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This is known as Jacobi's identity. We can also derive it from the fact that [Y, Z] is a natural operation and hence for any one parameter group φt of dif- feomorphisms we have

φt*([Y, Z])=[φt*Y, φt*Z].

If X is the infinitesimal generator of φt then differentiating the preceding equa- tion with respect to t at t=0 gives

[X, [Y, Z]]=[[X, Y], Z]+[Y, [X, Z

In other words, X acts as a derivation of the (mutliplication“ given by Lie bracket. This is just Jacobi's identity when we use the antisymmetry of the bracket. In the future we we will have occasion to take cyclic sums such as those which arise on the left of Jacobi's identity. So if F is a function of three vector fields (or of three elements of any set) with values in some vector space (for example in the space of vector fields) we will define the cyclic sum CycF by

CycF(X, Y, Z):=F(X, Y, Z)+F(Y, Z, X)+F(Z, X, Y) .

With this definition Jacobi's identity becomes

Cyc[X, [Y, Z]]=0. (2.6)


2.13 Left invariant forms.

Let G be a group and M be a set. A left action of G on M consists of a map


satisfying the conditions

φ(a, φ(b, m =φ(ab, m)

(an associativity law) and

φ(e, m)=m, mM

where e is the identity element of the group. When there is no risk of confusion we will write am for φ(a, m).(But in much of the beginning of the following exercises there will be a risk of confusion since there will be several different actions of the same group G on the set M). We think of an action as assigning to each element aG a transformation, φa, of M:

φa:MM, φa:mφ(a, m) .

So we also use the notation

φam=φ(a, m) .



For example, we may take M to be the group G itself and let the action be left multiplication, L, so

L(a, m)=am.

We will write

La:GG, Lam=am.

We may may also consider the (left) action of right multiplication:

R : G×GG, R(a, m)=ma-1.

(The inverse is needed to get the order right in R(a, R(b, m))=R(ab, m So we will write

Ra:GG, Ram=ma-1.

We will be interested in the case that G is a Lie group, which means that G is a manifold and the multiplication map G×GG and the inverse map GG, aa-1 are both smooth maps. Then the differential, (dLa)m maps the tangent space to G at m, to the tangent space to G at am:


and similarly


In particular,

dLa-1: TGaTGe.

Let G=Gl(n) be the group of all invertible n×n matrices. It is an open subset (hence a submanifold) of the n2 dimensional space Mat (n) of all n×n matrices. We can think of the tautological map which sends every AG into itself thought of as an element of Mat(n) as a matrix valued function on G. Put another way, A is a matrix of functions on G, each of the matrix entries Aij of A is a function on G. Hence dA=(dA) is a matrix of differential forms (or, we may say, a matrix valued differential form). So we may consider


which is also a matrix valued differential form on G. Let B be a fixed element of G.

1. Show that

LB*(A-1dA)=A-1dA. (2.7)

So each of the entries of A-1dA is left invariant.

2. Show that

RB*(A-1dA)=B(A-1dA)B-1. (2.8)

So the entries of A-1dA are not right invariant (in general), but (2.8) shows how they are transformed into one another by right multiplication.



For any two matrix valued differential forms R=(Rij) and S=(Sij) define their matrix exterior product RS by the usual formula for matrix product, but with exterior multiplication of the entries instead of ordinary multiplication, so


Also, if R=(Rij) is a matrix valued differential form, define dR by applying d to each of the entries. So

(dR)ij:=(dRij) .

Finally, if ψ : XY is a smooth map and R=(Rij) is a matrix valued form on Y then we define its pullback by pulling back each of the entries:

(ψ*R)ij:=(ψ*Rij) .

2.14 The Maurer Cartan equations.

3. In elementary calculus we have the formula d(1/x)=-dx/x2. What is the generalization of this formula for the matrix function A-1. In other words, what is the formula for d(A-1) ?

