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PRINCIPAL BUNDLES AND PART II; PRINCIPAL BUNDLES

1. INTRODUCTION

The1 plan for this is as follows. First we will present principal bundles through the example of frame bundles of vector bundles. This has the advantage of a canonical system of coordinates. Then we will look at the more general form of the theory. There are several good reasons for moving from vector bundles to principal bundles. 1. Change of the local of sections is a action on the vector but in the situation it is difficult to see what the is really doing. In principal bundles this knowledge becomes more systematic and we understand it better. 2. A on a vector bundle is associated with a particular local basis of sections. We have the formulas for changing to a new connection when we change to a new local basis of sections, but we understand this only in the most shallow way. In the Principle Bundle context we see that this is an action of the group on the Lie Algebra. 3. Once a connection is put on a it metastasizes through all the associated bundles in a natural way. This replaces what look like haphazard and ad hoc methods for constructing connections on bundles related to a vector bundle. 4. The role of ”preserving a structure” in relation to connections is clearly reflected in the group of the principal bundle. Various properties like skew symmetry of the connection for the using an in Riemannian Geometry become understandable as properties of the Lie Algebra of the group O(n, R). 5. It is a convenient mathematical setting in which to study . These are fairly compelling reasons for introducing principal bundles. For amusement I will offer an analogy. The role of principal bundles in differential geometry might be compared to the role of evolution in biology. It is perfectly possible to work in a specific area of biology, for example turtles, without evo- lution. But the big scale structure of turtle theory will never be clear without evolution. Each new turtle requires a specific investigation. For a specific example in Differential Geometry, consider the following. It is natural to require in the Riemannian case that the covariant derivative satisfy d(σ, ρ) = (Dσ,ρ) + (σ,Dρ)

111 April 2012

1 If we come at this from the principal bundle approach, we don’t have to worry about whether it’s natural; it will automatically fall out of the theory when the group is O(n, R), which is the natural choice of group in the Riemannian context.

2. REVIEW OF VECTOR BUNDLES

In this section we will briefly review what we know about Vector Bundles and digest the formulas for future use. All these were derived in Part I. Let π : E → M be a vector bundle. We will denote by σ = {σ1,...,σn} a local basis of sections over U ⊆ M and by {σ˜1,..., σ˜n} second such. We will denote individual sections over U by τ,ρ. Thus σα,τ,ρ ∈ Γ(U, E). Let {σα} be a local basis of sections over U. For any x ∈ U there is a vector −1 space π [x] isomorphic to E above x in E, and σ(x) = (σ1(x),...,σn(x)) will be a basis of this vector space. (The basis may be arbitrary or it may be a special kind of basis, for example orthonormal, depending on additional structures on the manifold, for example a Riemannian structure.) In general, if {σ˜α} is a second such local basis of sections, then they will be connected by a group α R element gβ (x) ∈ GL(n, ) so that α σ˜β = σαgβ (x)

∞ α We assume that everything is smooth (C ) so that we may regard (gβ (x)) as a local section of GL(n, R) over U. (Here we are thinking of GL(n, R) as being the trivial bundle U × GL(n, R) over U.) Now we wish to digest the material on Vector Bundles that we derived in Part I so it will be handy for reference.

Vector Bundle (E,M,π) where π : E → M Fibre is E U coordinate patch of M

{σ1,...,σn} fixed local basis of sections over U u1,...,ud local coordinates on U ρ is a local section of E and thus α 1 d ρ = σαρ (u ,...,u ) ρ has local coordinates u1,...,ud,ρ1,...,ρn (bundle coordinates) ∂ ∂ ∂ ∂ T (E) has a basis ,..., , ,..., ρ ∂u1 ∂ud ∂ρ1 ∂ρn ∂ ∂ v ∈ T (E) has a representation v = vi + V α ρ ∂ui ∂ρα ∂ w is in the VERTICAL space ⇔ w = W α ⇔ all wi =0 ∂ρα

2 The connection is characterized by its HORIZONTAL SPACE. We can define this by a projection Π at each point in E which projects the at that point of the bundle to the vertical space which is the tangent space of the fibre and a subset of the full tangent space. Given a connection, the projection Π is defined by

i ∂ α ∂ α α β i ∂ Π v i + V α = V +Γβiρ v α ∂u ∂ρ ∂ρ Note how Π is a projection onto the Vertical Space. The Horizontal Space is defined as the kernel of Π;

v is in the HORIZONTAL space ⇔ 0 = Πv ⇔

v1 1 . W Γ1 ρβ, Γ1 ρβ, 1, 0, 0  .  2 β 1 βd  W   Γ2 ρβ, Γ2 ρβ, 0, 1, 0   vd  0= . = β,1 βd   .  V 1   .       n   Γn ρβ, Γn ρβ, 0, 0, 1   .   W   β,1 βd   .       V n    α β i β ⇔ 0=Γβiρ v + V A vector field X on M has the form ∂ X = Xi ∂ui The LIFT of X to HE is ∂ ∂ X˜ = Xi + X˜ α ∂ui ∂ρα where X˜ is in the Horizontal Space and thus

0 = ΠX˜ ˜ α α β i 0 = X +Γβiρ X so ∂ X˜ = Xi − Γα ρβXi ∂ui βi We wish to lift the curve x(t) from M to E. We do this by lifting it’s tangent vectorx ˙(t) to HE⊆TE by using the lift equations and then integrating along the lift to get the curve in E. (Technically, it would be best to embedx ˙(t) in a vector field in a neighborhood of the curve x(t), lift the whole field, and then pursue the integral curve in E.) Let’s look at these equations explicitly.

Curve x(t) with coordinates ui(t) and

3 dui ∂ tangent vector X(t)= dt ∂ui dui ∂ dui ∂ Lift: X˜(t)= − Γα ρβ dt ∂ui βi dt ∂ρα Equations of the liftx ˜(t) of x(t) which is the integral curve of X˜(t) and has coordinatesu ˜i(t),ρα(t):

du˜i dui = dt dt dρα dui = X˜ α = −Γα ρβ dt βi dt whereu ˜i(0) = ui(0) are the coordinates of x(0) so that πx˜(t)= x(t) and where ρα(0) are the coordinates of an arbitrary vector lying over x(0) in E. To further understand what the lift of the curve x(t) means, recall that

α Dρ = σαDρ α α β = σα(dρ + ωβ ρ ) α α β i = σα(dρ +Γβiρ du )

Thus for a tangent vector v in T (E)

α α β i j ∂ β ∂ Dρ(v) = σα(dρ +Γβiρ du ) v j + V β ∂u ∂ρ α α β i = σα V +Γβiρ v Since the tangent vector to the lift is

dui ∂ dui ∂ x˜˙ = − Γα ρβ dt ∂ui βi dt ∂ρα we see Dρ(x˜˙ )=0 Just to be completely clear here, ρ is an element of the vector bundle E. It is determined by coordinates ui which determine the point of U over which ρ α i α α lies, and by the coordinates ρ . The u determine Γβi , and the Γβi and the ρα determine Dρ which is closely related to the projection Π from the Tangent Space of E to the Vertical Space of E. Note that ρ lives in E, not in the tangent space T (E). Now let’s review the equations for of a vector ρ along a curve x(t) in U. We recall from part I that the equations for parallel transport

α 0 = Dρ(x ˙) = σαDρ (x ˙) j α α β du ∂ = σα(dρ + ωβ ρ ) j dt ∂u

4 α j ∂ρ i α β i du ∂ = σα( i du +Γβiρ du ) j ∂u dt ∂u α i i ∂ρ du α β du = σα i +Γβiρ ∂u dt dt α i dρ α β du = σα +Γβiρ dt dt Thus we see that lifting x(t) tox ˜(t) is just another way of describing parallel transport of vectors. Finally, we recall that

∂ [X, Y ] − [X,˜ Y˜ ] = Ω α(X, Y )ρβ β ∂ρα ∂ = R α XiY j ρβ β ij ∂ρα

Thus if one knows the horizontal space, one knows the lifts of vector fields and thus one knows the curvature.

3. FROM VECTOR BUNDLES TO FRAME BUNDLES

Now we are ready to introduce the Principal Bundle. In the present context we have two pictures of the principle bundle, the first of which is called the . 1. In the first picture we get the principal bundle by replacing the fibre π−1[x] of E by a fibre consisting of all the different bases of the fibre E of E. An ele- ment of the new fibre over x will be denoted by σ(x)= {σ1(x),...,σn(x)}. A section over U ⊆ M will be a smooth choice of a basis in the fibre of each x ∈ U. 2. In the second picture we get the principal bundle by replacing the fibre π−1[x] of E by a fibre consisting of the group G of the vector bundle E. In general this group is GL(n, R) and this is what we are using in our examples but it could be, for example, O(n, R) or a number of other groups. The two pictures are equivalent but not canonically so. To transition be- tween the two pictures we must select a fixed local section σ0 over U. Then for any x ∈ U and any element σ(x) in P , we can find a group element k(x) ∈ G for which σ(x)= σ0(x)k(x). Thus the local section k(x) of U × GL(n, R) forms a local section in the second picture. We symbolize this as

