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Journal of Generalized Lie Theory and Applications
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A connection whose curvature is the Lie bracket


Department of Mathematics, California Polytechnic State University, San Luis Obispo, CA 93407, USA

E-mail: [email protected]

*Corresponding Author:
Department of Mathematics
California Polytechnic State University
San Luis Obispo, CA 93407, USA
E-mail: [email protected]

Received date: February 19, 2009; Revised date: March 06, 2009

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Let G be a Lie group with Lie algebra g. On the trivial principal G-bundle over g there is a natural connection whose curvature is the Lie bracket of g. The exponential map of G is given by parallel transport of this connection. If G is the di eomorphism group of a manifold M, the curvature of the natural connection is the Lie bracket of vector elds on M. In the case that G = SO(3) the motion of a sphere rolling on a plane is given by parallel transport of a pullback of the natural connection by a map from the plane to so(3). The motion of a sphere rolling on an oriented surface in R3 can be described by a similar connection.

A natural connection and its curvature

Samelson [6] has shown that the covariant derivative of a connection can be expressed as a Lie bracket. It is the purpose of this article to show that the Lie bracket of a Lie algebra can be expressed as the curvature form of a natural connection. Although it is plausible that this natural connection has been described before or is known as mathematical folklore, it does not appear in the standard references (e.g., Kobayashi and Nomizu [4, 5] or Sharpe [7])..

The setting for this result is the following. Let π: P → X be a right principal G-bundle with the Lie group G as the structure group. A connection on P is a smooth G-equivariant distribution of horizontal spaces in the tangent bundle TP complementary to the vertical tangent spaces of the bers. The curvature of a connection is a g-valued 2-form on the total space P. A good reference is Bleecker's book [2].

Let G be a Lie group and g its Lie algebra. Let imagebe the total space of the trivial right principal G-bundle with projection P → g :image The right action of G on P is given by (x; h)  g = (x; hg) and let Rg denote the action of g on P; that is, Rg : P → P : imageLet 1 image G be the identity element and letimage be the identity section of the bundle.

Let image and image'be the left and right multiplication by g. The context should make it clear whether Rg is acting on P or on G. The adjoint action of G on g is the derivative at the identity of the conjugation action of G on itself. That is, image

Finally, we recall the de nition of a fundamental vector eld on a principal G-bundle P. For image be the vector eld on P whose value at p is given by


The vector field iamge is called a fundamental vector field. In the case that P is the trivial bundle image it is routine to check that


Theorem 1.1. There is a natural connection on the trivial bundle image whose local curvature 2-form (with respect to the identity section ) is the constant g-valued 2-form on g whose value on a pair of tangent vectors image is the Lie bracketimage

Proof. We de ne the connection on P by choosing the horizontal spaces. For (x; g) image P let H(x;g) be the subspace of T (x;g)P de ned by


It is easy to check that the distribution is right equivariant and that H(x;g) is complementary to the vertical tangent space at (x; g).

Let α be the g-valued 1-form on P de ning this connection. It is characterized by having the horizontal space H(x;g) as the kernel of (x;g) and by satisfying the conditions


From these properties one can check that α is de ned by


The curvature of a connection is the g-valued 2-form on the total space P de ned by


Pulling α back to g by imagegives the local connection 1-form image

The local curvature form Ω is then


Computing image we see that


Hence, image is a constant form, andimageWe evaluate[image,image ] on a pair of tangent vectorsimage


Therefore, the local curvature form Ω is the constant g-valued 2-form that maps a pair of tangent vectors imagewhich are just elements of g, to the Lie bracket of iamgeand image


For a Lie group homomorphism image be the associated Lie algebra homomorphism. (Note that image when the Lie algebras are viewed as the tangent spaces at the identities of the groups.) Then the map


is a morphism of principal bundles, which means that it commutes with the right actions of G1 and G2:


Furthermore, it is a morphism that preserves the horizontal spaces of the natural connections de ned in Theorem 1.1. More precisely, image maps the horizontal subspaces inIMAGE to the horizontal subspaces in IMAGE ) as follows. LetIMAGE


which is an element of the horizontal space at image

We also note that for a composition of Lie group homomorphismsimage the principal bundle map image and thatimageis the identity on the principal bundle for the identity image Hence, we have proved the following theorem

Theorem 1.2. There is a covariant functor F rom the category of Lie groups to the category of principal bundles with connection, which is de ned on objects so that F(G) is the trivial principal bundle image with its natural connection and de ned on morphisms byimage

Next we consider the relationship between the connection 1-forms of the two bundles.

