cartan homotopy formula
play

Cartan Homotopy Formula Qingyuan Bai Last update: July 23, 2019 - PDF document

Cartan Homotopy Formula Qingyuan Bai Last update: July 23, 2019 Abstract We give a proof for Cartan homotopy formula concerning the Lie derivative on a smooth manifold M and then explain how it fits into the representation theory of


  1. Cartan Homotopy Formula Qingyuan Bai Last update: July 23, 2019 Abstract We give a proof for Cartan homotopy formula concerning the Lie derivative on a smooth manifold M and then explain how it fits into the representation theory of Gerstenhaber algebras. 1 Lie derivative on smooth manifold Recall that for any smooth manifold M , the global sections of its tangent bundle Γ( TM ) is a vector space over R . We can make it into a Lie algebra by the following defition: for any X, Y ∈ Γ( TM ), their Lie bracket is defined to be [ X, Y ] := XY − Y X To make sense of this expression, we need to identify Γ( TM ) with the space of derivations D := { T ∈ End R ( C ∞ ( M )) | T ( fg ) = ( Tf ) g + ( Tg ) f } Then for any two linear operators X and Y , XY − Y X is still a linear operator. Moreover, if they are both derivations, XY − Y X will also be a derivation. The fact that ( D, [ ]) is a Lie algebra follows from the fact that taking commutator makes End ( V ) into a Lie algebra for any vector space V . We can compute this under a coordinate chart ( U, u 1 , ..., u n ) � ( XY − Y X ) u i = x j ∂ j y i − y j ∂ j x i j so � � XY − Y X = ( x j ∂ j y i − y j ∂ j x i ) ∂ i i j Meanwhile, there is a geometric approach to this Lie bracket, or ‘Lie derivative’, to be more precise. We view a global section X as a vector field on M and integrate it. For each point p ∈ M , there is a neighbourhood U of p and ǫ > 0 such that there is a family of open embeddings: Φ t : U → M, ∀| t | ≤ ǫ 1

  2. and for any q ∈ U d dt Φ t ( q ) = X (Φ t ( q )) To define Lie derivatives, we use this flow of X to flow functions, vector fields, differential forms... etc. and differentiate. For example, we define Lie derivative of vector field Y along X at q to be L X Y ( q ) := d dtD Φ − 1 t ( Y (Φ t ( q )) This gives a global section of TM (think through why it is smooth). For any form ω on M , L X ω is defined likewise. To give an explicit formula for L X , we first consider the case of vector fields: Theorem 1. L X ( Y ) = [ X, Y ] . Sketch of proof. Carry out the computation under a local chart. Then for 0-forms (functions) Theorem 2. L X ( f ) = Xf Sketch of proof. This is more or less the deifinition of vector fields and flows (it might depend on how you define tangent vectors). To compute the Lie derivatives of general n -forms, we first introduce interior multiplication (sometimes called contraction) ι v . 1 For a vector space V and v ∈ V , we have a map ι v : ( V ∗ ) ⊗ k → ( V ∗ ) ⊗ k − 1 f ( · ) �→ f ( v, · ) Here we identified ( V ∗ ) ⊗ k with k -linear functions on V . More generally, given v ∈ � k V , we have ˆ l l − k V ∗ → � � V ∗ ι ˆ v : where ι ˆ v ( ω )(ˆ u ) := ω (ˆ v ∧ ˆ u ) v : � • V ∗ → � • V ∗ is called the inner multiplication with ˆ This map ι ˆ v . Here are some basic propositions of interior multiplication: • ι 2 v = 0 ˆ v ( ω ) ∧ γ + ( − 1) | ω || ˆ v | ω ∧ ι ˆ • ι ˆ v ( ω ∧ γ = ι ˆ v ( γ ) This construction can be carried out on manifolds by evaluating everything X ∈ Γ( � k TM ) and ω ∈ Ω l ( M ) = Γ( � l T ∗ M ), ι ˆ pointwise. For ˆ X ( ω ) ∈ Ω l ( M ) is given by X ( ω ) | p = ι ˆ X | p ( ω | p ) ι ˆ (again, think through why this is smooth.) 1 For a vector space V , we often make the useful identification � k V ∗ = ( � k V ) ∗ . 2

  3. Theorem 3. Let X ∈ Γ( TM ) be a vector field, the following holds: • L X : Ω • ( M ) → Ω • ( M ) is a derivation. • L X commutes with d . • (Cartan homotopy formula) L X = dι X + ι X d . • (Leibniz rule) For ω ∈ Ω k ( M ) and Y i ∈ Γ( TM ) , � L X ( ω ( Y 1 , ..., Y k )) = ( L X ω )( Y 1 , ..., Y k ) + ω ( Y 1 , ..., L X ( Y i ) , ...Y k ) . Sketch of proof. The 1st and the 4th claim can be checked directly by compu- tation under local charts. The ‘reason’ that they should hold is that wedge and pairing operation is realized by multiplication of functions under local charts, and we have Leibniz rule for multiplication of functions. The 2nd claim follows from the fact that d commutes with pullback. It remains to prove Cartan homotopy formula. The crucial observation is that both sides of the equation ‘commute’ with ∧ and d . Also note that the formula holds on C ∞ ( M ). Locally, if U is a coordinate chart. Ω k ( U ) is generated by { du i 1 ∧ ... ∧ du i k } over C ∞ ( U ), so it suffices to show that the formula holds for elements that take form fdu i 1 ∧ ... ∧ du i k . This follows from our observations above. Note that the Leibniz rule above in fact gives a global formula for Lie deriva- tive. As an application, we prove the global formula for exterior derivatives: Corollary 1. Assume k ≥ 1 , for ω ∈ Ω k ( M ) and Y i ∈ Γ( TM ) , we have: ( − 1) i Y i ω ( Y 0 , ..., ˆ � ( dω )( Y 0 , ..., Y k ) = Y i , ..., Y k ) i � ( − 1) i + j ω ([ Y i , Y j ] , Y 0 , ..., ˆ Y i , ..., ˆ + Y j , ..., Y k ) i<j Sketch of proof. First we prove it for 1-forms. We only need to check it locally. As in the proof for Cartan homotopy formula, fix a coordinate chart U , we may assume ω = fdg , where f, g ∈ ( U ). Compute directly: For general case: ( dω )( Y 0 , ..., Y k ) = ( ι Y 0 dω )( Y 1 , ..., Y k ) = ( L Y 0 ω − dι Y 0 ω )( Y 1 , ..., Y k ) The first term can be computed from global formula for Lie derivative, and the second term can be computed from global formula for d on k − 1 forms. We might conclude the proof by induction. 3

  4. 2 Cartan homotopy formula for Gerstenhaber algebra A Gerstenharber algebra is a graded-commutative algebra with a Lie bracket of degree − 1 satisfying Poisson identity. To be more precise, let A be a graded algebra, with its original multiplication written as ab and Lie bracket written as [ a, b ], we require the following: • | ab | = | a || b | (multiplication is of degree 0) • ab = ( − 1) | ab | ba (multiplication is graded-commutative) • ( ab ) c = a ( bc ) (multiplication is associatitve) • | [ a, b ] | = | a | + | b | − 1 (Lie bracket is of degree -1) • [ a, b ] = − ( − 1) ( | a |− 1)( | b |− 1) [ b, a ](Lie bracket is graded antisymmetric) • [ a, bc ] = [ a, b ] c + ( − 1) ( | a |− 1) | b | b [ a, c ](Poisson identity) • [ a, [ b, c ]] = [[ a, b ] , c ] + ( − 1) ( | a |− 1)( | b |− 1) [ b, [ a, c ]](Jacobi identity) Roughly speaking, these conditions say that the grading, multiplication and Lie bracket structures on A are compatible with each other. Example 1. Let ( l, [ , ]) be a Lie algebra. On � • l , there is a graded multipli- cation ∧ . Moreover, the Lie bracket extends to maps: m n m + n − 1 � � � l × l → l such that � ( − 1) i + j [ v i , v j ] ∧ v 1 ∧ ... ∧ u n [ v 1 ∧ ... ∧ v m , u 1 ∧ ... ∧ u n ] = i,j and we set [ k, ˆ v ] = 0 for any k ∈ R . Then ( � • l, ∧ , []) is a Gerstenhaber algebra. Example 2. For a smooth manifold M , we can extend [ , ] on Γ( TM ) to Γ( � • TM ) likewise. However, we require [ X, f ] := Xf and extend this expression to all of Γ( � • TM ) by Poisson identity. Then (Γ( � • TM ) , ∧ , []) is a Gerstenhaber algebra. Given a graded commutative algebra B , equipped with a multiplication and a bracket by elements of A . We say that it is a representation of Gerstenharber algebra if all the requirements above still holds for multiplication and bracket by elements of A . The question is that: if B also comes with a differential, how do we require compatibility of product and multiplication structure and the differential? 4

Download Presentation
Download Policy: The content available on the website is offered to you 'AS IS' for your personal information and use only. It cannot be commercialized, licensed, or distributed on other websites without prior consent from the author. To download a presentation, simply click this link. If you encounter any difficulties during the download process, it's possible that the publisher has removed the file from their server.

Recommend


More recommend