σ0 σ ←→ k

5 Naturally if we select a different local basis σ1 the correspondence will change. If σ1 = σ0h then σ0k = σ = σ1k˜ = σ0hk˜ and thus

k = hk˜ k˜ = h−1k so σ1 σ ←→ h−1k Since all these objects are local sections I have quietly dropped out the x. We are now going to perform some of the same activities in the Principal Bundle that we did in vector bundles. The new feature is that instead of the parallel translation of a vector field along a curve we are going to transform an entire frame. This will then produce a group element and eventually an element of the Lie Algebra of the group. Since the frame is made up of vectors, initially the change will require little more than bookkeeping. As before, we let x(t) be a curve in U ⊆ M where U is a coordinate patch with coordinates u1,...,ud. An element p ∈ P will be represented by a frame, and we will assume that the frame coincides in the fibre over x(0) with the base frame σ0. That is, we impose the initial condition that the transported frameσ ˜(t) coincides with the value of the section σ0 over x(0) when t =0 Now we transport each vector σα(0) of the base frame along the curve x(t) to get a frame fieldσ ˜α(t). For small t (at least) the vectorsσ ˜α(t) are linearly α independent sinceσ ˜α(0) are. Hence there are coefficients kβ (t) so that

α σ˜β(t)= σα(t)kβ (t)

Here and in the future we are using σ(t) for the value of the section σ0 over x(t). Since theσ ˜β(t) are linearly independent (for small t at least) the elements α α kβ (t) may be put into a matrix (kβ (t)) and regarded as an element of the R α group GL(n, ). For each β, the elements (kβ (t)) play the same role in the principal bundle that the coefficients ρα(t) played in the vector bundle with parallel transfer. Thus the equations for the transfer of the frame are the same;

α Dkβ (x ˙) = 0 for β =1,...,n α α γ (dkβ + ωγ kβ)x ˙ = 0 dkα dui β +Γα kγ = 0 dt γi β dt α The initial conditions for the kβ (t) are

α α kβ (0) = δβ

6 α reflecting thatσ ˜(0) = σ(0). Thus the connection coefficients Γβi for the principal bundle P are identical to those for the vector bundle E from which it arose. This α is important; the Γβi may be specified for the frame bundle P or for the vector bundle E, and E is our first example of an of the principal bundle P . It will turn out hat the connection on the principal bundle P will determine a connection on all its associated bundles. More on this later. Note that because parallel transfer from 0 to s+t is the same as from 0 to s and from s to s + t, we are dealing with a one parameter subgroup of GL(n, R). dk This then introduces a new feature, since dt |t=0 will then be an element of the Lie Algebra gl(n, R) of the group GL(n, R). Let’s look at what it will be. We have dkα dui β = −Γα kγ dt γi β dt so, letting t → 0, we have

dkα dui β | = [−Γα kγ ] dt t=0 γi β dt t=0 dui = −Γα δγ | γi β dt t=0 dui = −Γα | βi dt t=0 α = −ωβ (x ˙(0))

α Sincex ˙(0) can be any element of Tπ(p)(M), we can now interpret ωβ in a new α way; ωβ is a Lie Algebra valued one form. We input a tangent vector of M and it outputs a element in the Lie Algebra gl(n, R) of G. We have looked at this for the example G =GL(n, R) but it would work exactly the same way if G were some other matrix group. This is our first new insight; the connection on the Vector Bundle E functions as a connection ω in the Principal Bundle and thus forms a Lie Algebra valued one-form. This is one of the two properties we wish to have for connections on principal bundles. The second property is an invariance principle so that the covariant derivative D will not depend on the choice of basis. We will investigate this in a subsequent section. Next we need to discuss the analog for Principal Bundles of the Horizontal Space for vector bundles. While the coordinates in the principal bundle can α i be taken as kβ , u , it turns out in the sequel that the elements of the matrix α α −1 (ℓβ ) = (kβ ) are a better choice, as they match up better with out previous work on Vector Bundles. This is perhaps not obvious at this time but we will discuss it further in the next section. α i Using the coordinates (ℓβ , u ),the vectors in the tangent space TpP look like i ∂ α ∂ v = v i + Vβ α ∂u ∂ℓβ

7 Vectors in the vertical space Vp will look like

α ∂ w = Wβ α ∂ℓβ α The connection ωβ gives rise, just as in the vector bundle, to a projection Πp defined by

α ∂ i ∂ α ∂ Wβ α = Πp v = Πp v i + Vβ α ∂ℓβ ∂u ∂ℓβ α α γ i ∂ = Vβ +Γγiℓβv α ∂ℓβ α α α γ i Wβ = Vβ +Γγiℓβv

i It is obvious that if w ∈ Vp(P ) then Πp(w)= w since all v = 0. From this wesee 2 that Πp[Tp(P )] = Vp(P ). Hence Πp is a projection. Since dim Tp(P )= n + d 2 and dim Vp = n we have dim ker Πp = d = dim M. We define

Hp(P ) = ker Πp and thus dim Hp(P )= d. −1 Since Vp(P ) is the tangent space to the fibre π [x], we have dπ[Vp(P )] = π∗[Vp(P )] = 0. Because of our choice of coordinates, we then have

i ∂ α ∂ i ∂ π∗ v i + Vβ α = v i ∂u ∂ℓβ ∂u

Thus π∗ is a projection of Hp(P ) to Tp(M) and since they both have dimension d, dπ = π∗ is an of Hp(P ) to Tp(M). We note that if the structure group G is a subgroup of GL(n, R) instead of the whole group, the details would be only slightly different. We will consider this later. Next we note that the elements of Vp are essentially n × n matrices. Indeed

α ∂ α Wβ α ←→ Wβ ∂ℓβ α R and the (Wβ ) may be regarded as elements of the Lie Algebra gl(n, ).

4. COORDINATES ON THE PRINCIPAL BUNDLE AND INVARIANCE OF THE CONNECTION

It is critical that the Horizontal Space H(P ) be invariant under the right G−action. This means we must have

Hpg(P )= Rg∗Hp(P )

8 This will be a corollary of the formula

ΠpgRg∗(v)= Rg∗Π(pv) which we are going to derive in this section. This formula can be written in another way, as −1 Dpg = Ad(g ) Dp Formulas like these were derived for Vector Bundles and we need “only” replicate the process. However, there are some tricky aspects. As has often been the case in these notes I wish to do this proof in coordi- nates so we have the techniques and coordinate system available should we need them later. Having a “good” coordinate system will perhaps facilitate this and later adventures. What to choose for a good coordinate system is often a matter of experience, which one accumulates by trying coordinate systems which are less good. Recall that we are working over a single coordinate patch and that we have chosen a local basis of sections σ01,...,σ0n which we will keep in the shadows but is necessary for coordinates. All sections of P can then be written α 1 d as σ = σ0k where k = kβ (u ,...,u ) . We will work with two sections σ = σ0k andσ ˜ = σ0k˜ related byσ ˜ = σg α 1 d where g = gβ (u ,...,u ) . We then have σ = σ0k σ0k˜ =σ ˜ = σg = σ0kg k˜ = kg

α i Since σ = σ0k it would be possible to take (kβ ,u ) as coordinates for P . Possible α i but not such a good idea. The (kβ ,u ) transform in an inconvenient manner under the G−action because of the equation k˜ = kg. We are free to use any coordinates we like, so we will use a more convenient set. I now suggest how we might find a more convenient set. Recall that the coordinates {ρ1,...,ρn} for a section ρ of a vector bundle transformed under change of local basis of sections asρ ˜ = hρ (where h = g−1). We will be better able to utilize the insights of the vector bundles if we have coordinates the transform the same way. Set h = g−1, ℓ = k−1, ℓ˜ = k˜−1. Then since ˜ ˜−1 −1 −1 ˜ α k = kg, we have k = g k and thus ℓ = hℓ. Thus I propose to take {ℓβ } as ˜γ ˜ the coordinates of σ = σ0k (and thus {ℓδ } become the coordinates ofσ ˜ = σ0k). The equation ℓ˜= hℓ now mirrorsρ ˜ = hρ, as desired. Explicitly, if p ∈ P then p is represented by some σ on the fibre over some 1 d x ∈ U with coordinates (u ,...,u ) and σ = σ0k for some k ∈ GL(n, R) and, setting ℓ = k−1 we have the coordinates of p ∈ P are

α i p ←→ (ℓβ ,u )

Since ℓ˜= hℓ ˜α α γ ℓβ = hγ ℓβ

9 Now we recall that for a vector bundle if we have ∂ ∂ v = V α + vi ∂ρα ∂ui then α α i γ ∂ Πpv = V +Γγiv ρ α ∂ρ α i where we use coordinates (ρ ,u ). We define Πp by a similar formula where we α α replace ρ by ℓβ . Then we have

α ∂ i ∂ v = Vβ α + v i ∂ℓβ ∂u

α α i γ ∂ Πpv = Vβ +Γγiv ℓβ α ∂ℓβ Ten seconds thought shows that this is a projection onto the vertical space; if α ∂ i v = Vβ α (all v =0so v ∈ Vσ(P ))) then Πv = v. Thus the crucial question is ∂ℓβ whether it is invariant under the G−action. This is where the choice of coordi- nates is important; with these coordinates the calculation greatly resembles the calculation showing the covariant derivative in a vector bundle does not depend α α on the choice of basis. Formally one just replace ρ by ℓβ . Our goal is to prove the following Theorem

Πpg Rg∗v = Rg∗ Πpv ΠRg p ◦ Rg∗ = Rg∗ ◦ Πp

Proof The proof consists of two parts. As usual, we replace p by the σ = (σ1,...,σn) which represents it. First we find a formula for Rg∗ and second we verify the formula; neither is difficult. Since Rgσ = σg =σ ˜,

Rg∗ : Tσ(P ) → Tσ˜(P )

We set v ∈ Tσ(P ), w ∈ Tσ˜(P )

