Stability of Photon Orbits In the context of general relativity the photonsphere occurs in $r = 3M$ (schwarzschild), and it is a saddle (unstable) fixed point on the phase space. Is it possible for a saddle point to be an homoclinic point? If yes, how could an orbit in the phase space connect the stable and unstable manifold once the unstable orbit goes to the infinity?
 A: The existence of homoclinic (null) orbits depends on how the dynamical system is formulated. In particular, the choice of parameter along the orbit matters. If we would take an affine parameter $\lambda$ along the orbit then any null orbit that asymptotes to $r=3M$ (from above) as $\lambda\to -\infty$ will tend to $r=\infty$ as $\lambda\to \infty$. I.e. their exist no homoclinic orbits.
However, if we introduce an alternative (non-affine) "time" parameter $\tilde{\lambda}$ related to $\lambda$ by
$$\frac{d\tilde{\lambda}}{d\lambda} = \frac{L}{r^2},$$
where $L$ is the particle's angular momentum (this an affine version of the well-known Mino time parameter), and we introduce the coordinate $x := M/r$ then the equation of motion for $x$ is given by
$$ \left(\frac{dx}{d\tilde{\lambda}}\right)^2 = \frac{M^2E^2}{L^2} - x^2(1+2x). $$
For $\frac{M^2E^2}{L^2}=1/27$, this has as a solution,
$$ x = \frac{1}{2}\tanh^2 \frac{\tilde{\lambda}}{2}-\frac{1}{6}.$$
This is a homoclinic orbit that tends to the lightring at $x=1/3$ for $\tilde{\lambda}\to \pm\infty$. It also has a minimum value of $x=-1/6$ at $\tilde{\lambda}=0$. This last bit goes towards answering the second half of the question. The homoclinic orbit connects back to the equilibrium point by going "through" infinity (i.e. $x=0$) to the negative $x$ region and returning to the positive $x$ region. It should be noted that this negative $x$ region "beyond infinity" does not correspond to any physical region of space time, but is just a formal mathematical construct.
Update:
Although the introduction of $x$ is convenient. It is not necessary. One can also look for homoclinic orbits that tend to $r=3M$ from below. This exists and is given by:
$$ r = 6M\frac{\sinh^2\frac{\tilde{\lambda}}{2}}{2+\cosh\tilde{\lambda}}$$
The introduction of an alternative time parameter appears essential though.
