Intuition of curvature

Question

The only physical intuition of curvature that I know: parallel transport along a closed loop doesn't close (e.g. parallel transport of a tangent vector on a sphere).

The Ambrose-Singer theorem says "curvature equals holonomy". See, the intuition in the above paragraph is actually the "holonomy" part, not exactly the "curvature" part. Is there another physical intuition of curvature? Something like, what is the direct physical interpretation of ${\textrm{hor d}}\omega(X, Y)$? Why, intuitively, is this thing curvature (other than that it equals holonomy)?

Answer 1

By the way, if you find my note hard to read, I recommend that you read Ballmann's notes, whose link is in my post.

The idea is as follows: Although holonomy is usually described by an embedded square, I find it easier to use an embedded disk parameterized by a square. Given a point $p$ in a Riemannian manifold $M$ and a topologically trivial loop $c: [0,1]\rightarrow M$, where $c(0) = c(1) = p$, let $S \subset M$ be a parameterized square, where 3 of its sides map to $p$, and the 4th is the loop $c$. In other words, there is a map $$ C: [0,1] \times [0,1] \rightarrow M $$ such that $$C(\{0\}\times [0,1] \cup [0,1]\times\{0,1\}) = \{p\}$$ and $c(t) = C(1,t)$.

The idea now is to extend $X$ appropriately to the interior of the square, define a reference frame on the square, and measure the infinitesimal holonomy of $X$ at each point inside the square. Done properly, the integral of the infinitesimal holonomy will be equal to $P_CX - X$ at $p$. On the other hand, the infinitesimal holonomy at each point on the square is just the curvature tensor evaluated using two vectors tangent to the square, $X$, and one of the vectors in the reference frame.

First, parallel translate $X$ along the curve from $p$ back to $p$. Extend $X$ to the interior of the square as follows: For each $0 \le t \le 1$, observe that the curve $C(s,\cdot)$ is a loop from $p$ to $p$, where if $s= 0$, it is the trivial loop at $p$ and if $s=1$, it is the original loop we started with. Parallel translate $X$ along each loop, from $C(0,t)$ to $C(1,t)$, Observe that, even though $C(0,0) = C(1,1) = p$, $X(1,1) - X(0,0) \ne 0$. In fact, $$ P_C(X)-X = X(1,1)-X(0,0).$$

To calculate the infinitesimal holonomy in the interior of the square, it is necessary to measure $X$ relative to a reference frame. Let $Y$ denote one of the vectors in the frame at $p$. We then want to extend $Y$ appropriately to the square and measure how $g(X,Y)$ varies in the interior of the square. Here, we want $Y(1,1) = Y(0,0)$, so that $g(X,Y)$ really does measure the holonomy of $X$ relative to $Y$. So what we do is to parallel translate $Y$ from $p$ to each point $c(t) = C(1,t)$ along the curve $C(\cdot,t)$.

We now have vector fields $X(s,t)$ and $Y(s,t)$, where $$ g(P_C(X)-X,Y) = g(X(1,1),Y(1,1)) - g(X(0,0),Y(0,0)). $$ If you now calculate $\partial^2_{st}g(X(s,t),Y(s,t))$, the curvature tensor appears. The fundamental theorem of calculus says that the integral of this on the square is equal to $g(P_C(X)-X,Y)$.