4. Show that if we set ω=A-1dA then

dω+ωω=0. (2.9)

Here is another way of thinking about A-1dA: Since G=Gl(n) is an open subset of the vector space Mat (n) , we may identify the tangent space TGA with the vector space Mat (n) . That is we have an isomorphism between TGA and Mat (n) . If you think about it for a minute, it is the form dA which effects this isomorphism at every point. On the other hand, left multiplication by A-1 is a linear map. Under this identification, the differential of a linear map L looks just like L. So in terms of this identification, A-1dA, when evaluated at the tangent space TGA is just the isomorphism dLA-1 : TGATGI where I is the identity matrix.

2.15 Restriction to a subgroup

Let H be a Lie subgroup of G. This means that H is a subgroup of G and it is also a submanifold. In other words we have an embedding




which is a ( n injective) group homomorphism.Let


denote the tangent space to H at the identity element.

5. Conclude from the preceding discussion that if we now set


then ω takes values in h. In other words, when we evaluate ω on any tangent vector at any point of H we get a matrix belonging to the subspace h.

6. Show that on a group, the only transformations which commute with all the right multiplications, Rb, bG, are the left multiplications, La.

For any vector ξTHI, define the vector field X by


(Recall that RA-1 is right multiplication by A and so sends I into A.) For example, if we take H to be the full group G=Gl(n) and identify the tangent space at every point with Mat (n) then the above definition becomes


By construction, the vector field X is right invariant, i.e. is invariant under all the diffeomorphisms RB.

7. Conclude that the flow generated by X is left multiplication by a one param- eter subgroup. Also conclude that in the case H=Gl(n) the flow generated by X is left multiplication by the one parameter group


Finally conclude that for a general subgroup H, if ξh then all the exptξ lie in H.

8. What is the space h in the case that H is the group of Euclidean motions in three dimensional space, thought of as the set of all four by four matrices of the form

(Av01), AA=I, vR3?

2.16 Frames.

Let V be an n dimensional vector space. Recall that frame on V is, by defini- tion, an isomorphism f: RnV. Giving f is the same as giving each of the



vectors fi=f(δi) where the δi range over the standard basis of R. So giving a frame is the same as giving an ordered basis of V and we will sometimes write

f=(f1, ..., fn) .

If AGl(n) then A is an isomorphism of Rn with itself, so fo A-1 is another frame. So we get an action, R:Gl(n)×FF where F=F(V) denotes the space of all frames:

R(A, f)=fA-1. (2.10)

If f and g are two frames, then g-1 of =M is an isomorphism of Rn with itself, i.e. a matrix. So given any two frames, f and g, there is a unique MGl(n) so that g=fM-1. Once we fix an f, we can use this fact to identify F with Gl(n) , but the identification depends on the choice of f. But in any event the (non-unique) identification shows that F is a manifold and that (2.10) defines an action of Gl(n) on F. Each of the fi (the ith basis vector in the frame) can be thought of as a V valued function on F. So we may write

dfj=ωijfi (2.11)

where the ωij are ordinary (number valued) linear differential forms on F. We think of this equation as giving the expansion of an infinitesimal change in fj in terms of the basis f=(f1, ..., fn) . If we use the ""row“ representation of f as above, we can write these equations as

df=fω (2.12)

where ω=(ωij) .

9. Show that the ω defined by (2.12) satisfies

RB*ω=BωB-1. (2.13)

To see the relation with what went on before, notice that we could take V=Rn itself. Then f is just an invertible matrix, A and (2.12) becomes our old equation ω=A-1dA. So (2.13) reduces to (2.8).

If we take the exterior derivative of (2.12) we get


from which we conclude

dω+ωω=0. (2.14)

2.17 Euclidean frames.

We specialize to the case where V=Rn, n=d+1 so that the set of frames becomes identified with the group Gl(n) and restrict to the subgroup, H, of



Euclidean motions which consist of all n×m matrices of the form

(Av01), AO(d), vRd.

