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Unformatted text preview: Ma1c Analytic Recitation 4/16/09 1 The Fundamental Group I mentioned last time about topologies and how you could define your open sets to be strange things, as long as they satisfied some axioms. Well, that stuff is certainly interesting, but topologists are mostly concerned with differentiating spaces with their usual topologies. For example, the two topological spaces R 2 and R 2 \ (0 , 0) (with their usual topologies) are clearly intuitively not the same space. However, how do you tell them apart? One common thing that topologists do is to find the fundamental group of a space. If the fundamental groups are different, then the spaces must be different. To define the fundamental group, pick a point x in your space X . Then consider all loops in the space which start and end at x . Now this number is clearly hugely uncountable, but we dont really care if you wiggle a loop aroundthe important thing is how it wraps around the space. Therefore, consider all loops up to smooth deformation. This is called up to homotopy. Notice that in the case of R 2 , there is only one loop up to deformationthe constant loop at x . However, in R 2 \ (0 , 0) there are lots up loops. Any two loops which wrap a different number of times around (0 , 0) cannot be deformed into each other. Notice also that there are two directions in which loops can wrap. Therefore, there is a bijection between the integers and loops in R 2 \ (0 , 0). In fact, the map is more that a bijectionyou can compose loops (simply do one and then the other), and composition of loops corresponds to addition of integers. Therefore we say that the fundamental group of R 2 is trivial and the fundamental group of R 2 \ (0 , 0) is Z (the integers). 2 The Chain Rule Obviously, there should be some kind of chain rule for partial derivatives. Theorem 8.8 gives you a nice one: If r : R R n and f : R n R , and g = f r , then g ( t ) = f ( a ) r ( t ), where a = r ( t ). 2.1 Example Let r ( t ) = ( a cos( t ) ,a sin( t )), let f = x 2- y 2 , and g = f r . Then f ( r ( t )) = (2 x,- 2 y ) | r ( t ) = (2 a cos( t ) ,- 2 a sin( t )). We also have r ( t ) = (- a sin( t ) ,a cos( t )). Therefore, g ( t ) = f ( r ( t )) r ( t ) = (2 a cos( t ) ,- 2 a sin( t )) (- a sin( t ) ,a cos( t )) =- 2 a 2 cos( t )sin( t )- 2 a 2 cos( t )sin( t ) =- 2 a 2 sin(2 t ) If you picture in your head the saddle and then the path that g ( t ) takes, you will see that this answer is intuitively correct....
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