Maxwell tensor in spherical and static spacetime I am approaching Problem 6.3 of Wald’s General Relativity.
I have some issues in understanding why the most general form of the Maxwell tensor in a static and spherically symmetric spacetime is
$F_{ab}= 2A(r)(e_{0})_{[a}(e_{1})_{b]}+2B(r)(e_2)_{[a}(e_3)_{b]}$
where the tetrad $e_\mu$ is defined as
$(e_0)_a = f^{1/2}(r) (dt)_a$
$(e_1)_a = h^{1/2}(r) (dr)_a$
$(e_2)_a = r (d\theta)_a$
$(e_3)_a = r \sin(\theta) (d\phi)_a$
I know I am missing something very stupid here... 
The general form should be something like
$F_{ab}= 2A(r)(e_{0})_{[a}(e_{1})_{b]}+2B(r)(e_2)_{[a}(e_3)_{b]} + C(r)(e_{0})_{[a}(e_{2})_{b]}  + D(r) (e_{0})_{[a}(e_{3})_{b]} + E(r) (e_{1})_{[a}(e_{2})_{b]} + G(r) (e_{1})_{[a}(e_{3})_{b]}$
since the coefficients cannot depend on $\theta$ or $\phi$ for spherical symmetry, and $F_{ab}$ is antisymmetrical. 
If I apply a transformation $\theta^{'} = -\theta$ to the tetrad, I get
$(e_2)^{'}_a = -(d\theta)_a$ and  $(e_3)^{'}_a = -r\sin(\theta)(d\phi)_a$, while the others are left unchanged. Then, to have $F_{ab}$ unchanged under the transformation, I must have $C(r)=D(r)=E(r)=G(r)=0$, leading to the expected result.
Now, if a transformation $\theta^{'} = \theta + \alpha$ is applied, the only effect on the tetrad should be $(e_3)^{'}_a = r\sin(\theta+\alpha)(d\phi)_a$, while all the others should be left unchanged (or not? I think that the problem should be here...).
To have $F_{ab}$ left unchanged for every $\alpha$, it must be $B(r)=G(r)=D(r)=0$, which is different from expected (for $B(r) =0$).
What am I missing?
 A: You've made several mistakes here, though they're easy mistakes to make.  There's a reason that Wald didn't approach his derivation of the general static spherically symmetric metric in this way.  Among other reasons, things are more subtle than you think, because the pullbacks that define an isometry also require evaluation at the same geometric point in both coordinate systems, which are given by different coordinate values, so you can't directly compare components so easily.  This is a whole class of mistakes that even professional relativists make all the time.
To be a little more specific, neither of your transformations is a symmetry of the metric — an isometry.  Look back at the definition of an isometry.  It's a mapping, in this case, from the manifold to itself.  And in particular it's an invertible mapping, so every point gets moved to another point.  So let's take the point $(\theta, \phi) = (\pi, 0)$ for example.  Under your first mapping, that goes to $(\theta', \phi') = (-\pi, 0)$.  But that point didn't exist in the original coordinate system, so I'll just have to go ahead and guess that you mean for it to be a different name for the same point.  And now, you've gone and defined a different set of basis one-forms for that new coordinate system, and you've demanded that $F_{ab}$ in one system has the same components as in the other system.  This doesn't make much sense.
Your second example is an even clearer case of non-isometry.  Obviously, there are again some points whose new coordinates didn't exist in the old system and vice versa, but maybe you'll try to argue that you meant for it to be a "local" isometry in some way.  Okay, let's say $|\alpha|<\pi/2$ (and not zero) for the sake of argument, and let's look at the equator $\theta = \pi/2$.  I suppose you're saying that these points get moved to the circle of points whose old coordinate was $\theta + \alpha$.  But this circle is smaller than the equator, so you're not preserving distances — it's not an isometry!
So abandon that train of thought.  Instead, I encourage you to look at just one point in your
manifold and consider a transformation of the manifold that leaves that point invariant.  (This gets
rid of a lot of those awkward details of pullbacks that I mentioned above.)  The transformation I
have in mind is a rotation $\mathcal{R}$ about your chosen point — say through some angle $\gamma$.
