I'm currently trying to understand the sectional curvature of riemannian manifolds and I don't know if I'm thinking correctly.
So, say we have a riemannian manifold $(M,g)$ with constant sectional curvature $k$. As far as i know the curvature at any given point is completely determined by the metric $g$. After doing some calculations it seems true to me, that if we equip $M$ with a new metric
$$
\tilde{g} = \lambda \cdot g,
$$
where $\lambda$ is just some strictly positive scalar, the curvature of the manifold $(M,\tilde{g})$ is scaled in the same way, i.e. $(M,\tilde{g})$ has constant sectional curvature $\lambda \cdot k$.
Now, I know that the sectional curvature of the sphere with radius $1$ is equal to $1$ and the curvature of the sphere with radius $2$ is equal to $\frac{1}{4}$. I'm wondering if, from a riemannian geometry viewpoint, it's somehow the same to look at the radius $2$ sphere, embedded in euclidean space OR the radius $1$ sphere, equipped with the "compressed" metric $\frac{1}{4}g$, where $g$ is the usual scalar product of $\mathbb{R}^n$.
This is probably a rather soft question, but maybe someone has something to add or can correct my reasoning 🙂
Best Answer
You probably made a mistake when you computed the curvature of the scaled manifold. In local coordinates $\mathrm{Riem} \approx \partial \Gamma + \Gamma^2$ and $\Gamma \approx g^{-1} \partial g$. So the Riemann curvature as a $(1,3)$ tensor is scale invariant. This means that the sectional curvature
$$ K(u,v) = \frac{\langle \mathrm{Riem}(u,v)v,u\rangle}{\langle u,u\rangle\langle v,v\rangle - \langle u,v\rangle^2} $$
scales like $g * g^{-2}$ or that $\tilde{K} = \lambda^{-1} K$ if $\tilde{g} = \lambda g$.