Vertical bundle

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In mathematics, the vertical bundle of a smooth fiber bundle is the subbundle of the tangent bundle that consists of all vectors that are tangent to the fibers.

More precisely, if π : E → M is a smooth fiber bundle over a smooth manifold M and eE with π(e) = x ∈ M, then the vertical space VeE at e is the tangent space Te(Ex) to the fiber Ex containing e. That is, VeE = Te(Eπ(e)). The vertical space is therefore a vector subspace of TeE, and the union of the vertical spaces is a subbundle VE of TE: this is the vertical bundle of E.

Formal definition[edit]

Let π:EM be a smooth fiber bundle over a smooth manifold M. The vertical bundle is the kernel VE := ker(dπ) of the tangent map dπ : TE → TM.[1]

Since dπe is surjective at each point e, it yields a regular subbundle of TE. Furthermore the vertical bundle VE is also integrable.

An Ehresmann connection on E is a choice of a complementary subbundle to VE in TE, called the horizontal bundle of the connection.


A simple example of a smooth fiber bundle is a Cartesian product of two manifolds. Consider the bundle B1 := (M × N, pr1) with bundle projection pr1 : M × NM : (xy) → x. Applying the definition in the paragraph above to find the vertical bundle, we consider first a point (m,n) in M × N. Then the image of this point under pr1 is m. The preimage of m under this same pr1 is {m} × N, so that T(m,n) ({m} × N) ={m} × TN.The Vertical bundle is then VB1 = M × TN, which is a subbundle of T(M ×N). If we take the other projection pr2 : M × N → N : (xy) → y to define the fiber bundle B2 := (M × N, pr2) then the vertical bundle will be VB2 = TM × N.

In both cases, the product structure gives a natural choice of horizontal bundle, and hence an Ehresmann connection: the horizontal bundle of B1 is the vertical bundle of B2 and vice versa.

See also[edit]


  1. ^ Kolář, Ivan; Michor, Peter; Slovák, Jan (1993), Natural Operations in Differential Geometry (PDF), Springer-Verlag  (page 77)