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This is regarding the proof of Birkhoff's theorem. A part of the proof requires one to show that the most general spherically symmetric metric can be written in the form $$ ds^2 = -e^{2\alpha(t,r)} dt^2 + e^{2\beta(t,r)}dr^2 + r^2 d\Omega_2^2 $$ In order to show this, Carroll briefly describes how one would go about setting the coordinate system in which the metric takes the form above. For this, he begins by stating that "spherical symmetry implies existence of 3 Killing vectors $K_i$ satisfying $[K_i, K_j] = \epsilon_{ijk} K_k$" (this is the Lie algebra of $SO(3)$ which is the symmetry group of $S^2$). He then invokes Frobenius's theorem to say that starting from any point $q \in M$, we can construct a two-dimensional submanifold $S_q$ (he then uses the word "sphere" instead of submanifold due to the obvious relation of $S_q$ to our ordinary understanding of spheres). From here, I present the following quote from the book (p. 199-200 below eq. (5.28))

...Begin by considering a single point $q$ lying on a sphere $S_q$ (note that $q$ must not be a degenerate point at which all of the Killing vectors vanish). Put coordinates $(\theta, \phi)$ on this particular sphere only, not yet through the manifold. At each point $p$ on $S_q$, there will be a two-dimensional orthogonal subspace $O_p$, consisting of points along geodesics emanating from $p$ whose tangent vectors at $p$ are orthogonal to $S_q$. Note that there will be a one-dimensional subgroup $R_p$ of rotations that leave $p$ fixed; indeed, these rotations keep fixed any direction perpendicular to $S_q$ at $p$, and hence the entire two-surface $O_p$ is left invariant by $R_p$. Consider a point $r$ that is not on $S_q$, but on some other sphere $S_r$ in the foliation, and that lies in the two-surface $O_p$ orthogonal to $S_q$ at $p$. Since $p$ is arbitrary, this includes any possible point $r$ in a neighborhood of $S_q$. Note that $O_p$ will be orthogonal to $S_r$ as well as to $S_q$. To see this, consider the two-dimensional plane $V_r$ of vectors in the tangent space $T_rM$ that are orthogonal to the two-surface $O_p$. Since $O_p$ is left invariant by the rotations $R_p$, these rotations must take $V_r$ into itself, because they are an isometry, and hence preserve orthogonality. But $R_p$ also takes the set of vectors tangent to $S_q$ into itself, since these rotations leave the spheres invariant. In four dimensions, two planes that are both orthogonal to a given plane at the same point must be the same plane; hence, the vectors tangent to $S_r$ must be orthogonal to $O_p$.

I follow every statement up to the last few lines. The points of confusion are

But $R_p$ also takes the set of vectors tangent to $S_q$ into itself, since these rotations leave the spheres invariant. In four dimensions, two planes that are both orthogonal to a given plane at the same point must be the same plane; hence, the vectors tangent to $S_r$ must be orthogonal to $O_p$.

While I can still follow the above lines, I do not see how he reaches the final conclusion and I feel like the crux of the argument is there. I also feel like the following line is important, but I do not quite see how

Since $p$ is arbitrary, this includes any possible point $r$ in a neighborhood of $S_q$

Does anyone have a clearer explanation for why $S_r$ is also orthogonal to $O_p$?

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It's a lot easier to see this argument in three flat dimensions, where the spheres embed using the normal spherical coordinates. Then, your normal space is the normal line to the sphere at point $p$. Then, it should be easy enough to see that, since geodesics preserve angle, the tangent space of our first sphere is parallel translated along the geodesic to the tangent space of any concentric sphere in a neighborhood of the first. Then, by construction, the geodesic is normal to the second sphere, since it is also normal to the first.

Once you understand this, extending it to the case Carroll is talking about shouldn't be hard.

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