lecture1

LECTURE 1: DIFFERENTIAL FORMS 1.1-forms on Rn In calculus, you may have seen the differential orexterior derivative dfof a functionf(x,y,z ) defined to be df=∂f ∂xdx+∂f ∂ydy+∂f ∂zdz. The expression dfis called a 1-form. But what does this really mean? Definition: Asmooth 1-form φonRnis a real-valued function on the set of all tangent vectors to Rn, i.e., φ:TRn→R with the properties that 1.φis linear on the tangent space TxRnfor eachx∈Rn. 2. For any smooth vector field v=v(x), the function φ(v) :Rn→Ris smooth. Given a 1-form φ, for eachx∈Rnthe map φx:TxRn→R is an element of the dual space ( TxRn)∗. When we extend this notion to all ofRn, we see that the space of 1-forms on Rnis dual to the space of vector fields on Rn. In particular, the 1-forms dx1,... ,dxnare defined by the property that for any vector v= (v1,... ,vn)∈TxRn, dxi(v) =vi. Thedxi’s form a basis for the 1-forms on Rn, so any other 1-form φmay be expressed in the form φ=n/summationdisplay i=1fi(x)dxi. If a vector field vonRnhas the form v(x) = (v1(x),... ,vn(x)), 1 2 JEANNE NIELSEN CLELLAND then at any point x∈Rn, φx(v) =n/summationdisplay i=1fi(x)vi(x). 2.p-forms on Rn The 1-forms on Rnare part of an algebra, called the algebra of differential forms onRn. The multiplication in this algebra is called wedge product , and it is skew-symmetric: dxi∧dxj=−dxj∧dxi. One consequence of this is that dxi∧dxi= 0. If each summand of a differential form φcontainspdxi’s, the form is called ap-form . Functions are considered to be 0-forms, and any form on Rnof degreep>n must be zero due to the skew-symmetry. A basis for the p-forms on Rnis given by the set {dxi1∧···∧dxip: 1≤i1<i2<···<ip≤n}. Anyp-formφmay be expressed in the form φ=/summationdisplay |I|=pfIdxi1∧···∧dxip whereIranges over all multi-indices I= (i1,... ,i p) of length p. Just as 1-forms act on vector fields to give real-valued functions, so p-forms act onp-tuples of vector fields to give real-valued functions. For instance, if φ,ψare 1-forms and v,ware vector fields, then (φ∧ψ)(v,w) =φ(v)ψ(w)−φ(w)ψ(v). In general, if φ1,... ,φ pare 1-forms and v1... ,v pare vector fields, then (φ1∧···∧φp)(v1,... ,v p) =/summationdisplay σ∈Spsgn(σ)φ1(vσ(1))φ2(vσ(2))···φn(vσ(n)). 3.The exterior derivative The exterior derivative is an operation that takes p-forms to (p+ 1)-forms. We will first define it for functions and then extend this definition to higher degree forms. Definition: Iff:Rn→Ris differentiable, then the exterior derivative of fis the 1-form dfwith the property that for any x∈Rn, v∈TxRn, dfx(v) =v(f), i.e.,dfx(v) is the directional derivative of fatxin the direction of v. LIE GROUPS AND THE METHOD OF THE MOVING FRAME 3 It is not difficult to show that df=n/summationdisplay i=1∂f ∂xidxi. The exterior derivative also obeys the Leibniz rule d(fg) =gdf+fdg and the chain rule d(h(f)) =h/prime(f)df. We extend this definition to p-forms as follows: Definition: Given ap-formφ=/summationdisplay |I|=pfIdxi1∧···∧dxip, the exterior deriv- ativedφis the (p+ 1)-form dφ=/summationdisplay |I|=pdfI∧dxi1∧···∧dxip. Ifφis ap-form andψis aq-form, then the Leibniz rule takes the form d(φ∧ψ) =dφ∧ψ+ (−1)pφ∧dψ. Very Important Theorem: d2= 0. i.e., for any differential form φ, d(dφ) = 0. Proof: First suppose that