Proposition 1.3. The following diagram commutes:



Proof. Starting with the composition image we have


Proposition 1.4. The following diagram commutes:


Proof. Because imge is linear,image for allimage Thus, forimage


Also, the local curvature forms commute with the Lie algebra homomorphism image

Proposition 1.5. Let image be the local curvature 2-form for the natural connection on image

Proof. This is simply restating the fact that image is a homomorphism of Lie algebras:image

Remark 1.6. Let G be the di eomorphism group of a manifold M with g being the space of vector elds on M. Although G is not, strictly speaking, a Lie group, the natural connection on still makes sense, and so the curvature of this connection is given by the Lie bracket of vector fields.

Given a smooth curve imageparallel transport along c is horizontal lift (c(t); g(t)) with initial condition g(0) = 1. Therefore, g is a solution of the di erential equation


which simply says that imageis in the horizontal subspace at the point (c(t); g(t))

Theorem 1.7. Let iamgebe an element of the Lie algebra image and defineimage Then parallel transport along c is given by iamge

Proof. It suces to show that iamgesatis es the di erential equation imageimage with initial condition g(0) = 1. The derivative of expimage is given by


Remark 1.8. With this result there is another way to see that the Lie bracket is the curvature, since


is the in nitesimal parallel transport around the parallelogram spanned by ciamgeand image which, in turn, is the curvature tensor applied to iamgeand image

Corollary 1.9. Let image and iamge be elements of image and define image be the horizontal lift of c with image

Proof. Since image the value of  does not matter. Therefore, let  = 0. By the theoremimage is the horizontal lift of c with image The right-invariance of the horizontal spaces implies that image

Remark 1.10. Parallel transport along c is also known as the time-ordered (or path-ordered) exponential of image. One of the equivalent ways to de ne the time-ordered exponential of a curve image is to de ne it as the solution of the di erential equationimage withimage this is the di erential equation de ning parallel transport along c. To understand the use of the phrase \time-ordered exponential," we use a piecewise linear approximation to c starting at c(0) and consisting of the line segments connecting image and imageThen by repeated use of the corollary g(t) is approximated by


Therefore, g(t) is the limit as n goes to in nity of this product of exponentials, which are ordered according the parameter value.

Remark 1.11. The holonomy subgroup (at a point x0 in the base space) of a connection on a principal G-bundle is the subgroup of G consisting of the results of parallel transporting around closed curves starting and ending at x0. The null holonomy group is the subgroup resulting from transporting around null-homotopic curves. By the Ambrose-Singer theorem [1] the Lie algebra of the null holonomy group is generated by the values of the curvature tensor. Now image is simply connected and so the null holonomy group is the holonomy group and its Lie algebra is the derived algebra [image; image]. Assuming G is connected, the holonomy group is the derived group of G.



The special orthogonal group

Let G be the rotation group SO(3) with g = iamge(3), the Lie algebra of in nitesimal rotations. We identify iamge(3) with R3 in the standard way so that the vector v in R3 corresponds to the in nitesimal rotation with axis v that goes counterclockwise with respect to an observer with v pointing towards him. The angular velocity is the magnitude of v. If v = (v1; v2; v3), then the corresponding matrix image iniamge(3) is


Then imageso that the map  is an isomorphism of Lie algebras imageimage

Given a curve image parallel transport along c is given by the curveimage SO(3), which is the unique solution to the di erential equation


In other words, the in nitesimal rotation at t has for its axis of rotation the vector imageOne can visualize this as a sphere of radius 1 rotating at the head of a screw (with right-hand threads) that is tunneling through space following the trajectory given by the curve c. Note that the spherical head is not rigidly attached to the screw because the axis of rotation must be free to vary.

A rolling sphere on a plane

A variation of this natural connection can be used to describe the geometry of a sphere rolling without slipping on a horizontal plane. The plane is R2 embedded in R3 as the set of points imge and a sphere of radius 1 sits on top of the plane. Letimage be a smooth curve with initial pointimage Roll the sphere along the curve c until it reaches the endpoint imageAs the sphere rolls along, the point of contact is c(t) and the con guration of the sphere is given by a curve g(t) in SO(3). At each point c(t) the in nitesimal rotation is about the axis iamgewhere iamegrotation counterclockwise de ned by imageWe can formulate the di erential equation satis ed by g(t) as


(Briefly image from which the di erential equation follows.) In order to see this di erential equation as the parallel transport equation for the curve c with initial condition g(0) = I, we de ne a connection on IMAGEwith horizontal space at the point (x; g) in IMAGEgiven by


The connection 1-form is de ned by


Let iamge'be the local connection 1-form. Since


we see that is constant, i.e., Then the local curvature 2-form acts on a pair of tangent vectors imaeg


For the last step note that imagebecause  is an isomorphism of Lie algebras. Also, image from the geometric properties of the cross product. Finally, image

The derived algebra of image is image because each of the basis elementsimage is a cross product. The group SO(3) is connected, and so, by the Ambrose-Singer theorem, the holonomy subgroup is all of SO(3). In other words, any rotation of a sphere can be achieved by rolling the sphere around some closed path in the plane. An elementary proof (without the apparatus of modern di erential geometry) of this old result has recently appeared [3].

