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 $N = N_1 \oplus N_2.$ ?

  • 1
    $\begingroup$ 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 $P_2$. $\endgroup$ Nov 4, 2020 at 22:55

1 Answer 1


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. $P$.

  • $\begingroup$ Thanks a lot @Kajelad ! I really appreciate your help! $\endgroup$
    – Maria
    Nov 4, 2020 at 16:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .