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

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 $$T\Sigma$$?

• Any choice of a projection $\pi\colon \iota^*(TM) \to T\Sigma$ induces a connection $\nabla^{\pi} = \pi(\iota^*\nabla)$ on $T\Sigma$, but there is no canonical projection in the absence of any additional structure. For instance, a Riemannian metric on $\iota^*(TM)$ selects a particular projection: the orthogonal projection. Nov 20 at 19:41
• I see, so the point is that we can view $T\Sigma\subset \iota^\ast(TM)$. As a result, since $\iota^\ast\nabla_X$ acts on sections of $\iota^\ast(TM)$ it can also act on sections of $T\Sigma$, but then we need to make sure that the result lies again in $T\Sigma$ and therefore we just apply a projection? I think this discussion altogether avoids the need of discussing extensions of vector fields off $\Sigma$, which is very nice.
– Gold
Nov 20 at 20:02
• This is precisely this. You can answer your own question and accept it, so that it appears as answered and might help other people someday Nov 20 at 20:21

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.

• Nicely settled. Don't forget to accept your answer so that the question is marked as solved Nov 21 at 13:17
• Thanks for the help! I'm only allowed to accept it tomorrow, then I'll do it.
– Gold
Nov 21 at 15:59
• Oh, I didn't know that. By the way, both of your equations are labelled with (2) Nov 21 at 16:09
• Yes, looks like if you answer your own question you can only accept it after 2 days. Thanks for pointing about the eq. numbering
– Gold
Nov 21 at 22:41