Such a matrix, when applied to a vector


sends it into the vector


and Aw+v is the orthogonal transformation A applied to w followed by the translation by v. The corresponding Euclidean frames (consisting of the columns of the elements of H) are thus defined to be the frames of the form

fi=(ei0), i=1, ...d,

where the ei form an orthonormal basis of Rd and


where vRd is an arbitrary vector. The idea is that v represents a choice of origin in d dimensional space and e=(e1, ..., ed) is an orthonormal basis. We can write this in shorthand notation as


If ι denotes the embedding of H into G, we know from the exercise 5 that




So the pull back of (2.12) becomes

d(ev01)=(eΩeθ00) (2.15)

or, in more expanded notation,

dej=iΩijei, dv=iθiei.



Let (, ) denote the Euclidean scalar product. Then we can write

θi=(dv, ei) (2.16)


(dej, ei)=Ωij.

If we set


this becomes

(dei, ej)=Θij. (2.17)

Then (2.14) becomes


Or, in more expanded notation,

dΘ=ΘΘ. (2.18)

dθi=jΘijθj, dΘik=jΘijΘjk. (2.19)

Equations (2.16)-(2.18) or (2.19) are known as the structure equations of

Euclidean geometry.

2.18 Frames adapted to a submanifold.

Let M be a k dimensional submanifold of R. This determines a submanifold of the manifold, H, of all Euclidean frames by the following requirements:

i) vM and

ii) eiTMv for ik. We will usually write m instead of v to emphasize the first requirement -that the frames be based at points of M. The second requirement says that the first k vectors in the frame based at m be tangent to M (and hence that the last n-k vectors in the frame are normal to M). We will denote this manifold by O(M) . It has dimension


The first term comes from the point m varying on M, the second is the dimension of the orthogonal group O(k) corresponding to the choices of the first k vectors in the frame, and the third term is dimO(d-k) correspond to the last (n-k) vectors. We have an embedding of O(M) into H, and hence the forms θ and Θ pull back to O(M) . As we are running out of letters, we will continue to denote these pull backs by the same letters. So the pulled back forms satisfy the same structure equations (2.16)-(2.18) or (2.19) as above, but they are supplemented by

θi=0, i>k. (2.20)



2.19 Curves and surfaces-their structure equa- tions.

We will be particularly interested in curves and surfaces in three dimensional Euclidean space. For a curve, C, the manifold of frames is two dimensional, and we have

dC = θ1e1 (2.21)

de1 = Θ12e2+Θ13e3 (2.22)

de2 = Θ21e1+Θ23e3 (2.23)

de3 = Θ31e1+Θ32e2. (2.24)

One can visualize the manifold of frames as a sort of tube: about each point of the curve there is a circle in the plane normal to the tangent line corresponding the possible choices of e2.

For the case of a surface the manifold of frames is three dimensional: we can think of it as a union of circles each centered at a point of S and in the plane tangent to S at that point. Then equation (2.21) is replaced by

dX=θ1e1+θ2e2 (2.25)

but otherwise the equations are as above, including the structure equations (2.19). These become

dθ1 = Θ12θ2 (2.26)

dθ2 = -Θ12θ1 (2.27)

0 = Θ31θ1+Θ32θ2 (2.28)

dΘ12 = Θ13Θ32 (2.29)

dΘ13 = Θ12Θ23 (2.30)

dΘ23 = Θ21Θ13 (2.31)

Equation (2.29) is known as Gauss' equation, and equations (2.30) and (2.31) are known as the Codazzi-Mainardi equations.

2.20 The sphere as an example.

In computations with local coordinates, we may find it convenient to use a ""cross-section“ of the manifold of frames, that is a map which assigns to each point of neighborhood on the surface a preferred frame. If we are given a parametrization m=m(u, v) of the surface, one way of choosing such a cross- section is to apply the Gram-Schmidt orthogonalization procedure to the tan- gent vector fields mu and mv, and take into account the chosen orientation.