fis a function, i.e., a 0-form. Then d(df) =d(n/summationdisplay i=1∂f ∂xidxi) =/summationdisplay i,j∂2f ∂xi∂xjdxj∧dxi =/summationdisplay i<j(∂2f ∂xj∂xi−∂2f ∂xi∂xj)dxi∧dxj = 0 because mixed partials commute. Next, note that dxireally does mean d(xi), wherexiis theith coordinate function. So by the argument above, d(dxi) = 0. Now suppose that φ=/summationdisplay |I|=pfIdxi1∧···∧dxip. 4 JEANNE NIELSEN CLELLAND Then by the Leibniz rule, d(dφ) =d(/summationdisplay |I|=pdfI∧dxi1∧···∧dxip) =/summationdisplay |I|=p[d(dfI)∧dxi1∧···∧dxip−dfI∧d(dxi1)∧···∧dxip+...] = 0.2 Definition: Ap-formφisclosed ifdφ= 0. φisexact if there exists a (p−1)-formηsuch thatφ=dη. By the Very Important Theorem, every exact form is closed. The converse is only partially true: every closed form is locally exact. This means that given a closed p-formφon an open set U⊂Rn, any point x∈Uhas a neighborhood on which there exists a ( p−1)-formηwithdη=φ. 4.Differential forms on manifolds Given a smooth manifold M, asmooth 1-form φonMis a real-valued function on the set of all tangent vectors to Msuch that 1.φis linear on the tangent space TxMfor eachx∈M. 2. For any smooth vector field vonM, the function φ(v) :M→Ris smooth. So for each x∈M, the map φx:TxM→R is an element of the dual space ( TxM)∗. Wedge products and exterior derivatives are defined similarly as for Rn. If f:M→Ris a differentiable function, then we define the exterior derivative offto be the 1-form dfwith the property that for any x∈M, v∈TxM, dfx(v) =v(f). A local basis for the space of 1-forms on Mcan be described as before in terms of any local coordinate chart ( x1,... ,xn) onM, and it is possible to show that the coordinate-based notions of wedge product and exterior derivative are in fact independent of the choice of local coordinates and so are well-defined. More generally, suppose that M1,M2are smooth manifolds and that F: M1→M2is a differentiable map. For any x∈M1, the differential dF (also denoted F∗) :TxM1→TF(x)M2may be thought of as a vector-valued 1-form, because it is a linear map from TxM1to the vector space TF(x)M2. There is an analogous map in the opposite direction for differential forms, called the pullback and denoted F∗. It is defined as follows. LIE GROUPS AND THE METHOD OF THE MOVING FRAME 5 Definition: IfF:M1→M2is a differentiable map, then 1. Iff:M2→Ris a differentiable function, then F∗f:M1→Ris the function (F∗f)(x) = (f◦F)(x). 2. Ifφis ap-form onM2, thenF∗φis thep-form onM1defined as follows: ifv1,... ,v p∈TxM1, then (F∗φ)(v1,... ,v p) =φ(F∗(v1),... ,F ∗(vp)). In terms of local coordinates ( x1,... ,xn) onM1and (y1,... ,ym) onM2, suppose that the map Fis described by yi=yi(x1,... ,xn),1≤i≤m. Then the differential dFat each point x∈M1may be represented in this coordinate system by the matrix/bracketleftbigg ∂yi ∂xj/bracketrightbigg . Thedxj’s are forms on M1, thedyi’s are forms on M2, and the pullback mapF∗acts on the dyi’s by F∗(dyi) =n/summationdisplay j=1∂yi ∂xjdxj. The pullback map behaves as nicely as one could hope with respect to the various operations on differential forms, as described in the following theo- rem. Theorem: LetF:M1→M2be a differentiable map, and let φ,ηbe differential forms on M2. Then 1.F∗(φ+η) =F∗φ+F∗η. 2.F∗(φ∧η) =F∗φ∧F∗η. 3.F∗(dφ) =d(F∗φ). 