α ∂ i ∂ v = Vβ α + v i ∂ℓβ ∂u ∂ ∂ w = R (v) = W γ + vi g∗ δ ˜γ i ∂ℓδ ∂u α γ Then we have only to construct the Jacobian of Rg to connect Vβ to Wδ . A ˜γ α i tricky point is that ℓδ depends on both the ℓβ and the u . Indeed

˜γ γ α γ β α ℓδ = hαℓδ = hαδδ ℓβ

10 so ˜γ ˜γ γ ∂ℓδ γ β ∂ℓδ ∂hα α α = hαδδ and i = i ℓδ ∂ℓβ ∂u ∂u Then ˜γ ˜γ γ α ∂ℓδ i ∂ℓα Wδ = Vβ α + v i ∂ℓβ ∂u ∂hγ = V αhγ δβ + vi α ℓα β α δ ∂ui δ ∂hγ = hγ V α + α viℓα α δ ∂ui δ Comparison with the Vector Bundle calculation shows the results are for- α α mally identical (with ℓβ replacing ρ ) although the method of derivation was (superficially) different. Part two is now a little calculation. We have as usual

α ∂ i ∂ v = Vβ α + v i ∂ℓβ ∂u ∂ Π v = V α +Γα viℓ p β i β ∂ℓα β ∂ Note that in Πσv there are no terms with ∂ui . Hence ∂ R Π v = R V α +Γα viℓ g∗ p g∗ δ i δ ∂ℓα δ ∂ = hγ V α +Γα viℓ α δ i δ ˜γ ∂ℓδ using the formula previously derived. −1 −1 For ΠσgRg∗v we must use the formulaω ˜ = g dg + Ad(g )ω for trans- forming the connection one-form from Tσ to Tσg. This is automatic for the frame bundle, but for further development we should keep in mind that this would work in any Principal Bundle with a connection, (that is, a g−valued one form), satisfying this formula. The calculations would be identical. Hence the proof proves something more general than our original goal. Unpacking the formula we have

α α i ωβ (v) = Γiv γ ˜γ i ω˜ν (w) = Γνiv ∂gα = hγ ν vi + hγ Γα gρvi α ∂ui α ρi ν

γ α γ ∂ γ α Next we note hαgν = δν and ∂ui (hαgν ) = 0 so we have ∂hγ ∂gα α gα + hγ ν =0 ∂ui ν α ∂ui

11 Using this we finally have

ΠpgRg∗v = Πpgw ∂ ∂ = Π W γ + vi pg δ ˜γ ∂ui ∂ℓδ ∂ = W γ + Γ˜γ viℓ˜ν δ νi δ ˜γ ∂ℓδ ∂hγ ∂gα ∂ = hγ V α + α viℓα + (hγ ν vi + hγ Γα gρvi)hν ℓσ α δ i δ α i α ρi ν σ δ ˜γ ∂u ∂u ∂ℓδ ∂hγ ∂hγ ∂ = hγ V α + α viℓα − α gαvihν ℓσ + hγ Γα gρvihν ℓσ α δ i δ i ν σ δ α ρi ν σ δ ˜γ ∂u ∂u ∂ℓδ ∂hγ ∂hγ ∂ = hγ V α + α viℓα − α δαviℓσ + hγ Γα δρ viℓσ α δ i δ i σ δ α ρi σ δ ˜γ ∂u ∂u ∂ℓδ ∂hγ ∂hγ ∂ = hγ V α + α viℓα − α viℓα + hγ Γα viℓρ α δ i δ i δ α ρi δ ˜γ ∂u ∂u ∂ℓδ ∂ = hγ V α +Γα viℓρ α δ ρi δ ˜γ ∂ℓδ = Rg∗Πpv and the theorem is proved. Note carefully how the non-tensorial term g−1dg in γ ∂hα i α the transformation formula for ω cancels out the “extra” term ∂ui v ℓδ in the α transformation of Vδ in the seventh step. We will later have need of a formula which we can conveniently derive at this point. If we extract the next to the last step in the previous proof we have the formula ∂ Π R (v)= hγ V α +Γα viℓρ pg g∗ α δ ρi δ ˜γ ∂ℓδ From the formula ˜γ ∂ℓδ γ β α = hαδδ ∂ℓβ we see that ∂ℓα β = gαδδ ˜γ γ β ∂ℓδ and thus that ∂ ∂ℓα ∂ ∂ = β = gαδδ ˜γ ˜γ α γ β α ∂ℓδ ∂ℓδ ∂ℓβ ∂ℓβ ˜α We then put this formula in place of ∂/∂ℓβ in the previous equation to get ∂ Π R (v) = hγ V α +Γα viℓρ pg g∗ α δ ρi δ ˜γ ∂ℓδ γ α α i ρ α δ ∂ = hα Vδ +Γρiv ℓδ gγ δβ α ∂ℓβ

12 γ α α i ρ α ∂ = hα Vβ +Γρiv ℓβ gγ α ∂ℓβ

We then emphasize this result as a corollary.

γ α α i ρ α ∂ Corollary ΠpgRg∗(v)= hα Vβ +Γρiv ℓβ gγ ∂ℓα β 5. ALTERNATE WAYS OF LOOKING AT CONNECTIONS ON THE PRINCIPAL BUNDLE

We now wish to examine Πp from another point of view. Since the fibres are isomorphic to the group G, the vertical space Vp(P ) is isomorphic to the Lie Algebra g of G. We wish to develop this isomorphism explicitly, which is easy. Recall that an element X ∈ g generates a local one parameter subgroup k(t) of G which, for matrix groups can be written down explicitly as 1 1 k(t) = exp(Xt)= I + Xt + X2t2 + X3t3 + ... 2! 3! (This is often written as eXt.) We can use the right action of G on p to form a path in the fibre π−1[π(p)] pk(t) This is a path in the manifold P , so its derivative at t = 0 will yield a tangent vector to the fibre and thus be in Vp(P )

d pk(t)| ∈ V (P ) dt t=0 p More abstractly, we may obtain the vector field on P which corresponds to X ∈ g by setting d X˜ (f)= f(p exp(Xt))| X ∈ g p dt t=0 where f is any smooth function on a neighborhood of p. Since we are working with GL(n, R) and its subgroups all this can be carried α out extremely explicitly. A one parameter group k(t) = (kβ (t)) can be correlated with a Lie Algebra element d W α = kα(t)| β dt β t=0 α and W = (Wβ ) produces k(t) by

k(t) = exp(Wt)

13 The only limitation on k(t) is that det k(t) = 0 and g = gl(n, R) is the full matrix algebra over R. the action on P is given by

α σk(t) = (σαkβ (t)) and the element of Vp(P ) will be

d W˜ = (σ kα(t))| p dt α β t=0 α ∂ = Wβ α ∂kβ

(For the last equation, recall that if a path on a manifold has coordinates ui(t) dui ∂ then the tangent vector is dt ∂ui ) Using these methods we can now rewrite the projection Πp in a way which is abstractly preferable. We have a) the projection Πp : Tp(P ) → Vp(P ) b) the isomorphism W → W˜ p : g → Vp These combine, using the first and the inverse of the second, to form a map

Zp : Tp(P ) → g

Notice that these constructions are not dependent on the particular choice of G = GL(n, R); they are available as long as Πp is. The converse is also true; given a Zp it is easy to construct a Πp : Tp(P ) → Vp(P ) by composing Zp with the map W → W˜ p. Thus we see that a connection can be specified in four formally different ways:

α α i 1. Through the differential forms ωβ =Γβidu

2. Through the Horizontal Space Hp(P )

3. Through the projection Πp : Tp(P ) → Vp(P )

4. Through the Zp : Tp(P ) → g = Lie(G) The specification of any of these four will give a connection on a Principal Bundle in an abstract sense, but we do not wish to study the situation in this generality. We wish to restrict our consideration to connections on Principal Bundles that arise from connections on Vector Bundles. For this to be true one needs a property we call invariance which is related to base change in the Vector Bundle, which we will study in a subsequent section. In the more general situation we can even make a general definition for connection on a general fibre bundle, although we will not follow this trail. Given any fibre bundle (E,M,π,E,G) where G is the structure group, we set

14 Vp equal to the tangent space to the fibre at p. A on E is a vector valued differential form Πp from Tp(P ) (one for each p) onto Vp(P ) and which is the identity on Vp(P ). Then for t ∈ Tp(P )

t ∈ Hp(P ) ⇐⇒ Πp(t)=0 Such a differential form is called HORIZONTAL. I do not wish to pursue the subject at this level of generality. For Principal Bundles and Vector Bundles we wish to restrict our attention to connections which are well behaved under the right action of G. (There need be no such action in a general fibre bundle.) This will be treated in a subsequent section. Briefly, we will study connections that satisfy the additional condition that

Rg∗Πp = ΠpgRg∗ which is equivalent to −1 −1 ωpg = g dg + Ad(g )ωp

6. COVARIANT DERIVATIVE OF HIGHER ORDER VECTOR VALUED FORMS

Let π : E → M be a vector bundle and π : P → M the corresponding frame bundle, both with group G. The covariant derivative D acts on objects in the bundles

α α α β Dρ = σαDρ = σα(dρ + ωβ ρ ) Dσ = σω

In each case the output is a vector valued one form or frame valued 1-form. For the moment, let us concentrate on E. We will denote the algebra of differential forms on the tangent bundle of M by

∞ A(M)= Ai(M) i=0 where Ai(M) is the differential forms of degree i on M. A0(M) is just the functions on M. In a coordinate patch elements of Ap(M) look locally like

i1 ip fj1,...,jp du ∧ ... ∧ du

0≤j1,<...

