Vertical bundle

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 e ∈ E with π(e) = x ∈ M, then the vertical space V_eE at e is the tangent space T_e(E_x) to the fiber E_x containing e. That is, V_eE = T_e(E_π(e)). The vertical space is therefore a vector subspace of T_eE, and the union of the vertical spaces is a subbundle VE of TE: this is the vertical bundle of E.

Formal definition

Let π:E→M 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.

Since dπ_e 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.

Example

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 × N → M : (x, y) → 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 : (x, y) → 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

• Vertical vector field