Hello Ali! It would be helpful if you write the matrices in question elementwise. Then we are really not far away from the answer. In the moment is not clear if A* is matrix tranposition + complex conjugation, or only matrix transposition, and is also not so clear if the projections are refering to C2 or something different. So please give us the set in question as a parametrization with 2 variables (if it really has to be S2).

Dear Mihai, Thank you for your comment. Sorry for my unclear question. 

Yes by A* I mean the adjoint of a complex matrix the combination of transpoition and complex conjucate. the homeomorphism between S^2 and the space of projections is the following; (x,y,z) \in S^2 map to 1/2 \begin{pmatrix]}1-z&x+iy//x-iy&1+z \end{pmatrix}. this is homeomorphism between sphere and the space of nontrivial projections of M_{2}(C).  I learned this fact from page 15 of the printed version of book "Non commutative geometry " bu Alain Connes.

My first try is the following: The  unitary group act transitively on the space of projections.  The action of a unitary U on a projection A is U*AU. So it suffice to prove that each unitary matrix act isometrically on the space of projections. This would imply that the curvature is constant.  BUT I am not sure that a ll unitary matrices have an isometric action .

OK, Ali. So your environment is in fact R8 with the metric given by its euclidian norm sqrt( sum of xi2 ) . Let a matrix be (z1 z2  ;  z3 z4 ). Your set lies in an intersection of hyperplanes, which are Im z1 = Im z4 = 0 , Re z1 + Re z4= 2, Re z2 = Re z3, Im z2 + Im z3 = 0. Those are exactly 5 independent affine hyperplanes (prove independence by rank computation!), and their intersection is an  affine space T  of real dimension 8 - 5 = 3. On the other side the euclidian norm of R8 computed for points of your set is sqrt([ (1-z)2 + (1 + z)2 + 2x2+ 2y2 ] / 4 ) = sqrt ( [1 + x2 + y2 + z2 ] / 2 ) = 1 because (x, y, z) in S2 as embedded in R3. So your set (call it S) is the intersection of a 3-dimensional linear space T with the sphere S7 canonically embedded in R8 , and the euclidian distance in R8 generates (the same) metric on T. For me it is clear: your sphere is round, constant curvature, and canonically embedded. However, the affine space T does not contain 0 in R8 so the center of our sphere, which lies in T, is not 0 from R8. This center and the radius of the sphere, as sphere embedded in T, must be computed. As a result, we will see that the sphere has not curvature 1. It will have a different oneI

The fact that this sphere is not centered in the matrix (0 0 ; 0 0) explains also why your approach with the unitary group did not work. Most of them do not preserve the set!

The center is the middle of a diameter connecting opposite points. That is 1/2 x 1/2 x (2 0 ; 0 2) = 1/2 x I, where I is the unit matrix. It belongs to T, where we observe that one equation for T is Re z_1 + Re z_2 = 1, and not 2 as I wrote before, forgetting 1/2. The radius is the distance of every point to the center, so is sqrt(1/4 x [z^2 + z^2 + 2x^2 + 2y^2]) = sqrt(2) / 2. 

Dear Mihai thank you very much  for your very interesting answer. Regarding the action of unitary group: the action of a unitary matrix, a matrix U with UU*=Id, on a projection A is U*AU. this is again a projection. So why you think projections are not preserved under this action?

This approach in real case is usefull. It gives us a riemannian metric on real grassmanian with constant curvature, since the orthonormal matrices act transitively on grassmanian. This implies that the curvature is constant.More precisely: The space of all projections in M_{n}(R) which have trace k(equivalently rank k)  is homeomorphic to G(k,n), the real grassmanian. The natural inner product of M_{n}(R)  ~R^{n^2} is -trace(AB^tr). Obviously this inner product is invariant under orthonormal conjucation, so the metric (and its restriction to projections) is invariant under the action of O(n).

Dear Ali:

Only now I saw your question. Maybe you got an answer or you are no longer interested. In general, the Grassmannian (real or complex) embedded into the space of symmetric or hermitian nxn-matrices, is homogeneous but does not have constant curvature (however it has "constant" = parallel curvature tensor). But in dimension n=2, the complex Grassmannian is simply the projective line CP^1 = S^2. What is its precise curvature? There is a little difficulty: The projectors have trace 1, hence they are lying in the affine subspace  {A \in H : trace A = 1}  where  H  denotes the hermitian matrices. It is slightly easier and even more natural to pass from the space of projections to the space of reflections (which differ only by a constant multiple of the identity matrix and scaling: the eigenvalues are 1 and -1 rather than 0 and 1). Also it is more convenient to devide the trace metric by n, then the identity matrix I has norm 1 and all unitary (or orthogonal) matrices have norm 1. The full circle g(t) =  I \cos t + J \sin t  with  J = (0,-1\\1,0) has length 2\pi. Now you have to compute the length of  g(t) S g(t)^{-1} where S is the reflection (1,0\\0,-1), but you are back to  S  already at  t = \pi. So you can compute the length of a great circle and thus the curvature of the sphere. In the projection model you can work similarly.

Best regards

Jost