5.The Lie derivative The final operation that we will define on differential forms is the Lie de- rivative. This is a generalization of the notion of directional derivative of a function. 6 JEANNE NIELSEN CLELLAND Suppose that v(x) is a vector field on a manifold M, and letϕ:M×(−ε,ε)→ Mbe the flow ofv. This is the unique map that satisfies the conditions ∂ϕ ∂t(x,t) =v(ϕ(x,t)) ϕ(x,0) =x. In other words, ϕt(x) =ϕ(x,t) is the point reached at time tby flowing along the vector field v(x) starting from the point xat time 0. Recall that if f:M→Ris a smooth function, then the directional derivative offatxin the direction of vis v(f) = lim t→0f(ϕt(x))−f(x) t = lim t→0(ϕ∗ t(f)−f)(x) t. Similarly, given a differential form φwe define the Lie derivative ofφalong the vector field v(x) to be Lvφ= lim t→0ϕ∗ tφ−φ t. Fortunately there is a practical way to compute the Lie derivative. First we need the notion of the left-hook of a differential form with a vector field. Given ap-formφand a vector field v, the left-hookvφofφwithv(also called the interior product ofφwithv) is the (p−1)-form defined by the property that for any w1,... ,w p−1∈TxRn, (vφ)(w1,... ,w p−1) =φ(v,w1,... ,w p−1). For instance, ∂ ∂x(dx∧dy+dz∧dx) =dy−dz. Now according to Cartan’s formula , the Lie derivative of φalong the vector fieldvis Lvφ=vdφ+d(vφ). Exercises 1. Classical vector analysis avoids the use of differential forms on R3by converting 1-forms and 2-forms into vector fields by means of the following one-to-one correspondences. ( ε1,ε2,ε3will denote the standard basis ε1= [1,0,0], ε2= [0,1,0], ε3= [0,0,1].) LIE GROUPS AND THE METHOD OF THE MOVING FRAME 7 f1dx1+f2dx2+f3dx3←→f1ε1+f2ε2+f3ε3 f1dx2∧dx3+f2dx3∧dx1+f3dx1∧dx2←→f1ε1+f2ε2+f3ε3 Vector analysis uses three basic operations based on partial differentiation: 1.Gradient of a function f: grad(f) =3/summationdisplay i=1∂f ∂xiεi 2.Curl of a vector field v=3/summationdisplay i=1vi(x)εi: curl(v) =/parenleftbigg∂v3 ∂x2−∂v2 ∂x3/parenrightbigg ε1+/parenleftbigg∂v1 ∂x3−∂v3 ∂x1/parenrightbigg ε2+/parenleftbigg∂v2 ∂x1−∂v1 ∂x2/parenrightbigg ε3 3.Divergence of a vector field v=3/summationdisplay i=1vi(x)εi: div(v) =3/summationdisplay i=1∂vi ∂xi Prove that all three operations may be expressed in terms of exterior deriva- tives as follows: 1.df↔grad(f) 2. Ifφis a 1-form and φ↔v, thendφ↔curl(v). 3. Ifηis a 2-form and η↔v, thendη↔div(v)dx1∧dx2∧dx3. Show that the identities curl(grad(f)) = 0 div(curl(v)) = 0 follow from the fact that d2= 0. 2. Letfandgbe real-valued functions on R2. Prove that df∧dg=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle∂f ∂x∂f ∂y ∂g ∂x∂g ∂y/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingledx∧dy. (You may recognize this from the change-of-variables formula for double integrals.) 3. Suppose that φ,ψare 1-forms on Rn. Prove the Leibniz rule d(φ∧ψ) =dφ∧ψ−φ∧dψ. 8 JEANNE NIELSEN CLELLAND 4. Prove the statement above that if F:M1→M2is described in terms of local coordinates by yi=yi(x1,... ,xn),1≤i≤m then F∗(dyi) =n/summationdisplay j=1∂yi ∂xjdxj. 5. Let (r,θ) be coordinates on R2and (x,y,z ) coordinates on R3. Let F:R2→R3be defined by F(r,θ) = (cosθ,sinθ,r). Describe the differential dFin terms of these coordinates and compute the pullbacksF∗(dx),F∗(dy),F∗(dz).