A point from the comments that should be emphasised is that any finite-dimensional vector space is isomorphic to any other of the same dimension.
What defines a representation of a group is the action of the group elements on the vector space. Then it is often convenient to choose some particular manifestation of the representation space in which the action looks nice, or familiar.
In the case of the proper Lorentz group, to classify the representations we use the fact that its complexification is isomorphic (up to a $\mathbb{Z}_2$) to $SL(2,\mathbb{C})\times SL(2,\mathbb{C})$, so an element can be given by a pair $(A,B)$ of $SL(2,\mathbb{C})$ matrices. Then the $(\frac12,\frac12)$ representation can be conveniently described as acting on the space of $2\times2$ matrices as
$$
(A,B):M\mapsto A M B^\dagger.
$$
This is natural if you think of $M$ as a tensor product of vectors in the $(\frac12,0)$ and $(0,\frac12)$ representations, like $M=u v^\dagger$ (or a sum of such terms).
To see the relation to the vector representation, write the matrices as
$$
M=\begin{pmatrix}t+z & x-i y\\
x+iy & t-z
\end{pmatrix}
$$
and notice that the determinant is $t^2-x^2-y^2-z^2$, and further that this is preserved under the action of the group. Using the $(t,x,y,z)$ basis instead of these matrices would give the usual Lorentz transformation of vectors. Going from the complexification to the real section of $SO(3,1)_+$ restricts you to a single $SL(2,\mathbb{C})$, which corresponds to $A=B$ the way it's written here, so Hermiticity (or reality of $(t,x,y,z)$) is also preserved.
It's pretty annoying that P&S just give you
$$S^{\mu \nu} = \frac{i}{4} [\gamma^{\mu},\gamma^{\nu}]$$
from thin air, here is a way to derive it similar to Bjorken-Drell's derivation (who start from the Dirac equation) but from the Clifford algebra directly, assuming that products of the gamma matrices form a basis. Given a Clifford algebra of $\gamma^{\mu}$'s satisfying
\begin{align}
\{ \gamma^{\mu} , \gamma^{\mu} \} = 2 \eta^{\mu \nu} I
\end{align}
we note that for an invertible transformation $S$ we have
\begin{align}
2 \eta^{\mu \nu} I &= 2 \eta^{\mu \nu} S^{-1} S \\
&= S^{-1}(2 \eta^{\mu \nu}) S \\
&= S^{-1}\{ \gamma^{\mu} , \gamma^{\mu} \} S \\
&= \{ S^{-1} \gamma^{\mu} S, S^{-1}\gamma^{\mu} S \} \\
&= \{ \gamma'^{\mu} , \gamma'^{\mu} \}
\end{align}
showing us that the Clifford algebra of matrices
$$\gamma'^{\mu} = S^{-1} \gamma^{\mu} S$$
also satisfies the Clifford algebra, hence any set of matrices satisfying the Clifford algebra can be obtained from a given set $\gamma^{\mu}$ using a non-singular transformation $S$. Since the anti-commutation relations involve the metric $\eta_{\mu \nu}$, and we know the metric is left invariant under Lorentz transformations
$$\eta^{\mu \nu} = \Lambda^{\mu} \, _{\rho} \Lambda^{\nu} \, _{\sigma} \eta^{\rho \sigma} $$
this immediately implies
\begin{align}
2 \eta^{\mu \nu} I &= \Lambda^{\mu} \, _{\rho} \Lambda^{\nu} \, _{\sigma} 2 \eta^{\rho \sigma} I \\
&= \Lambda^{\mu} \, _{\rho} \Lambda^{\nu} \, _{\sigma} \{ \gamma^{\rho} , \gamma^{\sigma} \} \\
&= \{ \Lambda^{\mu} \, _{\rho} \gamma^{\rho} , \Lambda^{\nu} \, _{\sigma} \gamma^{\sigma} \} \\
&= \{ \gamma'^{\mu} , \gamma'^{\mu} \}
\end{align}
which shows that the Lorentz transformation of a gamma matrix also satisfies the Clifford algebra, and so is itself a gamma matrix, and hence can be expressed in terms of some non-singular transformation $S$
