# If $\nabla_1$ and $\nabla_2$ are Levi-Civita connections for a metric on the smooth sphere, then their curvature tensor would recover the radius…?

I am a little confused by an idea suggested to me: putting a connection on a sphere doesn't specify a metric geometry - it remembers notions like straightness of paths, but not length of paths.

Let me ask a specific question.

Let $\nabla$ be a connection on the smooth sphere $S^2$ (a topological manifold with a smooth atlas). Provided the connection $\nabla$ spherically symmetric (I capriciously define this to mean that it pulls back to itself for all elements in some conjugacy class of $SO(n)$ in the full diffeomorphism group of the zero set $x^2 + y^2 + z^2 = 1$ in $R^3$), torsion-free, and has constant scalar curvature $1/2(1/r^2)$, can I conclude that the connection is the Levi-Civita connection of the round sphere of radius $r$ with the usual spherical metric?

Thanks - I am just trying to understand better exactly what information is contained in the connection / covariant derivative.

• The scalar curvature of a connection is not defined without a choice of metric. – Jack Lee Aug 9 '15 at 14:53
• @JackLee Oh, makes sense. I'll write up your comment as an answer, thanks. (Unless you want points.) – Lorenzo Najt Aug 9 '15 at 15:21

(This answer was suggested by JackLee's comment. The dependence of scalar curvature on the metric is basically explained here: https://en.wikipedia.org/wiki/Scalar_curvature )

Let $R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{ [X,Y] } Z$ define the Riemann curvature tensor. Then for any pair of $\eta, \mu \in T_p$, the Riemann curvature tensor defines an endomorphism of the tangent space $v \to R(v, \eta) \mu$. The trace of this map is a $(0,2)$ tensor, since it takes two vectors, $\eta$ and $\mu$ in the tangent space, and spit out a number. The map $(\eta, \mu) \to Tr ( v \to R(v, \eta) \mu)$ is called the Ricci curvature tensor.

In order to compute the scalar curvature, one first has to convert the Ricci curvature into a $(1,1)$ tensor, so that it is an endomorphism of $T_pM$, and then take the trace. The operation of lowering the index requires the metric, hence the connection does not see this value.