This is a smooth, well-defined mapping, and we know that it is a symmetry of the metric as required.
We get new coordinates $(t', r', \theta', \phi')$ for each point, and in general $(t', r') = (t, r)$.  By construction, the coordinates
of our chosen point did not change, so just for this one point, we also have $(\theta', \phi') =
(\theta, \phi)$.  But we can also construct a new basis $(e'_\mu)_a$ in the same way as
before.  And we can take the pullbacks of our basis elements into our original tangent space, so
that we can compare them directly.  At our chosen point this pullback is deceptively simple, so
it's not hard to express the pullbacks $(\mathcal{R}_\ast e'_\mu)_a$ in the original basis.  We have
at our chosen point only
\begin{align}
  (\mathcal{R}_\ast e'_0)_a &= (e_0)_a, \\
  (\mathcal{R}_\ast e'_1)_a &= (e_1)_a, \\
  (\mathcal{R}_\ast e'_2)_a &= (e_2)_a \cos\gamma + (e_3)_a \sin\gamma, \\
  (\mathcal{R}_\ast e'_3)_a &= (e_3)_a \cos\gamma - (e_2)_a \sin\gamma.
\end{align}
Obviously $(\mathcal{R}_\ast e'_0)_{[a}(\mathcal{R}_\ast e'_1)_{b]} = (e_0)_{[a}(e_1)_{b]}$, and a
simple calculation shows that we also have $(\mathcal{R}_\ast e'_2)_{[a}(\mathcal{R}_\ast e'_3)_{b]}
= (e_2)_{[a}(e_3)_{b]}$.
Now, to say that $F_{ab}$ has this transformation as a symmetry is not just to say that it
transforms as any tensor would, but that the symmetry transformation in this coordinate system
leaves those coefficient functions invariant.  So, for example, expand the quantity
\begin{equation}
  C(r')\, (\mathcal{R}_\ast e'_0)_{[a}(\mathcal{R}_\ast e'_2)_{b]}
  +
  D(r')\, (\mathcal{R}_\ast e'_0)_{[a}(\mathcal{R}_\ast e'_3)_{b]}
\end{equation}
and take the component of $(e_0)_{[a}(e_2)_{b]}$ in the result.  You should find $C(r) = C(r')
\cos\gamma - D(r') \sin\gamma$.  But the functional forms of $C$ and $D$ cannot depend on the
transformation, and $r'=r$ at our chosen point, and since this must be true for any $\gamma$, we
must have $C = D = 0$.
There's a geometric interpretation to this.  Just as a vector represents a line (as well as a magnitude and a "sense" or direction along the line), a bivector represents a plane (as well as a magnitude and a "sense" or orientation in the plane).  Bivectors are represented algebraically as antisymmetric rank-2 tensors.  Of course, these terms like $(e_2)_{[a}(e_3)_{b]}$ represent the dual form of a bivector.  In particular, $(e_2)_{[a}(e_3)_{b]}$ represents the plane tangent to the sphere at a point.  If you rotate about that point, you are simply rotating this plane onto itself — in other words, you do nothing to it: you haven't changed the geometric plane it represents, or the orientation in the plane, or the associated magnitude.  Similarly, if you rotate about a different point, you rotate this tangent plane onto another tangent plane to the same sphere.  The associated magnitude and orientation is described by $2B(r)$, but that will be the same quantity for any tangent plane.  So it makes sense that the dual-bivector field $2B(r)(e_2)_{[a}(e_3)_{b]}$ should be symmetric under rotation.  Of course the elements of $(e_0)_{[a}(e_1)_{b]}$ are not changed by rotation, so they should also be symmetric.  On the other hand, a term like $(e_0)_{[a}(e_2)_{b]}$ represents the plane spanned by $dt$ and $d\theta$.  But of course, $d\theta$ changes when it's rotated, so the plane this term represents will be rotated — and so, not symmetric.
