MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose we have a smooth vector bundle $\pi: E \rightarrow B$ and two sub vector bundles $\pi_1: E_1 \rightarrow B_1$ and $\pi_2: E_2 \rightarrow B_2$ such that the bases $B_1$ and $B_2$ are submanifolds of $B$. Now suppose we would like to intersect both subbundles, that is we would like to define 'something like' $\pi_{1,2}: E_{12} \rightarrow B_1 \cap B_2$ where:

1.) $B_1 \cap B_2$ is a (smooth) submanifold of $B$.

2.) For each $b \in B_1 \cap B_2$ we define $\pi_{1,2}^{-1}(b):= \pi_{1}^{-1}(b) \cap \pi_{2}^{-1}(b)$ as the (standard set theoretic) intersection of the fibers of $\pi_1$ and $\pi_2$ and $E_{12}:=\bigcup_{b \in B_1 \cap B_2}\pi_{1,2}^{-1}(b)$.

3.) The dimension $dim(\pi_{1,2}^{-1}(b))$ is constant for all $b \in B_1 \cap B_2$.


Is this a vector bundle?

Is it a sub(vector) bundle of $\pi$?


P.S.: I know the discussion here:

About the intersection of two vector bundles

but since it doesn't solve the problem, I think its o.k. to ask my question anyway.

share|cite|improve this question
up vote 1 down vote accepted

For every vector space $V$ we have a difference map

$$ D: V\oplus V\to V,\;\; D(v_0,v_1)=v_1-v_0$$

whose kernel is the diagonal $\Delta_V\subset V\oplus V$. More generally, for vector bundles we have a bundle map $$D: E\oplus E\to E$$

whose kernel is the diagonal sub-bundle $\Delta_E$. Consider now the restriction $\bar{D}$ of $D$ to the subbundle $F:=E_1\oplus E_2 \subset E\oplus E$. For any $b\in B$ the kernel of $\bar{D}_b$ can be identified with the subspace $E_1(b)\cap E_2(b)$. This suggests a more general problem.

Suppose that $E,F\to B$ are smooth vector bundles over a compact smooth manifold $B$ and $T: F\to E$ is a smooth bundle morphism such that $\dim > \ker T_b$ is independent of $b$. Then the family of subspaces $ \ker T_b$ forms a smooth vector bundle.

This is certainly the case. To se this equip $E$ and $F$ with metrics and observe that

$$\ker T=\ker \left(T^*T: E\to E\right)$$

so we reduce the problem to the case when $E=F$ and $T$ is selfadjoint and nonnegative definite. This is what I will assume in the sequel.

For $b\in B$ we denote by $\lambda(b)$ the smallest nonzero eigenvalue of $T_b$. The fact that the dimension of $\ker T_b$ is independent of $b$ implies that

$$\lambda_0:=\inf_{b\in B} \lambda(b) >0. $$

Use the spectral theorem to represent the projection onto $\ker T_b$ as a contour integral along a circle in the plane centered at $0$ and of radius $\lambda_0/2$. This proves that this projection depends smoothly on $b$ and thus solves the above problem.

share|cite|improve this answer
nice answer. Unfortunately neither of $B$, $B_1$, $B_2$ nor $B_1 \cap B_2$ is assumed to be compact. – Mirco Jan 11 '12 at 14:28
I know that in the case of topological vector bundles the kernel of a vector bundle morphism is a vector bundle if the morphism is of constant rank. But I don't know if this holds in the smooth setting,too. I guess together with your beautiful diagonal construction, this will work, right? – Mirco Jan 11 '12 at 14:41
The smoothness comes form the representation of the projection as a contour integral. This integral depends smoothly on parameters. Note that in your problem you can assume from the very beginning that $E_1,E_2$ are defined on $B'=B_1\cap B_2$. Thus you can assume from the very beginning that $B=B_1=B_2$. Next, compactness is not needed. All we need is that for any compact $K$ the number $\inf_{b\in K}\lambda_b$ is positive. – Liviu Nicolaescu Jan 11 '12 at 21:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.