α i1 ip σα fj1,...,jp du ∧ ... ∧ du α 0≤j1,<...

15 Thus the operator D is

D : A0(M, E) → A1(M, E) α α α β σα ⊗ ρ → σα(dρ + ωβ ρ )

This is fine as far as it goes, but we want a D that works on Ap(M, E). We can define this inductively so that

D : Ap(M, E) → Ap+1(M, E) by means of Leibniz’ rule:

for η ∈ Ap(M) and ρ a section of E D(ρ ⊗ η) = ρ ⊗ dη + Dρ ∧ η

This works because Ap(M) is a free algebra. We will do a couple of examples. Let f ∈ A0(M) (a function). Then the rule says D(ρf)= ρ df + (Dρ) f However, we could also do this the old way:

α D(ρf) = σαD(ρ f) α α β = σα d(ρ f)+ ωβ ρ f α α α β = σα(ρ df)+ σα dρ + ωβ ρ f = ρ df + (Dρ)f so we have consistency. For our second example we compute D2ρ:

2 α D ρ = DD(σαρ ) α α β = D σα(dρ + ωβ ρ ) α α γ α α γ = σα d(dρ + ωγ ρ ) + (Dσα) ∧ (dρ + ωγ ρ ) α γ α γ β α α γ = σα dωγ ρ − ωγ dρ + σβ ωα ∧ (dρ + ωγ ρ ) β γ β γ β α β α γ = σβ dωγ ρ − ωγ dρ + ωα dρ + ωα ∧ ωγ ρ β β α γ = σβ dωγ + ωα ∧ ωγ ρ β γ = σβ Ωγ ρ Thus the second covariant derivative is intimately related to curvature, as second derivatives always are. Note the lack of any terms involving derivatives of ρ. This means that D2ρ is tensorial. We will discuss this later.

7. ASSOCIATED BUNDLES, Part I

In this section we show the algebraic construction of associated bundles and give some important examples. All our constructions here can be regarded as

16 taking place on a single fibre, so the constructions will actually work for any vector space, although we have set up the notation to be suitable for bundles. Let (P,M,π,G) be a Principal Bundle and let ρ : G → Aut(V ) be a representation of G on a vector space V . We will create a vector bundle with structure group G out of ρ and V . To do this we we first form the Cartesian product P × V . There is a natural right action of G on P × V that comes from the right action of G on P , namely

(σ, v)g = (σg,ρ(g−1)v)

This is indeed a right action:

((σ, v)g)h = (σg,ρ(g−1)v)h = (σg)h,ρ(h−1)ρ(g−1)v = (σ(gh),ρ(h−1g−1)v) = (σ(gh),ρ((gh)−1)v) = (σ, v)(gh)

We can now take the quotient of P × V by the G-action, which is to say we make an equivalence class out of each orbit of G, and these equivalence classes form the desired associated vector space to P . We denote the equivalence class by [σ, v]. Thus (σ, v) ∼ (σg,ρ(g−1)v) and we have Def Let g ∈ G, σ ∈ P and v ∈ V . Then

[σ, v] = [σg,ρ(g−1)v]

This is indeed an ; transitivity is checked in the previous calculation. Of course we could take a less sophisticated view and simply define the equivalence relation and form the classes without mentioning the G-action, but this way is more elegant. The equivalence classes now form a vector space, for let ξ1 = [σ1, v1] and ξ2 = [σ2, v2]. Since G is transitive on P there is a g ∈ G for which σ1g = σ2. We now define

ξ1 + ξ2 = [σ1, v1] + [σ2, v2] −1 = [σ1g,ρ(g )v1] + [σ2, v2] −1 = [σ2,ρ(g )v1] + [σ2, v2] −1 = [σ2,ρ(g )v1 + v2]

It is easy to show that with this definition that addition is well defined and with the obvious definition α[σ, v] = [σ, αv] for scalar multiplication that the equivalence classes form a vector space. It would be nice to have a notation for this new construction. The equiva- lence class lives in the space (P × V ) so one notation would be (P × V )/ ∼. We will use the standard notation which is

P ×ρ V

17 There are some standard for sections of these bundles which we will discuss after a few examples. The point of the construction is that a con- nection on a Vector Bundle E uploads to a connection on its principal bundle P which then metastasizes through all of P ’s associated bundles. We should also pause a minute and look at the situation from a higher point of view. We are interested in associated vector bundles, but the construction above will work in more general circumstances. Suppose that V , instead of being a vector space, is any object on which G acts from the LEFT. This means that for h,g ∈ G and v ∈ V we have, denoting the action by g.v, h.(g.v) = (hg).v In our case we have the action given by g.v = ρ(g)v and this is a left action since h.(g.v)= ρ(h)ρ(g)v = ρ(hg)v = (hg).v There is confusion lurking here because in the equivalence relation [σ, v] = [σg,ρ(g−1)v] the action g, v → ρ(g−1)v is a right action. This is irrelevant. To see things in the proper light, we rewrite the equivalence relations as follows: [σg,v] = [σg,ρ(g−1)ρ(g)v] = [σ, ρ(g)v] Then the right action of G on P and the left action of G on V are more clearly visible. Unfortunately the equation [σg,v] = [σ, ρ(g)v] does not cor- relate too well with the forms we usually use in writing down expressions in vector spaces. However, it can be very useful for calculation; for example to decode [σ(gh),ρ((gh)−1)v], we can use it as follows: [σ(gh),ρ((gh)−1)v] = [(σg)h,ρ(h−1)ρ(g−1)v] = [(σg),ρ(h)ρ(h−1)ρ(g−1)v] = [σg,ρ(g−1)v] = [σ, ρ(g)ρ(g−1)v] = [σ, v] This should also convince you that it is correct, even though for some people it intuitively feels wrong. Now suppose that P is the bundle of frames of a vector bundle W of di- mension n and fibre W and that G =GL(n, R). We then recover the bundle W from the construction in the following obvious way. We take V to be Rn ρ(g) = g ∈ GL(R,n) and correlate w ∈ W with [σ, v] as follows. Let σ be a i frame and w ∈ W . Then w = σiw and we correlate w1  .  w ↔ [σ, . ]  wn   

18 i If we write w = σiw as w1 w1  .   .  w = σ . = (σ1,...,σn) .  wn   wn      Then the familiar relations w1 w1 w˜1  .  −1  .   .  w = (σ1,...,σn) . w = (σ1,...,σn)gg . = (˜σ1,..., σ˜n) .  wn   wn   w˜n        correlate perfectly with w1 w1 w˜1  .  −1  .   .  [σ, . ] = [σg,ρ(g ) . ] = [(˜σ1,..., σ˜n), . ]  wn   wn   w˜n        so that we see we merely have two different notations for the same idea. Note that each equivalence class correlates to a unique vector and that different mem- bers of the same equivalence class correlate with representations of the same vector in different bases. This is very nice way to look at basis change. It is now easy to show that W ∗ and Hom(W, W ) are examples of the As- sociated Bundle construction. First we review how the dual basis works and construct a matrix notation for it. Recall that we may form a basis σ∗ for W ∗ by taking the dual basis of σ. We define σ∗i by

∗i i σ (σj )= δj It is convenient, as we will see, to represent the dual basis as a column σ∗1 ∗  .  σ = .  σ∗n    We shall refer to the elements λ of W ∗ as covectors. Then a covector λ may ∗ ∗i be expanded in terms of the basis σ as λ = λiσ , and the matrix notation for this is σ∗1 ∗i  .  λ = λiσ = (λ1,...,λn) .  σ∗n    This is convenient because we can now find λ(v) where v ∈ W by matrix mul- tiplication σ∗1 v1  .   .  λ(v) = (λ1,...,λn) . (σ1,...,σn) .  σ∗n   vn     

19 v1 ∗i  .  = (λ1,...,λn) σ (σj ) .  vn    v1 i  .  = (λ1,...,λn) δj .  vn    i = λiv as expected. This looks even better (and more symmetric) if we replace the notation λ(v) by the notation λ, v, which emphasizes the duality between W ∗ and W . Since λ(v) does not depend upon the basis, we should be able to incorporate the G action into the above equation. As before we writeσ ˜ = σg to indicate the G action on a basis of W . The corresponding dual basis of W ∗ will beσ ˜∗ and since both σ∗ andσ ˜∗ are bases there will be matrix h for which hσ∗ =σ ˜∗ The h is written to the left of σ∗ because of the way we write σ as a column. We now want to find h in terms of g, which is easy with our efficient notation. Using I for the identity matrix we have I =˜σ∗σ˜ = hσ∗σg = h I g = hg so that h = g−1 σ˜∗ = g−1σ∗ Because of the inverse, this is a right action of G on the set of bases of W ∗. In order to proceed further without typographical stress we introduce some additional notation. We will denote an element of Rn by v and an element of Rn∗ by λ; v1  .  v = . λ = (λ1,...,λn)  vn    and thus with w ∈ W and λ ∈ W ∗ we have v1  .  v = σv = (σ1,...,σn) .  vn    σ∗1 ∗  .  λ = λσ = (λ1,...,λn) .  σ∗n    This will simplify writing things down. Let’s practice with the new notation by checking that works as it should: v = σ v = σgg−1v =σ ˜ v˜

20 λ = λ σ∗ = λgg−1σ∗ = λ˜ σ˜∗

Note that we forgot to find out how the coordinates of λ change when we change the basis, but from the above we can read off immediately that the formula must be λ˜ = λg. We already knew that v˜ = g−1v which we can read off from the first line. Now let’s check that we can compute the value of a linear functional on a α vector in any coordinate system. Recall that we saw λ(v)= λαv .

