Associated vector bundles and frame bundles
Translate between a torsor of frames, representation-valued coordinates, and intrinsic vector-bundle sections.
On this page
The question for today
What you’ll understand
- Construct an associated bundle with a fixed quotient convention.
- Recover a vector bundle from its frame bundle.
- Relate section coordinates to equivariant functions.
A principal bundle stores symmetry, while a representation specifies which objects that symmetry acts on. Applying different representations to one frame bundle produces vectors, covectors, exterior powers, and many other familiar geometric quantities. The construction is economical, but only if the quotient relation agrees with the transition convention. We retain principal right actions and let be a smooth left representation on a finite-dimensional real or complex vector space. We will derive the coordinate law rather than guess an inverse, then use it to distinguish an actual section from a mere list of unrelated local vectors.
The associated quotient
Define by the relation . Equivalently, take the quotient of the diagonal right action . These are the same relation; quoting both prevents an inverse from disappearing between sources. Local principal sections identify the quotient with , so it is a vector bundle over the base. Addition and scalar multiplication are performed in one fiber and are independent of the chosen representative.
Write an associated section locally as . On an overlap, makes . Consequently the component column in chart equals the transition matrix times the column in chart . This is a relation between two coordinate descriptions of one vector, not an active rotation of two different vectors. The distinction later determines both the covariant derivative and the gauge-transformation law.
Frames reconstruct the original bundle
For a rank- vector bundle , a frame at is a linear isomorphism . The frame bundle has right action . Its standard associated bundle maps to by . The quotient relation makes this well-defined because . Local frames show that it is a smooth vector-bundle isomorphism. No global frame is presumed in this reconstruction.
A fiber metric reduces real frames to orthonormal frames and complex frames to unitary frames. Such metrics exist over our paracompact manifold bases by partitions of unity, but choosing one is additional geometric data. An orientation reduces a real structure group from to its positive-determinant subgroup. It does not generally reduce it to the identity group or produce a global frame. The tangent bundle of the sphere is an immediate warning that oriented bundles can remain nontrivial.
Equivariant functions and tensor operations
A section can also be represented by a smooth function satisfying . The point is then independent of the frame in a given fiber. Conversely, represent a section uniquely in each frame to obtain such a function. This formulation lets differentiation on the principal bundle produce tensorial operations downstairs. An arbitrary function lacking this equivariance does not define a section, even if its values look plausible in a chosen trivialization.
Dual representations act by inverse transpose, and determinant representations record volume scaling. Tensor products of representations produce tensor-product bundles. For a complex line bundle, tensor power multiplies clutching winding, while taking the dual reverses it. These constructions need not change the base space or the underlying principal bundle, but they change the representation and therefore the resulting associated object. A representation with a nontrivial kernel can even forget part of the principal topology, so reconstructing the principal bundle from one associated bundle is not automatic.
Work it through
The same vector in two frames
Take the standard representation and overlapping frames related by . A vector has column . Find its other column and test the dual pairing.
Use the associated-bundle law , giving . This calculation changes its coordinates, not the vector itself.
Choose a covector with column , interpreted through . The dual law gives .
Both pairings equal eleven: and . Direct equality confirms that the transpose and inverse conventions agree.
The vector has columns (5,3) and (1,3), while the covector has columns (1,2) and (2,3). Their scalar pairing is an intrinsic quantity.
The living notebook
Explore the model
Interactive lab · matrix
A coordinate change that preserves invertibility
Pause/Reset and set diagonal scale 2 and shear 1. Predict determinant 2 and image of (1,1) equal to (3,1). Reset and set both controls to scale 1 and shear 1: the determinant is 1 and that corner becomes (2,1). Final-matrix readouts are fixed while playback interpolates geometry. This is one invertible frame-change family, not a nontrivial bundle atlas.
- Area scale · det A
- 1.4
- Singular values · σmax, σmin
- 1.597, 0.876
- Condition number · κ₂(A)
- 1.823
The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forCoordinates and areas change according to the chosen representation; an intrinsic pairing uses compatible dual coordinates.
Pause & reason
Check your understanding
Your turn
Try it yourself
A complex line bundle has chosen clutching map . Find the clutching maps for its dual and third tensor power, then determine whether their direct sum has trivial determinant clutching.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Apply the inverse representation for the dual, multiplication for tensor powers, and multiply the two line transitions for the determinant.
Compare with the solution
The dual has transition and the third power has transition . The determinant of their direct sum is the product , with winding minus four in the specified equatorial convention. This determinant line is nontrivial over the sphere, so the rank-two direct sum cannot be trivial. A nonzero determinant obstruction suffices here; in general a trivial determinant would not prove a higher-rank bundle trivial.
Keep the idea
Take it with you
- The associated quotient fixes all component inverse conventions.
- A vector bundle is recovered by evaluating its standard associated frame bundle.
- Tensor operations can preserve, amplify, or forget different parts of twisting.
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