I want to solve this question:

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I feel like the previous question is similar to the one given in this link:

Natural projection of tangent bundle is submersion

Am I correct? but what does the Jacobian matrix look like in our situation here?



1 Answer 1


I don't recall all the notation from G&P so I'll try to just explain why the projection from the normal bundle of a manifold $M$ to the manifold is a submersion. Let's write this as $\pi:NM\to M$. A trivializing open neighborhood for $NM\to M$ is an open neighborhood $U\subseteq M$ so that $\pi^{-1}(U)\cong U\times \mathbb{R}^k$. If $U$ is a trivializing open neighborhood for $NM$ with $U$ also a coordinate chart neighborhood on $M$ with coordinates $(x^1,\ldots, x^m)$, then on $\pi^{-1}(U)$ the bundle is parametrized by $(x^1,\ldots, x^m, v^1,\ldots, v^k)$.

If we think about the projection $\pi^{-1}(U)\to U$ and put it in coordinates, we are just forgetting about the last $k$ data above, so the map is locally $(x^1,\ldots, x^m, v^1,\ldots, v^k)\to (x^1,\ldots, x^m).$ The associated Jacobian matrix is of the form $J=[I_m|0_k]$ where $I_m$ is the $m\times m$ identity matrix and $0_k$ is the $k\times k$ zero matrix. Notice that $J$ is a surjective linear map, so that $\pi: NM\to M$ is a submersion.

  • $\begingroup$ why $J$ is a surjective linear map? $\endgroup$
    – Idonotknow
    Commented Nov 27, 2018 at 16:39
  • $\begingroup$ Because its rank equals the dimension of the target space. $\endgroup$ Commented Nov 27, 2018 at 16:40
  • $\begingroup$ I understood you are saying that our map gives us the first coordinate only and the second is not included ..... this is why the jacobian is the identity for the first coordinates while zero for the last ..... is my understanding correct? $\endgroup$
    – Idonotknow
    Commented Nov 27, 2018 at 18:20
  • $\begingroup$ What about the question of what specifically is the preimage ? $\endgroup$
    – Emptymind
    Commented Nov 27, 2018 at 18:36
  • $\begingroup$ what is the dimension of the normal bundle (our domain)? $\endgroup$
    – Idonotknow
    Commented Nov 28, 2018 at 19:51

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