\begin{align}
\gamma'^{\mu} &= \Lambda^{\mu} \, _{\nu} \gamma^{\nu} \\
&= S^{-1} \gamma^{\mu} S
\end{align}
where $S$ is to be determined. Since the operators $S$ represent performing a Lorentz transformation on $\gamma^{\mu}$, and Lorentz transformations on fields expand as $I - \frac{i}{2}\omega_{\mu \nu} M^{\mu \nu}$, we expand $\Lambda$ and $S$ as
\begin{align}
\Lambda^{\mu} \, _{\nu} &= \delta ^{\mu} \, _{\nu} + \omega^{\mu} \, _{\nu} \\
S &= I - \frac{i}{2} \omega_{\mu \nu} \Sigma^{\mu \nu}
\end{align}
where $\Sigma^{\mu \nu}$ must be anti-symmetric and constructed from a basis of gamma matrices, hence from
\begin{align}
\gamma'^a &= \Lambda^a \, _{\mu} \gamma^{\mu} \\
&= (\delta^a \, _{\mu} + \omega^a \, _{\mu})\gamma^{\mu} \\
&= \gamma^a + \omega^a \, _{\mu} \gamma^{\mu} \\
&= \gamma^a + \omega_{b \mu} \eta^{a b} \gamma^{\mu} \\
&= \gamma^a + \omega_{b \mu} \eta^{a [b} \gamma^{\mu]} \\
&= \gamma^a + \frac{1}{2} \omega_{b \mu} (\eta^{a b} \gamma^{\mu} - \eta^{a \mu} \gamma^b) \\
&= \gamma^a + \frac{1}{2} \omega_{\nu} (\eta^{a \mu} \gamma^{\nu} - \eta^{a \nu} \gamma^{\mu}) \\
&= S^{-1} \gamma^a S \\
&= (I - \frac{i}{2} \omega_{\mu \nu} \Sigma^{\mu \nu}) \gamma^a (I + \frac{i}{2} \omega_{\mu \nu} \Sigma^{\mu \nu}) \\
&= \gamma^a - \frac{i}{2} \omega_{\mu \nu} [\gamma^a, \Sigma^{\mu \nu}]
\end{align}
we have the relation (which can be interpreted as saying that $\gamma^a$ transforms as a vector under spinor representations of Lorentz transformations, as e.g. in Tong's QFT notes)
\begin{align}
i (\eta^{a \mu} \gamma^{\nu} - \eta^{a \nu} \gamma^{\mu}) = [\gamma^a, \Sigma^{\mu \nu}]
\end{align}
and we know $\Sigma^{\mu \nu}$, since it is anti-symmetric, must involve a product's of $\gamma$ matrices (because of the 16-dimensional basis formed from Clifford algebra elements), only two by the left-hand side, and from
\begin{align}
\gamma^{\mu} \gamma^{\nu} &= - \gamma^{\nu} \gamma^{\mu} , \ \ \ \mu \neq \nu, \\
\gamma^{\mu} \gamma^{\mu} &= \gamma^{\nu} \gamma^{\mu} , \ \ \ \mu = \nu,
\end{align}
we expect that
\begin{align}
\Sigma^{\mu \nu} &= c [\gamma^{\mu},\gamma^{\nu}] \\
&= c (\gamma^{\mu} \gamma^{\nu} - \gamma^{\nu} \gamma^{\mu}) \\
&= 2 c ( \gamma^{\mu} \gamma^{\nu} - \eta^{\mu \nu})
\end{align}
for some $c$ which we constrain by the (vector) relation above
\begin{align}
i (\eta^{a \mu} \gamma^{\nu} - \eta^{a \nu} \gamma^{\mu}) &= [\gamma^a, \Sigma^{\mu \nu}] \\
&= c [\gamma^a, 2( \gamma^{\mu} \gamma^{\nu} - \eta^{\mu \nu})] \\
&= 2 c [\gamma^a, \gamma^{\mu} \gamma^{\nu}] \\
&= 2 c ( \gamma^{\mu} [\gamma^a,\gamma^{\nu}] + [\gamma^a, \gamma^{\mu}] \gamma^{\nu}) \\
&= 2 c [ \gamma^{\mu} 2( \gamma^a \gamma^{\nu} - \eta^{a \nu}) + 2( \gamma^a \gamma^{\mu} - \eta^{a \mu}) \gamma^{\nu}] \\
&= 4 c [ \gamma^{\mu} ( \gamma^a \gamma^{\nu} - \eta^{a \nu}) + ( \gamma^{\mu} \gamma^{a} + 2 \eta^{a \mu} - \eta^{a \mu}) \gamma^{\nu}] \\
&= 4 c (\eta^{a \mu} \gamma^{\nu} - \eta^{a \nu} \gamma^{\mu}).
\end{align}
This gives the result $c = i/4$. The generator of Lorentz transformations of gamma matrices is
\begin{align}
\Sigma^{\mu \nu} &= \dfrac{i}{4} [\gamma^{\mu},\gamma^{\nu}] \\
&= \dfrac{i}{2}(\gamma^{\mu} \gamma^{\nu} - \eta^{\mu \nu}) \ \ \text{i.e.} \\
S &= I - \frac{i}{2} \omega_{\mu \nu} (\frac{i}{4} [\gamma^{\mu},\gamma^{\nu}]) \\
&= I + \dfrac{1}{8} \omega_{\mu \nu} [\gamma^{\mu},\gamma^{\nu}].
\end{align}
Using the fact that the gamma matrices transform as a vector under the spinor representation of an infinitesimal Lorentz transformation,
\begin{align}
[\Sigma^{\mu \nu}, \gamma^{\rho}] = i (\gamma^{\mu} \eta^{\nu \rho} - \gamma^{\nu} \eta^{\mu \rho})
\end{align}
we can show the spinor representation of a Lorentz transformation satisfies the Lorentz algebra commutation relations, since for $\rho \neq \sigma$
\begin{align}
[\Sigma^{\mu \nu},\Sigma^{\rho \sigma}] &= \frac{i}{2}[\Sigma^{\mu \nu},\gamma^{\rho} \gamma^{\sigma}] \\
&= \frac{i}{2}([\Sigma^{\mu \nu},\gamma^{\rho} ] \gamma^{\sigma} + \gamma^{\rho} [\Sigma^{\mu \nu}, \gamma^{\sigma}]) \\
&= \frac{i}{2}\{ i (\gamma^{\mu} \eta^{\nu \rho} - \gamma^{\nu} \eta^{\mu \rho}) \gamma^{\sigma} + \gamma^{\rho} i (\gamma^{\mu} \eta^{\nu \sigma} - \gamma^{\nu} \eta^{\mu \sigma}) \} \\
&= - \frac{1}{2}\{ \gamma^{\mu} \eta^{\nu \rho} \gamma^{\sigma} - \gamma^{\nu} \eta^{\mu \rho} \gamma^{\sigma} + \gamma^{\rho} \gamma^{\mu} \eta^{\nu \sigma} - \gamma^{\rho} \gamma^{\nu} \eta^{\mu \sigma} \} \\
&= \frac{i}{2}\{ \eta^{\nu \rho} (2 \Sigma^{\mu \sigma} + \eta^{\mu \sigma}) - \eta^{\mu \rho} (2 \Sigma^{\nu \sigma} - \eta^{\nu \sigma}) + (2 \Sigma^{\rho \mu} - \eta^{\rho \mu}) \eta^{\nu \sigma} - (2 \Sigma^{\rho \nu}) - \eta^{\rho \nu}) \eta^{\mu \sigma} \} \\
&= i ( \eta^{\nu \rho} \Sigma^{\mu \sigma} - \eta^{\mu \rho} \Sigma^{\nu \sigma} + \Sigma^{\rho \mu} \eta^{\nu \sigma} - \Sigma^{\rho \nu} \eta^{\mu \sigma} ).
\end{align}
This method generalizes from $SO(3,1)$ to $SO(N)$, see e.g. Kaku QFT Sec. 2.6, and the underlying reason for doing any of this in the first place is that one seeks to find projective representations which arise due to the non-simple-connectedness of these orthogonal groups. Regarding your question about arbitrary metrics $g_{\mu \nu}$, this method applies to, and arises due to the non-simple-connectedness of, special orthogonal groups, you can't generalize to arbitrary metrics, this is a problem which can be circumvented in supergravity and superstring theory using veilbein's.
References:
- Bjorken, J.D. and Drell, S.D., 1964. Relativistic quantum mechanics; Ch. 2.
- Kaku, M., 1993. Quantum field theory: a modern introduction. Oxford Univ. Press; Sec. 2.6.
- Tong, Quantum Field Theory Notes http://www.damtp.cam.ac.uk/user/tong/qft.html.
- http://www.damtp.cam.ac.uk/user/examples/D18S.pdf
- Does $GL(N,\mathbb{R})$ own spinor representation? Which group is its covering group? (Kaku's QFT textbook)
Best Answer
Physically, the only thing that the electromagnetic field tensor and a Lorentz transformation generator have in common is that they both happen to be antisymmetric rank 2 tensors. The link doesn't go any farther than that.
However, this coincidence does lead to a few analogies. For example, if you know about Lorentz transformations, then you know that an antisymmetric rank 2 tensor contains two three-vectors inside it, namely $\boldsymbol{\zeta}$ and $\mathbf{K}$. Then if somebody tells you the electromagnetic field is the same kind of tensor, you'll automatically know that it can be broken down into two three-vectors, namely the electric and magnetic fields. But this is a purely mathematical analogy.
A more physical result comes from the equation of motion $$\frac{d u_\mu}{d\tau} = (q/m) F_{\mu\nu} u^\nu.$$ where $u^\mu$ is the four-velocity; you can expand this in components to verify it's just the Lorentz force law. Now, comparing this with an infinitesimal (active) Lorentz transformation $$\Delta u_\mu = \Lambda_{\mu\nu} u^\nu$$ we see that the Lorentz force is equivalent to an active Lorentz transformation acting on the four-velocity, with generator $(q/m) F_{\mu\nu}$.
We can do some quick sanity checks:
Two caveats to this result: