We define \u03b9 v \u03b2 \u03b2 v R for a 1 covector \u03b2 on V and \u03b9 v \u03b2 0 for a 0 covector \u03b2 a

# We define ι v β β v r for a 1 covector β on v and

• morvarid2191
• 430
• 50% (2) 1 out of 2 people found this document helpful

This preview shows page 246 - 248 out of 430 pages.

We define ι v β = β ( v ) R for a 1-covector β on V and ι v β = 0 for a 0-covector β (a constant) on V . Proposition 20.7. For 1 -covectors α 1 ,..., α k on a vector space V and v V, ι v ( α 1 ∧···∧ α k )= k i = 1 ( 1 ) i 1 α i ( v ) α 1 ∧···∧ hatwide α i ∧···∧ α k , where the caret hatwide over α i means that α i is omitted from the wedge product. Proof. parenleftBig ι v parenleftBig α 1 ∧···∧ α k parenrightBigparenrightBig ( v 2 ,..., v k ) = parenleftBig α 1 ∧···∧ α k parenrightBig ( v , v 2 ,..., v k ) = det α 1 ( v ) α 1 ( v 2 ) ··· α 1 ( v k ) α 2 ( v ) α 2 ( v 2 ) ··· α 2 ( v k ) . . . . . . . . . . . . α k ( v ) α k ( v 2 ) ··· α k ( v k ) ( Proposition 3.27 ) = k i = 1 ( 1 ) i + 1 α i ( v ) det [ α ( v j )] 1 k ,ℓ negationslash = i 2 j k ( expansion along first column ) = k i = 1 ( 1 ) i + 1 α i ( v ) parenleftBig α 1 ∧···∧ hatwide α i ∧···∧ α k parenrightBig ( v 2 ,..., v k ) (Proposition 3.27) . Proposition 20.8. For v in a vector space V, let ι v : logicalandtext ( V ) logicalandtext ∗− 1 ( V ) be interior multiplication by v. Then (i) ι v ι v = 0 , (ii) for β logicalandtext k ( V ) and γ logicalandtext ( V ) , ι v ( β γ )=( ι v β ) γ +( 1 ) k β ι v γ . In other words, ι v is an antiderivation of degree 1 whose square is zero. Proof. (i) Let β logicalandtext k ( V ) . By the definition of interior multiplication, ( ι v ( ι v β ))( v 3 ,..., v k )=( ι v β )( v , v 3 ,..., v k )= β ( v , v , v 3 ,..., v k )= 0 , because β is alternating and there is a repeated variable v among its arguments.
228 § 20 The Lie Derivative and Interior Multiplication (ii) Since both sides of the equation are linear in β and in γ , we may assume that β = α 1 ∧···∧ α k , γ = α k + 1 ∧···∧ α k + , where the α i are all 1-covectors. Then ι v ( β γ ) = ι v ( α 1 ∧···∧ α k + ) = parenleftBigg k i = 1 ( 1 ) i 1 α i ( v ) α 1 ∧···∧ hatwide α i ∧···∧ α k parenrightBigg α k + 1 ∧···∧ α k + +( 1 ) k α 1 ∧···∧ α k k i = 1 ( 1 ) i + 1 α k + i ( v ) α k + 1 ∧···∧ hatwidest α k + i ∧···∧ α k + (by Proposition 20.7) = ( ι v β ) γ +( 1 ) k β ι v γ . Interior multiplication on a manifold is defined pointwise. If X is a smooth vector field on M and ω Ω k ( M ) , then ι X ω is the ( k 1 ) -form defined by ( ι X ω ) p = ι X p ω p for all p M . The form ι X ω on M is smooth because for any smooth vector fields X 2 ,..., X k on M , ( ι X ω )( X 2 ,..., X k )= ω ( X , X 2 ,..., X k ) is a smooth function on M (Proposition 18.7(iii) (i)). Of course, ι X ω = ω ( X ) for a 1-form ω and ι X f = 0 for a function f on M . By the properties of interior multiplication at each point p M (Proposition 20.8), the map ι X : Ω ( M ) Ω ( M ) is an antiderivation of degree 1 such that ι X ι X = 0.

#### You've reached the end of your free preview.

Want to read all 430 pages?