma^\mu, \gamma^\nu]$ then the $\sigma$ matrices form a basis for a representation of the Lorentz Lie algebra.
Now we want to construct a Lorentz invariant inner product on the spinor space. An inner product on a finite-dimensional complex vector space always takes the form
$$ \tag{1} \langle u, v \rangle = v^\dagger H u $$
where $H$ is a Hermitian matrix. How can we construct $H$ in general without "guessing"?
Clearly such an $H$ must satisfy
$$\tag 2 S^\dagger H S = H$$
where $S$ is an arbitrary Lorentz transformation acting on spinors, and taking the derivative at the identity yields the equivalent condition
$$\tag{3} \sigma^\dagger H + H \sigma = 0$$
where $\sigma$ is an arbitrary linear combination of the $\sigma^{\mu\nu}$'s.
The usual approach I see is to choose a representation of the gamma matrices with $\gamma^0$ Hermitian and $\gamma^i$ anti-Hermitian for $i = 1, 2, 3$. One can easily check in that case that (3) is satisfied when we set $H = \gamma^0$. This gives the Lorentz-invariant inner product
$$\tag{4} \langle u, v \rangle = v^\dagger \gamma^0 u$$
which we can also write as
$$\tag{5} \langle u, v \rangle = \overline{v} u$$
with $\overline{v} = v^\dagger \gamma^0$. This is standard but it seems to require guessing that we should make $\gamma^0$ Hermitian and $\gamma^i$ anti-Hermitian. Moreover, I feel that this presentation is misleading because it appears to single out a privileged direction in space-time, namely the time direction in a fixed frame of reference.
Questions:


*

*Is it the case that $H$ always has to be a scalar multiple of $\gamma^0$ in order for (3) to hold? If so, why?

*If not, how can we construct an appropriate $H$ given an arbitrary representation $\gamma^\mu$ of the spacetime algebra?

*Can this be done without relying on the fact that $\eta$ is diagonal? If so, how? If not, why not?

 A: If we choose the signature of the metric $\eta$ to be $(1,-1,-1,-1)$ and choose the gamma matrices $\gamma^{\mu}$ to be unitary (as they can be because they form a representation of a finite group), then it would follow from the commutation relations $\{\gamma^{\mu},\gamma^{\nu}\}=2\eta^{\mu\nu}$ that $\gamma^{0}$ would be Hermitian and $\gamma^{i}\:,i=1,2,3$ would be antihermitian. In such a case we can take $H$ to be $\gamma^0$ or $\gamma^1\gamma^2\gamma^3$. On the other hand, if the signature of the metric $\eta$ is chosen to be $(-1,1,1,1)$ (and again gamma matrices are chosen to be unitary) then $\gamma^{0}$ would be antiheritian and $\gamma^i$ would be hermitian. In this case we can take $H$ to be either $i\gamma^0$ or $i\gamma^1\gamma^2\gamma^3$.
I think the source of nonuniqueness in the choice of $H$ is the fact that the space $\mathbb{C}^4$ is not an irreducible representation of the spin group (though it is an irreducible representation of the Clifford algebra). The two irreducible representation are eigenspaces of the matrix $\gamma_5$. This seems to be the reason that the two choices of $H$ above are related as $H_1=\text{const.}\gamma_5H_2$; however, I am not fully sure about this.  
About question 3, I think there is no loss of generality in choosing the matrix $\eta$ to be diagonal as one can always find a basis in which this is true. 