The connection just de ned is actually just a pullback of the natural connection on imageSO(3).

Theorem 2.1. Define


Then the connection on IAMGE associated to the rolling sphere is the pullback byIMAGEof the natural connection on imager

Proof. The pullback bundle of the trivial bundle is also trivial. Let imagef denote the bundle map image Then it suces to compute image, the pullback of the connection 1-form α on imageto see that it is the connection 1-form of the rolling sphere. At a point image


A rolling sphere on a surface

Now more generally, we consider a sphere rolling on a surface in R3. We will construct a connection on the trivial SO(3)-bundle over the surface whose parallel transport describes the rotation of the sphere. However, in this generality, the connection need not be a pullback of the natural connection on imageLet X be a smooth orientable surface in R3 and let n be the unit normal vector eld pointing to the side on which the sphere rolls. Let J be the automorphism of the tangent bundle TX that rotates each tangent space counterclockwise image with axis of rotation given by the unit normal n. With the natural identi cation ofimage withimageAs compared with the planar surface it is now more complicated to describe the in nitesimal rotation at a point IMAGEin the direction imagebecause of the twisting and turning of the tangent spaces of the surface.

The unit normal vector eld is a map image The derivative of n at x is a linear map imageDifferentiating the constant function image shows thatshows thatimage Therefore,image and henceimagealso lies in imageThen the vector image is the axis of the in nitesimal rotation of the sphere at x in the direction v. We de ne a connection on the trivial bundle IMAGE whose horizontal space at (x; g) is


The connection 1-form α is given by


Let image be the local connection 1-form. Thus,


A rolling sphere on a sphere.

When the surface X is itself a sphere, it is possible to explicitly compute the local curvature form. Let X be the sphere of radius r centered at the origin.Then n(x) = x/r, Dn(x)(v) = v/r, and v + Dn(x)(v) = (1 + 1/r)v. Recall that Jx(v) = n(x)  v, and so


In order to compute the local curvature form imagein coordinates we use the isomorphism image between R3 and so(3) in order to treat as an R3-valued 1-form. Hence


With coordinates image




Evaluating imaeon a pair of tangent vectors u; v gives


Next we evaluate imageRecall that image is de ned so that its value on a pair of tangent vectors u and v is


Therefore, imageIn this case the bracket operation is the cross product and so


For the last step above we use the geometry of the cross product with x=r being a unit vector normal to both u and v to conclude thatimage. Putting the pieces together we see that


As the radius goes to in nity, the curvature form approaches the curvature form for the sphere rolling on a plane|as one would expect. For the sphere rolling on the outside of a sphere of radius 1, the curvature vanishes and so the connection is at and the holonomy is trivial around any null-homotopic path and hence around any closed path because S2 is simply connected. Fix a basepoint imageThere is a global section of the trivial bundle image mappingimage to the holonomy along any path from x0 to x. Antipodal points take the same value as can be seen by rolling the sphere along a great circle from the north to south pole. This section is an integral surface for the horizontal distribution of the connection. It is possible to describe this map in coordinates explicitly and it is especially nice using unit quaternions, i.e., S3, to represent rotations. The quaternion q de nes a rotation by mapping imageThe Lie group homomorphism IMAGE has kernel {1}g. In fact, S3 is the universal cover of SO(3) and this homomorphism is the projection. Using (0; 0; 1) for the basepoint, the holonomy map from S2 to SO(3) lifts to a map to S3 given by IMAGEThis represents the counterclockwise rotation through the angle IMAGEabout the axis (-y; x; 0). One can easily check that rolling the unit sphere from the north pole (0; 0; 1) to (x; y; z) does indeed turn the sphere through twice the angle between the two points and with axis that is normal to (x; y; 0).

To roll the sphere on the inside of a sphere of radius r simply change the unit normal to IMAGE and follow the same calculations. The local curvature formIMAGE turns out to be exactly the same


Although the curvature forms are equal, the connections are not the same and parallel transport is not the same. This can be seen easily for r = 1. Rolling inside the sphere produces no movement at all; the rolling sphere stays xed. Rolling the sphere on the outside does change the con guration along nonconstant paths.


The author would like to thank Jim Delany for helpful discussions about the rolling sphere and for Mathematica code to compute holonomy for the plane and spheres.


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