For example, consider the sphere of radius R. We can parameterize the sphere with the north and south poles (and one longitudinal semi-circle) removed



by the (u, v)(0,2π)×(0, π) by X=X(u, v) where

X(u, v)=(RsinvcosuRsinusinvRcosv)

Here v denotes the angular distance from the north pole, so the excluded value v=0 corresponds to the north pole and the excluded value v=π corresponds to the south pole. Each constant value of v between 0 and π is a circle of latitude with the equator given by v=π2. The parameter u describes the longitude from the excluded semi-circle.

In any frame adapted to a surface in R3, the third vector e3 is normal to the surface at the base point of the frame. There are two such choices at each base point. In our sphere example let us choose the outward pointing normal, which at the point m(u, v) is

e3(m(u, v))=(sinvcosusinusinvcosv)

We will write the left hand side of this equation as e3(u, v) . The coordinates u, v are orthogonal, i.e. Xu and Xv are orthogonal at every point, so the or- thonormalization procedure amounts only to normalization: Replace each of these vectors by the unit vectors pointing in the same direction at each point. So we get

e1(u, v)=(-sinucosu0), e2(u, v)=(cosucosvsinucosv-sinv)

We thus obtain a map ψ from (0,2π)×(0, π) to the manifold of frames,

ψ(u, v)=(X(u, v), e1(u, v), e2, (u, v), e3(u, v

Since Xue1=Rsinv and Xve2=R we have

dX(u, v)=(Rsinvdu)e1(u, v)+(Rdv)e2(u, v) .

Thus we see from (2.25) that

ψ*θ1=Rsinvdu, ψ*θ2=Rdv

and hence that


Now R2sin vdudv is just the area element of the sphere expressed in u, v co- ordinates. The choice of e1, e2 determines an orientation of the tangent space to the sphere at the point X(u, v) and so ψ*(θ1θ2) is the pull-back of the corresponding oriented area form.



10. Compute ψ*Θ12, ψ*Θ13, and ψ*Θ23 and verify that


where K=1/R2 is the curvature of the sphere.

We will generalize this equation to an arbitrary surface in R3 in section ??.

2.21 Ribbons

The idea here is to study a curve on a surface, or rather a curve with an ""infinitesimal“ neighborhood of a surface along it. So let C be a curve and O(C) its associated two dimensional manifold of frames. We have a projection π : O(C)C sending every frame into its origin. By a ribbon based on C we mean a section n: CO(C) , so n assigns a unique frame to each point of the curve in a smooth way. We will only be considering curves with non-vanishing tangent vector everywhere. With no loss of generality we may assume that we have parametrized the curve by arc length, and the choice of e1 determines an orientation of the curve, so θ=ds. The choice of e2 at every point then de- termines e3 up to a±sign. So a good way to visualize s is to think of a rigid metal ribbon determined by the curve and the vectors e2 perpendicular to the curve (determined by n) at each point. The forms Θij all pull back under n to function multiples of ds:

n*Θ12=kds, n*Θ23=-τds, n*Θ13=wds (2.32)

where k, τ and w are functions of s. We can write equations (2.21)-(2.24) above as



de1ds=ke2+we3, de2ds=-ke1-τe3, de3ds=-we1+τe3. (2.33)

For later applications we will sometimes be sloppy and write Θij instead of n*Θij for the pull back to the curve, so along the ribbon we have Θ12=kds etc. Also it will sometimes be convenient in computations (as opposed to proving theorems) to use parameters other than arc length.

11. Show that two ribbons (defined over the same interval of s values) are congruent (that is there is a Euclidean motion carrying one into the other) if and only if the functions k, τ, and w are the same.

A ribbon is reallyjust a curve in the space, H, of all Euclidean frames, having the property that the base point, that is the v of the frame (v, e1, e2, e3) has non- vanishing derivative. The previous exercise says that two curves, i:IH and j : IH in H differ by an overall left translation (that is satisfy j=Lhi) if and only if the forms θ, Θ12, Θ13, Θ23 pull back to the same forms on I. The