This is a very interesting question.

Here is a paper that considers Riemann curvature as well as Ricci curvature at r = 2GM in a study of the effective photon mass:

Geodesic Deviation of Photons in Einstein and Higher Derivative Gravity

https://arxiv.org/pdf/gr-qc/9701009.pdf

And here is a very detailed 2018 paper that considers Riemann curvature:

QUILLEN TYPE BUNDLES AND GEOMETRIC QUANTIZATION OF VORTEX MODULI SPACES ON KAHLER SURFACES

https://arxiv.org/pdf/1808.02705.pdf

See Section 3, starting on page 4, on curvature.

Hello James,

The references you mentioned don't have a sortie answer what I am considering. I particularly want the answer like GM/r^3 (Abhas Mitra, 10.4236/ijaa.2012.24031) or of order 1/M^2 (Gravitation book, pp. 822). But how it is derived?

To calculate the curvature at the event horizon, you need a coordinate system that does not become singular at the event horizon.

One such coordinate system is Kruskal-Szekeres coordinates, which are nonsingular everywhere except for the physical singularity itself. If you have access to the computer algebra system Maxima, here is a short script that calculates the Riemann tensor in these coordinates and shows its nonzero values at X=T (which is the event horizon in Kruskal-Szekeres coordinates). It also shows that the spacetime is, in fact, a vacuum spacetime (Ricci-flat):

load(ctensor);

ct_coords:[T,X,u,v];

lg:zeromatrix(4,4);

r:2*G*M*(1+lambert_w((X^2-T^2)/%e));

lg[1,1]:32*G^3*M^3/r*exp(-r/2/G/M);

lg[2,2]:-lg[1,1];

lg[3,3]:-r^2;

lg[4,4]:-r^2*sin(u)^2;

cmetric();

ratfac:true;

matchdeclare(_X,true,_Y,true);

tellsimpafter(%e^(lambert_w(_X)+_Y)*lambert_w(_X),%e^_Y*_X);

lriemann(true);

ric:zeromatrix(4,4);

ricci(true);

lriem[2,2,1,1],X=T;

lriem[3,3,1,1],X=T;

lriem[3,3,2,2],X=T;

lriem[4,4,1,1],X=T;

lriem[4,4,2,2],X=T;

Hello V. T. Toth,

I think the preexistence of non-vanishing Riemann curvature tensor at r=2GM leaded transformation of coordinate system in terms of the Kruskal-Szekeres coordinates. It means both are distinct aspects of same thing but independent to each other rather the coordinate systems are guided by Riemann tensor. There are a lots of references those write this statement directly or indirectly. I am expecting an answer like what it is mentioned in my last comment above through terms of proper derivation. Anyway thank you for your effort to make coding like that..but I am searching from a theoretical point of view.

I found a document dealing the problem (@Gary Oas ) but it will result finally R11=0. See page 5 and insert all values thus derived in https://web.stanford.edu/~oas/SI/SRGR/notes/SchwarzschildSolution.pdf

Dear Bishal Banjara: Kruskal-Szekeres coordinates were developed not because the Riemann tensor is nonvanishing at the horizon, but because other coordinate systems are degenerate at the horizon. I offered to you a proper derivation. You can only derive components of the Riemann tensor at the horizon using a coordinate system that is defined at the horizon. The components, of course, are coordinate system dependent. So you cannot make a general statement like, say, R11 = 0, as that is dependent on the choice of coordinates. (Incidentally, R11 with its two indices looks like a component of the Ricci tensor, which vanishes in the Schwarzschild case, because it is a vacuum spacetime. So Rmn is zero everywhere, regardless of the choice of coordinate system. That, too, is demonstrated in the derivation that I offered.)

hello V. T. Toth, I always feel that I could not make people understand what I meant but you are that rare one who understood my point at once. Any way thank you for your sortie answer. I had yesterday got the point from https://physics.stackexchange.com/questions/295814/non-zero-components-of-the-riemann-tensor-of-the-schwarzschild-metric too. But i have a doubt about your last post as " To calculate the curvature at the event horizon, you need a coordinate system that does not become singular at the event horizon. " But although for r=2GM coordinate make the solution as the singular at event horizon, we have non-vanishing curvature of order 1/M^2 (Gravitation, pp. 822). How to take it?

Dear Bishal Banjara: The stackexchange answer shows the derivation in Schwarzschild coordinates, which are degenerate at the horizon. You can see for yourself, the Riemann tensor in these coordinates has components containing the factor 1/(c2r − 2GM), which diverges at the horizon. This is why it is important to use a coordinate system that is well-defined at the horizon, such that the Riemann tensor is expressed in terms of components that also are well-behaved at the horizon.

Hello V. T. Toth, I request you to read my post again as I have edited to make a additional question for you.

Hello Toth,

Take it as my innocence and reply me, does the Kruskal-Szekeres coordinate transformation passes all the test of Schwarzschild Solution like light bent, mercury precision,...?

Dear Bishal Banjara: For some reason, I missed your inquiry from last year (Researchgate's interface is not always user friendly.) I came across it just now. My apologies. To answer your question, physical observables such as the bending of light do NOT depend on the choice of coordinates. Which is why we have the freedom to pick whichever coordinate system suits the task at hand best. Actually calculating something in an ill-suited coordinate system (e.g., the bending of light in Kruskal-Szekeres coordinates) can be tedious if not impossible, which is why we rely on general theorems instead regarding general covariance. The basic idea is that so long as we express our physical observables using only quantities that transform as tensors do under general coordinate transformations, we are assured that the final results about physically observable phenomena are coordinate-independent results. This is also the case in everyday physics: e.g., the trajectory of a cannonball does not change just because, say, we opt to use Cartesian vs. polar coordinates to calculate it.

Thank you V.T. tooth.