Can we use the pullback connection to induce a connection on the tangent bundle of a submanifold?

Question

Let $M$ be a smooth manifold. Let also $\nabla$ be an affine connection on $TM$. If $\Sigma\subset M$ is an embedded sub manifold and $\iota:\Sigma\to M$ is the inclusion we can construct the pullback bundle $\iota^\ast(TM)$. Moreover we can define on it the pullback connection $\iota^\ast\nabla$ defined by its action on pullback sections $$(\iota^\ast\nabla)_{X}(\iota^\ast Y)=\iota^\ast(\nabla_{\iota_{\ast}X}Y)\tag{1}.$$

This defines a connection because the pullback sections $\iota^\ast Y$ span the space of sections of $\iota^\ast(TM)$. All that said, this is a connection on $\iota^\ast(TM)$. My question here is: does the pullback connection allows us to induce a connection on $T\Sigma$?

Naively it seems that one way one could proceed is: let $Y\in \Gamma(T\Sigma)$ and let $p\in \Sigma$. If we extend it a vector field $\mathscr{Y}\in\Gamma(TU)$ in an open subset $U\subset M$ containing $p$, we can then pull it back to $\iota^\ast(TM)$ and apply $\iota^\ast \nabla$ to it. The problem are in the details of this construction: (1) how can we be sure that the result is well-defined, i.e., independent of the extension? Can this really be formulated as a local question, i.e., can we do this construction focusing on a small neighborhood of each point as I have proposed and still get a globally defined connection on $T\Sigma$?

Answer 1

The pullback bundle is defined as $$\iota^\ast(TM)=\{(x,v)\in \Sigma\times TM : v\in T_xM\}.\tag{1}$$

As a result it contains a sub-bundle that can be identified naturally with $T\Sigma$, namely, the one consisting of all the pairs $(x,v)$ with $v\in T_x\Sigma\subset T_xM$. As a result, given $(\iota^\ast\nabla)_X$ for some fixed $X\in \Gamma(T\Sigma)$, since this map can act on any section of $\iota^\ast(TM)$ it can naturally act on sections of $T\Sigma$ as well by restriction.

The result, however, may not lie in $T\Sigma$. This can be remedied by fixing a projection map $\Pi:\iota^\ast(TM)\to T\Sigma$ and defining

$$\nabla^\Pi_X Y \equiv \Pi\left((\iota^\ast\nabla)_XY\right)\tag{2}$$

The choice of $\Pi$ is equivalent to the choice of a splitting $\iota^\ast(TM)\simeq T\Sigma\oplus N\Sigma$ where $N\Sigma$ is a choice of definition of "normal bundle". When $M$ has a Riemannian metric $g$ then indeed it gives rise to a canonical split, namely, the orthogonal split, in which $N\Sigma$ is indeed the normal bundle of $\Sigma$ with respect to $g$. In that case it gives a preferred choice of induced connection.