I see the scalar curvature R as an indicator of how a manifold curves locally (the easiest example is for a 2-dimensional manifold M, where the R=0 in a point means that it is flat there, R>0 that it makes like a hill and R<0 that it is a saddle point).

Are there analogous interpretations for the Riemann tensor Rm and for the Ricci curvature Rc? I tried to think about it but I really can't get anything making sense.

**Answer**

We consider the **Riemann tensor** first. A crucial observation is that if we parallel

transport a vector u at p to q along two different pathes vw and wv, the resulting

vectors at q are different in general (following figure). If, however, we parallel transport

a vector in a Euclidean space, where the parallel transport is defined in our

usual sense, the resulting vector does not depend on the path along which it

has been parallel transported. We expect that this non-integrability of parallel

transport characterizes the intrinsic notion of curvature, which does not depend

on the special coordinates chosen.

It is useful to say that in this sense visualization of the **first Bianchi identity** is very easy:

We can give a quantitative geometric interpretation to the **sectional curvature
tensor** in any dimension. Let M be a Riemannian n-manifold and p ∈ M. If Π is any 2-dimensional subspace of TpM, and V⊂TpM is any neighborhood

of zero on which expp is a diffeomorphism, then SΠ:=expp(Π∩V) is a

2-dimensional submanifold of M containing p (following figure), called the plane

section determined by Π. Note that SΠ is just the set swept out by geodesics

whose initial tangent vectors lie in Π. We define the sectional curvature of M associated with Π, denoted K(Π),

to be the Gaussian curvature of the surface SΠ at p with the induced metric. If (X,Y) is any basis for Π, we also use the notation K(X,Y) for K(Π).

Proposition: If (X,Y) is any basis for a 2-plane Π⊂TpM, then K(X,Y)=Rm(X,Y,Y,X)|X|2|Y|2−⟨X,Y⟩2

We can also give a geometric interpretation for the Ricci and scalar

curvatures. Given any unit vector V∈TpM, choose an orthonormal basis

{Ei} for TpM such that E1=V . Then Rc(V,V) is given by

Rc(V,V)=R11=Rkk11=n∑k=1Rm(Ek,E1,E1,Ek)=n∑k=2K(E1,Ek)

Therefore the Ricci tensor has the following interpretation: *For any unit
vector V∈TpM, Rc(V,V) is the sum of the sectional curvatures of planes
spanned by V and other elements of an orthonormal basis.* Since Rc is symmetric

and bilinear, it is completely determined by its values of the form

Rc(V,V) for unit vectors V .

Similarly, the scalar curvature is

S=Rjj=n∑j=1Rc(Ej,Ej)=n∑j,k=1Rm(Ek,Ej,Ej,Ek)=∑j≠kK(Ej,Ek)

Therefore the scalar curvature is *the sum of all sectional curvatures of
planes spanned by pairs of orthonormal basis elements.*

**Attribution***Source : Link , Question Author : Daniel Robert-Nicoud , Answer Author : Sepideh Bakhoda*