Elementary Riemann curvature calculation Suppose I have a surface like $f(x, y) = x^2 + y^2$. The gradient of this surface is $(2x, 2y)$. I understand the Riemann curvature of this surface is like the commutator of 2nd derivatives, which here is obviously $0$. A really naive question is, how does this map onto the actual idea of the curvature of this surface, which increases by $2$ in any $x$ or $y$ direction? There is an acceleration along curves, even if that kind of lengthening is in a "straight" direction ... so I'm a little confused about how there is no curvature.
What I'm really trying to get to the bottom of is how we can understand lack of commuting 2nd derivatives as a force in general relativity. I get failing to parallel transport around a curve, and there should definitely be loops like that on the surface of $f(x, y)$, which is kind of like a gently curving slope. Though I am still calculating it as $0$. Sorry for the elementary question, but any help in understanding why the calculated curvature of $f(x, y)$ is not the naive curving on the surface would be really great.
 A: You are asking (rightfully) what the second derivative of a function has to do with curvature in a geometric sense. Let's keep the following  principles in mind:

*

*Riemannian curvature measures the rate of change of the direction in the tangent space of a tangential vector to the manifold.


*when the manifold is at least $2D$ we can move between two points along different curves and that's what makes the concept of Riemannian curvature a bit difficult. Parallel transport is needed to understand this.


*the tangent spaces at a one-dimensional curve are all one dimensional. A vector cannot change its direction in a tangent space (there is no Riemannian curvature). However, we can ask how a vector changes its direction in the ambient space the curve lives in.
Let's look at this $1D$ case in more detail.
Example 1. $f(x)=x^2\,.$ This one-dimensional curve is a parabola in ambient space $\mathbb R^2\,.$ It can be parametrized by
$$
\left(\begin{matrix}x\\x^2\end{matrix}\right)\,,\quad\text{ point on curve }\,.
$$
The derivative w.r.t. $x$ of this vector is the tangent vector to the curve at its point with parameter $x\,:$
$$
\left(\begin{matrix}1\\2x\end{matrix}\right)\,,\quad\text{ tangent vector to the curve }\,.
$$
Because we are only interested in the direction of the tangent vector let's switch to the angle $\alpha$ between the vector and the $x$ axis:
$$
\alpha=\arctan(2x)\,.
$$
The rate of change of this direction when $x$ varies is (if I am not mistaken)
$$
\frac{d\alpha}{dx}=\cos^2(\arctan 2x)\,.
$$
At $x=0$ this naive curvature is one and decreses to zero when $x^2$ gets very large. That's intuitively clear as the parabola looks pretty much like a straight line when $x^2$ is very large.
Example 2. $f(x)=\sqrt{1-x^2}\,.$ This is the upper part of a circle with radius one. We expect this to have constant curvature. To see that this is in fact the case it is better to parametrize the curve in polar coordinates:
$$
\left(\begin{matrix}\cos\theta\\\sin\theta\end{matrix}\right)\,,\quad\text{ point on curve }\,,\quad
\left(\begin{matrix}-\sin\theta\\\cos\theta\end{matrix}\right)\,,\quad\text{ tangent vector to the curve }\,.
$$
The angle between the tangent vector and the $x$ axis is
$$
\alpha=\theta+\frac{\pi}{2}\,.
$$
Its rate of change is constant - as expected.
