LectureNotes14G

LectureNotes14G - Math 6455 Nov 1, 2006 1 Differential...

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Unformatted text preview: Math 6455 Nov 1, 2006 1 Differential Geometry I Fall 2006, Georgia Tech Lecture Notes 14 Connections Suppose that we have a vector field X on a Riemannian manifold M . How can we measure how much X is changing at a point p M in the direction Y p T p M ? The main problem here is that there exists no canonical way to compare a vector in some tangent space of a manifold to a vector in another tangent space. Hence we need to impose a new kind of structure on a manifold. To gain some insight, we first study the case where M = R n . 0.1 Differentiation of vector fields in R n Since each tangent space T p R n is canonically isomorphic to R n , any vector field on R n may be identified as a mapping X : R n R n . Then for any Y p T p R n we define the covariant derivative of X with respect to Y p as Y p X := ( Y p ( X 1 ) ,...,Y p ( X n ) ) . Recall that Y p ( X i ) is the directional derivative of X i at p in the direction of Y , i.e., if : (- , ) M is any smooth curve with (0) = p and (0) = Y , then Y p ( X i ) = ( X i ) (0) = grad X i ( p ) ,Y . The last equality is an easy consequence of the chain rule. Now suppose that Y : R n R n is a vector field on R n , p Y 7- Y p , then we may define a new vec- tor field on R n by ( Y X ) p := Y p X. Then the operation ( X,Y ) 7- X Y may be thought of as a mapping : X ( R n ) X ( R n ) X ( R n ), where X denotes the space of vector fields on R n . Next note that if X X ( R n ) is any vector field and f : M R is a function, then we may define a new vector field fX ( R n ) by setting ( fX ) p := f ( p ) X p (do not confuse fX , which is a vector field , with Xf which is a function defined by Xf ( p ) := X p ( f )). Now we observe that the covariant differentiation of vector fields on R n satisfies the following properties: 1. Y ( X 1 + X 2 ) = Y X 1 + Y X 2 1 Last revised: November 14, 2006 1 2. Y ( fX ) = ( Y f ) Y X + f Y X 3. Y 1 + Y 2 X = Y 1 X + Y 2 X 4. fY X = f Y X It is an easy exercise to check the above properties. Another good exercise to write down the pointwise versions of the above expressions. For instance note that item (2) implies that Y p ( fX ) = ( Y p f ) Y p X + f ( p ) Y p X, for all p M . 0.2 Definition of connection and Christoffel symbols Motivated by the Euclidean case, we define a connection on a manifold M as any mapping : X ( M ) X ( M ) X ( M ) which satisfies the four properties mentioned above. We say that is smooth if whenever X and Y are smooth vector fields on M , then Y X is a smooth vector field as well. Note that any manifold admits the trivial connection 0. In the next sections we study some nontrivial examples....
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LectureNotes14G - Math 6455 Nov 1, 2006 1 Differential...

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