α −1 β λ(v)= λαv = λv = λ˜g gv˜ = λ˜v˜ = λ˜βv˜

Next we want to describe W ∗ as an associated bundle to the bundle P of frames of W . To do this, we must find a representation ρ of G on some vector ∗ space V so that P ×ρ V will be isomorphic to W . This is only a matter of of getting the transformation under basis change correct. n∗ For V we take R and write elements of V as rows λ = (λ1,...,λn). Define ρ by g.λ = ρ(g)λ = λg−1 If λ ∈ W and λ = λ σ∗ is the representation of λ in the dual basis then the isomorphism is λ ↔ [σ, λ] Now we must check that the isomorphism correctly survives basis change:

[σ, λ] = [σg,ρ(g−1)λ]=[˜σ, λg]=[˜σ, λ˜] using λ˜ = λg and other formulas derived above. It is trivial to verify the homomorphism properties so

∗ n∗ W ↔ P ×ρ R is an isomorphism and we see that W ∗ is an associated vector bundle of P . Without going into the trivial details, it should now be clear that with n −1 n∗ ρ1(g)v = gv on R and ρ2(g)λ = λg on R we have a Tensor Product representation ρ = ρ1 ⊗⊗ ρ1 ⊗ ρ2 ⊗⊗ ρ2 of G on Rn ⊗⊗ Rn ⊗ Rn∗ ⊗⊗ Rn∗ (any finite number of Rn’s and Rn∗’s) and this shows that

W ⊗⊗ W ⊗ W ∗ ⊗⊗ W ∗ is an associated bundle of P . It will be instructive to examine W ⊗ W ∗, which is generated by elements of the form w ⊗ λ, in more detail. We have the isomorphism

w ⊗ λ ↔ [σ, w ⊗ λ]

21 (where w and λ are the coordinates of w and λ in the basis σ). Let’s look at the equivalence classes

[σ, w ⊗ λ] ∼ [σg, ρ(g−1)( w ⊗ λ)] −1 −1 ∼ [˜σ, ρ1(g ) w ⊗ ρ2(g )λ] ∼ [˜σ, g−1 w ⊗ λg] ∼ [˜σ, w˜ ⊗ λ˜] just as we would expect. We can extract a little more out of W ⊗W ∗. We define an action of W ⊗W ∗ on W by (w ⊗ λ)(v)= wλ(v) Moving this over to the matrix representation, any matrix may be counterfeited in Rn ⊗ Rn∗ by using its rows as in the following example.

1 −3 1 0 ↔ ⊗ (1, −3)+ ⊗ (2, 5) 2 5 0 1 1 −3 3 1 0 3 ↔ ⊗ (1, −3)+ ⊗ (2, 5) 2 5 2 0 1 2 1 0 −3 ↔ (−3) + (16) = 0 1 16

This immediately generalizes; Hom(W, W ) is isomorphic to W ⊗ W ∗ (with the α action described above). If T has matrix (tβ ) in the basis σ then we have an isomorphism Rn Rn∗ T ↔ [σ, t] ∈ P ×ρ1⊗ρ2 ( ⊗ ) α where t is formed from the matrix (tβ ) as shown above. Then T ↔ [σg, ρ(g−1)t] ∼ [˜σ, g−1tg] using the action of ρ defined above. From this we read off that if T has matrix α −1 α (tβ ) in the σ basis then it will have matrix g (tβ )g in theσ ˜ = σg basis. This indicates how our formalism can be a handy tool in linear algebra.

8. ASSOCIATED BUNDLES, Part II

Having defined Associated Bundles in the previous section, our next task is to use the connection on the principal bundle to generate a connection on each of the associated bundles. It is easy enough to give a formula for this, but to motivate the formula will take some time. While there are no doubt many ways to do this, we will come at the problem by means of parallel displacement. One reason for this approach is that parallel

22 displacement is a rather easy to visualize geometric process, which we hope will make the construction seem more natural and geometric. Let V = (P,V,ρ) be an associated vector bundle to the principal bundle P . We visualize P as a bundle of frames of of a vector bundle E and also as the bundle with fibre G as discussed above; there is a fixed local basis of sections σ which in this section will remain fixed. The vector bundle structure for V is defined as follows: take v = [σ, v] ∈V and define π(v) to be π(σ). Since σ and σg are in the same fibre over M, π(v) is independent of the choice of equivalence class for v. Next we show how to describe a section. Let (u1,...,un) be coordinates on some open set U ⊂ M. Then we can define a section by giving [σ(u1,...,ud), v(u1,...,ud)] in which it is required that π σ(u1,...,ud) = x(u1,...,ud) ∈ M π[σ(u1,...,ud), v(u1,...,ud)] = π σ(u1,...,ud)= x(u1,...,ud) ∈ M To see how to define parallel transport for V we will first review previous work in E and P . Let x(t) be a curve in M with 0 in its domain. We will concentrate our attention at t = 0. Recall parallel transport in E. Let ρ0 ∈ −1 π [x(0)]. Then the parallel transport ρ(t) of ρ0 along x(t) is governed by the equations

ρ(0) = ρ0 α dρ α β i D ρ = σα +Γ ρ u˙ = 0 x˙(t) dt βi For notational simplicity we will denote the frame σ along x(t) by σ(t) = σ(u1(t),...,un(t) . Next we remind ourselves that the transport can be done for each of the particular vectors σα(0) of the basis σ(0). The equations will α then be, withσ ˜γ (t)= σα(t)gγ (t)

σ˜α(0) = σα(0) α dgγ α β i D σ˜γ (t) = σα +Γ g (t)u ˙ (t) = 0 x˙ (t) dt βi γ α where the Γβi are the connection coefficients for the frame σ. The first condition α α shows that gγ (0) = δγ . Thus we can replace these equations by α α gγ (0) = δγ dgα γ +Γα gβ(t)u ˙ i(t) = 0 dt βi γ

We collect the transported vectorsσ ˜α(t) into a new frameσ ˜ = (˜σ1(t),..., σ˜n(t)). γ γ Now suppose we letρ ˜(t)=˜σγ (t)ρ0 where the ρ0 are constant. The we have γ γ ρ˜(0) =σ ˜γ (0)ρ0 = σγ (0)ρ0 = ρ0 and we also have γ γ Dx˙(0)ρ˜(t)= Dx˙(0) [˜σγ (t)ρ0 ]= Dx˙(0) [˜σγ (t)] ρ0 =0

23 Thus ρ(t) andρ ˜(t) are both solutions of the same linear system of differential equations and have he same initial conditions. Hence by uniqueness of solutions ρ˜(t)= ρ(t). Putting this into words, we can solve the parallel transport problem for each ρ0 or we can solve it by translating the whole frame and using the coordinates of ρ at t = 0 as the coefficients of the translated frame. Thus solving the parallel transport problem in P automatically solves it in E and vice versa. α α Next we note that the uniqueness of solution for gβ (t) implies that gβ (t) forms a one parameter group in G =GL(n, R). From this we see that d gα(t) | = − Γα gγ (t)u ˙ i(t) | dt β t=0 γi β t=0 α γ i = − Γγi δβ (t)u ˙ (0) α i α = − Γβi u˙ (0) := −ωβ (x ˙ 0)

But the derivative of a one-parameter group in a is an element of the α Lie Algebra of the Lie Group. Thus ωβ (x ˙ 0) is an element of the Lie Algebra. R α R α For example, if G = O(n, ) then ωβ (x ˙ 0) is in o(n ) which implies that ωβ (x ˙ 0) α is skew symmetric. Since ωβ (x ˙ 0) depends linearly on the tangent vectorx ˙ 0 ∈ α Tx˙ 0 (M) we see that ωβ is a Lie algebra valued one form. It is important that α the connection ωβ can be found from

d ω α(v)= Γα u˙ i(0) = − gα(t) β βi dt β t=0 α where the gβ are found by parallel transfer of σ(0) along any curve for which x˙(0) = v.

9. ASSOCIATED BUNDLES, Part III

Before we introduce connections into all the associated bundles it will be con- venient to introduce a concept we have neglected up till now. Another way to think of parallel transport in a bundle is that it sets up an isomorphism between π−1[x(0)] and π−1[x(t)]. Specifically, for the principal bundle P we have the isomorphism

−1 −1 θt : π [x(0)] → π [x(t)]

θt : σ0 → σ˜(t)

If we write this in terms of the local basis of sections (the frame) we have

θt : σ0 → σ(t)g(t)

24 α where we are using the same notation as in Part II, where g(t) = (gβ (t)) and α −1 σ˜(t)= σ(t)g(t) meansσ ˜β(t)= σα(t)gβ (t). We are really most interested in θt which moves the basis σ(t) in π−1[x(t)] to the basis in π−1[x(0)] which parallel translates into the basis σ(t). For convenience let h(t)= g(t)−1. Then

