# Kernel of a Constant Rank Bundle Map Using the Constant Rank Theorem

Let $M$ be a smooth manifold (without boundary) and $E$ and $E'$ be smooth vector bundles of over $M$.

Let $F:E\to E'$ be a bundle homomorphism.

For each $p\in M$, we define the rank of $F$ at $p$ as the rank of the linear map $F|E_p$, where $E_p$ denotes the fibre over $p$ in $E$.

We say that $F$ is of constant rank if the rank of $F$ is invariant with respect to $p$.

The kernel of a bundle homomorphism $F:E\to E'$ is defined as $\ker F=\bigsqcup_{p\in M}\ker F|E_p$.

THEROEM. The kernel of a constant rank bundle homomorpsim $F:E\to E'$ is a subbundle of $E$.

This theorem can be found in Lee's Introduction to Smooth Manifolds, II Edition, pg 266.

The proof given in Lee's book is via the concept of local frames.

I was wondering whether we can somehow apply the following fact: Let $M$ and $N$ be smooth manifolds and $F:M\to N$ be a constant rank smooth map. Let $q$ be any point in the image of $F$. Then $F^{-1}(q)$ is an embedded submanifold of $M$.

(Thanks to User 43687 for pointing out a mistake in the above.)

That would lead to a shorter proof.

Can somebody help?

Thanks.

• What do you mean by the kernel of a smooth map. If you mean it's derivative, then this would be a special case of the claim where $E$ is the tangent bundle. – user113529 Feb 24 '15 at 20:06
• @user43687 I am sorry. I made a blunder. Let me correct it. – caffeinemachine Feb 24 '15 at 20:09
• I don't think you'll be able to use this. Since the total spaces of each bundle is a smooth manifold, certainly $F^{-1}(s(p))$ is an embedded submanifold ($s$ is the zero section, $p$ a point on the base). In fact, it's a subspace of the fiber. The problem is stitching these sub manifolds together smoothly. – user113529 Feb 24 '15 at 20:26
• Also, the statement that $F$ has constant rank (as a smooth map between manifolds) means that the derivative $dF$ has constant rank as a map between tangent bundles. Hence, to fit it in to your context, $dF$ would need to have constant rank as a map of the tangent bundles over the TOTAL spaces $E$ and $E^{\prime}$. – user113529 Feb 24 '15 at 20:35
• I agree with your remark. I still have a feeling that we can still make it work. But it's just a feeling. – caffeinemachine Feb 24 '15 at 20:49