Vertical bundle

Vertical bundle

In mathematics, the vertical bundle of a smooth fiber bundle is the subbundle of the tangent bundle that consists of all vectors which are tangent to the fibers. More precisely, if "π":"E"→"M" is a smooth fiber bundle over a smooth manifold "M" and "e" ∈ "E" with "π"("e")="x" ∈ "M", then the vertical space V"e""E" at "e" is the tangent space T"e"("E""x") to the fiber "E""x" containing "e". That is, V"e""E" = T"e"(E"π"("e")). The vertical space is therefore a subspace of T"e""E", and the union of the vertical spaces is a subbundle V"E" of T"E": this is the vertical bundle of "E".

The vertical bundle is the kernel of the differential d"π":T"E"→"π"-1T"M"; where π-1T"M" is the pullback bundle; symbolically, Ve"E"=ker(dπe). Since dπe is surjective at each point "e", it yields a canonical identification of the quotient bundle T"E"/V"E" with the pullback "π"-1T"M".

An Ehresmann connection on "E" is a choice of a complementary subbundle to V"E" in T"E", called the horizontal bundle of the connection.

Example

A simple example of a smooth fiber bundle is a Cartesian product of two manifolds. Consider the bundle "B"1 := ("M" × "N", pr1) with bundle projection pr1 : "M" × "N" → "M" : ("x", "y") → "x". The vertical bundle is then V"B"1 = "M" × T"N", which is a subbundle of T("M"×"N"). If we take the other projection pr2 : "M" × "N" → "N" : ("x", "y") → "y" to define the fiber bundle "B"2 := ("M" × "N", pr2) then the vertical bundle will be V"B"2 = T"M" × "N".

In both cases, the product structure gives a natural choice of horizontal bundle, and hence an Ehresmann connection: the horizontal bundle of "B"1 is the vertical bundle of "B"2 and vice versa.

References

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