Decomposing a normal bundle

I'm reading a paper on symplectic geometry. I reached to a point where the author uses normal bundles ( which I know the definition but I've never work with). He actually decomposes a given normal bundle into a sum of two other normal bundles. I can't give my precise question and the normal bundles I'm talking about, because this will require to introduce a lot of stuff. But i think he uses the following argument:

Let M be a smooth manifold, and let C and P be two submanifold of M such that: $$C \subset P \subset M.$$

Let N be the normal bundle of C in M,

$$N_1:=$$ the normal bundle of C in P.

$$N_2:=$$ the normal bundle of P in M, restricted to C.

Then, we get $$N = N_1 \oplus N_2.$$

My questions are: first, what is the meaning of restricting a normal bundle to a submanifold (as is used to define the normal bundle $$N_2)$$, and the second one is, is the above argument true, namely is $$N = N_1 \oplus N_2.$$ ?

• In general, if you have a complete intersection, a submanifold $C$ that is the transverse intersection of two submanifolds $P_1$ and $P_2$ of $M$, then the normal bundle of $C$ is the direct sum of the normal bundles of $P_1$ and $P_2$. Nov 4, 2020 at 22:55

In general, given a vector bundle $$\pi:B\to M$$ and an embedded submanifold $$S\subseteq M$$, one can define the restricted bundle as $$R=\pi^{-1}(S)$$, with the projection given by $$\pi|_R:R\to S$$. This is also a vector bundle. This restriction can also be defined as the pullback bundle $$R=\iota^*B$$ where $$\iota:S\to M$$ is the inclusion map.
To your second qustion, the answer is yes; this can be seen by noting that, for $$p\in C$$, we have the decompositions $$T_pM=T_pP\oplus N_pP$$ and $$T_pP=T_pC\oplus N_pC$$ where $$N_pC$$ denotes the normal space w.r.t. $$P$$.