Differential Geometry – Intuition for Curvature in Riemannian Geometry

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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.

Best Answer

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.)

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