θt(σ0 h(t)) = θt(σ0)h(t)=˜σ(t)h(t)= σ(t)g(t)h(t)= σ(t) so we have −1 θt σ(t) = σ0 h(t) All of this works because parallel transport preserves linear combinations of vectors. In a similar way we can define θt on E. Let ρ be a section and let ρ0 = ρ(x(0) and ρ(t)= ρ(x(t)). Letρ ˜(t) be the parallel transport of ρ0 along x(t). Define −1 −1 θt : πE [x(0)] → πE [x(t)] θt : ρ0 → ρ˜(t)=˜σ(t)ρ0 = σ(t)g(t)ρ0 as we discussed in Part II. Now we want to consider taking the derivative of ρ. This is a special case of a result we will derive later so discussing it now is inefficient, but perhaps illuminating psychologically. The naive formula for the derivative is of course ρ(t) − ρ(0) ρ′(0) = lim t→0 t −1 −1 However, this is nonsense since ρ(t) ∈ πE [x(t)] and ρ(0) ∈ πE [x(0)] and these are different spaces and so the subtraction is impossible. We want to subtract −1 things in the same space, preferably πE [x(0)], and one way to do this is to −1 replace ρ(t) by θt (ρ(t)). So we are going to try this and compute θ−1ρ(t) − ρ(0) lim t t→0 t Our first job is to find the derivative of h(t)= g(t)−1 we have h(t)g(t) = Id dh dg g(t)+ h(t) = 0 dt dt dh dg = −h(t) h(t) dt dt We will have need of this below. −1 α Our second job is to compute θt ρ(t). Let θtρ = ρ(t) and ρ = σα(0)ρ and α ρ(t)= σα(t)ρ (t). If we set ρ1 ρ1(t)  .   .  ρ = . and ρ(t)= .  ρn   ρn(t)     

25 then we can write more briefly

θt(ρ) = θt(σ(0)ρ) = θt(σ(0))ρ =σ ˜(t)ρ = σ(t)g(t)ρ using the techniques we developed in Associated Bundles Part II. But since

ρ(t)= σ(t)ρ(t) we have

σ(t)g(t)ρ = ρ(t) = θt(ρ) = σ(t)ρ(t) g(t)ρ = ρ(t) ρ = h(t)ρ(t) where h(t) is the matrix inverse to g(t). Thus

ρ = σ(0)ρ = σ(0)h(t)ρ(t)

Now we have everything necessary to compute the derivative.

θ−1ρ(t) − ρ(0) σ(0)h(t)ρ(t) − σ(0)ρ(0) lim t = lim t→0 t t→0 t h(t)ρ(t) − h(0)ρ(0) = σ(0) lim t→0 t d = σ(0) h(t)ρ(t) dt t=0 dρ(t) dh = σ(0) h(t) + ρ(t) dt dt t=0 dρ(t) dg = σ(0) h(t) − h(t) h(t)ρ(t) dt dt t=0 dρ(t) = σ(0) h(t) + h(t)ω(x ˙(t))h(t)ρ(t) dt t=0 dρ(t) = σ(0) + ω(x ˙(0))ρ dt 0 = σ(0)Dx˙ (0)ρ

= Dx˙(0)ρ Thus our new methods have reinvented the covariant derivative, which is very reassuring. Also notice that the critical technical pieces were the knowledge of −1 the connection on P and the use of the map θt . With the above as a model we can now guess the proper formula for a connection on any associated bundle. First the setup. Let the bundle be V with representation space V and representation ρ of G on V . As before we have a curve x(t) and a section

v(u1,...,un) = [σ(u1,...,ud), v(u1,...,ud)]

26 with values along the curve

v(t)= v(u1(t),...,ud(t)) = [σ(u1(t),...,ud(t)), v(u1(t),...,ud)(t)] where v ∈ V . We define parallel transport of v(0) along x(t) in the obvious way −1 Def The parallel transport of v(0) ∈ πV [x(0] along x(t) is v˜(t)=[˜σ(t), v(0)] = [σ(t)g(t), v(0)] = [σ(t),ρ g(t) (v(0))] Then we have the usual parallel transport functions 2

θt[σ(0), v(0)] → [σ(t),ρ g(t) (v(0))] −1 θt [σ(t), v(t)] → [σ(0),ρ h(t) (v(t))] where h(t)= g(t)−1. Now we define Def The covariant derivative of a section v(u1,...,ud) = [σ(u1,...,ud), v(u1,...,ud) of the vector bundle V at v(0) is

−1 θt (v(t)) − v(0) d Dx˙(0)v = lim = σ(0), ρ h(t) v(t) t→0 t dt t=0 In detail,

θ−1(v(t)) − v(0) [σ(0),ρ h(t) (v(t))] − [σ(0),ρ h(0) (v(0))] lim t = lim t→0 t t→0 t ρ h(t) (v(t)) − ρ h(0) (v(0))] = σ(0), lim, t→0 t d = σ(0), ρ(h(t)v(t) dt t=0 We are now in the enviable position of having defined parallel transport in any associated bundle, and we have a formula for the covariant derivative of sections of an associated bundle. The formula will be effective if we have sufficient control of the representation ρ, as we will illustrate in the following section, where we find the covariant derivative in various popular bundles. We will also investigate some theoretical considerations involving tensor products which are important in the sequel.

10. EXAMPLES WITH TENSOR PRODUCTS

In this section we are going to calculate covariant derivatives using the technol- ogy of the previous section for some of our favorite bundles. Let ρ be a rep- resentation of the group G of P on some vector space V , V the corresponding 2 These θt go in the opposite direction from those in most of the literature. Beware.

27 associated bundle, and let x(t) be a curve on the base space M. The notations and formulas we will use are as follows.

g(t) = g(ui(t)) h(t) = g(t)−1 dh dh = −h(t) h(t) dt dt dg = − ω α(x ˙(t)) dt β d D v = σ(0), [ρ(h(t))v(t)] x˙ (0) dt t=0 where v = [σ, v] ∈ P ×ρ V . These formulas are intended to be generic, with V the associated bundle, V the representation space, whatever it happens to be, v ∈ V, and v the representation of v in the basis σ. Details will vary from example to example and complete consistency of notation would be burdensome. The Bundle E We begin with the simplest associated bundle. If P is the bundle of frames of a vector bundle E then we know E can be reconstructed from P using V = Rn and ρ(g)v = gv. We will now show how the covariant derivative and connection α α i on E conforms to our previous formulas. Let (ωβ )=(Γβidu ) be the connection on P for the local basis of sections σ = (σ1,...,σn). We regard E as

n E = P ×ρ R with typical element

v1  .  v = [σ, v] where v = .  vn   

Now let x(t) be a curve in M andx ˙ 0 =x ˙(0). Then d D v = [σ(0), ρ(h(t))(v(t) ] x˙ 0 dt t=0 d = [σ(0), h(t)(v(t) ] dt t=0 dv dh = [σ(0), h(t) + v(t)] dt dt t=0 dv dg = [σ(0), h(t) − h(t) h(t) v(t)] dt dt t=0 dv = [σ(0), h(t) + h(t) ω(x(t)) h(t) v(t)] dt t=0 dv = [σ(0), + ω(x ) v(0)] dt 0

28 Since there is nothing special about t = 0 we can translate this into more normal notation for any t as

α j dv α du β Dx˙(t)v = σα(t) +Γβj v (t) dt dt

Since for any w ∈ Tx(M) we can find a curve withx ˙(0) = w we can write the above formula (again with t = 0) as ∂vα duj D v = σ +Γα wj vβ(t) w α ∂uj dt βj α ∂v α β j = σα j +Γβjv (t) w ∂u and thus the covariant differential is α ∂v j α β j Dv = σα j du +Γβj v du ∂u α α β = σα dv + ωβ v ) α = σαD v which we recognize as our old formula for the covariant derivative. We also want to make contact with the ancient notation. We set ∂vα vα = +Γα vβ vβ |j ∂uj βj and then

α j Dv = σαv |j du duj D v = σ vα x˙ (t) α |j dt

The Bundle E∗ ∗ n∗ We obtain the bundle E by choosing V = R and ρ(g)(λ1,...,λn) = −1 (λ1,...,λn)g . (This is a left action on V .) With the same notations we have d D λ = [σ(0), ρ(h(t))λ(t) ] x˙ 0 dt t=0 d = [σ(0), λ(t)h(t)−1 ] dt t=0 d = [σ(0), λ(t)g(t) ] dt t=0 dλ dg = [σ(0), g(t)+ λ(t) ] dt dt t=0 dλ = [σ(0), g(t) − λ(t)(ω α(x ˙(t))] dt β t=0 dλ α = [σ(0), − λ(0)(ωβ (x ˙(0))] dt t=0 29 Since there is nothing special about t = 0 we may replace the 0 by t and then switch to a more normal notation. We then have

j β ∂λβ α du Dx˙ 0 λ = σ j − λαΓβj ∂u dt β ∂λβ α j Dwλ = σ j − λαΓβj w ∂u β ∂λβ α j Dλ = σ j − λαΓβj du ∂u and setting ∂λ λ = β − λ Γα β|j ∂uj α βj we have β j Dλ = σ λβ|j du which is the covariant differential. Linear Transformations; the bundle E⊗E∗ For the space V we choose M(n, R) and, for the representation for a ∈ M(n, R), we choose ρ(g)a = gag−1. Let the Linear Transformation T have the matrix a in the basis σ so that

T = [σ, a]

Then in the basisσ ˜ = σg, T will have the basis b = g−1ag. In the associated bundle notation we have

[σ, a] = [σgg−1, a] = [σg, ρ(g−1)a] = [σg, g−1ag]=[˜σ, g−1ag] so that everything is consistent. By our general formula, (with h = g−1)

d D T = [σ, ρ(h(t))a(t) ] x˙ 0 dt t=0 d = [σ, h(t)a(t)g(t) ] dt t=0 da dh dg = [σ, h(t) g(t)+ a(t)g(t)+ h(t)a(t) ] dt dt dt t=0 da dg dg = [σ, h(t) g(t) − h(t) h(t)a(t)g(t)+ h(t)a(t) ] dt dt dt t=0 da = [σ, h(t) g(t)+ h(t)ω(xh ˙ (t)a(t)g(t) − h(t)a(t)ω(x ˙] dt t=0 da = [σ, + ω(x ˙ 0)a(t) − a(t)ω(x ˙ 0)] dt t=0

