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](https://mathoverflow.net/questions/7926/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.



Intersection of subvector bundles
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.


