Axial symmetry constraints on the metric

I am reading the paper on Gravitational Waves in General Relativity. VII. Waves from Axi-Symmetric Isolated Systems by H. Bondi, M. G. J. van der Burg, A. W. K. Metzner. (link) Here is a quote(s) from that paper

Throughout this paper, we shall suppose that the 4-space is axially symmetric and reflexion symmetrical....

...... From the axial symmetry, the azimuth angle is readily defined. Suppose we now put a source of light at a point $O$ on the axis of symmetry and surround it by a small sphere on which we can produce the azimuth co-ordinate $\phi$ together with a colatitude $\theta$ and a time co-ordinate $u$. We then define the $u$, $\theta$, $\phi$ co-ordinates of an arbitrary event $E$ to be the $u$, $\theta$, $\phi$ co-ordinates of the event at which the light ray $OE$ intersects the small sphere. In other words, along an outward radial light ray the three co-ordinates $u$, $\theta$, $\phi$ are constant. If we wish to write down the metric for such a system of co-ordinates (in which the part referring to the azimuth angle $\phi$ appears separately) then we know that since only the co-ordinate $r$ varies along a light ray, the term $g_{11}$ of the metric tensor must vanish, the four co-ordinates $u$, $r$, $\theta$, $\phi$ being denoted by 0, 1, 2, 3 in that order. Moreover, we must have $$\Gamma^0_{11} = \Gamma^2_{11}$$

QUESTION: Why do the above Christoffel symbols vanish?

EDIT: QUESTION 2: What does one mean by reflexion symmetry?

I finally figured it out. It's simple really. He requires that we have radial null geodesics right? This implies that the geodesic equation $$\frac{d^2 x^\lambda}{d \tau^2} + \Gamma^\lambda_{\mu\nu} \frac{d x^\mu}{d \tau} \frac{d x^\nu}{d \tau} = 0$$ should be solved by $(u,\theta,\phi)$ constant. Plugging this into the equation above, we immediately find the constraints $$\Gamma^0_{11} = \Gamma^2_{11} = \Gamma^3_{11} = 0$$ He doesn't mention $\Gamma^3_{11}$ because it is trivially satisfied for the metric at hand, since $$\Gamma^3_{11} = \frac{1}{2} g^{3\rho} \left( 2 g_{\rho 1, 1} - g_{11,\rho} \right) = \frac{1}{2} g^{33} \left( 2 g_{3 1, 1} - g_{11,3} \right) = 0$$ where we have used $g_{11} = 0$ ($(u,\theta,\phi)$ constant along null geodesics), $\partial_3 g_{\mu\nu} = 0$ (axial symmetry) and $g_{03}=g_{13}=g_{23} = 0$(reflexion symmetry)