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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.

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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.

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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

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