More generally, Is intersection of two subbundles of a fibre bundle a subbundle? If not then under what conditions is it true.
No. If V is the intersection of two vector bundles, then the dimension of the fiber of V over a point could be non-constant and then it would not be a vector bundle.
Let V=R^3 (R=real numbers) and define
p:V--> R by p(x,y,t)=t. This is a vector bundle of rank 2.
Let V_1={(x,y,t)| x=0}. This is a subbundle of V of rank 1.
Let V_2={(x,y,t)| x=yt). This is another subbundle of V of rank 1.
The intersection of V_1 and V_2 is not a vector bundle over R.
The fiber over t=s is a point when s is nonzero, and a line if t=0.
Compactness isn't enough. The trivial bundle S^1 x R^2 over S^1 contains the trivial line bundle S^1 x R as well as the twisted Mobius bundle. No matter how you arrange them, some of the fibers will intersect at a point and others will intersect in a line (the entire fiber over a point).
Not offhand. Any time you have a non-stably trivial k-plane bundle and a trivial k-plane bundle over a space B embedded in the trivial 2k-plane bundle over B, as in the example above, it seems you have a similar situation. I think this can happen when B is a (real, complex, quaternionic) projective space. Finding (non-trivial) conditions under which the intersection is a subbundle seems difficult. You might be able to look at classifying maps to get something.
Let E be a vector bundle over (a connected manifold) M and V, W vector subbundle. Then V cap W is a vector subbundle iff V + W is a vector subbundle and this happens iff the dimension of either the sum or the intersection of the fibers is constant along M.
No, it is not possible to put any condition on the bundle to guarantee that, because it’s a local property. As for a reference, I don’t have one right now, but I guess that most books in differential geometry discuss this. In any case the proof is not hard and everyone with basic knowledge in differential geometry can afford it.
I think following is the proof of the claim by Prof. Vitagliano:
Note that only sufficient part requires proof.
Let E1 and E2 are subbundles of a vect. bundle E over B. Let E1+E2 be whitney sum of E1 and E2. Define f: E1+E2 --> E by f(u,v)=u-v . f is smooth bundle morphism and its kernel is equivalent to intersection of E1 and E2. So if dim of intersection of fibres is constant then E1cap E2 is a subundle. Note that dimension of intersection of fibres is constant iff dimension of sum of fibres is constant by rank nullity theorem in linear algebra.
Yes. The key property here is that kernel (and cokernel) of a constant rank vector bundle map are vector bundles.
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