# Pullback-connection

The standard connection on $\mathbb{R}^2$ is $\nabla_XY= \sum_{ij=1}^2 X^i \partial_i Y^j \partial_j.$

I was wondering how to calculate the induced connection of this on $\mathbb{S}^1$

More precisely, let $X,Y$ be vector fields on $\mathbb{R}^2$ that are also in $T\mathbb{S}^1,$ then we can write them as functions of $x$ on the upper and lower part of the circle such that $x \mapsto X(x,\sqrt{1-x^2}) \in T_{(x,\pm \sqrt{1-x^2})}\mathbb{S}^1.$

I would like to know how the induced connection on the sphere now looks-like.

• The usual way to think about the induced connection on a submanifold of $\Bbb R^n$ is to differentiate using the connection on $\Bbb R^2$ (in your case) and project the result onto the tangent space of the submanifold. (If you've not played around with this before, you might get some intuition from my (free) differential geometry text — see in particular section 2.4.) Commented Apr 10, 2017 at 21:00

In general, you don't have a notion of an "induced connection" on a submanifold. That is, if $N \subseteq M$ is a submanifold and you have a connection $\nabla$ on $M$, you can't get a connection on $N$ without some extra data (see here for details).
However, in your case the connection you work with on $\mathbb{R}^2$ is the Levi-Civita connection of the standard Euclidean metric on $\mathbb{R}^2$. In this case, you endow $S^1$ with the induced Riemannian metric and consider the associated Levi-Civita connection. If we work with the coordinate system $(\cos \theta, \sin \theta)$ on $S^1$, the metric has the form $d\theta^2$ and so the associated Levi-Civita connection has the form
$$\nabla_{f \partial_{\theta}}(g \partial_{\theta}) = f \partial_{\theta}(g) \partial_{\theta}.$$
• I don't fully agree with this. If you have some sort of split $TM|_\Sigma=T\Sigma\oplus S$, where $\Sigma$ is the submanifold, $TM|_\Sigma$ is the restricted tangent bundle and $S$ is some complementary subbundle then for a connection $\nabla$, we have an induced connection by the projection of $\nabla$ to $\Sigma$ (projection generated by the split). Of course, the most natural way to obtain such a split is to have a Riemannian metric and have $S$ be the normal bundle, but even then $\nabla$ doesn't have to be the Levi-Civita connection. Commented Apr 11, 2017 at 8:27
• @Uldreth: For me, such a splitting is "extra data". And if the connection on $M$ was the Levi-Civita connection of a metric $g$ then the splitting you get using the metric together with projecting gives you the Levi-Civita connection of the induced metric on $\Sigma$. Commented Apr 11, 2017 at 19:18