Why is the photon-sphere around a Kerr Black Hole spherical and not ellipsoid? Kerr Black Holes have usually (excluding extrema $a=0$, $a=1$) due to their spinning activity an ellipsoidal ergosphere. 
So why does the photon-sphere does not have an ellipsoidal form?

On the possibility of observation of the future for movement in the field of black holes of different types. Yu.V. Pavlov. Gen. Relativ. Gravit. 45, 17 (2013), arXiv:1203.4000.

 A: 
So why does the photon-sphere does not have an ellipsoidal form?

It does have an an ellipsoidal form, or, more exactly, that of an oblated sheroid; in Boyer Lindquist coordinates where
$${x} = \sqrt {r^2 + a^2} \sin\theta\cos\phi \ , \ \ {y} = \sqrt {r^2 + a^2} \sin\theta\sin\phi \ , \ \ {z} = r \cos\theta$$
the r-coordinate of all the possible photon orbits is constant, but if you transform the pseudospherical coordinate system into the cartesian background space the constant r does indeed transform into an ellipsoid. 
For comparison see the ergospheres and horizons of a rotating black hole in pseudospherical $r, \theta, \phi$ (left, for comparison see Fig. 3 in Nigel Sharp's paper "On embeddings of the Kerr geometry") and $x,y,z$ (right, comparison at Fig. 3 in Matt Visser's paper "The Kerr spacetime: A brief introduction" ) coordinates:

While the horizons have a constant $r$ in Boyer Lindquist coordinates, they don't have a constant $R$ in cartesian coordinates. That of course also goes for photon orbits:

As you can see, the shape of the photon trajectory in cartesian background space is not sperical, but an oblated spheroid. In Fig. 4 of Edward Teo's paper "Spherical photon orbits around a Kerr black hole" (pdf), you can find the same orbit, but in pseudospherical Boyer Lindquist coordinates, where the orbit looks spherical again.
A: I dont' see much discussion on the linked article about the photon sphere being a true sphere -- they talk about the photon orbit at $r=3M$, but this is only valid for an equatorial orbit (and I'd expect that it would depend on the value of $a$ and you'd have a different radius for corotation and anti-rotation), and when you go away from the equatorial plane, you'll have different values for the orbit. 
I'll admit that I"m talking from intuition here, I haven't done the calculation for closed null geodesics in the kerr spacetime.  
Short, non-rambly answer: I don't know if the kerr hole even has a photon-sphere in the Schwarzschild case, and if it did, my intuition screams that it will not be a sphere.  (also note, $r=const, t=const$ does not define a sphere in the Kerr spacetime)
A: Photon "spheres" of Kerr metric are not yet found.  That needs a lot of work. Two  closed photon circles are known.
