Pulling back a connection to a curve What does it mean to pull a connection back to a curve?
For example, if I take the connection $\nabla s = ds$ on the trivial bundle $\mathbb R^2 \times \mathbb R^2$ over $\mathbb R^2$, and the curve $\gamma=(a(t),b(t))$, what would pullback connection look like in coordinates?
I tried to say $(\gamma^*d)(f(t))=d(\gamma(f(t))$, but I don't think this is right. The second object is $d$ of a function to $\mathbb R^2$, not $\mathbb R$, so we end up with a vector of $1$-forms instead of a single $1$-form.
 A: For a map $\gamma: M \rightarrow N$ and a bundle $E \rightarrow N$ with connection $\nabla$, the pullback connection is a connection $^\gamma \nabla$ on the pullback bundle $\gamma^* E \rightarrow M$ satisfying $^\gamma \nabla_X (s \circ \gamma) |_p= \nabla_{d\gamma_p(X)} s|_{\gamma(p)}$ where $p \in M$ and $X$ and $s$ are sections of $TM \rightarrow M$ and $E \rightarrow N$ respectively.
In the example you describe above where $\gamma : I \rightarrow \mathbb{R}^2$ is a curve and $E$ is the trivial rank 2 bundle, the pullback bundle is the trivial rank 2 bundle over $I$.
We can identify sections of the pullback bundle with maps $f: I \rightarrow \mathbb{R}^2$ and any such map can locally be written as $s \circ \gamma$ for some $s: \mathbb{R}^2 \rightarrow \mathbb{R}^2$.
The equation for the pullback connection then yields that
$$^\gamma \nabla_{\partial_t} f |_t= ds_{\gamma(t)} (d\gamma_t \partial_t)=df_t (\partial_t)$$
In other words, the pullback connection is the exterior derivative for maps $I \rightarrow \mathbb{R}^2$.