30 Decoding, and replacingx ˙ 0 byx ˙(t) since there is nothing special about t = 0, we find the representation for Dx˙(t) to be

σ da D T ←→ + ω(x ˙)a(t) − a(t)ω(x ˙) x˙ (t) dt α i i i σ ∂a du du du ←→ β +Γα aγ − Γγ aα ∂ui dt γi β dt βi γ dt α i σ ∂a du ←→ β +Γα aγ − Γγ aα ∂ui γi β βi γ dt Thus setting, as the ancients did, ∂aα aα = β +Γα aγ − Γγ aα β|i ∂ui γi β βi γ we have

σ α i DT ←→ aβ|i du α i DT = [σ, aβ|i du ] The equation ∂aα aα = β +Γα aγ − Γγ aα β|i ∂ui γi β βi γ α α can be written in another way. With, as usual, a = (aβ ) and ω = (ωβ ) = α i (Γβi du ) we have, with ∂a a = + (ω ∧ a) − (a ∧ ω) |i ∂ui i i where

α γ (ω ∧ a)i = (ωβ ) ∧ (aβ) i α γ = (Γγiaβ) i ω ∧ a = (ω ∧ a)i du and similarly for a ∧ ω. (The wedges are strictly speaking unnecessary since a is a 0-form.) Then we have, using the Lie Bracket notation [ω,a] for ω ∧ a − a ∧ ω], ∂a a = + [ω,a] |i ∂ui i σ i DT ←→ a|idu σ ←→ da + [ω,a] DT = σ, da + [ω,a] In the literature they often say this as ”D acts on E⊗E∗ = End(E) via the Lie Bracket.”

31 Bilinear Forms This example is more difficult than the preceding ones because it is not possible to express what we need using matrices in a convenient way. I could make it more convenient if I were willing to use the transpose of matrices, but I feel that using transpose without the presence of a inner product is immoral, so I shall not do it. In addition, the techniques which I use here are perfectly general, and the reader is urged to watch the calculations with the idea of using these techniques for any tensor in mind, since I will later claim this is possible. Thus this is a proof by example. We we will use the notation B[u, v] ∈ E∗ ⊗E∗ for this treatment and we will take V = Rn∗ ⊗ Rn∗ ≃ M(n, R) but here it is natural to write the matrix with low indices. The representation will be the tensor product of the previ- ously introduced λ with itself, although we will allow ourselves some flexibility in the notation. We could follow the same trail that we trod with linear trans- formations but I want to use a related technique because it is more readily generalizable to more complicated tensor situations. The space of bilinear forms can be visualized as Rn∗ ⊗ Rn∗. Here is how we do this. First we define an action of Rn∗ ⊗ Rn∗ on Rn ⊗ Rn by

(λ ⊗ )(v ⊗ w)= λ(v)( w)

Next we define

bαβ = B[σα, σβ]

βα = (bα1,bα2,...,bαn) th σα = (0,..., 0, 1, 0,..., 0) 1 in α place Note that, with u = [σ; u] and other previously established conventions we have

α β B[u, v] = B[σαu , σβ v ] α β = B[σα, σβ]u v α β = bαβ u v

Then we can represent B[u, v] by

Rn∗ Rn∗ b = σα ⊗ βα ∈ ⊗ α We then have

b(u, v) = σα ⊗ βα(u, v) α

= σα(u)βα(v) α α β = u ( bαβv ) α β

32 α β = bαβu v αβ = B[u, v]

n∗ Now recall that ρ2 : GL(n, R) → Aut(R ) is given by −1 −1 ρ2(g)(λ1,...,λn) = (λ1,...,λn)g = λg and we can define ρ : GL(n, R) → Aut(Rn∗ ⊗ Rn∗) by

ρ = ρ2 ⊗ ρ2

ρ(g)(λ ⊗ ) = ρ2(g)(λ) ⊗ ρ2(g)() = λg−1 ⊗ g−1 Now using the formula d D b = [σ(0), ρ(h(t))( σ ⊗ β (t)) ] x˙ 0 dt α α t=0 α d = [σ(0), ( σ h(t)−1 ⊗ β (t)h(t)−1 ] dt α α t=0 α d = [σ(0), ( σ g(t) ⊗ β (t)g(t) ] dt α α t=0 α dσ dg = [σ(0), α g(t)+ σ ⊗ β g(t) dt α dt α α dβ dg + σ g(t) ⊗ α g(t)+ β ] α dt α dt dg dβ dg = [σ(0), σ ⊗ β g(t)+ σ g(t) ⊗ α g(t)+ β ] α dt α α dt α dt t=0 α because dσα/dt = 0 since σα is constant. Hence, since dg/dt = −ω(x ˙(t)), and keeping in mind that βα is a row vector and ω(x ˙(t) is a matrix, dβ D b = [σ(0), − σ ω(x ˙(t)) ⊗ β g(t)+ σ g(t) ⊗ α g(t) − β ω(x ˙(t) ] x˙ 0 α α α dt α t=0 α dβ = [σ(0), − σ ω(x ˙ ) ⊗ β + σ ⊗ α − β ω(x ˙ ) ] α 0 α α dt α 0 α dβ = [σ(0), σ ⊗ α − σ ω(x ˙ ) ⊗ β − σ ⊗ β ω(x ˙ ) ] α dt α 0 α α α 0 α The first and third terms are no problem but the second term is, because it does not match up well with matrix calculations. Hence we must do some unpleasant footwork. First we transform the term a bit.

σγ ω((x ˙ 0) ⊗ βγ = (0 ..., 0, 1, 0,..., 0)ω((x ˙ 0) ⊗ (bγ1,...,bγn) γ γ

33 γ γ = (ω1 (x ˙ 0),...,ωn(x ˙ 0)) ⊗ (bγ1,...,bγn) γ γ = ωβ (x ˙ 0) σβ ⊗ (bγ1,...,bγn) γ γ γ = σβ ⊗ (ωβ (x ˙ 0)bγ1,...,ωβ (x ˙ 0)bγn) γ

γ For the rest of the calculation we will suppress the argument for ωβ (x ˙ 0) writing γ it as ωβ . In the same way that

σ (bαβ) ←→ σα ⊗ βα α σ ←→ σα ⊗ (bα1,...,bαn) α we have

γ γ σ γ γ (ωα )(βγβ) = (ωα βγβ) ←→ σα ⊗ (ωα bγ1,...,ωα bγn) α Thus the second term in the above sum is

γ γ γ [σ(0), σα ⊗ (ωα bγ1,...,ωα bγn) ] = [σ(0), −(ωα βγβ)] α γ Thus if we write Dx˙ 0 b out in matrix form and give ωα its argumentx ˙ 0 back, we get

dbαβ γ γ Dx˙ 0 b = [σ0, − ωα bγβ(x ˙ 0) − bαγωβ (x ˙ 0) ] dt j ∂bαβ du γ j γ j = [σ0, j − Γαj bγβu˙ 0 − bαγ Γβju˙ 0 ] ∂u dt duj = [σ , b ] 0 αβ|j dt where ∂b b = αβ − Γγ b − b Γγ αβ|j ∂uj αj γβ αγ βj and thus j Db = bαβ|jdu I hope it is now clear that the preceding calculations are general enough so that we see the general pattern. That is

αβ...δ ρ σ υ T = [σ0,Tρ σ...υ σα ⊗ σβ ⊗ ... ⊗ σδ ⊗ σ ⊗ σ ⊗ ... ⊗ σ ] αβ...δ ρ σ υ j DT = [σ0,Tρ σ...υ|j σα ⊗ σβ ⊗ ... ⊗ σδ ⊗ σ ⊗ σ ⊗ ... ⊗ σ ⊗ du ]

34 where ∂T αβ...δ T αβ...δ = ρσ...υ +Γα T θβ...δ + ... +Γα T αβ...θ ρσ...υ|j ∂uj θj ρ σ...υ θj ρ σ...υ θ αβ...δ θ αβ...δ − ΓρjTθ σ...υ − ... − ΓυjTρ σ...θ Thus we have re-derived the formulas from the Classical Tensor Calculus. One more matter remains for this section. I wish to show that scalar func- tions form and associated bundle also. This is a totally trivial matter, merely a matter of chasing definitions, but it is important to show that from our defi- nitions we get Df = df. As you may already have guessed, the representation space will be V = R and the representation will be trivial; that is ρ(g) = Identity for all g ∈ G. Thus a function maybe be regarded as a section

f = [σ, f] f(x) = [σ(x), f(x)] and [σ, f] = [σg, ρ(g−1)f] = [σg, f] so the choice of basis has no effect on the scalar function. Let x(t) be a curve in U. Setting, for convenience, f(t)= f(x(t)) we have

θt[σ(0), f(0)] = [˜σ(t), f(0)] = [σ(t)g, f(0)] = [σ(t), f(0)] so −1 θt [σ(0), f(t)] = [σ(0), f(t)] and we have for the curve x(t)

−1 θt [σ(t), f(t)] − [σ(0, f(0)] Dx˙ 0 = lim t→0 t [σ(0), f(t)] − [σ(0, f(0)] = lim t→0 t d = [σ(0), f(x(t)] dt t=0 ∂f dui = [σ(0), ] ∂ui dt t=0 ∂f i = [σ(0), i u˙ (0)] ∂u t=0 ∂f = [σ(0), dui(x ˙ )] ∂ui 0 = [σ(0), df(x ˙ 0)]

35 Thus

Dx˙ 0 f = df(x ˙ 0) Df = df which is the result we wished to establish. It is important to realize here that we are taking covariant derivatives of real values functions on M. Later we will consider functions on P and the formula Df = df will not hold for this more general case. If the function f is constant on the fibres we will of course be back to this case and the formula will again hold.

