LectureNotes11G - Math 6455 Oct 2, 2006 1 Differential...

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Unformatted text preview: Math 6455 Oct 2, 2006 1 Differential Geometry I Fall 2006, Georgia Tech Lecture Notes 11 Orientability Any ordered basis ( b 1 ,...,b n ) of R n may be viewed as a matrix B GL ( n ) whose i th column is b i . Thus the set of ordered basis of R n are in one-to-one correspondence with elements of GL ( n ), and so we may partition them into two subsets: those whose corresponding matrices have positive determinant, GL ( n ) + , and those whose corresponding matrices have negative determinant, GL ( n )- . By an orientation for R n we mean the choice of GL ( n ) + or GL ( n )- as the preferred source for picking a basis for R n . The standard orientation for R n is that determined by GL ( n ) + . Lemma 0.0.1. GL ( n ) has exactly two path-connected components: GL ( n ) + and GL ( n )- . Proof. First note that since det: GL ( n ) R n- { } is continuous, and R- { } is not connected, then GL ( n ) is not connected. So it must have at least two compo- nents GL ( n ) + = det- 1 ((0 , )), and GL ( n )- = det- 1 ((- , 0)). Secondly note that multiplying the first row of any element of GL ( n ) + yields an element of GL ( n )- . This implies that GL ( n ) + and GL ( n )- are homeomorphic. So we just need to check that GL ( n ) + is path connected. This may be achieved in two steps: first we deform each element of GL ( n ) + to an element of SO ( n ) and then show that SO ( n ) is path connected. Step 1: This may be achieved with the aid of the Gram-Schmidt process. In particular recall that if ( b 1 ,...,b n ) is any basis of R n and ( b 1 ,...,b n ) is the corre- sponding orthonormalizaion of it, then b i is never antiparallel to b i and each pair b i and b i span a subspace which does not include any other of the elements of the basis. Thus tb i + (1- t ) b i yields a continuous deformation of ( b 1 ,...,b n ) to ( b 1 ,...,b n ) through a family of basis. Step 2: Let B SO ( n + 1), then we may write B = ( b 1 ,...,b n ) where b i are the columns of B . We may rotate B until b n coincides with (0 ,..., , 1). Then b 1 ,...,b n- 1 lie in R n- 1 { } , and by a rotation in R n- 1 we may bring b n- 1 in coincidence with (0 ,..., , 1 , 0). Continuing this procedure we may continuously deform ( b 1 ,...,b n ) through a family of orthonormal basis until b 2 ,...,b n coincide with the last n- 1 elements of the standard basis of R n . At that point we must either have b 1 = (1 , ,..., 0) or b 1 = (- 1 , ,..., 0). But the latter is impossible 1 Last revised: October 10, 2006 1 because B has positive determinant, and continuous deformations of it through a family of basis preserve the sign of the determinant. So b 1 = (1 , ,..., 0) and we are done. This immediately yields that Corollary 0.0.2. Two basis of R n can be continuously transformed to each other, through a family of basis, if and only if they belong to the same orientation class....
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This note was uploaded on 08/25/2011 for the course MATH 6456 taught by Professor Staff during the Spring '08 term at Georgia Institute of Technology.

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LectureNotes11G - Math 6455 Oct 2, 2006 1 Differential...

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