Is a normal connection indeed a connection?

Question

I am reading GTM176. In the first picture, it says that the second fundamental form is bilinear over $C^\infty(M)$, i.e., $(\tilde{\nabla}_XY)^{\bot}$ is a tensor. However, in the second picture, it defines a normal connection $$\nabla_X^\bot N=(\tilde{\nabla}_XN)^{\bot}.$$ We know that a connection is not a tensor, so my question is that are these definitions contradict? Plus, in the third picture of do carmo’s book, it says that a normal connection is a connection, so I am very confused.

The first picture in GTM176

The second picture in GTM176

The third picture in do carmo’s book

Answer 1

Recall that $T : \mathfrak{X}(M)\times\mathfrak{X}(M) \to \mathfrak{X}(M)$ is a tensor field if it is $C^{\infty}(M)$-linear over both arguments. While $\mathrm{II}$ is $C^{\infty}(M)$-linear over both arguments, it is not a tensor field because $\mathrm{II} : \mathfrak{X}(M)\times\mathfrak{X}(M) \to \Gamma(NM)$; i.e. it outputs a normal vector field, not a tangential one.

As the second image shows, $\nabla^{\perp} : \mathfrak{X}(M)\times\Gamma(NM) \to \Gamma(NM)$ so $\mathrm{II}$ and $\nabla^{\perp}$ are not defined on the same objects. For the first, you need a pair of tangential vector fields, while for the second, you need a tangential vector field and a normal vector field. In particular, the fact that $\mathrm{II}$ is $C^{\infty}(M)$-linear in both its arguments has no bearing on $\nabla^{\perp}$.