# Intuition for curvature in Riemannian geometry

Studying the various notion of curvature, I have not been able to get the intuition and deeper understanding beyond their definitions.

Let me first give the definitions I know. Throughout, I will consider a $m-$dimensional Riemannian manifold $(M,g)$ equipped with Levi-Civita connection $\nabla$.

We have defined Riemannian curvature tensor to be the collection of trilinear maps $$R_p:T_pM \times T_pM \times T_pM \to T_pM, \ (u,v,w)\mapsto R(X,Y)Z(p), \quad p\in M$$ where $X,Y,Z$ are vector fields defined on some neighbourhood of $p$ with $X(p)=u,Y(p)=v,Z(p)=w$, and $R(X,Y)Z:=\nabla_Y\nabla_XZ-\nabla_X\nabla_YZ+\nabla_{[X,Y]}Z.$ This seems to be a purely algebraic defintion; I don't see any geometry here.

Second is the sectional curvature. For $p \in M$, let us take a two-dimensional subspace $E \leq T_pM.$ Suppose $(u,v)$ be the basis of $E$. We then define the sectional curvature $K(E)$ of $M$ at $p$ with respect to $E$ as $$K(E)=K(u,v):=\frac{\langle R(u,v)v,u\rangle}{\langle u,u\rangle\langle v,v\rangle-\langle u,v\rangle^2}.$$ Third is the Ricci curvature. Let $p\in M$ and $x\in T_pM$ be a unit vector. Let $(z_1,\cdots,z_m)$ be an orthonormal basis of $T_pM$ s.t. $z_m=x.$ The Ricci curvature of $M$ at $p$ with respect of $x$ is $$Ric_p(x):=\frac{1}{m-1}\sum_{i=1}^{m-1}K(x,z_i)$$ where $K(x,z_i)$ is the sectional curvature defined above.

Finally the scalar curvature of $M$ at $p$ is defined to be $$K(p):=\frac{1}{m}\sum_{i=1}^m Ric_p(z_i).$$

My questions are about understanding these four notions beyond the defintions. How should I think of each of them? Are they related to one another in a sense that is one notion of curvature stronger than another? I am sorry if these questions are too much for one post.

• The geometry is in the derivative symbols $\nabla_X$, $\nabla_Y$, etc. Just as in ordinary calculus, where the derivative encodes important infinitesmal geometric information (the infinitesimal slope), those derivatives and expressions involving them encode important geometric information. Aug 11, 2016 at 21:53
• See the explanation of curvature in V. Arnol'd's Mathematical Methods of Classical Mechanics.
– avs
Aug 11, 2016 at 21:57

Here's a brutally terse (and slightly imprecise, though "morally correct") account that I hope conveys the geometric content.

In a Riemannian manifold, a tangent vector may be carried uniquely along a piecewise $C^{1}$ path by parallel transport, a geometric notion.

If $X$ and $Y$ are tangent vectors at a point, and if a tangent vector $Z$ is carried around "the small parallelogram with sides $tX$ and $tY$", the parallel-translated vector is, to second order, $Z + t^{2} R(X, Y) Z$. Qualitatively, curvature measures the failure of commutativity of the covariant differentiation operators along $X$ and along $Y$; parallel transporting an orthonormal frame $F$ around a small parallelogram causes $F$ to rotate by an amount linear in each of $X$ and $Y$.

If an (ordered) orthonormal pair $(e_{1}, e_{2})$ is parallel transported around a small square with sides $te_{1}$ and $te_{2}$, it rotates through an angle (approximately) equal to $t^{2}K(e_{1}, e_{2})$. Alternatively, if you're happy with Gaussian curvature as an intrinsic quantity, the sectional curvature of a $2$-plane $E$ at $p$ is the Gaussian curvature of the image of $E$ under the exponential map at $p$.

The Ricci curvature of a unit vector $u$ at $p$ is the average of the sectional curvatures of all $2$-planes at $p$ containing $u$.

The scalar curvature at $p$ is the average of all sectional curvatures of $2$-planes at $p$.

(As in the formulas you give, the averages in the Ricci and scalar curvatures are usually defined and computed as finite averages over suitable pairs of vectors from an orthonormal basis, but in fact the averages may be taken continuously, as described above.)

• Maybe to give an even better understand of the algebric form of the Riemann tensor you should also say that the last part of the tensor (the one with the covariant derivative respect of to the commutator of X and Y) is just added to make $R(X,Y)Z$ a tensor. The interesting part is just the difference between the first two terms as you already explained.
– Dac0
Aug 12, 2016 at 6:18
• @Dac0, on those lines, can I ask what does tensor mean? I am familiar with the notion only in linear algebra.
– user166467
Aug 12, 2016 at 7:10
• You can define a tensor in different ways: as a multilinear application (as in linear algebra), as a mathemathical object that has coefficient that change according a certain rule at changing coordinate (as in some old-fashioned differential geometry book), etc... The main thing here in differential Geometry is that a Tensor is something that is coordinate free, i.e. doesn't change changing coordinate. Here you want that the Riemann tensor being something that doesn't change its value changing coordinate frame. To do so you had to subtract the extra term at the end.
– Dac0
Aug 12, 2016 at 8:39
• @user166467 In response to "I am familiar with the notion only in linear algebra": a lot of the notions one defines when dealing with manifolds are just "smoothly varying versions" of the corresponding notions from linear algebra. For instance, vector fields and differential forms are respectively smoothly varying vectors and smooth varying functionals living on the tangent space of the manifold. Tensors on manifolds also work in the same way. Apr 30, 2020 at 22:14
• For the intuitive understanding, I think it would help to also suggest what information you lose when doing these averages that takes you from the Riemann tensor to the Ricci tensor to the Ricci scalar. For example, in GR the stress-energy tensor is of order 2 and so you need to match it to the Ricci tensor - but intuitively you might ask what does this mean, apart from the fact that this is how spacetime seems to work :) Jul 7, 2022 at 9:48