Take the 2-minute tour ×

Let $\pi_1:V_1 \rightarrow B_1$ and $\pi_2:V_2 \rightarrow B_2$ be smooth vector bundles and write $\phi_V: V_1 \rightarrow V_2$ as well as $\phi_B:B_1 \rightarrow B_2$ respectively for the total and base part of a smooth vector bundle morphism.

The Question is, what additional properties must we assume on $\phi_V$ / $\phi_B$ such that:

1.) The image is a smooth vector bundle?

2.) The image is a smooth sub(vector)bundle?

3.) The preimage $\phi_V^{-1}(W) \rightarrow \phi_B^{-1}(A)$ of a smooth sub(vector)bundle $W \subset V_2 \rightarrow A \subset B_2$ is a smooth sub(vector)bundle of $\pi_1$?

share|improve this question
What's the difference between (1) and (2)? What's your definition of a morphism of smooth vector bundles? –  Zhen Lin Dec 27 '11 at 2:20
A smooth vector bundle morphism is a pair of smooth maps that respects the linear and the bundle structure. (Look for example at Wikipeadia or are there others?) Are 1.) and 2.) different? Thats part of the question. If not why can't we assume a priory that the smooth structures are equal? –  Mark Neuhaus Dec 27 '11 at 2:58
Well, that's precisely the point: (1) is not interesting. We want the smooth structure the image to be induced from the smooth structure of $V_2$ – so we really only care about (2). –  Zhen Lin Dec 27 '11 at 3:16
Ok. I agree with that because otherwise we get problems with the smoothness of the maps. So question 1.) is equivalent to question 2.) –  Mark Neuhaus Dec 27 '11 at 3:19

1 Answer 1

As Zhen has noted then we need $\phi_B$ to be surjective for the image of $V_1$ to have any chance of being a vector bundle. Once this is satisfied then I think it is necessary and sufficient that the rank of $Im(V_1)_x$, as a subspace of $V_{2,\phi_B(x)}$ be the same for all $x$ in $B_1$.

The problem is that it is very hard to give any kind of non-trivial verifiable general condition on the map $\phi_V$ that is equivalent to the previous condition, reducing it to the usefulness of a tautology. Even in the category of analytic manifolds and maps, which is highly rigid, there is no such condition.

An example may help. Let's take as our base $B$ the complex plane $\mathbb C$ and let's consider the trivial vector bundle $V$ with fiber $\mathbb C^n$ over $B$. Then $V$ is just the product $\mathbb C^n \times B$ and $\pi$ is the projection onto the second factor.

Now consider the following endomorphism of $V$. We define $\phi : V \to V$ by setting $\phi(v,z) = (zv, z)$. At a point $z \not= 0$ the image of $\phi$ is all of $\mathbb C^n$. At the point $z = 0$ however, the image of $\phi$ is the trivial space $\{ 0 \}$. We see that the image is not of constant rank and thus not a vector bundle. In fact it isn't even a coherent sheaf in the analytic category, as the rank of the stalks of such objects only "jumps up" (we would need to "sheafify" the image to get such a coherent image sheaf).

share|improve this answer
I know that! In a sense this is all offtopic because it is not an answer to my question. –  Mark Neuhaus Dec 27 '11 at 16:12
@Mark: Then please consider rephrasing your question to make it more clear what you want. –  Zhen Lin Dec 27 '11 at 16:16
Question is logical and clear. The morphisms you assumed are not the naturals because they won't form 'the' category of smooth vector bundles. ... Maybe this is the wrong place to ask, because it seems that it is more a research level question. –  Mark Neuhaus Dec 27 '11 at 16:46
Dear @Mark, I fear I don't see how one would answer the question you asked in a way that isn't very similar to what Zhen and I have done. The moral of both answers is that there are no general criteria for what you want. If these are not the answers you wanted then it may be time to edit the question and provide more motivation and detail. –  Gunnar Magnusson Dec 28 '11 at 9:25

Your Answer