11 Tensors of type Ad and the Curvature Ten- sor

As we saw, linear transformations live in E⊗E∗ and are tensors that use the representation ρ(g)a = gag−1 for g ∈ M(n, R. Because this is the formula for the Adjoint (left) action of a Matrix Lie Group on it’s Lie Algebra, these tensors care called Tensors of type Ad. Recall that tensor of type Ad form an associated tensor bundle of the Principal Bundle P with the formula

[σ, a] = [σgg−1,a] = [σg,ρ(g−1a] = [σg,g−1ag]

The most important of these tensors is the curvature tensor Ω. (The cur- vature tensor Ω is actually an element of E⊗E∗ ⊗ T (M) ⊗ T (M)∗ but we are interested at this point only in the E⊗E∗ part of it. Recall that 1 Ω = R β e ⊗ eα ⊗ duk ∧ dul 2 α kl β β α k l = Rα kleβ ⊗ e ⊗ du ∧ du α,β,k

36 As part of our review we will again derive the change of basis formula for ω. For this we will use only

Dσ = σω and D(fv)= df v + fDv for σ = (σ1,...,σn) ∈ P (a basis of sections), v a section over M and f a function on M. The we have, for σ′ = σg,

σ′ω′ = Dσ′ = D(σg) = σ dg + (Dσ)g σgω′ = σ dg + (σω)g = σ(dg + ωg) gω′ = dg + ωg ′ −1 −1 −1 ω = g dg + g ωg = ωG + Ad(g )ω where ωG is the fundamental left invariant 1-form on G and Ad is the adjoint action of G on its Lie Algebra g introduced above. For the remainder of this section we will use only the formula

ω′ = g−1dg + g−1ωg relating the ω′ for σ′ to the ω for σ. It is interesting how much can be pulled out of this formula. Let’s rewrite the formula in the forms

gω′ = dg + ωg and dg = gω′ − ωg

Recall that for a wedge product of the forms ω ∈ Λr and η ∈ Λs we have

d(ω ∧ η)= dω ∧ η + (−1)rω ∧ dη

Using this on the previous formula (where g ∈ Λ0) we have

dg ∧ ω′ + g ∧ dω′ = ddg + (dω) ∧ g − ω ∧ dg

Note the critical sign shift on the right hand side, and note that it is due to ω′ coming second in the product gω′ but ω comes first in the product ωg. Next we have, using dg = gω′ − ωg

(gω′ − ωg)ω′ + g ∧ dω′ = 0+(dω)g − ω ∧ (gω′ − ωg) gω′ ∧ ω′ − ωg ∧ ω′ + gdω′ = (dω)g − ω ∧ gω′ + ω ∧ ωg

Now note that ωg ∧ ω′ = ω ∧ gω′ by the associativity of matrix multiplication and the fact that functions commute with wedge. Hence the terms cancel and we have

g(dω′ + ω′ ∧ ω′) = (dω + ω ∧ ω)g dω′ + ω′ ∧ ω′ = g−1(dω + ω ∧ ω)g so that dω +ω ∧ω is a tensor of type Ad. This suggests that dω +ω ∧ω might be something important, and of course we know it is the curvature tensor Ω. The

37 point is that one of the ways of knowing what is important is keeping an eye on the transformation properties. Notice also that we derived the transformation properties of dω + ω ∧ ω solely from ω′ = g−1dg + g−1ωg. Now we will develop some theory for the curvature tensor Ω. We will begin from Dσ = σω. Then by the definition of D we have

D(Dσ)= D(σω)= σD(ω)

But we have

D(Dσ) = D(σω) = σdω + (Dσ) ∧ ω = σdω + (σω) ∧ ω = σ(dσ + ω ∧ ω)

Comparing we see

D2σ = σD(ω)= σ(dω + ω ∧ ω)= σΩ and thus D(ω)= dω + ω ∧ ω =Ω None of these formulas are new be we have derived them very efficiently here. We also have

dΩ = d(dω + ω ∧ ω) = ddω + d(ω ∧ ω) = 0+(dω) ∧ ω − ω ∧ dω = (Ω − ω ∧ ω) ∧ ω − ω ∧ (Ω − ω ∧ ω) = Ω ∧ ω − (ω ∧ ω) ∧ ω − ω ∧ Ω+ ω ∧ (ω ∧ ω) = Ω ∧ ω − ω ∧ Ω = [Ω,ω] which is an important formula. Notice that there are no exterior differentials in this formula; they have cancelled out. To indicate what I mean more explicitly, α β ω is expressed in terms of Γβi and Ω in terms of Rα kl and dΩ is a complicated linear combination of products of these terms, with no additional derivatives. Let us now generalize to a tensor T whose expression in matrix form is an α n × n matrix of k-forms tσβ . Thus in terms of the basis σ, T corresponds to α α k tσ = (tσβ ) and tσβ ∈ Λ (T ∗ (M)). We now assume that T is a tensor of type Ad. This means that for σ′ = σg

−1 −1 tσ′ = Ad(g )tσ = g tσg

We are now going to duplicate the calculation we made with Ω which will work since both ω and T are tensors of type Ad and the k manages to slip through the calculation. We begin as before

gtσ′ = tσg

38 and take the exterior derivative of both sides

d(gtσ′ ) = d(tσg) k dg ∧ tσ′ + g dtσ′ = (dtσ)g + (−1) tσ ∧ dg using the d(η ∧ ω) = (dη) ∧ ω + (−1)deg η η ∧ dω formula again. Recalling that ω′ = g−1dg + g−1ωg we see that dg = gω′ − ωg and then we have ′ k ′ (gω − ωg) ∧ tσ′ + g dtσ′ = (dtσ )g + (−1) tσ ∧ (gω − ωg) ′ k+1 k ′ g(dtσ′ + ω ∧ tσ′ ) − ωg ∧ tσ′ = (dtσ + (−1) tσ ∧ ω)g + (−1) tσ ∧ gω Now, using the fact that T is of type Ad, ′ −1 k+1 k −1 ′ g(dtσ′ + ω ∧ tσ′ ) − ωg ∧ g tσg = (dtσ + (−1) tσ ∧ ω)g + (−1) gtσ′ g ∧ gω There is some cancellation of g and g−1 due to the associativity of matrix multiplication. We now move each of the second terms to the opposite side to get ′ k ′ k+1 g(dtσ′ + ω ∧ tσ′ − (−1) tσ′ ∧ ω ) = (dtσ + (−1) tσ ∧ ω + ω ∧ tσ)g ′ k ′ −1 k dtσ′ + ω ∧ tσ′ − (−1) tσ′ ∧ ω = g (dtσ + ω ∧ tσ − (−1) tσ ∧ ω)g k showing that dtσ + ω ∧ tσ − (−1) tσ ∧ ω is a tensor of type Ad. In fact it is Dtσ as can be easily shown.

As a first application, Ωσ is a tensor of type Ad. Hence 2 DΩσ = dΩσ + ω ∧ Ωσ − (−1) Ω ∧ ω

= dΩσ + [ωσ, Ωσ]

But we earlier showed that dΩσ = [Ωσ,ωσ] so

DΩσ = [Ωσ,ωσ] + [ωσ, Ωσ]

= Ωσ ∧ ωσ − ωσ ∧ Ωσ + ωσ ∧ Ωσ − Ωσ ∧ ωσ = 0 Hence DΩ= σDΩσ =0

Now we want a general formula for D2T , where T is a tensor of type Ad represented by a matrix of k-forms as before. Since the formulas involve tensors we may as well use an invariant notation. We have DT = dT + ω ∧ T − (−1)kT ∧ ω D(DT ) = d(DT )+ ω ∧ DT − (−1)k+1DT ∧ ω = d(dT + ω ∧ T − (−1)kT ∧ ω)+ ω ∧ (dT + ω ∧ T − (−1)kT ∧ ω) −(−1)k+1(dT + ω ∧ T − (−1)kT ∧ ω) ∧ ω = 0+ dω ∧ T − ω ∧ dT − (−1)kdT ∧ ω − (−1)2kT ∧ dω +ω ∧ dT + ω ∧ ω ∧ T − (−1)kω ∧ T ∧ ω −(−1)k+1dT ∧ ω − (−1)k+1ω ∧ T ∧ ω + (−1)k+1+kT ∧ ω ∧ ω

39 This gives us 10 terms (including the 0). Moving terms 8 and 9 into positions after terms 3 and 7 and simplifying the (-1) factors gives us

D2T = dω ∧ T − ω ∧ dT − (−1)kdT ∧ ω + (−1)kdT ∧ ω + ω ∧ dT + ω ∧ ω ∧ T − (−1)kω ∧ T ∧ ω + (−1)kω ∧ T ∧ ω −T ∧ ω ∧ ω

Removing the cancelling terms we have

D2T = (dω + ω ∧ ω) ∧ T − T ∧ (dω + ω ∧ ω) = Ω ∧ T − T ∧ Ω = [Ω,T ]

This shows how intricately the curvature Ω is intertwined with the second co- variant derivative of forms of type Ad.

40