# What does the Jacobian matrix of the projection mapping for Normal bundle look like? (2.3.14 G&P)

I want to solve this question:

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?

Thanks.

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.
• why $J$ is a surjective linear map? Commented Nov 27, 2018 at 16:39